Quantum Mechanics I

Peter S. Riseborough

August 29, 2013

Contents

1 Principles of Classical Mechanics 91.1 Lagrangian Mechanics . . . . . . . . . . . . . . . . . . . . . . . . 9

1.1.1 Exercise 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.1.2 Solution 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.1.3 The Principle of Least Action . . . . . . . . . . . . . . . . 121.1.4 The Euler-Lagrange Equations . . . . . . . . . . . . . . . 151.1.5 Generalized Momentum . . . . . . . . . . . . . . . . . . . 161.1.6 Exercise 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.1.7 Solution 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.2 Hamiltonian Mechanics . . . . . . . . . . . . . . . . . . . . . . . 191.2.1 The Hamilton Equations of Motion . . . . . . . . . . . . . 201.2.2 Exercise 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.2.3 Solution 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.2.4 Time Evolution of a Physical Quantity . . . . . . . . . . . 221.2.5 Poisson Brackets . . . . . . . . . . . . . . . . . . . . . . . 22

1.3 A Charged Particle in an Electromagnetic Field . . . . . . . . . . 251.3.1 The Electromagnetic Field . . . . . . . . . . . . . . . . . 251.3.2 The Lagrangian for a Classical Charged Particle . . . . . 261.3.3 Exercise 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.3.4 Solution 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.3.5 The Hamiltonian of a Classical Charged Particle . . . . . 281.3.6 Exercise 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . 291.3.7 Solution 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2 Failure of Classical Mechanics 332.1 Semi-Classical Quantization . . . . . . . . . . . . . . . . . . . . . 34

2.1.1 Exercise 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . 342.1.2 Solution 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . 342.1.3 Exercise 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.1.4 Solution 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

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3 Principles of Quantum Mechanics 383.1 The Principle of Linear Superposition . . . . . . . . . . . . . . . 403.2 Wave Packets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

3.2.1 Exercise 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . 463.2.2 Solution 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . 463.2.3 Exercise 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . 483.2.4 Solution 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

3.3 Probability, Mean and Deviations . . . . . . . . . . . . . . . . . . 523.3.1 Exercise 10 . . . . . . . . . . . . . . . . . . . . . . . . . . 543.3.2 Solution 10 . . . . . . . . . . . . . . . . . . . . . . . . . . 543.3.3 Exercise 11 . . . . . . . . . . . . . . . . . . . . . . . . . . 573.3.4 Solution 11 . . . . . . . . . . . . . . . . . . . . . . . . . . 573.3.5 Exercise 12 . . . . . . . . . . . . . . . . . . . . . . . . . . 583.3.6 Solution 12 . . . . . . . . . . . . . . . . . . . . . . . . . . 59

3.4 Operators and Measurements . . . . . . . . . . . . . . . . . . . . 603.4.1 Operator Equations . . . . . . . . . . . . . . . . . . . . . 613.4.2 Operator Addition . . . . . . . . . . . . . . . . . . . . . . 623.4.3 Operator Multiplication . . . . . . . . . . . . . . . . . . . 623.4.4 Commutators . . . . . . . . . . . . . . . . . . . . . . . . . 643.4.5 Exercise 13 . . . . . . . . . . . . . . . . . . . . . . . . . . 663.4.6 Solution 13 . . . . . . . . . . . . . . . . . . . . . . . . . . 663.4.7 Exercise 14 . . . . . . . . . . . . . . . . . . . . . . . . . . 673.4.8 Solution 14 . . . . . . . . . . . . . . . . . . . . . . . . . . 673.4.9 Exercise 15 . . . . . . . . . . . . . . . . . . . . . . . . . . 683.4.10 Solution 15 . . . . . . . . . . . . . . . . . . . . . . . . . . 683.4.11 Exercise 16 . . . . . . . . . . . . . . . . . . . . . . . . . . 703.4.12 Solution 16 . . . . . . . . . . . . . . . . . . . . . . . . . . 703.4.13 Exercise 17 . . . . . . . . . . . . . . . . . . . . . . . . . . 713.4.14 Solution 17 . . . . . . . . . . . . . . . . . . . . . . . . . . 713.4.15 Exercise 18 . . . . . . . . . . . . . . . . . . . . . . . . . . 723.4.16 Solution 18 . . . . . . . . . . . . . . . . . . . . . . . . . . 723.4.17 Eigenvalue Equations . . . . . . . . . . . . . . . . . . . . 733.4.18 Exercise 19 . . . . . . . . . . . . . . . . . . . . . . . . . . 753.4.19 Exercise 20 . . . . . . . . . . . . . . . . . . . . . . . . . . 753.4.20 Solution 20 . . . . . . . . . . . . . . . . . . . . . . . . . . 763.4.21 Exercise 21 . . . . . . . . . . . . . . . . . . . . . . . . . . 783.4.22 Solution 21 . . . . . . . . . . . . . . . . . . . . . . . . . . 783.4.23 Exercise 22 . . . . . . . . . . . . . . . . . . . . . . . . . . 793.4.24 Solution 22 . . . . . . . . . . . . . . . . . . . . . . . . . . 793.4.25 Adjoint or Hermitean Conjugate Operators . . . . . . . . 813.4.26 Hermitean Operators . . . . . . . . . . . . . . . . . . . . . 853.4.27 Exercise 23 . . . . . . . . . . . . . . . . . . . . . . . . . . 863.4.28 Exercise 24 . . . . . . . . . . . . . . . . . . . . . . . . . . 863.4.29 Solution 24 . . . . . . . . . . . . . . . . . . . . . . . . . . 863.4.30 Exercise 25 . . . . . . . . . . . . . . . . . . . . . . . . . . 893.4.31 Solution 25 . . . . . . . . . . . . . . . . . . . . . . . . . . 89

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3.4.32 Exercise 26 . . . . . . . . . . . . . . . . . . . . . . . . . . 913.4.33 Solution 26 . . . . . . . . . . . . . . . . . . . . . . . . . . 913.4.34 Eigenvalues and Eigenfunctions of Hermitean Operators . 923.4.35 Exercise 27 . . . . . . . . . . . . . . . . . . . . . . . . . . 943.4.36 Solution 27 . . . . . . . . . . . . . . . . . . . . . . . . . . 943.4.37 Exercise 28 . . . . . . . . . . . . . . . . . . . . . . . . . . 953.4.38 Solution 28 . . . . . . . . . . . . . . . . . . . . . . . . . . 963.4.39 Exercise 29 . . . . . . . . . . . . . . . . . . . . . . . . . . 993.4.40 Solution 29 . . . . . . . . . . . . . . . . . . . . . . . . . . 993.4.41 Exercise 30 . . . . . . . . . . . . . . . . . . . . . . . . . . 993.4.42 Solution 30 . . . . . . . . . . . . . . . . . . . . . . . . . . 1003.4.43 Exercise 31 . . . . . . . . . . . . . . . . . . . . . . . . . . 1023.4.44 Exercise 32 . . . . . . . . . . . . . . . . . . . . . . . . . . 1023.4.45 Solution 32 . . . . . . . . . . . . . . . . . . . . . . . . . . 1033.4.46 Hermitean Operators and Physical Measurements . . . . . 1043.4.47 Exercise 33 . . . . . . . . . . . . . . . . . . . . . . . . . . 1053.4.48 Solution 33 . . . . . . . . . . . . . . . . . . . . . . . . . . 1053.4.49 Exercise 34 . . . . . . . . . . . . . . . . . . . . . . . . . . 1073.4.50 Solution 34 . . . . . . . . . . . . . . . . . . . . . . . . . . 1073.4.51 Exercise 35 . . . . . . . . . . . . . . . . . . . . . . . . . . 1083.4.52 Solution 35 . . . . . . . . . . . . . . . . . . . . . . . . . . 109

3.5 Quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1123.5.1 Relations between Physical Operators . . . . . . . . . . . 1123.5.2 The Correspondence Principle . . . . . . . . . . . . . . . 1123.5.3 The Complementarity Principle . . . . . . . . . . . . . . . 1133.5.4 Coordinate Representation . . . . . . . . . . . . . . . . . 1143.5.5 Momentum Representation . . . . . . . . . . . . . . . . . 1173.5.6 Exercise 36 . . . . . . . . . . . . . . . . . . . . . . . . . . 1233.5.7 Exercise 37 . . . . . . . . . . . . . . . . . . . . . . . . . . 1253.5.8 Exercise 38 . . . . . . . . . . . . . . . . . . . . . . . . . . 1253.5.9 Solution 38 . . . . . . . . . . . . . . . . . . . . . . . . . . 1263.5.10 Exercise 39 . . . . . . . . . . . . . . . . . . . . . . . . . . 1273.5.11 Solution 39 . . . . . . . . . . . . . . . . . . . . . . . . . . 1273.5.12 Exercise 40 . . . . . . . . . . . . . . . . . . . . . . . . . . 1283.5.13 Solution 40 . . . . . . . . . . . . . . . . . . . . . . . . . . 1283.5.14 Exercise 41 . . . . . . . . . . . . . . . . . . . . . . . . . . 1293.5.15 Solution 41 . . . . . . . . . . . . . . . . . . . . . . . . . . 1293.5.16 Commuting Operators and Compatibility . . . . . . . . . 1303.5.17 Non-Commuting Operators . . . . . . . . . . . . . . . . . 1323.5.18 Exercise 42 . . . . . . . . . . . . . . . . . . . . . . . . . . 1323.5.19 Solution 42 . . . . . . . . . . . . . . . . . . . . . . . . . . 1323.5.20 The Uncertainty Principle . . . . . . . . . . . . . . . . . . 1333.5.21 Exercise 43 . . . . . . . . . . . . . . . . . . . . . . . . . . 1353.5.22 Solution 43 . . . . . . . . . . . . . . . . . . . . . . . . . . 1353.5.23 Exercise 44 . . . . . . . . . . . . . . . . . . . . . . . . . . 1363.5.24 Solution 44 . . . . . . . . . . . . . . . . . . . . . . . . . . 136

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3.5.25 Exercise 45 . . . . . . . . . . . . . . . . . . . . . . . . . . 1363.5.26 Solution 45 . . . . . . . . . . . . . . . . . . . . . . . . . . 1373.5.27 Exercise 46 . . . . . . . . . . . . . . . . . . . . . . . . . . 1373.5.28 Solution 46 . . . . . . . . . . . . . . . . . . . . . . . . . . 137

3.6 The Philosophy of Measurement . . . . . . . . . . . . . . . . . . 1403.6.1 Exercise 47 . . . . . . . . . . . . . . . . . . . . . . . . . . 1433.6.2 Solution 47 . . . . . . . . . . . . . . . . . . . . . . . . . . 1443.6.3 Exercise 48 . . . . . . . . . . . . . . . . . . . . . . . . . . 1453.6.4 Solution 48 . . . . . . . . . . . . . . . . . . . . . . . . . . 1463.6.5 Exercise 49 . . . . . . . . . . . . . . . . . . . . . . . . . . 1473.6.6 Solution 49 . . . . . . . . . . . . . . . . . . . . . . . . . . 148

3.7 Time Evolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1493.7.1 The Schrodinger Picture. . . . . . . . . . . . . . . . . . . 1503.7.2 Exercise 50 . . . . . . . . . . . . . . . . . . . . . . . . . . 1513.7.3 Solution 50 . . . . . . . . . . . . . . . . . . . . . . . . . . 1513.7.4 The Heisenberg Picture. . . . . . . . . . . . . . . . . . . . 1523.7.5 Exercise 51 . . . . . . . . . . . . . . . . . . . . . . . . . . 1533.7.6 Solution 51 . . . . . . . . . . . . . . . . . . . . . . . . . . 1533.7.7 Exercise 52 . . . . . . . . . . . . . . . . . . . . . . . . . . 1543.7.8 Solution 52 . . . . . . . . . . . . . . . . . . . . . . . . . . 1543.7.9 Exercise 53 . . . . . . . . . . . . . . . . . . . . . . . . . . 1553.7.10 Exercise 54 . . . . . . . . . . . . . . . . . . . . . . . . . . 1553.7.11 Solution 54 . . . . . . . . . . . . . . . . . . . . . . . . . . 1553.7.12 Exercise 55 . . . . . . . . . . . . . . . . . . . . . . . . . . 1563.7.13 Exercise 56 . . . . . . . . . . . . . . . . . . . . . . . . . . 1573.7.14 The Schrodinger Equation . . . . . . . . . . . . . . . . . . 1573.7.15 Exercise 57 . . . . . . . . . . . . . . . . . . . . . . . . . . 1603.7.16 Solution 57 . . . . . . . . . . . . . . . . . . . . . . . . . . 1603.7.17 Time Development of a Wave Packet . . . . . . . . . . . . 1613.7.18 Exercise 58 . . . . . . . . . . . . . . . . . . . . . . . . . . 1623.7.19 Solution 58 . . . . . . . . . . . . . . . . . . . . . . . . . . 1623.7.20 Time Evolution and Energy Eigenfunctions . . . . . . . . 1643.7.21 Exercise 59 . . . . . . . . . . . . . . . . . . . . . . . . . . 1663.7.22 Solution 59 . . . . . . . . . . . . . . . . . . . . . . . . . . 1663.7.23 The Correspondence Principle . . . . . . . . . . . . . . . 1683.7.24 The Continuity Equation and Particle Conservation . . . 169

4 Applications of Quantum Mechanics 1744.1 Exact Solutions in One Dimension . . . . . . . . . . . . . . . . . 174

4.1.1 Particle Confined in a Deep Potential Well . . . . . . . . 1744.1.2 Time Dependence of a Particle in a Deep Potential Well . 1824.1.3 Exercise 60 . . . . . . . . . . . . . . . . . . . . . . . . . . 1834.1.4 Particle Bound in a Shallow Potential Well . . . . . . . . 1834.1.5 Exercise 61 . . . . . . . . . . . . . . . . . . . . . . . . . . 1904.1.6 Solution 61 . . . . . . . . . . . . . . . . . . . . . . . . . . 1914.1.7 Scattering from a Shallow Potential Well . . . . . . . . . 194

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4.1.8 Exercise 62 . . . . . . . . . . . . . . . . . . . . . . . . . . 1994.1.9 Solution 62 . . . . . . . . . . . . . . . . . . . . . . . . . . 1994.1.10 Exercise 63 . . . . . . . . . . . . . . . . . . . . . . . . . . 2014.1.11 Solution 63 . . . . . . . . . . . . . . . . . . . . . . . . . . 2014.1.12 The Threshold Energy for a Bound State . . . . . . . . . 2034.1.13 Transmission through a Potential Barrier . . . . . . . . . 2044.1.14 Exercise 64 . . . . . . . . . . . . . . . . . . . . . . . . . . 2074.1.15 Solution 64 . . . . . . . . . . . . . . . . . . . . . . . . . . 2084.1.16 The Double Well Potential . . . . . . . . . . . . . . . . . 2094.1.17 The delta function Potential . . . . . . . . . . . . . . . . . 2134.1.18 Bound States of a delta function Potential . . . . . . . . . 2154.1.19 Exercise 65 . . . . . . . . . . . . . . . . . . . . . . . . . . 2244.1.20 Solution 65 . . . . . . . . . . . . . . . . . . . . . . . . . . 2244.1.21 Exercise 66 . . . . . . . . . . . . . . . . . . . . . . . . . . 2264.1.22 Solution 66 . . . . . . . . . . . . . . . . . . . . . . . . . . 2264.1.23 Exercise 67 . . . . . . . . . . . . . . . . . . . . . . . . . . 2324.1.24 Solution 67 . . . . . . . . . . . . . . . . . . . . . . . . . . 2324.1.25 Exercise 68 . . . . . . . . . . . . . . . . . . . . . . . . . . 2344.1.26 Solution 68 . . . . . . . . . . . . . . . . . . . . . . . . . . 235

4.2 The One-Dimensional Harmonic Oscillator . . . . . . . . . . . . . 2374.2.1 The Raising and Lowering Operators . . . . . . . . . . . . 2384.2.2 The Effect of the Lowering Operator . . . . . . . . . . . . 2384.2.3 The Ground State . . . . . . . . . . . . . . . . . . . . . . 2394.2.4 The Effect of The Raising Operator . . . . . . . . . . . . 2404.2.5 The Normalization . . . . . . . . . . . . . . . . . . . . . . 2414.2.6 The Excited States . . . . . . . . . . . . . . . . . . . . . . 2414.2.7 Exercise 69 . . . . . . . . . . . . . . . . . . . . . . . . . . 2434.2.8 Solution 69 . . . . . . . . . . . . . . . . . . . . . . . . . . 2444.2.9 Exercise 70 . . . . . . . . . . . . . . . . . . . . . . . . . . 2464.2.10 Solution 70 . . . . . . . . . . . . . . . . . . . . . . . . . . 2464.2.11 Exercise 71 . . . . . . . . . . . . . . . . . . . . . . . . . . 2464.2.12 Time Development of the Harmonic Oscillator . . . . . . 2474.2.13 Exercise 72 . . . . . . . . . . . . . . . . . . . . . . . . . . 2494.2.14 Solution 72 . . . . . . . . . . . . . . . . . . . . . . . . . . 2504.2.15 Hermite Polynomials . . . . . . . . . . . . . . . . . . . . . 2514.2.16 Exercise 73 . . . . . . . . . . . . . . . . . . . . . . . . . . 2564.2.17 Solution 73 . . . . . . . . . . . . . . . . . . . . . . . . . . 2564.2.18 Exercise 74 . . . . . . . . . . . . . . . . . . . . . . . . . . 2574.2.19 Solution 74 . . . . . . . . . . . . . . . . . . . . . . . . . . 2574.2.20 The Completeness Condition . . . . . . . . . . . . . . . . 258

4.3 Dual-symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2604.4 Bargmann Potentials . . . . . . . . . . . . . . . . . . . . . . . . . 263

4.4.1 Exercise 75 . . . . . . . . . . . . . . . . . . . . . . . . . . 2664.4.2 Solution 75 . . . . . . . . . . . . . . . . . . . . . . . . . . 2664.4.3 Exercise 76 . . . . . . . . . . . . . . . . . . . . . . . . . . 2704.4.4 Solution 76 . . . . . . . . . . . . . . . . . . . . . . . . . . 270

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4.4.5 Exercise 77 . . . . . . . . . . . . . . . . . . . . . . . . . . 2734.4.6 Solution 77 . . . . . . . . . . . . . . . . . . . . . . . . . . 273

4.5 Orbital Angular Momentum . . . . . . . . . . . . . . . . . . . . . 2764.5.1 Exercise 78 . . . . . . . . . . . . . . . . . . . . . . . . . . 2784.5.2 Solution 78 . . . . . . . . . . . . . . . . . . . . . . . . . . 2784.5.3 Exercise 79 . . . . . . . . . . . . . . . . . . . . . . . . . . 2804.5.4 Simultaneous Eigenfunctions. . . . . . . . . . . . . . . . . 2824.5.5 The Raising and Lowering Operators . . . . . . . . . . . . 2844.5.6 The Eigenvalues and Degeneracy . . . . . . . . . . . . . . 2854.5.7 The Effect of the Raising Operators. . . . . . . . . . . . . 2864.5.8 Explicit Expressions for the Eigenfunctions . . . . . . . . 2874.5.9 Legendre Polynomials . . . . . . . . . . . . . . . . . . . . 2904.5.10 Associated Legendre Functions . . . . . . . . . . . . . . . 2934.5.11 Spherical Harmonics . . . . . . . . . . . . . . . . . . . . . 2944.5.12 Exercise 80 . . . . . . . . . . . . . . . . . . . . . . . . . . 2984.5.13 Solution 80 . . . . . . . . . . . . . . . . . . . . . . . . . . 2984.5.14 Exercise 81 . . . . . . . . . . . . . . . . . . . . . . . . . . 3024.5.15 Solution 81 . . . . . . . . . . . . . . . . . . . . . . . . . . 3034.5.16 Exercise 82 . . . . . . . . . . . . . . . . . . . . . . . . . . 3044.5.17 Solution 82 . . . . . . . . . . . . . . . . . . . . . . . . . . 3054.5.18 The Addition Theorem . . . . . . . . . . . . . . . . . . . 3064.5.19 Finite-Dimensional Representations . . . . . . . . . . . . 3084.5.20 Exercise 83 . . . . . . . . . . . . . . . . . . . . . . . . . . 3124.5.21 Exercise 84 . . . . . . . . . . . . . . . . . . . . . . . . . . 3124.5.22 Solution 84 . . . . . . . . . . . . . . . . . . . . . . . . . . 3134.5.23 The Laplacian Operator . . . . . . . . . . . . . . . . . . . 3154.5.24 An Excursion into d-Dimensional Space . . . . . . . . . . 3174.5.25 Exercise 85 . . . . . . . . . . . . . . . . . . . . . . . . . . 3194.5.26 Solution 85 . . . . . . . . . . . . . . . . . . . . . . . . . . 320

4.6 Spherically Symmetric Potentials . . . . . . . . . . . . . . . . . . 3234.6.1 Exercise 86 . . . . . . . . . . . . . . . . . . . . . . . . . . 3234.6.2 Solution 86 . . . . . . . . . . . . . . . . . . . . . . . . . . 3244.6.3 The Free Particle . . . . . . . . . . . . . . . . . . . . . . . 3254.6.4 The Spherical Square Well . . . . . . . . . . . . . . . . . . 3344.6.5 Exercise 87 . . . . . . . . . . . . . . . . . . . . . . . . . . 3424.6.6 Solution 87 . . . . . . . . . . . . . . . . . . . . . . . . . . 3434.6.7 Exercise 88 . . . . . . . . . . . . . . . . . . . . . . . . . . 3454.6.8 Exercise 89 . . . . . . . . . . . . . . . . . . . . . . . . . . 3454.6.9 Solution 89 . . . . . . . . . . . . . . . . . . . . . . . . . . 3454.6.10 Exercise 90 . . . . . . . . . . . . . . . . . . . . . . . . . . 3464.6.11 Solution 90 . . . . . . . . . . . . . . . . . . . . . . . . . . 3464.6.12 Exercise 91 . . . . . . . . . . . . . . . . . . . . . . . . . . 3484.6.13 Solution 91 . . . . . . . . . . . . . . . . . . . . . . . . . . 3484.6.14 Exercise 92 . . . . . . . . . . . . . . . . . . . . . . . . . . 3494.6.15 Solution 92 . . . . . . . . . . . . . . . . . . . . . . . . . . 3494.6.16 Ladder operators for a free particle . . . . . . . . . . . . . 351

6

4.6.17 The Rayleigh Equation . . . . . . . . . . . . . . . . . . . 3554.6.18 The Isotropic Planar Harmonic Oscillator . . . . . . . . . 3584.6.19 The Spherical Harmonic Oscillator . . . . . . . . . . . . . 3604.6.20 Exercise 93 . . . . . . . . . . . . . . . . . . . . . . . . . . 3624.6.21 Solution 93 . . . . . . . . . . . . . . . . . . . . . . . . . . 3634.6.22 Exercise 94 . . . . . . . . . . . . . . . . . . . . . . . . . . 3654.6.23 The Bound States of the Coulomb Potential . . . . . . . . 3674.6.24 Exercise 95 . . . . . . . . . . . . . . . . . . . . . . . . . . 3744.6.25 Exercise 96 . . . . . . . . . . . . . . . . . . . . . . . . . . 3744.6.26 Solution 96 . . . . . . . . . . . . . . . . . . . . . . . . . . 3744.6.27 Exercise 97 . . . . . . . . . . . . . . . . . . . . . . . . . . 3754.6.28 Ladder Operators for the Hydrogen Atom . . . . . . . . . 3764.6.29 Rydberg Wave Packets . . . . . . . . . . . . . . . . . . . . 3804.6.30 Laguerre Polynomials . . . . . . . . . . . . . . . . . . . . 3834.6.31 Exercise 98 . . . . . . . . . . . . . . . . . . . . . . . . . . 3904.6.32 Solution 98 . . . . . . . . . . . . . . . . . . . . . . . . . . 3904.6.33 Exercise 99 . . . . . . . . . . . . . . . . . . . . . . . . . . 3924.6.34 Solution 99 . . . . . . . . . . . . . . . . . . . . . . . . . . 3934.6.35 Exercise 100 . . . . . . . . . . . . . . . . . . . . . . . . . . 3954.6.36 Solution 100 . . . . . . . . . . . . . . . . . . . . . . . . . 395

4.7 A Charged Particle in a Magnetic Field . . . . . . . . . . . . . . 3994.7.1 Exercise 101 . . . . . . . . . . . . . . . . . . . . . . . . . . 4004.7.2 Exercise 102 . . . . . . . . . . . . . . . . . . . . . . . . . . 4004.7.3 Solution 102 . . . . . . . . . . . . . . . . . . . . . . . . . 4004.7.4 The Degeneracy of the Landau Levels . . . . . . . . . . . 4014.7.5 Exercise 103 . . . . . . . . . . . . . . . . . . . . . . . . . . 4034.7.6 Solution 103 . . . . . . . . . . . . . . . . . . . . . . . . . 4044.7.7 The Aharonov-Bohm Effect . . . . . . . . . . . . . . . . . 406

4.8 The Pauli Spin Matrices . . . . . . . . . . . . . . . . . . . . . . . 4124.8.1 Exercise 104 . . . . . . . . . . . . . . . . . . . . . . . . . . 4154.8.2 Solution 104 . . . . . . . . . . . . . . . . . . . . . . . . . 4154.8.3 Exercise 105 . . . . . . . . . . . . . . . . . . . . . . . . . . 4174.8.4 Solution 105 . . . . . . . . . . . . . . . . . . . . . . . . . 4174.8.5 Exercise 106 . . . . . . . . . . . . . . . . . . . . . . . . . . 4194.8.6 Solution 106 . . . . . . . . . . . . . . . . . . . . . . . . . 4194.8.7 Exercise 107 . . . . . . . . . . . . . . . . . . . . . . . . . . 4204.8.8 Solution 107 . . . . . . . . . . . . . . . . . . . . . . . . . 4214.8.9 Exercise 108 . . . . . . . . . . . . . . . . . . . . . . . . . . 4234.8.10 Solution 108 . . . . . . . . . . . . . . . . . . . . . . . . . 4244.8.11 The Pauli Equation . . . . . . . . . . . . . . . . . . . . . 4264.8.12 Spin Dynamics . . . . . . . . . . . . . . . . . . . . . . . . 4284.8.13 Exercise 109 . . . . . . . . . . . . . . . . . . . . . . . . . . 4304.8.14 Solution 109 . . . . . . . . . . . . . . . . . . . . . . . . . 4304.8.15 Exercise 110 . . . . . . . . . . . . . . . . . . . . . . . . . . 4314.8.16 Solution 110 . . . . . . . . . . . . . . . . . . . . . . . . . 4324.8.17 The Berry Phase . . . . . . . . . . . . . . . . . . . . . . . 434

7

4.9 Transformations and Invariance . . . . . . . . . . . . . . . . . . . 4384.9.1 Time Translational Invariance . . . . . . . . . . . . . . . . 4394.9.2 Translational Invariance . . . . . . . . . . . . . . . . . . . 4404.9.3 Periodic Translational Invariance . . . . . . . . . . . . . . 4434.9.4 Exercise 111 . . . . . . . . . . . . . . . . . . . . . . . . . . 4504.9.5 Solution 111 . . . . . . . . . . . . . . . . . . . . . . . . . 4514.9.6 Rotational Invariance . . . . . . . . . . . . . . . . . . . . 4524.9.7 Exercise 112 . . . . . . . . . . . . . . . . . . . . . . . . . . 4614.9.8 Solution 112 . . . . . . . . . . . . . . . . . . . . . . . . . 4614.9.9 Exercise 113 . . . . . . . . . . . . . . . . . . . . . . . . . . 4634.9.10 Solution 113 . . . . . . . . . . . . . . . . . . . . . . . . . 4634.9.11 Exercise 114 . . . . . . . . . . . . . . . . . . . . . . . . . . 4684.9.12 Solution 114 . . . . . . . . . . . . . . . . . . . . . . . . . 4684.9.13 Gauge Invariance . . . . . . . . . . . . . . . . . . . . . . . 4714.9.14 Exercise 116 . . . . . . . . . . . . . . . . . . . . . . . . . . 4724.9.15 Solution 116 . . . . . . . . . . . . . . . . . . . . . . . . . 4734.9.16 Galilean Boosts . . . . . . . . . . . . . . . . . . . . . . . . 474

5 The Rotating Planar Oscillator 476

6 Dirac Formulation 4826.1 Dirac Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 482

6.1.1 Bracket Notation . . . . . . . . . . . . . . . . . . . . . . . 4836.1.2 Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . 4846.1.3 Adjoints and Hermitean Operators . . . . . . . . . . . . . 4856.1.4 Representation of Operators . . . . . . . . . . . . . . . . . 486

6.2 Representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4876.3 Gram-Schmidt Orthogonalization . . . . . . . . . . . . . . . . . . 489

7 Appendices 490

8

1 Principles of Classical Mechanics

Newton’s laws can be reformulated in a variety of different ways. These differ-ent formulations provide more powerful and elegant methods for solving prob-lems which involve many different variables and have a natural formulation innon-Cartesian coordinate systems, such as spherical polar coordinates. In non-Cartesian coordinate systems, vector formulations are complicated by the factthat the orthogonal directions associated with the variables depend upon thevalues of the generalized coordinates. The advantage of the alternate formula-tion of Newton’s laws is based on the fact that they involve scalar quantitiesrather than vector quantities, therefore, they do not require transforming theequations of motion between the Cartesian coordinates and the non-Cartesiansystem.

1.1 Lagrangian Mechanics

The Lagrangian approach to classical mechanics is based on a scalar quantity,the Lagrangian L, which depends upon the generalized coordinates and veloci-ties.

For a Cartesian coordinate system, the coordinates are the position of theparticle x, y and z. The generalized velocities are the time derivatives of thecoordinates, which we represent by x, y and z.

For a non-Cartesian coordinate system, such as spherical polar coordinates,the generalized coordinates for one particle are r, θ, and ϕ. The generalizedvelocities are the time derivatives of the coordinates, which are r, θ, and ϕ.

The Lagrangian is given by the difference of the kinetic energy T and thepotential energy V ,

L = T − V (1)

In Cartesian coordinates, we have

L =m

2x2 +

m

2y2 +

m

2z2 − V (x, y, z) (2)

In spherical polar coordinates, we have

L =m r2

2+

m r2 θ2

2+

m r2 sin2 θ ϕ2

2− V (r, θ, ϕ) (3)

9

ϕϕϕϕ

θθθθ

x

y

zr

r

Figure 1: The Spherical Polar Coordinate System. A general point is labelledby the coordinates (r, θ, ϕ).

——————————————————————————————————

1.1.1 Exercise 1

Find the Lagrangian for a particle in terms of spherical polar coordinates.

——————————————————————————————————

1.1.2 Solution 1

The Lagrangian for a particle in a potential is given by

L =m

2r2 − V (r) (4)

and with r ≡ (r, θ, ϕ) one has

r =∂r

∂r

dr

dt+

∂r

∂θ

dθ

dt+

∂r

∂ϕ

dϕ

dt(5)

but the three orthogonal unit vectors of spherical polar coordinates er, eθ andeϕ are defined as the directions of increasing r, increasing θ and increasing ϕ.

10

ϕϕϕϕ

θθθθ

x

y

z dr er

r

r dθ θ θ θ eθθθθ

r sinθθθθ dϕ ϕ ϕ ϕ eϕϕϕϕ

dθθθθ

dϕϕϕϕ

Figure 2: The Spherical Polar Coordinate System. An orthogonal set of unitvectors er, eθ and eϕ can be constructed which, respectively, correspond to thedirections of increasing r, θ and ϕ.

Thus,

∂r

∂r= er

∂r

∂θ= r eθ

∂r

∂ϕ= r sin θ eϕ

(6)

Hencer = er

dr

dt+ r eθ

dθ

dt+ r sin θ eϕ

dϕ

dt(7)

and as the unit vectors are orthogonal

L =m

2

[ (dr

dt

)2

+ r2(dθ

dt

)2

+ r2 sin2 θ

(dϕ

dt

)2 ]− V (r) (8)

——————————————————————————————————

11

For a problem involving N particles, we denote the generalized coordinatesby qi, where i runs over the 3N values corresponding to the 3 coordinates foreach of the N particles, and the generalized velocities by qi. The LagrangianL is a function of the set of qi and qi, and we shall write this as L(qi, qi) inwhich only one set of coordinates and velocities appears. However, L dependson all the coordinates and velocities. The Lagrangian is the sum of the kineticenergy of the particles minus the total potential energy, which is the sum of theexternal potentials acting on each of the particles together with the sum of anyinteraction potentials acting between pairs of particles.

1.1.3 The Principle of Least Action

The equations of motion originate from an extremum principle, often called theprinciple of least action. The central quantity in this principle is given by theaction S which is a number that depends upon the specific function which isa trajectory qi(t′). These trajectories run from the initial position at t′ = 0,which is denoted by qi(0), to a final position at t′ = t, denoted by qi(t).These two sets of values are assumed to be known, and they replace the twosets of initial conditions, qi(0) and qi(0), used in the solution of Newton’s laws.There are infinitely many arbitrary trajectories that run between the initial andfinal positions. The action for any one of these trajectories, qi(t′) is given by anumber which has the value of the integration

S =∫ t

0

dt′ L(qi(t′), qi(t′)) (9)

The value of S depends on the particular choice of trajectory qi(t′). The actionis an example of a functional S[qi(t′)] as it yields a number that depends uponthe choice of a function. The extremum principle asserts that the value of Sis an extremum, i.e. a maximum, minimum or saddle point, for the trajectorywhich satisfies Newton’s laws.

To elucidate the meaning of the extremal principle, we shall consider anarbitrary trajectory qi(t′) that goes between the initial and final position in atime interval of duration t. Since this trajectory is arbitrary, it is different fromthe trajectory that satisfies Newton’s laws, which as we shall show later is anextremal trajectory qex

i (t′). The difference or deviation between the arbitrarytrajectory and the extremal trajectory is defined by

δqi(t′) = qi(t′) − qexi (t′) (10)

An important fact is that this deviation tends to zero at the end points t′ = 0and t′ = t since our trajectories are defined to all run through the specificinitial qi(0) and final positions qi(t) at t′ = 0 and t′ = t. Let us considerthe variety of the plots of δqi(t′) versus t′. There are infinitely many differentcurves. Let us concentrate on a single shape of the curve, then we can generate

12

0

1

2

3

4

5

6

-1 0 1 2 3

t

q(t)

qex(t)

q(tf)

q(ti)

ti tf

q(t)

Figure 3: Arbitrary trajectories qi(t) originating from a specific initial pointqi(ti) at t = ti and ending up at a specific final point qi(tf ) at time t = tf .

a whole family of such curves by either increasing or decreasing the magnitudeof the deviation by a factor of λ. The family of trajectories is given by

qi(t′) = qexi (t′) + λ δqi(t′) (11)

When λ = 0 the original curve reduces to the extremal curve and when λ = 1we recover our initial choice for the arbitrary trajectory.

If we substitute this family of trajectories into the action, we would find anumber S(λ) that depends on λ. This function S(λ) should be extremal, i.e.either a maximum, minimum, or a saddle point as a function of λ at λ = 0 ifthe action is an extremum at the extremal trajectory. The condition that theaction is extremal is just that

∂S

∂λ= 0 (12)

at λ = 0, or the first order term in the Taylor series expansion of S(λ) in λ iszero.

Let us first look at a simple example of motion in one dimension, where theLagrangian is given by

L(x, x) =m

2x2 − V (x) (13)

13

0

1

2

3

4

5

6

-1 0 1 2 3

t

q(t)

qex(t)

q(tf)

q(ti)

ti tf

q(t)

δδδδq(t)

Figure 4: Arbitrary trajectories qi(t) going between specific initial and finalpoints, and the extremal trajectory qex

i (t). The deviation δqi(t) is defined asδqi(t) = qi(t) − qex

i (t).

and let us substitutex(t′) = xex(t′) + λ δx(t′) (14)

in S(λ) and expand in powers of λ,

S(λ) =∫ t

0

dt′ L(xex(t′) + λ δx(t′) , xex(t′) + λ δx(t′))

=∫ t

0

dt′ L(xex(t′), xex(t′))

+ λ

∫ t

0

dt′[∂

∂xL(x(t′), x(t′))

∣∣∣∣λ=0

δx(t′) +∂

∂xL(x(t′), x(t′))

∣∣∣∣λ=0

δx(t′)]

+ O(λ2) (15)

In the above expression, the partial derivatives of the Lagrangian are evaluatedwith the extremal trajectory. Since we are only concerned with the conditionthat S is extremal at λ = 0, the higher order terms in the Taylor expansionin λ are irrelevant. If S is to be extremal, then the extremum condition meansthat the term linear in λ must vanish, no matter what our particular choice of

14

δx(t′) is. Thus, we require that∫ t

0

dt′[∂

∂xL(x(t′), x(t′))

∣∣∣∣λ=0

δx(t′) +∂

∂xL(x(t′), x(t′))

∣∣∣∣λ=0

δx(t′)]

= 0

(16)

for any shape of δx(t′). Since this expression involves both δx(t′) and δx(t′), weshall eliminate the time derivative of the deviation in the second term. To dothis we integrate the second term by parts, that is∫ t

0

dt′∂

∂xL(x(t′), x(t′))

∣∣∣∣λ=0

δx(t′) =

∂

∂xL(x(t′), x(t′))

∣∣∣∣λ=0

δx(t′)∣∣∣∣t0

−∫ t

0

dt′(d

dt′∂

∂xL(x(t′), x(t′))

∣∣∣∣λ=0

)δx(t′)

(17)

The boundary terms vanish at the beginning and the end of the time intervalsince the deviations δx(t′) vanish at both these times. On substituting theexpression (17) back into the term of S(λ) linear in λ, one obtains∫ t

0

dt′[∂

∂xL(x(t′), x(t′))

∣∣∣∣λ=0

−(d

dt′∂

∂xL(x(t′), x(t′))

∣∣∣∣λ=0

) ]δx(t′) = 0

(18)This integral must be zero for all shapes of the deviation δx(t′) if xex(t′) is theextremal trajectory. This can be assured if the term in the square brackets isidentically zero. This gives the equation

∂

∂xL(x(t′), x(t′))

∣∣∣∣λ=0

−(d

dt′∂

∂xL(x(t′), x(t′))

∣∣∣∣λ=0

)= 0 (19)

which determines the extremal trajectory. Now using the form of the Lagrangiangiven in equation 13, one finds

∂V (xex(t′))∂x

+ md x(t′)dt′

= 0 (20)

which is identical to the equations found from Newton’s laws. Thus, the ex-tremal principle reproduces the results obtained from Newton’s laws.

1.1.4 The Euler-Lagrange Equations

Let us now go back to the more general case with N particles, and arbitrarycoordinates qi and arbitrary Lagrangian L. It is straight forward to repeat thederivation of the extremal condition and find that the equations of motion forthe extremal trajectory qi(t′) reduce to the 3N equations,

∂

∂qiL(qj(t′), qj(t′)) −

(d

dt′∂

∂qiL(qj(t′), qj(t′))

)= 0 (21)

15

0

1

2

3

4

5

6

-1 0 1 2 3

t

x(t)

xex(t)

x(tf)

x(ti)

ti tf

δδδδx(t)

Figure 5: Arbitrary trajectories x(t) going between specific initial and finalpoints, and the extremal trajectory xex(t). The deviation δx(t) is defined asδx(t) = x(t) − xex(t).

where there is one equation for each value of i. The value of j is just the dummyvariable which reminds us that L depends on all the coordinates and velocities.These equations are the Euler-Lagrange equations, and are a set of second orderdifferential equations which determine the classical trajectory.

1.1.5 Generalized Momentum

The angular momentum is an example of what we call a generalized momentum.We define a generalized momentum in the same way as the components ofmomentum are defined for a particle in Cartesian coordinates. The generalizedmomentum pi conjugate to the generalized coordinate qi is given by the equation

pi =(∂ L

∂qi

)(22)

Thus, in a Cartesian coordinate system, we find the x-component of a particle’smomentum is given by

px =(∂ L

∂x

)= mx (23)

16

Likewise, for the y and z components

py =(∂ L

∂y

)= my (24)

and

pz =(∂ L

∂z

)= mz (25)

The Euler-Lagrange equations of motion for the general case is re-written interms of the generalized momentum as

∂ L

∂qi−(d pi

dt′

)= 0 (26)

This equation has the same form as Newton’s laws involving the rate of changeof momentum on one side and the derivative of the Lagrangian w.r.t a coordi-nate on the other side.

——————————————————————————————————

1.1.6 Exercise 2

Find the classical equations of motion for a particle, in spherical polar coordi-nates.

——————————————————————————————————

1.1.7 Solution 2

The generalized momenta are found via

pr =∂L

∂r= m r

pθ =∂L

∂θ= m r2 θ

pϕ =∂L

∂ϕ= m r2 sin2 θ ϕ (27)

17

The Euler-Lagrange equations become

dpr

dt=

∂L

∂r= m r

(θ2 + sin2 θ ϕ2

)− ∂V

∂r

dpθ

dt=

∂L

∂θ= m r2 sin θ cos θ ϕ2 − ∂V

∂θdpϕ

dt=

∂L

∂ϕ= − ∂V

∂ϕ(28)

——————————————————————————————————

18

1.2 Hamiltonian Mechanics

Hamiltonian Mechanics formulates mechanics not in terms of the generalizedcoordinates and velocities, but in terms of the generalized coordinates and mo-menta. The Hamiltonian will turn out to be the equivalent of energy. SinceNewton’s laws give rise to a second order differential equation and require twoinitial conditions, to solve Newton’s laws we need to integrate twice. The firstintegration can be done with the aid of an integrating factor. For example, with

m x = − ∂V

∂x(29)

the integrating factor is the velocity, x. On multiplying the equation by theintegrating factor and then integrating, one obtains

m

2x2 = E − V (x) (30)

where the constant of integration is the energy E. Note that the solution isnow found to lie on the surface of constant energy in the two-dimensional spaceformed by x and x. The solution of the mechanical problem is found by in-tegrating once again. The point is, once we have obtained the energy, we arecloser to finding a solution of the equations of motion. Hamiltonian mechanicsresults in a set of first order differential equations.

The Hamiltonian, H(qi, pi, t), is a function of the generalized coordinates qiand generalized momentum pi. It is defined as a Legendre transformation ofthe Lagrangian

H(qi, pi, t) =∑

i

qi pi − L(qi, qi, t) (31)

The Legendre transformation has the effect of eliminating the velocity qi andreplacing it with the momentum pi.

The equations of motion can be determined from the Lagrangian equationsof motion. Since the Hamiltonian is considered to be a function of coordinatesand momenta alone, an infinitesimal change in H occurs either through aninfinitesimal change in the coordinates dqi, momenta dpi or, if the Lagrangianhas any explicit time dependence, through dt,

dH =∑

i

(∂H

∂qidqi +

∂H

∂pidpi

)+

∂H

∂tdt (32)

However, from the definition of H one also has

dH =∑

i

(qi dpi + pi dqi −

∂L

∂qidqi −

∂L

∂qidqi

)− ∂L

∂tdt (33)

19

The terms proportional to dqi cancel as, by definition, pi is the same as ∂L∂qi

.Thus, we have

dH =∑

i

(qi dpi −

∂L

∂qidqi

)− ∂L

∂tdt (34)

The cancellation of the terms proportional to the infinitesimal change dqi is aresult of the Legendre transformation, and confirms that the Hamiltonian is afunction of only the coordinates and momenta. We also can use the Lagrangianequations of motion to express ∂L

∂qias the time derivative of the momentum pi.

1.2.1 The Hamilton Equations of Motion

We can now compare the specific form of the infinitesimal change in H foundabove, with the infinitesimal differential found from its dependence on pi andqi. On equating the coefficients of dqi, dpi and dt, one has

qi =∂H

∂pi

pi = − ∂H

∂qi

− ∂L

∂t=

∂H

∂t(35)

The first two equations are the Hamiltonian equations of motion. The Hamilto-nian equations are two sets of first order differential equations, rather than theone set of second order differential equations given by the Lagrangian equationsof motion.

An example is given by motion in one dimension where

L =m

2x2 − V (x) (36)

the momentum is given by

p =∂L

∂x= m x (37)

Then, the Hamiltonian becomes

H = p x − L

= p x − m

2x2 + V (x)

=p2

2m+ V (x) (38)

20

which is the same as the energy.

For a single particle moving in a central potential, we find that the Hamil-tonian in spherical polar coordinates has the form

H =p2

r

2 m+

p2θ

2 m r2+

p2ϕ

2 m r2 sin2 θ+ V (r) (39)

which is the energy of the particle in spherical polar coordinates.

——————————————————————————————————

1.2.2 Exercise 3

Find the Hamiltonian and the Hamiltonian equations of motion for a particlein spherical polar coordinates.

——————————————————————————————————

1.2.3 Solution 3

Using the expression for the Lagrangian

L =m

2

[ (dr

dt

)2

+ r2(dθ

dt

)2

+ r2 sin2 θ

(dϕ

dt

)2 ]− V (r) (40)

one finds the generalized momenta

pr =∂L

∂r= m r

pθ =∂L

∂θ= m r2 θ

pϕ =∂L

∂ϕ= m r2 sin2 θ ϕ (41)

The Hamiltonian is given by

H = pr r + pθ θ + pϕ ϕ − L (42)

which on eliminating r, θ and ϕ in terms of the generalized momenta, leads to

H =p2

r

2 m+

p2θ

2 m r2+

p2ϕ

2 m r2 sin2 θ+ V (r) (43)

The equations of motion become

r =pr

m

θ =pθ

m r2

ϕ =pϕ

m r2 sin2 θ(44)

21

and

pr = −(

p2θ

m r3+

p2ϕ

m sin2 θ r3

)− ∂V

∂r

pθ = − cos θp2

ϕ

m sin3 θ r2− ∂V

∂θ

pϕ = − ∂V

∂ϕ(45)

——————————————————————————————————

1.2.4 Time Evolution of a Physical Quantity

Given any physical quantity A then it can be represented by a function of theall the coordinates and momenta, and perhaps explicitly on time t, but not onderivatives with respect to time. This quantity A is denoted by A(qi, pi, t). Therate of change of A with respect to time is given by the total derivative,

dA

dt=∑

i

(∂A

∂qiqi +

∂A

∂pipi

)+

∂A

∂t(46)

where the first two terms originate from the dynamics of the particle’s trajectory,the last term originates from the explicit time dependence of the quantity A.On substituting the Hamiltonian equations of motion into the total derivative,and eliminating the rate of change of the coordinates and momenta, one finds

dA

dt=∑

i

(∂A

∂qi

∂H

∂pi− ∂A

∂pi

∂H

∂qi

)+

∂A

∂t(47)

1.2.5 Poisson Brackets

The Poisson Brackets of two quantities, A and B, is given by the expression

[ A , B ]PB =∑

i

(∂A

∂qi

∂B

∂pi− ∂A

∂pi

∂B

∂qi

)(48)

The equation of motion for A can be written in terms of the Poisson Bracket,

dA

dt= [ A , H ]PB +

∂A

∂t(49)

22

From the definition, it can be seen that the Poisson Bracket is anti-symmetric

[ A , B ]PB = − [ B , A ]PB (50)

The Poisson Bracket of a quantity with itself is identically zero

[ A , A ]PB = 0 (51)

If we apply this to the Hamiltonian we find

dH

dt= [ H , H ]PB +

∂H

∂tdH

dt=

∂H

∂t(52)

Thus, if the Hamiltonian doesn’t explicitly depend on time, the Hamiltonian isconstant. That is, the energy is conserved.

Likewise, if A doesn’t explicitly depend on time and if the Poisson Bracketbetween H and A is zero,

[ A , H ]PB = 0 (53)

one finds that A is also a constant of motion

dA

dt= [ A , H ]PB

= 0 (54)

Another important Poisson Bracket relation is the Poisson Bracket of thecanonically conjugate coordinates and momenta, which is given by

[ pj , qj′ ]PB =∑

i

(∂pj

∂qi

∂qj′

∂pi− ∂pj

∂pi

∂qj′

∂qi

)= 0 −

∑i

δi,j δi,j′

= − δj,j′ (55)

The first term is zero as q and p are independent. The last term involves theKronecker delta function. The Kronecker delta function is given by

δi,j = 1 if i = j

δi,j = 0 if i 6= j (56)

and is zero unless i = j, where it is unity. Thus, the Poisson Bracket betweena generalized coordinate and its conjugate generalized momentum is − 1,

[ pj , qj′ ]PB = − δj,j′ (57)

23

while the Poisson Bracket between a coordinate and the momentum conjugateto a different coordinate is zero. We can also show that

[ pj , pj′ ]PB = [ qj , qj′ ]PB = 0 (58)

These Poisson Brackets shall play an important role in quantum mechanics,and are related to the commutation relations of canonically conjugate coordi-nate and momentum operators.

24

1.3 A Charged Particle in an Electromagnetic Field

In the classical approximation, a particle of charge q in an electromagnetic fieldrepresented by E(r, t) and B(r, t) is subjected to a Lorentz force

F = q

(E(r, t) +

1cr ∧ B(r, t)

)(59)

The Lorentz force acts as a definition of the electric and magnetic fields, E(r, t)and B(r, t) respectively. In classical mechanics, the fields are observable throughthe forces they exert on a charged particle.

In general, quantum mechanics is couched in the language of potentials in-stead of forces, therefore, we shall be replacing the electromagnetic fields bythe scalar and vector potentials. The electromagnetic fields are solutions ofMaxwell’s equations which not only express the fields in terms of the sourcesand but also form consistency conditions. The scalar and vector potentialssimplify Maxwell’s equations since they automatically satisfy the consistencyconditions. However, as their definitions specify that they are solutions of first-order partial differential equations, they are not unique. Despite the ambiguityin the potentials, the physical results that can be derived from them are unique.

1.3.1 The Electromagnetic Field

The electromagnetic field satisfies Maxwell’s eqns.,

∇ . B(r, t) = 0

∇ ∧ E(r, t) +1c

∂

∂tB(r, t) = 0

∇ . E(r, t) = ρ(r, t)

∇ ∧ B(r, t) − 1c

∂

∂tE(r, t) =

1cj(r, t) (60)

where ρ(r, t) and j(r, t) are the charge and current densities. The last twoequations describe the relation between the fields and the sources. The first twoequations are the source free equations and are automatically satisfied if oneintroduced a scalar φ(r, t) and a vector potential A(r, t) such that

B(r, t) = ∇ ∧A(r, t)

E(r, t) = − ∇ φ(r, t) − 1c

∂

∂tA(r, t) (61)

In classical mechanics, the electric and magnetic induction fields, E and B,are regarded as the physically measurable fields, and the scalar φ(r, t) and avector potential A(r, t) are not physically measurable. There is an arbitrarinessin the values of the potentials φ(r, t) and A(r, t) as they are defined to be thesolutions of differential equations which relate them to the physically measurable

25

E(r, t) and B(r, t) fields. This arbitrariness is formalized in the concept of agauge transformation, which means that the potentials are not unique and if onereplaces the potentials by new values which involve derivatives of any arbitraryscalar function Λ(r, t)

φ(r, t) → φ(r, t) − 1c

∂

∂tΛ(r, t)

A(r, t) → A(r, t) + ∇ Λ(r, t) (62)

the physical fields, E(r, t) and B(r, t), remain the same. This transformationis called a gauge transformation. Although the laws of physics are formulatedin terms of the gauge fields φ(r, t) and A(r, t), the physical results are gauge-invariant.

1.3.2 The Lagrangian for a Classical Charged Particle

The Lagrangian for a classical particle in an electromagnetic field is expressedas

L = − m c2

√ (1 − r2

c2

)− q φ(r, t) +

q

cA(r, t) . r (63)

The canonical momentum now involves a component originating from the fieldas well as the mechanical momentum

p =m r√ (

1 − r2

c2

) +q

cA(r, t) (64)

The Lagrangian equations of motion for the classical particle are

d

dt

(m r

1√1 − r2

c2

+q

cA(r, t)

)= − q ∇ φ(r, t) +

q

c∇(r . A(r, t)

)(65)

where the time derivative is a total derivative. The total derivative of the vectorpotential term is written as

d

dtA(r, t) =

∂

∂tA(r, t) + r . ∇ A(r, t) (66)

as it relates the change of vector potential experienced by a moving particle.The change of vector potential may occur due to an explicit time dependenceof A(r, t) at a fixed position, or may occur due to the particle moving to a newposition in a non-uniform field A(r, t).

——————————————————————————————————

26

1.3.3 Exercise 4

Show that these equations reduce to the relativistic version of the equations ofmotion with the Lorentz force law

d

dt

(m r

1√1 − r2

c2

)= q

(E(r, t) +

1cr ∧ B(r, t)

)(67)

——————————————————————————————————

1.3.4 Solution 4

The Euler-Lagrange equation is of the form

d

dt

(m r

1√1 − r2

c2

+q

cA(r, t)

)= − q ∇ φ(r, t) +

q

c∇(r . A(r, t)

)(68)

On substituting the expression for the total derivative of the vector potential

d

dtA(r, t) =

∂

∂tA(r, t) + r . ∇ A(r, t) (69)

one obtains the equation

d

dt

(m r

1√1 − r2

c2

)= − q ∇ φ(r, t) − q

c

∂

∂tA(r, t)

+q

c∇(r . A(r, t)

)− q

c

(r . ∇

)A(r, t) (70)

The last two terms can be combined to yield

d

dt

(m r

1√1 − r2

c2

)= − q ∇ φ(r, t) − q

c

∂

∂tA(r, t)

+q

cr ∧

(∇ ∧ A(r, t)

)(71)

The first two terms on the right hand side are identifiable as the expression forthe electric field E(r, t) and the last term involving the curl A(r, t) is recognizedas involving the magnetic induction field B(r, t) and, therefore, comprises themagnetic component of the Lorentz Force Law.

——————————————————————————————————

27

Despite the fact that the Lagrangian of a charged particle depends on thegauge fields and, therefore, changes form under a gauge transformation, theclassical equations of motion are gauge invariant. The equations of motion aregauge invariant since they only dependent on the electromagnetic fields E(r, t)and B(r, t). The gauge invariance of the equations of motion can be seen in adifferent way, that is by directly applying a gauge transformation to the actionS. Under a gauge transformation

φ(r, t) → φ(r, t) − 1c

∂

∂tΛ(r, t)

A(r, t) → A(r, t) + ∇ Λ(r, t) (72)

the Lagrangian L of a particle with charge q

L =m

2r2 − q φ(r, t) +

q

cA(r, t) . r (73)

transforms to

L → L′ = L +q

c

(∂

∂tΛ(r, t) + ∇ Λ(r, t) . r

)(74)

However, the term proportional to q is recognized as being a total derivativewith respect to time. That is, if one considers r to have the time-dependencer(t) of any trajectory connecting the initial and final position of the particle,then

L → L′ = L +q

c

d

dtΛ(r(t), t) (75)

where the derivative is evaluated over a trajectory of the particle. Hence, undera gauge transformation, the action changes from S to S′, where

S′ =∫ tf

ti

L(r, r) +q

c

∫ tf

ti

dtd

dtΛ(r(t), t)

= S +q

c

(Λ(rf , tf ) − Λ(ri, ti)

)(76)

which only depends on the end-points and not on the path that was chosen.Thus, the gauge transformation only adds a constant term to the action andthe trajectory identified by extremal principle is unaffected by the gauge trans-formation.

1.3.5 The Hamiltonian of a Classical Charged Particle

The Hamiltonian for a charged particle is found from

H = p . r − L (77)

28

which, with the relation between the momentum and the mechanical momen-tum,

p − q

cA(r, t) = m r

1√1 − r2

c2

(78)

leads to

H =

√(c2(p − q

cA(r, t)

)2

+ m2 c4)

+ q φ(r, t) (79)

The presence of the electromagnetic field results in the two replacements

p → p − q

cA(r, t)

H → H − q φ(r, t) (80)

An electromagnetic field is often incorporated in a Hamiltonian describing freecharged particles through these two replacements. These replacements retainrelativistic invariance as both the pairs E and p and φ(r, t) and A(r, t) formfour vectors. This procedure of including an electromagnetic field is based onwhat is called the minimal coupling assumption. The non-relativistic limit ofthe Hamiltonian is found by expanding the square root in powers of p2

m2 c2 andneglecting the rest mass energy m c2 results in the expression

H =1

2 m

(p − q

cA(r, t)

)2

+ q φ(r, t) (81)

which forms the basis of the Hamiltonian used in the Schrodinger equation.

——————————————————————————————————

1.3.6 Exercise 5

Derive the Hamiltonian for a charged particle in an electromagnetic field.

——————————————————————————————————

1.3.7 Solution 5

The Lagrangian is given by

L = − m c2

√1 − r2

c2− q φ(r, t) +

q

cr . A(r, t) (82)

so the generalized momentum p is given by

p − q

cA(r, t) =

m r√1 − r2

c2

(83)

29

and so on inverting this one finds

r

c=

(p − q

c A(r, t))

√m2 c2 +

(p − q

c A(r, t))2

(84)

and

1 − r

c

2

=m2 c2

m2 c2 +(p − q

c A(r, t))2 (85)

The Hamiltonian is then found from the Legendre transformation

H = p . r − L

=(p − q

cA(r, t)

). r + m c2

√1 − r2

c2+ q φ(r, t)

= c

√m2 c2 +

(p − q

cA(r, t)

)2

+ q φ(r, t) (86)

——————————————————————————————————

On expanding the quadratic kinetic energy term, one finds the Hamilto-nian has the form of a sum of the unperturbed Hamiltonian and an interactionHamiltonian Hint,

H =(

p2

2 m+ q φ(r, t)

)+ Hint (87)

The interaction Hint couples the particle to the vector potential

Hint = − q

2 m c

(p . A(r, t) + A(r, t) . p

)+

q2

2 m c2A2(r, t) (88)

where the first term linear in A is the paramagnetic coupling and the last termquadratic in A2 is known as the diamagnetic interaction. For a uniform staticmagnetic field B(r, t) = B, one possible form of the vector potential is

A(r) = − 12r ∧ B (89)

The interaction Hamiltonian has the form

Hint = +q

2 m c

( (r ∧ B

). p

)+

q2

8 m c2

(r ∧ B

)2

(90)

30

The first term can be written as the ordinary Zeeman interaction between theorbital magnetic moment and the magnetic field

Hint = − q

2 m c

(B . L

)+

q2

8 m c2

(r ∧ B

)2

(91)

where the orbital magnetic moment M is related to the orbital angular momen-tum via

M = +q

2 m cL (92)

Hence, the ordinary Zeeman interaction has the form of a dipole interaction

Lq

M = q / (2mc) L

Figure 6: A particle with charge q and angular momentum L, posseses a mag-netic dipole moment M given by M = + q

2 m c L.

with the magnetic fieldHZeeman = − M . B (93)

which has the tendency of aligning the magnetic moment parallel to the field.

Elementary particles such as electrons have another form of magnetic mo-ment and angular momentum which is intrinsic to the particle, and is not con-nected to any physical motion of the particle. The intrinsic angular momentumof elementary particles is known as spin.

There are two general approaches that can be taken to Quantum Mechanics.One is the path integral approach which was first developed by Dirac and then

31

popularized by Feynmann. This approach is based on the use of the Lagrangianformulation of classical mechanics. The other approach which is more common,and is the one that we shall follow exclusively, is based on the Hamiltonianformulation of classical mechanics.

32

2 Failure of Classical Mechanics

Classical Mechanics fails to correctly describe some physical phenomena. Thisfirst became apparent at the atomic level. Historically, the failure of classicalmechanics was first manifested after Rutherford’s discovery of the structure ofthe atom. Classically, an electron orbiting around a charged nucleus shouldcontinuously radiate energy according to Maxwell’s Electromagnetic Theory.The radiation leads to the electron experiencing a loss of energy, and therebycontinuously reducing the radius of the electrons orbit. Thus, the atom be-comes unstable as the electron spirals into the nucleus. However, it is a wellestablished experimental fact that atoms are stable and that the atomic energylevels have discrete values for the energy. The quantization of the energy levelsis seen through the Franck-Hertz experiment, which involves inelastic collisionsbetween atoms. Other experimental evidence for the quantization of atomic en-ergy levels is given by the emission and absorption of electromagnetic radiation.For example, the series of dark lines seen in the transmitted spectrum whenlight, with a continuous spectrum of wavelengths, falls incident on hydrogengas is evidence that the excitation spectrum consists of discrete energies. NielsBohr discovered that the excitation energy ∆E and the angular frequency ofthe light ω are related via

∆E = h ω (94)

The quantity h is known as Planck’s constant and has the value of

h = 1.0545 × 10−34 J s

= 0.65829 × 10−15 eV s (95)

The Balmer, Lyman and Paschen series of electromagnetic absorption by hy-drogen atoms establishes that ∆E takes on discrete values.

Another step in the development of quantum mechanics occurred when Louisde Broglie postulated wave particle duality, namely that entities which have theattributes of particles also posses attributes of waves. This is formalized by therelationships

E = h ω

p = h k (96)

The de Broglie relations were verified by Davisson and Germer in their exper-iments in which a beam of electrons were placed incident on the surface of acrystalline solid, and the reflected beam showed a diffraction pattern indicativeof the fact that the electrons have a wave length λ. The diffraction conditionrelates the angle of the diffracted beam to the ratio of the separation betweenthe planes of atoms and the wavelength λ. Furthermore, the diffraction condi-tion showed a variation with the particle’s energies which is consistent with thewavelength momentum relation.

33

2.1 Semi-Classical Quantization

The first insight into quantum phenomena came from Niels Bohr, who imposedan additional condition on Classical Mechanics1. This condition, when imposedon systems where particles undergo periodic orbits, reduces the continuous val-ues of allowed energies to a set of discrete energies. This semi-classical quanti-zation condition is written as ∮

pi dqi = ni h (97)

where pi and qi are the canonically conjugate momentum and coordinatesof Lagrangian or Hamiltonian Mechanics, and ni is an integer from the set(0,1,2,3,4,. . . ,∞) and h is a universal constant. The integration is over one pe-riod of the particle’s orbit in phase space2. The discrete values of the excitationenergies found for the hydrogen atom produces reasonably good agreement withthe excitation energies found for the absorption or emission of light from hydro-gen gas.

——————————————————————————————————

2.1.1 Exercise 6

Assuming that electrons move in circular orbits in the Coulomb potential dueto a positively charged nucleus, (of charge Z e ), find the allowed values of theenergy when the semi-classical quantization condition is imposed.

——————————————————————————————————

2.1.2 Solution 6

The Lagrangian L is given in terms of the kinetic energy T and the scalarelectrostatic potential V by L = T − V . Then in spherical polar coordinates(r, θ, ϕ), one finds the Lagrangian

L =m r2

2+

m r2 θ2

2+

m r2 sin2 θ ϕ2

2+

Z e2

r(98)

The Euler-Lagrange equation for the radius r is given by

∂L

∂r=

d

dt

(∂L

∂r

)1N. Bohr, Phil. Mag. 26, 1 (1913).2Einstein showed that this quantization rule can be applied to certain non-separable sys-

tems [A. Einstein, Deutsche Physikalische Gesellschaft Verhandlungen 19, 82 (1917)]. In anon-ergodic system, the accessible phase space defines a d-dimensional torus. The quantiza-tion condition can be applied to any of the d independent closed loops on the torus. Theseloops do not have to coincide with the system’s trajectory.

34

m r θ2 + m r sin2 θ ϕ2 − Z e2

r2= m r (99)

while the equation of motion for the polar angle θ is given by

∂L

∂θ=

d

dt

(∂L

∂θ

)m r2

2sin 2θ ϕ2 = m

d

dt

(r2 θ

)(100)

and finally we find

∂L

∂ϕ=

d

dt

(∂L

∂ϕ

)0 = m

d

dt

(r2 sin2 θ ϕ

)(101)

According to the assumption, the motion is circular, which we choose to be inthe equatorial plane (r = a, θ = π

2 ). Thus, one has

Z e2

a2= m a ϕ2

0 = m a2 ϕ (102)

Hence, we find that the angular velocity is a constant, ϕ = ω.

To impose the semi-classical quantization condition we need the canonicalmomentum. The canonical momenta are given

pr =∂L

∂r= 0

pθ =∂L

∂θ= 0

pϕ =∂L

∂ϕ= m r2 sin2 θ ϕ = m a2 ω (103)

The semi-classical quantization condition becomes∮pϕ dϕ = m a2 ω 2π = nϕh (104)

From the above one has the two equations

m a2 ω = nϕ h

Z e2

a2= m a ω2 (105)

On solving these for the Bohr radius a and angular frequency ω, one finds

ω =Z2 e4

n3ϕ h3

a =n2

ϕ h2

Z e2 m(106)

35

Substituting these equations in the expression for the energy E, one finds theexpression first found by Bohr

E =m a2 ω2

2− Z e2

a

E = − Z e2

2 a

E = − m Z2 e4

2 n2ϕ h2 (107)

This agrees with the exact (non-relativistic) quantum mechanical expression forthe energy levels of electrons bound to a H ion.

——————————————————————————————————

2.1.3 Exercise 7

Find the energy of a one-dimensional simple Harmonic oscillator, of mass m andfrequency ω, when the semi-classical quantization condition is imposed.

——————————————————————————————————

2.1.4 Solution 7

The Hamiltonian is given by

H(p, q) =p2

2m+

m ω2 q2

2(108)

Hamilton’s equations of motions are

p = − ∂H

∂q= − m ω2 q

q = +∂H

∂p=

p

m(109)

Thus, we have the equation of motion

q + ω2 q = 0 (110)

which has the solution in the form

q(t) = A sin(ωt+ φ)p(t) = m ω A cos(ωt+ φ) (111)

36

The semi-classical quantization condition becomes∮p(t) dq(t) = m ω A2

∫ 2π

0

dϕ cos2 ϕ

= m ω A2 2 π = n h (112)

Thus, we find A2 = 2 nhm ω where h = h

2π and the energy becomes

E = H(p, q) =m ω2 A2

2= n h ω (113)

This should be compared with the exact quantum mechanical result E =hω ( n + 1

2 ). The difference between these results become negligible forlarge n. That is for a fixed Energy E, the results approach each other in thelimit of large n or equivalently for small h. This is example is illustrative of thecorrespondence principle, which states that Quantum Mechanics, should closelyapproximate Classical Mechanics where Classical Mechanics is known to providean accurate description of nature, and that this in this limit h can be consideredto be small.

——————————————————————————————————

37

3 Principles of Quantum Mechanics

The Schrodinger approach to Quantum Mechanics is known as wave mechan-ics. The Schrodinger formulation is in terms of states of a system which arerepresented by complex functions defined in Euclidean space, Ψ(r) known aswave functions and measurements are represented by linear differential opera-tors. The Schrodinger approach, though most common is not unique. An alter-nate approach was pursued by Heisenberg, in which the state of the system arerepresented by column matrices and measurements are represented by squarematrices. This second approach is known as Heisenberg’s matrix mechanics.These two approaches were shown to be equivalent by Dirac, who developed anabstract formulation of Quantum Mechanics using an abstract representation ofstates and operators.

We shall first consider the system at a fixed time t, say t = 0. The wavefunction, Ψ(r), represents a state of a single particle at that instant of time,and has a probabilistic interpretation. Consider an ensemble of N identical andnon-interacting systems, each of which contains a measurement apparatus anda single particle. Each measurement apparatus has its own internal referenceframe with its own origin. A measurement of the position r (referenced tothe coordinate system attached to the measurement apparatus) is to be madeon each particle in the ensemble. Just before the measurements are made, eachparticle is in a state represented by Ψ(r). Measurements of the positions of eachparticle in the ensemble will result in a set of values of r that represents points inspace, referenced w.r.t. the internal coordinate system3. The probability that ameasurement of the position r of a particle will give a value in the infinitesimalvolume d3r containing the point r, is given by

P (r) d3r = | Ψ(r) |2 d3r (114)

Thus, P (r) d3r is the probability of finding the particle in the infinitesimalvolume d3r located at r. The probability is directly proportional to the size ofthe volume d3r, and P (r) = | Ψ(r) |2 is the probability density. Since theparticle is somewhere in three-dimensional space, the probability is normalizedsuch that ∫

| Ψ(r) |2 d3r = 1 (115)

This normalization condition will have to be enforced on the wave function ifit is to represent a single-particle state. The normalization condition must betrue for all times, if the particle number is conserved4.

3Alternatively, instead of considering an ensemble of identical systems, one could considerperforming N successive measurements on a single system. However, before each successivemeasurement is made, one would have to reset the initial condition. That is, the systemshould be prepared so that, just before each measurement is made, the particle is in the statedescribed by Ψ(r).

4For well-behaved functions, the normalization condition implies that | Ψ(r) |2 vanishes as|r| → ∞.

38

Given two states, Ψ(r) and Φ(r), one can define an inner product or overlapmatrix element as the complex constant given by∫

d3r Φ∗(r) Ψ(r) (116)

where the integration runs over all volume of three-dimensional space. It shouldbe noted that inner product of two wave functions depends on the order thatthe wave functions are specified. If the inner product is taken in the oppositeorder, one finds∫

d3r Ψ∗(r) Φ(r) =( ∫

d3r Φ∗(r) Ψ(r))∗

(117)

which is the complex conjugate of the original inner product. The normalizationof the wave function Ψ(r) just consists of the inner product of the wave functionwith itself. The interpretation of the squared modulus of the wave function asa probability density requires that the normalization of a state is unity.

We should note that if all observable quantities for a state always involve thewave function times its complex conjugate, then the absolute phase of the wavefunction is not observable. Only phase differences are measurable. Therefore,it is always possible to transform a wave function by changing its phase

Ψ(r) → Ψ′(r) = exp[i Λ(r)h

]Ψ(r) (118)

where Λ(r) is an arbitrary real function.

39

3.1 The Principle of Linear Superposition

In quantum mechanics, a measurement of a physical quantity of a single particlewhich is in a unique state will result in a value of the measured quantity that isone of a set of possible results an. Repetition of the measurement on a particlein exactly the same initial state may yield other values of the measured quantity(such as am). The probability distribution for the various results an is governedby the particular state of the system that the measurement is being performedon.

This suggests that a state of a quantum mechanical system, at any instant,can be represented as a superposition of states corresponding to the differentpossible results of the measurement. Let Φn(r) be a state such that a mea-surement of A on the state will definitely give the result an. The simplest wayof making a superposition of states is by linearly adding multiples of the wavefunctions Φn(r) corresponding to the possible results.

Thus, the principle of linear superposition can be stated as

Ψ(r) =∑

n

Cn Φn(r) (119)

where the expansion coefficients Cn are complex numbers. The expansion coef-ficients can be determined from a knowledge of the wave function of the stateΨ(r) and the set of functions, Φn(r), representing the states in which a mea-surement of A is known to yield the result an. The expansion coefficients Cn

are related to the probability that the measurement on the state Ψ(r) results inthe value an.

The principle of linear superposition of the wave function can lead to inter-ference in the results of measurements. For example, the results of a measure-ment of the position of the particle r, leads to a probability distribution P (r)according to

P (r) d3r = | Ψ(r) |2 d3r

=∑n,m

C∗m Cn Φ∗m(r) Φn(r) d3r (120)

On isolating the terms with n = m, one finds terms in which the phases ofthe wave functions Φn(r) and the phases of the complex numbers Cn separatelycancel. The remaining terms, in which n 6= m, represent the interferenceterms. Thus

P (r) d3r =∑

n

| Cn |2 | Φn(r) |2 +∑n 6=m

C∗m Cn Φ∗m(r) Φn(r) d3r (121)

The coefficient | Cn |2 is the probability that the state represented by Ψ(r) isin the state n.

40

As an example, consider a state which is in a superposition of two states eachof which represents a state of definite momentum, p = h k and p = − h k.The forward and backward travelling states are

Φk(r) = Ck exp[

+ i k . r

]Φ−k(r) = C−k exp

[− i k . r

](122)

These states are not normalizable and, therefore, each state must representbeams of particles with definite momentum in which the particles are uniformlydistributed over all space. Since the integral

∫| Ψ(r) |2 d3r → ∞, the beam

must be considered to contain an infinite number of particles.

The probability densities or intensities of the two independent beams aregiven by

Pk(r) = | Φk(r) |2 = | Ck |2

P−k(r) = | Φ−k(r) |2 = | C−k |2 (123)

We shall assume that these beams have the same intensities, that is | C−k |2 =| C+k |2. Then, in this case the wave function can be expressed in terms of thephases of C± = | C | exp[ i δ± ]. When the beams are superimposed, the stateis described by the wave function

Ψ(r) = | C | exp[i

(δ+ + δ−)2

]×

(exp

[+ i (k . r +

(δ+ − δ−)2

)]

+ exp[− i (k . r +

(δ+ − δ−)2

)])

(124)

This state is a linear superposition and has a probability density for finding theparticle at r given by P (r) where

P (r) = 4 | C |2 cos2(k . r +

(δ+ − δ−)2

)(125)

Thus, the backward and forward travelling beam interfere, the superpositiongives rise to consecutive planes of maxima and minima. The maxima are locatedat

k . r +(δ+ − δ−)

2= π n (126)

and the minima are located at

k . r +(δ+ − δ−)

2=

π

2( 2 n + 1 ) (127)

41

There is a strong analogy between Quantum Mechanics and Optics. Thewave function Ψ(r, t) plays the role corresponding to the electric field E(r, t).Both the wave function and the electric field obey the principle of linear super-position. The probability density of finding the particle at point r, | Ψ(r, t) |2plays the role of the intensity of light, I, which is proportional to | E(r, t) |2. Infact, the intensity of light is just proportional to the probability density of find-ing a photon at the point r. The phenomenon of interference occurs in QuantumMechanics and also in Optics. The analogy between Quantum Mechanics andOptics is not accidental, as Maxwell’s equations are intimately related to the”Schrodinger equation” for a massless particle with intrinsic spin S = 1.

42

3.2 Wave Packets

By combining momentum states with many different values of the momentum,one can obtain states in which the particle is essentially localized in a finitevolume. These localized states are defined to be wave packets. Wave packetsare the closest one can get to a classical state of a free particle, which has awell defined position and momentum. For a quantum mechanical wave packet,the distribution of results for measurements of the position and momentum aresharply peaked around the classical values. The wave function can be expressed

-1

-0.5

0

0.5

1

-3 -2 -1 0 1 2 3

x

Ψ(

Ψ(

Ψ(

Ψ(x

) )))

Re

Im

Figure 7: The real and imaginary parts of a wavefunction Ψ(x) representing aGaussian wave packet in one dimension.

as a Fourier transform

Ψ(r) =(

12 π

) 32∫

d3k Φ(k) exp[

+ i k . r

](128)

where Φ(k) is related to the momentum probability distribution function5. Sincethe exponential factor represents states of different momenta, this relation is anexample of how a wave function can be expanded in terms of states correspond-ing to the various possible results of a physical measurement. The momentum

5This is an example of the principle of linear superposition, in which the discrete variablen has been replaced by a Riemann sum over the continuous variable k and the expansioncoefficients Cn are proportional to Φ(k).

43

probability distribution function is related to the wave vector probability dis-tribution function Pk(k), defined by

Pk(k) d3k = | Φ(k) |2 d3k (129)

The function Φ(k) can be determined from knowledge of Ψ(r) from the inverserelation

Φ(k) =(

12 π

) 32∫

d3r Ψ(r) exp[− i k . r

](130)

These results are from the mathematical theory of Fourier Transformations.The consistency of the Fourier Transform with the Inverse Fourier Transformcan be seen by combining them via

Ψ(r) =(

12 π

) 32∫

d3k Φ(k) exp[

+ i k . r

]=

(1

2 π

)3 ∫d3k

∫d3r′ Ψ(r′) exp

[i k . ( r − r′ )

](131)

together with the representation of the three-dimensional Dirac delta function

δ3( r − r′ ) =(

12 π

)3 ∫d3k exp

[i k . ( r − r′ )

](132)

On inserting the integral representation of the Dirac delta function eqn(132)into eqn(131), one finds an equation which is the formal definition of the Diracdelta function

Ψ(r) =∫

d3r′ Ψ(r′) δ3( r − r′ ) (133)

That equation (132) provides a representation of the three-dimensional Diracdelta function can by seen directly by factorizing it into the product of threeindependent one-dimensional delta functions as

δ3( r − r′ ) = δ( x − x′ ) δ( y − y′ ) δ( z − z′ ) (134)

and then comparing with the right hand side which can also be factorizedinto three independent one-dimensional integrals. Each factor in the three-dimensional delta function of eqn(132) can be replaced by the representationof the one-dimensional delta function as a limit of a sequence of Lorentzianfunctions of width ε,

δ( x − x′ ) = limε → 0

1π

ε

( x − x′ )2 + ε2(135)

The sequence of function is shown in fig(8). Then on evaluating the integrationsover each of the one-dimensional variables in the right hand side of eqn(132)

44

Dirac delta dunction

0

0.5

1

1.5

2

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

x

δ ε(x

)

δε(x) = 1/π ε/(ε2+x2)

lim ε → 0

Figure 8: The Dirac delta function δ(x). The Dirac delta function is defined asthe limit, ε → 0 of a sequence of functions δε(x).

and using the replacement ( x − x′ ) → ( x− x′ ) ± i ε needed to keep theintegral convergent, one finds that each of the three factors have the same form

=1

2 πlim

L→∞

∫ +L

−L

dk exp[i k ( x − x′ )

]=

12 π

limL→∞

(∫ +L

0

dk exp[i k ( x − x′ ) − ε k

]+∫ 0

−L

dk exp[i k ( x − x′ ) + ε k

] )=

12 π

(− 1

i ( x − x′ ) − ε+

1i ( x − x′ ) + ε

)= lim

ε → 0

1π

ε

( x − x′ )2 + ε2(136)

which equals the corresponding representation of the delta function on the lefthand side. This completes the identification, and proves that the inverse Fouriertransform of the Fourier transform is the original function. It has also provedthat any square integrable function, i.e. normalizable function, can be expandedas a sum of momentum eigenfunctions. The momentum eigenstates interfere de-structively almost everywhere, except at the position where the wave packet is

45

peaked.

In general, the d-dimensional momentum distribution is related to the dis-tribution of k vectors, | Φ(k) |2 by the relation

Pp(p) =∫

ddk δd( p − h k )∣∣∣∣ Φ(k)

∣∣∣∣2=

1hd

∣∣∣∣ Φ(p

h

) ∣∣∣∣2 (137)

In future, we shall find it convenient to include factors of h−d2 into the definition

of the d-dimensional Fourier transform so that, on squaring the modulus, theygive the properly normalized momentum distribution function.

——————————————————————————————————

3.2.1 Exercise 8

Given a wave function Ψ(x) where

Ψ(x) = C exp[− ( x − x0 )2

4 δx2

]exp

[+ i k0 x

](138)

determine C, up to an arbitrary phase factor, and determine Φ(kx). Later, weshall see that | Φ(kx) |2 is proportional to the probability distribution of the xcomponent of momentum of the system. In three dimensions | Φ(k) |2 d3k isthe probability of finding the system with a wave vector k in an infinitesimalvolume d3k around the point k.

——————————————————————————————————

3.2.2 Solution 8

The magnitude of the coefficient C is determined from the normalization con-dition ∫ + ∞

− ∞dx | Ψ(x) |2 = 1 (139)

which is evaluated as

| C |2∫ + ∞

− ∞dx exp

[− ( x − x0 )2

2 δx2

]= 1 (140)

On changing the variable of integration from x to y where

y =x − x0√

2 δx(141)

46

the normalization condition becomes

1 = | C |2√

2 δx∫ + ∞

− ∞dy exp

[− y2

]1 = | C |2

√2 π δx (142)

Hence, the magnitude of C is found as

| C | =1

(2 π)14 δx

12

(143)

Real-space distribution function

0

0.1

0.2

0.3

0.4

0.5

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

x

l ψ(x

) l2

x0

Figure 9: The real space distribution function | Ψ(x) |2. The distribution func-tion is centered around x0.

The k-space wave function is found from

Φ(kx) =1√2 π

∫ + ∞

− ∞dx Ψ(x) exp[ − i kx x ]

=1

(2 π)34 δx

12

∫ + ∞

− ∞dx exp

[− ( x − x0 )2

4 δx2

]exp[ + i k0 x ] exp[ − i kx x ]

(144)

Combining the exponential factors and completing the square by changing vari-able from x to z where

z = x − x0 + 2 i ( kx − k0 ) δx2 (145)

47

one finds that

Φ(kx) =1

(2 π)34 δx

12

∫ + ∞

− ∞dz exp

[− ( z )2

4 δx2

]× exp

[− ( k0 − kx )2 δx2

]exp

[− i ( kx − k0 ) x0

]= δx

12

(2π

) 14

exp[− ( k0 − kx )2 δx2

]exp

[− i ( kx − k0 ) x0

](146)

Thus, we see that in kx space the modulus squared wave function | Φ(kx) |2 iscentered around k0 and has a k width which is proportional to δx−1.

Momentum-space distribution function

0

0.2

0.4

0.6

0.8

1

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

p

l Φ(p

) l2

p0

Figure 10: The momentum space distribution function | Φ(p) |2. The momen-tum distribution is centered about p0.

——————————————————————————————————

3.2.3 Exercise 9

Given the wave function

Ψ(x) = C exp[− λ | x |

](147)

48

where λ is a positive real number, find C and the properly normalized momen-tum distribution function.

Real-space distribution function

0

0.2

0.4

0.6

0.8

1

1.2

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

x

l ψ(x

) l2

Figure 11: The real space distribution function | Ψ(x) |2.

——————————————————————————————————

3.2.4 Solution 9

The normalization condition is given by∫ + ∞

− ∞dx | Ψ(x) |2 = 1 (148)

which is evaluated as

| C |2∫ + ∞

− ∞dx exp

[− 2 λ | x |

]= 1 (149)

The integration can be broken up into two parts, one over the range ( − ∞ , 0 )and the second over the range ( 0 , + ∞ ). On replacing | x | by ± x in theappropriate interval, we have

1 = | C |2( ∫ 0

− ∞dx exp

[+ 2 λ x

]+

∫ + ∞

0

dx exp[− 2 λ x

] )

49

1 = | C |2(

12 λ

+1

2 λ

)(150)

Thus, the normalization is given by

| C | =√λ (151)

Momentum-space distribution function

0

0.2

0.4

0.6

0.8

1

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

p

l Φ(p

) l2

Figure 12: The momentum space distribution function | Φ(p) |2.

The k-space wave function is given by

Φ(kx) =1√2 π

∫ + ∞

− ∞dx Ψ(x) exp

[− i kx x

]=

√λ

2 π

∫ + ∞

− ∞dx exp

[− λ | x |

]exp

[− i kx x

](152)

This is evaluated, as before, by breaking up the integral into two intervals. Thefirst interval is an integration over the range ( − ∞ , 0 ) and the second intervalhas x in the range ( 0 , + ∞ ). Thus, we find

Φ(kx) =

√λ

2 π

(1

λ − i kx+

1λ + i kx

)

50

=

√λ

2 π2 λ

λ2 + k2x

(153)

The width of the kx distribution is proportional to λ. The momentum distribu-tion Pp(px) is given by

Pp(px) dpx =∣∣∣∣ Φ(px

h

) ∣∣∣∣2 dpx

h(154)

Thus, we find that the momentum distribution function is given by

Pp(px) =2π

h3 λ3

( h2 λ2 + p2x )2

(155)

which is properly normalized to unity.

——————————————————————————————————

51

3.3 Probability, Mean and Deviations

In Quantum Mechanics one often deals with systems in a well defined state,however, the result of a particular measurement is indeterminate. That is, usu-ally, the result of the measurement cannot be predicted with absolute certainty.Therefore, we shall review some aspects of probabilities and averages.

Sequential measurements of A

0

0.5

1

1.5

2

0 5 10 15 20

i

Ai

< A >

∆A9

∆A4 ∆Arms

N=20

Figure 13: The set of results of N sequential measurements of A. The averagevalue < A > and the magnitude of the root mean squared deviation ∆Arms

of this set of measurements are indicated in red. Also shown are ∆Ai whichrepresent the deviations of some individual data points from the average value.

The average or mean value of a measured quantity A, denoted by A, isexpressed as an integral (and/or a sum) over all the possible values of themeasurement A, weighted with the probability distribution function P (A) via

A =∫

dA P (A) A (156)

The integration runs over all possible (real) values of A. The higher momentsAm have averages, Am, which are given by

Am =∫

dA P (A) Am (157)

52

Probability of measurements of A

0

0.05

0.1

0.15

0.2

0.1 0.3 0.5 0.7 0.9 1.1 1.3 1.5 1.7 1.9

A

P(A

)N →∞

Figure 14: The probability distribution P (A) is defined as the relative frequencyof the occurrence of result A, in the limit N → ∞ where N is the number ofmeasurements.

The deviation of the variable A from its mean value is defined by

∆A = A − A (158)

The average value of ∆A is zero, as can be seen from

∆A = A − A = 0 (159)

which just reflects the fact that the variable A fluctuates equally on both sidesof the average value A. A measure of the size of the fluctuations can be foundfrom the mean squared deviation ∆A2 via

∆A2 =∫dA P (A) ( A − A )2

=∫dA P (A) ( A2 − 2 A A + A

2)

= A2 − A2

(160)

The mean squared deviation is non-zero as ∆A2 is always positive definite. Thevariance or root mean squared (r.m.s.) deviation ∆Arms provides a measure ofthe fluctuations in A.

∆Arms =√

∆A2

53

=√

( A − A )2

=√

( A2 − A2

) (161)

The r.m.s. deviation is a measure of the uncertainty in the value of A.

From the examples of wave packets given in exercise 4, it can be found thatthe uncertainty in the i-th component of the momentum and the i-th componentof the particle’s position satisfies the inequality

∆ri rms ∆pi rms

≥ h

2(162)

This is an example of the Heisenberg uncertainty relation, applied to the vari-ables r and p.

——————————————————————————————————

3.3.1 Exercise 10

Given P (x) where

P (x) =(

12 π

) 12

δx−1 exp[− ( x − x0 )2

2 δx2

](163)

find x, x2, ∆x2 and ∆x4.

——————————————————————————————————

3.3.2 Solution 10

The average value of x, denoted as x is given by

x =∫ + ∞

− ∞dx P (x) x

=∫ + ∞

− ∞dx | Ψ(x) |2 x

=1√2 π

∫ + ∞

− ∞dx δx−1 exp

[− ( x − x0 )2

2 δx2

]x

(164)

Changing variable to y = x − x0, one has

x =1√2 π

∫ + ∞

− ∞dy δx−1 exp

[− y2

2 δx2

]( y + x0 )

(165)

54

The term linear in y is odd and, therefore, vanishes. The term proportional tox0 factors out of the integral to yield

x = x01√2 π

∫ + ∞

− ∞dy δx−1 exp

[− y2

2 δx2

]= x0

∫ + ∞

− ∞dx P (x)

= x0 (166)

as P (x) is normalized to unity. Thus, the average value of x is x0.

The average value of x2 is given by

x2 =∫ + ∞

− ∞dx P (x) x2

=∫ + ∞

− ∞dx | Ψ(x) |2 x2

=1√2 π

∫ + ∞

− ∞dx δx−1 exp

[− ( x − x0 )2

2 δx2

]x2

(167)

Changing variable to y = x − x0 one has

x2 =1√2 π

∫ + ∞

− ∞dy δx−1 exp

[− y2

2 δx2

]( y2 + 2 y x0 + x2

0 )

(168)

The term linear in y is odd and, therefore, vanishes. The term proportional tox2

0 factors out of the integral to yield

x2 =1√2 π

∫ + ∞

− ∞dy δx−1 y2 exp

[− y2

2 δx2

]+

+ x20

1√2 π

∫ + ∞

− ∞dy δx−1 exp

[− y2

2 δx2

]=

1√2 π

∫ + ∞

− ∞dy δx−1 y2 exp

[− y2

2 δx2

]+

+ x20

∫ + ∞

− ∞dx P (x)

= x20 +

1√2 π

∫ + ∞

− ∞dy δx−1 y2 exp

[− y2

2 δx2

](169)

as P (x) is normalized to unity. Since we have

1√π

∫ + ∞

− ∞dy exp

[− α y2

]=

1√α

(170)

55

then by differentiating with respect to α we find

1√π

∫ + ∞

− ∞dy y2 exp

[− α y2

]= − ∂

∂α

1√π

∫ + ∞

− ∞dy exp

[− α y2

]= − ∂

∂α

1√α

= +12

1√α3

(171)

Hencex2 = x2

0 + δx2 (172)

Thus, the average value of x2 is x20 + δx2.

