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Today’s Outline - February 01, 2018

• Problems from Chapter 6

Homework Assignment #04:Chapter 7:1,2,4,5,7,9due Thursday, February 08, 2018

Midterm Exam #1Thursday, February 22, 2018

C. Segre (IIT) PHYS 406 - Spring 2018 February 01, 2018 1 / 18

Problem 6.23

Consider the (eight) n = 2 states, |2 l ml ms〉. Find the energy of eachstate, under strong Zeeman splitting. Express each answer as the sum ofthree terms: the Bohr energy, the fine structure (proportional to α2), andthe Zeeman contribution (proportional to µBBext). If you ignore finestructure altogether, how many distinct levels are there and what are theirdegeneracies?

The Bohr term is identicalfor all of these states

the strong Zeeman termis

and the fine structure per-turbation for l > 0 is

E2 = −13.6 eV

22= −3.4 eV

EZ = µBBext(ml + 2ms)

Efs = (1.7 eV)α2

8−[l(l + 1)−mlms

l(l + 1/2)(l + 1)

Problem 6.23

E2 = −13.6 eV

22= −3.4 eV

Efs = (1.7 eV)α2

8−[l(l + 1)−mlms

l(l + 1/2)(l + 1)

Problem 6.23

E2 = −13.6 eV

= −3.4 eV

Efs = (1.7 eV)α2

8−[l(l + 1)−mlms

l(l + 1/2)(l + 1)

Problem 6.23

E2 = −13.6 eV

22= −3.4 eV

Efs = (1.7 eV)α2

8−[l(l + 1)−mlms

l(l + 1/2)(l + 1)

Problem 6.23

E2 = −13.6 eV

22= −3.4 eV

Efs = (1.7 eV)α2

8−[l(l + 1)−mlms

l(l + 1/2)(l + 1)

Problem 6.23

E2 = −13.6 eV

22= −3.4 eV

Efs = (1.7 eV)α2

8−[l(l + 1)−mlms

l(l + 1/2)(l + 1)

Problem 6.23

E2 = −13.6 eV

22= −3.4 eV

Efs = (1.7 eV)α2

8−[l(l + 1)−mlms

l(l + 1/2)(l + 1)

Problem 6.23

E2 = −13.6 eV

22= −3.4 eV

Efs = (1.7 eV)α2

8−[l(l + 1)−mlms

l(l + 1/2)(l + 1)

]}C. Segre (IIT) PHYS 406 - Spring 2018 February 01, 2018 2 / 18

Problem 6.23 (cont.)

Etot = −3.4 eV + (3.4 eV)α2

8−[l(l + 1)−mlms

l(l + 1/2)(l + 1)

]}+ µBBext(ml + 2ms)

|n l ml ms〉 (ml + 2ms) {. . . } Total Energy

|1〉 = |2 0 0 12〉 1 −58 −3.4 eV[1 + 5

16α2] + µBBext

|2〉 = |2 0 0 −12〉 −1 −58 −3.4 eV[1 + 516α

2]− µBBext

|3〉 = |2 1 1 12〉 2 −18 −3.4 eV[1 + 1

16α2] + 2µBBext

|4〉 = |2 1 −1 −12〉 −2 −18 −3.4 eV[1 + 116α

2]− 2µBBext

|5〉 = |2 1 0 12〉 1 − 7

24 −3.4 eV[1 + 748α

2] + µBBext

|6〉 = |2 1 1 −12〉 0 −1124 −3.4 eV[1 + 1148α

|7〉 = |2 1 0 −12〉 −1 − 724 −3.4 eV[1 + 7

48α2]− µBBext

|8〉 = |2 1 −1 12〉 0 −1124 −3.4 eV[1 + 11

48α2]

ignoring fine structure there are 5 distinct levels with two of them singlydegenerate and the other double degenerate

Etot = −3.4 eV + (3.4 eV)α2

8−[l(l + 1)−mlms

l(l + 1/2)(l + 1)

|1〉 = |2 0 0 12〉 1 −58 −3.4 eV[1 + 5

16α2] + µBBext

|2〉 = |2 0 0 −12〉 −1 −58 −3.4 eV[1 + 516α

2]− µBBext

|3〉 = |2 1 1 12〉 2 −18 −3.4 eV[1 + 1

16α2] + 2µBBext

|4〉 = |2 1 −1 −12〉 −2 −18 −3.4 eV[1 + 116α

2]− 2µBBext

|5〉 = |2 1 0 12〉 1 − 7

24 −3.4 eV[1 + 748α

2] + µBBext

|6〉 = |2 1 1 −12〉 0 −1124 −3.4 eV[1 + 1148α

|7〉 = |2 1 0 −12〉 −1 − 724 −3.4 eV[1 + 7

48α2]− µBBext

|8〉 = |2 1 −1 12〉 0 −1124 −3.4 eV[1 + 11

48α2]

Etot = −3.4 eV + (3.4 eV)α2

8−[l(l + 1)−mlms

l(l + 1/2)(l + 1)

|1〉 = |2 0 0 12〉 1 −58 −3.4 eV[1 + 5

16α2] + µBBext

|2〉 = |2 0 0 −12〉 −1 −58 −3.4 eV[1 + 516α

2]− µBBext

|3〉 = |2 1 1 12〉 2 −18 −3.4 eV[1 + 1

16α2] + 2µBBext

|4〉 = |2 1 −1 −12〉 −2 −18 −3.4 eV[1 + 116α

2]− 2µBBext

|5〉 = |2 1 0 12〉 1 − 7

24 −3.4 eV[1 + 748α

2] + µBBext

|6〉 = |2 1 1 −12〉 0 −1124 −3.4 eV[1 + 1148α

|7〉 = |2 1 0 −12〉 −1 − 724 −3.4 eV[1 + 7

48α2]− µBBext

|8〉 = |2 1 −1 12〉 0 −1124 −3.4 eV[1 + 11

48α2]

Etot = −3.4 eV + (3.4 eV)α2

8−[l(l + 1)−mlms

l(l + 1/2)(l + 1)

|1〉 = |2 0 0 12〉 1 −58 −3.4 eV[1 + 5

16α2] + µBBext

|2〉 = |2 0 0 −12〉 −1 −58 −3.4 eV[1 + 516α

2]− µBBext

|3〉 = |2 1 1 12〉 2 −18 −3.4 eV[1 + 1

16α2] + 2µBBext

|4〉 = |2 1 −1 −12〉 −2 −18 −3.4 eV[1 + 116α

2]− 2µBBext

|5〉 = |2 1 0 12〉 1 − 7

24 −3.4 eV[1 + 748α

2] + µBBext

|6〉 = |2 1 1 −12〉 0 −1124 −3.4 eV[1 + 1148α

|7〉 = |2 1 0 −12〉 −1 − 724 −3.4 eV[1 + 7

48α2]− µBBext

|8〉 = |2 1 −1 12〉 0 −1124 −3.4 eV[1 + 11

48α2]

Etot = −3.4 eV + (3.4 eV)α2

8−[l(l + 1)−mlms

l(l + 1/2)(l + 1)

|1〉 = |2 0 0 12〉 1 −58 −3.4 eV[1 + 5

16α2] + µBBext

|2〉 = |2 0 0 −12〉 −1 −58 −3.4 eV[1 + 516α

2]− µBBext

|3〉 = |2 1 1 12〉 2 −18 −3.4 eV[1 + 1

16α2] + 2µBBext

|4〉 = |2 1 −1 −12〉 −2 −18 −3.4 eV[1 + 116α

2]− 2µBBext

|5〉 = |2 1 0 12〉 1 − 7

24 −3.4 eV[1 + 748α

2] + µBBext

|6〉 = |2 1 1 −12〉 0 −1124 −3.4 eV[1 + 1148α

|7〉 = |2 1 0 −12〉 −1 − 724 −3.4 eV[1 + 7

48α2]− µBBext

|8〉 = |2 1 −1 12〉 0 −1124 −3.4 eV[1 + 11

48α2]

Etot = −3.4 eV + (3.4 eV)α2

8−[l(l + 1)−mlms

l(l + 1/2)(l + 1)

|1〉 = |2 0 0 12〉 1 −58 −3.4 eV[1 + 5

16α2] + µBBext

|2〉 = |2 0 0 −12〉 −1 −58 −3.4 eV[1 + 516α

2]− µBBext

|3〉 = |2 1 1 12〉 2 −18 −3.4 eV[1 + 1

16α2] + 2µBBext

|4〉 = |2 1 −1 −12〉 −2 −18 −3.4 eV[1 + 116α

2]− 2µBBext

|5〉 = |2 1 0 12〉 1 − 7

24 −3.4 eV[1 + 748α

2] + µBBext

|6〉 = |2 1 1 −12〉 0 −1124 −3.4 eV[1 + 1148α

|7〉 = |2 1 0 −12〉 −1 − 724 −3.4 eV[1 + 7

48α2]− µBBext

|8〉 = |2 1 −1 12〉 0 −1124 −3.4 eV[1 + 11

48α2]

