Quantum MechanicsLecture Notes forChemistry 6312

Quantum Chemistry

Eric R. BittnerUniversity of HoustonDepartment of Chemistry

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Lecture Notes on Quantum ChemistryLecture notes to accompany Chemistry 6321

Copyright @1997-2003, University of Houston and Eric R. BittnerAll Rights Reserved.

August 12, 2003

Contents

0 Introduction 80.1 Essentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100.2 Problem Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110.3 2003 Course Calendar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

I Lecture Notes 14

1 Survey of Classical Mechanics 151.1 Newton’s equations of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.1.1 Elementary solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.1.2 Phase plane analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.2 Lagrangian Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.2.1 The Principle of Least Action . . . . . . . . . . . . . . . . . . . . . . . . . 181.2.2 Example: 3 dimensional harmonic oscillator in spherical coordinates . . . . 20

1.3 Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231.4 Hamiltonian Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

1.4.1 Interaction between a charged particle and an electromagnetic field. . . . . 241.4.2 Time dependence of a dynamical variable . . . . . . . . . . . . . . . . . . . 261.4.3 Virial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2 Waves and Wavefunctions 292.1 Position and Momentum Representation of |ψ〉 . . . . . . . . . . . . . . . . . . . . 292.2 The Schrodinger Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.2.1 Gaussian Wavefunctions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322.2.2 Evolution of ψ(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

2.3 Particle in a Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.3.1 Infinite Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.3.2 Particle in a finite Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382.3.3 Scattering states and resonances. . . . . . . . . . . . . . . . . . . . . . . . 402.3.4 Application: Quantum Dots . . . . . . . . . . . . . . . . . . . . . . . . . . 43

2.4 Tunneling and transmission in a 1D chain . . . . . . . . . . . . . . . . . . . . . . 492.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 492.6 Problems and Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

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3 Semi-Classical Quantum Mechanics 553.1 Bohr-Sommerfield quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563.2 The WKB Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

3.2.1 Asymptotic expansion for eigenvalue spectrum . . . . . . . . . . . . . . . . 583.2.2 WKB Wavefunction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 603.2.3 Semi-classical Tunneling and Barrier Penetration . . . . . . . . . . . . . . 62

3.3 Connection Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 653.4 Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

3.4.1 Classical Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 703.4.2 Scattering at small deflection angles . . . . . . . . . . . . . . . . . . . . . . 733.4.3 Quantum treatment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 743.4.4 Semiclassical evaluation of phase shifts . . . . . . . . . . . . . . . . . . . . 753.4.5 Resonance Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

3.5 Problems and Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

4 Postulates of Quantum Mechanics 804.0.1 The description of a physical state: . . . . . . . . . . . . . . . . . . . . . . 854.0.2 Description of Physical Quantities: . . . . . . . . . . . . . . . . . . . . . . 854.0.3 Quantum Measurement: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 864.0.4 The Principle of Spectral Decomposition: . . . . . . . . . . . . . . . . . . . 864.0.5 The Superposition Principle . . . . . . . . . . . . . . . . . . . . . . . . . . 874.0.6 Reduction of the wavepacket: . . . . . . . . . . . . . . . . . . . . . . . . . 894.0.7 The temporal evolution of the system: . . . . . . . . . . . . . . . . . . . . 904.0.8 Dirac Quantum Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

4.1 Dirac Notation and Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . 944.1.1 Transformations and Representations . . . . . . . . . . . . . . . . . . . . . 944.1.2 Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 964.1.3 Products of Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 974.1.4 Functions Involving Operators . . . . . . . . . . . . . . . . . . . . . . . . . 98

4.2 Constants of the Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1004.3 Bohr Frequency and Selection Rules . . . . . . . . . . . . . . . . . . . . . . . . . . 1014.4 Example using the particle in a box states . . . . . . . . . . . . . . . . . . . . . . 1024.5 Time Evolution of Wave and Observable . . . . . . . . . . . . . . . . . . . . . . . 1034.6 “Unstable States” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1044.7 Problems and Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

5 Bound States of The Schrodinger Equation 1105.1 Introduction to Bound States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1105.2 The Variational Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

5.2.1 Variational Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1125.2.2 Constraints and Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . 1145.2.3 Variational method applied to Schrodinger equation . . . . . . . . . . . . . 1175.2.4 Variational theorems: Rayleigh-Ritz Technique . . . . . . . . . . . . . . . . 1185.2.5 Variational solution of harmonic oscillator ground State . . . . . . . . . . . 119

5.3 The Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

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5.3.1 Harmonic Oscillators and Nuclear Vibrations . . . . . . . . . . . . . . . . . 1245.3.2 Classical interpretation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1315.3.3 Molecular Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

5.4 Numerical Solution of the Schrodinger Equation . . . . . . . . . . . . . . . . . . . 1365.4.1 Numerov Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1365.4.2 Numerical Diagonalization . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

5.5 Problems and Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

6 Quantum Mechanics in 3D 1526.1 Quantum Theory of Rotations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1536.2 Eigenvalues of the Angular Momentum Operator . . . . . . . . . . . . . . . . . . 1576.3 Eigenstates of L2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1586.4 Eigenfunctions of L2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1596.5 Addition theorem and matrix elements . . . . . . . . . . . . . . . . . . . . . . . . 1626.6 Legendre Polynomials and Associated Legendre Polynomials . . . . . . . . . . . . 1646.7 Quantum rotations in a semi-classical context . . . . . . . . . . . . . . . . . . . . 1656.8 Motion in a central potential: The Hydrogen Atom . . . . . . . . . . . . . . . . . 170

6.8.1 Radial Hydrogenic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 1736.9 Spin 1/2 Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173

6.9.1 Theoretical Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1746.9.2 Other Spin Observables . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1756.9.3 Evolution of a state . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1756.9.4 Larmor Precession . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176

6.10 Problems and Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177

7 Perturbation theory 1807.1 Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1807.2 Two level systems subject to a perturbation . . . . . . . . . . . . . . . . . . . . . 182

7.2.1 Expansion of Energies in terms of the coupling . . . . . . . . . . . . . . . 1837.2.2 Dipole molecule in homogenous electric field . . . . . . . . . . . . . . . . . 184

7.3 Dyson Expansion of the Schrodinger Equation . . . . . . . . . . . . . . . . . . . . 1887.4 Van der Waals forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190

7.4.1 Origin of long-ranged attractions between atoms and molecules . . . . . . . 1907.4.2 Attraction between an atom a conducting surface . . . . . . . . . . . . . . 192

7.5 Perturbations Acting over a Finite amount of Time . . . . . . . . . . . . . . . . . 1937.5.1 General form of time-dependent perturbation theory . . . . . . . . . . . . 1937.5.2 Fermi’s Golden Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194

7.6 Interaction between an atom and light . . . . . . . . . . . . . . . . . . . . . . . . 1977.6.1 Fields and potentials of a light wave . . . . . . . . . . . . . . . . . . . . . 1977.6.2 Interactions at Low Light Intensity . . . . . . . . . . . . . . . . . . . . . . 1987.6.3 Photoionization of Hydrogen 1s . . . . . . . . . . . . . . . . . . . . . . . . 2027.6.4 Spontaneous Emission of Light . . . . . . . . . . . . . . . . . . . . . . . . 204

7.7 Time-dependent golden rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2097.7.1 Non-radiative transitions between displaced Harmonic Wells . . . . . . . . 2107.7.2 Semi-Classical Evaluation . . . . . . . . . . . . . . . . . . . . . . . . . . . 216

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7.8 Problems and Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219

8 Many Body Quantum Mechanics 2228.1 Symmetry with respect to particle Exchange . . . . . . . . . . . . . . . . . . . . . 2228.2 Matrix Elements of Electronic Operators . . . . . . . . . . . . . . . . . . . . . . . 2278.3 The Hartree-Fock Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . 229

8.3.1 Two electron integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2308.3.2 Koopman’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231

8.4 Quantum Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2318.4.1 The Born-Oppenheimer Approximation . . . . . . . . . . . . . . . . . . . . 232

8.5 Problems and Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241

A Physical Constants and Conversion Factors 247

B Mathematical Results and Techniques to Know and Love 249B.1 The Dirac Delta Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249

B.1.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249B.1.2 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250B.1.3 Spectral representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250

B.2 Coordinate systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253B.2.1 Cartesian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253B.2.2 Spherical . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254B.2.3 Cylindrical . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255

C Mathematica Notebook Pages 256

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List of Figures

1.1 Tangent field for simple pendulum with ω = 1. The superimposed curve is a linearapproximation to the pendulum motion. . . . . . . . . . . . . . . . . . . . . . . . 17

1.2 Vector diagram for motion in a central forces. The particle’s motion is along theZ axis which lies in the plane of the page. . . . . . . . . . . . . . . . . . . . . . . 21

1.3 Screen shot of using Mathematica to plot phase-plane for harmonic oscillator.Here k/m = 1 and our xo = 0.75. . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2.1 A gaussian wavepacket, ψ(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332.2 Momentum-space distribution of ψ(k). . . . . . . . . . . . . . . . . . . . . . . . . 332.3 Go for fixed t as a function of x. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352.4 Evolution of a free particle wavefunction. . . . . . . . . . . . . . . . . . . . . . . 362.5 Particle in a box states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382.6 Graphical solution to transendental equations for an electron in a truncated hard

well of depth Vo = 10 and width a = 2. The short-dashed blue curve correspondsto the symmetric case and the long-dashed blue curve corresponds to the asymetric

case. The red line is√

1− V o/E. Bound state solution are such that the red andblue curves cross. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

2.7 Transmission (blue) and Reflection (red) coefficients for an electron scattering overa square well (V = −40 and a = 1 ). . . . . . . . . . . . . . . . . . . . . . . . . . 42

2.8 Transmission Coefficient for particle passing over a bump. . . . . . . . . . . . . . 432.9 Scattering waves for particle passing over a well. . . . . . . . . . . . . . . . . . . . 442.10 Argand plot of a scattering wavefunction passing over a well. . . . . . . . . . . . . 452.11 Density of states for a 1-, 2- , and 3- dimensional space. . . . . . . . . . . . . . . . 462.12 Density of states for a quantum well and quantum wire compared to a 3d space.

Here L = 5 and s = 2 for comparison. . . . . . . . . . . . . . . . . . . . . . . . . . 472.13 Spherical Bessel functions, j0, j1, and j1 (red, blue, green) . . . . . . . . . . . . . 482.14 Radial wavefuncitons (left column) and corresponding PDFs (right column) for an

electron in a R = 0.5Aquantum dot. The upper two correspond to (n, l) = (1, 0)(solid) and (n, l) = (1, 1) (dashed) while the lower correspond to (n, l) = (2, 0)(solid) and (n, l) = (2, 1) (dashed) . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

3.1 Eckart Barrier and parabolic approximation of the transition state . . . . . . . . . 633.2 Airy functions, Ai(y) (red) and Bi(y) (blue) . . . . . . . . . . . . . . . . . . . . . 663.3 Bound states in a graviational well . . . . . . . . . . . . . . . . . . . . . . . . . . 693.4 Elastic scattering trajectory for classical collision . . . . . . . . . . . . . . . . . . 70

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3.5 Form of the radial wave for repulsive (short dashed) and attractive (long dashed)potentials. The form for V = 0 is the solid curve for comparison. . . . . . . . . . . 76

4.1 Gaussian distribution function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 814.2 Combination of two distrubitions. . . . . . . . . . . . . . . . . . . . . . . . . . . . 824.3 Constructive and destructive interference from electron/two-slit experiment. The

superimposed red and blue curves are P1 and P2 from the classical probabilities . 834.4 The diffraction function sin(x)/x . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

5.1 Variational paths between endpoints. . . . . . . . . . . . . . . . . . . . . . . . . . 1165.2 Hermite Polynomials, Hn up to n = 3. . . . . . . . . . . . . . . . . . . . . . . . . 1285.3 Harmonic oscillator functions for n = 0 to 3 . . . . . . . . . . . . . . . . . . . . . 1325.4 Quantum and Classical Probability Distribution Functions for Harmonic Oscillator.1335.5 London-Eyring-Polanyi-Sato (LEPS) empirical potential for the F+H2 → FH+H

chemical reaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1355.6 Morse well and harmonic approximation for HF . . . . . . . . . . . . . . . . . . . 1365.7 Model potential for proton tunneling. . . . . . . . . . . . . . . . . . . . . . . . . . 1375.8 Double well tunneling states as determined by the Numerov approach. . . . . . . . 1385.9 Tchebyshev Polynomials for n = 1− 5 . . . . . . . . . . . . . . . . . . . . . . . . 1405.10 Ammonia Inversion and Tunneling . . . . . . . . . . . . . . . . . . . . . . . . . . 147

6.1 Vector model for the quantum angular momentum state |jm〉, which is representedhere by the vector j which precesses about the z axis (axis of quantzation) withprojection m. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156

6.2 Spherical Harmonic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1606.3 Classical and Quantum Probability Distribution Functions for Angular Momentum.168

7.1 Variation of energy level splitting as a function of the applied field for an ammoniamolecule in an electric field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

7.2 Photo-ionization spectrum for hydrogen atom. . . . . . . . . . . . . . . . . . . . . 205

8.1 Various contributions to the H+2 Hamiltonian. . . . . . . . . . . . . . . . . . . . . 236

8.2 Potential energy surface for H+2 molecular ion. . . . . . . . . . . . . . . . . . . . . 238

8.3 Three dimensional representations of ψ+ and ψ− for the H+2 molecular ion. . . . . 238

8.4 Setup calculation dialog screen . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2438.5 HOMO-1, HOMO and LUMO for CH2 = O. . . . . . . . . . . . . . . . . . . . . . 2458.6 Transition state geometry for H2 +C = O → CH2 = O. The Arrow indicates the

reaction path. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246

B.1 sin(xa)/πx representation of the Dirac δ-function . . . . . . . . . . . . . . . . . . 251B.2 Gaussian representation of δ-function . . . . . . . . . . . . . . . . . . . . . . . . . 251

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List of Tables

3.1 Location of nodes for Airy, Ai(x) function. . . . . . . . . . . . . . . . . . . . . . . 68

5.1 Tchebychev polynomials of the first type . . . . . . . . . . . . . . . . . . . . . . . 1405.2 Eigenvalues for double well potential computed via DVR and Numerov approaches 143

6.1 Spherical Harmonics (Condon-Shortley Phase convention. . . . . . . . . . . . . . . 1606.2 Relation between various notations for Clebsch-Gordan Coefficients in the literature169

8.1 Vibrational Frequencies of Formaldehyde . . . . . . . . . . . . . . . . . . . . . . . 244

A.1 Physical Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248A.2 Atomic Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248A.3 Useful orders of magnitude . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248

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Chapter 0

Introduction

Nothing conveys the impression of humungous intellect so much as even the sketchiestknowledge of quantum physics, and since the sketchiest knowledge is all anyone willever have, never be shy about holding forth with bags of authority about subatomicparticles and the quantum realm without having done any science whatsoever.Jack Klaff –Bluff Your Way in the Quantum Universe

The field of quantum chemistry seeks to provide a rigorous description of chemical processesat its most fundamental level. For ordinary chemical processes, the most fundamental and un-derlying theory of chemistry is given by the time-dependent and time-independent version of theSchrodinger equation. However, simply stating an equation that provides the underlying theoryin now shape or form yields and predictive or interpretive power. In fact, most of what we doin quantum mechanics is to develop a series of well posed approximation and physical assump-tions to solve basic equations of quantum mechanics. In this course, we will delve deeply intothe underlying physical and mathematical theory. We will learn how to solve some elementaryproblems and apply these to not so elementary examples.

As with any course of this nature, the content reflects the instructors personal interests inthe field. In this case, the emphasis of the course is towards dynamical processes, transitionsbetween states, and interaction between matter and radiation. More “traditional” quantumchemistry courses will focus upon electronic structure. In fact, the moniker “quantum chemistry”typically refers to electronic structure theory. While this is an extremely rich topic, it is mypersonal opinion that a deeper understanding of dynamical processes provides a broader basisfor understanding chemical processes.

It is assumed from the beginning, that students taking this course have had some exposureto the fundamental principles of quantum theory as applied to chemical systems. This is usuallyin the context of a physical chemistry course or a separate course in quantum chemistry. Ialso assume that students taking this course have had undergraduate level courses in calculus,differential equations, and have some concepts of linear algebra. Students lacking in any of theseareas are strongly encouraged to sit through my undergraduate Physical Chemistry II course(offered in the Spring Semester at the Univ. of Houston) before attempting this course. Thiscourse is by design and purpose theoretical in nature.

The purpose of this course is to provide a solid and mathematically rigorous tour throughmodern quantum mechanics. We will begin with simple examples which can be worked out

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exactly on paper and move on to discuss various approximation schemes. For cases in whichanalytical solutions are either too obfuscating or impossible, computer methods will be intro-duced using Mathematica. Applications toward chemically relevant topics will be emphasizedthroughout.

We will primarily focus upon single particle systems, or systems in which the particles aredistinguishable. Special considerations for systems of indistinguishable particles, such as theelectrons in a molecule, will be discussed towards the end of the course. The pace of the courseis fairly rigorous, with emphasis on solving problems either analytically or using computer.

I also tend to emphasize how to approach a problem from a theoretical viewpoint. As youwill discover rapidly, very few of the problem sets in this course are of the “look-up the rightformula” type. Rather, you will need to learn to use the various techniques (perturbation theory,commutation relations, etc...) to solve and work out problems for a variety of physical systems.

The lecture notes in this package are really to be regarded as a work in progress and updatesand additions will be posted as they evolve. Lacking is a complete chapter on the Hydrogenatom and atomic physics and a good overview of many body theory. Also, I have not includeda chapter on scattering and other topics as these will be added over the course of time. Certainsections are clearly better than others and will be improved upon over time. Each chapter endswith a series of exercises and suggested problems Some of which have detailed solutions. Others,you should work out on your own. At the end of this book are a series of Mathematica notebooksI have written which illustrate various points and perform a variety of calculations. These can bedownloaded from my web-site (http://k2.chem.uh.edu/quantum/) and run on any recent versionof Mathematica. (≥ v4.n).

It goes entirely without saying (but I will anyway) that these notes come from a wide varietyof sources which I have tried to cite where possible.

9

0.1 Essentials

• Instructor: Prof. Eric. R. Bittner.

• Office: Fleming 221 J

• Email: bittner@uh.edu

• Phone: -3-2775

• Office Hours: Monday and Thurs. afternoons or by appointment.

• Course Web Page: http://k2.chem.uh.edu/quantum/ Solution sets, course news, classnotes, sample computer routines, etc...will be posted from time to time on this web-page.

• Other Required Text: Quantum Mechanics, Landau and Lifshitz. This is volume 3 ofL&L’s classical course in modern physics. Every self-respecting scientist has at least two orthree of their books on their book-shelves. This text tends to be pretty terse and uses theclassical phrase it is easy to show... quite a bit (it usually means just the opposite). Theproblems are usually worked out in detail and are usually classic applications of quantumtheory. This is a land-mark book and contains everything you really need to know to doquantum mechanics.

• Recommended Texts: I highly recommend that you use a variety of books since oneauthor’s approach to a given topic may be clearer than another’s approach.

– Quantum Mechanics, Cohen-Tannoudji, et al. This two volume book is very compre-hensive and tends to be rather formal (and formidable) in its approach. The problemsare excellent.

–

– Lectures in Quantum Mechanics, Gordon Baym. Baym’s book covers a wide range oftopics in a lecture note style.

– Quantum Chemistry, I. Levine. This is usually the first quantum book that chemistsget. I find it to be too wordy and the notation and derivations a bit ponderous. Levinedoes not use Dirac notation. However, he does give a good overview of elementaryelectronic structure theory and some if its important developments. Good for startingoff in electronic structure.

– Modern Quantum Mechanics, J. J. Sakurai. This is a real classic. Not good for a firstexposure since it assumes a fairly sophisticated understanding of quantum mechanicsand mathematics.

– Intermediate Quantum Mechanics, Hans Bethe and Roman Jackiw. This book is agreat exploration of advanced topics in quantum mechanics as illustrated by atomicsystems.

10

– What is Quantum Mechanics?, Transnational College of LEX. OK, this one I foundat Barnes and Noble and it’s more or less a cartoon book. But, it is really good. Itexplores the historical development of quantum mechanics, has some really interest-ing insights into semi-classical and ”old” quantum theory, and presents the study ofquantum mechanics as a unfolding story. This book I highly recommend if thisis the first time you are taking a course on quantum mechanics.

– Quantum Mechanics in Chemistry by George Schatz and Mark Ratner. Ratner andSchatz have more in terms of elementary quantum chemistry, emphasizing the use ofmodern quantum chemical computer programs than almost any text I have reviewed.

• Prequisites: Graduate status in chemistry. This course is required for all Physical Chem-istry graduate students. The level of the course will be fairly rigorous and I assume that stu-dents have had some exposure to quantum mechanics at the undergraduate level–typicallyin Physical Chemistry, and are competent in linear algebra, calculus, and solving elemen-tary differential equations.

• Tests and Grades: There are no exams in this course, only problem sets and participationin discussion. This means coming to lecture prepared to ask and answer questions. Mygrading policy is pretty simple. If you make an honest effort, do the assigned problems(mostly correctly), and participate in class, you will be rewarded with at least a B. Ofcourse this is the formula for success for any course.

0.2 Problem Sets

Your course grade will largely be determined by your performance on these problems as well asthe assigned discussion of a particular problem. My philosophy towards problem sets is that thisis the only way to really learn this material. These problems are intentionally challenging, butnot overwhelming, and are paced to correspond to what will be going on in the lecture.

Some ground rules:

1. Due dates are posted on each problem–usually 1 week or 2 weeks after they are assigned.Late submissions may be turned in up to 1 week later. All problems must be turned in byDecember 3. I will not accept any submissions after that date.

2. Handwritten Problems. If I can’t read it, I won’t grade it. Period. Consequently, I stronglyencourage the use of word processing software for your final submission. Problem solutionscan be submitted electronically as Mathematica, Latex, or PDF files to bittner@uh.edu withthe subject: QUANTUM PROBLEM SET. Do not send me a MSWord file as an emailattachment. I expect some text (written in compete and correct sentences) to explain yoursteps where needed and some discussion of the results. The computer lab in the basementof Fleming has 20 PCs with copies of Mathematica or you can obtain your own licensefrom the University Media Center.

3. Collaborations. You are strongly encouraged to work together and collaborate on problems.However, simply copying from your fellow student is not an acceptable collaboration.

11

4. These are the only problems you need to turn in. We will have additional exercises–mostlycoming from the lecture. Also, at the end of the lectures herein, are a set of suggestedproblems and exercises to work on. Many of these have solutions.

12

0.3 2003 Course Calendar

This is a rough schedule of topics we will cover. In essence we will follow the starting froma basic description of quantum wave mechanics and bound states. We will then move ontothe more formal aspects of quantum theory: Dirac notation, perturbation theory, variationaltheory, and the like. Lastly, we move onto applications: Hydrogen atom, many-electron systems,semi-classical approximations, and a semi-classical treatment of light absorption and emission.

We will also have a recitation session in 221 at 10am Friday morning. The purpose of thiswill be to specifically discuss the problem sets and other issues.

• 27-August: Course overview: Classical Concepts

• 3-Sept: Finishing Classical Mechanics/Elementary Quantum concepts

• 8-Sept: Particle in a box and hard wall potentials (Perry?)

• 10 Sept: Tunneling/Density of states (Perry)

• 15/17 Bohr-Sommerfield Quantization/Old quantum theory/connection to classical me-chanics (Perry)

• 22/24 Sept: Semiclassical quantum mechanics: WKB Approx. Application to scattering

• 29 Sept/1 Oct. Postulates of quantum mechanics: Dirac notation, superposition principle,simple calculations.

• 6/8 Oct: Bound States: Variational principle, quantum harmonic oscillator

• 13/15 Oct: Quantum mechanics in 3D: Angular momentum (Chapt 4.1-4.8)

• 20/22 Oct: Hydrogen atom/Hydrogenic systems/Atomic structure

• 27/29 Oct: Perturbation Theory:

• 3/5 Nov: Time-dependent Perturbation Theory:

• 10/12 Identical Particles/Quantum Statistics

• 17/19 Nov: Helium atom, hydrogen ion

• 24/26 Nov: Quantum Chemistry

• 3 Dec–Last day to turn in problem sets

• Final Exam: TBA

13

Part I

Lecture Notes

14

Chapter 1

Survey of Classical Mechanics

Quantum mechanics is in many ways the cumulation of many hundreds of years of work andthought about how mechanical things move and behave. Since ancient times, scientists havewondered about the structure of matter and have tried to develop a generalized and underlyingtheory which governs how matter moves at all length scales.

For ordinary objects, the rules of motion are very simple. By ordinary, I mean objects thatare more or less on the same length and mass scale as you and I, say (conservatively) 10−7m to106 m and 10−25g to 108g moving less than 20% of the speed of light. On other words, almosteverything you can see and touch and hold obey what are called “classical” laws of motion. Theterm “classical” means that that the basic principles of this class of motion have their foundationin antiquity. Classical mechanics is a extremely well developed area of physics. While you maythink that given that classical mechanics has been studied extensively for hundreds of yearsthere really is little new development in this field, it remains a vital and extremely active area ofresearch. Why? Because the majority of universe “lives” in a dimensional realm where classicalmechanics is extremely valid. Classical mechanics is the workhorse for atomistic simulationsof fluids, proteins, polymers. It provides the basis for understanding chaotic systems. It alsoprovides a useful foundation of many of the concepts in quantum mechanics.

Quantum mechanics provides a description of how matter behaves at very small length andmass scales: i.e. the realm of atoms, molecules, and below. It was developed over the last centuryto explain a series of experiments on atomic systems that could not be explained using purelyclassical treatments. The advent of quantum mechanics forced us to look beyond the classicaltheories. However, it was not a drastic and complete departure. At some point, the two theoriesmust correspond so that classical mechanics is the limiting behavior of quantum mechanics formacroscopic objects. Consequently, many of the concepts we will study in quantum mechanicshave direct analogs to classical mechanics: momentum, angular momentum, time, potentialenergy, kinetic energy, and action.

Much like classical music is in a particular style, classical mechanics is based upon the principlethat the motion of a body can be reduced to the motion of a point particle with a given massm, position x, and velocity v. In this chapter, we will review some of the concepts of classicalmechanics which are necessary for studying quantum mechanics. We will cast these in formwhereby we can move easily back and forth between classical and quantum mechanics. We willfirst discuss Newtonian motion and cast this into the Lagrangian form. We will then discuss theprinciple of least action and Hamiltonian dynamics and the concept of phase space.

15

1.1 Newton’s equations of motion

Newton’s Principia set the theoretical basis of mathematical mechanics and analysis of physicalbodies. The equation that force equals mass times acceleration is the fundamental equation ofclassical mechanics. Stated mathematically

mx = f(x) (1.1)

The dots refer to differentiation with respect to time. We will use this notion for time derivatives.We may also use x′ or dx/dt as well. So,

x =d2x

dt2.

For now we are limiting our selves to one particle moving in one dimension. For motion inmore dimensions, we need to introduce vector components. In cartesian coordinates, Newton’sequations are

mx = fx(x, y, z) (1.2)

my = fy(x, y, z) (1.3)

mz = fz(x, y, z) (1.4)

where the force vector ~f(x, y, z) has components in all three dimensions and varies with location.We can also define a position vector, ~x = (x, y, z), and velocity vector ~v = (x, y, z). We can alsoreplace the second-order differential equation with two first order equations.

x = vx (1.5)

vx = fx/m (1.6)

These, along with the initial conditions, x(0) and v(0) are all that are needed to solve for themotion of a particle with mass m given a force f . We could have chosen two end points as welland asked, what path must the particle take to get from one point to the next. Let us considersome elementary solutions.

1.1.1 Elementary solutions

First the case in which f = 0 and x = 0. Thus, v = x = const. So, unless there is an appliedforce, the velocity of a particle will remain unchanged.

Second, we consider the case of a linear force, f = −kx. This is restoring force for a springand such force laws are termed Hooke’s law and k is termed the force constant. Our equationsare:

x = vx (1.7)

vx = −k/mx (1.8)

or x = −(k/m)x. So we want some function which is its own second derivative multiplied bysome number. The cosine and sine functions have this property, so let’s try

x(t) = A cos(at) +B sin(bt).

16

Taking time derivativesx(t) = −aA sin(at) + bB cos(bt);

x(t) = −a2A cos(at)− b2B sin(bt).

So we get the required result if a = b =√k/m, leaving A and B undetermined. Thus, we need

two initial conditions to specify these coefficients. Let’s pick x(0) = xo and v(0) = 0. Thus,

x(0) = A = xo and B = 0. Notice that the term√k/m has units of angular frequency.

ω =

√k

m

So, our equation of motion are

x(t) = xo cos(ωt) (1.9)

v(t) = −xoω sin(ωt). (1.10)

-2 p -p 0 p 2 px

-3

-2

-1

0

1

2

3

v

-2 p -p 0 p 2 p

Figure 1.1: Tangent field for simple pendulum with ω = 1. The superimposed curve is a linearapproximation to the pendulum motion.

1.1.2 Phase plane analysis

Often one can not determine the closed form solution to a given problem and we need to turnto more approximate methods or even graphical methods. Here, we will look at an extremelyuseful way to analyze a system of equations by plotting their time-derivatives.

First, let’s look at the oscillator we just studied. We can define a vector s = (x, v) =(v,−k/mx) and plot the vector field. Fig. 1.3 shows how to do this in Mathematica. The

17

superimposed curve is one trajectory and the arrows give the “flow” of trajectories on the phaseplane.

We can examine more complex behavior using this procedure. For example, the simplependulum obeys the equation x = −ω2 sin x. This can be reduced to two first order equations:x = v and v = −ω2 sin(x).

We can approximate the motion of the pendulum for small displacements by expanding thependulum’s force about x = 0,

−ω2 sin(x) = −ω2(x− x3

6+ · · ·).

For small x the cubic term is very small, and we have

v = −ω2x = − k

mx

which is the equation for harmonic motion. So, for small initial displacements, we see that thependulum oscillates back and forth with an angular frequency ω. For large initial displacements,xo = π or if we impart some initial velocity on the system vo > 1, the pendulum does not oscillateback and forth, but undergoes librational motion (spinning!) in one direction or the other.

1.2 Lagrangian Mechanics

1.2.1 The Principle of Least Action

The most general form of the law governing the motion of a mass is the principle of least actionor Hamilton’s principle. The basic idea is that every mechanical system is described by a singlefunction of coordinate, velocity, and time: L(x, x, t) and that the motion of the particle is suchthat certain conditions are satisfied. That condition is that the time integral of this function

S =∫ tf

toL(x, x, t)dt

takes the least possible value give a path that starts at xo at the initial time and ends at xf atthe final time.

Lets take x(t) be function for which S is minimized. This means that S must increase forany variation about this path, x(t)+ δx(t). Since the end points are specified, δx(0) = δx(t) = 0and the change in S upon replacement of x(t) with x(t) + δx(t) is

δS =∫ tf

toL(x+ δx, x+ δx, t)dt−

∫ tf

toL(x, x, t)dt = 0

This is zero, because S is a minimum. Now, we can expand the integrand in the first term

L(x+ δx, x+ δx, t) = L(x, x, t) +

(∂L

∂xδx+

∂L

∂xδx

)

Thus, we have ∫ tf

to

(∂L

∂xδx+

∂L

∂xδx

)dt = 0.

18

Since δx = dδx/dt and integrating the second term by parts

δS =

[∂L

δxδx

]tf

to

+∫ tf

to

(∂L

∂x− d

dt

∂L

∂x

)δxdt = 0

The surface term vanishes because of the condition imposed above. This leaves the integral. Ittoo must vanish and the only way for this to happen is if the integrand itself vanishes. Thus wehave the

∂L

∂x− d

dt

∂L

∂x= 0

L is known as the Lagrangian. Before moving on, we consider the case of a free particle.The Lagrangian in this case must be independent of the position of the particle since a freelymoving particle defines an inertial frame. Since space is isotropic, L must only depend upon themagnitude of v and not its direction. Hence,

L = L(v2).

Since L is independent of x, ∂L/∂x = 0, so the Lagrange equation is

d

dt

∂L

∂v= 0.

So, ∂L/∂v = const which leads us to conclude that L is quadratic in v. In fact,

L =1

mv2,

which is the kinetic energy for a particle.

T =1

2mv2 =

1

2mx2.

For a particle moving in a potential field, V , the Lagrangian is given by

L = T − V.

L has units of energy and gives the difference between the energy of motion and the energy oflocation.

This leads to the equations of motion:

d

dt

∂L

∂v=∂L

∂x.

Substituting L = T − V , yields

mv = −∂V∂x

which is identical to Newton’s equations given above once we identify the force as the minus thederivative of the potential. For the free particle, v = const. Thus,

S =∫ tf

to

m

2v2dt =

m

2v2(tf − to).

19

You may be wondering at this point why we needed a new function and derived all this fromsome minimization principle. The reason is that for some systems we have constraints on the typeof motion they can undertake. For example, there may be bonds, hinges, and other mechanicalhinderances which limit the range of motion a given particle can take. The Lagrangian formalismprovides a mechanism for incorporating these extra effects in a consistent and correct way. Infact we will use this principle later in deriving a variational solution to the Schrodinger equationby constraining the wavefunction solutions to be orthonormal.

Lastly, it is interesting to note that v2 = (dl/d)2 = (dl)2/(dt)2 is the square of the elementof an arc in a given coordinate system. Thus, within the Lagrangian formalism it is easy toconvert from one coordinate system to another. For example, in cartesian coordinates: dl2 =dx2 + dy2 + dz2. Thus, v2 = x2 + y2 + z2. In cylindrical coordinates, dl = dr2 + r2dφ2 + dz2, wehave the Lagrangian

L =1

2m(r2 + r2φ2 + z2)

and for spherical coordinates dl2 = dr2 + r2dθ2 + r2 sin2 θdφ2; hence

L =1

2m(r2 + r2θ2 + r2 sin2 θφ2).

1.2.2 Example: 3 dimensional harmonic oscillator in spherical coor-dinates

Here we take the potential energy to be a function of r alone (isotropic)

V (r) = kr2/2.

Thus, the Lagrangian in cartesian coordinates is

L =m

2(x2 + y2 + z2) +

k

2r2

since r2 = x2 + y2 + z2, we could easily solve this problem in cartesian space since

L =m

2(x2 + y2 + z2) +

k

2(x2 + y2 + z2) (1.11)

=

(m

2x2 +

k

2x2

)+

(m

2y2 +

k

2y2

)+

(m

2z2 +

k

2z2

)(1.12)

and we see that the system is separable into 3 independent oscillators. To convert to sphericalpolar coordinates, we use

x = r sin(φ) cos(θ) (1.13)

y = r sin(φ) sin(θ) (1.14)

z = r cos(θ) (1.15)

and the arc length given above.

L =m

2(r2 + r2θ2 + r2 sin2 θφ2)− k

2r2

20

The equations of motion are

d

dt

∂L

∂φ− ∂L

∂φ=

d

dt(mr2 sin2 θφ = 0 (1.16)

d

dt

∂L

∂θ− ∂L

∂θ=

d

dt(mr2θ)−mr2 sin θ cos θφ = 0 (1.17)

d

dt

∂L

∂r− ∂L

∂r=

d

dt(mr)−mrθ2 −mr sin2 θφ2 + kr = 0 (1.18)

We now prove that the motion of a particle in a central force field lies in a plane containingthe origin. The force acting on the particle at any given time is in a direction towards the origin.Now, place an arbitrary cartesian frame centered about the particle with the z axis parallel tothe direction of motion as sketched in Fig. 1.2 Note that the y axis is perpendicular to the planeof the page and hence there is no force component in that direction. Consequently, the motionof the particle is constrained to lie in the zx plane, i.e. the plane of the page and there is noforce component which will take the particle out of this plane.

Let’s make a change of coordinates by rotating the original frame to a new one whereby thenew z′ is perpendicular to the plane containing the initial position and velocity vectors. In thesketch above, this new z′ axis would be perpendicular to the page and would contain the y axiswe placed on the moving particle. In terms of these new coordinates, the Lagrangian will havethe same form as before since our initial choice of axis was arbitrary. However, now, we havesome additional constraints. Because the motion is now constrained to lie in the x′y′ plane,θ′ = π/2 is a constant, and θ = 0. Thus cos(π/2) = 0 and sin(π/2) = 1 in the equations above.From the equations for φ we find

d

dtmr2φ = 0

ormr2φ = const = pφ.

This we can put into the r equation

d

dt(mr)−mrφ2 + kr = 0 (1.19)

d

dt(mr)−

p2φ

mr3+ kr = 0 (1.20)

o

F

Z

X

Y

a

Figure 1.2: Vector diagram for motion in a central forces. The particle’s motion is along the Zaxis which lies in the plane of the page.

21

ZZ'

X'

Y'Y

X

where we notice that −p2φ/mr

3 is the centrifugal force. Taking the last equation, multiplying byr and then integrating with respect to time gives

r2 = −p2

φ

m2r2− kr2 + b (1.21)

i.e.

r =

√−

p2φ

m2r2− kr2 + b (1.22)

Integrating once again with respect to time,

t− to =∫ rdr

r(1.23)

=∫ rdr√

− p2φ

m2 − kr4 + br2

(1.24)

=1

2

∫ dx√a+ bx+ cx2

(1.25)

where x = r2, a = −p2φ/m

2, b is the constant of integration, and c = −k This is a standardintegral and we can evaluate it to find

r2 =1

2ω(b+ A sin(ω(t− to))) (1.26)

where

A =

√b2 −

ω2p2φ

m2.

What we see then is that r follows an elliptical path in a plane determined by the initial velocity.

22

This example also illustrates another important point which has tremendous impact on molec-ular quantum mechanics, namely, the angular momentum about the axis of rotation is conserved.We can choose any axis we want. In order to avoid confusion, let us define χ as the angularrotation about the body-fixed Z ′ axis and φ as angular rotation about the original Z axis. Soour conservation equations are

mr2χ = pχ

about the Z ′ axis andmr2 sin θφ = pφ

for some arbitrary fixed Z axis. The angle θ will also have an angular momentum associatedwith it pθ = mr2θ, but we do not have an associated conservation principle for this term since itvaries with φ. We can connect pχ with pθ and pφ about the other axis via

pχdχ = pθdθ + pφdφ.

Consequently,mr2χ2dχ = mr2(φ sin θdφ+ θdθ).

Here we see that the the angular momentum vector remains fixed in space in the absence ofany external forces. Once an object starts spinning, its axis of rotation remains pointing in agiven direction unless something acts upon it (torque), in essence in classical mechanics we can

fully specify Lx, Ly, and Lz as constants of the motion since d~L/dt = 0. In a later chapter, wewill cover the quantum mechanics of rotations in much more detail. In the quantum case, wewill find that one cannot make such a precise specification of the angular momentum vector forsystems with low angular momentum. We will, however, recover the classical limit in end as weconsider the limit of large angular momenta.

1.3 Conservation Laws

We just encountered one extremely important concept in mechanics, namely, that some quantitiesare conserved if there is an underlying symmetry. Next, we consider a conservation law arisingfrom the homogeneity of time. For a closed dynamical system, the Lagrangian does not explicitlydepend upon time. Thus we can write

dL

dt=∂L

∂xx+

∂L

∂xx (1.27)

Replacing ∂L/∂x with Lagrange’s equation, we obtain

dL

dt= x

d

dt

(∂L

∂x

)+∂L

∂xx (1.28)

=d

dt

(x∂L

∂x

)(1.29)

Now, rearranging this a bit,

d

dt

(x∂L

∂x− L

)= 0. (1.30)

23

So, we can take the quantity in the parenthesis to be a constant.

E =

(x∂L

∂x− L

)= const. (1.31)

is an integral of the motion. This is the energy of the system. Since L can be written in formL = T − V where T is a quadratic function of the velocities, and using Euler’s theorem onhomogeneous functions:

x∂L

∂x= x

∂T

∂x= 2T.

This gives,E = T + V

which says that the energy of the system can be written as the sum of two different terms: thekinetic energy or energy of motion and the potential energy or the energy of location.

One can also prove that linear momentum is conserved when space is homogeneous. That is,when we can translate our system some arbitrary amount ε and our dynamical quantities mustremain unchanged. We will prove this in the problem sets.

1.4 Hamiltonian Dynamics

Hamiltonian dynamics is a further generalization of classical dynamics and provides a crucial linkwith quantum mechanics. Hamilton’s function, H, is written in terms of the particle’s positionand momentum, H = H(p, q). It is related to the Lagrangian via

H = xp− L(x, x)

Taking the derivative of H w.r.t. x

∂H

∂x= −∂L

∂x= −p

Differentiation with respect to p gives∂H

∂p= q.

These last two equations give the conservation conditions in the Hamiltonian formalism. If His independent of the position of the particle, then the generalized momentum, p is constant intime. If the potential energy is independent of time, the Hamiltonian gives the total energy ofthe system,

H = T + V.

1.4.1 Interaction between a charged particle and an electromagneticfield.

We consider here a free particle with mass m and charge e in an electromagnetic field. TheHamiltonian is

H = pxx+ pyy + pz z − L (1.32)

= x∂L

∂x+ y

∂L

∂y+ z

∂L

∂z− L. (1.33)

24

Our goal is to write this Hamiltonian in terms of momenta and coordinates.For a charged particle in a field, the force acting on the particle is the Lorenz force. Here it

is useful to introduce a vector and scaler potential and to work in cgs units.

~F =e

c~v × (~∇× ~A)− e

c

∂ ~A

∂t− e~∇φ.

The force in the x direction is given by

Fx =d

dtmx =

e

c

(y∂Ay

∂x+ z

∂Az

∂x

)− e

c

(y∂Ax

∂y+ z

∂Ax

∂z+∂Ax

∂t

)− e

∂φ

∂x

with the remaining components given by cyclic permutation. Since

dAx

dt=∂Ax

∂t+ x

∂Ax

∂x+ y

∂Ax

∂y+ z

∂Ax

∂z,

Fx =e

c

(+x

∂Ax

∂x+ y

∂Ax

∂y+ z

∂Ax

∂z

)− e

c~v · ~A− eφ.

Based upon this, we find that the Lagrangian is

L =1

2mx2 1

2my2 1

2mz2 +

e

c~v · ~A− eφ

where φ is a velocity independent and static potential.Continuing on, the Hamiltonian is

H =m

2(x2 + y2 + z2) + eφ (1.34)

=1

2m((mx)2 + (my)2 + (my)2) + eφ (1.35)

The velocities, mx, are derived from the Lagrangian via the canonical relation

p =∂L

∂x

From this we find,

mx = px −e

cAx (1.36)

my = py −e

cAy (1.37)

mz = pz −e

cAz (1.38)

and the resulting Hamiltonian is

H =1

2m

[(px −

e

cAx

)2

+(py −

e

cAy

)2

+(pz −

e

cAz

)2]

+ eφ.

We see here an important concept relating the velocity and the momentum. In the absence of avector potential, the velocity and the momentum are parallel. However, when a vector potentialis included, the actual velocity of a particle is no longer parallel to its momentum and is in factdeflected by the vector potential.

25

1.4.2 Time dependence of a dynamical variable

On of the important applications of Hamiltonian mechanics is in the dynamical evolution of avariable which depends upon p and q, G(p, q). The total derivative of G is

dG

dt=∂G

∂t+∂G

∂qq +

∂G

∂pp

From Hamilton’s equations, we have the canonical definitions

q =∂H

∂p, p = −∂H

∂q

Thus,

dG

dt=

∂G

∂t+∂G

∂q

∂H

∂p− ∂G

∂p

∂H

∂q(1.39)

dG

dt=

∂G

∂t+ G,H, (1.40)

where A,B is called the Poisson bracket of two dynamical quantities, G and H.

G,H,= ∂G

∂q

∂H

∂p− ∂G

∂p

∂H

∂q

We can also define a linear operator L as generating the Poisson bracket with the Hamiltonian:

LG =1

iH,G

so that if G does not depend explicitly upon time,

G(t) = exp(iLt)G(0).

where exp(iLt) is the propagator which carried G(0) to G(t).Also, note that if G,H = 0, then dG/dt = 0 so that G is a constant of the motion. This

too, along with the construction of the Poisson bracket has considerable importance in the realmof quantum mechanics.

1.4.3 Virial Theorem

Finally, we turn our attention to a concept which has played an important role in both quan-tum and classical mechanics. Consider a function G that is a product of linear momenta andcoordinate,

G = pq.

The time derivative is simply.G

dt= qp+ pq

26

Now, let’s take a time average of both sides of this last equation.⟨d

dtpq

⟩= lim

T→∞

1

T

∫ T

0

(d

dtpq

)dt (1.41)

= limT→∞

1

T

∫ T

0d(pq) (1.42)

= limT→∞

1

T((pq)T − (pq)0) (1.43)

If the trajectories of system are bounded, both p and q are periodic in time and are thereforefinite. Thus, the average must vanish as T →∞ giving

〈pq + qp〉 = 0 (1.44)

Since pq = 2T and p = −F , we have

〈2T 〉 = −〈qF 〉. (1.45)

In cartesian coordinates this leads to

〈2T 〉 = −⟨∑

i

xiFi

⟩. (1.46)

For a conservative system F = −∇V . Thus, if we have a centro-symmetric potential givenby V = Crn, it is easy to show that

〈2T 〉 = n〈V 〉.

For the case of the Harmonic oscillator, n = 2 and 〈T 〉 = 〈V 〉. So, for example, if we have atotal energy equal to kT in this mode, then 〈T 〉+ 〈V 〉 = kT and 〈T 〉 = 〈V 〉 = kT/2. Moreover,for the interaction between two opposite charges separated by r, n = −1 and

〈2T 〉 = −〈V 〉.

27

Figure 1.3: Screen shot of using Mathematica to plot phase-plane for harmonic oscillator. Herek/m = 1 and our xo = 0.75.

28

Chapter 2

Waves and Wavefunctions

In the world of quantum physics, no phenominon is a phenominon until it is a recordedphenominon.– John Archibald Wheler

The physical basis of quantum mechanics is

1. That matter, such as electrons, always arrives at a point as a discrete chunk, but that theprobibility of finding a chunk at a specified position is like the intensity distribution of awave.

2. The “quantum state” of a system is described by a mathematical object called a “wave-function” or state vector and is denoted |ψ〉.

3. The state |ψ〉 can be expanded in terms of the basis states of a given vector space, |φi〉as

|ψ〉 =∑

i

|φi〉〈φi|ψ〉 (2.1)

where 〈φi|ψ〉 denotes an inner product of the two vectors.

4. Observable quantities are associated with the expectation value of Hermitian operators andthat the eigenvalues of such operators are always real.

5. If two operators commute, one can measure the two associated physical quantities simul-taneously to arbitrary precision.

6. The result of a physical measurement projects |ψ〉 onto an eigenstate of the associatedoperator |φn〉 yielding a measured value of an with probability |〈φn|ψ〉|2.

2.1 Position and Momentum Representation of |ψ〉1 Two common operators which we shall use extensively are the position and momentum operator.

1The majority of this lecture comes from Cohen-Tannoudji Chapter 1, part from Feynman & Hibbs

29

The position operator acts on the state |ψ〉 to give the amplitude of the system to be at agiven position:

x|ψ〉 = |x〉〈x|ψ〉 (2.2)

= |x〉ψ(x) (2.3)

We shall call ψ(x) the wavefunction of the system since it is the amplitude of |ψ〉 at point x. Herewe can see that ψ(x) is an eigenstate of the position operator. We also define the momentumoperator p as a derivative operator:

p = −ih ∂∂x

(2.4)

Thus,

pψ(x) = −ihψ′(x). (2.5)

Note that ψ′(x) 6= ψ(x), thus an eigenstate of the position operator is not also an eigenstate ofthe momentum operator.

We can deduce this also from the fact that x and p do not commute. To see this, first consider

∂

∂xxf(x) = f(x) + xf ′(x) (2.6)

Thus (using the shorthand ∂x as partial derivative with respect to x.)

[x, p]f(x) = ih(x∂xf(x)− ∂x(xf(x))) (2.7)

= −ih(xf ′(x)− f(x)− xf ′(x)) (2.8)

= ihf(x) (2.9)

What are the eigenstates of the p operator? To find them, consider the following eigenvalueequation:

p|φ(k)〉 = k|φ(k)〉 (2.10)

Inserting a complete set of position states using the idempotent operator

I =∫|x〉〈x|dx (2.11)

and using the “coordinate” representation of the momentum operator, we get

− ih∂xφ(k, x) = kφ(k, x) (2.12)

Thus, the solution of this is (subject to normalization)

φ(k, x) = C exp(ik/h) = 〈x|φ(k)〉 (2.13)

30

We can also use the |φ(k)〉 = |k〉 states as a basis for the state |ψ〉 by writing

|ψ〉 =∫dk|k〉〈k|ψ〉 (2.14)

=∫dk|k〉ψ(k) (2.15)

where ψ(k) is related to ψ(x) via:

ψ(k) = 〈k|ψ〉 =∫dx〈k|x〉〈x|ψ〉 (2.16)

= C∫dx exp(ikx/h)ψ(x). (2.17)

This type of integral is called a “Fourier Transfrom”. There are a number of ways to definethe normalization C when using this transform, for our purposes at the moment, we’ll set C =1/√

2πh so that

ψ(x) =1√2πh

∫dkψ(k) exp(−ikx/h) (2.18)

and

ψ(x) =1√2πh

∫dxψ(x) exp(ikx/h). (2.19)

Using this choice of normalization, the transform and the inverse transform have symmetric formsand we only need to remember the sign in the exponential.

2.2 The Schrodinger Equation

Postulate 2.1 The quantum state of the system is a solution of the Schrodinger equation

ih∂t|ψ(t)〉 = H|ψ(t)〉, (2.20)

where H is the quantum mechanical analogue of the classical Hamiltonian.

From classical mechanics, H is the sum of the kinetic and potential energy of a particle,

H =1

2mp2 + V (x). (2.21)

Thus, using the quantum analogues of the classical x and p, the quantum H is

H =1

2mp2 + V (x). (2.22)

To evaluate V (x) we need a theorem that a function of an operator is the function evaluatedat the eigenvalue of the operator. The proof is straight forward, Taylor expand the functionabout some point, If

V (x) = (V (0) + xV ′(0) +1

2V ′′(0)x2 · · ·) (2.23)

31

then

V (x) = (V (0) + xV ′(0) +1

2V ′′(0)x2 · · ·) (2.24)

Since for any operator

[f , fp] = 0∀ p (2.25)

Thus, we have

〈x|V (x)|ψ〉 = V (x)ψ(x) (2.26)

So, in coordinate form, the Schrodinger Equation is written as

ih∂

∂tψ(x, t) =

(− h

2m

∂2

∂x2+ V (x)

)ψ(x, t) (2.27)

2.2.1 Gaussian Wavefunctions

Let’s assume that our initial state is a Gaussian in x with some initial momentum k.

ψ(x, 0) =(

2

πa2

)1/4

exp(ikox) exp(−x2/a2) (2.28)

The momentum representation of this is

ψ(k, 0) =1

2πh

∫dxe−ikxψ(x, 0) (2.29)

= (πa)1/2e−(k−ko)2a2/4) (2.30)

In Fig.2.1, we see a gaussian wavepacket centered about x = 0 with ko = 10 and a = 1.For now we will use dimensionaless units. The red and blue components correspond to the realand imaginary components of ψ and the black curve is |ψ(x)|2. Notice, that the wavefunction ispretty localized along the x axis.

In the next figure, (Fig. 2.2) we have the momentum distribution of the wavefunction, ψ(k, 0).Again, we have chosen ko = 10. Notice that the center of the distribution is shifted about ko.

So, for f(x) = exp(−x2/b2), ∆x = b/√

2. Thus, when x varies form 0 to ±∆x, f(x) isdiminished by a factor of 1/

√e. (∆x is the RMS deviation of f(x).)

For the Gaussian wavepacket:

∆x = a/2 (2.31)

∆k = 1/a (2.32)

or

∆p = h/a (2.33)

Thus, ∆x∆p = h/2 for the initial wavefunction.

32

-3 -2 -1 1 2 3

-0.75

-0.5

-0.25

0.25

0.5

0.75

Figure 2.1: Real (red), imaginary (blue) and absolute value (black) of gaussian wavepacket ψ(x)

6 8 10 12 14k

0.5

1

1.5

2

2.5

yè@kD

Figure 2.2: Momentum-space distribution of ψ(k).

33

2.2.2 Evolution of ψ(x)

Now, let’s consider the evolution of a free particle. By a “free” particle, we mean a particlewhose potential energy does not change, I.e. we set V (x) = 0 for all x and solve:

ih∂

∂tψ(x, t) =

(− h

2m

∂2

∂x2

)ψ(x, t) (2.34)

This equation is actually easier to solve in k-space. Taking the Fourier Transform,

ih∂tψ(k, t) =k2

2mψ(k, t) (2.35)

Thus, the temporal solution of the equation is

ψ(k, t) = exp(−ik2/(2m)t/h)ψ(k, 0). (2.36)

This is subject to some initial function ψ(k, 0). To get the coordinate x-representation of thesolution, we can use the FT relations above:

ψ(x, t) =1√2πh

∫dkψ(k, t) exp(−ikx) (2.37)

=∫dx′〈x| exp(−ip2/(2m)t/h)|x′〉ψ(x′, 0) (2.38)

=

√m

2πiht

∫dx′ exp

(im(x− x′)2

2ht

)ψ(x′, 0) (2.39)

=∫dx′Go(x, x

′)ψ(x′, 0) (2.40)

(homework: derive Go and show that Go is a solution of the free particle schrodinger equationHGo = i∂tGo.) The function Go is called the “free particle propagator” or “Green’s Function”and tells us the amplitude for a particle to start off at x′ and end up at another point x at timet.