The mean squared deviation is then found as

∆x2rms = ∆x2 =

∫ + ∞

− ∞dx P (x) ( x − x0 )2

=1√2 π

∫ + ∞

− ∞dy δx−1 y2 exp

[− y2

2 δx2

](173)

The integration was previously evaluated, and yields

∆x2rms = δx2 (174)

Finally, we can show that

∆x4 =∫ + ∞

− ∞dx P (x) ( x − x0 )4

=1√2 π

∫ + ∞

− ∞dy δx−1 y4 exp

[− y2

2 δx2

](175)

The integration can be performed by taking the second derivative with respectto α of

1√π

∫ + ∞

− ∞dy exp

[− α y2

]=

1√α

(176)

Then, by differentiating with respect to α, we find

1√π

∫ + ∞

− ∞dy y4 exp

[− α y2

]= +

∂2

∂α2

1√π

∫ + ∞

− ∞dy exp

[− α y2

]= +

∂2

∂α2

1√α

= +34

1√α5

(177)

56

Thus, we have

∆x4 =∫ + ∞

− ∞dx P (x) ( x − x0 )4

=1√2 π

∫ + ∞

− ∞dy δx−1 y4 exp

[− y2

2 δx2

]= 3 δx4 (178)

——————————————————————————————————

3.3.3 Exercise 11

Find the r.m.s. deviation in the particle’s position ∆xrms and momentum∆px rms for the wave packet

Ψ(x) =(

12 π

) 14

δx−12 exp

[− ( x − x0 )2

4 δx2

]exp

[+ i k0 x

](179)

——————————————————————————————————

3.3.4 Solution 11

From the results of Exercise 10 we immediately find that ∆xrms = δx.

The kx space wave function is given by

Φ(kx) = δx12

(2π

) 14

exp[− ( k0 − kx )2 δx2

]exp

[− i ( kx − k0 ) x0

](180)

Thus, with px = h kx one has the momentum probability distribution P (px)given by

P (px) dpx =∣∣∣∣ Φ(px

h

) ∣∣∣∣2 dpx

h

=(

2π

) 12

exp[− 2 ( k0 − kx )2 δx2

]δx

hdpx (181)

Hence, this is a Gaussian distribution centered at px = h k0, so the averagevalue of the momentum is h k0. The width of the Gaussian is = h

2 δx , so

∆px rms ∆xrms =h

2(182)

57

which is an example of the Heisenberg uncertainty principle in which the equal-ity holds.

——————————————————————————————————

3.3.5 Exercise 12

Given that a particle moving in one dimension is in a state given by(1

2 L

) 12

if | x | < L

Ψ(x) =0 otherwise (183)

find the probability that it is found in a momentum eigenstate, and verify that

0

0.2

0.4

0.6

0.8

-2 -1 0 1 2

x/L

Ψ ΨΨΨ(x

)

Figure 15: The wavefunction Ψ(x) given in Exercise 12.

the probability distribution is properly normalized.

——————————————————————————————————

58

3.3.6 Solution 12

The real space wave function is expressed in terms of the momentum space wavefunction via

Ψ(x) =∫

dp√2 π h

exp[

+ ip x

h

]Φ(p) (184)

and the inverse relation is

Φ(p) =∫

dx√2 π h

exp[− i

p x

h

]Ψ(x) (185)

The properly normalized momentum space wave function Φ(p) is evaluated as

Φ(p) =∫ + L

− L

dx√2 π h

1√2 L

exp[− i

p x

h

]=

1√4 π L h

∫ + L

− L

dx exp[− i

p x

h

]

=

√L

π h

sin(

p Lh

)p Lh

(186)

The normalized momentum probability density P (p) is given by the modulussquared of the momentum space wave function

P (p) =∣∣∣∣ Φ(p)

∣∣∣∣2

=L

π h

sin2

(p Lh

)(

p Lh

)2 (187)

The distribution P (p) is properly normalized since∫ + ∞

− ∞dx

sin2 x

x2= π (188)

——————————————————————————————————

59

Momentum Probability Distribution

0

0.1

0.2

0.3

0.4

-8 -6 -4 -2 0 2 4 6 8

pL/!

P(p

) ! /

L

Figure 16: The momentum probability distribution function P (p) for the Ψ(x)given in Exercise 12.

3.4 Operators and Measurements

In Schrodinger’s wave mechanics, a physical or measurable quantity is repre-sented by a (linear) differential operator A. An operator A is defined by itsaction on the states represented by arbitrary wave functions Ψ(r). The opera-tor A transforms the state Ψ(r) into another state Φ(r) via

A Ψ(r) = Φ(r) (189)

A measurement on a quantum mechanical system generally disturbs the sys-tem thereby causing a change in the state of the system. As the measurementresults in a change in the state of a system, the measurement is represented byan operator.

A linear operator A is defined by its action on a state Ψ(r) that is a linearsuperposition,

Ψ(r) =∑

n

Cn Φn(r) (190)

An operator A acting on a state Ψ(r) formed as a linear superposition is defined

60

to be a linear operator if, and only if, it produces the result

A∑

n

Cn Φn(r) =∑

n

Cn A Φn(r) (191)

That is, the resulting state must be equivalent to the superposition of the statesformed by A acting on the components Φn(r).

The position operator r is given by the vector r. Its operation on the stateΨ(r) is just that of multiplication by r.

r Ψ(r) = r Ψ(r) (192)

The momentum operator p is given by − i h ∇. The operator ∇ is given inCartesian coordinates as

∇ = ex∂

∂x+ ey

∂

∂x+ ez

∂

∂z(193)

where ex, ey and ez are the three orthogonal unit vectors used to define aCartesian coordinate system. The action of the momentum operator on a stateΨ(r) is given by

p Ψ(r) = − i h ∇Ψ(r) (194)

The kinetic energy operator is given by T = − h2

2m ∇2. Its action on astate Ψ(r) is given by

T Ψ(r) = − h2

2m∇2Ψ(r) (195)

where the Laplacian is given in Cartesian coordinates as

∇2 =∂2

∂x2+

∂2

∂y2+

∂2

∂z2(196)

The operators have no mathematical meaning except when they act upon awave function.

3.4.1 Operator Equations

Just as in Hamiltonian Mechanics, the operators corresponding to physicallymeasurable quantities can be expressed in terms of the momentum and positionoperators. This requires compounding operators, via addition and multiplica-tion. The results of compounding operators can lead to operator equations.Operator equations, if valid, must be true if the same expression results whenthe operators on both side of the equality sign act on the same arbitrary wavefunction. Usually in writing an operator equation the wave function is sup-pressed, but it is implicitly assumed to be present.

61

3.4.2 Operator Addition

Given two differential operators A and B, each defined by their action on everywave function Ψ(r) by

A Ψ(r) = ΦA(r)B Ψ(r) = ΦB(r) (197)

where the final wave functions ΦX(r) are dependent on the particular choice ofthe wave function Ψ(r). The sum of two operators is defined as ( A + B ) andis given by the linear sum

( A + B ) Ψ(r) = ΦA(r) + ΦB(r)= A Ψ(r) + B Ψ(r) (198)

An example is given by the differential operator

∂2

∂x2+ 3

∂

∂x+ x4 (199)

Another example is given by the kinetic energy operator for a particle inCartesian coordinates

T = − h2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)(200)

where the kinetic energies for motion along the x, y and z axis add.

3.4.3 Operator Multiplication

Given two differential operators A and B, each defined by their action on everyconceivable wave function Ψ(r) by

A Ψ(r) = ΦA(r)B ΦA(r) = ΦBA(r) (201)

then the product of two operators is defined by their successive actions.

B A Ψ(r) = B ΦA(r)= ΦBA(r) (202)

62

Two examples are given by multiplying two operators ∂n

∂xn and ∂m

∂xm whichyields the compound operator ∂n+m

∂xn+m and similarly by multiplying the opera-tors xn and xm yields xn+m.

Important examples of the multiplication of two operators are given by :

The kinetic energy for motion in one dimension.

The kinetic energy operator is derived from the classical expression

Tx =p2

x

2 m

=px . px

2 m

=1

2 m

(− i h

∂

∂x

) (− i h

∂

∂x

)= − h2

2 m∂2

∂x2(203)

The kinetic energy for motion in three dimensions.

The kinetic energy operator is derived as

T =p2

2 m

=p . p

2 m

=1

2 m

(− i h ex

∂

∂x− i h ey

∂

∂y− i h ez

∂

∂z

)2

= − h2

2 m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)(204)

where we have used the representation of the vector momentum p in terms ofits components and the unit vectors ex, ey and ez.

The (vector) orbital angular momentum operator L is given by the vectorproduct of two vector operators

L = r ∧ p (205)

and is expressed asL = ex Lx + ey Ly + ez Lz (206)

63

where

Lz = − i h

(x∂

∂y− y

∂

∂x

)Ly = − i h

(z∂

∂x− x

∂

∂z

)Lx = − i h

(y∂

∂z− z

∂

∂y

)(207)

This is actually a pseudo-vector, as it transforms like a vector under every sym-metry operation except inversion.

It should be noted that in general, unlike in the previous two examples, theproduct of two operators depends on the order in which they act on the wavefunction. If two operators have the effect

B Ψ(r) = ΦB(r)A ΦB(r) = ΦAB(r) (208)

then the product of the two operators is defined by their successive actions

A B Ψ(r) = A ΦB(r)= ΦAB(r)6= ΦBA(r) (209)

For example, in multiplying A = ∂∂x and B = xm one has

A B Ψ(x) =∂

∂x

(xm Ψ(x)

)= m xm−1 Ψ(x) + xm ∂

∂xΨ(x) (210)

whereas on taking the product in the opposite order one obtains a differentresult

B A Ψ(x) = xm ∂

∂x

(Ψ(x)

)(211)

3.4.4 Commutators

The commutator of two operators A and B is defined as the difference of theproducts of two operators taken in different orders

[ A , B ] = A B − B A (212)

64

In this equation, like other operator equations, it should always be rememberedthat the operators are assumed to act on an arbitrary wave function. Note thatthe commutator is anti-symmetric

[ A , B ] = − [ B , A ] (213)

From the previous example, where A = ∂∂x and B = xm, one has

[ A , B ] Ψ(x) = A B Ψ(x)− B A Ψ(x)

=∂

∂x

(xm Ψ(x)

)− xm ∂

∂x

(Ψ(x)

)= m xm−1 Ψ(x) (214)

Since the above equation is true for every arbitrary Ψ(x), one has the operatorequation

∂

∂xxm − xm ∂

∂x= m xm−1 (215)

where the presence of the wave function is implicitly assumed. An importantexample is given by the commutator of x = x and px = − i h ∂

∂x then

[ x , px ] = [ x , − i h∂

∂x] = i h (216)

The commutators of the x, y, and z components of the momentum and positionoperators are given by

[ x , px ] = [ y , py ] = [ z , pz ] = i h (217)

If the commutator of two operators is zero, then the two operators are saidto commute. An example of two commuting operators is given by x = x andpy = − i h ∂

∂y . In general, the commutators between different components ofthe particle’s coordinates and momenta are zero.

[ x , py ] = [ y , px ] = 0[ y , pz ] = [ z , py ] = 0[ x , pz ] = [ z , px ] = 0 (218)

The commutator of an operator A and the sum of two operators B and Cis given by

[ A , B + C ] = [ A , B ] + [ A , C ] (219)

65

whereas the commutator of an operator and a product of two operators B andC is given by

[ A , B C ] = [ A , B ] C + B [ A , C ] (220)

and[ A B , C ] = A [ B , C ] + [ A , C ] B (221)

Repeated application of the above relations allows one to express the com-mutator of [ A , Bn ] as

[ A , Bn ] = [ A , B ] Bn−1 + B [ A , B ] Bn−2 + B2 [ A , B ] Bn−3

+ ..... + Bn−1 [ A , B ] B + Bn−1 [ A , B ] (222)

which involves the sum of n commutators.

If the commutator [ A , B ] is not an operator but is a constant instead,the above expression simplifies to

[ A , Bn ] = n [ A , B ] Bn−1 (223)

This can be used to prove that if the commutator [ A , B ] is a constant then,for any function f(B) which has a power series expansion, one has

[ A , f(B) ] = [ A , B ] f ′(B) (224)

where f ′(x) is the derivative of f(x), that is f ′(x) = ∂f∂x . Examination of this

operator equation gives justification for representing any pair of operators thathave a constant commutator as a pair which consists of the variable and thecommutation constant times a derivative w.r.t. the variable.

——————————————————————————————————

3.4.5 Exercise 13

Determine the commutator [ p , V (r) ]. This expression occurs in the equationwhich determines the time derivative of the average value of the momentum, p.

——————————————————————————————————

3.4.6 Solution 13

The commutators between the components of momentum and the componentsof the position are given by

[ pi , xj ] = − i h δi,j (225)

66

These commutators are constants. Thus, on defining the potential as a Taylorseries expansion in r one has

[ pi , V (r) ] = − i h∂V (r)∂xi

(226)

which is the i-th component of the commutator

[ p , V (r) ] = − i h ∇ V (r) (227)

The commutator of the momentum and the potential is related to the gradientof the potential and thus is related to the force.

——————————————————————————————————

3.4.7 Exercise 14

Determine the three commutators [ x , Lz ] , [ y , Lz ] and [ z , Lz ] of thecomponents of the position vector and the z component of the angular momen-tum operator.

Also find the commutator of Lz with the components of the momentum p ,[ px , Lz ] , [ py , Lz ] and [ pz , Lz ] .

——————————————————————————————————

3.4.8 Solution 14

The z component of the angular momentum is given by

Lz = x py − y px (228)

Then, the commutator with the x component of the position is given by

[ x , Lz ] = [ x , x py ] − [ x , y px ]= − y [ x , px ]= − i h y (229)

The commutator with the y component of the position is given by

[ y , Lz ] = [ y , x py ] − [ y , y px ]= + x [ y , py ]= + i h x (230)

The third commutator is given by

[ z , Lz ] = [ z , x py ] − [ x , y px ]= 0 (231)

67

Hence, we obtain

[ x , Lz ] = − i h y

[ y , Lz ] = + i h x

[ z , Lz ] = 0 (232)

The remaining commutators can be obtained from the two non-zero commuta-tors by cyclically permuting x, y and z.

Similarly, the commutation relations between the components of the angularmomentum and the components of the momentum are evaluated from

[ px , Lz ] = [ px , x py ] − [ px , y px ]= [ px , x ] py

= − i h py (233)

Hence, the commutators of the z component of angular momentum with thecomponents of the momentum are found as

[ px , Lz ] = − i h py

[ py , Lz ] = + i h px

[ pz , Lz ] = 0 (234)

The other commutators can be found by cyclic permutation of the subscripts x,y and z.

——————————————————————————————————

3.4.9 Exercise 15

Determine the double commutator [ x , [ x , H ] ], where the Hamiltonian is

given by H =p2

2 m + V (r). This commutator forms the basis of the Thomas-Reiche-Kuhn Sum rule for the intensity of optical absorption by atoms.

——————————————————————————————————

3.4.10 Solution 15

The commutator between the x component of the position and the HamiltonianH can be decomposed into the sum of the commutators between x and thepotential V (r) and the commutator of x with the potential energy p2

2m

[ x , H ] = [ x , V (r) ] + [ x ,p2

2 m] (235)

68

The commutator between the x component of the position and the potentialvanishes

[ x , V (r) ] = 0 (236)

The only non-zero part of the commutator is the commutator between the xcomponent of the kinetic energy and the position. This non-zero commutatoris evaluated as

[ x ,p2

x

2 m] =

12 m

([ x , px ] px + px [ x , px ]

)=

i h

2 m2 px (237)

Hence, the double commutator is given by

[ x , [ x , H ] ] = − h2

m(238)

——————————————————————————————————

Another special example of an operator equation that we shall extensivelyuse is in the definition of an exponential of an operator

exp[λ A

]=∑

n

λn An

n!(239)

where λ is any complex number.

Consider the operator which is a function of λ and defined via,

f(λ) = eλ A B e− λ A (240)

then it is easy to prove that the derivatives are given by the series of expressions,(df

dλ

)= A f(λ) − f(λ) A

= [ A , f(λ) ](d2f

dλ2

)= [ A ,

df

dλ]

= [ A , [ A , f(λ) ] ] (241)

This infinite set of equations can be used in the Taylor expansion of f(λ) toyield

eλ A B e− λ A = B + λ [ A , B ] +λ2

2![ A , [ A , B ] ]

+λ3

3![ A , [ A , [ A , B ] ] ] + . . .

(242)

69

where we have used the fact that f(0) = B.

——————————————————————————————————

3.4.11 Exercise 16

Given the first order differential equation

i hdU(t)dt

= H U(t) (243)

and U(0) = 1, find an expression for U(t). This equation and operator isusually seen in the context of the time evolution of a wave function, and theinitial condition expresses the fact that the wave function is continuous at t = 0.

——————————————————————————————————

3.4.12 Solution 16

The solution for the time evolution operator is found by integrating the equation

U(t) = U(0) − i

h

∫ t

0

dt′ H U(t′) (244)

and then iterating

U(t) = U(0) − i

h

∫ t

0

dt′ H U(0) − 1h2

∫ t

0

dt′∫ t′

0

dt” H H U(t”)

= U(0) − i

h

∫ t

0

dt′ H U(0) − 1h2

∫ t

0

dt′∫ t′

0

dt” H H U(0) + . . .

= U(0) − i

ht H U(0) − 1

h2

t2

2H H U(0) + . . .

= exp[− i t

hH

]U(0) (245)

In the last line the series is recognized as the expansion of the exponential. Sincethe initial condition corresponds to U(0) = 1, we obtain the solution as

U(t) = exp[− i t

hH

](246)

——————————————————————————————————

70

3.4.13 Exercise 17

Given the first order linear differential operator equation

dB(λ)dλ

= [ A , B(λ) ] (247)

with the boundary condition B(0) = C, find an expression for B(λ).

——————————————————————————————————

3.4.14 Solution 17

Since∂B(λ)∂λ

= [ A , B(λ) ] (248)

one can integrate the equation to yield

B(λ) − B(0) =∫ λ

0

dλ′ [ A , B(λ′) ]

B(λ) − C =∫ λ

0

dλ′ [ A , B(λ′) ] (249)

This equation can be iterated to yield

B(λ) = C +∫ λ

0

dλ′ [ A , C ] +∫ λ

0

dλ′∫ λ′

0

dλ” [ A , [ A , B(λ”) ] ]

B(λ) = C +∫ λ

0

dλ′ [ A , C ] +∫ λ

0

dλ′∫ λ′

0

dλ” [ A , [ A , C ] ] + . . .

(250)

which can be evaluated as

B(λ) = C + λ [ A , C ] +λ2

2![ A , [ A , C ] ] + . . . (251)

which is the Taylor series of B(λ).

From eqn(242) of the previous example, the Taylor series can be recognizedas the expansion of

B(λ) = exp[

+ λ A

]C exp

[− λ A

](252)

Alternatively, the series can be re-arranged as

B(λ) =(

1 + λ A +λ2

2!A2 + . . .

)C

(1 − λ A +

λ2

2!A2 + . . .

)= exp

[+ λ A

]C exp

[− λ A

](253)

71

giving the same result.

——————————————————————————————————

3.4.15 Exercise 18

The Baker - Campbell - Hausdorf relation6

exp[A

]exp

[B

]= exp

[( A + B )

]exp

[12

[ A , B ]]

(254)

is only valid for operators A and B that have commutator which is a constant,i.e.

[ A , B ] = Const. (255)

Derive the Baker - Cambell - Hausdorff relation.

——————————————————————————————————

3.4.16 Solution 18

First we note that, from the series expansion of the exponential, one can show[exp

[λ B

], A

]= λ exp

[λ B

][ B , A ] (256)

Now let us consider the operator function f(λ) defined by

f(λ) = exp[λ A

]exp

[λ B

]exp

[− λ ( A + B )

](257)

Then the derivative with respect to λ is given by

∂f

∂λ= exp

[λ A

]A exp

[λ B

]exp

[− λ ( A + B )

]+ exp

[λ A

]exp

[λ B

]B exp

[− λ ( A + B )

]− exp

[λ A

]exp

[λ B

]( A + B ) exp

[− λ ( A + B )

](258)

6J.E. Campbell, Proc. London Math. Soc. 29, 14 (1898).H.F. Baker, London Math. Soc. Ser. 3, 24 (1904).F. Hausdorf, Ber. Vehr. Sachs. Wissen. Leipzig, Math.-Naturwiss. Kl. 58, 19 (1906).

72

On commuting exp[ λ B ] with A in the first term, we find

∂f

∂λ= λ exp

[λ A

]exp

[λ B

][ A , B ] exp

[− λ ( A + B )

](259)

as the other terms cancel identically. Furthermore as the commutator is aconstant it may be factored out, thus we have

∂f∂λ

f= λ [ A , B ] (260)

which on integrating and re-arranging yields

exp[λ A

]exp

[λ B

]= exp

[λ ( A + B )

]exp

[λ2

2[ A , B ]

](261)

and on setting λ = 1 we have proved the relationship.

——————————————————————————————————

The Taylor-MacLaurin expansion of a function f(x) is given by

f(x+ a) = exp[a∂

∂x

]f(x) (262)

which has the three-dimensional analogue

f(r + a) = exp[a . ∇

]f(r) (263)

This is equivalent to a shift of the origin by + a, or a shift of the function by − a.

3.4.17 Eigenvalue Equations

Given a differential operator A one can find eigenfunctions ϕa(r) and eigenval-ues a. The eigenfunctions and eigenvalues are determined from the eigenvalueequation,

A ϕa(r) = a ϕa(r) (264)

An example is given by the eigenfunctions of the x component of the mo-mentum, px, defined on the interval (−∞,∞). The eigenvalue equation is

px ϕp(r) = px ϕp(r)

− i h∂

∂xϕp(r) = px ϕp(r) (265)

73

The eigenfunctions are of the form

ϕp(r) ∝ exp[ipx x

h

](266)

and the eigenvalue px is any real number. Thus, the eigenvalues form a continu-ous spectra with values in the interval (−∞,∞). The momentum eigenfunctionsare also eigenfunctions of the kinetic energy operator Tx, since Tx = p2

x

2 m . Weshould note that two eigenfunctions correspond to the same eigenvalue of Tx.These two eigenfunctions are those which correspond to the momentum eigen-values + px and − px. When an eigenvalue corresponds to more than oneeigenfunction, the eigenvalue is said to be degenerate. The above eigenfunc-tions of Tx are doubly degenerate. The degeneracy is the number of linearlyindependent7 eigenfunctions corresponding to the same eigenvalue, and this de-pends on the space of the functions being considered. For example, if the wavefunctions depend on x, y and z, one can find eigenfunctions of px of the form

Ψ(x, y, z) =(

12 π h

) 32

exp[ipx x

h

]exp

[ipy y

h

]exp

[ipz z

h

](267)

the eigenvalue px is infinitely degenerate.

Note that a factor of h−d2 has been introduced into the normalization of the

d-dimensional momentum eigenstates, or equivalently the d-dimensional Fouriertransform. This factor was introduced in order that the momentum distributionfunction is properly normalized.

In general, given an eigenfunction Ψa(r) of an operator A with eigenvaluea, then any function of the operator f(A) satisfies the eigenvalue equation

f(A) Ψa(r) = f(a) Ψa(r) (268)

where the eigenvalue is f(a).

Another example is given by the energy eigenfunctions of the one-dimensionalharmonic oscillator, which has the Hamiltonian operator H

H = − h2

2 m∂2

∂x2+

m ω2

2x2 (269)

7A set of functions φn is said to be linearly independent if any function of the set cannotbe expressed as a linear superposition of the other members of the set. That is, if the set ofφn is linearly independent, the only solution of the equation∑

n

Cn φn = 0

is the trivial solution in which all the coefficients are zero Cn = 0.

74

The energy eigenvalue equation for the harmonic oscillator is

H Ψn(x) =(− h2

2 m∂2

∂x2+

m ω2

2x2

)Ψn(x)

= En Ψn(x) (270)

which has the energy eigenvalues En = h ω ( n + 12 ) where n is any positive

integer ( including 0 ). The energy eigenfunctions are Ψn(x), where

Ψn(x) =( − 1 )n(2n n!

) 12

(m ω

hπ

) 14

exp[ +m ω

2 hx2 ]

(h

m ω

)n2 dn

dxnexp[ − m ω

hx2 ]

(271)In this case the eigenvalues are in the form of a discrete set of numbers, andthe eigenfunctions are localized about the origin. The harmonic oscillator, clas-sically, corresponds to a situation in which a particle would be bound to theorigin by the quadratic potential, for any value of the energy E.

——————————————————————————————————

3.4.18 Exercise 19

Show that if

Ψ(r) = exp[− φ(r)

](272)

satisfies appropriate boundary conditions then it represents a bound state wavefunction with energy eigenvalue E for a particle moving in the potential

V (r) =h2

2 m

[( ∇φ(r) )2 − ∇2 φ(r)

]+ E (273)

——————————————————————————————————

3.4.19 Exercise 20

An electron is moving in one dimension and is constrained to the region outsidea perfect conductor x > 0, where the potential is given by the image potential

V (x) = − e2

2 x(274)

where e is the charge of the electron. Find a bound state energy eigenfunctionφ(x) and the average value of x.

75

Image potential for electrons near a metal surface

-2

-1.5

-1

-0.5

0

0.5

1

-2 0 2 4 6 8 10

x / x0

V(x

) x 0

/ e2

V(x) = - e2 / 2 x

Figure 17: The one-dimensional image potential V (x) which confines an electronclose to the surface of a metal. The bound state energy is marked by a horizontalline.

Hint: Try a wave function of the form x exp[ − α x ] for x > 0.

——————————————————————————————————

3.4.20 Solution 20

The trial wave function is chosen as

φ(x) = A x exp[− α x

](275)

for x > 0 and is zero for negative values of x, and is shown in fig(18). Thus,the wave function is localized as x → ∞ and vanishes at the surface at x = 0,representing a particle trapped in the attractive image potential. The electrondoes not penetrate into the metal. On substituting the wave function into theenergy eigenvalue equation[

− h2

2 m∂2

∂x2− e2

2 x

]φ(x) = E φ(x) (276)

and on noting that∂φ

∂x= (

1x− α ) φ(x) (277)

76

Bound state wave function

0

0.1

0.2

0.3

0.4

0.5

0.6

-2 0 2 4 6 8 10x / x0

Ψ0(

x)

Figure 18: The wave function Ψ(x) of an electron bound to the surface of ametal.

and the second derivative is given by

∂2φ

∂x2= − (

2x− α ) α φ(x) (278)

one finds that the eigenvalue equation reduces to[− h2

2 m( α2 − 2

α

x) − e2

2 x

]φ(x) = E φ(x) (279)

This is an algebraic equation, which is solved by choosing α such that

α =m e2

2 h2 (280)

and the energy eigenvalue E is then given by

E = − m e4

8 h2 (281)

The eigenvalue E is negative, corresponding to the fact that classically the elec-tron is trapped in the potential well and does not have enough energy to reachx → ∞. The threshold energy for the particle to reach infinity is E = 0, andhence the particle must have a negative energy.

——————————————————————————————————

77

3.4.21 Exercise 21

A particle of mass m is confined to the region x > 0 and is subjected to apotential V (x) given by

V (x) = V0

[a2

x2− a

x

](282)

where V0 and a are constants. The potential is sketched in fig(19). Derive the

-0.5

0

0.5

1

0 2 4 6 8 10

x/a

V(x

)

V(x) = V0 [ (a/x)2 - (a/x) ]

Figure 19: The potential V (x) for a one-dimensional hydrogen-like atom.

bound state energy and wave function.

Hint: Try a wave function of the form xs exp[ − α x ] for x > 0.

——————————————————————————————————

3.4.22 Solution 21

We shall consider the trial wave function

φ(x) = C xs exp[− α x

](283)

78

which has the second derivative given by

∂2

∂x2φ(x) =

(s ( s − 1 )

x2− 2 α

s

x+ α2

)φ(x) (284)

On substituting into the energy eigenvalue equation one finds the consistencyequations

s ( s − 1 ) =2 m V0

h2 a2

2 s α =2 m V0

h2 a

α2 = − 2 m E

h2 (285)

Thus, we find the exponent s as

s =12±√

14

+2 m V0

h2 a2 (286)

Since the wave function should be square integrable we take the positive root.Then from the second equation we find

α =2 m V0

h2 a

1 +√

1 + 8 m V0h2 a2

(287)

from which one can find the energy E.

——————————————————————————————————

3.4.23 Exercise 22

Prove that the energy eigenfunctions of the harmonic oscillator are given byΨn(x) given above.

——————————————————————————————————

3.4.24 Solution 22

The wave function Ψn(x) should satisfy the energy eigenvalue equation,[− h2

2 m∂2

∂x2+

m ω2

2x2

]Ψn(x) = hω ( n +

12

) Ψn(x) (288)

First let us transform to the dimensionless variable y where

y =√m ω

hx (289)

79

Then, the energy eigenvalue equation can be simplified to[− ∂2

∂y2+ y2

]Φn(y) = ( 2 n + 1 ) Φn(y) (290)

We have to show that Φn(y) given by

Φn(y) = exp[

+y2

2

] (∂

∂y

)n

exp[− y2

](291)

satisfies the above dimensionless eigenvalue equation. We shall prove this byinduction.

We first note that the above trial wave function with n = 0 satisfies theeigenvalue equation as[

− ∂2

∂y2+ y2

]exp

[− y2

2

]= exp

[− y2

2

](292)

So the trial wave function satisfies the equation for n = 0, and is thus a solu-tion for this particular n.

Then we shall prove that, if the trial wave function satisfies the eigenvalueequation for an arbitrary n, then the function also satisfies the equation for( n + 1 ).

Let us first note that

exp[

+y2

2

] (∂

∂y

)n

exp[− y2

2

]=(

exp[

+y2

2

] (∂

∂y

)exp

[− y2

2

] )n

=(

+∂

∂y− y

)n

(293)

The trial wave function is assumed to satisfy the equation[− ∂2

∂y2+ y2

] (+

∂

∂y− y

)n

exp[− y2

2

]= ( 2 n + 1 )

(+

∂

∂y− y

)n

exp[− y2

2

](294)

and we aim to show that if the above equation is true then the equation withn + 1 [

− ∂2

∂y2+ y2

] (+

∂

∂y− y

)n+1

exp[− y2

2

]= ( 2 n + 3 )

(+

∂

∂y− y

)n+1

exp[− y2

2

](295)

80

is also true. We note that the operator identity[− ∂2

∂y2+ y2

] (+

∂

∂y− y

)=(

+∂

∂y− y

) [− ∂2

∂y2+ y2 + 2

](296)

is always valid. On substituting this identity into the eigenvalue equation withthe n + 1 th trial function, we have(

+∂

∂y− y

) [− ∂2

∂y2+ y2 + 2

] (+

∂

∂y− y

)n

exp[− y2

2

]=(

+∂

∂y− y

)( 2 n + 3 )

(+

∂

∂y− y

)n

exp[− y2

2

](297)

and on cancelling the term(+

∂

∂y− y

)2(

+∂

∂y− y

)n

exp[− y2

2

](298)

from both sides of the equation, we obtain(+

∂

∂y− y

) [− ∂2

∂y2+ y2

] (+

∂

∂y− y

)n

exp[− y2

2

]=(

+∂

∂y− y

)( 2 n + 1 )

(+

∂

∂y− y

)n

exp[− y2

2

](299)

This is true if eqn(294) holds true, as we are only taking a derivative of eqn(294).Equation(294) is true for n = 0 and so from the above we have shown that itis true for n = 1. By induction this equation is also true for all higher integervalues of n.

——————————————————————————————————

Given an operator, A, which corresponds to a physical measurement A anda system that has a wave function that is an eigenfunction of the operator cor-responding to the eigenvalue a then a measurement of that physical quantity Adefinitely results in the value a.

3.4.25 Adjoint or Hermitean Conjugate Operators

Given an operator A one can construct its adjoint or Hermitean conjugate, A†.The adjoint operator or Hermitean conjugate operator is defined by the matrixelements between two arbitrary wave functions∫

d3r Φ∗(r) A Ψ(r) =( ∫

d3r Ψ∗(r) A† Φ(r))∗

(300)

81

This definition involves the inner product of the wave function A Ψ(r), whichis the state Ψ(r) after it is transformed by the operator A, and Φ(r). The defi-nition relates this inner product to the complex conjugate of the inner productof the state A† Φ(r) with Ψ(r).

From the definition of the adjoint or Hermitean conjugate, it is easy to provethat the Hermitean conjugate of the Hermitean conjugate of an operator A isthe same as the original operator A,(

A†)†

= A (301)

This is proved by using the definition of the adjoint twice∫d3r Φ∗(r) A Ψ(r) =

( ∫d3r Ψ∗(r) A† Φ(r)

)∗=

( ( ∫d3r Φ∗(r)

(A†)†

Ψ(r))∗ )∗

(302)

which, since (z∗)∗ = z for any complex number, yields the identification(A†)† = A.

Also a complex constant a, when regarded as an operator, has a Hermiteanconjugate which is just its complex conjugate a∗.

An example of finding the Hermitean conjugate of an operator is given byconsidering the differential operator in one dimension.

A =∂

∂x(303)

The Hermitean conjugate or self adjoint operator, A†, is found from∫ +∞

−∞dx Φ∗(x)

∂

∂xΨ(x) =

= Φ∗(x) Ψ(x)∣∣∣∣∞−∞

−∫ +∞

−∞dx

∂

∂x

(Φ∗(x)

)Ψ(x)

= −∫ +∞

−∞dx

∂

∂x

(Φ∗(x)

)Ψ(x) (304)

where the second line involves integration by parts and the third line utilizesthe boundary conditions on the wave functions

lim|x| → ∞

Ψ(x) = 0 (305)

82

which are implied by the normalization condition. From this, one identifies theHermitean conjugate operator as

A† = − ∂

∂x(306)

Another example of a Hermitean conjugate of an operator is given by trans-lation operator, which translates the wave function by the vector a.

S(a) = exp[− a . ∇

](307)

This, when acting on a wave function Ψ(r) generates the Taylor MacLaurinseries expansion of the function Ψ(r − a). The Hermitean conjugate is foundfrom the series expansion of the exponential operator, and integrating by partsto yield

S†(a) = exp[

+ a . ∇]

(308)

which coincides with the inverse of S(a), as

S†(a) S(a) = S(a) S†(a) = 1 (309)

Thus, we haveS†(a) = S−1(a) (310)

Operators that have their Hermitean conjugate operators equal to their inversesare called unitary operators. Thus, an operator A is a unitary operator if

A† = A−1 (311)

Hence, S(a) is a unitary operator.

Another unitary operator is given by the time translation operator U(t)which is found to be

U(t) = exp[− i

H t

h

](312)

where H is the Hamiltonian operator

H = − h2

2m∂2

∂x2+ V (x) (313)

and V (x) is a real function. Then it can be shown that the Hermitean conjugateof U is given by

U†(t) = exp[

+ iH t

h

](314)

83

Translation through a distance a

0

0.2

0.4

0.6

0.8

1

φ(x) φ(x-a)

x0 x0+a0 01 2 23 1 3

x x

φ'(x)=φ(T-1x)=φ(x-a)

x'=Tx=x+a

Figure 20: The unitary operator S(a) produces a translation of the wave func-tion by a. The translation operator T acts on the coordinates and transformsthe point x to the point x′ = x + a, i.e. x′ = T x. Under the translation,the wave function φ(x) is transformed to φ′(x), such that the value of φ′ at x′

is the same as the value of φ at the point x. Therefore, φ′(x′) = φ(x) so,φ′(x) = φ(T−1x). The unitary operator which transforms the wave functionis then specified as φ′(x) = S(a) φ(x) = φ(T−1x).

In this case, the Hermitean conjugate is just the complex conjugate of the op-erator. Also we have that the Hermitean conjugate coincides with the inverseoperator

U†(t) U(t) = U(t) U†(t) = 1 (315)

This is the condition that has to be satisfied for an operator to be a unitaryoperator. The unitary condition for the time evolution operator ensures thatthe normalization of the states is independent of time.

The Hermitean conjugate of a product of two operators ( A B )† is givenby the product of the Hermitean conjugates taken in reverse order, ( B† A† ).This can be proved by starting with the definition of the Hermitean conjugateof the product(∫

d3r Ψ∗(r)(A B

)Φ(r)

)∗=∫

d3r Φ∗(r)(A B

)†Ψ(r)

(316)

The product operator is defined in terms of the state which results when the

84

operator A acts on the state B Φ(r). On using the definition of the Hermiteanconjugate twice, successively, one finds

=(∫

d3r Ψ∗(r) A(B Φ(r)

) )∗=

∫d3r

(B Φ(r)

)∗ (A† Ψ(r)

)=

(∫d3r

(A† Ψ(r)

)∗ (B Φ(r)

) )∗=

∫d3r Φ∗(r) B†

(A†Ψ(r)

)=

∫d3r Φ∗(r) B† A† Ψ(r) (317)

which completes the proof.

3.4.26 Hermitean Operators

A Hermitean operator (or self Adjoint operator) is any operator that satisfiesthe condition

A† = A (318)

That is, if an operator is Hermitean, the Hermitean conjugate of the operatoris equal to the operator.

Examples of Hermitean operators are given by, the components of the mo-mentum operators px = − i h ∂

∂x , py = − i h ∂∂y and pz = − i h ∂

∂z , andalso the single-particle Hamiltonian defined by

H = − h2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)+ V (r) (319)

is Hermitean.

In spherical polar coordinates (r,θ,ϕ), the canonical generalized momentacould correspond to the Hermitean operators

pr = − i h

r

∂

∂rr

pθ = − i h

sin12 θ

∂

∂θsin

12 θ

pϕ = − i h∂

∂ϕ(320)

These operators are Hermitean operators. Note that when these operators aretaken between two wave functions, and integrated over the volume r2 dr sin θ dθ dϕ

85

the resulting factors of r and sin θ in the expressions are symmetrically placedwith respect to the differential operator.

——————————————————————————————————

3.4.27 Exercise 23

Prove that the canonical momenta in spherical polar coordinates given in eqn(320)are Hermitean operators.

——————————————————————————————————

3.4.28 Exercise 24

The expression for the kinetic energy in the Hamiltonian of eqn(39) when quan-tized involves the sum of the three operators

p2r = − h2

r2∂

∂rr2

∂

∂r

p2θ = − h2

sin θ∂

∂θsin θ

∂

∂θ

p2ϕ = − h2 ∂2

∂ϕ2(321)

Show that these three operators are Hermitean.

——————————————————————————————————

3.4.29 Solution 24

The Hermitean conjugate of p2ϕ is calculated from the matrix elements∫ 2 π

0

dϕ

∫ π

0

dθ sin θ∫ ∞

0

dr r2 Φ∗(r, θ, ϕ) p2ϕ Ψ(r, θ, ϕ)

= − h2

∫ 2 π

0

dϕ

∫ π

0

dθ sin θ∫ ∞

0

dr r2 Φ∗(r, θ, ϕ)∂2

∂ϕ2Ψ(r, θ, ϕ)

(322)

by integrating by parts twice with respect to ϕ. On integrating by parts for thefirst time, one obtains

− h2

∫ 2 π

0

dϕ

∫ π

0

dθ sin θ∫ ∞

0

dr r2 Φ∗(r, θ, ϕ)∂2

∂ϕ2Ψ(r, θ, ϕ)

86

= − h2

∫ π

0

dθ sin θ∫ ∞

0

dr r2 Φ∗(r, θ, ϕ)(∂

∂ϕΨ(r, θ, ϕ)

)∣∣∣∣2 π

0

+ h2

∫ 2 π

0

dϕ

∫ π

0

dθ sin θ∫ ∞

0

dr r2(∂

∂ϕΦ∗(r, θ, ϕ)

) (∂

∂ϕΨ(r, θ, ϕ)

)(323)

The boundary term vanishes since the wave functions are 2π periodic in ϕ, i.e.,Ψ(r, θ, ϕ) = Ψ(r, θ, ϕ+ 2π). On integrating by parts once again, one obtains

− h2

∫ 2 π

0

dϕ

∫ π

0

dθ sin θ∫ ∞

0

dr r2∂2

∂ϕ2Φ∗(r, θ, ϕ) Ψ(r, θ, ϕ)

= − h2

( ∫ 2 π

0

dϕ

∫ π

0

dθ sin θ∫ ∞

0

dr r2 Ψ∗(r, θ, ϕ)∂2

∂ϕ2Φ(r, θ, ϕ)

)∗(324)

where the boundary terms have vanished once again. Thus, we have found that(p2

ϕ

)†= − h2 ∂2

∂ϕ2(325)

Hence, p2ϕ is Hermitean.

The Hermitean conjugate of p2θ is calculated from the matrix elements∫ 2 π

0

dϕ

∫ π

0

dθ sin θ∫ ∞

0

dr r2 Φ∗(r, θ, ϕ) p2θ Ψ(r, θ, ϕ)

= − h2

∫ 2 π

0

dϕ

∫ π

0

dθ

∫ ∞

0

dr r2 Φ∗(r, θ, ϕ)∂

∂θ

(sin θ

∂

∂θΨ(r, θ, ϕ)

)(326)

by integrating by parts twice with respect to θ. Note that the factor sin θ in themetric for the volume integral has cancelled with a factor from the componentof the kinetic energy. On integrating by parts for the first time, one obtains

− h2

∫ 2 π

0

dϕ

∫ π

0

dθ

∫ ∞

0

dr r2 Φ∗(r, θ, ϕ)∂

∂θ

(sin θ

∂

∂θΨ(r, θ, ϕ)

)= − h2

∫ 2 π

0

dϕ

∫ ∞

0

dr r2 Φ∗(r, θ, ϕ) sin θ(∂

∂θΨ(r, θ, ϕ)

)∣∣∣∣π0

+ h2

∫ 2 π

0

dϕ

∫ π

0

dθ sin θ∫ ∞

0

dr r2(∂

∂θΦ∗(r, θ, ϕ)

) (∂

∂θΨ(r, θ, ϕ)

)(327)

The boundary term vanishes since the wave functions are finite at the poleswhere sin θ = 0 . On integrating by parts once again, one obtains

− h2

∫ 2 π

0

dϕ

∫ π

0

dθ

∫ ∞

0

dr r2∂

∂θ

(sin θ

∂

∂θΦ∗(r, θ, ϕ)

)Ψ(r, θ, ϕ)

87

= − h2

( ∫ 2 π

0

dϕ

∫ π

0

dθ

∫ ∞

0

dr r2 Ψ∗(r, θ, ϕ)∂

∂θ

(sin θ

∂

∂θΦ(r, θ, ϕ)

) )∗(328)

where the boundary terms have vanished once again. Thus, we have found that(p2

ϕ

)†= − h2 1

sin θ∂

∂θ

(sin θ

∂

∂θ

)(329)

Hence, p2ϕ is Hermitean. Note that the weight factor sin θ in the integral over

the solid angle played a crucial role in the proof of Hermiticity.

The radial component of the kinetic energy have the matrix elements∫ 2 π

0

dϕ

∫ π

0

dθ sin θ∫ ∞

0

dr r2 Φ∗(r, θ, ϕ) p2r Ψ(r, θ, ϕ)

= − h2

∫ 2 π

0

dϕ

∫ π

0

dθ sin θ∫ ∞

0

dr Φ∗(r, θ, ϕ)∂

∂r

(r2

∂

∂rΨ(r, θ, ϕ)

)(330)

by integrating by parts twice with respect to r. On integrating by parts for thefirst time, one obtains

− h2

∫ 2 π

0

dϕ

∫ π

0

dθ sin θ∫ ∞

0

dr Φ∗(r, θ, ϕ)∂

∂r

(r2

∂

∂rΨ(r, θ, ϕ)

)= − h2

∫ 2 π

0

dϕ

∫ π

0

dθ sin θ r2 Φ∗(r, θ, ϕ)(∂

∂rΨ(r, θ, ϕ)

)∣∣∣∣∞0

+ h2

∫ 2 π

0

dϕ

∫ π

0

dθ sin θ∫ ∞

0

dr r2(∂

∂rΦ∗(r, θ, ϕ)

) (∂

∂rΨ(r, θ, ϕ)

)(331)

The boundary term vanishes since the wave function vanishes as r → ∞, i.e.,limr → ∞ Ψ(r, θ, ϕ) = 0 and the weight function vanishes at r = 0 . Onintegrating by parts once again, one obtains

− h2

∫ 2 π

0

dϕ

∫ π

0

dθ sin θ∫ ∞

0

dr∂

∂r

(r2

∂

∂rΦ∗(r, θ, ϕ)

)Ψ(r, θ, ϕ)

= − h2

( ∫ 2 π

0

dϕ

∫ π

0

dθ sin θ∫ ∞

0

dr Ψ∗(r, θ, ϕ)∂

∂r

(r2

∂

∂rΦ(r, θ, ϕ)

) )∗(332)

where the boundary terms have vanished once again. Thus, we have found that(p2

r

)†= − h2

r2∂

∂r

(r2

∂

∂r

)(333)

88

is Hermitean. In the proof, the weight factor r2 played a crucial role in showingthe radial part of the kinetic energy is Hermitean.

——————————————————————————————————

3.4.30 Exercise 25

Using the expression for the momentum operator p = − i h ∇ in sphericalpolar coordinates

p = − i h

(er

∂

∂r+ eθ

1r

∂

∂θ+ eϕ

1r sin θ

∂

∂ϕ

)(334)

where the unit vectors are given in terms of the Cartesian unit vectors via

er = sin θ cosϕ ex + sin θ sinϕ ey + cos θ ez

eθ = cos θ cosϕ ex + cos θ sinϕ ey − sin θ ez

eϕ = − sinϕ ex + cosϕ ey (335)

show that the resulting expression for the kinetic energy agrees with the expres-sion inferred from the previous exercise.

——————————————————————————————————

3.4.31 Solution 25

First we shall note that the unit vectors are independent of r and are onlyfunctions of the angles, so

∂er

∂θ= eθ

∂er

∂ϕ= sin θ eϕ

∂eθ

∂θ= − er

∂eθ

∂ϕ= cos θ eϕ

∂eϕ

∂θ= 0

∂eϕ

∂ϕ= −

(sin θ er + cos θ eθ

)(336)

Then, we have

p2 = − h2

[er

∂

∂r. er

∂

∂r+ er

∂

∂r. eθ

1r

∂

∂θ

89

+ er∂

∂r. eϕ

1r sin θ

∂

∂ϕ+ eθ

1r

∂

∂θ. er

∂

∂r

+ eθ1r

∂

∂θ. eθ

1r

∂

∂θ+ eθ

1r

∂

∂θ. eϕ

1r sin θ

∂

∂ϕ

+ eϕ1

r sin θ∂

∂ϕ. er

∂

∂r+ eϕ

1r sin θ

∂

∂ϕ. eθ

1r

∂

∂θ

+ eϕ1

r sin θ∂

∂ϕ. eϕ

1r sin θ

∂

∂ϕ

](337)

which on evaluating the derivatives and evaluating the scalar products of theorthogonal unit vectors becomes

p2 = − h2

[∂2

∂r2+

2r

∂

∂r

+1r2

∂2

∂θ2+

cos θr2 sin θ

∂

∂θ

+1

r2 sin2 θ

∂2

∂ϕ2

](338)

The terms containing the first derivatives come from the change in the unitvectors of the fourth, seventh and eight terms of the previous equation. Thenthis can be simplified to yield

p2 = − h2

[1r2

∂

∂r

(r2

∂

∂r

)+

1r2 sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1r2 sin2 θ

∂2

∂ϕ2

](339)

which involves the appropriate expression for the Laplacian in spherical polarcoordinates.

——————————————————————————————————

Given an arbitrary operator B which is not Hermitean, i.e. B 6= B†, thenone can always construct a Hermitean operator A as

A =12

(B + B†

)(340)

since

A† =12

(B† + ( B† )†

)=

12

(B† + B

)= A (341)

The operator A is the Hermitean part of the arbitrary operator B.

——————————————————————————————————

90

3.4.32 Exercise 26

Construct the Hermitean operators corresponding to the operators Br = − i ∂∂r

and Bθ = − i ∂∂θ , expressed in spherical polar coordinates r, θ and ϕ.

——————————————————————————————————

3.4.33 Solution 26

The Hermitean conjugate of Br is found by considering the matrix elements andby integrating by parts with respect to r,

− i

∫ ∞

0

dr r2∫ π

0

dθ sin θ∫ 2 π

0

dϕ Φ∗(r, θ, ϕ)∂

∂rΨ(r, θ, ϕ)

= +(i

∫ ∞

0

dr

∫ π

0

dθ sin θ∫ 2 π

0

dϕ Ψ∗(r, θ, ϕ)∂

∂rr2 Φ(r, θ, ϕ)

)∗(342)

where we have assumed that the wave function satisfies the boundary conditions

limr → ∞

Ψ(r, θ, ϕ) → 0 (343)

Thus, the Hermitean conjugate, B†r , is given by

B†r = − i1r2

∂

∂rr2 (344)

From this we can construct a Hermitean operator as the Hermitean part of B,via

Ar = − i

2

(∂

∂r+

1r2

∂

∂rr2)

= − i

(∂

∂r+

1r

)= − i

(1r

∂

∂rr

)(345)

Similarly, the Hermitean conjugate of Bθ is found from

− i

∫ ∞

0

dr r2∫ π

0

dθ sin θ∫ 2 π

0

dϕ Φ∗(r, θ, ϕ)∂

∂θΨ(r, θ, ϕ)

= +(i

∫ ∞

0

dr r2∫ π

0

dθ

∫ 2 π

0

dϕ Ψ∗(r, θ, ϕ)∂

∂θsin θ Φ(r, θ, ϕ)

)∗(346)

91

The boundary term vanishes at θ = 0 and θ = π due to the presence of thesin θ term. Thus, B†θ is given by

B†θ = − i1

sin θ∂

∂θsin θ (347)

From this we can construct a Hermitean operator Aθ via

Aθ = − i

2

(∂

∂θ+

1sin θ

∂

∂θsin θ

)= − i

(∂

∂θ+

cos θ2 sin θ

)= − i

1

sin12 θ

(∂

∂θsin

12 θ

)(348)

——————————————————————————————————

3.4.34 Eigenvalues and Eigenfunctions of Hermitean Operators

Hermitean operators have eigenvalues and eigenfunctions that have three im-portant properties. These are :

(i) The eigenvalues are real numbers.

(ii) Any two eigenfunctions corresponding to different eigenvalues have ma-trix elements which are zero. Wavefunctions that have this property are said tobe orthogonal.

(iii) Any wave function can be expanded in terms of a complete set of theeigenfunctions of a Hermitean operator.

We shall now prove the first two statements. Consider a set of eigenfunctionsof the operator A. These satisfy the eigenvalue equation

A φa(r) = a φa(r) (349)

Then take the matrix elements with an eigenfunction corresponding to the eigen-value b, ∫

d3r φ∗b(r) A φa(r) = a

∫d3r φ∗b(r) φa(r) (350)

but since A is Hermitean

=( ∫

d3r φ∗a(r) A† φb(r))∗

=( ∫

d3r φ∗a(r) A φb(r))∗

(351)

92

However, φb(r) is an eigenstate of A with eigenvalue b, so

=( ∫

d3r φ∗a(r) b φb(r))∗

= b∗∫

d3r φ∗b(r) φa(r) (352)

Thus, we find the equality

( a − b∗ )( ∫

d3r φ∗b(r) φa(r))

= 0 (353)

Thus, if the two eigenvalues are the same, b = a, and φa(r) is normalizedto unity, one must have a = b∗ = a∗. Hence, we have proved that theeigenvalues of a Hermitean operator are real.

If the eigenvalues are different, b 6= a, one then has∫d3r φ∗b(r) φa(r) = 0 (354)

Thus, eigenfunctions of the Hermitean operator corresponding to different eigen-values are orthogonal.

If an eigenvalue is degenerate, then the eigenfunctions corresponding to thiseigenvalue are not necessarily orthogonal. However, by choosing appropriatelinear combinations of these eigenfunctions one can construct a set of eigen-functions which are orthogonal. This procedure is known as Gram-Schmidtorthogonalization. We shall implicitly assume that this procedure has alwaysbeen performed on any set of eigenfunctions.

We shall delay the proof of the expansion of an arbitrary wave functiontill later. However, we shall show how the orthonormality properties of theeigenfunctions allow a simple evaluation of the expansion coefficients Cn of anarbitrary wave function

Ψ(r) =∑

n

Cn φn(r) (355)

Taking the matrix elements with the eigenfunction φm(r), one finds∫d3r φ∗m(r) Ψ(r) =

∑n

Cn

∫d3r φ∗m(r) φn(r)

=∑

n

Cn δn,m

= Cm (356)

93

Therefore, the expansion of an arbitrary wave function Ψ(r) can be written inthe form

Ψ(r) =∑

n

∫d3r′ φ∗n(r′) Ψ(r′) φn(r) (357)

After re-arranging, one recognizes that the sum over the eigenfunctions has thesame property as the Dirac delta function. Thus, we have found the complete-ness condition ∑

n

φ∗n(r′) φn(r) = δ3( r − r′ ) (358)

This condition has to be satisfied if any wave function can be expanded in termsof a complete set of eigenfunctions. On going back to the example of momen-tum eigenfunctions and wave packets, one recognizes that the sum over theeigenvalues n is to be replaced by an integral over the continuous eigenvaluesk. Furthermore, the completeness condition is just the representation of theDirac delta function given by eqn(132). In the case of degenerate eigenvalues,an appropriate orthonormal set of eigenfunctions should be constructed.

——————————————————————————————————

3.4.35 Exercise 27

Express the x, y and z components of the orbital angular momentum oper-ator L in terms of the spherical polar coordinate variables (r,θ,ϕ). Find allthe eigenfunctions and eigenvalues of the z component of the (Hermitean) or-bital angular momentum operator Lz. Show that they form an orthonormal set.

By using the theorems of discrete Fourier transforms for fixed r and θ, showthat any wave function can be expanded in terms of the complete set of eigen-functions of Lz, φm(ϕ) as

Ψ(r, θ, ϕ) =∑m

Φm(r, θ) φm(ϕ) (359)

where the expansion coefficients Φm(r, θ) are functions of r and θ.

——————————————————————————————————

3.4.36 Solution 27

Since the momentum is given by

p = − i h

(er

∂

∂r+ eθ

1r

∂

∂θ+ eϕ

1r sin θ

∂

∂ϕ

)(360)

94

and r = r er then with L = r ∧ p, one finds that the angular momentumsolely has eθ and eϕ components. The total orbital angular momentum can thenbe resolved into the Cartesian components as

Lx = − i h

(− sin ϕ

∂

∂θ− cosϕ cot θ

∂

∂ϕ

)Ly = − i h

(+ cos ϕ

∂

∂θ− sinϕ cot θ

∂

∂ϕ

)Lz = − i h

∂

∂ϕ(361)

The eigenfunctions of the z component of the orbital angular momentum oper-ator are

φm(ϕ) =(

12 π

) 12

exp[i m ϕ

](362)

where, because of the condition that φm(ϕ) is single valued

φm( ϕ ) = φm( ϕ + 2 π ) (363)

then one finds exp[ i m 2 π ] = 1 which implies that m can only have integervalues. Since these eigenfunctions only involve ϕ, and not r and θ, the normal-ization is only due to the integration over ϕ between 0 and 2 π. Any functionof ϕ which has period 2 π can be expanded in terms of this set of orthonormaleigenfunctions.

——————————————————————————————————

3.4.37 Exercise 28

Find the expectation value of the nested commutation relation [ x , [ x , H ] ]between the Hermitean operators x and the energy H between the lowest energyeigenfunctions φ0, i.e., evaluate∫

d3r φ∗0(r) [ x , [ x , H ] ] φ0(r) (364)

and then use the completeness relation to prove the Thomas-Reiche-Kuhn sumrule ∑

n

∣∣∣∣ ∫ d3r φ∗0(r) x φn(r)∣∣∣∣2 ( En − E0 ) = +

h2

2 m(365)

which relates the intensities of optical dipole allowed transitions |∫d3r φ∗0(r) x φn(r) |2

and the energies of the photon involved in the transitions ( En − E0 ). If allthese quantities are measured experimentally, multiplied and summed over allpossible transitions, one should obtain unity. If the sum is different from unity

95

the experiment has failed to identify some transitions or the experimental ap-paratus has not been calibrated correctly.

——————————————————————————————————

3.4.38 Solution 28

Starting from the commutation relation

[ x , [ x , H ] ] = − h2

m(x2 H + H x2 − 2 x H x

)= − h2

m(366)

and taking the matrix elements between Ψ∗0(x) and Ψ0(x), one has∫ + ∞

− ∞dx Ψ∗

0(x)(x2 H + H x2 − 2 x H x

)Ψ0(x)

= − h2

m

∫ + ∞

− ∞dx Ψ∗

0(x) Ψ0(x)

= − h2

m(367)

since the wave functions are normalized to unity. The right hand side of thecommutator consists of four terms. The first and second term will turn out tobe identical, although it may not seem obvious at first. The third and fourthare obviously identical and have already been combined and these give rise tothe term which contains a factor of 2.

We shall now examine the first term which is given by∫ + ∞

− ∞dx Ψ∗

0(x) x x H Ψ0(x) (368)

and introduce a factor of 1 between the two x’s. This 1 can be written as anintegral over a Dirac delta function δ(x− x′) with respect to x′. Thus, we have∫ + ∞

− ∞dx

∫ + ∞

− ∞dx′ Ψ∗

0(x) x δ( x − x′ ) x′ H Ψ0(x′) (369)

where everything to the right of the delta function has been expressed in termsof the primed variable x′. Since Ψ0(x′) is an energy eigenfunction it satisfies

H Ψ0(x′) = E0 Ψ0(x′) (370)

96

This can be substituted in the expression for the term we are trying to evaluate,to yield

=∫ + ∞

− ∞dx

∫ + ∞

− ∞dx′ Ψ∗

0(x) x δ( x − x′ ) x′ E0 Ψ0(x′) (371)

On writing the delta function in terms of the completeness relation

δ( x − x′ ) =∑

n

Ψn(x) Ψ∗n(x′) (372)

one obtains

= E0

∫ + ∞

− ∞dx

∫ + ∞

− ∞dx′ Ψ∗

0(x) x∑

n

Ψn(x) Ψ∗n(x′) x′ Ψ0(x′)

= E0

∑n

∫ + ∞

− ∞dx

∫ + ∞

− ∞dx′ Ψ∗

0(x) x Ψn(x) Ψ∗n(x′) x′ Ψ0(x′)

= E0

∑n

∫ + ∞

− ∞dx Ψ∗

0(x) x Ψn(x)∫ + ∞

− ∞dx′ Ψ∗

n(x′) x′ Ψ0(x′)

= E0

∑n

( ∫ + ∞

− ∞dx Ψ∗

0(x) x Ψn(x)) ( ∫ + ∞

− ∞dx′ Ψ∗

n(x′) x′ Ψ0(x′))

= E0

∑n

∣∣∣∣ ∫ + ∞

− ∞dx Ψ∗

0(x) x Ψn(x)∣∣∣∣2 (373)

where we have interchanged the order of summation and integrations.

The second term can be immediately relate it to the first term. The secondterm is given by ∫ + ∞

− ∞dx Ψ∗

0(x) H x x Ψ0(x) (374)

Let us note that the Hermitean conjugate of the operator H x x is just x x H.Using the definition of the Hermitean conjugate of an operator, one finds∫ + ∞

− ∞dx Ψ∗

0(x) H x x Ψ0(x) =( ∫ + ∞

− ∞dx Ψ∗

0(x) x x H Ψ0(x))∗

(375)

Thus, the second term is just the complex conjugate of the first term, which wehave seen is real. The first two terms are thus identical.

The remaining terms are evaluated in the same way as the first term. Westart with the expression∫ + ∞

− ∞dx Ψ∗

0(x) x H x Ψ0(x) (376)

97

and introduce a factor of 1 between the H and an x. This 1 can be written asan integral over a Dirac delta function δ(x − x′) with respect to x′. Thus, wehave ∫ + ∞

− ∞dx

∫ + ∞

− ∞dx′ Ψ∗

0(x) x H δ( x − x′ ) x′ Ψ0(x′) (377)

where everything to the right of the delta function has been expressed in termsof the primed variable x′. On writing the delta function in terms of the com-pleteness relation

δ( x − x′ ) =∑

n

Ψn(x) Ψ∗n(x′) (378)

one obtains

=∫ + ∞

− ∞dx

∫ + ∞

− ∞dx′ Ψ∗

0(x) x H∑

n

Ψn(x) Ψ∗n(x′) x′ Ψ0(x′)

(379)

Since Ψn(x′) is an energy eigenfunction it satisfies

H Ψn(x′) = En Ψn(x′) (380)

This can be substituted in the expression for the term we are trying to evaluate,to yield

=∫ + ∞

− ∞dx

∫ + ∞

− ∞dx′ Ψ∗

0(x) x H∑

n

Ψn(x) Ψ∗n(x′) x′ Ψ0(x′)

=∫ + ∞

− ∞dx

∫ + ∞

− ∞dx′ Ψ∗

0(x) x∑

n

En Ψn(x) Ψ∗n(x′) x′ Ψ0(x′)

=∑

n

En

∫ + ∞

− ∞dx

∫ + ∞

− ∞dx′ Ψ∗

0(x) x Ψn(x) Ψ∗n(x′) x′ Ψ0(x′)

=∑

n

En

∫ + ∞

− ∞dx Ψ∗

0(x) x Ψn(x)∫ + ∞

− ∞dx′ Ψ∗

n(x′) x′ Ψ0(x′)

=∑

n

En

( ∫ + ∞

− ∞dx Ψ∗

0(x) x Ψn(x)) ( ∫ + ∞

− ∞dx′ Ψ∗

n(x′) x′ Ψ0(x′))

=∑

n

En

∣∣∣∣ ∫ + ∞

− ∞dx Ψ∗

0(x) x Ψn(x)∣∣∣∣2 (381)

where we have interchanged the order of summation and integrations.

Combining the four term leads to the equation

2∑

n

( E0 − En )∣∣∣∣ ∫ + ∞

− ∞dx Ψ∗

0(x) x Ψn(x)∣∣∣∣2 = − h2

m(382)

which was to be proved.——————————————————————————————————

98

3.4.39 Exercise 29

Evaluate all the 6 commutation relations between the various pairs of the dif-ferent Cartesian components of the orbital angular momentum operators L.

——————————————————————————————————

3.4.40 Solution 29

The components of the angular momentum are given by

Lx = y pz − z py

Ly = z px − x pz

Lz = x py − y px (383)

The commutators can be evaluated by expanding them

[ Lx , Ly ] = [ y pz , Ly ] − [ z py , Ly ]= [ y pz , z px ] + [ z py , x pz ]= − i h y px + i h x py

= + i h Lz (384)

so one has the result[ Lx , Ly ] = i h Lz (385)

The set of commutators can be summarized as

[ Li , Lj ] = i h εi,j,k Lk (386)

where εi,j,k is the antisymmetric Levi-Civita symbol, defined as

εi,j,k = 1 (387)

if (i, j, k) is an even permutation of (x, y, z), and

εi,j,k = − 1 (388)

if (i, j, k) is an odd permutation of (x, y, z) or εi,j,k is zero otherwise.

——————————————————————————————————

3.4.41 Exercise 30

In our discussion of the classical mechanics of a charged particle in an electro-magnetic field, we have seen that the momentum of the particle is changed due

99

to the presence of the vector potential A(r, t) and that the Hamiltonian is ofthe form

H =p2

2 m− q

m cp . A(r, t) +

q2

2 m c2A2(r, t) + q φ(r, t) (389)

In the radiation or Coulomb gauge, defined by the condition ∇ . A(r, t) = 0,electromagnetic waves are solely represented by the vector potential. Thus, tolowest order in A, light couples to the particle via the interaction Hint

Hint = − q

2 m c

(p . A(r, t) + A(r, t) . p

)(390)

Show that the momentum operator can be expressed as a commutator of the po-sition operator and the Hamiltonian for a charged particle in the scalar potentialrepresented by the Hamiltonian H0,

− ih

mp = [ H0 , r ] (391)

Hence, by using the energy eigenfunctions Ψn(r) and eigenvalues En of theunperturbed Hamiltonian H0 , show that if A(r, t) only varies over long lengthscales, i.e. A(r, t) ' A(t), then∫

d3r Ψ∗n(r) p . A(r, t) Ψm(r) =

i m

h

(En − Em

) ∫d3r Ψ∗

n(r) r . A(r, t) Ψm(r)

(392)This provides the usual basis for discussing optical transitions involving the ab-sorption or emission of light in terms of dipole transitions, when the wave lengthof light is much longer than the scale associated with the atomic or electronicstructure.

——————————————————————————————————

3.4.42 Solution 30

The matrix elements of the interaction term can be expressed as

− q

m c

∫d3r φ∗m(r) p . A φn(r) (393)

where we are using the radiation gauge

∇ . A = 0 (394)

in which the vector potential is transverse to the direction of propagation. Wecan substitute the expression for p given by

p = − im

h[ r , H0 ] (395)

100

in the matrix elements, to find

iq

h c

∫d3r φ∗m(r) A . [ r , H0 ] φn(r) (396)

The matrix elements of the commutator can be expanded as two terms

+ iq

h c

∫d3r φ∗m(r) A . r H0 φn(r)

− iq

h c

∫d3r φ∗m(r) H0 A . r φn(r) (397)

Now on using the energy eigenvalue equation

H0 φn(r) = En φn(r) (398)

and the approximate r independence of the vector potential A, the first termcan be evaluated, yielding

+ iq

h cEn A .

( ∫d3r φ∗m(r) r φn(r)

)− i

q

h cA .

( ∫d3r φ∗m(r) H0 r φn(r)

)(399)

The matrix elements of r H0 are related to the complex conjugate of its Her-mitean conjugate H0 r, thus( ∫

d3r φ∗m(r) H0 r φn(r))

=( ∫

d3r φ∗n(r) r H0 φm(r))∗

= Em

( ∫d3r φ∗n(r) r φm(r)

)∗(400)

Hence, we find

+ iq

h c

(En − Em

)A .

( ∫d3r φ∗m(r) r φn(r)

)(401)

which can be re-written as

+ iq

cωn,m A .

( ∫d3r φ∗m(r) r φn(r)

)(402)

where En − En = h ωn,m is related to the frequency of the photon involvedin the transition. The matrix elements of q r can be considered to provide anelectric dipole for the transition, and hence the approximation that A is almostr independent is called the dipole approximation.

——————————————————————————————————

101

3.4.43 Exercise 31

Extend the proof of the Thomas-Reiche-Kuhn sum rule to show∑n

( En − Em )∣∣∣∣ ∫ d3r φ∗n(r) exp

[i k . r

]φm(r)

∣∣∣∣2 =h2 k2

2 m(403)

——————————————————————————————————

3.4.44 Exercise 32

The scattering of light is encapsulated by the Kramers - Heisenberg formula.This involves the square of matrix elements of the interaction between a chargedparticle and the field, which is of the order of the fourth power of the vector po-tential A. The contributions to the matrix elements for this process are secondorder in the vector potential, and occurs as a combination of processes involvingthe paramagnetic interaction squared and the lowest order diamagnetic interac-tion. At low frequencies, one has Rayleigh scattering of light, which involves apartial cancellation between the contribution of the diamagnetic interaction andthe process involving the paramagnetic interaction. In this exercise, we shallexpress the matrix elements of the diamagnetic interaction in a form similarto that of the second order process involving the paramagnetic interaction, asneeded to demonstrate the partial cancellation.

The diamagnetic interaction term can be written as

Hdia = +q2

2 m c2A(r, t) . A(r, t) (404)

The matrix elements of this interaction can be related to the scalar product oftwo unit vectors

eα . eβ (405)

Show that,

eα . eβ =1m

∑m

1Em − En

×

×[ ∫

d3r φ∗m(r) ( p . eα ) φn(r)∫

d3r′ φ∗n(r′) ( p . eβ ) φm(r′)

+∫

d3r φ∗m(r) ( p . eβ ) φn(r)∫

d3r′ φ∗n(r′) ( p . eα ) φm(r′)]

(406)

where φm(r) are energy eigenstates with eigenvalues Em.——————————————————————————————————

102

3.4.45 Solution 32

We shall first write

eα . eβ =∑i,j

( eα . ei ) δi,j ( ej . eβ )

=∑i,j

( eα . ei )[ pi , xj ]

i h( ej . eβ )

= eα .[ p , r ]i h

. eβ (407)

It shall be understood that the scalar product of the operator p is taken witheα, in both terms of the commutator. On taking the matrix elements betweenφ∗n(r) and φn(r) and inserting the completeness relation, we have

eα . eβ = eα .

∫d3r φ∗n(r)

[ p , r ]− i h

φn(r) . eβ

=i

heα .

∫d3r φ∗n(r) p r φn(r) . eβ

− i

heβ .

∫d3r φ∗n(r) r p φn(r) . eα

=i

heα .

∫d3r φ∗n(r) p

∫d3r′ δ( r − r′ ) r′ φn(r′) . eβ

− i

heβ .

∫d3r φ∗n(r) r

∫d3r′ δ( r − r′ ) p′ φn(r′) . eα

=i

h

∑m

[eα .

∫d3r φ∗n(r) p φm(r)

∫d3r′ φ∗m(r′) r′ φn(r′) . eβ

− eβ .

∫d3r φ∗n(r) r φm(r)

∫d3r′ φ∗m(r′) p′ φn(r′) . eα

](408)

On using the equality

( Em − En ) φ∗m(r) r φn(r) = φ∗m(r) [ H , r ] φn(r)

= − ih

mφ∗m(r) p φn(r) (409)

one obtains the result

eα . eβ =1m

∑m

[1

Em − En

] [ ∫d3r φ∗n(r) ( eα . p ) φ∗m(r)

×∫

d3r′ φm(r′) ( p′ . eβ ) φn(r′)]

+1m

∑m

[1

Em − En

] [ ∫d3r φ∗n(r) ( eβ . p ) φ∗m(r)

103

×∫

d3r′ φm(r′) ( p′ . eα ) φn(r′)]

(410)

as was to be proved.

——————————————————————————————————

3.4.46 Hermitean Operators and Physical Measurements

A physical measurements of a physical quantity A can result in a set of realnumbers a that are the possible results of the measurement. It is reasonable torepresent the operator corresponding to A as a Hermitean operator A, wherethe eigenvalues correspond to the possible set of results a.

Given a state represented by Ψ(r) which has the expansion in terms of theeigenfunctions of A

Ψ(r) =∑

a

Ca φa(r) (411)

then the probability of finding a specific result a in the measurement of A isgiven by

P (a) = | Ca |2 (412)

The orthonormality of the set of eigenfunctions ensures that the probabilitydistribution P (a) is properly normalized.∫

d3r | Ψ(r) |2 = 1

=∫

d3r

∣∣∣∣ ∑a

Ca φa(r)∣∣∣∣2

=∑a,b

C∗b Ca δa,b

=∑

a

| Ca |2 (413)

Thus, the sum over probabilities P (a) is also normalized to unity.

The moments of the probability distribution are given by the matrix elementsof powers of the operator

An =∫

d3r Ψ∗(r) An Ψ(r)

=∫

d3r∑m′

C∗m′Φ∗m′(r) An∑m

Cm Φm(r)

=∫

d3r∑m′

C∗m′Φ∗m′(r)∑m

Cm anmΦm(r)

104

=∑

m,m′

C∗m′ Cm δm,m′ anm

=∑m

| Cm |2 anm (414)

and in particular the average value of A is given by the matrix elements

A =∫

d3r Ψ∗(r) A Ψ(r) (415)

This is often called the expectation value of A in the state Ψ(r).

——————————————————————————————————

3.4.47 Exercise 33

Find the average values of the energy and the momentum of a particle in thestate represented by the wave function

Ψ(r) = C exp[− α ( r − r0 )2

](416)

where α2 = m ω2 h , and the Hamiltonian is given by

H =p2

2 m+

m ω2 r2

2(417)

Also find the mean squared deviation of the (vector) position operator.

——————————————————————————————————

3.4.48 Solution 33

The magnitude of the normalization constant is found from the condition

1 = | C |2∫

d3r exp[− 2 α ( r − r0 )2

](418)

The integral is evaluated to yield

1 = | C |2(

π

2 α

) 32

(419)

Thus, we have the normalization

| C | =(

2 απ

) 34

(420)

105

up to an arbitrary choice of phase.

The average value of the energy is given by the expectation value of theHamiltonian

E =∫

d3r Ψ∗(r) H Ψ(r)

=∫

d3r Ψ∗(r)[

p2

2 m+

m ω2 r2

2

]Ψ(r)

=(

2 απ

) 32∫

d3r

[h2

2 m

(6 α − 4 α2 ( r − r0 )2

)+

m ω2 r2

2

]×

× exp[− 2 α ( r − r0 )2

](421)

The integral is evaluated as

E =3 h2 α

m+

m ω2 r202

− 3 h2 α

2 m+

3 m ω2

8 α

=m ω2 r20

2+

3 h2 α

2 m+

3 m ω2

8 α(422)

where the first term is the energy of a displaced classical harmonic oscillator.On substituting α, one finds

E =m ω2 r20

2+

32h ω (423)

where the last term represents the zero-point energy for the three independentdegrees of freedom.

The average value of the momentum is zero, due to the fact that the wavefunction has no spatially varying phase. Hence, the integrand of the expectationvalue of p is antisymmetric around r0

p = 2 i h α | C |2∫

d3r ( r − r0 ) exp[− 2 α ( r − r0 )2

](424)

Thus, the expectation value of the momentum vanishes.

The average value of r2 is given by

r2 = r20 +3

4 α(425)

Hence, the mean squared deviation is given by

∆r2rms =3

4 α

=3 h

2 m ω(426)

106

which corresponds to the sum of the mean square displacements along the threeCartesian axes.

——————————————————————————————————

3.4.49 Exercise 34

Find the average value of the z component of the orbital angular momentumand square of the z component of the angular momentum for the following threewave functions

Ψ1a(r) = C1 r f(r) sin θ cosϕΨ1b(r) = C1 r f(r) sin θ sinϕΨ1c(r) = C1 r f(r) cos θ (427)

Do you need to evaluate the radial part of the integrations?

——————————————————————————————————

3.4.50 Solution 34

The operator corresponding to the z component of the angular momentum isgiven by

Lz = − i h∂

∂ϕ(428)

Thus, the expectation value of the angular momentum in a state Ψ(r, θ, ϕ) isgiven by the expression

Lz =∫ ∞

0

dr r2∫ π

0

dθ

∫ 2 π

0

dϕ Ψ∗(r, θ, ϕ) Lz Ψ(r, θ, ϕ)

= − i h

∫ ∞

0

dr r2∫ π

0

dθ

∫ 2 π

0

dϕ Ψ∗(r, θ, ϕ)∂

∂ϕΨ(r, θ, ϕ)

(429)

The expectation value for the first two states involve the integration∫ 2 π

0

dϕ sinϕ cosϕ = 0 (430)

whereas the third state is an eigenstate of the z component of the angular mo-mentum operator with eigenvalue zero. Hence, the average value of Lz is zeroin all three states.

107

The mean squared value of the z component of the angular momentum isgiven by

L2z =

∫ ∞

0

dr r2∫ π

0

dθ

∫ 2 π

0

dϕ Ψ∗(r, θ, ϕ) L2z Ψ(r, θ, ϕ)

= − h2

∫ ∞

0

dr r2∫ π

0

dθ

∫ 2 π

0

dϕ Ψ∗(r, θ, ϕ)∂2

∂ϕ2Ψ(r, θ, ϕ)

(431)

In the first two states one has the factors

− h2

∫ 2 π

0

dϕ cosϕ∂2

∂ϕ2cosϕ = h2

∫ 2 π

0

dϕ cos2 ϕ

= h2 π

− h2

∫ 2 π

0

dϕ sinϕ∂2

∂ϕ2sinϕ = h2

∫ 2 π

0

dϕ sin2 ϕ

= h2 π (432)

and the corresponding factor for the third state is identically zero as it is aneigenstate with zero eigenvalue. One recognizes that the expectation valuesare proportional to the normalization integral and, as the state are properlynormalized, one has

L2z = h2 (433)

for the first two states and the expectation value is zero for the last state.

——————————————————————————————————

3.4.51 Exercise 35

Find the average values of the x, y and z component of the orbital angularmomentum and square of the component of the angular momentum for thefollowing three wave functions

Ψ1a(r) = C1 ( x + i y ) f(r)Ψ1b(r) = C1 ( x − i y ) f(r)Ψ1c(r) = C1 z f(r) (434)

where f(r) is a function of the radial distance, defined via r2 = x2 + y2 + z2.What can one conclude about the magnitude of the angular momentum of theseexamples?

——————————————————————————————————

108

3.4.52 Solution 35

We shall first note that the components of the angular momentum operator are

Lx = − i h

(y∂

∂z− z

∂

∂y

)Ly = − i h

(z∂

∂x− x

∂

∂z

)Lz = − i h

(x∂

∂y− y

∂

∂x

)(435)

and that any radially symmetric function f(r) is an eigenfunction of the compo-nents of the angular momentum operator with a zero eigenvalue. For example,on considering the action of the operator Lz on an arbitrary radially symmetricfunction f(r), one has

Lz f(r) = − i h

(x∂f

∂y− y

∂f

∂x

)= − i h

(xy

r− y

x

r

)∂f

∂r

= 0 (436)

as long as r 6= 0. Likewise, the other components also produce zero whenacting on an arbitrary radial function f(r).

The three functions given by Ψ1(r) are seen to be eigenfunctions of Lz witheigenvalues ± h and 0, since

Lz Ψ1a/b(r) = − i h C1 f(r)(x∂

∂y− y

∂

∂x

)( x ± i y )

= − i h C1 f(r)(± x i − y 1

)= ± h C1 f(r) ( x ± i y )= ± h Ψ1a/b (437)

while

Lz Ψ1c(r) = − i h C1 f(r)(x∂

∂y− y

∂

∂x

)z

= 0 (438)

Thus, these states are eigenstates of Lz and the average value of Lz is equal toeither ± h or 0 respectively. Also the average values of L2

z are just 1 and 0.

The average value of Lx and Ly in states Ψ1a/b(r) can be found by notingthat

Lx Ψ1a/b(r) = ∓ h Ψ1c(r)

Ly Ψ1a/b(r) = − i h Ψ1c(r) (439)

109

and the expectation values of Lx and Ly are zero in these states since the origi-nal states and the transformed states are all eigenstates of Lz corresponding todifferent eigenvalues, and the eigenstates of a Hermitean operator with differenteigenvalues are orthogonal.

The expectation value of Lx and Ly in the state Ψ1c(r) can be found bynoting that

Lx Ψ1c(r) = − i h C1 y f(r)

= − h

2

(Ψ1a(r) − Ψ1b(r)

)Ly Ψ1c(r) = + i h C1 x f(r)

= ih

2

(Ψ1a(r) + Ψ1b(r)

)(440)

which are just linear combinations of the eigenstates of Lz. The expectationvalue is zero as the overlap between eigenstates of Lz with different eigenvaluesis zero.

We recognize that the set of functions Ψ1a(r), Ψ1b(r) and Ψ1c(r) form aclosed set under the angular momentum operators, as they operators acting onany member of the set just transform them into linear combinations of the set.

By repeated use of the above relations one can show that∫d3r Ψ∗

1a(r) L2x Ψ1a(r) = − h

∫d3r Ψ∗

1a(r) Lx Ψ1c(r)

=h2

2

∫d3r Ψ∗

1a(r)(

Ψ1a(r) − Ψ1c(r))

=h2

2(441)

and also that∫d3r Ψ∗

1b(r) L2x Ψ1b(r) = + h

∫d3r Ψ∗

1b(r) Lx Ψ1c(r)

= − h2

2

∫d3r Ψ∗

1b(r)(

Ψ1a(r) − Ψ1c(r))

=h2

2(442)

Likewise, for the expectation values of L2y one finds that∫

d3r Ψ∗1a(r) L2

y Ψ1a(r) = − i h

∫d3r Ψ∗

1a(r) Ly Ψ1c(r)

110

=h2

2

∫d3r Ψ∗

1a(r)(

Ψ1a(r) + Ψ1c(r))

=h2

2(443)

and also that∫d3r Ψ∗

1b(r) L2y Ψ1b(r) = − i h

∫d3r Ψ∗

1b(r) Ly Ψ1c(r)

= − h2

2

∫d3r Ψ∗

1b(r)(

Ψ1a(r) + Ψ1c(r))

=h2

2(444)

The expectation value of L2x and L2

y in the state Ψ1c(r) are found from∫d3r Ψ∗

1c(r) L2x Ψ1c(r) = − h

2

∫d3r Ψ∗

1c(r) Lx

(Ψ1a(r) − Ψ1b(r)

)= h2

∫d3r Ψ∗

1c(r) Ψ1c(r)

= h2 (445)

and also that∫d3r Ψ∗

1c(r) L2y Ψ1c(r) = + i

h

2

∫d3r Ψ∗

1c(r) Ly

(Ψ1a(r) + Ψ1b(r)

)= h2

∫d3r Ψ∗

1c(r) Ψ1c(r)

= h2 (446)

The average value of L2 = L2x + L2

y + L2z is 2 h2 for all these states. In fact

one can show that the states Ψ1a(r), Ψ1b(r), Ψ1c(r) are not only eigenstates ofLz but are are all eigenstates of the operator

L2 = L2x + L2

y + L2z (447)

with eigenvalues of 2 h2. This operator corresponds to the squared length of thevector angular momentum. These states can be thought of as having angularmomentum of the same magnitude, but with different orientations.

——————————————————————————————————

111

3.5 Quantization

We have emphasized that because a physical measurement results in a changeof the state of the system, measurements should be represented by operators.We have also asserted that a measurement of a physical quantity A yields areal value a as the result of the measurement, and that immediately after themeasurement has been made it is certain that the resulting state correspondsto the state in which A has the definite value a. This means that, immediatelyafter the measurement, the system is in an eigenstate of the operator A whichhas an eigenvalue a. Furthermore, we have made a choice that the physicaloperators should be Hermitean so that we know that the eigenvalues are real.Thus far, we have been mainly examining the mathematics of operators andwave functions using several physical examples. It remains to discuss how theseoperators are chosen, and what the relationships between them are.

3.5.1 Relations between Physical Operators

A physical operator that appears in Quantum Mechanics is associated with thename of an analogous quantity or variable in Classical Mechanics. That is,we associate a classical variable with a quantum mechanical operator. In theHamiltonian formulation of classical mechanics, the classical quantities may beexpressed in terms of coordinates and momenta and perhaps time, but not interms of derivatives. These relationships are called constitutive or definitive re-lationships. In quantum mechanics, the various operators representing physicalquantities satisfy the same constitutive relationships as the classical variablesdo. That is, the classical quantities in a definitive relationship are replaced bythe quantum operators. However, the order in which the quantum operatorsappear in the definition may have to be chosen judiciously, as they may not becommuting operators.

3.5.2 The Correspondence Principle

The relationships between the various quantum operators are chosen to corre-spond to the relations between variables in classical mechanics, for a specificpurpose. Classical Mechanics has a well defined regime in which it provides anexcellent and accurate description of nature. This is often called the regime ofvalidity of Classical Mechanics. Quantum Mechanics, if it is to provide a betterdescription of nature, should produce results which are identical to the resultsof classical mechanics (to within the experimental accuracy) within the regimeof validity of classical mechanics. By assuming that the same relationships holdbetween classical variables and quantum operators, it is ensured that if thequantum operators can, to a good approximation, be replaced by values, i.e.real variables, then Quantum Mechanics will reduce to classical mechanics. Weshall examine the conditions that are required for a quantum state to lie withinthe regime of validity of Classical Mechanics. One condition that is needed to be

112

satisfied, if quantum operators may be approximated by real variables, is thatthe value associated with the commutator [ A , B ] between two operators A andB should be negligibly small when compared to the products of the values asso-ciated with A and B. The values or variables associated with the operators arenaturally identified with the expectation values for the quantum state in ques-tion. Since, in Hamiltonian mechanics, all quantities are expressed in terms ofthe canonically conjugate momenta and coordinates qi and pj , the commutatorsof all operators can be reduced to the non-zero combinations of the fundamentalcommutator [ qi , pj ]. The Complementarity Principle associates the value ofthe fundamental commutators with the corresponding classical Poisson Brack-ets, up to a constant proportional to h. If the commutators, hence h, can beconsidered to be negligibly small when compared with the average values of pi

and qj in the physical state, then Quantum Mechanics will reduce to ClassicalMechanics in this limit. The Correspondence Principle can be embodied in theconcise statement that Quantum Mechanics should reduce to Classical Mechan-ics in the limit h → 0.

3.5.3 The Complementarity Principle

The complementarity principle has the effect that the Poisson Brackets betweentwo classical variables are to be replaced by the commutator between two quan-tum operators

[ A , B ]PB → [ A , B ]i h

(448)

where h is a universal constant. The equivalence can be seen by examiningthe Poisson Bracket algebra between two products of quantum operators, whilerespecting the order

[ A1 A2 , B1 B2 ]PB = [ A1 , B1 ]PB B2 A2 + B1 [ A1 , B2 ]PB A2

+ A1 [ A2 , B1 ]PB B2 + A1 B1 [ A2 , B2 ]PB

= [ A1 , B1 ]PB A2 B2 + A1 [ A2 , B1 ]PB B2

+ B1 [ A1 , B2 ]PB A2 + A1 B1 [ A2 , B2 ]PB

(449)

Thus, on equating these two results we find

[ A1 , B1 ]PB [ A2 , B2 ] = [ A2 , B3 ]PB [ A1 , B1 ] (450)

which, since the pair of operators A1 and B1 are independent of the pair A2 andB2, leads to the discovery that the quantum Poisson bracket is proportional tothe commutator, with a universal constant of proportionality

[ A1 , B1 ] = i h [ A1 , B1 ]PB (451)

On retaining the classical value of the Poisson bracket between canonically con-jugate coordinates qi and momentum pj , one finds that the quantum operators

113

must satisfy the commutation relations

[ pj , qi ] = − i h δj,i

[ pj , pi ] = [ qj , qi ] = 0 (452)

Having found the commutation relationships between the momenta and coor-dinate operators, the basis for quantization is complete as any operator can beexpressed in terms of the canonical conjugate momenta and coordinates. How-ever, it is usual to find explicit representations for the operators.

The procedure of quantization is simplest and free from ambiguities, if onequantizes in a Cartesian coordinate system and then transforms to other coor-dinate systems.

The classical limit is obtained when the operators can be replaced by theireigenvalues and this implies that the commutators are negligible. As the com-mutation relation between two operators ultimately involves a commutationrelation between position and momentum, the commutation relation is propor-tional to h. The classical limit is approached when h is negligible comparedwith the products of appropriate expectation values in a state.

3.5.4 Coordinate Representation

The coordinate representation is that usually used in Schrodinger’s wave me-chanics. In this case, the position operator r is diagonal in that it is representedby the vector variable r. The momentum operator p is represented by a Her-mitean first order differential operator. The wave function is a function of theposition Ψ(r), as r is diagonal. The momentum operator is represented as thesum of a gradient term and a real function

r = r

p = − i h ∇ + ∇ Λ(r) (453)

where Λ(r) is an arbitrary real function. The gradient of the real function, ∇ Λ,is to be regarded as a function and not as an operator. Usually the arbitraryfunction is removed from the momentum operator by an appropriate choice ofthe phase of the wave function.

p → p′ = − i h ∇

Ψ(r) → Ψ′(r) = exp[

+ iΛ(r)h

]Ψ(r) (454)

This representation is the one we shall mostly use in this class.

For example, in generalized coordinate systems, the infinitesimal displace-ment ds is given in terms of the metric gi,j and the infinitesimal changes in the

114

generalized coordinates dqj via

ds2 =∑i,j

gi,j dqi dqj (455)

The metric is a symmetric tensor gi,j = gj,i. Since the kinetic energy is definedin terms of the square of the infinitesimal displacement per unit time, (ds

dt )2, the

classical Lagrangian is given by

L(q, q) =m

2

∑i,j

qi gi,j qj − V (q) (456)

The canonical momenta are defined by the derivatives of the Lagrangian w.r.t.the generalized velocities

pi =∂L

∂qi= m gi,j q

j (457)

The associated quantum mechanical operators have to be Hermitean with theinner product ∫ ∏

j

dqj √g Φ∗(r) Ψ(r) (458)

where g is given by the determinant of the covariant metric tensor

g = det(gi,j(q)

)(459)

The quantum mechanical operators must satisfy the canonical commutationrules

[ pi , qj ] = − i h δj

i

[ pi , pj ] = 0

[ qi , qj ] = 0 (460)

In the coordinate representation, the generalized momenta are given by

pj = − i h g−14

∂

∂qjg

14 (461)

or

pj = − i h

(∂

∂qj+

12

Γj

)(462)

where

Γj = g−12

(∂

∂qjg

12

)(463)

The position operators are given by

qj = qj (464)

115

As shown by Podolsky8, the Hamiltonian can be written in the symmetric Her-mitean form

H =1

2 m

∑i,j

g−14 pi g

14 gi,j g

14 pj g

− 14 + V (q) (465)

where the contra-variant metric tensor is given by gi,j = g−1i,j .