Etot = −3.4 eV + (3.4 eV)α2

8−[l(l + 1)−mlms

l(l + 1/2)(l + 1)

|1〉 = |2 0 0 12〉 1 −58 −3.4 eV[1 + 5

16α2] + µBBext

|2〉 = |2 0 0 −12〉 −1 −58 −3.4 eV[1 + 516α

2]− µBBext

|3〉 = |2 1 1 12〉 2 −18 −3.4 eV[1 + 1

16α2] + 2µBBext

|4〉 = |2 1 −1 −12〉 −2 −18 −3.4 eV[1 + 116α

2]− 2µBBext

|5〉 = |2 1 0 12〉 1 − 7

24 −3.4 eV[1 + 748α

2] + µBBext

|6〉 = |2 1 1 −12〉 0 −1124 −3.4 eV[1 + 1148α

|7〉 = |2 1 0 −12〉 −1 − 724 −3.4 eV[1 + 7

48α2]− µBBext

|8〉 = |2 1 −1 12〉 0 −1124 −3.4 eV[1 + 11

48α2]

Etot = −3.4 eV + (3.4 eV)α2

8−[l(l + 1)−mlms

l(l + 1/2)(l + 1)

|1〉 = |2 0 0 12〉 1 −58 −3.4 eV[1 + 5

16α2] + µBBext

|2〉 = |2 0 0 −12〉 −1 −58 −3.4 eV[1 + 516α

2]− µBBext

|3〉 = |2 1 1 12〉 2 −18 −3.4 eV[1 + 1

16α2] + 2µBBext

|4〉 = |2 1 −1 −12〉 −2 −18 −3.4 eV[1 + 116α

2]− 2µBBext

|5〉 = |2 1 0 12〉 1 − 7

24 −3.4 eV[1 + 748α

2] + µBBext

|6〉 = |2 1 1 −12〉 0 −1124 −3.4 eV[1 + 1148α

|7〉 = |2 1 0 −12〉 −1 − 724 −3.4 eV[1 + 7

48α2]− µBBext

|8〉 = |2 1 −1 12〉 0 −1124 −3.4 eV[1 + 11

48α2]

Etot = −3.4 eV + (3.4 eV)α2

8−[l(l + 1)−mlms

l(l + 1/2)(l + 1)

|1〉 = |2 0 0 12〉 1 −58 −3.4 eV[1 + 5

16α2] + µBBext

|2〉 = |2 0 0 −12〉 −1 −58 −3.4 eV[1 + 516α

2]− µBBext

|3〉 = |2 1 1 12〉 2 −18 −3.4 eV[1 + 1

16α2] + 2µBBext

|4〉 = |2 1 −1 −12〉 −2 −18 −3.4 eV[1 + 116α

2]− 2µBBext

|5〉 = |2 1 0 12〉 1 − 7

24 −3.4 eV[1 + 748α

2] + µBBext

|6〉 = |2 1 1 −12〉 0 −1124 −3.4 eV[1 + 1148α

|7〉 = |2 1 0 −12〉 −1 − 724 −3.4 eV[1 + 7

48α2]− µBBext

|8〉 = |2 1 −1 12〉 0 −1124 −3.4 eV[1 + 11

48α2]

Etot = −3.4 eV + (3.4 eV)α2

8−[l(l + 1)−mlms

l(l + 1/2)(l + 1)

|1〉 = |2 0 0 12〉 1 −58 −3.4 eV[1 + 5

16α2] + µBBext

|2〉 = |2 0 0 −12〉 −1 −58 −3.4 eV[1 + 516α

2]− µBBext

|3〉 = |2 1 1 12〉 2 −18 −3.4 eV[1 + 1

16α2] + 2µBBext

|4〉 = |2 1 −1 −12〉 −2 −18 −3.4 eV[1 + 116α

2]− 2µBBext

|5〉 = |2 1 0 12〉 1 − 7

24 −3.4 eV[1 + 748α

2] + µBBext

|6〉 = |2 1 1 −12〉 0 −1124 −3.4 eV[1 + 1148α

|7〉 = |2 1 0 −12〉 −1 − 724 −3.4 eV[1 + 7

48α2]− µBBext

|8〉 = |2 1 −1 12〉 0 −1124 −3.4 eV[1 + 11

48α2]

Etot = −3.4 eV + (3.4 eV)α2

8−[l(l + 1)−mlms

l(l + 1/2)(l + 1)

|1〉 = |2 0 0 12〉 1 −58 −3.4 eV[1 + 5

16α2] + µBBext

|2〉 = |2 0 0 −12〉 −1 −58 −3.4 eV[1 + 516α

2]− µBBext

|3〉 = |2 1 1 12〉 2 −18 −3.4 eV[1 + 1

16α2] + 2µBBext

|4〉 = |2 1 −1 −12〉 −2 −18 −3.4 eV[1 + 116α

2]− 2µBBext

|5〉 = |2 1 0 12〉 1 − 7

24 −3.4 eV[1 + 748α

2] + µBBext

|6〉 = |2 1 1 −12〉 0 −1124 −3.4 eV[1 + 1148α

|7〉 = |2 1 0 −12〉 −1 − 724 −3.4 eV[1 + 7

48α2]− µBBext

|8〉 = |2 1 −1 12〉 0 −1124 −3.4 eV[1 + 11

48α2]

Problem 6.25

Work out the matrix elements of H ′Z and H ′fs and construct the W-matrixgiven in the text, for n = 2.

The derivation in the book used the 8 distinct j , and mj eigenstates as thebasis upon which to build the interaction matrix

ψ1 ≡∣∣1212

⟩= |00〉

∣∣1212

⟩ψ2 ≡

∣∣12 −

⟩= |00〉

∣∣12 −

ψ3 ≡∣∣3232

⟩= |11〉

∣∣1212

⟩ψ4 ≡

∣∣32 −

⟩= |1−1〉

∣∣12 −

⟩ψ5 ≡

∣∣3212

⟩=√

23 |10〉

∣∣1212

⟩+√

13 |11〉

∣∣12 −

⟩ψ6 ≡

∣∣1212

⟩= −

√13 |10〉

∣∣1212

⟩+√

23 |11〉

∣∣12 −

⟩ψ7 ≡

∣∣32 −

⟩=√

13 |1−1〉

∣∣1212

⟩+√

23 |10〉

∣∣12 −

⟩ψ8 ≡

∣∣12 −

⟩= −

√23 |1−1〉

∣∣1212

⟩+√

13 |10〉

∣∣12 −

Problem 6.25

ψ1 ≡∣∣1212

⟩= |00〉

∣∣1212

⟩ψ2 ≡

∣∣12 −

⟩= |00〉

∣∣12 −

ψ3 ≡∣∣3232

⟩= |11〉

∣∣1212

⟩ψ4 ≡

∣∣32 −

⟩= |1−1〉

∣∣12 −

⟩ψ5 ≡

∣∣3212

⟩=√

23 |10〉

∣∣1212

⟩+√

13 |11〉

∣∣12 −

⟩ψ6 ≡

∣∣1212

⟩= −

√13 |10〉

∣∣1212

⟩+√

23 |11〉

∣∣12 −

⟩ψ7 ≡

∣∣32 −

⟩=√

13 |1−1〉

∣∣1212

⟩+√

23 |10〉

∣∣12 −

⟩ψ8 ≡

∣∣12 −

⟩= −

√23 |1−1〉

∣∣1212

⟩+√

13 |10〉

∣∣12 −

Problem 6.25

{ψ1 ≡

∣∣1212

⟩= |00〉

∣∣1212

⟩ψ2 ≡

∣∣12 −

⟩= |00〉

∣∣12 −

ψ3 ≡∣∣3232

⟩= |11〉

∣∣1212

⟩ψ4 ≡

∣∣32 −

⟩= |1−1〉

∣∣12 −

⟩ψ5 ≡

∣∣3212

⟩=√

23 |10〉

∣∣1212

⟩+√

13 |11〉

∣∣12 −

⟩ψ6 ≡

∣∣1212

⟩= −

√13 |10〉

∣∣1212

⟩+√

23 |11〉

∣∣12 −

⟩ψ7 ≡

∣∣32 −

⟩=√

13 |1−1〉

∣∣1212

⟩+√

23 |10〉

∣∣12 −

⟩ψ8 ≡

∣∣12 −

⟩= −

√23 |1−1〉

∣∣1212

⟩+√

13 |10〉

∣∣12 −

⟩C. Segre (IIT) PHYS 406 - Spring 2018 February 01, 2018 4 / 18

We will use the ml , ms choice of basis set wavefunctions instead

The l = 0 states are the same asbefore but now the other 6 stateshave mixed j , and mj represen-tations using the same Clebsch-Gordan table