The sketch tells me that in order to got far away from the initial point in time t I need tohave a lot of energy (wiggles get closer together implies higher Fourier component )

Here we see that the probability to find a particle at the initial point decreases with time.Since the period of oscillation (T ) is the time required to increase the phase by 2π.

2π =mx2

2ht− mx2

2h(t+ T )(2.41)

=mx2

2ht2

(T 2

1 + T/t

)(2.42)

Let ω = 2π/T and take the long time limit t T , we can estimate

ω ≈ m

2h

(x

t

)2

(2.43)

34

Figure 2.3: Go for fixed t as a function of x.

-10 -5 5 10

-0.4

-0.2

0.2

0.4

Since the classical kinetic energy is given by E = m/2v2, we obtain

E = hω (2.44)

Thus, the energy of the wave is proportional to the period of oscillation.We can evaluate the evolution in x using either the Go we derived above, or by taking the

FT of the wavefunction evolving in k-space. Recall that the solution in k-space was

ψ(k, t) = exp(−ik2/(2m)t/h)ψ(k, 0) (2.45)

Assuming a Gaussian form for ψ(k) as above,

ψ(x, t) =

√a

(2π)3/4

∫dke−a2/4(k−ko)2ei(kx−ω(k)t) (2.46)

where ω(k) is the dispersion relation for a free particle:

ω(k) =hk2

2m(2.47)

Cranking through the integral:

ψ(x, t) =

(2a2

π

)1/4eiφ(

a4 + 4h2t2

m2

)1/4eikox exp

[(x− hko/mt)

2

a2 + 2iht/m

](2.48)

where φ = −θ − hk2o/(2m)t and tan 2θ = 2ht/(ma2).

Likewise, for the amplitude:

|ψ(x, t)|2 =

√1

2π∆x(t)2exp

[−(x− vt)

2

2∆x(t)2

](2.49)

35

Figure 2.4: Evolution of a free particle wavefunction. In this case we have given the initial statea kick in the +x direction. Notice that as the system moves, the center moves at a constant ratewhere as the width of the packet constantly spreads out over time.

Where I define

∆x(t) =a

2

√1 +

4h2t2

m2a4(2.50)

as the time dependent RMS width of the wave and the group velocity:

vo =hko

m. (2.51)

Now, since ∆p = h∆k = h/a is a constant for all time, the uncertainty relation becomes

∆x(t)∆p ≥ h/2 (2.52)

corresponding to the particle’s wavefunction becoming more and more diffuse as it evolves intime.

2.3 Particle in a Box

2.3.1 Infinite Box

The Mathematica handout shows how one can use Mathematica to set up and solve some simpleproblems on the computer. (One good class problem would be to use Mathematica to carryout the symbolic manipulations for a useful or interesting problem and/or to solve the problemnumerically.)

The potential we’ll work with for this example consists of two infinitely steep walls placedat x = ` and x = 0 such that between the two walls, V (x) = 0. Within this region, we seeksolutions to the differential equation

∂2xψ(x) = −2mE/h2ψ(x). (2.53)

The solutions of this are plane waves traveling to the left and to the right,

ψ(x) = A exp(−ikx) +B exp(+ikx) (2.54)

The coefficients A and B we’ll have to determine. k is determined by substitution back into thedifferential equation

ψ′′(x) = −k2ψ(x) (2.55)

Thus, k2 = 2mE/h2, or hk =√

2mE. Let’s work in units in which h = 1 and me = 1. Energy inthese units is the Hartree (≈ 27.eV.) Posted on the web-page is a file (c-header file) which has anumber of useful conversion factors.

36

Since ψ(x) must vanish at x = 0 and x = `

A+B = 0 (2.56)

A exp(ik`) +B exp(−ik`) = 0 (2.57)

We can see immediately that A = −B and that the solutions must correspond to a family of sinefunctions:

ψ(x) = A sin(nπ/`x) (2.58)

Just a check,

ψ(`) = A sin(nπ/``) = A sin(nπ) = 0. (2.59)

To obtain the coefficient, we simply require that the wavefunctions be normalized over the rangex = [0, `]. ∫ `

0sin(nπx/`)2dx =

`

2(2.60)

Thus, the normalized solutions are

ψn(x) =

√2

`sin(nπ/`x) (2.61)

The eigenenergies are obtained by applying the Hamiltonian to the wavefunction solution

Enψn(x) = − h2

2m∂2

xψn(x) (2.62)

=h2n2π2

2a2mψn(x) (2.63)

Thus we can write En as a function of n

En =h2π2

2a2mn2 (2.64)

for n = 0, 1, 2, .... What about the case where n = 0? Clearly it’s an allowed solution ofthe Schrodinger Equation. However, we also required that the probability to find the particleanywhere must be 1. Thus, the n = 0 solution cannot be permitted.

Note also that the cosine functions are also allowed solutions. However, the restriction ofψ(0) = 0 and ψ(`) = 0 discounts these solutions.

In Fig. 2.5 we show the first few eigenstates for an electron trapped in a well of length a = π.The potential is shown in gray. Notice that the number of nodes increases as the energy increases.In fact, one can determine the state of the system by simply counting nodes.

What about orthonormality. We stated that the solution of the eigenvalue problem form anorthonormal basis. In Dirac notation we can write

〈ψn|ψm〉 =∫dx〈ψn|x〉〈x|ψm〉 (2.65)

=∫ `

0dxψ∗n(x)ψm(x) (2.66)

=2

`

∫ `

0dx sin(nπx/`) sin(mπx/`) (2.67)

= δnm. (2.68)

37

-1 1 2 3 4

2

4

6

8

10

12

14

Figure 2.5: Particle in a box states

Thus, we can see in fact that these solutions do form a complete set of orthogonal states onthe range x = [0, `]. Note that it’s important to specify “on the range...” since clearly the sinfunctions are not a set of orthogonal functions over the entire x axis.

2.3.2 Particle in a finite Box

Now, suppose our box is finite. That is

V (x) =

−Vo if −a < x < a0 otherwise

(2.69)

Let’s consider the case for E < 0. The case E > 0 will correspond to scattering solutions. Inside the well, the wavefunction oscillates, much like in the previous case.

ψW (x) = A sin(kix) +B cos(kix) (2.70)

where ki comes from the equation for the momentum inside the well

hki =√

2m(En + Vo) (2.71)

We actually have two classes of solution, a symmetric solution when A = 0 and an antisym-metric solution when B = 0.

Outside the well the potential is 0 and we have the solutions

ψO(x) = c1eρxandc2e

−ρx (2.72)

38

We will choose the coefficients c1 and c2 as to create two cases, ψL and ψR on the left and righthand sides of the well. Also,

hρ =√−2mE (2.73)

Thus, we have three pieces of the full solution which we must hook together.

ψL(x) = Ceρxfor x < −a (2.74)

ψR(x) = De−ρxfor x > −a (2.75)

(2.76)

ψW (x) = A sin(kix) +B cos(kix)for inside the well (2.77)

To find the coefficients, we need to set up a series of simultaneous equations by applying theconditions that a.) the wavefunction be a continuous function of x and that b.) it have continuousfirst derivatives with respect to x. Thus, applying the two conditions at the boundaries:

ψL(−a)− ψW (−a) = 0 (2.78)

(2.79)

ψR(a)− ψW (a) = 0 (2.80)

(2.81)

ψ′L(−a)− ψ′W (−a) = 0 (2.82)

(2.83)

ψ′R(a)− ψ′W (a) = 0 (2.84)

The matching conditions at x = aThe final results are (after the chalk dust settles):

1. For A = 0. B = D sec(aki)e−aρ and C = D. (Symmetric Solution)

2. For B = 0, A = C csc(aki)e−aρ and C = −D. (Antisymmetric Solution)

So, now we have all the coefficients expressed in terms of D, which we can determine by normal-ization (if so inclined). We’ll not do that integral, as it is pretty straightforward.

For the energies, we substitute the symmetric and antisymmetric solutions into the Eigenvalueequation and obtain:

ρ cos(aki) = ki sin(ki) (2.85)

39

or

ρ

ki

= tan(aki) (2.86)

√E

Vo − E= tan(a

√2m(Vo − E)/h) (2.87)

for the symmetric case and

ρ sin(aki) = −ki cos(aki) (2.88)

for the anti-symmetric case, or

ρ

ki

= cot(aki) (2.89)

(2.90)

√E

Vo − E= cot(a

√2m(Vo − E)/h) (2.91)

Substituting the expressions for ki and ρ into final results for each case we find a set ofmatching conditions: For the symmetric case, eigenvalues occur when ever the two curves√

1− Vo/E = tan(a√

2m(E − Vo)/h) (2.92)

and for the anti-symmetric case,√1− Vo/E = cot(a

√2m(E − Vo)/h) (2.93)

These are called “transcendental” equations and closed form solutions are generally impossibleto obtain. Graphical solutions are helpful. In Fig. ?? we show the graphical solution to thetransendental equations for an electron in a Vo = −10 well of width a = 2. The black dotsindicate the presence of two bound states, one symmetric and one anti-symmetric at E = 2.03and 3.78 repectively.

2.3.3 Scattering states and resonances.

Now let’s take the same example as above, except look at states for which E > 0. In this case, wehave to consider where the particles are coming from and where they are going. We will assumethat the particles are emitted with precise energy E towards the well from −∞ and travel fromleft to right. As in the case above we have three distinct regions,

1. x > −a where ψ(x) = eik1x +Re−ik1x = ψL(x)

2. −a ≤ x ≤ +a where ψ(x) = Ae−ik2x +Be+ik2x = ψW (x)

40

2 4 6 8 10E

-4

-3

-2

-1

1

2

3

4symêasym

Figure 2.6: Graphical solution to transendental equations for an electron in a truncated hardwell of depth Vo = 10 and width a = 2. The short-dashed blue curve corresponds to thesymmetric case and the long-dashed blue curve corresponds to the asymetric case. The red line

is√

1− V o/E. Bound state solution are such that the red and blue curves cross.

3. x > +a where ψ(x) = Te+ik1x = ψR(x)

where k1 =√

2mE/h is the momentum outside the well, k2 =√

2m(E − V )/h is the momentuminside the well, and A, B, T , and R are coefficients we need to determine. We also have thematching conditions:

ψL(−a)− ψW (−a) = 0

ψ′L(−a)− ψ′W (−a) = 0

ψR(a)− ψW (a) = 0

ψ′R(a)− ψ′W (a) = 0

This can be solved by hand, however, Mathematica make it easy. The results are a series of ruleswhich we can use to determine the transmission and reflection coefficients.

T → −4e−2iak1+2iak2k1k2

−k12 + e4iak2k1

2 − 2k1k2 − 2e4iak2k1k2 − k22 + e4iak2k2

2 ,

A → 2e−iak1+3iak2k1 (k1 − k2)

−k12 + e4iak2k1

2 − 2k1k2 − 2e4iak2k1k2 − k22 + e4iak2k2

2 ,

B → −2e−iak1+iak2k1 (k1 + k2)

−k12 + e4iak2k1

2 − 2k1k2 − 2e4iak2k1k2 − k22 + e4iak2k2

2 ,

R →

(−1 + e4iak2

) (k1

2 − k22)

e2iak1

(−k1

2 + e4iak2k12 − 2k1k2 − 2e4iak2k1k2 − k2

2 + e4iak2k22)

41

10 20 30 40En HhartreeL

0.2

0.4

0.6

0.8

1

R,T

Figure 2.7: Transmission (blue) and Reflection (red) coefficients for an electron scattering overa square well (V = −40 and a = 1 ).

The R and T coefficients are related to the rations of the reflected and transimitted flux tothe incoming flux. The current operator is given by

j(x) =h

2mi(ψ∗∇ψ − ψ∇ψ∗) (2.94)

Inserting the wavefunctions above yields:

jin =hk1

m

jref = − hk1R2

m

jtrans =hk1T

2

m

Thus, R2 = −jref/jin and T 2 = jtrans/jin. In Fig. 2.7 we show the transmitted and reflectioncoefficients for an electron passing over a well of depth V = −40, a = 1 as a function of incidentenergy, E.

Notice that the transmission and reflection coefficients under go a series oscillations as theincident energy is increased. These are due to resonance states which lie in the continuum. Thecondition for these states is such that an integer number of de Broglie wavelength of the wave inthe well matches the total length of the well.

λ/2 = na

Fig. 2.8,show the transmission coefficient as a function of both incident energy and the welldepth and (or height) over a wide range indicating that resonances can occur for both wells andbumps. Figures 2.9 show various scattering wavefunctions for on an off-resonance cases. Lastly,Fig. ?? shows an Argand plot of both complex components of ψ.

42

10

20

30

40

En

-10

-5

0

5

10

V

0.85

0.9

0.95

1

T

10

20

30

40

En

Figure 2.8: Transmission Coefficient for particle passing over a bump. Here we have plotted T asa function of V and incident energy En. The oscillations correspond to resonance states whichoccur as the particle passes over the well (for V < 0) or bump V > 0.

2.3.4 Application: Quantum Dots

One of the most active areas of research in soft condensed matter is that of designing physicalsystems which can confine a quantum state in some controllable way. The idea of engineering aquantum state is extremely appealing and has numerous technological applications from smalllogic gates in computers to optically active materials for biomedical applications. The basicphysics of these materials is relatively simple and we can use the basic ideas presented in thischapter. The basic idea is to layer a series of materials such that electrons can be trapped in ageometrically confined region. This can be accomplished by insulator-metal-insulator layers andetching, creating disclinations in semiconductors, growing semi-conductor or metal clusters, andso on. A quantum dot can even be a defect site.

We will assume through out that our quantum well contains a single electron so that we cantreat the system as simply as possible. For a square or cubic quantum well, energy levels aresimply those of an n-dimensional particle in a box. For example for a three dimensional system,

Enx,ny ,nz =h2π2

2m

(nx

Lx

)2

+

(ny

Ly

)2

+(nz

Lz

)2 (2.95)

where Lx, Ly, and Lz are the lengths of the box and m is the mass of an electron.The density of states is the number of energy levels per unit energy. If we take the box to be

43

-10 -5 5 10

-1.5

-1

-0.5

0.5

1

1.5

-10 -5 5 10

-1

-0.5

0.5

1

Figure 2.9: Scattering waves for particle passing over a well. In the top graphic, the particle ispartially reflected from the well (V < 0) and in the bottom graphic, the particle passes over thewell with a slightly different energy than above, this time with little reflection.

a cube Lx = Ly = Lz we can relate n to a radius of a sphere and write the density of states as

ρ(n) = 4π2n2 dn

dE= 4π2n2

(dE

dn

)−1

.

Thus, for a 3D cube, the density of states is

ρ(n) =

(4mL2

πh2

)n

i.e. for a three dimensional cube, the density of states increases as n and hence as E1/2.Note that the scaling of the density of states with energy depends strongly upon the dimen-

44

-10

0

10

x-5

0

5

Re@yD

-4

-2

0

2

4

Im@yD

-10

0

10

x

Figure 2.10: Argand plot of a scattering wavefunction passing over a well. (Same parameters asin the top figure in Fig. 2.9).

sionality of the system. For example in one dimension,

ρ(n) =2mL2

h2π2

1

n

and in two dimensionsρ(n) = const.

The reason for this lies in the way the volume element for linear, circular, and spherical integrationscales with radius n. Thus, measuring the density of states tells us not only the size of the system,but also its dimensionality.

We can generalize the results here by realizing that the volume of a d dimensional sphere ink space is given by

Vd =kdπd/2

Γ(1 + d/2)

where Γ(x) is the gamma-function. The total number of states per unit volume in a d-dimensionalspace is then

nk = 21

2π2Vd

and the density is then the number of states per unit energy. The relation between energy andk is

Ek =h2

2mk2.

i.e.

k =

√2Ekm

h

45

0.2 0.4 0.6 0.8 1energy HauL0.5

1

1.5

2

2.5

3

DOS

STMNotebook.nb 1

Figure 2.11: Density of states for a 1-, 2- , and 3- dimensional space.

which gives

ρd(E) =2−1+ d

2 d π−2+ d2

(√m εh

)d

εΓ(1 + d2)

A quantum well is typically constructed so that the system is confined in one dimensionand unconfined in the other two. Thus, a quantum well will typically have discrete state onlyin the confined direction. The density of states for this system will be identical to that of the3-dimensional system at energies where the k vectors coincide. If we take the thickness to be s,then the density of states for the quantum well is

ρ =L

sρ2(E)

⌊L

ρ3(E)

Lρ2(E)/s

⌋

where bxc is the ”floor” function which means take the largest integer less than x. This is plottedin Fig. 2.12 and the stair-step DOS is indicative of the embedded confined structure.

Next, we consider a quantum wire of thickness s along each of its 2 confined directions. TheDOS along the unconfined direction is one-dimensional. As above, the total DOS will be identical

46

0.005 0.01 0.015 0.02 0.025 0.03e HauL

10

20

30

DOS Quantum well vs. 3d body

0.05 0.1 0.15 0.2 0.25 0.3e HauL20

40

60

80

100

120

DOS Quantum wire vs. 3d body

Figure 2.12: Density of states for a quantum well and quantum wire compared to a 3d space.Here L = 5 and s = 2 for comparison.

to the 3D case when the wavevectors coincide. Increasing the radius of the wire eventually leadsto the case where the steps decrease and merge into the 3D curve.

ρ =(L

s

)2

ρ1(E)

⌊L2ρ2(E)

L2ρ2(E)/s

⌋

For a spherical dot, we consider the case in which the radius of the quantum dot is smallenough to support discrete rather than continuous energy levels. In a later chapter, we will derivethis result in more detail, for now we consider just the results. First, an electron in a sphericaldot obeys the Schrodinger equation:

− h2

2m∇2ψ = Eψ (2.96)

where ∇2 is the Laplacian operator in spherical coordinates

∇2 =1

r

∂2

∂r2r +

1

r2 sin θ

∂

∂θsin θ

∂

∂θ+

1

r2 sin2 θ

∂2

∂φ2.

The solution of the Schrodinger equation is subject to the boundary condition that for r ≥ R,ψ(r) = 0, where R is the radius of the sphere and are given in terms of the spherical Besselfunction, jl(r) and spherical harmonic functions, Ylm.

ψnlm =21/2

R3/2

jl(αr/R)

jl+1(α)Ylm(Ω), (2.97)

with energy

E =h2

2m

α2

R2(2.98)

Note that the spherical Bessel functions (of the first kind) are related to the Bessel functions via,

jl(x) =

√π

2xJl+1/2(x). (2.99)

47

5 10 15 20x

-0.2

0.2

0.4

0.6

0.8

1jlHxL

Figure 2.13: Spherical Bessel functions, j0, j1, and j1 (red, blue, green)

The first few of these are

j0(x) =sin x

x(2.100)

j1(x) =sin x

x2− cosx

x(2.101)

j2(x) =(

3

x3− 1

x

)sin x− 3

x2cosx (2.102)

jn(x) = (−1)nxn

(1

x

d

dx

)n

j0(x) (2.103)

where the last line provides a way to generate jn from j0.The α’s appearing in the wavefunction and in the energy expression are determined by the

boundary condition that ψ(R) = 0. Thus, for the lowest energy state we require

j0(α) = 0, (2.104)

i.e. α = π. For the next state (l = 1),

j1(α) =sinα

α2− cosα

α= 0. (2.105)

This can be solved to give α = 4.4934. These correspond to where the spherical Bessel functionspass through zero. The first 6 of these are 3.14159, 4.49341, 5.76346, 6.98793, 8.18256, 9.35581.These correspond to where the first zeros occur and give the condition for the radial quantization,n = 1 with angular momentum l = 0, 1, 2, 3, 4, 5. There are more zeros, and these correspond tothe case where n > 1.

In the next set of figures (Fig. 2.14), we look at the radial wavefunctions for an electron ina 0.5Aquantum dot. First, the case where n = 1, l = 0 and n = 0, l = 1. In both cases, thewavefunctions vanish at the radius of the dot. The radial probability distribution function (PDF)is given by P = r2|ψnl(r)|2. Note that increasing the angular momentum l from 0 to 1 causesthe electron’s most probable position to shift outwards. This is due to the centrifugal force dueto the angular motion of the electron. For the n, l = (2, 0) and (2, 1) states, we have 1 node inthe system and two peaks in the PDF functions.

48

0.1 0.2 0.3 0.4 0.5r

2

4

6

8

10

12

y

0.1 0.2 0.3 0.4 0.5r

1

2

3

4

P

0.1 0.2 0.3 0.4 0.5r

-25

-20

-15

-10

-5

5

y

0.1 0.2 0.3 0.4 0.5r

1

2

3

4

P

Figure 2.14: Radial wavefuncitons (left column) and corresponding PDFs (right column) for anelectron in a R = 0.5Aquantum dot. The upper two correspond to (n, l) = (1, 0) (solid) and(n, l) = (1, 1) (dashed) while the lower correspond to (n, l) = (2, 0) (solid) and (n, l) = (2, 1)(dashed) .

2.4 Tunneling and transmission in a 1D chain

In this example, we are going to generalize the ideas presented here and look at what happens ifwe discretize the space in which a particle can move. This happens physically when we considerwhat happens when a particle (eg. an electron) can hop from one site to another. If an electronis on a given site, it has a certain energy ε to be there and it takes energy β to move the electronfrom the site to its neighboring site. We can write the Schrodinger equation for this system as

ujε+ βuj+1 + βuj−1 = Euj.

for the case where the energy depends upon where the electron is located. If the chain is infinite,we can write uj = Teikdj and find that the energy band goes as E = ε + 2β cos(kd) where k isnow the momentum of the electron.

2.5 Summary

We’ve covered a lot of ground. We now have enough tools at hand to begin to study some physicalsystems. The traditional approach to studying quantum mechanics is to progressively a seriesof differential equations related to physical systems (harmonic oscillators, angular momentum,

49

hydrogen atom, etc...). We will return to those models in a week or so. Next week, we’re going tolook at 2 and 3 level systems using both time dependent and time-independent methods. We’lldevelop a perturbative approach for computing the transition amplitude between states. We willalso look at the decay of a state when its coupled to a continuum. These are useful models fora wide variety of phenomena. After this, we will move on to the harmonic oscillator.

2.6 Problems and Exercises

Exercise 2.1 1. Derive the expression for

Go(x, x′) = 〈x| exp(−ihot/h)|x′〉 (2.106)

where ho is the free particle Hamiltonian,

ho = − h2

2m

∂2

∂x2(2.107)

2. Show that Go is a solution of the free particle Schrodinger Equation

ih∂tGo(t) = hoGo(t). (2.108)

Exercise 2.2 Show that the normalization of a wavefunction is independent of time.

Solution:

i∂t〈ψ(t)|ψ(t)〉 = (i〈ψ(t)|)(|ψ(t)〉) + (〈ψ(t)|)(i|ψ(t)〉) (2.109)

= −〈ψ(t)|H†|ψ(t)〉+ 〈ψ(t)|H|ψ(t)〉 (2.110)

= −〈ψ(t)|H|ψ(t)〉+ 〈ψ(t)|H|ψ(t)〉 = 0 (2.111)

Exercise 2.3 Compute the bound state solutions (E < 0) for a square well of depth Vo where

V (x) =

−Vo −a/2 ≤ x ≤ a/20 otherwise

(2.112)

1. How many energy levels are supported by a well of width a.

2. Show that a very narrow well can support only 1 bound state, and that this state is an evenfunction of x.

3. Show that the energy of the lowest bound state is

E ≈ mV 2o a

2

2h2 (2.113)

4. Show that as

ρ =

√−2mE

h2 → 0 (2.114)

the probability of finding the particle inside the well vanishes.

50

Exercise 2.4 Consider a particle with the potential

V (x) =

0 for x > a−Vo for 0 ≤ x ≤ a∞ for x < 0

(2.115)

1. Let φ(x) be a stationary state. Show that φ(x) can be extended to give an odd wavefunctioncorresponding to a stationary state of the symmetric well of width 2a (i.e the one studiedabove) and depth Vo.

2. Discuss with respect to a and Vo the number of bound states and argue that there is alwaysat least one such state.

3. Now turn your attention toward the E > 0 states of the well. Show that the transmissionof the particle into the well region vanishes as E → 0 and that the wavefunction is perfectlyreflected off the sudden change in potential at x = a.

Exercise 2.5 Which of the following are eigenfunctions of the kinetic energy operator

T = − h2

2m

∂2

∂x2(2.116)

ex, x2, xn,3 cos(2x), sin(x) + cos(x), e−ikx,

f(x− x′) =∫ ∞

−∞dke−ik(x−x′)e−ik2/(2m) (2.117)

.

Solution Going in order:

1. ex

2. x2

3. xn

4. 3 cos(2x)

5. sin(x) + cos(x)

6. e−ikx

Exercise 2.6 Which of the following would be acceptable one dimensional wavefunctions for abound particle (upon normalization): f(x) = e−x, f(x) = e−x2

, f(x) = xe−x2, and

f(x) =

e−x2

x ≥ 0

2e−x2x < 0

(2.118)

Solution In order:

51

1. f(x) = e−x

2. f(x) = e−x2

3. f(x) = xe−x2

4.

f(x) =

e−x2

x ≥ 0

2e−x2x < 0

(2.119)

Exercise 2.7 For a one dimensional problem, consider a particle with wavefunction

ψ(x) = Nexp(ipox/h)√

x2 + a2(2.120)

where a and po are real constants and N the normalization.

1. Determine N so that ψ(x) is normalized.∫ ∞

−∞dx|ψ(x)|2 = N2

∫ ∞

−∞dx

1

x2 + a2(2.121)

= N2π

a(2.122)

Thus ψ(x) is normalized when

N =

√a

π(2.123)

2. The position of the particle is measured. What is the probability of finding a result between−a/

√3 and +a/

√3?

a

π

∫ +a/√

3

−a/√

3dx|ψ(x)|2 =

∫ +a/√

3

−a/√

3dx

1

x2 + a2(2.124)

=1

πtan−1(x/a)

∣∣∣∣+a/√

3

−a/√

3(2.125)

=1

3(2.126)

3. Compute the mean value of a particle which has ψ(x) as its wavefunction.

〈x〉 =a

π

∫ ∞

−∞dx

x

x2 + a2(2.127)

= 0 (2.128)

52

Exercise 2.8 Consider the Hamiltonian of a particle in a 1 dimensional well given by

H =1

2mp2 + x2 (2.129)

where x and p are position and momentum operators. Let |φn〉 be a solution of

H|φn〉 = En|φn〉 (2.130)

for n = 0, 1, 2, · · ·. Show that

〈φn|p|φm〉 = αnm〈φn|x|φm〉 (2.131)

where αnm is a coefficient depending upon En − Em. Compute αnm. (Hint: you will need to usethe commutation relations of [x, H] and [p, H] to get this). Finally, from all this, deduce that

∑m

(En − Em)2|φn|x|φm〉|2 =h2

2m〈φn|p2|φn〉 (2.132)

Exercise 2.9 The state space of a certain physical system is three-dimensional. Let |u1〉, |u2〉,and |u3〉 be an orthonormal basis of the space in which the kets |ψ1〉 and |ψ2〉 are defined by

|ψ1〉 =1√2|u1〉+

i

2|u2〉+

1

2|u3〉 (2.133)

|ψ2〉 =1√3|u1〉+

i√3|u3〉 (2.134)

1. Are the states normalized?

2. Calculate the matrices, ρ1 and ρ2 representing in the |ui〉〉 basis, the projection operatorsonto |ψ1〉 and |ψ2〉. Verify that these matrices are Hermitian.

Exercise 2.10 Let ψ(r) = ψ(x, y, z) be the normalized wavefunction of a particle. Express interms of ψ(r):

1. A measurement along the x-axis to yield a result between x1 and x2.

2. A measurement of momentum component px to yield a result between p1 and p2.

3. Simultaneous measurements of x and pz to yield x1 ≤ x ≤ x2 and pz > 0.

4. Simultaneous measurements of px, py, and pz, to yield

p1 ≤ px ≤ p2 (2.135)

(2.136)

p3 ≤ py ≤ p4 (2.137)

(2.138)

p5 ≤ pz ≤ p6 (2.139)

Show that this result is equal to the result of part 2 when p3, p5 → −∞ and p4, p6 → +∞.

53

Exercise 2.11 Consider a particle of mass m whose potential energy is

V (x) = −α(δ(x+ l/2) + δ(x− l/2))

1. Calculate the bound states of the particle, setting

E = − h2ρ2

2m.

Show that the possible energies are given by

e−ρl = ±(

1− 2ρ

µ

)

where µ = 2mα/h2. Give a graphic solution of this equation.

(a) The Ground State. Show that the ground state is even about the origin and that it’senergy, Es is less than the bound state of a particle in a single δ-function potential,−EL. Interpret this physically. Plot the corresponding wavefunction.

(b) Excited State. Show that when l is greater than some value (which you need to deter-mine), there exists an odd excited state of energy EA with energy greater than −EL.Determine and plot the corresponding wavefunction.

(c) Explain how the preceeding calculations enable us to construct a model for an ionizeddiatomic molecule, eg. H+

2 , whose nuclei are separated by l. Plot the energies of thetwo states as functions of l, what happens as l→∞ and l→ 0?

(d) If we take Coulombic repulsion of the nuclei into account, what is the total energyof the system? Show that a curve which gives the variation with respect to l of theenergies thus obtained enables us to predict in certain cases the existence of boundstates of H+

2 and to determine the equilibrium bond length.

2. Calculate the reflection and transmission coefficients for this system. Plot R and T asfunctions of l. Show that resonances occur when l is an integer multiple of the de Brogliewavelength of the particle. Why?

54

Chapter 3

Semi-Classical Quantum Mechanics

Good actions ennoble us, and we are the sons of our own deeds.–Miguel de Cervantes

The use of classical mechanical analogs for quantum behavour holds a long and proud tradition inthe development and application of quantum theory. In Bohr’s original formulation of quantummechanics to explain the spectra of the hydrogen atom, Bohr used purely classical mechanicalnotions of angular momentum and rotation for the basic theory and imposed a quantizationcondition that the angular momentum should come in integer multiples of h. Bohr worked underthe assumption that at some point the laws of quantum mechanics which govern atoms andmolecules should correspond to the classical mechanical laws of ordinary objects like rocks andstones. Bohr’s Principle of Correspondence states that quantum mechanics was not completelyseparate from classical mechanics; rather, it incorporates classical theory.

From a computational viewpoint, this is an extremely powerful notion since performing aclassical trajectory calculation (even running 1000’s of them) is simpler than a single quantumcalculation of a similar dimension. Consequently, the development of semi-classical methods hasand remains an important part of the development and untilization of quantum theory. In facteven in the most recent issues of the Journal of Chemical Physics, Phys. Rev. Lett, and otherleading physics and chemical physics journals, one finds new developments and applications ofthis very old idea.

In this chapter we will explore this idea in some detail. The field of semi-classical mechanicsis vast and I would recommend the following for more information:

1. Chaos in Classical and Quantum Mechanics, Martin Gutzwiller (Springer-Verlag, 1990).Chaos in quantum mechanics is a touchy subject and really has no clear-cut definition thatanyone seems to agree upon. Gutzwiller is one of the key figures in sorting all this out. Thisis very nice and not too technical monograph on quantum and classical correspondence.

2. Semiclassical Physics, M. Brack and R. Bhaduri (Addison-Wesley, 1997). Very interestingbook, mostly focusing upon many-body applications and Thomas-Fermi approximations.

3. Computer Simulations of Liquids, M. P. Allen and D. J. Tildesley (Oxford, 1994). Thisbook mostly focus upon classical MD methods, but has a nice chapter on quantum methodswhich were state of the art in 1994. Methods come and methods go.

There are many others, of course. These are just the ones on my bookshelf.

55

3.1 Bohr-Sommerfield quantization

Let’s first review Bohr’s original derivation of the hydrogen atom. We will go through this a bitdifferently than Bohr since we already know part of the answer. In the chapter on the Hydrogenatom we derived the energy levels in terms of the principle quantum number, n.

En = −me4

2h2

1

n2(3.1)

In Bohr’s correspondence principle, the quantum energy must equal the classical energy. Sofor an electron moving about a proton, that energy is inversely proportional to the distance ofseparation. So, we can write

− me4

2h2

1

n2= − e

2

2r(3.2)

Now we need to figure out how angular momentum gets pulled into this. For an orbiting bodythe centrifugal force which pulls the body outward is counterbalenced by the inward tugs of thecentripetal force coming from the attractive Coulomb potential. Thus,

mrω2 =e2

r2, (3.3)

where ω is the angular frequency of the rotation. Rearranging this a bit, we can plug this intothe RHS of Eq. 3.2 and write

− me4

2h2

1

n2= −mr

3ω2

2r(3.4)

The numerator now looks amost like the classical definition of angular momentum: L = mr2ω.So we can write the last equation as

− me4

2h2

1

n2= − L2

2mr2. (3.5)

Solving for L2:

L2 =me4

2h2

2mr2

n2. (3.6)

Now, we need to pull in another one of Bohr’s results for the orbital radius of the H-atom:

r =h2

me2n2. (3.7)

Plug this into Eq.3.6 and after the dust settles, we find

L = hn. (3.8)

But, why should electrons be confined to circular orbits? Eq. 3.8 should be applicable toany closed path the electron should choose to take. If the quantization condition only holds

56

for circular orbits, then the theory itself is in deep trouble. At least that’s what Sommerfieldthought.

The numerical units of h are energy times time. That is the unit of action in classicalmechanics. In classical mechanics, the action of a mechanical system is given by the integral ofthe classical momentum along a classical path:

S =∫ x2

x1

pdx (3.9)

For an orbit, the initial point and the final point must coincide, x1 = x2, so the action integralmust describe some the area circumscribed by a closed loop on the p−x plane called phase-space.

S =∮pdx (3.10)

So, Bohr and Sommerfield’s idea was that the circumscribed area in phase-space was quantizedas well.

As a check, let us consider the harmonic oscillator. The classical energy is given by

E(p, q) =p2

2m+k

2q2.

This is the equation for an ellipse in phase space since we can re-arrange this to read

1 =p2

2mE+

k

2Eq2

=p2

a2+q2

b2(3.11)

where a =√

2mE and b =√

2E/k describe the major and minor axes of the ellipse. The area ofan ellipse is A = πab, so the area circumscribed by a classical trajectory with energy E is

S(E) = 2Eπ√m/k (3.12)

Since√k/m = ω, S = 2πE/ω = E/ν. Finally, since E/ν must be an integer multiple of h, the

Bohr-Sommerfield condition for quantization becomes∮pdx = nh (3.13)

where p is the classical momentum for a path of energy E, p =√

2m(V (x)− E. Taking this a

bit farther, the de Broglie wavelength is p/h, so the Bohr-Sommerfield rule basically states thatstationary energies correspond to classical paths for which there are an integer number of deBroglie wavelengths.

Now, perhaps you can see where the problem with quantum chaos. In classical chaos, chaotictrajectories never return to their exact staring point in phase-space. They may come close, butthere are no closed orbits. For 1D systems, this is does not occur since the trajectories are thecontours of the energy function. For higher dimensions, the dimensionality of the system makesit possible to have extremely complex trajectories which never return to their starting point.

Exercise 3.1 Apply the Bohr-Sommerfield proceedure to determine the stationary energies fora particle in a box of length l.

57

3.2 The WKB Approximation

The original Bohr-Sommerfield idea can be imporoved upon considerably to produce an asymp-totic (h→ 0) approximation to the Schrodinger wave function. The idea was put forward at aboutthe same time by three different theoreticians, Brillouin (in Belgium), Kramers (in Netherlands),and Wentzel (in Germany). Depending upn your point of origin, this method is the WKB (US& Germany), BWK (France, Belgium), JWKB (UK), you get the idea. The original referencesare

1. “La mecanique odularatoire de Schrodinger; une methode generale de resolution par ap-proximations successives”, L. Brillouin, Comptes rendus (Paris). 183, 24 (1926).

2. “Wellenmechanik und halbzahlige Quantisierung”, H. A. Kramers, Zeitschrift fur Physik39, 828 (1926).

3. “Eine Verallgemeinerung der Quantenbedingungen fur die Zwecke der Wellenmechanik”,Zeitschrift fur Physik 38, 518 (1926).

We will first go through how one can use the approach to determine the eigenvalues of theSchrodinger equation via semi-classical methods, then show how one can approximate the actualwavefunctions themselves.

3.2.1 Asymptotic expansion for eigenvalue spectrum

The WKB proceedure is initiated by writing the solution to the Schodinger equation

ψ′′ +2m

h2 (E − V (x))ψ = 0

as

ψ(x) = exp(i

h

∫χdx

). (3.14)

We will soon discover that χ is the classical momentum of the system, but for now, let’s considerit to be a function of the energy of the system. Substituting into the Schrodinger equationproduces a new differential equation for χ

h

i

dχ

dx= 2m(E − V )− χ2. (3.15)

If we take h→ 0, it follows then that

χ = χo =√

2m(E − V ) = |p| (3.16)

which is the magnitude of the classical momentum of a particle. So, if we assume that this issimply the leading order term in a series expansion in h we would have

χ = χo +h

iχ1 +

(h

i

)2

χ2 . . . (3.17)

58

Substituting Eq. 3.17 into

χ =h

i

1

ψ

∂ψ

x(3.18)

and equating to zero coefficients with different powers of h, one obtains equations which determinethe χn corrections in succession:

d

dxχn−1 = −

n∑m=0

χn−mχm (3.19)

for n = 1, 2, 3 . . .. For example,

χ1 = −1

2

χ′oχo

=1

4

V ′

E − V(3.20)

χ2 = −χ21 + χ′12χo

= − 1

2χo

V ′2

16(E − V )2+

V ′2

4(E − V )2+

V ′′

4(E − V )

.

= − 5V ′2

32(2m)1/2(E − V )5/2− V ′′

8(2m)1/2(E − V )3/2(3.21)

and so forth.

Exercise 3.2 Verify Eq. 3.19 and derive the first order correction in Eq.3.20.

Now, to use these equations to determine the spectrum, we replace x everywhere by a complexcoordinate z and suppose that V (z) is a regular and analytic function of z in any physicallyrelevant region. Consequently, we can then say that ψ(z) is an analytic function of z. So, wecan write the phase integral as

n =1

h

∫Cχ(z)dz

=1

2πi

∫C

ψ′n(z)

ψn(z)dz (3.22)

where ψn is the nth discrete stationary solution to the Schrodinger equation and C is somecontour of integration on the z plane. If there is a discrete spectrum, we know that the numberof zeros, n, in the wavefunction is related to the quantum number corresponding to the n + 1energy level. So if ψ has no real zeros, this is the ground state wavefunction with energy Eo, onereal zero corresponds to energy level E1 and so forth.

Suppose the contour of integration, C is taken such that it include only these zeros and noothers, then we can write

n =1

h

∫Cχodz +

1

2πi

∫c−h

∫Cχ2dz + . . . (3.23)

59

Each of these terms involves E − V in the denominator. At the classical turning points whereV (z) = E, we have poles and we can use the residue theorem to evaluate the integrals. Forexample, χ1 has a pole at each turnining point with residue −1/4 at each point, hence,

1

2πi

∫Cχ1dz = −1

2. (3.24)

The next term we evaluate by integration by parts∫C

V ′′

(E − V (z))3/2dz = −3

2

∫C

V ′2

(E − V (z))5/2dz. (3.25)

Hence, we can write ∫Cχ2(z)dz =

1

32(2m)1/2

∫C

V ′2

(E − V (z))5/2dz. (3.26)

Putting it all together

n+ 1/2 =1

h

∫c

√2m(E − V (z))dz

− h

128π2(2m)1/2

∫c

V ′2

(E − V (z))5/2dz + . . . (3.27)

Granted, the above analysis is pretty formal! But, what we have is something new. Notice thatwe have an extra 1/2 added here that we did not have in the original Bohr-Sommerfield (BS)theory. What we have is something even more general. The original BS idea came from thenotion that energies and frequencies were related by integer multiples of h. But this is reallyonly valid for transitions between states. If we go back and ask what happens at n = 0 inthe Bohr-Sommerfield theory, this corresponds to a phase-space ellipse with major and minoraxes both of length 0–which violates the Heisenberg Uncertainly rule. This new quantizationcondition forces the system to have some lowest energy state with phase-space area 1/2.

Where did this extra 1/2 come from? It originates from the classical turning points whereV (x) = E. Recall that for a 1D system bound by a potential, there are at least two such points.Each contributes a π/4 to the phase. We will see this more explicitly in the next section.

3.2.2 WKB Wavefunction

Going back to our original wavefunction in Eq. 3.14 and writing

ψ = eiS/h

where S is the integral of χ, we can derive equations for S:

1

2m

(∂S

∂x

)− ih

2m

∂2S

∂x2+ V (x) = E. (3.28)

Again, as above, one can seek a series expansion of S in powers of h. The result is simply theintegral of Eq. 3.17.

S = So +h

iS1 + . . . (3.29)

60

If we make the approximation that h = 0 we have the classical Hamilton-Jacobi equation for theaction, S. This, along with the definition of the momentum, p = dSo/dx = χo, allows us to makea very firm contact between quantum mechanics and the motion of a classical particle.

Looking at Eq. 3.28, it is clear that the classical approximation is valid when the second termis very small compared to the first. i.e.

hS ′′

S ′2 1

hd

dx

(dS

dx

)(dx

dS

)2

1

hd

dx

1

p 1 (3.30)

where we equate dS/dx = p. Since p is related to the de Broglie wavelength of the particleλ = h/p , the same condition implies that∣∣∣∣∣ 1

2π

dλ

dx

∣∣∣∣∣ 1. (3.31)

Thus the semi-classical approximation is only valid when the wavelength of the particle as de-termined by λ(x) = h/p(x) varies slightly over distances on the order of the wavelength itself.Written another way by noting that the gradiant of the momentum is

dp

dx=

d

dx

√2m(E − V (x)) = −m

p

dV

dx.

Thus, we can write the classical condition as

mh|F |/p3 1 (3.32)

Consequently, the semi-classical approximation can only be used in regions where the momentumis not too small. This is especially important near the classical turning points where p → 0. Inclassical mechanics, the particle rolls to a stop at the top of the potential hill. When this happensthe de Broglie wavelength heads off to infinity and is certainly not small!

Exercise 3.3 Verify the force condition given by Eq. 3.32.

Going back to the expansion for χ

χ1 = −1

2

χ′oχo

=1

4

V ′

E − V(3.33)

or equivalently for S1

S ′1 = − S ′′o2S ′

= − p′

2p(3.34)

So,

S1(x) = −1

2log p(x)

61

If we stick to regions where the semi-classical condition is met, then the wavefunction becomes

ψ(x) ≈ C1√p(x)

eih

∫p(x)dx +

C2√p(x)

e−ih

∫p(x)dx (3.35)

The 1/√p prefactor has a remarkably simple interpretation. The probability of find the particle

in some region between x and x+dx is given by |ψ|2 so that the classical probability is essentiallyproportional to 1/p. So, the fast the particle is moving, the less likely it is to be found in somesmall region of space. Conversly, the slower a particle moves, the more likely it is to be found inthat region. So the time spend in a small dx is inversely proportional to the momentum of theparticle. We will return to this concept in a bit when we consider the idea of time in quantummechanics.

The C1 and C2 coefficients are yet to be determined. If we take x = a to be one classicalturning point so that x > a corresponds to the classically inaccessible region where E < V (x),then the wavefunction in that region must be exponentially damped:

ψ(x) ≈ C√|p|

exp(−1

h

∫ x

a|p(x)|dx

)(3.36)

To the left of x = a, we have a combination of incoming and reflected components:

ψ(x) =C1√p

exp(i

h

∫ a

xpdx

)+C2√p

exp(− i

h

∫ a

xpdx

)(3.37)

3.2.3 Semi-classical Tunneling and Barrier Penetration

Before solving the general problem of how to use this in an arbitrary well, let’s consider the casefor tunneling through a potential barrier that has some bumpy top or corresponds to some simplepotential. So, to the left of the barrier the wavefunction has incoming and reflected components:

ψL(x) = Aeikx +Be−ikx. (3.38)

Inside we have

ψB(x) =C√|p(x)|

e+ih

∫|p|dx +

D√|p(x)|

e−ih

∫|p|dx (3.39)

and to the right of the barrier:

ψR(x) = Fe+ikx. (3.40)

If F is the transmitted amplitude, then the tunneling probability is the ratio of the transmittedprobability to the incident probability: T = |F |2/|A|2. If we assume that the barrier is high orbroad, then C = 0 and we obtain the semi-classical estimate for the tunneling probability:

T ≈ exp

(−2

h

∫ b

a|p(x)|dx

)(3.41)

62

where a and b are the turning points on either side of the barrier.Mathematically, we can “flip the potential upside down” and work in imaginary time. In this

case the action integral becomes

S =∫ b

a

√2m(V (x)− E)dx. (3.42)

So we can think of tunneling as motion under the barrier in imaginary time.There are a number of useful applications of this formula. Gamow’s theory of alpha-decay is

a common example. Another useful application is in the theory of reaction rates where we wantto determine tunneling corrections to the rate constant for a particular reaction. Close to the topof the barrier, where tunneling may be important, we can expand the potential and approximatethe peak as an upside down parabola

V (x) ≈ Vo −k

2x2

where +x represents the product side and −x represents the reactant side. See Fig. 3.1 Set thezero in energy to be the barrier height, Vo so that any transmission for E < 0 corresponds totunneling. 1

-4 -2 2 4x

-0.8

-0.6

-0.4

-0.2

0

e

Figure 3.1: Eckart Barrier and parabolic approximation of the transition state

At sufficiently large distances from the turning point, the motion is purely quasi-classical andwe can write the momentum as

p =√

2m(E + kx2/2) ≈ x√mk + E

√m/k/x (3.43)

and the asymptotic for of the Schrodinger wavefunction is

ψ = Ae+iξ2/2ξ+iε−1/2 +Be−iξ2/2ξ−iε−1/2 (3.44)

where A and B are coefficients we need to determine by the matching condition and ξ and ε are

dimensionless lengths and energies given by ξ = x(mk/h)1/4, and ε = (E/h)√m/k.

1The analysis is from Kembel, 1935 as discussed in Landau and Lifshitz, QM

63

The particular case we are interested in is for a particle coming from the left and passing tothe right with the barrier in between. So, the wavefunctions in each of these regions must be

ψR = Be+iξ2/2ξiε−1/2 (3.45)

and

ψL = e−iξ2/2(−ξ)−iε−1/2 + Ae+iξ2/2(−ξ)iε−1/2 (3.46)

where the first term is the incident wave and the second term is the reflected component. So,|A|2| is the reflection coefficient and |B|2 is the transmission coefficient normalized so that

|A|2 + |B|2 = 1.

Lets move to the complex plane and write a new coordinate, ξ = ρeiφ and consider what happensas we rotate around in φ and take ρ to be large. Since iξ2 = ρ2(i cos 2φ− sin 2φ), we have

ψR(φ = 0) = Beiρ2

ρ+iε−1/2

ψL(φ = 0) = Aeiρ2

(−ρ)+iε−1/2 (3.47)

and at φ = π

ψR(φ = π) = Beiρ2

(−ρ)+iε−1/2

ψL(φ = π) = Aeiρ2

ρ+iε−1/2 (3.48)

So, in otherwords, ψR(φ = π) looks like ψL(φ = 0) when

A = B(eiπ)iε−1/2

So, we have the relation A = −iBe−πε. Finally, after we normalize this we get the transmissioncoefficient:

T = |B|2 =1

1 + e−2πε

which must hold for any energy. If the energy is large and negative, then

T ≈ e−2πε.

Also, we can compute the reflection coefficient for E > 0 as 1−D,

R =1

1 + e+2πε.

Exercise 3.4 Verify these last relationships by taking the ψR and ψL, performing the analyticcontinuation.

64

This gives us the transmission probabilty as a function of incident energy. But, normalchemical reactions are not done at constant energy, they are done at constant temperature.To get the thermal transmission coefficient, we need to take a Boltzmann weighted average oftransmission coefficients

Tth(β) =1

Z

∫dEe−EβT (E) (3.49)

where β = 1/kT and Z is the partition function. If E represents a continuum of energy statesthen

Tth(β) = −βωh(ψ(0)(βωh

4π)− ψ(0)(1

4(βωh

π+ 2)))

4π(3.50)

where ψ(n)(z) is the Polygamma function which is the nth derivative of the digamma function,ψ(0)(z), which is the logarithmic derivative of Eulers gamma function, ψ(0)(z) = Γ(z)/Γ(z).2

3.3 Connection Formulas

In what we have considered thus far, we have assumed that up until the turning point thewavefunction was well behaved and smooth. We can think of the problem as having two domains:an exterior and an interior. The exterior part we assumed to be simple and the boundaryconditions trivial to impose. The next task is to figure out the matching condition at the turningpoint for an arbitrary system. So far what we have are two pieces, ψL and ψR, in the notationabove. What we need is a patch. To do so, we make a linearizing assumption for the force atthe classical turning point:

E − V (x) ≈ Fo(x− a) (3.51)

where Fo = −dV/dx evaluated at x = a. Thus, the phase integral is easy:

1

h

∫ x

apdx =

2

3h

√2mFo(x− a)3/2 (3.52)

But, we can do better than that. We can actually solve the Schrodinger equation for the linearpotential and use the linearized solutions as our patch. The Mathematica Notebook AiryFunc-tions.nb goes through the solution of the linearized Schrodinger equation

− h2

2m

dψ

dx2+ (E + V ′)ψ = 0 (3.53)

which can be re-written as

ψ′′ = α3xψ (3.54)

with

α =(

2m

h2 V′(0)

)1/3

.

2See the Mathematica Book, sec 3.2.10.

65

Absorbing the coefficient into a new variable y, we get Airy’s equation

ψ′′(y) = yψ.

The solutions of Airy’s equation are Airy functions, Ai(y) and Bi(y) for the regular and irregularcases. The integral representation of the Ai and Bi are

Ai(y) =1

π

∫ ∞

0cos

(s3

3+ sy

)ds (3.55)

and

Bi(y) =1

π

∫ ∞

0

[e−s3/3+sy + sin

(s3

3+ sy

)]ds (3.56)

Plots of these functions are shown in Fig. 3.2.

-10 -8 -6 -4 -2 2y

-0.5

0.5

1

1.5

Ai@yD, Bi@yD

Figure 3.2: Airy functions, Ai(y) (red) and Bi(y) (blue)

Since both Ai and Bi are acceptible solutions, we will take a linear combination of the twoas our patching function and figure out the coefficients later.

ψP = aAi(αx) + bBi(αx) (3.57)

We now have to determine those coefficients. We need to make two assumptions. One,that the overlap zones are sufficiently close to the turning point that a linearized potential isreasonable. Second, the overlap zone is far enough from the turning point (at the origin) thatthe WKB approximation is accurate and reliable. You can certainly cook up some potentialfor which this will not work, but we will assume it’s reasonable. In the linearized region, themomentum is

p(x) = hα3/2(−x)3/2 (3.58)

So for +x, ∫ x

0|p(x)|dx = 2h(αx)3/2/3 (3.59)

66

and the WKB wavefunction becomes:

ψR(x) =D√

hα3/4x1/4e−2(αx)3/2/3. (3.60)

In order to extend into this region, we will use the asymptotic form of the Ai and Bi functionsfor y 0

Ai(y) ≈ e−2y3/2/3

2√πy1/4

(3.61)

Bi(y) ≈ e+2y3/2/3

√πy1/4

. (3.62)

Clearly, the Bi(y) term will not contribute, so b = 0 and

a =

√4π

αhD.