In planar polar coordinates, (r, ϕ), the infinitesimal length element is givenby

ds2 = dr2 + r2 dϕ2 (466)

Thus, the co-variant metric has the diagonal form

gi,j =(

1 00 r2

)(467)

and the contra-variant metric is

gi,j =(

1 00 1

r2

)(468)

The infinitesimal volume element is given by

dr dϕ√g = dr dϕ r (469)

The generalized momenta operators are given by

pr = − i h r−12∂

∂rr

12

pϕ = − i h r−12∂

∂ϕr

12 (470)

The Hamiltonian is given by

H =1

2 m

[r−

12 pr r pr r

− 12 + r−

12 pϕ

1rpϕ r−

12

]+ V (r)

= − h2

2 m

[1r

∂

∂r

(r∂

∂r

)+

1r2

∂2

∂ϕ2

]+ V (r) (471)

In spherical polar coordinates, (r, θ, ϕ), the infinitesimal length ds is givenby

ds2 = dr2 + r2 dθ2 + r2 sin2 θ dϕ2 (472)

Thus, the co-variant metric has the diagonal form

gi,j =

1 0 00 r2 00 0 r2 sin2 θ

(473)

8B. Podolsky, Phys. Rev. 32, 812 (1928).

116

and the contra-variant tensor is given by the inverse matrix

gi,j =

1 0 00 1

r2 00 0 1

r2 sin2 θ

(474)

Thus, the infinitesimal volume element is given by

dr dθ dϕ√g = dr dθ dϕ r2 sin θ (475)

and the canonical momentum pj are given by

pr = − i h1

r√

sin θ∂

∂rr√

sin θ

= − i h1r

∂

∂rr

pθ = − i h1

r√

sin θ∂

∂θr√

sin θ

= − i h1√sin θ

∂

∂θ

√sin θ

pϕ = − i h1

r√

sin θ∂

∂ϕr√

sin θ

= − i h∂

∂ϕ(476)

Hence, the Hamiltonian has the form

H =1

2 m

[1rpr r

2 pr1r

+1

r√

sin θpθ sin θ pθ

1r√

sin θ+ pϕ

1r2 sin2 θ

pϕ

]+ V (r)

= − h2

2 m

[1r2

∂

∂r

(r2

∂

∂r

)+

1r2 sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1r2 sin2 θ

∂2

∂ϕ2

]+ V (r)

(477)

3.5.5 Momentum Representation

The momentum representation is the representation in which the momentumoperator is diagonal, and the wave function is given as a function of p, Φ(p).The momentum and position operators are given by

p = p

r = + i h ∇p − ∇p Λ(p) (478)

where the gradients are derivatives w.r.t. p . This representation of the opera-tors satisfies the required commutation relations between the components of r

117

and p . In complete analogy to the real-space representation, the momentumdistribution function is given by

P (p) d3p = | Φ(p) |2 d3p (479)

and is normalized to unity ∫P (p) d3p = 1 (480)

The transform between the real space and momentum representation is givenby the integral Fourier Transform,

Ψ(r) =(

12 π h

) 32∫

d3p exp[

+ ip . r

h

]Φ(p) (481)

and its inverse

Φ(p) =(

12 π h

) 32∫

d3r exp[− i

p . r

h

]Ψ(r) (482)

The components of momentum p and r position operators in the coordinaterepresentation are defined by their action on an arbitrary state Ψ(r), whichproduces transformed states ψpx and ψx

ψpx(r) = − i h ∇x Ψ(r)ψx(r) = rx Ψ(r) (483)

The corresponding operators in the momentum representation are found byFourier transforming these equations. On Fourier Transforming and integratingthe first equation by parts, and with appropriate boundary conditions, one findsthe expression for the momentum operator in the momentum representation.Likewise, on differentiating the integral Fourier transform with respect to px ,one obtains a representation of the position operator rx in momentum space.The resulting equations are

φpx(p) = p

xΦ(p)

φx(p) = + i h ∇pxΦ(p) (484)

This confirms the identification of the operators in the momentum representa-tion.

——————————————————————————————————

Example

118

The momentum space representation provides a simple way for finding thebound states of a particle moving in one dimension, in the presence of a partic-ular potential

V (x) → ∞ for x < 0

V (x) = − e2

2 xfor x > 0 (485)

Image potential for electrons near a metal surface

-2

-1.5

-1

-0.5

0

0.5

1

-2 0 2 4 6 8 10

x / x0

V(x

) x 0

/ e2

V(x) = - e2 / 2 x

Figure 21: The one-dimensional image potential V (x) which confines a particleclose to an in-penetrable region for x < 0. The bound state energy is markedby a horizontal line.

——————————————————————————————————

Solution

On multiplying by x, the energy eigenvalue equation becomes[x

p2

2 m− e2

2

]φn(p) = x En φn(p) (486)

A simplification has occurred for our particular choice of potential V (x), sincethe energy eigenvalue equation has been transformed so that it is now linear inx. Therefore,

i h∂

∂p

[p2

2 m− En

]φn(p) =

e2

2φn(p) (487)

119

This is a first order differential equation(i h

p

m− e2

2

)φn(p) = − i h

(p2

2 m− En

) (∂φn

∂p

)(488)

The equation can be put in the form of an integral

∫ p

0

dp′

(∂φn

∂p′

)φn(p′)

= −∫ p

0

dp′( p′

m + ih

e2

2p′2

2 m − En

)(489)

which can be evaluated as

lnφn(p)φn(0)

= − ln(

1 − p2

2 m En

)− i e2

2 h

√2 m− En

tan−1 p√−2 m En

(490)

Hence, one finds

φn(p) =(

φn(0)

1 − p2

2 m En

)exp

[− i e2

2 h

√2 m− En

tan−1 p√− 2 m En

](491)

Thus, φn(p) vanishes as p → ± ∞ and is normalizable. This indicates thatψn(x) is also normalizable for En < 0 and hence satisfies the boundary condi-tion at x → ∞.

The energy eigenvalues are found by insisting that the real space wave func-tion ψn(x) satisfy the boundary condition

ψn(0) = 0 (492)

at x = 0, since V (0) → ∞. The wave function in real space is given in termsof the momentum space wave function through

ψn(x) =1√

2 π h

∫ ∞

−∞dp φn(p) exp

[ip x

h

](493)

so the boundary condition at x = 0 becomes

0 = ψn(0) =1√

2 π h

∫ ∞

−∞dp φn(p) (494)

The boundary condition has the explicit form

0 =1√

2 π h

∫ ∞

−∞dp

(φn(0)

1 − p2

2 m En

)exp

[− i e2

2 h

√2 m− En

tan−1 p√− 2 m En

](495)

120

The imaginary part of the integral is odd in p and, thus, vanishes identically.Therefore, the boundary condition is satisfied, if one requires that the real partvanishes

0 =1√

2 π h

∫ ∞

−∞dp

(φn(0)

1 − p2

2 m En

)cos[e2

2 h

√2 m− En

tan−1 p√− 2 m En

](496)

On integrating this, one finds the condition

sin(π e2

4 h

√2 m− En

)= 0 (497)

Hence, the energy eigenvalues are determined by the boundary condition

π e2

4 h

√2 m− En

= n π (498)

and are given by

En = − 2 m e4

16 h2 n2(499)

which has the form of a Rydberg series.

——————————————————————————————————Example: A Particle Confined in a Uniform Force Field.

Another problem which can be solved in the momentum representation isthat of a particle of mass m moving in one dimension in the presence of apotential

V (x) = − F x if x < 0→ ∞ if x > 0 (500)

The potential is depicted in fig(22). A potential of this type confines electronsnear the interface of semiconductors with different levels of doping9. The infinitepotential excludes the particle from the region x > 0 and introduces theboundary condition at x = 0

φ(0) = 0 (501)

The energy eigenvalue equation becomes

− h2

2 m∂2φ

∂x2− F x φ(x) = E φ(x) (502)

for x < 0. The classical turning point − a is given by

a =E

F(503)

9T.S. Rahman, D.L. Mills and P.S. Riseborough, Phys. Rev. B 23, 4081, (1981).

121

Confining Potential

0

5

10

15

-15 -10 -5 0 5

x/ξξξξ

V(x

) [

unit

s of

(h2 F

2 /2m

)1/3 ]

En

- F x

E0

E1

Figure 22: The potential of the uniform force and the infinite potential confinesthe particle in the region x < 0. The energy eigenvalues En are depicted bythe dashed lines.

On eliminating the energy in the eigenvalue equation in favor of the turningpoint, one finds

− h2

2 m∂2φ

∂x2− F ( x + a ) φ = 0 (504)

A characteristic length scale ξ is defined by

ξ =(

h2

2 m F

) 13

(505)

Hence, after introducing the dimensionless variable z = x + aξ , one finds that

the eigenvalue equation for z ξ < a has the form

∂2φ

∂z2+ z φ = 0 (506)

This equation is related to Airy’s equation and has solutions which are linearcombinations Ai(−z) and Bi(−z). The function Bi(z) diverges exponentiallyas z → ∞. Hence, a solution containing Bi(−z) is not a physically acceptablesolution in the region x < 0. The function Bi(−z) is discarded, so the solutionis of the form

φ(x) = C Ai

(− x+ a

ξ

)(507)

122

for x < 0. The solution has the integral representation

Ai

(− x+ a

ξ

)=

ξ

π

∫ ∞

0

dk cos(

13k3 ξ3 − k ( x + a )

)(508)

The asymptotic form of the wave function can be found by the method ofstationary phase and is given by

φ(x) ∼ C√π

(x+ a

ξ

)− 14

sin[

23

(x+ a

ξ

) 32

+π

4

](509)

within the classically accessible region (0 > x > − a), and decays like

φ(x) ∼ C

2√π

(− x+ a

ξ

)− 14

exp[− 2

3

(− x+ a

ξ

) 32]

(510)

in the classically forbidden region (x < − a). However, to be an eigenfunction,the wave function must also satisfy the boundary condition of eqn(501) at x = 0.Hence, one requires

Ai

(− an

ξ

)= 0 (511)

The boundary condition determines the allowed values of the turning pointa which are labeled by an, or equivalently, by using eqn(503), the boundarycondition determines the energy eigenvalue En. The asymptotic large n valuesof the ratio is given by the formula

an

ξ=(

3 π8

( 4 n − 1 )) 2

3

(512)

This approximate formula is quite accurate even for n = 1, as can be seen fromTable(1). For large n, the zeros vary sub-linearly as an ∼ n

23 . Hence, the

asymptotic form of the energy eigenvalues is given by

En =(h2 F 2

2 m

) 13(

3 π8

( 4 n − 1 )) 2

3

(513)

so the level spacing decreases with increasing n. The first few wave functionsare shown in fig(23).

——————————————————————————————————

3.5.6 Exercise 36

Find the momentum representation of the kinetic energy operator T =p2

2 mstarting from the known expression in the real space or position representation

T = − h2

2 m∇2 (514)

123

Table 1: The first few zeros of the Airy function.

n an / ξ an / ξ(exact value) (asymptotic)

1 2.338 2.3202 4.088 4.0823 5.521 5.5174 6.787 6.7845 7.944 7.9426 9.023 9.0217 10.040 10.0398 11.009 11.0089 11.936 11.935

The lowest energy eigenfunctions

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

-10 -8 -6 -4 -2 0x/ξ

φ n(x

)

φ1(x)φ2(x)φ3(x)

Figure 23: The x dependence of the first few eigenfunctions. The wave functionsare non-zero in the region x < 0. The wave functions decay beyond the classicalturning points and are zero for x > 0.

——————————————————————————————————

124

3.5.7 Exercise 37

Starting from the real space representation of the pseudo-vector angular mo-mentum operator L = r ∧ p, acting on a wave function, find the momentumrepresentation of the operator through integration by parts.

——————————————————————————————————

3.5.8 Exercise 38

Find a momentum space representation for the Hamiltonian of a particle mov-ing in three dimensions, in the presence of an electrostatic potential φ(r). Alsowrite the explicit expression that occurs when the potential produces a uniformelectric field E(r, t) = E0. Find the momentum space eigenfunctions for thisparticular Hamiltonian.

Potential due to a Uniform Force Field

-12

-6

0

6

-4 -2 0 2 4 6 8 10 12

x

V(x

)

V(x) = - F x

E

Figure 24: The potential V (x) = − F x, which gives rise to a unform forcefield F . The energy E is denoted by a horizontal line.

——————————————————————————————————

125

Real-space energy eigenfunctionof a particle moving in a uniform field

-2

-1

0

1

2

-4 -2 0 2 4 6 8 10 12

x

Ψ(x

)

Figure 25: The real space wave function Ψ(x) of a particle of mass m and energyE moving in a uniform force field.

3.5.9 Solution 38

The Hamiltonian in the momentum representation has the form

H =p2

2 m− i h F

∂

∂p(515)

The energy eigenvalue equation is(p2

2 m− i h F

∂

∂p

)Φ(p) = E Φ(p) (516)

and has the solution

Φ(p) = C exp[− i

h F

(p3

6 m− p E

) ](517)

whence the real space wave function is found as

Ψ(x) =2 C√2 π h

∫ ∞

0

dp cos(

p3

6 m h F− p ( F x + E )

h F

)(518)

——————————————————————————————————

126

3.5.10 Exercise 39

Find the two lowest energy eigenvalues and eigenfunctions for a one-dimensionalharmonic oscillator, in the momentum representation.

——————————————————————————————————

3.5.11 Solution 39

The normalized ground state wave function for the one-dimensional Harmonicoscillator is given by

ψ0(x) =(m ω

π h

) 14

exp[− m ω x2

2 h

](519)

The momentum space wave function φ0(p) is given by the Fourier transform

φ0(p) =(

1√2 π h

) ∫ ∞

−∞dx exp

[− i

p x

h

]ψ0(x) (520)

which is evaluated by completing the square and integrating

φ0(p) =(

1π h m ω

) 14

exp[− p2

2 h m ω

](521)

The momentum space wave function is properly normalized, thus P (p) =| φ0(p) |2 is the probability density for finding the particle with momentump. It is notable that the most probable value of the momentum is zero, as theclassical particle would be at rest.

The normalized first excited state wave function is given by

ψ1(x) =(

2√π

) 12(m ω

h

) 34

x exp[− m ω x2

2 h

](522)

The momentum space wave function is evaluated as

φ1(p) =(

2√π

) 12(

1h m ω

) 34

p exp[− p2

2 h m ω

](523)

which is also properly normalized. It should be noted that the most probablevalues of p is non-zero, as in the excited state the classical particle has finitevalues of the momentum.

——————————————————————————————————

127

3.5.12 Exercise 40

Prove that the overlap matrix between two different states Ψ(r) and ψ(r) inposition representation are related via∫

d3r Ψ∗(r) ψ(r) =∫

d3p Φ∗(p) φ(p) (524)

to the corresponding wave functions, Φ(p) and ψ(p), in the momentum spacerepresentation.

——————————————————————————————————

All observable physical quantities can be expressed in terms of matrix ele-ments of operators.

——————————————————————————————————

3.5.13 Solution 40

On evaluating the integral ∫d3p Φ∗(p) φ(p) (525)

by using the transform

φ(p) =(

12 π h

) 32∫

d3r exp[− i

p . r

h

]ψ(r) (526)

and the complex conjugate

Φ∗(p) =(

12 π h

) 32∫

d3r′ exp[

+ ip . r′

h

]Ψ∗(r′) (527)

one has(1

2 π h

)3 ∫d3r

∫d3r′

∫d3p exp

[i

( r′ − r ) . ph

]Ψ∗(r′) ψ(r) (528)

Since the integral over all space leads to the three-dimensional Dirac delta func-tion (

12 π h

)3 ∫d3p exp

[i

( r′ − r ) . ph

]= δ3( r′ − r ) (529)

the overlap between the states is given by∫d3p Φ∗(p) φ(p) =

∫d3r

∫d3r′ Ψ∗(r′) δ( r′ − r ) ψ(r)

=∫

d3r Ψ∗(r) ψ(r) (530)

128

as was to be shown.

——————————————————————————————————

3.5.14 Exercise 41

Consider the following matrix elements in the position representation∫d3r Ψ∗(r)

(− i h

)∇ ψ(r) (531)

and ∫d3r Ψ∗(r) r ψ(r) (532)

Find the equivalent expressions in the momentum space representation.

——————————————————————————————————

3.5.15 Solution 41

Starting with the expression∫d3r Ψ∗(r)

(− i h

)∇ ψ(r) (533)

and substituting the expressions for the Fourier Transformed wave functions

− i h

∫d3r

(2πh)3

∫d3p′ exp[ − i

p′ . r

h] Φ∗(p′) ∇

∫d3p exp[ + i

p . r

h] φ(p)

=∫

d3r

(2πh)3

∫d3p′ exp[ − i

p′ . r

h] Φ∗(p′)

∫d3p p exp[ + i

p . r

h] φ(p)

(534)

Then on interchanging the orders of integration and on recognizing the expres-sion for the Dirac delta function

δ3( p − p′ ) =∫

d3r

( 2 π h )3exp

[i

( p − p′ ) . rh

](535)

one has∫d3r Ψ∗(r)

(− i h

)∇ ψ(r) =

∫d3p Φ∗(p) p φ(p) (536)

129

The procedure for evaluating the matrix elements of the position operatoris quite analogous. Starting with the expression∫

d3r Ψ∗(r) r ∇ ψ(r) (537)

and substituting the expressions for the Fourier Transformed wave functions∫d3r

(2πh)3

∫d3p′ exp[ − i

p′ . r

h] Φ∗(p′) r

∫d3p exp[ + i

p . r

h] φ(p)

= − i h

∫d3r

(2πh)3

∫d3p′ exp[ − i

p′ . r

h] Φ∗(p′)

∫d3p φ(p) ∇p exp[ + i

p . r

h]

(538)

Integrating by parts with respect to p, and as the boundary term is zero, wefind

= + i h

∫d3r

(2πh)3

∫d3p′ exp[ − i

p′ . r

h] Φ∗(p′)

∫d3p exp[ + i

p . r

h] ∇pφ(p)

(539)

Then on interchanging the orders of integration and on recognizing the expres-sion for the Dirac delta function

δ3( p − p′ ) =∫

d3r

( 2 π h )3exp

[i

( p − p′ ) . rh

](540)

one has ∫d3r Ψ∗(r) r ψ(r) = + i h

∫d3p Φ∗(p) ∇p φ(p) (541)

which completes the solution. We see that the matrix elements between thewave functions and operators in the real space representation have identical val-ues to the equivalent expressions in the momentum representation.

——————————————————————————————————

3.5.16 Commuting Operators and Compatibility

If two operators A and B commute, [ A , B ] = 0, these operators are com-patible in the sense that it is possible to find functions that are simultaneouslyeigenfunctions of A and B.

The proof is simple if the eigenfunctions of the operator A are non-degenerate.In this case, the eigenfunctions of A satisfy

A φa(r) = a φa(r) (542)

130

Taking the matrix elements of the commutator of A and B, one has∫d3r φ∗a′(r) [ A , B ] φa(r) = 0 (543)

which is evaluated as

( a′ − a )∫

d3r φ∗a′(r) B φa(r) = 0 (544)

Thus, the operator B only has non-zero matrix elements between states of thesame eigenvalue a = a′. Since by assumption the eigenfunctions of A are non-degenerate and form a complete set, the eigenfunctions of A are simultaneouslyeigenfunctions of B.

The converse is that, if there are simultaneous eigenfunctions of the twooperators A and B, then the operators commute. By definition one has,

A φn(r) = an φn(r)B φn(r) = bn φn(r) (545)

Then, the action of the commutator on the wave function is zero as

A B φn(r) = A bn φn(r)= an bn φn(r)= B an φn(r)= B A φn(r) (546)

and since an arbitrary wave function Ψ(r) can be expanded in terms of φn(r)the commutator is identically zero.

Clearly an example of two commuting operators is given by any operator Aand any function f(A) with a Taylor expansion,

[ A , f(A) ] = 0 (547)

The operators A and f(A) do not form an independent set, as one can beexpressed in terms of the other members of the set. Alternatively if A and Bcommute then

[ g(A) , f(B) ] = 0 (548)

An example of an independent set of commuting operators is given by any ofthe Hermitean operators p

z, p

yor p

x, or alternatively rz, ry or rx. The degen-

eracy of simultaneous eigenvalues of a set of operators is equal to the numberof eigenfunctions corresponding to the set of eigenvalues.

A complete set of commuting operators is defined to be a maximal set ofindependent commuting Hermitean operators. Any state can be uniquely ex-panded in terms of the eigenfunctions of a complete set of commuting Hermitean

131

operators. What exactly constitutes a complete set of operators depends uponthe physical space, i.e. whether it is one-dimensional, two-dimensional or three-dimensional, and whether there is an intrinsic space associated with the particlesuch as spin. If the set is not complete then one can add another independent op-erator to the set until the set becomes complete. A state which corresponds to aneigenfunction of a complete set of commuting operators is uniquely determinedand is completely defined without any remaining arbitrariness. By successivelyadding operators to a set of independent commuting operators, one has reducedthe degeneracy until the simultaneous eigenvalues are non-degenerate and theset of operators is complete.

3.5.17 Non-Commuting Operators

It is impossible to find a set of simultaneous eigenfunctions of non-commutingoperators. Thus, if

[ A , B ] = C (549)

and if φ(r) is a simultaneous eigenfunction then

[ A , B ] φ(r) = C φ(r) = 0 (550)

This would require that φ(r) is also an simultaneous eigenfunction of C witheigenvalue zero, which is very restrictive and very unlikely. Thus, it is impossibleto know the values of A and B at the same time if they are non-commuting. Anexample of this is given by the momentum and position operators correspondingto the same coordinate, p

xand rx.

——————————————————————————————————

3.5.18 Exercise 42

Show that if A and B are Hermitean operators and

[ A , B ] = i C (551)

then C is Hermitean.

——————————————————————————————————

3.5.19 Solution 42

Consider the matrix elements of the commutator∫d3r Φ∗(r) [ A , B ] Ψ(r) =

∫d3r Φ∗(r) A B Ψ(r) −

∫d3r Φ∗(r) B A Ψ(r)

(552)

132

On using the definition of the Hermitean conjugate of the products, we find

=( ∫

d3r Ψ∗(r) B† A† Φ(r))∗

−( ∫

d3r Φ∗(r) A† B† Ψ(r))∗

(553)

and as A and B are Hermitean, one has

= −( ∫

d3r Ψ∗(r) [ A , B ] Φ(r))∗

(554)

Hence, the Hermitean conjugate of the commutator is

[ A , B ]† = − [ A , B ] (555)

so (i C

)†= −

(i C

)(556)

and as the Hermitean conjugate of i is − i , we have

− i C† = − i C (557)

Thus, we recognize C as being Hermitean.

——————————————————————————————————

3.5.20 The Uncertainty Principle

The Heisenberg uncertainty relation provides a quantification of the degree towhich one can determine the values of two non-commuting Hermitean opera-tors, A and B in a specific state Ψ(r).

Consider the deviations ∆A and ∆B and form the positive definite quantity∑n

∣∣∣∣ ∫ d3r Ψ∗(r)(

∆A + i µ ∆B)φn(r)

∣∣∣∣2 ≥ 0 (558)

where µ is a real variable, and φn(r) form a complete set of eigenfunctions. Onusing the completeness relation, one finds

=∫

d3r Ψ∗(r)(

∆A + i µ ∆B) (

∆A − i µ ∆B)

Ψ(r)

=∫

d3r Ψ∗(r)(

∆A2 + i µ [ ∆B , ∆A ] + µ2 ∆B2

)Ψ(r) (559)

and since this has no zeros as a function of µ, the discriminant of this quadraticequation in µ must be greater than zero. Hence,

4∫

d3r Ψ∗(r) ∆A2 Ψ(r)∫

d3r′ Ψ∗(r′) ∆B2 Ψ(r′)

≥( ∫

d3r Ψ∗(r) i [ ∆A , ∆B ] Ψ(r))2

(560)

133

Thus, we have the Heisenberg uncertainty relation

2 ∆Arms ∆Brms ≥∣∣∣∣ ∫ d3r Ψ∗(r) i [ A , B ] Ψ(r)

∣∣∣∣ (561)

The equality only holds if the effects of the operators on the state Ψ(r) areproportional to each other. This can be seen by examining eqn(558) with theequality sign holding. This implies that, if the equality holds, the sums of thesquares of the matrix elements∫

d3r φ∗n(r)(

∆A − i µ ∆B)

Ψ(r) (562)

must be zero for all n. If we expand the transformed state(

∆A − i µ ∆B)

Ψ

in terms of φn (∆A − i µ ∆B

)Ψ(r) =

∑n

Cn φn(r) (563)

then we recognize that we just proved that if the equality is to hold Cn = 0for all n. The completeness condition then implies that we have the expansionof the zero state, so (

∆A − i µ ∆B)

Ψ(r) = 0 (564)

when the equality holds. The value of the constant, µ, can be obtained fromthe inequality by noting that if the equality holds, then as ∆B2

rms > 0 thevalue of µ must be a repeated root. On solving the quadratic equation for therepeated root, we find

µ = i

∫d3r Ψ∗(r) [ A , B ] Ψ(r)

2 ∆B2rms

(565)

Hence, on substituting the value of µ back we have the equation

∆A Ψ(r) =(∫

d3r Ψ∗(r) [ A , B ] Ψ(r)2 ∆B2

rms

)∆B Ψ(r) (566)

This equation is satisfied for the state Ψ(r) which produces the minimum un-certainty in ∆A and ∆B.

Applying the uncertainty relation to the canonically conjugate coordinatesand momenta, we have the inequality

∆pi rms

∆rj rms ≥ h

2δi,j (567)

134

in which the right hand side is independent of the state of the system. Theminimum uncertainty equation takes the form(

− i h∂

∂x− px

)Ψ(x) =

i h

2 ∆x2rms

(x − x

)Ψ(x) (568)

which has the solution

Ψ(x) =(

2 π ∆x2rms

)− 14

exp[−

(x − x

)2

4 ∆x2rms

+ ipx x

h

](569)

——————————————————————————————————

3.5.21 Exercise 43

Consider a particle moving in one dimension in the presence of a potential V (x).Show that the uncertainty in the energy and position are related via

∆Erms ∆xrms =h

2 mp (570)

——————————————————————————————————

3.5.22 Solution 43

The Hamiltonian is given by

H =p2

2 m+ V (x) (571)

The commutator of H and x is found as

[ H , x ] = − ih

mp (572)

Then on using the generalized uncertainty principle, one has

∆Erms ∆xrms =h

2 mp (573)

as was to be proved.

——————————————————————————————————

135

3.5.23 Exercise 44

Derive uncertainty relations for the various components of the orbital angularmomenta. For what states is it possible to measure Lx and Ly simultaneouslywith minimum uncertainty?

——————————————————————————————————

3.5.24 Solution 44

From the commutation relation

[ Lx , Ly ] = i h Lz (574)

one has the uncertainty relation

( ∆Lx )rms ( ∆Ly )rms ≥ h

2Lz (575)

Thus, it is possible to measure Lx and Ly with minimum uncertainty instates Ψ(r) where the average value of Lz is zero

Lz =∫

d3r Ψ∗(r) Lz Ψ(r) = 0 (576)

In particular, the state with total angular momentum zero is a simultaneouseigenstate of Lx, Ly and Lz, with all three eigenvalues being zero.

——————————————————————————————————

3.5.25 Exercise 45

If the Hermitean operators A , B and C satisfy the commutation relations

[ B , C ] = i A

[ A , C ] = i B (577)

then show that the uncertainty relation

∆( A B )rms ∆Crms ≥ 12

(A2 + B2

)(578)

is satisfied.

——————————————————————————————————

136

3.5.26 Solution 45

This exercise is a simple example of the uncertainty principle. The commutatorof the composite operator A B with C is evaluated as

[ A B , C ] = A [ B , C ] + [ A , C ] B

= i

(A2 + B2

)(579)

Hence, on substituting into the general statement of the uncertainty principle,one obtains the result

∆( A B )rms ∆Crms ≥ 12

(A2 + B2

)(580)

which was to be proved.

——————————————————————————————————

3.5.27 Exercise 46

Show that the z component of the angular momentum Lz and the azimuthalangle ϕ satisfy the commutation relations

[ Lz , cosϕ ] = + i h sinϕ[ Lz , sinϕ ] = − i h cosϕ (581)

Hence, derive the uncertainty relations

∆(cosϕ) ∆Lz ≥ h

2sinϕ

∆(sinϕ) ∆Lz ≥ h

2cosϕ (582)

——————————————————————————————————

3.5.28 Solution 46

The uncertainty relations applied to the azimuthal angle are problematic, sinceit is not clear whether the angle ϕ is really only well defined modulo 2π. An angleϕ cannot be distinguished by a single physical measurement from ϕ + m 2 πwhere m is an integer. If, on the other hand, ϕ is considered to be defined onlyon the restricted interval (0, 2π) then the uncertainty principle for ϕ and Lz

would read(∆ϕ)rms (∆Lz)rms >

h

2(583)

137

-2

-1

0

1

2

-2 -1 0 1 2

x

y

sin φφφφ

cos φφφφ

φφφφ

Figure 26: The azimuthal angle ϕ of a point on a unit circle.

which would imply that either there are no eigenstates of Lz, which is false, orthat the uncertainty of ϕ can be infinite, which is contrary to our assumption.The only way out of this quandry is to assert that an angle is not measurable,but that only the coordinates sinϕ or cosϕ are measurable. These functionsuniquely define ϕ modulo 2 π. We need to restrict our attention to functionsand operators which are periodic in ϕ since Lz is only Hermitean on the spaceof functions which are periodic in ϕ.

The commutation relations for the coordinates and angular momentum areevaluated as

[ Lz , cosϕ ] = i h sinϕ[ Lz , sinϕ ] = − i h cosϕ (584)

Simple substitution into the uncertainty relation leads to the equations

( ∆Lz )2 ( ∆ cosϕ )2 ≥ h2

4

(sinϕ

)2

( ∆Lz )2 ( ∆ sinϕ )2 ≥ h2

4

(cosϕ

)2

(585)

138

The above equation does not produce any contradiction with the fact that

Φm(ϕ) =1√2 π

exp[i m ϕ

](586)

are eigenstates of Lz since the expectation value of the trigonometric functionscosϕ and sinϕ are both zero. In this case, one finds that the uncertainty rela-tions reduce to the equality 0 = 0.

——————————————————————————————————

139

3.6 The Philosophy of Measurement

In Quantum Mechanics the measurement process disturbs the system and is,therefore, represented by an operator. The measurement of A produces instan-taneous changes from one state, say Ψ(r) of the system to another. The stateof the system immediately after the measurement is that which corresponds toan eigenstate of the operator A which has the eigenvalue that corresponds tothe result of the measurement a. Prior to the measurement, and the discoveryof the result a, it is impossible to predict which eigenstate the system will take.All that can be predicted is the probability that the measurement will resultin a particular eigenvalue a. This probability is proportional to the squaredmodulus of the expansion coefficient Ca

Ψ(r) =∑

a

Ca φa(r) (587)

Thus, one can view quantum mechanical measurement as a sort of filter, whichinstantaneously filters out the state φa(r) from the initial state of the systemΨ(r). Since this resulting state after the measurement is represented as a pureeigenfunction of A, it has only one expansion coefficient. Therefore, there is aprobability of one for finding the result a in a further measurement of A, as longas this measurement is made very soon after the first measurement was made.

——————————————————————————————————

Example

The Hermitean operators A and B do not commute. The operator A hastwo non-degenerate eigenstates and eigenvalues given by

A Φi(r) = ai Φi(r) (588)

for i = 1 , 2. The operator B also has two non-degenerate eigenstates

B Θi(r) = bi Θi(r) (589)

for i = 1 , 2. The state Φ1(r) can be expressed as a linear superposition ofthe eigenstates Θi(r), as

Φ1(r) = C1 Θ1(r) + C2 Θ2(r) (590)

where C1 and C2 are known complex constants.

(i) Find an expression for Φ2(r) in terms of the Θi(r), and determine theexpansion coefficients in terms of the known quantities Ci.

A quantum mechanical particle is in a state Ψ(r) given by

Ψ(r) =

√16

Φ1(r) +

√56

Φ2(r) (591)

140

(ii) Determine the probabilities of finding the results a1 and a2 in a mea-surement of A. The quantity B is then measured immediately after A has beenmeasured. What is the probability that the measurement of B yields the resultb1?

(iii) If the measurement of A had not been performed, what is the probabil-ity that the measurement of B will result in the measured value b1?

——————————————————————————————————

Solution

Since B is Hermitean the eigenstates of A can be expanded in terms of theeigenstates of B. The eigenstate Φ1(r) is expanded as

Φ1(r) = C1 Θ1(r) + C2 Θ2(r) (592)

and Φ2(r) can be expanded in terms of the complete set of Θi(r) as

Φ2(r) = D1 Θ1(r) + D2 Θ2(r) (593)

where D1 and D2 are to be determined. As Φ2(r) must be orthogonal to Φ1(r),then

0 =∫

d3r Φ∗1(r) Φ2(r)

= C∗1 D1

∫d3r Θ∗

1(r) Θ1(r) + C∗1 D2

∫d3r Θ∗

1(r) Θ2(r)

+ C∗2 D1

∫d3r Θ∗

2(r) Θ1(r) + C∗2 D2

∫d3r Θ∗

2(r) Θ2(r)

(594)

Since Θ1(r) and Θ2(r) form an orthonormal set, one has

0 = C∗1 D1 + C∗2 D2 (595)

Also, as the states are normalized to unity, one has

| C1 |2 + | C2 |2 = | D1 |2 + | D2 |2 = 1 (596)

and so the unknown constants Di are found as

D1 = − C∗2

D2 = C∗1 (597)

up to an arbitrary common phase. Hence, we have

Φ2(r) = − C∗2 Θ1(r) + C∗1 Θ2(r) (598)

141

A measurement of A on the state

Ψ(r) =

√16

Φ1(r) +

√56

Φ2(r) (599)

yields the result a1 with probability P (a1) = 16 and the result a2 occurs with

probability P (a2) = 56 .

If the measurement A has been performed and the result a1 has been found,the system then is in the state Φ1(r). The conditional probability that a subse-quent measurement of B yields the result b1 is given by

P (a1|b1) = | C1 |2 (600)

and the conditional probability that the measurement of B yields the result b2is given by

P (a1|b2) = | C2 |2 (601)

However, if on the other hand the result of the first measurement is a2,the system is in state Φ2(r) just after the first measurement. The conditionalprobability that a subsequent measurement of B yields the result b1 is given by

P (a2|b1) = | C2 |2 (602)

and the conditional probability that the measurement of B yields the result b2is given by

P (a2|b2) = | C1 |2 (603)

The total probability that the second measurement of B will result in thevalue b1, no matter what value of A is measured, is given by

P (A|b1) =16| C1 |2 +

56| C2 |2 (604)

If the measurement of B is performed directly on the state Ψ, then as

Ψ(r) =( √

16C1 −

√56C∗2

)Θ1(r) +

( √56C2 +

√16C∗1

)Θ2(r) (605)

the result b1 occurs with probability

P (b1) =∣∣∣∣√

16C1 −

√56C∗2

∣∣∣∣2 (606)

142

The probability P (b1) is not the same as P (A|b1) as the measurement of Adisturbs the system.

——————————————————————————————————

The time evolution of a system between two successive instants of time isgoverned by the Schrodinger equation, as long as no measurements are madeon the system in the time interval between the two times. In order to solvethe Schrodinger equation, which is a first order differential equation in the timevariable, it is necessary to have one initial condition for the wave function. Thisis usually given by the state of the system at the initial time t0, Ψ(r, t0). Oncethe initial condition is known, the Schrodinger equation will predict how thestate evolves with time until a measurement (say a measurement of A) is per-formed on the system. Let us assume that the measurement occurs at time t1.The measurement at t1 disrupts the evolution process, and sends the systeminto an eigenstate of the measurement operator with a definite eigenvalue. Thevalue of the eigenvalue is the result of the measurement. The probability ofgetting the result a, depends on the expansion coefficient of the state Ψ(r, t1).After the measurement, the system has an new initial state at t1 which is φa(r).This new initial state should be used in the Schrodinger equation to predictthe subsequent evolution of the system from t1 to later times t2, as long asno other measurements occur in this time interval. If there is a second mea-surement (such as a measurement represented by B), the process repeats. Thesystem evolves forward from the time of the second measurement according tothe Schrodinger equation, and the result of the second measurement providesthe new initial condition at the time of the latest measurement.

——————————————————————————————————

3.6.1 Exercise 47

A particle moving in three dimensions is in a quantum mechanical state de-scribed by a wave function

Ψ(r) = C exp[− α r2

](607)

in the presence of a spherically symmetric potential. The Hamiltonian is givenby the operator

H = − h2

2 m∇2 +

m ω2

2r2 (608)

The energy of the state is measured. Find the probability that the system is inthe ground state E0 = 3

2 h ω, where the ground state wave function φ0(r) isgiven by

φ0(r) =(m ω

h π

) 34

exp[− m ω r2

2 h

](609)

143

Determine the probability that the system is found to be in an excited state.

What is the probability that the above system is found to be in a state φ1(r)corresponding to E1 = 5

2 h ω, where

φ1(r) = ( 2 π )12

(m ω

h π

) 54

r cos θ exp[− m ω r2

2 h

](610)

which is also an eigenstate of angular momentum l = 1.

——————————————————————————————————

3.6.2 Solution 47

The normalization constant for Ψ(r) is found from the equation

1 =∫

d3r | Ψ(r) |2

= | C |2∫

d3r exp[− 2 α r2

]= | C |2

∫ 2 π

0

dϕ

∫ π

0

dθ sin θ∫ ∞

0

dr r2 exp[− 2 α r2

]= | C |2 4 π

∫ ∞

0

dr r2 exp[− 2 α r2

](611)

Hence,

| C | =(

2 απ

) 34

(612)

The wave function Ψ(r) can be expanded in terms of the eigenfunctions of H,which are the φn(r). Since the wave functions are normalized, the probabilitythat a measurement of H will result in the value E0 is given by the modulussquare of the expansion coefficient. The expansion coefficient for the state φ0(r)is found from the overlap∫

d3r φ∗0(r) Ψ(r) = 4 π | C |2(m ω

h π

) 32∫ ∞

0

dr r2 exp[− ( α +

m ω

2 h) r2

]=

(4 α m ω

2 h

( α + m ω2 h )2

) 34

(613)

The probability that the system is found in the ground state is given by

P0 =(

4 α m ω2 h

( α + m ω2 h )2

) 32

(614)

144

and the probability that the system ends up in an excited state is given by

Pexc = 1 − P0

= 1 −(

4 α m ω2 h

( α + m ω2 h )2

) 32

(615)

Note that if α has the valueα =

m ω

2 h(616)

then the probability that the system will undergo a transition to a differenteigenstate when H is measured is zero. That is, the system is already in aneigenstate of H with eigenvalue E0 and any subsequent measurements of H(but only if infinitesimally small times have elapsed since Ψ(r) was created) willyield the same result E0.

The probability that the system ends up in the excited state φ1(r) is zero,since on evaluating the expansion coefficient one finds∫

d3r φ∗0(r) Ψ(r)

= ( 2 π )32 | C |

(m ω

π h

) 54∫ π

0

dθ sin θ cos θ∫ ∞

0

dr r3 exp[− ( α +

m ω

2 h) r2

]= 0 (617)

since the integral over θ vanishes identically. The vanishing of this matrix ele-ment is a natural consequence of the conservation of angular momentum for aspherically symmetric potential. The initial state is an eigenstate of the squareof the angular momentum, L2, with eigenvalue zero, and is also an eigenstateof the z component of the angular momentum operator with eigenvalue zero.The state φ1(r) is an eigenstate of the square of the angular momentum L2

with eigenvalue 2 h2 and is an eigenstate of the z component of the angularmomentum with angular momentum zero. Thus, these states correspond toeigenstates of L2 with different eigenvalues. The overlap is zero because statescorresponding to different eigenvalues of Hermitean operators are orthogonal.

——————————————————————————————————

3.6.3 Exercise 48

The initial state of a system is represented by the wave function

Ψ(r) =(m ω

h π

) 34

exp[− m ω ( r − a )2

2 h

](618)

where a = ez a. What is the probability that after an energy measurementthe system will be found in the energy eigenstates φ0(r) and φ1(r) given below?

145

The state φ0(r) is given by

φ0(r) =(m ω

h π

) 34

exp[− m ω r2

2 h

](619)

which is a simultaneous eigenstate of energy E0 = 3/2 h ω and of angularmomentum with l = 0. The state φ1(r) is given by

φ1(r) = ( 2 π )12

(m ω

h π

) 54

r cos θ exp[− m ω r2

2 h

](620)

which is an eigenstate of energy E1 = 5/2 hω and is an eigenstate of angularmomentum with l = 1 and m = 0.

——————————————————————————————————

3.6.4 Solution 48

The probability that the system initially in the state Ψ(r) is found in an eigen-state with energy En is found from the eigenstate expansion

Ψ(r) =∑

n

Cn φn(r) (621)

asP (En) = | Cn |2 (622)

whereCn =

∫d3r φ∗n(r) Ψ(r) (623)

The expansion coefficient C0 is found as

C0 =∫

d3r

(m ω

π h

) 32

exp[− m ω

2 h(r − a

2)2]

exp[− m ω

2 h(r +

a

2)2]

= exp[− m ω

4 ha2

](624)

Thus, we find that the probability of finding the result E0 if the energy of stateΨ(r) is measured is given by

P (E0)l=0 = exp[− m ω

2 ha2

](625)

which decreases exponentially with increasing a.

146

The expansion coefficient C1 is determined from

C1 = ( 2 π )12

∫d3r

(m ω

π h

)2

( z +az

2) exp

[− m ω

h( r2 +

a2

4)]

=(m ω a2

z

2 h

) 12

exp[− m ω

4 ha2

](626)

where az is the z component of the vector displacement a. Hence, one hasthe probability of finding a simultaneous eigenstate of energy E1 and angularmomentum l = 1 and m = 0 is given by

P (E1)l=1,m=0 =(m ω a2

z

2 h

)exp

[− m ω

2 ha2

](627)

——————————————————————————————————

3.6.5 Exercise 49

A particle is confined to move in a one-dimensional interval between x = 0and x = L. The initial state is given by

Ψ(x) =

√30L5

x ( x − L ) (628)

for L > x > 0, and 0 otherwise. Find the probability that, after a measurementof the energy, the system will be found in the eigenstate

φn(x) =

√2L

sinn π x

L(629)

corresponding to energy En = h2 π2 n2

2 m L2

If the energy is measured and found to be equal 9 h2π2

2 m L2 , what is the prob-ability density that in a second measurement the particle will then be found inan interval of width δx around x = L

2 ?

If the measurement of En had not taken place, what would the probabilitydensity for finding the particle at x = L

2 have been?

——————————————————————————————————

147

3.6.6 Solution 49

First we shall evaluate the indefinite integrals∫ x

dx′ x′ sinα x′ = − ∂

∂α

∫ x

dx′ cosα x′

=sinα x − α x cosα x

α2(630)

and ∫ x

dx′ x′2 sinα x′ = − ∂2

∂α2

∫ x

dx′ sinα x′

=( 2 − α2 x2 ) cosα x + 2 α x sinα x

α3

(631)

Then the coefficients Cn in the expansion of Ψ(x) in terms of the eigenfunctionsφn(x) are given by

Cn =∫ L

0

dx′ φ∗n(x) Ψ(x)

=√

60L3

∫ L

0

dx′ x ( x − L ) sinn π x

L

=√

60n3 π3

2 ( cosn π − 1 ) (632)

Hence, only terms with odd values of n appear in the expansion. The probabilityof observing the eigenvalue En is only finite for odd n and is given by

P (n) = | C2n |

P (n) =480n6 π6

( 1 − cosn π ) (633)

Since the energy measurement picks out the non-degenerate eigenfunctioncorresponding to n = 3, the probability density for finding the particle atx = L

2 is given by

P (L/2) =2L

(634)

whereas if the measurement had not occurred then we would have anticipatedthat the probability density would be

P (L/2) =158 L

(635)

Thus, the measurement of the energy has changed the probabilities of the mea-surement of the particle’s position made at an instant later.

——————————————————————————————————

148

3.7 Time Evolution

The evolution equation of a physical quantity B(p, q, t) in Classical Mechanicsis given in terms of the Poisson Brackets as,

dB

dt= [ B , H ]PB +

∂B

∂t(636)

where H is the Hamiltonian, which is a function of the canonically conjugatemomenta p, coordinates q and possibly t. We regard this as an equation ofmotion for the expectation value. To obtain the quantum mechanical equationof motion from the classical equation, we replace the classical quantity B(p, q, t)by the operator B = B(p, q, t), and the classical Hamiltonian H(p, q, t) by theHamiltonian operator H = H(p, q, t). In the non-relativistic limit, the classicalHamiltonian for a particle, of charge q, in an electromagnetic field is given by

H =1

2 m

(p − q

cA(r, t)

)2

+ q φ(r, t) (637)

where we are now representing the position vector as r. The Hamiltonian oper-ator is

H =1

2 m

(p − q

cA(r, t)

)2

+ q φ(r, t) (638)

which in the coordinate representation becomes

H(r, t) =1

2 m

(− i h ∇ − q

cA(r, t)

)2

+ q φ(r, t) (639)

On substituting operators for the classical variables in the classical equation ofmotion, replacing the Poisson Bracket with the commutator divided by the i h,and taking the expectation value in a state Ψ(r, t), one has

i hd

dt

∫d3r Ψ∗(r, t) B(r, t) Ψ(r, t) =

∫d3r Ψ∗(r, t) [ B(r, t) , H(r, t) ] Ψ(r, t)

+ i h

∫d3r Ψ∗(r, t)

∂

∂tB(r, t) Ψ(r, t)

(640)

This should be regarded as an equation which governs the time dependence ofthe expectation value of B(t)

B(t) =∫

d3r Ψ∗(r, t) B(r, t) Ψ(r, t) (641)

The time dependence of the expectation value B(t) has a formal solution givenby

B(t) =∫

d3r Ψ∗(r, 0) U†(t, 0) B(r, t) U(t, 0) Ψ(r, 0) (642)

149

where U(t, t′) is the time evolution operator which describes the implicit timedependence of the system and satisfies the first order equation

i hd

dtU(t, t′) = H(r, t) U(t, t′) (643)

with the boundary condition U(t′, t′) = 1 and U†(t, t′) is its Hermitean conju-gate.

From the solution given in eqn(642) one can derive a number of “pictures”of quantum mechanics. The two most popular are :

(i) The Schrodinger picture.

(ii) The Heisenberg picture.

3.7.1 The Schrodinger Picture.

In this picture, the operators are independent of time (the only exception isif there is some externally imposed time dependence, like a time-dependentelectromagnetic field etc.). That is, the operators only change through theirexplicit time dependence. The implicit time dependence of the system is carriedin the wave function. Thus, in the Schrodinger picture

Ψ(r, t) = U(t, t′) Ψ(r, t′)B(r, t) = B(r, t′) (644)

The time evolution is usually referenced with respect to an initial time t′ whichis usually chosen as t′ = 0. If the time evolution operator U(t, t′) is unitary,the wave function will retain its initial normalization. From this, one finds thatthe implicit time dependence is contained in the wave function and, is governedby the first order differential equation

i hd

dtΨ(r, t) = H(r, t) Ψ(r, t) (645)

with one initial condition. In the momentum space representation, the Schrodingerequation has the form

i hd

dtΦ(p, t) =

(p2

2 m+ q φ( i h ∇p , t)

)Φ(p, t) (646)

——————————————————————————————————

150

3.7.2 Exercise 50

Derive the momentum space form the Schrodinger equation from the real spaceform.

——————————————————————————————————

3.7.3 Solution 50

The time-dependent Schrodinger equation is

i h∂Ψ(r, t)∂t

= − h2

2 m∇2 Ψ(r, t) + q φ(r, t) Ψ(r, t) (647)

in position representation, where Ψ(r, t) is the time-dependent wave function.The time-dependent momentum space wave function Φ(p, t) is related to Ψ(r, t)via the generalized Fourier transform

Ψ(r, t) =(

12 π h

) 32∫

d3p′ Φ(p′, t) exp[

+ ip′ . r

h

](648)

On substituting this relation in the first term on the right hand side, one obtainsthe contribution

− h2

2 m∇2 Ψ(r, t) =

h2

2 m

(1

2 π h

) 32∫

d3p′ Φ(p′, t)p′2

h2 exp[

+ ip′ . r

h

]=

(1

2 π h

) 32∫

d3p′ Φ(p′, t)p′2

2 mexp

[+ i

p′ . r

h

](649)

On performing the inverse Fourier Transform

Φ(p, t) =(

12 π h

) 32∫

d3r Ψ(r, t) exp[− i

p . r

h

](650)

on the time-dependent Schrodinger equation, one has

i h∂Φ(p, t)∂t

=(

12 π h

)3 ∫d3r

∫d3p′ exp

[i

( p′ − p ) . rh

]p′2

2 mΦ(p′, t)

+ q

(1

2 π h

) 32∫

d3r Ψ(r, t) φ(r, t) exp[− i

p . r

h

](651)

where we have switched the order of the wave function and the scalar potentialin the last term. On recognizing the integral over r as being proportional to the

151

integral representation of the delta function δ3(p − p′), and then performingthe integration over p, one obtains

i h∂Φ(p, t)∂t

=p2

2 mΦ(p, t)

+ q

(1

2 π h

) 32∫

d3r Ψ(r, t) φ(r, t) exp[− i

p . r

h

](652)

The last term can be manipulated as

i h∂Φ(p, t)∂t

=p2

2 mΦ(p, t)

+ q

(1

2 π h

) 32∫

d3r Ψ(r, t) φ(i h ∇p, t) exp[− i

p . r

h

]=

p2

2 mΦ(p, t)

+ q φ(i h ∇p, t)(

12 π h

) 32∫

d3r Ψ(r, t) exp[− i

p . r

h

]=

p2

2 mΦ(p, t) + q φ(i h ∇p, t) Φ(p, t) (653)

In the last line we have used the definition of the inverse Fourier transform.Thus, we have derived the equation that governs the time dependence of themomentum space wave function

i h∂Φ(p, t)∂t

=[

p2

2 m+ q φ(i h ∇p, t)

]Φ(p, t) (654)

analogous to the Schrodinger equation in the position representation.

——————————————————————————————————

3.7.4 The Heisenberg Picture.

In the Heisenberg picture, the wave function is chosen to be independent oftime and the operators carry the implicit time dependence which represents thedynamics of the system. Thus, in this picture we have

Ψ(r, t) = Ψ(r, t′) = Ψ(r, 0)B(r(t), t) = U†(t, t′) B(r, t) U(t, t′) (655)

Hence, the equation of motion for the implicit time dependence comes from theequation of motions for the operators and is given by

i hd

dtB(r(t), t) = [ B(r(t), t) , H(r(t), t) ] + i h

∂

∂tB(r(t), t) (656)

152

——————————————————————————————————

3.7.5 Exercise 51

Find the equation of motion for the position operator r(t) in the Heisenbergpicture for a system with the Hamiltonian given by

H(t) =p(t)2

2 m+ V (r(t)) (657)

——————————————————————————————————

3.7.6 Solution 51

We shall examine the Heisenberg equation of motion

i hdr

dt= [ r(t) , H(t) ] + i h

∂r

∂t(658)

at the instant of time, say t = 0. Then in the position representation withr(0) = r and p(0) = − i h ∇ one has no explicit time dependence for r andthe implicit dependence is governed by

i hdr

dt= [ r , H ]

= [ r ,p2

2 m] + [ r , V (r) ]

= 2 i hp

2 m

Thus, we find the operator equation

dr

dt=

p

m(659)

This operator equation has the same form as the momentum velocity relationin classical mechanics.

——————————————————————————————————

153

3.7.7 Exercise 52

Find the Heisenberg picture equation of motion for the momentum operatorp(t) for a system with the Hamiltonian given by

H =p2

2 m+ V (r) (660)

——————————————————————————————————

3.7.8 Solution 52

We shall examine the Heisenberg equation of motion

i hdp

dt= [ p(t) , H(t) ] + i h

∂p

∂t(661)

at the instant of time, say t = 0. Then in the position representation withr(0) = r and p(0) = − i h ∇, one has no explicit time dependence for p andthe implicit dependence is governed by

i hdp

dt= [ p , H ]

= [ p ,p2

2 m] + [ p , V (r) ]

= − i h ∇ V (r) (662)

Thus, we find the operator equation

dp

dt= − ∇ V (r) (663)

This equation has the same form as Newton’s laws, except the classical quanti-ties are replaced by operators.

——————————————————————————————————

The rate of change with respect to time of the operators found in Exercises51 and 52 should look exactly like Hamilton’s equations of motion in classicalmechanics.

——————————————————————————————————

154

3.7.9 Exercise 53

Prove that in the Heisenberg picture the rate of change of the angular momen-tum operator satisfies the equation

d

dtL(t) = − q r(t) ∧ ∇φ(r(t), t) (664)

which is similar to the corresponding equation involving the torque classicalmechanics.

If an operator B does not explicitly depend on time t, then the Heisenbergequation of motion reduces to

i hd

dtB(r(t)) = [ B(r(t)) , H(r(t), t) ] (665)

and so we find that the operator does not change with time if B commutes withthe Hamiltonian H. As the wave function is time independent in the Heisenbergpicture, the expectation values of the operator will also be time independent.Thus, B will be a constant of motion, or a conserved quantity, if

[ B , H ] = 0 (666)

Since any operator commutes with itself, when one substitutes H for B, onefinds that the Hamiltonian commutes with itself. Thus, if the Hamiltonian hasno explicit dependence on time then the energy is a constant of motion.

——————————————————————————————————

3.7.10 Exercise 54

Find the time dependence of a Hermitean operator B which has no explicit timedependence, if the commutator with the Hamiltonian is a complex constant C,i.e.

[ H , B ] = C (667)

——————————————————————————————————

3.7.11 Solution 54

The time dependence of the operator B is governed by the Heisenberg equationof motion

i hdB

dt= [ B , H ] (668)

155

Hence, on using the commutation relation

i hdB

dt= − C (669)

which leads toB(t) = B(0) + i

C

ht (670)

This has an application for the rate of change of momentum p for a particlein a uniform applied field F ,

H =p2

2 m− F . r (671)

which leads to the commutation relation

[ p , H ] = + i h F (672)

Thus, we have the time dependence of the momentum operator

p(t) = p(0) + F t (673)

which is analogous to the classical solution for the momentum of a particle inan applied uniform field.

——————————————————————————————————

3.7.12 Exercise 55

Consider a free particle moving in one dimension, calculate the r.m.s. position∆xrms(t) as a function of time, by using the Heisenberg equations of motionrepeatedly. Do not assume any particular form for the wave packet. Show that

( ∆xrms(t) )2 = ( ∆xrms(0) )2 +

+2m

[12x p + p x (0) − x (0) p (0)

]t +

+( ∆prms(0) )2

m2t2

( ∆prms(t) )2 = ( ∆prms(0) )2 (674)

The average value of the momentum and the r.m.s. momenta are time indepen-dent as a consequence of p being conserved.

——————————————————————————————————

156

3.7.13 Exercise 56

Consider a one-dimensional harmonic oscillator with the Hamiltonian.

H =p2

2 m+

m ω2

2x2 (675)

Derive the equations of motion for the expectation value of x. Solve this equationto show that the mean position oscillates

x(t) = x(0) cos ω t +p(0)m ω

sin ω t (676)

and that the r.m.s. position is given by

∆x2(t) = ∆x

2(0) cos2 ω t +

∆p2(0)

m2 ω2sin2 ω t +

+[

12x p + p x (0) − x (0) p (0)

]sin 2 ω t

m ω(677)

Show that this result reduces to that of the previous exercise when ω → 0.

——————————————————————————————————

The Schrodinger and Heisenberg pictures are equivalent since the only phys-ically important quantities are measurable and, therefore, are in the form of anexpectation value, as displayed in eqn(642). We shall mainly be concerned withthe Schrodinger picture, in the position space representation.

3.7.14 The Schrodinger Equation

In the Schrodinger picture, the operators are time independent and the wavefunctions evolve with time according to

Ψ(r, t) = U(t, t0) Ψ(r, t0) (678)

The time-dependent wave functions satisfy the Schrodinger equation,

i h∂

∂tΨ(r, t) = H(r, t) Ψ(r, t)

i h∂

∂tΨ(r, t) =

[1

2 m

(− i h ∇ − q

cA(r, t)

)2

+ q φ(r, t)]

Ψ(r, t)

(679)

This linear partial differential equation is first order in time and, therefore,requires one initial condition. At the initial time t0 = 0 the initial condition isgiven by knowledge of Ψ(r, 0). For the case where the Hamiltonian contains a

157

time-independent scalar electrostatic potential φ(r), the Schrodinger equationhas the form

i h∂

∂tΨ(r, t) =

[− h2

2 m∇2 + q φ(r)

]Ψ(r, t) (680)

In this case, where the Hamiltonian H is time independent, one has a formalsolution

Ψ(r, t) = U(t, 0) Ψ(r, 0)

= exp[− i

H t

h

]Ψ(r, 0) (681)

which contains the wave function at the initial time. It should be noted that,since the time evolution operator U(t, 0) satisfies the equation

U†(t, 0) U(t, 0) = I (682)

it is a unitary operator. The unitarity condition ensures that the normalizationof the wave function Ψ(r, t) = U(t, 0) Ψ(r, 0) is independent of time.

Let us now consider a Hamiltonian which contains a time-dependent externalfield, due to some classical source. For example,

H(t) =p2

2 m+ V (r) − ( F . r ) sinω t (683)

We should be aware that the Hamiltonian at different times does not commute

[ H(t1) , H(t2) ] = − i h ( F . p )(

sinω t1 − sinω t2

)(684)

When the Hamiltonian has an explicit time dependence, H(t), due to anexternal field then the solution of the Schrodinger equation can be expressed as

Ψ(r, t) = U(t, 0) Ψ(r, 0) (685)

where the time dependence of the evolution operator is governed by the equation

i h∂

∂tU(t, 0) = H(t) U(t, 0) (686)

On integrating the time evolution equation, one obtains

U(t, 0) = 1 − i

h

∫ t

0

dt1 H(t1) U(t1, 0) (687)

158

This involves the time evolution operator U on both sides of the equation. Weshall iterate this equation, that is, substitute the expression for U(t, 0) with thetwo terms on the right hand side as an expression for U(t1, 0) in the right handside. After the substitution, the equation now possesses two completely knownterms, one of order zero in H and the second linear in H and the unknown terminvolving U appears in a third term and has a coefficient which is proportional toH2. On further iterating this equation an infinite number of times, one obtainsan infinite series in the operator H,

U(t, 0) = 1 − i

h

∫ t

0

dt1 H(t1) −1h2

∫ t

0

dt1

∫ t′

0

dt2 H(t1) H(t2) + . . . (688)

where one recognizes that the Hamiltonian at earlier times always appears tothe right of Hamiltonian at later times. The general higher order term in thisinfinite series is given by

+(− i

h

)n ∫ t

0

dt1

∫ t1

0

dt2 . . .

∫ tn−1

0

dtn H(t1) H(t2) . . . H(tn) (689)

which we again note has a time ordered structure, in that the time tm+1 isalways earlier than the time tm, and H(tm+1) is on the right of H(tm). In thegeneral term in this expansion, the intermediate times in the integration arealways ordered and the Hamiltonian operators are also time ordered.

We could permute the dummy variables t1 to tn in many ways, in fact thereare n! such permutations. Each permutation would result in a different timeordering of the labels, and each permutation gives the same result as long asthe Hamiltonians are ordered such that the Hamiltonians at consecutive timesare adjacent to each other with the earlier time to the right. This procedure iscalled time ordering and was first introduced by the Gian-Carlo Wick. We shalldefine an operator W which time orders a string of Hamiltonians according tothe time label. We note that each of the n! permutations involve integrationsover different sectors of the “space” defined by the set of times (t1, t2, . . ., tn) where t > tm > 0, and that the n! sectors completely run over theentire “volume”, tn, of this space. Hence, we can re-write the general term asan integration over the entire volume, as long as we keep the Hamiltonian in thecorrect time ordered places, and this is done by following the instruction W ,

+1n!

(− i

h

)n ∫ t

0

dt1

∫ t

0

dt2 . . .

∫ t

0

dtn W H(t1) H(t2) . . . H(tn)

(690)

On summing the series, we find

U(t, 0) = W exp[− i

h

∫ t

0

dt′ H(t′)]

(691)

This is the best we can do for an arbitrary time-dependent Hamiltonian. How-ever, if H has no explicit time dependence, then the effect of W can be neglected

159

since all the H(tm) are the same. In this case we recover our previous result,namely

U(t, 0) = exp[− i

h

∫ t

0

dt′ H

]= exp

[− i

H t

h

](692)

——————————————————————————————————

3.7.15 Exercise 57

Find the time dependence of a state representing a particle in a field free envi-ronment, with the initial condition that at t = 0 the particle is in momentumeigenstate given by

Ψ(r, 0) =(

12 π h

) 32

exp[

+ ip0. r

h

](693)

——————————————————————————————————

3.7.16 Solution 57

The time dependence of the wave function is given by the Schrodinger equation

i hd

dtΨ(r, t) = H Ψ(r, t) (694)

The solution can found first by integrating

Ψ(r, t) = Ψ(r, 0) − i

h

∫ t

0

dt′ H Ψ(r, t′) (695)

and then by iterating

Ψ(r, t) = Ψ(r, 0) − i

h

∫ t

0

dt′ H Ψ(r, 0) − 1h2

∫ t

0

dt′∫ t′

0

dt” H2 Ψ(r, t”)

(696)The equation simplifies, since the initial wave function Ψ(r, 0) is also an eigen-state of the free-particle Hamiltonian H

H =p2

2 m(697)

160

so thatH Ψ(r, 0) = E(p

0) Ψ(r, 0) (698)

where

E(p0) =

p20

2 m(699)

is the energy eigenvalue. On substituting the eigenvalue for H in eqn(696) anditerating an infinite number of times, one finds that

Ψ(r, t) = exp[− i

E(p0) t

h

]Ψ(r, 0)

=(

12 π h

) 32

exp[

+ i

(p0. r − E(p

0) t

h

) ](700)

——————————————————————————————————

3.7.17 Time Development of a Wave Packet

A wave packet Ψ(r, t) representing a free particle evolves according to

i h∂

∂tΨ(r, t) = − h2

2 m∇2 Ψ(r, t) (701)

since the potential is a constant q φ(r, t) = V0 which can be absorbed into thereference energy, and can be set to zero.

The evolution equation can be solved by Fourier decomposition, since for freeparticles the momentum eigenfunctions are also eigenfunctions of the Hamilto-nian. By Fourier decomposition of the initial wave function Ψ(r, 0), one findsΦ(p, 0) as the coefficient of the momentum eigenfunction, φp(r). Then

φp(r) ∝ exp[ip . r

h

](702)

the wave function in momentum representation evolves with time according to

Φ(p, t) = exp[− i

E t

h

]Φ(p, 0) (703)

where E =p2

2 m . Thus, as the initial value of the wave function in momentumrepresentation is given by

Φ(p, 0) =(

12 π h

) 32∫

d3r′ Ψ(r′, 0) exp[− i

p . r′

h

](704)

161

we find that the time development of the wave function in momentum represen-tation is governed by

Φ(p, t) =(

12 π h

) 32∫

d3r′ Ψ(r′, 0) exp[− i

(p . r′ + E t

)h

](705)

On performing the inverse Fourier transform of Φ(p, t), we obtain the wavefunction in the position representation at time t as, Ψ(r, t) where

Ψ(r, t) =(

12 π h

) 32∫

d3p Φ(p, 0) exp[

+ i

(p . r −

p2

2 mt

)/ h

](706)

Thus, all one needs to do is to evaluate the Fourier transform Φ(p, 0) of the ini-tial condition, eqn(704), and the inverse transform, given in eqn(706), to obtainthe time dependence of the free particle wave function.

——————————————————————————————————

3.7.18 Exercise 58

Given the initial wave function

Ψ(r, 0) =(

12 π δx2

) 34

exp[− ( r − r0 )2

4 δx2

]exp

[+ i

p0. r

h

](707)

find the time evolution of the wave function for a free particle, the time de-pendence of the average position, r(t), and the mean squared deviation of theparticle’s position, ∆r(t)2.

——————————————————————————————————

3.7.19 Solution 58

The momentum space wave function, at time t = 0 is found from the initialwave function in real space via the Fourier transform

Φ(p, 0) =(

12 π h

) 32∫

d3r Ψ(r, 0) exp[− i

p . r

h

]=

(2 δx2

π h2

) 34

exp[−

( p − p0

)2 δx2

h2

]exp

[− i

r0 . ( p − p0

)h

](708)

162

Dispersion of a Gaussian wave packet

0

0.5

1

1.5

-1.5 -1 -0.5 0 0.5 1 1.5x

l Ψ(x

,t) l2

t = 0t = 2t0

t = 4t0

t = t0

t = 3t0

Figure 27: The time dependence of the probability density for an initial Gaussianwavepacket, Ψ(x, t). The wavepacket moves with velocity p

m and disperses withincreasing t.

Then, the momentum space wave function at time t is given by

Φ(p, t) =(

2 δx2

π h2

) 34

exp[−

( p − p0

)2 δx2

h2

]× exp

[− i

r0 . ( p − p0

)h

]exp

[− i

p2 t

2 m h

](709)

The time dependence only enters via the phase. Since the momentum probabil-ity distribution involves the modulus squared wave function | Φ(p, t) |2, we findthat the momentum distribution is time independent. Furthermore the averagemomentum is p

0and the distribution has a r.m.s. width, in each of the three

orthogonal directions, which is equal to h2 δx , independent of time. The wave

packet does not spread in momentum space.

The real space wave function at time t is given by the inverse Fourier trans-form of Φ(p, t)

Ψ(r, t) =(

12 π h

) 32∫

d3p Φ(p, t) exp[

+ ip . r

h

](710)

163

This has the explicit form

Ψ(r, t) =(

12 π h

) 32∫

d3p

(2 δx2

π h2

) 34

exp[−

( p − p0

)2 δx2

h2

]× exp

[− i

r0 . ( p − p0

)h

]exp

[− i

p2 t

2 m h

]exp

[+ i

p . r

h

](711)

Combining the exponential factors and completing the square by changing vari-able from p to z where

z = p − p0− i

2( r − r0 − t

p0

m )( δx2 + i t h

2 m )(712)

one finds that

Ψ(r, t) =(

12 π ( δx2 + i t h

2 m )

) 34

exp[−

( r − r0 − 1m p

0t )2

4 ( δx2 + i t h2 m )

]

× exp[

+ i( p

0. r − p2

2 m t )h

](713)

From this we see that the wave packet corresponds to a real space probabilitydistribution function in which the maximum moves along the classical trajectory,

r = r0 +p0t

m(714)

The width of the distribution function, however, increases with increasing t,corresponding to a spreading of the wave packet. The root mean squared widthis given by

∆x2(t)rms =

√δx4 + t2

h2

4 m2(715)

which increases linearly with t for sufficiently large t. This corresponds to thespread in the positions of a set of classical particles which have a momentumdistribution of width h

2 δx .

——————————————————————————————————

3.7.20 Time Evolution and Energy Eigenfunctions

The analysis of the time development of a wave packet for a free particle can beextended to motion in an arbitrary potential, with the knowledge of the energyeigenfunctions of the time-independent Hamiltonian operator.

H φn(r) = En φn(r) (716)

164

The energy eigenstates evolve according to the Schrodinger equation

i h∂

∂tφn(r, t) = H φn(r, t)

= En φn(r, t) (717)

since they are eigenstates. The solution of the Schrodinger equation yields thetime dependence of the energy eigenstates as

φn(r, t) = exp[− i

En t

h

]φn(r) (718)

The energy eigenstates are often called stationary states, as the only time de-pendence occurs through a phase factor, which is generally unobservable.

Given that the eigenstates of the Hamiltonian form a complete set, we canexpand the initial wave function in terms of the energy eigenstates

Ψ(r, 0) =∑

n

Cn φn(r) (719)

where the expansion coefficients are given by

Cn =∫

d3r′ φ∗n(r′) Ψ(r′, 0) (720)

Hence, we have the formal solution

Ψ(r, t) =∑

n

Cn φn(r) exp[− i

En t

h

]Ψ(r, t) =

∑n

∫d3r′ φn(r, 0) φ∗n(r′) exp

[− i

En t

h

]Ψ(r′, 0)

(721)

This expression for the time-dependent wave function involves an infinite sumof energy eigenfunctions, and satisfies the time-dependent Schrodinger equationas can be seen by direct substitution. We are implicitly assuming that thesummation converges for all times, including t = 0 where the sum reduces tothe Dirac delta function

δ( r′ − r ) (722)

due to the completeness relation. Thus, at t = 0, the integration over r′ yieldsthe initial wave function Ψ(r, 0).

——————————————————————————————————

165

3.7.21 Exercise 59

A two-level system has energy eigenfunctions φ1 and φ2 with energy eigenvaluesE1 and E2 respectively. The operator A does not commute with the Hamiltonianand has eigenfunctions θ1 and θ2 corresponding to the eigenvalues a1 and a2. Ifthe system is in the eigenstate θ1 at time t = 0 (i.e. Ψ(r, 0) = θ1(r)), showthat it is possible for the expectation value of A to obey the equation

a(t) =(a1 + a2

2

)+(a1 − a2

2

)cos

( E1 − E2 ) th

(723)

where the expectation value is defined by

a(t) =∫d3r Ψ∗(r, t) A Ψ(r, t) (724)

Also find the probabilities that a measurement of A at time t will give the resulta1 or a2, respectively.

——————————————————————————————————

3.7.22 Solution 59

The principle of linear superposition, together with the Schrodinger equationfor the time-independent Hamiltonian yields, the wave function of the systemas

Ψ(r, t) = C1 φ1(r) exp[− i

E1 t

h

]+ C2 φ2(r) exp

[− i

E2 t

h

](725)

At t = 0 the system is in an eigenstate of A with eigenvalue a1. Thus,

θ1(r) = C1 φ1(r) + C2 φ2(r) (726)

By orthogonality of the eigenstates of A we have

θ2(r) = C∗2 φ1(r) − C∗1 φ2(r) (727)

Hence, on solving for the energy eigenstates in terms of the eigenstates of A, wefind that the state of the system is given by

Ψ(r, t) =(| C1 |2 exp

[− i

E1 t

h

]+ | C2 |2 exp

[− i

E2 t

h

] )θ1(r)

+ C1 C2

(exp

[− i

E1 t

h

]− exp

[− i

E2 t

h

] )θ2(r)

(728)

166

The expectation value of A in this state is given by

a(t) =∫

d3r Ψ∗(r, t) A Ψ(r, t)

= a1

(| C1 |4 + | C2 |4 + 2 | C1 |2 | C2 |2 cos

( E1 − E2 ) th

)+ a2 2 | C1 |2 | C2 |2

(1 − cos

( E1 − E2 ) th

)(729)

This yields the required result if

| C1 |4 + | C2 |4 =12

(730)

and2 | C1 |2 | C2 |2 =

12

(731)

On using the normalization condition in the last equation, one finds a quadraticequation. The solution of the quadratic equation leads to

| C1 |2 = | C2 |2 =12

(732)

The probability of finding the result a1 is given by

P (a1) =(| C1 |4 + | C2 |4 + 2 | C1 |2 | C2 |2 cos

( E1 − E2 ) th

)(733)

and the probability of finding the result a2 is

P (a2) = 2 | C1 |2 | C2 |2(

1 − cos( E1 − E2 ) t

h

)(734)

Note that, due to the normalization, the sum of the probabilities is unity

P (a1) + P (a2) =(| C1 |2 + | C2 |2

)2

= 1 (735)

and independent of t.

——————————————————————————————————

167

3.7.23 The Correspondence Principle

The correspondence principle, as previously stated, implies that quantum me-chanics should reduce to classical mechanics in the limit h→ 0. We shall showhow quantum mechanics reproduces the Hamilton-Jacobi equations of classicalmechanics, in this limit.

Starting from the Schrodinger equation in the case where the vector potentialis absent

i h∂

∂tΨ(r, t) =

(− h2

2 m∇2 + q φ(r, t)

)Ψ(r, t) (736)

one can write the complex wave function in terms of two real functions A(r, t)and S(r, t) representing the amplitude and phase

Ψ(r, t) = A(r, t) exp[

+i S(r, t)

h

](737)

Then, on substituting this form in the Schrodinger equation and taking the realparts and the imaginary parts, one has two equations. The first represents thereal part of the Schrodinger equation

−(∂S

∂t

)=

12 m

(∇ S(r, t)

)2

+ q φ(r, t) − h2

2 m∇2 A

A(738)

and the second represents the imaginary part

i h∂A

∂t= − i

h

2 mA ∇2 S − h

m

(∇ A

).

(∇ S

)(739)

On taking the limit h→ 0, one finds that the equation for the phase S satisfiesthe Hamilton - Jacobi equation of classical mechanics,

−(∂S

∂t

)=

12 m

(∇ S(r, t)

)2

+ q φ(r, t) (740)

In the Hamilton - Jacobi formulation of classical mechanics one identifies theclassical momentum with the gradient of S

p = ∇ S (741)

The equation for the amplitude A can be reduced to(∂A2

∂t

)= − ∇ .

(A2 ∇ S

m

)(742)

by multiplying by an integrating factor of A(r, t). This last equation is iden-tified as a continuity equation which relates the probability density ρ(r, t) =

168

| Ψ(r, t) |2 = A(r, t)2, to the probability current density j(r, t) defined byj(r, t) = ∇ S

m ρ. The continuity equation is

∂

∂tρ(r, t) + ∇ . j(r, t) = 0 (743)

Thus, to lowest order in h, the phase of the wave function satisfies the Hamilton- Jacobi equation and the amplitude satisfies a continuity equation.

Applying this analysis to a wave packet, one finds that the wave packetfollows a trajectory which is governed by the classical action S that appearsin the exponent. The phase factor has the slowest variation along the classicaltrajectory. The continuity condition shows that the probability density ρ willvanish outside the region that moves along the classical trajectory. Furthermore,the value of any physical quantity can be approximated by the classical valueif the wave function is dominated by the exponential factor. In this case onecan consider the unitary transformation, which when applied to the momentumoperator results in

exp[− i S

h

]p exp

[+

i S

h

]= p + ∇ S (744)

This leads to the operator being represented by the classical value of the momen-tum and a quantum correction. Thus, for an arbitrary operator B = B(r, p)one has

exp[− i S

h

]B(r, p) exp

[+

i S

h

]= B(r, p+∇S) (745)

If the amplitude A(r, t) has a sufficiently slow variation, then the expectationvalue of B can be replaced by its classical value, since∫

d3r Ψ∗(r) B Ψ(r) =∫

d3r A exp[− i S

h

]B(r, p) exp

[+

i S

h

]A

=∫

d3r A B(r, p+∇S) A

≈ B(r,∇S) (746)

where the real function B is evaluated on the classical trajectory on which A isnon-vanishing.

Let us examine the conservation of probability density that is inherent inthe exact quantum mechanical Schrodinger equation.

3.7.24 The Continuity Equation and Particle Conservation

An exact continuity equation can be derived from the Schrodinger equation,which ensures that the number of particles (i.e. the probability of finding the

169

particle anywhere), is conserved. The form of the equation is

∂

∂tρ(r, t) + ∇ . j(r, t) = 0 (747)

where ρ(r, t) = Ψ∗(r, t) Ψ(r, t) is the probability density and (in the absenceof a vector potential) the probability current density j(r, t) is given by

j(r, t) =h

2 m i

[Ψ∗(r, t) ∇ Ψ(r, t) − Ψ(r, t) ∇ Ψ∗(r, t)

](748)

The expression for the probability current density, j(r, t), corresponds to the realfunction given by the velocity operator sandwiched between the wave functions

j(r, t) = Real[

Ψ∗(r)p

mΨ(r)

](749)

The current density is real, and the real part is found by adding the expressionand its complex conjugate, and then dividing by two.

The proof that j(r, t) and ρ(r, t) satisfy the continuity equation starts fromthe Schrodinger equation

i h∂

∂tΨ(r, t) =

(− h2

2 m∇2 + q φ(r, t)

)Ψ(r, t) (750)

and its complex conjugate

− i h∂

∂tΨ∗(r, t) =

(− h2

2 m∇2 + q φ(r, t)

)Ψ∗(r, t) (751)

Multiplying the first equation by Ψ∗(r, t) and the complex conjugate equationby Ψ(r, t) and then subtracting it, one obtains

i h∂

∂tΨ∗(r, t) Ψ(r, t) = − h2

2 m

(Ψ∗(r, t) ∇2 Ψ(r, t) − Ψ(r, t) ∇2 Ψ∗(r, t)

)= − h2

2 m∇ .

(Ψ∗(r, t) ∇ Ψ(r, t) − Ψ(r, t) ∇ Ψ∗(r, t)

)(752)

which was to be proved.

Thus, the total probability of finding the particle is unity, and remains unityfor all times. This can be seen by integrating the continuity equation over alarge fixed volume of space∫

d3r∂

∂tρ(r, t) +

∫d3r

(∇ . j(r, t)

)= 0 (753)

170

On using Green’s theorem to express the integral over the divergence of thecurrent as an integral over the surface of the volume S, one has∫

d3r∂

∂tρ(r, t) +

∫d2S . j(r, t) = 0 (754)

If the current falls to zero at the boundary of the surface, one is left with∫d3r

∂

∂tρ(r, t) = 0 (755)

which, as the volume of integration is fixed, becomes

d

dt

∫d3r ρ(r, t) = 0 (756)

Thus, one finds that the wave function that satisfies the Schrodinger equationremains normalized ∫

d3r Ψ∗(r, t) Ψ(r, t) = 1 (757)

for all times.

——————————————————————————————————

Example: The Probability Current Density of a Spherical Wave.

Spherical Waves have wave functions of the form

Ψ(r) =exp

[± i k r

]r

(758)

where k is a constant. This wave function is not normalizable and is independentof time. It should be interpreted as representing a steady state of a large numberof beams of particles which are either diverging from or converging on the origin.

The probability current density j(r) is calculated from

j(r) =h

2 m i

[Ψ∗(r) ∇ Ψ(r) − Ψ(r) ∇ Ψ∗(r)

](759)

which involves the gradient of the spherical wave. The gradient is evaluated as

∇Ψ(r) = er

(± i k − 1

r

) exp[± i k r

]r

(760)

which is purely radial. Hence, the current density is purely radial

j = ± h k

m

1r2

er (761)

171

and falls off as r−2, with increasing r. The exponential phase factor in the wavefunctions cancel in the expression for the probability current density. The r−2

dependence of the probability current density is a consequence of continuityin the steady state, since it shows that the same number of particles, per unittime, pass through any spherical shell surrounding the origin, irrespective of theradius r of the shell. This is seen as follows: The number of particles passingthrough an infinitesimal area d2S per unit time is given by the product

j . d2S = ± h k

m

er

R2. d2S (762)

where R is the radial distance of the surface element from the origin. Onintegrating this over the surface of a sphere of radius R, for which d2S is alsoradially directed, one finds

− dN

dt=∫d2S . j = ± 4 π

h k

m(763)

since, the surface area of the sphere is given by 4 π R2. Therefore, the numberof particles passing through the spherical surface is independent of R.

Spherical Waves

Figure 28: Outgoing Spherical Waves.

The continuity condition is not satisfied at the origin and, depending on thesign of the probability current, the origin either acts as a source or sink where

172

particles are created or annihilated.

——————————————————————————————————

173

4 Applications of Quantum Mechanics

4.1 Exact Solutions in One Dimension

One-dimensional problems are instructive, as they are often exactly soluble, andare relatively easy to solve. They also provide good illustrations of the princi-ples of quantum mechanics. However, some phenomena found in one dimensionmay lead to conclusions that do not hold in higher dimensions. For example,an attractive potential in one dimension always leads to the formation of abound state, but this does not remain true in higher dimensions. Nevertheless,we shall now examine some exactly soluble one-dimensional eigenstate problems.

4.1.1 Particle Confined in a Deep Potential Well

Let us consider the problem of a quantum mechanical particle, moving in onedimension. The particle is confined by a potential to move in a region of spacewhere 0 < x < L, because the potential satisfies

0 if L > x > 0V (x) =

∞ otherwise (764)

A classical particle with energy E < ∞ is forbidden to exist in the region ofspace outside the interval of length L. The particle’s motion consists of an orbitwhich moves back and forth with momentum p = ±

√2 m E . The average

position of the particle is given by x = L2 and the root mean squared deviation

of the particle’s position is given by L√12

. The average value of the momentumis zero because it spends half the time moving forward with momentum + p andhalf the time moving backward with momentum − p. However, the root meansquared value of the particle’s momentum is exactly equal to p.

Inside the interval L > x > 0, the energy eigenvalue equation takes theform

− h2

2 m∂2

∂x2ΨE(x) = E ΨE(x) (765)

which corresponds to the motion of a free particle. This equation is satisfied bythe form

ΨE(x) = A exp[

+ i k x

]+ B exp

[− i k x

](766)

with arbitrary A and B, and p = h k =√

2 m E . Within the region definedby L > x > 0, the form of the solution appears to correspond to a linearsuperposition of momentum eigenstates, with momentum ± p.

174

Infinite Potential Well

-20

0

20

40

60

80

100

120

-1 -0.5 0 0.5 1 1.5 2

x / L

V(x

)V0

V0 >> E > 0 E

Figure 29: The potential of a deep potential well of depth V0, where V0 → ∞.The zero of energy is chosen to be the bottom of the potential well.

Outside the interval, the energy eigenvalue equation has the form[− h2

2 m∂2

∂x2+ V0

]ΨE(x) = E ΨE(x) (767)

where the limit V0 → ∞ should be taken. In this range only the formΨE(x) = 0 satisfies this equation. This corresponds to the classical expec-tation that the particle should only be found within the interval of length L.

The solution must be continuous for all values of x, even at the boundariesx = 0 and x = L. Thus, we require ΨE(x) to satisfy the boundary conditions

ΨE(0) = ΨE(L) = 0 (768)

That is, the wave function inside the interval should match smoothly onto thesolution outside the interval. This yields the two conditions

A + B = 0 (769)

at x = 0 and

A exp[

+ i k L

]+ B exp

[− i k L

]= 0 (770)

175

at x = L. The first condition yields A = − B and so inside the potentialwell the wave function is just

ΨE(x) = A

(exp

[+ i k x

]− exp

[− i k x

] )= 2 i A sin k x (771)

for L > x > 0. The continuity condition at x = L then becomes

ΨE(L) = 2 i A sin k L

= 0 (772)

Since we require A 6= 0, this places a condition on the allowed values of k whichmust satisfy k L = n π for integer values of n. The negative integer valuesof n give the same solution as the positive integers, and thus are not needed.The value n = 0 gives a trivial solution which doesn’t describe a particle atall. The solution n = 0 is not normalizable, and there is no probability forfinding a particle in an interval ∆x even if L > x > 0. Thus, the allowedvalues of k are given by kn = n π

L where n is a positive integer greater thanzero. Therefore, the allowed values of the energy eigenvalue are also given bythe discrete values

En =h2 k2

n

2 m

=h2 n2 π2

2 m L2(773)

The energy eigenvalues are discrete as the particle is trapped in a region of finitespatial extent. The corresponding energy eigenfunctions are given by

Ψn(x) = C sinn π x

L(774)

where C is an arbitrary complex number. The normalization condition∫ + ∞

− ∞dx | Ψn(x) |2 = 1 (775)

reduces to ∫ L

0

dx | Ψn(x) |2 = 1 (776)

since Ψn(x) vanishes outside the region of integration. Using the form of Ψn(x)one finds that | C | is given by

1 = | C |2∫ L

0

dx sin2 n π x

L

= | C |2∫ L

0

dx12

(1 − cos

2 n π xL

)

176

= | C |2 12

(x − L

2 n πsin

2 n π xL

) ∣∣∣∣x = L

x = 0

= | C |2 L

2(777)

Thus, the properly normalized wave functions are,

Ψn(x) =

√2L

sinn π x

L

[Θ(x) − Θ(x− L)

](778)

where Θ(x) is the Heaviside step function defined by

Θ(x) = 1 if x > 0Θ(x) = 0 if x < 0 (779)

Eigenfunctions of the infinite potential well

0 0.2 0.4 0.6 0.8 1x / L

Ψ ΨΨΨn(

x)

ΨΨΨΨ1(x)

ΨΨΨΨ2(x)

ΨΨΨΨ5(x)

ΨΨΨΨ3(x)

ΨΨΨΨ4(x)0

0

0

0

0

Figure 30: The energy eigenfunctions Ψn(x) of a particle in a deep potentialwell.

These wave functions are the eigenfunctions of a Hermitean operator H. Wesee that the eigenvalues are real

En =n2 π2 h2

2 m L2(780)

177

and are non-degenerate. The eigenfunctions are orthogonal, as can be seen bydirectly evaluating the integral∫ + ∞

− ∞dx Ψ∗

m(x) Ψn(x) =2L

∫ L

0

dx sinm π x

Lsin

n π x

L

=1L

∫ L

0

dx

(cos

(m − n ) π xL

− cos(m + n ) π x

L

)=

1π

(sin( m − n ) π

m − n− sin( m + n ) π

m + n

)= δm − n (781)

The overlap integral is zero for m 6= n and for m = n the wave functionsare normalized to unity, as can be seen via invoking l’Hopital’s rule. Also onerecognizes that functions in the finite interval L > x > 0 can, by the theoryof finite Fourier series, be expanded in terms of the set of eigenfunctions. Thechoice of interval for the function which is to be expanded in a Fourier series,usually extends from − L to + L, and the series contains terms of cos n π x

L inaddition to the terms sin n π x

L . However, if we insist that our wave functionsΨ(x) over the enlarged interval are odd, i.e. Ψ( − x ) = − Ψ( x ) then onlythe sin terms remain in the Fourier series expansion. That is, the eigenfunctionsform a complete set, only in the interval L > x > 0 .

The average values of the position of the particle in these energy eigenstatesis given by

x =∫ L

0

dx Ψ∗n(x) x Ψn(x)

=2L

∫ L

0

dx x sin2 n π x

L

=1L

∫ L

0

dx x

(1 − cos

2 n π xL

)(782)

However, the integral can be evaluated via∫dx x cosα x =

∂

∂α

∫dx sinα x

= − ∂

∂α

(cosα x

α

)=

(cosα x + x α sinα x

α2

)(783)

Thus, we find that the average position is given by

x =L

2(784)

178

Probability Density

0 0.2 0.4 0.6 0.8 1

x / L

l Ψ Ψ Ψ Ψn(

x) l2

n = 5

n = 1

n = 2

n = 3

n = 4

0

0

0

0

0

Figure 31: The quantum mechanical probability density P (x) = | Ψn(x) |2 fora particle in an energy eigenstate with energy eigenvalue E is compared withthe classically calculated probability density for a particle with energy E. Theclassical probability density is calculated by averaging over the initial conditionsfor classical trajectories with energy E. In accordance with the correspondenceprinciple, the probability densities should agree in the limit h → 0.

which is exactly the same as the classical average.

The mean squared deviation in the particle’s position is given by

∆x2 =∫ L

0

dx Ψ∗n(x)

(x − L

2

)2

Ψn(x)

=2L

∫ L

0

dx

(x2 − L2

4

)sin2 n π x

L

=1L

∫ L

0

dx

(x2 − L2

4

) (1 − cos

2 n π xL

)=

1L

∫ L

0

dx

(x2 − L2

4

) (1 − cos

2 n π xL

)(785)

This integral is evaluated with the aid of the equality∫dx x2 cosα x = − ∂2

∂α2

∫dx cosα x

179

= − ∂2

∂α2

(sinα x

α

)=

(− 2 sinα x + 2 α x cosα x + α2 x2 sinα x

α3

)(786)

Thus, the mean squared deviation becomes

∆x2 =L2

12− L2

2 n2 π2(787)

which for large enough n becomes identical to the classical value. This is anothermanifestation of the correspondence principle at work. These results correspondto a particle which has an average position half way along the allowed interval,and fluctuates back and forth with excursions proportional to the root meansquared deviation.

The average value of the momentum is given by

p = − i h

∫ ∞

−∞dx Ψ∗

n(x)d

dxΨn(x)

= − i2 hL

∫ L

0

dx sinn π x

L

d

dxsin

n π x

L

− i2 hL

∫ L

0

dx sin2 n π x

L

d

dx

[Θ(x) − Θ(x− L)

]= − i

2 π n hL2

∫ L

0

dx sinn π x

Lcos

n π x

L

− i2 hL

∫ L

0

dx sin2 n π x

L

[δ(x) − δ(x− L)

]= − i

π n h

L2

∫ L

0

dx sin2 n π xL

= iπ n h

L

(cos 2 n π − 1

)2 n π

= 0 (788)

The average value of the momentum of the quantum mechanical particle is zero,like the classical average of p. The root mean squared momentum is just givenby

p2 = − h2

∫ ∞

−∞dx Ψ∗

n(x)d2

dx2Ψn(x)

= − 2 h2

L

∫ L

0

dx sinn π x

L

d2

dx2sin

n π x

L

180

− 22 h2

L

∫ L

0

dx sinn π x

L

(d

dxsin

n π x

L

)d

dx

[Θ(x) − Θ(x− L)

]− 2 h2

L

∫ L

0

dx sin2 n π x

L

d2

dx2

[Θ(x) − Θ(x− L)

]= − 2 h2

L

∫ L

0

dx sinn π x

L

d2

dx2sin

n π x

L

− 22 h2

L

∫ L

0

dx sinn π x

L

(d

dxsin

n π x

L

) [δ(x) − δ(x− L)

]− 2 h2

L

∫ L

0

dx sin2 n π x

L

d

dx

[δ(x) − δ(x− L)

]= +

2 π2 n2 h2

L3

∫ L

0

dx sinn π x

Lsin

n π x

L

=π2 n2 h2

L2(789)

In these calculations, the contributions from the discontinuities in the first andsecond derivatives of the Heaviside step functions vanish identically. Thus, theaverage momentum is zero just like for the classical particle, and the root meansquared deviation of the momentum is equal to p = h n π

L . This is similar tothe classical particle, except that the r.m.s. momentum is quantized, that is, itonly has discrete values.