ψ3 ≡ |11〉∣∣1212

⟩=∣∣3232

⟩ψ4 ≡ |1−1〉

∣∣12 −

⟩=∣∣32 −

⟩ψ5 ≡ |10〉

∣∣1212

⟩=√

∣∣3212

⟩−√

∣∣1212

⟩ψ6 ≡ |11〉

∣∣12 −

⟩=√

∣∣3212

⟩+√

∣∣1212

⟩ψ7 ≡ |10〉

∣∣12 −

⟩=√

∣∣32 −

⟩+√

∣∣12 −

⟩ψ8 ≡ |1−1〉

∣∣1212

⟩=√

∣∣32 −

⟩−√

∣∣12 −

ψ3 ≡ |11〉∣∣1212

⟩=∣∣3232

⟩ψ4 ≡ |1−1〉

∣∣12 −

⟩=∣∣32 −

⟩ψ5 ≡ |10〉

∣∣1212

⟩=√

∣∣3212

⟩−√

∣∣1212

⟩ψ6 ≡ |11〉

∣∣12 −

⟩=√

∣∣3212

⟩+√

∣∣1212

⟩ψ7 ≡ |10〉

∣∣12 −

⟩=√

∣∣32 −

⟩+√

∣∣12 −

⟩ψ8 ≡ |1−1〉

∣∣1212

⟩=√

∣∣32 −

⟩−√

∣∣12 −

ψ3 ≡ |11〉∣∣1212

⟩=∣∣3232

⟩ψ4 ≡ |1−1〉

∣∣12 −

⟩=∣∣32 −

⟩ψ5 ≡ |10〉

∣∣1212

⟩=√

∣∣3212

⟩−√

∣∣1212

⟩ψ6 ≡ |11〉

∣∣12 −

⟩=√

∣∣3212

⟩+√

∣∣1212

⟩ψ7 ≡ |10〉

∣∣12 −

⟩=√

∣∣32 −

⟩+√

∣∣12 −

⟩ψ8 ≡ |1−1〉

∣∣1212

⟩=√

∣∣32 −

⟩−√

∣∣12 −

ψ3 ≡ |11〉∣∣1212

=∣∣3232

⟩ψ4 ≡ |1−1〉

∣∣12 −

⟩=∣∣32 −

⟩ψ5 ≡ |10〉

∣∣1212

⟩=√

∣∣3212

⟩−√

∣∣1212

⟩ψ6 ≡ |11〉

∣∣12 −

⟩=√

∣∣3212

⟩+√

∣∣1212

⟩ψ7 ≡ |10〉

∣∣12 −

⟩=√

∣∣32 −

⟩+√

∣∣12 −

⟩ψ8 ≡ |1−1〉

∣∣1212

⟩=√

∣∣32 −

⟩−√

∣∣12 −

ψ3 ≡ |11〉∣∣1212

⟩=∣∣3232

ψ4 ≡ |1−1〉∣∣12 −

⟩=∣∣32 −

⟩ψ5 ≡ |10〉

∣∣1212

⟩=√

∣∣3212

⟩−√

∣∣1212

⟩ψ6 ≡ |11〉

∣∣12 −

⟩=√

∣∣3212

⟩+√

∣∣1212

⟩ψ7 ≡ |10〉

∣∣12 −

⟩=√

∣∣32 −

⟩+√

∣∣12 −

⟩ψ8 ≡ |1−1〉

∣∣1212

⟩=√

∣∣32 −

⟩−√

∣∣12 −

ψ3 ≡ |11〉∣∣1212

⟩=∣∣3232

⟩ψ4 ≡ |1−1〉

∣∣12 −

=∣∣32 −

⟩ψ5 ≡ |10〉

∣∣1212

⟩=√

∣∣3212

⟩−√

∣∣1212

⟩ψ6 ≡ |11〉

∣∣12 −

⟩=√

∣∣3212

⟩+√

∣∣1212

⟩ψ7 ≡ |10〉

∣∣12 −

⟩=√

∣∣32 −

⟩+√

∣∣12 −

⟩ψ8 ≡ |1−1〉

∣∣1212

⟩=√

∣∣32 −

⟩−√

∣∣12 −

ψ3 ≡ |11〉∣∣1212

⟩=∣∣3232

⟩ψ4 ≡ |1−1〉

∣∣12 −

⟩=∣∣32 −

ψ5 ≡ |10〉∣∣1212

⟩=√

∣∣3212

⟩−√

∣∣1212

⟩ψ6 ≡ |11〉

∣∣12 −

⟩=√

∣∣3212

⟩+√

∣∣1212

⟩ψ7 ≡ |10〉

∣∣12 −

⟩=√

∣∣32 −

⟩+√

∣∣12 −

⟩ψ8 ≡ |1−1〉

∣∣1212

⟩=√

∣∣32 −

⟩−√

∣∣12 −

ψ3 ≡ |11〉∣∣1212

⟩=∣∣3232

⟩ψ4 ≡ |1−1〉

∣∣12 −

⟩=∣∣32 −

⟩ψ5 ≡ |10〉

∣∣1212

∣∣3212

⟩−√

∣∣1212

⟩ψ6 ≡ |11〉

∣∣12 −

⟩=√

∣∣3212

⟩+√

∣∣1212

⟩ψ7 ≡ |10〉

∣∣12 −

⟩=√

∣∣32 −

⟩+√

∣∣12 −

⟩ψ8 ≡ |1−1〉

∣∣1212

⟩=√

∣∣32 −

⟩−√

∣∣12 −

ψ3 ≡ |11〉∣∣1212

⟩=∣∣3232

⟩ψ4 ≡ |1−1〉

∣∣12 −

⟩=∣∣32 −

⟩ψ5 ≡ |10〉

∣∣1212

⟩=√

∣∣3212

⟩−√

∣∣1212

ψ6 ≡ |11〉∣∣12 −

⟩=√

∣∣3212

⟩+√

∣∣1212

⟩ψ7 ≡ |10〉

∣∣12 −

⟩=√

∣∣32 −

⟩+√

∣∣12 −

⟩ψ8 ≡ |1−1〉

∣∣1212

⟩=√

∣∣32 −

⟩−√

∣∣12 −

ψ3 ≡ |11〉∣∣1212

⟩=∣∣3232

⟩ψ4 ≡ |1−1〉

∣∣12 −

⟩=∣∣32 −

⟩ψ5 ≡ |10〉

∣∣1212

⟩=√

∣∣3212

⟩−√

∣∣1212

⟩ψ6 ≡ |11〉

∣∣12 −

∣∣3212

⟩+√

∣∣1212

⟩ψ7 ≡ |10〉

∣∣12 −

⟩=√

∣∣32 −

⟩+√

∣∣12 −

⟩ψ8 ≡ |1−1〉

∣∣1212

⟩=√

∣∣32 −

⟩−√

∣∣12 −

ψ3 ≡ |11〉∣∣1212

⟩=∣∣3232

⟩ψ4 ≡ |1−1〉

∣∣12 −

⟩=∣∣32 −

⟩ψ5 ≡ |10〉

∣∣1212

⟩=√

∣∣3212

⟩−√

∣∣1212

⟩ψ6 ≡ |11〉

∣∣12 −

⟩=√

∣∣3212

⟩+√

∣∣1212

ψ7 ≡ |10〉∣∣12 −

⟩=√

∣∣32 −

⟩+√

∣∣12 −

⟩ψ8 ≡ |1−1〉

∣∣1212

⟩=√

∣∣32 −

⟩−√

∣∣12 −

ψ3 ≡ |11〉∣∣1212

⟩=∣∣3232

⟩ψ4 ≡ |1−1〉

∣∣12 −

⟩=∣∣32 −

⟩ψ5 ≡ |10〉

∣∣1212

⟩=√

∣∣3212

⟩−√

∣∣1212

⟩ψ6 ≡ |11〉

∣∣12 −

⟩=√

∣∣3212

⟩+√

∣∣1212

⟩ψ7 ≡ |10〉

∣∣12 −

∣∣32 −

⟩+√

∣∣12 −

⟩ψ8 ≡ |1−1〉

∣∣1212

⟩=√

∣∣32 −

⟩−√

∣∣12 −

ψ3 ≡ |11〉∣∣1212

⟩=∣∣3232

⟩ψ4 ≡ |1−1〉

∣∣12 −

⟩=∣∣32 −

⟩ψ5 ≡ |10〉

∣∣1212

⟩=√

∣∣3212

⟩−√

∣∣1212

⟩ψ6 ≡ |11〉

∣∣12 −

⟩=√

∣∣3212

⟩+√

∣∣1212

⟩ψ7 ≡ |10〉

∣∣12 −

⟩=√

∣∣32 −

⟩+√

∣∣12 −

ψ8 ≡ |1−1〉∣∣1212

⟩=√

∣∣32 −

⟩−√

∣∣12 −

ψ3 ≡ |11〉∣∣1212

⟩=∣∣3232

⟩ψ4 ≡ |1−1〉

∣∣12 −

⟩=∣∣32 −

⟩ψ5 ≡ |10〉

∣∣1212

⟩=√

∣∣3212

⟩−√

∣∣1212

⟩ψ6 ≡ |11〉

∣∣12 −

⟩=√

∣∣3212

⟩+√

∣∣1212

⟩ψ7 ≡ |10〉

∣∣12 −

⟩=√

∣∣32 −

⟩+√

∣∣12 −

⟩ψ8 ≡ |1−1〉

∣∣1212

∣∣32 −

⟩−√

∣∣12 −

ψ3 ≡ |11〉∣∣1212

⟩=∣∣3232

⟩ψ4 ≡ |1−1〉

∣∣12 −

⟩=∣∣32 −

⟩ψ5 ≡ |10〉

∣∣1212

⟩=√

∣∣3212

⟩−√

∣∣1212

⟩ψ6 ≡ |11〉

∣∣12 −

⟩=√

∣∣3212

⟩+√

∣∣1212

⟩ψ7 ≡ |10〉

∣∣12 −

⟩=√

∣∣32 −

⟩+√

∣∣12 −

⟩ψ8 ≡ |1−1〉

∣∣1212

⟩=√

∣∣32 −

⟩−√

∣∣12 −

Complete n = 2, l = 0, 1 degenerate set

The complete set of 8 n = 2, l = 0, 1 degenerate states using quantumnumbers l , s, ml , and ms can now be listed