Now, for the other side, we do the same proceedure. Except this time x < 0 so the phaseintegral is ∫ 0

xpdx = 2h(−αx)3/2/3. (3.63)

Thus the WKB wavefunction on the left hand side is

ψL(x) =1√p

(Be2i(−αx)3/2/3 + Ce−2i(−αx)3/2/3

)(3.64)

=1√

hα3/4(−x)1/4

(Be2i(−αx)3/2/3 + Ce−2i(−αx)3/2/3

)(3.65)

That’s the WKB part, to connect with the patching part, we again use the asymptotic forms fory 0 and take only the regular solution,

Ai(y) ≈ 1√π(−y)1/4

sin(2(−y)3/2/3 + π/4

)≈ 1

2i√π(−y)1/4

(eiπ/4ei2(−y)3/2/3 − e−iπ/4e−i2(−y)3/2/3

)(3.66)

Comparing the WKB wave and the patching wave, we can match term-by-term

a

2i√πeiπ/4 =

B√hα

(3.67)

−a2i√πe−iπ/4 =

C√hα

(3.68)

Since we know a in terms of the normalization constant D, B = ieiπ/4D and C = ie−iπ/4. Thisis the connection! We can write the WKB function across the turning point as

ψWKB(x) =

2D√p(x)

sin[

1h

∫ 0x pdx+ π/4

]x < 0

2D√|p(x)|

e−1h

∫ 0

xpdx x > 0

(3.69)

67

Table 3.1: Location of nodes for Airy, Ai(x) function.node xn

1 -2.338112 -4.087953 -5.520564 -6.786715 -7.944136 -9.022657 -10.0402

Example: Bound states in the linear potential

Since we worked so hard, we have to use the results. So, consider a model problem for a particlein a gravitational field. Actually, this problem is not so far fetched since one can prepare trappedatoms above a parabolic reflector and make a quantum bouncing ball. Here the potential isV (x) = mgx where m is the particle mass and g is the graviational constant (g = 9.80m/s).We’ll take the case where the reflector is infinite so that the particle cannot penetrate into it.The Schrodinger equation for this potential is

− h2

2mψ′′ + (E −mgx)ψ = 0. (3.70)

The solutions are the Airy Ai(x) functions. Setting, β = mg and c = h2/2m, the solutions are

ψ = CAi(

(−βc

)1/3

(x− E/β)) (3.71)

However, there is one caveat: ψ(0) = 0, thus the Airy functions must have their nodes at x = 0.So we have to systematically shift the Ai(x) function in x until a node lines up at x = 0. Thenodes of the Ai(x) function can be determined and the first 7 of them are To find the energylevels, we systematically solve the equation(

−βc

)1/3En

β= xn

So the ground state is where the first node lands at x = 0,

E1 =2.33811β

(β/c)1/3

=2.33811mg

(2m2g/h2)1/3(3.72)

and so on. Of course, we still have to normalize the wavefunction to get the correct energy.We can make life a bit easier by using the quantization condition derived from the WKB

approximation. Since we require the wavefunction to vanish exactly at x = 0, we have:

1

h

∫ xt

0p(x)dx+

π

4= nπ. (3.73)

68

2 4 6 8 10

-10

-5

5

10

15

Figure 3.3: Bound states in a graviational well

This insures us that the wave vanishes at x = 0, xt in this case is the turning point E = mgxt.(See Figure 3.3) As a consequence,∫ xt

0p(x)dx = (n− 1/4)π

Since p(x) =√

2m(En −mgx), The integral can be evaluated

∫ xt

0

√2m(E −mghdx =

√2

2En

√Enm

3gm+

2√m (En − gmxt) (−En + gmxt)

3gm

(3.74)

Since xt = En/mg for the classical turning point, the phase intergral becomes

2√

2En2

3g√Enm

= (n− 1/4)πh.

Solving for En yields the semi-classical approximation for the eigenvalues:

En =g

23 m

13

((1− 4n)2

) 13 (3π)

23 h

23

4 213

(3.75)

In atomic units, the gravitional constant is g = 1.08563 × 10−22bohr/au2 (Can you guess whywe rarely talk about gravitational effects on molecules?). For n = 0, we get for an electronEsc

o = 2.014× 10−15hartree or about 12.6 Hz. So, graviational effects on electrons are extremelytiny compared to the electron’s total energy.

69

rc

θc

χb r

θ

Figure 3.4: Elastic scattering trajectory for classical collision

3.4 Scattering

The collision between two particles plays an important role in the dynamics of reactive molecules.We consider here the collosion between two particles interacting via a central force, V (r). Work-ing in the center of mass frame, we consider the motion of a point particle with mass µ withposition vector ~r. We will first examine the process in a purely classical context since it isintuitive and then apply what we know to the quantum and semiclassical case.

3.4.1 Classical Scattering

The angular momentum of the particle about the origin is given by

~L = ~r × ~p = µ(~r × ~r) (3.76)

we know that angular momentum is a conserved quantity and it is is easy to show that ~L = 0viz.

~L =d

dt~r × ~p = (~r × ~r + (~r × ~p). (3.77)

Since r = p/µ, the first term vanishes; likewise, the force vector, ~p = −dV/dr, is along ~r so thatthe second term vanishes. Thus, L = const meaning that angular momentum is a conservedquantity during the course of the collision.

In cartesian coordinates, the total energy of the collision is given by

E =µ

2(x2 + y2) + V. (3.78)

To convert from cartesian to polar coordinates, we use

x = r cos θ (3.79)

70

y = r sin θ (3.80)

x = r cos θ − rθ sin θ (3.81)

y = r sin θ + rθ cos θ. (3.82)

Thus,

E =mu

2r2 + V (r) +

L2

2µr2(3.83)

where we use the fact that

L = µr2θ2 (3.84)

where L is the angular momentum. What we see here is that we have two potential contributions.The first is the physical attraction (or repulsion) between the two scattering bodies. The second isa purely repulsive centrifugal potential which depends upon the angular momentum and ultimatlyupon the inpact parameters. For cases of large impact parameters, this can be the dominantforce. The effective radial force is given by

µr =L2

2r3µ− ∂V

∂r(3.85)

Again, we note that the centrifugal contribution is always repulsive while the physical interactionV(r) is typically attractive at long range and repulsive at short ranges.

We can derive the solutions to the scattering motion by integrating the velocity equations forr and θ

r = ±(

2

µ

(E − V (r)− L2

2µr2

))1/2

(3.86)

θ =L

µr2(3.87)

and taking into account the starting conditions for r and θ. In general, we could solve theequations numerically and obtain the complete scatering path. However, really what we areinterested in is the deflection angle χ since this is what is ultimately observed. So, we integratethe last two equations and derive θ in terms of r.

θ(r) =∫ θ

0dθ = −

∫ r

∞

dθ

drdr (3.88)

= −∫ r

∞

L

µr2

1√(2/µ)(E − V − L2/2µr2)

dr (3.89)

where the collision starts at t = −∞ with r = ∞ and θ = 0. What we want to do is derive thisin terms of an impact parameter, b, and scattering angle χ. These are illustrated in Fig. 3.4 andcan be derived from basic kinematic considerations. First, energy is conserved through out, so if

71

we know the asymptotic velocity, v, then E = µv2/2. Secondly, angular momentum is conserved,so L = µ|r × v| = µvb. Thus the integral above becomes

θ(r) = −b∫ r

∞

dθ

drdr (3.90)

= −∫ r

∞

dr

r2√

1− V/E − b2/r2. (3.91)

Finally, the angle of deflection is related to the angle of closest approach by 2θc + χ = π; hence,

χ = π − 2b∫ ∞

rc

dr

r2√

1− V/E − b2/r2(3.92)

The radial distance of closest approach is determined by

E =L2

2µr2c

+ V (rc) (3.93)

which can be restated as

b2 = r2c

(1− V (rc)

E

)(3.94)

Once we have specified the potential, we can compute the deflection angle using Eq.3.94. IfV (rc) < 0 , then rc < b and we have an attractive potential, if V (rc) > 0 , then rc > b and thepotential is repulsive at the turning point.

If we have a beam of particles incident on some scattering center, then collisions will occurwith all possible impact parameter (hence angular momenta) and will give rise to a distributionin the scattering angles. We can describe this by a differential cross-section. If we have someincident intensity of particles in our beam, Io, which is the incident flux or the number of particlespassing a unit area normal to the beam direction per unit time, then the differential cross-section,I(χ), is defined so that I(χ)dΩ is the number of particles per unit time scattered into some solidangle dΩ divided by the incident flux.

The deflection pattern will be axially symmetric about the incident beam direction due thespherical symmetry of the interaction potential; thus, I(χ) only depends upon the scattering an-gle. Thus, dΩ can be constructed by the cones defining χ and χ+dχ, i.e. dΩ = 2π sinχdχ. Evenif the interaction potential is not spherically symmetric, since most molecules are not spherical,the scattering would be axially symmetric since we would be scattering from a homogeneous dis-tribution of al possible orientations of the colliding molecules. Hence any azimuthal dependencymust vanish unless we can orient or the colliding species.

Given an initial velocity v, the fraction of the incoming flux with impact parameter betweenb and b + db is 2πbdb. These particles will be deflected between χ and χ + dχ if dχ/db > 0 orbetween χ and χ− dχ if dχ/db < 0. Thus, I(χ)dΩ = 2πbdb and it follows then that

I(χ) =b

sinχ|dχ/db|. (3.95)

72

Thus, once we know χ(b) for a given v, we can get the differential cross-section. The totalcross-section is obtained by integrating

σ = 2π∫ π

0I(χ) sinχdχ. (3.96)

This is a measure of the attenuation of the incident beam by the scattering target and has theunits of area.

3.4.2 Scattering at small deflection angles

Our calculations will be greatly simplified if we consider collisions that result in small deflectionsin the forward direction. If we let the initial beam be along the x axis with momentum p, thenthe scattered momentum, p′ will be related to the scattered angle by p′ sinχ = p′y. Taking χ tobe small

χ ≈p′yp′

=momentum transfer

momentum. (3.97)

Since the time derivative of momentum is the force, the momentum transfered perpendicular tothe incident beam is obtained by integrating the perpendicular force

F ′y = −∂V

∂y= −∂V

∂r

∂r

∂y= −∂V

∂r

b

r(3.98)

where we used r2 = x2 + y2 and y ≈ b. Thus we find,

χ =p′y

µ(2E/µ)1/2(3.99)

= −b(2µE)−1/2∫ +∞

−∞

∂V

∂r

dt

r(3.100)

= −b(2µE)−1/2

(2E

µ

)−1/2 ∫ +∞

−∞

∂V

∂r

dx

r(3.101)

= − b

E

∫ ∞

b

∂V

∂r(r2 − b2)−1/2dr (3.102)

where we used x = (2E/µ)1/2t and x varies from −∞ to +∞ as r goes from −∞ to b and back.Let us use this in a simple example of the V = C/rs potential for s > 0. Substituting V into

the integral above and solving yields

χ =sCπ1/2

2bsE

Γ((s+ 1)/2)

Γ(s/2 + 1). (3.103)

This indicates that χE ∝ b−s and |dχ/db| = χs/b. Thus, we can conclude by deriving thedifferential cross-section

I(χ) =1

sχ−(2+2/s)

(sCπ1/2

2E

Γ((s+ 1)/2)

Γ(s/2 + 1)

)2/s

(3.104)

for small values of the scattering angle. Consequently, a log-log plot of the center of massdifferential cross-section as a function of the scattering angle at fixed energy should give a straightline with a slope −(2+2/s) from which one can determine the value of s. For the van der Waalspotential, s = 6 and I(χ) ∝ E−1/3χ−7/3.

73

3.4.3 Quantum treatment

The quantum mechanical case is a bit more complex. Here we will develop a brief overviewof quantum scattering and move onto the semiclassical evaluation. The quantum scattering isdetermined by the asymptotic form of the wavefunction,

ψ(r, χ)r→∞−→ A

(eikz +

f(χ)

reikr

)(3.105)

where A is some normalization constant and k = 1/λ = µv/h is the initial wave vector alongthe incident beam direction (χ = 0). The first term represents a plane wave incident uponthe scatterer and the second represents an out going spherical wave. Notice that the outgoingamplitude is reduced as r increases. This is because the wavefunction spreads as r increases. Ifwe can collimate the incoming and out-going components, then the scattering amplitude f(χ) isrelated to the differential cross-section by

I(χ) = |f(χ)|2. (3.106)

What we have is then that asymptotic form of the wavefunction carries within it informationabout the scattering process. As a result, we do not need to solve the wave equation for all ofspace, we just need to be able to connect the scattering amplitude to the interaction potential.We do so by expanding the wave as a superposition of Legendre polynomials

ψ(r, χ) =∞∑l=0

Rl(r)Pl(cosχ). (3.107)

Rl(r) must remain finite as r = 0. This determines the form of the solution.When V (r) = 0, then ψ = A exp(ikz) and we can expand the exponential in terms of spherical

waves

eikz =∞∑l=0

(2l + 1)eilπ/2 sin(kr − lπ/2)

krPl(cosχ) (3.108)

=1

2i

∞∑l=0

(2l + 1)ilPl(cosχ)

(ei(kr−lπ/2)

kr+e−i(kr−lπ/2)

kr

)(3.109)

We can interpret this equation in the following intuitive way: the incident plane wave is equiv-alent to an infinite superposition of incoming and outgoing spherical waves in which each termcorresponds to a particular angular momentum state with

L = h√l(l + 1) ≈ h(l + 1/2). (3.110)

From our analysis above, we can relate L to the impact parameter, b,

b =L

µv≈ l + 1/2

kλ. (3.111)

In essence the incoming beam is divided into cylindrical zones in which the lth zone containsparticles with impact parameters (and hence angular momenta) between lλ and (l + 1)λ.

74

Exercise 3.5 The impact parameter, b, is treated as continuous; however, in quantum mechanicswe allow only discrete values of the angular momentum, l. How will this affect our results, sinceb = (l + 1/2)λ from above.

If V (r) is short ranged (i.e. it falls off more rapidly than 1/r for large r), we can derive ageneral solution for the asymptotic form

ψ(r, χ) −→∞∑l=0

(2l + 1) exp

(i

(lπ

2+ ηl

))sin(kr − lπ/2 + ηl

krPl(cosχ). (3.112)

The significant difference between this equation and the one above for the V (r) = 0 case is theaddition of a phase-shift ηl. This shift only occurs in the outgoing part of the wavefunction and sowe conclude that the primary effect of a potential in quantum scattering is to introduce a phasein the asymptotic form of the scattering wave. This phase must be a real number and has thefollowing physical interpretation illustrated in Fig. 3.5 A repulsive potential will cause a decreasein the relative velocity of the particles at small r resulting in a longer de Broglie wavelength. Thiscauses the wave to be “pushed out” relative to that for V = 0 and the phase shift is negative.An attractive potential produces a positive phase shift and “pulls” the wavefunction in a bit.Furthermore, the centrifugal part produces a negative shift of −lπ/2.

Comparing the various forms for the asymptotic waves, we can deduce that the scatteringamplitude is given by

f(χ) =1

2ik

∞∑l=0

(2l + 1)(e2iηl − 1)Pl(cosχ). (3.113)

From this, the differential cross-section is

I(χ) = λ2

∣∣∣∣∣∞∑l=0

(2l + 1)eiηl sin(ηl)Pl(cosχ)

∣∣∣∣∣2

(3.114)

What we see here is the possibility for interference between different angular momentum com-ponents

Moving forward at this point requires some rather sophisticated treatments which we reservefor a later course. However, we can use the semiclassical methods developed in this chapter toestimate the phase shifts.

3.4.4 Semiclassical evaluation of phase shifts

The exact scattering wave is not so important. What is important is the asymptotic extent ofthe wavefunction since that is the part which carries the information from the scattering centerto the detector. What we want is a measure of the shift in phase between a scattering with andwithout the potential. From the WKB treatment above, we know that the phase is related tothe classical action along a given path. Thus, in computing the semiclassical phase shifts, weare really looking at the difference between the classical actions for a system with the potentialswitched on and a system with the potential switched off.

ηSCl = lim

R→∞

(∫ R

rc

dr

λ(r)−∫ R

b

dr

λ(r)

)(3.115)

75

5 10 15 20 25 30

-1

-0.5

0.5

1

Figure 3.5: Form of the radial wave for repulsive (short dashed) and attractive (long dashed)potentials. The form for V = 0 is the solid curve for comparison.

R is the radius a sphere about the scattering center and λ(r) is a de Broglie wavelength

λ(r) =h

p=

1

k(r)=

h

µv(1− V (r)− b2/r2)1/2(3.116)

associated with the radial motion. Putting this together:

ηSCl = lim

R→∞k

∫ R

rc

(1− V (r)

E− b2

r2)1/2dr −

∫ R

b

(1− b2

r2

)1/2

dr

(3.117)

= limR→∞

∫ R

rc

k(r)dr − k∫ R

b

(1− b2

r2

)1/2

dr

(3.118)

(k is the incoming wave-vector.) The last integral we can evaluate:

k∫ R

b

(r2 − b2))1/2

rdr = k

((r2 − b2)− b cos−1 b

r

)∣∣∣∣∣R

b

= kR− kbπ

2(3.119)

Now, to clean things up a bit, we add and subtract an integral over k. (We do this to get rid ofthe R dependence which will cause problems when we take the limit R→∞.)

ηSCl = lim

R→∞

(∫ R

rc

k(r)dr −∫ R

rc

kdr +∫ R

rc

kdr − (kR− kbπ

2)

)(3.120)

=∫ R

rc

(k(r)− k)dr − k(rc − bπ/2) (3.121)

=∫ R

rc

(k(r)− k)dr − krcπ(l + 1/2)/2 (3.122)

This last expression is the standard form of the phase shift.The deflection angle can be determined in a similar way.

χ = limR→∞

[(π − 2

∫actual path

dθ)−(π −

∫V =0 path

dθ)]

(3.123)

76

We transform this into an integral over r

χ = −2b

∫ ∞

rc

(1− V (r)

E− b2

r2

)−1/2dr

r2−∫ ∞

b

(1− b2

r2

)−1/2dr

r2

(3.124)

Agreed, this is weird way to express the scattering angle. But let’s keep pushing this forward.The last integral can be evaluated

∫ ∞

b

(1− b2

r2

)−1/2dr

r2=

1

bcos−1 b

r

∣∣∣∣∣∞

b

= − π

2b. (3.125)

which yields the classical result we obtained before. So, why did we bother? From this we canderive a simple and useful connection between the classical deflection angle and the rate of changeof the semiclassical phase shift with angular momentum, dηSC

l /dl. First, recall the Leibnitz rulefor taking derivatives of integrals:

d

dx

∫ b(x)

a(x)f(x, y)dy =

b

dxf(b(x), y)− da

dxf(a(x), y) +

∫ b(x)

a(x)

∂f(x, y)

∂xdy (3.126)

Taking the derivative of ηSCl with respect to l, using the last equation a and the relation that

(∂b/∂l)E = b/k, we find that

dηSCl

dl=χ

2. (3.127)

Next, we examine the differential cross-section, I(χ). The scattering amplitude

f(χ) =λ

2i

∞∑l=0

(2l + 1)e2iηlPl(cosχ). (3.128)

where we use λ = 1/k and exclude the singular point where χ = 0 since this contributes nothingto the total flux.

Now, we need a mathematical identity to take this to the semiclassical limit where the po-tential varies slowly with wavelength. What we do is to first relate the Legendre polynomial,Pl(cos θ) to a zeroth order Bessel function for small values of θ (θ 1).

Pl(cos θ) = J0((l + 1/2)θ). (3.129)

Now, when x = (l + 1/2)θ 1 (i.e. large angular momentum), we can use the asymptoticexpansion of J0(x)

J0(x) →√

2

πxsin

(x+

π

4

). (3.130)

Pulling this together,

Pl(cos θ) →[

2

π(l + 1/2)θ

]1/2

sin ((l + 1/2)θ + π/4) ≈[

2

π(l + 1/2)

]1/2sin ((l + 1/2)θ + π/4)

(sin θ)1/2(3.131)

77

for θ(l + 1/2) 1. Thus, we can write the semi-classical scattering amplitude as

f(χ) = −λ∞∑l=0

((l + 1/2)

2π sinχ

)1/2 (eiφ+

+ eiφ−)

(3.132)

where

φ± = 2ηl ± (l + 1/2)χ± π/4. (3.133)

The phases are rapidly oscillating functions of l. Consequently, the majority of the terms mustcancel and the sum is determined by the ranges of l for which either φ+ or φ− is extremized.This implies that the scattering amplitude is determined almost exclusively by phase-shifts whichsatisfy

2dηl

dl± χ = 0, (3.134)

where the + is for dφ+/dl = 0 and the − is for dφ−/dl = 0. This demonstrates that the only thephase-shifts corresponding to impact parameter b can contribute significantly to the differentialcross-section in the semi-classical limit. Thus, the classical condition for scattering at a givendeflection angle, χ is that l be large enough for Eq. 3.134 to apply.

3.4.5 Resonance Scattering

3.5 Problems and Exercises

Exercise 3.6 In this problem we will (again) consider the ammonia inversion problem, this timewe will proceed in a semi-classical context.

Recall that the ammonia inversion potential consists of two symmetrical potential wells sepa-rated by a barrier. If the barrier was impenetrable, one would find energy levels corresponding tomotion in one well or the other. Since the barrier is not infinite, there can be passage betweenwells via tunneling. This causes the otherwise degenerate energy levels to split.

In this problem, we will make life a bit easier by taking

V (x) = α(x4 − x2)

as in the examples in Chapter 5.Let ψo be the semi-classical wavefunction describing the motion in one well with energy Eo.

Assume that ψo is exponentially damped on both sides of the well and that the wavefunctionis normalized so that the integral over ψ2

o is unity. When tunning is taken into account, thewavefunctions corresponding to the new energy levels, E1 and E2 are the symmetric and anti-symmetric combinations of ψo(x) and ψo(−x)

ψ1 = (ψo(x) + ψo(−x)/√

2

ψ2 = (ψo(x)− ψo(−x)/√

2

where ψo(−x) can be thought of as the contribution from the zeroth order wavefunction in theother well. In Well 1, ψo(−x) is very small and in well 2, ψo(+x) is very small and the productψo(x)ψo(−x) is vanishingly small everywhere. Also, by construction, ψ1 and ψ2 are normalized.

78

1. Assume that ψo and ψ1 are solutions of the Schrodinger equations

ψ′′o +2m

h2 (Eo − V )ψo = 0

and

ψ′′1 +2m

h2 (E1 − V )ψ1 = 0,

Multiply the former by ψ1 and the latter by ψo, combine and subtract equivalent terms, andintegrate over x from 0 to ∞ to show that

E1 − Eo = − h2

mψo(0)ψ′o(0),

Perform a similar analysis to show that

E2 − Eo = +h2

mψo(0)ψ′o(0),

2. Show that the unperturbed semiclassical wavefunction is

ψo(0) =

√ω

2πvo

exp[−1

h

∫ a

0|p|dx

]

andψ′o(0) =

mvo

hψo(0)

where vo =√

2(Eo − V (0))/m and a is the classical turning point at Eo = V (a).

3. Combining your results, show that the tunneling splitting is

∆E =hω

πexp

[−1

h

∫ +a

−a|p|dx

].

where the integral is taken between classical turning points on either side of the barrier.

4. Assuming that the potential in the barrier is an upside-down parabola

V (x) ≈ Vo − kx2/2

what is the tunneling splitting.

5. Now, taking α = 0.1 expand the potential about the barrier and compute determine theharmonic force constant for the upside-down parabola. Using the equations you derived andcompute the tunneling splitting for a proton in this well. How does this compare with thecalculations presented in Chapter 5.

79

Chapter 4

Postulates of Quantum Mechanics

When I hear the words “Schrodinger’s cat”, I wish I were able to reach for my gun.Stephen Hawkings.

The dynamics of physical processes at a microscopic level is very much beyond the realm ofour macroscopic comprehension. In fact, it is difficult to imagine what it is like to move about onthe length and timescales for whcih quantum mechanics is important. However, for molecules,quantum mechanics is an everyday reality. Thus, in order to understand how molecules move andbehave, we must develop a model of that dynamics in terms which we can comprehend. Makinga model means developing a consistent mathematical framework in which the mathematicaloperations and constructs mimic the physical processes being studied.

Before moving on to develop the mathematical framework required for quantum mechanics,let us consider a simple thought experiment. WE could do the experiment, however, we wouldhave to deal with some additional technical terms, like funding. The experiment I want toconsider goes as follows: Take a machine gun which shoots bullets at a target. It’s not a veryaccurate gun, in fact, it sprays bullets randomly in the general direction of the target.

The distribution of bullets or histogram of the amount of lead accumulated in the target isroughly a Gaussian, C exp(−x2/a). The probability of finding a bullet at x is given by

P (x) = Ce−x2/a. (4.1)

C here is a normalization factor such that the probability of finding a bullet anywhere is 1. i.e.∫ ∞

−∞dxP (x) = 1 (4.2)

The probability of finding a bullet over a small interval is∫ b

adxP (x)〉0. (4.3)

Now suppose we have a bunker with 2 windows between the machine gun and the target suchthat the bunker is thick enough that the bullets coming through the windows rattle around afew times before emerging in random directions. Also, let’s suppose we can “color” the bulletswith some magical (or mundane) means s.t. bullets going through 1 slit are colored “red” and

80

Figure 4.1: Gaussian distribution function

-10 -5 5 10

0.2

0.4

0.6

0.8

1

81

Figure 4.2: Combination of two distrubitions.

-10 -5 5 10

0.2

0.4

0.6

0.8

1

bullets going throught the other slit are colored “blue”. Thus the distribution of bullets at atarget behind the bunker is now

P12(x) = P1(x) + P2(x) (4.4)

where P1 is the distribution of bullets from window 1 (the blue bullets) and P2 the “red” bullets.Thus, the probability of finding a bullet that passed through either 1 or 2 is the sum of theprobabilies of going through 1 and 2. This is shown in Fig. ??

Now, let’s make an “electron gun” by taking a tungsten filiment heated up so that electronsboil off and can be accellerated toward a phosphor screen after passing through a metal foil witha pinhole in the middle We start to see little pin points of light flicker on the screen–these arethe individual electron “bullets” crashing into the phosphor.

If we count the number of electrons which strike the screen over a period of time–just as inthe machine gun experiment, we get a histogram as before. The reason we get a histogram isslightly different than before. If we make the pin hole smaller, the distribution gets wider. Thisis a manifestation of the Heisenberg Uncertainty Principle which states:

∆x · δp ≥ h/2 (4.5)

In otherwords, the more I restrict where the electron can be (via the pin hole) the more uncertainI am about which direction is is going (i.e. its momentum parallel to the foil.) Thus, I wind upwith a distribution of momenta leaving the foil.

Now, let’s poke another hole in the foil and consider the distribution of electrons on the foil.Based upon our experience with bullets, we would expect:

P12 = P1 + P2 (4.6)

BUT electrons obey quantum mechanics! And in quantum mechanics we represent a particlevia an amplitude. And one of the rules of quantum mechanics is that we first add amplitudesand that probabilities are akin to the intensity of the combinded amplitudes. I.e.

P = |ψ1 + ψ2|2 (4.7)

82

Figure 4.3: Constructive and destructive interference from electron/two-slit experiment. Thesuperimposed red and blue curves are P1 and P2 from the classical probabilities

-10 -5 5 10

0.2

0.4

0.6

0.8

1

where ψ1 and ψ2 are the complex amplitudes associated with going through hole 1 and hole 2.Since they are complex numbers,

ψ1 = a1 + ib1 = |psi1|eiφ1 (4.8)

ψ2 = a2 + ib2 = |psi2|eiφ2 (4.9)

Thus,

ψ1 + ψ2 = |ψ1|eiφ1 + |ψ2|eiφ2 (4.10)

|ψ1 + ψ2|2 = (|ψ1|eiφ1 + |ψ2|eiφ2)

× (|ψ1|e−iφ1 + |ψ2|e−iφ2) (4.11)

P12 = |ψ1|2 + |ψ2|2 + 2|ψ1||ψ2| cos(φ1 − φ2) (4.12)

P12 = P1 + P2 + 2√P1P2 cos(φ1 − φ2) (4.13)

In other words, I get the same envelope as before, but it’s modulated by the cos(φ1 − φ2)“interference” term. This is shown in Fig. 4.3. Here the actual experimental data is shown as adashed line and the red and blue curves are the P1 and P2. Just as if a wave of electrons struckthe two slits and diffracted (or interfered) with itself. However, we know that electrons come indefinite chunks–we can observe individual specks on the screen–only whole lumps arrive. Thereare no fractional electrons.

Conjecture 1 Electrons–being indivisible chunks of matter–either go through slit 1 or slit 2.

Assuming Preposition 1 is true, we can divide the electrons into two classes:

1. Those that go through slit 1

83

2. Those that go through slit 2.

We can check this preposition by plugging up hole 1 and we get P2 as the resulting distribution.Plugging up hole 2, we get P1. Perhaps our preposition is wrong and electrons can be split inhalf and half of it went through slit 1 and half through slit 2. NO! Perhaps, the electron wentthrough 1 wound about and went through 2 and through some round-about way made its wayto the screen.

Notice that in the center region of P12, P12 > P1+P2, as if closing 1 hole actually decreased thenumber of electrons going through the other hole. It seems very hard to justify both observationsby proposing that the electrons travel in complicated pathways.

In fact, it is very mysterious. And the more you study quantum mechanics, the more mys-terious it seems. Many ideas have been cooked up which try to get the P12 curve in terms ofelectrons going in complicated paths–all have failed.

Surprisingly, the math is simple (in this case). It’s just adding complex valued amplitudes.So we conclude the following:

Electrons always arrive in discrete, indivisible chunks–like particles. However, theprobability of finding a chunk at a given position is like the distribution of the intensityof a wave.

We could conclude that our conjecture is false since P12 6= P1 + P2. This we can test.Let’s put a laser behind the slits so that an electron going through either slit scatters a bitof light which we can detect. So, we can see flashes of light from electrons going through slit1, flashes of light from electrons going through slit 2, but NEVER two flashes at the sametime. Conjecture 1 is true. But if we look at the resulting distribution: we get P12 = P1 + P2.Measuring which slit the electon passes through destroys the phase information. When we makea measurement in quantum mechanics, we really disturb the system. There is always the sameamount of disturbance because electrons and photons always interact in the same way every timeand produce the same sized effects. These effects “rescatter” the electrons and the phase info issmeared out.

It is totally impossible to devise an experiment to measure any quantum phenomina withoutdisturbing the system you’re trying to measure. This is one of the most fundimental and perhapsmost disturbing aspects of quantum mechanics.

So, once we have accepted the idea that matter comes in discrete bits but that its behavouris much like that of waves, we have to adjust our way of thinking about matter and dynamicsaway from the classical concepts we are used to dealing with in our ordinary life.

These are the basic building blocks of quantum mechanics. Needless to say they are stated ina rather formal language. However, each postulate has a specific physical reason for its existance.For any physical theory, we need to be able to say what the system is, how does it move, andwhat are the possible outcomes of a measurement. These postulates provide a sufficient basis forthe development of a consistent theory of quantum mechanics.

84

4.0.1 The description of a physical state:

The state of a physical system at time t is defined by specifying a vector |ψ(t)〉 belonging to astate space H. We shall assume that this state vector can be normalized to one:

〈ψ|ψ〉 = 1

4.0.2 Description of Physical Quantities:

Every measurable physical quantity, A, is described by an operator acting in H; this operator isan observable.

A consequence of this is that any operator related to a physical observable must be Hermitian.This we can prove. Hermitian means that

〈x|O|y〉 = 〈y|O|x〉∗ (4.14)

Thus, if O is a Hermitian operator and 〈O〉 = 〈ψ|O|ψ〉 = λ〈ψ|ψ〉,

〈O〉 = 〈x|O|x〉+ 〈x|O|y〉+ 〈y|O|x〉+ 〈y|O|y〉. (4.15)

Likewise,

〈O〉∗ = 〈x|O|x〉∗ + 〈x|O|y〉∗ + 〈y|O|x〉∗ + 〈y|O|y〉∗

= 〈x|O|x〉+ 〈y|O|x〉∗ + 〈x|O|y〉+ 〈y|O|y〉= 〈O〉 (4.16)

= λ (4.17)

If O is Hermitian, we can also write

〈ψ|O = λ〈ψ|. (4.18)

which shows that 〈ψ| is an eigenbra of O with real eigenvalue λ. Therefore, for an arbitrary ket,

〈ψ|O|φ〉 = λ〈ψ|φ〉 (4.19)

Now, consider eigenvectors of a Hermitian operator, |ψ〉 and |φ〉. Obviously we have:

O|ψ〉 = λ|ψ〉 (4.20)

O|φ〉 = µ|φ〉 (4.21)

Since O is Hermitian, we also have

〈ψ|O = λ〈ψ| (4.22)

〈φ|O = µ〈φ| (4.23)

Thus, we can write:

〈φ|O|ψ〉 = λ〈φ|ψ〉 (4.24)

〈φ|O|ψ〉 = µ〈φ|ψ〉 (4.25)

Subtracting the two: (λ− µ)〈φ|ψ〉 = 0. Thus, if λ 6= µ, |ψ〉 and |φ〉 must be orthogonal.

85

4.0.3 Quantum Measurement:

The only possible result of the measurement of a physical quantity is one of the eigenvalues ofthe corresponding observable. To any physical observable we ascribe an operator, O. The resultof a physical measurement must be an eigenvalue, a. With each eigenvalue, there correspondsan eigenstate of O, |φa〉. This function is such that the if the state vector, |ψ(t)〉 = |φa〉 wheret corresponds to the time at which the measurement was preformed, O|ψ〉 = a|ψ〉 and themeasurement will yield a.

Suppose the state-function of our system is not an eigenfunction of the operator we areinterested in. Using the superposition principle, we can write an arbitrary state function as alinear combination of eigenstates of O

|ψ(t)〉 =∑a

〈φa|ψ(t)〉|φa〉

=∑a

ca|φa〉. (4.26)

where the sum is over all eigenstates of O. Thus, the probability of observing answer a is |ca|2.IF the measurement DOES INDEED YIELD ANSWER a, the wavefunction of the system

at an infinitesmimal bit after the measurement must be in an eigenstate of O.

|ψ(t+ )〉 = |φa〉. (4.27)

4.0.4 The Principle of Spectral Decomposition:

For a non-discrete spectrum: When the physical quantity, A, is measured on a system in anormalized state |ψ〉, the probability P(an) of obtaining the non-degenerate eigenvalue an of thecorresponding observable is given by

P(an) = |〈un|ψ〉|2 (4.28)

where |un〉 is a normalized eigenvector of A associated with the eigenvalue an. i.e.

A|un〉 = an|un〉

For a discrete spectrum: the sampe principle applies as in the non-discrete case, except wesum over all possible degeneracies of an

P(an) =gn∑i=1

|〈un|ψ〉|2

Finally, for the case of a continuous spectrum: the probability of obtaining a result betweenα and α+ dα is

dPα = |〈α|ψ〉|2 dα

86

4.0.5 The Superposition Principle

Let’s formalize the above discussion a bit and write the electron’s state |ψ〉 = a|1〉+ b|2〉 where|1〉 and |2〉 are “basis states” corresponding to the electron passing through slit 1 or 2. Thecoefficients, a and b, are just the complex numbers ψ1 and ψ2 written above. This |ψ〉 is a vectorin a 2-dimensional complex space with unit length since ψ2

1 + ψ21 = 1. 1

Let us define a Vector Space by defining a set of objects |ψ〉, an addition rule: |φ〉 =|ψ〉+ |ψ′ > which allows us to construct new vectors, and a scaler multiplication rule |φ〉 = a|ψ〉which scales the length of a vector. A non-trivial example of a vector space is the x, y plane.Adding two vectors gives another vector also on the x, y plane and multiplying a vector by aconstant gives another vector pointed in the same direction but with a new length.

The inner product of two vectors is written as

〈φ|ψ〉 = (φxφy)

(ψx

ψy

)(4.29)

= φ∗xψx + φ∗yψy (4.30)

= 〈ψ|φ〉∗. (4.31)

The length of a vector is just the inner product of the vector with itself, i.e. ψ|ψ〉 = 1 for thestate vector we defined above.

The basis vectors for the slits can be used as a basis for an arbitrary state |ψ〉 by writing itas a linear combination of the basis vectors.

|ψ〉 = ψ1|1〉+ ψ1|1〉 (4.32)

In fact, any vector in the vector space can always be written as a linear combination of basisvectors. This is the superposition principle.

The different ways of writing the vector |ψ〉 are termed representations. Often it is easier towork in one representation than another knowing fully that one can always switch back in forth atwill. Each different basis defines a unique representation. An example of a representation are theunit vectors on the x, y plane. We can also define another orthonormal representation of the x, yplane by introducing the unit vectors |r〉, |θ〉, which define a polar coordinate system. One canwrite the vector v = a|x〉+ b|y > as v =

√a2 + b2|r〉+ tan−1(b/a)|θ〉 or v = r sin θ|x〉+ r cos θ|y〉

and be perfectly correct. Usually experience and insight is the only way to determine a prioriwhich basis (or representation) best suits the problem at hand.

Transforming between representations is accomplished by first defining an object called anoperator which has the form:

I =∑

i

|i〉〈i|. (4.33)

The sum means “sum over all members of a given basis”. For the xy basis,

I = |x〉〈x|+ |y〉〈y| (4.34)

1The notation we are introducing here is known as “bra-ket” notation and was invented by Paul Dirac. Thevector |ψ〉 is called a “ket”. The corresponding “bra” is the vector 〈ψ| = (ψ∗xψ

∗y), where the ∗ means complex

conjugation. The notation is quite powerful and we shall use is extensively throughout this course.

87

This operator is called the “idempotent” operator and is similar to multiplying by 1. For example,

I|ψ〉 = |1〉〈1|ψ〉+ |2〉〈2|ψ〉 (4.35)

= ψ1|1〉+ ψ2|2〉 (4.36)

= |ψ〉 (4.37)

We can also write the following:

|ψ〉 = |1〉〈1|ψ〉+ |2〉〈2|ψ〉 (4.38)

The state of a system is specified completely by the complex vector |ψ〉 which can be writtenas a linear superposition of basis vectors spanning a complex vector space (Hilbert space). Innerproducts of vectors in the space are as defined above and the length of any vector in the spacemust be finite.

Note, that for state vectors in continuous representations, the inner product relation can bewritten as an integral:

〈φ|ψ〉 =∫dqφ∗(q)φ(q) (4.39)

and normalization is given by

〈ψ|ψ〉 =∫dq|φ(q)|2 ≤ ∞. (4.40)

The functions, ψ(q) are termed square integrable because of the requirement that the innerproduct integral remain finite. The physical motivation for this will become apparent in amoment when ascribe physical meaning to the mathematical objects we are defining. The classof functions satisfying this requirement are also known as L2 functions. (L is for Lebesgue,referring to the class of integral.)

The action of the laser can also be represented mathematically as an object of the form

P1 = |1〉〈1|. (4.41)

and

P2 = |2〉〈2| (4.42)

and note that P1 + P2 = I.When P1 acts on |ψ〉 it projects out only the |1〉 component of |ψ〉

P1|ψ〉 = ψ1|1〉. (4.43)

The expectation value of an operator is formed by writing:

〈P1〉 = 〈ψ|P1|ψ〉 (4.44)

Let’s evaluate this:

〈Px〉 = 〈ψ|1〉〈1|ψ〉= ψ2

1 (4.45)

88

Similarly for P2.Part of our job is to insure that the operators which we define have physical counterparts.

We defined the projection operator, P1 = |1〉〈1|, knowing that the physical polarization filterremoved all “non” |1〉 components of the wave. We could have also written it in another basis,the math would have been slightly more complex, but the result the same. |ψ1|2 is a real numberwhich we presumably could set off to measure in a laboratory.

4.0.6 Reduction of the wavepacket:

If a measurement of a physical quantity A on the system in the state |ψ〉 yields the result, an,the state of the physical system immediately after the measurement is the normalized projectionPn|ψ〉 onto the eigen subspace associated with an.

In more plain language, if you observe the system at x, then it is at x. This is perhapsthe most controversial posulate since it implies that the act of observing the system somehowchanges the state of the system.

Suppose the state-function of our system is not an eigenfunction of the operator we areinterested in. Using the superposition principle, we can write an arbitrary state function as alinear combination of eigenstates of O

|ψ(t)〉 =∑a

〈φa|ψ(t)〉|φa〉

=∑a

ca|φa〉. (4.46)

where the sum is over all eigenstates of O. Thus, the probability of observing answer a is |ca|2.IF the measurement DOES INDEED YIELD ANSWER a, the wavefunction of the system

at an infinitesmimal bit after the measurement must be in an eigenstate of O.

|ψ(t+ )〉 = |φa〉. (4.47)

This is the only postulate which is a bit touchy deals with the reduction of the wavepacketas the result of a measurement. On one hand, you could simply accept this as the way onegoes about business and simply state that quantum mechanics is an algorithm for predictingthe outcome of experiments and that’s that. It says nothing about the inner workings of theuniverse. This is what is known as the “Reductionist” view point. In essence, the Reductionistview point simply wants to know the answer: “How many?”, “How wide?”, “How long?”.

On the other hand, in the Holistic view, quantum mechanics is the underlying physical theoryof the universe and say that the process of measurement does play an important role in how theuniverse works. In otherwords, in the Holist wants the (w)hole picture.

The Reductionist vs. Holist argument has been the subject of numerous articles and booksin both the popular and scholarly arenas. We may return to the philosophical discussion, butfor now we will simply take a reductionist view point and first learn to use quantum mechanicsas a way to make physical predictions.

89

4.0.7 The temporal evolution of the system:

The time evolution of the state vector is given by the Schrodinger equation

ih∂

∂t|ψ(t)〉 = H(t)|ψ(t)〉

where H(t) is an operator/observable associated withthe total energy of the system.As we shall see, H is the Hamiltonian operator and can be obtained from the classical Hamil-

tonian of the system.

4.0.8 Dirac Quantum Condition

One of the crucial aspects of any theory is that we need to be able to construct physical observ-ables. Moreover, we would like to be able to connect the operators and observables in quantummechanics to the observables in classical mechanics. At some point there must be a correspon-dence. This connection can be made formally by relating what is known as the Poisson bracketin classical mechanics:

f(p, q), g(p, q) =∂f

∂q

∂g

∂p− ∂g

∂q

∂f

∂p(4.48)

which looks a lot like the commutation relation between two linear operators:

[A, B] = AB − BA (4.49)

Of course, f(p, q) and g(p, q) are functions over the classical position and momentum of thephysical system. For position and momentum, it is easy to show that the classical Poissonbracket is

q, p = 1.

Moreover, the quantum commutation relation between the observable x and p is

[x, p] = ih.

Dirac proposed that the two are related and that this relation defines an acceptible set ofquantum operations.

The quantum mechanical operators f and g, which in classical theory replace theclassically defined functions f and g, must always be such that the commutator of fand g corresponds to the Poisson bracket of f and g according to

ihf, g = [f , g] (4.50)

To see how this works, we write the momentum operator as

p =h

i

∂

∂x(4.51)

90

Thus,

pψ(x) =h

i

∂ψ(x)

∂x(4.52)

Let’s see if x and p commute. First of all

∂

∂xxf(x) = f(x) + xf ′(x) (4.53)

Thus,

[x, p]f(x) =h

i(x

∂

∂xf(x)− ∂

∂xxf(x)

=h

i(xf ′(x)− f(x)− xf ′(x))

= ihf(x) (4.54)

The fact that x and p do not commute has a rather significant consequence:In other words, if two operators do not commute, one cannot devise and experiment to

simultaneously measure the physical quantities associated with each operator. This in fact limitsthe precision in which we can preform any physical measurement.

The principle result of the postulates is that the wavefunction or state vector of the systemcarries all the physical information we can obtain regarding the system and allows us to makepredictions regarding the probable outcomes of any experiment. As you may well know, if onemake a series of experimental measurements on identically prepared systems, one obtains adistribution of results–usually centered about some peak in the distribution.

When we report data, we usually don’t report the result of every single experiment. Fora spectrscopy experiment, we may have made upwards of a million individual measurement,all distributed about some average value. From statistics, we know that the average of anydistribution is the expectation value of some quantity, in this case x:

E(x) =∫P(x)xdx (4.55)

for the case of a discrete spectra, we would write

E[h] =∑n

hnPn (4.56)

where hn is some value and Pn the number of times you got that value normalized so that∑n

Pn = 1

. In the language above, the hn’s are the possible eigenvalues of the h operator.A similar relation holds in quantum mechanics:

Postulate 4.1 Observable quantities are computed as the expectation value of an operator 〈O〉 =〈ψ|O|ψ〉. The expectation value of an operator related to a physical observable must be real.

91

For example, the expectation value of x the position operator is computed by the integral

〈x〉 =∫ +∞

−∞ψ∗(x)xψ(x)dx.

or for the discrete case:〈O〉 =

∑n

on|〈n|ψ〉|2.

Of course, simply reporting the average or expectation values of an experiment is not enough,the data is usually distributed about either side of this value. If we assume the distribution isGaussian, then we have the position of the peak center xo = 〈x〉 as well as the width of theGaussian σ2.

The mean-squared width or uncertainty of any measurement can be computed by taking

σ2A = 〈(A− 〈A〉)〉.

In statistical mechanics, this the fluctuation about the average of some physical quantity, A. Inquantum mechanics, we can push this definition a bit further.

Writing the uncertainty relation as

σ2A = 〈(A− 〈A〉)(A− 〈A〉)〉 (4.57)

= 〈ψ|(A− 〈A〉)(A− 〈A〉)|ψ〉 (4.58)

= 〈f |f〉 (4.59)

where the new vector |f〉 is simply short hand for |f〉 = (A − 〈A〉)|ψ〉. Likewise for a differentoperator B

σ2B = = 〈ψ|(B − 〈B〉)(B − 〈B〉)|ψ〉 (4.60)

= 〈g|g〉. (4.61)

We now invoke what is called the Schwartz inequality

σ2Aσ

2B = 〈f |f〉〈g|g〉 ≥ |〈f |g〉|2 (4.62)

So if we write 〈f |g〉 as a complex number, then

|〈f |g〉|2 = |z|2

= <(z)2 + =(z)2

≥ =(z)2 =(

1

2i(z − z∗)

)2

≥(

1

2i(〈f |g〉 − 〈g|f〉)

)2

(4.63)

So we conclude

σ2Aσ

2B ≥

(1

2i(〈f |g〉 − 〈g|f〉)

)2

(4.64)

92

Now, we reinsert the definitions of |f〉 and |g〉.

〈f |g〉 = 〈ψ|(A− 〈A〉)(B − 〈B〉)|ψ〉= 〈ψ|(AB − 〈A〉B − A〈B〉+ 〈A〉〈B〉)|ψ〉= 〈ψ|AB|ψ〉 − 〈A〉〈ψ|B|ψ〉 − 〈B〉〈ψ|A|ψ〉+ 〈A〉〈B〉= 〈AB〉 − 〈A〉〈B〉 (4.65)

Likewise

〈g|f〉 = 〈BA〉 − 〈A〉〈B〉. (4.66)

Combining these results, we obtain

〈f |g〉 − 〈g|f〉 = 〈AB〉 − 〈BA〉 = 〈AB −BA〉 = 〈[A,B]〉. (4.67)

So we finally can conclude that the general uncertainty product between any two operators isgiven by

σ2Aσ

2B =

(1

2i〈[A,B]〉

)2

(4.68)

This is commonly referred to as the Generalized Uncertainty Principle. What is means is thatfor any pair of observables whose corresponding operators do not commute there will always besome uncertainty in making simultaneous measurements. In essence, if you try to measure twonon-commuting properties simultaneously, you cannot have an infinitely precise determination ofboth. A precise determination of one implies that you must give up some certainty in the other.

In the language of matrices and linear algebra this implies that if two matrices do not com-mute, then one can not bring both matrices into diagonal form using the same transformationmatrix. in other words, they do not share a common set of eigenvectors. Matrices which docommute share a common set of eigenvectors and the transformation which diagonalizes one willalso diagonalize the other.

Theorem 4.1 If two operators A and B commute and if |ψ〉 is an eigenvector of A, then B|ψ〉is also an eigenvector of A with the same eigenvalue.

Proof: If |ψ〉 is an eigenvector of A, then A|ψ〉 = a|ψ〉. Thus,

BA|ψ〉 = aB|ψ〉 (4.69)

Assuming A and B commute, i.e. [A,B] = AB −BA = 0,

AB|ψ〉 = a(B|ψ〉) (4.70)

Thus, (B|ψ〉) is an eigenvector of A with eigenvalue, a.

Exercise 4.1 1. Show that matrix multiplication is associative, i.e. A(BC) = (AB)C, butnot commutative (in general), i.e. BC 6= CB

2. Show that (A+B)(A−B) = A2 +B2 only of A and B commute.

3. Show that if A and B are both Hermitian matrices, AB + BA and i(AB − BA) are alsoHermitian. Note that Hermitian matrices are defined such that Aij = A∗

ji where ∗ denotescomplex conjugation.

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4.1 Dirac Notation and Linear Algebra

Part of the difficulty in learning quantum mechanics comes from fact that one must also learn anew mathematical language. It seems very complex from the start. However, the mathematicalobjects which we manipulate actually make life easier. Let’s explore the Dirac notation and therelated mathematics.

We have stated all along that the physical state of the system is wholly specified by thestate-vector |ψ〉 and that the probability of finding a particle at a given point x is obtained via|ψ(x)|2. Say at some initial time |ψ〉 = |s〉 where s is some point along the x axis. Now, theamplitude to find the particle at some other point is 〈x|s〉. If something happens between thetwo points we write

〈x|operator describing process|s〉 (4.71)

The braket is always read from right to left and we interpret this as the amplitude for“startingoff at s, something happens, and winding up at i”. An example of this is the Go function in thehomework. Here, I ask “what is the amplitude for a particle to start off at x and to wind up atx′ after some time interval t?”

Another Example: Electrons have an intrinsic angular momentum called “spin” . Accordingly,they have an associated magnetic moment which causes electrons to align with or against animposed magnetic field (eg.this gives rise to ESR). Lets say we have an electron source whichproduces spin up and spin down electrons with equal probability. Thus, my initial state is:

|i〉 = a|+〉+ b|−〉 (4.72)

Since I’ve stated that P (a) = P (b), |a|2 = |b|2. Also, since P (a) + P (b) = 1, a = b = 1/√

2.Thus,

|i〉 =1√2(|+〉+ |−〉) (4.73)

Let’s say that the spin ups can be separated from the spin down via a magnetic field, B and wefilter off the spin down states. Our new state is |i′〉 and is related to the original state by

〈i′|i〉 = a〈+|+〉+ b〈+|−〉 = a. (4.74)

4.1.1 Transformations and Representations

If I know the amplitudes for |ψ〉 in a representation with a basis |i〉 , it is always possible tofind the amplitudes describing the same state in a different basis |µ〉. Note, that the amplitudebetween two states will not change. For example:

|a〉 =∑

i

|i〉〈i|a〉 (4.75)

also

|a〉 =∑µ

|µ〉〈µ|a〉 (4.76)

94

Therefore,

〈µ|a〉 =∑

i

〈µ|i〉〈i|a〉 (4.77)

and

〈i|a〉 =∑µ

〈i|µ〉〈µ|a〉. (4.78)

Thus, the coefficients in |µ〉 are related to the coefficients in |i〉 by 〈µ|i〉 = 〈i|µ〉∗. Thus, we candefine a transformation matrix Sµi as

Sµi =

〈µ|i〉 〈µ|j〉 〈µ|k〉〈ν|i〉 〈ν|j〉 〈ν|k〉〈λ|i〉 〈λ|j〉 〈λ|k〉

(4.79)

and a set of vectors

ai =

〈i|a〉〈j|a〉〈k|a〉

(4.80)

aµ =

〈µ|a〉〈ν|a〉〈λ|a〉

(4.81)

Thus, we can see that

aµ =∑

i

Sµiai (4.82)

Now, we can also write

Siµ =

〈µ|i〉∗ 〈µ|j〉∗ 〈µ|k〉∗

〈ν|i〉 〈ν|j〉 〈ν|k〉∗〈λ|i〉∗ 〈λ|j〉∗ 〈λ|k〉∗

= S∗µi (4.83)

Thus,

ai =∑µ

Siµaµ (4.84)

Since 〈i|µ〉 = 〈µ|i〉∗, S is the Hermitian conjugate of S. So we write

S = S† (4.85)

(4.86)

95

S† = (ST )∗ (4.87)

(4.88)

(S†)ij = S∗ji (4.89)

So in short;

a = Sa = SSa = S†Sa (4.90)

thus

S†S = 1 (4.91)

and S is called a unitary transformation matrix.

4.1.2 Operators

A linear operator A maps a vector in space H on to another vector in the same space. We canwrite this in a number of ways:

|φ〉 A7→ |χ〉 (4.92)

or

|χ〉 = A|φ〉 (4.93)

Linear operators have the property that

A(a|φ〉+ b|χ〉) = aA|φ〉+ bA|χ〉 (4.94)

Since superposition is rigidly enforced in quantum mechanics, all QM operators are linear oper-ators.

The Matrix Representation of an operator is obtained by writing

Aij = 〈i|A|j〉 (4.95)

For example: Say we know the representation of A in the |i〉 basis, we can then write

|χ〉 = A|φ〉 =∑

i

A|i〉〈i|φ〉 =∑j

|j〉〈j|χ〉 (4.96)

Thus,

〈j|χ〉 =∑

i

〈j|A|i〉〈i|φ〉 (4.97)

We can keep going if we want by continuing to insert 1’s where ever we need.

96

The matrix A is Hermitian if A = A†. If it is Hermitian, then I can always find a basis |µ〉 inwhich it is diagonal, i.e.

Aµν = aµδµν (4.98)

So, what is A|µ〉 ?

A|µ〉 =∑ij

|i〉〈i|A|j〉〈j|µ〉 (4.99)

(4.100)

=∑ij

|i〉Aijδjµ (4.101)

(4.102)

=∑

i

|i〉Aiµ (4.103)

(4.104)

=∑

i

|i〉aµδiµ (4.105)

(4.106)

= aµ|µ〉 (4.107)

An important example of this is the “time-independent” Schroedinger Equation:

H|ψ〉 = E|ψ〉 (4.108)

which we spend some time in solving above.Finally, if A|φ〉 = |χ〉 then 〈φ|A† = 〈χ|.

4.1.3 Products of Operators

An operator product is defined as

(AB)|ψ〉 = A[B|ψ〉] (4.109)

where we operate in order from right to left. We proved that in general the ordering of theoperations is important. In other words, we cannot in general write AB = BA. An example ofthis is the position and momentum operators. We have also defined the “commutator”

[A, B] = AB − BA. (4.110)

Let’s now briefly go over how to perform algebraic manipulations using operators and commu-tators. These are straightforward to prove

97

1. [A, B] = −[B, A]

2. [A, A] = −[A, A] = 0

3. [A, BC] = [A, B]C + B[A, C]

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

5. [A, [B, C]] + [B, [C, A]] + [C, [A, B]] = 0 (Jacobi Identity)

6. [A, B]† = [A†, B†]

4.1.4 Functions Involving Operators

Another property of linear operators is that the inverse operator always can be found. I.e. if|χ〉 = A|φ〉 then there exists another operator B such that |φ〉 = B|χ〉. In other words B = A−1.