The momentum space wave function Φn(p) is given by

Φn(p) =1√

2 π h

∫ L

0

dx

√2L

sinn π x

Lexp

[− i

p x

h

](790)

which yields

Φn(p) =2 i L√π h L

n π sin(

p L2 h + n π

2

)(n π

)2

−(

p Lh

)2 exp[− i

2

(p L

h+ n π

) ](791)

Hence, the momentum distribution function P (p) is given by

P (p) = | Φn(p) |2

=4 Lπ h

(n π

)2

sin2

(p L2 h + n π

2

)[ (

n π

)2

−(

p Lh

)2]2 (792)

which is peaked at p = ± n π hL .

181

Momentum distribution function

0

0.02

0.04

0.06

0.08

0.1

-40 -30 -20 -10 0 10 20 30 40

p L / h

Pn(p)

n = 4

Figure 32: The momentum distribution function Pn(p) = | Φn(p) |2 of aparticle in a deep potential well, with n = 4.

The uncertainty in the momentum is given by ∆prms = h n πL and the

uncertainty in the position is

∆xrms = L

√ (112

− 12 n2 π2

)(793)

The uncertainties satisfy the uncertainty principle,

∆xrms ∆prms ≥ h

2(794)

4.1.2 Time Dependence of a Particle in a Deep Potential Well

Consider a particle in an initial state given by an arbitrary wave function Ψ(x; 0),which satisfies the boundary conditions Ψ(0) = Ψ(L) = 0. Then on decom-posing this into energy eigenstates via the expansion

Ψ(x; 0) =∑

n

Cn

√2L

sinn π x

L(795)

182

one finds that the expansion coefficients are given by

Cn =

√2L

∫ L

0

dx sinn π x

LΨ(x; 0) (796)

Then, the solution of the Schrodinger equation at time t is given by the expres-sion

Ψ(x; t) =∑

n

Cn exp[− i

h π2 n2

2 m L2t

] √2L

sinn π x

L(797)

in which each term has a different time dependence.

4.1.3 Exercise 60

Find the time development of the initial wave function Ψ(x; 0) = δ( x − x0 ), ofa particle in a box. Plot the time dependence of the probability density P (x; t)for finding the particle at position x at various times. Find the time dependenceof the average value of the particle’s position x(t).

4.1.4 Particle Bound in a Shallow Potential Well

Let us consider the problem of a quantum mechanical particle moving in onedimension, in the presence of a potential given by

− V0 if L > x > 0V (x) =

0 otherwise (798)

The energy of our particle is E < 0.

Classically, the particle would confined by the potential to move in a regionof space where 0 < x < L and the energy of the particle would be restrictedby 0 > E > − V0. Hence, the classical particle would be confined to theregion near the origin.

Inside the interval L > x > 0, the energy eigenvalue equation takes theform

− h2

2 m∂2

∂x2ΨE(x) = ( E + V0 ) ΨE(x) (799)

which corresponds to the motion of a free particle. This equation is satisfied bythe form

ΨE(x) = A exp[

+ i k x

]+ B exp

[− i k x

](800)

183

Shallow Potential Well

-1.5

-1

-0.5

0

0.5

-1 -0.5 0 0.5 1 1.5 2

x / L

V(x

) / V

0

- V0

- V0 < E < 0E

Figure 33: A potential well V (x) of width L and depth − V0. The bound stateenergy E is indicated by a horizontal line.

with arbitrary A and B, and p = h k =√

2 m ( E + V0 ) . Within thisregion, the form of the solution appears to correspond to a linear superpositionof momentum eigenstates, with momentum ± p.

Outside the interval L > x > 0, the particle should satisfy the equation,

− h2

2 m∂2

∂x2ΨE(x) = E ΨE(x) (801)

which corresponds to the motion of a free particle but with negative energy. Thekinetic energy instead of being positive must be negative, so this implies thatwe should consider imaginary values for the momentum. That is, we should trylooking for solutions in which the momentum is given by p = i h κ. In factthe energy eigenvalue equation is satisfied by the general form

ΨE(x) = Cα exp[

+ κ x

]+ Dα exp

[− κ x

](802)

where h κ =√

2 m | E | . The coefficients Cα and Dα are arbitrary andhave to be separately determined in each region. The form of the solution cor-responds to a linear superposition of momentum eigenstates, with imaginaryvalue of the momentum ± i h κ.

184

In the region where x > L the solution must be of the form of a decreasingexponential

ΨE(x) = D> exp[− κ x

]for x > L (803)

The term proportional to the coefficient C> has been discarded as the wave func-tion must be square integrable, and since the term proportional to C> woulddiverge when x → ∞. The above wave function falls to zero at the boundary,x → ∞.

In the region where x < 0, the solution must be of the form

ΨE(x) = C< exp[

+ κ x

]for x < 0 (804)

as this tends to zero as x → − ∞. The term proportional to the coefficientD< has been discarded as the wave function must be square integrable.

The above forms of the solution involve four (as yet) unknown constants A,B, C and D. These constant must be determined by the boundary conditions atx = 0 and x = L, and the normalization condition. As we shall presently see,the wave function and its derivative must be continuous at each of these points.Thus, there appears to be five conditions and only four unknowns. However,if we also regarded the energy E as a variable, then the above conditions mayrestrict the energy to discrete allowed values10.

The wave function should be continuous and also the first derivative shouldbe continuous at both the points x = 0 and x = L. The need for the derivativeof the wave function be continuous at these points can be seen by writing thepotential as

V (x) = + V0

(Θ( x − L ) − Θ( x )

)(805)

and integrating the energy eigenvalue equation over a small interval of width2 ε around either x = L or x = 0. Then

− h2

2 m∂

∂xΨE(x)

∣∣∣∣L + ε

L − ε

+∫ L + ε

L − ε

dx V (x′) ΨE(x′) = E

∫ L + ε

L − ε

dx ΨE(x′)

(806)Since the wave function is continuous, the term proportional to the energy hasa magnitude of 2 E ΨE(L) ε. Also, as we are considering potentials V (x) whichonly have finite discontinuities and the wave function is continuous, the term

10More precisely, the boundary conditions at | x | → ∞ are responsible for quantizingthe energy. This can be shown by solving the eigenvalue equation for ΨE(x) (up to thenormalization) for an arbitrary energy. If one enforces one boundary condition (say at x →− ∞) and integrates the equation, then the solution will diverge at the other boundary(x → + ∞) unless E is restricted to have certain discrete values.

185

Energy-dependence of trial energy-eigenfunctions of the attractive square well potential

-0.5

0

0.5

1

1.5

2

-1 -0.5 0 0.5 1 1.5 2

x/L

ΨE(x

)

E > E0

E < E0

E = E0

Figure 34: The energy-dependence of trial energy eigenfunctions ΨE(x) for ashallow potential well. The solution is chosen to satisfy the boundary conditionat x→ −∞ and is normalized to unity at x = 0. It is seen that the solution doesnot satisfy the boundary condition at x→∞ unless E = E0. This suggests thatthe discreteness of the bound state energy is caused by the boundary conditionsat infinity.

involving the integral of the potential has a magnitude of − V0 ε ΨE(L). In thelimit ε → 0, one finds

∂

∂xΨE(x)

∣∣∣∣x = L + ε

=∂

∂xΨE(x)

∣∣∣∣x = L − ε

(807)

Thus, the first derivative of the wave function must be continuous at the bound-ary.

The continuity of the wave function at x = L yields the condition

ΨE(L) = A exp[

+ i k L

]+ B exp

[− i k L

]= D exp

[− κ L

](808)

and the continuity of the first derivative yields

∂

∂xΨE(L) = i k

(A exp

[+ i k L

]− B exp

[− i k L

] )

186

= − κ D exp[− κ L

](809)

The continuity of the wave function at x = 0 results in the equation

ΨE(0) = A + B = C (810)

and the continuity of the first derivative yields

∂

∂xΨE(0) = i k

(A − B

)= + κ C (811)

Solving for A and B in terms of C, we find

A =i k + κ

2 i kC

B =i k − κ

2 i kC (812)

The two continuity conditions at x = L can be used to eliminate D andyields another equation, involving A and B

− κ

(A exp

[+ i k L

]+ B exp

[− i k L

] )= i k

(A exp

[+ i k L

]− B exp

[− i k L

] )(813)

However, since A and B have already been determined in terms of C, and sinceC can be factored out (the magnitude of C is determined by the normalizationcondition), the above condition restricts the possible values of the energy E.Hence, we obtain the secular equation which determines the eigenvalues

exp[

+ 2 i k L]

=

(( i k − κ )2

( i k + κ )2

)(814)

On taking the square root, we can rewrite this as

exp[i k L

]= ± ( i k − κ )2

( k2 + κ2 )(815)

which can be rewritten as an equation for the ratio of the imaginary and realparts, in which case the ± sign drops out

tan k L =2 k κ

k2 − κ2(816)

187

which we shall refer to later on. What we really need to do is to take the fourthroot of the equation (814), which yields the two equations

exp[i k L / 2

]= ± i

( i k − κ )√( k2 + κ2 )

(817)

and

exp[i k L / 2

]= ± ( i k − κ )√

( k2 + κ2 )(818)

which correspond to the four fourth roots of unity. On taking the ratio ofthe imaginary to the real parts, these two equations can be re-written as twoconditions

k tank L

2= + κ

k cotk L

2= − κ (819)

The above two conditions determine the bound state energies E as the wavevector k and the value κ are related to the energy through

κ2 = − 2 m E

h2

k2 =2 m ( E + V0 )

h2 (820)

The conditions on k can be solved graphically for the bound states, by squaringthe equation.

For the first condition, as both k and κ are positive, squaring introducesspurious solutions, corresponding to regions where k tan k L

2 is negative. Wethen plot

κ2 =2 m V0

h2 − k2 (821)

as a function of k, on the same graph as the function

k2 tan2 k L

2(822)

For positive V0 we have a parabola that intersects the k axis at k =√

2 m V0h2 .

The function k2 tan2 k L2 is never negative, but falls to zero at k = 2 n π

L forinteger n and diverges to infinity at k = ( 2 n + 1 ) π

L . Thus, for k in the range

0 < k <√

2 m V0h2 , one has roots near k = ( 2 n + 1 ) π

L . Since tan k L2

is negative in each alternate segment 2 n πL > k > ( 2 n − 1 ) π

L where theslope of our function is negative, the solution in these regions are spurious. Thesolutions of our condition corresponds to the intersections of our function with

188

Self Consistency Equation

0

50

100

150

200

0 1 2 3 4 5 6 7 8

k L

k2 cot2kL/2

2mV0/h2- k2

k2tan2kL/2

Figure 35: Graphical solution of the equations determining the bound stateenergies of the shallow potential well. The k values where the inverted parabolaintersects the curves k2 tan2 kL

2 or k2 cot2 kL2 represents possible allowed wave

vectors k. As explained in the text, there is a maximum of one allowed solutionfor k in each interval of π

L , as the second solution is spurious.

the parabola, but only in the intervals ( 2 n + 1 ) πL > k > 2 n π

L where ourfunction has a positive slope.

The energy eigenstates satisfying the second condition in eqn(819) can befound by a similar procedure. However, we should note that in this case k isreal and − κ is negative. Thus, the roots of the second condition should liein the intervals where tan k L

2 is negative, and the intersections in the regions( 2 n + 1 ) π

L > k > 2 n πL are spurious and should be discarded. Thus, the

regions of k which contain bound states are interchanged for the two conditions.

Let us note that for large enough V0 the bound states correspond to the setof k values given by

k ≈ n π

L(823)

if the values of k are sufficiently small and are limited by

h2 k2

2 m V0 (824)

189

In this case, the eigenvalues are related to the eigenvalues of the infinite poten-tial well.

Since the values of k are found to be discrete, the energy eigenvalues are alsodiscrete. The allowed values of E are found to be in the range 0 ≥ E ≥ − V0.The only remaining undetermined constant is the value of C which can be foundby normalizing the wave function.

Bound state wave functions for a shallow potential well

-1.5

-1

-0.5

0

0.5

1

1.5

-1 -0.5 0 0.5 1 1.5 2

x / L

Ψ ΨΨΨn(

x)

ΨΨΨΨ2(x)

ΨΨΨΨ1(x)

ΨΨΨΨ0(x)

Figure 36: Bound state wave functions Ψn(x) for a shallow potential well.

——————————————————————————————————

4.1.5 Exercise 61

A particle of mass m is confined to move in a one-dimensional potential V (x)where

V (x) → ∞ x < 0V (x) = − V0 0 < x < a

V (x) = 0 a < x (825)

Derive the equation for the bound state energy.Find the minimum value of V0 which will produce a bound state.

190

An asymmetric potential well

-1.5

-1

-0.5

0

0.5

-1 -0.5 0 0.5 1 1.5 2

x / L

V(x

) / V

0

- V0

- V0 < E < 0E

Figure 37: A one-dimensional potential V (x), which excludes the particle fromthe region x < 0. The minimum value of the potential is − V0. The potentialV0 has to exceed a minimum value if the particle is to be bound.

——————————————————————————————————

4.1.6 Solution 61

The bound state wave function is given by

Ψ(x) = 0 for x < 0

where V (x) → ∞. In the region of positive x, the energy eigenvalue equationis given by

− h2

2 m∂2Ψ∂x2

+ V0 Ψ(x) = E Ψ(x) (826)

inside the potential well where a > x > 0, and

− h2

2 m∂2Ψ∂x2

= E Ψ(x) (827)

outside the potential well where x > a. This has the solution

Ψ(x) = A exp[

+ i k x

]+ B exp

[− i k x

](828)

191

Bound state wave function with minimal energy

0

0.05

0.1

0.15

-1 0 1 2

x / L

Ψ ΨΨΨ0(

x)

ΨΨΨΨ0(x)

Figure 38: The form of the bound state wave function, for a value of the potentialV0 which is just sufficently strong enough to create one bound state.

where the energy eigenvalue is given by

E = − V0 +h2

2 mk2 (829)

Since the wave function has to be continuous at x = 0, one has

A + B = 0 (830)

Thus, the solution inside the well is of the form

Ψ(x) = 2 i A sin k x (831)

Outside the potential well, the eigenvalue equation has the solution of the form

Ψ(x) = C exp[− κ x

]+ D exp

[+ κ x

](832)

However, as the wave function must vanish as x → ∞ then one has D = 0.Thus, the wave function for x > a is simply given by

Ψ(x) = C exp[− κ x

](833)

192

and the energy eigenvalue is given by

E = − h2

2 mκ2 (834)

Continuity at x = a yields

2 i A sin k a = C exp[− κ a

](835)

and continuity of the derivative yields

2 i k A cos k a = − κ C exp[− κ a

](836)

Hence on dividing the above equations, one has

k cot k a = − κ (837)

On squaring this equation, one has

k2 cot2 k a =2 m V0

h2 − k2 (838)

ork2 =

2 m V0

h2 sin2 k a (839)

Thus,2 m ( V0 + E )

h2 =2 m V0

h2 sin2 k a (840)

If a bound state is just formed, it has the maximum bound state energywhich is zero, E = 0. Hence, in this case, the above equation simplifies to

1 = sin2 k a (841)

Thus, the minimum value of k is found as

k =π

2 a(842)

and since E = 0, one has

0 = − V0 +h2

2 mπ2

4 a2(843)

Hence, the minimum value of V0 needed to produce a bound state is given by

V0 =h2 π2

8 m a2(844)

193

Alternatively, the result can be derived from the condition that the kineticenergy inside the well and outside the well must be minimized. The wave func-tion inside the well is oscillatory, but must have the minimum number of oscil-lations. The wave function outside the well must decay slowly, therefore, thedecay constant for the wave function for x > a is given by κ = 0. Thus, thematching condition for the derivative of the wave function shows that the slopemust be zero at the edge of the potential well x = a and, since E = 0. Sincethe wave function has a maximum at x = a, one has

k cos k a = 0 (845)

from which one has k = π2 a . Hence, as

E = − h2

2 mκ2

= − V0 +h2

2 mk2 (846)

the minimum value of the potential needed to produce a bound state is givenby

V0 =h2 π2

8 m a2(847)

——————————————————————————————————

4.1.7 Scattering from a Shallow Potential Well

The shallow potential well is described by

V (x) = − V0 for 0 < x < L

V (x) = 0 otherwise (848)

Classically, the energy E = 0 separates the bound states, which have E < 0,and the scattering states, which have E > 0. We shall now consider stateswhere the energy E is greater than zero so that the particle may travel toinfinity.

The allowable forms for the wave function in the three regions are

Ψ(x) = A exp[

+ i k x

]+ B exp

[− i k x

]0 > x

Ψ(x) = C exp[

+ i k′ x

]+ D exp

[− i k′ x

]L > x > 0

Ψ(x) = F exp[

+ i k x

]+ G exp

[− i k x

]x > L

(849)

194

Scattering from a Potential Well

-1.5

-1

-0.5

0

0.5

-1 -0.5 0 0.5 1 1.5 2

x / L

V(x

) / V

0

- V0

E > 0

Figure 39: Scattering from a shallow potential well V (x) of depth − V0. Theenergy of the scattering states E is > 0, so classically the particle is free tomove to infinity.

where the energy eigenvalue is given by E = h2 k2

2 m and E + V0 = h2 k′2

2 mHowever, as we only have four matching conditions, the six coefficients are notdetermined, and k is a continuous real variable. We can choose one coefficient toset the normalization of the wave function. The other coefficient can be chosenas desired. We shall choose the coefficient G to be zero. This represents a beamof particles being produced at x → − ∞ with momentum p falling incident onthe potential at x = 0. Some fraction of the beam is reflected back towardsx → − ∞, with momentum − p and a certain fraction being transmitted tox → + ∞ with momentum p. The vanishing of G corresponds to the absenceof a source of particles, with momentum − p, at x → ∞.

The matching conditions at x = 0 yield

A + B = C + D

k ( A − B ) = k′ ( C − D ) (850)

The matching conditions at x = L are given by

C exp[

+ i k′ L

]+ D exp

[− i k′ L

]= F exp

[+ i k L

]

195

k′(C exp

[+ i k′ L

]− D exp

[− i k′ L

] )= k F exp

[+ i k L

](851)

This pair of equations can be solved for C and D in terms of the amplitudeof the transmitted beam F ,

C exp[

+ i k′ L

]=

(k′ + k

2 k′

)F exp

[i k L

]D exp

[− i k′ L

]=

(k′ − k

2 k′

)F exp

[i k L

](852)

whereas the amplitudes of the incident beam A and reflected beam B are givenby

A =(k + k′

2 k

)C +

(k − k′

2 k

)D

B =(k − k′

2 k

)C +

(k + k′

2 k

)D (853)

Thus, we find

A exp[− i k L

]=

(2 k k′ cos k′ L − i ( k2 + k′2 ) sin k′ L

2 k k′

)F

B exp[− i k L

]=

(( k′2 − k2 ) i sin k′ L

2 k k′

)F (854)

The amplitude of the incident beam is assumed to be known, as it can becontrolled by the particle accelerator. Thus, we shall express the amplitudesof the transmitted and reflected beams as ratios with respect to the incidentbeam. The ratio of the transmitted beam to the incident beam is given by

F

A=( 2 k k′ exp

[− i k L

]2 k k′ cos k′ L − i ( k2 + k′2 ) sin k′ L

)(855)

The ratio of the amplitudes of the reflected beam to the incident beam isgiven by

B

A=(

( k′2 − k2 ) i sin k′ L2 k k′ cos k′ L − i ( k2 + k′2 ) sin k′ L

)(856)

The amplitude of the reflected beam is proportional to k′2 − k2 and thus, isproportional to V0, and the denominator varies as E for large E. The denom-inator is non-zero when both k and k′ are real. This is because the magnitude

196

of the denominator is given by the sum of two squares

4 k2 k′2 cos2 k′L + ( k2 + k′2 )2 sin2 k′L (857)

and when the sine factor is zero, the cosine factor is positive. Thus, there areno simultaneous zeros.

In the above expressions k and k′ are continuous variables and so the energyeigenvalues have a continuous spectrum in the range E ≥ 0.

The intensities of the reflected and transmitted beams add to equal theintensity of the incident beam,

| A |2 = | B |2 + | F |2 (858)

This is a necessary condition for the probability to be conserved. As the proba-bility does not build up with time at any point in space, but is in a steady state,the probability currents entering any region of space must be equal to thoseleaving. As the magnitude of the velocities outside the potential well are simplygiven by h k

m , the equality of incoming flux with the outgoing flux reduces tothe above statement about the intensities of the beams.

The reflection coefficient is defined as the ratio

R =| B |2

| A |2(859)

which takes into account that the magnitude of the velocity of the incidentbeam is always the same as that of the reflected beam, as they are moving inthe same region of space, and thus experience the same potential. The reflectioncoefficient is given by

R(k) =(

( k′2 − k2 )2 sin2 k′ L

4 k2 k′2 cos2 k′ L + ( k2 + k′2 )2 sin2 k′ L

)(860)

We see that the reflection coefficient has a numerator proportional to V 20 , and

the denominator grows like E2 for sufficiently large E. It should be noted that,in this case, quantum mechanics produces results which are completely differentfrom classical mechanics as, in one dimension, a classical particle is not reflectedby an attractive potential well.

The transmission coefficient is defined as the ratio

T =kt | F |2

ki | A |2(861)

in which kt is the magnitude of the wave vector of the transmitted beamF exp[ i kt x ] and ki is the magnitude of the wave vector for the incident

197

Reflection Coefficient for a beam of particles incident on a shallow potential well

0

0.2

0.4

0.6

0.8

1

0 2 4 6 8 10 12

k L

R(k

)

Figure 40: The reflection coefficient R(k) for a beam of particles of momentumh k incident on a shallow potential well of depth − V0 and width L. Note that,due to the Ramsauer effect, the reflection coefficient falls to zero periodically.

beam A exp[ i ki x ]. In our examples ki and kt have the same value and,therefore, cancel. In the more general case, the velocities have to be included,and then the reflection and transmission coefficients will satisfy the equation,

R + T = 1 (862)

In the above example, the transmission coefficient is evaluated as

T (k) =(

4 k2 k′2

4 k2 k′2 cos2 k′ L + ( k2 + k′2 )2 sin2 k′ L

)(863)

It is seen that the beam is totally transmitted whenever sin k′ L = 0. Thetransmission coefficient shows oscillations with increasing k′. The minimum inthe reflection coefficient and the corresponding maximum in the transmissioncoefficient is known as the Ramsauer effect11. It is seen in the scattering oflow-energy electrons from inert gas atoms.

——————————————————————————————————

11C. Ramsauer, Ann. Phys. (Leipzig) 64, 546 (1921).

198

4.1.8 Exercise 62

Consider a particle of mass m moving in one dimension in a potential given by

0 if x < 0V (x) =

− V0 if x > 0 (864)

Consider the energy eigenstate which corresponds to a beam of particles of mo-mentum − p incident on the potential step from x → ∞, find the amplitudeof the reflected and transmitted waves, for E > 0. Also find the reflection andtransmission coefficients and show that they add up to unity.

——————————————————————————————————

4.1.9 Solution 62

The incident beam is in the region x > 0 and is represented by the wavetravelling along the negative x direction

A exp[− i k′ x

](865)

In addition, for positive x, one expects a reflected wave of amplitude B thattravels along the positive x direction. The form of the solution for x > 0 isgiven by

Ψ(x) = A exp[− i k′ x

]+ B exp

[+ i k′ x

](866)

On substituting the trial form of the solution in the energy eigenvalue equationfor x > 0

− h2

2 m∂2

∂x2Ψ(x) − V0 Ψ(x) = E Ψ(x) (867)

where we have used V (x) = − V0 for x > 0, one obtains a relation betweenthe energy eigenvalue and k′,

h2

2 mk′2 − V0 = E (868)

The solution for x < 0 corresponds to the transmitted beam

Ψ(x) = C exp[− i k x

](869)

199

which on substituting into the energy eigenvalue equation

− h2

2 m∂2

∂x2Ψ(x) = E Ψ(x) (870)

leads toh2

2 mk2 = E (871)

The matching conditions at x = 0 yield

C = A + B

k C = k′ ( A − B ) (872)

These can be solved to yield the amplitude of the reflected and transmittedwaves

B

A=

(k′ − k

k′ + k

)C

A=

(2 k′

k′ + k

)(873)

as ratios to the amplitude of the incident wave. The transmission coefficient, T ,is given by

T =k

k′

∣∣∣∣ CA∣∣∣∣2

=4 k k′

( k + k′ )2(874)

where the factor k / k′ represents the change in the flux of particles due tothe difference in the velocities of the transmitted incident beam. The reflectioncoefficient R is given by

R =∣∣∣∣ BA

∣∣∣∣2=

( k′ − k )2

( k + k′ )2(875)

Hence, we find that reflection and transmission coefficients add

R + T =( k′ − k )2

( k + k′ )2+

4 k k′

( k + k′ )2

=( k + k′ )2

( k + k′ )2

= 1 (876)

200

to yield unity.

——————————————————————————————————

4.1.10 Exercise 63

By considering the solution of the Schrodinger equation with the initial statedescribed by the energy eigenstate found in the previous example, show thatthe system is in a steady state. Find the probability current density in the tworegions, and show that it satisfies the continuity condition, when E > 0. Also,show that the continuity condition is satisfied in the two regions when E < 0.

——————————————————————————————————

4.1.11 Solution 63

The probability density ρ(x, t) for finding a particle in the region x > 0 isgiven by

ρ(x, t) =∣∣∣∣ Ψ(x, t)

∣∣∣∣2=

∣∣∣∣ A exp[− i k′ x

]+ B exp

[+ i k′ x

] ∣∣∣∣2∣∣∣∣ exp[− i

hE t

] ∣∣∣∣2= | A |2 + | B |2 + A B∗ exp

[− 2 i k′ x

]+ A∗ B exp

[+ 2 i k′ x

]= | A |2

[2(

k2 + k′2

( k + k′ )2

)+ 2

(k′ − k

k′ + k

)cos 2 k′ x

](877)

which is time independent, but exhibits oscillations due to the interference be-tween the incident and reflected wave. The probability density ρ(x, t) for findinga particle in the region x < 0 is given by

ρ(x, t) =∣∣∣∣ Ψ(x, t)

∣∣∣∣2=

∣∣∣∣ C exp[− i k′ x

]exp

[− i

hE t

] ∣∣∣∣2= | C |2

= | A |2 4 k′2

( k + k′ )2(878)

which is spatially uniform and time independent.

201

The probability current density jx(x, t) is given by

jx(x, t) = Real

[Ψ∗(x, t)

(− i h

m

∂

∂x

)Ψ(x, t)

](879)

For x > 0, one finds the

jx(x, t) = Real

[− h k′

m( | A |2 − | B |2 )

]

+ Real

[− h k′

m( A B∗ exp

[− 2 i k′ x

]− B A∗ exp

[+ 2 i k′ x

])

]

= Real

[− h k′

m( | A |2 − | B |2 )

](880)

as the last term is purely imaginary. This represents the net flux along thepositive direction, which is the difference between the transmitted flux and theincident flux. Note that the ratio of these two fluxes is (apart from a sign) justthe reflection coefficient R. The total probability current density for x > 0 isevaluated as

jx(x, t) = − h

mk′(

4 k k′

( k + k′ )2

)| A |2 (881)

For x < 0 one finds

jx(x, t) = − h k

m| C |2 (882)

On evaluating the probability current density for x < 0, one finds that thecurrent density is time independent and spatially uniform. The current densityhas a value of

jx(x, t) = − h

mk

(4 k′2

( k + k′ )2

)| A |2 (883)

Note that this just represents the flux corresponding to the transmitted beam.The ratio of the transmitted flux to the incident flux is just the transmissioncoefficient T .

Since, the probability current density has the same value for x > 0 andx < 0 it is independent of x. The probability density is also independent of t.Thus, the continuity condition

∂ρ(x, t)∂t

+∂jx(x, t)∂x

= 0 (884)

202

is trivially satisfied.

——————————————————————————————————

4.1.12 The Threshold Energy for a Bound State

We have examined the shallow potential well and found energy eigenstates inthe energy regions, E > 0, which corresponded to scattering states with con-tinuous energy, and the energy region 0 > E > − V0 which had boundstates with discrete energy eigenvalues. This raises the question as to whetherthere are any states with energy E less than − V0. Classically, it is impossibleto have states with an energy lower than the minimum of the potential energyas the only other contribution to the total energy, apart from the potential,is the kinetic energy which is positive definite. We shall now prove that it isimpossible to have quantum states with energy eigenvalues that are lower thanthe minimum value of the potential.

If a bound state exists, then | Ψ(x) |2 must decay exponentially as | x | → ∞in order that the wave function be normalized. This means that | Ψ(x) |2 musthave at least one maximum. This means that

∂

∂x| Ψ(x) |2 = 0 (885)

and∂2

∂x2| Ψ(x) |2 < 0 (886)

at the value of x = x0 where the maximum is located.

We shall start with the energy eigenvalue equation

− h2

2 m∂2Ψ∂x2

+(V (x) − E

)Ψ = 0 (887)

and pre-multiply by the complex conjugate wave function Ψ∗ and add this tothe complex conjugate of the energy eigenvalue equation post-multiplied by Ψ.The result is

− h2

2 m

(Ψ∗ ∂

2Ψ∂x2

+ Ψ∂2Ψ∗

∂x2

)+ 2

(V (x) − E

)| Ψ |2 = 0 (888)

We can re-write the above equation as

− h2

2 m

(∂2

∂x2| Ψ |2

)+ 2

(V (x) − E

)| Ψ |2 = − h2

m

∣∣∣∣ ∂Ψ∂x

∣∣∣∣2 (889)

This must hold at all values of x, including x0. But at the maximum we have

∂

∂x| Ψ(x0) | = 0 (890)

203

and∂2

∂x2| Ψ(x0) |2 < 0 (891)

Hence, we find that the equality

− h2

2 m

(∂2

∂x2| Ψ(x0) |2

)= 2

(E − V (x0)

)| Ψ(x0) |2 (892)

holds at x0. Since the first term is positive, the second term must also bepositive. Furthermore, as | Ψ(x0) |2 ≥ 0, we must have

E ≥ V (x0) (893)

By definition, V (x0) is greater than or equal to the minimum value of V (x).Therefore, the bound state energy E must be greater than or equal to the min-imum value of the potential.

4.1.13 Transmission through a Potential Barrier

Let us now consider a particle with energy E > 0 moving in a potential whichnow has the form

V (x) = + V0 if L > x > 0V (x) = 0 otherwise (894)

We shall consider states in the energy range where V0 > E > 0. In this energyrange, a classical particle would be forbidden to exist in the region L > x > 0but could be found in the other regions of space.

A specific form of the solution of the Schrodinger equation can be written as

Ψ(x) = A exp[

+ i k x

]+ B exp

[− i k x

]x < 0

Ψ(x) = C exp[

+ κ x

]+ D exp

[− κ x

]L > x > 0

Ψ(x) = F exp[

+ i k x

]x > L (895)

where k is related to the energy eigenvalue E via

k2 = +2 m E

h2

κ2 =2 m ( V0 − E )

h2 (896)

This form is appropriate to the case where a beam of particles of momentump = h k is incident from x → − ∞ onto the potential barrier. A portion of

204

Potential Barrier

-0.5

0

0.5

1

1.5

-1 -0.5 0 0.5 1 1.5 2

x / L

V(x

) / V

0

V0

V0 > E > 0 E

Figure 41: A potential V (x) with a barrier of height V0 and width L.

the beam will be reflected back to x → − ∞ and a portion will be transmittedto x → ∞. The continuity condition at x = 0 can be solved to yield

A =(

+ κ + i k

2 i k

)C +

(− κ + i k

2 i k

)D

B =(− κ + i k

2 i k

)C +

(+ κ + i k

2 i k

)D (897)

The continuity condition at x = L yields C and D in terms of F . This leadsto

C =(κ + i k

2 κ

)F exp

[i k L

]exp

[− κ L

]D =

(κ − i k

2 κ

)F exp

[i k L

]exp

[+ κ L

](898)

On substituting the expressions for C and D into the previous expressions forA and B, one obtains

A =exp

[i k L

]4 i k κ

(4 i k κ coshκ L + 2 ( k2 − κ2 ) sinhκ L

)F

205

B =exp

[i k L

]4 i k κ

(2 ( k2 + κ2 ) sinhκ L

)F (899)

The ratio of the amplitude of the transmitted wave to the incident wave is F / Awhich is found as

F

A=

4 i k κ exp[− i k L

](

4 i k κ coshκ L + 2 ( k2 − κ2 ) sinhκ L) (900)

whereas the ratio of the amplitude of the reflected wave to the incident wave isB / A where

B

A=

(2 ( k2 + κ2 ) sinhκ L

)(

4 i k κ coshκ L + 2 ( k2 − κ2 ) sinhκ L) (901)

These expressions are obviously related to the amplitudes for reflection fromand transmission through a shallow potential well, via the analytic continuation

k′ → i κ (902)

The transmission coefficient T (k) is evaluated as

T (k) =∣∣∣∣ FA

∣∣∣∣2=

4 k2 κ2(4 k2 κ2 cosh2 κ L + ( k2 − κ2 )2 sinh2 κ L

) (903)

For large κ L, we have coshκ L ' 12 exp

[+ κ L

]and sinhκ L '

12 exp

[+ κ L

]. Thus, we find that transmission coefficient decreases exponen-

tially with increasing L. The amplitude of the transmitted wave is exponentiallysuppressed, and the leading exponential term is

F

A≈ 4 i k κ

( k + i κ )2exp

[− κ L

]exp

[− i k L

](904)

The phenomenon in which particles are transmitted through a potential barrier(known as quantum tunnelling) is due to the exponentially small probability offinding the particles in the region of space where they are classically forbidden toexist due to energetic reasons. In the classically forbidden regions, the classical

206

Transmission through a potential barrier

-2 -1 0 1 2 3 4

x/a

Rea

l [ Φ

k(x)

] R T

Figure 42: Transmission Through a Potential Barrier. The real part of the wavefunction Φk(x) is plotted as a function of x. A beam of particles is emitted froma source located at x → − ∞ and falls incident on the potential barrier. Thebarrier extends from x = 0 up to x = a, and its extremities are marked bythe red vertical lines. The height of the barrier V0 is greater than the kineticenergy E of the incident particles, and hence is impenetrable to a beam ofclassical particles of energy E. Most of the incident beam is reflected from thepotential barrier. The remaining part of the beam tunnels through the potentialbarrier and is transmitted to infinity. Therefore, some of the particles quantummechanically tunnel through the classically forbidden region.

kinetic energy would have to be negative. In the limit of large κ L, within theapproximation of leading exponential terms, the reflected wave has a relativeamplitude of unity

B

A≈ ( k2 + κ2 )

( k + i κ )2(905)

as is needed if the probability current is to satisfy the continuity equation.

——————————————————————————————————

4.1.14 Exercise 64

Construct the probability current density in the three regions and show thatthe continuity equations is satisfied in each region, including the region where

207

Transmission Coefficient for a Potential Barrier

0

0.2

0.4

0.6

0.8

1

1.2

0 2 4 6 8 10 12

k L

T(k

)

Figure 43: The transmission coefficient T (k) of a beam of particles, of wavevector k, incident on a potential barrier of height V0. For energies in the rangeV0 > E > 0, where a beam of classical particles would be totally reflected, thequantum mechanical particles have an exponentially small probability of beingtransmitted.

tunnelling occurs.

——————————————————————————————————

4.1.15 Solution 64

Since the states are energy eigenstates, the probability density is time indepen-dent. For x < 0, the probability density

ρ(x, t) = | A |2 + | B |2 +(A B∗ exp

[i 2 k x

]+ B A∗ exp

[− i 2 k x

] )(906)

is oscillatory. For L > x > 0, the probability density is also non-uniform

ρ(x, t) = | C |2 exp[

+ 2 κ x]

+ | D |2 exp[− 2 κ x

]+(C D∗ + D C∗

)(907)

208

For x > L, the probability density is spatially uniform

ρ(x, t) = | F |2 (908)

The probability current density jx(x, t) in the region x < 0 is given by

jx(x, t) =h k

m

(| A |2 − | B |2

)(909)

In the barrier region, the probability current density is given by

jx(x, t) = − ih κ

m

(C D∗ − C∗ D

)(910)

for L > x > 0. This quantity is only finite if the coefficients C and Dare complex. Since the coefficients are determined via the matching conditions,the probability current density is only finite if the wave function shows periodicoscillations in some regions of space. In the last region, the probability currentdensity is also spatially uniform

jx(x, t) =h k

m| F |2 (911)

for x > L.

On evaluating the probability current density, it is found to be given by

jx(x, t) =h k

m

4 k2 κ2

4 k2 κ2 cosh2 κ L + ( k2 − κ2 )2 sinh2 κ L| A |2 (912)

and is uniform throughout space. Since the probability current density is con-stant and the probability density is time independent, the continuity equation

∂ρ

∂t+

∂jx∂x

= 0 (913)

is trivially satisfied.

——————————————————————————————————

4.1.16 The Double Well Potential

Consider a particle of mass m moving in one dimension in the presence of adouble well potential

V (x) = ∞ for | x | > ( a + b )V (x) = 0 for ( a + b ) > | x | > ( b − a )V (x) = V0 otherwise (914)

209

The double well potential

x

V(x

)

(b-a) (b+a)-(b+a) -(b-a)

En

En

V0

Figure 44: The double well potential.

The potential V (x) is a symmetric function of x. The parity operator P isdefined as the operator which, when acting on an arbitrary function ψ(x), hasthe effect

P ψ(x) = ψ(−x) (915)

On applying the parity operator twice, it is seen that it satisfies the relation

P 2 = I (916)

The eigenfunctions φp(x) of the parity operator satisfy the equations

P φp(x) = p φp(x) (917)

where p is the parity eigenvalue. Because of the relation expressed by eqn(916),the eigenvalues of the parity operator are restricted to have values ± 1. Sincethe potential is an even function of x, the Hamiltonian commutes with the parityoperator

[ P , H ] = 0 (918)

so that the parity is a conserved quantity. In other words, parity is a constant ofmotion. Since the Hamiltonian commutes with the parity operator, one can findsimultaneous eigenstates of H and P . We shall seek such solutions for whichE < V0. In this case, classical particles are excluded from traversing the regioncontaining the potential barrier.

210

The even parity eigenfunctions can be expressed in the form

φ(x) = A sin k ( x − a − b ) for ( a + b ) > x > ( b − a )φ(x) = B cosh κ x for ( b − a ) > | x | (919)

where the energy eigenvalue is related to k and κ via

E =h2 k2

2 m= V0 −

h2 κ2

2 m(920)

Since φ(x) is an even parity eigenfunction φ(−x) = φ(x), the form of thesolution for − ( b − a ) > x > − ( b + a ) is given by

φ(x) = − A sin k ( x + a + b ) for − ( b − a ) > x > − ( b + a )(921)

The above form of the solution φ(x) identically satisfies the boundary conditionsat the hard walls x = ± ( b + a ). The remaining matching conditions atx = ± ( b − a ) lead to the self-consistency equation

k cot 2 k a = − κ tanh κ ( b − a ) (922)

This equation together with the relation

k2 + κ2 =2 m V0

h2 (923)

can be solved graphically. The solutions determine the quantized values of k.

The odd parity eigenfunctions can be expressed as

φ(x) = A sin k ( x − a − b ) for ( a + b ) > x > ( b − a )φ(x) = B sinh κ x for ( b − a ) > | x | (924)

The form of the solution for negative values of x can be obtained from thoseof positive values of x by noting that they have odd parity φ(−x) = − φ(x).If the above forms of φ(x) are to satisfy the differential equation, the energyeigenvalue must be related to k and κ via eqn(920). The above form of thesolution φ(x) automatically satisfies the boundary conditions at the hard wallsx = ± ( b + a ). The matching conditions for the odd parity eigenfunctionsat x = ± ( b − a ) lead to the equation

k cot 2 k a = − κ coth κ ( b − a ) (925)

The quantized values of k are given by the solutions of the above self-consistencyequation.

211

Consistency conditions

-10

-5

0

5

0 0.5 1 1.5 2

2ka /π

k cot 2ka

−κ coth κ(b-a)

−κ tanh κ(b-a)

2(2mV0a2/h2π2)

1/2

Figure 45: The graphical solution of the self consistency conditions for thedouble potential well. The right hand side for odd parity solutions is shown bythe dashed line, the right hand side for even parity solutions is shown by thesolid line.

Graphical solution of the even and odd parity self consistency equationsshows that, for large barrier heights or large barrier widths such that

tanhκ( b − a ) ∼ 1 (926)

the eigenvalues occur in pairs which are almost degenerate. The pairs of energieshave magnitudes which are close to the bound state energies of the isolated wells,so

k cot 2 k a ∼ − κ (927)

The splitting ∆E between the lowest pairs of eigenvalues is exponentially small,and is approximated by

∆E ∼ h2 π2

2 m a2

2κ a

exp[− 2 κ ( b − a )

](928)

The tunnel splitting between the pairs of energy levels increases for the higherenergy levels.

212

Energy eigenfunctions of the double well potential

x

Ψn(

x)

-(b+a) (b+a)-(b-a) (b-a)

Ψ0(x)

Ψ1(x)

Figure 46: The lowest pair of energy eigenfunctions for the double well potential.

4.1.17 The delta function Potential

Let us consider a beam of particles of mass m and momentum h k scatteringfrom an attractive potential V (x) localized at the origin

V (x) = − V0 a δ(x) (929)

where V0 a is a measure of the strength of the potential. In the strength, V0 hasunits of potential and a has units of length.

The energy eigenvalue equation becomes[− h2

2 m∂2

∂x2− V0 a δ(x)

]φk(x) = E φk(x) (930)

We shall look for a solution which is of the form

φk(x) = A exp[

+ i k x

]+ B exp

[− i k x

](931)

in the region 0 > x and has the form

φk(x) = F exp[

+ i k x

](932)

213

Delta function potential

-14

-12

-10

-8

-6

-4

-2

0

2

-2 -1 0 1 2x / a

V(x

)V(x) = - V0 a δ(x)

E0

Figure 47: The attractive delta function potential. The bound state energy isshown by a horizontal dashed line.

for x > 0. These forms satisfy the energy eigenvalue equation in these tworegions if

E =h2 k2

2 m(933)

since, the potential vanishes identically in these two separate regions. Again, theform is appropriate for discussion of the scattering of a beam of particles withenergy E > 0 and momentum h k incident on the potential from x → − ∞.

The matching conditions at x = 0 are affected by the presence of thenon-zero potential. The continuity of the wave function yields

A + B = F (934)

The discontinuity in the first derivative is produced by the infinite potential atthe origin. Let us integrate the energy eigenvalue equation from x = − ε tox = + ε and let ε → 0. Then, as E and φk(0) are finite, in the limit ε → 0one has∫ + ε

− ε

dx

[− h2

2 m∂2

∂x2− V0 a δ(x)

]φk(x) = E

∫ + ε

− ε

dx φk(x)

− h2

2 m∂

∂xφk(x)

∣∣∣∣+ ε

− ε

− V0 a φk(0) = E 2 ε φk(0) (935)

214

which reduces to

− h2

2 m∂

∂xφk(x)

∣∣∣∣+ ε

− ε

= V0 a φk(0) (936)

This yields the second condition

− h2

2 m

(i k F

)+

h2

2 m

(i k A − i k B

)= V0 a F (937)

These two equations can be solved to yield B and F in terms of A,

B

A=

− 11 + i k h2

m a V0

F

A=

i k h2

m a V0

1 + i k h2

m a V0

(938)

Thus, we again have

| B |2

| A |2+| F |2

| A |2= 1

T + R = 1 (939)

and E is a continuous variable, in the range E > 0. We note that for large E,R → 0 and T → 1, as the potential becomes ineffective in scattering particleswith sufficiently high energies.

4.1.18 Bound States of a delta function Potential

We shall now consider the bound states of a delta function potential which musthave E < 0 and as a result

E = − h2 κ2

2 m(940)

where κ is a positive real number. Accordingly, the wave function in the tworegions must have the forms given by

φκ(x) = A exp[− κ x

]+ B exp

[+ κ x

](941)

when 0 > x and

φκ(x) = F exp[− κ x

](942)

when x > 0, corresponding to the analytic continuation k → i κ. In orderfor the wave function to be normalizable at x → − ∞, one must have A = 0.The matching conditions then yield

B = F (943)

215

and

− h2

2 m

(− κ F

)+

h2

2 m

(+ κ B

)= V0 a F (944)

which yields the energy eigenvalue condition

h2

mκ = V0 a (945)

Thus, the energy eigenvalue is given by

E = − ( V0 )2(2 h2

m a2

) (946)

which has the dimensions of energy.

This result could have been obtained directly by noting that the reflectionand transmission coefficients have poles at

h2

mi k = − V0 a (947)

which is satisfied for imaginary values of k. Furthermore, since A can be ne-glected in comparison with B and F for this imaginary value of k, and sincek = i κ, the wave function decays away from the origin

φκ(x) = B exp[

+ κ x

](948)

in the region 0 > x and

φκ(x) = B exp[− κ x

](949)

for x > 0. The amplitudes of the decaying wave functions are equal (B = F ),as can be seen from eqn(938) by noting that at the pole i k h2

m a V0= − 1. Thus,

we have recovered the bound state energy and the form of the wave function byanalytic continuation to negative energies.

The normalization of the wave function yields the amplitude, up to an arbi-trary phase, as

B =√κ (950)

Solution in the Momentum Representation

216

Bound state wave function

0

0.5

1

1.5

-2 -1 0 1 2x / a

Ψ0(

x )

Figure 48: The bound state wave function of an attractive delta function poten-tial. Note that the wave function has a discontinuous first derivative at x = 0.

The bound states of the attractive one-dimensional delta function potentialcan be found by transforming to the momentum representation. The energyeigenvalue equation is[

− h2

2 m∂2

∂x2− V0 a δ(x)

]ψ0(x) = E0 ψ0(x) (951)

which after performing the Fourier Transform

φ0(k) =(

12 π

) 12∫ ∞

−∞dx ψ0(x) exp

[− i k x

](952)

becomesh2 k2

2 mφ0(k) −

V0 a√2 π

ψ0(0) = E0 φ0(k) (953)

On expressing the energy eigenvalue E0 in terms of κ

E0 = − h2 κ2

2 m(954)

the equation is solved for φ0(k) as

φ0(k) =1√2 π

(2 m V0 a

h2

)ψ0(0)

( k2 + κ2 )(955)

217

The real space wave function ψ0(x) is given by the inverse Fourier Transform ofφ0(k)

ψ0(x) =(

12 π

) 12∫ ∞

−∞dk φ0(k) exp

[+ i k x

](956)

On performing the integral via Cauchy’s method, in which the contour is com-pleted with a semi-circle in either the upper or lower half of the complex planedepending on the sign of x, one finds that the real space wave function is givenby

ψ0(x) =(m V0 a

h2 κ

)ψ0(0) exp

[− κ | x |

](957)

If the form of the solution for ψ0(x) is consistent with the value of ψ0(0) atx = 0, the value of κ must satisfy the equation

κ =m V0 a

h2 (958)

This condition determines the allowed value of the bound state energy.

——————————————————————————————————

Example: The Bound States of a Triple delta function Potential.

We shall examine the bound states of a particle of mass m moving in onedimension, in the presence of a potential V (x) given by

V (x) = − V0 a

[δ(x− a) + δ(x) + δ(x+ a)

](959)

where V0 is a positive constant with units of potential and a is a length scale.The potential is an even function of x. The parity operator P is defined as theoperator which, when acting on an arbitrary function ψ(x), has the effect

P ψ(x) = ψ(−x) (960)

Since the potential is an even function of x, the Hamiltonian commutes withthe parity operator

[ P , H ] = 0 (961)

so that the parity is a conserved quantity. In other words, parity is a constantof motion.

We shall search for bound states that are simultaneous eigenstates of theparity operator and the Hamiltonian. We shall look for eigenfunctions whichare of the form

φn(x) = A exp[− κ x

](962)

218

Triple delta function potential

-20

-15

-10

-5

0

5

-2 -1 0 1 2

x/a

V(x

)/V

0 a

Figure 49: The potential V (x) for three equally spaced attractive delta func-tions.

in the region x > a and which are of the form

φn(x) = B exp[− κ x

]+ C exp

[+ κ x

](963)

in the region a > x > 0. The forms of the solution for x < 0 are found fromthe above forms by requiring that the eigenfunctions either have even or oddparities. The above forms satisfy the energy eigenvalue equation in the regionswhere the potential is zero, as long as the energy eigenvalue of the bound stateEn is expressed in terms of κ via

En = − h2 κ2

2 m(964)

The above form of the eigenfunction is a solution of the differential equationas long as it satisfies the differential equation at the boundary points x = 0 andx = ± a. The boundary conditions at x = a are the condition of continuity ofthe wave function and the condition that the discontinuity of its first derivativeis related to the integral of the potential. These boundary conditions can beexpressed as

A = B + C exp[

2 κ a]

(965)

219

and

A

(1 − 2 m V0 a

h2 κ

)= B − C exp

[2 κ a

](966)

The above two equations can be combined to yield

A

(1 − m V0 a

h2 κ

)= B (967)

andm V0 a

h2 κA = C exp

[2 κ a

](968)

The eigenfunctions also have to satisfy boundary conditions at x = 0. Theexplicit forms of the boundary conditions at x = 0 depend on the parity of theeigenfunctions, therefore, we shall discuss the two cases separately.

Odd Parity

For odd parity solutions, φn(−x) = − φn(x), so the boundary condition atx = 0 reduces to

φn(0) = 0 (969)

orB = − C (970)

Hence, one finds that B and C are completely determined by A, and A is deter-mined, up to the phase, by the normalization condition. The above conditionsimply that for odd parity solutions, κ must satisfy the equation(

m V0 a2

h2 − κ a

)=

m V0 a2

h2 exp[− 2 κ a

](971)

This equation can be solved graphically by plotting the left hand side and theright hand side as a function of κ a on the same graph, as seen in fig(50). Thesolutions for κ correspond to the points where the two curves cross. Both curvesstart at the same value

m V0 a2

h2 (972)

when κ a = 0, and are both decreasing functions of κ a. The left hand sidedecreases linearly with increasing κ and becomes negative at

κ a =m V0 a

2

h2 (973)

and the right hand side approaches the value of 0 exponentially as κ a → ∞.By considering the initial slopes of both curves, one finds that if

2m V0 a

2

h2 > 1 (974)

220

Odd Parity Consistency Equation

-1

0

1

2

0 1 2 3

κ a

(mV0a2/h2) exp [ -2κa ]

(mV0a2/h2-κa)

Figure 50: The graphical solution for the odd parity bound states, shown fortwo different strengths of the potential V0. The left hand side of eqn(971) isshown by a red line, and the right hand side by a blue line. For large values ofthe potential or the separation, the equation (represented by the pair of solidlines) has one solution at a non-zero value of κ. For small values of V0, the lefthand and right hand sides of the equation are depicted by the dashed lines. Thedashed curves only intersect at κ = 0.

then the two curves also cross at a finite value of κ a. The value of κ a at thecrossing yields a bound state solution with odd parity. The solution at κ a = 0represents an odd parity zero energy resonance.

Even Parity

For even parity solutions, φn(−x) = φn(x). Therefore, the matching condi-tion at the x = 0 boundary leads to(

B − C

)=

m V0 a

h2 κ

(B + C

)(975)

or

B

(1 − m V0 a

h2 κ

)= C

(1 +

m V0 a

h2 κ

)(976)

On combining these conditions with the boundary conditions at x = a, onefinds that κ a must satisfy the equation(

κ a − m V0 a2

h2

)2

=m V0 a

2

h2

(κ a +

m V0 a2

h2

)exp

[− 2 κ a

](977)

221

This equation determines the energy eigenvalues and energy eigenfunctions witheven parity. The above transcendental equation can be solved graphically, asseen in fig(51). The left hand side and right hand side take on the same value

Even Parity Consistency Equations

0

1

2

3

4

0 1 2 3

κ a

(mV0a2/h2)(κa+mV0a

2/h2) exp [ -2κa ]

(κa-mV0a2/h2)2

Figure 51: The graphical solution of eqn(977) for the even parity bound states,shown for two different strengths of the potential. The left hand side of theequation is shown by a red line, and the right hand side by a blue line. Forlarge values of V0 ma2

h2 , the equation (represented by the solid lines) has twosolutions at non-zero values of κ. For small values of V0, the equation only hasone solution at a non-zero value of κ, located at the intersection of the dashedlines.

at κ a = 0. The left hand side is always positive but touches zero at

κ a =m V0 a

2

h2 (978)

and varies quadratically as κ a → ∞. The right hand side is monotonicallydecreasing and approaches zero as κ a → ∞. From consideration of the initialslope, it is found that there are three solutions of this equation if

2m V0 a

2

h2 > 3 (979)

and, otherwise, only has two solutions. The solution at κ a = 0 correspondsto a zero energy bound state or resonance. The solutions at finite values of κ ahave finite binding energies.

222

Bound State Energies

-3

-2

-1

0

0 0.5 1 1.5 2

mV0a2/h2

E n m

a2 /h

2

n=1

n=2

n=3

Figure 52: The bound state energy eigenvalues En, for the triple delta functionpotential.

Bound state wave functions

-2 -1 0 1 2

x/a

ψn(

x)

n=3

n=2

n=1

mV0a2/h2 = 1.8

0

0

0

Figure 53: The bound state wave functions φn(x), for the triple delta functionpotential.

223

The three bound state energies En are plotted as a function of κ a in fig(52).It is seen that the number of bound states increases as the value of V0 increases.As the separation between the wells is increased, all the bound state energiesapproach the asymptotic expression

En → − m V 20 a2

2 h2 (980)

expected for three isolated delta function potentials. The three bound statewave functions φn(x) are shown in fig(53). It is seen that the number of nodesincreases with increasing energy.

——————————————————————————————————

4.1.19 Exercise 65

Find the reflection and transmission coefficients for a potential

V (x) = V0 Θ(x)− V0 a δ(x) (981)

where Θ(x) is the Heaviside step function,

1 for x > 0Θ(x) =

0 for x < 0 (982)

The potential is shown in fig(54). Show that the reflection and transmissioncoefficients add up to unity.

——————————————————————————————————

4.1.20 Solution 65

The energy eigenfunction corresponding to a scattering experiment where theincident beam is travelling towards the right is given by

φk(x) = A exp[i k x

]+ B exp

[− i k x

]x < 0

φk(x) = C exp[i k′ x

]x > 0 (983)

where E = h2 k2

2 m and E = V0 + h2 k′2

2 m

The continuity condition leads to

C = A + B (984)

224

-4

-3

-2

-1

0

1

2

-1 -0.5 0 0.5 1

x/a

V(x

)

V(x) = -V0 a δδδδ(x) + V0 ΘΘΘΘ(x)

Figure 54: The potential V (x) used in Exercise 65.

whereas the discontinuity in the first derivative satisfies the equation(i k′ +

2 m V0 a

h2

)C = i k ( A − B ) (985)

ThereforeA

C=(k + k′ − i 2 m V0 a

h2

2 k

)(986)

andB

C=(k − k′ + i 2 m V0 a

h2

2 k

)(987)

Thus, we have the transmission coefficient

T =k′

k

| C |2

| A |2

T =4 k k′

( k + k′ )2 + ( 2 m V0 ah2 )2

(988)

and the reflection coefficient is given by

R =| B |2

| A |2

R =( k − k′ )2 + ( 2 m V0 a

h2 )2

( k + k′ )2 + ( 2 m V0 ah2 )2

(989)

225

0

0.2

0.4

0.6

0.8

1

1 1.5 2 2.5 3

k/k0

T(k

)T(k)

R(k)

Figure 55: The transmission coefficient T (k) and the reflection coefficient R(k)calculated in Exercise 65.

——————————————————————————————————

4.1.21 Exercise 66

Find the reflection and transmission coefficients for a beam of particles incidenton the potential V (x) given by

V (x) = − V0 a

(δ( x − a ) + δ( x + a )

)(990)

Also, find the bound states.

——————————————————————————————————

4.1.22 Solution 66

The energy eigenfunctions corresponding to the scattering states of energy

E =h2 k2

2 m(991)

226

-8

-6

-4

-2

0

2

4

-3 -2 -1 0 1 2 3

x / a

V(x

)

V(x) = - V0 a δδδδ(x-a) - V0 a δδδδ(x+a)

E0

E1

Figure 56: The potential for two attractive delta function potentials.

can be expressed in terms of a steady state in which a particle beam of momen-tum h k is incident from x → − ∞ or where a particle beam is incident fromx → + ∞. We shall consider the solution corresponding to a beam incidentfrom the left travelling to the right. The energy eigenfunction can be expressedas

φk(x) = A exp[i k x

]+ B exp

[− i k x

]x < − a

φk(x) = C exp[i k x

]+ D exp

[− i k x

]− a < x < a

φk(x) = F exp[i k x

]a < x (992)

where A is the amplitude of the incident beam and B and F are the amplitudesof the reflected and transmitted beams, respectively.

The continuity conditions at x = − a is given by

A exp[− i k a

]+ B exp

[+ i k a

]= C exp

[− i k a

]+ D exp

[+ i k a

](993)

227

whereas, from integrating the energy eigenvalue equation, one finds the discon-tinuity in the first derivative satisfies(

A exp[− i k a

]− B exp

[+ i k a

] )=

(C exp

[− i k a

]− D exp

[+ i k a

] )+

2 m V0 a

i k h2

(C exp

[− i k a

]+ D exp

[+ i k a

] )(994)

This pair of equations can be solved for A and B in terms of C and D.

A = C

(1 +

m V0 a

i k h2

)+ D exp

[+ 2 i k a

] (m V0 a

i k h2

)B = D

(1 − m V0 a

i k h2

)− C exp

[− 2 i k a

] (m V0 a

i k h2

)(995)

The boundary conditions at x = a are similar, the continuity equation is

C exp[

+ i k a

]+ D exp

[− i k a

]= F exp

[+ i k a

](996)

whereas, from integrating the energy eigenvalue equation, one finds the discon-tinuity in the first derivative satisfies(

C exp[

+ i k a

]− D exp

[− i k a

] )= F exp

[+ i k a

] (1 +

2 m V0 a

i k h2

)(997)

This pair of equations can be solved to yield

C = F

(1 +

m V0 a

i k h2

)D = −

(m V0 a

i k h2

)F exp

[+ 2 i k a

](998)

The expressions for C and D can be substituted in the expressions for A andB, yielding the amplitudes of the reflected and transmitted beam

A

F=

(1 +

m V0 a

i k h2

)2

−(m V0 a

i k h2

)2

exp[

+ 4 i k a]

228

B

F= −

(m V0 a

i k h2

) (exp

[+ 2 i k a

]+ exp

[− 2 i k a

] )(999)

The transmission coefficient is given by the inverse of the squared modulus ofthe first equation and the reflection coefficient is found by dividing the one equa-tion by the other and then multiplying be its complex conjugate.

Self Consistency Equation

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2κ a

( m V0 a /h2)2 exp[ - 4 κ a]

( m V0 a /h2 - κ )2

Figure 57: A graphical solution for the bound state energies.

The bound states are given by the analytic continuation k → i κ and thevalues of κ are determined from the poles of the transmission and reflectioncoefficients, i.e. A = 0. Hence, we find(

κ − m V0 a

h2

)2

=(m V0 a

h2

)2

exp[− 4 κ a

](1000)

For large κ a the solution reduces to the solution for two independent deltafunction potentials, but these are subject to the exponentially small bondingand anti-bonding splitting

κ ≈(m V0 a

h2

) (1 ∓ exp

[− 2

m V0 a2

h2

] )(1001)

For small κ a the splitting becomes larger and one obtains the values of κ as

κ ≈ 2(m V0 a

h2

)

229

Bound state wave functions

-1

-0.5

0

0.5

1

-4 -3 -2 -1 0 1 2 3 4

x / a

Ψn(

x)

Ψ0(x)

Ψ1(x)

Figure 58: The bound state wave functions for the double delta function poten-tial.

κ ≈ 0 (1002)

This potential may serve as a one-dimensional model for a molecule, in whichthe levels are doubly degenerate. If the atoms are far apart, the energies justcorrespond to the energies of the individual atoms. However, when the atomsare brought closer together, there is an energy gain for doubly occupying thelowest energy state. This gain occurs through the bonding - anti-bonding split-ting and results in the molecule being stabilized.

Alternate Derivation of the Bound State Solutions.

We shall find the bound states of particles moving in one dimension, in thepresence of the potential V (x) given by

V (x) = − V0 a

(δ( x − a ) + δ( x + a )

)(1003)

The energy eigenvalue equation[− h2

2 m∂2

∂x2− V (x)

]ψα(x) = Eα ψα(x) (1004)

230

will be solved in the momentum space representation. On Fourier Transforming

φα(k) =(

12 π

) 12∫ ∞

−∞dx ψα(x) exp

[− i k x

](1005)

the energy eigenvalue equation, one obtains

h2 k2

2 mφα(k)− V0 a√

2 π

(ψα(a) exp

[− i k a

]+ ψα(−a) exp

[+ i k a

] )= Eα φα(k)

(1006)On expressing the energy eigenvalue Eα in terms of κ

Eα = − h2 κ2

2 m(1007)

the momentum space form of the energy eigenvalue equation is solved for φα(k),yielding

φα(k) =1√2 π

(2 m V0 a

h2

)1

( k2 + κ2 )

(ψα(a) exp

[− i k a

]+ ψα(−a) exp

[+ i k a

] )(1008)

The real space wave function ψα(x) is given by the inverse Fourier Transformof φα(k)

ψα(x) =(

12 π

) 12∫ ∞

−∞dk φα(k) exp

[+ i k x

](1009)

The integrals are performed using Cauchy’s method, in which the contours arecompleted with semi-circles at infinity. One finds that the real space wavefunction is given by

ψα(x) =(m V0 a

h2 κ

) (ψα(a) exp

[− κ | x− a |

]+ ψα(−a) exp

[− κ | x+ a |

] )(1010)

This solution must be consistent at the two points x = ± a. This yields thetwo consistency conditions

ψα(±a)[

1 −(m V0 a

h2 κ

) ]= ψα(∓a)

(m V0 a

h2 κ

)exp

[− 2 κ a

](1011)

These equations determine the allowed values of κ. On combining these equa-tions, one has[

1 −(m V0 a

h2 κ

) ]2=(m V0 a

h2 κ

)2

exp[− 4 κ a

](1012)

which has two non-trivial solutions if

2(m V0 a

2

h2

)> 1 (1013)

231

Otherwise, the above equation only has one non-trivial solution, and the trivialsolution κ = 0.

——————————————————————————————————

4.1.23 Exercise 67

A particle of mass m moves in one dimension under the influence of an attrac-tive delta function potential centered at the origin, and of strength V0 a. Theparticle is in the bound state at t = 0 and then the strength of the attractivepotential is suddenly changed to V ′0 a. What is the probability that the particlewill remain bound to the potential?

——————————————————————————————————

4.1.24 Solution 67

The energy eigenstate Ψ0(x) of the initial Hamiltonian H is governed by theeigenvalue equation

− h2

2 m∂2

∂x2Ψ0(x) − V0 a δ(x) Ψ0(x) = E0 Ψ0(x) (1014)

For x 6= 0 the eigenvalue equation reduces to

− h2

2 m∂2

∂x2Ψ0(x) = E0 Ψ0(x) (1015)

and has a solution of the form

Ψ0(x) = A exp[− κ x

]x > 0

Ψ0(x) = B exp[

+ κ x

]x < 0 (1016)

where

E0 = − h2

2 mκ2 (1017)

The matching condition at x = 0 yields

Ψ0( + ε ) = Ψ0( − ε ) (1018)

orA = B (1019)

232

On integrating the differential equation over the infinitesimal interval between( − ε , + ε ), one obtains∫ +ε

−ε

dx

[− h2

2 m∂2

∂x2Ψ0(x) − V0 a δ(x) Ψ0(x)

]= E0

∫ +ε

−ε

dx Ψ0(x)

− h2

2 m∂

∂xΨ0(x)

∣∣∣∣+ε

−ε

− V0 a Ψ0(0) = E0 2 ε Ψ0(0)

(1020)

or in the limit ε → 0

+h2

mκ A − V0 a A = 0 (1021)

Hence,

κ =m a V0

h2 (1022)

thus we have

E0 = − m a2 V0

2 h2 (1023)

and the bound state wave function is given by

Ψ0(x) = A exp[− m a V0

h2 | x |]

(1024)

The magnitude of the constant A is given by the normalization condition∫ ∞

−∞dx | Ψ0(x) |2 = 1

| A |2∫ ∞

−∞dx exp

[− 2

m a V0

h2 | x |]

= 1

2 | A |2∫ ∞

0

dx exp[− 2

m a V0

h2 x

]= 1

| A |2 h2

m a V0= 1 (1025)

Hence,

A =

√m a V0

h2 (1026)

up to an arbitrary phase. The initial ground state wave function is given by

Ψ0(x) =

√m a V0

h2 exp[− m a V0

h2 | x |]

(1027)

233

The initial state wave function Ψ0(x) can be expressed as a linear superpo-sition of the eigenstates of the final Hamiltonian, H ′, φn(x), via

Ψ0(x) =∑

n

Cn φn(x) (1028)

where the expansion coefficients are given by

Cn =∫ ∞

−∞dx φ∗n(x) Ψ0(x) (1029)

The probability that a measurement of H ′ will result in a value for the energyof E′n is given by

P (n) = | Cn |2 (1030)

The overlap between the initial bound state and the final bound state of poten-tial V ′0 , is given by∫ ∞

−∞dx φ∗0(x) Ψ0(x) = 2

m a√

V0 V ′0h2

∫ ∞

0

dx exp[− m a ( V0 + V ′0 )

h2 x

]=

2√

V0 V ′0V0 + V ′0

(1031)

The probability that the particle remains in the bound state, P (0), is given by

P (0) =∣∣∣∣ ∫ ∞

−∞dx φ∗0(x) Ψ0(x)

∣∣∣∣2=

4 V0 V′0

( V0 + V ′0 )2(1032)

Thus, the probability that the particle ends up in an excited state is given by

Pexc =∑n 6=0

P (n)

= 1 − P (0)

=( V0 − V ′0 )2

( V0 + V ′0 )2(1033)

since the probabilities are normalized to unity. Note that if V0 = V ′0 no tran-sitions take place.

——————————————————————————————————

4.1.25 Exercise 68

A particle moves in one dimension in a potential of the form

V (x) = ∞ for x < 0V (x) = − V0 a δ( x − a ) for x > 0 (1034)

234

-15

-10

-5

0

5

10

15

-1 0 1 2

x/a

V(x

)V(x) = - V0 a δ δ δ δ(x-a) + V1 ΘΘΘΘ(-x)

Figure 59: The potential V (x) used in Exercise 68. The limit V1 → ∞ shouldbe taken.

The potential is shown in fig(59). Find the condition that determines the boundstates. What is the minimum value of V0 for which a bound state appears?

——————————————————————————————————

4.1.26 Solution 68

We need to consider the bound state solutions of the energy eigenvalue equation[− h2

2 m∂2

∂x2+ V (x)

]φ0(x) = E0 φ0(x) (1035)

In the region when x > a the bound state solution must have the form ofa decaying exponential

φ0(x) = C exp[− κ x

](1036)

where the bound state energy is

E = − h2 κ2

2 m(1037)

235

Whereas in the region where 0 < x < a the wave function must have theform

φ0(x) = A exp[− κ x

]+ B exp

[+ κ x

](1038)

and vanish for x < 0.

To satisfy the boundary condition at x = 0 one must have A = − B.

The continuity condition at x = a yields

A

(exp

[− κ a

]− exp

[+ κ a

] )= C exp

[− κ a

](1039)

and on integrating the energy eigenvalue equation in an infinitesimal regionabout x = a to obtain the discontinuity of the first derivative in terms of thestrength of the delta function potential,

− κ C exp[− κ a

]− κ A

(exp

[− κ a

]+ exp

[+ κ a

] )= − 2 m V0 a

h2 C exp[− κ a

](1040)

These two equations can be solved to yield

coth κ a = −(

1 − 2 m V0 a

h2 κ

)(1041)

The bound state only just exists if κ → 0 in which case the above equationcan be expanded in powers of κ. The solution at κ → 0 only exists if

2 m V0 a2

h2 = 1 (1042)

——————————————————————————————————

236

4.2 The One-Dimensional Harmonic Oscillator

We shall find all the energy eigenfunctions of the one-dimensional Harmonic os-cillator in a systematic way. The Hamiltonian of the one-dimensional harmonicoscillator can be written as

H =[− h2

2 m∂2

∂x2+

m ω2 x2

2

](1043)

The Harmonic Oscillator Potential

-1

0

1

2

3

4

5

-3 -2 -1 0 1 2 3

x

V(x

)

V(x) = 1/2 m ω2 x2

E > 0 E0

E1

E2

E3

Figure 60: The Harmonic Oscillator Potential V (x) = m ω2

2 x2.

The Hamiltonian can be re-written in terms of the dimensionless variable

ξ =√m ω

hx (1044)

Then, the Hamiltonian has the form

H =h ω

2

[− ∂2

∂ξ2+ ξ2

](1045)

which shows that h ω provides the natural energy scale for the quantum system.The Hamiltonian operator can be expressed in terms of its classical factorizationand a constant either as

H =h ω

2

[ (− ∂

∂ξ+ ξ

) (+

∂

∂ξ+ ξ

)+ 1

](1046)

237

or as

H =h ω

2

[ (+

∂

∂ξ+ ξ

) (− ∂

∂ξ+ ξ

)− 1

](1047)

since the factors do not commute. In fact, the commutator of the factors isgiven by [ (

+∂

∂ξ+ ξ

),

(− ∂

∂ξ+ ξ

) ]= 2 (1048)

4.2.1 The Raising and Lowering Operators

Let us denote the factors by the operators a† and a since these factors are Her-mitean conjugates. We shall also introduce a factor of

√2 into their definition

a =1√2

(+

∂

∂ξ+ ξ

)a† =

1√2

(− ∂

∂ξ+ ξ

)(1049)

to simplify the commutation relations. Thus, these operators have the commu-tation relations

[ a , a† ] = + 1 (1050)

4.2.2 The Effect of the Lowering Operator

Let us consider the energy eigenvalue equation

H Ψn(x) = En Ψn(x) (1051)

with wave function Ψn(x) and energy eigenvalue En. The Hamiltonian whenexpressed in terms of the new operators is just

H =hω

2

(2 a† a + 1

)H =

hω

2

(2 a a† − 1

)(1052)

The effect of the operators a on the eigenvalue equation can be found as

a H Ψn(x) = En a Ψn(x) (1053)

but since the commutator of a with the Hamiltonian is

a H = ah ω

2

(2 a† a + 1

)=

h ω

2

(2 a† a + 3

)a

=(H + h ω

)a (1054)

238

one finds eqn(1053) can be re-written as(H + h ω

)a Ψn(x) = En a Ψn(x) (1055)

Thus, on rewriting this we find

H a Ψn(x) =(En − h ω

)a Ψn(x) (1056)

Hence, the wave functiona Ψn(x) (1057)

is an eigenfunction of the Hamiltonian with an eigenvalue of En − h ω. Thus,the operator a when acting on an energy eigenfunction produces another energyeigenfunction with a lower energy eigenvalue. In other words,

a Ψn(x) = Cn Ψn−1(x) (1058)

where Cn is a constant of proportionality. This property of a justifies namingit as the lowering operator.

4.2.3 The Ground State

The lowest energy state has eigenvalue E0 and, therefore, as there is no lowerenergy eigenfunction, one must have

a Ψ0(x) = 0 (1059)

This leads to the equation for the ground state wave function(+

∂

∂ξ+ ξ

)Ψ0(x) = 0 (1060)

This has the solution

Ψ0(x) = C exp[− ξ2

2

](1061)

where C is the normalization constant that still has to be determined. In termsof the original variables, the ground state wave function is found as

Ψ0(x) = C exp[− m ω x2

2 h

](1062)

The normalization condition yields

| C | =(m ω

h π

) 14

(1063)

239

Since the lowest energy state satisfies eqn(1059) and due to the first form of theHamiltonian given in eqn(1052), we deduce that the lowest energy eigenvalue ish ω2 . The method implies that there exists higher-energy eigenstates that have

energy eigenvalues which are larger by multiples of h ω, that is

En = h ω

(n +

12

)(1064)

4.2.4 The Effect of The Raising Operator

The effect of the Hermitean conjugate of a can be found by considering its effecton the energy eigenvalue equation

a† H Ψn(x) = a† En Ψn(x) (1065)

On using the commutation relations, one finds

a† H = a†h ω

2

(2 a† a + 1

)=

h ω

2

(2 a† a − 1

)a†

=(H − h ω

)a† (1066)

so one finds that eqn(1065) becomes(H − h ω

)a† Ψn(x) = En a† Ψn(x) (1067)

Thus, on re-writing the above equation, we find

H a† Ψn(x) =(En + h ω

)a† Ψn(x) (1068)

Hence, the wave functiona† Ψn(x) (1069)

is an eigenfunction of the Hamiltonian with an eigenvalue of En + h ω, i.e.En+1. Thus, the operator a† when acting on an energy eigenfunction producesanother energy eigenfunction with a higher-energy eigenvalue.

a† Ψn(x) = C∗n+1 Ψn+1(x) (1070)

where C∗n+1 is a constant of proportionality related to that in eqn(1058). Thisproperty of a† justifies naming it as the raising operator.

240

4.2.5 The Normalization

The constant of proportionality for the raising operator is related to the constantof proportionality for the lowering operator as can be found by considering thedefinition of the Hermitean conjugate of an operator∫ + ∞

− ∞dx Ψ∗

n−1(x) a Ψn(x) =( ∫ + ∞

− ∞dx Ψ∗

n(x) a† Ψn−1(x))∗

(1071)

This implies that, with properly normalized eigenfunctions and on using thedefinition in eqn(1068) and its consequence eqn(1070), the coefficients of pro-portionality satisfy

Cn =(C∗n

)∗(1072)

as expected. The constant of proportionality can be found, up to an arbitraryphase, from the energy eigenvalue equation

h ω

2

(2 a† a + 1

)Ψn(x) = h ω

(n +

12

)Ψn(x) (1073)

Thus, on using the equations for the raising and lowering operators togetherwith the constants of proportionality, one finds

| Cn |2 = n (1074)

Hence, on choosing a real phase for the Cn, we have the equations

a Ψn(x) =√n Ψn−1(x)

a† Ψn(x) =√n+ 1 Ψn+1(x) (1075)

4.2.6 The Excited States

All the energy eigenfunctions can be obtained from the ground state wave func-tion by successive action of the raising operator, Hence, we have

Ψ1(x) = a† Ψ0(x)

Ψ1(x) =1√2

(−√

h

m ω

∂

∂x+√m ω

hx

)Ψ0(x)

Ψ1(x) =(

m ω

4 h π

) 14(−√

h

m ω

∂

∂x+√m ω

hx

)exp

[− m ω x2

2 h

](1076)

241

The higher-energy eigenfunctions are given by iteration

Ψn+1(x) =a†√

( n + 1 )Ψn(x)

Ψn+1(x) =1√

2 ( n + 1 )

(−√

h

m ω

∂

∂x+√m ω

hx

)Ψn(x)

(1077)

Then the n-th excited state wave function is given in terms of the ground statewave function Ψ0(x) by acting on it by the raising operator n times

Ψn(x) =( a† )n

√n!

Ψ0(x)

=( a† )n

√n!

(m ω

h π

) 14

exp[− m ω x2

2 h

]=

1√2n n!

(−√

h

m ω

∂

∂x+√m ω

hx

)n (m ω

h π

) 14

exp[− m ω x2

2 h

](1078)

Furthermore, using the operator identity(−√

h

m ω

∂

∂x+√m ω

hx

)≡

≡ ( − 1 ) exp[

+m ω x2

2 h

] √h

m ω

∂

∂xexp

[− m ω x2

2 h

](1079)

one finds the eigenstates are given by the expression

Ψn(x) =( − 1 )n

√2n n!

exp[

+m ω x2

2 h

] ( √h

m ω

∂

∂x

)n

exp[− m ω x2

2 h

]×(m ω

h π

) 14

exp[− m ω x2

2 h

](1080)

The solution is recognized as involving the n-th order Hermite polynomialHn(x)given by

Hn

(√m ω

hx

)= (− 1 )n exp

[+m ω x2

h

] (√h

m ω

∂

∂x

)n

exp[− m ω x2

h

](1081)

Thus, we have found the expression for the general energy eigenfunction for theone-dimensional harmonic oscillator as

Ψn(x) =1√

2n n!

(m ω

h π

) 14

exp[− m ω x2

2 h

]Hn

(√m ω

hx

)(1082)

242

These states are all properly normalized to unity, as the ground state wave func-tion was properly normalized.

Harmonic Oscillator wave functions

-4 -3 -2 -1 0 1 2 3 4

x

Ψ ΨΨΨn(

x)

ΨΨΨΨ7(x)

ΨΨΨΨ1(x)

ΨΨΨΨ2(x)

ΨΨΨΨ3(x)

ΨΨΨΨ4(x)

ΨΨΨΨ5(x)

ΨΨΨΨ6(x)

Figure 61: The ground state Ψ0(x) and excited state wave functions Ψn(x) ofthe one-dimensional harmonic oscillator.

The quantum probability density for finding a particle with energy E atposition x should be compared with the classical probability density if the ini-tial position is unknown. Unlike the classical probability density, the quantumprobability density shows spatial oscillations. If the spatial oscillations are aver-aged over, then it is seen that as the energy increases, the quantum probabilitydensity approaches the classical density.

——————————————————————————————————

4.2.7 Exercise 69

Find an expression for the expectation value of x4, in the n-th excited state ofthe harmonic oscillator.

——————————————————————————————————

243

0

0.2

0.4

0.6

0.8

-4.5 -3.5 -2.5 -1.5 -0.5 0.5 1.5 2.5 3.5 4.5

x

Pn(

x) =

l Ψ ΨΨΨ

n(x)

l 2

n = 4

Figure 62: Comparison of the quantum mechanical probability densities forfinding a particle at position x, P (x) = | Ψn(x) |2 with the classical probabilitydensities for n = 4.

4.2.8 Solution 69

The average value of x4 in the n-th energy eigenstate of the harmonic oscillatorcan be found be expressing x in terms of the raising and lowering operators

x =

√h

2 m ω( a† + a ) (1083)

Hence

x4 =∫ +∞

−∞dx φ∗n(x) x4 φn(x)

=(

h

2 m ω

)2 ∫ +∞

−∞dx φ∗n(x) ( a† + a )4 φn(x)

(1084)

The only non-zero terms are those involving the same number of raising andlowering operators as terms where these numbers are different can be related tothe matrix elements of energy states with different values of n which, because ofthe orthogonality of non-degenerate energy eigenstates, are zero. The non-zero

244

0

0.2

0.4

0.6

0.8

-4.5 -3.5 -2.5 -1.5 -0.5 0.5 1.5 2.5 3.5 4.5

x

Pn(

x) =

l Ψ ΨΨΨ

n(x)

l 2

n = 8

Figure 63: Comparison of the quantum mechanical probability densities forfinding a particle at position x, P (x) = | Ψn(x) |2 with the classical probabilitydensities for n = 8.

terms are

x4 =(

h

2 m ω

)2 ∫ +∞

−∞dx φ∗n(x)

(a† a† a a + a† a a† a

+ a† a a a† + H.c.

)φn(x)

=(

h

2 m ω

)2 (n ( n − 1 ) + n n + n ( n + 1 )

+ ( n + 1 ) n + ( n + 1 ) ( n + 1 ) + ( n + 2 ) ( n + 1 ))

x4 =(

h

2 m ω

)2

3 ( 2 n2 + 2 n + 1 ) (1085)

——————————————————————————————————

245

4.2.9 Exercise 70

Show that for a harmonic oscillator in the n-th energy eigenstates, the uncer-tainty in the position and momentum satisfy the equation

∆xrms ∆prms = h2 n + 1

2(1086)

——————————————————————————————————

4.2.10 Solution 70

The expectation value of the ∆x2 is equal to the expectation value of x2

∆x2 =h

2 m ω

∫dx φ∗n(x) ( a† + a )2 φn(x)

=h

2 m ω( 2 n + 1 )

In the first line we have used the fact that the term∫dx φ∗n(x) ( a† )2 φn(x) (1087)

and the Hermitean conjugate term both vanish identically. In the second line, wehave used the fact that the state φn(x) is an eigenstate of the number operatora† a = n with eigenvalue n. Likewise

∆p2 = − h2 m ω

2 h

∫dx φ∗n(x) ( a† − a )2 φn(x)

=h m ω

2( 2 n + 1 )

Combining these, we find the equality

∆prms ∆xrms =h

2( 2 n + 1 ) (1088)

——————————————————————————————————

4.2.11 Exercise 71

A particle is moving in one dimension in a potential V (x) which is given by

V (x) =m ω2

2x2 x > 0

V (x) → ∞ x < 0 (1089)

Find the energy eigenfunctions and eigenvalues.

246

4.2.12 Time Development of the Harmonic Oscillator

Given an initial condition Ψ(x, 0), one can expand this initial wave function interms of the energy eigenstates φn(x) of the harmonic oscillator via

Ψ(x; 0) =∑

n

Cn φn(x) (1090)

where the expansion coefficients may be found from

Cn =∫ + ∞

− ∞dx φ∗n(x) Ψ(x; 0) (1091)

The wave function at future times t can be found from the solution of theSchrodinger equation, and as the Hamiltonian is time independent the solutioncan be expressed as

Ψ(x; t) = exp[− i

H t

h

]Ψ(x; 0) (1092)

which gives rise to the expression for the time-dependent wave function as

Ψ(x; t) =∑

n

Cn exp[− i

H t

h

]φn(x)

Ψ(x; t) =∑

n

Cn exp[− i

En t

h

]φn(x) (1093)

since φn(x) satisfy the energy eigenvalue equation

H φn(x) = En φn(x) (1094)

The motion of the particle can be viewed through examination of the averageposition x(t) or momentum p(t).

The average position is given by the expectation value

x(t) =∫ + ∞

− ∞dx Ψ∗(x; t) x Ψ(x; t) (1095)

which, on using the expression in eqn(1093), leads to

x(t) =∑n,m

C∗m Cn exp[

+ i( Em − En ) t

h

] ∫ + ∞

− ∞dx φ∗m(x) x φn(x)

(1096)

247

The matrix elements of x between the two energy eigenfunctions can be evalu-ated by expressing the x in terms of the raising and lowering operators,

x =

√h

m ω

12

(ξ +

∂

∂ξ+ ξ − ∂

∂ξ

)=

√h

2 m ω

(a + a†

)(1097)

Then the matrix elements can be found to be∫ + ∞

−∞dx φ∗m(x) x φn(x) =

√h

2 m ω

( √n+ 1 δm,n+1 +

√n δm,n−1

)(1098)

On substituting this expression back into eqn(1096) and performing the doublesummation and then using En+1 − En = h ω, one finds

x(t) =∫ + ∞

−∞dx Ψ∗(x; t) x Ψ(x; t)

=

√h

2 m ω

∑n

( √n+ 1 C∗n+1 Cn exp

[+ i ω t

]+

+√n C∗n−1 Cn exp

[− i ω t

] )=

√h

2 m ω

∑n

√n+ 1

(C∗n+1 Cn exp

[+ i ω t

]+

+ C∗n Cn+1 exp[− i ω t

] )(1099)

where we have shifted the summation index by one in the second term. Since thefirst term is equal to the complex conjugate of the second term, the expectationvalue is real. This is as it must be, because x is a Hermitean operator. Let usdenote the expansion coefficients in terms of an amplitude and phase

Cn = | Cn | exp[i ϕn

](1100)

then we have

x(t) =

√2 hm ω

∑n

√n+ 1 | Cn+1 | | Cn | cos

(ω t + ϕn − ϕn+1

)(1101)

Hence, the position oscillates with frequency ω, just like the classical value.However, if the phase differences ϕn+1 − ϕn are randomly distributed, destruc-tive interference may mask the oscillations.

248

One can evaluate the expectation value of the momentum via a similar pro-cedure. The expectation value is given by

p(t) = − i h

∫ + ∞

− ∞dx Ψ∗(x; t)

∂

∂xΨ(x; t) (1102)

which yields

p(t) = − i h∑n,m

C∗m Cn exp[

+ i( Em − En ) t

h

] ∫ + ∞

− ∞dx φ∗m(x)

∂

∂xφn(x)

(1103)The matrix elements can be evaluated by expressing the derivative in terms ofthe raising and lowering operators

∂

∂x=

√m ω

h

12

(ξ +

∂

∂ξ− ξ +

∂

∂ξ

)=

√m ω

2 h

(a − a†

)(1104)

Hence, we have the matrix elements∫ + ∞

− ∞dx φ∗m(x)

∂

∂xφn(x) =

√m ω

2h

( √n δm,n−1 −

√n+ 1 δm,n+1

)(1105)

On substituting the above matrix elements into the expectation value of themomentum, one finds

p(t) = − h

√2 m ω

h

∑n

| Cn+1 | | Cn |√n+ 1 sin ( ω t + ϕn − ϕn+1 )

(1106)Thus, we see that the average momentum oscillates out of phase with the averageposition, and

p(t) = m∂

∂tx(t) (1107)

which is just the same as the classical mechanical expression.

——————————————————————————————————

4.2.13 Exercise 72

Consider a harmonic oscillator in a state given by

Ψ(x) =1√2 s

n=N+s∑n=N−s

exp[− i n ϕ

]φn(x) (1108)

249

where φn(x) is the n-th excited state of the harmonic oscillator and N s 1. Find the time dependence of x(t) and p(t). Compare these expectation valueswith the classical expressions.

——————————————————————————————————

4.2.14 Solution 72

This type of state is a coherent state as it is a superposition of a very largenumber of energy eigenstates, where the exists a definite phase relation betweenthe component states. The time dependence of the coherent states is given by

Ψ(x, t) =1√2 s

n=N+s∑n=N−s

exp[− i n ϕ

]exp

[− i n ω t

]φn(x) (1109)

As a result, the component states always have a simple phase relationship be-tween them.

The expectation value of x is given by

x(t) =1

2 s

∑n,n′

exp[i ( n′ − n ) ( ϕ + ω t )

] ∫ ∞

−∞dx φ∗n′(x) x φn(x) (1110)

However, the operator x can be expressed in terms of the creation and annihi-lation operators via

x =

√h

2 m ω( a† + a ) (1111)

Then, the matrix elements of x between the energy eigenstates φn(x) and φ∗n(x)can be evaluated as∫ ∞

−∞dx φ∗n′(x) x φn(x) =

√h

2 m ω

(δn′,n+1

√n + 1 + δn′,n−1

√n

)(1112)

Thus, the expectation value is calculated as

x(t) =1

2 s

√h

2 m ω

∑n,n′

( √n + 1 exp

[i ( ϕ + ω t )

]δn′,n+1

+√n exp

[− i ( ϕ + ω t )

]δn′,n−1

)(1113)

If one only retains the terms of leading order in 1N and 1

s , one finds that theexpectation value of the position is given by

x(t) =

√h N

2 m ωcos( ω t + ϕ ) (1114)

250

Likewise, for the expectation values of the momentum, one can show that

p(t) = −√m ω N

2 hsin( ω t + ϕ ) (1115)

The expectation value of the momentum is related to the expectation value ofthe position via the classical relation

p(t) = m∂x(t)∂t

(1116)

Schrodinger12 determined that the coherent state

Ψλ(x) = exp[− 1

2λ2

] ∞∑n=0

λn exp[− i n ϕ

]√n!

φn(x) (1117)

is an exact eigenstate of the lowering operator. This can be seen as

a Ψλ(x) = λ exp[− i ϕ

]Ψλ(x) (1118)

Coherent states play an important role in laser optics as they approximate clas-sical states, containing many quanta, that have definite phases.

——————————————————————————————————

4.2.15 Hermite Polynomials

The Hermite Polynomials Hn(z) can be defined as

Hn(z) =(− 1

)n exp[

+ z2

]∂n

∂znexp

[− z2

](1119)

From this definition, we can see that they satisfy the differential equation

∂2

∂z2Hn − 2 z

∂

∂zHn + 2 n Hn(z) = 0 (1120)

This is proved by evaluating the first and second derivatives of Hn(z). The firstderivative is given by

∂

∂zHn = 2 z

(− 1

)n exp[

+ z2

]∂n

∂znexp

[− z2

]−(− 1

)n exp[

+ z2

]∂n

∂zn2 z exp

[− z2

](1121)

12E. Schrodinger, Naturwissenschaften, 14, 664 (1926).