ψ1 ≡ |00〉∣∣1212

⟩=∣∣1212

⟩ψ2 ≡ |00〉

∣∣12 −

⟩=∣∣12 −

ψ3 ≡ |11〉∣∣1212

⟩=∣∣3232

⟩ψ4 ≡ |1−1〉

∣∣12 −

⟩=∣∣32 −

⟩ψ5 ≡ |10〉

∣∣1212

⟩=√

∣∣3212

⟩−√

∣∣1212

⟩ψ6 ≡ |11〉

∣∣12 −

⟩=√

∣∣3212

⟩+√

∣∣1212

⟩ψ7 ≡ |10〉

∣∣12 −

⟩=√

∣∣32 −

⟩+√

∣∣12 −

⟩ψ8 ≡ |1−1〉

∣∣1212

⟩=√

∣∣32 −

⟩−√

∣∣12 −

now build the W matrix for the H ′ = HZ + Hfs perturbation

{ψ1 ≡ |00〉

∣∣1212

⟩=∣∣1212

⟩ψ2 ≡ |00〉

∣∣12 −

⟩=∣∣12 −

ψ3 ≡ |11〉∣∣1212

⟩=∣∣3232

⟩ψ4 ≡ |1−1〉

∣∣12 −

⟩=∣∣32 −

⟩ψ5 ≡ |10〉

∣∣1212

⟩=√

∣∣3212

⟩−√

∣∣1212

⟩ψ6 ≡ |11〉

∣∣12 −

⟩=√

∣∣3212

⟩+√

∣∣1212

⟩ψ7 ≡ |10〉

∣∣12 −

⟩=√

∣∣32 −

⟩+√

∣∣12 −

⟩ψ8 ≡ |1−1〉

∣∣1212

⟩=√

∣∣32 −

⟩−√

∣∣12 −

now build the W matrix for the H ′ = HZ + Hfs perturbation

{ψ1 ≡ |00〉

∣∣1212

⟩=∣∣1212

⟩ψ2 ≡ |00〉

∣∣12 −

⟩=∣∣12 −

ψ3 ≡ |11〉∣∣1212

⟩=∣∣3232

⟩ψ4 ≡ |1−1〉

∣∣12 −

⟩=∣∣32 −

⟩ψ5 ≡ |10〉

∣∣1212

⟩=√

∣∣3212

⟩−√

∣∣1212

⟩ψ6 ≡ |11〉

∣∣12 −

⟩=√

∣∣3212

⟩+√

∣∣1212

⟩ψ7 ≡ |10〉

∣∣12 −

⟩=√

∣∣32 −

⟩+√

∣∣12 −

⟩ψ8 ≡ |1−1〉

∣∣1212

⟩=√

∣∣32 −

⟩−√

∣∣12 −

⟩now build the W matrix for the H ′ = HZ + Hfs perturbation

Matrix element calculations

The Zeeman and finestructure correctionsare

taking n = 2 and re-placing the constantswith β and γ for sim-plification

EZ = µBBext(ml + 2ms) = β(ml + 2ms)

Efs = −13.6α2

j + 12

]= −γ

j + 12

now compute the various matrix elements usingthe appropriate basis set for each correction⟨

ψ1|H ′|ψ1

⟩=⟨1212 |Hfs |12

⟩+⟨001

212 |HZ |12

1200⟩

= −γ[8− 3] + β

= −5γ + β⟨ψ2|H ′|ψ2

⟩=⟨12 −

12 |Hfs |12 −

⟩+⟨001

2 −12 |HZ |12 −

1200⟩

= −γ[8− 3]− β= −5γ − β⟨

ψ3|H ′|ψ3

⟩=⟨3232 |Hfs |32

⟩+⟨111

212 |HZ |12

1211⟩

= −γ[4− 3] + 2β

= −γ + 2β⟨ψ4|H ′|ψ4

⟩=⟨32 −

32 |Hfs |32 −

⟩+⟨1−11

2 −12 |HZ |12 −

121−1

⟩= −γ[4− 3]− 2β

= −γ − 2β

= β(ml + 2ms)

Efs = −13.6α2

j + 12

]= −γ

j + 12

ψ1|H ′|ψ1

⟩=⟨1212 |Hfs |12

⟩+⟨001

212 |HZ |12

1200⟩

= −γ[8− 3] + β

= −5γ + β⟨ψ2|H ′|ψ2

⟩=⟨12 −

12 |Hfs |12 −

⟩+⟨001

2 −12 |HZ |12 −

1200⟩

= −γ[8− 3]− β= −5γ − β⟨

ψ3|H ′|ψ3

⟩=⟨3232 |Hfs |32

⟩+⟨111

212 |HZ |12

1211⟩

= −γ[4− 3] + 2β

= −γ + 2β⟨ψ4|H ′|ψ4

⟩=⟨32 −

32 |Hfs |32 −

⟩+⟨1−11

2 −12 |HZ |12 −

121−1

⟩= −γ[4− 3]− 2β

= −γ − 2β

= β(ml + 2ms)

Efs = −13.6α2

j + 12

= −γ

j + 12

ψ1|H ′|ψ1

⟩=⟨1212 |Hfs |12

⟩+⟨001

212 |HZ |12

1200⟩

= −γ[8− 3] + β

= −5γ + β⟨ψ2|H ′|ψ2

⟩=⟨12 −

12 |Hfs |12 −

⟩+⟨001

2 −12 |HZ |12 −

1200⟩

= −γ[8− 3]− β= −5γ − β⟨

ψ3|H ′|ψ3

⟩=⟨3232 |Hfs |32

⟩+⟨111

212 |HZ |12

1211⟩

= −γ[4− 3] + 2β

= −γ + 2β⟨ψ4|H ′|ψ4

⟩=⟨32 −

32 |Hfs |32 −

⟩+⟨1−11

2 −12 |HZ |12 −

121−1

⟩= −γ[4− 3]− 2β

= −γ − 2β

= β(ml + 2ms)

Efs = −13.6α2

j + 12

= −γ

j + 12

ψ1|H ′|ψ1

⟩=⟨1212 |Hfs |12

⟩+⟨001

212 |HZ |12

1200⟩

= −γ[8− 3] + β

= −5γ + β⟨ψ2|H ′|ψ2

⟩=⟨12 −

12 |Hfs |12 −

⟩+⟨001

2 −12 |HZ |12 −

1200⟩

= −γ[8− 3]− β= −5γ − β⟨

ψ3|H ′|ψ3

⟩=⟨3232 |Hfs |32

⟩+⟨111

212 |HZ |12

1211⟩

= −γ[4− 3] + 2β

= −γ + 2β⟨ψ4|H ′|ψ4

⟩=⟨32 −

32 |Hfs |32 −

⟩+⟨1−11

2 −12 |HZ |12 −

121−1

⟩= −γ[4− 3]− 2β

= −γ − 2β

Efs = −13.6α2

j + 12

= −γ

j + 12

ψ1|H ′|ψ1

⟩=⟨1212 |Hfs |12

⟩+⟨001

212 |HZ |12

1200⟩

= −γ[8− 3] + β

= −5γ + β⟨ψ2|H ′|ψ2

⟩=⟨12 −

12 |Hfs |12 −

⟩+⟨001

2 −12 |HZ |12 −

1200⟩

= −γ[8− 3]− β= −5γ − β⟨

ψ3|H ′|ψ3

⟩=⟨3232 |Hfs |32

⟩+⟨111

212 |HZ |12

1211⟩

= −γ[4− 3] + 2β

= −γ + 2β⟨ψ4|H ′|ψ4

⟩=⟨32 −

32 |Hfs |32 −

⟩+⟨1−11

2 −12 |HZ |12 −

121−1

⟩= −γ[4− 3]− 2β

= −γ − 2β

Efs = −13.6α2

j + 12

]= −γ

j + 12

ψ1|H ′|ψ1

⟩=⟨1212 |Hfs |12

⟩+⟨001

212 |HZ |12

1200⟩

= −γ[8− 3] + β

= −5γ + β⟨ψ2|H ′|ψ2

⟩=⟨12 −

12 |Hfs |12 −

⟩+⟨001

2 −12 |HZ |12 −

1200⟩

= −γ[8− 3]− β= −5γ − β⟨

ψ3|H ′|ψ3

⟩=⟨3232 |Hfs |32

⟩+⟨111

212 |HZ |12

1211⟩

= −γ[4− 3] + 2β

= −γ + 2β⟨ψ4|H ′|ψ4

⟩=⟨32 −

32 |Hfs |32 −

⟩+⟨1−11

2 −12 |HZ |12 −

121−1

⟩= −γ[4− 3]− 2β

= −γ − 2β

Efs = −13.6α2

j + 12

]= −γ

j + 12

now compute the various matrix elements usingthe appropriate basis set for each correction