We also need to know how to evaluate functions of operators. Say we have a function, F (z)which can be expanded as a series

F (z) =∞∑

n=0

fnzn (4.111)

Thus, by analogy

F (A) =∞∑

n=0

fnAn. (4.112)

For example, take exp(A)

exp(x) =∞∑

n=0

xn

n!= 1 + x+ x2/2 + · · · (4.113)

thus

exp(A) =∞∑

n=0

An

n!(4.114)

If A is Hermitian, then F (A) is also Hermitian. Also, note that

[A, F (A)] = 0.

Likewise, if

A|φa〉 = a|φa〉 (4.115)

then

An|φa〉 = an|φa〉. (4.116)

98

Thus, we can show that

F (A)|φa〉 =∑n

fnAn|φa〉 (4.117)

(4.118)

=∑n

fnan|φa〉 (4.119)

(4.120)

= F (a) (4.121)

Note, however, that care must be taken when we evaluate F (A + B) if the two operatorsdo not commute. We ran into this briefly in breaking up the propagator for the SchroedingerEquation in the last lecture (Trotter Product). For example,

exp(A+ B) 6= exp(A) exp(B) (4.122)

unless [A, B] = 0. One can derive, however, a useful formula (Glauber)

exp(A+ B) = exp(A) exp(B) exp(−[A, B]/2) (4.123)

Exercise 4.2 Let H be the Hamiltonian of a physical system and |φn〉 the solution of

H|φn〉 = En|φn〉 (4.124)

1. For an arbitrary operator, A, show that

〈φn|[A,H]|φn〉 = 0 (4.125)

2. Let

H =1

2mp2 + V (x) (4.126)

(a) Compute [H, p], [H, x], and [H, xp].

(b) Show 〈φn|p|φn〉 = 0.

(c) Establish a relationship between the average of the kinetic energy given by

Ekin = 〈φn|p2

2m|φn〉 (4.127)

and the average force on a particle given by

F = 〈φn|x∂V (x)

∂x|φn〉. (4.128)

Finally, relate the average of the potential for a particle in state |φn〉 to the averagekinetic energy.

99

Exercise 4.3 Consider the following Hamiltonian for 1d motion with a potential obeying a sim-ple power-law

H =p2

2m+ αxn (4.129)

where α is a constant and n is an integer. Calculate

〈A〉 = 〈ψ|[xp,H]||ψ〉 (4.130)

and use the result to relate the average potential energy to the average kinetic energy of thesystem.

4.2 Constants of the Motion

In a dynamical system (quantum, classical, or otherwise) a constant of the motion is any quantitysuch that

∂tA = 0. (4.131)

For quantum systems, this means that

[A,H] = 0 (4.132)

(What’s the equivalent relation for classical systems?) In other words, any quantity whichcommutes with H is a constant of the motion. Furthermore, for any conservative system (inwhich there is no net flow of energy to or from the system),

[H,H] = 0. (4.133)

From Eq.4.131, we can write that

∂t〈A〉 = ∂t〈ψ(t)|A|ψ(t)〉 (4.134)

Since [A,H] = 0, we know that if the state |φn〉 is an eigenstate of H,

H|φn〉 = En|φn〉 (4.135)

then

A|φn〉 = an|φn〉 (4.136)

The an are often referred to as “good quantum numbers”. What are some constants of themotion for systems that we have studied thus far? (Bonus: how are constants of motion relatedto particular symmetries of the system?)

A state which is in an eigenstate ofH it’s also in an eigenstate ofA. Thus, I can simultaneouslymeasure quantities associated with H and A. Also, after I measure with A, the system remainsin a the original state.

100

4.3 Bohr Frequency and Selection Rules

What if I have another operator, B, which does not commute with H? What is 〈B(t)〉? This wecan compute by first writing

|ψ(t)〉 =∑n

cne−iEnt/h|φn〉. (4.137)

Then

〈B(t)〉 = 〈ψ|B|ψ(t)〉 (4.138)

=∑n

cnc∗me

−i(En−Em)t/h〈φm|B|φn〉. (4.139)

Let’s define the “Bohr Frequency” as ωnm = (En − Em)/h.

〈B(t)〉 =∑n

cnc∗me

−iωnmt〈φm|B|φn〉. (4.140)

Now, the observed expectation value of B oscillates in time at number of frequencies corre-sponding to the energy differences between the stationary states. The matrix elements Bnm =〈φm|B|φn〉 do not change with time. Neither do the coefficients, cn. Thus, let’s write

B(ω) = cnc∗m〈φm|B|φn〉δ(ω − ωnm) (4.141)

and transform the discrete sum into an continuous integral

〈B(t)〉 =1

2π

∫ ∞

0e−iωtB(ω) (4.142)

where B(ω) is the power spectral of B. In other words, say I monitor < B(t) > with myinstrument for a long period of time, then take the Fourier Transform of the time-series. I getthe power-spectrum. What is the power spectrum for a set of discrete frequencies: If I observethe time-sequence for an infinite amount of time, I will get a series of discretely spaced sticksalong the frequency axis at precisely the energy difference between the n and m states. Theintensity is related to the probability of making a transition from n to m under the influenceof B. Certainly, some transitions will not be allowed because 〈φn|B|φm〉 = 0. These are the“selection rules”.

We now prove an important result regarding the integrated intensity of a series of transitions:

Exercise 4.4 Prove the Thomas-Reiche-Kuhn sum rule:2∑n

2m|xn0|2

h2 (En − Eo) = 1 (4.143)

where the sum is over a compete set of states, |ψn〉 of energy En of a particle of mass m whichmoves in a potential; |ψo〉 represents a bound state, and xn0 = 〈ψn|x|ψo〉. (Hint: use the com-mutator identity: [x, [x,H]] = h2/m)

2This is perhaps one of the most important results of quantum mechanics since it is gives the total spectralintensity for a series of transitions. c.f Bethe and Jackiw for a great description of sum-rules.

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Figure 4.4: The diffraction function sin(x)/x

-20 -10 10 20

-0.2

0.2

0.4

0.6

0.8

1

4.4 Example using the particle in a box states

What are the constants of motion for a particle in a box?Recall that the energy levels and wavefunctions for this system are

En =n2π2h2

2ma2(4.144)

φn(x) =

√2

asin(

nπ

ax) (4.145)

Say our system in in the nth state. What’s the probability of measuring the momentum andobtaining a result between p and p+ dp?

Pn(p)dp = |φn(p)|2dp (4.146)

where

φn(p) =1√2πh

∫ a

0dx

√2

asin(nπx/a)e−ipx/h (4.147)

=1

2i

1√2πh

[ei(nπ/a−p/h)a − 1

i(nπ/a− p/h)− e−i(nπ/a−p/h)a − 1

−i(nπ/a− p/h)

](4.148)

=1

2i

√a

πhexp i(nπ/2− pa/(2h))[F (p− nπh/2) + (−1)n+1F (p+ nπh/a)] (4.149)

Where the F (p) are “diffraction functions”

F (p) =sin(pa/(2h))

pa/(2h)(4.150)

Note that the width 4πh/a does not change as I change n. Nor does the amplitude. However,note that (F (x+ n)± FF (x− n))2 is always an even function of x. Thus, we can say

〈p〉n =∫ +∞

−∞Pn(p)pdp = 0 (4.151)

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We can also compute:

〈p2〉 = h2∫ ∞

0dx

∣∣∣∣∣∂φn(x)

∂x

∣∣∣∣∣2

(4.152)

= h2∫ a

0

2

a

(nπ

a

)2

cos(nπx/a)dx (4.153)

=

(nπh

a

)2

= 2mEn (4.154)

Thus, the RMS deviation of the momentum:

∆pn =√〈p2〉n − 〈p〉2n =

nπh

a(4.155)

Thus, as n increases, the relative accuracy at which we can measure p increases due the factthat we can resolve the wavefunction into two distinct peaks corresponding to the particle eithergoing to the left or to the right. ∆p increases due to the fact that the two possible choices forthe measurement are becoming farther and farther apart and hence reflects the distance betweenthe two most likely values.

4.5 Time Evolution of Wave and Observable

Now, suppose we put our system into a superposition of box-states:

|ψ(0)〉 =1√2(|φ1〉+ |φ2〉) (4.156)

What is the time evolution of this state? We know the eigen-energies, so we can immediatelywrite:

|ψ(t)〉 =1√2(exp(−iE1t/h)|φ1〉+ exp(−iE2t/h)|φ2〉) (4.157)

Let’s factor out a common phase factor of e−iE1t/h and write this as

|ψ(t)〉 ∝ 1√2(|φ1〉+ exp(−i(E2 − E1)t/h)|φ2〉) (4.158)

and call (E2 − E1)/h = ω21 the Bohr frequency.

|ψ(t)〉 ∝ 1√2(|φ1〉+ exp(−iω21t)|φ2〉) (4.159)

where

ω21 =3π2h

2ma2. (4.160)

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The phase factor is relatively unimportant and cancels out when I make a measurement. Eg.the prob. density:

|ψ(x, t)|2 = |〈x|ψ(t)〉|2 (4.161)

(4.162)

=1

2φ2

1(x) +1

2φ2

2(x) + φ1(x)φ2(x) cos(ω21t) (4.163)

Now, let’s compute 〈x(t)〉 for the two state system. To do so, let’s first define x′ = x − a/2as the center of the well to make the integrals easier. The first two are easy:

〈φ1|x′|φ1〉 ∝∫ a

0dx(x− a/2) sin2(πx/a) = 0 (4.164)

〈φ2|x′|φ2〉 ∝∫ a

0dx(x− a/2) sin2(2πx/a) = 0 (4.165)

which we can do by symmetry. Thus,

〈x′(t)〉 = Ree−iω21t〈φ1|x′|φ2〉 (4.166)

〈φ1|x′|φ2〉 = 〈φ1|x|φ2〉 − (a/2)〈φ1|φ2〉 (4.167)

=2

a

∫ a

0dxx sin(πx/a) sin(2πx/a) (4.168)

= −16a

9π2(4.169)

Thus,

〈x(t)〉 =a

2− 16a

9π2cos(ω21t) (4.170)

Compare this to the classical trajectory. Also, what about 〈E(t)〉?

4.6 “Unstable States”

So far in this course, we have been talking about systems which are totally isolated from therest of the universe. In these systems, there is no influx or efflux of energy and all our dynamicsare governed by the three principle postulates I mentioned a the start of the lecture. IN essence,if at t = 0 I prepare the system in an eigenstate of H, then for all times later, it’s still in thatstate (to within a phase factor). Thus, in a strictly conservative system, a system prepared inan eigenstate of H will remain in an eigenstate forever.

However, this is not exactly what is observed in nature. We know from experience that atomsand molecules, if prepared in an excited state (say via the absorption of a photon) can relax

104

back to the ground state or some lower state via the emission of a photon or a series of photons.Thus, these eigenstates are “unstable”.

What’s wrong here? The problem is not so much to do with what is wrong with our descriptionof the isolated system, it has to do with full description is not included. A isolated atom ormolecule can still interact with the electro-magnetic field (unless we do some tricky confinementexperiments). Thus, there is always some interaction with an outside “environment”. Thus,while it is totally correct to describe the evolution of the global system in terms of some global“atom” + “environment” Hamiltonian, it it NOT totally rigorous to construct a Hamiltonianwhich describes only part of the story. But, as the great Prof. Karl Freed (at U. Chicago) oncetold me “Too much rigor makes rigor mortis”.

Thankfully, the coupling between an atom and the electromagnetic field is pretty weak. Eachphoton emission probability is weighted by the fine-structure constant, α ≈ 1/137. Thus a 2photon process is weighted by α2. Thus, the isolated system approximation is pretty good. Also,we can pretty much say that most photon emission processes occur as single photon events.

Let’s play a bit “fast and loose” with this idea. We know from experience that if we preparethe system in an excited state at t = 0, the probability of finding it still in the excited state atsome time t later, is

P (t) = e−t/τ (4.171)

where τ is some time constant which we’ll take as the lifetime of the state. One way to “prove”this relation is to go back to Problem Set 0. Let’s say we have a large number of identical systemsN , each prepared in the excited state at t = 0. At time t, there are

N(t) = N e−t/τ (4.172)

systems in the excited state. Between time t and t + dt a certain number, dn(t) will leave theexcited state via photon emission.

dn(t) = N(t)−N(t+ dt) = −dN(t)

dtdt = N(t)

dt

τ(4.173)

Thus,

dn(t)

N(t)=dt

τ(4.174)

Thus, 1/τ is the probability per unit time for leaving the unstable state.The Avg. time a system spends in the unstable state is given by:

1

τ

∫ ∞

0dtte−t/τ = τ (4.175)

For a stable state P (t) = 1 thus, τ →∞.The time a system spends in the state is independent of its history. This is a characteristic of

an unstable state. (Also has to do with the fact that the various systems involved to not interactwith each other. )

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Finally, according to the time-energy uncertainty relation:

∆Eτ ≈ h. (4.176)

Thus, an unstable system has an intrinsic “energy width” associated with the finite time thesystems spends in the state.

For a stable state:

|ψ(t)〉 = e−iEnt/h|φn〉 (4.177)

and

Pn(t) = |e−iEnt/h|2 = 1 (4.178)

for real energies.What if I instead write: E ′

n = En − ihγn/h? Then

Pn(t) = |e−iEnt/he−γn/2t|2 = e−γnt (4.179)

Thus,

γn = 1/τn (4.180)

is the “Energy Width” of the unstable state.The surprising part of all this is that in order to include dissipative effects (photon emission,

etc..) the Eigenvalues of H become complex. In other words, the system now evolves under anon-hermitian Hamiltonian! Recall the evolution operator for an isolated system:

U(t) = e−iHt/h (4.181)

(4.182)

U †(t) = eiHt/h (4.183)

where the first is the forward evolution of the system and the second corresponds to the back-wards evolution of the system. Thus, Unitarity is thus related to the time-reversal symmetryof conservative systems. The inclusion of an “environment” breaks the intrinsic time-reversalsymmetry of an isolated system.

4.7 Problems and Exercises

Exercise 4.5 Find the eigenvalues and eigenvectors of the matrix:

M =

0 0 0 10 0 1 00 1 0 01 0 0 0

(4.184)

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Solution: You can either do this the hard way by solving the secular determinant and thenfinding the eigenvectors by Gramm-Schmidt orthogonalization, or realize that since M = M−1

and M = M †, M is a unitary matrix, this, its eigenvalues can only be ± 1. Furthermore, sincethe trace of M is 0, the sum of the eigenvalues must be 0 as well. Thus, λ = (1, 1,−1,−1) arethe eigenvalues. To get the eigenvectors, consider the following. Let φmu be an eigenvector ofM , thus,

φµ =

x1

x2

x3

x4

. (4.185)

Since Mφµ = λµφmu, x1 = λµx4 and x2 = λµx3 Thus, 4 eigenvectors are

−1001

,

01−10

,

1001

,

0110

(4.186)

for λ = (−1,−1, 1, 1).

Exercise 4.6 Let λi be the eigenvalues of the matrix:

H =

2 −1 −3−1 1 2−3 2 3

(4.187)

calculate the sums:

3∑i

λi (4.188)

and

3∑i

λ2i (4.189)

Hint: use the fact that the trace of a matrix is invariant to choice representation.

Solution: Using the hint,

trH =∑

i

λi =∑

i

Hii = 2 + 3 + 1 = 6 (4.190)

and ∑i

λ2i =

∑ij

HijHji =∑ij

H2ij = 42 (4.191)

107

Exercise 4.7 1. Let |φn〉 be the eigenstates of the Hamiltonian, H of some arbitrary systemwhich form a discrete, orthonormal basis.

H|φn〉 = En|φn〉.

Define the operator, Unm asUnm = |φn〉〈φm|.

(a) Calculate the adjoint U †nm of Unm.

(b) Calculate the commutator, [H,Unm].

(c) Prove: UmnU†pq = δnqUmp

(d) For an arbitrary operator, A, prove that

〈φn|[A,H]|φn〉 = 0.

(e) Now consider some arbitrary one dimensional problem for a particle of mass m andpotential V (x). For here on, let

H =p2

2m+ V (x).

i. In terms of p, x, and V (x), compute: [H, p], [H, x], and [H, xp].

ii. Show that 〈φn|p|φn〉 = 0.

iii. Establish a relation between the average value of the kinetic energy of a state

〈T 〉 = 〈φn|p2

2m|φn〉

and

〈φn|xdV

dx|φn〉.

The average potential energy in the state φn is

〈V 〉 = 〈φn|V |φn〉,

find a relation between 〈V 〉 and 〈T 〉 when V (x) = Voxλ for λ = 2, 4, 6, . . ..

(f) Show that〈φn|p|φm〉 = α〈φn|x|φm〉

where α is some constant which depends upon En − Em. Calculate α, (hint, considerthe commutator [x,H] which you computed above.

(g) Deduce the following sum-rule for the linear -response function.

〈φ0|[x, [H, x]]|φ0〉 = 2∑n>0

(En − E0)|〈φ0|x|φn〉|2

Here |φ0〉 is the ground state of the system. Give a physical interpretation of this lastresult.

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Exercise 4.8 For this section, consider the following 5× 5 matrix:

H =

0 0 0 0 10 0 0 1 00 0 1 0 00 1 0 0 01 0 0 0 0

(4.192)

1. Using Mathematica determine the eigenvalues, λj, and eigenvectors, φn, of H using theEigensystem[] command. Determine the eigenvalues only by solving the secular determi-nant.

|H − Iλ| = 0

Compare the computational effort required to perform both calculations. Note: in enteringH into Mathematica, enter the numbers as real numbers rather than as integers (i.e. 1.0vs 1 ).

2. Show that the column matrix of the eigenvectors of H,

T = φ1, . . . , φ5,

provides a unitary transformation of H between the original basis and the eigenvector basis.

T †HT = Λ

where Λ is the diagonal matrix of the eigenvalues λj. i.e. Λij = λiδij.

3. Show that the trace of a matrix is invarient to representation.

4. First, without using Mathematica, compute: Tr(H2). Now check your result with Mathe-matica.

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Chapter 5

Bound States of The SchrodingerEquation

A #2 pencil and a dream can take you anywhere.– Joyce A. Myers

Thus far we have introduced a series of postulates and discussed some of their physical impli-cations. We have introduced a powerful notation (Dirac Notation) and have been studying howwe describe dynamics at the atomic and molecular level where h is not a small number, but isof order unity. We now move to a topic which will serve as the bulk of our course, the studyof stationary systems for various physical systems. We shall start with some general principles,(most of which we have seen already), and then tackle the following systems in roughly thisorder:

1. Harmonic Oscillators: Molecular vibrational spectroscopy, phonons, photons, equilibriumquantum dynamics.

2. Angular Momentum: Spin systems, molecular rotations.

3. Hydrogen Atom: Hydrogenic Systems, basis for atomic theory

5.1 Introduction to Bound States

Before moving on to these systems, let’s first consider what is meant by a “bound state”. Saywe have a potential well which has an arbitrary shape except that at x = ±a, V (x) = 0 andremains so in either direction. Also, in the range of −a ≤ x ≤ a, V (x) < 0. The SchrodingerEquation for the stationary states is:[

−h2

2m

∂2

∂x2+ V (x)

]φn(x) = Enφn(x) (5.1)

Rather than solve this exactly (which we can not do since we haven’t specified more aboutV ) let’s examine the topology of the allowed bound state solutions. As we have done with thesquare well cases, let’s cut the x axis into three domains: Domain 1 for x < −a, Domain 2 for−a ≤ x ≤ a, Domain 3 for x > a. What are the matching conditions that must be met?

110

For Domain 1 we have:

∂2

∂x2φI(x) = −2mE

h2 φI(x) ⇒ (+)φI(x) (5.2)

For Domain 2 we have:

∂2

∂x2φII(x) =

2m(V (x)− E)

h2 φII(x) ⇒ (−)φII(x) (5.3)

For Domain 3 we have:

∂2

∂x2φIII(x) = −2m(E)

h2 φIII(x) ⇒ (+)φIII(x) (5.4)

At the rightmost end of each equation, the (±) indicates the sign of the second derivative ofthe wavefunction. (i.e. the curvature must have the same or opposite sign as the function itself.)For the + curvature functions, the wavefunctions curve away from the x-axis. For − curvature,the wavefunctions are curved towards the x-axis.

Therefore, we can conclude that for regions outside the well, the solutions behave much likeexponentials and within the well, the behave like superpositions of sine and cosine functions.Thus, we adopt the asymptotic solution that

φ(x) ≈ exp(+αx)for x < a as x→ −∞ (5.5)

and

φ(x) ≈ exp(−αx)for x > a as x→ +∞ (5.6)

Finally, in the well region, φ(x) oscillates about the x-axis. We can try to obtain a more completesolution by combining the solutions that we know. To do so, we must find solutions which areboth continuous functions of x and have continuous first derivatives of x.

Say we pick an arbitrary energy, E, and seek a solution at this energy. Define the righthandpart of the solution to within a multiplicative factor, then the left hand solution is then acomplicated function of the exact potential curve and can be written as

φIII(x) = B(E)e+ρx +B′(E)e−ρx (5.7)

where B(E) and B′(E) are both real functions of E and depend upon the potential function.Since the solutions must be L2, the only appropriate bound states are those for which B(E) = 0.Any other value of B(E) leads to diverging solutions.

Thus we make the following observations concerning bound states:

1. They have negative energy.

2. They vanish exponentially outside the potential well and oscillate within.

3. They form a discrete spectrum as the result of the boundary conditions imposed by thepotential.

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5.2 The Variational Principle

Often the interaction potential is so complicated that an exact solution is not possible. Thisis often the case in molecular problems in which the potential energy surface is a complicatedmultidimensional function which we know only at a few points connected by some interpola-tion function. We can, however, make some approximations. The method, we shall use is the“Variational method”.

5.2.1 Variational Calculus

The basic principle of the variational method lies at the heart of most physical principles. Theidea is that we represent a quantity as a stationary integral

J =∫ x2

x1

f(y, yx, x)dx (5.8)

where f(y, yx, x) is some known function which depened upon three variables, which are alsofunctions of x, y(x), yx = dy/dx, and x itself. The dependency of y on x uis generally unknown.This means that while we have fixed the end-points of the integral, the path that we actuallytake between the endpoints is not known.

Picking different paths leads to different values of J . However ever certain paths will minimize,maximize, or find the saddle points of J . For most cases of physical interest, its the extremathat we are interested. Lets say that there is one path, yo(t) which minimizes J (See Fig. 5.1).If we distort that path slightly, we get another path y(x) which is not too unlike yo(x) and wewill write it as y(x) = yo(x) + η(x) where η(x1) = η(x2) = 0 so that the two paths meet at theterminal points. If η(x) differs from 0 only over a small region, we can write the new path as

y(x, α) = yo(x) + αη(x)

and the variation from the minima as

δy = y(x, α)− yo(x, 0) = αη(x).

Since yo is the path which minimizes J , and y(x, α) is some other path, then J is also afunction of α.

J(α) =∫ x2

x1

f(y(x, α), y′(x, α), x)dx

and will be minimized when (∂J

∂α

)α=0

= 0

Because J depends upon α, we can examine the α dependence of the integral(∂J

∂α

)α=0

=∫ x2

x1

(∂f

∂y

∂y

∂α+∂f

∂y′∂y′

∂α

)dx

Since∂y

∂α= η(x)

112

and∂y′

∂α=∂η

∂x

we have (∂J

∂α

)α=0

=∫ x2

x1

(∂f

∂yη(x) +

∂f

∂y′∂η

∂x

)dx.

Now, we need to integrate the second term by parts to get η as a common factor. Rememberintegration by parts? ∫

udv = vu−∫vdu

From this ∫ x2

x1

(∂f

∂y′∂η

∂x

)dx = η(x)

∂f

∂x

∣∣∣∣∣x2

x1

−∫ x2

x1

η(x)d

dx

∂f

∂y′dx

The boundaty term vanishes since η vanishes at the end points. So putting it all together andsetting it equal to zero: ∫ x2

x1

η(x)

(∂f

∂y− d

dx

∂f

∂y′

)dx. = 0

We’re not done yet, since we still have to evaluate this. Notice that α has disappeared fromthe expression. In effect, we can take an arbitrary variation and still find the desired path thaminimizes J . Since η(x) is arbitrary subject to the boundary conditions, we can make it have thesame sign as the remaining part of the integrand so that the integrand is always non-negative.Thus, the only way for the integral to vanish is if the bracketed term is zero everywhere.(

∂f

∂y− d

dx

∂f

∂y′

)= 0 (5.9)

This is known as the Euler equation and it has an enormous number of applications. Perhapsthe simplest is the proof that the shortest distance between two points is a straight line (oron a curved space, a geodesic). The straightline distance between two points on the xy plane

is s =√x2 + y2 and the differential element of distance is ds =

√(dx)2 + (dy)2 =

√1 + y2

xdx.Thus, we can write a distance along some line in the xy plane as

J =∫ x2y2

x1y1

ds =∫ x2y2

x1y1

√1 + y2

xdx.

If we knew y(x) then J would be the arclength or path-length along the function y(x) betweentwo points. Sort of like, how many steps you would take along a trail between two points. Thetrail may be curvy or straight and there is certainly a single trail which is the shortest. So,setting

f(y, yx, x) =√

1 + y2x

and substituting it into the Euler equation one gets

d

dx

∂f

∂yx

= − d

dx

1√1 + y2

x

= 0. (5.10)

113

So, the only way for this to be true is if

1√1 + y2

x

= constant. (5.11)

Solving for yx produces a second constant: yx = a, which immediatly yields that y(x) = ax+ b.In other words, its a straight line! Not too surprising.

An important application of this principle is when the integrand f is the classical Lagrangianfor a mechanical system. The Lagrangian is related to Hamiltonian and is defined as the differencebetween the kinetic and potential energy.

L = T − V (5.12)

where as H is the sum of T +V . Rather than taking x as the independent variable, we take time,t, and position and velocity oa a particle as the dependent variables. The statement of δJ = 0is a mathematical statement of Hamilton’s principle of least action

δ∫ t2

t1L(x, x, t)dt = 0. (5.13)

In essence, Hamilton’s principle asserts that the motion of the system from one point to anotheris along a path which minimizes the integral of the Lagrangian. The equations of motion for thatpath come from the Euler-Lagrange equations,

d

dt

∂L

∂x− ∂L

∂x= 0. (5.14)

So if we write the Lagrangian as

L =1

2mx2 − V (x) (5.15)

and substitute this into the Euler-Lagarange equation, we get

mx = −∂V∂x

(5.16)

which is Newton’s law of motion: F = ma.

5.2.2 Constraints and Lagrange Multipliers

Before we can apply this principle to a quantum mechanical problem, we need to ask our selveswhat happens if there is a constraint on the system which exclues certain values or paths so thatnot all the η’s may be varied arbitrarily? Typically, we can write the constraint as

φi(y, x) = 0 (5.17)

For example, for a bead on a wire we need to constrain the path to always lie on the wire or fora pendulum, the path must lie on in a hemisphere defined by the length of the pendulum from

114

the pivot point. In any case, the general proceedure is to introduce another function, λi(x) andintegrate ∫ x2

x1

λi(x)φi(y, x)dx = 0 (5.18)

so that

δ∫ x2

x1

λi(x)φi(y, x)dx = 0 (5.19)

as well. In fact it turns out that the λi(x) can be even be taken to be a constant, λi for thiswhole proceedure to work.

Regardless of the case, we can always write the new stationary integral as

δ∫

(f(y, yx, x) +∑

i

λiφi(y, x))dx = 0. (5.20)

The multiplying constants are called Lagrange Mulipliers. In your statistical mechanics course,these will occur when you minimize various thermodynamic functions subject to the variousextensive constraints, such as total number of particles in the system, the average energy ortemperature, and so on.

In a sence, we have redefined the original function or Lagrangian to incorporate the constraintinto the dynamics. So, in the presence of a constraint, the Euler-Lagrange equations become

d

dt

∂L

∂x− ∂L

∂x=∑

i

∂φi

∂xλi (5.21)

where the term on the right hand side of the equation represents a force due to the constraint.The next issue is that we still need to be able to determine the λi Lagrange multipliers.

115

Figure 5.1: Variational paths between endpoints. The thick line is the stationary path, yo(x)and the dashed blue curves are variations y(x, α) = yo(x) + αη(x).

-1.5 -1 -0.5 0.5 1 1.5x

-2

-1

1

2

fHxL

116

5.2.3 Variational method applied to Schrodinger equation

The goal of all this is to develop a procedure for computing the ground state of some quantummechanical system. What this means is that we want to minimize the energy of the systemwith respect to arbitrary variations in the state function subject to the constraint that the statefunction is normalized (i.e. number of particles remains fixed). This means we want to constructthe variation:

δ〈ψ|H|ψ〉 = 0 (5.22)

with the constraint 〈ψ|ψ〉 = 0.In the coordinate representation, the integral involves taking the expectation value of the

kinetic energy operator...which is a second derivative operator. That form is not too convenientfor our purposes since it will allow us to write Eq.5.22 in a form suitable for the Euler-Lagrangeequations. But, we can integrate by parts!

∫ψ∗∂2ψ

∂x2dx = ψ∗

∂ψ

∂x

∣∣∣∣∣−∫ (

∂ψ∗

∂x

)(∂ψ

∂x

)dx (5.23)

Assuming that the wavefunction vanishes at the limits of the integration, the surface term van-ishes leaving only the second term. We can now write the energy expectation value in termsof two dependent variables, ∇ψ and ψ. OK, they’re functions, but we can still treat them asdependent variables just like we treated the y(x)′s above.

E =∫ (

h2

2m(∇ψ∗)(∇ψ) + V ψ∗ψ

)dx (5.24)

Adding on the constraint and defining the Lagrangian as

L =

(h2

2m(∇ψ∗)(∇ψ) + V ψ∗ψ

)− λψ∗ψ, (5.25)

we can substitute this into the Euler-Lagrange equations

∂L

∂ψ∗−∇∂x ∂L

∂(∇ψ∗= 0. (5.26)

This produces the result

(V − λ)ψ =h2

2m∇2ψ, (5.27)

which we immediately recognize as the Schrodinger equation.While this may be a rather academic result, it gives us the key to recognize that we can

make an expansion of ψ in an arbitrary basis and take variations with respect to the coeffientsof that basis to find the lowest energy state. This is the basis of a number of powerful numericalmethods used solve the Schrodinger equation for extremely complicated systems.

117

5.2.4 Variational theorems: Rayleigh-Ritz Technique

We now discuss two important theorems:

Theorem 5.1 The expectation value of the Hamiltonian is stationary in the neighborhood of itseigenstates.

To demonstrate this, let |ψ〉 be a state in which we compute the expectation value of H. Also,let’s modify the state just a bit and write

|ψ〉 → |ψ〉+ |δψ〉. (5.28)

Expectation values are computed as

〈H〉 =〈ψ|H|ψ〉〈ψ|ψ〉

(5.29)

(where we assume arb. normalization). In other words

〈ψ|ψ〉〈H〉 = 〈ψ|H|ψ〉 (5.30)

Now, insert the variation

〈ψ|ψ〉δ〈H〉+ 〈δψ|ψ〉〈H〉+ 〈ψ|δψ〉〈H〉 = 〈δψ|H|ψ〉+ 〈ψ|H|δψ〉 (5.31)

or

〈ψ|ψ〉δ〈H〉 = 〈δψ|H − 〈H〉|ψ〉 − 〈ψ|H − 〈H〉|δψ〉 (5.32)

If the expectation value is to be stationary, then δ〈H〉 = 0. Thus the RHS must vanish for anarbitrary variation in the wavefunction. Let’s pick

|δψ〉 = ε|ψ〉. (5.33)

Thus,

(H − 〈H〉)|ψ〉 = 0 (5.34)

That is to say that |ψ〉 is an eigenstate of H. Thus proving the theorem.The second theorem goes:

Theorem 5.2 The Expectation value of the Hamiltonian in an arbitrary state is greater than orequal to the ground-state energy.

The proof goes as this: Assume that H has a discrete spectrum of states (which we demonstratedthat it must) such that

H|n〉 = En|n〉 (5.35)

118

Thus, we can expand any state |ψ〉 as

|ψ〉 =∑n

cn|n〉. (5.36)

Consequently

〈ψ|ψ〉 =∑n

|cn|2, (5.37)

and

〈ψ|H|ψ〉 =∑n

|cn|2En. (5.38)

Thus, (assuming that |ψ〉 is normalized)

〈H〉 =∑n

En|cn|2 ≥∑n

Eo|cn|2 ≥ Eo (5.39)

quid est demonstrato.Using these two theorems, we can estimate the ground state energy and wavefunctions for a

variery of systems. Let’s first look at the Harmonic Oscillator.

Exercise 5.1 Use the variational principle to estimate the ground-state energy of a particle inthe potential

V (x) =

Cx x > 0+∞ x ≤ 0

(5.40)

Use xe−ax as the trial function.

5.2.5 Variational solution of harmonic oscillator ground State

The Schrodinger Equation for the Harmonic Osc. (HO) is[− h2

2m

∂2

∂x2+k2

2x2

]φ(x)− Eφ(x) = 0 (5.41)

Take as a trial function,

φ(x) = exp(−λx2) (5.42)

where λ is a positive, non-zero constant to be determined. The variational principle states thatthe energy reaches a minimum

∂〈H〉∂λ

= 0. (5.43)

when φ(x) is the ground state solution. Let us first derive 〈H〉(λ).

〈H〉(λ) =〈φ|H|φ〉〈φ|φ〉

(5.44)

119

To evaluate this, we break the problem into a series of integrals:

〈φ|φ〉 =∫ ∞

−∞dx|φ(x)|2 =

√π

2λ(5.45)

〈φ|p2|φ〉 =∫ ∞

−∞dxφ′′(x)φ(x) = −2λ〈φ|φ〉+ 4λ2〈φ|x2|φ〉 (5.46)

and

< φ|x2|φ〉 =∫ ∞

−∞dxx2|φ(x)|2 =

1

4λ〈φ|φ〉. (5.47)

Putting it all together:

〈φ|H|φ〉〈φ|φ〉

=

(− h2

2m

)(−2λ+ 4λ2 1

4λ

)+k

2

1

4λ(5.48)

〈φ|H|φ〉〈φ|φ〉

=

(h2

2m

)λ+

k

8λ(5.49)

Taking the derivative with respect to λ:

∂〈H〉∂λ

=h2

2m− k

8λ2= 0 (5.50)

Thus,

λ = ±√mk

2h(5.51)

Since only positive values of λ are allowed.

λ =

√mk

2h(5.52)

Using this we can calculate the ground state energy by substituting λ back into 〈H〉(λ).

〈H〉(λ) =

√mk

2h

(h2

2m+k

8

4h2

mk

)=h

2

√k

m(5.53)

Now, define the angular frequency: ω =√k/m.

〈H〉(λ) =h

2ω (5.54)

which ( as we can easily prove) is the ground state energy of the harmonic oscillator.Furthermore, we can write the HO ground state wavefunction as

120

φo(x) =1√〈φ|φ〉

φ(x) (5.55)

φo(x) =

(2λ

π

)1/4

exp

(−√mk

2hx2

)(5.56)

φo(x) =

(√mk

hπ

)1/4

exp

(−√mk

2hx2

)(5.57)

To compute the “error” in our estimate, let’s substitute the variational solution back into theSchrodinger equation:

[− h2

2m

∂2

∂x2+k2

2x2

]φo(x) = − h2

2mφ′′o(x) +

k2

2φo(x) (5.58)

− h2

2mφ′′o(x) +

k2

2φo(x) = − h2

2m

(kmx2 − h

√km

h2

)φo(x) +

k

2x2φo(x) (5.59)

− h2

2mφ′′o(x) +

k2

2φo(x) =

h

2

√k

mφo(x) (5.60)

Thus, φo(x) is in fact the correct ground state wavefunction for this system. If it were notthe correct function, we could re-introduce the solution as a new trial function, re-compute theenergy, etc... and iterate through until we either find a solution, or run out of patience! (Usuallyit’s the latter than the former.)

5.3 The Harmonic Oscillator

Now that we have the HO ground state and the HO ground state energy, let us derive the wholeHO energy spectrum. To do so, we introduce “dimensionless” quantities: X and P related tothe physical position and momentum by

X = (mω

2h)1/2x (5.61)

P = (1

2hmω)1/2p (5.62)

This will save us from carrying around a bunch of coefficients. In these units, the HO Hamiltonianis

H = hω(P 2 +X2). (5.63)

121

The X and P obey the canonical comutational relation:

[X,P ] =1

2h[x, p] =

i

2(5.64)

We can also write the following:

(X + iP )(X − iP ) = X2 + P 2 + 1/2 (5.65)

(X − iP )(X + iP ) = X2 + P 2 − 1/2. (5.66)

Thus, I can construct the commutator:

[(X + iP ), (X − iP )] = (X + iP )(X − iP )− (X − iP )(X + iP )

= 1/2 + 1/2

= 1 (5.67)

Let’s define the following two operators:

a = (X + iP ) (5.68)

a† = (X + iP )† = (X − iP ). (5.69)

Therefore, the a and a† commute as

[a, a†] = 1 (5.70)

Let’s write H in terms of the a and a† operators:

H = hω(X2 + P 2) = hω(X − iP )(X + iP ) + hω/2 (5.71)

or in terms of the a and a† operators:

H = hω(a†a+ 1/2) (5.72)

Now, consider that |φn〉 is the nth eigenstate of H. Thus, we write:

hω(a†a+ 1/2)|φn〉 = En|φn〉 (5.73)

What happens when I multiply the whole equation by a? Thus, we write:

ahω(a†a+ 1/2)|φn〉 = aEn|φn〉 (5.74)

hω(aa† + 1/2)(a|φn〉) = En(a|φn〉) (5.75)

Now, since aa† − a†a = 1,

hω(a†a+ 1/2− 1)(a|φn〉) = En(a|φn〉) (5.76)

In other words, a|φn〉 is an eigenstate of H with energy E = En − hω.

122

What happends if I do the same procedure, this time using a†? Thus, we write:

a†hω(a†a+ 1/2)|φn〉 = a†En|φn〉 (5.77)

Since

[a, a†] = aa† − a†a (5.78)

we have

a†a = aa† − 1 (5.79)

we can write

a†a†a = a†(aa† − 1) (5.80)

= (a†a− 1)a†. (5.81)

Thus,

a†hω(a†a+ 1/2)|φn〉 = hω((a†a− 1 + 1/2)a†)|φn〉 (5.82)

or,

hω(a†a− 1/2)(a†|φn〉) = En(a†|φn〉). (5.83)

Thus, a†|φn〉 is an eigenstate of H with energy E = En + hω.Since a† and a act on harmonic oscillator eigenstates to give eigenstates with one more or one

less hω quanta of energy, these are termed “creation” and “annihilation” operators since theyact to create additional quata of excitation or decrease the number of quanta of excitation inthe system. Using these operators, we can effectively “ladder” our way up the energy scale anddetermine any eigenstate one we know just one.

Well, we know the ground state solution. That we got via the variational calculation. Whathappens when I apply a† to the φo(x) we derived above? In coordinate form:

(X − iP )φo(x) =

((mω

2h

)1/2

x+(

1

2mωh

)1/2 ∂

∂x

)φo(x)

(5.84)

=

((mω

2h

)1/2

x+(

1

2mωh

)1/2 ∂

∂x

)(mω

hπ

)1/4

e(−x2 mω2h ) (5.85)

X acting on φo is:

Xφo(x) =

√mω

2hxφo(x) (5.86)

iP acting on φo is

iPφo(x) = −h√

1

mω2h

∂

∂xφo(x) (5.87)

123

iPφo(x) = −x√mω

2hφo(x) (5.88)

= −Xφo(x) (5.89)

After cleaning things up:

(X − iP )φo(x) = 2Xφo(x) (5.90)

= 2

√mω

2hxφo(x) (5.91)

= 2Xφo(x) (5.92)

= 2

√mω

2hx(mω

2h

)1/4

exp(−x2mω

2h

)(5.93)

5.3.1 Harmonic Oscillators and Nuclear Vibrations

We introduced one of the most important applications of quantum mechanics...the solution ofthe Schrodinger equation for harmonic systems. These are systems in which the amplitude ofmotion is either small enough so that the physical potential energy operator can be expandedabout its minimum to second order in the displacement from the minima. When we do so, theHamiltonian can be written in the form

H = hω(P 2 +X2) (5.94)

where P and X are dimensionless operators related to the physical momentum and positionoperators via

X =

√mω

2hx (5.95)

and

P =

√1

2hmωp. (5.96)

We also used the variational method to deduce the ground state wavefunction and demonstratedthat the spectrum of H is a series of levels separated by hω and that the ground-state energy ishω/2 above the energy minimum of the potential.

We also defined a new set of operators by taking linear combinations of the X and P .

a = X + iP (5.97)

a† = X − iP. (5.98)

We also showed that the commutation relation for these operators is

[a, a†] = 1. (5.99)

These operators are non-hermitian operators, and hence, do not correspond to a physical ob-servable. However, we demonstrated that when a acts on a eigenstate of H, it produces another

124

eigenstate withe energy En − hω. Also, a† acting on an eigenstate of H produces another eigen-state with energy En + hω. Thus,we called a the destruction or annihilation operator since itremoves a quanta of excitation from the system and a† the creation operator since it adds aquanta of excitation to the system. We also wrote H using these operators as

H = hω(a†a+1

2) (5.100)

Finally, ω is the angular frequency of the classical harmonic motion, as obtained via Hooke’slaw:

x = − k

mx. (5.101)

Solving this produces

x(t) = xo sin(ωt+ φ) (5.102)

and

p(t) = po cos(ωt+ φ). (5.103)

Thus, the classical motion in the x, p phase space traces out the circumference of a circle every1/ω regardless of the initial amplitude.

The great advantage of using the a, and a† operators is that they we can replace a differentialequation with an algebraic equation. Furthermore, since we can represent any Hermitian operatoracting on the HO states as a combination of the creation/annihilation operators, we can replacea potentially complicated series of differentiations, integrations, etc... with simple algebraicmanipulations. We just have to remember a few simple rules regarding the commutation of thetwo operators. Two operators which we may want to construct are:

• position operator:(

2hmω

)1/2(a† + a)

• momentum operator: i(

hmω2

)1/2(a† − a).

Another important operator is

N = a†a. (5.104)

and

H = hω(N + 1/2). (5.105)

Since [H,N ] = 0, eigenvalues of N are “good quantum numbers” and N is a constant of themotion. Also, since

H|φn〉 = En|φn〉 = hω(N + 1/2)|φn〉 (5.106)

then if

N |φn〉 = n|φn〉, (5.107)

then n must be an integer n = 0, 1, 2, · · · corresponding to the number of quanta of excitation inthe state. This gets the name “Number Operator”.

Some useful relations (that you should prove )

125

1. [N, a] = [a†a, a] = −a

2. [N, a†] = [a†a, a†] = a†

To summarize, we have the following relations using the a and a† operators:

1. a|φn〉 =√n|φn−1〉

2. a†|φn〉 =√n+ 1|φn+1〉

3. 〈φn|a =√n+ 1〈φn+1 = (a|φn〉)†

4. 〈φn|a† =√n+ 1〈φn−1|

5. N |φn〉 = n|φn〉

6. 〈φn|N = n〈φn|

Using the second of these relations we can write

|φn+1〉 =a†√n+ 1

|φn〉 (5.108)

which can be iterated back to the ground state to produce

|φn〉 =(a†)n

√n!|φo〉 (5.109)

This is the “generating relation” for the eigenstates.Now, let’s look at x and p acting on |φn〉.

x|φn〉 =

√h

2mω(a† + a)|φn〉 (5.110)

=

√h

2mω(√n+ 1|φn+1〉+

√n|φn−1〉) (5.111)

Also,

p|φn〉 = i

√mhω

2(a† − a)|φn〉 (5.112)

= i

√mhω

2(√n+ 1|φn+1〉 −

√n|φn−1〉) (5.113)

Thus,the matrix elements of x and p in the HO basis are:

〈φm|x|φn〉 =

√h

2mω

(√n+ 1δm,n+1 +

√nδm,n−1

)(5.114)

126

〈φm|p|φn〉 = i

√mωh

2

(√n+ 1δm,n+1 −

√nδm,n−1

)(5.115)

The harmonic oscillator wavefunctions can be obtained by solving the equation:

〈x|a|φo〉 = (X + iP )φo(x) = 0 (5.116)

(mω

hx+

∂

∂x

)φo(x) = 0 (5.117)

The solution of this first order differential equation is easy:

φo(x) = c exp(−mω2x2) (5.118)

where c is a constant of integration which we can obtain via normalization:∫dx|φo(x)|2 = 1 (5.119)

Doing the integration produces:

φo(x) =(mω

hπ

)1/4

e−mω2h

x2

(5.120)

127

Figure 5.2: Hermite Polynomials, Hn up to n = 3.

-3 -2 -1 1 2 3x

-10

-7.5

-5

-2.5

2.5

5

7.5

10HnHxL

128

Since we know that a† acting on |φo〉 gives the next eigenstate, we can write

φ1(x) =

(mω

hx− ∂

∂x

)φo(x) (5.121)

Finally, using the generating relation, we can write

φn(x) =1√n!

(mω

hx− ∂

∂x

)n

φo(x). (5.122)

Lastly, we have the “recursion relations” which generates the next solution one step higher orlower in energy given any other solution.

φn+1(x) =1√n+ 1

(mω

hx− ∂

∂x

)φn(x) (5.123)

and

φn−1(x) =1√n

(mω

hx+

∂

∂x

)φn(x). (5.124)

These are the recursion relationships for a class of polynomials called Hermite polynomials, afterthe 19th French mathematician who studied such functions. These are also termed “Gauss-Hermite” and form a set of orthogonal polynomials. The first few Hermite Polynomials, Hn(x)are 1, 2x,−2 + 4x2,−12x+ 8x3, 12− 48x2 + 16x4 for n = 0 to 4. Some of these are plottedin Fig. 5.2

The functions themselves are defined by the generating function

g(x, t) = e−t2+2tx =∞∑

n=0

Hn(x)tn

n!. (5.125)

Differentiating the generating function n times and setting t = 0 produces the nth Hermitepolynomial

Hn(x) =dn

dtng(x, t)

∣∣∣∣∣ = (−1)nex2 dn

dxne−x2

(5.126)

Another useful relation is the Fourier transform relation:

1√2π

∫ ∞

−∞eitxe−x2/2Hn(x)dx = −ine−t2/2Hn(t) (5.127)

which is useful in generating the momentum space representation of the harmonic oscillatorfunctions. Also, from the generating function, we can arrive at the recurrence relation:

Hn+1 = 2xHn − 2nHn−1 (5.128)

and

H ′n(x) = 2nHn−1(x). (5.129)

129

Consequently, the hermite polynomials are solutions of the second-order differental equation:

H ′′n − 2xH ′

n + 2nHn = 0 (5.130)

which is not self-adjoint! To put this into self-adjoint form, we multiply by the weighting functionw = e−x2

, which leads to the orthogonality integral∫ ∞

−∞Hn(x)Hm(x)e−x2

dx = δnm. (5.131)

For the harmonic oscillator functions, we absorb the weighting function into the wavefunctionitself

ψn(x) = e−x2/2Hn(x).

When we substitute this function into the differential equation for Hn we get

ψ′′n + (2n+ 1− x2)ψn = 0. (5.132)

To normalize the functions, we first multipy g by itself and then multiply by w

e−x2

e−s2+2sxe−t2+2tx =∑mn

e−x2

Hn(x)Hm(x)smtn

n!m!(5.133)

When we integrate over −∞ to ∞ the cross terms drop out by orthogonality and we are leftwith ∑

n=0

(st)n

n!n!

∫ ∞

−∞e−x2

H2n(x)dx =

∫ ∞

−∞e−x2−s2+2sx−t2+2xtdx

=∫ ∞

−∞e−(x−s−t)2dx

= π1/2e2st =∑n=0

2n(st)n

n!. (5.134)

Equating like powers of st we obtain,∫ ∞

−∞e−x2

H2n(x)dx = 2nπ1/2n!. (5.135)

When we apply this technology to the SHO, the solutions are

ψn(z) = 2−n/2π−1/4(n!)−1/2e−z2

Hn(z) (5.136)

where z = αx andα2 =

mω

h.

A few gratuitous solutions:

φ1(x) =

[4

π

(mω

h

)3]1/4

x exp(−1

2mωx2) (5.137)

φ2(x) =(mω

4πh

)1/4 (2mω

hx2 − 1

)exp(−1

2mωx2) (5.138)

Fig. 5.3 shows the first 4 of these functions.

130

5.3.2 Classical interpretation.

In Fig. 5.3 are a few of the lowest energy states for the harmonic oscillator. Notice that as thequantum number increases the amplitude of the wavefunction is pushed more and more towardslarger values of ±x. This becomes more pronounced when we look at the actual probabilitydistribution functions, |ψn(x)|2| for the same 4 states as shown in Fig 5.4.

Here, in blue are the actual quantum distributions for the ground-state through n = 3. Ingray are the classical probability distrubutions for the corresponding energies. The gray curvestell us the probabilty per unit length of finding the classical particle at some point x and anypoint in time. This is inversely proportional to how long a particle spends at a given point...i.e.Pc(x) ∝ 1/v(x). Since E = mv2/2 + V (x),

v(x) =√

2(E − V (x))/m

and

P (x) ∝√

m

2(E − V (x))

For the Harmonic Oscillator:

Pn(x) ∝√

m

2(hω(n+ 1/2)− kx2/2).

Notice that the denominator goes to zero at the classical turning points, in other words,the particle comes to a dead-stop at the turning point and consequently we have the greatestlikelyhood of finding the particle in these regions. Likewise in the quantum case, as we increasethe quantum number, the quantum distrubution function becomes more and more like its classicalcounterpart. This is shown in the last four frames of Fig. 5.4 where we have the same plots as inFig. 5.4, except we look at much higher quantum numbers. For the last case, where n = 19 theclassical and quantum distributions are nearly identical. This is an example of the correspondenceprinciple. As the quantum number increases, we expect the quantum system to look more andmore like its classical counter part.

131

Figure 5.3: Harmonic oscillator functions for n = 0 to 3

-4 -2 2 4

-2

-1

1

2

-4 -2 2 4

-4

-2

2

4

-4 -2 2 4

0.2

0.4

0.6

0.8

1

-4 -2 2 4

-1

-0.5

0.5

1

132

Figure 5.4: Quantum and Classical Probability Distribution Functions for Harmonic Oscillatorfor n = 0, 1, 2, 3, 4, 5, 9, 14, 19

-4 -2 2 4

0.25

0.5

0.75

1

1.25

1.5

-4 -2 2 4

0.25

0.5

0.75

1

1.25

1.5

-4 -2 2 4

0.2

0.4

0.6

0.8

1

1.2

1.4

-4 -2 2 4

0.2

0.4

0.6

0.8

1

1.2

1.4

-4 -2 2 4

0.5

1

1.5

2

2.5

-4 -2 2 4

0.5

1

1.5

2

2.5

-4 -2 2 4

0.25

0.5

0.75

1

1.25

1.5

1.75

-4 -2 2 4

0.25

0.5

0.75

1

1.25

1.5

1.75

-7.5 -5 -2.5 2.5 5 7.5

0.1

0.2

0.3

0.4

0.5

0.6

-7.5 -5 -2.5 2.5 5 7.5

0.1

0.2

0.3

0.4

0.5

0.6

-7.5 -5 -2.5 2.5 5 7.5

0.1

0.2

0.3

0.4

0.5

0.6

-7.5 -5 -2.5 2.5 5 7.5

0.1

0.2

0.3

0.4

0.5

0.6

-7.5 -5 -2.5 2.5 5 7.5

0.2

0.4

0.6

0.8

-7.5 -5 -2.5 2.5 5 7.5

0.2

0.4

0.6

0.8

-7.5 -5 -2.5 2.5 5 7.5

0.10.20.30.40.50.60.7

-7.5 -5 -2.5 2.5 5 7.5

0.10.20.30.40.50.60.7

133

5.3.3 Molecular Vibrations

The fully quantum mechanical treatment of both the electronic and nuclear dynamics of even adiatomic molecule is a complicated affair. The reason for this is that we are forced to find thestationary states for a potentially large number of particles–all of which interact, constrainedby a number of symmetry relations (such as the fact that no two electrons can be in the samestate at the same time.) In general, the exact solution of a many-body problem (such as this)is impossible. (In fact, believe it it is rigorously impossible for even three classically interactingparticles..although many have tried. ) However, the mass of the electron is on the order of 103 to104 times smaller than the mass of a typical nuclei. Thus, the typical velocities of the electronsis much larger than the typical nuclear velocities. We can then assume that the electronic cloudsurrounding the nuclei will respond instantaneously to small and slow changes to the nuclearpositions. Thus, to a very good approximation, we can separate the nuclear motion from theelectonic motion. This separation of the nuclear and electronic motion is called the Born-Oppenheimer Approximation or the Adiabatic Approximation. This approximation isone of the MOST important concepts in chemical physics and is covered in more detail in Section8.4.1.