251

and the second derivative is given by

∂2

∂z2Hn = ( 4 z2 + 2 )

(− 1

)n exp[

+ z2

]∂n

∂znexp

[− z2

]− 8 z

(− 1

)n exp[

+ z2

]∂n

∂znz exp

[− z2

]+(− 1

)n exp[

+ z2

]∂n

∂zn( 4 z2 − 2 ) exp

[− z2

](1122)

The terms + 2 and − 2 in the round brackets cancel. On forming the combina-tion

∂2

∂z2Hn − 2 z

∂

∂zHn (1123)

we have

∂2

∂z2Hn − 2 z

∂

∂zHn = − 4 z

(− 1

)n exp[

+ z2

]∂n

∂znz exp

[− z2

]+(− 1

)n exp[

+ z2

]∂n

∂zn4 z2 exp

[− z2

](1124)

The righthand side can be simplified as

= + 2 z(− 1

)n exp[

+ z2

]∂n+1

∂zn+1exp

[− z2

]−(− 1

)n exp[

+ z2

]∂n

∂zn2 z

∂

∂zexp

[− z2

](1125)

which after commuting the z in the last term to the front, we have

= − 2 n(− 1

)n exp[

+ z2

]∂n

∂znexp

[− z2

]= − 2 n Hn(z) (1126)

Thus, we have found that the Hermite polynomials satisfy the equation

∂2

∂z2Hn − 2 z

∂

∂zHn + 2 n Hn(z) = 0 (1127)

as was to be proved.

This differential equation also has an integral representation of its solution.The integral representation of the solution is given by

Hn(z) =2n

√π

∫ + ∞

− ∞du exp

[− u2

] (z + i u

)n

(1128)

252

This can be shown by first evaluating the terms

∂2

∂z2Hn(z) = n ( n − 1 )

2n

√π

∫ + ∞

− ∞du exp

[− u2

] (z + i u

)n−2

∂

∂zHn(z) = n

2n

√π

∫ + ∞

− ∞du exp

[− u2

] (z + i u

)n−1

(1129)

and then by forming the expression

2 n Hn(z) − 2 z∂

∂zHn(z)

= i n2n

√π

∫ + ∞

− ∞du 2 u exp

[− u2

] (z + i u

)n−1

= − i n2n

√π

∫ + ∞

− ∞du

(∂

∂uexp

[− u2

] ) (z + i u

)n−1

(1130)

On integrating by parts, one finds this simplifies to

= i n2n

√π

∫ + ∞

− ∞du exp

[− u2

]∂

∂u

(z + i u

)n−1

= − n ( n − 1 )2n

√π

∫ + ∞

− ∞du exp

[− u2

] (z + i u

)n−2

= − ∂2

∂z2Hn(z) (1131)

Thus, the integral expression of eqn(1128) satisfies the differential equation

∂2

∂z2Hn − 2 z

∂

∂zHn + 2 n Hn(z) = 0 (1132)

and, therefore, is a representation of the Hermite polynomials. The integralrepresentation can be used to yield the explicit forms of low-order Hermitepolynomials. Expressions for first few lowest order Hermite polynomials Hn(z)are given in Table(2).

We can show that

∂

∂zHn(z) = 2 n Hn−1(z) (1133)

This is seen by examining the first derivative

∂

∂zHn = 2 z

(− 1

)n exp[

+ z2

]∂n

∂znexp

[− z2

]−(− 1

)n exp[

+ z2

]∂n

∂zn2 z exp

[− z2

](1134)

253

Table 2: The Lowest Order Hermite Polynomials Hn(z).

n Hn(z)

0 H0(z) 11 H1(z) 2 z2 H2(z) 4 z2 − 23 H3(z) 8 z3 − 12 z4 H4(z) 16 z4 − 48 z2 + 125 H5(z) 32 z5 − 160 z3 + 120 z

On commuting the term proportional to z in the second term to the front, wehave

∂

∂zHn = − 2 n

(− 1

)n exp[

+ z2

]∂n−1

∂zn−1exp

[− z2

]= 2 n Hn−1(z) (1135)

Thus, the derivative of the n-th order Hermite polynomial is related to the Her-mite polynomial of (n− 1)-th order.

From this relationship, we can derive the generating function for the Hermitepolynomials

exp[− t2 + 2 t z

]=

∞∑n = 0

Hn(z)n!

tn (1136)

The proof starts with examining F (z, t), which is defined via

F (z, t) =∞∑

n = 0

Hn(z)n!

tn (1137)

On taking the derivative with respect to z, one has

∂

∂zF (z, t) =

∞∑n = 0

∂

∂z

Hn(z)n!

tn

=∞∑

n = 0

2 nHn−1(z)

n!tn

= 2∞∑

n = 0

Hn−1(z)(n− 1)!

tn

= 2 t F (z, t) (1138)

254

The equation∂

∂zF (z, t) = 2 t F (z, t) (1139)

can be integrated to yield

lnF (z, t)F (0, t)

= 2 t z (1140)

Thus, we have

F (z, t) = F (0, t) exp[

2 z t]

(1141)

We can evaluate F (0, t) from Hn(0) as

F (0, t) =∞∑

n = 0

Hn(0)n!

tn (1142)

We shall re-write the expression for Hn(z) in terms of a power series and thenset z = 0,

Hn(z) =(− 1

)n exp[

+ z2

]∂n

∂zn

∑m

(− 1

)m ( z2 )m

m!(1143)

The only term that remains on setting z = 0 is the term with n = 2 m.Hence, only the even order Hermite polynomials remain finite at z = 0. Wefind that these are given by

H2m(0) =(− 1

)m ( 2 m )!m!

(1144)

Thus, we have

F (0, t) =∞∑

n = 0

Hn(0)n!

tn

=∞∑

m = 0

(− 1

)mm!

t2m

= exp[− t2

](1145)

Hence, we obtain

F (z, t) = exp[

+ 2 z t − t2]

=∞∑

n = 0

Hn(z)n!

tn (1146)

which is the generating function expansion for the Hermite polynomials.

——————————————————————————————————

255

4.2.16 Exercise 73

The initial wave function of a particle of mass m in a harmonic potential offrequency ω is given by

Ψ(x, 0) =(m ω

π h

) 14

exp[− m ω

2 h( x − a )2

](1147)

Find the probability that a measurement of the energy will give the result

En = h ω ( n +12

) (1148)

——————————————————————————————————

4.2.17 Solution 73

Using the generating function expansion, we can decompose the initial wavefunction as

Ψ(x, 0) =(m ω

π h

) 14

exp[− m ω

2 h( x − a )2

]=

(m ω

π h

) 14

exp[− m ω

2 hx2

]× exp

[− m ω

2 ha2 +

m ω

hx a

]=

(m ω

π h

) 14

exp[− m ω

2 hx2

]× exp

[− m ω

4 ha2

]exp

[− m ω

4 ha2 + 2

m ω

hxa

2

]=

(m ω

π h

) 14

exp[− m ω

4 ha2

]×∑

n

exp[− m ω

2 hx2

]Hn(

√m ω

h x )n!

(m ω a2

4 h

)n2

= exp[− m ω

4 ha2

] ∑n

φn(x)√n!

(m ω a2

2 h

)n2

(1149)

where φn(x) is the n-th energy eigenfunction given by

φn(x) =(m ω

π h

) 14

exp[− m ω

2 h x2

]2

n2 ( n! )

12

Hn(√m ω

hx) (1150)

256

Thus, as the probability P (n) is given by the square of the expansion coefficient,we have

P (n) =

(m ω a2

2 h

)n

n!exp

[− m ω

2 ha2

](1151)

——————————————————————————————————

4.2.18 Exercise 74

Find the time dependence of the state which has the initial wave function

Ψ(x, 0) =(m ω

π h

) 14

exp[− m ω

2 h( x − a )2

](1152)

and, hence, find the time dependence of the probability density P (x, t) of find-ing the particle at position x.

——————————————————————————————————

4.2.19 Solution 74

From the previous example, we have the expansion of the initial wave functionin terms of energy eigenstates,

Ψ(x, 0) =(m ω

π h

) 14

exp[− m ω

2 h( x − a )2

]= exp

[− m ω

4 ha2

]∑n

φn(x)√n!

(m ω a2

2 h

)n2

(1153)

The time dependence of the wave function is then found from the time depen-dence of the energy eigenstates

Ψ(x, t) = exp[− i

H t

h

]Ψ(x, 0)

= exp[− i

ω t

2

]exp

[− m ω

4 ha2

]×∑

n

exp[− i n ω t

]φn(x)√n!

(m ω a2

2 h

)n2

(1154)

which can be re-summed to yield

Ψ(x, t) =(m ω

π h

) 14

exp[− i

ω t

2

]exp

[− m ω

2 hx2

]

257

× exp[− m ω

4 ha2 ( 1 + e − 2 i ω t ) +

m ω

hx a e − i ω t

](1155)

Hence, we find the probability density P (x, t) as

P (x, t) = | Ψ(x, t) |2

=(m ω

π h

) 12

exp[− m ω

h( x − a cos ω t )2

](1156)

Hence we see that the state, which was initially displaced from the equilibriumstate by a distance a, performs oscillations of amplitude a and frequency ω justlike a classical particle in the same potential.

——————————————————————————————————

4.2.20 The Completeness Condition

As the Hamiltonian operator for the harmonic oscillator is a Hermitean operator,the energy eigenfunctions form a complete set. The completeness condition forthe harmonic oscillator energy eigenfunctions can be proved by using the integralrepresentation and the generating function. We take the generating function

F (x, t) = exp[

+ 2 x t − t2]

=∞∑

n = 0

Hn(x)n!

tn (1157)

then let t = y + i u, so

exp[

+ 2 x y − y2

]exp

[2 i ( x − y ) u + u2

]=

∞∑n = 0

Hn(x)n!

( y + i u )n (1158)

On multiplying this equation by exp[− u2

]and then integrating over u from

− ∞ to + ∞, we find∫ + ∞

− ∞du exp

[+ 2 x y − y2

]exp

[2 i ( x − y ) u

]=

∞∑n = 0

Hn(x)n!

∫ + ∞

− ∞du exp

[− u2

]( y + i u )n (1159)

258

so we have

π exp[

+ 2 x y − y2

]δ( x − y ) =

√π

2n

∞∑n = 0

Hn(x)n!

Hn(y) (1160)

The above equation can be rearranged to yield

δ( x − y ) =∞∑

n = 0

12n n!

√π

exp[− x2

2

]Hn(x) exp

[− y2

2

]Hn(y)

(1161)On identifying the normalized energy eigenfunctions φn(x) with the productof the exponential and the Hermite polynomials, one finds the completenessrelation for the energy eigenfunctions of the Hermitean Hamilton operator

δ( x − y ) =∑

n

φ∗n(x) φn(y) (1162)

The completeness condition allows an arbitrary function to be expanded in termsof the energy eigenstates.

259

4.3 Dual-symmetry

Let us assume that we know the ground state solution of the energy eigenvalueequation for a particle moving in one dimension in the presence of a specificpotential V (x). Let the ground state energy be denoted by E0 and the groundstate wave function be denoted by φ0(x). Using these definitions, the energyeigenvalue equation has the form

H φ0(x) = E0 φ0(x)[− h2

2m∂2

∂x2+ V (x)

]φ0(x) = E0 φ0(x) (1163)

Alternatively, we can identify H − E0 with an operator

H − E0 =h2

2m

[− ∂2

∂x2+(φ′′0(x)φ0(x)

) ](1164)

This (energy-shifted) Hamiltonian of eqn(1164) can be factorized as the productof a pair of Hermitean conjugate operators. The first operator is defined by

A =h√2 m

[∂

∂x−(φ′0(x)φ0(x)

) ](1165)

and its Hermitean conjugate is then found to be

A† =h√2 m

[− ∂

∂x−(φ′0(x)φ0(x)

) ](1166)

The Hamiltonian (up to the definition of the ground state energy) is given bythe product

H − E0 = A† A (1167)

The pair of Hermitean conjugate operators have a commutation relation givenby

[ A , A† ] = 2(h2

2 m

) [ (φ′0(x)φ0(x)

)2

−(φ′′0(x)φ0(x)

) ](1168)

For the case of the Harmonic Oscillator, A and A† respectively are proportionalto the lowering and raising operators, and their commutator is merely the con-stant h ω.

The dual partner Hd of the Hamiltonian H is given by the product of thepair of operators taken in the reverse order

Hd = A A† (1169)

The dual potential Vd(x) is defined in terms of the dual Hamiltonian. The dualpotential is found to be

Vd(x) = V (x) − E0 + [ A , A† ]

= V (x) − E0 −h2

m

∂

∂x

(φ′0(x)φ0(x)

)(1170)

260

The pair of potentials V (x) and Vd(x) are dual partner potentials. For the Har-monic Oscillator, the potential and the dual potential are the same except fora constant shift of the energy.

In general, Vd(x) and V (x) have the same energy level spectrum, En − E0 =Ed

n′ . The exceptional case is the zero energy eigenvalue for the ground state ofV (x), and this exception occurs since the operator A annihilates the groundstate. The equality between the shifted eigenvalues is proved by noting that, if

H φn = (En − E0) φn (1171)

then

Hd

(A φn

)= A A†

(A φn

)= A

(H φn

)= (En − E0)

(A φn

)(1172)

Hence, Edn′ = (En − E0), unless A φn = 0. Likewise, one can establish the

inverse relationship between the eigenstates of Hd with the eigenstates of H.That is, if φd

n(x) is an eigenstate of Hs with eigenvalue Edn, then A† φd

n(x) is aneigenstate of H with eigenvalue Ed

n. Thus, A and A† connect states of differentHamiltonians that have the same energies.

Duality and the Infinite Square Well

The infinite square well has an infinite number of bound states with energyeigenvalues given by

En =h2

2m

(π n

L

)2

(1173)

and energy eigenfunctions are given by

φn(x) =

√2L

sin(n π x

L

)(1174)

where n is a positive integer, excluding zero. Since the logarithmic derivativeof the ground state wave function is given by(

φ′1(x)φ1(x)

)=(π

L

)cot(π x

L

)(1175)

the dual partner potential is found to be

Vd(x) =h2

2m

(π

L

)2 [csc2

(πx

L

)− 1

](1176)

261

Eigenfunctions of the Infinite Square Well

0 0.2 0.4 0.6 0.8 1

x/L

φ n(x

)n=3

n=2

n=1

0

0

0

Figure 64: The first three eigenfunctions φn(x) for the infinite square well po-tential.

The dual energy eigenstates are given, up to an arbitrary multiplicative con-stant, by

φdn(x) ∝ A φn(x)

∝[∂

∂x−(π

L

)cot(π x

L

) ]sin(n π x

L

)

∝ n cos(n π x

L

)− cos

(π x

L

) [ sin(

n π xL

)(

π xL

) ](1177)

where φd1(x) and Ed

1 are absent.

262

Dual Potential

0

2

4

6

8

10

0 0.2 0.4 0.6 0.8 1

x/L

Vd (x

) [ 2

m/h

2 (L/ π

)2 ]

Ed2

Ed3

Figure 65: The potential V d(x) dual to the infinite square well potential. Theenergy eigenvalues are marked by the dashed horizonatl lines.

4.4 Bargmann Potentials

Bargmann potentials are an interesting class of one-dimensional potentials.Bargmann potentials have the very unusual property that they are reflectionlesspotentials, which means that a plane wave incident on the potential does notproduce a reflected wave. The potential produces bound states and does havean effect on the states with E > 0 in that the transmitted wave experiences aphase shift relative to the incident wave.

We shall consider solutions of the family of energy eigenvalue equations gov-erned by the integer parameter n. The eigenvalue equation is given by

− h2

2 m∂2

∂x2Ψα,n(x) − V0 n ( n + 1 ) sech2

(x

ξ

)Ψα,n(x) = Eα Ψα,n(x)

(1178)

where the length scale ξ is given by

ξ2 =h2

2 m V0(1179)

263

Dual wave functions

0 0.2 0.4 0.6 0.8 1x/L

φn(

x)n=3

n=2

0

0

Figure 66: The lowest two energy eigenstates φdn(x) of the dual potential.

For n = 0, the Bargmann energy eigenvalue equation reduces to the freeparticle problem. The general solution of the free particle problem is simply

Ψα,0(x) = A exp[

+ i k x

]+ B exp

[− i k x

](1180)

where the energy eigenvalue is given by

Eα,0 =h2 k2

2 m(1181)

We can relate the general scattering solution for arbitrary n to the generalsolution for the free particle problem (where n = 0) by an iterative method.This relation involves dual-symmetry. We shall show that

Ψα,n(x) =[ξ∂

∂x− n tanh

(x

ξ

) ]Ψα,n−1(x) (1182)

First let us change variable from x to the dimensionless variable z = xξ . Then

the Bargmann equation takes the form

∂2

∂z2Φα,n(z) + n ( n + 1 ) sech2z Φα,n(z) = − 2 m Eα ξ2

h2 Φα,n(z)

(1183)

264

Bargmann Potentials

-15

-10

-5

0

5

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

x

V(x

) / V

0

V(x) = - n ( n + 1 ) V0 sech2 x

n = 1

n = 3

n = 2

Figure 67: Bargmann potentials, Vn(x) for various values of n.

Let us assume that Φα,n−1(z) satisfies the Bargmann equation of order n − 1.Then on pre-multiplying the eigenvalue equation by

Mn =[∂

∂z− n tanh z

](1184)

we have[∂

∂z− n tanh z

] [∂2

∂z2Φα,n−1(z) + n ( n − 1 ) sech2z Φα,n−1(z)

]= − 2 m Eα ξ2

h2

[∂

∂z− n tanh z

]Φα,n−1(z)

(1185)

The Hamiltonian Hn depends on n via

Hn = − h2

2 m ξ2

[∂2

∂z2+ n ( n + 1 ) sech2z

](1186)

On commuting Mn with Hn−1, we obtain

Mn Hn−1 = Hn−1 Mn − nh2

m ξ2sech2z

[∂

∂z− n tanh z

]= Hn Mn (1187)

265

Thus, we find that

Hn Mn Φα,n−1 = Eα Mn Φα,n−1 (1188)

This shows that the wave function

Φα,n(z) = Mn Φα,n−1(z)

=[∂

∂z− n tanh z

]Φα,n−1(z) (1189)

is an eigenfunction of Hn with the same eigenvalue Eα.

——————————————————————————————————

4.4.1 Exercise 75

Show that the raising operator Mn satisfies the relation

Mn Hn−1 = Hn Mn (1190)

——————————————————————————————————

4.4.2 Solution 75

In dimensionless units, the raising operator is given by

Mn =∂

∂z− n tanh z (1191)

and the Hamiltonian Hn containing the n-th order Bargmann potential can beput into the form

Hn = −(

h2

2 m ξ2

) [∂2

∂z2+ n ( n + 1 ) sech2z

](1192)

The identity can be proved by starting with the product Mn Hn−1 andre-writing it as

Mn Hn−1 = Hn−1 Mn + [ Mn , Hn−1 ] (1193)

The commutator can be evaluated as

[ Mn , Hn−1 ] = −(

h2

2 m ξ2

)[∂

∂z, n ( n − 1 ) sech2z ] +

(h2

2 m ξ2

)n [ tanh z ,

∂2

∂z2]

266

= 2 n ( n − 1 )(

h2

2 m ξ2

)sech2 z tanh z − n

(h2

2 m ξ2

)[∂2

∂z2, tanh z ]

=(

h2

2 m ξ2

)2 n ( n − 1 ) sech2 z tanh z

− n

(h2

2 m ξ2

)∂

∂z[∂

∂z, tanh z ] − n

(h2

2 m ξ2

)[∂

∂z, tanh z ]

∂

∂z

= 2 n ( n − 1 )(

h2

2 m ξ2

)sech2 z tanh z

− n

(h2

2 m ξ2

)∂

∂zsech2z − n

(h2

2 m ξ2

)sech2z

∂

∂z

= 2 n2

(h2

2 m ξ2

)sech2 z tanh z − 2 n

(h2

2 m ξ2

)sech2z

∂

∂z

= − 2 n(

h2

2 m ξ2

)sech2z

[∂

∂z− n tanh z

]= − 2 n

(h2

2 m ξ2

)sech2z Mn (1194)

Inserting this in the expression (1193), one finds

Mn Hn−1 = Hn−1 Mn − 2 n(

h2

2 m ξ2

)sech2z Mn

= Hn Mn (1195)

which proves the identity.

——————————————————————————————————

Starting from n = 0, one can generate all the scattering states by iteration,

Φk,n(z) = Mn Mn−1 . . . M2 M1 Φk,0(z) (1196)

Let us note that at z → + ∞ the solution takes the asymptotic form

Φk,n(z) = A exp[

+ i k z

] m=n∏m=1

(+ i k − m

)

+ B exp[− i k z

] m=n∏m=1

(− i k − m

)(1197)

whereas at z → − ∞ the solution takes the form

Φk,n(z) = A exp[

+ i k z

] m=n∏m=1

(+ i k + m

)

+ B exp[− i k z

] m=n∏m=1

(− i k + m

)(1198)

267

On setting B = 0, a case which corresponds to an incident beam with momen-tum p = h k, we see that there is no reflected beam and the transmitted beamhas a phase which is different from that of the incident beam. The change ofphase 2 δ(k) is calculated from

δ(k) =m=n∑m=1

tan−1

(k

m

)(1199)

The phase shift δ(k) varies as k is varied. The total variation of the pase shift

Phase shift for the n=4 Bargmann potential

0

0.5

1

1.5

2

0 1 2 3 4 5 6 7 8

k a / ππππ

δ(δ(δ(δ(k )

/π )/π)/π)/π

n=4

Figure 68: The k dependence of the phase shift δ(k) for the scattering states ofthe n = 4 Bargmann potential.

is defined asδ(∞) − δ(0) (1200)

and is equal to n π2 . The total change in the phase shift is related to the number

of bound states of the equation through Levinson’s theorem13.

On increasing n by unity, the number of bound states of the Bargmannincreases by unity. Thus, the Bargmann equation of order n has n bound states.The lowest energy bound state is given by the un-normalized wave function

Φn,n(z) = sechnz (1201)

13N. Levinson, Kgl. Danske Videnskab. Selskab, Mat.-fys. Medd., 25, 1 (1949).

268

Raising Operators for the Bargmann Potentials

-5

-4

-3

-2

-1

0

1

2E

/ V

0

E =0E1

E2

E3

n = 0 n = 1 n = 2 n = 3

M3

M1 M2

M3

M3Ek > 0

M2

Figure 69: A graphical description of the raising operators for the Bargmannequations. The raising operators Mn transform the solutions of the (n − 1)-thBargmann potential to solutions of the n-th potential, but with the same energyeigenvalue.

and has a bound state energy given by

En = − h2

2 m ξ2n2 (1202)

Then the higher energy bound states Φm,n(z) can be created from the corre-sponding bound states Φm,m(z) which have energies Em = − h2

2 m ξ2 m2. The

higher-energy bound states are found by operating on Φm,m(z) with the sequen-tially ordered product of operators Mn Mn−1 . . . Mn′ . . . Mm+2 Mm+1 (wheren ≥ n′ ≥ m+ 1). Thus, for example, the second lowest energy (un-normalized)eigenfunction of the n-th order Bargmann equation is

Φn−1,n(z) = Mn Φn−1,n−1(z)

=[∂

∂z− n tanh z

]sechn−1z

= − ( 2 n − 1 ) sechn−1z tanh z (1203)

which has the energy eigenvalue

En−1 = − h2

2 m ξ2( n − 1 )2 (1204)

269

——————————————————————————————————

4.4.3 Exercise 76

Find the eigenfunctions and eigenvalues of the Bargmann equation with indexn = 2. Also show that the bound states are orthogonal to the scattering states.

The n=2 Bargmann Potential

-8

-6

-4

-2

0

2

4

-3 -2 -1 0 1 2 3x

V(x

) / V

0

E0

E1

n = 2

Figure 70: The Bargmann potential for n = 2. There are two bound statesand the bound state energies are denoted by horizontal lines.

——————————————————————————————————

4.4.4 Solution 76

In terms of the dimensionless variable z, lowest energy bound state of the n-thBargmann potential has the wave function

φn,n(z) = C sechn z (1205)

and the dimensionless energy eigenvalue is En = − n2. The higher energybound state is found by using the raising operator Mn acting on the bound

270

states of the (n− 1)-th Bargmann potential, i.e.,

φm,n(z) = Mn φm,n−1(z) (1206)

The resulting bound state φm,n(z) has energy Em. In this manner, we find that

φn−1,n(z) = Mn φn−1,n−1(z)

=[∂

∂z− n tanh z

]sechn−1z

= − ( 2 n − 1 ) sechn−1z tanh z (1207)

The eigenfunction φn−1,n(z) corresponds to the energy eigenvalue En−1 =− ( n − 1 )2. Thus, for n = 2, the two bound states are given by

φ2(z) = sech2z

φ1(z) = sech z tanh z (1208)

The bound state wave functions are shown in fig(71).

Bound state wave functions for the n = 2 potential

-1

-0.5

0

0.5

1

-4 -3 -2 -1 0 1 2 3 4

x

Ψn(

x)

n = 2 Ψ0(x)

Ψ1(x)

Figure 71: The bound states of Bargmann potential for n = 2. For n = 2there are two bound states, the two bound state wave functions are denoted byΨ0(x) and Ψ1(x).

271

The scattering states are given by φk(z) which can be determined from thescattering states of the n = 0 potential, by using the raising operator twice.The scattering states for n = 2 are given by

φk(z) = M2 M1

(A exp

[+ i k z

]+ B exp

[− i k z

] )=

[∂

∂z− 2 tanh z

] [∂

∂z− tanh z

] (A exp

[+ i k z

]+ B exp

[− i k z

] )=

(2 − k2 − 3 i k tanh z − 3 sech2z

)A exp

[+ i k z

]+(

2 − k2 + 3 i k tanh z − 3 sech2z

)B exp

[− i k z

](1209)

Note that on analytically continuing from k to i κ, one finds that the asymptoticexponentially growing term, for both positive and negative z, has a vanishingcoefficient in the asymptotic limit if κ is equal to either 2 or 1. The solutions forthese special values of κ correspond to the bound states that we have alreadyfound.

It remains to show that the bound states are orthogonal to the scatteringstates. That is, we have to show that∫ +∞

−∞dz φm(z) φk(z) = 0 (1210)

For the lowest bound state where m = 2, this requires that coefficients of Aand B vanish separately. These both vanish if the integral∫ ∞

0

dz cos k z(

2− k2 − 3 sech2z

)sech2z − 3 i k

∫ ∞

0

dz sin k z tanh z sech2z

(1211)also vanishes. On integrating the second term by parts, we establish that thetwo terms can be combined as

=∫ ∞

0

dz cos k z(

2 +12k2 − 3 sech2z

)sech2z (1212)

As the integrals are evaluated as∫ ∞

0

dz cos k z sech2z =π k2

sinh π k2

(1213)

and ∫ ∞

0

dz cos k z sech4z =13!

π k2

sinh π k2

(4 + k2

)(1214)

272

The overlap matrix element vanishes. The orthogonality of the scattering statesand the m = 1 bound state is treated similarly.

——————————————————————————————————

4.4.5 Exercise 77

Find the normalization for the two lowest energy bound states of the Bargmannpotential for arbitrary n.

——————————————————————————————————

4.4.6 Solution 77

The lowest energy solution of the n-th order Bargmann eigenvalue equation

− h2

2 m

[∂2

∂x2+

n ( n + 1 )ξ2

sech2

(x

ξ

) ]φm(x) = Em φm(x) (1215)

is seen to be

φn(x) = A sechn

(x

ξ

)(1216)

This can be most easily seen by changing to the scaled variable

z =x

ξ(1217)

so the eigenvalue equation becomes[∂2

∂z2+ n ( n + 1 )

]φn(z ξ) = − 2 m ξ2

h2 E φn(z ξ) (1218)

The function φn(zξ) is an eigenfunction, as can be seen by direct substitutionof

φn(z ξ) = A sechnz (1219)

into the eigenvalue equation and then by noting that

∂2

∂z2sechnz = n ( n + 1 ) sechnz tanh2 z − n sechnz

= n2 sechnz − n ( n + 1 ) sech(n+2)z

(1220)

Thus, sechnz is a bound state with the bound state energy given by

En = − h2

2 m ξ2n2 (1221)

273

Hence, the second bound state φn−1 can be found by applying Mn on sechn−1z.

φn−1(z) ∝[∂

∂z− n tanh z

]sechn−1z (1222)

which leads to the identification of

φn−1 = B sechn−1z tanh z (1223)

as the bound state with energy eigenvalue

En−1 = − h2

2 m ξ2( n − 1 )2 (1224)

The magnitude of the normalization constant A can be determined from

1 = | A |2∫ ∞

−∞dx sech2nz

= | A |2 ξ∫ ∞

−∞dz sech2nz (1225)

The integration can be evaluated by integration by parts∫ ∞

−∞dz sech2n dz =

∫ ∞

−∞dz sech2n−2z

∂

∂ztanh z

= sech2(n−1)z tanh z∣∣∣∣∞−∞

−∫ ∞

−∞dz tanh z

∂

∂zsech2(n−1)z

(1226)

The boundary term is zero for n ≥ 2 and for n = 1 the boundary term isjust 2. Since,

∂

∂zsech2(n−1)z = − 2 ( n − 1 ) sech2(n−1)z tanh z (1227)

andtanh2 z = 1 − sech2z (1228)

one has the recursion relation∫ ∞

−∞dz sech2n dz =

2 ( n − 1 )( 2 n − 1 )

∫ ∞

−∞dz sech2(n−1)z

=2n ( n − 1 )!( 2 n − 1 )!!

(1229)

for n ≥ 1. Thus, the normalized wave function for the lowest energy boundstate is given by

φn(x) =

√( 2 n − 1 )!!ξ 2n ( n − 1 )!

sechn

(x

ξ

)(1230)

274

The normalization of the next lowest-energy bound state is found from

1 = | B |2 ξ∫ ∞

−∞dz sech2(n−1)z tanh2 z

= | B |2 ξ( ∫ ∞

−∞dz sech2(n−1)z −

∫ ∞

−∞dz sech2nz

)= | B |2 ξ

(2(n−1) ( n − 2 )!

( 2 n − 3 )!!− 2n ( n − 1 )!

( 2 n − 1 )!!

)= | B |2 ξ

(2(n−1) ( n − 2 )!

( 2 n − 1 )!!

)(1231)

Thus, the normalized bound state wave function is given by

φn−1(x) =

√( 2 n − 1 )!!

ξ 2n−1 ( n − 2 )!sechn

(x

ξ

)tanh

(x

ξ

)(1232)

——————————————————————————————————

275

4.5 Orbital Angular Momentum

In three dimensions, the orbital angular momentum operator is a pseudo-vectorand is defined by the vector product

L = r ∧ p (1233)

therefore, it can be decomposed in terms of Cartesian unit vectors via

L = ex Lx + ey Ly + ez Lz (1234)

where the components are given by the expressions

Lx = y pz − z py

Ly = z px − x pz

Lz = x py − y px (1235)

From the commutation relations between position and momentum operators,one finds that the components of the angular momentum operator satisfy thefollowing type of commutation relations with the components of the positionand momentum

[ Lx , y ] = i h z

[ Lx , py ] = i h pz

[ Lx , x ] = 0[ Lx , px ] = 0 (1236)

The above expressions can be used together with the rules for the commutatorsof sums and products of operators, to prove that

[ Lx , Ly ] = i h Lz

[ Ly , Lz ] = i h Lx

[ Lz , Lx ] = i h Ly (1237)

Since the components of the orbital angular momentum operators do not com-mute, the uncertainty principle asserts that, in general, it is not possible to findan a simultaneous eigenstate of more than one component. Since the angularmomentum operators satisfy the above commutation relations, they provide anexample of a Lie algebra14.

14The operators of a Lie algebra satisfy commutation relations of the form

[ Ai , Aj ] =∑

k

Cki,j Ak

The set of operators Ai which form a Lie algebra are linearly related to the infinitesimalgenerators of transformation groups, known as Lie Groups. In the case of orbital angularmomentum, the Lie group is the group of three-dimensional rotations. It is also noteworthy

276

The commutation relations between the various components of the angularmomentum can be evaluated through

[ Lx , Ly ] = [ Lx , z px ] − [ Lx , x pz ]

= [ Lx , z ] px + z [ Lx , px ] − [ Lx , x ] pz − x [ Lx , pz ]= − i h y px + i h x py

= i h Lz (1238)

The others can be obtained by cyclic permutations, corresponding to the differ-ent choices of Cartesian coordinate axes.

The operator expressing the squared magnitude of the angular momentumvector follows from the operators representing the Cartesian components andthe Pythagorean theorem

L2 = L2x + L2

y + L2z (1239)

That is, the squared magnitude of the angular momentum is merely the sumof the squares of the operator components. The magnitude of L2 is a scalarquantity. It can be proved that

[ Lx , L2 ] = 0

[ Ly , L2 ] = 0

[ Lz , L2 ] = 0 (1240)

Thus, it is possible to find simultaneous eigenfunctions of any one componentof the angular momentum and the magnitude squared.

The commutation relation between the z component of the angular momen-tum Lz and the magnitude of the angular momentum L2 is proved by considering

[ Lz , L2 ] = [ Lz , L

2x ] + [ Lz , L

2y ] + [ Lz , L

2z ] (1241)

that the set of commutation relations between the angular momentum operators and anyvector quantity with components vi satisfy commutation relations of the form

[ Li , vj ] = i h∑

k

εi,j,k vk

where εi,j,k is the Levi-Civita symbol. The Levi-Civita symbol is defined as

εi,j,k =

1 if i, j, k is a cyclic permutation of 1, 2, 3−1 if i, j, k is not a cyclic permutation of 1, 2, 30 otherwise

The above commutation relations imply that all vectors transform in the same way under theset of rotation operations.

277

which becomes

[ Lz , L2 ] = [ Lz , L

2x ] + [ Lz , L

2y ]

= [ Lz , Lx ] Lx + [ Lz , Lx ] Lx + [ Lz , Ly ] Ly + Ly [ Lz , Ly ]

= i h

(Ly Lx + Lx Ly

)− i h

(Lx Ly + Ly Lx

)= 0 (1242)

Thus, Lz and L2 commute. By invariance under the permutation of the coor-dinate axes, the commutators of L2 with the other components of L are alsofound to be zero.

——————————————————————————————————

4.5.1 Exercise 78

Find the expression for the operator L2 in terms of the Cartesian componentsof the position and derivatives w.r.t position. Express your final result entirelyin terms of invariant quantities, such as scalar products.

——————————————————————————————————

4.5.2 Solution 78

The components of the angular momentum can be written in terms of the Levi-Civita symbol εi,j,k as

Lk = − i h∑i,j

εi,j,k xi∂

∂xj(1243)

whereεi,j,k = 1 (1244)

if i, j, k are an even permutation of the indices 1, 2, and 3, and

εi,j,k = − 1 (1245)

if i, j, k are an odd permutation of the indices 1, 2, and 3, and

εi,j,k = 0 (1246)

if any index is repeated.

278

The magnitude of the angular momentum is given by

L2 =∑

k

L2k

= − h2∑

k

∑i,j

εi,j,k xi∂

∂xj

∑l,m

εl,m,k xl∂

∂xm

(1247)

But on using the identity∑k

εi,j,k εl,m,k = δi,l δj,m − δi,m δj,l (1248)

one obtains

L2 = − h2∑i,j

(xi

∂

∂xjxi

∂

∂xj− xi

∂

∂xjxj

∂

∂xi

)

= − h2∑i,j

(xi xi

∂

∂xj

∂

∂xj+ xi

∂xi

∂xj

∂

∂xj

− xi xj∂

∂xj

∂

∂xi− xi

∂xj

∂xj

∂

∂xi

)= − h2

∑i,j

(x2

i

∂2

∂x2j

+ xi δi,j∂

∂xj

− xi xj∂

∂xj

∂

∂xi− xi

∂

∂xi

)(1249)

Thus, we have

L2 = − h2

( ∑i

x2i

∑j

∂2

∂x2j

+∑

i

xi∂

∂xi

−∑i,j

xi xj∂

∂xj

∂

∂xi− 3

∑i

xi∂

∂xi

)(1250)

Hence, the magnitude of the angular momentum can be written as

L2 = − h2

( ∑i

x2i

∑j

∂2

∂x2j

+∑

i

xi∂

∂xi

−∑i,j

xi xj∂

∂xj

∂

∂xi− 3

∑i

xi∂

∂xi

)

= − h2

( ∑i

x2i

∑j

∂2

∂x2j

−∑i,j

xi xj∂

∂xj

∂

∂xi− 2

∑i

xi∂

∂xi

)(1251)

279

which can be expressed as

L2 = − h2

( ∑i

x2i

∑j

∂2

∂x2j

−∑i,j

xj∂

∂xjxi

∂

∂xi− 1

∑i

xi∂

∂xi

)(1252)

When put in vector notation, the magnitude of the angular momentum becomes

L2 = − h2

(r2 ∇2 − ( r . ∇ )2 − ( r . ∇ )

)(1253)

which is seen to be a scalar operator.

——————————————————————————————————

4.5.3 Exercise 79

Show that the following functions are eigenfunctions of Lz and L2 and find thecorresponding eigenvalues.

Ψ0(r) = g0(r)Ψ1(r) = z g1(r)Ψ2(r) = ( x + i y ) g2(r)Ψ3(r) = ( x − i y ) g3(r)Ψ4(r) = ( 3 z2 − r2 ) g4(r)Ψ5(r) = ( x + i y )2 g5(r)Ψ6(r) = ( x − i y )2 g6(r)Ψ7(r) = z ( x + i y ) g7(r)Ψ8(r) = z ( x − i y ) g8(r) (1254)

where the functions gn(r) are arbitrary functions of the radial distance.

——————————————————————————————————

The Cartesian components of the angular momentum operators can be ex-pressed in terms of spherical polar coordinates. The Cartesian components arefound to be given by the expressions

Lx = − i h

(− sinϕ

∂

∂θ− cosϕ cot θ

∂

∂ϕ

)Ly = − i h

(+ cosϕ

∂

∂θ− sinϕ cot θ

∂

∂ϕ

)Lz = − i h

∂

∂ϕ(1255)

280

and these components can be shown to satisfy the same commutation relationsthat were derived from a purely Cartesian formulation.

We shall commence the derivation of eqns(1255) by expressing the angularmomentum in terms of the unit vectors of the polar coordinate system. Theorbital angular momentum operator is expressed as

L = r ∧ p

= − i h r ∧ ∇ (1256)

and on using the representation

∇ = er∂

∂r+ eθ

1r

∂

∂θ+ eϕ

1r sin θ

∂

∂ϕ(1257)

one finds

L = − i h r

(er ∧ eθ

1r

∂

∂θ+ er ∧ eϕ

1r sin θ

∂

∂ϕ

)= − i h

(eϕ

∂

∂θ− eθ

1sin θ

∂

∂ϕ

)(1258)

In the last line we have used the fact that the unit vectors er, eθ and eϕ forman orthogonal coordinate system. Furthermore, as

r = er r

= r cos θ ez + r sin θ(

sinϕ ey + cosϕ ex

)(1259)

one can express the spherical polar coordinate unit vectors in terms of theCartesian unit vectors by using the definition of the unit vectors

er =∂

∂rr

= cos θ ez + sin θ(

sinϕ ey + cosϕ ex

)eθ =

1r

∂

∂θr

= − sin θ ez + cos θ(

sinϕ ey + cosϕ ex

)eϕ =

1r sin θ

∂

∂ϕr

=(

cosϕ ey − sinϕ ex

)(1260)

On substituting the above expressions for the unit vectors into the equation forthe angular momentum vector L in eqn(1258), the angular momentum is found

281

in the form

L = − i h

(− ex sinϕ

∂

∂θ+ ey cosϕ

∂

∂θ

− ex cot θ cosϕ∂

∂ϕ− ey cot θ sinϕ

∂

∂ϕ+ ez

∂

∂ϕ

)(1261)

which has three components given by the expressions of eqn(1255).

The square of the angular momentum L2 is given by

L2 = − h2

[1

sin2 θ

∂2

∂ϕ2+

1sin θ

∂

∂θ

(sin θ

∂

∂θ

) ](1262)

4.5.4 Simultaneous Eigenfunctions.

In spherical polar coordinates, the angular momentum operators act on thewave functions Ψ(r, θ, ϕ), but since r does not appear in the operators, it is

redundant. The simultaneous eigenfunctions of the pair of operators Lz and L2

are functions of θ and ϕ alone, and are written as Y lm(θ, ϕ). These eigenfunctions

satisfy the pair of eigenvalue equations,

L2 Y lm(θ, ϕ) = − h2

[1

sin2 θ

∂2

∂ϕ2+

1sin θ

∂

∂θ

(sin θ

∂

∂θ

) ]Y l

m(θ, ϕ)

= λl Ylm(θ, ϕ) (1263)

Lz Ylm(θ, ϕ) = − i h

∂

∂ϕY l

m(θ, ϕ)

= µ Y lm(θ, ϕ) (1264)

where λl is the eigenvalue of L2 and µ is the eigenvalue of Lz. The eigenfunctionsY l

m(θ, ϕ) are usually factorized according to

Y lm(θ, ϕ) = Θl

m(θ) Φm(ϕ) (1265)

From the form of the Lz operator as a derivative w.r.t. ϕ, one can easily findthat the ϕ dependence of Y l

m(θ, ϕ) must be given by

Y lm(θ, ϕ) = Θl

m(θ)(

12 π

) 12

exp[iµ ϕ

h

](1266)

The modulus of the wave function is single valued, as it must represent a uniqueprobability density for each point in space. Therefore, we expect that the wave

282

function at the point (r,θ,ϕ) must have the same values at (r,θ,ϕ + 2π) sincethis represents the same point. In this case, we have

exp[iµ ϕ

h

]= exp

[iµ ( ϕ + 2 π )

h

](1267)

which is satisfied if µ = h m for any positive or negative integer m.

Classically, the magnitude of a component of the angular momentum shouldbe smaller than the magnitude of the pseudo-vector L. From this observation,one expects that an inequality should exist between the eigenvalues of λl andm h. The maximum value of m will be denoted by l. Later, we shall show thatthe eigenvalue λl is related to l via λl = h2 l ( l + 1 ).

This suggests a picture of the orbital angular momentum eigenstates as be-ing states where L has a definite magnitude

√λl and a definite z component

m h. The picture is that in which the vector L has a definite z component andmagnitude, but has an uncertain direction due to precession around the z axis.

-3

-2

-1

0

1

2

3

m

[ l ( l + 1 ) ]1/2

Figure 72: A semi-classical picture of a state with angular momentum l andz component m. The picture of the angular momentum is a vector of lengthh√

l ( l + 1 ) which has a projection h m along the z axis. The angularmomentum vector can be thought of as precessing around the z axis, so thatthe x and y components are indeterminate.

283

4.5.5 The Raising and Lowering Operators

Two useful operators are given by the raising and lowering operators, L+ andL− defined by

L+ = Lx + i Ly

L− = Lx − i Ly (1268)

The operators L+ and L− are Hermitean conjugates. These operators also

commute with L2.[ L2 , L+ ] = [ L2 , L− ] = 0 (1269)

as both Lx and Ly commute with L2.

In spherical polar coordinates, the raising and lowering operators are givenby

L+ = + h exp[

+ i ϕ

] (∂

∂θ+ i cot θ

∂

∂ϕ

)L− = − h exp

[− i ϕ

] (∂

∂θ− i cot θ

∂

∂ϕ

)(1270)

The raising and lowering operators satisfy the commutation relations

[ Lz , L+ ] = + h L+

[ Lz , L− ] = − h L−

[ L+ , L− ] = 2 h Lz (1271)

On operating with the commutators [ L2 , L± ] on the eigenfunctions of L2,one finds

L2 L± Y lm(θ, ϕ) = L± L2 Y l

m(θ, ϕ)= λl L± Y l

m(θ, ϕ) (1272)

and recognizes that L± Y lm(θ, ϕ) is also an eigenfunction of L2 with the same

eigenvalue λl as found for Y lm(θ, ϕ). Thus, L± acting on an eigenfunction of the

magnitude of the orbital angular momentum does not change the eigenvalue λl.

On operating with the commutator [ Lz , L± ] = ± h L± on the eigen-functions Y l

m(θ, ϕ), one finds

Lz L± Y lm(θ, ϕ) =

(L± Lz ± h L±

)Y l

m(θ, ϕ)

=(m h ± h

)L± Y l

m(θ, ϕ) (1273)

284

Thus, the raising and lowering operators when acting on a simultaneous eigen-function of L2 and Lz produce other simultaneous eigenfunctions with Lz eigen-values that are either raised or lowered by h. The action of L± on the eigen-function Y l

m(θ, ϕ) is to produce a function proportional to Y lm±1(θ, ϕ). Thus,

we haveL± Y l

m(θ, ϕ) = C±(l,m) Y lm±1(θ, ϕ) (1274)

The constants of proportionality, C±(l,m), have yet to be determined.

4.5.6 The Eigenvalues and Degeneracy

First we shall note the two equalities,

L− L+ = L2 − L2z − h Lz

L+ L− = L2 − L2z + h Lz (1275)

We shall first consider the effect of the raising operator. On operating L− L+

on the eigenfunction Y lm(θ, ϕ), on using the first equality of eqn(1275), one finds

L− L+ Y lm(θ, ϕ) =

(L2 − L2

z − h Lz

)Y l

m(θ, ϕ)

=(λl − h2 m2 − h2 m

)Y l

m(θ, ϕ) (1276)

If the value of m is the maximum value l, then we have

L+ Y ll (θ, ϕ) = 0 (1277)

and so, on substituting m = l in the previous equation, we find

λl = h2 l

(l + 1

)(1278)

Thus, we have found the eigenvalue of the square of the angular momentum interms of the maximum eigenvalue of the z component of the angular momentumoperator.

Now we shall determine explicit expressions for the products of the coeffi-cients C±(l,m). On using the properties of the raising and lowering operatorsgiven by eqn(1274), one finds

L− L+ Y lm(θ, ϕ) = C−(l,m+ 1) C+(l,m) Y l

m(θ, ϕ) (1279)

but we also have

L− L+ Y lm(θ, ϕ) =

(λl − h2 m2 − h2 m

)Y l

m(θ, ϕ) (1280)

285

Therefore, we can determine the products of the coefficient C±(l,m) by equatingthe above two expressions and then by substituting the expression for λl. Thus,we find

C−(l,m+ 1) C+(l,m) = λl − h2

(m2 + m

)= h2

(l ( l + 1 ) − m2 − m

)(1281)

The effect of the lowering operator is found by considering the other equality,and considering the minimum value of l. On operating L+ L− on the eigenfunc-tion Y l

m(θ, ϕ) and on using the second equality of eqn(1275), one finds

L+ L− Y lm(θ, ϕ) =

(L2 − L2

z + h Lz

)Y l

m(θ, ϕ)

= h2

(l ( l + 1 ) − m2 + m

)Y l

m(θ, ϕ) (1282)

If m is the minimum value l′, then we have

L− Y ll′(θ, ϕ) = 0 (1283)

therefore

h2

(l ( l + 1 ) − l′2 + l′

)= 0 (1284)

so we find that the minimum value of m is given by l′ = − l. Thus, thepossible values of m run through the set of numbers − l , − ( l − 1 ) , − ( l −2 ) , . . . , ( l − 2 ) , ( l − 1 ) , l , which includes 0. There are ( 2 l + 1 )

different eigenfunctions of m for fixed l. Thus, the eigenvalues of L2 correspond-ing to the value l have a degeneracy of ( 2 l + 1 ). Since we have proved thatthe eigenvalues m for the z component of the orbital angular momentum areinteger, we note that l must also be an integer.

4.5.7 The Effect of the Raising Operators.

The coefficients C±(l,m) are found by noting that as L+ and L− are Hermiteanconjugates ∫ π

0

dθ sin θ∫ 2 π

0

dϕ Y lm+1(θ, ϕ)∗ L+ Y l

m(θ, ϕ)

= C+(l,m)

=(∫ π

0

dθ sin θ∫ 2 π

0

dϕ Y lm(θ, ϕ)∗ L− Y l

m+1(θ, ϕ))∗

= C∗−(l,m+ 1) (1285)

286

Thus, we have

| C+(l,m) |2 = h2

(l ( l + 1 ) − m ( m + 1 )

)= h2 ( l − m ) ( l + m + 1 ) (1286)

and, by considering the lowering operator, one finds the analogous equation

| C−(l,m) |2 = h2

(l ( l + 1 ) − m ( m − 1 )

)= h2 ( l + m ) ( l − m + 1 ) (1287)

Thus, the effect of the raising and lowering operators are given by

L+ Y lm(θ, ϕ) = h

√( l − m ) ( l + m + 1 ) Y l

m+1(θ, ϕ)

L− Y lm(θ, ϕ) = h

√( l + m ) ( l − m + 1 ) Y l

m−1(θ, ϕ) (1288)

where we have chosen the phase of the constants to be zero.

4.5.8 Explicit Expressions for the Eigenfunctions

Explicit expressions for the simultaneous eigenfunctions Y lm(θ, ϕ) can be found

starting from the effect of the raising operator on the state with maximum m,i.e. m = l. This results in the equation

L+ Y ll (θ, ϕ) = 0 (1289)

which has the explicit form

h exp[

+ i ϕ

] (∂

∂θ+ i cot θ

∂

∂ϕ

)Y l

l (θ, ϕ) = 0 (1290)

As the eigenfunction has a ϕ dependence of exp[ i l ϕ ], and as the eigenfunctionis factorized as

Y lm(θ, ϕ) = Θl

m(θ)(

12 π

) 12

exp[i m ϕ

](1291)

one finds∂

∂θΘl

l(θ) = l cot θ Θll(θ) (1292)

This equation can be integrated to yield

Θll(θ) = Bl

l sinl θ (1293)

where Bll is a constant of proportionality. The magnitude of Bl

l is determinedby the normalization condition

| Bll |2

∫ π

0

dθ sin θ sin2l θ = 1 (1294)

287

Thus, the magnitude of the normalization is found as

| Bll | =

(( 2 l + 1 )!22l+1 ( l! )2

) 12

(1295)

The normalized eigenfunction Y ll (θ, ϕ) is found as

Y ll (θ, ϕ) =

(1

4 π( 2 l + 1 )!22l ( l! )2

) 12

sinl θ exp[i l ϕ

](1296)

up to an arbitrary constant phase.

The lowering operator acting on the state with minimum m, i.e. m = − l,results in the equation

L− Y l−l(θ, ϕ) = 0 (1297)

which has the explicit form

h exp[− i ϕ

] (− ∂

∂θ+ i cot θ

∂

∂ϕ

)Y l

l (θ, ϕ) = 0 (1298)

As the eigenfunction has a ϕ dependence of exp[ − i l ϕ ], and as the eigenfunc-tion is factorized as

Y lm(θ, ϕ) = Θl

m(θ)(

12 π

) 12

exp[i m ϕ

](1299)

one finds∂

∂θΘl−l(θ) = l cot θ Θl

−l(θ) (1300)

which can be integrated to yield

Θl−l(θ) = Bl

−l sinl θ (1301)

where Bl−l is a constant of proportionality. The magnitude of Bl

−l is determinedby the normalization condition

| Bl−l |2

∫ π

0

dθ sin θ sin2l θ = 1 (1302)

Thus, the magnitude of the normalization is found as

| Bl−l | =

(( 2 l + 1 )!22l+1 ( l! )2

) 12

(1303)

The normalized eigenfunction Y l−l(θ, ϕ) is found as

Y l−l(θ, ϕ) =

(1

4 π( 2 l + 1 )!22l ( l! )2

) 12

sinl θ exp[− i l ϕ

](1304)

288

up to an arbitrary constant phase.

The eigenfunctions corresponding to larger values of m can be found bysuccessive operation with the raising operators,

exp[

+ i ϕ

] (∂

∂θ+ i

∂

∂ϕcot θ

)Y l

m(θ, ϕ)

=√

( l − m ) ( l + m + 1 ) Y lm+1(θ, ϕ) (1305)

Thus, on factorizing the eigenfunctions, we have(+

∂

∂θ− m cot θ

)Θl

m(θ) =√

( l − m ) ( l + m + 1 ) Θlm+1(θ)

sinm θ

(∂

∂θ

)1

sinm θΘl

m(θ) =√

( l − m ) ( l + m + 1 ) Θlm+1(θ)

sinm+1 θ

(∂

∂ cos θ

)1

sinm θΘl

m(θ) =√

( l − m ) ( l + m + 1 ) Θlm+1(θ)

(1306)

Starting with m = − l, after l +m iterations, one finds

Θlm(θ) =

( − 1 )l+m

2l l!

[( 2 l + 1 ) ( l − m )!

2 ( l + m )!

] 12

sinm θ

[∂

∂ cos θ

]l+m

sin2l θ

(1307)Thus, the spherical harmonics are given by

Y lm(θ) =

( − 1 )l+m

2l l!

[( 2 l + 1 ) ( l − m )!

4 π ( l + m )!

] 12

×

× sinm θ

[∂

∂ cos θ

]l+m

sin2l θ exp[i m ϕ

](1308)

These eigenfunctions are normalized via∫ 2 π

0

dϕ

∫ π

0

dθ sin θ | Y lm(θ, ϕ) |2 = 1 (1309)

The eigenfunction for positive m and negative m are related via

Y l−m(θ, ϕ) = ( − 1 )m Y l

m(θ, ϕ)∗ (1310)

The spherical harmonics also satisfy eigenvalue equation for L2. This canbe seen explicitly from the representation in spherical polar coordinates

− h2

[1

sin2 θ

∂2

∂ϕ2+

1sin θ

∂

∂θ

(sin θ

∂

∂θ

) ]Y l

m(θ, ϕ) = h2 l ( l + 1 ) Y lm(θ, ϕ)

(1311)

289

which on factorizing the eigenfunctions into polar and azimuthal parts, yields[+

m2

sin2 θ− 1

sin θ∂

∂θ

(sin θ

∂

∂θ

) ]Θl

m(θ) = l ( l + 1 ) Θlm(θ) (1312)

which identifies Θlm(θ) with the associated Legendre Polynomials P l

m(cos θ).This identification proceeds through changing variables x = cos θ, so one finds

∂

∂x

[( 1 − x2 )

∂Θ∂x

]− m2

1 − x2Θ = − l ( l + 1 ) Θ (1313)

which can then be put into the standard form

( 1 − x2 )∂2Θ∂x2

− 2 x∂Θ∂x

− m2

1 − x2Θ = − l ( l + 1 ) Θ (1314)

The spherical harmonics are given in terms of the associated Legendre Polyno-mials via

Y lm(θ, ϕ) =

√2 l + 1

4 π( l − m )!( l + m )!

( − 1 )m P lm( cos θ ) exp

[i m ϕ

](1315)

4.5.9 Legendre Polynomials

The generating function F (z, t) for the Legendre polynomials has an expansiongiven by

F (z, t) =1√

1 + t2 − 2 z t

=∞∑

l = 0

tl Pl(z) (1316)

for t < 1.

The Legendre polynomials satisfy recurrence relations, which can be derivedfrom the generating function expansion. The first is obtained by differentiatingthe generating function with respect to t(

∂F (z, t)∂t

)=

( z − t )( 1 + t2 − 2 z t )

32

=l=∞∑l=0

l tl−1 Pl(z) (1317)

After multiplying the above equation with a factor of ( 1 + t2 − 2 z t ) andthen substituting the generating function expansion, one obtains the relation

( 1 + t2 − 2 z t )l=∞∑l=0

l tl−1 Pl(z) = ( z − t )l=∞∑l=0

tl Pl(z) (1318)

290

-1

-0.5

0

0.5

1

-1 -0.5 0 0.5 1

z

|Ylm(θ,φ)|2

θ

Figure 73: Since the spherical harmonics Y lm(θ, ϕ) only depend on ϕ through

a phase factor, their θ dependence can be conveniently represented by a polarplot. In a polar plot, points on a curve are represented by the radial and angular

coordinates(|Y l

m(θ, ϕ)|2, θ)

.

On equating like powers of t, one has

( l + 1 ) Pl+1(z) − ( 2 l + 1 ) z Pl(z) + l Pl−1(z) = 0 (1319)

A second recurrence relation can be found by differentiating the generatingfunction expansion with respect to z(

∂F (z, t)∂z

)=

t

( 1 + t2 − 2 z t )32

=l=∞∑l=0

tl P ′l (z) (1320)

On multiplying by ( 1 + t2 − 2 z t ) and then substituting the generatingfunction expansion, one obtains

t∑

l

tl Pl(z) = ( 1 + t2 − 2 z t )∑

l

tl P ′l (z) (1321)

which yields the relation

P ′l+1(z) + P ′l−1(z) = 2 z P ′l (z) + Pl(z) (1322)

291

involving the derivatives of the polynomials.

Numerous other relations can be obtained from the above recurrence rela-tions. If we take the derivative of the first recurrence relation, eqn(1319), thenwe have

( l + 1 ) P ′l+1(z) − ( 2 l + 1 ) z P ′l (z) + l P ′l−1(z) = ( 2 l + 1 ) Pl(z) (1323)

The term proportional to z P ′l (z) can be eliminated from the above equation.This is achieved by multiplying eqn(1323) by two and then subtracting ( 2 l + 1 )times eqn(1322), which leads to

P ′l−1(z) − P ′l+1(z) = − ( 2 l + 1 ) Pl(z) (1324)

A more useful set of equations can be obtained by relating the Legendre poly-nomials of one order to the next. For example, adding eqn(1322) and eqn(1324)one eliminates P ′l+1(z)

P ′l−1(z) = z P ′l (z) − l Pl(z) (1325)

On subtracting the same pair of the equations, one has

P ′l+1(z) = ( l + 1 ) Pl(z) + z P ′l (z) (1326)

Another pair of equations can be obtained from eqn(1325) and eqn(1326) byshifting the index of eqn(1325) from l − 1 to l and then eliminating the termproportional to P ′l+1(z), leading to the equation

( 1 − z2 ) P ′l (z) = ( l + 1 )[z Pl(z) − Pl+1(z)

](1327)

Another recurrence relation can be derived from the above equation by usingeqn(1319) to eliminate the term proportional to ( l + 1 ) Pl+1(z). This processresults in

( 1 − z2 ) P ′l (z) = l Pl−1(z) − l z Pl(z) (1328)

From these relations we can find the differential equation for the Legendrepolynomials Pl(z). By differentiating eqn(1328), one obtains an equation in-volving P ′l−1(z) which can be eliminated using eqn(1325). This procedure leadsto Legendre’s equation

( 1 − z2 ) P ′′l (z) − 2 z P ′l (z) = − l ( l + 1 ) Pl(z) (1329)

292

On setting z = cos θ, the differential equation takes the form

1sin θ

∂

∂θ

(sin θ

∂

∂θPl( cos θ )

)+ l ( l + 1 ) Pl( cos θ ) = 0 (1330)

This is recognized as the eigenvalue equation for the total angular momentum,in the case where there is a simultaneous zero eigenvalue for the z component ofthe angular momentum, m = 0. In the general case, where m 6= 0, the simul-taneous eigenfunctions were denoted by Θl

m(θ). For finite m, the θ dependenceof the angular momentum eigenvalues is expressed in terms of the associatedLegendre polynomials.

4.5.10 Associated Legendre Functions

The Associated Legendre functions, P lm(z) for positive m, are defined as

P lm(z) = ( 1 − z2 )

m2

∂m

∂zmPl(z) (1331)

Clearly, the associated Legendre function with m = 0 is identical with theLegendre polynomial. Furthermore, since Pl(z) is a polynomial of order l, theassociated Legendre functions vanish for m > l.

The associated Legendre functions satisfy a differential equation obtainedby differentiating Legendre’s equation m times. This leads to

( 1 − z2 ) v′′ − 2 z ( m + 1 ) v′ + ( l − m ) ( l + m + 1 ) v = 0 (1332)

wherev =

∂m

∂zmPl(z) (1333)

The associated Legendre function P lm(z) and v are related via

v =1

( 1 − z2 )m2P l

m(z) (1334)

The first and second derivatives of v are evaluated as

v′ = mz

1 − z2v +

1( 1 − z2 )

m2

∂

∂zP l

m(z) (1335)

and

v′′ = m

[( m + 2 )

z2

( 1 − z2 )2v +

11 − z2

v

]+

2 m z

1 − z2

1( 1 − z2 )

m2

∂

∂zP l

m(z) +1

( 1 − z2 )m2

∂2

∂z2P l

m(z)

(1336)

293

Substitution of the expressions for v, v′ and v′′ into eqn(1332) leads to

( 1 − z2 )∂2

∂z2P l

m(z) − 2 z∂

∂zP l

m(z) +[l ( l + 1 ) − m2

1 − z2

]P l

m(z) = 0

(1337)which is the differential equation satisfied by the associated Legendre functions.On changing variables from z to cos θ, one finds that the associated Legendrepolynomials satisfy

1sin θ

∂

∂θ

(sin θ

∂

∂θP l

m( cos θ ))

+[l ( l + 1 ) − m2

sin2 θ

]P l

m( cos θ ) = 0

(1338)The above equation occurs as part of eigenvalue equation for the magnitude ofthe orbital angular momentum, when the ϕ dependence has been separated outby the introduction of the quantum numberm corresponding to the z componentof the orbital angular momentum. The normalized part of the wave functionwhich depends on the polar angle θ is given by

Θlm(θ) =

√( 2 l + 1 )

2( l − m )!( l + m )!

P lm(cos θ) (1339)

which involves the associated Legendre function in the variable cos θ.

4.5.11 Spherical Harmonics

Explicit expressions for the first few spherical harmonics Y lm(θ, ϕ) are given in

Tables(3-4). The spherical harmonics are normalized so that∫ 2 π

0

dϕ

∫ π

0

dθ Y l′

m′(θ, ϕ)∗ Y lm(θ, ϕ) = δl,l′ δm,m′ (1340)

294

Table 3: The Lowest Order Spherical Harmonics Y lm(θ, ϕ).

l m Y lm(θ, ϕ)

l = 0 m = 0 Y 00 (θ, ϕ) 1√

4 π

l = 1 m = 0 Y 10 (θ, ϕ)

√3

4 π cos θ

m = ±1 Y 1±1(θ, ϕ) ∓

√3

8 π e± i ϕ sin θ

l = 2 m = 0 Y 20 (θ, ϕ)

√5

16 π

(3 cos2 θ − 1

)

m = ±1 Y 2±1(θ, ϕ) ∓

√158 π e± i ϕ cos θ sin θ

m = ±2 Y 2±2(θ, ϕ)

√15

32 π e± i 2 ϕ sin2 θ

Table 4: The Spherical Harmonics with l = 3, Y 3m(θ, ϕ).

l m Y lm(θ, ϕ)

l = 3 m = 0 Y 30 (θ, ϕ)

√7

16 π

(5 cos3 θ − 3 cos θ

)

m = ±1 Y 3±1(θ, ϕ) ∓

√21

64 π e± i ϕ ( 5 cos2 θ − 1 ) sin θ

m = ±2 Y 3±2(θ, ϕ)

√10532 π e± i 2 ϕ cos θ sin2 θ

m = ±3 Y 3±3(θ, ϕ) ∓

√35

64 π e± i 3 ϕ sin3 θ

295

Figure 74: The angular dependence of the squared modulus of the sphericalharmonic | Y 0

0 (θ, ϕ) |2. Since the spherical harmonic does not depend on (θ, ϕ)via the exponential term exp[ i m ϕ ], the angular distribution is sphericallysymmetric. The θ dependence for l = 0 is shown as a polar plot.

l = 0

m = 0

l ΘΘΘΘ00(θθθθ) l2

296

Figure 75: The angular dependence of the squared modulus of the sphericalharmonics | Y l

m(θ, ϕ) |2. Since the spherical harmonics only depend on ϕ viathe exponential term exp[ i m ϕ ], the probability density is independent ofϕ. The angular dependence is solely determined by the associated Legendrefunctions Θl

m(θ). The θ dependence for l = 1, m = ± 1 , 0 are shown aspolar plots.

l = 1

m = 0

l ΘΘΘΘ10(θθθθ) l2

l = 1

m = 1

l ΘΘΘΘ11(θθθθ) l2

297

Under inversion r → − r, then as ϕ → ϕ + π and θ → π − θ, onefinds that the spherical harmonics transform as

Y lm(θ, ϕ) → Y l

m(π − θ, π + ϕ)→ ( − 1 )l Y l

m(θ, ϕ) (1341)

Thus, all orbital angular momentum wave functions with the same value of lmust have the same parity.

The spherical harmonics are simultaneous eigenstates of two Hermitean op-erators, therefore, they form a complete set. The completeness relation in thespace of (θ,ϕ) can be expressed as

∞∑l=0

l∑m=−l

Y lm(θ′, ϕ′)∗ Y l

m(θ, ϕ) = δ( r′ − r ) (1342)

where the delta function expresses the condition that the directions r and r′

must be the same.

——————————————————————————————————

4.5.12 Exercise 80

Calculate the vector probability current density j(r, θ, ϕ) for a wave function ofthe form

Ψ(r, θ, ϕ) = f(r) Y lm(θ, ϕ) (1343)

which has an orbital angular momentum of l = 1, for the various eigenvaluesof the z component of the angular momentum, m, if f(r) is a real function.

——————————————————————————————————

4.5.13 Solution 80

The velocity is expressed as

v =p

me

= − ih

me∇ (1344)

The gradient is written as

∇ = er∂

∂r+ eθ

1r

∂

∂θ+ eϕ

1r sin θ

∂

∂ϕ(1345)

298

l = 2

m = 0l ΘΘΘΘ2

0(θθθθ) l2

Figure 76: The angular dependence of the squared modulus of the sphericalharmonics | Y l

m(θ, ϕ) |2. Since the spherical harmonics only depend on ϕ viathe exponential term exp[ i m ϕ ], the probability density is independent ofϕ. The angular dependence is solely determined by the associated Legendrefunctions Θl

m(θ). The θ dependence for l = 2 , m = ± 2 , ± 1 , 0 are shownas polar plots.

l = 2

m = 1

l ΘΘΘΘ21(θθθθ) l2

l = 2

m = 2l ΘΘΘΘ2

2(θθθθ) l2

299

l = 3

m = 0

l ΘΘΘΘ30(θθθθ) l2

Figure 77: The angular dependence of the squared modulus of the sphericalharmonics | Y l

m(θ, ϕ) |2. Since the spherical harmonics only depend on ϕ viathe exponential term exp[ i m ϕ ], the probability density is independent ofϕ. The angular dependence is solely determined by the associated Legendrefunctions Θl

m(θ). The θ dependence for l = 3 , m = ± 3 , ± 2 , ± 1 , 0are shown as polar plots. The probability density is peaked up in the equatorialplane θ = π

2 for m = ± l.

l = 3m = 1 l ΘΘΘΘ3

1(θθθθ) l2

l = 3m = 2 l ΘΘΘΘ3

2(θθθθ) l2

l = 3m = 3

l ΘΘΘΘ33(θθθθ) l2

300

Inversion

θ

ϕx

y

z r

- r

π−θ

π+ϕ

r

r

Figure 78: In spherical polar coordinates, the inversion or parity operator pro-duces the transformation ϕ → ϕ + π and θ → π − θ.

The probability current density is expressed in terms of the wave function Ψ(r, t)as the imaginary part of a vector quantity

j(r, t) =h

me=m

[Ψ∗(r, t) ∇ Ψ(r, t)

](1346)

which in the case of a pure energy eigenvalue is time independent. In sphericalpolar coordinates, the probability current is expressed as

j(r, t) = erh

2 me i

(Ψ∗(r)

∂

∂rΨ(r) − Ψ(r)

∂

∂rΨ∗(r)

)+ eθ

h

2 me i r

(Ψ∗(r)

∂

∂θΨ(r) − Ψ(r)

∂

∂θΨ∗(r)

)+ eϕ

h

2 me i r sin θ

(Ψ∗(r)

∂

∂ϕΨ(r) − Ψ(r)

∂

∂ϕΨ∗(r)

)(1347)

On expressing the wave function as

Ψ(r) = g(r) Θlm(θ) exp

[i m ϕ

](1348)

301

where both g(r) and the associated Legendre functions Θlm(θ) are real, one

obtains

j(r, t) = eϕh m

me r sin θ

(g(r) Θl

m(θ))2

= eϕh m

me r sin θ

∣∣∣∣ Ψ(r)∣∣∣∣2 (1349)

where m is the azimuthal or magnetic quantum number. Hence, if the stateis an eigenstate of the z component of the angular momentum, the probabilitycurrent orbits around the atom in orbits parallel to the x - y plane.

——————————————————————————————————

4.5.14 Exercise 81

In the dipole approximation, the average power radiated when the hydrogenatom decays from the n, l th excited state to the n′, l′ th excited state is givenby

Pn,l;n′,l′ =43ω4

n,l;n′,l′e2

c3

∣∣∣∣ ∫ d3r φ∗n′,l′,m′(r) r φn,l,m(r)∣∣∣∣2 (1350)

By considering the angular integration, show that the only non-zero matrixelements occur when the angular momentum quantum numbers l and m for thetwo states are related via

∆l = l′ − l = ± 1∆m = m′ − m = 0 , ± 1 (1351)

These are the selection rules for dipole radiation. From conservation of angularmomentum, and since ∆l = ± 1, one sees that the photon must have spinone. The photon has a spin which is an integer multiple of h and, therefore,it is a boson. This contrasts with particles which have spins that are half-oddinteger multiples of h, such as the electron which has 1

2 h, that are fermions15.Transitions which do not satisfy the dipole selection rules are forbidden. Gen-erally, forbidden transitions still can occur, but require going beyond the dipoleapproximation and have intensities which are lower by factors of the order ofmagnitude a2

0λ2 where a0 is the Bohr radius and λ is the wave length of light.

——————————————————————————————————

15The relationship between spin and statistics was first discussed by W. Pauli in the article,Physical Review, 58, 716 (1940).

302

Energy levels of Hydrogen

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

En

[ Ryd

berg

s ]

n=1

n=2n=3n=4

l=0 l=1 l=2 l=3 l=4

Balmer

Lyman

Figure 79: The energy levels for Hydrogen (with 4 ≥ l). The energies areplotted in units of Rydbergs. Some of the transitions allowed by the electricdipole selection rule are shown.

4.5.15 Solution 81

The vector r is expressed in terms of the Cartesian unit vectors through

r = r

(cos θ ez + sin θ ( sinϕ ey + cosϕ ex )

)(1352)

and substituted into the matrix elements. To evaluate the angular integrations,we start by finding the two recurrence relations

( 2 l + 1 ) cos θ P lm(cos θ) = ( l + m ) P l−1

m (cos θ) + ( l − m + 1 ) P l+1m (cos θ)

( 2 l + 1 ) sin θ P lm(cos θ) = P l−1

m+1(cos θ) + P l+1m+1(cos θ) (1353)

and then show that

cos θ Y lm(θ, ϕ) =

=(

(l −m+ 1) (l +m+ 1)(2l + 1) (2l + 3)

) 12

Y l+1m +

((l −m) (l +m)(2l − 1) (2l + 1)

) 12

Y l−1m

(1354)

exp[

+ i ϕ

]sin θ Y l

m(θ, ϕ) =

303

=(

(l +m+ 1) (l +m+ 2)(2l + 1) (2l + 3)

) 12

Y l+1m+1 +

((l −m) (l −m− 1)

(2l − 1) (2l + 1)

) 12

Y l−1m+1

(1355)

exp[− i ϕ

]sin θ Y l

m(θ, ϕ) =

=(

(l −m+ 1) (l −m+ 2)(2l + 1) (2l + 3)

) 12

Y l+1m−1 +

((l +m) (l +m− 1)

(2l − 1) (2l + 1)

) 12

Y l−1m−1

(1356)

The expected result then follows immediately from the orthogonality of thespherical harmonics.

——————————————————————————————————

4.5.16 Exercise 82

Consider the commutation relations

[ Lz , x ] = + i h y

[ Lz , y ] = − i h x

[ Lz , z ] = 0 (1357)

Take the matrix elements of the commutator involving z between states withangular momentum (l′,m′) and (l,m) and use the eigenvalue equation to showthat

( m − m′ )∫ 2π

0

dϕ

∫ π

0

dθ sin θ Y l′

m′(θ, ϕ)∗ cos θ Y lm(θ, ϕ) = 0 (1358)

Likewise, by considering the commutation rules involving x and y, show thatthey can be combined to yield

( m − m′ )2∫ 2π

0

dϕ

∫ π

0

dθ sin θ Y l′

m′(θ, ϕ)∗ sin θ cosϕ Y lm(θ, ϕ)

=∫ 2π

0

dϕ

∫ π

0

dθ sin θ Y l′

m′(θ, ϕ)∗ sin θ cosϕ Y lm(θ, ϕ) (1359)

Hence, deduce the selection rule that the matrix element∫d3r φ∗n′,l′,m′(r) r φn,l,m(r) = 0 (1360)

vanishes unless m′ − m = ± 1 or m′ − m = 0.

304

Following a similar procedure with the double commutator16

[ L2 , [ L2 , r ] ] = 2 h2

(r L2 + L2 r

)(1362)

deduce that the matrix elements∫d3r φ∗n′,l′,m′(r) r φn,l,m(r) (1363)

are only non-zero if l and l′ satisfy the equation(( l′ + l + 1 )2 − 1

) (( l′ − l )2 − 1

)= 0 (1364)

Since the first factor is always positive (unless l′ = l = 0), the electric dipoleselection rule becomes l′ − l = ± 1.

——————————————————————————————————

4.5.17 Solution 82

The selection rules for the z-component of the electron’s orbital angular mo-mentum can be proved by considering the commutation relations

[ Lz , x ] = i h y

[ Lz , y ] = − i h x

[ Lz , z ] = 0 (1365)

On taking the matrix elements between states with definite z-components of theangular momenta, one finds

< n′l′m′ | [ Lz , x ] | nlm > = i h < n′l′m′ | y | nlm >

< n′l′m′ | [ Lz , y ] | mlm > = − i h < n′l′m′ | x | nlm >

< n′l′m′ | [ Lz , z ] | nlm > = 0 (1366)

which reduce to

( m′ − m ) < n′l′m′ | x | nlm > = i < n′l′m′ | y | nlm >

( m′ − m ) < n′l′m′ | y | mlm > = − i < n′l′m′ | x | nlm >

( m′ − m ) < n′l′m′ | z | nlm > = 0 (1367)

16As was shown by Dirac (P. A. M. Dirac, Proc. Roy. Soc. A111, 251 (1926).), this doublecommutation relation is a specific example of the more general relation

[ J2 , [ J2 , A ] ] = 2 h2

(A J2 + J2 A

)− 4 h2 ( A . J ) J (1361)

valid for any arbitrary vector operator A.

305

From the last equation, it follows that either m′ = m or that

< n′l′m′ | z | nlm > = 0 (1368)

On combining the first two equations, one finds that

( m′ − m )2 < n′l′m′ | x | nlm > = i ( m′ − m ) < n′l′m′ | y | nlm >

= < n′l′m′ | x | nlm > (1369)

This equation is solved by requiring that either

( m′ − m )2 = 1 (1370)

or< n′l′m′ | x | nlm > = 0 (1371)

Hence, we have derived the selection rules ∆m = ± 1, 0.

The selection rules for the magnitude of the electrons orbital angular mo-mentum are found by considering the double commutator

[ L2 , [ L2 , r ] ] = 2 h2

(r L2 + L2 r

)(1372)

On taking the matrix elements of this equation between eigenstates of the mag-nitude of the orbital angular momentum, one finds

h4

(l′ ( l′ + 1 )− l ( l + 1 )

)2

< n′l′m′ | r | nlm >=(l′ ( l′ + 1 ) + l ( l + 1 )

)< n′l′m′ | r | nlm >

(1373)Hence, either

< n′l′m′ | | nlm > = 0 (1374)

or (( l′ + l + 1 )2 − 1

) (( l′ − l )2 − 1

)= 0 (1375)

Since the first factor is always positive for non-zero l and l′, the electric dipoleselection rule becomes ∆l = ± 1.

——————————————————————————————————

4.5.18 The Addition Theorem

The addition theorem for spherical harmonics states that

Pl( cos γ ) =4 π

2 l + 1

m = l∑m = − l

Y lm(θ1, ϕ1) Y l

m(θ2, ϕ2)∗ (1376)

306

where γ is the angle between the directions (θ1, ϕ1) and (θ2, ϕ2), i.e.,

cos γ = cos θ1 cos θ2 + sin θ1 sin θ2 cos ( ϕ1 − ϕ2 ) (1377)

We shall assume that we can expand Pl( cos γ ) as a Laplace series

Pl( cos γ ) =∑m1

alm1

Y lm1

(θ1, ϕ1) (1378)

whereal

m1=∫

dΩ1 Pl( cos γ ) Y lm1

(θ1, ϕ1) (1379)

But, on noting that the associated Legendre polynomials are the same as theLegendre polynomials when m = 0 and on using eqn(1315), one obtains

Pl( cos γ ) =(

4 π2 l + 1

) 12

Y l0 (γ, 0) (1380)

The choice of the azimuthal angle as 0 is irrelevant, as m = 0. Hence, we have

alm1

=(

4 π2 l + 1

) 12∫

dΩ1 Yl0 (γ, 0) Y l

m1(θ1, ϕ1)∗ (1381)

The choice of the orientation of the axes is irrelevant, as the integration over dΩ1

runs over all 4π solid angle. We shall choose the direction (θ2, ϕ2) as our polaraxis. In this case, we need to re-express the spherical harmonic Y l

m1(θ1, ϕ1) in

terms of our new variable of integration

Y lm1

(θ1, ϕ1)∗ =∑m2

blm2

∗ Y lm2

(γ, 0)∗ (1382)

where, as we are dealing with an eigenfunction of the scalar magnitude of theangular momentum with eigenvalue governed by l, the eigenvalue l is completelyunaltered by a different choice of Cartesian coordinates. Thus,

alm1

=(

4 π2 l + 1

) 12 ∑

m2

∫dΩγ Y

l0 (γ, 0) blm2

∗ Y lm2

(γ, 0)∗

=(

4 π2 l + 1

) 12 ∑

m2

blm2

∗ δm2,0

=(

4 π2 l + 1

) 12

bl0∗ (1383)

where we have used the orthogonality property of the spherical harmonics.Therefore, we see that the coefficients al

m which appear in the Laplace ex-pansion are related to only one non-zero coefficient bl0 in their expansion given

307

by eqn(1382). The coefficient bl0 can be obtained directly when γ = 0 as theexpansion only contains the one term

Y lm1

(θ1, ϕ1)∗ = bl0∗ Y l

0 (0, 0)∗

= bl0∗(

2 l + 14 π

) 12

(1384)

since the spherical harmonics with m 6= 0 vanish when γ = 0. In the above equa-tion the variables (θ1, ϕ1) take on the fixed numerical values (θ2, ϕ2), becauseγ ≡ 0. Hence,

bl0∗ =

(4 π

2 l + 1

) 12

Y lm1

(θ2, ϕ2)∗ (1385)

Inserting this into eqn(1383), we have

alm1

=(

4 π2 l + 1

)Y l

m1(θ2, ϕ2)∗ (1386)

On substituting the coefficient alm1

back into the Laplace series expansion, onehas

Pl( cos γ ) =4 π

2 l + 1

m1=l∑m1=−l

Y lm1

(θ1, ϕ1) Y lm1

(θ2, ϕ2)∗ (1387)

This concludes the proof of the Addition Theorem for Spherical Harmonics.

4.5.19 Finite-Dimensional Representations

One can find a representation of the angular momentum operators Li withinthe manifold of eigenstates of L2 with the eigenvalue λl. The states are tobe represented by column vectors. Since the eigenvalues are ( 2 l + 1 ) folddegenerate, the space is spanned by unit column vectors with ( 2 l + 1 ) rows.Any arbitrary column vector can be expressed as a linear superposition of the( 2 l + 1 ) basis column vectors. The angular momentum operators act on thestates, and transform the states into other states. The rules of transformationcorrespond to the laws of matrix multiplication, in which the operators are( 2 l + 1 ) × ( 2 l + 1 ) dimensional square matrices. For convenience, thebasis states may be taken to be the ( 2 l + 1 ) eigenstates of Lz. For definiteness,we shall restrict our discussion to l = 1. The three basis eigenstates may berepresented by the unit column vectors;

Y 11 =

( 100

)(1388)

with eigenvalue + h,

Y 10 =

( 010

)(1389)

308

with eigenvalue 0, and

Y 1−1 =

( 001

)(1390)

with eigenvalue − h. Any arbitrary state Ψ with eigenvalue λl can be expressedas the linear superposition of the basis states via

Ψ =m=1∑

m=−1

Ψm Y lm (1391)

in which Ψm are the expansion coefficients. This equation can be expressedmore concretely as( Ψ1

Ψ0

Ψ−1

)= Ψ1

( 100

)+ Ψ0

( 010

)+ Ψ−1

( 001

)(1392)

The inner product of two states Ψ and Φ is defined to be a complex number,obtained via

Φ† Ψ =m=1∑

m=−1

Φ∗m Ψm (1393)

The inner product is a generalization of the scalar product to arbitrary dimen-sional (in this case ( 2 l + 1 ) dimensional) column matrices with complexcomponents. It involves a row vector Φ† which is the complex conjugate andtranspose of the column vector Φ. In general, it is convenient to label the com-ponents of the basis vectors by an abstract index i, instead of the quantumnumber m which corresponds to the z component of angular momentum. Weshall use this more general labelling below. The inner product is just the resultof the matrix multiplication of the row matrix Φ† with the column matrix Ψ,i.e.,

Φ† Ψ =(

Φ∗1 Φ∗2 Φ∗3

) ( Ψ1

Ψ2

Ψ3

)(1394)

where we are labelling the three components of our states Φ and Ψ by the indexi. We shall assume that our states Ψ are normalized to unity, that is, we require

Ψ† Ψ =3∑

i=1

Ψ∗i Ψi = 1 (1395)

This just corresponds to the condition that the sum over all possibilities yieldsa probability of unity.

The effect of an operator A on a state Ψ transforms it into another state Φ,according to the laws of matrix multiplication

Φ = A Ψ (1396)

309

where A is a three by three matrix. More concretely, the action of the operatoris given by ( Φ1

Φ2

Φ3

)=

(A1,1 A1,2 A1,3

A2,1 A2,2 A2,3

A3,1 A3,2 A3,3

) ( Ψ1

Ψ2

Ψ3

)(1397)

which results in the state Φ having the components

Φi =j=3∑j=1

Ai,j Ψj (1398)

The operator A is specified by specifying the matrix elements Ai,j . Operatorsrepresented by matrices can be compounded by the laws of matrix additionand multiplication. Matrix multiplication is generally non-commutative, and issuitable for representing non-commuting operators.

The Hermitean conjugate of the operator is defined by the equation(Φ† A Ψ

)∗= Ψ† A† Φ (1399)

which must be true for any arbitrary states Φ and Ψ. From this, one can seethat the Hermitean conjugate of a matrix is just the complex conjugate of thetransposed matrix. Thus, if the matrix A has matrix elements Ai,j , then theHermitean conjugate A† has matrix elements given by A∗j,i. A Hermitean ma-trix is defined to be a matrix for which A = A†.

Since the basis states are eigenfunctions of the operator Lz, one has

Lz Y1m = m h Y 1

m (1400)

The three linear equations can be trivially interpreted as a matrix equation, inwhich the operator Lz is diagonal in the chosen basis. Thus, one can representthe operator Lz by the diagonal three by three matrix

Lz = h

( 1 0 00 0 00 0 − 1

)(1401)

where the diagonal matrix elements are just the eigenvalues of Lz. The effectof the raising operator L+ on the basis states is to raise the angular momentumaccording to the law

L+ Y lm = h

√( l − m ) ( l + m + 1 ) Y l

m+1 (1402)

Since the effect of L+ on the state with maximal m, m = + 1, is to producezero, one has one trivial and two non-trivial algebraic equations, which can be

310

used to represent the raising operator as

L+ = h

( 0√

2 00 0

√2

0 0 0

)(1403)

Likewise, from considering the equation

L− Y lm = h

√( l + m ) ( l − m + 1 ) Y l

m−1 (1404)

one can construct a representation of the lowering operator as

L− = h

( 0 0 0√2 0 0

0√

2 0

)(1405)

The x and y components of the angular momentum are found from the defini-tions

L± = Lx ± i Ly (1406)

asLx =

12

( L+ + L− ) (1407)

andLy =

12 i

( L+ − L− ) (1408)

From which, one has

Lx =h√2

( 0 1 01 0 10 1 0

)(1409)

and

Ly =h√2

( 0 − i 0i 0 − i0 i 0

)(1410)

It can be shown that these matrices, when compounded according to the lawsof matrix multiplication and addition, satisfy the commutation relations

[ Li , Lj ] = i h∑

k

εi,j,k Lk (1411)

where εi,j,k is the anti-symmetric Levi-Civita symbol. The anti-symmetric Levi-Civita symbol is defined so that εi,j,k = 1 if i, j, k correspond to an even numberof permutations of x, y, z, or is − 1, if i, j, k corresponds to an odd permutationof x, y, z. Otherwise, εi,j,k = 0, when one or more index is repeated. Fur-thermore, since the Hermitean conjugate of the operator just corresponds tothe complex conjugate of the transposed matrix, we see that the componentsLk not only satisfy the angular momentum commutation relations, but are also

311

Hermitean operators.

It can then be easily shown that the operator

L2 = L2x + L2

y + L2z (1412)

is a diagonal matrix, with non-zero matrix elements λl = 1 ( 1 + 1 ) h2. Thisis as expected, since the basis states are all eigenstates of the magnitude of theangular momentum. Hence, we have

L2 = 2 h2

( 1 0 00 1 00 0 1

)(1413)

Since the operator corresponding to the magnitude of the angular momentum isproportional to the unit operator, it commutes with all the components of theangular momentum Li. The operator L

2is sometimes referred to as a Casimir

operator.

——————————————————————————————————

4.5.20 Exercise 83

Evaluate the commutators of the three by three matrices representing the com-ponents of the angular momentum, and show that they satisfy the commutationrelations, such as,

[ Lx , Ly ] = i h Lz (1414)

Also evaluate the matrix corresponding to the square of the angular momentumL2 and show that it commutes with the various components of L, and find theeigenvalues of L2.

——————————————————————————————————

4.5.21 Exercise 84

Find the eigenvalues and eigenvectors of the component of the angular momen-tum along the unit vector η, where

η = sin θ cosϕ ex + sin θ sinϕ ey + cos θ ez (1415)

Work in the space of l = 1 with the basis in which Lz is diagonal.

Hence, find the unitary matrix Ue(θ) which produces a rotation of the statesthrough an angle θ, about the axis

e = − sinϕ ex + cosϕ ey (1416)

312

The operator Ue(θ) should bring the eigenstates of Lz into coincidence with theeigenstates of Lη. It is also required that Ue(θ) does not transform the eigen-states of Le.

——————————————————————————————————

4.5.22 Solution 84

We seek the eigenvalues and eigenfunctions of the operator

η . L = sin θ cosϕ Lx + sin θ sinϕ Ly + cos θ Lz (1417)

which has the matrix representation

η . L = h

cos θ 1√2

sin θ exp[ − i ϕ ] 01√2

sin θ exp[ + i ϕ ] 0 1√2

sin θ exp[ − i ϕ ]0 1√

2sin θ exp[ + i ϕ ] − cos θ

(1418)

in the basis where Lz is diagonal. The eigenvalues µ h are determined from thesecular equation

0 =

∣∣∣∣∣cos θ − µ 1√

2sin θ exp[ − i ϕ ] 0

1√2

sin θ exp[ + i ϕ ] − µ 1√2

sin θ exp[ − i ϕ ]0 1√

2sin θ exp[ + i ϕ ] − cos θ − µ

∣∣∣∣∣(1419)

and, thus, are found to be µ = 0 , ± h. The eigenfunction Y ηµ corresponding

to the eigenvalue µ, are denoted by

Y ηµ =

αηµ

βηµ

γηµ

(1420)

The components of the µ = 0 eigenfunction are found from the eigenvalueequation 0

00

=

√2 cos θ sin θ exp[ − i ϕ ] 0

sin θ exp[ + i ϕ ] 0 sin θ exp[ − i ϕ ]0 sin θ exp[ + i ϕ ] −

√2 cos θ

αη0

βη0

γη0

(1421)

Thus, the normalized eigenfunction of the operator in eqn(1417) correspondingto the zero eigenvalue is found as

Y η0 =

1√2

− sin θ exp[ − i ϕ ]√2 cos θ

sin θ exp[ + i ϕ ]

(1422)

313

Likewise, one finds the eigenfunction corresponding to µ = + 1 is given by

Y η1 =

12

( 1 + cos θ ) exp[ − i ϕ ]√2 sin θ

( 1 − cos θ ) exp[ + i ϕ ]

(1423)

and the eigenfunction corresponding to µ = − 1 is found as

Y η−1 =

12

( 1 − cos θ ) exp[ − i ϕ ]−√

2 sin θ( 1 + cos θ ) exp[ + i ϕ ]

(1424)

The phases of the eigenfunctions are completely arbitrary.