⟨ψ1|H ′|ψ1

⟩=⟨1212 |Hfs |12

⟩+⟨001

212 |HZ |12

1200⟩

= −γ[8− 3] + β

= −5γ + β⟨ψ2|H ′|ψ2

⟩=⟨12 −

12 |Hfs |12 −

⟩+⟨001

2 −12 |HZ |12 −

1200⟩

= −γ[8− 3]− β= −5γ − β⟨

ψ3|H ′|ψ3

⟩=⟨3232 |Hfs |32

⟩+⟨111

212 |HZ |12

1211⟩

= −γ[4− 3] + 2β

= −γ + 2β⟨ψ4|H ′|ψ4

⟩=⟨32 −

32 |Hfs |32 −

⟩+⟨1−11

2 −12 |HZ |12 −

121−1

⟩= −γ[4− 3]− 2β

= −γ − 2β

Efs = −13.6α2

j + 12

]= −γ

j + 12

ψ1|H ′|ψ1

=⟨1212 |Hfs |12

⟩+⟨001

212 |HZ |12

1200⟩

= −γ[8− 3] + β

= −5γ + β⟨ψ2|H ′|ψ2

⟩=⟨12 −

12 |Hfs |12 −

⟩+⟨001

2 −12 |HZ |12 −

1200⟩

= −γ[8− 3]− β= −5γ − β⟨

ψ3|H ′|ψ3

⟩=⟨3232 |Hfs |32

⟩+⟨111

212 |HZ |12

1211⟩

= −γ[4− 3] + 2β

= −γ + 2β⟨ψ4|H ′|ψ4

⟩=⟨32 −

32 |Hfs |32 −

⟩+⟨1−11

2 −12 |HZ |12 −

121−1

⟩= −γ[4− 3]− 2β

= −γ − 2β

Efs = −13.6α2

j + 12

]= −γ

j + 12

ψ1|H ′|ψ1

⟩=⟨1212 |Hfs |12

⟩+⟨001

212 |HZ |12

1200⟩

= −γ[8− 3] + β

= −5γ + β⟨ψ2|H ′|ψ2

⟩=⟨12 −

12 |Hfs |12 −

⟩+⟨001

2 −12 |HZ |12 −

1200⟩

= −γ[8− 3]− β= −5γ − β⟨

ψ3|H ′|ψ3

⟩=⟨3232 |Hfs |32

⟩+⟨111

212 |HZ |12

1211⟩

= −γ[4− 3] + 2β

= −γ + 2β⟨ψ4|H ′|ψ4

⟩=⟨32 −

32 |Hfs |32 −

⟩+⟨1−11

2 −12 |HZ |12 −

121−1

⟩= −γ[4− 3]− 2β

= −γ − 2β

Efs = −13.6α2

j + 12

]= −γ

j + 12

ψ1|H ′|ψ1

⟩=⟨1212 |Hfs |12

⟩+⟨001

212 |HZ |12

1200⟩

= −γ[8− 3] + β

= −5γ + β⟨ψ2|H ′|ψ2

⟩=⟨12 −

12 |Hfs |12 −

⟩+⟨001

2 −12 |HZ |12 −

1200⟩

= −γ[8− 3]− β= −5γ − β⟨

ψ3|H ′|ψ3

⟩=⟨3232 |Hfs |32

⟩+⟨111

212 |HZ |12

1211⟩

= −γ[4− 3] + 2β

= −γ + 2β⟨ψ4|H ′|ψ4

⟩=⟨32 −

32 |Hfs |32 −

⟩+⟨1−11

2 −12 |HZ |12 −

121−1

⟩= −γ[4− 3]− 2β

= −γ − 2β

Efs = −13.6α2

j + 12

]= −γ

j + 12

ψ1|H ′|ψ1

⟩=⟨1212 |Hfs |12

⟩+⟨001

212 |HZ |12

1200⟩

= −γ[8− 3] + β

= −5γ + β

⟨ψ2|H ′|ψ2

⟩=⟨12 −

12 |Hfs |12 −

⟩+⟨001

2 −12 |HZ |12 −

1200⟩

= −γ[8− 3]− β= −5γ − β⟨

ψ3|H ′|ψ3

⟩=⟨3232 |Hfs |32

⟩+⟨111

212 |HZ |12

1211⟩

= −γ[4− 3] + 2β

= −γ + 2β⟨ψ4|H ′|ψ4

⟩=⟨32 −

32 |Hfs |32 −

⟩+⟨1−11

2 −12 |HZ |12 −

121−1

⟩= −γ[4− 3]− 2β

= −γ − 2β

Efs = −13.6α2

j + 12

]= −γ

j + 12

ψ1|H ′|ψ1

⟩=⟨1212 |Hfs |12

⟩+⟨001

212 |HZ |12

1200⟩

= −γ[8− 3] + β

= −5γ + β⟨ψ2|H ′|ψ2

=⟨12 −

12 |Hfs |12 −

⟩+⟨001

2 −12 |HZ |12 −

1200⟩

= −γ[8− 3]− β= −5γ − β⟨

ψ3|H ′|ψ3

⟩=⟨3232 |Hfs |32

⟩+⟨111

212 |HZ |12

1211⟩

= −γ[4− 3] + 2β

= −γ + 2β⟨ψ4|H ′|ψ4

⟩=⟨32 −

32 |Hfs |32 −

⟩+⟨1−11

2 −12 |HZ |12 −

121−1

⟩= −γ[4− 3]− 2β

= −γ − 2β

Efs = −13.6α2

j + 12

]= −γ

j + 12

ψ1|H ′|ψ1

⟩=⟨1212 |Hfs |12

⟩+⟨001

212 |HZ |12

1200⟩

= −γ[8− 3] + β

= −5γ + β⟨ψ2|H ′|ψ2

⟩=⟨12 −

12 |Hfs |12 −

⟩+⟨001

2 −12 |HZ |12 −

1200⟩

= −γ[8− 3]− β= −5γ − β⟨

ψ3|H ′|ψ3

⟩=⟨3232 |Hfs |32

⟩+⟨111

212 |HZ |12

1211⟩

= −γ[4− 3] + 2β

= −γ + 2β⟨ψ4|H ′|ψ4

⟩=⟨32 −

32 |Hfs |32 −

⟩+⟨1−11

2 −12 |HZ |12 −

121−1

⟩= −γ[4− 3]− 2β

= −γ − 2β

Efs = −13.6α2

j + 12

]= −γ

j + 12

ψ1|H ′|ψ1

⟩=⟨1212 |Hfs |12

⟩+⟨001

212 |HZ |12

1200⟩

= −γ[8− 3] + β

= −5γ + β⟨ψ2|H ′|ψ2

⟩=⟨12 −

12 |Hfs |12 −

⟩+⟨001

2 −12 |HZ |12 −

1200⟩

= −γ[8− 3]− β

= −5γ − β⟨ψ3|H ′|ψ3

⟩=⟨3232 |Hfs |32

⟩+⟨111

212 |HZ |12

1211⟩

= −γ[4− 3] + 2β

= −γ + 2β⟨ψ4|H ′|ψ4

⟩=⟨32 −

32 |Hfs |32 −

⟩+⟨1−11

2 −12 |HZ |12 −

121−1

⟩= −γ[4− 3]− 2β

= −γ − 2β

Efs = −13.6α2

j + 12

]= −γ

j + 12

ψ1|H ′|ψ1

⟩=⟨1212 |Hfs |12

⟩+⟨001

212 |HZ |12

1200⟩

= −γ[8− 3] + β

= −5γ + β⟨ψ2|H ′|ψ2

⟩=⟨12 −

12 |Hfs |12 −

⟩+⟨001

2 −12 |HZ |12 −

1200⟩

= −γ[8− 3]− β= −5γ − β

⟨ψ3|H ′|ψ3

⟩=⟨3232 |Hfs |32

⟩+⟨111

212 |HZ |12

1211⟩

= −γ[4− 3] + 2β

= −γ + 2β⟨ψ4|H ′|ψ4

⟩=⟨32 −

32 |Hfs |32 −

⟩+⟨1−11

2 −12 |HZ |12 −

121−1

⟩= −γ[4− 3]− 2β

= −γ − 2β

Efs = −13.6α2

j + 12

]= −γ

j + 12

ψ1|H ′|ψ1

⟩=⟨1212 |Hfs |12

⟩+⟨001

212 |HZ |12

1200⟩

= −γ[8− 3] + β

= −5γ + β⟨ψ2|H ′|ψ2

⟩=⟨12 −

12 |Hfs |12 −

⟩+⟨001

2 −12 |HZ |12 −

1200⟩

= −γ[8− 3]− β= −5γ − β⟨

ψ3|H ′|ψ3

=⟨3232 |Hfs |32

⟩+⟨111

212 |HZ |12

1211⟩

= −γ[4− 3] + 2β

= −γ + 2β⟨ψ4|H ′|ψ4

⟩=⟨32 −

32 |Hfs |32 −

⟩+⟨1−11

2 −12 |HZ |12 −

121−1

⟩= −γ[4− 3]− 2β

= −γ − 2β

Efs = −13.6α2

j + 12

]= −γ

j + 12

ψ1|H ′|ψ1

⟩=⟨1212 |Hfs |12

⟩+⟨001

212 |HZ |12

1200⟩

= −γ[8− 3] + β

= −5γ + β⟨ψ2|H ′|ψ2

⟩=⟨12 −

12 |Hfs |12 −

⟩+⟨001

2 −12 |HZ |12 −

1200⟩

= −γ[8− 3]− β= −5γ − β⟨

ψ3|H ′|ψ3

⟩=⟨3232 |Hfs |32

⟩+⟨111

212 |HZ |12

1211⟩

= −γ[4− 3] + 2β

= −γ + 2β⟨ψ4|H ′|ψ4

⟩=⟨32 −

32 |Hfs |32 −

⟩+⟨1−11

2 −12 |HZ |12 −

121−1

⟩= −γ[4− 3]− 2β

= −γ − 2β

Efs = −13.6α2

j + 12

]= −γ

j + 12

ψ1|H ′|ψ1

⟩=⟨1212 |Hfs |12

⟩+⟨001

212 |HZ |12

1200⟩

= −γ[8− 3] + β

= −5γ + β⟨ψ2|H ′|ψ2

⟩=⟨12 −

12 |Hfs |12 −

⟩+⟨001

2 −12 |HZ |12 −

1200⟩

= −γ[8− 3]− β= −5γ − β⟨

ψ3|H ′|ψ3

⟩=⟨3232 |Hfs |32

⟩+⟨111

212 |HZ |12

1211⟩

= −γ[4− 3] + 2β

= −γ + 2β⟨ψ4|H ′|ψ4

⟩=⟨32 −

32 |Hfs |32 −

⟩+⟨1−11

2 −12 |HZ |12 −

121−1

⟩= −γ[4− 3]− 2β

= −γ − 2β

Efs = −13.6α2

j + 12

]= −γ

j + 12

ψ1|H ′|ψ1

⟩=⟨1212 |Hfs |12

⟩+⟨001

212 |HZ |12

1200⟩

= −γ[8− 3] + β

= −5γ + β⟨ψ2|H ′|ψ2

⟩=⟨12 −

12 |Hfs |12 −

⟩+⟨001

2 −12 |HZ |12 −

1200⟩

= −γ[8− 3]− β= −5γ − β⟨

ψ3|H ′|ψ3

⟩=⟨3232 |Hfs |32

⟩+⟨111

212 |HZ |12

1211⟩

= −γ[4− 3] + 2β

= −γ + 2β

⟨ψ4|H ′|ψ4

⟩=⟨32 −

32 |Hfs |32 −

⟩+⟨1−11

2 −12 |HZ |12 −

121−1

⟩= −γ[4− 3]− 2β

= −γ − 2β

Efs = −13.6α2

j + 12

]= −γ

j + 12

ψ1|H ′|ψ1

⟩=⟨1212 |Hfs |12

⟩+⟨001

212 |HZ |12

1200⟩

= −γ[8− 3] + β

= −5γ + β⟨ψ2|H ′|ψ2

⟩=⟨12 −

12 |Hfs |12 −

⟩+⟨001

2 −12 |HZ |12 −

1200⟩

= −γ[8− 3]− β= −5γ − β⟨

ψ3|H ′|ψ3

⟩=⟨3232 |Hfs |32

⟩+⟨111

212 |HZ |12

1211⟩

= −γ[4− 3] + 2β

= −γ + 2β⟨ψ4|H ′|ψ4

=⟨32 −

32 |Hfs |32 −

⟩+⟨1−11

2 −12 |HZ |12 −

121−1