Fundimental notion is that the nuclear motion of a molecule occurs in the average field ofthe electrons. In other words, the electronic charge distribution acts as an extremely complexmulti-dimensional potential energy surface which governs the motion and dynamics of the atomsin a molecule. Consequently, since chemistry is the science of chemical structure, changes, anddynamics, nearly all chemical reactions can be described in terms of nuclear motion on one (ormore) potential energy surface. In Fig. ?? is the London-Eyring-Polanyi-Sato (LEPS) [1]surfacefor the F + H2 → HF + H reaction using the Mukerman V set of parameters.[3] The LEPSsurface is an empirical potential energy surface based upon the London-Heitler valance bondtheory. Highly accurate potential functions are typically obtained by performing high level abinitio electronic structure calculations sampling over numerous configurations of the molecule.[2]

For diatomic molecules, the nuclear stretching potential can be approximated as a Morsepotential curve

V (r) = De(1− e−α(r−req)2 −De (5.139)

where De is the dissociation energy, α sets the range of the potential, and req is the equilibriumbond length. The Morse potential for HF is shown in Fig. 5.6 and is parameterized by De =

591.1kcal/mol, α = 2.2189A−1

, and req = 0.917A.Close to the very bottom of the potential well, where r − re is small, the potential is nearly

harmonic and we can replace the nuclear SE with the HO equation by simply writing that theangular frequancy is

ω =

√V ′′(re)

m(5.140)

So, measuring the vibrational spectrum of the well will give us the curvature of the well since(En −Em)/h is always an integer multiple of ω for harmonic systems. The red curve in Fig. 5.6is a parabolic approximation for the bottom of the well.

V (r) = De(−1 + α2(r − re)2/2 + . . .) (5.141)

134

Figure 5.5: London-Eyring-Polanyi-Sato (LEPS) empirical potential for the F +H2 → FH +Hchemical reaction

0.5 1 1.5 2 2.5 3 3.5 4

0.5

1

1.5

2

2.5

3

3.5

4

rFH

rHH

135

Figure 5.6: Morse well and harmonic approximation for HF

1 2 3 4rFH

-600

-400

-200

200

400

600

V HkCalêmolL

Clearly, Deα2 is the force constant. So the harmonic frequency for the well is ω =

√k/µ, where

µ is the reduced mass, µ = m1m2/(m1 +m2) and one would expect that the vibrational energylevels would be evenly spaced according to a harmonic progression. Deviations from this are dueto anharmonic effects introduced by the inclusion of higher order terms in the Taylor expansionof the well. As one might expect, the harmonic expansion provides a descent estimate of thepotential energy surface close to the equilibrium geometry.

5.4 Numerical Solution of the Schrodinger Equation

5.4.1 Numerov Method

Clearly, finding bound state soutions for the Schrodinger equation is an important task. Un-fortunately, we can only solve a few systems exactly. For the vast majority of system whichwe cannot handle exactly, we need to turn to approximate means to finde solutions. In laterchapters, we will examine variational methods and perturbative methods. Here, we will look ata very simple scheme based upon propagating a trial solution at a given point. This methods iscalled the ”Numerov” approach after the Russian astronomer who developed the method. It canbe used to solve any differential equation of the form:

f ′′(r) = f(r)u(r) (5.142)

where u(r) is simply a function of r and f ′′ is the second derivative of f(r) with respect to r andwhich is the solution we are looking for. For the Schrodinger equation we would write:

ψ′′ =2m

h2 (V (r)− E)ψ (5.143)

136

Figure 5.7: Model potential for proton tunneling.

-1 -0.5 0.5 1x HbohrL

-4000

-2000

2000

4000

6000V Hcm-1L

The basic proceedure is to expand the second derivative as a finite difference so that if we knowthe solution at one point, xn, we can get the solution at some small point a distance h away,xn+1 = xn + h.

f [n+ 1] = f [n] + hf ′[n] +h2

2!f ′′[n] +

h3

3!f ′′′[n] +

h4

4!f (4)[n] . . . (5.144)

f [n− 1] = f [n]− hf ′[n] +h2

2!f ′′[n]− h3

3!f ′′′[n] +

h4

4!f (4)[n] (5.145)

If we combine these two equations and solve for f [n+ 1] we get a result:

f [n+ 1] = −f [n− 1] + 2f [n] + f ′′[n]h2 +h4

12f (4)[n] +O[h6] (5.146)

Since f is a solution to the Schrodinger equation, f ′′ = Gf where

G =2m

h2 (V − E)

we can get the second derivative of f very easily. However, for the higher order terms, we haveto work a bit harder. So let’s expand f ′′

f ′′[n+ 1] = −2f ′′[n]− f ′′[n− 1] + s2f (4)[n]

and truncate at order h6. Now, solving for f (4)[n] and substituting f ′′ = Gf we get

f [n+ 1] =

(2f [n]− f [n− 1] + h2

12(G[n− 1]f [n− 1] + 10G[n]f [n])

)(1− h2

12G[n+ 1]

) (5.147)

which is the working equation.Here we take a case of proton tunneling in a double well potential. The potential in this case

is the V (x) = α(x4 − x2) function shown in Fig. 5.7. Here we have taken the parameter α = 0.1

137

-2 -1 1 2

-1

-0.5

0.5

1

-2 -1 1 2

-1

1

2

3

Figure 5.8: Double well tunneling states as determined by the Numerov approach. On the leftis the approximate lowest energy (symmetric) state with no nodes and on the rigth is the nextlowest (antisymmetric) state with a single node. The fact that the wavefunctions are heading offtowards infinity indicated the introduction of an additional node coming in from x = ∞.

and m = 1836 as the proton and use atomic units throughout. Also show in Fig. 5.7 are effectiveharmonic oscillator wells for each side of the barrier. Notice that the harmonic approximationis pretty crude since the harminic well tends to over estimate the steepness of the inner portionand under estimate the steepness of the outer portions. Nonetheless, we can use the harminicoscillator ground states in each well as starting points.

To use the Numerov method, one starts by guessing an initial energy, E, and then propagatinga trial solution to the Schrodinger equation. The Curve you obtain is in fact a solution to theequation, but it will ususally not obey the correct boundary conditions. For bound states, theboundary condition is that ψ must vanish exponentially outside the well. So, we initialize themethod by forcing ψ[1] to be exactly 0 and ψ[2] to be some small number. The exact valuesreally make no difference. If we are off by a bit, the Numerov wave will diverge towards ±∞ as xincreases. As we close in on a physically acceptible solution, the Numerov solution will begin toexhibit the correct asymptotic behavour for a while before diverging. We know we have hit uponan eigenstate when the divergence goes from +∞ to −∞ or vice versa, signeling the presenceof an additional node in the wavefunction. The proceedure then is to back up in energy a bit,change the energy step and gradually narrow in on the exact energy. In Figs. 5.8a and 5.8b arethe results of a Numerov search for the lowest two states in the double well potential. One at-3946.59cm−1 and the other at -3943.75cm−1. Notice that the lowest energy state is symmetricabout the origin and the next state is anti-symmetric about the origin. Also in both cases, theNumerov function diverges since we are not precisely at a stationary solution of the Schrodingerequation...but we are within 0.01cm−1 of the true eigenvalue.

The advantage of the Numerov method is that it is really easy to code. In fact you caneven code it in Excell. Another advantage is that for radial scattering problems, the out goingboundary conditions occur naturally, making it a method of choice for simple scattering problems.In the Mathematica notebooks, I show how one can use the Numerov method to compute thescattering phase shifts and locate resonance for atomic collisions. The disadvantage is that youhave to search by hand for the eigenvalues which can be extremely tedious.

138

5.4.2 Numerical Diagonalization

A more general approach is based upon the variational principle (which we will discuss later)and the use of matrix representations. If we express the Hamiltonian operator in matrix form insome suitable basis, then the eigenfunctions of H can also be expressed as linear combinationsof those basis functions, subject to the constraint that the eigenfunctions be orthonormal. So,what we do is write:

〈φn|H|φm〉 = Hnm

and|ψj〉 =

∑n

〈φn|ψj〉|φn〉

The 〈φn|ψj〉 coefficients are also elements of a matrix, Tnj which transforms a vector in the φbasis to the ψ basis. Conseqently, there is a one-to-one relation between the number of basisfunctions in the ψ basis and the basis functions in the φ basis.

If |ψn〉 is an eigenstate of H, then

H|ψj〉 = Ej|ψj〉.

Multiplying by 〈φm| and resolving the identity,∑n

〈φm|H|φn〉〈φn|ψj〉 = 〈φm|ψj〉Ej∑n

HmnTnj = EjTmj (5.148)

Thus, ∑mn

TmjHmnTnj = Ej (5.149)

or in more compact formT †HT = EI

where I is the identity matrix. In otherwords, the T -matrix is simply the matrix which bringsH to diagonal form.

Diagonalizing a matrix by hand is very tedeous for anything beyond a 3×3 matrix. Since thisis an extremely common numerical task, there are some very powerful numerical diagonalizationroutines available. Most of the common ones are in the Lapack package and are included as partof the Mathematica kernel. So, all we need to do is to pick a basis, cast our Hamiltonian intothat basis, truncate the basis (usually determined by some energy cut-off) and diagonalize away.Usually the diagonalization part is the most time consuming. Of course you have to be prudentin choosing your basis.

A useful set of basis functions are the trigonmetric forms of the Tchebychev polynomials.1

These are a set of orthogonal functions which obey the following recurrence relation

Tn+1(x)− 2xTn(x) + Tn−1(x) = 0 (5.150)

139

-1 -0.5 0.5 1

-1

-0.75

-0.5

-0.25

0.25

0.5

0.75

1

Figure 5.9: Tchebyshev Polynomials for n = 1− 5

Table 5.1: Tchebychev polynomials of the first type

To = 1

T1 = x

T2 = 2x2 − 1

T3 = 4x3 − 3x

T4 = 8x4 − 8x2 − 1

T5 = 16x5 − 20x3 + 5x

140

Table 5.1 lists a the first few of these polynomials as functions of x and a few of these are plottedin Fig. 5.9

It is important to realize that these functions are orthogonal on a finite range and thatintegrals over these functions must include a weighting function w(x) = 1/

√1− x2. The orthog-

onality relation for the Tn polynomials is

∫ +1

−1Tm(x)Tn(x)w(x)dx =

0 m 6= mπ2

m = n 6= 0π m = n = 0

(5.151)

Arfkin’s Mathematical Methods for Physicists has a pretty complete overview of these specialfunctions as well as many others. As usual, These are encorporated into the kernel of Mathe-matica and the Mathematica book and on-line help pages has some useful information regardingthese functions as well as a plethera of other functions.

From the recurrence relation it is easy to show that the Tn(x) polynomials satisfy the differ-ential equation:

(1− x2)T ′′n − xT ′

x + n2Tn = 0 (5.152)

If we make a change of variables from x = cos(θ) and dx = − sin θdθ, then the differentialequation reads

dTn

dθ+ n2Tn = 0 (5.153)

This is a harmonic oscillator and has solutions sinnθ and cosnθ. From the boundary conditionswe have two linearly independent solutions

Tn = cosnθ = cosn(arccosx)

andVn = sinnθ.

The normalization condition then becomes:∫ +1

−1Tm(x)Tn(x)w(x)dx =

∫ π

0cos(mθ) cos(nθ)dθ (5.154)

and ∫ +1

−1Vm(x)Vn(x)w(x)dx =

∫ π/2

−π/2sin(mθ) sin(nθ)dθ (5.155)

which is precisely the normalization integral we perform for the particle in a box state assumingthe width of the box was π. For more generic applications, we can scale θ and its range to anyrange.

1 There are at least 10 ways to spell Tchebychev’s last name Tchebychev, Tchebyshev, Chebyshev are themost common, as well as Tchebysheff, Tchebycheff, Chebysheff, Chevychef, . . .

141

The way we use this is to use the φn = N sinnx basis functions as a finite basis and truncateany expansion in this basis at some point. For example, since we are usually interested in lowlying energy states, setting an energy cut-off to basis is exactly equivalent to keeping only thelowest ncut states. The kinetic energy part of the Hamiltonian is diagonal in this basis, so we getthat part for free. However, the potential energy part is not diagonal in the φn = N sinnx basis,so we have to compute its matrix elements:

Vnm =∫φn(x)V (x)φm(x)dx (5.156)

To calculate this integral, let us first realize that [V, x] = 0, so the eigenstates of x are alsoeigenstates of the potential. Taking matrix elements in the finite basis,

xnm = N2∫φn(x)xφm(x)dx,

and diagonalizing it yields a finite set of ”position” eigenvalues, xi and a transformation forconverting between the ”position representation” and the ”basis representation”,

Tin = 〈xi|φn〉,

which is simply a matrix of the basis functions evaluated at each of the eigenvalues. The specialset of points defined by the eigenvalues of the position operator are the Gaussian quadraturepoints over some finite range.

This proceedure termed the ”discrete variable representation” was developed by Light andcoworkers in the 80’s and is a very powerful way to generate coordinate representations of Hamil-tonian matrixes. Any matrix in the basis representation (termed the FBR for finite basis represen-tation) can be transformed to the discrete variable representation (DVR) via the transformationmatrix T . Moreover, there is a 1-1 correspondency between the number of DVR points and thenumber of FBR basis functions. Here we have used only the Tchebychev functions. One cangenerate DVRs for any set of orthogonal polynomial function. The Mathematica code below gen-erates the required transformations, the points, the eigenvalues of the second-derivative operator,and a set of quadrature weights for the Tchebychev sine functions over a specified range:

dv2fb[DVR_, T_] := T.DVR.Transpose[T];

fb2dv[FBR_, T_] := Transpose[T].FBR.T;

tcheby[npts_, xmin_, xmax_] := Module[pts, fb, del,

del = xmax - xmin;

pts = Table[i*del*(1/(npts + 1)) + xmin, i, npts] // N;

fbrke = Table[(i*(Pi/del))^2, i, npts] // N;

w = Table[del/(npts + 1), i, npts] // N;

T = Table[

Sqrt[2.0/(npts + 1)]*Sin[(i*j)*Pi/(npts + 1)],

i, npts, j, npts] // N;

Return[pts, T, fbrke, w]

]

To use this, we first define a potential surface, set up the Hamiltonian matrix, and simplydiagonalize. For this example, we will take the same double well system described above andcompare results and timings.

142

V[x_] := a*(x^4 - x^2);

cmm = 8064*27.3;

params = a -> 0.1, m -> 1836;

x, T, K, w = tcheby[100, -1.3, 1.3];

Kdvr = (fb2dv[DiagonalMatrix[K], T]*m)/2 /. params;

Vdvr = DiagonalMatrix[V[x]] /. params;

Hdvr = Kdvr + Vdvr;

tt = Timing[w, psi = Transpose[

Sort[Transpose[Eigensystem[Hdvr]]]]];

Print[tt]

(Select[w*cmm , (# < 3000) &]) // TableForm

This code sets up the DVR points x, the transformation T and the FBR eigenvalues K usingthe tcheby[n,xmin,xmax] Mathematica module defined above. We then generate the kineticenergy matrix in the DVR using the transformation

KDV R = T †KFBRT

and form the DVR Hamiltonian

HDV R = KDV R + VDV R.

The eigenvalues and eigenvectors are computed via the Eigensystem[] routine. These are thensorted according to their energy. Finally we print out only those states with energy less than3000 cm−1 and check how long it took. On my 300 MHz G3 laptop, this took 0.3333 seconds tocomplete. The first few of these are shown in Table 5.2 below. For comparison, each Numeroviteration took roughly 1 second for each trial function. Even then, the eigenvalues we found areprobabily not as accurate as those computed here.

Table 5.2: Eigenvalues for double well potential computed via DVR and Numerov approaches

i ωi (cm−1) Numerov1 -3946.574 -3946.592 -3943.7354 -3943.753 -1247.09744 -1093.52045 591.3666 1617.424

143

5.5 Problems and Exercises

Exercise 5.2 Consider a harmonic oscillator of mass m and angular frequency ω. At timet = 0, the state of this system is given by

|ψ(0)〉 =∑n

cn|φn〉 (5.157)

where the states |φn〉 are stationary states with energy En = (n+ 1/2)hω.

1. What is the probability, P , that at a measurement of the energy of the oscillator at somelater time will yield a result greater than 2hω. When P = 0, what are the non-zero coeffi-cients, cn?

2. From now on, let only co and c1 be non zero. Write the normalization condition for |ψ(0)〉and the mean value 〈H〉 of the energy in terms of co and c1. With the additional requirementthat 〈H〉 = hω, calculate |co|2 and |c1|2.

3. As the normalized state vector |ψ〉 is defined only to within an arbitrary global phase factor,as can fix this factor by setting co to be real and positive. We set c1 = |c1|eiφ. We assumealso that 〈H〉 = hω and show that

〈x〉 =1

2

√h

mω. (5.158)

Calculate φ.

4. With |ψ〉 so determined, write |ψ(t)〉 for t > 0 and calculate the value of φ at time t.Deduce the mean of 〈x〉(t) of the position at time t.

Exercise 5.3 Find 〈x〉, 〈p〉, 〈x2〉 and 〈p2〉 for the ground state of a simple harmonic oscillator.What is the uncertainty relation for the ground state.

Exercise 5.4 In this problem we consider the the interaction between molecule adsorbed on asurface and the surface phonons. Represent the vibrational motion of the molecule (with reducedmass µ) as harmonic with force constant K

Ho =−h2

2µ

∂2

∂x2+K

2x2 (5.159)

and the coupling to the phonons as

H ′ = −x∑k

Vk cos(Ωkt) (5.160)

where Vk is the coupling between the molecule and phonon of wavevector k and frequency Ωk.

144

1. Express the total Hamiltonian as a displaced harmonic well. What happens to the well asa function of time?

2. What is the Golden-Rule transition rate between the ground state and the nth excited stateof the system due to phonon interactions? Are there any restrictions as to which final statecan be reached? Which phonons are responsible for this process?

3. From now on, let the perturbing force be constant in time

H ′ = x∑k

Vk (5.161)

where Vk is the interaction with a phonon with wavevector k. Use the lowest order levelof perturbation theory necessary to construct the transition probability between the groundstate and the second-excited state.

Exercise 5.5 Let

X = (mω

2h)1/2x (5.162)

P = (1

2hmω)1/2p (5.163)

Show that the Harmonic Oscillator Hamiltonian is

H = hω(P 2 +X2) (5.164)

Now, define the operator: a† = X − iP . Show that a† acting on the harmonic oscillator groundstate is also an eigenstate of H. What is the energy of this state? Use a† to define a generatingrelationship for all the eigenstates of H.

Exercise 5.6 Show that if one expands an arbitrary potential, V (x) about its minimum at xmin,and neglects terms of order x3 and above, one always obtains a harmonic well. Show that aharmonic oscillator subject to a linear perturbation can be expressed as an unperturbed harmonicoscillator shifted from the origin.

Exercise 5.7 Consider the one-dimensional Schrodinger equation with potential

V (x) =

m2ω2x2 x > 0

+∞ x ≤ 0(5.165)

Find the energy eigenvalues and wavefunctions.

Exercise 5.8 An electron is contained inside a hard sphere of radius R. The radial componentsof the lowest S and P state wavefunctions are approximately

ψS(x) ≈ sin(kr)

kr(5.166)

ψP (x) ≈ cos(kr)

kr− sin(kr)

(kr)2=∂ψS(kr)

∂(kr). (5.167)

145

1. What boundary conditions must each state obey?

2. Using E = k2h2/(2m) and the above boundary conditions, what are the energies of eachstate?

3. What is the pressure exerted on the surface of the sphere if the electron is in the a.) S state,b.) the P state. (Hint, recall from thermodynamics: dW = PdV = −(dE(R)/dR)dR.)

4. For a solvated e− in water, the S to P energy gap is about 1.7 eV. Estimate the size ofthe the hard-sphere radius for the aqueous electron. If the ground state is fully solvated,the pressure of the solvent on the electron must equal the pressure of the electron on thesolvent. What happens to the system when the electron is excited to the P -state from theequilibrated S state? What happens to the energy gap between the S and P as a result ofthis?

Exercise 5.9 A particle moves in a three dimensional potential well of the form:

V (x) =

∞ z2 > a2

mω2

2(x2 + y2), otherwise

(5.168)

Obtain an equation for the eigenvalues and the associated eigenfunctions.

Exercise 5.10 A particle moving in one-dimension has a ground state wavefunction (not-normalized)of the form:

ψo(x) = e−α4x4/4 (5.169)

where α is a real constant with eigenvalue Eo = h2α2/m. Determine the potential in which theparticle moves. (You do not have to determine the normalization.)

Exercise 5.11 A two dimensional oscillator has the Hamiltonian

H =1

2(p2

x + p2y) +

1

2(1 + δxy)(x2 + y2) (5.170)

where h = 1 and δ << 1. Give the wavefunctions for the three lowest energy levels when δ = 0.Evaluate using first order perturbation theory these energy levels for δ 6= 0.

Exercise 5.12 For the ground state of a harmonic oscillator, find the average potential andkinetic energy. Verify the virial theorem, 〈T 〉 = 〈V 〉, holds in this case.

Exercise 5.13 In this problem we will explore the use of two computational tools to calculate thevibrational eigenstates of molecules. The the Spartan structure package is available on one of thePCs in the computer lab in the basement of Fleming. First, we will generate a potential energysurface using ab initio methods. We will then fit that surface to a functional form and try tocalculate the vibrational states. The system we will consider is the ammonia molecule undergoinginversion from its degenerate C3v configuration through a D3h transition state.

146

Figure 5.10: Ammonia Inversion and Tunneling

1. Using the Spartan electronic structure package (or any other one you have access to), builda model of NH3, and determine its ground state geometry using various levels of ab initiotheory. Make a table of N − H bond lengths and θ = 6 H − N − H bond angles for theequilibrium geometries as a function of at least 2 or 3 different basis sets. Looking in theliterature, find experimental values for the equilibrium configuration. Which method comesclosest to the experimental values? Which method has the lowest energy for its equilibriumconfiguration.

2. Using the method which you deamed best in part 1, repeat the calculations you performedabove by systematically constraining the H − N − H bond angle to sample configurationsaround the equilibrium configuration and up to the planar D3h configuration. Note, it maybe best to constrain two H − N − H angles and then optimize the bond lengths. Sampleenough points on either side of the minimum to get a descent potential curve. This is yourBorn-Oppenheimer potential as a function of θ.

3. Defining the orgin of a coordinate system to be the θ = 120o D3h point on the surface, fityour ab initio data to the ”W”-potential

V (x) = αx2 + βx4 (5.171)

What are the theoretical values of α and β?

4. We will now use perturbation theory to compute the tunneling dynamics.

(a) Show that the points of minimum potential energy are at

xmin = ±(α

2β

)1/2

and that the energy difference between the top of the barrier and the minimum energyis given by

V = V (0)− V (xmin) (5.172)

=α2

4β(5.173)

147

(b) We first will consider the barrier to be infinitely high so that we can expand the poten-tial function around each xmin. Show that by truncating the Taylor series expansionabove the (x − xmin)2 terms that the potential for the left and right hand sides aregiven by

VL = 2α (x+ xmin)2 − V

andVR = 2α (x− xmin)2 − V.

What are the vibrational energy levels for each well?

(c) The wavefunctions for the lowest energy states in each well are given by

ψ(x) =γ1/2

π1/4exp[−γ

2

2(x± xmin)2]

with

γ =

((4µα)1/2

h

)1/2

.

The energy levels for both sides are degenerate in the limit that the barrier height isinfinite. The total ground state wavefunction for this case is

Ψ(x) =

(ψL(x)ψR(x)

).

However, as the barrier height decreases, the degenerate states begin to mix causingthe energy levels to split. Define the ”high barrier” hamiltonian as

H = − h2

2µ

∂2

∂x+ VL(x)

for x < 0 and

H = − h2

2µ

∂2

∂x+ VR(x)

for x > 0. Calculate the matrix elements of H which mix the two degenerate left andright hand ground state wavefunctions: i.e.

〈Ψ|H|Ψ〉 =

(HRR HLR

HRL HLL

)

whereHRR = 〈ψR|H|ψR〉

, with similar definitions for HRL, HLL and HLR. Obtain numerical values of eachmatrix element using the values of α and β you determined above (in cm−1). Use themass of a H atom for the reduced mass µ.

148

(d) Since the ψL and ψR basis functions are non-orthogonal, you will need to consider theoverlap matrix, S, when computing the eigenvalues of H. The eigenvalues for thissystem can be determined by solving the secular equation∣∣∣∣∣ α− λ β − λS

β − λS α− λ

∣∣∣∣∣ = 0 (5.174)

where α = HRR = HLL and β = HLR = HRL (not to be confused with the potentialparameters above). Using Eq. 5.174, solve for λ and determine the energy splittingin the ground state as a function the unperturbed harmonic frequency and the barrierheight, V . Calculate this splitting using the parameters you computed above. What isthe tunneling frequency? The experimental results is ∆E = 0.794cm−12.

Exercise 5.14 Consider a system in which the Lagrangian is given by

L(qi, qi) = T (qi, qi)− V (qi) (5.175)

where we assume T is quadratic in the velocities. The potential is independent of the velocityand neither T nor V carry any explicit time dependency. Show that

d

dt

∑j

qj∂L

∂qj− L

= 0

The constant quantity in the (. . .) defines a Hamiltonian, H. Show that under the assumedconditions, H = T + V

Exercise 5.15 The Fermat principle in optics states that a light ray will follow the path, y(x)which minimizes its optical length, S, through a media

S =∫ x2,y2

x1,y1

n(y, x)ds

where n is the index of refraction. For y2 = y1 = 1 and −x1 = x2 = 1 find the ray-path for

1. n = exp(y)

2. n = a(y − yo) for y > yo

Make plots of each of these trajectories.

Exercise 5.16 In a quantum mechanical system there are gi distinct quantum states betweenenergy Ei and Ei + dEi. In this problem we will use the variational principle and Lagrangemultipliers to determine how ni particles are distributed amongst these states subject to the con-straints

1. The number of particles is fixed:n =

∑i

ni

2From Molecular Structure and Dynamics, by W. Flygare, (Prentice Hall, 1978)

149

2. the total energy is fixed ∑i

niEi = E

We consider two cases:

1. For identical particles obeying the Pauli exclusion principle, the probability of a given con-figuration is

WFD =∏i

gi

ni!(gi − ni)!(5.176)

Show that maximizing WFD subject to the constraints above leads to

ni =gi

eλ1+λ2Ei + 1

with the Lagrange multipliers λ1 = −Eo/kT and λ2 = 1/kT . Hint: try working with thelogW and use Sterling’s approximation in the limit of a large number of particles .

2. In this case we still consider identical particles, but relax the restriction on the fixed numberof particles in a given state. The probability for a given distribution is then

WBE =∏i

(ni + gi − 1)!

ni!(gi − 1)!.

Show that by minimizing WBE subject to the constraints above leads to the occupationnumbers:

ni =gi

eλ1+λ2Ei − 1

where again, the Lagrange multipliers are λ1 = −Eo/kT and λ2 = 1/kT . This yields theBose-Einstein statistics. Note: assume that gi 1

3. Photons satisfy the Bose-Einstein distribution and the constraint that the total energy isconstant. However, there is no constrain regarding the total number of photons. Show thatby eliminating the fixed number constraint leads to the foregoing result with λ1 = 0.

150

Bibliography

[1] C. A. Parr and D. G. Trular, J. Phys. Chem. 75, 1844 (1971).

[2] H. F. Schaefer III, J. Phys. Chem. 89, 5336 (1985).

[3] P. A. Whitlock and J. T. Muckermann, J. Chem. Phys. 61, 4624 (1974).

151

Chapter 6

Quantum Mechanics in 3D

In the next few lectures, we will focus upon one particular symmetry, the isotropy of free space.As a collection of particles rotates about an arbitrary axis, the Hamiltonian does not change. Ifthe Hamoltonian does in fact depend explicitly upon the choice of axis, the system is “gauged”,meaning all measurements will depend upon how we set up the coordinate frame. A Hamiltonianwith a potential function which depends only upon the coordinates, e.g. V = f(x, y, z), is gaugeinvarient, meaning any measurement that I make will not depend upon my choice of referenceframe. On the other hand, if our Hamiltonian contains terms which couple one reference frameto another (as in the case of non-rigid body rotations), we have to be careful in how we selectthe “gauge”. While this sounds like a fairly specialized case, it turns out that many ordinaryphenimina depend upon this, eg. figure skaters, falling cats, floppy molecules. We focus uponrigid body rotations first.

For further insight and information into the quantum mechanics of angular momentum, Irecommend the following texts and references:

1. Theory of Atomic Structure, E. Condon and G. Shortley. This is the classical book onatomic physics and theory of atomic spectroscopy and has inspired generations since itcame out in 1935.

2. Angular Momentum–understanding spatial aspects in chemistry and physics, R. N. Zare.This book is the text for the second-semester quantum mechanics at Stanford taught byZare (when he’s not out looking for Martians). It’s a great book with loads of examples inspectroscopy.

3. Quantum Theory of Angular Momentum, D. A. Varshalovich, A. Moskalev, and V. Kher-sonskii. Not to much physics in this book, but if you need to know some relation betweenWigner-D functions and Racah coefficients, or how to derive 12j symbols, this book is foryou.

First, we need to look at what happens to a Hamiltonian under rotation. In order to show thatH is invariant to any rotations, we need only to show that it is invarient under an infinitesimalrotation.

152

6.1 Quantum Theory of Rotations

Let δ~φ by a vector of a small rotation equal in magnitude to the angle δφ directed long anarbitrary axis. Rotating the system by δ~φ changes the direction vectors ~rα by δ~rα.

δ~rα = δ~φ× ~rα (6.1)

Note that the × denotes the vector “cross” product. Since we will be using cross-productsthrough out these lectures, we pause to review the operation.

A cross product between two vectors is computed as

~c = ~a×~b

=

∣∣∣∣∣∣∣i j kai aj ak

bi bj bk

∣∣∣∣∣∣∣= i(ajbk − bjak)− j(aibk − biak) + k(aibj − biaj)

= εijkajbk (6.2)

Where εijk is the Levi-Cevita symbol or the “anti-symmetric unit tensor” defined as

εijk =

0 if any of the indices are the same1 for even permuations of the indices−1 for odd permutations of the indices

(6.3)

(Note that we also have assumed a “summation convention” where by we sum over all repeatedindices. Some elementary properties are εiklεikm = δlm and εiklεikl = 6.)

So, an arbitrary function ψ(r1, r2, · · ·) is transformed by the rotation into:

ψ1(r1 + δr1, r2 + δr2, · · ·) = ψ(r1, r2, · · ·) +∑a

δra · ~∇ψa

= ψ(r1, r2, · · ·) +∑a

δ~φ× ra · ~∇aψa

=

(1 + δ~φ ·

∑a

~ra × ~∇a

)ψa (6.4)

Thus, we conclude, that the operator

1 + δ~φ ·∑a

~ra × ~∇a (6.5)

is the operator for an infintesimal rotation of a system of particles. Since δφ is a constant, wecan show that this operator commutes with the Hamiltonian

[

(∑a

~ra × ~∇a

), H] = 0 (6.6)

This implies then a particular conservation law related to the isotropy of space. This is of courseangular momentum so that (∑

a

~ra × ~∇a

)(6.7)

153

must be at least proportional to the angular momentum operator, L. The exact relation is

hL = ~r × ~p = −ih~r × ~∇ (6.8)

which is much like its classical counterpart

L =1

m~r × ~v. (6.9)

The operator is of course a vector quantity, meaning that is has direction. The components ofthe angular momentum vector are:

hLx = ypz − zpy (6.10)

hLy = zpx − zpz (6.11)

hLz = xpy − ypx (6.12)

hLi = εijkxjpk (6.13)

For a system in a external field, antgular momentum is in general not conserved. However, ifthe field posesses spherical symmetry about a central point, all directions in space are equivalentand angular momentum about this point is conserved. Likewise, in an axially symmetric field,motion about the axis is conserved. In fact all the conservation laws which apply in classicalmechanics have quantum mechanical analogues.

We now move on to compute the commutation rules between the Li operators and the x andp operators First we note:

[Lx, x] = [Ly, y] = [Lz, z] = 0 (6.14)

[Lx, y] =1

h((ypz − zpy)y − y(ypz − zpy)) = −z

h[py, y] = iz (6.15)

In short hand:

[Li, xk] = iεiklxl (6.16)

We need also to know how the various components commute with one another:

h[Lx, Ly] = Lx(zpx− xpz)− (zpx − xpz)Lx (6.17)

= (Lxz − zLx)px − x(Lxpz − pzLx) (6.18)

= −iypx + ixpy (6.19)

154

= ihLz (6.20)

Which we can summarize as

[Ly, Lz] = iLx (6.21)

[Lz, Lx] = iLy (6.22)

[Lx, Ly] = iLz (6.23)

[Li, Lj] = iεijkLk (6.24)

Now, denote the square of the modulus of the total angular momentum by L2, where

L2 = L2x + L2

y + L2z (6.25)

Notice that this operator commutes with all the other Lj operators,

[L2, Lx] = [L2, Ly] = [L2, Lz] = 0 (6.26)

For example:

[L2x, Lz] = Lx[Lx, Lz] + [Lx, Lz]Lx = −i(LxLy + LyLx) (6.27)

also,

[L2y, Lz] = i(LxLy + LyLx) (6.28)

Thus,

[L2, Lz] = 0 (6.29)

Thus, I can measure L2 and Lz simultaneously. (Actually I can measure L2 and any one com-ponent Lk simultaneously. However, we ueually pick this one as the z axis to make the matheasier, as we shall soon see.)

A consequence of the fact that Lx, Ly, and Lz do not commute is that the angular momentum

vector ~L can never lie exactly along the z axis (or exactly along any other axis for that matter).

We can interpret this in a classical context as a vector of length |L| = h√L(L+ 1) with the Lz

component being hm. The vector is then constrained to lie in a cone as shown in Fig. ??. Wewill take up this model at the end of this chapter in the semi-classical context.

It is also convienent to write Lx and Ly as a linear combination

L+ = Lx + iLyL− = Lx − iLy (6.30)

(Recall what we did for Harmonic oscillators?) It’s easy to see that

[L+, L−] = 2Lz (6.31)

155

Figure 6.1: Vector model for the quantum angular momentum state |jm〉, which is representedhere by the vector j which precesses about the z axis (axis of quantzation) with projection m.

|j|=(j (j + 1))1/2

Z

X

Y

m θ

[Lz, L+] = L+ (6.32)

[Lz, L−] = −L− (6.33)

Likewise:

L2 = L+L− + L2z − Lz = L−L+ + L2

z + Lz (6.34)

We now give some frequently used expressions for the angular momentum operator for asingle particle in spherical polar coordinates (SP). In SP coordinates,

x = r sin θ cosφ (6.35)

y = r sin θ sinφ (6.36)

z = r cos θ (6.37)

It’s easy and straightforward to demonstrate that

Lz = −i ∂∂φ

(6.38)

and

L± = e±φ

(± ∂

∂θ+ i cot θ

∂

∂φ

)(6.39)

156

Thus,

L2 = − 1

sin θ

[1

sin θ

∂2

∂φ2+

∂

∂θsin θ

∂

∂θ

](6.40)

which is the angular part of the Laplacian in SP coordinates.

∇2 =1

r2 sin θ

[sin θ

∂

∂rr2 ∂

∂r+

∂

∂θsin θ

∂

∂θ+

1

sin θ

∂2

∂φ2

](6.41)

=1

r2

∂

∂rr2 ∂

∂r− 1

r2L2 (6.42)

In other words, the kinetic energy operator in SP coordinates is

− h2

2m∇2 = − h2

2m

(1

r2

∂

∂rr2 ∂

∂r− 1

r2L2

)(6.43)

6.2 Eigenvalues of the Angular Momentum Operator

Using the SP form

Lzψ = i∂ψ

∂φ= lzψ (6.44)

Thus, we conclude that ψ = f(r, θ)eilzφ. This must be single valued and thus periodic in φ withperiod 2π. Thus,

lz = m = 0,±1,±2, · · · (6.45)

Thus, we write the azimuthal solutions as

Φm(φ) =1√2πeimφ (6.46)

which are orthonormal functions:∫ 2π

0Φ∗

m(φ)Φm′(φ)dφ = δmm′ (6.47)

In a centrally symmetric case, stationary states which differ only in their m quantum numbermust have the same energy.

We now look for the eigenvalues and eigenfunctions of the L2 operator belonging to a set ofdegenerate energy levels distinguished only by m. Since the +z−axis is physically equivalent tothe −z−axis, for every +m there must be a −m. Let L denote the greatest possible m for agiven L2 eigenstate. This upper limit must exist because of the fact that L2−L2

z = L2x +L2

y is aoperator for an essentially positive quantity. Thus, its eigenvalues cannot be negative. We nowapply LzL± to ψm.

Lz(L±ψm) = (Lz ± 1)(L±ψm) = (m± 1)(L±ψm) (6.48)

157

(note: we used [Lz, L±] = ±L± ) Thus, L±ψm is an engenfunction of Lz with eigenvalue m± 1.

ψm+1 ∝ L+ψm (6.49)

ψm−1 ∝ L−ψm (6.50)

If m = l then, we must have L+ψl = 0. Thus,

L−L+ψl = (L2 − L2z − Lz)ψl = 0 (6.51)

i.e.

L2ψl = (L2z + Lz)ψl = l(l + 1)ψl (6.52)

Thus, the eigenvalues of L2 operator are l(l + 1) for l any positive integer (including 0). For agiven value of l, the component Lz can take values

l, l − 1, · · · , 0,−l (6.53)

or 2l + 1 different values. Thus an energy level with angular momentum l has 2l + 1 degeneratestates.

6.3 Eigenstates of L2

Since l ansd m are the good quantum numbers, we’ll denote the eigenstates of L2 as

L2|lm〉 = l(l + 1)|lm〉. (6.54)

This we will often write in short hand after specifying l as

L2|m〉 = l(l + 1)|m〉. (6.55)

Since L2 = L+L− + L2z − Lz, we have

〈m|L2|m〉 = m2 −m−∑m′〈m|L+|m′〉〈m′|L−|m〉 = l(l + 1) (6.56)

Also, note that

〈m− 1|L−|m〉 = 〈m|L+|m− 1〉∗, (6.57)

thus we have

|〈m|L+|m− 1〉|2 = l(l + 1)−m(m− 1) (6.58)

Choosing the phase (Condon and Shortly phase convention) so that

〈m− 1|L−|m〉 = 〈m|L+|m− 1〉 (6.59)

〈m|L+|m− 1〉 =√l(l + 1)−m(m− 1) =

√(l +m)(l −m+ 1) (6.60)

Using this relation, we note that

〈m|Lx|m− 1〉 = 〈m− 1|Lx|m〉 =1

2

√(l +m)(l −m+ 1) (6.61)

〈m|Ly|m− 1〉 = 〈m− 1|Lx|m〉 =−i2

√(l +m)(l −m+ 1) (6.62)

Thus, the diagonal elements of Lx and Ly are zero in states with definite values of 〈Lz〉 = m.

158

6.4 Eigenfunctions of L2

The wavefunction of a particle is not entirely determined when l and m are presribed. We stillneed to specify the radial component. Thus, all the angular momentum operators (in SP coords)contain and explicit r dependency. For the time, we’ll take r to be fixed and denote the angularmomentum eigenfunctions in SP coordinates as Ylm(θ, φ) with normalization∫

|Ylm(θ, φ)|2dΩ (6.63)

where dΩ = sin θdθdφ = d(cos θ)dφ and the integral is over all solid angles. Since we candetermine common eigenfunctions for L2 and Lz, there must be a separation of variables, θ andφ, so we seek solutions of the form:

Ylm(θ, φ) = Φm(φ)Θlm(θ) (6.64)

The normalization requirement is that∫ π

0|Θlm(θ)|2 sin θdθ = 1 (6.65)

and I require ∫ 2π

0

∫ π

0Y ∗

l′m′YlmdΩ = δll′δmm′ . (6.66)

I thus seek solution of(1

sin θ

∂

∂θsin θ

∂

∂θ+

1

sin2 θ

∂2

∂φ2

)ψ + l(l + 1)ψ = 0 (6.67)

i.e. (1

sin θ

∂

∂θsin θ

∂

∂θ− m2

sin2 θ+ l(l + 1)

)Θlm(θ) = 0 (6.68)

which is well known from the theory of spherical harmonics.

Θlm(θ) = (−1)mil

√√√√(2l + 1)(l −m)!

2(l −m)!Pm

l (cos θ) (6.69)

for m > 0. Where Pml are associated Legendre Polynomials. For m < 0 we get

Θl,−|m| = (−1)mΘl,|m| (6.70)

Thus, the angular momentum eigenfunctions are the spherical harmonics, normalized so thatthe matrix relations defined above hold true. The complete expression is

Ylm = (−1)(m+|m|)/2il[2l + 1

4π

(l − |m|)!(l + |m|)!

]1/2

P|m|l (cos θ)eimφ (6.71)

159

Table 6.1: Spherical Harmonics (Condon-Shortley Phase convention.

Y00 =1√4π

Y1,0 =(

3

4π

)1/2

cos(θ)

Y1,±1 = ∓(

3

8π

)1/2

sin(θ)e±iφ

Y2,±2 = 3

√5

96πsin2 θe∓iφ

Y2,±1 = ∓3

√5

24πsin θ cos θeiφ

Y2,0 =

√5

4π

(3

2cos2 θ − 1

2

)These can also be generated by the SphericalHarmonicY[l,m,θ,φ] function in Mathematica.

Figure 6.2: Spherical Harmonic Functions for up to l = 2. The color indicates the phase of thefunction.

160

For the case of m = 0,

Yl0 = il(

2l + 1

4π

)1/2

Pl(cos θ) (6.72)

Other useful relations are in cartesian form, obtained by using the relations

cos θ =z

r, (6.73)

sin θ cosφ =x

r, (6.74)

and

sin θ sinφ =y

r. (6.75)

Y1,0 =(

3

4π

)1/2 z

r(6.76)

Y1,1 =(

3

8π

)1/2 x+ iy

r(6.77)

Y1,−1 =(

3

8π

)1/2 x− iy

r(6.78)

The orthogonality integral of the Ylm functions is given by∫ 2π

0

∫ π

0Y ∗

lm(θ, φ)Yl′m′(θ, φ) sin θdθdφ = δll′δmm′ . (6.79)

Another useful relation is that

Yl,−m = (−1)mY ∗lm. (6.80)

This relation is useful in deriving real-valued combinations of the spherical harmonic functions.

Exercise 6.1 Demonstrate the following:

1. [L+, L2] = 0

2. [L−, L2] = 0

Exercise 6.2 Derive the following relations

ψl,m(θ, φ) =

√√√√ (l +m)!

(2l!)(l −m)!(L−)l−mψl,l(θ, φ)

and

ψl,m(θ, φ) =

√√√√ (l −m)!

(2l!)(l +m)!(L+)l+mψl,−l(θ, φ)

where ψl,m = Yl,m are eigenstates of the L2 operator.

161

6.5 Addition theorem and matrix elements

In the quantum mechanics of rotations, we will come across integrals of the general form∫Y ∗

l1m1Yl2m2Yl3m3dΩ

or ∫Y ∗

l1m1Pl2Yl3m3dΩ

in computing matrix elements between angular momentum states. For example, we may beasked to compute the matrix elements for dipole induced transitions between rotational states ofa spherical molecule or between different orbital angular momentum states of an atom. In eithercase, we need to evaluate an integral/matrix element of the form

〈l1m1|z|l2m2〉 =∫Y ∗

l1m1zYl2m2dΩ (6.81)

Realizing that z = r cos θ = r√

4π/3Y10(θ, φ), Eq. 6.87 becomes

〈l1m1|z|l2m2〉 =

√4π

3r∫Y ∗

l1m1Y10Yl2m2dΩ (6.82)

Integrals of this form can be evaluated by group theoretical analysis and involves the introduc-tion of Clebsch-Gordan coefficients, CLM

l1m1l2m2

1 which are tabulated in various places or can becomputed using Mathematica. In short, some basic rules will always apply.

1. The integral will vanish unless the vector sum of the angular momenta sums to zero.i.e.|l1 − l3| ≤ l2 ≤ (l1 + l3). This is the “triangle” rule and basically means you have to beable make a triangle with length of each side being l1, l2, and l3.

2. The integral will vanish unless m2 + m3 = m1. This reflects the conservation of the zcomponent of the angular momentum.

3. The integral vanishes unless l1 + l2 + l3 is an even integer. This is a parity conservationlaw.

So the general proceedure for performing any calculation involving spherical harmonics is to firstcheck if the matrix element violates any of the three symmetry rules, if so, then the answer is 0and you’re done. 2

To actually perform the integration, we first write the product of two of the Ylm’s as aClebsch-Gordan expansion:

Yl1m1Yl2m2 =∑LM

√√√√(2l1 + 1)(2l2 + 1)

4π(2L+ 1)CL0

l10l20CLMl1m1l2m2

YLM . (6.83)

1Our notation is based upon Varshalovich’s book. There at least 13 different notations that I know of forexpressing these coefficients which I list in a table at the end of this chapter.

2In Mathematica, the Clebsch-Gordan coefficients are computed using the functionClebschGordan[j1,m1,j2,m2,j,m] for the decomposition of |jm〉 in to |j1,m1〉 and |j2,m2〉.

162

We can use this to write

∫Y ∗

lmYl1m2Yl2m2dΩ =∑LM

√√√√(2l1 + 1)(2l2 + 1)

4π(2L+ 1)CL0

l10l20CLMl1m1l2m2

∫Y ∗

lmYLMdΩ

=∑LM

√√√√(2l1 + 1)(2l2 + 1)

4π(2L+ 1)CL0

l10l20CLMl1m1l2m2

δlLδmM

=

√√√√(2l1 + 1)(2l2 + 1)

4π(2l + 1)C l0

l10l20Clml1m1l2m2

(6.84)

In fact, the expansion we have done above for the product of two spherical harmonics can beinverted to yield the decomposition of one angular momentum state into a pair of coupled angularmomentum states, such as would be the case for combining the orbital angular momentum of aparticle with, say, its spin angular momentum. In Dirac notation, this becomes pretty apparent

|LM〉 =∑

m1m2

〈l1m1l2m2|LM〉|l1m1l2m2〉 (6.85)

where the state |l1m1l2m2〉 is the product of two angular momentum states |l1m1〉 and |l2m2〉.The expansion coefficients are the Clebsch-Gordan coefficients

CLMl1m1l2m2

= 〈l1m1l2m2|LM〉 (6.86)

Now, let’s go back the problem of computing the dipole transition matrix element betweentwo angular momentum states in Eq. 6.87. The integral we wish to evaluate is

〈l1m1|z|l2m2〉 =∫Y ∗

l1m1zYl2m2dΩ (6.87)

and we noted that z was related to the Y10 spherical harmonic. So the integral over the angularcoordinates involves: ∫

Y ∗l1m1

Y10Yl2m2dΩ. (6.88)

First, we evaluate which matrix elements are going to be permitted by symmetry.

1. Clearly, by the triangle inequality, |l1 − l2| = 1. In other words, we change the angularmomentum quantum number by only ±1.

2. Also, by the second criteria, m1 = m2

3. Finally, by the third criteria: l1 + l2 + 1 must be even, which again implies that l1 and l2differ by 1.

Thus the integral becomes

∫Y ∗

l+1,mY10Ylm =

√√√√(2l + 1)(2 + 1)

4π(2l + 3)C l+1,0

l010 C1ml+1,ml0 (6.89)

163

From tables,

C l+1,0l010 = −

( √2 (1 + l)√

2 + 2 l√

3 + 2 l

)

C1ml+1,ml0 = −

(√2√

1 + l −m√

1 + l +m√2 + 2 l

√3 + 2 l

)So

C l+1,0l010 C1m

l+1,ml0 =2 (1 + l)

√1 + l −m

√1 + l +m

(2 + 2 l) (3 + 2 l)

Thus,

∫Y ∗

l+1,mY10YlmdΩ =

√3

4π

√√√√(l +m+ 1)(l −m+ 1)

(2l + 1)(2l + 3)(6.90)

Finally, we can construct the matrix element for dipole-transitions as

〈l1m1|z|l2m2〉 = r

√√√√(l +m+ 1)(l −m+ 1)

(2l + 1)(2l + 3)δl1±1,l2δm1,m2. (6.91)

Physically, this make sense because a photon carries a single quanta of angular momentum. Soin order for molecule or atom to emit or absorb a photon, its angular momentum can only changeby ±1.

Exercise 6.3 Verify the following relations

∫Y ∗

l+1,m+1Y11Ylmdω =

√3

8π

√√√√(l +m+ 1)(l +m+ 2)

2l + 1)(2l + 3)(6.92)

∫Y ∗

l−1,m−1Y11Ylmdω = −√

3

8π

√√√√(l −m)(l −m− 1)

2l − 1)(2l + 1)(6.93)

∫Y ∗

lmY00YlmdΩ =1√4π

(6.94)

6.6 Legendre Polynomials and Associated Legendre Poly-

nomials

Ordinary Legendre polynomials are generated by

Pl(cos θ) =1

2ll!

dl

(d cos θ)l(cos2 θ − 1)l (6.95)

164

i.e. (x = cos θ)

Pl(x) =1

2ll!

∂l

∂xl(x2 − 1)l (6.96)

and satisfy (1

sin θ

∂

∂θsin θ

∂

∂θ+ l(l + 1)

)Pl = 0 (6.97)

The Associated Legendre Polynomials are derived from the Legendre Polynomials via

Pml (cos θ) = sinm θ

∂m

(∂ cos θ)mPl(cos θ) (6.98)

6.7 Quantum rotations in a semi-classical context

Earlier we established the fact that the angular momentum vector can never exactly lie on asingle spatial axis. By convention we take the quantization axis to be the z axis, but this isarbitrary and we can pick any axis as the quantization axis, it is just that picking the z axismake the mathematics much simpler. Furthermore, we established that the maximum lengththe angular momentum vector can have along the z axis is the eigenvalue of Lz when m = l,

so 〈lz〉 = l which is less than√l(l + 1). Note, however, we can write the eigenvalue of L2 as

l2(1+1/l2) . As l becomes very large the eigenvalue of Lz and the eigenvalue of L2 become nearlyidentical. The 1/l term is in a sense a quantum mechanical effect resulting from the uncertainty

in determining the precise direction of ~L.We can develop a more quantative model for this by examining both the uncertainty product

and the semi-classical limit of the angular momentum distribution function. First, recall, that ifwe have an observable, A then the spread in the measurements of A is given by the varience.

∆A2 = 〈(A− 〈A〉)2〉 = 〈A2〉 − 〈A〉2. (6.99)

In any representation in which A is diagonal, ∆A2 = 0 and we can determine A to any level ofprecision. But if we look at the sum of the variances of lx and ly we see

∆L2x + ∆L2

y = l(l + 1)−m2. (6.100)

So for a fixed value of l and m, the sum of the two variences is constant and reaches its minimumwhen |m| = l corresponding to the case when the vector points as close to the ±z axis as itpossible can. The conclusion we reach is that the angular momentum vector lies somewhere in acone in which the apex half-angle, θ satisfies the relation

cos θ =m√

l(l + 1)(6.101)

which we can varify geometrically. So as l becomes very large the denominator becomes for m = l

l√l(l + 1)

=1

1√

1 + 1/l2→ 1 (6.102)

165

and θ = 0,cooresponding to the case in which the angular momentum vector lies perfectly alongthe z axis.

Exercise 6.4 Prove Eq. 6.100 by writing 〈L2〉 = 〈L2x〉+ 〈L2

y〉+ 〈L2z〉.

To develop this further, let’s look at the asymptotic behavour of the Spherical Harmonics atlarge values of angular momentum. The angular part of the Spherical Harmonic function satisfies(

1

sin θ

∂

∂θsin θ

∂

∂θ+ l(l + 1)− m2

sin2 θ

)Θlm = 0 (6.103)

For m = 0 this reduces to the differential equation for the Legendre polynomials(∂2

∂θ2+ cot θ

∂

∂θ+ l(l + 1)

)Pl(cos θ) = 0 (6.104)

If we make the substitution

Pl(cos θ) =χlθ

(sin θ)1/2(6.105)

then we wind up with a similar equation for χl(θ)(∂2

∂θ2+ (l + 1/2)2 +

csc2θ

4

)χl = 0. (6.106)

For very large l, the l+ 1/2 term dominates and we can ignore the cscθ term everywhere exceptfor angles close to θ = 0 or θ = π. If we do so then our differential equation becomes(

∂2

∂θ2+ (l + 1/2)2

)χl = 0, (6.107)

which has the solution

χl(θ) = Al sin ((l + 1/2)θ + α) (6.108)

where Al and α are constants we need to determine from the boundary conditions of the problem.For large l and for θ l−1 and π − θ l−1 one obtains

Pl(cos θ) ≈ Alsin((l + 1/2)θ + α)

(sin θ)1/2. (6.109)

Similarly,

Ylo(θ, φ) ≈(l + 1/2

2π

)1/2

Alsin((l + 1/2)θ + α)

(sin θ)1/2. (6.110)

so that the angular probability distribution is

|Ylo|2 =

(l + 1/2

2π

)A2

l

sin2((l + 1/2)θ + α)

sin θ. (6.111)

166

When l is very large the sin2((l+ 1/2)θ) factor is extremely oscillatory and we can replace itby its average value of 1/2. Then, if we require the integral of our approximation for |Yl0|2 to benormalized, one obtains

|Yl0|2 =1

2π2 sin(θ)(6.112)

which holds for large values of l and all values of θ except for theta = 0 or θ = π.We can also recover this result from a purely classical model. In classical mechanics, the

particle moves in a circular orbit in a plane perpendicular to the angular momentum vector. Form = 0 this vector lies in the xy plane and we will define θ as the angle between the particle andthe z axis, φ as the azimuthal angle of the angular momentum vector in the xy plane. Since theparticles speed is uniform, its distribution in θ is uniform. Thus the probability of finding theparticle at any instant in time between θ and θ+ dθ is dθ/π. Furthermore, we have not specifiedthe azimuthal angle, so we assume that the probability distribution is also uniform over φ andthe angular probability dθ/π must be smeared over some band on the uniform sphere defined bythe angles θ and θ + dθ. The area of this band is 2π sin θdθ. Thus, we can define the “classicalestimate” of the as a probability per unit area

P (θ) =dθ

π

1

2π sin θ=

1

2π2 sin θ

which is in agreement with the estimate we made above.For m 6= 0 we have to work a bit harder since the angular momentum vector is tilted out of

the plane. For this we define two new angles γ which is the azimuthal rotation of the particle’sposition about the L vector and α with is constrained by the length of the angular momentumvector and its projection onto the z axis.

cosα =m√l(l + 1

≈ m

l

The analysis is identical as before with the addition of the fact that the probability in γ (takento be uniform) is spread over a zone 2π sin θdθ. Thus the probability of finding the particle withsome angle θ is

P (θ) =dγ

dθ

1

2π2 sin θ.