The required rotation is about an axis in the plane perpendicular to the unitvectors ez and eη. Since the rotation is specified by three parameters (the axisof the rotation and the angle of rotation), the transformation matrix containsthree parameters that have to be determined. The rotation matrix Ue(θ) ispartly determined from its action on the eigenstates of Lz

Y ηµ = Ue(θ) Y z

µ (1425)

There is also the subsidiary requirement that the component of the spins alongthe rotation axis are unaffected, which means that the eigenstates of Le shouldbe mapped onto themselves. Thus, we have αη

µ

βηµ

γηµ

= 1

2

( 1 + cos θ ) exp[ − i ϕ ] −√

2 sin θ exp[ − i ϕ ] ( 1 − cos θ ) exp[ − i ϕ ]√2 sin θ 2 cos θ −

√2 sin θ

( 1 − cos θ ) exp[ + i ϕ ]√

2 sin θ exp[ + i ϕ ] ( 1 + cos θ ) exp[ + i ϕ ]

×

αzµ

βzµ

γzµ

(1426)

The columns of the transformation matrix Ue(ϕ) are the eigenvectors that wehave just found. The phases of the columns can be altered to fit our require-ment that a state with angular momentum directed along the axis of rotation eremains unchanged (apart from an over all phase factor). The phase factors ofthe Y η

µ are chosen as exp[ i µ ϕ ] so that the diagonal elements of the unitarytransformation are real17. This choice of phase factors produces the desiredrotation since the eigenstates of Le are unchanged by it. More precisely, under

17The change of phase can be understood as being the result of a rotation through ϕ aboutthe η axis

314

this transformation, the eigenstates of Le only change by a θ dependent phasefactor. Also, with the above choice of phase, the eigenstates Y z

µ are completelyunaffected by the transformation, if the angle of rotation θ is set to zero.

The above result can be re-written as follows. On recognizing that thecomponent of the angular momentum operator parallel to the axis of rotationis given by(

e . L

)=

h√2

0 − i exp[ − i ϕ ] 0i exp[ + i ϕ ] 0 − i exp[ − i ϕ ]

0 i exp[ + i ϕ ] 0

(1427)

and that the square is given by(e . L

)2

=h2

2

1 0 − exp[ − 2 i ϕ ]0 2 0

− exp[ + 2 i ϕ ] 0 1

(1428)

we find that the rotation operator may be written in the form

Ue(θ) = 1 +(

cos θ − 1) (

e . L

h

)2

− i sin θ(e . L

h

)(1429)

This is an example of Cayley’s theorem, as the rotation about the axis e can bewritten as the exponential of a matrix

Ue(θ) = exp[− i

hθ ( e . L )

](1430)

——————————————————————————————————

The angular momentum operator corresponding to larger l values can alsobe constructed by the same method, but with a larger number of basis states.In particular, one requires ( 2 l + 1 ) basis states to span the space with totalangular momentum l. In fact, almost any operator can be represented by aN × N square matrix acting on a N -dimensional column vector, however, thisis only useful for bases where the dimension N is finite.

4.5.23 The Laplacian Operator

The magnitude of the angular momentum can be expressed in terms of theLaplacian by using the vector identity(A ∧ B

).

(C ∧ D

)=(A . C

) (B . D

)−(A . D

) (B . C

)(1431)

315

which, for our operators, can be adapted as

L2 = ( r ∧ p ) . ( r ∧ p )= − ( r ∧ p ) . ( p ∧ r )

= − r .

[p ( p . r ) − p2 r

](1432)

In manipulating the expression for the square of the angular momentum, onehas to carefully respect the ordering of r and p. In the second line we have usedthe identity

r ∧ p = − p ∧ r (1433)

which is valid because the different components of the position and angularmomentum commute. The last term in eqn(1432) can be written in terms ofr2 p2 by using the commutator

[ r , p2 ] = 2 i h p (1434)

Also the first term in eqn(1432) can be re-written by using the commutator

p . r − r . p = − 3 i h (1435)

in the last factor. Hence, we obtain

L2 = r2 p2 + i h r . p − ( r . p ) ( r . p ) (1436)

The term r . p is evaluated as

r . p = − i h r∂

∂r(1437)

Putting these together, we find that the square of the angular momentum isgiven by

L2 = r2 p2 + h2 ∂

∂r

(r2

∂

∂r

)(1438)

Thus, we find that the Laplacian operator is expressed in terms of the angularmomentum by

p2 =L2

r2− h2

r2∂

∂r

(r2

∂

∂r

)(1439)

This results in the usual expression for the kinetic energy in spherical polarcoordinates

T =p2

2 m

=L2

2 m− h2

2 m r2∂

∂r

(r2

∂

∂r

)(1440)

316

4.5.24 An Excursion into d-Dimensional Space

As angular momentum is associated with a rotation in a two-dimensional planeand since in d dimensions there is no unique normal to a two-dimensional plane,the angular momentum is not a vector. The angular momentum is a two-formand has d(d − 1)/2 components that are labelled by two indices denoting theplane of rotation18. In a Cartesian coordinate system, the components of Lj,k

are defined via

Lj,k = − i h

(xj

∂

∂xk− xk

∂

∂xj

)(1441)

The quantity Lj,k is antisymmetric in the indices j and k. The square of theangular momentum no longer has a simple geometric significance, but is stilldefined as the operator

L2 =d∑

j>k

L2j,k

= − h2d∑

j>k

(xj

∂

∂xk− xk

∂

∂xj

)2

(1442)

The above form can be expanded as

L2 = − h2d∑

j,k

(x2

j

∂2

∂x2k

− xj xk∂

∂xj

∂

∂xk

)+ ( d − 1 ) h2

d∑j=1

xj∂

∂xj

= r2 p2 + h2d∑

j,k

xj xk∂

∂xj

∂

∂xk+ ( d − 1 ) h2

d∑j=1

xj∂

∂xj(1443)

where the radius is defined by the d-dimensional version of the Pythagoreantheorem

r2 =d∑

j=1

x2j (1444)

A point in a d-dimensional space can be described by hyper-spherical polar co-ordinates involving the hyper-radius r and (d−1) angle variables (θ1, . . . , θd−1).The hyper-spherical polar coordinates are related to the Cartesian coordinatesvia the set of equations

xd = r cos θd−1

xd−1 = r sin θd−1 cos θd−2

xd−2 = r sin θd−1 sin θd−2 cos θd−3

......

...x2 = r sin θd−1 sin θd−2 . . . sin θ2 cos θ1x1 = r sin θd−1 sin θd−2 . . . sin θ2 sin θ1 (1445)

18In three dimensions, the angular momentum is an antisymmetric second rank tensor thatis only masquerading as a vector.

317

The set of equations can be written symbolically as

xj = r fj(θi) (1446)

Hence, we can evaluated the partial derivative w.r.t. r as(∂

∂r

)θi

=∑

j

(∂xj

∂r

)θi

∂

∂xj

=∑

j

(fj(θi)

)∂

∂xj

=∑

j

(xj

r

)∂

∂xj(1447)

and the second derivative is evaluated as(∂2

∂r2

)θi

=∑j,k

(∂xj

∂r

)θi

∂

∂xj

(∂xk

∂r

)θi

∂

∂xk

=∑j,k

(fj(θi)

) (fk(θi)

)∂

∂xj

∂

∂xk

=∑j,k

xj xk

r2∂

∂xj

∂

∂xk(1448)

Hence, we find that the magnitude of the angular momentum is related to themagnitude the momentum and some radial derivatives

L2 = r2 p2 + h2 r2∂2

∂r2+ ( d − 1 ) h2 r

∂

∂r(1449)

or

p2 = − h2 1rd−1

∂

∂r

(rd−1 ∂

∂r

)+

L2

r2(1450)

This equation can be used to express the kinetic energy in terms of the angularmomentum in situations where there is spherical symmetry. The above identityreduces to the usual expression found when d = 3.

The eigenvalues of the angular momentum are the hyper-spherical harmon-ics. Harmonic functions hλ

µ are defined as homogeneous polynomials of order λ(homogeneous polynomials in the Cartesian coordinates) that satisfy Laplace’sequation

∇2 hλµ = 0 (1451)

The hyper-spherical harmonics Y λµ with angular momentum λ can be expressed

in terms of the complete set of harmonic functions of order λ via

rλ Y λµ =

∑ν

Cµ,ν hλν (1452)

318

where the Cµ,ν are expansion coefficients19. The above equation implies thatthe hyper-spherical harmonics only depend on the direction of the vector x.Operating with p2 on rλ Y λ

µ yields

p2 rλ Y λµ = 0 (1453)

since rλ Y λµ is a combination of harmonic functions. Furthermore, the above

equation together with eqn(1450) implies that[− h2 1

rd−1

∂

∂r

(rd−1 ∂

∂r

)+

L2

r2

]rλ Y λ

µ = 0 (1454)

Since the Y λµ only depend on the set of angles θi and do not depend on r, we

find that the above equation reduces to

L2 Y λµ = h2 λ ( λ + d − 2 ) Y λ

µ (1455)

Thus, the hyper-spherical harmonics Y λµ are eigenfunctions L2 and the eigen-

values are h2 λ ( λ + d −2 ). The eigenvalue equation reduces to the standardeigenvalue equation in three dimensions.

——————————————————————————————————

4.5.25 Exercise 85

Show that the radial momentum pr, which in classical mechanics is defined by

pr =1r

( r . p ) (1456)

must be represented by the symmetrized form

pr =12

[1r

( r . p ) + ( p . r )1r

](1457)

in order to yield the Hermitean operator

pr = − i h

(∂

∂r+

1r

)(1458)

Show that in d dimensions this leads to the expression

pr = − i h

(∂

∂r+d − 1

2 r

)(1459)

which is a Hermitean operator.

——————————————————————————————————

19The expansion coefficients Cµ,ν are dependent on the choice of the set of basis functionshλ

µ, the choice of the set of quantum numbers µ and an overall phase factor.

319

4.5.26 Solution 85

The quantum operator corresponding to the un-symmetrized form of the clas-sical radial momentum is found from

1r

( r . p ) =1r

( r er . p )

= − i h1r

(r er . ∇

)= − i h

1r

(r∂

∂r

)= − i h

∂

∂r(1460)

where we have used the expression for ∇ in d-dimensional spherical polar coor-dinates

∇ = er∂

∂r+ eθ1

1r

∂

∂θ1+ eθ2

1r sin θ1

∂

∂θ2+ . . . + eθd

1r sin θ1 . . . sin θd−2

∂

∂θd−1

= er∂

∂r+

d−1∑i=1

1r hi

eθi

∂

∂θi(1461)

In this expression, we have introduced the unit vectors eθiin the orthogonal

directions defined by

eθi=

1hi

∂

∂θier (1462)

and the product hi of the dimensionless angular factors

hi =i−1∏j=1

(sin θj

)(1463)

This operator is not Hermitean, as can be seen by constructing the Her-mitean conjugate. The Hermitean conjugate is evaluated by considering theeffect of the matrix elements with two different arbitrary but normalizable ra-dial wave functions Rn(r) and Rm(r),∫ ∞

0

dr rd−1 R∗m(r)(− i h

∂

∂r

)Rn(r) (1464)

The factor rd−1 represents the r dependence of the surface area of a d-dimensionalsphere, and dr rd−1 is proportional to the volume of an infinitesimal shell ofthickness dr. By integration by parts, and using the fact that the wave functionsare normalizable to eliminate the boundary terms, one obtains the result

=∫ ∞

0

dr Rn(r)(

+ i h∂

∂r

)rd−1 R∗m(r)

320

=∫ ∞

0

dr rd−1 Rn(r)(

+ i h∂

∂r+ i h

d − 1r

)R∗m(r)

(1465)

On using the definition of the Hermitean conjugate, one identifies the Hermiteanconjugate operator as(

1r

( r er . p ))†

= − i h

(∂

∂r+

d − 1r

)(1466)

The symmetric partner

( p . r )1r

(1467)

is quantized as(p . r

)1r

= − i h

(∇ . r er

)1r

= − i h

(∇ . er

)= − i h

∂

∂r− i h

d−1∑i=1

1r hi

eθi .

(∂

∂θier

)(1468)

The factor of r cancels with 1r in the second line, as the differential operator

acts on both factors. In the last line, we have utilized the identity∂

∂rer = 0 (1469)

The derivative of the unit vector er with respect to the angle θi is evaluated as∂er

∂θi= hi eθi

(1470)

As the unit vectors are orthogonal, one has

( p . r )1r

= − i h∂

∂r− i h

d−1∑i=1

1r hi

eθi.

(∂

∂θier

)

= − i h∂

∂r− i h

d−1∑i=1

1r hi

eθi .d−1∑j=1

hj eθj

= − i h∂

∂r− i h

d−1∑i=1

1r hi

d−1∑j=1

hj δi−j

= − i h∂

∂r− i h

d−1∑i=1

1r

= − i h∂

∂r− i h

d − 1r

(1471)

321

Hence, the quantized symmetric partner is equal to the Hermitean conjugate.

The Hermitean operator corresponding to the radial component of the mo-mentum is given by the Hermitean part of the operator

pr =12

[1r

( r . p ) +(

1r

( r . p ))† ]

=12

[1r

( r . p ) + ( p . r )1r

]= − i h

[∂

∂r+

d − 12 r

](1472)

——————————————————————————————————

322

4.6 Spherically Symmetric Potentials

The Hamiltonian for a particle of mass m and charge q moving in a sphericallysymmetric potential is given by

H =p2

2 m+ q φ(r) (1473)

where the potential φ(r) only depends on the radial coordinate r and is inde-pendent of (θ, ϕ). Since the Hamiltonian commutes with the orbital angularmomentum operator,

[ H , L ] = 0 (1474)

and, therefore, also commutes with the magnitude

[ H , L2 ] = 0 (1475)

one may find simultaneous eigenfunctions of H, Lz and L2. These eigenfunctionsmust be of the separable form

ΨE,l,m(r) = RE,l(r) Y lm(θ, ϕ) (1476)

Thus, the energy eigenvalue equation becomes[− h2

2 m r2∂

∂r

(r2

∂

∂r

)+

L2

2 m r2+ q φ(r)

]ΨE,l,m(r) = E ΨE,l,m(r)

(1477)

After substituting eqn(1476) in the above equation, and after using the fact thatY l

m(θ, ϕ) is an eigenstate of L2, with eigenvalue h2 l ( l + 1 ), one finds thatthe spherical harmonic is a common factor that can be cancelled. The radialpart of the wave function must satisfy the radial equation[− h2

2 m r2∂

∂r

(r2

∂

∂r

)+

h2 l ( l + 1 )2 m r2

+ q φ(r)]RE,l(r) = E RE,l(r)

(1478)This is an ordinary differential equation, instead of a partial differential equa-tion, in which the centrifugal potential appears as an addition to the electrostaticpotential. The radial equation is independent of the azimuthal quantum num-ber m, and hence, so are the radial eigenfunctions and eigenvalues.

——————————————————————————————————

4.6.1 Exercise 86

A particle of mass m moves in a three-dimensional spherically symmetric po-tential V (r) which vanishes at r → ∞. It has an eigenstate which is givenby

Φ(r) = C rs exp[− α r

]sinl θ exp

[i l ϕ

](1479)

323

Effective Radial Potential

-0.4

-0.2

0

0.2

0.4

0 5 10 15 20

r

Vef

f(r)

l=0

l=1

l=2

Veff(r) = V(r) + 2 l ( l + 1 ) / 2 m r2

Centrifugal Barrier

Figure 80: The effective radial potential Veff (r) is the sum of the potential V (r)and the centrifugal potential.

What are the values of the angular momentum and energy quantum numbersfor this state? Find the potential V (r).

——————————————————————————————————

4.6.2 Solution 86

The state is an eigenstate of angular momentum with l = l and m = l. Theenergy eigenvalue and the potential can be determined from[

V (r) − E

]Φ(r) =

h2

2 m∇2 Φ(r) (1480)

where

∇2 Φ(r) =[

1r

∂2

∂r2r − l ( l + 1 )

r2

]C rs exp

[− α r

]sinl θ exp

[i l ϕ

]=

[α2 − 2 α

( 1 + s )r

+s ( s + 1 ) − l ( l + 1 )

r2

]Φ(r)

(1481)

324

Thus, at r → ∞ where V (r) → 0, we find

E = − h2 α2

2 m(1482)

and

V (r) =h2

2 m

[( s − l ) ( s + l + 1 )

r2− 2 α ( 1 + s )

r

](1483)

——————————————————————————————————

4.6.3 The Free Particle

The free particle corresponds to the case of a vanishing electrostatic potentialφ(r) = 0. The time-independent Schrodinger equation for the free particleseparates in spherical polar coordinates. The radial wave function is given bythe solution of[

− h2

2 m r2∂

∂r

(r2

∂

∂r

)+

h2 l ( l + 1 )2 m r2

]RE,l(r) = E RE,l(r)

(1484)

which, with h k =√

2 m E, becomes[− 1

r2∂

∂r

(r2

∂

∂r

)+

l ( l + 1 )r2

]RE,l(r) = k2 RE,l(r) (1485)

It is also useful to introduce a dimensionless variable ρ = k r to obtain thescaled equation[

− 1ρ2

∂

∂ρ

(ρ2 ∂

∂ρ

)+

l ( l + 1 )ρ2

]RE,l(ρ/k) = RE,l(ρ/k) (1486)

This differential equation reduces to Bessel’s equation. This can be seen bymaking the transformation

RE,l(r) =yl(ρ)√ρ

(1487)

which leads to

d2 yl

d ρ2+

1ρ

d yl

dρ+(

1 −( l + 1

2 )2

ρ2

)yl = 0 (1488)

This is Bessel’s equation for half-integer indices, and the solutions we seek areproportional to the Bessel functions Jl+ 1

2(ρ). The Bessel functions are finite at

325

the origin ρ = 0. The solutions for RE,l are, therefore, known as the sphericalBessel functions

RE,l(r) = jl(ρ) =√π

2

Jl+ 12(ρ)

√ρ

(1489)

Equation (1486) actually has two independent solutions jl(z) and ηl(z) whichare the spherical Bessel functions and the spherical Neumann functions. Forvalues of ρ close to the origin, the solutions have leading terms in the powerseries expansion which vary as

Rl(r) ∝ ρα + O(ρα+2) (1490)

where α has to be determined. On substituting the leading term into the dif-ferential equation, and noting that the term of order unity can be neglectedcompared with the centrifugal barrier near r = 0, one finds that the differen-tial equation reduces to(

− α ( α + 1 ) + l ( l + 1 ))ρα−2 + O(ρα) = 0 (1491)

for small ρ. The values of α, for which this is satisfied, satisfy a quadraticequation and are given by either of the solutions α = l or α = − ( l + 1 ).The spherical Bessel functions jl(ρ) vary as ρl close to the origin, which is aconsequence of the high centrifugal barrier that prevents classical particles withfixed l approaching the origin. The spherical Neumann functions ηl(ρ) vary asρ−(l+1) as ρ → 0. The spherical Neumann functions are to be discarded asr2 | RE,l(r) |2 dr is proportional to the probability of finding a particle in theinterval dr at a radial distance r. The inclusion of the spherical Neumann func-tions would lead to the probability diverging like r−2l as r → 0. Therefore,the spherical Neumann functions represent unphysical solutions near the origin.

Since the spherical Bessel functions and spherical Neumann functions playan important role in the description of scattering by short ranged potentials, weshall digress to examine the general properties of these functions in more detail.

The solution of equation(1486) with α = l, is the spherical Bessel functionjl(z), and can be written as

jl(z) = D zl

∫ +1

−1

ds exp[ i s z ] ( 1 − s2 )l (1492)

where the usual normalization is given as

Dl =1

2l+1 l!(1493)

It can be shown that jl(z) satisfies the differential equation for Rl,E by directsubstitution. We first need to evaluate the various derivatives in the equation.

326

The first derivative of jl(z) is given by

∂

∂zjl(z) = D l zl−1

∫ +1

−1

ds exp[ i s z ] ( 1 − s2 )l

+ D i zl

∫ +1

−1

ds exp[ i s z ] s ( 1 − s2 )l (1494)

and the second derivative is given by

∂2

∂z2jl(z) = D l ( l − 1 ) zl−2

∫ +1

−1

ds exp[ i s z ] ( 1 − s2 )l

+ D 2 i l zl−1

∫ +1

−1

ds exp[ i s z ] s ( 1 − s2 )l

− D zl

∫ +1

−1

ds exp[ i s z ] s2 ( 1 − s2 )l (1495)

On substituting in the differential equation number (1486) and collecting liketerms, after some cancellations, we find

∂2

∂z2jl(z) +

2z

∂

∂zjl(z) +

(1 − l ( l + 1 )

z2

)jl(z)

= D 2 i ( l + 1 ) zl−1

∫ +1

−1

ds exp[ i s z ] s ( 1 − s2 )l

+ D zl

∫ +1

−1

ds exp[ i s z ] ( 1 − s2 )l+1

= − D i zl−1

∫ +1

−1

ds exp[ i s z ]d

ds( 1 − s2 )l+1

+ D zl

∫ +1

−1

ds exp[ i s z ] ( 1 − s2 )l+1 (1496)

In the last expression, we have rewritten the integrand in the first term as a dif-ferential with respect to s. On integrating by parts in the first term, and notingthe boundary term is zero, we find that the right hand side of the differentialequation is given by

= − D zl

∫ +1

−1

ds exp[ i s z ] ( 1 − s2 )l+1

+ D zl

∫ +1

−1

ds exp[ i s z ] ( 1 − s2 )l+1

= 0 (1497)

Thus, jl(z) satisfies the radial differential equation.

327

From the above integral representation expression for, jl and its derivative,we can easily verify that the recursion relation

jl+1(z) =l

zjl(z) −

∂

∂zjl(z) (1498)

holds true. A second useful recursion relation that can be derived is

( 2 l + 1 )∂

∂zjl(z) = l jl−1(z) − ( l + 1 ) jl+1(z) (1499)

A second solution ηl(z) is given by the integral

ηl(z) = C1

zl+1

∫ +1

−1

ds exp[ i s z ]1

( 1 − s2 )l+1(1500)

since the differential equation is invariant under the transformation l → − ( l +1 ).

From the integral expressions for jl(z) and ηl(z), one can show that thespherical Bessel functions are regular at the origin

limz → 0

jl(z) ∼2l l!

( 2 l + 1 )!zl (1501)

while the spherical Neumann functions are divergent at the origin,

limz → 0

ηl(z) ∼ − ( 2 l )!2l l!

z−(l+1) (1502)

The asymptotic large z variation of the spherical Bessel function is foundfrom the integral representation

jl(z) =zl

2l+1 l!

∫ +1

−1

ds exp[ i s z ] ( 1 − s2 )l (1503)

by integrating by parts l times. Each integration by parts reduces the power ofthe mononomial in z in front of the integral by unity, yielding

jl(z) =il

2l+1 l!

∫ +1

−1

ds exp[ i s z ](d

ds

)l

( 1 − s2 )l (1504)

since the boundary terms all vanish. The vanishing of the boundary terms occursas there are more factors of ( 1 − s2 ) than there are derivatives with respect tos. However, on further integration by parts, the higher order derivatives remainnon-zero at the boundaries s = ± 1. Repeated integration by parts yields an

328

expansion where the non-zero terms are ordered in increasing powers of z−1.The first non-zero term in the asymptotic expansion is given by

jl(z) ∼ − il+1

2l+1 l! zexp[ i s z ]

(d

ds

)l

( 1 − s2 )l

∣∣∣∣s=+1

s=−1

+ O(z−2) (1505)

Thus, the leading term in the asymptotic large z expansion is given by

limz → ∞

jl(z) ∼cos[ z − ( l + 1 ) π

2 ]z

(1506)

The asymptotic expansion for the spherical Neumann function is given by ananalogous expression which is most easily found from eqn(1506) via the substi-tution l → − ( l + 1 ), and is given by

limz → ∞

ηl(z) ∼sin[ z − ( l + 1 ) π

2 ]z

(1507)

The explicit forms of the first few spherical Bessel and Neumann functions aregiven in Table(5).

The eigenfunctions for the free particle have to be square integrable overany finite volume of space, as the integral of the modulus squared of the wavefunction represents the probability that a particle is found in the volume. Sinceonly the spherical Bessel functions are integrable over the region near the ori-gin, they can represent physical wave functions near the origin. The sphericalNeumann functions have to be discarded near the origin.

The orthogonality condition for the spherical Bessel functions is given by∫ ∞

0

dr r2 jl(kr) jl(k′r) =π

2 k2δ( k − k′ ) (1508)

The normalization condition corresponds to the continuous eigenvalue spectrumassociated with the delocalized wave functions. As the wave functions are pro-portional to the spherical harmonics, they are orthogonal for different values ofl. Therefore, we only require that the Bessel functions with the same values ofl but different energy eigenvalues are orthogonal.

The solution of the free particle energy eigenvalue equation, with angularmomentum l, also has uses in cases where there is a spherically symmetricpotential φ(r) which is short ranged. In this case, the energy eigenvalue equationcan be separated into radial and angular parts

Ψ(r, θ, ϕ) = Rl,k(r) Y lm(θ, ϕ) (1509)

since the potential is spherically symmetric. The wave functions of the spatiallyextended states with energy eigenvalues Ek > 0 have asymptotic large r

329

Table 5: The Spherical Bessel jl(z) and Neumann ηl(z) Functions.

l jl(z)

l = 0 j0(z) sin zz

l = 1 j1(z) sin z − z cos zz2

l = 2 j2(z)( 3 − z2 ) sin z − 3 z cos z

z3

l = 3 j3(z)( 15 − 6 z2 ) sin z − ( 15 z − z3 ) cos z

z4

l ηl(z)

l = 0 η0(z) − cos zz

l = 1 η1(z) − cos z + z sin zz2

l = 2 η2(z) − ( 3 − z2 ) cos z + 3 z sin zz3

l = 3 η3(z) − ( 15 − 6 z2 ) cos z + ( 15 z − z3 ) sin zz4

330

Spherical Bessel Functions

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

0 2 4 6 8 10 12

j1(x)

j0(x)

j2(x)

Figure 81: The spherical Bessel functions jl(z) as a function of z for various l.

limits which are expressible as linear superpositions of the spherical Bessel andNeumann functions, since φ(r) ≈ 0 in this region. Thus,

Rl,k(r) = Al jl(kr) + Bl ηl(kr) (1510)

The ratio of the amplitude of the two terms contains information about the shortranged potential, and is usually expressed in terms of the phase shift δl(k),

Bl

Al= − tan δl(k) (1511)

On using this definition of the phase shift, and the asymptotic large r formsof the spherical Bessel and Neumann functions, one finds that the asymptoticform of the radial wave function is given by

Rl,k(r) ∼ Al

cos δl(k)

sin(k r − l π

2 + δl(k))

k r(1512)

Hence, the large r form of the radial wave function resembles that of a free par-ticle. The main difference is that the phase of the radial oscillations is shiftedby δl(k).

The normalization of the extended states, for an arbitrary short rangedpotential, can be inferred from the asymptotic large r behavior. This can be

331

Spherical Neumann Functions

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0 2 4 6 8 10 12

x

η1(x)η2(x)

η0(x)

Figure 82: The spherical Neumann functions ηl(z) as a function of z for variousl. The spherical Neumann functions diverge at the origin.

shown by re-writing radial part of the energy eigenvalue equation in the alternateform

1r

∂2

∂r2

(r Rl,k(r)

)+

2 m E

h2 Rl,k(r) =[l ( l + 1 )

r2+

2 m q φ(r)h2

]Rl,k(r)

(1513)A generalized continuity equation can be derived from the time-independentSchrodinger equation. The continuity equation has the form

( k2 − k′2 )∫ R

0

dr r2 R∗l,k′(r) Rl,k(r) =

= −[r R∗l,k′(r)

∂

∂rr Rl,k(r) − r Rl,k

∂

∂rr R∗l,k′

]R

0

(1514)

We note that as r → 0, Rl,k(r) vanishes for non-zero l. Hence, the normaliza-tion is governed solely by the asymptotic behavior. As previously mentioned,the radial wave function has the asymptotic large r form

Rl,k(r) ∼ 1r

sin(k r − l π

2+ δl(k)

)(1515)

where δl(k) is the phase shift and k2 = 2 m Eh2 . Then, on substituting the

332

asymptotic form in the expression for the normalization, one finds∫ R

0

dr r2 R∗l,k′(r) Rl,k(r) =

=1

k′2 − k2

[k sin

(k′ R − l π

2+ δl(k′)

)cos(k R − l π

2+ δl(k)

)− k′ sin

(k R − l π

2+ δl(k)

)cos(k′ R − l π

2+ δl(k′)

) ](1516)

The above expression can be re-written in the form

=π

2

[ sin(

(k′ − k) R+ δl(k′)− δl(k))

( k′ − k )

+sin(

(k′ + k) R+ δl(k′) + δl(k)− lπ)

( k′ + k )

](1517)

The region of physical interest is that in which both k and k′ are both positive.In the limit r → ∞, the first term dominates the above expression and so thenormalization integral can be expressed as

=π

2δ( k′ − k ) (1518)

where we have used the representation of the delta function

limN→∞

sin N x

x→ δ(x) (1519)

This shows that the orthonormality condition of the continuum states mustinvolve the Dirac delta function∫ ∞

0

dr r2 R∗l,k′(r) Rl,k(r) =π

2δ( k′ − k ) (1520)

The othonormality condition of the spherical Bessel functions provides one con-crete example of this relation. For the free particle, the radial wave function isidentified with the spherical Bessel function by

Rl(r) = k jl(kr) (1521)

since we require that, when r →∞, the radial wave function has the same nor-malization as the asymptotic form given in eqn(1515). Therefore, one recoversthe identity ∫ ∞

0

dr r2 jl(k′r) jl(kr) =π

2 k2δ( k′ − k ) (1522)

where k and k′ are both restricted to have positive values.

333

The Dirac delta Function

-20

-10

0

10

20

30

40

50

60

70

-0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5x

sin

N x

/ π πππ

x

200

100

50

25

N

Figure 83: The Dirac delta function δ(x) can be regarded as the limit N →∞of the sequence of functions sin N x

π x .

4.6.4 The Spherical Square Well

The spherically symmetric potential represents a short ranged potential, inwhich q φ(r) = − V0 for r < a and q φ(r) = 0 for r > a. Thestates of angular momentum l have radial wave functions which satisfy

− h2

2 m r2d

dr

(r2

d R

dr

)+

h2 l ( l + 1 )2 m r2

R = ( E + V0 ) R (1523)

for r < a and

− h2

2 m r2d

dr

(r2

d R

dr

)+

h2 l ( l + 1 )2 m r2

R = E R (1524)

for r > a. The condition for regularity at the origin leads us to select thesolution

R(r) = Cl jl(k′r) (1525)

for r < a, where k′ is given by

h2 k′2 = 2 m ( E + V0 ) (1526)

If E > 0, the solutions outside the well are oscillatory and correspond to linearcombinations of the spherical Bessel and spherical Neumann functions. These

334

solutions correspond to scattering states. On the other hand, the bound stateshave energies E < 0 which correspond to imaginary values of k. The boundstate radial wave functions ought to vanish as r → ∞.

Scattering States

Inside the well, the radial wave function with angular momentum l must beproportional to the spherical Bessel function with wave vector k′

Rl,E(r) = Cl jl(k′r) (1527)

for r < a. As previously mentioned, the spherical Neumann function divergesat the origin and, therefore, is not an physically acceptable solution inside thewell. Outside the well, r > a, the scattering states can be written as the linearcombination

Rl,E(r) = Al jl(kr) + Bl ηl(kr) (1528)

The ratio of the coefficients Al and Bl can be obtained from the continuityconditions at r = a. These conditions are that the radial wave function iscontinuous at r = a,

limε → 0

Rl,E(a+ ε) = limε → 0

Rl,E(a− ε) (1529)

and that the derivative is also continuous at r = a

limε → 0

∂

∂rRl,E(r)

∣∣∣∣a+ε

= limε → 0

∂

∂rRl,E(r)

∣∣∣∣a−ε

(1530)

These equations can be divided, yielding

k′ j′l(k′a)

k jl(k′a)=

Al j′l(ka) + Bl η

′l(ka)

Al jl(ka) + Bl ηl(ka)(1531)

Thus, the ratio of Bl to Al is determined as

Bl

Al=

k′ j′l(k′a) jl(ka) − k j′l(ka) jl(k

′a)k η′l(ka) jl(k′a) − k′ j′l(k′a) ηl(ka)

(1532)

This ratio is usually written in terms of the tangent of the phase shift δl(k). Thephase shift (modulo 2 π) can be inferred from experiments in which a beam ofparticles of energy E falls incident on the potential. For low energies, the scat-tering is obviously dominated by the l = 0 phase shift, since classically theangular momentum is given by l = k r. Hence, only the l = 0 contributionis expected to be important at low energies where k → 0.

For the case l = 0, the matching procedure leads to

B0

A0=

k′ cos k′a sin ka − k cos ka sin k′ak′ cos k′a cos ka + k sin ka sin k′a

(1533)

335

The ratio of amplitudes can be expressed in terms of the l = 0 phase shift δ0(k),via the definition

− tan δ0(k) =B0

A0(1534)

Hence, the phase shift is given by

tan δ0(k) =(

( kk′ ) tan k′ a − tan k a

1 + ( kk′ ) tan k′ a tan k a

)(1535)

or

δ0(k) = tan−1

(k

k′tan k′ a

)− k a (1536)

where h2 k′2 = h2 k2 + 2 m V0. The phase shift vanishes identically fork′ = k and, therefore, is a non-linear measure of the strength of the potential.The phase shift δ0(k) (in units of π) is shown in fig(84) as a function of k′a

π forfixed small values of the energy. The phase shifts are only slightly dependent onk at low energies. It is seen that the phase shift rapidly increases by π at various

Phase Shift for the attractive spherical potential well

0

1

2

3

4

5

6

0 1 2 3 4 5 6

k'a/ππππ

δδδδ 0(k

)/π πππ

Figure 84: The l = 0 phase shift δ0(k) for low-energy particles incident on anattractive potential well, as a function of k′a.

values of k′a. These values are tabulated in Table(6). At these particular valuesof k′a, the potential is almost strong enough to produce another bound statewith l = 0. This can be demonstrated by noting that these special values of k′acorrespond to k′a = (n+ 1

2 ) π, so the radial wave function inside the potential

336

Table 6: The values of (k′aπ ) near which the l = 0 phase shift changes by π and

resonant scattering occurs.

l = 0 bound state(

k′aπ

)

n = 1 0.48n = 2 1.50n = 3 2.51n = 4 3.50n = 5 4.51n = 6 5.51

well has a node at the origin and has zero slope at the edge of the well. Thewave function with zero slope inside the well only matches to a wave functionwith zero slope outside the well. Outside the well, the eigenfunction is constant(since k = 0) and corresponds to having a zero value of the decay constantκ and, therefore, represents a zero energy bound state. As seen in fig(85), thescattering cross-sections are large for low-energy particles incident on potentialswhere V0 is close to these critical values. This type of scattering is known as

Scattering cross-section

0 1 2 3 4 5 6

k'a/π

d σ/dΩ

Figure 85: The l = 0 scattering cross-section for low-energy particles incidenton an attractive potential well, as a function of k′a.

337

resonant scattering.

In the limit k → 0, the matching condition reduces to

B0

A0= k a

(1 − tan k′a

k′a

)(1537)

Hence, the ratio vanishes proportionally to k, as k → 0. This means that theasymptotic, large r, radial wave function has the form of

limk →0

R0,k(r) = A0

(sin krkr

− B0

A0

cos krkr

)= A0

(1 − asc

r

)(1538)

where

asc = a

(1 − tan k′a

k′a

)(1539)

is the scattering length20. In the limit of low energies, the differential scatteringcross-section is given by

dσ

dΩ= | asc |2 (1540)

which is isotropic, that is, independent of the scattering angle θ.

——————————————————————————————————Example: l = 0 Scattering from a Repulsive Spherical Well Potential.

In this case, we consider the energy eigenstates with l = 0 of the sphericallysymmetric potential

V (r) = V0 if r < a

V (r) = 0 if r > a (1541)

We shall consider the case where E < V0. By analytic continuation from thecase where V0 > E, the radial wave function on the inside of the well has theform

R0(r) = C0sinhκ rκ r

(1542)

which is proportional to the l = 0 spherical Bessel function for imaginary ar-gument. This function satisfies the radial equation, if the energy eigenvalue isrelated to the potential V0 and κ via

E = V0 −h2 κ2

2 m(1543)

20Note that if the magnitude of the potential is such that k′a is close to an odd multipleof π

2, the scattering length can be anomalously large. When this occurs, low energy particles

are resonantly scattered.

338

Outside the well, the radial wave function is given by a linear combination ofthe l = 0 spherical Bessel and Neumann functions

R0(r) = A0sin k r

k r− B0

cos k rk r

(1544)

where k is related to the energy eigenvalue by

E =h2 k2

2 m(1545)

The matching conditions for the radial wave function at r = a lead to theequation

tan( k a + δ0(k) )k a

=tanhκ aκ a

(1546)

The matching condition leads to the phase shift being given by

δ0(k) = tan−1

(k

κtanh κ a

)− k a (1547)

In the limit of a hard sphere, V0 → ∞, so κ a → ∞. In the hard sphere limit,the above expression for the phase shift simplifies to

limV0→∞

δ0(k) → − k a (1548)

The k → 0 limit of the scattering amplitude is found as

asc = − limk→0

tan δ0(k)k

= a (1549)

Therefore, at low energies where the scattering is isotropic and independent ofk, the total scattering cross-section of the hard sphere is found as

σT =∫

dΩ(dσ

dΩ

)=

∫dΩ a2

sc

= 4 π a2 (1550)

This derivation is only valid for small k. The total cross-section should be com-pared with geometrical cross-section, which is defined as the area of (the fronthalf of) the target’s surface projected onto the plane of the beam. As seen infig(86), the geometrical cross-section for the hard sphere is given by π a2, thearea of a disk of radius a. Therefore, as the total scattering cross-section of thehard sphere potential is 4 π a2, the total cross-section is four times greater thanthe geometrical cross-section.

339

Geometric cross-section

Target

Incident BeamWave Front

Figure 86: The geometrical scattering cross-section for particles incident on ahard sphere.

——————————————————————————————————Bound States

In order to determine the bound states it is necessary to examine the be-havior of the radial wave function for imaginary wave vectors as r → ∞. Itis useful to define certain combinations of the spherical Bessel and Neumannfunctions that prove to be useful. From the asymptotic variation of the Besseland Neumann functions, one recognizes that the combinations

h(1)l (ρ) = jl(ρ) + i ηl(ρ)

h(2)l (ρ) = jl(ρ) − i ηl(ρ) (1551)

known as the spherical Hankel functions of the first and second kind, have simpleexponential variations

h(1)l (ρ) =

1ρ

exp[

+ i

(ρ − ( l + 1 )

π

2

) ]h

(2)l (ρ) =

1ρ

exp[− i

(ρ − ( l + 1 )

π

2

) ](1552)

For imaginary k, such that ρ = + i κ r, only the spherical Hankel functionsof the first kind decay exponentially at r → ∞. Thus, when discussing bound

340

Table 7: The spherical Hankel Functions h(m)l (z)

h(1)l (z) h

(2)l (z)

l = 0 − iz exp[ + i z ] + i

z exp[ − i z ]

l = 1 −i (1−iz)z2 exp[ + i z ] i (1+iz)

z2 exp[ − i z ]

l = 2 −i (3−3iz−z2)z3 exp[ + i z ] i (3+3iz−z2)

z3 exp[ − i z ]

l = 3 −i (15−15iz−6z2+iz3)z4 exp[ + i z ] i (15+15iz−6z2−iz3)

z4 exp[ − i z ]

states it is useful to express the radial wave function outside the well in termsof the spherical Hankel function of the first kind.

The wave functions in the interior and exterior regions should be matched atr = a. For the bound states, since E < 0, the radial wave functions exteriorto the well have the form h

(1)l (i κ r) while the wave functions interior to the

well have the form jl(k′ r). Therefore, on equating the logarithmic derivatives,one finds the condition

k′ j′l(k′ a)

jl(k′ a)=

i κ h′(1)l (i κ a)

h(1)l (i κ a)

(1553)

which, together with the relation

k′2 − 2 m V0

h2 = − κ2 (1554)

leads to discrete values of κ and, hence, discrete energy eigenvalues Eκ,l.

This procedure for graphically determining the allowed values of κ illustratedfor the l = 0 bound states in fig(87). The dashed curves shows the relationshipbetween κ and k′ of eqn(1554), for different values of V0. The intersection ofthe solid lines and the dashed lines yield the allowed values of κ for the l = 0bound states. The number of intersections and the bound states increase as V0

increases. The wave function for the lowest energy l = 0 bound state is shownin fig(88), for various values of V0.

——————————————————————————————————

341

Graphical Solution for Bound State Energies

-4

-2

0

2

4

6

8

0 0.5 1 1.5 2

k' a / ππππ

- co

t k'

a / k

' aκ κ κ κ a

Figure 87: The graphical solution of the l = 0 version of the self consistencycondition (eqn(1553)), for various values of V0.

4.6.5 Exercise 87

The deuteron consists of a neutron which is bound in an l = 0 state to a protonby a short-ranged attractive potential. The neutron and proton in the deuteronare in a spin triplet state. Assume that the potential can be represented by aspherically symmetric “square” well of depth V0 and radius a. The experimen-tally determined binding energy E0 is 2.21 MeV. The scattering length at

sc for aneutron-proton system in a triplet spin state is inferred, from low-energy elasticscattering experiments21, to have a magnitude of 5.38 F. Using this data andthe reduced mass, estimate the depth of the potential V0 and the radius a. Theestimated value of a can be compared with the value of 2.1 F for the rms radiusof the charge distribution deuteron and 0.8 F for the rms radius for the chargedistribution of the proton, obtained by the diffraction of high-energy electrons22.

The neutron-proton interaction is highly spin-dependent. For example, theexperimentally determined singlet neutron-proton scattering length is given byas

sc = − 23.7 F.

——————————————————————————————————21L.J. Rainwater, W.W. Havens, J.R. Dunning and C.S. Wu, Phys. Rev. 73, 733 (1948).22E.E. Chambers and R. Hofstadter, Phys. Rev. 103, 1454 (1956), and J.A. McIntyre and

R. Hofstadter, Phys. Rev. 98, 158 (1955).

342

The radial wave function for the l=0 bound state

0

0.5

1

1.5

0 0.5 1 1.5 2

r / a

rR0(

r)

Figure 88: The lowest energy l = 0 bound state wave function, for variousvalues of V0.

4.6.6 Solution 87

The triplet scattering length atsc = 5.38 F, was measured in experiments

involving the scattering of very low-energy particles. For the model potential,the scattering length is given by

asc = a

(1 − tan k′′a

k′′a

)(1555)

where the wave vector k′′ is associated with a zero-energy particle within thewell. Therefore,

h2 k′′2

2 µ= V0 (1556)

If the potential only has one bound state and the scattering is resonant, thenthe product k′′ a must be slightly greater than π

2 .

The bound state energy E0 is related to the imaginary wave vector κ via

E0 = − h2 κ2

2 µ(1557)

where µ = m2 is the reduced mass of the neutron-proton system. The bound

343

state energy is determined from the condition

k′ cot k′a = − κ (1558)

where

V0 =h2

2 µ

(k′2 + κ2

)(1559)

ork′′2 = k′2 + κ2 (1560)

There are two parameters that need to be determined a and k′′. We shalleliminate a in terms of k′, via

κ a =κ√

k′′2 − κ2tan−1

[−√k′′2 − κ2

κ

](1561)

where the inverse tangent has to be understood as giving an angle greater thanπ2 . The equation involves the dimensionless variable x = k′′

κ . The resultingvalue of a is to be substituted in the equation for the scattering length, yieldingthe equation

κ atsc =

1√x2 − 1

tan−1

[−√x2 − 1

]− 1x

tan( x tan−1

[−√x2 − 1

]√x2 − 1

)(1562)

This equation involves the dimensionless product κ atsc which for triplet scat-

tering has a value κ atsc ∼ 1.23. This leads to the value of x = 4.05. The

depth of the potential is then given by

V0 = E0 x2 (1563)

which is estimated to be V t0 = 36.25 MeV. The width of the well is given by

a = 2.01 F.

Since the experimentally determined singlet neutron-proton scattering lengthis given by as

sc = − 23.7 F, then using the value of a just calculated, one de-termines that the depth of the singlet potential is V s

0 = 23.7 MeV. The singletpotential is not deep enough to bind the neutron and proton. Similar resultswere obtained by using a more realistic form of the potential23.

——————————————————————————————————

23P.M. Morse, J.B. Fisk and L.I. Schiff, Phys. Rev. 50, 748 (1936).

344

4.6.7 Exercise 88

Find the energy eigenvalues of the l = 0 states of the spherically symmetricinfinite square well potential,

∞ if r > a

V (r) =0 if r < a (1564)

——————————————————————————————————

4.6.8 Exercise 89

Show that the spherical square well has no bound states unless

V0 a2 >

h2 π2

8 m(1565)

——————————————————————————————————

4.6.9 Solution 89

The bound state solution outside the well has to have the form of a decayingexponential and, therefore, is given by a spherical Hankel function in whichk → i κ, and is given by a spherical Bessel function inside the well, with wavevector k′, where

h2 k2

2 m− V0 = − h2 κ2

2 m(1566)

On matching the exponentially decaying spherical Hankel functions with imag-inary k with the spherical Bessel functions of argument k′r at r = a oneobtains a transcendental equation for the bound state. The largest value of κoccurs, for fixed V0, when l = 0 as this minimizes the effect of the rotationalkinetic energy or centrifugal potential and, therefore, maximizes the effect ofthe attractive potential. For l = 0, the bound state condition simplifies to

k′ cot k′ a = − κ (1567)

On squaring this equation, one obtains

k′2 cot2 k′ a = + κ2 =2 m V0

h2 − k′2 (1568)

345

which contains some spurious solutions. The left hand side is greater or equalto zero, and only falls to zero for k′ equal to every odd multiple of π

2 a . Theright hand side is an inverse parabola, which becomes negative for k′2 greaterthan 2 m V0

h2 . The solutions of the squared equation are, therefore, restricted tothe region of relatively small k′ where the inverted parabola is positive. Thesolutions of the squared equation correspond to the solutions of the originalequation only in the restricted range where cot k′ a is negative. Thus, thelowest energy bound states occur for π

2 a < k′ < πa . Hence, on increasing the

strength of V0 from zero, the lowest energy bound state first occurs when

2 m V0

h2 ≥ π2

4 a2(1569)

——————————————————————————————————

4.6.10 Exercise 90

Show that if a square well just binds an energy level of angular momentum l,the parameters satisfy the equation

jl−1

( √2 m V0 a2

h2

)= 0 (1570)

which utilized the recurrence formulas for the Bessel functions.

——————————————————————————————————

4.6.11 Solution 90

Inside the well of radius a, a > r, the solution for the radial wave function,with angular momentum l, is proportional to the spherical Bessel function

Rl(r) = Al jl(kr) (1571)

The wave vector k is related to the energy eigenvalue by

E =h2 k2

2 m− V0 (1572)

Since the spherical Neumann function ηl(kr) diverges at r = 0, it yields anunphysical probability density and, therefore, is discarded for r < a.

Outside the well, r > a, the solution is proportional to the spherical Hankelfunction

Rl(r) = Bl h(1)l (iκr) (1573)

346

whereh

(1)l (iκr) = jl(iκr) + i ηl(iκr) (1574)

Since E ≤ 0, the momentum is imaginary and is given by

E = − h2 κ2

2 m(1575)

The spherical Hankel function h(1)l (iκr) represents the combination of the pos-

sible solutions which decays exponentially as r → ∞. The other combinationh

(2)l (iκr) blows up as r → ∞ and so is not normalizable and, therefore, the

unphysical solution is discarded.

The bound state wave functions must satisfy the continuity conditions atr = a, which can be combined to yield

k j′l(ka)jl(ka)

=i κ h

′(1)l (iκa)

h(1)l (iκa)

(1576)

If the potential is so weak that it just binds the particle, then E ∼ 0, andhence κ = 0. The spherical Hankel functions are dominated by the sphericalNeumann functions as they diverge like ( κ a )−(l+1) in the limit κ → 0. Hence,in the limit κ → 0, the matching condition becomes

k j′l(ka)jl(ka)

= − ( l + 1 )a

(1577)

and we have the condition

k a j′l(ka) + ( l + 1 ) jl(ka) = 0 (1578)

The recurrence relations for the spherical Bessel functions can be combined toyield

( l + 1 ) jl(ka) + ka j′l(ka) = ka jl−1(ka) (1579)

Hence, on using the above recurrence relation, we find that the condition thatthe spherical square potential well is just sufficiently strong that it starts havea bound state of angular momentum l is given by

jl−1(ka) = 0 (1580)

in whichk2 =

2 m V0

h2 (1581)

since E = 0. Thus, we have derived the required condition.

——————————————————————————————————

347

4.6.12 Exercise 91

Find the bound states corresponding to angular momentum l = 0 for theattractive delta function spherical shell potential

V (r) = − V0 a δ( | r | − a ) (1582)

Find the minimum value of V0 for which a bound state occurs.

——————————————————————————————————

4.6.13 Solution 91

The energy eigenfunction function for a spherically symmetric Hamiltonian canbe separated in terms of a radial function R(r) and an angular function Y l

m(θ, ϕ),

Φ(r) = Rn,l(r) Y lm(θ, ϕ) (1583)

The solution for the radial function inside the potential shell is given by theanalytic continuation of the spherical Bessel function, of angular momentum l,

Rn,l(r) = Al jl( i κ r ) for r < a (1584)

since the wave function must be normalizable at the origin. However, outsidethe shell it is given by a linear combination of the spherical Bessel and Neumannfunctions that decay as r → ∞ exponentially. This combination defines thespherical Hankel functions,

Rn,l(r) = Bl jl( i κ r ) + Cl ηl( i κ r ) for r > a (1585)

The wave function is continuous at r = a. Thus,

Rn,l(a+ ε) = Rn,l(a− ε) (1586)

The delta function only depends on r and so the first derivative has a discon-tinuous slope at r = a, given by integrating the radial equation∫ a+ε

a−ε

dr r2(

h2

2 m1r2

∂

∂rr2

∂

∂r+ V0 a δ( r − a ) + E

)Rn,l(r) = 0 (1587)

Taking the limit ε → 0, one obtains

R′n,l(a+ ε) − R′n,l(a− ε) +2 m V0 a

h2 R(a) = 0 (1588)

The energy eigenvalue is given by

E = − h2 κ2

2 m(1589)

348

For l = 0, one has

Rn,0(r) = A0sinh κ r

rfor r < a

Rn,0(r) = B0

exp[− κ r

]r

for r > a (1590)

On eliminating B and A, we find the eigenvalue equation for α

exp[κ a

]κ =

2 m V0 a

h2 sinh κ a (1591)

The bound state just forms when the solution κ → 0, where sinh κ a → κ a.In this case we have

V0 =h2

2 m a2(1592)

——————————————————————————————————

4.6.14 Exercise 92

A particle is confined to move in the volume between two concentric spheres ofradius R0 and radius R1, where R1 > R0. The potential is radially symmetricand has the form

V (r) = 0 if R1 > r > R0

V (r) → ∞ elsewhere (1593)

(a) Find the values of the energy eigenvalues if the particle has angular mo-mentum given by l = 0.

(b) Consider the limit R1 → R0, which resembles a particle rigidly boundat radius R1, but is free to rotate. Assume that the particle is in the groundstate. The excitation energy of any energy state is defined as the differencebetween its energy eigenvalue and the energy of the ground state. In this limit,show that the excitation energies for fixed l are independent of the value of l.Also, evaluate all of the remaining finite excitation energies.

——————————————————————————————————

4.6.15 Solution 92

The general solution for the energy eigenstates of angular momentum l is of theform

φl(r) =(Al jl(kr) + Bl ηl(kr)

)Y l

m(θ, ϕ) (1594)

349

for R1 > r > R0, and zero elsewhere. The wave function has to satisfy theboundary conditions (

Al jl(kR1) + Bl ηl(kR1))

= 0 (1595)

at r = R1 and (Al jl(kR0) + Bl ηl(kR0)

)= 0 (1596)

at r = R0. These equations can be used to eliminate the ratio of a to B,yielding a transcendental equation for k

jl(kR0)ηl(kR0)

=jl(kR1)ηl(kR1)

(1597)

For l = 0, this can be re-written as

sin k ( R0 − R1 ) = 0 (1598)

which yields the allowed values of k as

kn =n π

R1 − R0(1599)

and has the energy eigenvalue

En,l=0 =h2 n2 π2

2 m ( R1 − R0 )2(1600)

In the limit R1 → R0, the energy differences between the eigenstates ofdifferent n (with fixed l) are independent of l as the radial equation becomes

− h2

2 m R20

∂

∂r

(r2∂Rl(r)∂r

)Rl(r) =

(E − h2 l ( l + 1 )

2 m R20

)Rl(r) (1601)

Thus, the differences in the eigenvalues, for fixed l, are identical to the differenceswith l = 0, which we have shown tend to infinity. The only remaining finiteexcitation energies correspond to states of the same n values but with differentrotational energies. That is we have a rigid rotor, with excitation energies

El′−l =h2

2 m R20

(l′ ( l′ + 1 ) − l ( l + 1 )

)(1602)

This corresponds to the excitation energies of a three-dimensional rigid rotor,with moment of inertia I = m R2

0. The difference between successive excita-tion energies increase with increasing l as h2

I ( l + 1 ). These differences inexcitation energies are frequently observed in the optical absorption spectra ofmolecules.

——————————————————————————————————

350

4.6.16 Ladder operators for a free particle

The radial wave function Rl(r) for a free particle with angular momentum lsatisfies the eigenvalue equation

0 = −(∂2Rl

∂ρ2

)− 2

ρ

(∂Rl

∂ρ

)+[l ( l + 1 )

ρ2− 1

]Rl

0 = − 1ρ

(∂2ρRl

∂ρ2

)+[l ( l + 1 )

ρ2− 1

]Rl (1603)

where the dimensionless variable ρ is defined by

ρ = k r (1604)

and k is related to the energy eigenvalue E via

E =h2k2

2 m(1605)

We shall introduce ladder operators that transform radial wave functions Rl(r)into radial wave functions with the same value of the energy but with differentvalues of the angular momentum 24. Since the ladder operators that are tobe introduced keep the energy E unchanged, the raising and lowering operatorscan be expressed in terms of the variable ρ and the derivative with respect to ρ.

The raising operator A+l is defined via

A+l = − 1

ρ

∂

∂ρρ +

l

ρ(1606)

The lowering operator A−l is the Hermitean conjugate of the raising operatorA+

l . In three dimensions, the lowering operator A−l is calculated as

A−l = +1ρ

∂

∂ρρ +

l

ρ(1607)

The products of the raising and lowering operators are given by

A+l A−l = − 1

ρ

∂2

∂ρ2ρ +

l ( l + 1 )ρ2

(1608)

or, when taken in the opposite order

A−l A+l = − 1

ρ

∂2

∂ρ2ρ +

l ( l − 1 )ρ2

(1609)

On comparing the above two forms, we find the relation

A−l A+l = A+

l−1 A−l−1 (1610)

24L. Infeld, Phys. Rev. 59, 737 (1941).

351

Radial distribution function

0

0.5

1

1.5

0 1 2 3 4 5 6

ρ / πρ / πρ / πρ / π

j l2 ( ρ) ρ)ρ)ρ) ρ ρρρ2

l-0l=1l=2l=3

Figure 89: The (un-normalized) radial distribution for a free particle with an-gular momentum l. The centrifugal barrier excludes particles with finite valuesof l from a region surrounding the origin.

Thus, we have found two equivalent forms of the operator that appears in theradial equation with angular momentum (l − 1).

When acting on Rl, the lowering operator A−l produces the radial wavefunction Rl−1 which is an energy eigenstate with the same energy. This can beseen by first using eqn(1608) to write the radial equation in the form(

A+l A−l − 1

)Rl = 0 (1611)

and then acting on this with A−l . This yields the equation

A−l

(A+

l A−l − 1)Rl = 0 (1612)

which can be re-written as(A−l A+

l − 1)A−l Rl = 0 (1613)

Hence, on using eqn(1610), one finds(A+

l−1 A−l−1 − 1

)A−l Rl = 0 (1614)

352

which indicates that A−l Rl is proportional to the radial wave function Rl−1

with angular momentum (l − 1) and the same energy E

Rl−1 ∝ A−l Rl (1615)

Since the angular momentum is reduced by unity, the above relation justifiesnaming A−l as the lowering operator.

The raising operator A+l+1 acting on Rl produces a radial wave function Rl+1

where the angular momentum is raised by one unit. This can be shown by firstsubstituting l by (l + 1) in eqn(1610), to produce

A−l+1 A+l+1 = A+

l A−l (1616)

The radial equation for Rl can be written as(A−l+1 A

+l+1 − 1

)Rl = 0 (1617)

If the raising operator A+l+1 acts on the radial equation as expressed in eqn(1617),

one has

0 = A+l+1

(A−l+1 A

+l+1 − 1

)Rl

0 =(A+

l+1 A−l+1 − 1

)A+

l+1 Rl (1618)

The factor in the parentheses is identified as the radial equation operator withangular momentum (l + 1), therefore the product A+

l+1 Rl must be the radialwave function with angular momentum (l + 1). Hence, we have

Rl+1 ∝ A+l+1 Rl (1619)

so the raising operator A+l+1 increases the angular momentum of the radial wave

function Rl by one unit, but keeps the energy unchanged.

The radial wave function with l = 0 can be determined from the radialequation, which we write in the form(

∂2ρR0

∂ρ2

)= − ρR0 (1620)

The general solution is easily found and is given by

ρ R0(r) = A0 sin ρ + B0 cos ρ (1621)

The physically acceptable solution has B0 = 0 since the probability of findingthe particle in the neighborhood of ρ→ 0 must be finite. Hence, the radial wavefunction with l = 0 is given by

R0(r) = A0

(sin ρρ

)(1622)

353

which is proportional to the zero-th order spherical Bessel function j0(ρ) whichis given by

j0(ρ) =(

sin ρρ

)(1623)

The radial wave functions with l 6= 0 can be found by successive use of theraising operators. For example, the radial wave function with l = 1 is foundfrom j0(ρ) by

R1(r) ∝[− 1

ρ

∂

∂ρρ +

1ρ

]R0(r)

= − A11ρ

∂

∂ρ

(sin ρ

)+ A0

1ρ

sin ρρ

= A1

(sin ρ − ρ cos ρ

ρ2

)(1624)

The radial wave function R1(r) is proportional to the spherical Bessel functionj1(ρ), which is given by

j1(ρ) =(

sin ρ − ρ cos ρρ2

)(1625)

It is seen that the spherical Bessel function of order one tends to zero like ρ whenρ → 0. Repeated applications of the raising operators yield the un-normalizedradial functions in the form

Rl(r) = Al jl(ρ) (1626)

where the coefficient Al is an arbitrary complex number. These solutions, likethe spherical Bessel functions, are all well-behaved at the origin. Furthermore,the probability of finding the particle near the origin must decrease as l in-creases, due to the increase of the centrifugal barrier in the effective potential.

The spherical Neumann functions ηl(ρ) of order l are independent solutionsof the radial equation with angular momentum l. However, these solutions aresingular at the origin, as can be seen by examining the spherical Neumannfunction of order zero

η0(ρ) = −(

cos ρρ

)(1627)

which diverges like ρ−1 when ρ→ 0. Other independent solutions of the radialequation can be obtained by successive actions of the raising operators. Forexample, the spherical Neumann function of order one is obtained from η0(ρ)by

η1(ρ) =(− 1

ρ

∂

∂ρρ +

1ρ

)η0(ρ)

354

= −(− 1

ρ

∂

∂ρρ +

1ρ

) (cos ρρ

)= −

(cos ρ + ρ sin ρ

ρ2

)(1628)

The function η1(ρ) diverges like ρ−2 as ρ → 0. The general form of ηl(ρ) asρ→ 0 can be determined by induction. One can show that if

limρ→0

ηl(ρ) ∼ − 1ρl+1

(1629)

then since

ηl+1(ρ) =(− 1

ρ

∂

∂ρρ +

(l + 1)ρ

)ηl(ρ) (1630)

one must have

limρ→0

ηl+1(ρ) ∼ −(− 1

ρ

∂

∂ρ

1ρl

+(l + 1)ρl+2

)= − 1

ρl+2(1631)

Since the asymptotic form holds true for l = 0, one has shown that

limρ→0

ηl(ρ) ∼ − 1ρl+1

(1632)

is true for all values of l. The asymptotic ρ→ 0 behavior of the spherical Besselfunctions is given by

limρ→0

jl(ρ) ∼ ρl (1633)

The above two asymptotic forms of the independent solutions can be simplyobtained from the radial equation, if one assumes that the leading terms of thesolutions are of the form R ∼ ρα for some unknown values of α.

4.6.17 The Rayleigh Equation

A plane wave can be expanded in a series of spherical waves, via the Rayleighequation

exp[i k r cos θ

]=∑

l

il ( 2 l + 1 ) jl( kr ) Pl( cos θ ) (1634)

The expansion can be thought of in terms as the decomposition of a wavefront (a plane of constant phase) of a beam in terms of its angular momentumcomponents, as is indicated schematically in fig(90). The above expansion is tobe expected since the eigenstates of the Laplacian with eigenvalue − k2, whichare simultaneous eigenfunctions of the linear momentum must be related to

355

Classical decomposition of a beam into angular momentum eigenstates

l = k r

l = 1l = 2

l = 3l = 4

l = 0

Cross section of the Beam

Figure 90: A semi-classical description of the decomposition of a plane wave interms of states with angular momentum l. If a point on the wave front at adistance r from the axis with ( l + 1

2 ) h > p r > ( l − 12 ) h has angular

momentum l.

the eigenstates of the Laplacian which are simultaneous eigenfunctions of theangular momentum. The Rayleigh expansion can be verified by differentiatingthe above relation with respect to k r leading to

i cos θ exp[i k r cos θ

]=∑

l

il ( 2 l + 1 ) j′l( kr ) Pl( cos θ )

cos θ∑

l

il+1 ( 2 l + 1 ) jl( kr ) Pl( cos θ ) =∑

l

il ( 2 l + 1 ) j′l( kr ) Pl( cos θ )

(1635)

The recurrence relation

( 2 l + 1 ) z Pl(z) = ( l + 1 ) Pl+1(z) + l Pl−1(z) (1636)

can be used to write∑l

il+1 jl( kr )[

( l + 1 ) Pl+1( cos θ ) + l Pl−1( cos θ )]

=∑

l

il ( 2 l + 1 ) j′l( kr ) Pl( cos θ ) (1637)

356

The recurrence relation for j′l( kr ) given by

( 2 l + 1 ) j′l( kr ) = l jl−1( kr ) − ( l + 1 ) jl+1( kr ) (1638)

can be used in eqn(1637) to obtain∑l

il+1 jl( kr )[

( l + 1 ) Pl+1( cos θ ) + l Pl−1( cos θ )]

=∑

l

il[l jl−1( kr ) − ( l + 1 ) jl+1( kr )

]Pl( cos θ ) (1639)

which, on changing the index from l + 1 to l in the first term and l − 1 to l inthe second term, yields the equality∑

l

il[jl−1( kr ) l Pl( cos θ ) − jl+1( kr ) ( l + 1 ) Pl( cos θ )

]=∑

l

il[l jl−1( kr ) − ( l + 1 ) jl+1( kr )

]Pl( cos θ ) (1640)

This concludes the proof of the Rayleigh expansion.

An alternate proof can be found by using the orthogonality of the Legendrepolynomials and forming the matrix elements with the plane wave expansion

exp[i k r cos θ

]=∑l′

al′ Pl′( cos θ ) (1641)

with unknown expansion coefficients al′ . Then, on multiplying by Pl(cos θ) sin θand integrating w.r.t θ, and using the orthogonality condition, one finds∫ 1

−1

d cos θ Pl( cos θ ) exp[i k r cos θ

]=∑l′

al′2 δl,l′

2 l + 1(1642)

Thus, the expansion coefficients are evaluated as

al =( 2 l + 1 )

2

∫ 1

−1

ds exp[i s k r

]Pl(s)

= ( 2 l + 1 ) il jl( kr ) (1643)

where we have used the Rodriguez formula expression for the Legendre polyno-mial

Pl(s) =1

2l l!∂l

∂sl( s2 − 1 )l (1644)

and integrated by parts l times and then used an integral representation of thespherical Bessel function

jl(z) =zl

2l+1 l!

∫ +1

−1

ds exp[i z s

]( 1 − s2 )l (1645)

357

The above two expressions allowed us to express the integral involving the Leg-endre polynomial in terms of the spherical Bessel function. The Rayleigh ex-pansion is often used in considerations of scattering.

4.6.18 The Isotropic Planar Harmonic Oscillator

The potential for the Isotropic Planar Harmonic Oscillator is given by

V (r) =m ω2

2r2 (1646)

which only depends on the radial distance r. In planar polar coordinates (r, ϕ),the energy eigenvalue equation for the isotropic two-dimensional Harmonic Os-cillator is of the form[− h2

2 m1r

∂

∂rr∂

∂r− h2

2 m r2∂2

∂ϕ2+

m ω2 r2

2− E

]Ψ(r, ϕ) = 0 (1647)

The Laplacian in the eigenvalue equation is expressed in the form appropriatefor two-dimensional motion. In particular, it can be seen that this operator isHermitean if one uses the two-dimensional measure of “volume”, dr r dϕ, inthe definition of the inner product. The potential only depends on the radialdistance r, and so the two-dimensional angular momentum operator Lz com-mutes with the Hamiltonian. Thus, the energy and angular momentum areconserved. Also, it is possible to find simultaneous eigenfunctions of H and Lz.The eigenvalue of Lz is denoted by h mz. On introducing the separable formfor the eigenfunction

Ψ(r, ϕ) = R(r)1√2 π

exp[i mz ϕ

](1648)

the differential equation for the radial wave function becomes[− h2

2 m1r

∂

∂rr∂

∂r+

h2 m2z

2 m r2+

m ω2 r2

2− E

]R(r) = 0 (1649)

On introducing the dimensionless variable, ρ, defined by

ρ =m ω r2

h(1650)

it is found that the derivative with respect to r is given in terms of the differentialwith respect to ρ, via

∂

∂r=

∂ρ

∂r

∂

∂ρ

= 2 ρ12

√m ω

h

∂

∂ρ(1651)

358

Thus, the radial differential equation takes the dimensionless form

h ω

[− 2 ρ

∂2

∂ρ2− 2

∂

∂ρ+

m2z

2 ρ+

ρ

2− E

h ω

]R(r) = 0 (1652)

Furthermore, the radial function is parameterize in the form

R(r) = ρ|mz2 | exp

[− ρ

2

]L(ρ) (1653)

which has the physically acceptable asymptotic variation at the boundaries,ρ → 0 and ρ → ∞. The term proportional to ρ|

mz2 | indicates that we seek

the solution which is normalizable at ρ = 0. The exponential factor indicatesthat we seek the solution which decays exponentially as ρ → ∞. The firstorder derivative of R(r) is evaluated as

∂R

∂ρ= ρ|

mz2 | exp

[− ρ

2

] (∂L

∂ρ+|mz|2 ρ

L − 12L

)(1654)

and the second order derivative is given by a lengthier, but analogous, expres-sion. The differential equation becomes

− 2 ρ∂2L

∂ρ2+ 2 ( ρ − |mz| − 1 )

∂L

∂ρ+(|mz| + 1 − E

h ω

)L = 0 (1655)

which is the differential equation for the associated Laguerre polynomials. Thedifferential equation can be solved by the Frobenius method, in which we expandthe solution in powers of ρ

L(ρ) =∞∑

n=0

an ρn (1656)

On substituting the series into the differential equation, and demanding thatthe coefficient of ρn should vanish, one finds the recurrence relation

( n + 1 ) ( n + |mz| + 1 ) an+1 =(n +

12

( |mz| + 1 − E

h ω))an (1657)

which relates an+1 to an. Thus, all the coefficients in the series are determinedin terms of a0. The value of a0 is only determined from the normalization of thewave function. If the series did not truncate, then one can show that the solutionwould diverge exponentially as ρ → ∞, and would not be normalizable. Thus,in order to satisfy the boundary conditions the series must truncate. The seriestruncates at the nr-th term, so anr+1 = anr+2 = anr+3 = . . . = 0, if theenergy eigenvalue is given by

E = h ω ( 2 nr + |mz| + 1 ) (1658)

359

In Cartesian coordinates, the energy eigenvalue equation separates into twoordinary differential equations, and each differential equation represents a one-dimensional Harmonic Oscillator. The energy eigenvalue corresponds to thesum of the energy eigenvalues for the two one-dimensional problems. Therefore,the energy is given by

E = h ω ( nx +12

+ ny +12

) (1659)

which contains a contribution of h ω from the energy of the zero-point motion.

4.6.19 The Spherical Harmonic Oscillator

The potential for the spherically symmetric Harmonic Oscillator is given by

V (r) =m ω2

2r2 (1660)

As the potential is only a function of r, the angular momentum operators L2

and Lz commute with the Hamiltonian. Thus, angular momentum is conserved,and one can find simultaneous eigenstates of the operators H, L

2and Lz.

The energy eigenvalue equation for the Spherically Symmetric HarmonicOscillator is separable in spherical polar coordinates. If the wave function isassumed to be of the form

Ψ(r, θ, ϕ) = Rl(r) Y lm(θ, ϕ) (1661)

then, after using the properties of the spherical harmonics, it is found that theradial wave function R(r) satisfies an equation independent of the variables, θand ϕ. The radial equation for the spherically symmetric harmonic oscillator isgiven by

− h2

2 m r2d

dr

(r2

d R

dr

)+[h2 l ( l + 1 )

2 m r2+

m ω2 r2

2

]R = E R

(1662)

The solution can be written in the form

Rl(r) = rl exp(− m ω

2 hr2)f(r) (1663)

which minimizes the effect of the centrifugal potential barrier at the origin, andthe exponential term takes care of the binding at asymptotic large values ofr. It should be noted that a second solution exists which, at large r, variesasymptotically as

R(r) ∼ exp(

+m ω

2 hr2)

(1664)

360

This solution has been discarded as it is not normalizable, and hence, unphysical.On changing variable to

ρ =m ω r2

h(1665)

and on substituting the wave function into eqn(1662), one finds

ρ∂2f

∂ρ2+(

32

+ l − ρ

)∂f

∂ρ+

12

(E

h ω− (

32

+ l ))f = 0 (1666)

which leads to f(√

hρmω ) being recognized as being proportional to an associated

Laguerre polynomial in the variable ρ. The associated Laguerre polynomialLc

a(z) is a solution of the differential equation

z∂2

∂z2L(z) + ( c + 1 − z )

∂

∂zL(z) + a L(z) = 0 (1667)

The solutions of this equation only satisfy the appropriate boundary conditionsat z → ∞ for specific values of a. This discrete set of a values give rise tothe discrete eigenvalues of the energy E of the harmonic oscillator. This can beseen by substituting the Frobenius series for L(z)

L(z) =∑

n

an zn (1668)

into the radial equation and combining like powers of zn. The form of L(z)satisfies the differential equation if the coefficients of each power of z vanishidentically. This procedure results in the recursion relations for the coefficientsan,

( n + 1 ) ( n + c + 1 ) an+1 = ( n − a ) an (1669)

If the series did not terminate, then the asymptotic large n behavior wouldresult in the sum of the series L(ρ) having a positive exponential variation(L(ρ) ∼ exp[ + ρ ]) since an ∼ an−1

n . If this hypothetical form of theexponential factor for the series sum was combined with the explicit exponentialfactor which we extracted from the radial wave function via the substitution

R(ρ) = ρl2 exp

[− ρ

2

]L(ρ) (1670)

this would lead to a solution which has the second form of asymptotic variation(R(ρ) ∼ exp[ + ρ

2 ]) as r → ∞. If this were the case, the radial wavefunction would not be square integrable and would not represent a bound state.Therefore, the series must terminate. For the series to terminate, one requiresthat n = a for some n. For an integer value of a, say equal to nr, the Laguerrepolynomials truncate after the first ( nr + 1 ) terms, and all the higher ordercoefficients vanish, anr+1 = anr+2 = . . . = 0. The series expansion alwaysconverges and the wave function satisfies appropriate boundary conditions atr → 0 and at r → ∞. For the isotropic three-dimensional harmonic oscillator,

361

enforcing the requirement that the boundary conditions are satisfied, i.e. theseries truncates, yields the eigenvalues as

E = h ω

(32

+ l + 2 nr

)(1671)

The factor of 32 is the sum of the three independent zero-point motions, and

2 nr + l are the number of excited quanta. Since each state with given l has adegeneracy of 2 l + 1, and states withN quanta have a degeneracy correspondingto the range of values of l from 0 to N , the states with N quanta are highlydegenerate. The degeneracy of a state with energy E = h ω ( 3

2 + N ) isequal to ( N + 2 ) ( N + 1 )

2 .

Radial wave functions for the spherical harmonic oscillator

-1

0

1

2

3

0 1 2 3 4

ρρρρ1/21/21/21/2

Rn,

l

1s2s3s

nr=0,1,2l=0

Figure 91: The radial wavefunction Rn,l(r) for the simple harmonic oscillatorfor l = 0 and nr = 0 , 1 , 2. The radial quantum number nr determines thenumber of nodes in the radial wave function.

——————————————————————————————————

4.6.20 Exercise 93

Show that the substitution of

R(r) = ρl2 exp

[− ρ

2

]L(ρ) (1672)

362

Radial wave functions for the spherical harmonic oscillator

0

0.5

1

1.5

2

0 1 2 3 4

ρρρρ1/21/21/21/2

Rnl

1s

1p

1d

nr=0l=0,1,2

Figure 92: The radial wavefunction Rn,l(r) for the simple harmonic oscillatorfor nr = 0 and l = 0 , 1 , 2. For non-zero values of l, the centrifugal barrierprevents particles from reaching the origin.

whereρ =

m ω

hr2 (1673)

into the radial equation

− h2

2 m1r2

∂

∂rr2

∂R

∂r+

h2 l ( l + 1 )2 m r2

R +m ω2

2r2 R = E R (1674)

leads to the differential equation for the associated Laguerre polynomials.

——————————————————————————————————

4.6.21 Solution 93

On defining the dimensionless variable ρ through

ρ =m ω

hr2 (1675)

one finds that∂

∂r=

∂ρ

∂r

∂

∂ρ= 2

√m ω

hρ

12∂

∂ρ(1676)

363

The radial differential equation becomes

0 = − h ω

2

[4ρρ

12∂

∂ρρ

32∂

∂ρ− l ( l + 1 )

ρ− ρ +

2 Eh ω

]R

= − h ω

2

[4 ρ

∂2

∂ρ2+ 6

∂

∂ρ− l ( l + 1 )

ρ− ρ +

2 Eh ω

]R

(1677)

The derivatives of

R(r) = ρl2 exp

[− ρ

2

]L(ρ) (1678)

are evaluated as

∂R

∂ρ=

l

2ρ

l2−1 exp

[− ρ

2

]L(ρ)− 1

2ρ

l2 exp

[− ρ

2

]L(ρ) + ρ

l2 exp

[− ρ

2

]∂

∂ρL(ρ)

(1679)and

∂2R

∂ρ2=

l

2(l

2− 1 ) ρ

l2−2 exp

[− ρ

2

]L(ρ) − l

4ρ

l2−1 exp

[− ρ

2

]L(ρ)

+l

2ρ

l2−1 exp

[− ρ

2

]∂

∂ρL(ρ)

− l

4ρ

l2−1 exp

[− ρ

2

]L(ρ) +

14ρ

l2 exp

[− ρ

2

]L(ρ)

− 12ρ

l2 exp

[− ρ

2

]∂

∂ρL(ρ)

+l

2ρ

l2−1 exp

[− ρ

2

]∂

∂ρL(ρ) − 1

2ρ

l2 exp

[− ρ

2

]∂

∂ρL(ρ)

+ ρl2 exp

[− ρ

2

]∂2

∂ρ2L(ρ) (1680)

Hence, we have

4 ρ∂2R

∂ρ2= ρ

l2 exp

[− ρ

2

] [4 ρ

∂2

∂ρ2+ 4 ( l − ρ )

∂

∂ρ+l ( l − 2 )

ρ+ ρ− 2 l

]L(ρ)

(1681)and

6∂R

∂ρ= ρ

l2 exp

[− ρ

2

] [6∂

∂ρ− 3 +

3 lρ

]L(ρ) (1682)

which on substituting into the radial equation leads to

− h ω

2ρ

l2 exp

[− ρ

2

] [4 ρ

∂2

∂ρ2+ ( 4 l + 6− 4 ρ )

∂

∂ρ+

2 Eh ω

− ( 2 l + 3 )]L(ρ) = 0

(1683)

364

On cancelling the common factor and dividing by 4, one obtains[ρ∂2

∂ρ2+ ( l +

32− ρ )

∂

∂ρ+

E

2 h ω− 1

4( 2 l + 3 )

]L(ρ) = 0 (1684)

which is the associated Laguerre equation, as was to be shown.

——————————————————————————————————

4.6.22 Exercise 94

Consider the set of energy eigenstates of the isotropic harmonic oscillator withE

h ω = 32 , 5

2 and 72 . Show that the eigenfunctions of the lowest energy eigen-

states are degenerate and that the wave functions and degeneracy found in thespherical polar representation coincide with the representation in terms of threeindependent one-dimensional harmonic oscillators. Evaluate the degeneracy ofthe energy eigenvalue with N quanta present.

——————————————————————————————————

The symmetry group of the spherical harmonic oscillator.

The degeneracy of a spherical harmonic oscillator can be discussed in termsof operators which, when acting on one state, produce other states degeneratewith it. These operators only have finite matrix elements between degeneratestates and must commute with the Hamiltonian. The commutators of these op-erators must also commute with the Hamiltonian and, therefore, the operatorsshould generate a Lie algebra. Since the angular momentum must commutewith the Hamiltonian, the Lie algebra of the three-dimensional harmonic oscil-lator should be greater than, and include, the angular momentum Lie algebra.

The Hamiltonian of the three-dimensional harmonic oscillator can be writtenin terms of the creation and annihilation operators of the three one-dimensionalharmonic oscillators via

H =hω

2

3∑i=1

( a†i ai + ai a†i ) (1685)

Using the expressions for the position and momentum operators in terms of thecreation and annihilation operators

xj =(

h

2 m ω

) 12[a†j + aj

](1686)

and

pk = i

(h m ω

2

) 12[a†k − ak

](1687)

365

one can show that the operators representing the components of the angularmomentum are given by

Li = − i h∑j,k

εi,j,k a†j ak (1688)

where εi,j,k is the anti-symmetric Levi-Civita symbol. The states which corre-spond to the same total number of quanta are degenerate. The Lie algebra ofthe two-dimensional x− y sub-space is generated by the diagonal operator

τz =12

( a†1 a1 − a†2 a2 ) (1689)

and the ladder operators

τ+ = a†1 a2

τ− = a†2 a1 (1690)

These bi-linear τ operators do not change the total number of quanta and trans-form an energy eigenstate into eigenstates that are degenerate with it. The setof τ operators are linearly independent and obey similar commutation rela-tions to the angular momentum operators. Hence, the degeneracy of the two-dimensional harmonic oscillator is describable by a SU(2) symmetry group andis a Lie algebra of rank unity25. The symmetry group of the three-dimensionalharmonic oscillator must contain the SU(2) group as a sub-group. For thethree-dimensional harmonic oscillator, the Lie algebra is supplemented by thediagonal operator

λ =1

2√

3( a†1 a1 + a†2 a2 − 2 a†3 a3 ) (1691)

and, therefore, is a Lie algebra of rank 2. The Lie algebra of the three-dimensional harmonic oscillator also involves the following set of ladder op-erators

β+ = a†1 a3

β− = a†2 a3

γ+ = a†3 a2

γ− = a†3 a1 (1692)

Hence, the Lie algebra contains eight generators and corresponds to the SU(3)Lie group26.

25The rank of a Lie algebra is given by the maximal number of (linearly independent)simultaneously commuting operators in the algebra, excluding the Casimir operators. TheCasimir operators are non-linear combinations of the generators that commute with everygenerator of the Lie algebra. For semi-simple Lie groups of rank r, there are r Casimiroperators, and their eigenvalues uniquely characterize the multiplets of the group.

26The special unitary group in N dimensions SU(N) contains N2−1 generators. The SU(N)symmetry groups are discussed by S. Sternberg in Group Theory and Physics, CambridgeUniversity Press, (1994).

366

4.6.23 The Bound States of the Coulomb Potential

The time-independent Schrodinger equation27 for the attractive Coulomb po-tential is[− h2

2 m r2∂

∂r

(r2

∂

∂r

)+

h2 l ( l + 1 )2 m r2

− Ze2

r

]RE,l(r) = E RE,l(r)

(1693)where Z is the atomic number.

-0.75

-0.5

-0.25

0

0.25

-30 -20 -10 0 10 20 30

r/r0

V(r

)/V

(r0)

= -

r 0/r

E0

E1

E2E3

Figure 93: The radially symmetric potential, V (r), for the hydrogen atom. Theenergy eigenvalues are denoted by horizontal lines.

We introduce a dimensionless parameter

ρ =

√− 2 m E

h2 r (1694)

and a dimensionless constant ρ0 such that

V (r)E

=ρ0

ρ(1695)

so that

ρ0 =

√2 m− E

Z e2

h(1696)

27E. Schrodinger, Ann. Phys. (Leipzig) 79, 361 (1926).

367

Then, the radial wave function is expressed in the form

RE,l(r) = ρl exp[ − ρ ] f(ρ) (1697)

which explicitly displays the limiting behavior needed at the boundaries. Thefactor proportional to ρl is introduced so that the wave function satisfies theSchrodinger equation near the origin, where the centrifugal barrier dominatesthe potential. The exponential term describes the localization of the boundparticle to the region around the origin. It should be noted that there is asecond independent solution for which R would diverge at large ρ as exp[ + ρ ]but this second solution is to be discarded as it is not normalizable. Substitutingthe expression (1697) in the Schrodinger equation yields

∂2f

∂ρ2+ 2

(l + 1ρ

− 1)∂f

∂ρ−[

2 ( l + 1 ) − ρ0

ρ

]f = 0 (1698)

——————————————————————————————————

Mathematical details

In terms of the dimensionless variables, the equation for the radial wavefunction R(r) takes the form

− ∂2R

∂ρ2− 2

ρ

∂R

∂ρ+(l ( l + 1 )

ρ2+ 1

)R − ρ0

ρR = 0 (1699)

We shall substitute the radial wave function in the form

R(r) = ρl exp[ − ρ ] f(ρ) (1700)

into the above equation. The detailed form of R was chosen to satisfy ourboundary conditions. The first derivative of R is calculated as

∂R

∂ρ= ρl exp[ − ρ ]

[l

ρ− 1 +

∂

∂ρ

]f(ρ) (1701)

The second derivative of R is calculated as

∂2R

∂ρ2= ρl exp[ − ρ ]

[l ( l − 1 )

ρ2− 2 l

ρ+ 1

]f(ρ)

+ ρl exp[ − ρ ] 2(l

ρ− 1

)∂f

∂ρ

+ ρl exp[ − ρ ]∂2f

∂ρ2(1702)

On substituting these expressions in eqn(1699), one finds

∂2f

∂ρ2+ 2

(l + 1ρ

− 1)∂f

∂ρ−[

2 ( l + 1 ) − ρ0

ρ

]f = 0 (1703)

368

——————————————————————————————————

On multiplying by ρ, this equation reduces to the differential equation for theassociated Laguerre polynomials. On multiplying by ρ, this equation reducesto the equation for the associated Laguerre polynomials L(z) in the variablez = 2 ρ,

z∂2L

∂z2+(

2 ( l + 1 ) − z

)∂L

∂z+[ρ0

2− l − 1

]L = 0 (1704)

The solution has to satisfy the boundary condition at the origin ρ = 0 and atρ → ∞. The solution for f(ρ) can be obtained as a Frobenius series

f(ρ) =∑n=0

an ρn (1705)

On introducing this power series ansatz for the solution into the differentialequation (1698) and combining terms of similar order, one finds a polynomialequation. The coefficients of the terms of various powers of ρ in the resultingpolynomial must vanish identically. This procedure leads to a recursion relationwhich relates the coefficients an with different n. The recursion relation is foundas

an+1 ( n + 1 ) ( n + 2 l + 2 ) = an

(2 ( n + l + 1 ) − ρ0

)(1706)

If the series did not terminate, then one can show that the sum of the infiniteseries would be of order exp[ + 2 ρ ] for large ρ. This would result in theappearance of the second solution with the unphysical asymptotic behavior atlarge r

limr → ∞

R(r) ∼ ρl exp[

+ ρ

](1707)

but this would not satisfy our boundary conditions. Therefore, one insists thatthe series must truncate if it is to represent a bound state. The series truncatesat an integer value of n, say nr, if

ρ0 = 2 ( nr + l + 1 ) (1708)

and represents a bound state with radial quantum number nr. On using

ρ0 =

√2 m− E

Z e2

h(1709)

one finds the non-relativistic formula for the energy eigenvalues

E = − Z2 e4 m

2 h2 ( nr + l + 1 )2(1710)

369

Furthermore, on defining a characteristic length scale r0 in terms of ρ0 andeqn(1694), one has

r0 =

√h2

− 2 m Eρ0 (1711)

Therefore, one has

r0 = 2 ( nr + l + 1 )2h2

Z m e2(1712)

and one identifies the characteristic length scale with twice the Bohr radius.The principal quantum number n is usually defined as

n =ρ0

2= ( nr + l + 1 ) (1713)

where l is the quantum number corresponding to the orbital angular momentumand nr is the radial quantum number.

The lowest energy state or ground state corresponds to l = 0 and nr =0. The states with higher values of E are degenerate since for these energies,different values of nr and l can be combined to give the same value of ρ0 or n.The possible values of l for a fixed ρ0 are given by l = 0, 1, 2, ..... , ρ0

2 − 1.Since each l value has a degeneracy of 2 l + 1 due to the different m values,the total degeneracy is

ρ02 −1∑l=0

( 2 l + 1 ) =(ρ0

2

)2

= n2 (1714)

Thus, the energy eigenfunctions corresponding to the bound state energy En

are n2-fold degenerate, if we neglect spin.

Note that if we had deformed the Coulomb potential slightly, the degeneracybetween the states with principal quantum number n but with different orbitalangular momentum l would be lifted. This occurs in H, if one considers theeffect of the finite size of the nucleus.

The “accidental degeneracy” of the energy eigenvalues En of the hydrogenatom, corresponding to the states with principal quantum number n but withdifferent angular momentum quantum numbers l, involves the group SO(4).More specifically, it can be shown that the energy eigenstates of the hydrogenatom can be transformed into the eigenstates of a free rotor in a four-dimensionalspace28.

——————————————————————————————————

28V.A. Fock, Z. Phys. 98, 145 (1935).

370

-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0 2 4 6 8 10 12 14 16 18 20

r / a

Rn,

0(r)

l = 0 n = 1

n = 2

n = 3

Figure 94: The radial wave functions for the hydrogen atom Rn,l(r), for l =0 , 1 , 2. The radial quantum number nr determines the number of nodes inthe wave function.

-0.05

0

0.05

0.1

0.15

0.2

0 2 4 6 8 10 12 14 16 18 20

r / a

Rn,

1(r)

l = 1

n = 2

n = 3

0

0.01

0.02

0.03

0.04

0.05

0.06

0 2 4 6 8 10 12 14 16 18 20

r / a

Rn,

2(r)

l = 2

n = 3

371

0

0.2

0.4

0.6

0 1 2 3 4r/a0

r2 R

1,02 (r

)

n = 1

l = 0

Figure 95: The probability density P (r) = r2 | Rn,l(r) |2 for finding an electronat a distance r from the nucleus, for n = 1 , 2 , 3 and different l values.

0

0.05

0.1

0.15

0.2

0.25

0 2 4 6 8 10 12

r/a0

r2 R

2,l2 (r

)

n = 2

l = 1 l = 0

0

0.03

0.06

0.09

0.12

0 4 8 12 16 20

r/a0

r2 R

3,l2 (r

)

n = 3 l = 0

l = 1 l = 2

372

0

0.1

0.2

0.3

0.4

0.5

0 1 2 3 4 5 6 7 8 9 10

r / a0

P(r

)n = 2

l = 1 l = 0

Figure 96: The probability density P (r) = r2 | Rn,l(r) |2 for finding an electronat a distance r from the nucleus, for n = 2 , 3 and different l values. Theclassical probability densities are also shown. The classical probability densitiesare calculated with a fixed energy E and angular momentum l, but unknowninitial position r. Only the classical particle with l = 0 reaches the origin,however, it has a vanishing probability density there as the speed tends toinfinity as r → 0.