⟩= −γ[4− 3]− 2β

= −γ − 2β

Efs = −13.6α2

j + 12

]= −γ

j + 12

ψ1|H ′|ψ1

⟩=⟨1212 |Hfs |12

⟩+⟨001

212 |HZ |12

1200⟩

= −γ[8− 3] + β

= −5γ + β⟨ψ2|H ′|ψ2

⟩=⟨12 −

12 |Hfs |12 −

⟩+⟨001

2 −12 |HZ |12 −

1200⟩

= −γ[8− 3]− β= −5γ − β⟨

ψ3|H ′|ψ3

⟩=⟨3232 |Hfs |32

⟩+⟨111

212 |HZ |12

1211⟩

= −γ[4− 3] + 2β

= −γ + 2β⟨ψ4|H ′|ψ4

⟩=⟨32 −

32 |Hfs |32 −

⟩+⟨1−11

2 −12 |HZ |12 −

121−1

= −γ[4− 3]− 2β

= −γ − 2β

Efs = −13.6α2

j + 12

]= −γ

j + 12

ψ1|H ′|ψ1

⟩=⟨1212 |Hfs |12

⟩+⟨001

212 |HZ |12

1200⟩

= −γ[8− 3] + β

= −5γ + β⟨ψ2|H ′|ψ2

⟩=⟨12 −

12 |Hfs |12 −

⟩+⟨001

2 −12 |HZ |12 −

1200⟩

= −γ[8− 3]− β= −5γ − β⟨

ψ3|H ′|ψ3

⟩=⟨3232 |Hfs |32

⟩+⟨111

212 |HZ |12

1211⟩

= −γ[4− 3] + 2β

= −γ + 2β⟨ψ4|H ′|ψ4

⟩=⟨32 −

32 |Hfs |32 −

⟩+⟨1−11

2 −12 |HZ |12 −

121−1

⟩= −γ[4− 3]− 2β

= −γ − 2β

Efs = −13.6α2

j + 12

]= −γ

j + 12

ψ1|H ′|ψ1

⟩=⟨1212 |Hfs |12

⟩+⟨001

212 |HZ |12

1200⟩

= −γ[8− 3] + β

= −5γ + β⟨ψ2|H ′|ψ2

⟩=⟨12 −

12 |Hfs |12 −

⟩+⟨001

2 −12 |HZ |12 −

1200⟩

= −γ[8− 3]− β= −5γ − β⟨

ψ3|H ′|ψ3

⟩=⟨3232 |Hfs |32

⟩+⟨111

212 |HZ |12

1211⟩

= −γ[4− 3] + 2β

= −γ + 2β⟨ψ4|H ′|ψ4

⟩=⟨32 −

32 |Hfs |32 −

⟩+⟨1−11

2 −12 |HZ |12 −

121−1

⟩= −γ[4− 3]− 2β

= −γ − 2βC. Segre (IIT) PHYS 406 - Spring 2018 February 01, 2018 7 / 18

EZ = β(ml + 2ms) Efs = −γ

j + 12

⟨ψ5|H ′|ψ5

⟩= 2

⟨3212 |Hfs |32

⟩+ 1

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3] + β = −73γ + β⟨

ψ6|H ′|ψ6

⟩= 1

⟨3212 |Hfs |32

⟩+ 2

⟨1212 |Hfs |12

⟩+⟨111

2 −12 |HZ |12 −

1211⟩

= −γ 13 [4− 3]− γ 2

3 [8− 3] = −113 γ⟨

ψ7|H ′|ψ7

⟩= 2

⟨32 −

12 |Hfs |32 −

⟩+ 1

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12 −

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3]− β = −73γ − β⟨

ψ8|H ′|ψ8

⟩= 1

⟨32 −

12 |Hfs |32 −

⟩+ 2

⟨12 −

12 |Hfs |12 −

⟩+⟨1−11

212 |HZ |12

121−1

⟩= −γ 1

3 [4− 3]− γ 23 [8− 3] = −11

j + 12

⟨ψ5|H ′|ψ5

⟨3212 |Hfs |32

⟩+ 1

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3] + β = −73γ + β⟨

ψ6|H ′|ψ6

⟩= 1

⟨3212 |Hfs |32

⟩+ 2

⟨1212 |Hfs |12

⟩+⟨111

2 −12 |HZ |12 −

1211⟩

= −γ 13 [4− 3]− γ 2

3 [8− 3] = −113 γ⟨

ψ7|H ′|ψ7

⟩= 2

⟨32 −

12 |Hfs |32 −

⟩+ 1

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12 −

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3]− β = −73γ − β⟨

ψ8|H ′|ψ8

⟩= 1

⟨32 −

12 |Hfs |32 −

⟩+ 2

⟨12 −

12 |Hfs |12 −

⟩+⟨1−11

212 |HZ |12

121−1

⟩= −γ 1

3 [4− 3]− γ 23 [8− 3] = −11

j + 12

⟨ψ5|H ′|ψ5

⟩= 2

⟨3212 |Hfs |32

⟩+ 1

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3] + β = −73γ + β⟨

ψ6|H ′|ψ6

⟩= 1

⟨3212 |Hfs |32

⟩+ 2

⟨1212 |Hfs |12

⟩+⟨111

2 −12 |HZ |12 −

1211⟩

= −γ 13 [4− 3]− γ 2

3 [8− 3] = −113 γ⟨

ψ7|H ′|ψ7

⟩= 2

⟨32 −

12 |Hfs |32 −

⟩+ 1

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12 −

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3]− β = −73γ − β⟨

ψ8|H ′|ψ8

⟩= 1

⟨32 −

12 |Hfs |32 −

⟩+ 2

⟨12 −

12 |Hfs |12 −

⟩+⟨1−11

212 |HZ |12

121−1

⟩= −γ 1

3 [4− 3]− γ 23 [8− 3] = −11

j + 12

⟨ψ5|H ′|ψ5

⟩= 2

⟨3212 |Hfs |32

⟩+ 1

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3] + β

= −73γ + β⟨

ψ6|H ′|ψ6

⟩= 1

⟨3212 |Hfs |32

⟩+ 2

⟨1212 |Hfs |12

⟩+⟨111

2 −12 |HZ |12 −

1211⟩

= −γ 13 [4− 3]− γ 2

3 [8− 3] = −113 γ⟨

ψ7|H ′|ψ7

⟩= 2

⟨32 −

12 |Hfs |32 −

⟩+ 1

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12 −

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3]− β = −73γ − β⟨

ψ8|H ′|ψ8

⟩= 1

⟨32 −

12 |Hfs |32 −

⟩+ 2

⟨12 −

12 |Hfs |12 −

⟩+⟨1−11

212 |HZ |12

121−1

⟩= −γ 1

3 [4− 3]− γ 23 [8− 3] = −11

j + 12

⟨ψ5|H ′|ψ5

⟩= 2

⟨3212 |Hfs |32

⟩+ 1

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3] + β = −73γ + β

⟨ψ6|H ′|ψ6

⟩= 1

⟨3212 |Hfs |32

⟩+ 2

⟨1212 |Hfs |12

⟩+⟨111

2 −12 |HZ |12 −

1211⟩

= −γ 13 [4− 3]− γ 2

3 [8− 3] = −113 γ⟨

ψ7|H ′|ψ7

⟩= 2

⟨32 −

12 |Hfs |32 −

⟩+ 1

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12 −

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3]− β = −73γ − β⟨

ψ8|H ′|ψ8

⟩= 1

⟨32 −

12 |Hfs |32 −

⟩+ 2

⟨12 −

12 |Hfs |12 −

⟩+⟨1−11

212 |HZ |12

121−1

⟩= −γ 1

3 [4− 3]− γ 23 [8− 3] = −11

j + 12

⟨ψ5|H ′|ψ5

⟩= 2

⟨3212 |Hfs |32

⟩+ 1

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3] + β = −73γ + β⟨

ψ6|H ′|ψ6

⟨3212 |Hfs |32

⟩+ 2

⟨1212 |Hfs |12

⟩+⟨111

2 −12 |HZ |12 −

1211⟩

= −γ 13 [4− 3]− γ 2

3 [8− 3] = −113 γ⟨

ψ7|H ′|ψ7

⟩= 2

⟨32 −

12 |Hfs |32 −

⟩+ 1

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12 −

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3]− β = −73γ − β⟨

ψ8|H ′|ψ8

⟩= 1

⟨32 −

12 |Hfs |32 −

⟩+ 2

⟨12 −

12 |Hfs |12 −

⟩+⟨1−11

212 |HZ |12

121−1

⟩= −γ 1

3 [4− 3]− γ 23 [8− 3] = −11

j + 12

⟨ψ5|H ′|ψ5

⟩= 2

⟨3212 |Hfs |32

⟩+ 1

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3] + β = −73γ + β⟨

ψ6|H ′|ψ6

⟩= 1

⟨3212 |Hfs |32

⟩+ 2

⟨1212 |Hfs |12

⟩+⟨111

2 −12 |HZ |12 −

1211⟩

= −γ 13 [4− 3]− γ 2

3 [8− 3] = −113 γ⟨

ψ7|H ′|ψ7

⟩= 2

⟨32 −

12 |Hfs |32 −

⟩+ 1

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12 −

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3]− β = −73γ − β⟨

ψ8|H ′|ψ8

⟩= 1

⟨32 −

12 |Hfs |32 −

⟩+ 2

⟨12 −

12 |Hfs |12 −

⟩+⟨1−11

212 |HZ |12

121−1

⟩= −γ 1

3 [4− 3]− γ 23 [8− 3] = −11

j + 12

⟨ψ5|H ′|ψ5

⟩= 2

⟨3212 |Hfs |32

⟩+ 1

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3] + β = −73γ + β⟨

ψ6|H ′|ψ6

⟩= 1

⟨3212 |Hfs |32

⟩+ 2

⟨1212 |Hfs |12

⟩+⟨111

2 −12 |HZ |12 −

1211⟩

= −γ 13 [4− 3]− γ 2

3 [8− 3]