Since γ is the dihedral angle between the plane containing z and l and the plane containingl and r (the particle’s position vector), we can relate γ to θ and α by

cos θ = cosα cosπ

2+ sinα sin

π

2cos γ = sinα cos γ

Thus,sin θdθ = sinα sin γdγ.

This allows us to generalize our probability distribution to any value of m

|Ylm(θ, φ)|2 =1

2π2 sinα sin γ(6.113)

=1

2π2(sin2 α− cos2 θ)1/2(6.114)

167

Figure 6.3: Classical and Quantum Probability Distribution Functions for Angular Momentum.

0.5 1 1.5 2 2.5 3

0.2

0.4

0.6

0.8

1»Y4,4»2

0.5 1 1.5 2 2.5 3

0.2

0.4

0.6

0.8

1»Y10,10»2

0.5 1 1.5 2 2.5 3

0.2

0.4

0.6

0.8

1»Y4,2»2

0.5 1 1.5 2 2.5 3

0.2

0.4

0.6

0.8

1»Y10,2»2

0.5 1 1.5 2 2.5 3

0.2

0.4

0.6

0.8

1»Y4,0»2

0.5 1 1.5 2 2.5 3

0.25

0.5

0.75

1

1.25

1.5

»Y10,0»2

which holds so long as sin2 α > cos2 θ. This corresponds to the spatial region (π/2 − α) <θ < π/2 + α. Outside this region, the distribution blows up and corresponds to the classicallyforbidden region.

In Fig. 6.3 we compare the results of our semi-classical model with the exact results for l = 4and l = 10. All in all we do pretty well with a semi-classical model, we do miss some of thewiggles and the distribution is sharp close to the boundaries, but the generic features are allthere.

168

Table 6.2: Relation between various notations for Clebsch-Gordan Coefficients in the literature

Symbol Author

Cjmj1m1j2m2

Varshalovich a

Sj1j2jm1jm2

Wignerb

Ajj1j2mm1m2

Eckartc

Cm1mj2

Van der Weardend

(j1j2m1m2|j1j2jm) Condon and Shortley e

Cj1j2jm (m1m2) Fockf

X(j,m, j1, j2,m1) Boysg

C(jm;m1m2) Blatt and Weisskopfh

Cj1j2jm1m2m Beidenharni

C(j1j2j,m1m2) Rosej[j1 j2 jm1 m2 m

]Yutsis and Bandzaitisk

〈j1m1j2m2|(j1j2)jm〉 Fanol

a.) D. A. Varschalovich, et al. Quantum Theory of Angular Momentum, (World Scientific, 1988).b.) E. Wigner, Group theory, (Academic Press, 1959).c.) C. Eckart “The application of group theory to the quantum dynamics of monatomic systems”,Rev. Mod. Phys. 2, 305 (1930).d.) B. L. Van der Waerden, Die gruppentheorische methode in der quantenmechanik, (Springer,1932).e.)E. Condon and G. Shortley, Theory of Atomic Spectra, (Cambridge, 1932).f.) V. A. Fock, “ New Deduction of the Vector Model”, JETP 10,383 (1940).g.)S. F. Boys, “Electronic wave functions IV”, Proc. Roy. Soc., London, A207, 181 (1951).h.) J. M. Blatt and V. F. Weisskopf, Theoretical Nuclear Physics, (McGraw-Hill, 1952).i.) L. C. Beidenharn, ”Tables of Racah Coefficients”, ONRL-1098 (1952).j.) M. E. Rose, Multipole Fields, (Wiley 1955).k.) A. P. Yusis and A. A. Bandzaitit, The Theory of Angular Momentum in Quanutm Mechanics,(Mintus, Vilinus, 1965).l.) U. Fano, “Statistical matrix techniques and their application to the directional correlation ofradiation,” US Nat’l Bureau of Standards, Report 1214 (1951).

169

6.8 Motion in a central potential: The Hydrogen Atom

(under development)The solution of the Schrodinger equation for the hydrogen atom was perhaps the most signif-

icant developments in quantum theory. Since it is one of the few problems in nature in which wecan derive an exact solution to the equations of motion, it deserves special attention and focus.Perhaps more importantly, the hydrogen atomic orbitals form the basis of atomic physics andquantum chemistry.

The potential energy function between the proton and the electron is the centrosymmetricCoulombic potential

V (r) = −Ze2

r.

Since the potential is centrosymmetric and has no angular dependency the Hydrogen atom Hamil-tonian separates in to radial and angular components.

H = − h2

2µ

(1

r2

∂

∂rr2 ∂

∂r− L2

h2r2

)+e2

r(6.115)

where L2 is the angular momentum operator we all know and love by now and µ is the reducedmass of the electron/proton system

µ =memp

me +mp

≈ me = 1

Since [H,L] = 0, angular momentum and one component of the angular momentum must beconstants of the motion. Since there are three separable degrees of freedom, we have one otherconstant of motion which must correspond to the radial motion. As a consequence, the hydrogenwavefunction is separable into radial and angular components

ψnlm = Rnl(r)Ylm(θ, φ). (6.116)

Using the Hamiltonian in Eq. 6.115 and this wavefunction, the radial Schrodinger equation reads(in atomic units)

− h2

2

[1

r2

∂

∂rr2 ∂

∂r− l(l + 1)

r2

]− 1

r

Rnl(R) = ERnl(r) (6.117)

At this point, we introduce atomic units to make the notation more compact and drasticallysimplify calculations. In atomic units, h = 1 and e = 1. A list of conversions for energy, length,etc. to SI units is listed in the appendix. The motivation is so that all of our numbers are oforder 1.

The kinetic energy term can be rearranged a bit

1

r2

∂

∂rr2 ∂

∂r=

∂2

∂r2+

2

r

∂

∂r(6.118)

and the radial equation written as− h

2

2

[∂2

∂r2+

2

r

∂

∂r− l(l + 1)

r2

]− 1

r

Rnl(R) = ERnl(r) (6.119)

170

To solve this equation, we first have to figure out what approximate form the wavefunction musthave. For large values of r, the 1/r terms disappear and the asymptotic equation is

− h2

2

∂2

∂r2Rnl(R) = ERnl(r) (6.120)

or

∂2R

∂r2= α2R (6.121)

where α = −2mE/h2. This differential equation we have seen before for the free particle, so thesolution must have the same form. Except in this case, the function is real. Furthermore, forbound states with E < 0 the radial solution must go to zero as r → ∞, so of the two possibleasymptotic solutions, the exponentially damped term is the correct one.

R(r) ≡ e−αr (6.122)

Now, we have to check if this is a solution everywhere. So, we take the asymptotic solution andplug it into the complete equation:

α2e−αr +2

r(−αe−αr) +

2m

h2

(e2

r+ E

)e−αr = 0. (6.123)

Eliminating e−αr

(α2 +

2mE

h2

)+

1

r

(2me2

h2 − 2α

)= 0 (6.124)

For the solution to hold everywhere, it must also hold at r = 0, so two conditions must be met

α2 = −2mE/h2 (6.125)

which we defined above, and (2me2

h2 − 2α

)= 0. (6.126)

If these conditions are met, then e−αr is a solution. This last equation also sets the length scaleof the system since

α = me2/h2 = 1/ao (6.127)

where ao is the Bohr radius. In atomic units, ao = 1. Likewise, the energy can be determined:

E = − h2

2ma2o

= − h2

me2e2

2ao

= − e2

2ao

. (6.128)

In atomic units the ground states energy is E = −1/2hartree.

171

Finally, we have to normalize R∫d3re−2αr = 4π

∫ ∞

0r2e−2αrdr (6.129)

The angular normalization can be absorbed into the spherical harmonic term in the total wave-function since Y00 = 1/

√4π. So, the ground state wavefunction is

ψn00 = Ne−r/aoY00 (6.130)

The radial integral can be evaluated using Leibnitz’ theorem for differentiation of a definiteintegral

∂

∂β

∫ b

af(β, x)dx =

∫ b

a

∂f(β, x)

∂βdx (6.131)

Thus,

∫ ∞

0r2e−βrdr =

∫ ∞

0

∂2

∂β2e−βrdr

=∂2

∂β2

∫ ∞

0e−βrdr

= − ∂2

∂β2

1

β

=2

β3(6.132)

Exercise 6.5 Generalize this result to show that∫ ∞

0rne−βrdr =

n!

βn+1(6.133)

Thus, using this result and putting it all together, the normalized radial wavefunction is

R10 = 2(

1

ao

)3/2

e−r/ao . (6.134)

For the higher energy states, we examine what happens at r → 0. Using a similar analysisas above, one can show that close in, the radial solution must behave like a polynomial

R ≡ rl+1

which leads to a general solution

R = rl+1e−αr∞∑

s=0

asrs.

172

The proceedure is to substitute this back into the Schrodinger equation and evaluate term byterm. In the end one finds that the energies of the bound states are (in atomic units)

En = − 1

2n2

and the radial wavefunctions

Rnl =(

2r

nao

)l ( 2

nao

)((n− l − 1)!

2n((n+ l)!)3

)e−r/naoL2l+1

n+1

(2r

nao

)(6.135)

where the Lba are the associated Laguerre polynomials.

6.8.1 Radial Hydrogenic Functions

The radial wavefunctions for nuclei with atomic number Z are modified hydrogenic wavefunctionswith the Bohr radius scaled by Z. I.e a = ao/Z. The energy for 1 electron about a nucleus withZ protons is

En = −Z2

n2

1

2ao

= −Z2

n2

Ry

2(6.136)

Some radial wavefunctions are

R1s = 2(Z

ao

)3/2

e−Zr/ao (6.137)

R2s =1√2

(Z

ao

)3/2 (1− Zr

2ao

)e−Zr/2ao (6.138)

R2p =1

2√

6

(Z

ao

)5/2

re−Zr/2ao (6.139)

6.9 Spin 1/2 Systems

In this section we are going to illustrate the various postulates and concepts we have beendeveloping over the past few weeks. Rather than choosing as examples problems which arepedagogic (such as the particle in a box and its variations) or or chosen for theor mathematicalsimplicity, we are going to focus upon systems which are physically important. We are going toexamine, with out much theoretical introduction, the case in which the state space is limited totwo states. The quantum mechanical behaviour of these systems can be varified experimentallyand, in fact, were and still are used to test various assumptions regarding quantum behaviour.

Recall from undergraduate chemistry that particles, such as the electron, proton, and so forth,possess an intrinsic angular momentum, ~S, called spin. This is a property which has no analoguein classical mechanics. Without going in to all the details of angular momentum and how it getsquantized (don’t worry, it’s a coming event!) we are going to look at a spin 1/2 system, such as

173

a neutral paramagnetic Ag atom in its ground electronic state. We are going to dispense withtreating the other variables, the nuclear position and momentum,the motion of the electrons,etc... and focus only upon the spin states of the system.

The paramagnetic Ag atoms possess an electronic magnetic moment, ~M . This magneticmoment can couple to an externally applied magnetic field, ~B, resulting on a net force beingapplied to the atom. The potential energy in for this is

W = − ~M. ~B. (6.140)

We take this without further proof. We also take without proof that the magnetic moment andthe intrinsic angular momentum are proportional.

~M = γ~S (6.141)

The proportionality constant is the gyromagnetic ratio of the level under consideration. Whenthe atoms traverse through the magnetic field, they are deflected according to how their angularmomentum vector is oriented with the applied field.

~F = ~∇( ~M. ~B) (6.142)

Also, the total moment relative to the center of the atom is

~Γ = ~M × ~B. (6.143)

Thus, the time evolution of the angular momentum of the particle is

∂

∂t~S = ~Γ (6.144)

that it to say

∂

∂t~S = γ~S × ~B. (6.145)

Thus, the velocity of the angular momentum is perpendicular to ~S and the angular momentumvector acts like a gyroscope.

We can also show that for a homogeneous field the force acts parallel to z and is proportionaltoMz. Thus, the atoms are deflected according to how their angular momentum vector is orientedwith respect to the z axis. Experimentally, we get two distributions. Meaning that measurementof Mz can give rise to two possible results.

6.9.1 Theoretical Description

We associate an observable, Sz, with the experimental observations. This has 2 eigenvalues, at±h/2 We shall assume that the two are not degenerate. We also write the eigenvectors of Sz as|±〉 corresponding to

Sz|+〉 = +h

2|+〉 (6.146)

174

Sz|−〉 = +h

2|−〉 (6.147)

with

〈+|+〉 = 〈−|−〉 = 1 (6.148)

and

〈+|−〉 = 0. (6.149)

The closure, or idempotent relation is thus

|+〉〈+|+ |−〉〈−| = 1. (6.150)

The most general state vector is

|ψ〉 = α|+〉+ β|−〉 (6.151)

with

|α|2 + |β|2 = 1. (6.152)

In the |±〉 basis, the matrix representation of Sz is diagonal and is written as

Sz =h

2

(1 00 −1

)(6.153)

6.9.2 Other Spin Observables

We can also measure Sx and Sy. In the |±〉 basis these are written as

Sx =h

2

(0 11 0

)(6.154)

and

Sy =h

2

(0 i−i 0

)(6.155)

You can verify that the eigenvalues of each of these are ±h/2.

6.9.3 Evolution of a state

The Hamiltonian for a spin 1/2 particle in a B-field is given by

H = −γ|B|Sz. (6.156)

175

Where B is the magnitude of the field. This operator is time-independent, thus, we can solve theSchrodinger Equation and see that the eigenvectors of H are also the eigenvectors of Sz. (Thisthe eigenvalues of Sz are “good quantum numbers”.) Let’s write ω = −γ|B| so that

H|+〉 = +hω

2|+〉 (6.157)

H|−〉 = − hω2|−〉 (6.158)

Therefore there are two energy levels, E± = ±hω/2. The separation is proportional to themagnetic field. They define a single “Bohr Frequency”.

6.9.4 Larmor Precession

Using the |±〉 states, we can write any arb. angular momentum state as

|ψ(0)〉 = cos(θ

2)e−iφ/2|+〉+ sin(

θ

2)e+iφ/2|−〉 (6.159)

where θ and φ are polar coordinate angles specifing the directrion of the angular momentumvector at a given time. The time evolution under H is

|ψ(0)〉 = cos(θ

2)e−iφ/2e−iE+t/h|+〉+ sin(

θ

2)e+iφ/2e−iEmt/h|−〉, (6.160)

or, using the values of E+ and E−

|ψ(0)〉 = cos(θ

2)e−i(φ+ωt)/2|+〉+ sin(

θ

2)e+i(φ+ωt)/2|−〉 (6.161)

In other words, I can write

θ(t) = θ (6.162)

φ(t) = φ+ ωt. (6.163)

This corresponds to the precession of the angular momentum vector about the z axis at an angularfrequency of ω. More over, the expectation values of Sz, Sy, and Sx can also be computed:

〈Sz(t)〉 = h/2 cos(θ) (6.164)

〈Sx(t)〉 = h/2 sin(θ/2) cos(φ+ ωt) (6.165)

〈Sy(t)〉 = h/2 sin(θ/2) sin(φ+ ωt) (6.166)

Finally, what are the “populations” of the |±〉 states as a function of time?

|〈+|ψ(t)〉|2 = cos2(θ/2) (6.167)

|〈−|ψ(t)〉|2 = sin2(θ/2) (6.168)

Thus, the populations do not change, neither does the normalization of the state.

176

6.10 Problems and Exercises

Exercise 6.6 A molecule (A) with orbital angular momentum S = 3/2 decomposes into twoproducts: product (B) with orbital angular momentum 1/2 and product (C) with orbital angularmomentum 0. We place ourselves in the rest frame of A) and angular momentum is conservedthroughout.

A3/2 → B1/2 + C0 (6.169)

1. What values can be taken on by the relative orbital angular momentum of the two finalproducts? Show that there is only one possible value of the parity of the relative orbitalstate is fixed. Would this result remain the same if the spin of A was 3/2?

2. Assume that A is initially in the spin state characterized by the eigenvalue mah of its spincomponent along the z-axis. We know that the final orbital state has a definite parity. Is itpossible to determine this parity by measuring the probabilities of finding B in either state|+〉 or in state |−〉?

Exercise 6.7 The quadrupole moment of a charge distribution, ρ(r), is given by

Qij =1

e

∫3(xixj − δijr

2)ρ(r)d3r (6.170)

where the total charge e =∫d3rρ(r). The quantum mechanical equivalent of this can be written

in terms of the angular momentum operators as

Qij =1

e

∫r2[3

2(JiJj + JjJi)− δijJ

2]ρ(r)d3r (6.171)

The quadrupole moment of a stationary state |n, j〉, where n are other non-angular momentumquantum numbers of the system, is given by the expectation value of Qzz in the state in whichm = j.

1. Evaluate

Qo = 〈Qzz〉 = 〈njm = j|Qzz|njm = j〉 (6.172)

in terms of j and 〈r2〉 = 〈nj|r2|nj〉.

2. Can a proton (j = 1/2) have a quadrupole moment? What a bout a deuteron (j = 1)?

3. Evaluate the matrix element

〈njm|Qxy|nj′m′〉 (6.173)

What transitions are induced by this operator?

4. The quantum mechanical expression of the dipole moment is

po = 〈njm = j|reJz|njm = j〉 (6.174)

Can an eigenstate of a Hamiltonian with a centrally symmetric potential have an electricdipole moment?

177

Exercise 6.8 The σx matrix is given by

σx =

(0 11 0

), (6.175)

prove that

exp(iασx) = I cos(α) + iσx sin(α) (6.176)

where α is a constant and I is the unit matrix.

Solution: To solve this you need to expand the exponential. To order α4 this is

eiασx = I + iασx −α2

2σ2

x −iα3

3!σ3 +

α4

4!σ4

x + · · · (6.177)

Also, note that σx.σx = I, thus, σ2nx = I and σ2n+1

x = σx. Collect all the real terms and all theimaginary terms:

eiασx =

(I + I

α2

2+ I

α4

4!+ · · ·

)+ iσx

(α−−iα

3

3!+ · · ·

)(6.178)

These are the series expansions for cos and sin.

eiασx = I cos(α) + iσx sin(α) (6.179)

Exercise 6.9 Because of the interaction between the proton and the electron in the ground stateof the Hydrogen atom, the atom has hyperfine structure. The energy matrix is of the form:

H =

A 0 0 00 −A 2A 00 2A −A 00 0 0 A

(6.180)

in the basis defined by

|1〉 = |e+, p+〉 (6.181)

|2〉 = |e+, p−〉 (6.182)

|3〉 = |e−, p+〉 (6.183)

|4〉 = |e−, p−〉 (6.184)

where the notation e+ means that the electron’s spin is along the +Z-axis, and e− has the spinpointed along the −Z axis. i.e. |e+, p+〉 is the state in which both the electron spin and protonspin is along the +Z axis.

1. Find the energy of the stationary states and sketch an energy level diagram relating theenergies and the coupling.

178

2. Express the stationary states as linear combinations of the basis states.

3. A magnetic field of strength B applied in the +Z direction and couples the |e+, p+〉 and|e−, p−〉 states. Write the new Hamiltonian matrix in the |e±, p±〉 basis. What happens tothe energy levels of the stationary states as a result of the coupling? Add this informationto the energy level diagram you sketched in part 1.

Exercise 6.10 Consider a spin 1/2 particle with magnetic moment ~M = γ~S. The spin space isspanned by the basis of |+〉 and |−〉 vectors, which are eigenvectors of Sz with eigenvalues ±h/2.At time t = 0, the state of the system is given by

|ψ(0)〉 = |+〉

1. If the observable Sx is measured at time t = 0, what results can be found and with whatprobabilities?

2. Taking |ψ(0)〉 as the initial state, we apply a magnetic field parallel to the y axis withstrength Bo. Calculate the state of the system at some later time t in the |±〉 basis.

3. Plot as a function of time the expectation values fo the observables Sx, Sy, and Sz. Whatare the values and probabilities? Is there a relation between Bo and t for the result of oneof the measurements to be certain? Give a physical interpretation of this condition.

4. Again, consider the same initial state, this time at t = 0, we measure Sy and find +h/2What is the state vector |ψ(0+)〉 immediately after this measurement?

5. Now we take |ψ(0+)〉 and apply a uniform time-dependent field parallel to the z-axis. TheHamiltonian operator of the spin is then given by

H(t) = ω(t)Sz

Assume that prior to t = 0, ω(t) = 0 and for t > 0 increases linearly from 0 to ωo at timet = T . Show that for 0 ≤ t ≤ T , the state vector can be written as

|ψ(t)〉 =1√2

(eiθ(t)|+〉+ ie−iθ(t)|−〉

)where θ(t) is a real function of t (which you need to determine).

6. Finally, at time t = τ > T , we measure Sy. What results can we find and with whatprobabilities? Determine the relation which must exist between ωo and T in order for us tobe sure of the result. Give the physical interpretation.

179

Chapter 7

Perturbation theory

If you perturbate to much, you will go blind.– T. A. Albright

In previous lectures, we discusses how , say through application of and external driving force,the stationary states of a molecule or other quantum mechanical system can be come coupledso that the system can make transitions from one state to another. We can write the transitionamplitude exactly as

G(i→ j, t) = 〈j| exp(−iH(tj − ti))/h)|i〉 (7.1)

where H is the full Hamiltonian of the uncoupled system plus the applied perturbation. Thus, Gtells us the amplitude for the system prepared in state |i〉 at time ti and evolve under the appliedHamiltonian for some time tj − ti and be found in state |j〉. In general this is a complicatedquantity to calculate. Often, the coupling is very complex. In fact, we can only exactly determineG for a few systems: linearly driven harmonic oscillators, coupled two level systems to name themore important ones.

In today’s lecture and following lectures, we shall develop a series of well defined and system-atic approximations which are widely used in all applications of quantum mechanics. We startwith a general solution of the time-independent Schrodinger equation in terms and eventuallyexpand the solution to infinite order. We will then look at what happens if we have a pertur-bation or coupling which depends explicitly upon time and derive perhaps the most importantrule in quantum mechanics which is called: “Fermi’s Golden Rule”. 1

7.1 Perturbation Theory

In most cases, it is simply impossible to obtain the exact solution to the Schrodinger equation.In fact, the vast majority of problems which are of physical interest can not be resolved exactlyand one is forced to make a series of well posed approximations. The simplest approximation isto say that the system we want to solve looks a lot like a much simpler system which we can

1During a seminar, the speaker mentioned Fermi’s Golden Rule. Prof. Wenzel raised his arm and in Germanspiked English chided the speaker that it was in fact HIS golden rule!

180

solve with some additional complexity (which hopefully is quite small). In other words we wantto be able to write our total Hamiltonian as

H = Ho + V

where Ho represents that part of the problem we can solve exactly and V some extra part whichwe cannot. This we take as a correction or perturbation to the exact problem.

Perturbation theory can be formuated in a variery of ways, we begin with what is typicallytermed Rayleigh-Schrodinger perturbation theory. This is the typical approach and used mostcommonly. Let Ho|φn〉 = Wn|φn〉 and (Ho + λV )|ψ〉 = En|ψ〉 be the Schrodinger equations forthe uncoupled and perturbed systems. In what follows, we take λ as a small parameter andexpand the exact energy in terms of this parameter. Clearly, we write En as a function of λ andwrite:

En(λ) = E(0)n + λE(1)

n + λ2E(2)n . . . (7.2)

Likewise, we can expand the exact wavefunction in terms of λ

|ψn〉 = |ψ(0)n 〉+ λ|ψ(1)

n 〉+ λ2|ψ(2)n 〉 . . . (7.3)

Since we require that |ψ〉 be a solution of the exact Hamiltonian with energy En, then

H|ψ〉 = (Ho + λV )(|ψ(0)

n 〉+ λ|ψ(1)n 〉+ λ2|ψ(2)

n 〉 . . .)

(7.4)

=(E(0)

n + λE(1)n + λ2E(2)

n . . .) (|ψ(0)

n 〉+ λ|ψ(1)n 〉+ λ2|ψ(2)

n 〉 . . .)

(7.5)

Now, we collect terms order by order in λ

• λ0 : Ho|ψ(0)n 〉 = E(0)

n |ψ(0)n 〉

• λ1 : Ho|ψ(1)n 〉+ V |ψ(0)

n 〉 = E(0)n |ψ(1)〉+ E(1)

n |ψ(0)n 〉

• λ2 : Ho|ψ(2)n 〉+ V |ψ(1)〉 = E(0)

n |ψ(2)n 〉+ E(1)

n |ψ(1)n 〉+ E(2)

n |ψ(0)n 〉

and so on.The λ0 problem is just the unperturbed problem we can solve. Taking the λ1 terms and

multiplying by 〈ψ(0)n | we obtain:

〈ψ(0)n |Ho|ψ(0)

n 〉+ 〈ψ(0)n |V |ψ(0)〉 = E(0)

n 〈ψ(0)n |ψ(1)

n 〉+ E(1)n 〈ψ(0)

n |ψ(0)n 〉 (7.6)

In other words, we obtain the 1st order correction for the nth eigenstate:

E(1)n = 〈ψ(0)

n |V |ψ(0)〉.

Note to obtain this we assumed that 〈ψ(1)n |ψ(0)

n 〉 = 0. This is easy to check by performing asimilar calculation, except by multiplying by 〈ψ(0)

m | for m 6= n and noting that 〈ψ(0)n |ψ(0)

m 〉 = 0 areorthogonal state.

〈ψ(0)m |Ho|ψ(0)

n 〉+ 〈ψ(0)m |V |ψ(0)〉 = E(0)

n 〈ψ(0)m |ψ(1)

n 〉 (7.7)

181

Rearranging things a bit, one obtains an expression for the overlap between the unperturbed andperturbed states:

〈ψ(0)m |ψ(1)

n 〉 =〈ψ(0)

m |V |ψ(0)n 〉

E(0)n − E

(0)m

(7.8)

Now, we use the resolution of the identity to project the perturbed state onto the unperturbedstates:

|ψ(1)n 〉 =

∑m

|ψ(0)m 〉〈ψ(0)

m |ψ(1)n 〉

=∑m6=n

〈ψ(0)m |V |ψ(0)

n 〉E

(0)n − E

(0)m

|ψ(0)m 〉 (7.9)

where we explictly exclude the n = m term to avoid the singularity. Thus, the first-ordercorrection to the wavefunction is

|ψn〉 ≈ |ψ(0)n 〉+

∑m6=n

〈ψ(0)m |V |ψ(0)

n 〉E

(0)n − E

(0)m

|ψ(0)m 〉. (7.10)

This also justifies our assumption above.

7.2 Two level systems subject to a perturbation

Let’s say that in the |±〉 basis our total Hamiltonian is given by

H = ωSz + V Sx. (7.11)

In matrix form:

H =

(ω VV −ω

)(7.12)

Diagonalization of the matrix is easy, the eigenvalues are

E+ =√ω2 + V 2 (7.13)

E− = −√ω2 + V 2 (7.14)

We can also determine the eigenvectors:

|φ+〉 = cos(θ/2)|+〉+ sin(θ/2)|−〉 (7.15)

|φ−〉 = − sin(θ/2)|+〉+ cos(θ/2)|−〉 (7.16)

where

tan θ =|V |ω

(7.17)

For constant coupling, the energy gap ω between the coupled states determines how the statesare mixed as the result of the coupling.

plot splitting as a function of unperturbed energy gap

182

7.2.1 Expansion of Energies in terms of the coupling

We can expand the exact equations for E± in terms of the coupling assuming that the couplingis small compared to ω. To leading order in the coupling:

E+ = ω(1 +1

2

∣∣∣∣∣ |V |ω∣∣∣∣∣2

· · ·) (7.18)

E− = ω(1− 1

2

∣∣∣∣∣ |V |ω∣∣∣∣∣2

· · ·) (7.19)

On the otherhand, where the two unperturbed states are identical, we can not do this expan-sion and

E+ = |V | (7.20)

and

E− = −|V | (7.21)

We can do the same trick on the wavefunctions: When ω |V | (strong coupling) , θ ≈ π/2,Thus,

|ψ+〉 =1√2(|+〉+ |−〉) (7.22)

|ψ−〉 =1√2(−|+〉+ |−〉). (7.23)

In the weak coupling region, we have to first order in the coupling:

|ψ+〉 = (|+〉+|V |ω|−〉) (7.24)

|ψ−〉 = (|−〉+|V |ω|+〉). (7.25)

In other words, in the weak coupling region, the perturbed states look a lot like the unperturbedstates. Where as in the regions of strong mixing they are a combination of the unperturbedstates.

183

7.2.2 Dipole molecule in homogenous electric field

Here we take the example of ammonia inversion in the presence of an electric field. From theproblem sets, we know that the NH3 molecule can tunnel between two equivalent C3v config-urations and that as a result of the coupling between the two configurations, the unperturbedenergy levels Eo are split by an energy A. Defining the unperturbed states as |1〉 and |2〉 we candefine the tunneling Hamiltonian as:

H =

(Eo −A−A Eo

)(7.26)

or in terms of Pauli matrices:H = Eoσo − Aσx

Taking ψ to be the solution of the time-dependent Schrodinger equation

H|ψ(t)〉 = ih|ψ〉

we can insert the identity |1〉〈1|+ |2〉〈2| = 1 and re-write this as

ihc1 = Eoc1 − Ac2 (7.27)

ihc2 = Eoc2 − Ac1 (7.28)

where c1 = 〈1|ψ〉 and c2 = 〈2|ψ〉. are the projections of the time-evolving wavefunction onto thetwo basis states. Taking these last two equations and adding and subtracting them from eachother yields two new equations for the time-evolution:

ihc+ = (Eo − A)c+ (7.29)

ihc− = (Eo + A)c− (7.30)

where c± = c1 ± c2 (we’ll normalize this later). These two new equations are easy to solve,

c±(t) = A± exp(i

h(Eo ∓ A)t

).

Thus,

c1(t) =1

2eiEot/h

(A+e

−iAt/h + A−e+iAt/h

)and

c2(t) =1

2eiEot/h

(A+e

−iAt/h − A−e+iAt/h

).

Now we have to specify an initial condition. Let’s take c1(0) = 1 and c2(0) = 0 corresponding tothe system starting off in the |1〉 state. For this initial condition, A+ = A− = 1 and

c1(t) = eiEot/h cos(At/h)

andc2(t) = eiEot/h sin(At/h).

184

So that the time evolution of the state vector is given by

|ψ(t)〉 = eiEot/h [|1〉 cos(At/h) + |2〉 sin(At/h)]

So, left alone, the molecule will oscillate between the two configurations at the tunneling fre-quency, A/h.

Now, we apply an electric field. When the dipole moment of the molecule is aligned parallelwith the field, the molecule is in a lower energy configuration, whereas for the anti-parrallel case,the system is in a higher energy configuration. Denote the contribution to the Hamiltonian fromthe electric field as:

H ′ = µeEσz

The total Hamiltonian in the |1〉, |2〉 basis is thus

H =

(Eo + µeE −A−A Eo − µeE

)(7.31)

Solving the eigenvalue problem:|H − λI| = 0

we find two eigenvalues:

λ± = Eo ±√A2 + µ2

eE2.

These are the exact eigenvalues.In Fig. 7.1 we show the variation of the energy levels as a function of the field strength.

Figure 7.1: Variation of energy level splitting as a function of the applied field for an ammoniamolecule in an electric field

Weak field limit

If µeE/A 1, then we can use the binomial expansion

√1 + x2 ≈ 1 + x2/2 + . . .

to write

√A2 + µ2

eE2 = A

(1 +

(µeEA

)2)1

/2

≈ A

(1 +

1

2

(µeEA

)2)

(7.32)

Thus in the weak field limit, the system can still tunnel between configurations and the energysplitting are given by

E± ≈ (Eo ∓ A)∓ µ2eE2

A

185

To understand this a bit further, let us use perturbation theory in which the tunnelingdominates and treat the external field as a perturbing force. The unperturbed hamiltonian canbe diagonalized by taking symmetric and anti-symmetric combinations of the |1〉 and |2〉 basisfunctions. This is exactly what we did above with the time-dependent coefficients. Here thestationary states are

|±〉 =1√2

(|1〉 ± |2〉)

with energies E± = Eo ∓ A. So that in the |±〉 basis, the unperturbed Hamiltonian becomes:

H =

(Eo − A 0

0 Eo + A

).

The first order correction to the ground state energy is given by

E(1) = E(0) + 〈+|H ′|+〉

To compute 〈+|H ′|+〉 we need to transform H ′ from the |1〉, |2〉 uncoupled basis to the new |±〉coupled basis. This is accomplished by inserting the identity on either side of H ′ and collectingterms:

〈+|H ′|+〉 = 〈+|(|1〉 < 1|+ |2〉〈2|)H ′(|1〉 < 1|+ |2〉〈2|)〉 (7.33)

=1

2(〈1|+ 〈2|)H ′(|1〉+ |2〉) (7.34)

= 0 (7.35)

Likewise for 〈−|H ′|−〉 = 0. Thus, the first order correction vanish. However, since 〈+|H ′|−〉 =µeE does not vanish, we can use second order perturbation theory to find the energy correction.

W(2)+ =

∑m6=i

H ′miH

′im

Ei − Em

(7.36)

=〈+|H ′|−〉〈−|H ′|+〉

E(0)+ − E

(0)−

(7.37)

=(µeE)2

Eo − A− Eo − A(7.38)

= −µ2eE2

2A(7.39)

Similarly for W(2)− = +µ2

eE2/A. So we get the same variation as we estimated above by expandingthe exact energy levels when the field was week.

Now let us examine the wavefunctions. Remember the first order correction to the eigenstatesis given by

|+(1)〉 =〈−|H ′|−〉E+ − E−

|−〉 (7.40)

= −µE2A|−〉 (7.41)

186

Thus,

|+〉 = |+(0)〉 − µE2A|−〉 (7.42)

|−〉 = |−(0)〉+µE2A|+〉 (7.43)

So we see that by turning on the field, we begin to mix the two tunneling states. However, sincewe have assumed that µE/A 1, the final state is not too unlike our initial tunneling states.

Strong field limit

In the strong field limit, we expand the square-root term such that(

AµeE

)2 1.

√A2 + µ2

eE2 = Eµe

( A

µeE

)2

+ 1

1/2

= Eµe

1 +1

2

(A

µeE

)2 . . .

≈ Eµe +1

2

A2

µeE(7.44)

For very strong fields, the first term dominates and the energy splitting becomes linear in thefield strength. In this limit, the tunneling has been effectively suppressed.

Let us analyze this limit using perturbation theory. Here we will work in the |1, 2〉 basis andtreat the tunneling as a perturbation. Since the electric field part of the Hamiltonian is diagonalin the 1,2 basis, our unperturbed strong-field hamiltonian is simply

H =

(Eo − µeE 0

0 Eo − µeE

)(7.45)

and the perturbation is the tunneling component. As before, the first-order corrections to theenergy vanish and we are forced to resort to 2nd order perturbation theory to get the lowestorder energy correction. The results are

W (2) = ± A2

2µE

which is exactly what we obtained by expanding the exact eigenenergies above. Likewise, thelowest-order correction to the state-vectors are

|1〉 = |10〉 − A

2µE|20〉 (7.46)

|2〉 = |20〉+A

2µE|10〉 (7.47)

So, for large E the second order correction to the energy vanishes, the correction to the wave-function vanishes and we are left with the unperturbed (i.e. non-tunneling) states.

187

7.3 Dyson Expansion of the Schrodinger Equation

The Rayleigh-Schrodinger approach is useful for discrete spectra. However, it is not very usefulfor scattering or systems with continuous spectra. On the otherhand, the Dyson expansion of thewavefunction can be applied to both cases. Its development is similar to the Rayleigh-Schrodingercase, We begin by writing the Schrodinger Equation as usual:

(Ho + V )|ψ〉 = E|ψ〉 (7.48)

where we define |φ〉 and W to be the eigenvectors and eigenvalues of part of the full problem.We shall call this the “uncoupled” problem and assume it is something we can easily solve.

Ho|φ〉 = W |φ〉 (7.49)

We want to write the solution of the fully coupled problem in terms of the solution of theuncoupled problem. First we note that

(Ho − E)|ψ〉 = V |ψ〉. (7.50)

Using the “uncoupled problem” as a “homogeneous” solution and the coupling as an inhomoge-neous term, we can solve the Schrodinger equation and obtain |ψ〉 EXACTLY as

|ψ〉 = |φ〉+1

Ho − EV |ψ〉 (7.51)

This may seem a bit circular. But we can iterate the solution:

|ψ〉 = |φ〉+1

Ho − EV |φ〉+

1

Ho − EV

1

Ho −WV |ψ〉. (7.52)

Or, out to all orders:

|ψ〉 = |φ〉+∞∑

n=1

(1

Ho − EV)n

|φ〉 (7.53)

Assuming that the series converges rapidly (true for V << Ho weak coupling case), we cantruncate the series at various orders and write:

|ψ(0)〉 = |φ〉 (7.54)

|ψ(1)〉 = |φ〉+(

1

Ho − EV)|φ〉 (7.55)

|ψ(2)〉 = |ψ(1)〉+(

1

Ho − EV)2

|φ〉 (7.56)

and so on. Let’s look at |ψ(1)〉 for a moment. We can insert 1 in the form of∑

n |φn〉〈φn|

|ψ(1)n 〉 = |φn〉+

∑n

(1

Ho −Wm

)|φm〉〈φn|V |φm〉 (7.57)

188

i.e.

|ψ(1)n 〉 = |φn〉+

∑n

(1

Wn −Wm

)|φm〉〈φn|V |φm〉 (7.58)

Likewise:

|ψ(2)n 〉 = |ψ(1)

n 〉+∑lm

(1

(Wm −Wl)(Wn −Wm)

)2

VlmVmn|φn〉 (7.59)

where

Vlm = 〈φl|V |φm〉 (7.60)

is the matrix element of the coupling in the uncoupled basis. These last two expressions are thefirst and second order corrections to the wavefunction.

Note two things. First that I can actually solve the perturbation series exactly by noting thatthe series has the form of a geometric progression, for x < 1 converge uniformly to

1

1− x= 1 + x+ x2 + · · · =

∞∑n=0

xn (7.61)

Thus, I can write

|ψ〉 =∞∑

n=0

(1

Ho − EV)n

|φ〉 (7.62)

=∞∑

n=0

(GoV )n|φ〉 (7.63)

=1

1−GoV|φ〉 (7.64)

where Go = (Ho−E)−1 (This is the “time-independent” form of the propagator for the uncoupledsystem). This particular analysis is particularly powerful in deriving the propagator for the fullycoupled problem.

We now calculate the first order and second order corrections to the energy of the system.To do so, we make use of the wavefunctions we just derived and write

E(1)n = 〈ψ(0)

n |H|ψ(0)n 〉 = Wn + 〈φn|V |φn〉 = Wn + Vnn (7.65)

So the lowest order correction to the energy is simply the matrix element of the perturbationin the uncoupled or unperturbed basis. That was easy. What about the next order correction.Same procedure as before: (assuming the states are normalized)

E(2)n = 〈ψ(1)

n |H|ψ(1)n 〉

= 〈φn|H|φn〉

+∑m6=n

〈φn|H|φm〉(

1

Wn −Wm

)〈φm|V |φn〉+O[V 3]

= Wn + Vnn +∑m6=n

|Vnm|2

Wn −Wm

(7.66)

189

Notice that I am avoiding the case where m = n as that would cause the denominator tobe zero leading to an infinity. This must be avoided. The so called “degenerate case” mustbe handled via explicit matrix diagonalization. Closed forms can be obtained for the doublydegenerate case easily.

Also note that the successive approximations to the energy require one less level of approxi-mation to the wavefunction. Thus, second-order energy corrections are obtained from first orderwavefunctions.

7.4 Van der Waals forces

7.4.1 Origin of long-ranged attractions between atoms and molecules

One of the underlyuing principles in chemistry is that molecules at long range are attractivetowards each other. This is clearly true for polar and oppositely charges species. It is also truefor non-polar and neutral species, such as methane, noble gases, and etc. These forces are due topolarization forces or van der Waals forces, which is is attractive and decreases as 1/R7, i.e. theattractive part of the potential goes as −1/R6. In this section we will use perturbation theoryto understand the origins of this force, restricting our attention to the interaction between twohydrogen atoms separated by some distance R.

Let us take the two atoms to be motionless and separated by distance R with ~n being thevector pointing from atom A to atom B. Now let ~ra be the vector connecting nuclei A to itselectron and likewise for ~rB. Thus each atom has an instantaneous electric dipole moment

~µa = q ~Ra (7.67)

~µb = q ~Rb. (7.68)

(7.69)

We will assume that R ra & rb so that the electronic orbitals on each atom do not come intocontact.

Atom A creates an electrostatic potential, U , for atom B in which the charges in B caninteract. This creates an interaction energyW . Since both atoms are neutral, the most importantsource for the interactions will come from the dipole-dipole interactions. This, the dipole of Ainteracts with an electric field E = −∇U generated by the dipole field about B and vice versa. Tocalculate the dipole-dipole interaction, we start with the expression for the electrostatic potentialcreated by µa at B.

U(R) =1

4πεo

µa ·RR3

Thus,

~E = −∇U = − q

4πεo

1

R3(~ra − 3(~ra · ~n)~n) .

Thus the dipole-dipole interaction energy is

W = −~µb · ~E

=e2

R3(~ra · ~rb − 3(~ra · ~n)(~rb · ~n)) (7.70)

190

where e2 = q2/4πεo. Now, let’s set the z axis to be along ~n so we can write

W =e2

R3(xaxb + yayb − 2zazb).

This will be our perturbing potential which we add to the total Hamiltonian:

H = Ha +Hb +W

where Ha are the unperturbed Hamiltonians for the atoms. Let’s take for example wo hydrogenseach in the 1s state. The unperturbed system has energy

H|1s1; 1s2〉 = (E1 + E2)|1s1; 1s2〉 = −2EI |1s1; 1s2〉,

where EI is the ionization energy of the hydrogen 1s state (EI = 13.6eV). The first order vanishessince it involves integrals over odd functions. This we can anticipate since the 1s orbitals arespatially isotropic, so the time averaged valie of the dipole moments is zero. So, we have to looktowards second order corrections.

The second order energy correction is

E(2) =∑nlm

∑n′l′m′ =

|〈nlm;n′l′m′|W |1sa; 1sb〉|2

−2EI − En − En′

where we restrict the summation to avoid the |1sa; 1ab〉 state. SinceW ∝ 1/R3 and the deminatoris negative, we can write

E(2) = − C

R6

which explains the origin of the 1/R6 attraction.Now we evaluate the proportionality constant C Written explicitly,

C = e4∑nml

∑n′l′m′

|〈nlm′n′l′m′|(xaxb + yayb − 2zazb)|1sa; 1sb〉|2

2EI + En + En′(7.71)

Since n and n′ ≥ 2 and |En| = EI/n2 < EI , we can replace En and En′ with 0 with out

appreciable error. Now, we can use the resolution of the identity

1 =∑nml

∑n′l′m′

|nlm;n′l′m′〉〈nlm;n′l′m′|

to remove the summation and we get

C =e4

2EI

〈1sa; 1ab|(xaxb + yayb − 2zazb)2|1sa; 1sb〉 (7.72)

where EI is the ionization potential of the 1s state (EI = 1/2). Surprisingly, this is simpleto evaluate since we can use symmetry to our advantage. Since the 1s orbitals are sphericallysymmetric, any term involving cross-terms of the sort

〈1sa|xaya|1s〉 = 0

191

vanish. This leaves only terms of the sort

〈1s|x2|1s〉.

all of which are equal to 1/3 of the mean value of RA = x2a + y2

a + z2a. Thus,

C = 6e2

2EI

∣∣∣∣〈1s|R3 |1s〉∣∣∣∣2 = 6e2ao

where ao is the Bohr radius. Thus,

E(2) = −6e2ao

R6

What does all this mean. We stated at the beginning that the average dipole moment of a H1s atom is zero. That does not mean that every single measurement of µa will yield zero. What ismeans is that the probability of finding the atom with a dipole moment µa is the same for findingthe dipole vector pointed in the opposite direction. Adding the two together produces a net zerodipole moment. So its the fluctuations about the mean which give the atom an instantaneousdipole field. Moreover, the fluctuations in A are independent of the fluctuations in B, so firstorder effects must be zero since the average interaction is zero.

Just because the fluctuations are independent does not mean they are not correlated. Considerthe field generated by A as felt by B. This field is due to the fluctuating dipole at A. This fieldinduces a dipole at B. This dipole field is in turn felt by A. As a result the fluctuations becomecorrelated and explains why this is a second order effect. In a sense, A interacts with is owndipole field through “reflection” off B.

7.4.2 Attraction between an atom a conducting surface

The interaction between an atom or molecule and a surface is a fundimental physical processin surface chemistry. In this example, we will use perturbation theory to understand the long-ranged attraction between an atom, again taking a H 1s atom as our species for simplicity, anda conducting surface. We will take the z axis to be normal to the surface and assume that theatom is high enough off the surface that its altitude is much larger than atomic dimensions.Furthermore, we will assume that the surface is a metal conductor and we will ignore any atomiclevel of detail in the surface. Consequently, the atom can only interact with its dipole image onthe opposite side of the surface.

We can use the same dipole-dipole interaction as before with the following substitutions

e2 −→ −e2 (7.73)

R −→ 2d (7.74)

xb −→ x′a = xa (7.75)

yb −→ y′a = ya (7.76)

zb −→ z′a = −za (7.77)

where the sign change reflects the sign difference in the image charges. So we get

W = − e2

8d3(x2

a + y2a + 2z2

a)

192

as the interaction between a dipole and its image. Taking the atom to be in the 1s ground state,the first order term is non-zero:

E(1) = 〈1s|W |1s〉.

Again, using spherical symmetry to our advantage:

E(1) = − e2

8d34〈1s|r2|1s〉 = −e

2a2o

2d3.

Thus an atom is attracted to the wall with an interaction energy which varries as 1/d3. This isa first order effect since there is perfect correlation between the two dipoles.

7.5 Perturbations Acting over a Finite amount of Time

Perhaps the most important application of perturbation theory is in cases in which the couplingacts for a finite amount of time. Such as the coupling of a molecule to a laser field. The laser fieldimpinges upon a molecule at some instant in time and some time later is turned off. Alternativelywe can consider cases in which the perturbation is slowly ramped up from 0 to a final value.

7.5.1 General form of time-dependent perturbation theory

In general, we can find the time-evolution of the coefficients by solving

ihcn(t) = Encn(t) +∑k

λWnk(t)ck(t) (7.78)

where λWnk are the matrix elements of the perturbation. Now, let’s write the cn(t) as

cn(t) = bn(t)e−iEnt

and assume that bn(t) changes slowly in time. Thus, we can write a set of new equations for thebn(t) as

ihbn =∑k

eiωnktλWnkbk(t) (7.79)

Now we assume that the bn(t) can be written as a perturbation expansion

bn(t) = b(0)n (t) + λb(1)

n (t) + λ2b(2)n (t) + · · ·

where as before λ is some dimensionless number of order unity. Taking its time derivative andequating powers of λ one finds

ihb(i)n =∑k

eiωnktWnk(t)b(i−1)k

and that b(0)n (t) = 0.

193

Now, we calculate the first order solution. For t < 0 the system is assumed to be in somewell defined initial state, |φi〉. Thus, only one bn(t < 0) coefficient can be non-zero and must beindependent of time since the coupling has not been turned on. Thus,

bn(t = 0) = δni

At t = 0, we turn on the coupling and λW jumps from 0 to λW (0). This must hold for all orderin λ. So, we immediately get

b(0)n (0) = δni (7.80)

b(i)n (0) = 0 (7.81)

Consequently, for all t > 0, b(0)n (t) = δni which completely specifies the zeroth order result. This

also gives us the first order result.

ihb(1)n (t) =

∑k

eiωnktWnkδki

= eiωnitWni(t) (7.82)

which is simple to integrate

b(1)n (t) = − i

h

∫ t

0eiωnisWni(s)ds.

Thus, our perturbation wavefunction is written as

|ψ(t)〉 = e−iEot/h|φo〉+∑n6=0

λb(1)n (t)e−iEnt/h|φn〉 (7.83)

7.5.2 Fermi’s Golden Rule

Let’s consider the time evolution of a state under a small perturbation which varies very slowlyin time. The expansion coefficients of the state in some basis of eigenstates of the unperturbedHamiltonian evolve according to:

ihcs(t) =∑n

Hsncn(t) (7.84)

where Hsn is the matrix element of the full Hamiltonian in the basis.

Hsn = Esδns + Vsn(t) (7.85)

Assuming that V (t) is slowly varying and that Vsn << Es − En for all time. we can write theapproximate solution as

cs(t) = As(t) exp(−i/hEst) (7.86)

where As(t) is a function with explicit time dependence. Putting the approximate solution backinto the differential equation:

ihcs(t) = ihAs(t)cs(t) + Escs(t) (7.87)

(7.88)

194

= Escs(t) +∑n

VsnAn(t)e−i/hEnt (7.89)

We now proceed to solve this equation via series of well defined approximations. Our firstassumption is that Vsn << En − Es (weak coupling approximation). We can also write

As << 1 (7.90)

since we have assumed that A(t) varies slowly in time. For s 6= i we have the following initialconditions:

Ai(0) = 1 (7.91)

As(0) = 0 (7.92)

Thus, at t = 0 we can set all the coefficients in zero except for Ai, which is 1. Thus,

As(t) = −i/hVsie−i/h(Es−Ei)t (7.93)

which can be easily integrated

As(t) = − i

h

∫ t

0dt′Vsie

−i/h(Es−Ei)t′

(7.94)

where all the As(t) are assumed to be smaller than unity. Of course, to do the integral we needto know how the perturbation depends upon time. Let’s assume that

V (t) = 2V cos(ωt) = V(e+iωt + e−iωt

)(7.95)

where V is a time independent quantity. Thus we can determine A(t) as

As(t) = − i

h

∫ t

0dt′〈s|V |ı〉

[ei/h(Es−Ei+hω)t′ei/h(Es−Ei−hω)t′

](7.96)

(7.97)

= 〈s|V |ı〉[1− ei/h(Es−Ei+hω)t

Es − En + hω+

1− ei/h(Es−Ei−hω)t

Es − En − hω

](7.98)

Since the coupling matrix element is presumed to be small. The only significant contributioncomes when the denominator is very close to zero. i.e. when

hω ≈ |Es − En| (7.99)

For the case when Es > En we get only one term making a significant contribution and thus thetransition probability as a function of time is

Psn(t) = |cs(t)|2 = |As(t)|2 (7.100)

(7.101)

195

= 4|〈s|V |n〉|2[sin2((Es − En − hω)t/(2h))

(Es − En − hω)2

]. (7.102)

This is the general form for a harmonic perturbation. The function

sin2(ax)

x2(7.103)

is called the sinc function or the “Mexican Hat Function”. At x = 0 the peak is sharply peaked(corresponding to Es − En − hω = 0). Thus, the transition is only significant when the energydifference matches the frequency of the applied perturbation. As t→∞ (a = t/(2h)), the peakbecome very sharp and becomes a δ-function. The width of the peak is

∆x =2πh

t(7.104)

Thus, the longer we measure a transition, the more well resolved the measurement will become.This has profound implications for making measurements of meta-stable systems.

We have an expression for the transition probability between two discrete states. We havenot taken into account the the fact that there may be more than one state close by nor have wetaken into account the finite “width” of the perturbation (instrument function). When there amany states close the the transition, I must take into account the density of nearby states. Thus,we define

ρ(E) =∂N(E)

∂E(7.105)

as the density of states close to energy E where N(E) is the number of states with energy E.Thus, the transition probability to any state other than my original state is

Ps(t) =∑s

Pns(t) =∑s

|As(t)|2 (7.106)

to go from a discrete sum to an integral, we replace

∑n

→∫ dN

dEdE =

∫ρ(E)dE (7.107)

Thus,

Ps(t) =∫dE|As(t)|2ρ(E) (7.108)

Since |As(t)|2 is peaked sharply at Es = Ei + hω we can treat ρ(E) and Vsn as constant and write

Ps(t) = 4|〈s|V |n〉|2ρ(Es)∫ sin2(ax)

x2dx (7.109)

Taking the limits of integration to ±∞∫ ∞

−∞dx

sin2(ax)

x= πa =

πt

2h(7.110)

196

In other words:

Ps(t) =2πt

h|〈s|V |n〉|2ρ(Es) (7.111)

We can also define a transition rate as

R(t) =P (t)

t(7.112)

thus, the Golden Rule Transition Rate from state n to state s is

Rn→s(t) =2π

h|〈s|V |n〉|2ρ(Es) (7.113)

This is perhaps one of the most important approximations in quantum mechanics in that ithas implications and applications in all areas, especially spectroscopy and other applications ofmatter interacting with electro-magnetic radiation.