0

0.05

0.1

0.15

0.2

0 2 4 6 8 10 12 14 16 18 20

r / a0

P(r

)

n = 3

l = 1

l = 0

l = 2

373

4.6.24 Exercise 95

Derive the recursion relation for the series expansion of the associated Laguerrepolynomials that occur in the radial wave functions of electrons in the boundstates of Hydrogen atoms and determine the condition that the series terminatesafter n terms.

——————————————————————————————————

4.6.25 Exercise 96

An atom of tritium is in its ground state when the nucleus suddenly decaysinto a helium nucleus along with the emission of a fast electron (and an anti-neutrino) which leaves the atom without perturbing the extra-nuclear electron.Find the probability that the remaining He+ atom will be left in the state withn = 1, l = 0 and n = 2, l = 0. What is the selection rule for the l quantumnumber?

——————————————————————————————————

4.6.26 Solution 96

The electronic state immediately before the decay is identical to that of then = 1, l = 0 state of H, i.e. Z = 1. Then,

ΨZ=1,n=1,l=0(r) =1√πa− 3

20 exp

[− r

a0

](1715)

The probability that the electron falls into the energy levels of He+ is given bythe overlap with the respective Z = 2 wave functions

ΨZ=2,n=1,l=0(r) =1√π

(2a0

) 32

exp[− 2

r

a0

](1716)

and

ΨZ=2,n=2,l=0(r) =1√π

(1a0

) 32(

1 − r

a0

)exp

[− r

a0

](1717)

The matrix elements between the Z = 1 and Z = 2 states are evaluated viascaling and repeated integration by parts as

Cn=1,l=0 = 4 π∫ ∞

0

dr r2 Ψ∗Z=1,n=1,l=0(r) ΨZ=2,n=1,l=0(r)

= 4 π8

12

π a30

∫ ∞

0

dr r2 exp[− 3

r

a0

]

374

= 272

∫ ∞

0

dx x2 exp[− 3 x

]=

292

33(1718)

Thus, the probability of the electron falling into the ground state is equal to

P (0) =29

36(1719)

Similarly,

Cn=2,l=0 = 4 π∫ ∞

0

dr r2 Ψ∗Z=1,n=1,l=0(r) ΨZ=2,n=2,l=0(r)

= 4 π1

π a30

∫ ∞

0

dr r2(

1 − r

a0

)exp

[− 2

r

a0

]= 4

∫ ∞

0

dx x2

(1 − x

)exp

[− 2 x

]= − 1

2(1720)

Thus, the probability of the electron falling into the first excited state is equalto

P (1) =122

(1721)

The selection rule is ∆l = 0. This follows as the potential is spherically symmet-ric, and so angular momentum is conserved. Mathematically this is manifestedby the angular part of the wave function being given by spherical harmonics,and also by the fact that the spherical harmonics with different l are orthogonal.

——————————————————————————————————

4.6.27 Exercise 97

A tritium atom decays via the nuclear reaction

3H → 3He + e− + ν (1722)

and a fast electron leaves the atom almost instantaneously. Calculate the ex-pectation value of the energy for the remaining electron. The probability thatthe remaining energy is in the n-th bound state is denoted by P (n). It is foundthat

∞∑n=1

P (n) = 0.9755 (1723)

and∞∑

n=1

P (n)1n2

= 0.76660 (1724)

375

What is the probability that the Helium ion is doubly ionized and what is theaverage energy of the second emitted electron?

——————————————————————————————————

4.6.28 Ladder Operators for the Hydrogen Atom

The radial equation for the energy eigenstates of Hydrogen atom can be writtenin a dimensionless form. This is achieved by introducing a length scale b definedby

b =h2

Z e2 m(1725)

and a dimensionless variable ρ defined by

ρ =r

b(1726)

The energy eigenvalue can then be expressed in dimensionless form as

ε = E

(2 m b2

h2

)= E

(2 h2

Z2 e4 m

)=

(E

R

)(1727)

Thus, the dimensionless energy ε corresponds to the value of E when expressedin units of Rydbergs. In dimensionless units, the radial equation takes the form(

− ∂2

∂ρ2+

l ( l + 1 )ρ2

− 2ρ

)un,l = ε un,l (1728)

whereun,l(ρ) = ρ Rn,l(ρ) (1729)

is the reduced radial wave function. The bound state wave functions have nega-tive eigenvalues, are normalizable, satisfy the boundary condition limρ→0 un,l(ρ) →ρl+1 at ρ = 0, and vanish asymptotically as

limρ→∞

un,l(ρ) → exp[−√−ε ρ

](1730)

when ρ → ∞. The radial equation can be solved for the bound states usingappropriate raising and lowering operators.

The lowering operator A−l is defined as

A−l =∂

∂ρ+

l + 1ρ

− 1l + 1

(1731)

376

and the raising operator A+l is defined as the Hermitean conjugate of the low-

ering operator

A+l = − ∂

∂ρ+

l + 1ρ

− 1l + 1

(1732)

It immediately follows that

A−l A+l =

(− ∂2

∂ρ2+

l ( l + 1 )ρ2

− 2ρ

+1

( l + 1 )2

)(1733)

so the radial Schrodinger equation can be written as(A−l A+

l − 1( l + 1 )2

)un,l = ε un,l (1734)

The product of the raising and lowering operators, taken in the reverse order,can be expressed as

A+l A−l =

(− ∂2

∂ρ2+

( l + 2 ) ( l + 1 )ρ2

− 2ρ

+1

( l + 1 )2

)(1735)

Hence, one has the relation

A+l A−l − 1

( l + 1 )2= A−l+1 A

+l+1 −

1( l + 2 )2

(1736)

On multiplying the radial Schrodinger equation (as expressed in the form ofeqn(1734)) by the raising operator A+

l , one has

A+l

(A−l A+

l − 1( l + 1 )2

)un,l = ε A+

l un,l(A+

l A−l − 1( l + 1 )2

)A+

l un,l = ε A+l un,l (1737)

On substitution of the relation expressed by eqn(1736) in the above equation,one finds (

A−l+1 A+l+1 −

1( l + 2 )2

)A+

l un,l = ε A+l un,l (1738)

Hence, we see that A+l un,l is an eigenstate of the radial Schrodinger equation

with angular momentum l′ = l + 1 and energy eigenvalue ε. Thus, A+l is a

raising operator for the radial wave function.

The eigenstates with angular momentum l′ = l−1 can be obtained with theaid of the lowering operator. This can be shown by considering the Schrodingerequation written in the form(

A+l−1 A

−l−1 −

1l2

)un,l = ε un,l (1739)

377

On applying A−l−1 to this equation, one finds

A−l−1

(A+

l−1 A−l−1 −

1l2

)un,l = ε A−l−1 un,l(

A−l−1 A+l−1 −

1l2

)A−l−1 un,l = ε A−l−1 un,l(

A+l−2 A

−l−2 −

1( l − 1 )2

)A−l−1 un,l = ε A−l−1 un,l (1740)

where we have used eqn(1736) in the second line. On comparison of eqn(1740)with eqn(1739), one finds that A−l−1 un,l is an eigenstate with eigenvalue ε andangular momentum l′ = l − 1.

The energy eigenvalue ε satisfies an inequality which can be found directlyfrom the Radial Schrodinger equation

A−l A+l un,l =

(1

( l + 1 )2+ ε

)un,l (1741)

by multiplying by u∗n,l and integrating over ρ 29∫ ∞

0

dρ u∗n,l A−l A+

l un,l =∫ ∞

0

dρ u∗n,l

(1

( l + 1 )2+ ε

)un,l (1742)

However, un,l is normalized to unity and A−l is the Hermitean conjugate of A+l ,

so ∫ ∞

0

dρ | A+l un,l |2 =

(1

( l + 1 )2+ ε

)(1743)

Since the left-hand side is positive, one finds the inequality

ε ≥ − 1( l + 1 )2

(1744)

The inequality

ε ≥ − 1( l + 1 )2

(1745)

involving the (negative) bound state energy eigenvalue ε and the angular mo-mentum indicates that the angular momentum raising process must terminate.If this terminates for some value of l, say lmax, then the termination conditionis expressed as

A+lmax

un,lmax= 0 (1746)

29Due to the factor of ρ which appears in the definition of un,l in terms of the radial func-tion Rn,l(ρ), the three-dimensional radial equation is reduced to an effective one-dimensionalSchrodinger equation. The factor of ρ in un,l also reduces the appropriate measure for theradial integration is reduced from ρ2 to unity, as expected for an effective one-dimensionalgeometry.

378

On using the termination condition in the eigenvalue equation(A−lmax

A+lmax

− 1( lmax + 1 )2

)un,lmax

= ε un,lmax(1747)

so that the first term in the parenthesis is zero, one finds

ε = − 1( lmax + 1 )2

(1748)

orε = − 1

n2(1749)

where n = ( lmax + 1 ) is the principal quantum number.

The radial functions un,l can be obtained by first solving the first order dif-ferential equation for un,lmax

that expresses the termination and then, using thelowering operator successively to find all the eigenfunctions with lower l values.

The termination condition

A+lmax

un,lmax= 0 (1750)

is expressed as∂un,lmax

∂ρ=(n

ρ− 1n

)un,lmax (1751)

This can be integrated to yield

ln un,lmax= A + n ln ρ − ρ

n(1752)

where A is a constant of integration. On exponentiating, one finds the radialwave function has the form

un,lmax= C ρn exp

[− ρ

n

](1753)

where the constant C has to be determined from the normalization condition.The above form of un,l(ρ) yields a solution for the radial wave function whichdoes satisfy the correct boundary conditions.

The radial wave function with angular momentum l = lmax − 1 is obtainedfrom un,lmax by the action of the lowering operator

un,lmax−1 ∝ A−lmax−1 un,lmax(1754)

or, more explicitly

un,lmax−1 ∝(

∂

∂ρ+

( n − 1 )ρ

− 1( n − 1 )

)un,lmax

∝(

1 − ρ

n ( n − 1 )

)ρn−1 exp

[− ρ

n

](1755)

379

Table 8: Radial wave functions Rn,l(ρ) for the Coulomb potential, where ρ =Z ra . The Radial wave functions are normalized so that

∫∞0

dρ ρ2 Rn,l(ρ)2 = 1.

n = 1 l = 0 2 exp[− ρ

]

n = 2 l = 0 1√2

(1 − ρ

2

)exp

[− ρ

2

]

l = 1 12√

6ρ exp

[− ρ

2

]

n = 3 l = 0 2

332

(1 − 2

3 ρ + 227 ρ

2

)exp

[− ρ

3

]

l = 1 252

372

(1 − ρ

6

)ρ exp

[− ρ

3

]

l = 2 232

392√

5ρ2 exp

[− ρ

3

]

Therefore, the radial functions corresponding to smaller values of l can be ob-tained by sequentially operating with the appropriate A−l−1.

4.6.29 Rydberg Wave Packets

The correspondence between the classical picture of the orbital motion of elec-trons in the hydrogen atom and the quantum mechanical picture can be seen byexamining the time dependence of the peak in the probability density |Ψ(r, θ, ϕ; t) |2.For convenience, we shall consider motion in the plane θ = π

2 , and so shallset m to the maximal value m = l. In this case, due to the θ dependencegiven by Θl

l(θ) ∝ sinl θ, the probability density is maximal at θ = π2 . To

obtain a time dependence of the probability, it is necessary to superimpose en-ergy eigenstates to produce a wave packet. In particular, to obtain a probabilitydensity that shows a time-dependent variation in the ϕ dependence, eigenstatescorresponding to different energy eigenvalues must be superimposed. We shallsuperimpose the energy eigenstates with l = n − 1, which are given by

Ψn,n−1,n−1(r, θ, ϕ) = ℵ−1

(r

a0

)n−1

exp[− r

n a0

]sinn−1 θ exp

[i ( n− 1 ) ϕ

](1756)

380

where a0 = h2

m e2 is the Bohr radius, and ℵ is the normalization. This set ofenergy eigenstates correspond to circular orbits, as the radial quantum numbernr = 0. For the circular orbits, the principal quantum number n is just givenby l + 1. The energy eigenvalues are given by

En = − m e4

2 h2 n2(1757)

These states have radii defined by the maxima in

r2∣∣∣∣ Ψn,n−1,n−1(r, θ, ϕ)

∣∣∣∣2 (1758)

which occur at r = n2 a0. The products of the uncertainties in the radialposition and the radial momentum are given by

( ∆r )rms ( ∆pr )rms =h

2

(1 +

12 n

+ . . .

)(1759)

and also for the angular coordinates

( ∆θ )rms ( ∆pθ )rms =h

2

(1 +

14 n

+ . . .

)(1760)

A particular wave packet is represented by the wave function

Ψ(r; t) =∑

n

Cn Ψn,n−1,n−1(r, θ, ϕ) exp[− i t

hEn

](1761)

where the distribution of Cn produces a mean value of n denoted by n, and awidth given by ∆nrms.

The probability density shows that the wave packet describes periodic cir-cular orbits, with period given by the Kepler period TK . This is seen in fig(98).This can be derived from the expansion of the exponential factor in the wavepacket about n

exp[i ( n ϕ − t

hEn )

]exp

[i ( n − n )

(ϕ − t

h

(∂En

∂n

)∣∣∣∣n=n

) ]. . . (1762)

The first factor being a common overall phase factor does not appear in theprobability density. If the terms of order ( n − n )2 can be neglected, thesum over n of the second factor produces a finite Fourier series. The resultingapproximation to the wave packet is 2 π periodic in the variable

ϕ − t

h

(∂En

∂n

)∣∣∣∣n=n

(1763)

381

0

0.01

0.02

0.03

0.04

0.05

0 10 20 30 40 50 60 70 80 90 100

r / a0

P(r

)n = 6

l = 5 l = 0

Figure 97: A comparison of the quantum mechanical probability densities of thecircular orbit with l = n − 1 (blue) and the linear orbit l = 0 (red). For thecircular orbit with n = l + 1, the zero-point radial motion becomes negligiblefor large values of l and the radial wave function becomes highly peaked at theradius of the orbit. The classical probability densities are denoted by the brokenlines.

Thus, the Kepler period is given by

2 πTK

=1h

(∂En

∂n

)∣∣∣∣n=n

=m e4

h3 n3(1764)

or

TK =2 π h3 n3

m e4

= 2 π

(h2 n2

m e2

)(

e2

h n

)=

2 π a0 n2

v(1765)

which is the size of the circular orbit divided by the average speed v. The

382

average speed is given in terms of the fine structure constant and the velocityof light c via

v =e2

h n

=c

n

e2

h c(1766)

The terms of order ( n − n )2 have the effect of dephasing the oscillations. Atfirst, the ϕ dependence of the wave packets distort by spreading slowly in time.This increases the range of ϕ in which the wave packets have an appreciablemagnitude. The initial spreading is similar to the spreading of a wave packetfor a free particle. However, at large times, the wave packet recurs or re-forms asis shown in fig(99). The time scale for this phenomenon is found by expandingthe energy about the average value of n,

En = En + ( n − n )∂E

∂n

∣∣∣∣n=n

+12

( n − n )2∂2E

∂n2

∣∣∣∣n=n

(1767)

The recurrence time TR is defined via

− 1h

∂2E

∂n2

∣∣∣∣n=n

=2 πTR

(1768)

which yields

TR =n

3TK (1769)

The recurrence time causes the wave function to dephase and recur at timet = TR but with a phase shift of π. The phenomenon of recurrence is a quan-tum effect involving the non-linear dependence of the energy eigenvalues30.

4.6.30 Laguerre Polynomials

Laguerre’s differential equation is

z∂2φn

∂z2+ ( 1 − z )

∂φn

∂z+ n φn = 0 (1770)

The solution can be represented by a contour integral

φn(z) =∮

dt

2 π iexp

[− z t

1 − t

]1

( 1 − t ) tn+1(1771)

where the contour runs around a circle centered on the origin, with radius lessthan unity. We shall show that this is a solution of the differential equation.First we shall evaluate the derivatives

∂φn(z)∂z

= −∮

dt

2 π iexp

[− z t

1 − t

]t

( 1 − t )2 tn+1(1772)

30C. R. Stroud Jr. Physics and Probability, Essays in honor of E. T. Jaynes, Grandy andMilonini editors, Cambridge University Press (1993).

383

Figure 98: The time dependence of the probability density for an electron in acircular orbit, at times of order TK . The wave packet rotates with period Tk.[After C. R. Stroud Jr. (1993).]

and∂2φn(z)∂z2

= +∮

dt

2 π iexp

[− z t

1 − t

]t2

( 1 − t )3 tn+1(1773)

384

Figure 99: The time dependence of the probability density for an electron ina circular orbit, at large times. At large times the wave packet recurs. [AfterC. R. Stroud Jr. (1993).]

Then, on substituting the derivatives into the differential equation, one obtains

0 =∮

dt

2 π iexp

[− z t

1 − t

]1

( 1 − t )3 tn+1

[z t2 − t ( 1− t ) ( 1− z ) + n ( 1− t )2

](1774)

The right hand side is identified as the integral of a perfect differential

0 = −∮

dt

2 π i∂

∂t

(exp

[− z t

1 − t

]1

( 1 − t ) tn

)(1775)

On integrating the perfect differential around a closed contour without crossinga branch point, we find zero. Thus, we have verified that the expression givenby the contour integral does satisfy Laguerre’s equation.

We shall denote the solution of Laguerre’s equation by

Ln(z) =∮

dt

2 π iexp

[− z t

1 − t

]1

( 1 − t ) tn+1(1776)

The generating function for the Laguerre polynomials, G(z, t) can be founddirectly from this representation. Let

G(z, t) =∞∑

n=0

Ln(z) tn (1777)

385

then divide by tn+1 and integrate around a closed contour containing the origin.This procedure yields

Contour of Integration

-2

-1

0

1

2

-2 -1 0 1 2

Re tIm

t

Figure 100: The contour for the integration in eqn(1776). The circular contourencloses the pole at the origin but excludes the pole at t = 1. The two poles ofthe integrand are marked by the crosses.

∮dt

2 π iG(z, t)tn+1

=∑m

Lm(z)∮

dt

2 π itm

tn+1

= Ln(z) (1778)

since Cauchy’s theorem implies only the term with n = m is non-zero. Oncomparing the first line with the integral expression for Ln(z), we identify

G(z, t) =exp

[− z t

1 − t

]( 1 − t )

(1779)

Thus, we have the generating function series expansion

exp[− z t

1 − t

]( 1 − t )

=∞∑

n=0

Ln(z) tn (1780)

for t < 1.

386

A Rodriguez formula for the solution can be found by performing the non-linear transformation

t z

1 − t= s − z (1781)

which can be solved for t ast =

s − z

s(1782)

In this variable, the contour integral in eqn(1776) takes on the form

Ln(z) = exp [ z ]∮

ds

2 π isn exp[ − s ]( s − z )n+1

(1783)

where the new contour encloses the point s = z which corresponds to t = 0.By Cauchy’s theorem, this integral can be seen to be the n-th order derivativeof the numerator evaluated at the pole

Ln(z) =exp [ z ]

n!∂n

∂zn

(zn exp[ − z ]

)(1784)

for integral n.

The Laguerre differential operator is not Hermitean, but if the solution isexpressed as

Φn(z) = exp[− z

2

]Ln(z) (1785)

then the corresponding operator is Hermitean, and the differential equationbecomes

z∂2Φn

∂z2+

∂Φn

∂z+ ( n +

12− z

4) Φn(z) = 0 (1786)

Therefore, the Laguerre polynomials satisfy the orthogonality condition∫ ∞

0

dz exp[− z

]Lm(z) Ln(z) = δm−n (1787)

The Associated Laguerre polynomials are defined by

Lkn(z) = ( − 1 )k ∂k

∂zkLn+k(z) (1788)

For fixed k, one can find a generating function for Lkn(z) from the generating

function for the Laguerre polynomials

exp[− z t

1 − t

]( 1 − t )

=∞∑

n=−k

Ln+k(z) tn+k (1789)

387

by differentiating k times, with respect to z, obtaining

exp[− z t

1 − t

]tk

( 1 − t )k+1=

∞∑n=−k

( − 1 )k ∂k

∂zkLn+k(z) tn+k (1790)

Hence, since Ln+k(z) is a polynomial with highest order term of zn+k, the sumhas its first non-zero term when n = 0 and

exp[− z t

1 − t

]( 1 − t )k+1

=∞∑

n=0

Lkn(z) tn (1791)

which is the generating function expansion for the associated Laguerre poly-nomials. The expansion also shows that the associated Laguerre polynomialsreduce to the Laguerre polynomials when k = 0.

The associated Laguerre polynomials satisfy the differential equation

z∂2

∂z2Lk

n + ( k + 1 − z )∂

∂zLk

n + n Lkn = 0 (1792)

The associated Laguerre polynomials have the Rodriguez representation

Lkn(z) =

1n!

exp[

+ z

]z− k ∂n

∂zn

(exp

[− z

]zn + k

)(1793)

The differential equation for the associated Laguerre polynomials can be writtenin the form of a Hermitean operator eigenvalue equation by using the substitu-tion

φkn(z) = z

k2 exp

[− z

2

]Lk

n(z) (1794)

The functions φkn(z) form a complete orthonormal set because they satisfy the

differential equation

z∂2φk

n

∂z2+

∂φkn

∂z−(z

4− 2 n + k + 1

2+

k2

4 z

)φk

n = 0 (1795)

The normalization integral is∫ ∞

0

dz zk exp[− z

]Lk

n(z) Lkm(z) =

( n + k )!n!

δn−m (1796)

The normalized hydrogen atom wave function is given by

Ψn,m,l(r, θ, ϕ) = Cl,n,m ρl exp[− ρ

]L2l+1

n−l−1( 2 ρ ) Y lm(θ, ϕ) (1797)

388

Table 9: The Associated Laguerre Polynomials Lkn(z)

k = 0 k = 1

n L0n(z) L1

n(z)

n = 0 1 1

n = 1 1− z 2− z

n = 2 12 (2− 4z + z2) 1

2 (6− 6z + z2)

n = 3 16 (6− 18z + 9z2 − z3) 1

6 (24− 36z + 12z2 − z3)

k = 2 k = 3

n L2n(z) L3

n(z)

n = 0 1 1

n = 1 3− z 4− z

n = 2 12 (12− 8z + z2) 1

2 (20− 10z + z2)

n = 3 16 (60− 60z + 15z2 − z3) 1

6 (120− 90z + 18z2 − z3)

389

where we have written

ρ =Z r

n a0(1798)

and a0 is the Bohr radius. The normalization constant is given by

| Cl,n,m |2 =(

2 Zn a0

)3

22l ( n − l − 1 )!2 n ( n + l )!

(1799)

——————————————————————————————————

4.6.31 Exercise 98

Show that for the hydrogen atom, the average values of the moments of theradial distance of the electron from the proton satisfy the following relations,

4 ( s + 1 )ρ20

rs − ( 2 s + 1 ) a0 rs−1 +s

4

[( 2 l + 1 )2 − s2

]a20 r

s−2 = 0

s > − ( 2 l + 1 )(1800)

where a0 is the Bohr radius.

Also show that

r = a0ρ20

4

[1 +

12

(1 − 4

l ( l + 1 )ρ20

) ]r−1 =

4a0 ρ2

0

(1801)

——————————————————————————————————

4.6.32 Solution 98

We shall use the dimensionless variable ρ where

ρ = κ r =

√− 2 m E

h2 r (1802)

and write the Radial part of the wave function as R(ρ) = f(ρ)ρ . We also

introduce a constant ρ0 by

− Z e2

E r=

ρ0

ρ(1803)

390

Then, the radial part of the energy eigenvalue equation becomes[∂2f

∂ρ2+(− l ( l + 1 )

ρ2+

ρ0

ρ− 1

)f

]= 0 (1804)

We shall multiply this equation by

ρs+1 ∂f

∂ρ− C ρs f (1805)

and integrate over ρ. Then, as the average of the powers of ρ are given by

ρs =∫ ∞

0

dρ f(ρ) ρs f(ρ) (1806)

this will give us terms that are equal to the required expectation values.

The terms proportional to C are evaluated as

C

(l ( l + 1 ) ρs−2 − ρ0 ρs−1 + ρs

)− C

∫ ∞

0

dρ ρs f∂2f

∂ρ2(1807)

The last term can be integrated by parts, and the boundary terms vanish forappropriately large l. Then we obtain

C

(l ( l + 1 ) ρs−2 − ρ0 ρs−1 + ρs

)+ C

∫ ∞

0

dρ ρs f

(∂f

∂ρ

)2

− Cs ( s − 1 )

2ρs−2 (1808)

The remaining terms are re-written as∫ ∞

0

dρ ρs+1 ∂f

∂ρ

[∂2f

∂ρ2+(− l ( l + 1 )

ρ2+

ρ0

ρ− 1

)f

=12

∫ ∞

0

dρ ρs+1

(− l ( l + 1 )

ρ2+

ρ0

ρ− 1

) (∂f2

∂ρ

)+∫ ∞

0

dρ ρs+1

(∂f

∂ρ

) (∂2f

∂ρ2

)(1809)

These terms are evaluated as

12

(l ( l + 1 ) ( s − 1 ) ρs−2 − s ρ0 ρs−1 + ( s + 1 ) ρs

)− ( s + 1 )

2

∫ ∞

0

dρ ρs

(∂f

∂ρ

)2

(1810)

391

The terms involving(

∂f∂ρ

)2

vanish if we choose the constant C as

C =( s + 1 )

2(1811)

As the sum of the two terms are equal to zero, due to F satisfying the differentialequation, one has the equation relating the expectation values of powers of ρ

s

(l ( l + 1 ) − 1

4( s2 − 1 )

)ρs−2 − 2 s + 1

2ρ0 ρs−1 + ( s + 1 ) ρs = 0

(1812)

On putting s = 0 and noting ρ0 = 1 we have

ρ−1 =2ρ0

(1813)

and then with s = 1 we have

2 ρ =32ρ0 −

14

[( 2 l + 1 )2 − 1

]ρ−1

=32ρ0 −

12

[( 2 l + 1 )2 − 1

]ρ−10 (1814)

On converting back from dimensionless variables, we find the result given.

——————————————————————————————————

4.6.33 Exercise 99

Find the momentum space wave functions for the two lowest m = 0 energyeigenstates of the hydrogen atom, given by

Ψ1,0,0(r) =1√π

(Z

a0

) 32

exp[− Z r

a0

](1815)

and

Ψ2,1,0(r) =

√3

4 π 4!

(Z

a0

) 32(Z r

a0

)exp

[− Z r

2 a0

]cos θ (1816)

——————————————————————————————————

392

4.6.34 Solution 99

The momentum space wave function is given by the Fourier transform

Φ(p) =(

12 π h

) 32∫

d3r exp[− i

hp . r

]Ψ(r) (1817)

For the (1, 0, 0) wave function, the angular integration can be easily performedleading to

Φ1,0,0(p) =(

Z

2 π h a0

) 32

2√π h

×∫ ∞

0

dr r2( exp

[+ i

h p r

]− exp

[− i

h p r

]i p r

)exp

[− Z r

a0

](1818)

The first factor in the integrand is recognized as being proportional to the l = 0spherical Bessel function. The integral is easily evaluated to yield

Φ1,0,0(p) =(

Z h

2 π a0

) 32

2√πi

p

[ (1

p + i Z ha0

)2

−(

1p − i Z h

a0

)2]

=(

Z h

2 π a0

) 32

8√π

[Z ha0(

p2 + (Z ha0

)2)2

](1819)

The second momentum space wave function is found by using the Rayleighexpansion of the exponential plane wave. The Legendre polynomial is given interms of the angle between the momentum and the position vector. The spher-ical harmonic addition theorem is used to expand the Legendre polynomials interms of the products of the spherical harmonics, each separately involving thedirections of the momentum and position. On integrating over the directionsof the position vector and using orthonormality, one finds that the momentumspace wave function depends upon the spherical harmonic with the same (l,m)as the real space wave function, except that it depends on the direction of themomentum, (θp, ϕp), with respect to the Cartesian axes. The coefficient of thespherical harmonic Y l

m(θp, ϕp) is given by an integration involving the productof the spherical Bessel function jl(kr) and the radial dependence of the realspace wave function. The result entails evaluating integrals of the form∫ ∞

0

dz z(2+m) exp[− α z

]jl(z) (1820)

which for l = 2 simplifies to∫ ∞

0

dz z(2+m) exp[− α z

] (sin z − z cos z

z2

)

393

Table 10: The Gegenbauer Functions Cµν (x).

µ = 1 µ = 2 µ = 3 µ = 4 µ = 5

ν = 0 1ν = 1 2x 1ν = 2 4x2 − 1 4x 1ν = 3 8x3 − 4x 12x2 − 2 6x 1ν = 4 16x4 − 12x2 + 1 32x3 − 12x 24x2 − 3 8x 1ν = 5 32x5 − 32x2 + 6x 80x4 − 48x2 + 3 80x3 − 24x 40x2 − 4 10x

= ( − 1 )m ∂m

∂αm

∫ ∞

0

dz exp[− α z

] (sin z − z cos z

)= ( − 1 )m ∂m

∂αm

[2

( 1 + α2 )2

](1821)

The momentum space wave function of a general energy eigenstate of aHydrogen-like atom has been calculated by Podolsky and Pauling31. The mo-mentum eigenfunctions Φn,l,m(p, θ, ϕ) are found to be given by the general ex-pression

Φn,l,m(p, θ, ϕ) = Y lm(θ, ϕ)

(n a0

Z 2 π h

) 32[− ( − i )l π 22l+4 l!

(n(n− l − 1)!

(n+ l)!

) 12]

× ξl

(ξ2 + 1)l+2Cl+1

n−l−1

(ξ2 − 1ξ2 + 1

)(1822)

whereξ =

n p a0

Z h(1823)

and Cµν (x) are the Gegenbauer functions. Some of the Gegenbauer functions are

given in Table(10), and the others can be generated from the recursion relation

Cµν (x) =

2 µν

[x Cµ+1

ν−1 (x) − Cµ+1ν−2 (x)

](1824)

The momentum space distribution of the electrons in Hydrogen was determinedexperimentally32. The experiments involved the ionization of atomic hydrogenby a beam of high-energy electrons.

——————————————————————————————————

31B. Podolsky and L. Pauling, Phys. Rev. 34, 109 (1929).32B. Lohman and E. Weigold, Phys. Letts. 86 A, 139 (1981).

394

Momentum Space Probability Distribution

0

1

2

3

4

0 0.5 1 1.5 2

k a

4 π

P n,l(

k) k

2

n=2,l=0

n=2,l=1

n=1,l=0

(spherically averaged)

Figure 101: The spherically averaged momentum-space distribution function forthe few lowest energy eigenstates of Hydrogen.

4.6.35 Exercise 100

Find the scalar potential and vector potential produced by an electron in al = 1 state of a hydrogen atom by first calculating the charge and currentdensities and then by using Maxwell’s equations.

——————————————————————————————————

4.6.36 Solution 100

The hydrogen atom in a 2p orbital has a charge distribution

ρ(r, t) = | Ψ(r, t) |2

=e

64 π a50

r2 exp[− r

a0

]sin2 θ (1825)

where a0 is the Bohr radius h2

m e2 . In this case, the charge density is independentof time. The electrostatic potential φ(r) is given by

φ(r) =1

4 π

∫d3r′

ρ(r′)| r − r′ |

(1826)

395

We shall first perform the angular integral. The denominator is expandedin terms of the Legendre Polynomials via the expansion

1| r − r′ |

=1r>

∑l

(r<r>

)l

Pl(cos γ) (1827)

where γ is the angle between r and r′. Then using the addition theorem onehas

1| r − r′ |

=1r>

∑l

(r<r>

)l 4 π2 l + 1

m=+l∑m=−l

Y ∗ lm (θ′, ϕ′) Y l

m(θ, ϕ) (1828)

The angular dependence of the numerator can be expressed in terms of thespherical harmonics

sin2 θ′ = ( 1 − cos2 θ′ )

=23

√4 π Y 0

0 (θ′, ϕ′) −√

4 π5

23Y 2

0 (θ′, ϕ′) (1829)

The angular integration can be performed using the orthogonality of the spher-ical harmonics. Hence,

φ(r) =(

e

64 π a50

) ∫ ∞

0

dr′ r′41r>

exp[− r′

a0

]×∑

l

(r<r>

)l 12 l + 1

Y l0 (θ, ϕ)

(23

√4 π δl,0 −

√4 π5

23δl,2

)=

23

(e

64 π a50

) [I0 −

15I2

(3 cos2 θ − 1

2

) ](1830)

where

I0 =∫ ∞

0

dr′ r′4(

1r>

)exp

[− r′

a0

](1831)

and

I2 =∫ ∞

0

dr′ r′4(r2<r3>

)exp

[− r′

a0

](1832)

The radial integrations are broken into two parts, one for r > r′ and the otherfor r < r′. One finds

I0 =a50

r

[24 −

(24 + 24

(r

a0

)+ 12

(r

a0

)2

+ 4(r

a0

)3

+(r

a0

)4 )exp

[− r

a0

]+(

6(r

a0

)+ 6

(r

a0

)2

+ 3(r

a0

)3

+(r

a0

)4 )exp

[− r

a0

] ](1833)

396

and

I2 =a70

r3

[6! −

(6! + 6!

(r

a0

)+

6!2

(r

a0

)2

+

+ 5!(r

a0

)3

+ 30(r

a0

)4

+ 6(r

a0

)5

+(r

a0

)6 )exp

[− r

a0

]+( (

r

a0

)5

+(r

a0

)6 )exp

[− r

a0

] ](1834)

Hence, for r a0 where all the exponential terms are suppressed, one findsthat the electrostatic potential reduces to

φ(r) =e

4 π

(1r− 6

a20

r33 cos2 θ − 1

2

)(1835)

which is interpreted as the sum of a charge monopole and a quadrupole term.The deviation of the electron wave function from spherical symmetry has set upa quadrupolar electric field. When r < a0, the potential is reduced from thatof the monopole of charge e, since a Gaussian surface enclosing the origin onlycontains a fraction of the total electron charge.

To find the vector potential, first we find the current density j. The currentdensity is found from the expression

j(r, t) =e h

2 m i

(Ψ∗(r, t) ∇Ψ(r, t) − Ψ(r, t) ∇Ψ∗(r, t)

)(1836)

In spherical polar coordinates, the gradient is written as

∇ = er∂

∂r+ eθ

1r

∂

∂θ+ eϕ

1r sin θ

∂

∂ϕ(1837)

Therefore, the current density of the p electron is evaluated as

j(r, t) = eϕ

(e h

64 π m a50

)exp

[− r

a0

]r sin θ (1838)

The current density is resolved into Cartesian components, via

eϕ′ = cosϕ′ ey − sinϕ′ ex (1839)

and is then expressed in terms of spherical harmonics. The result is

j(r, t) = i

(e h

64 π m a50

)exp

[− r

a0

]r ×

×√

8 π3

[Y 1

1 (θ, ϕ)ex − i ey

2− Y 1

−1(θ, ϕ)ex + i ey

2

](1840)

397

which is time independent. The vector potential is given by

A(r) =1

4 π

∫d3 r′

j(r′)| r − r′ |

(1841)

Then the Legendre polynomial generating function expansion and the sphericalharmonic addition theorem are used to perform the angular integral over thevariables (θ′, ϕ′). The integrations utilize the orthonormality of the sphericalharmonics. The end result of the integrations over the angles (θ′, ϕ′) is

A(r) = i

(e h

64 π m a50

) ∫ ∞

0

dr′ r′3(r<r2>

)exp

[− r′

a0

]×

×√

8 π3

13

[Y 1

1 (θ, ϕ)ex − i ey

2− Y 1

−1(θ, ϕ)ex + i ey

2

]= eϕ

13

(e h

64 π m a50

) ∫ ∞

0

dr′ r′3(r<r2>

)exp

[− r′

a0

]sin θ

(1842)

The vector potential has a directional dependence which is similar to the cur-rent density. The radial integration can be performed, leading to the vectorpotential being dominated by a dipole contribution at large distances.

——————————————————————————————————

398

4.7 A Charged Particle in a Magnetic Field

Consider a particle of mass m and charge q moving in a uniform magnetic fieldB oriented along the z axis.

B = ez Bz (1843)

The vector potential A(r) can be found as a solution of

B = ∇ ∧ A(r) (1844)

One solution is given byA(r) = − y Bz ex (1845)

which corresponds to a particular choice of the gauge. The Hamiltonian of thecharged particle in the field can be written as

H =1

2 m

(px − q

cAx(r)

)2

+1

2 m

(p2

y + p2z

)=

12 m

(px +

q

cy Bz

)2

+1

2 m

(p2

y + p2z

)(1846)

We note that the x and z components of the momentum commute with theHamiltonian, as x and z do not appear in H . Thus,

[ H , px ] = 0[ H , py ] = 0 (1847)

which means that they are constants of motion. As the energy eigenstates arealso eigenstates of px and pz, we shall write the energy eigenfunction as

φn,px,py (r) =1

2 π hexp

[ipx x + pz z

h

]φn(y) (1848)

The energy eigenvalue equation can then be written in terms of the unknownfunction φn(y), as[

p2y

2 m+m

2

(q Bz

m c

)2 (y +

c px

q Bz

)2 ]φn(y) =

(E − p2

z

2 m

)φn(y) (1849)

This equation looks like the energy eigenvalue equation for the one-dimensionalharmonic oscillator, in which we have shifted the origin of the y coordinatethrough a distance c px

q Bz. The frequency of the oscillator corresponds to the

Larmor precession frequency ωL, which is given by

ω2L =

(q Bz

m c

)2

(1850)

Using our previously gained knowledge of the one-dimensional harmonic oscil-lator, we find that the energy eigenvalues are then given by

En,pz= h ωL

(n +

12

)+

p2z

2 m(1851)

399

We note that these energy levels (known as Landau levels33) are degenerate, asthe introduction of the B field causes states with different values of the (quasi-continuous quantum numbers) px and py to collapse onto states where the valueof the energy is determined by the integer n. The wave functions with the dif-ferent values of px have the “oscillations” in the y coordinate centered arounddifferent points.

——————————————————————————————————

4.7.1 Exercise 101

Show that both x0 and y0 commute with the Hamiltonian, where

x0 = x +c py

q Bz

y0 = − c px

q Bz(1852)

but x0 and y0 do not commute,

[ x0 , y0 ] 6= 0 (1853)

Show that in the classical limit, these variables correspond to the projection ofthe particle’s orbit on the x - y plane. Hence, x0 and y0 cannot be simultane-ously known for the quantum system.

——————————————————————————————————

4.7.2 Exercise 102

Consider a particle of mass m and charge e in perpendicular uniform electricand magnetic fields, E and B. Find the eigenfunctions and eigenvalues. Findthe average velocity in the x direction for any eigenstate.

——————————————————————————————————

4.7.3 Solution 102

Let the vector potential be written in the gauge

A(r) = B z ex (1854)

and the scalar potential is also a function of z

φ(z) = − | E | z (1855)33L.D. Landau, Z. f. Physik, 64, 629 (1930).

400

The Hamiltonian is given by

H =1

2 m

[ (px − e

cB z

)2

+ p2y + p2

z

]− e | E | z (1856)

which only depends on z. The momentum components px and py are goodquantum numbers. The eigenfunctions can be written as

Φkx,ky,n(r) = exp[i ( kx x + ky y )

]φn(z) (1857)

and φn(z) satisfies(E −

h2 k2y

2 m

)φn(z) =

(1

2 m

[ (h kx −

e

cB z

)2

+ p2y + p2

z

]− e | E | z

)φn(z)

(1858)This is recognized as the eigenvalue equation for a linear harmonic oscillatorcentered on z0

z0 = h kxc

e B+ m c2

| E |e B2

(1859)

Thus, the energy eigenvalue is given by

E = h ωc ( n +12

) +h2 k2

y

2 m− m c2

| E |2

2 B2− h kx c

| E |B

(1860)

The group velocity is given by

vx =∂E

∂px

= − c| E |B

(1861)

——————————————————————————————————

4.7.4 The Degeneracy of the Landau Levels

We shall consider a particle in a uniform magnetic field aligned along the zdirection. The particle is confined to move within a large volume. Let thevolume be bounded by surfaces, so that the accessible volume is described by

Lx > x > 0Ly > y > 0Lz > z > 0 (1862)

and the wave function should be zero outside this volume. Instead of apply-ing these boundary conditions, we shall impose Born-von Karman or periodic

401

boundary conditions.

The vector potential A(r) is defined to be the solution of

B = ∇ ∧ A(r) (1863)

One solution is given byA(r) = − y Bz ex (1864)

which corresponds to a particular choice of the gauge. The Hamiltonian of thecharged particle in the field can be written as

H =1

2 m

(px − q

cAx(r)

)2

+1

2 m

(p2

y + p2z

)=

12 m

(px +

q

cy Bz

)2

+1

2 m

(p2

y + p2z

)(1865)

in the region where the confining potential is zero.

The wave function that satisfies the boundary conditions, can be written as

Ψ(r) =1√

Lx Lz

exp[i

2 π nx x

Lx

]exp

[i

2 π nz z

Lz

]f(y) (1866)

where nx and nz are integers, nx = 0 , ± 1 , ± 2 , . . . etc. It is then foundthat the function f(y) satisfies the eigenvalue equation[

12 m

(2 π h nx

Lx+q

cy Bz

)2

− h2

2 m∂2

∂y2

]f(y) =

(E − 2 π2 h2 n2

z

m L2z

)f(y)

(1867)or[− h2

2 m∂2

∂y2+

q2 B2z

2 m c2

(2 π h c nx

q Bz Lx+ y

)2]f(y) =

(E − 2 π2 h2 n2

z

m L2z

)f(y)

(1868)which can be re-written in the form of a displaced harmonic oscillator equation,by writing

ωc =q Bz

m c(1869)

Then, the eigenvalue equation becomes[− h2

2 m∂2

∂y2+m ω2

c

2

(y +

nx Ly Φ0

Φ

)2]f(y) =

(E − 2 π2 h2 n2

z

m L2z

)f(y)

(1870)where we have introduced the flux Φ, defined by

Φ = Lx Ly Bz (1871)

402

and the fundamental flux quantum Φ0 defined by

Φ0 =2 π h c

q(1872)

Therefore, the eigenvalue E is just given by

E = h ωc ( n +12

) (1873)

and the eigenfunctions f(y) are the harmonic oscillator wave functions, φn(y)shifted through − nx

Φ0Φ Ly. That is, the normalized wave function is given by

f(y) = φn

(y + nx Ly

Φ0

Φ

)(1874)

The degeneracy can be found, in the semi-classical approximation, by as-suming that the deviation of the position of the harmonic oscillator from theequilibrium value y0 = − nx Ly

Φ0Φ is minimal. Under these circumstances,

one has 0 < y0 < Ly. Hence, nx is restricted to be such that

Ly > − nx LyΦ0

Φ> 0 (1875)

Hence, the degeneracy is the given by the maximum number of values that nx

can take. The degeneracy N is given by

N =(

ΦΦ0

)(1876)

which is controlled by the number of flux quanta, threading through the areaLx Ly.

——————————————————————————————————

4.7.5 Exercise 103

Find the energy eigenvalues and eigenfunctions of a particle of charge q andmass m moving in two dimensions, in the presence of a uniform magnetic field.Calculate the degeneracy of the lowest Landau level. Use the symmetric gaugewhere

A =Φ

2 π R20

r eϕ (1877)

and Φ is the total flux enclosed in the circular area π R20.

——————————————————————————————————

403

4.7.6 Solution 103

The energy eigenvalue equation is[− h2

2 m1r

∂

∂r

(r∂φ

∂r

)+

12 m

(− i

h

r

∂

∂ϕ− q

c

Φ2 π R2

0

r

)2

φ

]= E φ

(1878)The z component of the angular momentum Lz commutes with the Hamiltonian,and so one can find simultaneous eigenfunctions which are of the form

φ(r, ϕ) =1√2 π

exp[i µ ϕ

]R(r) (1879)

Hence, the radial wave function R(r) is given by[− h2

2 m1r

∂

∂r

(r∂R

∂r

)+

12 m

(h µ

r− q

c

Φ2 π R2

0

r

)2

R

]= E R (1880)

The radial wave function must satisfy boundary conditions at r = 0 andr → ∞. The physically acceptable form of the solution is

R(r) = ρµ2 exp

[− ρ

]f(ρ) (1881)

whereρ =

q Φc 4 π h R2

0

r2 (1882)

and f(ρ) is a finite order polynomial. The derivatives w.r.t. r can be expressedas derivatives with respect to ρ via

∂

∂r=

∂ρ

∂r

∂

∂ρ

=q Φ

c 2 π h R20

r∂

∂ρ(1883)

and

∂2

∂r2=

(q Φ

c 2 π h R20

)∂

∂ρ+(

q Φc 2 π h R2

0

)2

r2∂2

∂ρ2

=(

q Φc 2 π h R2

0

) [∂

∂ρ+ 2 ρ

∂2

∂ρ2

](1884)

Thus, the eigenvalue equation becomes

h q Φm c 2 π R2

0

[− ρ

∂2R

∂ρ2− ∂R

∂ρ+

1ρ

(µ

2− ρ

)2

R

]= E R (1885)

On substituting the form

R(r) = ρµ2 exp

[− ρ

]f(ρ) (1886)

404

into the differential equation, we find(h q Φ

m c 2 π R20

) [− ρ ∂2

∂ρ2+ ( 2 ρ− µ− 1 )

∂f

∂ρ+(

1− E m c 2 π R20

h q Φ

)f

]= 0

(1887)This differential equation can be solved by the Frobenius method, by expandingthe solution in the form

f(ρ) =n=nr∑n=0

an ρn (1888)

On substituting the series into the differential equation, one finds the recursionrelation

an+1 ( n + 1 ) ( n + µ + 1 ) =(

2 n + 1 − E m c 2 π R20

h q Φ

)an (1889)

The series for the polynomial f(ρ) truncates if the energy is given by

E =(

h q Φm c 2 π R2

0

)( 2 nr + 1 ) (1890)

where nr is an integer. In this case the boundary conditions are satisfied, andf(ρ) is a polynomial of degree nr.

The lowest energy eigenstates corresponds to nr = 0 and the eigenfunctioncan be written as

φ(r, ϕ) ∝ rµ exp[i µ ϕ

]exp

[− q Φ

c 4 π h R20

r2]

∝ ( x + i y )µ exp[− q Φ

c 4 π h R20

( x2 + y2 )]

∝ ( x + i y )µ exp[−(

ΦΦ0

) (x2 + y2

2 R20

) ](1891)

The probability density is peaked on a circle around the origin. The radius ofthe circle, rm, is given by

r2m = µc 2 π h R2

0

q Φ

= µ

(Φ0

Φ

)R2

0 (1892)

All the states with different µ are degenerate. The degeneracy is given by themaximum number of different allowed µ values. For a disk of area π R2

0 theallowed values of rµ satisfy rµ < R0, thus, the degeneracy is given by

µmax =(

ΦΦ0

)(1893)

——————————————————————————————————

405

4.7.7 The Aharonov-Bohm Effect

The Aharonov-Bohm effect34 provides a conclusive demonstration that a chargedquantum mechanical particle is sensitive to the vector potential, and not to themagnetic induction field B. In the Aharonov-Bohm effect, a charged particlemoves in a region of space where the magnetic field is zero. The magnetic fieldfree region is multiply connected, as a magnetic field threads through regionswhere the particle is excluded from35.

In order to provide a simple example of the Aharonov-Bohm effect, considera charged particle which is confined to move on ring of radius R. A magneticflux of strength Φ threads through the center of the ring and is contained withina cylinder of radius a.

Flux contained in a cylinder of radius a

BR

a

Figure 102: In the version of the Aharonov-Bohm effect being considered, theapplied magnetic flux Φ is confined within a cylinder of radius a. The chargedparticle is confined to move on a ring of radius R which encircles the cylinder.

The magnetic induction is in the z direction, so B = ez Bz. The magneticflux through the center of the ring is given by

Φ =∫

dx dy Bz

34Y. Aharonov and D. Bohm, Phys. Rev. 115, 485 (1959).35A. Tonomura, N. Osakabe, T. Matsuda, T. Kawasaki, J. Endo, S. Yano, and H. Yamada,

Phys. Rev. Lett. 56, 792 (1986).

406

=∫

d2S . B (1894)

in which the integral runs over the area inside the ring, where B is finite. Sincethe magnetic field is related to the vector potential via

B = ∇ ∧ A (1895)

then the flux is given by

Φ =∫

d2S .

(∇ ∧ A

)(1896)

On using Stoke’s theorem, one finds the total flux is given by

Φ =∮

dr . A (1897)

where the integral is around the perimeter of a ring of radius r and r > a.If the tangential component of the vector potential is denoted by Aϕ, then theloop integral is given by

Φ = 2 π r Aϕ (1898)

Hence, on considering the symmetry of the problem, one finds the vector po-tential is given by

A =Φ

2 π reϕ (1899)

There is no magnetic field present at this radius, since the definition

B = ∇ ∧ A (1900)

yields the z-component as

Bz = ( er∂

∂r+ eϕ

1r

∂

∂ϕ) ∧ Φ

2 π reϕ

= er ∧ eϕ∂

∂r

(Φ

2 π r

)+ eϕ

1r∧ ∂eϕ

∂ϕ

Φ2 π r

= er ∧ eϕ∂

∂r

(Φ

2 π r

)− 1

reϕ ∧ er

Φ2 π r

= 0 (1901)

The magnetic induction is zero since the total flux Φ contained within a loop isconstant, if r ≥ a. The vector potential in the region where the magnetic fieldis finite 0 < r < a is given by

A =r Φ

2 π a2eϕ (1902)

407

The energy eigenvalue equation for the particle confined on the ring of radiusR is

12 m

(− i

h

R

∂

∂ϕ− q Φ

c 2 π R

)2

Ψ(ϕ) = E Ψ(ϕ) (1903)

This has normalized eigenfunctions given by

Ψ(ϕ) =1√

2 π Rexp

[i µ ϕ

](1904)

where the energy eigenvalue E is determined by µ from

12 m R2

(µ h − q Φ

2 π c

)2

= E (1905)

Thus, the energy is given by

h2

2 m R2

(µ − q Φ

2 π h c

)2

= E (1906)

where 2 π h cq = Φ0 is the fundamental flux quantum. The modulus of the

wave function must be single valued, therefore

Ψ(ϕ) = Ψ(ϕ+ 2π) (1907)

or

1 = exp[i µ 2 π

](1908)

Hence, µ must be an integer, m = 0 , ± 1 , ± 2 . . . etc. The energy is givenby

E =h2

2 m R2

(m − Φ

Φ0

)2

(1909)

and the ground state energy and wave function is determined by the integer

value of m which minimizes E and, therefore, depends on the ratio of(

ΦΦ0

).

The ground state corresponds to the value of m such that

m − 12

<

(ΦΦ0

)<

m + 12

(1910)

Thus, the energy depends on the vector potential but not on the magnetic fieldB in the region where the particle moves. Furthermore, the ground state energyis a periodic function of Φ with periodicity Φ0, as the value m changes discon-tinuously from m to m+ 1 as Φ increases by an amount Φ0.

The current I produced by the charged particle flowing around the ring isdefined classically as

I = c∂E

∂Φ(1911)

408

Energy eigenvalues

0

0.1

0.2

0.3

0.4

-4 -3 -2 -1 0 1 2 3 4

Φ/ΦΦ/ΦΦ/ΦΦ/Φ0000

E( Φ ΦΦΦ

)

Figure 103: The energy eigenvalues Em in the Aharonov-Bohm effect, as afunction of the applied magnetic flux Φ threading the ring.

The quantum mechanical current operator is defined as

I = c∂H

∂Φ

=q h

2 π m R2

(i∂

∂ϕ+

ΦΦ0

)(1912)

Thus, in the ground state the current has an expectation value I , given by

I = − q h

2 π m R2

(m − Φ

Φ0

)(1913)

The current jumps discontinuously as the flux through the loop is increased byΦ0.

——————————————————————————————————

Example

Find the energy eigenvalues and eigenfunctions of a particle of charge q andmass m moving in two dimensions, in the presence of a uniform magnetic field.

409

Flux induced current

-0.75

-0.5

-0.25

0

0.25

0.5

0.75

-4 -3 -2 -1 0 1 2 3 4

Φ/ΦΦ/ΦΦ/ΦΦ/Φ0000

I(Φ ΦΦΦ

)/I 0

Figure 104: The dependence of the current I in the ground state on the appliedflux Φ.

Use the circularly symmetric gauge where

A =Φ

2 π reϕ (1914)

where Φ is the magnetic flux penetrating a ring of radius R0 r.

——————————————————————————————————

Solution

The energy eigenvalue equation is[− h2

2 m1r

∂

∂r

(r∂φ

∂r

)+

12 m

(− i

h

r

∂

∂ϕ− q

c

Φ2 π r

)2

φ

]= E φ (1915)

The z component of the angular momentum Lz commutes with the Hamiltonian,and so one can find simultaneous eigenfunctions which are of the form

φ(r, ϕ) =1√2 π

exp[i µ ϕ

]R(r) (1916)

Hence, the radial wave function R(r) is given by[− h2

2 m1r

∂

∂r

(r∂R

∂r

)+

12 m r2

(h µ − q

c

Φ2 π

)2

R

]= E R (1917)

410

The radial wave function must satisfy boundary conditions at r = 0 andr → ∞. The acceptable form for the solution at r = R0 → 0 vanishes atthe origin as

R(r) = rα (1918)

which minimizes the effect of the centrifugal potential. On introducing a di-mensionless variable

ρ = k r (1919)

where

E =h2 k2

2 m(1920)

and the elemental flux quantum Φ0, defined by

Φ0 =c 2 π h

q(1921)

one finds that the equation can be put in the dimensionless form

1ρ

∂

∂ρ

(ρ∂R

∂ρ

)+[

1 − 1ρ2

(µ − Φ

Φ0

)2 ]R = 0 (1922)

This is Bessel’s equation. The physically acceptable solution is finite in the limitas r → 0 and, therefore, the required solution is proportional to the Besselfunction of order ν where

ν = µ −(

ΦΦ0

)(1923)

The eigenfunction is given by

φ(r, ϕ) =1√2 π

exp[i µ ϕ

]Jν(kr) (1924)

The Bessel function has the asymptotic variation

Jν(ρ) ∼ ρν (1925)

as ρ → 0 and varies as

Jν(ρ) ∼√

2π ρ

cos(ρ − ( ν +

12

)π

2

)(1926)

when ρ → ∞. Note that, if the flux quantum threading the origin is changed sothat ν increases, the maxima of the probability density moves radially outward,however, there is no change in energy. The absence of any energy change is notsurprising in view of the fact that B = 0 in the region where the particle ismoving and so, the effect of the vector potential can be absorbed into the phaseof the wave function by a Gauge transformation.

——————————————————————————————————

411

4.8 The Pauli Spin Matrices

The Pauli Matrices are three two by two matrices, which have commutationrelations similar to the commutation relations of the three components of theangular momentum operators. The Pauli matrices are represented by

σx =

(0 11 0

)(1927)

σy =

(0 − ii 0

)(1928)

σz =

(1 00 − 1

)(1929)

These matrices are traceless, and have the properties that their square is equalto the unit matrix, σ0,

σ2x = σ2

y = σ2z = σ0 (1930)

where the unit matrix is given by

σ0 =

(1 00 1

)(1931)

The three Pauli matrices and the unit matrix are linearly independent, andform a basis for the two by two matrices. That is, any two by two matrixcan be expressed as a linear combination of the identity and Pauli matrices.Furthermore, they also satisfy the commutation relations

[ σx , σy ] = 2 i σz

[ σz , σx ] = 2 i σy

[ σy , σz ] = 2 i σx (1932)

When the Pauli matrices are multiplied by a factor of h2 , these commutation

relations become identical to the commutation relations of the components ofangular momentum, for fixed l. Thus, if one multiplies the Pauli matrices byh2 , one can identify these matrices as representing angular momentum operatorswhere the total angular momentum corresponds to l = 1/2. The spin angularmomentum operators are defined in terms of the Pauli matrices via

Si =h

2σi (1933)

where the index i can take on the values x, y and z.

The Pauli spin operators act on the space of two-component column vectors,Ψ,

Ψ =

(Ψ+

Ψ−

)(1934)

412

An arbitrary operator A given by

A =

(A1,1 A1,2

A2,1 A2,2

)(1935)

transforms the state Ψ according to the laws of matrix multiplication. That is,A acting on the state Ψ produces another state given by

A Ψ =

(A1,1 A1,2

A2,1 A2,2

) (Ψ+

Ψ−

)

=

(A1,1 Ψ+ + A1,2 Ψ−A2,1 Ψ+ + A2,2 Ψ−

)(1936)

A vector Ψ† which is the dual of the column vector Ψ is defined as the complexconjugate of the row vector

Ψ† =

(Ψ∗

+ Ψ∗−

)(1937)

An inner product can be defined for any two vectors on this space as the complexnumber formed from the components of the vectors

Φ† Ψ =

(Φ∗+ Φ∗−

) (Ψ+

Ψ−

)

=(

Φ∗+ Ψ+ + Φ∗− Ψ−

)(1938)

which, again, uses the laws of matrix multiplication. Physical states Ψ are tobe normalized such that

Ψ† Ψ = 1 (1939)

which results in the normalization condition(| Ψ+ |2 + | Ψ− |2

)= 1 (1940)

The existence of the inner product allows one to define the adjoint or Her-mitean conjugate of the operator A as the operator A† which has the effect

Φ† A Ψ =

(Φ∗+ Φ∗−

) (A1,1 A1,2

A2,1 A2,2

) (Ψ+

Ψ−

)

=

(Ψ+ Ψ−

) (A1,1 A2,1

A1,2 A2,2

) (Φ∗+Φ∗−

)

=(

Ψ† A† Φ)∗

(1941)

413

Hence, the Hermitean conjugate of A is given by the transpose of the complexconjugate of A

A† =

(A∗1,1 A∗1,2

A∗2,1 A∗2,2

)(1942)

A Hermitean matrix is a matrix for which A = A†.

The Pauli spin operators are Hermitean, and hence can represent physicalquantities such as the components of the angular momentum. The vector spinangular momentum is defined as

S =h

2σ

= exh

2σx + ey

h

2σy + ez

h

2σz (1943)

and the magnitude is given by

S2 = S2x + S2

y + S2z

=h

4

2

σ2x +

h

4

2

σ2y +

h

4

2

σ2z

=34h2 σ0 (1944)

The unit matrix σ0 is the identity operator as, when it acts on an arbitrarystate, Ψ, it has the effect

σ0 Ψ = Ψ (1945)

and so the eigenvalues of the unit matrix are unity. As the eigenvalues of themagnitude of the spin angular momentum are s ( s + 1 ) h2, this means thats = 1

2 . The Pauli spin operators only act on the space formed by s = 12 .

This space is two-dimensional as the eigenvalues of S2 have a degeneracy of( 2 s + 1 ) = 2. Thus, there are two independent basis states which can bechosen as the two column vectors that are the eigenvectors of σz

σz χ± = ± χ± (1946)

The eigenvector corresponding to the up-spin state with spin eigenvalue of + h2

is

χ+ =

(10

)(1947)

while the eigenvector corresponding to the down-spin state, with spin eigenvalue− h

2 is represented by

χ− =

(01

)(1948)

414

It should be noted that any pair of Pauli matrices, with different indices i andj, anti-commute. That is

σi σj + σj σi = 2 σ0 δi,j (1949)

The raising and lowering operators are defined as

σ± = σx ± i σy (1950)

which yields the representations

σ+ =

(0 20 0

)(1951)

and

σ− =

(0 02 0

)(1952)

——————————————————————————————————

4.8.1 Exercise 104

Prove that the above two column vectors of eqn(1947) and eqn(1948) are eigen-states of the z component of the spin operator Sz = h

2 σz, and find the effectof the raising and lowering operators on these states.

——————————————————————————————————

4.8.2 Solution 104

The column vector χ+ is an eigenvalue of σz, since it satisfies the eigenvalueequation

σz χ+ =

(1 00 − 1

) (10

)

=

(10

)= χ+ (1953)

with eigenvalue 1. Also, the column vector χ− is an eigenvalue of σz, since itsatisfies the eigenvalue equation

σz χ− =

(1 00 − 1

) (01

)

415

=

(0− 1

)

= −

(01

)= − χ− (1954)

with eigenvalue − 1. Hence, χ± are eigenstates of Sz with eigenvalues of ± h2 .

The effect of the raising operator, S+ = Sx + i Sy on the eigenstates ofSz with eigenvalue − h

2 is found from the effect of σ+

S+ =h

2σ+ =

h

2

(σx + i σy

)(1955)

acting on the eigenstates. This leads to

σ+ χ− =

(0 20 0

) (01

)

=

(20

)= 2 χ+ (1956)

This shows that the raising operator increases the eigenvalue from − h2 to + h

2 .When the raising operator acts on the state χ+ it has the effect

σ+ χ+ =

(0 20 0

) (10

)

=

(00

)= 0 (1957)

The raising operator annihilates the state with the maximal eigenvalue of Sz,since by definition it cannot raise it any further.

The effect of the lowering operator, σ− = σx − i σy on the eigenstates ofSz are given by

σ− χ+ =

(0 02 0

) (10

)

= 2

(01

)(1958)

416

which lowers the eigenvalue. However, since the state χ− has the lowest eigen-value of σz one finds

σ− χ− =

(0 02 0

) (01

)

=

(00

)= 0 (1959)

Thus, the lowering operator produces zero when it acts on the state with thelowest eigenvalue.

These equations can be summarized as

S+ χ− = h χ+

S+ χ+ = 0S− χ+ = h χ−

S− χ− = 0 (1960)

——————————————————————————————————

4.8.3 Exercise 105

Show that any two by two matrix can be expressed as a linear combination of thePauli matrices and the unit matrix. Also show that any two-component columnvector Ψ can be represented as the linear superposition of the two eigenstatesof σz, where

Ψ =

(Ψ+

Ψ−

)(1961)

——————————————————————————————————

4.8.4 Solution 105

Consider an arbitrary two by two matrix(A BC D

)(1962)

417

This can be expressed as the sum of four two by two matrices each with justone non-zero element(

A BC D

)= A

(1 00 0

)+ B

(0 10 0

)

+ C

(0 01 0

)+ D

(0 00 1

)(1963)

The four by four matrices with components on the diagonal can be expressedin terms of the sum of the unit matrix σ0 and σz via(

1 00 0

)=

12

[ (1 00 1

)+

(1 00 − 1

) ](1964)

while the matrix(0 00 1

)=

12

[ (1 00 1

)−

(1 00 − 1

) ](1965)

can be expressed as the difference.

The off-diagonal matrices can be expressed in terms of σx and σy via(0 10 0

)=

12

[ (0 11 0

)+ i

(0 − ii 0

) ](1966)

while the matrix(0 01 0

)=

12

[ (0 11 0

)− i

(0 − ii 0

) ](1967)

can be expressed as the difference.

Hence, we have the result that the arbitrary two by two matrix can beexpressed as(

A BC D

)=

12

(A+D ) σ0 +12

(A−D ) σz +12

(B + C ) σx +i

2(B − C ) σy

(1968)This proves that the Pauli matrices and the identity span the linear space formedby two by two matrices.

The states χ+ and χ− are eigenstates of σz. Since σz is a Hermitean operator,the eigenstates form a complete set. Thus, χ+ and χ− spans the space of the

418

column vectors as an arbitrary vector can be expressed as

Ψ =

(Ψ+

Ψ−

)

=

(Ψ+

0

)+

(0

Ψ−

)

= Ψ+

(10

)+ Ψ−

(01

)= Ψ+ χ+ + Ψ− χ− (1969)

as was expected.

——————————————————————————————————

4.8.5 Exercise 106

Prove the identity(σ . a

) (σ . b

)= σ0

(a . b

)+ i σ .

(a ∧ b

)(1970)

in which the spin vector is written in terms of the Pauli matrices as

σ = ex σx + ey σy + ez σz (1971)

——————————————————————————————————

4.8.6 Solution 106

First we note that the Pauli matrices satisfy the commutation relations

[ σi , σj ] = 2 i∑

k

εi,j,k σk (1972)

where εi,j,k is the anti-symmetric Levi-Civita symbol.

Also we note that the Pauli matrices satisfy

σi σj = − σj σi (1973)

if i 6= j, whereas if i = j then

σi σi = σ0 (1974)

419

where σ0 is the unit matrix. These relations can be combined in the anti-commutation relation

σi , σj + = 2 δi,j σ0 (1975)

in which the anti-commutator of two operators A and B is defined as A , B +,where

A , B + = A B + B A (1976)

Hence, on adding the commutation and anti-commutation relations, one hasthe identity

σi σj = δi,j σ0 + i∑

k

εi,j,k σk (1977)

The identity to be proved follows immediately by expressing the scalar prod-uct in terms of the sum of Cartesian components(

σ . a

) (σ . b

)=

∑i

σi ai

∑j

σj bj

=∑i,j

σi σj ai bj

=∑i,j

(δi,j σ0 + i

∑k

εi,j,k σk

)ai bj

=∑i,j

δi,j σ0 ai bj + i∑

k

σk

∑i,j

εi,j,k ai bj

= σ0

(a . b

)+ i

∑k

σk

(a ∧ b

)k

= σ0

(a . b

)+ i σ .

(a ∧ b

)(1978)

This completes the proof.

——————————————————————————————————

4.8.7 Exercise 107

Find all the eigenfunctions and eigenvalues for the Pauli matrix σx, and showthat they form a complete orthonormal set.

——————————————————————————————————

420

4.8.8 Solution 107

The matrix σx is given by

σx =

(0 11 0

)(1979)

and the eigenfunctions satisfy the eigenvalue equation

σx χx = λ χx (1980)

The eigenfunctions are represented by column vectors of the form

χx =

(αβ

)(1981)

which are normalized when

| α |2 + | β |2 = 1 (1982)

The eigenvalue equations are expressed as the coupled algebraic equations forthe components of Ψ as(

0 11 0

) (αβ

)= λ

(αβ

)(1983)

or (− λ 11 − λ

) (αβ

)= 0 (1984)

This equation only has a solution if the inverse matrix does not exist. Sincethe inverse matrix is inversely proportional to the determinant, the determinantmust vanish for there to be a non-trivial solution for the components. Therefore,the eigenvalue λ is given by the solution of∣∣∣∣∣ − λ 1

1 − λ

∣∣∣∣∣ = 0 (1985)

orλ2 = 1 (1986)

which is consistent with the relation

σ2x = σ0 (1987)

The eigenvalues of σx, are given by

λ = ± 1 (1988)

421

The eigenfunction χx+1 corresponding to λ = 1 is found from(

0 11 0

) (αβ

)=

(αβ

)(1989)

which yields α = β. Thus, the normalized eigenfunction is given by

χx+1 =

1√2

(11

)(1990)

up to an arbitrary phase.

The eigenfunction corresponding to λ = − 1 is found from(0 11 0

) (αβ

)= −

(αβ

)(1991)

which yields α = − β. Thus, the normalized eigenfunction is given by

χx−1 =

1√2

(1− 1

)(1992)

up to an arbitrary phase.

The eigenstates of σx are orthogonal as

χx+1

† χx−1 =

12

(1 1

)(1− 1

)

=12

(1 − 1

)= 0 (1993)

Furthermore, since the space of column vectors is two-dimensional, the set ofeigenfunctions are complete. That is, any arbitrary vector

Ψ =(

αβ

)(1994)

can be expressed as(αβ

)=

( α + β )√2

1√2

(11

)+

( α − β )√2

1√2

(1− 1

)(1995)

422

where the expansion coefficients were determined from the inner products withthe eigenstates

χx+1

† Ψ =( α + β )√

2(1996)

and

χx−1

† Ψ =( α − β )√

2(1997)

Thus, we have shown that an arbitrary state can be expanded in terms of theeigenstates of σx as

Ψ =( α + β )√

2χx

+1 +( α − β )√

2χx−1 (1998)

——————————————————————————————————

Thus, the Pauli matrices form a representation of half-integer angular mo-mentum, called spin. The spin quantum numbers are intrinsic to the particleand, therefore, the operators cannot be represented in terms of the position andmomentum of the particle. The spin of a particle first became manifest in theanomalous Zeeman effect, in which there is a coupling between the spin stateand an external magnetic field.

——————————————————————————————————

4.8.9 Exercise 108

Find the eigenvalues and eigenvectors of the component of the spin along theunit vector η, where

η = sin θ cosϕ ex + sin θ sinϕ ey + cos θ ez (1999)

Work in the basis in which σz is diagonal.

Hence, find the unitary matrix Ue(θ) which produces a rotation of the spinstates through an angle θ, about the axis

e = − sinϕ ex + cosϕ ey (2000)

perpendicular to the plane of ez and η.

Hint: In addition to producing the transformation of the spin state “aligned”parallel to the axis z to the spin state “aligned” with η, it is also required thatthe rotation does not change the “direction” of spin states aligned parallel tothe rotation axis. This is indicated schematically in figs(105) and (106) for a

423

Rotation of a set of Spin States

Rotation angle θη

φ

z

x

y

ε axis of rotn.

Sη = 0

Sη = 1

Sη = - 1

Figure 105: A spin state with eigenvalue of Sz equal to +h is to be transformedto a spin state which an eigenstate of Sη, with eigenvalue of +h.

spin one particle.

——————————————————————————————————

4.8.10 Solution 108

We seek the eigenvalues and eigenfunctions of the operator

η . σ = sin θ cosϕ σx + sin θ sinϕ σy + cos θ σz (2001)

which has the matrix representation

η . σ = h

(cos θ sin θ exp[ − i ϕ ]

sin θ exp[ + i ϕ ] − cos θ

)(2002)

in the basis where σz is diagonal. The eigenvalues µ h are determined from thesecular equation

0 =

∣∣∣∣∣ cos θ − µ sin θ exp[ − i ϕ ]sin θ exp[ + i ϕ ] − cos θ − µ

∣∣∣∣∣ (2003)

424

Rotation of a set of Spin States

θη

φ

z

x

y

ε

Sε eigenstates are invariant

Sε = 1

Sε = - 1

Sε = 0

Figure 106: It is also required that a spin rotation around the axis e does notchange the eigenstates of Se.

and, thus, are found to be µ = ± h. The eigenfunction χηµ corresponding to

the eigenvalue µ, is expressed in terms of its components through

χηµ =

(αη

µ

βηµ

)(2004)

The components for µ = + 1 are found from the eigenvalue equation(00

)=(

( cos θ − 1 ) sin θ exp[ − i ϕ ]sin θ exp[ + i ϕ ] − ( cos θ + 1 )

) (αη

+

βη+

)(2005)

Thus, the normalized eigenfunction of the operator in eqn(2001) correspondingto the µ = 1 eigenvalue is found as

χη+ =

(cos θ

2

sin θ2 exp[ + i ϕ ]

)(2006)

Likewise, one finds the eigenfunction corresponding to µ = − 1 is given by

χη− =

(− sin θ

2 exp[ − i ϕ ]cos θ

2

)(2007)

425

The rotation is about an axis in the plane perpendicular to the unit vectorsez and eη. The matrix Ue(θ) is determined from its action on the two eigenstatesof σz

χηµ = Ue(θ) χz

µ (2008)

Thus, we have(αη

µ

βηµ

)=

(cos θ

2 − sin θ2 exp[ − i ϕ ]

sin θ2 exp[ + i ϕ ] cos θ

2

) (αz

µ

βzµ

)(2009)

On recognizing that the component of the spin operator along the rotation axisis given by(

e . σ

)=(

0 − i exp[ − i ϕ ]+ i exp[ + i ϕ ] 0

)(2010)

and that the square is the unit matrix(e . σ

)2

=(

1 00 1

)= σ0 (2011)

we find that the spin rotation operator may be written in the form

Ue(θ) = cosθ

2σ0 − i sin

θ

2

(e . σ

)(2012)

Hence, the spin rotation operator is given by

Ue(θ) = exp[− i

θ

2e . σ

]= exp

[− i θ

he . S

](2013)

where the spin is defined as

S =h

2σ (2014)

——————————————————————————————————

4.8.11 The Pauli Equation

The existence of spin was first inferred by Uhlenbeck and Goudschmit36 as aresult of experiments where atoms were placed in a strong magnetic field and

36G.E. Uhlenbeck and S.A. Goudsmit, Naturwissenschaften, 13, 953 (1925) and Nature,117, 264 (1926).

426

the absorption lines were measured. In the ordinary Zeeman effect, the orbitalangular momentum of a charged electron produces a magnetic moment whichcouples to the field and, therefore, raises the degeneracy of the electronic levels.This results in a splitting of the absorption lines, when the field is applied. Thenumber of lines that are seen is equal to the degeneracy of the atomic levels inthe absence of the field which are labeled by n and l. In general, the degeneracyis expected to be given by 2 l + 1 corresponding to the different m values.However, due to the presence of spin, there are more lines seen than predictedby the spin-less version of the Schrodinger equation. Even the l = 0 state issplit into two levels corresponding to a degeneracy of 2. On introducing a spinoperator S representing spin angular momentum and equating the degeneracy2 S + 1 = 2, one finds that the spin must be half-integer S = 1

2 .

The Pauli equation37 assumes that the wave function for a spin half particlehas two components

Ψ(r) =

(Ψ+(r)Ψ−(r)

)(2015)

and expresses the kinetic energy operator of the free particle in terms of theproduct of a Laplacian and the unit two by two matrix,

T =p2

2 m(2016)

which can be re-written as

T =( p . σ ) ( p . σ )

2 m(2017)

by using the identity(σ . a

) (σ . b

)= σ0

(a . b

)+ i σ .

(a ∧ b

)(2018)

with a = b = p. Thus, the time-dependent Schrodinger equation for thetwo-component (column vector) wave function Ψ(r) representing a free particlecan be written in the form of the field-free Pauli equation

i h∂

∂tΨ(r, t) =

[− h2

2 m( σ . ∇ )2

]Ψ(r, t) (2019)

The corresponding equation for a charged particle in an electromagnetic field isobtained by using the minimum coupling assumption pµ → pµ − q

c Aµ. Theresulting equation is found as

i h∂

∂tΨ(r, t) =

[1

2 m

(σ . ( − i h ∇ − q

cA(r, t) )

)2

+ σ0 q φ(r, t)]

Ψ(r, t)

(2020)37W. Pauli, Z. f. Physik, 43, 601 (1927).

427

which, on using the identity, becomes

i h∂

∂tΨ(r, t) =

[1

2 mσ0

(− i h ∇ − q

cA(r, t)

)2

− h q

2 m cσ .

(∇ ∧ A(r, t) + A(r, t) ∧ ∇

)+ σ0 q φ(r, t)

]Ψ(r, t) (2021)

Furthermore, since

∇ ∧ A Ψ(r) = Ψ(r)(∇ ∧ A

)− A ∧ ∇ Ψ(r) (2022)

and B = ∇ ∧ A, one finds

i h∂

∂tΨ(r, t) =

[1

2 mσ0

(− i h ∇ − q

cA(r, t)

)2

− h q

2 m cσ . B(r, t) + σ0 q φ(r, t)

]Ψ(r, t) (2023)

Since the spin angular momentum S is given by h2 σ, one finds that the spin

couples to the magnetic field giving rise to the anomalous Zeeman effect, butthe relation between spin angular momentum and magnetic moment M is

M = 2(

q

2 m c

)S (2024)

which is different from the relation between the orbital angular momentum Land the magnetic moment found in the ordinary Zeeman effect

M =(

q

2 m c

)L (2025)

The extra factor of 2 found for the spin half particle is known as the gyro-magnetic ratio, g, and the magnitude of the factor q h

2 m c is known as the Bohrmagneton µB .

4.8.12 Spin Dynamics

We shall consider the dynamics of a particle with spin S = h2 that is localized

at some point in space, when a time-independent magnetic field B is applied tothe system. The direction of the field may be taken to define the direction ofthe z axis of our Cartesian coordinate system. We shall only be concerned withthe dynamics of the spin and, therefore, suppress any mention of the coordinatedependence of the wave function.

428

The wave function is then represented as a linear superposition of spin-upand spin-down states

Ψ(t) =

(Ψ+(t)Ψ−(t)

)(2026)

where Ψ±(t) are the time-dependent expansion coefficients. The Pauli equationis then given by the matrix equation

i h∂

∂tΨ(t) = − µB Bz σz Ψ(t) (2027)

where the Hamiltonian only consists of the Zeeman term. Since the energyeigenstates correspond to eigenstates of the z component of the spin, Sz = h

2σz,one may decompose the time-dependent wave function into energy eigenstates

Ψ(t) = Ψ+(t) χ+ + Ψ−(t) χ− (2028)

On substituting this into the time-dependent Schrodinger equation, one findsthat

i h∂

∂tΨ±(t) = ∓ µB B Ψ±(t) (2029)

which have the solutions

Ψ±(t) = exp[± i

µB B t

h

]Ψ±(0) (2030)

Hence, we have found the time dependence of our arbitrary initial spin wavefunction Ψ(0) as

Ψ(t) = exp[

+ iµB B t

h

]Ψ+(0) χ+ + exp

[− i

µB B t

h

]Ψ−(0) χ− (2031)

If the initial state was known at t = 0 to be an eigenstate of, say Sx with theeigenvalue + h

2 , then one has

Ψ(0) =1√2

(χ+ + χ−

)(2032)

or

Ψ(0) =1√2

(11

)(2033)

Thus, we have found that the time dependence of the wave function at latertimes, t, is given by

Ψ(t) =1√2

(exp

[+ i

µB B t

h

]χ+ + exp

[− i

µB B t

h

]χ−

)(2034)

429

From this one can see that the probability of finding the particle in an eigenstateof Sx with eigenvalue + h

2 is going to oscillate with time, as is the probabilitythat the particle is in the Sx eigenstate with eigenvalue − h

2 . This oscillationcorresponds to the classical precession of a spin around the z axis, with a fre-quency given by ω = 2 µB B

h , known as the Larmor precession frequency.

B Μ

Figure 107: A classical magnetic moment M in an applied magnetic field Bexperiences a torque τ = M ∧ B and, hence, precesses around the direction ofthe field.

——————————————————————————————————

4.8.13 Exercise 109

Evaluate the time dependence of the probability that a spin 12 in a magnetic

field B aligned along the z - axis is found in the eigenstates of Sy, if the initialstate is an eigenstate of Sx with eigenvalue + h

2 .

——————————————————————————————————

4.8.14 Solution 109

The initial state is an eigenstate of Sx with eigenvalue + h2 . The initial state

can be represented as

Ψ(0) =1√2

(11

)(2035)

430

which evolves with time as

Ψ(t) = exp[− i t

hH

]Ψ(0) (2036)

where the Hamiltonian is given by

H = − µB B σz (2037)

The time-dependent state is given by

Ψ(t) =1√2

exp[i t µB B

h

] (10

)+

1√2

exp[− i t µB B

h

] (01

)(2038)

The matrix σy has eigenvalues of ± 1 and the eigenvalues are given by

χy±1 =

1√2

(1± i

)(2039)

respectively. The state Ψ(t) is expanded in terms of the eigenstates χy± as

Ψ(t) =∑±

C±(t) χy± (2040)

where

C±(t) =12

(exp

[i t µB B

h

]∓ i exp

[− i t µB B

h

] )(2041)

Thus, we have the time-dependent probability given by

P y±(t) =

12

(1 ∓ sin

2 µB B t

h

)(2042)

——————————————————————————————————

4.8.15 Exercise 110

A spin half nucleus is placed in a static magnetic field B0 aligned along the zaxis and a smaller rotating magnetic field B1 in the x - y plane. The frequencyof the a.c. field is ω. If the nucleus is initially pointing in the + z direction att = 0, what is the probability that it will be aligned with the + z axis at latertimes?

431

Experiments were suggested and preformed by I.I. Rabi and co-workers38

which were based on calculations of the above type.

——————————————————————————————————

4.8.16 Solution 110

The spin vector can be written as

Ψ(t) =

(α(t)β(t)

)(2043)

and the Schrodinger equation is

i h∂

∂tΨ(t) = − µN

(B0 σz + B1 σx cosωt − B1 σy sinωt

)Ψ(t)

(2044)

or

i∂

∂tΨ(t) = −

(µN B0

hµN B1

h exp[ + i ω t ]µN B1

h exp[ − i ω t ] − µN B0h

)Ψ(t)

(2045)We should note that the sense of the rotation of B1 about the z-axis is suchthat the axis of rotation is anti-parallel to B0. We are also implicitly assumingthat µN is positive. The equations for the components of the spinors are

i∂

∂tα(t) = −

(µN B0

h

)α(t) −

(µN B1

h

)exp

[+ i ω t

]β(t)

(2046)

and

i∂

∂tβ(t) = +

(µN B0

h

)β(t) −

(µN B1

h

)exp

[− i ω t

]α(t)

(2047)

On solving the first equation for β and substituting this into the second, oneobtains a second order differential equation for α

∂2

∂t2α(t) − i ω

∂

∂tα(t) +

( (µN

h

)2

( B20 + B2

1 ) − ωµN B0

h

)α(t) = 0

(2048)

38I.I. Rabi, Phys. Rev. 51, 652 (1937), I.I. Rabi, J.R. Zacharias, S. Millman and P. Kusch,Phys. Rev. 53, 318 (1938).

432

This is solved by assuming that α(t) has the form

α(t) = α(0) exp[i λ t

](2049)

The characteristic equation is a quadratic equation in λ which can be solved,yielding

λ± = +ω

2±

√(µN B0

h− ω

2

)2

+(g µN B1

h

)2

(2050)

The general solution can be written as

α(t) = a+ exp[i λ+ t

]+ a− exp

[i λ− t

](2051)

The components of β can be written as

β(t) =(b+ exp

[i λ+ t

]+ b− exp

[i λ− t

] )exp

[− i ω t

](2052)

However, b+ and b− are given in terms of the a’s via the original differentialequations. Thus, b+ is linear in a+ and b− is linear in a−

b±a±

=

[( h ω

2 − µN B0 ) ∓

√(h ω2 − µN B0

)2

+(µN B1

)2 ]µN B1

(2053)To satisfy the initial conditions one must have α(0) = a+ + a− = 1 andβ(0) = b+ + b− = 0. These conditions allows one to solve for α(t). Theprobability of finding the spin-up eigenvalue is given by | α(t) |2 and

| α(t) |2 = 1 −

[ (µN B1

)2

(µN B0 − hω

2

)2

+(µN B1

)2

]×

× sin2

√(µN B0

h− ω

2

)2

+(µN B1

h

)2

t (2054)

As the spin one half wave function is normalized and has an overall phase factor,the spin state only depends on two variables. Therefore, the spin one half wavefunction can always be interpreted in terms of a classical state in which thespin has a definite direction. When the frequency ω is away from the resonancefrequency

ω = 2(µN B0

h

)(2055)

433

the spin precesses around the z-axis with frequency ω and the z component alsomakes makes small amplitude oscillations. That is, the spin motion is confinedwithin a small ring parallel the equator of the unit sphere. However, for fre-quencies near the resonance frequency, the z component of the spin performslarge amplitude oscillations and almost flips to the other pole with frequency

2(

µN B1h

).

——————————————————————————————————

4.8.17 The Berry Phase

In the presence of a magnetic field, an isolated spin will find an equilibriumstate which is the ground state. In the ground state the spin is aligned parallelto the field. If the direction of the field is subsequently changed, sufficientlyslowly, one expects that the direction of the spin will continue to align with themagnetic field at every instant of time. That is, one does not expect the slowlytime varying field to cause the spin to make a transition to states of higherenergy. This expectation is borne out by detailed calculations on systems ofvarious spin magnitudes. However, although the spin state does follow the fielddirection, the spin wave function does acquire an additional phase. If the field isreturned to the original direction, the additional phase factor is determined bythe topology of the spin’s motion39. The motion of the spin can be visualizedby projecting the spin direction onto the unit sphere. The direction of the spincan be specified by specifying the polar coordinates (θ(t), ϕ(t)). As the fieldis changed, the spins direction will map out a path on the unit sphere. Theadditional phase, or Berry phase δ, acquired by the spin is determined by thesolid angle Ω enclosed by the spin’s orbit. The Berry phase is determined bythe solid angle and the magnitude of the spin, in units of h. The path of thespin is parameterized as θ(ϕ). Since the infinitesimal solid angle dΩ is given by

dΩ = dϕ dθ sin θ (2056)

on integrating from the pole θ = 0 to θ(ϕ), the solid angle enclosed by theinfinitesimal wedge is

dΩ = dϕ ( 1 − cos θ(ϕ) ) (2057)

Then, the solid angle enclosed by the complete orbit is given by

Ω =∮

dϕ ( 1 − cos θ(ϕ) ) (2058)

where the integral over ϕ runs over 2 π.

39M.V. Berry, Proc. R. Soc. Lond. A 392, 45 (1984).

434

(θ(t),ϕ(t))Ω

Figure 108: The Berry Phase is determined by the solid angle Ω swept out bythe spins orbit on the unit sphere.

The Berry phase can be illustrated by considering a spin one half in a mag-netic field of constant magnitude, B, oriented along the direction (θ, ϕ). In thiscase, the Zeeman Hamiltonian is given by

HZ = − µB ( B . σ )

= − µB B

(cos θ σz + sin θ ( sinϕ σy + cosϕ σx )

)(2059)

which, on using the Pauli-spin matrices, can be expressed as

HZ = − µB B

(cos θ sin θ exp[− i ϕ ]

sin θ exp[ + i ϕ ] − cos θ

)(2060)

For fixed (θ, ϕ), the time-independent Hamiltonian HZ has an eigenstate witheigenfunction given by

χ+ =(

cos θ2

sin θ2 exp[ + i ϕ ]

)(2061)

corresponding to the eigenvalue

E0 = − µB B (2062)

435

Thus, in this eigenstate the spin is aligned parallel to the applied field. For astatic field one has the time-dependent wave function given by

χ+(t) =(

cos θ2

sin θ2 exp[ + i ϕ ]

)exp

[+ i

µB B

ht

](2063)

where the time dependence is given purely by the exponential phase factor.

If the direction of the field (θ(t), ϕ(t)) is changed very slowly, one expects thespin will adiabatically follow the field direction. That is, if the field is rotatedsufficiently slowly, one does not expect the spin to make a transition to thestate with energy E = + µB B where the spin is aligned anti-parallel to thefield. However, the wave function may acquire a phase which is different fromthe time and energy dependent phase factor expected for a static field. Thisextra phase is the Berry phase δ, which can be calculated from the Schrodingerequation

i h∂

∂t

(α(t)β(t)

)= − µB B

(cos θ(t) sin θ(t) exp[− i ϕ(t) ]

sin θ(t) exp[ + i ϕ(t) ] − cos θ(t)

) (α(t)β(t)

)(2064)

We shall assume that the wave function takes the adiabatic form(α(t)β(t)

)=

(cos θ(t)

2

sin θ(t)2 exp[ + i ϕ(t) ]

)exp

[+ i

(µB B

ht − δ(t)

) ](2065)

which instantaneously follows the direction of the field, but is also modifiedby the inclusion of the Berry phase δ(t). On substituting this ansatz into theSchrodinger equation, one finds that the non-adiabatic terms satisfy

− ∂δ

∂t

(cos θ(t)

2

sin θ(t)2 exp[ + i ϕ(t) ]

)+

∂ϕ

∂t

(0

sin θ(t)2 exp[ + i ϕ(t) ]

)

=i

2∂θ

∂t

(− sin θ(t)

2

cos θ(t)2 exp[ + i ϕ(t) ]

)(2066)

The above equation is then projected onto the adiabatic state by multiplying itby the row matrix (

cos θ(t)2 sin θ(t)

2 exp[ − i ϕ(t) ])

(2067)

One finds that the derivative of θ w.r.t. t cancels and that the equation simplifiesto

− ∂δ

∂t+

∂ϕ

∂tsin2 θ

2= 0 (2068)

436

Hence, the Berry phase is given by integrating w.r.t. to t,

δ(t) =∫ t

0

dt′∂ϕ

∂t′sin2 θ(t

′)2

=∫

dϕ sin2 θ(ϕ)2

=12

∫dϕ ( 1 − cos θ(ϕ) ) (2069)

On completing one orbit in spin space, the extra phase is given by

δ =12

Ω (2070)

as was claimed.

437

4.9 Transformations and Invariance

The assumption underlying the application of quantum mechanics to physicalsystems is that the basic structure and predictions of experimental results onquantum mechanical systems are independent of the inertial coordinate systemused. If we consider space-time which is governed by Galilean invariance, timebeing absolute and space being isotropic and governed by the laws of Euclideangeometry, we have non-relativistic quantum mechanics. Transformations be-tween different coordinate systems (passive transformations) leave the physicalsystem unaltered and merely provide relations between reference systems of dif-ferent observers. Alternately, two different physical systems may be related toeach other by a transformation of the physical position of the physical systemand measuring devices. Under an active transformation such as a physical ro-tation of one system, the two systems may be brought into coincidence.