= −113 γ⟨

ψ7|H ′|ψ7

⟩= 2

⟨32 −

12 |Hfs |32 −

⟩+ 1

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12 −

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3]− β = −73γ − β⟨

ψ8|H ′|ψ8

⟩= 1

⟨32 −

12 |Hfs |32 −

⟩+ 2

⟨12 −

12 |Hfs |12 −

⟩+⟨1−11

212 |HZ |12

121−1

⟩= −γ 1

3 [4− 3]− γ 23 [8− 3] = −11

j + 12

⟨ψ5|H ′|ψ5

⟩= 2

⟨3212 |Hfs |32

⟩+ 1

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3] + β = −73γ + β⟨

ψ6|H ′|ψ6

⟩= 1

⟨3212 |Hfs |32

⟩+ 2

⟨1212 |Hfs |12

⟩+⟨111

2 −12 |HZ |12 −

1211⟩

= −γ 13 [4− 3]− γ 2

3 [8− 3] = −113 γ

⟨ψ7|H ′|ψ7

⟩= 2

⟨32 −

12 |Hfs |32 −

⟩+ 1

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12 −

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3]− β = −73γ − β⟨

ψ8|H ′|ψ8

⟩= 1

⟨32 −

12 |Hfs |32 −

⟩+ 2

⟨12 −

12 |Hfs |12 −

⟩+⟨1−11

212 |HZ |12

121−1

⟩= −γ 1

3 [4− 3]− γ 23 [8− 3] = −11

j + 12

⟨ψ5|H ′|ψ5

⟩= 2

⟨3212 |Hfs |32

⟩+ 1

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3] + β = −73γ + β⟨

ψ6|H ′|ψ6

⟩= 1

⟨3212 |Hfs |32

⟩+ 2

⟨1212 |Hfs |12

⟩+⟨111

2 −12 |HZ |12 −

1211⟩

= −γ 13 [4− 3]− γ 2

3 [8− 3] = −113 γ⟨

ψ7|H ′|ψ7

⟨32 −

12 |Hfs |32 −

⟩+ 1

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12 −

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3]− β = −73γ − β⟨

ψ8|H ′|ψ8

⟩= 1

⟨32 −

12 |Hfs |32 −

⟩+ 2

⟨12 −

12 |Hfs |12 −

⟩+⟨1−11

212 |HZ |12

121−1

⟩= −γ 1

3 [4− 3]− γ 23 [8− 3] = −11

j + 12

⟨ψ5|H ′|ψ5

⟩= 2

⟨3212 |Hfs |32

⟩+ 1

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3] + β = −73γ + β⟨

ψ6|H ′|ψ6

⟩= 1

⟨3212 |Hfs |32

⟩+ 2

⟨1212 |Hfs |12

⟩+⟨111

2 −12 |HZ |12 −

1211⟩

= −γ 13 [4− 3]− γ 2

3 [8− 3] = −113 γ⟨

ψ7|H ′|ψ7

⟩= 2

⟨32 −

12 |Hfs |32 −

⟩+ 1

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12 −

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3]− β = −73γ − β⟨

ψ8|H ′|ψ8

⟩= 1

⟨32 −

12 |Hfs |32 −

⟩+ 2

⟨12 −

12 |Hfs |12 −

⟩+⟨1−11

212 |HZ |12

121−1

⟩= −γ 1

3 [4− 3]− γ 23 [8− 3] = −11

j + 12

⟨ψ5|H ′|ψ5

⟩= 2

⟨3212 |Hfs |32

⟩+ 1

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3] + β = −73γ + β⟨

ψ6|H ′|ψ6

⟩= 1

⟨3212 |Hfs |32

⟩+ 2

⟨1212 |Hfs |12

⟩+⟨111

2 −12 |HZ |12 −

1211⟩

= −γ 13 [4− 3]− γ 2

3 [8− 3] = −113 γ⟨

ψ7|H ′|ψ7

⟩= 2

⟨32 −

12 |Hfs |32 −

⟩+ 1

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12 −

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3]− β

= −73γ − β⟨

ψ8|H ′|ψ8

⟩= 1

⟨32 −

12 |Hfs |32 −

⟩+ 2

⟨12 −

12 |Hfs |12 −

⟩+⟨1−11

212 |HZ |12

121−1

⟩= −γ 1

3 [4− 3]− γ 23 [8− 3] = −11

j + 12

⟨ψ5|H ′|ψ5

⟩= 2

⟨3212 |Hfs |32

⟩+ 1

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3] + β = −73γ + β⟨

ψ6|H ′|ψ6

⟩= 1

⟨3212 |Hfs |32

⟩+ 2

⟨1212 |Hfs |12

⟩+⟨111

2 −12 |HZ |12 −

1211⟩

= −γ 13 [4− 3]− γ 2

3 [8− 3] = −113 γ⟨

ψ7|H ′|ψ7

⟩= 2

⟨32 −

12 |Hfs |32 −

⟩+ 1

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12 −

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3]− β = −73γ − β

⟨ψ8|H ′|ψ8

⟩= 1

⟨32 −

12 |Hfs |32 −

⟩+ 2

⟨12 −

12 |Hfs |12 −

⟩+⟨1−11

212 |HZ |12

121−1

⟩= −γ 1

3 [4− 3]− γ 23 [8− 3] = −11

j + 12

⟨ψ5|H ′|ψ5

⟩= 2

⟨3212 |Hfs |32

⟩+ 1

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3] + β = −73γ + β⟨

ψ6|H ′|ψ6

⟩= 1

⟨3212 |Hfs |32

⟩+ 2

⟨1212 |Hfs |12

⟩+⟨111

2 −12 |HZ |12 −

1211⟩

= −γ 13 [4− 3]− γ 2

3 [8− 3] = −113 γ⟨

ψ7|H ′|ψ7

⟩= 2

⟨32 −

12 |Hfs |32 −

⟩+ 1

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12 −

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3]− β = −73γ − β⟨

ψ8|H ′|ψ8

⟨32 −

12 |Hfs |32 −

⟩+ 2

⟨12 −

12 |Hfs |12 −

⟩+⟨1−11

212 |HZ |12

121−1

⟩= −γ 1

3 [4− 3]− γ 23 [8− 3] = −11

j + 12

⟨ψ5|H ′|ψ5

⟩= 2

⟨3212 |Hfs |32

⟩+ 1

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3] + β = −73γ + β⟨

ψ6|H ′|ψ6

⟩= 1

⟨3212 |Hfs |32

⟩+ 2

⟨1212 |Hfs |12

⟩+⟨111

2 −12 |HZ |12 −

1211⟩

= −γ 13 [4− 3]− γ 2

3 [8− 3] = −113 γ⟨

ψ7|H ′|ψ7

⟩= 2

⟨32 −

12 |Hfs |32 −

⟩+ 1

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12 −

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3]− β = −73γ − β⟨

ψ8|H ′|ψ8

⟩= 1

⟨32 −

12 |Hfs |32 −

⟩+ 2

⟨12 −

12 |Hfs |12 −

⟩+⟨1−11

212 |HZ |12

121−1

= −γ 13 [4− 3]− γ 2

3 [8− 3] = −113 γ

j + 12

⟨ψ5|H ′|ψ5

⟩= 2

⟨3212 |Hfs |32

⟩+ 1

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3] + β = −73γ + β⟨

ψ6|H ′|ψ6

⟩= 1

⟨3212 |Hfs |32

⟩+ 2

⟨1212 |Hfs |12

⟩+⟨111

2 −12 |HZ |12 −

1211⟩

= −γ 13 [4− 3]− γ 2

3 [8− 3] = −113 γ⟨

ψ7|H ′|ψ7

⟩= 2

⟨32 −

12 |Hfs |32 −

⟩+ 1

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12 −

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3]− β = −73γ − β⟨

ψ8|H ′|ψ8

⟩= 1

⟨32 −

12 |Hfs |32 −

⟩+ 2

⟨12 −

12 |Hfs |12 −

⟩+⟨1−11

212 |HZ |12

121−1

⟩= −γ 1