7.6 Interaction between an atom and light

What I am going to tell you about is what we teach our physics students in the thirdor fourth year of graduate school... It is my task to convince you not to turn awaybecause you don’t understand it. You see my physics students don’t understand it...That is because I don’t understand it. Nobody does.Richard P. Feynman, QED, The Strange Theory of Light and Matter

Here we explore the basis of spectroscopy. We will consider how an atom interacts with aphoton field in the low intensity limit in which dipole interactions are important. We will thenexamine non-resonant excitation and discuss the concept of oscillator strength. Finally we willlook at resonant emission and absorption concluding with a discussion of spontaneous emission.In the next section, we will look at non-linear interactions.

7.6.1 Fields and potentials of a light wave

An electromagnetic wave consists of two oscillating vector field components which are perpen-dicular to each other and oscillate at at an angular frequency ω = ck where k is the magnitudeof the wavevector which points in the direction of propagation and c is the speed of light. Forsuch a wave, we can always set the scalar part of its potential to zero with a suitable choice ingauge and describe the fields associate with the wave in terms of a vector potential, ~A given by

~A(r, t) = Aoezeiky−iωt + A∗

oeze−iky+iωt

Here, the wave-vector points in the +y direction, the electric field, E is polarized in the yz planeand the magnetic field B is in the xy plane. Using Maxwell’s relations

~E(r, t) = −∂A∂t

= iωez(Aoei(ky−ωt) − A∗

oe−i(ky−ωt))

197

and~B(r, t) = ∇× ~A = ikex(Aoe

i(ky−ωt) − A∗oe

−i(ky−ωt)).

We are free to choose the time origin, so we will choos it as to make Ao purely imaginary andset

iωAo = E/2 (7.114)

ikAo = B/2 (7.115)

where E and B are real quantities such that

EB

=ω

k= c.

Thus

E(r, t) = Eez cos(ky − ωt) (7.116)

B(r, t) = Bez sin(ky − ωt) (7.117)

where E and B are the magnitudes of the electric and magnetic field components of the planewave.

Lastly, we define what is known as the Poynting vector (yes, it’s pronounced pointing) whichis parallel to the direction of propagation:

~S = εoc2 ~E × ~B. (7.118)

Using the expressions for ~E and ~B above and averaging over several oscillation periods:

~S = εoc2E2ey (7.119)

7.6.2 Interactions at Low Light Intensity

The electromagnetic wave we just discussed can interact with an atomic electron. The Hamilto-nian of this electron can be given by

H =1

2m(P− qA(r, t))2 + V (r)− q

mS ·B(r, t)

where the first term represents the interaction between the electron and the electrical field ofthe wave and the last term represents the interaction between the magnetic moment of theelectron and the magnetic moment of the wave. In expanding the kinetic energy term, we haveto remember that momentum and position do not commute. However, in the present case, A isparallel to the z axis and Pz and y commute. So, we wind up with the following:

H = Ho +W

where

Ho =P2

2m+ V (r)

198

is the unperturbed (atomic) hamiltonian and

W = − q

mP ·A− q

mS ·B +

q2

2mA2.

The first two depend linearly upon A and the second is quadratic in A. So, for low intensity wecan take

W = − q

mP ·A− q

mS ·B.

Before moving on, we will evaluate the relative importance of each term by orders of magnitudefor transitions between bound states. In the second term, the contribution of the spin operatoris on the order of h and the contribution from B is on the order of kA. Thus,

WB

WE

=qmS ·B

qmP ·A

≈ hk

p

h/p is on the order of an atomic radius, ao and k = 2π/λ where λ is the wavelength of the light,typically on the order of 1000ao. Thus,

WB

WE

≈ ao

λ 1.

So, the magnetic coupling is not at all important and we focus only upon the coupling to theelectric field.

Using the expressions we derived previously, the coupling to the electric field component ofthe light wave is given by:

WE = − q

mpz(Aoe

ikye−iωt + A∗oe

−ikye+iωt).

Now, we expand the exponential in powers of y

e±iky = 1± iky − 1

2k2y2 + . . .

since ky ≈ ao/λ 1, we can to a good approximation keep only the first term. Thus we get thedipole operator

WD =qEmω

pz sin(ωt).

In the electric dipole approximation, W (t) = WD(t).Note, that one might expect that WD should have been written as

WD = −qEz cos(ωt)

since we are, after all, talking about a dipole moment associated with the motion of the electronabout the nucleus. Actually, the two expressions are identical! The reason is that I can alwayschoose a differnent gauge to represent the physical problem without changing the physical result.To get the present result, we used

A =Eωez sin(ωt)

199

andU(r) = 0

as the scalar potential. A gauge transformation is introduced by taking a function, f and defininga new vector potential and a new scalar potential as

A′ = A +∇f

U ′ = U − ∂f

∂t

We are free to choose f however we do so desire. Let’s take f = zE sin(ωt)/ω. Thus,

A′ = ezEω

(sin(ky − ωt) + sin(ωt))

andU ′ = −zE cosωt

is the new scalar potential. In the electric dipole approximation, ky is small, so we set ky = 0everywhere and obtain A′ = 0. Thus, the total Hamiltonian becomes

H = Ho + qU ′(r, t)

with perturbationW ′

D = −qzE cos(ωt).

This is the usual form of the dipole coupling operator. However, when we do the gauge trans-formation, we have to transform the state vector as well.

Next, let us consider the matrix elements of the dipole operator between two stationary statesof Ho: |ψi〉 and |ψf〉 with eigenenergy Ei and Ef respectively. The matrix elements of WD aregiven by

Wfi(t) =qEmω

sin(ωt)〈ψf |pz|ψi〉

We can evaluate this by noting that

[z,Ho] = ih∂Ho

∂pz

= ihpz

m.

Thus,〈ψf |pz|ψi〉 = imωfi〈ψf |z|ψi〉.

Consequently,

Wfi(t) = iqEωfisin(ωt)

ωzfi.

Thus, the matrix elements of the dipole operator are those of the position operator. This deter-mines the selection rules for the transition.

Before going through any specific details, let us consider what happens if the frequency ωdoes not coincide with ωfi. Soecifically, we limit ourselves to transitions originating from theground state of the system, |ψo〉. We will assume that the field is weak and that in the field the

200

atom acquires a time-dependent dipole moment which oscillates at the same frequency as thefield via a forced oscillation. To simplify matters, let’s assume that the electron is harmonicallybound to the nucleus in a classical potential

V (r) =1

2mωor

2

where ωo is the natural frequency of the electron.The classical motion of the electron is given by the equations of motion (via the Ehrenfest

theorem)

z + ω2z =qEm

cos(ωt).

This is the equation of motion for a harmonic oscillator subject to a periodic force. This inho-mogeneous differential equation can be solved (using Fourier transform methods) and the resultis

z(t) = A cos(ωot− φ) +qE

m(ω2o − ω2)

cos(ωt)

where the first term represents the harmonic motion of the electron in the absence of the drivingforce. The two coefficients, A and φ are determined by the initial condition. If we have a veryslight damping of the natural motion, the first term dissappears after a while leaving only thesecond, forced oscillation, so we write

z =qE

m(ω2o − ω2)

cos(ωt).

Thus, we can write the classical induced electric dipole moment of the atom in the field as

D = qz =q2E

m(ω2o − ω2)

cos(ωt).

Typically this is written in terms of a susceptibility, χ, where

χ =q2

m(ω2o − ω2)

.

Now we look at this from a quantum mechanical point of view. Again, take the initial stateto be the ground state and H = Ho +WD as the Hamiltonian. Since the time-evolved state canbe written as a superposition of eigenstates of Ho,

|ψ(t)〉 =∑n

cn(t)|φn〉

To evaluate this we can us the results derived previously in our derivation of the golden rule,

|ψ(t)〉 = |φo〉+∑n6=0

qE2imhω

〈n|pz|φo〉

×e−iωnot − eiωt

ωno + ω− e−iωnot − e−iωt

ωno − ω

|φn〉 (7.120)

where we have removed a common phase factor. We can then calculate the dipole momentexpectation value, 〈D(t)〉 as

〈D(t)〉 =2q2

hE cos(ωt)

∑n

ωon|〈φn|z|φo〉|2

ω2no − ω2

(7.121)

201

Oscillator Strength

We can now notice the similarity between a driven harmonic oscillator and the expectation valueof the dipole moment of an atom in an electric field. We can define the oscillator strength as adimensionless and real number characterizing the transition between |φo and |φn〉

fno =2mωno

h|〈φn|z|φo〉|2

In Exercise 2.4, we proved the Thomas-Reiche-Kuhn sum rule, which we can write in terms ofthe oscillator strengths, ∑

n

fno = 1

This can be written in a very compact form:

m

h2 〈φo|[x, [H, x]]|φo〉 = 1.

7.6.3 Photoionization of Hydrogen 1s

Up until now we have considered transitions between discrete states inrtoduced via some externalperturbation. Here we consider the single photon photoionization of the hydrogen 1s orbital toillustrate how the golden rule formalism can be used to calculate photoionization cross-sectionsas a function of the photon-frequency. We already have an expression for dipole coupling:

WD =qEmω

pz sin(ωt) (7.122)

and we have derived the golden rule rate for transitions between states:

Rif =2π

h|〈f |V |i〉|2δ(Ei − Ef + hω). (7.123)

For transitions to the continuum, the final states are the plane-waves.

ψ(k) =1

Ω1/2eik·r. (7.124)

where Ω is the volume element. Thus the matrix element 〈1s|V |k〉 can be written as

〈1s|pz|k〉 =hkz

Ω1/2

∫ψ1s(r)e

ik·rdr. (7.125)

To evaluate the integral, we need to transform the plane-wave function in to spherical coordinates.This can be done vie the expansion;

eik·r =∑

l

il(2l + 1)jl(kr)Pl(cos(θ)) (7.126)

where jl(kr) is the spherical Bessel function and Pl(x) is a Legendre polynomial, which we canalso write as a spherical harmonic function,

Pl(cos(θ)) =

√4π

2l + 1Yl0(θ, φ). (7.127)

202

Thus, the integral we need to perform is

〈1s|k〉 =1√πΩ

∑l

(∫Y ∗

00Yl0dΩ)il√

4π(2l + 1)∫ ∞

0r2e−rjl(kr)dr. (7.128)

The angular integral we do by orthogonality and produces a delta-function which restricts thesum to l = 0 only leaving,

〈1s|k〉 =1√Ω

∫ ∞

0r2e−rj0(kr)dr. (7.129)

The radial integral can be easily performed using

j0(kr) =sin(kr)

kr(7.130)

leaving

〈1s|k〉 =4

k

1

Ω1/2

1

(1 + k2)2. (7.131)

Thus, the matrix element is given by

〈1s|V |k〉 =qEhmω

1

Ω1/2

2

(1 + k2)2(7.132)

This we can indert directly in to the golden rule formula to get the photoionization rate to agiven k-state.

R0k =2πh

Ω

(qEmω

)2 4

(1 + k2)4δ(Eo − Ek + hω). (7.133)

which we can manipulate into reading as

R0k =16π

hΩ

(qEmω

)2

mδ(k2 −K2)

(1 + k2)4(7.134)

where we write K2 = 2m(EI + hω)/h2 to make our notation a bit more compact. Eventually,we want to know the rate as a function of the photon frequency, so let’s put everything exceptthe frequency and the volume element into a single constant, I, which is related to the intensityof the incident photon.

R0k =IΩ

1

ω2

δ(k2 −K2)

(1 + k2)4. (7.135)

Now, we sum over all possible final states to get the total photoionization rate. To do this, weneed to turn the sum over final states into an integral, this is done by

∑k

=Ω

(2π)34π∫ ∞

0k2dk (7.136)

203

Thus,

R =IΩ

1

ω2

Ω

(2π)34π∫ ∞

0k2 δ(k

2 −K2)

(1 + k2)4dk

=Iω2

1

2π2

∫ ∞

0k2 δ(k

2 −K2)

(1 + k2)2dk

Now we do a change of variables: y = k2 and dy = 2kdk so that the integral become

∫ ∞

0k2 δ(k

2 −K2)

(1 + k2)2dk =

1

2

∫ ∞

0

y1/2

(1 + y2)4δ(y −K2)dy

=K

2(1 +K2)4(7.137)

Pulling everything together, we see that the total photoionization rate is given by

R =Iω2

1

2π2

K

(1 +K2)4

=I√

mh2

√ω h− εo

√2π2 ω2

(1 + 2 m (ω h−εo)

h2

)4

= I√

2ω − 1

32π2 ω6(7.138)

where in the last line we have converted to atomic units to clean things up a bit. This expressionis clearly valid only when hω > EI = 1/2hartree (13.6 eV) and a plot of the photo-ionizationrate is given in Fig. 7.2

7.6.4 Spontaneous Emission of Light

The emission and absorption of light by an atom or molecule is perhaps the most spectacularand important phenomena in the universe. It happens when an atom or molecule undergoes atransition from one state to another due to its interaction with the electro-magnetic field. Becausethe electron-magnetic field can not be entirely eliminated from any so called isolated system(except for certain quantum confinement experiments), no atom or molecule is ever really isolated.Thus, even in the absence of an explicitly applied field, an excited system can spontaneously emita photon and relax to a lower energy state. Since we have all done spectroscopy experimentsat one point in our education or another, we all know that the transitions are between discreteenergy levels. In fact, it was in the examination of light passing through glass and light emittedfrom flames that people in the 19th century began to speculate that atoms can absorb and emitlight only at specific wavelengths.

We will use the GR to deduce the probability of a transition under the influence of an appliedlight field (laser, or otherwise). We will argue that the system is in equilibrium with the electro-magnetic field and that the laser drives the system out of equilibrium. From this we can deducethe rate of spontaneous emission in the absence of the field.

204

0.5 1 1.5 2Ñw HauL

0.2

0.4

0.6

0.8

R HarbL

Figure 7.2: Photo-ionization spectrum for hydrogen atom.

The electric field associated with a monochromatic light wave of average intensity I is

〈I〉 = c〈ρ〉 (7.139)

= c

εo 〈~E2o 〉2

+1

µo

〈B2o〉

2

(7.140)

=

(εoµo

)1/2 E2o

2(7.141)

= cεoE2

o

2(7.142)

where ρ is the energy density of the field, |~E| and |Bo| = (1/c)|~E| are the maximum amplitudesof the E and B fields of the wave. Units are MKS units.

The em wave in reality contains a spread of frequencies, so we must also specify the intensitydensity over a definite frequency interval:

dI

dωdω = cu(ω)dω (7.143)

where u(ω) is the energy density per unit frequency at ω.Within the “semi-classical” dipole approximation, the coupling between a molecule and the

light wave is

~µ · ~E(t) = ~µ · ~εEo

2cos(ωt) (7.144)

205

where ~µ is the dipole moment vector, and ~ε is the polarization vector of the wave. Using thisresult, we can go back to last week’s lecture and plug directly into the GR and deduce that

Pfi(ω, t) = 4|〈f |~µ · ~ε|i〉|2E2o

4

sin2 ((Ef − Ei − hω)t/(2h))

(Ef − Ei − hω)2(7.145)

Now, we can take into account the spread of frequencies of the em wave around the resonantvalue of ωo = (Ef − Ei)/h. To do this we note:

E2o = 2

〈I〉cεo

(7.146)

and replace 〈I〉 with (dI/dω)dω.

Pfi(t) =∫ ∞

0dωPfi(t, ω) (7.147)

=2

cεo

(dI

dω

)ωo

|〈f |~µ · ~ε|i〉|2∫ ∞

0

sin2((hωo − hω)(t/(2h)))

(hωo − hω)2dω (7.148)

To get this we assume that dI/dω and the matrix element of the coupling vary slowly withfrequency as compared to the sin2(x)/x2 term. Thus, as far as doing integrals are concerned,they are both constants. With ωo so fixed, we can do the integral over dw and get πt/(2h2). andwe obtain the GR transition rate:

kfi =π

cεoh2 |〈f |~µ · ~ε|i〉|

2

(dI

dω

)ωo

(7.149)

Notice also that this equation predicts that the rate for excitation is identical to the rate forde-excitation. This is because the radiation field contains both a +ω and −ω term (unless thefield is circularly polarized), this the transition rate to a state of lower energy to a higher energyis the same as that of the transition from a higher energy state to a lower energy state.

However, we know that systems can emit spontaneously in which a state of higher energycan go to a state of lower energy in the absence of an external field. This is difficult to explainin the presence frame-work since we have assumed that |i〉 is stationary.

Let’s assume that we have an ensemble of atoms in a cavity containing em radiation and thesystem is in thermodynamic equilibrium. (Thought you could escape thermodynamics, eh?) LetE1 and E2 be the energies of two states of the atom with E2 > E1. When equilibrium has beenestablished the number of atoms in the two states is determined by the Boltzmann equation:

N2

N1

=Ne−E2β

Ne−E1β= e−β(E2−E1) (7.150)

where β = 1/kT The number of atoms (per unit time) undergoing the transition from 1 to 20 isproportional to k21 induced by the radiation and to the number of atoms in the initial state, N1.

dN

dt(1 → 2) = N1k21 (7.151)

206

The number of atoms going from 2 to 1 is proportional to N2 and to k12 + A where A is thespontaneous transition rate

dN

dt(2 → 1) = N2(k21 + A) (7.152)

At equilibrium, these two rates must be equal. Thus,

k21 + A

k21

=N1

N2

= ehωβ (7.153)

Now, let’s refer to the result for the induced rate k21 and express it in terms of the energy densityper unit frequency of the cavity, u(ω).

k21 =π

εoh2 |〈2|~µ · ~ε|1〉|

2u(ω) = B21u(ω) (7.154)

where

B21 =π

εoh2 |〈2|~µ · ~ε|1〉|

2. (7.155)

For em radiation in equilibrium at temperature T the energy density per unit frequency is givenby Planck’s Law:

u(ω) =1

π2c3hω3

ehωβ − 1(7.156)

Combining the results we obtain

B12

B21

+A

B21

1

u(ω)= ehωβ (7.157)

B21

B12

+A

B21

π2c3

hω3(ehωβ − 1) = ehωβ (7.158)

(7.159)

which must hold for all temperatures. Since

B21

B12

= 1. (7.160)

we get

A

B21

π2c3

hω3= 1 (7.161)

and Thus, the spontaneous emission rate is

A =hω3

π2c3B12 (7.162)

207

=ω3

εoπhc3|〈2|~µ · ~ε|1〉|2 (7.163)

This is a key result in that it determines the probability for the emission of light by atomicand molecular systems. We can use it to compute the intensity of spectral lines in terms of theelectric dipole moment operator. The lifetime of the excited state is then inversely proportionalto the spontaneous decay rate.

τ =1

A(7.164)

To compute the matrix elements, we can make a rough approximation that 〈µ〉 ∝ 〈x〉e wheree is the charge of an electron and 〈x〉 is on the order of atomic dimensions. We also must includea factor of 1/3 for averaging over all orientations of (~µ ·~ε), since at any given time,the momentsare not all aligned.

1

τ= A =

4

3

ω3

hc3e2

4πεo|〈x〉|2 (7.165)

The factor

e2

4πεohc= α ≈ 1

137(7.166)

is the fine structure constant. Also, ω/c = 2π/λ. So, setting 〈x〉 ≈ 1 A

A =4

3

1

137c(

2π

λ

)3

(1A)2 ≈ 6× 1018

[λ(A)]3sec−1 (7.167)

So, for a typical wavelength, λ ≈ 4× 103A.

τ = 10−8sec (7.168)

which is consistent with observed lifetimes.We can also compare with classical radiation theory. The power radiated by an accelerated

particle of charge e is given by the Larmor formula (c.f Jackson).

P =2

3

e2

4πεo

(v)2

c3(7.169)

where v is the acceleration of the charge. Assuming the particle moves in a circular orbit ofradius r with angular velocity ω, the acceleration is v = ω2r. Thus, the time required to radiateenergy hω/2 is equivalent to the lifetime τ .

1

τclass

=2P

hω(7.170)

=1

hω

4

3

e2

4πεo

ω4r2

c3(7.171)

208

=4

3

ω3

hc3e2

4πεor2. (7.172)

This qualitative agreement between the classical and quantum result is a manifestation of thecorrespondence principle. However, it must be emphasized that the MECHANISM for radiationis entirely different. The classical result will never predict a discrete spectrum. This was in facta very early indication that something was certainly amiss with the classical electro-magneticfield theories of Maxwell and others.

7.7 Time-dependent golden rule

In the last lecture we derived the the Golden Rule (GR) transition rate as

k(t) =2π

h|〈s|V |n〉|2ρ(Es) (7.173)

This is perhaps one of the most important approximations in quantum mechanics in that ithas implications and applications in all areas, especially spectroscopy and other applications ofmatter interacting with electro-magnetic radiation. In today’s lecture, I want to show how wecan use the Golden Rule to simplify some very complex problems. Moreover, to show how weused the GR to solve a real problem. The GR has been used by a number of people in chemistryto look at a wide variery of problems. In fact, most of this lecture comes right out out of theJournal of Chemical Physics. Some papers you may be interested in knowing about include:

1. B. J. Schwartz, E. R. Bittner, O. V. Prezhdo, and P. J. Rossky, J. Chem. Phys. 104, 5242(1996).

2. E. Neria and A. Nitzan, J. Chem. Phys. 99, 1109 (1993).

3. A. Stiab and D. Borgis, J. Chem. Phys. 103, 2642 (1995)

4. E. J. Heller, J. Chem. Phys. 75, 2923 (1981).

5. W. Gelbart, K. Spears, K. F. Freed, J. Jortner, S. A. Rice, Chem. Phys. Lett. 6 345(1970).

The focus of the lecture will be to use the GR to calculate the transition rate from oneadiabatic potential energy surface to another via non-radiative decay. Recall in a previous lecture,we talked about potential energy curves of molecule and that these are obtained by solving theSchrodinger equation for the electronic degrees of freedom assuming that the nuclei move veryslowly. We defined the adiabatic or Born-Oppenheimer potential energy curves for the nuclei in amolecule by solving the Schrodinger equation for the electrons for fixed nuclear positions. Thesepotential curves are thus the electronic eigenvalues parameterized by the nuclear positions.

Vi(R) = Ei(R) (7.174)

209

Under the BO approximation, the nuclei move about on a single energy surface and the electronicwavefunction is simply |Ψi(R)〉 paramterized by the nuclear positions. However, when the nucleiare moving fast, this assumption is no longer true since,

ihd

dt|Ψi(R)〉 =

(ih∂R

∂t

∂

∂R+ Ei(R)

)|Ψi(R)〉. (7.175)

That is to say, when the nuclear velocities are large in a direction that the wavefunction changesa lot with varying nuclear position, the Born-Oppenheimer approximation is not so good andthe electronic states become coupled by the nuclear motion. This leads to a wide variety ofphysical phenomina including non-radiative decay and intersystem crossing and is an importantmechanism in electron transfer dynamics.

THe picture I want to work with today is a hybred quantum/classical picture (or semi-classical). I want to treat the nuclear dynamics as being mostly “classical” with some “quantumaspects”. IN this picture I will derive a semi-classical version of the Golden-Rule transition ratewhich can be used in concert with a classical Molecular dynamics simulation.

We start with the GR transition rate we derived the last lecture. We shall for now assumethat the states we are coupling are the vibrational-electronic states of the system, written as

|ψi〉 = |αi(R)I(R)〉 (7.176)

where R denotes the nuclear positions, |α(R)〉 is the adiabatic electronic eigenstate obtained atposition R and |I(R)〉 is the initial nuclear vibrational state on the α(R) potential energy surface.Let this denote the initial quantum state and denote by

|ψf〉 = |αf (R)F (R)〉 (7.177)

the final quantum state. The GR transition rate at nuclear position R is thus

kif =2π

h

∑f

|〈ψi|V |ψf〉|2δ(Ei − Ef ) (7.178)

where the sum is over the final density of states (vibrational) and the energys in the δ-functionis the electronic energy gap measured with respect to a common origin. We can also define a“thermal” rate constant by ensemble averaging over a collective set of initial states.

7.7.1 Non-radiative transitions between displaced Harmonic Wells

An important application of the GR comes in evaluating electron transfer rates between electronicstate of a molecule. Let’s approximate the diabatic electronic energy surfaces of a molecule asharmonic wells off-set by by energy and with the well minimums displaced by some amount xo.Let the curves cross at xs and assume there is a spatial dependent coupling V (x) which couplesthe diabatic electronic states. Let T1 denote the upper surface and So denote the ground-statesurface. The diabatic coupling is maximized at the crossing point and decays rapidly as we moveaway. Because these electronic states are coupled, the vibrational states on T1 become coupledto the vibrational states on So and vibrational amplitude can tunnel from one surface to the

210

other. The tunneling rate can be estimated very well using the Golden Rule (assuming that theamplitude only crosses from one surface to another once.)

kTS =2π

h|〈ΨT |V |ΨS〉|2ρ(Es) (7.179)

The wavefunction on each surface is the “vibronic” function mentioned above. This we will writeas a product of an electronic term |ψT 〉 (or |ψS〉) and a vibrational term |nT 〉 (or |nS〉). Forshorthand, let’s write the electronic contribution as

VTS = 〈ψT |V |ψS〉 (7.180)

Say I want to know the probability of finding the system in some initial vibronic state after sometime. The rate of decay of this state is thus the sum over all possible decay channels. So, I mustsum over all final states that I can decay into. The decay rate is thus.

kTS =2π

h

∑mS

|〈nT |VTS|mS〉|2ρ(Es) (7.181)

This equation is completely exact (within the GR approximation) and can be used in this form.However, let’s make a series of approximations and derive a set of approximate rates and comparethe various approximations.

Condon Approximation

First I note that the density of states can be rewritten as

ρ(Es) = Tr(Ho − Es)−1 (7.182)

so that

kTS =2π

h

∑mS

|〈nT |VTS|mS〉|2

En − Es

(7.183)

where En−Es is the energy gap between the initial and final states including the electronic energygap. What this means is that if the energy difference between the bottoms of the respective wellsis large, then the initial state will be coupled to the high-lying vibrational states in So. Next Imake the “Condon Approximation” that

〈nT |VTS|mS〉 ≈ VTS〈nT |mS〉 (7.184)

where 〈nT |mS〉 is the overlap between vibrational state |nT 〉 in the T1 well and state |mS〉 in theSo well. These are called Franck-Condon factors.

Evaluation of Franck Condon Factors

Define the Franck-Condon factor as

〈nT |mS〉 =∫dxϕ(T )∗

m (x)ϕ(S)n (x) (7.185)

211

where ϕ(T )n (x) is the coordinate representation of a harmonic osc. state. We shall assume that

the two wells are off-set by xs and each has a freq. ω1 and ω2. We can write the HO state foreach well as a Gauss-Hermite Polynomial (c.f. Compliment Bv in the text.)

ϕn(x) =

(β2

π

)1/41√2nn!

exp

(−β

2

2x2

)Hn(βx) (7.186)

where β =√mω/h and Hn(z) is a Hermite polynomial,

Hn(z) = (−1)nez2 ∂n

∂zne−z2

(7.187)

Thus the FC factor is

〈nT |mS〉 =

(β2

T

π

)1/4 (β2

S

π

)1/41√2nn!

1√2mm!

×∫dx exp

(−β

2T

2x2

)exp

(−β

2S

2(x− xs)

2

)× Hni

(βix)Hmf(βf (x− xs)) (7.188)

This integral is pretty difficult to solve analytically. In fact Mathematica even choked on thisone. Let’s try a simplification: We can expand the Hermite Polynomials as

Hn(z) = 2n√π

[1

Γ(12− n

2)− 2x

Γ(−n2)− nx3

Γ(12− n

2)

+ · · ·]

(7.189)

Thus, any integral we want to do involves doing an integral of the form

In =∫dxxn exp

(−β

21

2x2 − β2

2

2(x− xs)

2

)

=∫dxxn exp

(−β

21

2x2 − β2

2

2(x2 − 2xxs + x2

s)

)

=∫dxxn exp

(−β

21 + β2

2

2x2

)exp

(β2

2

2(2xxs − x2

s)

)

= exp

(−β

22

2x2

s

)∫dxxn exp

(−β

21 + β2

2

2x2

)exp

(β2

2(xxs))

(7.190)

= exp

(−β

22

2x2

s

)∫dxxn exp

(−a

2x2)

exp (bx) (7.191)

where I defined a = β21 + β2

2 and b = β22xs .

Performing the integral:

In = 2−1+n

21

a

2+n2

(1 + (−1)n)

√1

aaΓ(

1 + n

2) 1F1(

1 + n

2,1

2,b2

2 a)

−√

2 (−1 + (−1)n) bΓ(2 + n

2) 1F1(

2 + n

2,3

2,b2

2 a)

)(7.192)

212

where 1F1(a, b, z) is the Hypergeometric function (c.f Landau and Lifshitz, QM) and Γ(z) is theGamma function. Not exactly the easist integral in the world to evaluate. (In other words, don’tworry about having to solve this integral on an exam!) To make matters even worse, this is onlyone term. In order to compute, say the FC factor between ni = 10 and mf = 12 I would need tosum over 120 terms! However, Mathematica knows how to evaluate these functions, and we canuse it to compute FC factors very easily.

If the harmonic frequencies are the same in each well, life gets much easier. Furthermore, If Imake the initial vibrational state in the T1 well be the ground vibrational state, we can evaluatethe overlap exactly for this case. The answer is (see Mathematica hand out)

M [n] =βnxn

s√2nn!

(7.193)

Note that this is different than the FCF calculated in Ref. 6 by Gelbart, et al. who do not havethe square-root factor (their denominator is my denominator squared.) 2

Finally, we can evaluate the matrix element as

〈nT |VTS|mS〉 ≈ VTSβnxn

s√2nn!

(7.194)

Thus, the GR survival rate for the ground vibrational state of the T1 surface is

k =2π

hVTS

∑m

1

hΩ− hωm

βnxns√

2nn!(7.195)

where Ω is the energy difference between the T1 and So potential minimuma.

“Steep So Approximation”

In this approximation we assume that potential well of the So state is very steep and intesectsthe T1 surface at xs. We also assume that the diabatic coupling is a function of x. Thus, the GRsurvival rate is

k =2π

h〈V 2

TS〉ρS (7.196)

where

〈VTS〉 =∫dxψ∗T (x)ψS(x)V (x) (7.197)

When the So surface is steeply repulsive, the wavefunction on the So surface will be very oscillitoryat the classical turning point, which is nearly identical with xs for very steep potentials. Thus,for purposes of doing integrations, we can assume that

ψS(x) = Cδ(x− xs) + · · · (7.198)

2I believe this is may be a mistake in their paper. I’ll have to call Karl Freed about this one.

213

where xs is the classical turning point at energy E on the So surface. The justification for thisis from the expansion of the “semi-classical” wavefunction on a linear potential, which are theAiry functions, Ai(x).

Ai(−ζ) =1

2π

∫−∞

+∞ds exp(is2/3− iζs) (7.199)

Which can be expanded as

aAi(ax) = δ(x) + · · · (7.200)

Expansions of this form also let us estimate the coefficient C 3

Using the δ-function approximation∫dxψ∗T (x)ψS(x)V (x) = C

∫dxψ∗T (x)δ(x− xs)V (x) = Cψ∗T (xs)V (xs) (7.201)

Now, again assuming that we are in the ground vibrational state on the T1 surface,

ψT (x) =

(β2

π

)1/4

e−β2x2/2 (7.202)

∫dxψ∗T (x)ψS(x)V (x) = C

(β2

π

)1/4

e−β2x2s/2V (xs) (7.203)

So, we get the approximation:

k =2π

hCV (xs)

(β2

π

)1/4

e−β2x2s/2 1

hΩ(7.204)

where C remains to be determined. For that, refer to the Heller-Brown paper.

Time-Dependent Semi-Classical Evaluation

We next do something tricky. There are a number of ways one can represent the δ-function. Wewill use the Fourier representation of the function and write:

δ(Ei − Ef ) =h

2π

∫ ∞

−∞dtei/h(Ei−Ej)t (7.205)

Thus, we can write

kif =∫ ∞

−∞dt∑f

〈αi(R)I(R)|V |αf (R)F (R)〉

× 〈αf (R)F (R)|e+iHf t/hV e−iHit/h|αi(R)I(R)〉 (7.206)

3(see Heller and Brown, JCP 79 3336 (1983). )

214

Integrating over the electronic degrees of freedom, we can define

Vif (R) = 〈αi(R)|V |αf (R)〉 (7.207)

and thus write

kif =∫ ∞

−∞dt∑f

〈I(R)|Vif (R)|F (R)〉

× 〈F (R)|e+iHf t/hVif (R)e−iHit/h|I(R)〉 (7.208)

whereHi(R) is the nuclear Hamiltonian for the initial state andHf (R) is the nuclear Hamiltonianfor the final state.

At this point I can remove the sum over f and obtain

kif =∫ ∞

−∞dt〈I(R)|Vif (R)e+iHf t/hVif (R)e−iHit/h|I(R)〉. (7.209)

Next, we note that

Vif (R) = R · 〈αf (R)|ih∇R|αi(R)〉= R(0) ·Dif (R(0)) (7.210)

is proportional to the nuclear velocity at the initial time. Likewise, the term in the middlerepresents the non-adiabatic coupling at some later time. Thus, we can re-write the transitionrate as

kif =∫ ∞

−∞dt(R(0) ·Dif (R(0)))(R(t) ·Dif (R(t))) (7.211)

× 〈I(R)|e+iHf t/he−iHit/h|I(R)〉. (7.212)

Finally, we do an ensemble averate over the initial positions of the nuclei and obtain almost thefinal result:

kif =∫ ∞

−∞dt⟨(R(0) ·Dif (R(0)))(R(t) ·Dif (R(t)))Jif (t)

⟩. (7.213)

where I define

Jif (t) = 〈I(R)|e+iHf t/he−iHit/h|I(R)〉 (7.214)

This term represents the evolution of the initial nuclear vibrational state moving forward in timeon the initial energy surface

|I(R(t))〉 = e−i/hHit|I(R(0))〉 (7.215)

and backwards in time on the final energy surface.

〈I(R(t))| = 〈I(R(0))|ei/hHf t (7.216)

215

So, J(t) represents the time-dependent overlap integral between nuclear wavefunctions evolvingon the different potential energy surfaces.

Let us assume that the potential wells are Harmonic with the centers off set by some amountxs. We can define in each well a set of Harmonic Oscillator eigenstates, which we’ll write in shorthand as |ni〉 where the subscript i denotes the electronic state. At time t = 0, we can expandthe initial nuclear wavefunction as a superposition of these states:

|I(R(0))〉 =∑n

γn|ni〉 (7.217)

where γn = 〈ni|I(R(0))〉. The time evolution in the well is

|I(R(t))〉 =∑n

γn exp(−i/2(n+ 1)ωit)|ni〉. (7.218)

We can also express the evolution of the ket as a superposition of states in the other well.

〈I(R(t))| =∑m

ξ∗m exp(+i/2(m+ 1)ωf t)〈mf | (7.219)

where ξm = 〈mf |I(R)〉 are the coefficients. Thus, J(t) is obtained by hooking the two resultstogether:

J(t) =∑mn

ξ∗mγne+i/2(m+1)ωf te−i/2(n+1)ωit〈mf |ni〉 (7.220)

Now, we must compute the overlap between harmonic states in one well with harmonic states inanother well. This type of overlap is termed a Franck-Condon factor (FC). We will evaluate theFC factor using two different approaches.

7.7.2 Semi-Classical Evaluation

I want to make a series of simplifying assumptions to the nuclear wavefunction. Many of theseassumptions follow from Heller’s paper referenced at the beginning. The assumptions are asfollows:

1. At the initial time, 〈x|I(R)〉 can be written as a Gaussian of width β centered about R.

〈x|I(R)〉 =

(β2

π

)1/4

exp

(−β

2

2(x−R(t))2 +

i

hp(t)(x−R(t))

)(7.221)

where p(t) is the classical momentum. (c.f Heller)

2. We know that for a Gaussian wavepacket (especially one in a Harmonic Well), the centerof the Gaussian tracks the classical prediction. Thus, we can write that R(t) is the centerof the Gaussian evolves under Newton’s equation

mR(t) = Fi(R) (7.222)

where Fi(R) is the force computed as the derivative of the ith energy surface w.r.t. R,evaluated at the current nuclear position (i.e. the force we would get using the Born-Oppenheimer approximation.

Fi(R) = − ∂

∂RE(R). (7.223)

216

3. At t = 0, the initial classical velocities and positions of the nuclear waves evolving on thei and f surface are the same.

4. For very short times, we assume that the wave does not spread appreciably. We can fixthis assuption very easily if need be.

Using these assumptions, we can approximate the time-dependent FC factor as4

J(t) ≈ exp

[−β

2

4(Rf (t)−Ri(t))

2

]

× exp

[− 1

4β2h2 (pf (t)− pi(t))2

]

× exp[+i

2h(Rf (t)−Ri(t)) · (pj(t)− pi(t))

](7.224)

Next, we expand Ri(t) and pi(t) as a Tayler series about t = 0.

Ri(t) = Ri(0) + tRi(0) + t2Ri(0)

2!+ · · · (7.225)

Using Newton’s Equation:

Ri(t) = Ri(0) + tp(0)

m− t2

Fi(0)

m2!+ · · · (7.226)

pi(t) = pi(0) + Fi(0)t+1

2

∂Fi

∂tt2 · · · (7.227)

Thus the difference between the nuclear positions after a short amount of time will be

Ri(t)−Rf (t) ≈ −t2(Fi(0)

2m− Fj(0)

2m

)+ · · · (7.228)

Also, the momentum difference is

pf (t)− pi(t) ≈ (Ff (0)− Fi(0))t+ · · · (7.229)

Thus,

J(t) ≈ exp

[− β2t4

16m2(Fi(0)− Fj(0))

2

]

× exp

[− t2

4β2h2 ((Ff (0)− Fi(0))2

]

× exp[+

i

4mh(Ff (0)− Fi(0))t

3]

(7.230)

If we include the oscillitory term, the integral does not converge (so much for a short timeapprox.) However, when we do the ensemble average, each member of the ensemble contributes

4B. J. Schwartz, E. R. Bittner, O. V. Prezhdo, and P. J. Rossky, J. Chem. Phys. 104, 5242 (1996).

217

a slightly different phase contribution, so, we can safely ignore it. Furthermore, for short times,the decay of the overlap will be dominated by the term proportional to t2. Thus, we defined theapproximate decay curve as

J(t) = exp

(− t2

4β2h2 (Fj(0)− Fi(0))2

)(7.231)

Now, pulling everything together, we write the GR rate constant as

kif =∫ ∞

−∞dt⟨(R(0) ·Dif (R(0)))(R(t) ·Dif (R(t)))

× exp

(− t2

4β2h2 (Fj(0)− Fi(0))2

)⟩. (7.232)

The assumptions are that the overlap decays more rapidly than the oscillations in the autocor-relation of the matrix element. This actually bears itself out in reality.

Let’s assume that the overlap decay and the correlation function are un-correlated. (CondonApproximation) Under this we can write:

kif =∫ ∞

−∞dt⟨(R(0) ·Dif (R(0)))(R(t) ·Dif (R(t)))

⟩×

⟨exp

(− t2

4β2h2 (Fj(0)− Fi(0))2

)⟩. (7.233)

or defining

Cif (t) =⟨(R(0) ·Dif (R(0)))(R(t) ·Dif (R(t)))

⟩(7.234)

and using (c.f. Chandler, Stat. Mech.) ⟨eA⟩

= e〈A〉, (7.235)

the desired result is

kif =∫ ∞

−∞dtCif (t) exp

(−t2

⟨(Fj(0)− Fi(0))

2

4β2h2

⟩). (7.236)

Now, let’s assume that the correlation function is an oscillitory function of time.

Cif (t) = |Vif |2 cos(Ωt) (7.237)

Then

kif =∫ ∞

−∞dt|Vif |2 cos(Ωt) exp

(−t2

⟨(Fj(0)− Fi(0))

2

4β2h2

⟩). (7.238)

=

√π

be−Ω2/(4b)|Vif |2 (7.239)

218

where

b =

⟨(Fj(0)− Fi(0))

2

4β2h2

⟩(7.240)

In Ref 1. we used this equation (actually one a few lines up) to compute the non-radiativerelaxation rates between the p to s state of an aqueous electron in H20 and in D20 to estimatethe isotopic dependency of the transition rate. Briefly, the story goes as this. Paul Barbara’sgroup at U. Minnisota and Yann Gauduel’s group in France measured the fluorescence decay ofan excited excess electron in H20 and in D20 and noted that there was no resolvable differencebetween the two solvents. I.e. the non-radiative decay was not at all sensitive to isotopic changesin the solvent. (The experimental life-times are roughly 310 fs for both sovents with resolution ofabout 80 fs ) This is very surprising since, looking at the non-adiabatic coupling operator above,you will notice that the matrix element coupling states is proportional to the nuclear velocities.(The electronic matrix element is between the s and p state of the electron.) Thus, since thevelocity of a proton is roughly

√2 times faster than that of a deuteron of the same kinetic energy,

The non-adiabatic coupling matrix element between the s and p states in water should be twicethat as in heavy-water. Thus, the transition rate in water should be roughtly twice that as inheavy-water. It turns out that since the D’s move slower than the H’s, the nuclear overlap decaysroughly twice as slowly. Thus we get competing factors of two which cancel out.

7.8 Problems and Exercises

Exercise 7.1 A one dimensional harmonic oscillator, with frequency ω, in its ground state issubjected to a perturbation of the form

H ′(t) = Cpe−α|t| cos(Ωt) (7.241)

where p is the momentum operator and C, α, and Ω are constants. What is the probability thatas t → ∞ the oscillator will be found in its first excited state in first-order perturbation theory.Discuss the result as a function of Ω, ω, and α.

Exercise 7.2 A particle is in a one-dimensional infinite well of width 2a. A time-dependentperturbation of the form

H ′(t) = ToVo sin(πx

a)δ(t) (7.242)

acts on the system, where To and Vo are constants. What is the probability that the system willbe in the first excited state afterwards?

Exercise 7.3 Because of the finite size of the nucleus, the actual potential seen by the electronis more like:

219

0.02 0.04 0.06 0.08 0.1r(Bohr)

-120

-100

-80

-60

-40

-20

V(hartree)

1. Calculate this effect on the ground state energy of the H atom using first order perturbationtheory with

H ′ =

e2

r− e3

Rfor r ≤ R

0 otherwise(7.243)

2. Explain this choice for H ′.

3. Expand your results in powers of R/ao 1. (Be careful!)

4. Evaluate numerically your result for R = 1 fm and R = 100 fm.

5. Give the fractional shift of the energy of the ground state.

6. A more rigorous approach is to take into account the fact that the nucleus has a homoge-neous charge distribution. In this case, the potential energy experienced by the electron goesas

V (r) = −Ze2

r

when r > R and

V (r) = −Ze2

r

(1

2R

((r

R

)2

+ 2R

r− 3

)− 1

)for r ≤ R. What is the perturbation in this case? Calculate the energy shift for the H (1s)

220

energy level for R = 1fm and compare to the result you obtained above.

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Note that this effect is the “isotope shift” and can be observed in the spectral lines of the heavyelements.

221

Chapter 8

Many Body Quantum Mechanics

It is often stated that of all the theories proposed in this century, the silliest isquantum theory. In fact, some say that the only thing that quantum theory hasgoing for it is that it is unquestionably correct.–M. Kaku (Hyperspace, Oxford University Press, 1995)

8.1 Symmetry with respect to particle Exchange

Up to this point we have primarily dealt with quantum mechanical system for 1 particle or withsystems of distinguishable particles. By distinguishable we mean that one can assign a uniquelabel to the particle which distinguishes it from the other particles in the system. Electrons inmolecules and other systems, are identical and cannot be assigned a unique label. Thus, we mustconcern our selves with the consequences of exchanging the labels we use. To establish a firmformalism and notation, we shall write the many-particle wavefunction for a system as

〈ψN |ψN〉 =∫· · ·

∫d3r1 · · · d3rN |ψN(r1, r2, · · · , rN)|2 < +∞ (8.1)

=∫· · ·

∫d1d2 · · · dN |ψN(1, 2, · · · , N)|2. (8.2)

We will define the N -particle state space as the product of the individual single particle statespaces thusly

|ψN〉 = |a1a2 · · · aN) = |a1〉 ⊗ |a2〉 ⊗ · · · ⊗ |aN〉 (8.3)

For future reference, we will write the multiparticle state as with a curved bracket: | · · ·). Thesestates have wavefunctions

〈r|ψN〉 = (r1 · · · rn|a1a2 · · · aN) = 〈r1|a1〉〈r2|a2〉 · · · 〈rN |aN〉 (8.4)

= φa1(r1)φa1(r2) · · ·φaN(rN) (8.5)

222

These states obey analogous rules for constructing overlap (projections) and idempotent relations.They form a complete set of states (hence form a basis) and any multi-particle state in the statespace can be constructed as a linear combination of the basis states.

Thus far we have not taken into account the symmetry property of the wavefunction. Thereare a multitude of possible states which one can construct using the states we defined above.However, only symmetric and antisymmetric combinations of these state are actually observed innature. Particles occurring in symmetric or anti-symmetric states are called Bosons and Fermionsrespectively.

Let’s define the permutation operator Pαβ which swaps the positions of particles α and β.e.g.

P12|1, 2) = |2, 1) (8.6)

Also,

P12P12ψ(1, 2) = P 212ψ(1, 2) = ψ(1, 2) (8.7)

thus ψ(1, 2) is an eigenstate of P12 with eigenvalue ±1. In other words, we can also write

P12ψ(1, 2) = ζψ(1, 2) (8.8)

where ζ = ±1.A wavefunction of N bosons is totally symmetric and thus satisfies

ψ(P1, P2, · · · , PN) = ψ(1, 2, · · · , N) (8.9)

where (P1, P2, · · · , PN) represents any permutation P of the set (1, 2, · · · , N). A wavefunctionof N fermions is totally antisymmetric and thus satisfies

ψ(P1, P2, · · · , PN) = (−1)Pψ(1, 2, · · · , N). (8.10)

Here, (−1)P denotes the sign or parity of the permutation and is defined as the number of binarytranspositions which brings the permutation (P1, P2, ...) to its original from (1, 2, 3...).

For example: what is the parity of the permutation (4,3,5,2,1)? A sequence of binary trans-positions is

(4, 3, 5, 2, 1) → (2, 3, 5, 4, 1) → (3, 2, 5, 4, 1) → (5, 2, 3, 4, 1) → (1, 2, 3, 4, 5) (8.11)

So P = 4. Thus, for a system of 5 fermions

ψ(4, 3, 5, 2, 1) = ψ(1, 2, 3, 4, 5) (8.12)

In cases where we want to develop the many-body theory for both Fermions and Bosonssimultaneously, we will adopt the notation that ζ = ±1 and any wavefunction can be written as

ψ(P1, P2, · · · , PN) = (ζ)Pψ(1, 2, · · · , N). (8.13)

where ζ = −1 for fermions and +1 for bosons.

223

While these symmetry requirement are observed in nature, they can also be derived in thecontext of quantum field theory that given general assumptions of locality, causality and Lorentzinvariance, particles with half integer spin are fermions and those with integer spin are bosons.Some examples of bosons are photon, photons, pions, mesons, gluons and the 4He atom. Someexamples of fermions are protons, electrons, neutrons, muons, neutrinos, quarks, and the 3Heatom. Composite particles composes of any number of bosons and even or odd numbers offermions behave as bosons or fermions respectively at temperatures low compared to their bindingenergy. An example of this is super-conductivity where electron-phonon coupling induces thepairing of electrons (Cooper pairs) which form a Bose-condensate.

Now, consider what happens if I place two fermion particles in the same state:

|ψ(1, 2)〉 = |α(1)α(2)) (8.14)

where α(1) is a state with the “spin up” quantum number. This state must be an eigenstate ofthe permutation operator with eigenvalue ζ = −1.

P12|ψ(1, 2)〉 = −|α(2)α(1)) (8.15)

However, |α(1)α(2)) = |α(2)α(1)), thus the wavefunction of the state must vanish everywhere.For the general case of a system with N particles, the normalized wavefunction is

ψ =

(N1!N2!...

N !

)1/2∑ψp1(1)ψp2(2) · · ·ψpN(N) (8.16)

where the sum is over all permutations of different p1, p2... and the numbers Ni indicate howmany of these have the same value (i.e. how many particles are in each state) with

∑Ni = N .

For a system of 2 fermions, the wavefunction is

ψ(1, 2) = (ψp1(1)ψp2(2)− ψp1(2)ψp2(1))/√

2 (8.17)

Thus, in the example above:

ψ(1, 2) = (α(1)α(2)− α(2)α(1))/√

2 = 0 (8.18)

Likewise,

ψ(1, 2) = (β(1)α(2)− β(2)α(1))/√

2

= (β(1)α(2)− P12β(1)α(2))/√

2

= (β(1)α(2))/√

2 (8.19)

We will write such symmetrized states using the curly brackets

|ψ = |a1a2 · · · aN =

(N1!N2!...

N !

)1/2∑ψp1(1)ψp2(2) · · ·ψpN(N) (8.20)

For the general case of N particles, the fully anti-symmetrized form of the wavefunction takesthe form of a determinant

ψ =1√N !

∣∣∣∣∣∣∣φa(1) φa(1) φa(1)φb(1) φb(2) φb(2)φc(1) φc(3) φc(3)

∣∣∣∣∣∣∣ (8.21)

224

where the columns represent the particles and the rows are the different states. The interchangeof any two particles corresponds to the interchange of two columns, as a result, the determinantchanges sign. Consequently, if two rows are identical, corresponding to two particles occupyingthe same state, the determinant vanishes.

Another example, let’s consider the possible states for the He atom ground state. Let’sassume that the ground state wavefunction is the product of two single particle hydrogenic 1sstates with a spin wavefunction written thus

|ψ〉 = |1s(1)α(1), 1s(2)β(2)) (8.22)

Let’s denote |α〉 as the spin up state and |β〉 as the spin down state. We have the followingpossible spin combinations:

α(1)α(2) ↑↑ symmetricα(1)β(2) ↑↓ netherβ(1)α(2) ↓↑ neitherβ(1)β(2) ↓↓ symmetric

(8.23)

The ↑↑ and the ↓↓ states are clearly symmetric w.r.t particle exchanges. However, note thatthe other two are neither symmetric nor anti-symmetric. Since we can construct linear combina-tions of these states, we can use the two allowed spin configurations to define the combined spinstate Thus, we get two possible total spin states:

1√2(|α(1)β(2))± |β(1)α(2))) (8.24)

Thus, the possible two particle spin states are

α(1)α(2) ↑↑ symmetricβ(1)β(2) ↓↓ symmetric

1√2(α(1)β(2) + β(1)α(2)) ↑↓ + ↑↓ symmetric

1√2(α(1)β(2)− β(1)α(2)) ↑↓ − ↑↓ anti-symmetric

(8.25)

These spin states multiply the spatial state and the full wavefunction must be anti-symmetricw.r.t. exchange. For example, for the ground state of the He atom, the zero-th order spatialstate is |1s(1)1s(2)). This is symmetric w.r.t. exchange. Thus, the full ground-state wavefunctionmust be the product

|ψ〉 = |1s(1)1s(2))1√2(α(1)β(2)− β(1)α(2) (8.26)

The full state is an eigenstate of P12 with eigenvalue -1, which is correct for a system of fermions.What about the other states, where can we use them? What if we could construct a spatial

wavefunction that was anti-symmetric w.r.t particle exchange. Consider the first excited stateof He. The electron configuration for this state is

|1s(1)2s(2)) (8.27)

225

However, we could have also written

|1s(2)2s(1)) (8.28)

Taking the symmetric and anti-symmetric combinations

|ψ±〉 =1√2(|1s(1)2s(2))± |1s(2)2s(1))) (8.29)

The + state is symmetric w.r.t. particle exchange. Thus, the full state (including spin) must be

|ψ1〉 =1

2(|1s(1)2s(2)) + |1s(2)2s(1)))(α(1)β(2)− β(1)α(2)). (8.30)

The other three states must be

|ψ2〉 =1

2(|1s(1)2s(2))− |1s(2)2s(1)))(α(1)β(2) + β(1)α(2)) (8.31)

|ψ3〉 =1√2(|1s(1)2s(2))− |1s(2)2s(1)))(α(1)α(2)) (8.32)

|ψ4〉 =1√2(|1s(1)2s(2))− |1s(2)2s(1)))(β(1)β(2)). (8.33)

These states can also be constructed using the determinant wavefunction. For example, theground state configurations are generated using

|ψg =1√2

∣∣∣∣∣ 1s(1)α(1) 1s(1)β(1)1s(2)α(2) 1s(2)β(2)

∣∣∣∣∣ (8.34)

=1√2|1s(1)1s(2))[α(1)β(2)− α(2)β(1)] (8.35)

Likewise for the excited states, we have 4 possible determinant states

|ψ1 =1√2

∣∣∣∣∣ 1s(1)α(1) 2s(1)α(1)1s(2)α(2) 2s(2)α(2)

∣∣∣∣∣|ψ2 =

1√2

∣∣∣∣∣ 1s(1)α(1) 2s(1)β(1)1s(2)α(2) 2s(2)β(2)

∣∣∣∣∣|ψ3 =

1√2

∣∣∣∣∣ 1s(1)β(1) 2s(1)α(1)1s(2)β(2) 2s(2)α(2)

∣∣∣∣∣|ψ4 =

1√2

∣∣∣∣∣ 1s(1)β(1) 2s(1)β(1)1s(2)β(2) 2s(2)β(2)

∣∣∣∣∣ (8.36)

The |ψm) are related to the determinant states as follows

|ψ1 =1√2

[1s(1)α(1)2s(2)α(2)− 1s(2)α(2)2s(1)α(1)]

=1√2

[1s(1)2s(2)− 1s(2)2s(1)]α(1)α(2)

= |ψ3

|ψ4 =1√2

[1s(1)2s(2)− 1s(2)2s(1)] β(1)β(2) = |ψ4 (8.37)

226

The remaining two must be constructed from linear combinations of the determinant states:

|ψ2 =1√2

[1s(1)α(1)2s(2)β(2)− 1s(2)α(2)2s(1)β(1)] (8.38)

|ψ3 =1√2

[1s(1)β(1)2s(2)α(2)− 1s(2)β(2)2s(1)α(1)] (8.39)

|ψ2〉 =1√2

[|ψ2+ |ψ3] (8.40)

|ψ1〉 =1√2

[|ψ2 − |ψ3] (8.41)

When dealing with spin functions, a short hand notation is often used to reduced the notationa bit. The notation

|1s〉 ≡ |1sα〉 (8.42)

is used to denote a spin up state and

|1s〉 ≡ |1sβ〉 (8.43)

Using these, the above determinant functions can be written as

|ψ1 =1√2

∣∣∣∣∣ 1s(1) 2s(1)1s(2) 2s(2)

∣∣∣∣∣ (8.44)

=1√2

[|1s(1)2s(2))− |1s(2)2s(1))

](8.45)

|ψ2 =1√2

∣∣∣∣∣ 1s(1) 2s(1)1s(2) 2s(2)

∣∣∣∣∣ (8.46)

=1√2

[|1s(1)2s(2))− |1s(2)2s(1))

](8.47)

The symmetrization principal for fermions is often expressed as the Pauli exclusion principlewhich states: no two fermions can occupy the same same state at the same time. This, as we allwell know gives rise to the periodic table and is the basis of all atomic structure.