Equivalent quantum mechanical systems can be transformed into each other,via a unitary transformation associated with a unitary operator U . The unitaryoperator is defined as an operator that has a Hermitean conjugate which isidentical to the inverse operator

U† = U−1 (2071)

This has a consequence that the eigenvalues of a unitary operator are complexnumbers of magnitude unity, and the eigenfunctions form an orthonormal set.This can be seen by considering the matrix elements of U−1 U between twodifferent eigenfunctions φλ1 and φλ2∫

d3r φ∗λ2(r) U−1 U φλ1(r) = λ∗2 λ1

∫d3r φ∗λ2

(r) φλ1(r)

=∫

d3r φ∗λ2(r) φλ1(r) (2072)

since U is unitary. Thus, we have(λ∗2 λ1 − 1

) ∫d3r φ∗λ2

(r) φλ1(r) = 0 (2073)

Hence, the eigenvalues have modulus unity as

λ∗1 λ1 = 1 (2074)

and eigenfunctions corresponding to different eigenvalues are orthogonal.

Physically equivalent states are related via unitary transformations. Undera unitary transformation the states are transformed according to

Ψ(r) → Ψ′(r) = U Ψ(r) (2075)

438

which preserves the normalization of Ψ′ as∫d3r Ψ′∗(r) Ψ′(r) =

∫d3r Ψ∗(r) U† U Ψ(r)

=∫

d3r Ψ∗(r) Ψ(r)

= 1 (2076)

The operators are transformed according to

A → A′ = U A U† (2077)

as the expectation values of the transformed operators sandwiched between thetransformed states leads to the same expectation values as the untransformedoperators in the untransformed states∫

d3r Ψ′∗(r) A′ Ψ′(r) =∫

d3r Ψ∗(r) A Ψ(r) (2078)

4.9.1 Time Translational Invariance

The time translational operator U(t, t0) is a unitary operator which is definedby its action on a wave function

U(t, t0) Ψ(r, t0) = Ψ(r, t) (2079)

Since the wave function Ψ(r, t) has the same normalization as Ψ(r, t0), the timeevolution operator is a unitary operator as it is norm conserving.

The time translational operator can be expressed in terms of the time deriva-tive, since

exp[

( t − t0 )∂

∂t0

]Ψ(r, t0) = Ψ(r, t) (2080)

Thus, we find that the operator∂

∂t0(2081)

is the infinitesimal generator of time translations and the time translationaloperator can be expressed as

U(t, t0) = exp[

( t − t0 )∂

∂t0

](2082)

The time dependence of the Schrodinger equation is governed by the Hamilto-nian H, and if this is time independent then

i h∂

∂t≡ H (2083)

439

which gives the explicit term for the time translational operator

U(t, t0) = exp[− i

( t − t0 )h

H

](2084)

The time translational operator only depends on the time difference t − t0.

The time translation operators can be compounded, thereby forming a groupoperation

U(t2 − t1) U(t1 − t0) = U(t2 − t0) (2085)

since U only depends on the time differences t2 − t1. The identity operator is

U( 0 ) = 1 (2086)

The inverse of U( t ) isU−1( t ) = U( − t ) (2087)

such thatU( t ) U( − t ) = U( 0 ) (2088)

If a physical quantity is represented by a time-independent operator A inthe Schrodinger picture, and its expectation values are given by∫

d3r Ψ∗(r, t) A Ψ(r, t) =∫

d3r Ψ∗(r, t0) U†(t, t0) A U(t, t0) Ψ(r, t0)

(2089)this is independent of time if

U†(t, t0) A U(t, t0) = A (2090)

which is satisfied if the operator A commutes with the Hamiltonian H,

[ H , A ] = 0 (2091)

In particular, since time is homogeneous, and hence the Hamiltonian of aclosed system is independent of time, the energy is conserved as H commuteswith itself.

4.9.2 Translational Invariance

A translation of a system by a distance a changes the wave function of thesystem

Ψ(r) → Ψ′(r) = S(a) Ψ(r) (2092)

440

The unitary operator that represents a displacement of a system from r to r + ais given by

S(a) = exp[− a . ∇

](2093)

so

Ψ′(r) = S(a) Ψ(r)

= exp[− a . ∇

]Ψ(r)

= Ψ(r − a) (2094)

Thus, the wave function Ψ′ evaluated at r+a has the same value as Ψ evaluatedat r. The translational operators S(a) can be combined to yield other operatorsof the same type

S(a) S(b) = S(a+ b) (2095)

The translation operators form a representation of a group. The group isAbelian and it is a continuous group.

Translation through distance a

0

0.2

0.4

0.6

0.8

1

Ψ(x) Ψ(x-a)

x0 x0+a0 01 2 23 1 3

x x

Ψ'(x)=Ψ(T-1x)=Ψ(x-a)

x'=Tx=x+a

Figure 109: The translation of an arbitrary state Ψ(r) through a distance a.

For an infinitesimally small displacement ξ, one has

S(ξ) = 1 − i ξ .

(− i ∇

)+ O(ξ2)

= 1 − i

hξ . p + O(ξ2) (2096)

441

so − i ∇ is the generator of infinitesimal transformations. The generator ofinfinitesimal transformations corresponds to the momentum operator.

If a system is invariant under displacements then the expectation values ofphysical operators in these states should be the same for Ψ as Ψ′. In particular,the Hamiltonian is invariant under the transformation

S(ξ)† H S(ξ) = H (2097)

From the infinitesimal translation, we find that p and H must commute

[ p , H ] = 0 (2098)

Hence, we find p is a constant of motion. Thus, for translationally invariantsystems the momentum is conserved.

If we translate the state of the system and the measuring devices then, asspace is homogeneous, this is equivalent to a passive transformation in which thesystem remains unchanged but is represented in a different coordinate system.The measuring devices in this new coordinate system are represented by theoperators C ′ which are given in terms of the operators of the old coordinatesystem via

C ′ = S(a) C S†(a) (2099)

The momentum and coordinate operators are transformed accordingly as

p′ = S(a) p S†(a)

= exp[− a . ∇

]p exp

[+ a . ∇

]= p

and

r′ = S(a) r S†(a)

= exp[− a . ∇

]r exp

[+ a . ∇

]= r − a (2100)

For an active transformation, in which only the system is translated and themeasuring systems are kept in place, the corresponding operators are those ofthe original system.

When the homogeneity of space is disturbed, for example by the applicationof a magnetic field, the translation operators take a different form. For a uniformmagnetic field and no electrostatic potential, the system should be uniform and

442

so the translation operator should commute with the Hamiltonian. In this case,the correct generator of the translation is given by the pseudo-momentum

K = p − q

cA +

q

cB ∧ r (2101)

where the sum of the first two terms is recognized as the canonical momentum.The translation through a distance a is given by

S(a) = exp[− i

ha . K

](2102)

The components of the pseudo-momenta do not commute, as

[ Ki , Kj ] = − i hq

cεi,j,k Bk (2103)

and so the translation operators also do not commute

S(a) S(b) = S(b) S(a) exp[

+ iq

h cB . ( a ∧ b )

](2104)

The non-commutivity of the translational operators is a manifestation of thepresence of an Aharonov - Bohm phase.

4.9.3 Periodic Translational Invariance

Consider a potential V (x) which is periodic in translations through any integermultiple of a discrete repeat distance a. That is,

V (x+ na) = V (x) (2105)

This potential is periodic in x with period a. We seek eigenvalues of the energyeigenvalue equation

H φ(x) = E φ(x)[p2

2 m+ V (x)

]φ(x) = E φ(x) (2106)

The unitary operator S(na) defined by

S(na) = exp[− i n a

hp

](2107)

produces a translation of the wave function by a distance n a.

φ(x− na) = S(na) φ(x)

= exp[− i n a

hp

]φ(x) (2108)

443

since the operator generates a Taylor-MacLaurin expansion in the displacementn a. Furthermore, a displacement of (n+m) a can be generated by successivedisplacements of n a and m a. The unitary transformation corresponding tothe combined displacement can be expressed as

S((n+m)a) = S(na) S(ma) = S(ma) S(na) (2109)

Thus, the unitary operators corresponding to translations through any integermultiple of the basic periodicity a commute.

Since under the displacement the potential transforms as

V (x− na) = S(na) V (x) S†(na) (2110)

and as the potential is periodic

V (x− na) = V (x) (2111)

one finds thatS(na) V (x) S†(na) = V (x) (2112)

and so S(na) commutes with the Hamiltonian, as

H = S(na) H S†(na) (2113)

or, on post multiplying by S(na), one has

H S(na) = S(na) H (2114)

Thus, the set of operators H, and S(na), form a set of mutually commutingoperators and can be diagonalized.

We seek a simultaneous solution of the set of eigenvalue equations

H φ(x) = E φ(x)S(na) φ(x) = λn φ(x) (2115)

for all n. The eigenvalues of a unitary operator must have modulus of unity, soone has

| λn |2 = 1 (2116)

Since, the discrete translation operator satisfies the equation

S((n+m)a) = S(na) S(ma) (2117)

the eigenvalues must be related by the multiplicative relation

λn+m = λn λm (2118)

444

Hence, the eigenvalues may be written as

λn = exp[i n ϕ

](2119)

where ϕ is a real phase. This is a one-dimensional version of Bloch’s theorem,which states that eigenfunctions of a Hamiltonian with a periodic potential,with periodicity a, can be chosen such that

φ(x− na) = exp[i n ϕ

]φ(x) (2120)

Therefore, by demanding that φ(x) is an eigenfunction of S(na) we have ensuredthat the solution is finite at x → ± ∞.

——————————————————————————————————

Example

As an example, consider a periodic potential composed of repulsive deltafunctions of strength V0 a,

V (x) =n=+∞∑n=−∞

V0 a δ( x − n a ) (2121)

Classically, a particle of energy E > 0 would be free to move in the regionsbetween the repulsive delta functions. Quantum mechanically, we expect theparticle to be able to tunnel through the narrow potential barrier.

——————————————————————————————————

Solution

The differential equation can be solved in the region between the barriers,say, in the region n a < x < ( n + 1 ) a the eigenfunction has the form

φ(x) = An exp[i k x

]+ Bn exp

[− i k x

](2122)

corresponding to a linear superposition of a forward and backward travellingwaves. The wave vector k is related to the energy through

E =h2 k2

2 m(2123)

The wave function in the region ( n + 1 ) a < x < ( n + 2 ) a also has thesame form

φ(x) = An+1 exp[i k x

]+ Bn+1 exp

[− i k x

](2124)

445

V(x)

0

5

10

15

20

-4 -3 -2 -1 0 1 2 3 4

x/a

Figure 110: A periodic array of repulsive delta function potentials.

The pairs of coefficients (An, Bn) and (An+1, Bn+1) on each side of the deltafunction barrier at x = ( n + 1 ) a are related by demanding that the formsalso satisfy the eigenvalue equation at the boundary. Continuity of the wavefunction at x = ( n + 1 ) a yields the equation

(An+1 −An ) exp[i k ( n+ 1 ) a

]+ (Bn+1 −Bn ) exp

[− i k ( n+ 1 ) a

]= 0

(2125)The discontinuity in the derivative at the boundary is related to the strength ofthe delta function barrier through

( An+1 − An ) exp[i k ( n + 1 ) a

]− ( Bn+1 − Bn ) exp

[− i k ( n + 1 ) a

]=

(2 m V0 a

i k h2

) (An exp

[i k ( n + 1 ) a

]+ Bn exp

[− i k ( n + 1 ) a

] )(2126)

This pair of equations can be solved to yield (An+1, Bn+1) in terms of (An, Bn).Thus, on adding the pair of equations one has

2 ( An+1 − An ) exp[i k ( n + 1 ) a

]=

(2 m V0 a

i k h2

) (An exp

[i k ( n + 1 ) a

]+ Bn exp

[− i k ( n + 1 ) a

] )

446

(2127)

while on subtracting them one obtains

2 ( Bn+1 − Bn ) exp[− i k ( n + 1 ) a

]= −

(2 m V0 a

i k h2

) (An exp

[i k ( n + 1 ) a

]+ Bn exp

[− i k ( n + 1 ) a

] )(2128)

The pair of equations may be re-written as

An+1 =(

1 +m V0 a

i k h2

)An +

(m V0 a

i k h2

)exp

[− 2 i k ( n + 1 ) a

]Bn

(2129)and

Bn+1 =(

1 − m V0 a

i k h2

)Bn −

(m V0 a

i k h2

)exp

[+ 2 i k ( n + 1 ) a

]An

(2130)These two equations are not sufficient to determine a solution uniquely (up toan undetermined phase factor) since the ratio of the coefficients An and Bn areunknown. However, Bloch’s theorem gives the additional relation

φ(x− a) = exp[i ϕ

]φ(x) (2131)

which uniquely specifies the simultaneous eigenfunctions.

On using Bloch’s theorem, one finds

φ(x− a) = exp[i ϕ

]φ(x)

= exp[i ϕ

] (An+1 exp

[i k x

]+ Bn+1 exp

[− i k x

] )= An exp

[i k ( x − a )

]+ Bn exp

[− i k ( x − a )

](2132)

which, on equating the coefficients of the independent exponential functions,yields

An+1 = exp[− i ( ϕ + k a )

]An (2133)

and

Bn+1 = exp[− i ( ϕ − k a )

]Bn (2134)

447

This insures that the wave functions do not diverge in the limits x → ±∞. Onusing these relations to eliminate Bn+1 and An+1, one finds the two equations(

m V0 a

i k h2

) (Bn

An

)exp

[− 2 i k ( n + 1 ) a

]= exp

[− i ( k a + ϕ )

]−(

1 +m V0 a

i k h2

)(2135)

(m V0 a

i k h2

)(An

Bn

)exp

[+ 2 i k ( n + 1 ) a

]=

[ (1 − m V0 a

i k h2

)− exp

[+ i ( k a − ϕ )

] ](2136)

The above two equations can be combined to eliminate the ratio of An / Bn,which leads to the equation[

exp[− i (ka+ ϕ)

]−(

1 +m V0 a

i k h2

) ] [ (1 − m V0 a

i k h2

)− exp

[+ i (ka− ϕ)

] ]

=(m V0 a

i k h2

)2

(2137)

On re-arranging this equation, one finds

exp[− i 2 ϕ

]− 2 exp

[− i ϕ

] (cos k a + sin k a

m V0 a

k h2

)+ 1 = 0

(2138)

Multiplying this equation by exp[

+ i ϕ

]leads to

cosϕ =(

cos k a + sin k am V0 a

k h2

)= 1 (2139)

This equation only has real solutions for ϕ, when the right hand side has amagnitude less than unity. In these ranges of k, the above equation can besolved to yield k as a function of ϕ. We note that ϕ is only defined up tomultiples of 2 π. We shall examine the equation graphically, the right-hand sideis plotted as a function of k a in fig(111). As a function of k, the function

F (k) =(

cos k a + sin k am V0 a

k h2

)(2140)

has a maximum at k = 0 with the value

F (0) = 1 +m V0 a

2

h2 (2141)

448

and behaves as cos ka in the asymptotic large k limit. The allowed values of kare those for which cosϕ exists and, therefore, the function F (k) has a magni-tude less than unity. Hence, not all values of k will give rise to a solution. Inparticular, the values of k for which | F (k) | > 1 do not yield a solution. Therange of forbidden k values (at large k) are near k a = µ π, for integer µ.

-2

-1

0

1

2

3

4

5

6

0 1 2 3 4 5 6

k a / π

F(k)

Figure 111: The graphical determination of the allowed ranges of k.

Since, E and k are related via

E =h2 k2

2 m(2142)

and not all k values are allowed, we find that the spectrum of energy eigenvalueshave “gaps” centered near the energies

E =h2 µ2 π2

2 m a2(2143)

The allowed energies form bands, which are separated by band gaps. As k → ∞the band gaps become very narrow. Since the discontinuities become negligibleat high energies, the band energies closely follow the parabolic dispersion rela-tion. Since k is not a good quantum number and due to the gaps in the E(k)relation, the energies are not plotted as E(k). However, ϕ is a good quantumnumber as it determines the eigenvalue of the S(a), thus the bands are usuallyplotted as E(ϕ), against ϕ. Typical energy bands are shown in fig(112). It is

449

seen that there are energy gaps between the bands at the origin ϕ = 0 andat the boundary ϕ = π. The sizes of the gaps diminish and the widths of thebands increase as one successively inspects higher energy intervals.

Dispersion Relation

0

6

12

18

24

30

36

0 0.25 0.5 0.75 1

φ/πφ/πφ/πφ/π

E( φ φφφ

)

Figure 112: The band dispersion relation E(ϕ) plotted as a function of theBloch eigenvalue ϕ.

——————————————————————————————————

4.9.4 Exercise 111

Calculate the energy eigenvalues associated with a potential constructed froma periodic array of attractive delta function potentials

V (x) = −n=∞∑

n=−∞V0 a δ( x − n a ) (2144)

Explicitly, consider the eigenfunctions corresponding to E > 0 and E < 0.

——————————————————————————————————

450

4.9.5 Solution 111

The solution for the states with energy E > 0 and wave vector k can beexpressed as

φ(x) = An exp[i k x

]+ Bn exp

[− i k x

](2145)

in the intervals between the delta functions. The relation between the coeffi-cients An and Bn in successive intervals can be expressed in terms of the Blochlabel ϕ. The Bloch label is given by the solution of the equation

cosϕ =(

cos k a − sin k am V0 a

k h2

)≤ 1 (2146)

Not all values of k yield solutions for ϕ, and as the energy eigenvalues are givenby

E =h2 k2

2 m(2147)

one finds band gaps, for E > 0.

-4

-3

-2

-1

0

1

2

0 1 2 3 4 5 6

k a / ππππ

F(k

)

Figure 113: The graphical determination of the allowed ranges of k.

In addition to the E > 0 solutions, one has solutions which can be writtenas

φ(x) = An exp[

+ κ x

]+ Bn exp

[− κ x

](2148)

451

in each interval. These solutions correspond to negative energy eigenvalues

E = − h2 κ2

2 m(2149)

The eigenvalue of the translation operator can be expressed in terms of ϕ where

cosϕ =(

coshκ a − sinhκ am V0 a

κ h2

)≤ 1 (2150)

This only produces solutions for one finite range of κ values. Hence, we find asingle band with negative energies. The dispersion relation is shown in fig(115).

-4

-3

-2

-1

0

1

2

3

4

0 0.5 1 1.5 2 2.5 3

κκκκ a / ππππ

F( κ κκκ

)

Figure 114: The graphical determination of the allowed ranges of κ.

——————————————————————————————————

4.9.6 Rotational Invariance

A rotation of a system (through an angle θ about an arbitrary oriented axisrepresented by the unit vector e) can be expressed in terms of a unitary operation

Ue(θ) = exp[− i

θ

h

(e . J

) ](2151)

where J is the total angular momentum operator

J = L + S (2152)

452

-5

0

5

10

15

20

25

30

35

0 0.25 0.5 0.75 1

φ/πφ/πφ/πφ/π

E( φ φφφ

)

Figure 115: The band dispersion relation E(ϕ) for the periodic array of at-tractive delta function potentials, plotted as a function of the Bloch eigenvalueϕ.

composed of the sum of the orbital angular momentum and the spin angularmomentum operators.

That Ue(θ) affects a rotation can be seen by its effect on the wave functionof a spinless particle Ψ(r). Consider a rotation about the z axis through aninfinitesimal angle δϕ = ϕ0

N then

Ψ′(r) = Uz(δϕ) Ψ(r)

= exp[− i

δϕ

hLz

]Ψ( z ez + ey y + ex x )

=(

1 − δϕ ( x∂

∂y− y

∂

∂x) + O(δϕ2)

)Ψ(r)

= Ψ(r) + δϕ ( x ey − y ex ) . ∇ Ψ(r) + O(δϕ2)(2153)

and to lowest order in δϕ, we recognize this as a Taylor expansion of

= Ψ(r + δϕ x ey − δϕ y ex) (2154)

When the position vector is represented as

x = r sin θ cosϕ

453

Rotation of a scalar

Ψ(r)

Ψ(R-1r)Ψ

x-axis

y-axis

R

r'=R-1r

Figure 116: The unitary operator Uz(θ) produces a rotation of the scalar wavefunction through an angle of θ around the z-axis. The rotation operator R actson the coordinates and transforms the point r to the point r′, i.e. r′ = R r.Under the rotation, the wave function Ψ(r) is transformed to Ψ′(r), such that thevalue of Ψ′ at the point r′ is the same as the value of Ψ at the point r. Therefore,Ψ′(r′) = Ψ(r) so, Ψ′(r) = Ψ(R−1r). The unitary operator which transformsthe wave function is then determined by Ψ′(r) = U(θ) Ψ(r) = Ψ(R−1r).

y = r sin θ sinϕz = r cos θ (2155)

we find that the transformed wave function is

Ψ′(r) = Uz(δϕ) Ψ(r)= Ψ( z ez + r sin θ (sinϕ − cosϕδϕ) ey + r sin θ (cosϕ + sinϕδϕ) ex )= Ψ( z ez + r sin θ sin(ϕ− δϕ) ey + r sin θ cos(ϕ− δϕ) ex ) (2156)

which has suffered an infinitesimal rotation of δϕ about the z axis. Thus, we findthat the angular momentum operator is the generator for infinitesimal rotations.

A finite rotation through ϕ0 can be built up from N successive infinitesimaltransformations through δϕ = ϕ0/N by taking the limit N → ∞ via

Uz(ϕ0) = limN → ∞

UNz (δϕ)

= limN → ∞

(1 − i

ϕ0

N hLz

)N

454

= exp[− i

ϕ0

hLz

](2157)

This could have been immediately recognized had we used the representation ofboth the operator Lz and the wave function in spherical polar coordinates

Ψ′(r, θ, ϕ) = Uz(ϕ0) Ψ(r, θ, ϕ)

= exp[− i

ϕ0

hLz

]Ψ(r, θ, ϕ)

= exp[− ϕ0

∂

∂ϕ

]Ψ(r, θ, ϕ)

= Ψ(r, θ, ϕ− ϕ0) (2158)

Rotation of a vector through ϕ ϕ ϕ ϕ

r

e x r

e x ( e x r )

ϕ

e

r ( e . r )

Figure 117: The rotation of an arbitrary vector r about an angle ϕ about theaxis e.

An alternate derivation of the effect of a finite rotation through an angleϕ0 about an arbitrary unit vector e can be expressed in terms of its effect onthe position vector. By composition, one can build up the effect of a rotationon an arbitrary wave function. Under a finite rotation, an arbitrary vector rhas simple transformational properties as it behaves like a state with angular

455

momentum l = 1. After the rotation, the vector is given by40

r′ = ( r . e ) e − sinϕ0 e ∧ r + cosϕ0 e ∧(r ∧ e

)= ( r . e ) e − sinϕ0 e ∧ r + cosϕ0

(r − ( e . r ) e

)= r − sinϕ0 e ∧ r + ( cosϕ0 − 1 )

(r − ( e . r ) e

)(2159)

We shall use the identity

e ∧ r =(e ∧ ( r ∧ ∇ )

)r (2160)

and a second identity

e ∧ ( r ∧ e ) = −(e . ( r ∧ ∇ )

)2

r (2161)

Using these two identities, the rotated vector can be expressed as

r′ = r − sinϕ0

(e . ( r ∧ ∇ )

)r + ( 1 − cosϕ0 )

(e . ( r ∧ ∇ )

)2

r

=[

1 − sinϕ0

(e . ( r ∧ ∇ )

)+ ( 1 − cosϕ0 )

(e . ( r ∧ ∇ )

)2 ]r

=[

1 − i sinϕ0

(e . L

h

)− ( 1 − cosϕ0 )

(e . L

h

)2 ]r

=[

1 +∞∑

n=0

( − i ϕ0 )(2n+1)

(2n+ 1)!

(e . L

h

)+

∞∑n=1

( − i ϕ0 )(2n)

(2n)!

(e . L

h

)2 ]r

(2162)

where we have introduced the angular momentum operator. Since the compo-nent (e . r) of r is an eigenstate of ( e . L ) with eigenvalue 0, and since theperpendicular components are eigenstates of ( e . L )2 with eigenvalues + h2,we have (

e . L

h

)3

r =(e . L

h

)r (2163)

40This can be seen by introducing a basis consisting of three orthogonal unit vectors

e

eφ =1

sin θe ∧ r

eθ =1

sin θe ∧ ( e ∧ r )

where cos θ = ( e . r ). In this basis, the vector r has components (cos θ, 0,− sin θ). Therotation through ϕ0 can then be performed in this coordinate system.

456

This identity can be used to insert extra factors of(e . L

h

)2n

(2164)

into the general terms of both the even and odd series so that the powers of Lmatch the powers of ϕ0. The even and odd terms can be combined to form oneseries which exponentiates

r′ =[ ∞∑

n=0

( − i ϕ0 )n

n!

(e . L

h

)n ]r

= exp[− i

ϕ0

he . L

]r (2165)

giving the desired result.

Rotations R can be combined to give other rotations. The rotations forma group. If a rotation is denoted by its axis e and the angle of rotation θ, therotation Re(θ) can be represented on a unit sphere. The rotation axes can belabeled by their points of intersection with the unit sphere. The combinationof two rotations RP1(θ1) and RP2(θ2) can be found from Euler’s constructionshown in figs (118) and (119), and is neatly expressed algebraically in terms ofmultiplication of quaternions. Euler’s construction shows that rotations do notcommute if the rotations do not share the same axis, and so groups of rotationscan be non-Abelian.

We note that if the Hamiltonian of a system is invariant under rotations,then the angular momentum operator commutes with the Hamiltonian. If thisis the case, then angular momentum is conserved.

For a particle with spin, the rotation operator consists of a product of twofactors. The first factor acts on the spatial components and is identical to therotation operator for the spinless particle. The second factor acts (locally) onthe components of the spinor.

For a spin one half particle, the part of the unitary operator that acts onthe spin which represents a rotation through ϕ0 about the axis e is given by

Ue(ϕ0) = exp[− ϕ0

i

2e . σ

]= cos

ϕ0

2σ0 − i sin

ϕ0

2e . σ (2166)

This unitary operator41 acts on the two-component spinor wave function. Weshould note that a rotation by 2 π does not leave the spinor invariant but changes

41An arbitrary spin one half rotation operator can be represented by unitary 2×2 matrices,with determinant +1. Hence, the matrices are the special unitary matrices of SU(2).

457

Euler Construction

O

P1

P2

P3P4 θθθθ3/2

θθθθ2/2

θθθθ1/2

R1π−θπ−θπ−θπ−θ3/2

θθθθ2/2

θθθθ1/2

Figure 118: The Euler construction. The rotation RP1(θ1) through an angleθ1 about the axis OP1 is to be combined with the rotation RP2(θ2) through anangle θ2 about the axis OP2. Two auxiliary great circles are constructed fromP1 which subtend angles θ1

2 to the great circle connecting P1 and P2. Likewise,two more auxiliary great circles are constructed emanating from point P2. Thepoints of intersection of the pairs of auxiliary great circles are labeled as P3 andP4. It is seen that RP1(θ1) shifts P3 to position P4, and is shifted back to P3 bythe subsequent application of RP2(θ2). Hence, P3 is on the axis of the combinedrotation. Also P4 is on the axis of a combined rotation when the componentrotations are combined in the opposite order.

its phase by π. Thus, a spin half particle should be rotated through 4 π beforethe initial state is recovered42.

Thus, for example, a rotation of a spin by ϕ0 about the z-axis is producedby

Uz(ϕ0) =(

exp[ − i ϕ02 ] 0

0 exp[ + i ϕ02 ]

)(2167)

That this operator produces a rotation of the spin state can be verified byletting it act on the eigenstate of the operator η . S with eigenvalue + h

2 . In this

42In a recent experiment, spin one half particles were sent through a two-channel interfer-ometer. A magnetic field was used to rotate the particles in one channel. The interferencepattern was periodic in the rotation angle, with period 4 π. The experiment was described inA.W. Overhauser, A.R. Collela and S.A. Werner, Phys. Rev. Lett. 33, 1237 (1974).

458

Euler Construction

P2

P1P3

θθθθ1/2

θθθθ2222/2

θθθθ3/2

Rπ−θπ−θπ−θπ−θ3/2

θθθθ2222/2

O

Figure 119: The Euler construction. The angle of the combined rotation can befound by considering the effect of the successive rotations on point P1. Underthe first transformation P1 is invariant, since it is on the axis of rotation. Underthe second transformation P1 is swept to point R. Thus, under the combinedrotation around P3, the great circle segment P3P1 is swept to P3R. Since P2P3

bisects the segments P2P1 and P2R, the angle of the combined rotation is foundto be 2(π − θ3

2 ).

expression, η is a unit vector that has the Cartesian components

η =

sin θ cosϕsin θ sinϕ

cos θ

(2168)

The η component of the spin operator is given by

η . S =h

2

(cos θ sin θ exp[ − i ϕ ]

sin θ exp[ + i ϕ ] − cos θ

)(2169)

and the eigenstate with eigenvalue + h2 is given by

Ψη =(

cos θ2 exp[ − i ϕ

2 ]sin θ

2 exp[ + i ϕ2 ]

)(2170)

This eigenstate describes a spin pointing along the positive η direction. Therotated spin state is then given by

Ψ′η′ = Uz(ϕ0) Ψη

459

=(

cos θ2 exp[ − i ϕ+ϕ0

2 ]sin θ

2 exp[ + i ϕ+ϕ02 ]

)(2171)

Hence, we deduce that Ψ′η′ is the eigenstate of η′ . S with eigenvalue + h

2 , wherethe unit vector η′ is obtained by rotating η around the z axis through the angleϕ0. That is, the spins now point along the positive η′ direction. The operatorUz(ϕ0) produced a rotation of the spin about the z-axis through an angle of ϕ0.

The combined rotation on the spin and spatial wave function of the spin onehalf particle

Ψ(r) =(

Ψ+(r)Ψ−(r)

)(2172)

produces the transformed state

Ψ′(r) =

(Ψ

′

+(r)Ψ

′

−(r)

)=

(exp[+iϕ0

2 ] 00 exp[−iϕ0

2 ]

) (Ψ+(R−1

z r)Ψ−(R−1

z r)

)(2173)

In this expression, the spinor wave function is transferred from the point R−1z r

to the rotated point r and the direction of the spin is adjusted locally.

For a particle with spin one, the corresponding (local) spin rotation operatorcan be represented as

Ue(ϕ0) = exp[− i ϕ0

he . S

]= I − i sinϕ0

(e . S

h

)+(

cosϕ0 − 1) (

e . S

h

)2

(2174)

For integer spins, a rotation of 2 π does not produce a change in the phase ofspinor wave function. This is in contrast to the case of half integer spins wherethe phase of the wave function changes by π.

As an example, a spin is rotated through an angle ϕ0 about the z-axis bythe transformation Uz(ϕ0) where

Uz(ϕ0) =

exp[ − i ϕ0 ] 0 00 1 00 0 exp[ + i ϕ0 ]

(2175)

The state in which the direction of the classical spin is known to be along anarbitrary direction η corresponds to the quantum state which is an eigenstate

460

of the component of the spin along η is h. The relevant operator is given by

(η . S

h

)=

cos θ 1√2

sin θ exp[ − i ϕ ] 01√2

sin θ exp[ + i ϕ ] 0 1√2

sin θ exp[ − i ϕ ]0 1√

2sin θ exp[ + i ϕ ] − cos θ

(2176)

where θ and ϕ are the polar coordinates that specify the direction of the unitvector η. The operator has an eigenstate with eigenvalue h which is given by

Ψη =12

( 1 + cos θ ) exp[ − i ϕ ]√2 sin θ

( 1 − cos θ ) exp[ + i ϕ ]

(2177)

Hence, this eigenstate describes a spin pointing in the direction of positive η.This result is consistent with the calculated expectation values of Sx, Sy andSz for this state. It is simple to see that the transformation Uz(ϕ0) results in aspin state Uz(ϕ0) Ψη where the spin’s direction is rotated by ϕ0.

——————————————————————————————————

4.9.7 Exercise 112

Show that the spin rotation operator Ue(ϕ0), for a spin one half particle, satisfiesthe identity

Ue(ϕ0) = exp[− ϕ0

i

2e . σ

]= cos

ϕ0

2σ0 − i sin

ϕ0

2e . σ (2178)

——————————————————————————————————

4.9.8 Solution 112

First we expand the operator, and rearrange the series as a sum of even andodd terms

Ue(ϕ0) = exp[− i

ϕ0

2e . σ

]=

∞∑n=0

1n!

(− i

ϕ0

2e . σ

)n

=∞∑

n=0

12n!

(− i

ϕ0

2e . σ

)2n

+∞∑

n=0

1(2n+ 1)!

(− i

ϕ0

2e . σ

)(2n+1)

(2179)

461

Then prove the identity (e . σ

)2

= σ0 (2180)

which proceeds from the expansion of the scalar product(e . σ

)2

=( ∑

i

e . ei σi

)2

=∑i,j

( e . ei ) ( e . ei ) σi σj

=∑

i

( e . ei )2 σ2i +

∑i>j

( e . ei ) ( e . ei )(σi σj + σj σi

)=

∑i

( e . ei )2 σ0 +∑i>j

( e . ei ) ( e . ei ) δi,j

= ( e . e ) σ0

= σ0 (2181)

since the squares of the Pauli matrices are unity and the different Pauli matricesanti-commute. The last line follows as e is a unit vector. On substituting theidentity in the series expansion, one finds

Ue(ϕ0) = exp[− i

ϕ0

2e . σ

]=

∞∑n=0

( − 1 )n

2n!

(ϕ0

2

)2n

σ0 − i ( e . σ )∞∑

n=0

( − 1 )n

(2n+ 1)!

(ϕ0

2

)2n+1

(2182)

Since, the trigonometric functions are defined as

cosϕ0

2=

∞∑n=0

( − 1)n

(2n)!

(ϕ0

2

)2n

sinϕ0

2=

∞∑n=0

( − 1)n

(2n+ 1)!

(ϕ0

2

)(2n+1)

(2183)

one has proved the identity

Ue(ϕ0) = exp[− ϕ0

i

2e . σ

]= cos

ϕ0

2σ0 − i sin

ϕ0

2e . σ (2184)

——————————————————————————————————

462

4.9.9 Exercise 113

Consider the rotation of an arbitrary (spatially uniform) spin one half state Ψ.The rotation is represented by the unitary transformation Uz(ϕ0). Express theexpectation value of the x-component of spin in the transformed state Ψ′ interms of the untransformed states. Hence, show that

( Ψ′† Sx Ψ

′) = cosϕ0 ( Ψ† Sx Ψ ) − sinϕ0 ( Ψ† Sy Ψ ) (2185)

——————————————————————————————————

4.9.10 Solution 113

The transformed state is given by Ψ′where

Ψ′

= Uz(ϕ0) Ψ (2186)

Hence, the expectation value is given by

Ψ′† Sx Ψ

′= Ψ† U†z (ϕ0) Sx Uz(ϕ0) Ψ (2187)

In matrix form, we have

Ψ′† Sx Ψ

′=

h

2Ψ†

(exp[+iϕ0

2 ] 00 exp[−iϕ0

2 ]

) (0 11 0

) (exp[−iϕ0

2 ] 00 exp[+iϕ0

2 ]

)Ψ

= Ψ†(

0 exp[+iϕ0]exp[−iϕ0] 0

)Ψ

= cosϕ0 ( Ψ† Sx Ψ ) − sinϕ0 ( Ψ† Sy Ψ ) (2188)

The expectation value of Sy also changes like the y-component of a classicalvector under rotation, as does the expectation value of Sz. The expectationvalue of Sz doesn’t change since the rotation is about the z-axis. Hence, theexpectation value of S of any arbitrary state transforms like a vector.

——————————————————————————————————

The Vector Character of Spin One Particles

For a particle with spin one, the corresponding (local) spin rotation operatorcan be represented as

Ue(ϕ0) = exp[− i ϕ0

he . S

]= I − i sinϕ0

(e . S

h

)+(

cosϕ0 − 1) (

e . S

h

)2

(2189)

463

Therefore, the transformation Uz(ϕ0) which produces a rotation of the spin ofa spin one particle through the angle ϕ0 about the z-axis is given by

Uz(ϕ0) =

exp[ − i ϕ0 ] 0 00 1 00 0 exp[ + i ϕ0 ]

(2190)

in the basis formed by the set of eigenstates of Sz. Likewise, the forms of theoperators Uy and Ux that produce spin rotations about the y and x axes aregiven by

Uy(ϕ0) =

12 ( 1 + cosϕ0 ) − 1√

2sinϕ0

12 ( 1 − cosϕ0 )

1√2

sinϕ0 cosϕ0 − 1√2

sinϕ012 ( 1 − cosϕ0 ) 1√

2sinϕ0

12 ( 1 + cosϕ0 )

(2191)

and

Ux(ϕ0) =

12 ( 1 + cosϕ0 ) − i√

2sinϕ0

12 ( cosϕ0 − 1 )

− i√2

sinϕ0 cosϕ0 − i√2

sinϕ012 ( cosϕ0 − 1 ) − i√

2sinϕ0

12 ( 1 + cosϕ0 )

(2192)

The operator Q given by

Q =1√2

− 1 0 1− i 0 − i

0√

2 0

(2193)

represents a unitary transformation. This operator transforms states from thebasis formed by the eigenstates of Sz (Ψ) to a new basis (Ψ) via Ψx

Ψy

Ψz

= Q

Ψ+1

Ψ0

Ψ−1

(2194)

The z-component of the transformed spin operator is given by˜Sz = Q Sz Q

†

=h

2

− 1 0 1− i 0 − i

0√

2 0

1 0 00 0 00 0 − 1

− 1 i 00 0

√2

1 i 0

= h

0 − i 0i 0 00 0 0

(2195)

On applying the unitary transform to the y-component of the spin operator

Sy =h√2

0 − i 0i 0 − i0 i 0

(2196)

464

one obtains the transformed operator as˜Sy = Q Sy Q

†

=h

2

− 1 0 1− i 0 − i

0√

2 0

0 − i√

20

i√2

0 − i√2

0 i√2

0

− 1 i 0

0 0√

21 i 0

= h

0 0 i0 0 0− i 0 0

(2197)

Likewise, applying the transformation to the x-component

Sx =h√2

0 1 01 0 10 1 0

(2198)

one obtains the transformed operator˜Sx = Q Sx Q

†

=h

2

− 1 0 1− i 0 − i

0√

2 0

0 1√

20

1√2

0 1√2

0 1√2

0

− 1 i 0

0 0√

21 i 0

= h

0 0 00 0 − i0 i 0

(2199)

Therefore, the (j, k) matrix element of the transformed spin operator ˜Si is sim-ply given by ( ˜

Si

)j,k

= − i h εi,j,k (2200)

where εi,j,k is the Levi-Civita symbol. Since the transformed and untransformedoperators are related via a unitary transformation, the transformed operatorsobey the same commutation relations as the untransformed operators. In par-ticular, the squared magnitude of the transformed spin operator˜

S2 = ˜S2

x + ˜S2

y + ˜S2

z

= 2 h2

1 0 00 1 00 0 1

(2201)

is diagonal and, therefore, commutes with all other spin operators.

The form of the spin rotation operator ˜Ue(ϕ0) in the new basis is given by

˜Ue(ϕ0) = Q Ue(ϕ0) Q† (2202)

465

where˜Ue(ϕ0) = exp

[− i ϕ0

he .˜S

]

= I − i sinϕ0

(e .˜S

h

)+(

cosϕ0 − 1) (

e .˜S

h

)2

(2203)

In particular, the matrix which describes a rotation about the z-axis is given by

˜Uz(ϕ0) =

cosϕ0 − sinϕ0 0sinϕ0 cosϕ0 0

0 0 1

(2204)

The matrix which describes a rotation about the y-axis becomes

˜Uy(ϕ0) =

cosϕ0 0 sinϕ0

0 1 0− sinϕ0 0 cosϕ0

(2205)

and a rotation about the x-axis is produced by

˜Ux(ϕ0) =

1 0 00 cosϕ0 − sinϕ0

0 sinϕ0 cosϕ0

(2206)

These transformations are recognized as being identical to the expressions forthe transformations of vectors under rotations through a finite angle ϕ0 aboutthe Cartesian axes. A more general (local) rotation of the spin one wave functioncan be built up from infinitesimal rotations about an arbitrary axis. Under aninfinitesimal rotation δϕ0 about an arbitrary axis e, the spin one wave functionchanges (locally) according to Ψ′

x

Ψ′y

Ψ′z

=[

1 − i δϕ0

(e .˜S

h

) ] Ψx

Ψy

Ψz

(2207)

In terms of the components, the transformation is expressed as

Ψ′i =

∑j

[δi,j − i

∑k

δϕ0

(ek .

˜Sk

h

)i,j

]Ψj

=∑

j

[δi,j −

∑k

δϕ0 ek εk,i,j

]Ψj (2208)

If one interprets the components of wave function of the spin one particle asa vector in the same space as ordinary vectors, then on re-labelling and re-arranging the indices, one finds

−∑j,k

ek εk,i,j Ψj =

∑j,k

ej εi,j,k Ψk

466

Rotation of a Vector Field

x

yΨ(r)

Ψ'(r)=exp[-iSzφ] Ψ(Rz-1r)

Rz

Figure 120: The unitary operator Uz(ϕ) produces a rotation of the spin onewave function through an angle of ϕ around the z-axis. The rotation operatorRz acts on the coordinates and transforms the point r to the point r′, i.e.r′ = Rz r. Under the rotation, the vector wave function Ψ(r) is transformed toΨ′(r), such that the value of Ψ′ at the point r′ is the same as the value of Ψ atthe point r. However, the rotation also rotates the direction of the vector fieldby ϕ. Therefore, Ψ′(r′) = exp[−iϕ Sz] Ψ(r) so, Ψ′(r) = exp[−iϕ Sz] Ψ(R−1r).The unitary operator which transforms the wave function is then determinedby Ψ′(r) = U(ϕ) Ψ(r) = exp[−iϕ Sz] Ψ(R−1r).

=(e ∧ Ψ

)i

(2209)

Therefore, the spin one wave function transforms under rotations via

Ψ′

=[

Ψ + δϕ0

(e ∧ Ψ

) ](2210)

This is the same transformation law obeyed by a vector under a rotation throughan infinitesimal angle δϕ0 about the axis e. This suggests that the wave functionof a spin one particle should be considered as a vector field.

If the spin one wave function Ψ has a vector representation, then its decom-position in terms of Cartesian unit vectors ei and the Cartesian components Ψi

is given byΨ = ex Ψx + ey Ψy + ez Ψz (2211)

467

Sometimes it may be convenient to express the vector field in terms of compo-nents Ψm which have definite values of Sz. The components of the vector in thetwo basis sets are related by

Ψm = Q† Ψi (2212)

or, equivalently, in matrix form by Ψ+1

Ψ0

Ψ−1

=1√2

− 1 i 00 0

√2

1 i 0

Ψx

Ψy

Ψz

(2213)

——————————————————————————————————

4.9.11 Exercise 114

Find the spin states that are the simultaneous eigenstates of the operators ˜Sz

and ˜S2.

Show that, with an appropriate choice of phase, these states can be mappedonto the circularly polarized states Ψm which are expressed in terms of theCartesian basis by

Ψ+1 = − 1√2

1i0

Ψ0 =

001

Ψ−1 =

1√2

1− i0

(2214)

——————————————————————————————————

4.9.12 Solution 114

On expressing the eigenvectors in the form

Ψm =

αβγ

(2215)

468

one finds that the form of the eigenstates trivially satisfy the eigenvalue equation

for ˜S2. The eigenvalue equation for the operator ˜Sz is non-trivial and can bewritten as

h

0 − i 0i 0 00 0 0

αβγ

= λm

αβγ

(2216)

The eigenvalues λm are determined from the zeros of the determinant∣∣∣∣ − λ − i 0i − λ 00 0 − λ

∣∣∣∣ = 0 (2217)

which yield the solutions for λm as (1, 0,−1). The eigenvectors for λ = ± 1are determined from the equations

h

0 − i 0i 0 00 0 0

α±1

β±1

γ±1

= ±

α±1

β±1

γ±1

(2218)

which yields the equations

− i β±1 = ± α±1

i α±1 = ± β±1

0 = ± γ±1 (2219)

Since these are homogeneous linear equations, the (un-normalized) solutions canbe found by choosing α±1 = 1. The eigenvector corresponding to λm = 0 isgiven by

− i β0 = 0i α0 = 0

0 = 0 (2220)

which can be solved by choosing γ0 = 1. On normalizing the above set ofeigenstates, one finds that they can be expressed in forms

Ψ+1 =1√2

1i0

Ψ−1 =

1√2

1− i0

Ψ0 =

001

(2221)

469

These correspond to the given set of eigenstates, except for the difference inthe choice of phase for Ψ1. The desired phase could have been arrived at bychoosing α1 = − 1.

——————————————————————————————————

Exercise 115

Consider a general rotation of a vector through an angle θ in d dimensions.Show that trace of the rotation operator U(θ) is given by

Trace U(θ) = ( d − 2 ) + 2 cos θ (2222)

——————————————————————————————————

When expressed in terms of the angular momentum basis, an arbitrary vectorΨ has the decomposition

Ψ =m=+1∑m=−1

em Ψm (2223)

which defines the appropriate angular momentum basis vectors em. The basisvectors em are given in terms of the Cartesian basis vectors via

e+1 = − 1√2

( ex + i ey )

e0 = ez

e−1 =1√2

( ex − i ey ) (2224)

These complex basis vectors are the circular polarization vectors for a spinone particle. The set of complex basis vectors em for the angular momentumrepresentation satisfy

e∗m = (−1)m e−m (2225)

and their scalar products are given by

e∗m . em′ = (−1)m e−m . em′ = δm,m′ (2226)

The components of the vector Ψ would then be found from

Ψm = e∗m . Ψ (2227)

A similar type of decomposition in terms of circularly polarized basis vectors isfrequently applied to the vector wave function of the spin one photon.

470

4.9.13 Gauge Invariance

The Schrodinger equation for a charged particle in an electromagnetic field isgiven by

i h∂

∂tΨ(r, t) =

[1

2 m

(− i h ∇ − q

cA(r, t)

)2

+ q φ(r, t)]

Ψ(r, t)

(2228)

The vector and scalar potentials are invariant under the gauge transformations

A(r, t) → A′(r, t) = A(r, t) + ∇ Λ(r, t)

φ(r, t) → φ′(r, t) = φ(r, t) − 1c

∂

∂tΛ(r, t) (2229)

The primed and unprimed potentials are in different gauges, but represent thesame physical system. The Schrodinger equation in the primed gauge is

i h∂

∂tΨ′(r, t) =

[1

2 m

(− i h ∇ − q

cA′(r, t)

)2

+ q φ′(r, t)]

Ψ′(r, t)

(2230)

where the primed wave function is given in terms of the unprimed wave functionby

Ψ(r, t) → Ψ′(r, t) = Ψ(r, t) exp[iq

h cΛ(r, t)

](2231)

and where the scalar field involved in the gauge transformation has been ab-sorbed into the phase of the wave function.

The gauge transformation is produced by the unitary transformation U suchthat

Ψ′(r, t) = U Ψ(r, t)

U = exp[iq

h cΛ(r, t)

](2232)

Then the Hamiltonians in the two gauges are connected via

H ′ = U H U† − i h U∂

∂tU† (2233)

as can be seen by examining

i h∂

∂tΨ(r, t) = H Ψ(r, t) (2234)

which on substitutingΨ(r, t) = U† Ψ′(r, t) (2235)

471

becomes

U† i h∂

∂tΨ′(r, t) + i h

(∂

∂tU†)

Ψ′(r, t) = H U† Ψ′(r, t) (2236)

On utilizing the primed Schrodinger equation

i h∂

∂tΨ′(r, t) = H ′ Ψ′(r, t) (2237)

one has

U† H ′ Ψ′(r, t) + i h

(∂

∂tU†)

Ψ′(r, t) = H U† Ψ′(r, t) (2238)

This equation determines the time evolution of an arbitrary initial wave functionΨ′(r, t), and so we have an operator equation

U† H ′ + i h

(∂

∂tU†)

= H U† (2239)

which on pre-multiplying by U is identical to the relation between H and H ′.

We note that this gauge transformation leaves r and the velocity p − qc A(r, t)

invariant. This can be seen as

r′ = U r U† = r (2240)

and

p′ − q

cA′(r, t) = U

(p − q

cA(r, t)

)U†

= p − q

cA(r, t) − q

c∇ Λ(r, t)

= p − q

cA′(r, t) (2241)

Finally, as p′ = p = − i h ∇, the above quantity is gauge invariant as was tobe proved. Thus, the Schrodinger equation is gauge invariant.

——————————————————————————————————

4.9.14 Exercise 116

Find the continuity equation for a charged particle in an electromagnetic field,and show that the appropriate probability current density j(r, t) and the chargedensity ρ(r, t) are gauge invariant.

——————————————————————————————————

472

4.9.15 Solution 116

The equation of motion for the wave function Ψ is

i h∂

∂tΨ =

[1

2 m

(p − q

cA

)2

+ q φ

]Ψ (2242)

or

i h∂

∂tΨ =

[1

2 m

(− i h ∇ − q

cA

)2

+ q φ

]Ψ (2243)

and the complex conjugate Ψ∗ satisfies

− i h∂

∂tΨ∗ =

[1

2 m

(+ i h ∇ − q

cA

)2

+ q φ

]Ψ∗ (2244)

Multiplying the first equation by Ψ∗ and the second by Ψ, then subtracting thesecond from the first, one obtains

i h∂

∂t

(Ψ∗ Ψ

)= − h2

2 m

[Ψ∗ ∇2 Ψ − Ψ ∇2 Ψ∗

]+ i h ∇

(q

m cΨ∗ A(r, t) Ψ

)(2245)

Thus, one has the continuity equation

∂ρ

∂t+ ∇ j = 0 (2246)

but

ρ(r, t) = Ψ∗(r, t) Ψ(r, t)

j(r, t) =h

m=m

[Ψ∗ ∇ Ψ − Ψ ∇ Ψ∗

]− q

m cΨ∗(r, t) A(r, t) Ψ(r, t)

(2247)

The charge density ρ(r, t) is gauge invariant, as under the gauge transfor-mation one has

Ψ(r, t) → Ψ′(r, t) = exp[iq

c hΛ(r, t)

]Ψ(r, t) (2248)

and the phase factor drops out of the density.

In the current density, the gradient factor transforms as

Ψ∗ ∇ Ψ → Ψ′∗ ∇ Ψ′ = Ψ∗ ∇ Ψ + iq

c hΨ∗(∇ Λ

)Ψ

(2249)

473

and the term involving the vector potential transform as

Ψ∗ A(r, t) Ψ → Ψ′∗ A′(r, t) Ψ′ (2250)

where

Ψ′∗ A′(r, t) Ψ′ = Ψ∗ A(r, t) Ψ + Ψ∗ ∇Λ(r, t) Ψ (2251)

Thus, the factor in the current density involving ∇ Λ cancels, leaving j(r, t)gauge invariant.

——————————————————————————————————

4.9.16 Galilean Boosts

A transformation to a uniformly moving reference frame is represented by atime-dependent unitary transform. The (passive) transformation from a sta-tionary reference frame to one moving with a velocity - u is represented by theunitary matrix U defined by

U = exp[− i

hu . ( m r − p t )

](2252)

The above Galilean transformation is derived as the non-relativistic limit of aLorentz boost, which is just a (static) rotation in space-time. However, in thenon-relativistic limit the operator H is replaced by the rest mass energy mc2.The above transformation represents a Galilean boost since it has the effect that

r′ = U r U† = r + u t

p′ = U p U† = p + m u (2253)

as expected for a transformation to a reference frame moving with constant ve-locity.

The Hamiltonian is the generator of time-translations and, therefore, a stateΨ evolves according to the prescription

Ψ∗(i h

∂

∂t

)Ψ = Ψ∗ H Ψ (2254)

on transforming from Ψ to Ψ′ via

Ψ′ = U Ψ (2255)

one finds that

Ψ∗′(i h

∂

∂t

)Ψ′ = Ψ∗′ U H U† Ψ′ − i h Ψ∗′ U

∂

∂tU† Ψ′ (2256)

474

which defines the transformed Hamiltonian H ′ as

H ′ = U H U† − i h U∂

∂tU†

= U H U† − u . p (2257)

Thus, the energy eigenvalues are related via

E′ = E − u . p (2258)

which is recognized as the non-relativistic limit of the Lorentz transformationof the time-like component of a four-vector

E′ =E − u . p√

1 − (uc )2

(2259)

For a Hamiltonian in the stationary reference frame of the form

H =1

2 m

(p − q

cA(r)

)2

+ V (r) (2260)

the Hamiltonian in the moving reference frame has the form

H ′ =1

2 m

(p + m u − q

cA(r + ut)

)2

+ q φ(r + ut) − u . p

=1

2 m

(p − q

cA(r + ut)

)2

+ q φ(r + ut) − q

cu . A(r + ut) +

m u2

2(2261)

when expressed in terms of the untransformed position and momentum opera-tors. In the above equations we have ignored any explicit time-dependence ofthe electromagnetic potentials. The transformed Hamiltonian is increased by anadditive constant equal to the kinetic energy of the boost. Since we are workingin the non-relativistic limit, this additive constant has no physical consequence.The vector potential is transformed according to

A(r) → A′(r′) = A(r + ut) (2262)

which is the same as in the static reference frame (except that it is evaluated atthe transformed position r′ = r + ut), but the scalar potential is transformedaccording to

φ(r) → φ′(r′) = φ(r′) − 1cu . A(r′) (2263)

This has the implication that the magnetic field in the moving frame defined by

B′(r′) = ∇ ∧ A(r′) (2264)

is simply given by the magnetic field at the transformed position

B′(r′) = B(r + ut) (2265)

475

On the other hand, the transformed electric field defined by

E′(r′) = − ∇ φ′(r′) − 1c

∂

∂tA′(r′) (2266)

is evaluated as

E′(r′) = − ∇ φ(r′) +1c∇(u . A(r′)

)− 1

c

∂

∂tA(r + ut)

= − ∇ φ(r′) +1c∇(u . A(r′)

)− 1

c

(u . ∇

)A(r′) − 1

c

∂

∂tA(r′)

(2267)

where the last term is a partial derivative with respect to t at constant r′ = r+ut.On using the identity

∇(u . A

)−(u . ∇

)A = u ∧

(∇ ∧ A

)(2268)

one obtains the transformed electric field as

E′(r′) = E(r′) +1cu ∧ B(r′) (2269)

which is as might have been expected.

5 The Rotating Planar Oscillator

To end this first semester of quantum mechanics, we shall look at a simplesystem and show that there are still some surprises to be found in separablesystems. Usually systems are neither separable, nor can their excitations befound analytically. Most systems have to be solved numerically and it expectedthat their dynamics may sometimes show chaotic motion.

We shall consider a two-dimensional simple harmonic oscillator, that isslowly being rotated about the origin with frequency Ω. The classical La-grangian is given by

L =m

2

(r2 + r2 ( ϕ − Ω )2

)− m

2ω2 r2 (2270)

The generalized momenta are given by

pr = m r

pϕ = m r2(ϕ − Ω

)(2271)

Therefore, the classical Hamiltonian can be expressed as

H = pr r + pϕ ϕ − L

=p2

r

2 m+

p2ϕ

2 m r2+ pϕ Ω +

m

2ω2 r2 (2272)

476

In two dimensions, the angular momentum pϕ (which we shall now denote byL) is masquerading as a scalar. Hence, we shall write the Hamiltonian as

H =p2

r

2 m+

L2

2 m r2+ L Ω +

m

2ω2 r2 (2273)

We shall quantize this classical Hamiltonian and shall determine the energyeigenfunctions in the form

Ψnr,l(r, ϕ) = Rnr,l(r)1√2 π

exp[i l ϕ

](2274)

where l is a positive or negative integer. The radial part of the energy eigenvalueequation takes the form

− h2

2 m1r

∂

∂r

(r∂R

∂r

)+(

h2 l2

2 m r2+ h l Ω +

m

2ω2 r2

)R = Enr,l R (2275)

We shall introduce a dimensionless variable ρ via the definition

ρ =√m ω

hr (2276)

and re-write the radial equation as

1ρ

∂

∂ρ

(ρ∂Rnr,l

∂ρ

)+(l2

ρ2+ 2 l

Ωω

+ ρ2

)Rnr,l =

2 Enr,l

h ωRnr,l (2277)

For Ω = 0, the radial equation can easily be solved by using the operatoralgebra of raising and lowering operators. The raising operators are expressedas

A†l = −(

∂

∂ρ+

12 ρ

)+(ρ − 2 l + 1

2 ρ

)(2278)

and the lowering operators are defined as the Hermitean conjugate operators43

Al =(

∂

∂ρ+

12 ρ

)+(ρ − 2 l + 1

2 ρ

)(2279)

The commutator of the raising and lowering operator is evaluated as

[ A†l , Al ] = − 2 − 2 l + 1ρ2

(2280)

43The operators contain terms that depend on the dimensionality. In d dimensions, thecorresponding pair of Hermitean conjugate operators are given by

A†l

= −(

∂

∂ρ+

d − 1

2 ρ

)+

(ρ −

2 l + d − 1

2 ρ

)and

Al =

(∂

∂ρ+

d − 1

2 ρ

)+

(ρ −

2 l + d − 1

2 ρ

)

477

The product of the lowering and raising operators is evaluated as

A†l Al = − ∂2

∂ρ2− 1

ρ

∂

∂ρ+

l2

ρ2+ ρ2 − 2 ( l + 1 ) (2281)

which has a form similar to the left-hand side of the radial equation44. On usingthe commutator in eqn(2280), one finds that the product of the lowering andraising operators taken in opposite order is given by

Al A†l = − ∂2

∂ρ2− 1

ρ

∂

∂ρ+

( l + 1 )2

ρ2+ ρ2 − 2 l (2283)

The dimensionless form of radial equation is expressed as

Hl Rnr,l =2 Enr,l

h ωRnr,l (2284)

where the dimensionless effective Hamiltonian for angular momentum l is givenby

Hl =(A†l Al + 2 ( l + 1 )

)(2285)

and the effective Hamiltonian for angular momentum l + 1 is given by

Hl+1 =(Al A

†l + 2 l

)(2286)

By using the radial equation in the form

Hl Rnr,l = 2Enr,l

h ωRnr,l(

A†l Al + 2 ( l + 1 ))Rnr,l = 2

Enr,l

h ωRnr,l (2287)

and then on pre-multiplying by Al, one finds

Al

(A†l Al + 2 ( l + 1 )

)Rnr,l = 2

Enr,l

h ωAl Rnr,l(

Al A†l + 2 ( l + 1 )

)Al Rnr,l = 2

Enr,l

h ωAl Rnr,l(

Hl+1 + 2)Al Rnr,l = 2

Enr,l

h ωAl Rnr,l (2288)

44In d dimensions, the commutator of the raising and lowering operators is evaluated

[ A†l

, Al ] = − 2 −2 l + d − 1

ρ2

and the product of the operators is given by

A†l

Al = −∂2

∂ρ2−

d− 1

ρ

∂

∂ρ+

l ( l + d − 2 )

ρ2+ ρ2 − ( 2 l + d )

(2282)

478

Hence, we have shown that Al Rnr,l is an eigenstate of Hl+1 since

Hl+1 Al Rnr,l = 2(Enr,l

h ω− 1

)Al Rnr,l (2289)

Therefore, the lowering operator Al acting on Rnr,l produces another energyeigenfunction with indices (nr − 1, l + 1)

Al Rnr,l ∝ Rnr−1,l+1 (2290)

and with energy eigenvalue given by

Enr−1,l+1 = Enr,l − h ω (2291)

Since the energy is bounded from below there is a minimum value of nr (definedas nr = 0), so that the lowering operator acting on this state with nr = 0vanishes

Al R0,l = 0 (2292)

On using this condition in eqn(2287), one finds the energy eigenvalue equationsimplifies to

( l + 1 ) R0,l =E0,l

h ωR0,l (2293)

Hence, the energy eigenvalues for the states with nr = 0 are given by

E0,l = h ω

(l + 1

)(2294)

The radial eigenfunctions are found from the condition

Al R0,l = 0(∂

∂ρ+ ρ − l

ρ

)R0,l = 0 (2295)

which can be re-expressed as(∂R0,l

∂ρ

)R0,l

= − ρ +l

ρ(2296)

This can be integrated to yield

R0,l = C ρl exp[− ρ2

2

](2297)

The above form of the energy eigenfunction is also found for the isotropic d-dimensional harmonic oscillator, and the energy eigenvalue corresponding tothis eigenfunction is given by

E0,l = h ω

(l +

d

2

)(2298)

479

The radial eigenfunctions for larger values of nr can be obtained from theaction of the raising operators. This can be seen by considering the radialequation with angular momentum l + 1

Hl+1 Rnr,l+1 = 2Enr,l+1

h ωRnr,l+1(

Al A†l + 2 l

)Rnr,l+1 = 2

Enr,l+1

h ωRnr,l+1 (2299)

and then pre-multiplying by the raising operator A†l

A†l

(Al A

†l + 2 l

)Rnr,l+1 = 2

Enr,l+1

h ωA†l Rnr,l+1(

A†l Al + 2 l)A†l Rnr,l+1 = 2

Enr,l+1

h ωA†l Rnr,l+1(

Hl − 2)A†l Rnr,l+1 = 2

Enr,l+1

h ωA†l Rnr,l+1 (2300)

Hence, we have

Hl A†l Rnr,l+1 = 2

(Enr,l+1

h ω+ 1

)A†l Rnr,l+1 (2301)

so the raising operator acting on a radial eigenfunction with quantum num-bers (nr, l + 1) produces an eigenstate with quantum numbers (nr + 1, l) andeigenvalue

Enr+1,l = Enr,l+1 + h ω (2302)

Therefore, the eigenfunctions with higher values of n are found from

Rnr+1,l ∝ A†l Rnr,l+1 (2303)

The energy eigenvalues are found as

Enr,l = h ω

(2 nr + | l | + 1

)(2304)

This completely solves for the energy spectrum and the energy eigenfunctionswhen Ω = 0. The classical states can be recovered as coherent states, since theratios of the rotational frequencies to the vibrational frequencies are rational45.

In the more general case, one finds

Enr,l = h ω

(2 nr + | l | + l

Ωω

+ 1)

(2305)

45R. Balian and C. Bloch, Ann. Phys. N.Y. 69, 76 (1972).

480

In the general case, the ratios of the frequencies are not rational. The energyspectrum can be plotted as a function of the de-tuning parameter ν defined by

ν =(

Ω − ω

2 ω

)(2306)

The spectrum is shown in fig(121). For negative rational values of ν the spec-

Figure 121: The energy spectrum for the rotated planar harmonic oscillator asa function of the de-tuning parameter ν.

trum shows gaps46. These gaps are related to the occurrence of periodic orbits.The gaps are easily seen in the higher energy portions of the spectrum, shown infig(122). For ν = 0 a catastrophe occurs, all the energy levels become infinitelydegenerate and each group of levels is separated by the energy difference 2 h ω.

46R.K. Bhaduhri, S. Li, K. Tanaka and J.D. Waddington, J. Phys. A 27, L663 (1994).

481

Figure 122: The higher-energy eigenvalues for the rotated planar harmonic os-cillator as a function of the de-tuning.

6 Dirac Formulation

Schrodinger introduced wave mechanics in which a quantum state was describedin terms of a wave function and physical measurements were described by differ-ential operators. On the other hand, Heisenberg introduced matrix mechanicsin which states were represented by column vectors, similar to the column vec-tors used in our discussion of spin, and measurements were treated as matricessimilar to the spin matrices. These two formulations of quantum mechanics areidentical in physical content. The equivalence was first seen by Dirac and thisled to his abstract formulation.

6.1 Dirac Notation

Dirac introduced an abstract notation for states and their duals, and also forthe operators.

482

6.1.1 Bracket Notation

In the Dirac formulation, physical states are represented by the so called kets,which are written as

| Ψ > (2307)

These kets obey the principle of linear superposition

| Ψ > =∑

n

Cn | φn > (2308)

where Cn are complex numbers. Generally, if the discrete index n is supple-mented by a continuous index47 denoted by ξn, the superposition can be ex-pressed in terms of a sum and an integral

| Ψ > =∑

n

Cn | φn > +∫

dξ Cξ | φξ > (2309)

Dirac also introduced a dual space of bras. The bras are denoted as < Ψ |and are defined as the duals to the kets | Ψ >. They are the mirror image ofthe ket states. More precisely, they are defined by the scalar products with thekets defined as a complex number given by(

< Φ |)| Ψ > = < Φ | Ψ > (2310)

The scalar product is linear, so that

< Φ |(C1 | Ψ1 > + C2 | Ψ2 >

)= C1 < Φ | Ψ1 > + C2 < Φ | Ψ2 >

(2311)A bra vector is completely defined when the scalar product with every ket isknown. Thus, if the scalar product with every ket vector is zero, then the bravector is also zero.

< Φ | Ψ > = 0 for all | Ψ > (2312)

then

< Φ | = 0 (2313)

We assume that there is a one to one correspondence between the bras andthe kets. That is, the bra corresponding to the sum of the kets | φ1 > + | φ2 >

47For example, if the set of states | φn > corresponds to the eigenstates of some operator,the eigenvalue may have both discrete and continuous portions of its spectrum.

483

is the sum of the equivalent bras < φ1 | + < φ2 |. Also, the bra correspond-ing to the ket C | φ > is given by C∗ < φ |, which involves complex conjugation.

Because of this assumption, every physical state can also be represented bya bra as much as it can be represented by a ket.

The scalar product is defined such that the complex number equal to thescalar product

< Φ | Ψ > (2314)

is related to the number given by the scalar product of the dual vectors

< Ψ | Φ > (2315)

via complex conjugation. That is, we require that the scalar product satisfies

< Φ | Ψ > = < Ψ | Φ >∗ (2316)

In particular, the normalization of a state can be defined as the scalar prod-uct with itself,

< Ψ | Ψ > = < Ψ | Ψ >∗ (2317)

which must be real. We shall require that the scalar product be defined suchthat the normalization is both real number and positive

< Ψ | Ψ > ≥ 0 (2318)

Two states | Φ > and | Ψ > are defined to be orthogonal if their scalarproduct is zero

< Φ | Ψ > = 0 (2319)

6.1.2 Operators

A physical measurement is described by an operator A which is defined by itseffect on a ket, and in general, transforms a ket to another ket,

A | Ψ > = | Φ > (2320)

484

The operators are linear in that its effect on a linear superposition is expressedas a linear combination on the individual states

A | Ψ > =∑

n

Cn A | φn > (2321)

The effect of an operator on a bra can be defined via the scalar product

< Φ |(A | Ψ >

)= < Φ | A | Ψ >

=(< Φ | A

)| Ψ > (2322)

This defines the effect of the operator A on the bra < Φ |,

< Φ | A (2323)

6.1.3 Adjoints and Hermitean Operators

The Hermitean conjugate of an operator A is defined as the operator A† suchthat the bra

< Ψ | A (2324)

is equivalent to the ketA† | Ψ > (2325)

for any state | Ψ >. In particular, this means that

< Ψ | A | Φ > = < Φ | A† | Ψ >∗ (2326)

Physical operators are represented by Hermitean operators such that

A† = A (2327)

The eigenvalue equation of an operator A is

A | φn > = an | φn > (2328)

where an are the eigenvalues and | φn > are the eigenstates. It can be provedthat the eigenstates of a Hermitean operator form a complete orthogonal setand their eigenvalues are real.

485

An arbitrary state can be represented as a linear superposition of a completeorthonormal set of states | φn > . That is, an arbitrary state | Ψ > is re-writtenin terms of the expansion

| Ψ > =∑

n

Cn | φn > (2329)

where, due to orthonormality of the complete set, the expansion coefficient isevaluated as

Cn = < φn | Ψ > (2330)

Thus, we have| Ψ > =

∑n

< φn | Ψ > | φn > (2331)

which can be re-arranged as

| Ψ > =∑

n

| φn > < φn | Ψ > (2332)

Since | Ψ > is completely arbitrary, the relation is independent of | Ψ > .Omitting the arbitrary state leads to the completeness relation, in which theidentity operator I is written as

I =∑

n

| φn > < φn | (2333)

6.1.4 Representation of Operators

An arbitrary operator A can be represented in much the same way. We definethe operator via its effect on an arbitrary state | Ψ >

A | Ψ > = | Φ > (2334)

Then, with the analogous expression for | Φ > with expansion coefficients Bm,we find that the effect of the operator is given by

A∑

n

Cn | φn > =∑m

Bm | φm > (2335)

Taking the scalar product of the above equation with the bra < φm | , onefinds that the matrix elements of the operator < φm | A | φn > satisfy theequation

< φm | A | Ψ > =∑

n

< φm | A | φn > Cn

= Bm (2336)

486

Inserting the expression for Cn, one finds that the effect of an operator on anarbitrary state | Ψ > can be expressed directly in terms of the matrix elements

< φm | A | Ψ > =∑

n

(< φm | A | φn >

)< φn | Ψ >

= Bm (2337)

Hence, we can use the expression for Bm in the equation defining the effect ofthe operator

A | Ψ > =∑m

Bm | φm >

=∑m,n

| φm >

(< φm | A | φn >

)< φn | Ψ >

(2338)

Thus, as | Ψ > is arbitrary the above equation is independent of the choice of| Ψ > , so we may omit it. Hence, the operator A can be decomposed as

A =∑m,n

| φm >

(< φm | A | φn >

)< φn | (2339)

The operator is completely defined by its matrix elements between a completeset of states.

6.2 Representations

The position space states | r > are eigenstates of the Hermitean position oper-ator r,

r | r > = r | r > (2340)

Due to the Hermiticity, the eigenvalues of r are real. The orthogonality of theeigenstates are expressed as

< r | r′ > = δ3( r − r′ ) (2341)

The completeness condition is expressed in terms of the identity operator I,where

I =∫d3r | r > < r | (2342)

The position space representation associates a complex number Ψ(r) (thewave function) with every scalar product of a state represented by a ket | Ψ >and the bra representing a position state < r |

< r | Ψ > = Ψ(r) (2343)

487

alternatively, one obtains the complex conjugate of the wave function from

< Ψ | r > = Ψ∗(r) (2344)

A general scalar product can be evaluated using the completeness relationfor the position states, which involves the identity operator

I =∫

d3r | r > < r | (2345)

as

< Φ | Ψ > =∫

d3r < Φ | r > < r | Ψ >

=∫

d3r Φ∗(r) Ψ(r) (2346)

Thus, the scalar product has a position space representation as the overlap op-erator.

The position space representation of an operator A can also be found fromthe matrix elements between two arbitrary states, by inserting complete sets ofstates

< Φ | A | Ψ > =∫

d3r < Φ | r > < r | A | Ψ >

=∫

d3r

∫d3r′ < Φ | r > < r | A | r′ > < r′ | Ψ >

(2347)

The position space representation of the operator A is given by

< r | A | r′ > (2348)

For a local operator, the operator only depends on the position variable and thederivative at one point in space, so we have

< r | A | r′ > = δ3( r − r′ ) A( r′ ; − i h ∇′ ) (2349)

Thus, on integrating over r one obtains the usual expression

< Φ | A | Ψ > =∫

d3r′ < Φ | r′ > A( r′ ; − i h ∇′ ) < r′ | Ψ >

=∫

d3 r Φ∗(r) A( r ; − i h ∇ ) Ψ(r) (2350)

In this expression, the differential operator A( r ; − i h ∇ ) is that which ap-pears in the usual formulation of Schrodinger’s wave mechanics.

488

6.3 Gram-Schmidt Orthogonalization

Given a complete set of normalized states | φm > , that are not orthogonal,one can construct an orthonormal set | Ψn > via the process of Gram-Schmidtorthogonalization.

This is achieved by taking the first state as

| Ψ0 > = | φ0 > (2351)

which is normalized to unity. The second state in our orthogonal set is con-structed as

| Ψ1 > = N1

(| φ1 > − | Ψ0 > < Ψ0 | φ1 >

)(2352)

This second state is orthogonal to | Ψ0 > as can be seen by forming the scalarproduct, and noting that the first state is normalized to unity. The normaliza-tion of the second state N1 is chosen such that this state is also normalized tounity.

The higher order states are found by orthogonalizing to the properly nor-malized states found earlier, i.e.,

| Ψn > = Nn

(| φn > −

m=n−1∑m=0

| Ψm > < Ψm | φn >

)(2353)

and finding the normalization Nn constants before proceeding to find the nextstate. The normalization constants are given by

N−2n = 1 −

m=n−1∑m=0

| < Ψm | φn > |2 (2354)

Hence, given any complete set of states, it is possible to transform them into anorthonormal set.

489

7 Appendices

A: Non-spreading probability densities

The spreading of a wave packet (such as a Gaussian) is a feature which is of-ten found in quantum mechanics. Normalizable wave functions are interpretablein terms of the motion of a single particle. Non-normalizable wave functionsshould be thought of describing an ensemble of particles. There are only twonon-spreading wave functions, the plane wave which corresponds to a uniformbeam of particles with momentum h k, and the Airy wave function which weshall discuss below.

The Airy wave function is given at t = 0 by

Ψ(x; 0) = Ai[B x

h23

](2355)

which has the Fourier transform

Φ(k; 0) =1√2 π

h23

Bexp

[ih2 k3

3 B3

](2356)

Hence, in momentum representation, the time-dependent wave function for freeparticles is given by

Φ(k; t) =1√2 π

h23

Bexp

[i

(h2 k3

3 B3− h k2

2 mt

) ](2357)

Therefore, we find that the real-space wave function is given by the inverseFourier transform

Ψ(x; t) =1

2 πh

23

B

∫ ∞

−∞dk exp

[i

(k x +

h2 k3

3 B3− h k2

2 mt

) ](2358)

On changing the variable of integration from k to k′

k → k′ = k − B3

2 m ht (2359)

one finds that the integration results in another Airy function. The result is

Ψ(x; t) = Ai[B

h23

(x − B3 t2

4 m2

) ]exp

[iB3 t

2 m h

(x − B3 t2

6 m2

) ](2360)

Therefore, the probability density is given by

| Ψ(x; t) |2 = Ai2[B

h23

(x − B3 t2

4 m2

) ](2361)

490

Probability density for the Airy wave packet

0

0.5

1

1.5

2

-12 -8 -4 0 4

x

| Ψ(x

;0)|2

Figure 123: The initial probability density |Ψ(x; 0)|2 for an Airy wavepacket.The wavepacket does nor disperse with increasing t, but it does accelerate.

which propagates without distortion and has an uniform acceleration given by

B3

2 m2(2362)

This peculiar result can be understood on the basis of a comparison of thesemi-classical form of the wave function with a family of classical trajectories48.The semi-classical limit of the Airy function is given by

Ψ(x; 0) ∼ 1√π

(h

23

− B x

) 14

sin[

23

(− B x

h23

) 32

+π

4

](2363)

The semi-classical wave function is interpreted as a superposition of forwardand backward travelling particles with actions S± given by

S± = ± 23

(− B x

) 32

(2364)

From Hamilton-Jacobi theory, the momenta are given by

p± =∂S±∂x

= ±(− B x

) 12

(2365)

48M.V. Berry and N.L. Balazs, Am. J. Phys. 47, 264 (1979).

491

This implies that there is a phase-space relationship given by

X0(p) = − p2

B3(2366)

The Airy function can be considered as resembling an infinite ensemble of clas-sical particles, with initial positions given by X0(p) and each particle is param-eterized by its momentum p. These particles map out straight-line trajectoriesgiven by

X(p; t) = X0(p) +p

mt (2367)

The envelope of the family of trajectories is a parabola. This can be seen by

Envelope of a family of trajectories

-6

-4

-2

0

2

4

6

-5 -4 -3 -2 -1 0 1 2 3 4 5

t

x

x(t)=B3 t2/(4 m2)

X(p,t)=-p2/B3+pt/m

Figure 124: The family of trajectories represented by an Airy wave packet, andits parabolic envelope.

noting that at fixed value of t, the maximum value of X(p, t) corresponds to thevalue pt determined from

∂X(p, t)∂p

∣∣∣∣pt

= 0 (2368)

which yields the relation

pt =B3

2 mt (2369)

Since the family of trajectories touch the envelope at pt, the envelope x(t) isgiven by

x(t) = X(pt, t) = − p2t

B3+

pt

mt (2370)

492

or

x(t) = − B3

4 m2t2 +

B2

2 m2t2 =

B3

4 m2t2 (2371)

Hence the envelope is parabolic. This parabolic envelope represents the bound-aries of classical motion and closely follows the motion of the maximum valueof the quantum mechanical probability density.

——————————————————————————————————

B: Bound States within the Continuum

von Neumann and Wigner found that it is possible to construct potentialswhich have bound states at energies that are higher than the lowest energy ofthe continuum spectra49. However, in these unusual examples, the potentialcontains infinitely many oscillations. The method used is based on the solutionof the field-free energy eigenvalue equation, with any fixed values of the angularmomentum and energy. The energy eigenvalue equation for a free particle withangular momentum l has the form

− 1r2

∂

∂r

(r2

∂Rl

∂r

)+

l ( l + 1 )r2

Rl = k2 Rl (2372)

where the energy eigenvalue E is positive and is related to k via

E =h2 k2

2 m> 0 (2373)

The radial equation has the non-normalizable solution

Rl(r) = jl(kr) (2374)

corresponding to the spherical Bessel function. This energy eigenstate is ex-tended and corresponds to the continuous portion of the energy spectrum.

von Neumann and Wigner suggested that any continuum state may be mod-ified so as to produce a localized or normalizable state, by multiplying by somefunction f(r) which vanishes faster than r−

12 as r → ∞. As they showed, it

is possible to construct a potential such that the modified state is an eigenstateof the Hamiltonian with the same value of the energy. Extending these resultsto finite values of l, Stillinger and Herrick50 expressed the radial function as

Rl(r) = jl(kr) f(r) (2375)

Since this wave function is required to satisfy the energy eigenvalue equationwith fixed quantum numbers l and E

− 1r2

∂

∂r

(r2

∂Rl

∂r

) (l ( l + 1 )

r2+

2 mh2 V (r)

)Rl = k2 Rl (2376)

49J. von Neumann and E. Wigner, Phys. Z. 30, 465 (1929).50F.H. Stillinger and D.R. Herrick, Phys. Rev. A, 11 446 (1976).

493

the potential is specified in terms of derivatives of the eigenfunction

2 mh2 V (r) = 2 k

(j′l(kr)jl(kr)

+1k r

)f ′(r)f(r)

+f”(r)f(r)

(2377)

The potential falls to zero at infinity, if both f ′(r)f(r) and f”(r)

f(r) vanish as r →∞. For an arbitrary decaying form of f(r), the potential would have polesat the zeros of jl(kr). If the potential is required to be finite, then the polesoriginating from the zeros of jl(kr) should be suppressed by the zeros of f ′(r)

f(r) ,and furthermore f(r) should have no zeros. This suggests that the function f(r)should be an analytic function of the variable s(r), where

s(r) = k

∫ r

0

dr′ ( kr′ )2l+3 j2l (kr′)

=1

4 ( l + 1 )( kr )2l+4

[j2l (kr) + j2l+1(kr)

](2378)

This has the effect that derivative of s(r) vanishes at the zeros of jl(kr), buts(r) never decreases. The function f(r) may be chosen as

f(r) =1

A2 + s(r)(2379)

so that f(r) vanish at infinity and Rl(r) is normalizable. Due to the multiplica-tive factor of jl(kr) the wave function has an infinite number of nodes. Thepotential is then found as

V (r) = k2

[( kr )4l+6 j4l (kr)( A2 + s(r) )2

−2 ( kr )2l+3 jl(kr) j′l(kr) + ( l + 5

2 ) ( kr )2l+2 j2l (kr)A2 + s(r)

](2380)

The long-ranged behavior of the potential is then found to be given by

V (r) ∼ ( − 1 )l 8 ( l + 1 ) k2 sin 2 kr2 kr

(2381)

The absence of an outgoing “spherical” wave at infinity can be thought of asbeing caused by the infinite sequence of reflections by subsequent maxima inthe potential. At each maximum, there is always some reflection, even thoughthe particle may have an energy greater than the potential barriers. Indeed, theenergy of the bound state may even exceed the maximum value of the potential.In this truly exceptional case, even a classical particle would never be localized.

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C: The Symmetric Rotor

We shall consider the quantum mechanics describing the rotation of a rigidbody. Initially, the body is assumed to have non-identical moments of inertia

494

Radial wave function for l = 0

-0.1

0

0.1

0.2

0.3

0 0.5 1 1.5 2

k r / π

R0(

r)

Figure 125: The radial wave function R0(r) for an l = 0 bound state withpositive energy considered by von Neumann and Wigner.

I1, I2 and I3. We shall describe the orientation of the rigid body by the Eulerangles (α, β, γ) which describe the orientation of a set of axes embedded in thebody with respect to an external fixed frame of reference.

The Euler angles are angles which describe three successive rotations of thebody, which bring it from a configuration in which the axes embedded in thebody coincide with the axes of the fixed reference frame. The first angle αdescribes the first rotation of the body about the z-axis. This rotation rotatesthe x and y axes embedded in the body into new positions, x′ and y′. Thesecond rotation usually is prescribed as a rotation about the y′-axis, throughthe angle β. The second rotation brings the z-axis into the position z′′. Thez′′-axis has polar coordinates (β, α). The last rotation is a rotation through anangle γ about the z′′-axis.

The rotation can be represented by three by three matrices which acts onthe Cartesian coordinates (x1, x2, x3) of a vector. Under the rotation, the pointr rigidly attached to the body is transformed to the point r′, by

r′ = R(α, β, γ) r= Rz′′(γ) Ry′(β) Rz(α) r (2382)

The rotation through the angle α about the z-axis is represented by the three

495

Oscillating Potential

-4

-2

0

2

4

6

0 0.5 1 1.5 2

k r / π

V(r

) [un

its o

f h2 k2 /m

]

E

Figure 126: The oscillating potential V (r) of von Neumann and Wigner thatproduces an l = 0 bound state with positive energy.

by three matrix

Rz(α) =

cosα − sinα 0sinα cosα 0

0 0 1

(2383)

The rotation through an angle β about the y′-axis is described by

Ry′(β) = Rz(α) Ry(β) Rz(α)−1 (2384)

where

Ry(β) =

cosβ 0 sinβ0 1 0

− sinβ 0 cosβ

(2385)

The final rotation is the rotation about the angle γ about the z′′ axis. This isrepresented by

Rz′′(γ) = Ry′(β) Rz(α) Rz(γ) Rz(α)−1 Ry′(β)−1

= Rz(α) Ry(β) Rz(γ) Ry(β)−1 Rz(α)−1 (2386)

where

Rz(γ) =

cos γ − sin γ 0sin γ cos γ 0

0 0 1

(2387)

496

Hence, the combined rotation is given by the product of the rotations about thefixed axes as

R(α, β, γ) = Rz(α) Ry(β) Rz(γ) (2388)

and has the representation

R(α, β, γ) =

cosα − sinα 0sinα cosα 0

0 0 1

cosβ cos γ − cosβ sin γ sinβsin γ cos γ 0

− sinβ cos γ sinβ sin γ cosβ

=

cosα cosβ cos γ − sinα sin γ − cosα cosβ sin γ − sinα cos γ cosα sinβsinα cosβ cos γ + cosα sin γ − sinα cosβ sin γ + cosα cos γ sinα sinβ

− sinβ cos γ sinβ sin γ cosβ

The angle of rotation ψ can be found by evaluating the trace. The trace is foundas

cosβ + cos(α+ γ) ( 1 + cosβ ) (2389)

which is equal to 1 + 2 cosψ. Hence,

cosψ = cos2β

2cos(α+ γ) − sin2 β

2(2390)

The kinetic energy of the symmetric rotor is described by the Lagrangian

L =12

(I3 ω

23 + I2 ω

22 + I1 ω

21

)(2391)

where ω1, ω2 and ω3 are the components of the angular velocity of rotation aboutthe three body-fixed principal axes. We shall first express this Lagrangian interms of the Euler angles and then determine the Hamiltonian. The angularvelocity vector ω can be expressed as

ω = α eα + β eβ + γ eγ (2392)

The unit vectors eα, eβ , eγ are to be expressed in terms of the unit vectorsin the body-fixed coordinate system. Since the angular velocity γ takes placearound the body fixed z-axis (z′′), one has

eγ = e3 (2393)

The angular velocity β occurs around the y′-axis. The y′-axis is in the e1 − e2plane, since the rotation γ is around e3 and keeps the e1 − e2 plane invariant.The azimuthal angle of the y′ axis is π

2 − γ. Hence

eβ = e1 cos(π

2− γ) + e2 sin(

π

2− γ)

= e1 sin γ + e2 cos γ (2394)

497

The final component of the rotation is around the α axis, which is aligned alongthe original z-axis. The eα axis has polar angle β and azimuthal angle π − γrelative to the fixed body axes.

eα = sinβ(− cos γ e1 + sin γ e2

)+ e3 cosβ (2395)

Hence, the components of the angular velocity with respect to the body-fixedaxes are expressed in terms of the Euler angles via

ω1 = − sinβ cos γ α + sin γ βω2 = sinβ sin γ α + cos γ βω3 = cosβ α + γ (2396)

Hence, the Lagrangian is given as

L =12

[I1

(− sinβ cos γ α+sin γ β

)2

+ I2

(sinβ sin γ α+ cos γ β

)2

+ I3

(cosβ α+ γ

)2 ](2397)

For the symmetric top, one has I1 = I2, therefore, the Lagrangian simplifiesto

L =12

[I1

(sin2 β α2 + β2

)+ I3

(cosβ α + γ

)2 ](2398)

The momenta are given by

pα =∂L

∂α

= I1 sin2 β α + I3 cosβ(

cosβ α + γ

)(2399)

andpβ =

∂L

∂β= I1 β (2400)

and finally

pγ =∂L

∂β= I3

(cosβ α + γ

)(2401)

The Hamiltonian H is given by

H = pα α + pβ β + pγ γ − L

=1

2 I1p2

β +1

2 I1 sin2 β

(pα − cosβ pγ

)2

+1

2 I3p2

γ (2402)

This Hamiltonian is quantized as

H =1

2 I1p2

β +1

2 I1 sin2 β

(pα − cosβ pγ

)2

+1

2 I3p2

γ (2403)

498

It is seen that pα and pγ commute with the Hamiltonian and, therefore, one canfind simultaneous eigenstates of pα, pγ and H. The simultaneous eigenfunctionsare written in the form

Ψ(α, β, γ) =(

12 π

)exp

[i

h( α pα + γ pγ )

]φ(β) (2404)

Since the wave functions are required to be single-valued, the quantum numbersare determined to be pα = mα h and pγ = mγ h where (mα,mγ) are eitherboth integers or they are both half-integers. On substituting the above form,the expression for the eigenvalue equation reduces to

12 I1

[p2

β +1

sin2 β

(pα − cosβ pγ

)2 ]φ(β) =

(E − 1

2 I3p2

γ

)φ(β) (2405)

Since β is a polar angle, the above equation takes the explicit form

− h2

2 I11

sinβ∂

∂β

(sinβ

∂φ

∂β

)+

h2

2 I1 sin2 β

(mα − cosβ mγ

)2

φ(β) =(E −

h2 m2γ

2 I3

)φ(β)

(2406)or

− sinβ∂

∂β

(sinβ

∂φ

∂β

)+[ (

mα − cosβ mγ

)2

+ sin2 β

(I1I3m2

γ −2 I1 Eh2

) ]φ(β) = 0

(2407)Clearly, this can be rewritten as an eigenvalue equation in terms of the variablez = cosβ, where z is restricted to the range 1 ≥ z ≥ − 1. Therefore, we have

− ( 1− z2 )∂

∂z

(( 1− z2 )

∂φ(β)∂z

)+[m2

α +m2γ − λ− 2mα mγ z + λ z2

]φ(β) = 0

(2408)where λ is related to the energy eigenvalue via

λ =2 I1 Eh2 +

(1− I1

I3

)m2

γ (2409)

The eigenvalue equation has regular singular points at z = ± 1. Near thesepoints, the solution has a non-analytic form, and is expressed as

φ(β) = ( 1 − z )µ ( 1 + z )ν F (z) (2410)

where F (z) is an analytic function, and the indices µ and ν are to be determined.The indices are determined by considering the most singular terms, and arefound to be given by

µ = ±(mα −mγ

2

)(2411)

and

ν = ±(mα +mγ

2

)(2412)

499

The solutions with the negative values for the exponents are discarded since thesolution is required to be finite and normalizable. The function F (z) then mustsatisfy an equation of the form

− ( 1− z2 )∂2F

∂z2+ 2

[( 1 +mα ) z −mγ

]∂F

∂z+(mα (mα + 1 )− λ

)F = 0

(2413)For convenience, we have assumed that mα and mγ are both positive and thatmα > mγ . The eigenvalue λ is found by using the Frobenius method, expand-ing in powers about a singular point and requiring that the series expansionterminates, yielding a polynomial for F (z). This requires that

( mα +12

) ±√

14

+ λ (2414)

is a positive integer. Hence, we find that

λ = j ( j + 1 ) (2415)

and the energy eigenvalue is given by

E =h2

2 I1j (j + 1 ) +

h2 m2α

2 I3− h2 m2

α

2 I1(2416)

as could have been expected.

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“ c©2005 by Peter S. Riseborough”

500