8.2 Matrix Elements of Electronic Operators

We can write the Hamiltonian for a N -body problem as follows. Say our N -body Hamiltonianconsists of a sum of N single particle Hamiltonian, Ho, and two body interactions.

H =N∑n

H(n)o +

∑i6=j

V (ri − rj) (8.48)

Using the field operators, the expectation values of the Ho terms are∑α

∫dxφ∗α(x)Hoφα(x) =

∑α

nαWα (8.49)

227

since φα(x) is an eigenstate of Ho with eigenvalue Eα. α should be regards as a collection of allquantum numbers used to describe the Ho eigenstate.

For example, say we want a zeroth order approximation to the ground state of He and weuse hydrogenic functions,

|ψo〉 = |1s(2)1s(2)). (8.50)

This state is symmetric w.r.t electron exchange, so the spin component must be anti-symmetric.For now this will not contribute the the calculation. The zero-th order Schroedinger Equation is

(Ho(1) +Ho(2))|ψo〉 = Eo|ψo〉 (8.51)

Where Ho(1) is the zero-th order Hamiltonian for particle 1. This is easy to solve

(Ho(1) +Ho(2))|ψo〉 = −Z2|ψo〉 (8.52)

Z = 2, so the zero-th order guess to the He ground state energy is −4 (in Hartree units, recall 1Hartree = 27.6 eV). The correct ground state energy is more like -2.90 Hartree. Let’s now evaluateto first order in perturbation theory the direct Coulombic interaction between the electrons.

E(1) = (1s(1)1s(2)| 1

r12

|1s(1)1s(2)) (8.53)

The spatial wavefunction for the |1s(1)1s(2)) state is the product of two hydrogenic functions

(r1r2|1s(1)1s(2)) =Z3

πe−Z(r1+r2) (8.54)

Therefore,

E(1) =∫dV1

∫dV2

1

r12

Z3

πe−2Z(r1+r2) (8.55)

where dV = 4πr2dr is the volume element. This integral is most readily solved if we insteadwrite it as the energy of a charge distribution, ρ(2) = |ψ(2)|2, in the field of a charge distributionρ(1) = |ψ(1)|2 for r2 > r1.

E(1) = 2Z3

π

∫dV2e

−2Zr21

r2

∫ r2

0dV1e

−2Zr1 (8.56)

The factor of 2 takes into account that we get the same result for when r1 > r2. Doing theintegral, (See subsection 8.3.1.)

E(1) =5Z

8(8.57)

or 1.25 Hartee. Thus, for the He atom

E = Eo + E(1) = −Z2 +5Z

8= −4 + 1.25 = −2.75 (8.58)

228

Not too bad, the actual result is −2.90 Hartee.What we have not taken into consideration that there is an additional contribution to the

energy from the exchange interaction. In other words, we need to compute the Coulomb integralexchanging electron 1 with electron 2. We really need to compute the perturbation energy withrespect the determinant wavefunction.

E(1) = 1s(1)1s(2)|v|1s(1)1s(2) (8.59)

=1

2

((1s(1)1s(2)| − (1s(1)1s(2)|)

)v(|1s(1)1s(2))− |1s(1)1s(2))

)(8.60)

=1

2

((1s(1)1s(2)|v|1s(1)1s(2))− (1s(1)1s(2)|v|1s(1)1s(2))

− (1s(1)1s(2)|v|1s(1)1s(2)) + (1s(1)1s(2)|v|1s(1)1s(2)))

(8.61)

= 2(1s(1)1s(2)|v|1s(1)1s(2))− (1s(1)1s(2)|v|1s(1)1s(2)) (8.62)

However, the potential does not depend upon spin. Thus any matrix element which exchangesa spin from must vanish. This, we have no exchange contribution to the energy. We can in factmove on to higher orders in perturbation theory and solve accordingly.

8.3 The Hartree-Fock Approximation

We just saw how to estimate the ground state energy of a system in the presence of interactionsusing first order perturbation theory. To get this result, we assumed that the zeroth-orderwavefunctions were pretty good and calculated our results using these wavefunctions. Of course,the true ground state energy is obtained by summing over all diagrams in the perturbationexpansion:

E −W = 〈ψo|E −W |ψo〉 (8.63)

The second order term contains explicit two-body correlation interactions. i.e. the motion of oneelectron affects the motion of the other electron.

Let’s make a rather bold assumption that we can exclude connected two body interactionsand treat the electrons as independent but moving in an averaged field of the other electrons.

First we make some standard definitions:

Jαβ = (αβ|v|αβ) (8.64)

Kαβ = (αβ|v|βα) (8.65)

And write that

EHF −W =∑αβ

(2Jαβ −Kαβ) (8.66)

where sum over all occupied (n/2) spatial orbitals of a n−electron system. The J integral is thedirect interaction (Coulomb integral) and the K is the exchange interaction.

229

We now look for a a set of orbitals which minimize the variational integral EHF , subject to theconstrain that the wavefunction solutions be orthogonal. One can show (rather straightforwardly)that if we write the Hamiltonian as a functional of the electron density, ρ,

H[ρ] = Ho[1, 2] +∫ ∫

d3d412|v|34ρ(3, 4) (8.67)

= Ho[1] + 2J(1)−K(1) (8.68)

where ρ(1, 2) = φ(1)∗φ(2),

J(1) =∫d2|β(2)|2v(12) (8.69)

K(1) =∫d2α(2)∗β(2)v(12) (8.70)

The Hartree-Fock wavefunctions satisfy

H[ρ]ψ(1) = E(1)ψ(1) (8.71)

In other words, we diagonalize the Fock-matrix H[ρ] given an initial guess of the electron density.This gives a new set of electron orbitals, which we use to construct a new guess for the electrondensities. This procedure is iterated to convergence.

8.3.1 Two electron integrals

One of the difficulties encountered is in evaluating the Jαβ and Kαβ two electron integrals. Let’stake the case that the φα and φβ orbitals are centred on the same atom. Two centred terms canbe evaluated, but the analysis is more difficult. Writing the J integral in Eq. 8.65 out explicitlywe have:

Jαβ = (φα(1)φβ(2)|v(12)|φα(1)φβ(1))

=∫ ∫

φ∗α(1)φ∗β(2)v(1− 2)φα(1)φ2βd1d2

=∫φ∗α(r1)φα(r1)

∫φ∗β(r2)φβ(r2)v(r1 − r2)dr2dr1 (8.72)

If we can factor the single particle orbitals as φα(r, θ, φ) = Rnl(r)Ylm(θ, φ), then we can separatethe radial and angular integrals. Before we do that, we have to resolve the pair-interaction intoradial and angular components as well. For the Coulomb potential, we can use the expansion

1

|r1 − r2|=∑l=0

+l∑m=−l

4π

2l + 1

rl<

rl+1<

Y ∗lm(θ1φ1)Ylm(θ2φ2) (8.73)

where the notation r< denotes which is the smaller of r1 and r2 and r> the greater. For thehydrogen 1s orbitals (normalized and in atomic units)

φ1s =

√Z3

πe−Zr, (8.74)

230

the J integral for the 1s1s configuration is

J =Z6

π2

∫d3r1

∫d3r2e

−2Zr1e−2Zr21

r12

(8.75)

Inserting the expansion and using Y00 = 1/√

4π,

J = 16Z6∑

l

l∑m=−l

1

2l + 1

∫ ∞

0

∫ ∞

0e−2Zr1e−2Zr2

rl<

rl+1<

r21dr1r

22dr2

×∫Y ∗

lm(1)Y00(1)dΩ1

∫Ylm(2)Y ∗

00(2)dΩ2. (8.76)

The last two integrals are easy due to the orthogonality of the spherical harmonics. This leavesthe double integral,

J = 16Z6∫ ∞

0

∫ ∞

0e−2Zr1e−2Zr2

1

r>

r21dr1r

22dr2 (8.77)

which we evaluate by splitting into two parts,

J = 16Z2∫ ∞

0e−2Zr1r1

(∫ r1

0e−2Zr2r2

2dr2

)dr1

+∫ ∞

0e−2Zr1r2

1

(∫ ∞

r1

e−2Zr2r2dr2

)dr1

(8.78)

In this case the integrals are easy to evaluate and

J =5Z

8(8.79)

8.3.2 Koopman’s Theorem

Koopman’s theorem states that if the single particle energies are not affected by adding orremoving a single electron, then the ionization energy is energy of the highest occupied singleparticle orbital (the HOMO) and the electron affinity is the energy of the lowest unoccupiedorbital (i.e. the LUMO). For the Hartree-Fock orbitals, this is theorem can be proven to beexact since correlations cancel out at the HF level. For small molecules and atoms, the theoremfails miserably since correlations play a significant role. On the other hand, for large polyatomicmolecules, Koopman’s theorem is extremely useful in predicting ionization energies and spectra.From a physical point of view, the theorem is never exact since it discounts relaxation of boththe electrons and the nuclei.

8.4 Quantum Chemistry

Quantum chemical concepts play a crucial role in how we think about and describe chemicalprocesses. In particular, the term quantum chemistry usually denotes the field of electronicstructure theory. There is no possible way to cover this field to any depth in a single course andthis one section will certainly not prepare anyone for doing research in quantum chemistry. Thetopic itself can be divided into two sub-fields:

231

• Method development: The development and implementation of new theories and com-putational strategies to take advantage of the increaseing power of computational hardware.(Bigger, stronger, faster calculations)

• Application: The use of established methods for developing theoretical models of chemicalprocesses

Here we will go in to a brief bit of detail into various levels of theory and their implementationin standard quantum chemical packages. For more in depth coverage, refer to

1. Quantum Chemistry, Ira Levine. The updated version of this text has a nice overview ofmethods, basis sets, theories, and approaches for quantum chemistry.

2. Modern Quanutm Chemistry, A. Szabo and N. S. Ostlund.

3. Ab Initio Molecular Orbital Theory, W. J. Hehre, L. Radom, P. v. R. Schleyer, and J. A.Pople.

4. Introduction to Quantum Mechanics in Chemistry, M. Ratner and G. Schatz.

8.4.1 The Born-Oppenheimer Approximation

The fundimental approximation in quantum chemistry is the Born Oppenheimer approxima-tion we discussed earlier. The idea is that because the mass of an electron is at least 10−4 that ofa typical nuclei, the motion of the nuclei can be effectively ignored and we can write an electronicSchrodinger equation in the field of fixed nuclei. If we write r for electronic coordinates and Rfor the nuclear coordinates, the complete electronic/nuclear wavefunction becomes

Ψ(r, R) = ψ(r;R)χ(R) (8.80)

where ψ(r;R) is the electronic part and χ(R) the nuclear part. The full Hamiltonian is

H = Tn + Te + Ven + Vnn + Vee (8.81)

Tn is the nuclear kinetic energy, Te is the electronic kinetic energy, and the V ’s are the electron-nuclear, nuclear-nuclear, and electron-electron Coulomb potential interactions. We want Ψ to bea solution of the Schrodinger equation,

HΨ = (Tn + Te + Ven + Vnn + Vee)ψχ = Eψχ. (8.82)

So, we divide through by ψχ and take advantage of the fact that Te does not depend upon thenuclear component of the total wavefunction

1

ψχTnψχ+

1

ψTeψ + Ven + Vnn + Vee = E.

On the other hand, Tn operates on both components, and involves terms which look like

Tnψχ =∑n

− 1

2Mn

(ψ∇2nχ+ χ∇2

nψ + 2∇χ · ∇nψ)

232

where the sum is over the nuclei and ∇n is the gradient with respect to nuclear position n. Thecrudest approximation we can make is to neglect the last two terms–those which involve thederivatives of the electronic wave with respect to the nuclear coordinate. So, when we neglectthose terms, the Schrodinger equation is almost separable into nuclear and electronic terms

1

χ(Tn + Vn)χ+

1

ψ(Te + Ven + Vee)ψ = E. (8.83)

The equation is not really separtable since the second term depends upon the nuclear position.So, what we do is say that the electronic part depends parametrically upon the nuclear positiongiving a constant term, Eel(R), that is a function of R.

(Te + Ven(R) + Vee)ψ(r;R) = Eel(R)ψ(r;R). (8.84)

The function, Eel, depends upon the particular electronic state. Since it an eigenvalue of Eq. 8.84,there may be a manifold of these functions stacked upon each other.

Turning towards the nuclear part, we have the nuclear Schrodinger equation

(Tn + Vn(R) + E(α)el (R))χ = Wχ. (8.85)

Here, the potential governing the nuclear motion contains the electronic contribution, E(α)el (R),

which is the αth eigenvalue of Eq. 8.84 and the nuclear repulsion energy Vn. Taken together,these form a potential energy surface

V (R) = Vn + E(α)el (R)

for the nuclear motion. Thus, the electronic energy serves as the interaction potential betweenthe nuclei and the motion of the nuclei occurs on an energy surface generated by the electronicstate.

Exercise 8.1 Derive the diagonal non-adiabatic correction term 〈ψ|Tn|ψ > to produce a slightlymore accurate potential energy surface

V (R) = Vn + E(α)el + 〈ψα|Tn|ψα〉

.

The BO approximation breaks down when the nuclear motion becomes very fast and theelectronic states can become coupled via the nuclear kinetic energy operator. (One can see byinspection that the electonic states are not eigenstates of the nuclear kinetic energy since Hele

does not commute with ∇2N . )

Let’s assume that the nuclear motion is a time dependent quantity, R(t). Now, take the timederivative of |Ψe

n〉

d

dt|Ψe

n(R(t))〉 =∂R(t)

∂t

∂

∂R|Ψe

n(R)〉+∂

∂t|Ψe

n(R(t))〉 (8.86)

Now, multiply on the left by < Ψem(R(t))| where m 6= n

〈Ψem(R(t))| d

dt|Ψe

n(R(t))〉 =∂R(t)

∂t〈Ψe

m(R(t))| ∂∂R

|Ψen(R)〉 (8.87)

233

Cleaning things up,

〈Ψem(R(t))| d

dt|Ψe

n(R(t))〉 = R(t)· < Ψem(R(t))|∇N |Ψe

n(R)〉 (8.88)

we see that the nuclear motion couples electronic states when the nuclear velocity vector R islarge in a direction in which the electronic wavefunction changes most rapidly with R.

For diatomic molecules, we can separate out the center of mass motion and write m as thereduced mass

m =m1m2

m1 +m2

(8.89)

and write the nuclear Schrodinger equation (in one dimension)(−h2

2m

∂2

∂x2+ V (r)

)φ(r) = Eφ(r) (8.90)

where V (r) is the adiabatic or Born-Oppenheimer energy surface discussed above. Since V (r) isa polyomial of r, we can do a Taylor expansion of r about its minimum at re

V (r) = −Vo +1

2V ′′(re)(r − re)

2 +1

6V ′′′(re)(r − re)

3+ (8.91)

As an example of molecular bonding and how one computes the structure and dynamics ofa simple molecule, we turn towards the simplest molcular ion, H+

2 . For a fixed H −H distance,R, the electronic Schrodinger equation reads (in atomic units)(

−1

2∇2 − 1

r1− 1

r2

)ψ(r1, r2) = Eψ(r1, r2) (8.92)

The problem can be solved exactly in elliptical coordinates, but the derivation of the result isnot terribly enlightining. What we will do, however, is use a variational approach by combininghydrogenic 1s orbitals centered on each H nuclei. This proceedure is termed the Linear Combi-nation of Atomic Orbitals and is the underlying idea behind most quantum chemical calculations.

The basis functions are the hydrogen 1s orbitals. The rationalle for this basis is that as Rbecomes large, we have a H atom and a proton–a system we can handle pretty easily. Since theelectron can be on either nuclei, we take a linear combination of the two choices.

|ψ〉 = c1|φ1〉+ c2|φ2〉 (8.93)

We then use the variational proccedure to find the lowest energy subject to the constraint that〈ψ|ψ〉 = 1. This is an eigenvalue problem which we can write as∑

j=1,2

〈φi|H|φj〉 = E∑

j=1,2

cj〈φi|φj〉. (8.94)

or in Matrix form (H11 H12

H21 H22

)(c1c2

)= E

(S11 S12

S21 S22

)(8.95)

234

where Sij is the overlap between the two basis functions.

S12 = 〈φ1|φ2〉 =∫d3rφ∗1(r)φ2(r)

and assuming the basis functions are normalized to begin with:

S11 = S22 = 1.

For the hydrogenic orbitals

ψ1(r1) =e−(r1

√π

and

ψ2(r2) =e−(r2)

√π.

A simple calculation yields1

S = e−R

(1 +R +

R2

3

).

The matrix elements of the Hamiltonian need also to be computed. The diagonal terms areeasy and correspond to the hydrogen 1s energies plus the internuclear repulsion plus the Coulombinteraction between nuclei 2 and the electron distribution about nuclei 1.

H11 = −EI +1

R− J11 (8.96)

where

J11 = 〈φ1|1

r2|φ1〉 (8.97)

=∫d3r

1

r12

|φ1(r)|2 (8.98)

This too, we evaluate in elliptic coordinates and the result reads

J11 =2EI

R

(1− e−2R(1 +R)

). (8.99)

1 To derive this result, you need to first transform to elliptic coordinates u, v where

r1 =u+ v

2R

r2 =u− v

2R

the volume element is then d3r = R3(u2 − v2)/8dudvdφ where φ is the azimuthal angle for rotation about theH −H axis. The resulting integral reads

S =1π

∫ ∞

1

du

∫ +1

−1

dv

∫ 2π

0

dφR3

8(u2 − v2)e−uR.

235

Figure 8.1: Various contributions to the H+2 Hamiltonian.

2 4 6 8 10RHbohrL

0.2

0.4

0.6

0.8

1

S,J,A

By symmetry, H11 = H22 and we have the diagonal elements.We can think of J as being a modification of the nuclear repulsion due to the screening of the

electron about one of the atoms. |φ1(r)|2 is the charge density of the hydrogen 1s orbital and isspherically symmetric about nuclei 1. For large internuclear distances,

J =1

R(8.100)

and positive charge of nuclei 1 is completly counterballenced by the negative charge distributionabout it. At shorter ranges,

1

R− J > 0. (8.101)

However, screening along cannot explain a chemical bond since J does not go through a minimumat some distance R. Figure shows the variation of J , H11, and S as functions of R.

We now look at the off diagonal elements, H12 = H21. Written explicitly

H12 = 〈φ1|h(2)|φ2 > +1

RS12 − 〈φ1|

1

r12|φ2〉

= (−EI +1

R)S12 − A (8.102)

where

A = 〈φ1|1

r12

|φ2〉 =∫d3rφ1(r)

1

r12

φ2(r), (8.103)

which can also be evaluated using elliptical coordinates.

A = R2EI

∫ ∞

1du2ue−uR (8.104)

= 2EIe−R(1 +R) (8.105)

236

Exercise 8.2 Verify the expressions for J , S, and A by performig the transformation to ellipticcoordinates and performing the integrations.

A is termed the resonance integral and gives the energy for moving an electron from onenuclei to the other. When H12 6= 0, there is a finite probability for the electron to hop from onesite to the other and back. This oscillation results in the electron being delocalized between thenuclei andis the primary contribution to the formation of a chemical bond.

To wrap this up, the terms in the Hamiltonian are

S11 = S22 = 1 (8.106)

S12 = S21 = S (8.107)

H11 = H22 = −EI +1

R− C (8.108)

H12 = H21 = (−EI +1

R)S − A (8.109)

Since EI appears in each, we use that as our energy scale and set

E = εEI (8.110)

A = αEI (8.111)

J = γEI (8.112)

so that the secular equation now reads∣∣∣∣∣∣ −1 + 2R− γ − ε

(−1 + 2

R

)S − α− εS(

−1 + 2R

)S − α− εS −1 + 2

R− γ − ε

∣∣∣∣∣∣ = 0 (8.113)

Solving the secular equation yields two eigenvalues:

ε± = −12

R± α− γ

1∓ S. (8.114)

For large internuclear separations, ε± ≈ −1, or −EI which is the ground state of an isolated Hatom EI = 1/2. Choosing this as the energy origin, and putting it all back together:

E± = EI

2

R± 2e−R(1 +R)∓ 2(1− e−2R(1 +R))/R

1∓ e−R(1R +R2/3

(8.115)

Plots of these two energy surfaces are shown in Fig. 8.2. The energy minimum for the lower state(E−) is at E− = −0.064831hartree when Req = 2.49283ao (or -0.5648 hartree if we don’t set ourzero to be the dissociation limit). These results are qualitatively correct, but are quantitativelyway off the mark. The experimental values are De = 0.1025 hartree and Re = 2.00ao. Theresults can be improved upon by using improved basis function, using the charge as a variationalparameter and so forth. The important point is that even at this simple level of theory, we canget chemical bonds and equilibrium geometries.

For the orbitals, we have a symmetric and anti-symmetric combination of the two 1s orbitals.

ψ± = N±(φ1 ± φ2). (8.116)

237

Figure 8.2: Potential energy surface for H+2 molecular ion.

2 4 6 8 10R HbohrL

-0.05

0.05

0.1

0.15

0.2

0.25

e+ ,e- HhartreeL

Figure 8.3: Three dimensional representations of ψ+ and ψ− for the H+2 molecular ion generated

using the Spartan ab initio quantum chemistry program.

In Fig. 8.3, we show the orbitals from an ab initio calculation using the 6-31∗ set of basisfunctions. The first figure corresponds to the occupoled ground state orbital which forms a σbond between the two H atoms. The second shows the anti-bonding σ∗ orbital formed by theanti-symmetric combination of the 1s basis functions. The red and blue mesh indicates the phaseof the wavefunction.

238

Appendix: Creation and Annihilation Operators

Creation and annihilation operators are a convenient way to represent many-particle states andmany-particle operators. Recall, from the harmonic oscillator, a creation operator, a†, acts onthe ground state to produce 1 quanta of excitation in the system. We can generalize this to manyparticles by saying that a†λ creates a particle in state λ. e.g.

a†λ|λ1...λN〉 =√nλ + 1|λλ1...λN〉 (8.117)

where nλ is the occupation of the |λ〉 state. Physically, the operator αλ creates a particle in state|λ〉 and symmetries or antisymmetrizes the state as need be. For Bosons, the case is simple sinceany number of particles can occupy a given state. For Fermions, the operation takes a simplerform:

a†λ|λ1...λN〉 =

|λλ1...λN〉 if the state |λ〉 is not present in |λ1...λN〉

0 otherwise(8.118)

The anti-symmetrized basis vectors can be constructed using the a†j operators as

|λ1...λN = a†λ1a†λ2

· · · a†λN|0〉 (8.119)

Note that when we write the | states we do not need to keep track of the normalization factors.We do need to keep track of them them when we use the |〉 or |) vectors.

|λ1...λN〉 = a†λ1a†λ2

· · · a†λN|0〉 (8.120)

=

√1∏

λ nλ!a†λ1

a†λ2· · · a†λN

|0〉 (8.121)

The symmetry requirement places certain requirement on the commutation of the creationoperators. For example,

a†µa†ν |0〉 = |µν (8.122)

= ζ|νµ (8.123)

= ζa†νa†µ|0〉 (8.124)

Thus,

a†µa†ν − ζa†νa

†µ = 0 (8.125)

In other words, for ζ = 1 (bosons) the two operators commute for ζ = −1 (fermions) theoperators anti-commute.

239

We can prove similar results for the adjoint of the a† operator. In sort, for bosons we have acommutation relation between a†λ and aλ:

[aµ, a†ν ] = δµν (8.126)

While for fermions, we have a anti-commutation relation:

aµ, a†ν = aµa

†ν + a†µaν = δµν (8.127)

Finally, we define a set of field operators which are related to the creation/annihilation oper-ators as

ψ†(x) =∑α

〈α|x〉a†α =∑α

φ∗α(x)a†α (8.128)

ψ(x) =∑α

〈α|x〉aα =∑α

φα(x)aα (8.129)

These particular operators are useful in deriving various tight-binding approximations.Say that a†α places a particle in state α and aα(1) deletes a particle from state α. The

occupation of state |α〉 is thus

nα|α〉 = a†αaα|α〉 = nα|α〉 (8.130)

if the state is unoccupied, nα = 0 and if the state is occupied, the first operation removes theparticle and the second replaces, and nα = 1.

One-body operators can be evaluated as

U =∑α

nαUα =∑α

〈α|U |α〉a†αaα (8.131)

Likewise,

Uα = 〈α|U |α〉nα (8.132)

Two body operators are written as

V =1

2

∑αβγδ

∫dx∫dyφ∗α(x)φ∗β(y)V (x− y)φγ(x)φδ(y) = (αβ|v|γδ)a†αa

†βaγaδ (8.133)

Occasionally, it is useful to write the symmetrized variant of this operator

V =1

4

∑αβγδ

(αβ|v[|γδ)− |δγ)]a†αa†βaγaδ (8.134)

=1

4

∑αβγδ

αβ|v|γδa†αa†βaγaδ (8.135)

240

8.5 Problems and Exercises

Exercise 8.3 Consider the case of two identical particles with positions r1 and r2 trapped in acenteral harmonic well with a mutially repulsive harmonic interaction. The Hamiltonian for thiscase can be written as

H = − h2

2m

(∇2

1 +∇22

)+

1

2mω2(r2

1 + r22)−

λ

4mω2|r1 − r2|2 (8.136)

where λ is a dimensionalless scaling factor. This can be a model for two Bosons or Fermionstrapped in an optical trap and λmω2 simply tunes the s-wave scattering cross-section for the twoatoms.

1. Show that upon an appropriate change of variables

u = (r1 + r2)/√

2 (8.137)

v = (r1 − r2)/√

2 (8.138)

the Hamiltonian simplifies to two separable three-dimensional harmonic oscillators:

H =

[− h2

2m∇2

u +1

2mω2u2

]+

[− h2

2m∇2

v +1

2(1− λ)mω2v2

](8.139)

2. What is the exact ground state of this system?

3. Assuming the particles are spin 1/2 fermions. What are the lowest energy triplet and singletstates for this system?

4. What is the average distance of separation between the two particles in both the singlet andtriplet configurations.

5. Now, solve this problem via the variational approach by taking your trial wavefunction tobe a Slater determinant of the two lowest single particle states:

ψ(r1, r2) =

∣∣∣∣∣ α(1) β(2)φ(1) φ(2)

∣∣∣∣∣ (8.140)

Where φ(1) and φ(2) are the lowest energy 3D harmonic oscillator states modified suchthat we can take the width as a variational parameter:

φ(r) = N(ζ) exp(−r2/(2ζ2)

where N(ζ) is the normalization factor. Construct the Hamiltonan and determine thelowest energy state by taking the variation

δ〈ψ|H|ψ〉 = 0.

How does your variational estimate compare with the exact value for the energy?

241

Exercise 8.4 In this problem we consider an electron in a linear tri-atomic molecule formed bythree equidistant atoms. We will denote by |A〉, |B〉 , and |C〉 the three orthonormal state ofthe electron, corresponding to three wavefunctions localized about the three nuclei, A, B, and C.Whele there may be more than these states in the physical system, we will confine ourselves thesubspace spanned by these three vectors.

If we neglect the transfer of an electron from one site to an other site, its energy is described bythe Hamiltonian, Ho. The eigenstates of Ho are the three orthonormal states above with energiesEA, EB, and EC. For now, take EA = EB = EC = Eo. The coupling (i.e. electron hopping)between the states is described by an additional term W defined by its action on the basis vectors.

W |A〉 = −a|A〉

W |B〉 = −a(|A〉+ |C〉)

W |C〉 = −a|B〉

where a is a real positive constant.

1. Write both Ho and W in matrix form in the orthonormal basis and determine the eigenval-ues E1, E2, E3 and eigenvectors |1〉, |2〉, |3〉 for H = Ho +W . To do this numerically,pick your energy scale in terms of Eo/a.

2. Using the eigenvectors and eigenvalues you just determined, calculate the unitary timeevolution operator in the original basis. Eg.

〈A|U(t)|B〉 = 〈A| exp (−iH/ht) |B〉

3. If at time t = 0 the electron is localized on site A (in state |A〉), calculate the probabilityof finding the electron in any other state at some later time, t (i.e. PA, PB, and PC). Plotyour results. Is there some later time at which the probability of finding the electron backin the original state is exactly 1. Give a physical interpretation of this result.

4. Repeat your calculation in part 1 and 2, this time set EA = EC = Eo but set EB = 3Eo.Again, Plot your results for PA, PB, and PC and give a physical interpretation of yourresults.

In the next series of exercises, we will use Spartan ’02 for performing some elementary quan-tum chemistry calculations.

Exercise 8.5 Formaldehyde

1. Using the Builder on Spartan, build formaldehyde H2CO and perform an energy minimiza-tion. Save this. When this is done, use Geometry > Measure Distance and Geometry >Measure Angle to measure the C-H and C-O bond lengths and the H-C-O bond angle.

242

Figure 8.4: Setup calculation dialog screen

2. Set up a Geometry Optimization calculation (Setup ¿ Calculation). This will open up adialog screen that looks like Fig. 8.4. This will set up a Hartree-Fock calculation usingthe 6-31G** basis and print out the orbitals and their energies. It also forces Spartan topreserve the symmetry point-group of the initial configuration. After you do this, also setupsome calculations for generating some pictures of orbitals. Setup ¿ Graphics will open adialog window for adding graphics calculations. Add the following: HOMO, LUMO, andpotential. Close the dialog, and submit the job (Setup> Submit). Open the Spartan Monitorand wait until the job finishes. When this is done, use Geometry > Measure Distance andGeometry > Measure Angle to measure the C-H and C-O bond lengths and the H-C-O bondangle. This is the geometry predicted by a HF/6-31G** calculation.

3. Open Display> Surfaces and plot the HOMO and LUMO orbitals. Open up the text output(Display>Output ) generated by the calculation and figure out which molecular orbitalscorrespond to the HOMO and LUMO orbitals you plotted. What are their energies andirreducible representations? Are these σ or π orbitals? Considering the IRREPS of eachorbital, what is the lowest energy optical transition for this molecule? What are the atomicorbitals used for the O and C atoms in each of these orbitals?

4. Repeat the calculation you did above, this time including a calculation of the vibrationalfrequencies for both the ground state and first excited states (using CIS). Which vibrational

243

states undergo the largest change upon electronic excitation? Offer an explanation of thisresult noting that the excited state is pyramidal and that the So → S1 electronic transitionis an n→ π∗ transition. (This calculation will take some time.)

Table 8.1: Vibrational Frequencies of FormaldehydeSymmetry Description S0 S1

Sym CH strCO str

CH2 bendout-of-pland bendanti-sym CH str

CH2 rock

5. H2 + C=O → CH2=O transition state. Using the builder, make a model of the H2 + C=O→ CH2=O transition state. For this you will need to make a model that looks somethinglike what’s shown in Fig. 8.6. Then, go to the Search > Transition State menu. For thisyou will need to click on the H −H bond (head) and then the C for the tail of the reactionpath. Once you have done this, open Setup > Calculations and calculate Transition StateGeometry at the Ground state with Hartree Fock/6-31G**. Also compute the Frequencies.Close and submit the calculation. When the calculation finishes, examine the vibrationalfrequencies. Is there at least one imaginary frequency? Why do you expect only one suchfrequency? What does this tell you about the topology of the potential energy surface at thispoint. Record the energy at the transition state. Now, do two separate calculations of theisolated reactants. Combine these with the calculation you did above for the formaldehydeequilbrium geometry and sketch a reaction energy profile.

244

Figure 8.5: HOMO-1, HOMO and LUMO for CH2 = O.

245

Figure 8.6: Transition state geometry for H2 + C = O → CH2 = O. The Arrow indicates thereaction path.

246

Appendix A

Physical Constants and ConversionFactors

247

Table A.1: Physical ConstantsConstant Symbol SI ValueSpeed of Light c 299792458 m/s (exact)Charge of proton e 1.6021764 ×10−19 CPermitivity of Vacuum ε 8.8541878 −12J−1C2m−1

Avagadro’s Number NA 6.022142 ×1023 mol−1

Rest mass of electron me 9.109382 ×10−31kg

Table A.2: Atomic Units:In Atomic Units, the following quantities are unitary: h, e, me, ao.Quantity symbol or expression CGS or SI equiv.Mass me 9.109382 ×10−31kgCharge e 1.6021764 ×10−19 CAngular Momentum h 1.05457×10−34JsLength (bohr) ao = h2/(mee

2) 0.5291772 −10mEnergy (hartree) Eh = e2/ao 4.35974 ×10−18Jtime to = h3/(mee

4) 2.41888 ×10−17svelocity e2/h 2.18770 ×106m/sForce e2/a2

o 8.23872 ×10−8NElectric Field e/a2

o 5.14221 ×1011V/mElectric Potential e/ao 27.2114 V

Fine structure constant α = e2

hc1/137.036

Magnetic moment βe = eh/(2me) 9.27399 ×10−24J/TPermitivity of vacuum εo = 1/4π 8.8541878 −12J−1C2m−1

Hydrogen Atom IP −α2mec2/2 = −Eh/2 -13.60580 eV

Table A.3: Useful orders of magnitudeQuantity approximate value exact value

Electron rest mass mec2 ≈ 0.5 MeV 0.511003 MeV

Proton rest mass mpc2 ≈ 1000 MeV 938.280 MeV

neutron rest mass Mnc2 ≈ 1000MeV 939.573 MeV

Proton/Electron mass ratio mp/me ≈ 2000 1836.1515

One electron volt corresponds to a:Quantity symbol /relation exactfrequency: ν = 2.4× 1014Hz E = hν 2.417970 ×1014Hzwavelength: λ = 12000A λ = c/ν 12398.52 Awave number: 1/λ = 8000cm−1 8065.48 cm−1

temperature: T = 12000K E = kT 11604.5 K

248

Appendix B

Mathematical Results and Techniquesto Know and Love

B.1 The Dirac Delta Function

B.1.1 Definition

The Dirac delta-function is not really a function, per se, it is really a generalized function definedby the relation

f(xo) =∫ +∞

−∞dxδ(x− xo)f(x). (B.1)

The integral picks out the first term in the Taylor expansion of f(x) about the point xo and thisrelation must hold for any function of x. For example, let’s take a function which is zero only atsome arbitrary point, xo. Then the integral becomes:∫ +∞

−∞dxδ(x− xo)f(x) = 0. (B.2)

For this to be true for any arbitrary function, we have to conclude that

δ(x) = 0, for x 6= 0. (B.3)

Furthermore, from the Reimann-Lebesque theory of integration∫f(x)g(x)dx = lim

h→0

(a∑n

f(xn)g(xn)

), (B.4)

the only way for the defining relation to hold is for

δ(0) = ∞. (B.5)

This is a very odd function, it is zero every where except at one point, at which it is infinite. Soit is not a function in the regular sense. In fact, it is more like a distrubution function which isinfinitely narrow. If we set f(x) = 1, then we can see that the δ-function is normalized to unity.∫ +∞

−∞dxδ(x− xo) = 1. (B.6)

249

B.1.2 Properties

Some useful properties of the δ-function are as follows:

1. It is real: δ∗(x) = δ(x).

2. It is even: δ(x) = δ(−x).

3. δ(ax) = δ(x)/a for a > 0

4. ∫δ′(x)f(x)dx = f ′(0)

5. δ′(−x) = −δ′(x)

6. xδ(x) = 0

7.

δ(x2 − a2) =1

2a(δ(x+ a) + δ(x− a))

8. f(x)δ(x− a) = f(a)δ(x− a)

9. ∫δ(x− a)δ(x− b)dx = δ(a− b)

Exercise B.1 Prove the above relations.

B.1.3 Spectral representations

The δ function can be thought of as the limit of a sequence of regular functions. For example,

δ(x) = lima→∞

1

π

sin(ax)

x

This is the ”sinc” function or diffraction function with a width proportional to 1/a. For anyvalue of a, the function is regular. As we make a larger, the width increases and focuses aboutx = 0. This is shown in the Fig B.1, for increasing values of a. Notice that as a increases, thepeak increases and the function itself becomes extremely oscillitory.

Another extremely useful representation is the Fourier representation

δ(x) =1

2π

∫eikxdk. (B.7)

We used this representation in Eq. 7.205 to go from an energy representation to an integral overtime.

Finally, an other form is in terms of Gaussian functions as shown in Fig. B.2.

δ(x) = lima→∞

√a

πe−ax2

. (B.8)

250

Figure B.1: sin(xa)/πx representation of the Dirac δ-function

-20 -10 10 20

0.1

0.2

0.3

0.4

Figure B.2: Gaussian representation of δ-function

-4 -2 2 4

0.2

0.4

0.6

0.8

1

1.2

251

Here the height is proportional to a and the width to the standard deviation, 1/√

2a.Other representations include: Lorentzian form,

δ(x) = lima→

1

π

a

x2 + a2,

and derivative form

δ(x) =d

dxθ(x)

where θ(x) is the Heaviside step function

θ(x) =

0, x ≤ 01 x ≥ 0

(B.9)

This can be understood as the cumulative distribution function

θ(x) =∫ x

−∞δ(y)dy. (B.10)

252

B.2 Coordinate systems

In each case U is a function of coordinates and ~A is a vector.

B.2.1 Cartesianz

y

x

k

ji

• U = U(x, y, z)

• ~A = Axi+ Ay j + Azk

• Volume element: dV = dxdydz

• Vector product: A · B = AxBx + AyBy + AzBz

• Gradient~∇U =

∂U

∂xi+

∂U

∂yj +

∂U

∂zk

• Laplacian

∇2U =∂2U

∂x2+∂2U

∂y2+∂2U

∂z2

• Divergence

∇ · ~A =∂Ax

∂x+∂Ay

∂y+∂Az

∂z

• Curl

∇× ~A =

(∂Az

∂y− ∂Ay

∂z

)i−

(∂Ax

∂z− ∂Az

∂x

)j +

(∂Ay

∂x− ∂Ax

∂y

)k

253

B.2.2 Spherical

z

y

x

k

j

i

φ ρ

r

θ

• Coordinates:

• Transformation to cartesian:

x = r cosφ sin θ, y = r sin θ sinφ, z = r cos θ

• U = U(r, θ, φ)

• ~A = Arr + Aθθ + Aφφ

Ar = Aρ sin θ + Az cos θ (B.11)

Aθ = Aρ cos θ − Az sin θ (B.12)

Aφ = −Ax sinφ+ Ay cosφ (B.13)

Aρ = Ax cosφ+ Ay sinφ (B.14)

• Arc Length

• Volume element: dV =

• Vector product: A · B =

• Gradient~∇U =

• Laplacian∇2U =

• Divergence∇ · ~A =

• Curl∇× ~A =

254

B.2.3 Cylindrical

z

y

x

k

j

i

φ ρ

z

• Coordinates:

• Transformation to cartesian:

• U = U(x, y, z)

• ~A = Axi+ Ay j + Azk

• Volume element: dV = dxdydz

• Vector product: A · B = AxBx + AyBy + AzBz

• Gradient~∇U =

∂U

∂xi+

∂U

∂yj +

∂U

∂zk

• Laplacian

∇2U =∂2U

∂x2+∂2U

∂y2+∂2U

∂z2

• Divergence

∇ · ~A =∂Ax

∂x+∂Ay

∂y+∂Az

∂z

• Curl

∇× ~A =

(∂Az

∂y− ∂Ay

∂z

)i−

(∂Ax

∂z− ∂Az

∂x

)j +

(∂Ay

∂x− ∂Ax

∂y

)k

255

Appendix C

Mathematica Notebook Pages

256

3

1s

2s

2

Li

Lithium6.9

41

5.3

917

10N

eN

eon20.1

797

21.5

646

2

1s

2

He

Helium

4.0

0260

24.5

874

9

OOxygen

15.9

994

13.6

181

8

FFluorine

18.9

9840

17.4

228

7

NNitrogen

14.0

0674

14.5

341

6

CCarbon

12.0

107

11.2

603

5

1s

2s

2p

22

BBoron

10.8

11

8.2

980

57L

aLanthanum138.9

055

5.5

769

89A

cA

ctinium(2

27)

5.1

7

71L

uLutetium174.9

67

5.4

259

103Lr

Lawrencium

(262)

4.9

?

87

[Rn

]7s

Fr

Francium(2

23)

4.0

727

88R

aR

adium(2

26)

5.2

784

104R

fR

utherfordium(2

61)

6.0

?

72H

fH

afnium178.4

9

6.8

251

40Z

rZ

irconium91.2

24

6.6

339

39YY

ttrium88.9

0585

6.2

171

38S

rS

trontium87.6

2

5.6

949

56B

aB

arium137.3

27

5.2

117

73T

aTantalum180.9

479

7.5

496

54X

eX

enon131.2

9

12.1

298

19

[Ar]4

s

KP

otassium39.0

983

4.3

407

20C

aC

alcium40.0

78

6.1

132

21S

cS

candium44.9

5591

6.5

615

22T

iT

itanium47.8

67

6.8

281

30Z

nZinc

65.3

9

9.3

942

31G

aG

allium69.7

23

5.9

993

32G

eG

ermanium

72.6

1

7.8

994

33A

sA

rsenic74.9

2160

9.7

886

34S

eS

elenium78.9

6

9.7

524

35B

rB

romine

79.9

04

11.8

138

36K

rK

rypton83.8

0

13.9

996

23VVanadium50.9

415

6.7

462

24C

rC

hromium

51.9

961

6.7

665

25M

nM

anganese54.9

3805

7.4

340

26F

eIron55.8

45

7.9

024

27C

oC

obalt58.9

3320

7.8

810

28N

iN

ickel58.6

934

7.6

398

29C

uC

opper63.5

46

7.7

264

11

[Ne

]3s

Na

Sodium

22.9

8977

5.1

391

12M

gM

agnesium24.3

050

7.6

462

13[N

e]3

s3

p2

Al

Alum

inum26.9

8154

5.9

858

14S

iS

ilicon28.0

855

8.1

517

15P

Phosphorus30.9

7376

10.4

867

16SS

ulfur32.0

66

10.3

600

17C

lC

hlorine35.4

527

12.9

676

18A

rA

rgon39.9

48

15.7

596

1

HHydrogen

1.0

0794

13.5

984

4

1s

2s

22

Be

Beryllium

9.0

1218

9.3

227

37

[Kr]5

s

Rb

Rubidium

85.4

678

4.1

771

55

[Xe

]6s

Cs

Cesium

132.9

0545

3.8

939

42M

oM

olybdenum95.9

4

7.0

924

41N

bN

iobium92.9

0638

6.7

589

86R

nR

adon(2

22)

10.7

485

74WTungsten183.8

4

7.8

640

43T

cTechnetium

(98)

7.2

8

75R

eR

henium186.2

07

7.8

335

44R

uR

uthenium101.0

7

7.3

605

76O

sO

smium

190.2

3

8.4

382

45R

hR

hodium102.9

0550

7.4

589

77IrIridium

192.2

17

8.9

670

46P

dP

alladium106.4

2

8.3

369

78P

tP

latinum195.0

78

8.9

587

47A

gS

ilver107.8

682

7.5

762

79A

uGold

196.9

6655

9.2

255

48C

dC

admium

112.4

11

8.9

938

80H

gM

ercury200.5

9

10.4

375

60N

dN

eodymium

144.2

4

5.5

250

62S

mS

amarium

150.3

6

5.6

436

63E

uE

uropium151.9

64

5.6

704

64G

dG

adolinium157.2

5

6.1

501

65T

bTerbium

158.9

2534

5.8

638

61P

mP

romethium

(145)

5.5

82

66D

yD

ysprosium162.5

0

5.9

389

67H

oH

olmium

164.9

3032

6.0

215

68E

rE

rbium167.2

6

6.1

077

69T

mT

hulium168.9

3421

6.1

843

49InIndium114.8

18

5.7

864

50S

nTin

118.7

10

7.3

439

51S

bA

ntimony

121.7

60

8.6

084

52T

eTellurium127.6

0

9.0

096

53

IIodine126.9

0447

10.4

513

81T

lT

hallium204.3

833

6.1

082

82P

bLead207.2

7.4

167

83B

iB

ismuth

208.9

8038

7.2

856

84P

oP

olonium(2

09)

8.4

17

?

85A

tA

statine(2

10)

58C

eC

erium140.1

16

5.5

387

59P

rP

raseodymium

140.9

0765

5.4

73

70Y

bY

tterbium173.0

4

6.2

542

90T

hT

horium232.0

381

6.3

067

92UU

ranium238.0

289

6.1

941

93N

pN

eptunium(2

37)

6.2

657

94P

uP

lutonium(2

44)

6.0

262

95A

mA

mericium

(243)

5.9

738

96C

mC

urium(2

47)

5.9

915

91P

aP

rotactinium231.0

3588

5.8

9

97B

kB

erkelium(2

47)

6.1

979

98C

fC

alifornium(2

51)

6.2

817

99E

sE

insteinium(2

52)

6.4

2

100

Fm

Ferm

ium(2

57)

6.5

0

101Md

Mendelevium

(258)

6.5

8

102No

Nobelium(2

59)

6.6

5

1s

2s

2p

22

21

s2

s2

p2

23

1s

2s

2p

22

41

s2

s2

p2

25

1s

2s

2p

22

6

[Ne

]3s

2[N

e]3

s3

p2

2[N

e]3

s3

p2

3[N

e]3

s3

p2

4[N

e]3

s3

p2

5[N

e]3

s3

p2

6

[Ar]4

s2

[Ar]3

d4

s2

[Ar]3

d4

s2

2[A

r]3d

4s

32

[Ar]3

d4

s5

[Ar]3

d4

s5

2[A

r]3d

4s

62

[Ar]3

d4

s7

2[A

r]3d

4s

82

[Ar]3

d4

s1

0[A

r]3d

4s

10

2[A

r]3d

4s

4p

10

2[A

r]3d

4s

4p

10

22

[Ar]3

d4

s4

p1

02

3[A

r]3d

4s

4p

10

24

[Ar]3

d4

s4

p1

02

5[A

r]3d

4s

4p

10

26

[Kr]5

s2

[Kr]4

d5

s2

[Kr]4

d5

s2

2[K

r]4d

5s

4[K

r]4d

5s

5[K

r]4d

5s

52

[Kr]4

d5

s7

[Kr]4

d5

s8

[Kr]4

d1

0[K

r]4d

5s

10

[Kr]4

d5

s1

02

[Kr]4

d5

s5

p1

02

2[K

r]4d

5s

5p

10

23

[Kr]4

d5

s5

p1

02

4[K

r]4d

5s

5p

10

25

[Kr]4

d5

s5

p1

02

6

[Xe

]6s

2

[Xe

]5d

6s

2[X

e]4

f5d

6s

2[X

e]4

f6

s3

2[X

e]4

f6

s4

2[X

e]4

f6

s5

2[X

e]4

f6

s7

2[X

e]4

f6

s6

2[X

e]4

f5

d6

s7

2[X

e]4

f6

s9

2[X

e]4

f6

s1

02

[Xe

]4f

6s

11

2[X

e]4

f6

s1

22

[Xe

]4f

6s

13

2[X

e]4

f6

s1

42

[Xe

]4f

5d

6s

14

2

[Xe

]4f

5d

6s

14

22

[Xe

]4f

5d

6s

14

32

[Xe

]4f

5d

6s

14

42

[Xe

]4f

5d

6s

14

52

[Xe

]4f

5d

6s

14

26

[Xe

]4f

5d

6s

14

72

[Xe

]4f

5d

6s

14

9[X

e]4

f5

d6

s1

41

0[X

e]4

f5

d6

s1

41

02

[Hg

]6p

[Hg

]6p

2[H

g]6

p3

[Hg

]6p

4[H

g]6

p5

[Hg

]6p

6

[Rn

]6d

7s

2[R

n]6

d7

s2

2[R

n]5

f6

d7

s2

2[R

n]5

f6

d7

s3

2[R

n]5

f6

d7

s4

2[R

n]5

f7

s6

2[R

n]5

f7

s7

2[R

n]5

f6

d7

s7

2[R

n]5

f7

s9

2[R

n]5

f7

s1

02

[Rn

]5f

7s

11

2[R

n]5

f7

s1

22

[Rn

]5f

7s

13

2[R

n]5

f7

s1

42

[Rn

]5f

7s

7p

?1

42

[Rn

]5f

6d

7s

?1

42

2

[Kr]4

d5

s5

p1

02

Gro

up

IA

IIA

IIIAIV

AV

AV

IAV

IIAV

IIIAIB

IIB

IIIBIV

BV

BV

IBV

IIB

VIII

58C

eC

erium140.1

16

5.5

387 1G

°4

[Xe

]4f5

d6

s2

Ato

mic

Nu

mb

er

Sym

bo

l

Na

me

Gro

un

d-s

tate

Co

nfig

ura

tion

Gro

un

d-s

tate

Le

ve

l

Ion

iza

tion

En

erg

y(e

V)

105

107

106

108

109

111

110

112

Db

Dubnium(2

62)

Sg

Seaborgium

(263)

Hs

Hassium(2

65)

Bh

Bohrium(2

64)

Mt

Meitnerium(2

68)

Uu

nU

nunnilium(2

69)

Uu

bU

nunbium

Uu

uU

nununium(2

72)

Period

16 5 4 3 27

PE

RI

OD

IC

TA

BL

E

Ato

mic

Pro

pe

rties

of

the

Ele

me

nts

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For

them

ostaccuratevalues

oftheseand

otherconstants,visit

be

twe

en

C

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)

0.5

11

0M

eV

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89

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h=

h

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physics.nist.gov/constants

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!!!!

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444

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lids

Artific

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Pre

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1S0

1S0

3P2

3P0

2D3

2/

2D3

2/

3F2

?

2S1

2/

2D3

2/

3F2

1S0

4F3

2/

7S3

6S5

2/

5D4

4F9

2/

3F4

2S1

2/

2S1

2/

1S0

3P0

1S0

2S1

2/

6D1

2/

5D0

5F5

5D4

3D3

2S1

2/

2S1

2/

7F0

3H6

1S0

3F2

2P°1

2/

4S°3

2/

2P°3

2/

2P°1

2/

2P°3

2/

2P°1

2/

2P°3

2/

2P°1

2/

2P°1

2/

1G°4

4-I- °92/

6H°5

2/

8S°7

2/

9D°2

6H°1

52/

2F°7

2/

2P°1

2 ?/

4S°3

2/

3P2

1S0

1S0

3P0

4S°3

2/

3P2

1S0

1S0

2D3

2/

3F2

7S3

6S5

2/

4F9

2/

1S0

1S0

3P0

4S°3

2/

3P2

2P°3

2/

1S0

1S0

3F2

4F3

2/

6S5

2/

4F9

2/

1S0

3P0

4S°3

2/

3P2

2P°3

2/

1S0

1S0

2D3

2/

7F0

8S°7

2/

9D°2

6H°1

52/

3H6

2F°7

2/

1S0

5L°6

4K11

2/

6L11

2/

5-I-4

4-I- °15

2/

5-I-8

5-I-8

4-I- °15

2/

2S1

2/

2S1

2/

2S1

2/

257