3 [4− 3]− γ 23 [8− 3]

= −113 γ

j + 12

⟨ψ5|H ′|ψ5

⟩= 2

⟨3212 |Hfs |32

⟩+ 1

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3] + β = −73γ + β⟨

ψ6|H ′|ψ6

⟩= 1

⟨3212 |Hfs |32

⟩+ 2

⟨1212 |Hfs |12

⟩+⟨111

2 −12 |HZ |12 −

1211⟩

= −γ 13 [4− 3]− γ 2

3 [8− 3] = −113 γ⟨

ψ7|H ′|ψ7

⟩= 2

⟨32 −

12 |Hfs |32 −

⟩+ 1

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12 −

1210⟩

= −γ 23 [4− 3]− γ 1

3 [8− 3]− β = −73γ − β⟨

ψ8|H ′|ψ8

⟩= 1

⟨32 −

12 |Hfs |32 −

⟩+ 2

⟨12 −

12 |Hfs |12 −

⟩+⟨1−11

212 |HZ |12

121−1

⟩= −γ 1

3 [4− 3]− γ 23 [8− 3] = −11

j + 12

because of orthonormality of the basis sets, the only cross terms which arenon-zero are the ones where the same two original 〈jmj | terms arecombined: 〈ψ5|H ′|ψ6〉 and 〈ψ7|H ′|ψ8〉⟨

ψ5|H ′|ψ6

⟩=√

⟨3212 |Hfs |32

⟩−√

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12 −

1211⟩

= −γ√

29 [4− 3] + γ

√29 [8− 3] = 4

3 γ =⟨ψ6|H ′|ψ5

⟩⟨ψ7|H ′|ψ8

⟩=√

⟨32 −

12 |Hfs |32 −

⟩−√

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12

121−1

⟩= −γ

√29 [4− 3] + γ

√29 [8− 3] = 4

3 γ =⟨ψ8|H ′|ψ7

j + 12

]because of orthonormality of the basis sets, the only cross terms which arenon-zero are the ones where the same two original 〈jmj | terms arecombined: 〈ψ5|H ′|ψ6〉 and 〈ψ7|H ′|ψ8〉

⟨ψ5|H ′|ψ6

⟩=√

⟨3212 |Hfs |32

⟩−√

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12 −

1211⟩

= −γ√

29 [4− 3] + γ

√29 [8− 3] = 4

3 γ =⟨ψ6|H ′|ψ5

⟩⟨ψ7|H ′|ψ8

⟩=√

⟨32 −

12 |Hfs |32 −

⟩−√

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12

121−1

⟩= −γ

√29 [4− 3] + γ

√29 [8− 3] = 4

3 γ =⟨ψ8|H ′|ψ7

j + 12

]because of orthonormality of the basis sets, the only cross terms which arenon-zero are the ones where the same two original 〈jmj | terms arecombined: 〈ψ5|H ′|ψ6〉 and 〈ψ7|H ′|ψ8〉⟨

ψ5|H ′|ψ6

⟨3212 |Hfs |32

⟩−√

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12 −

1211⟩

= −γ√

29 [4− 3] + γ

√29 [8− 3] = 4

3 γ =⟨ψ6|H ′|ψ5

⟩⟨ψ7|H ′|ψ8

⟩=√

⟨32 −

12 |Hfs |32 −

⟩−√

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12

121−1

⟩= −γ

√29 [4− 3] + γ

√29 [8− 3] = 4

3 γ =⟨ψ8|H ′|ψ7

j + 12

ψ5|H ′|ψ6

⟩=√

⟨3212 |Hfs |32

⟩−√

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12 −

1211⟩

= −γ√

29 [4− 3] + γ

√29 [8− 3] = 4

3 γ =⟨ψ6|H ′|ψ5

⟩⟨ψ7|H ′|ψ8

⟩=√

⟨32 −

12 |Hfs |32 −

⟩−√

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12

121−1

⟩= −γ

√29 [4− 3] + γ

√29 [8− 3] = 4

3 γ =⟨ψ8|H ′|ψ7

j + 12

ψ5|H ′|ψ6

⟩=√

⟨3212 |Hfs |32

⟩−√

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12 −

1211⟩

= −γ√

29 [4− 3] + γ

√29 [8− 3]

= 4√2

3 γ =⟨ψ6|H ′|ψ5

⟩⟨ψ7|H ′|ψ8

⟩=√

⟨32 −

12 |Hfs |32 −

⟩−√

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12

121−1

⟩= −γ

√29 [4− 3] + γ

√29 [8− 3] = 4

3 γ =⟨ψ8|H ′|ψ7

j + 12

ψ5|H ′|ψ6

⟩=√

⟨3212 |Hfs |32

⟩−√

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12 −

1211⟩

= −γ√

29 [4− 3] + γ

√29 [8− 3] = 4

=⟨ψ6|H ′|ψ5

⟩⟨ψ7|H ′|ψ8

⟩=√

⟨32 −

12 |Hfs |32 −

⟩−√

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12

121−1

⟩= −γ

√29 [4− 3] + γ

√29 [8− 3] = 4

3 γ =⟨ψ8|H ′|ψ7

j + 12

ψ5|H ′|ψ6

⟩=√

⟨3212 |Hfs |32

⟩−√

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12 −

1211⟩

= −γ√

29 [4− 3] + γ

√29 [8− 3] = 4

3 γ =⟨ψ6|H ′|ψ5

⟨ψ7|H ′|ψ8

⟩=√

⟨32 −

12 |Hfs |32 −

⟩−√

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12

121−1

⟩= −γ

√29 [4− 3] + γ

√29 [8− 3] = 4

3 γ =⟨ψ8|H ′|ψ7

j + 12

ψ5|H ′|ψ6

⟩=√

⟨3212 |Hfs |32

⟩−√

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12 −

1211⟩

= −γ√

29 [4− 3] + γ

√29 [8− 3] = 4

3 γ =⟨ψ6|H ′|ψ5

⟩⟨ψ7|H ′|ψ8

⟨32 −

12 |Hfs |32 −

⟩−√

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12

121−1

⟩= −γ

√29 [4− 3] + γ

√29 [8− 3] = 4

3 γ =⟨ψ8|H ′|ψ7

j + 12

ψ5|H ′|ψ6

⟩=√

⟨3212 |Hfs |32

⟩−√

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12 −

1211⟩

= −γ√

29 [4− 3] + γ

√29 [8− 3] = 4

3 γ =⟨ψ6|H ′|ψ5

⟩⟨ψ7|H ′|ψ8

⟩=√

⟨32 −

12 |Hfs |32 −

⟩−√

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12

121−1

= −γ√

29 [4− 3] + γ

√29 [8− 3] = 4

3 γ =⟨ψ8|H ′|ψ7

j + 12

ψ5|H ′|ψ6

⟩=√

⟨3212 |Hfs |32

⟩−√

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12 −

1211⟩

= −γ√

29 [4− 3] + γ

√29 [8− 3] = 4

3 γ =⟨ψ6|H ′|ψ5

⟩⟨ψ7|H ′|ψ8

⟩=√

⟨32 −

12 |Hfs |32 −

⟩−√

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12

121−1

⟩= −γ

√29 [4− 3] + γ

√29 [8− 3]

= 4√2

3 γ =⟨ψ8|H ′|ψ7

j + 12

ψ5|H ′|ψ6

⟩=√

⟨3212 |Hfs |32

⟩−√

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12 −

1211⟩

= −γ√

29 [4− 3] + γ

√29 [8− 3] = 4

3 γ =⟨ψ6|H ′|ψ5

⟩⟨ψ7|H ′|ψ8

⟩=√

⟨32 −

12 |Hfs |32 −

⟩−√

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12

121−1

⟩= −γ

√29 [4− 3] + γ

√29 [8− 3] = 4

=⟨ψ8|H ′|ψ7

j + 12

ψ5|H ′|ψ6

⟩=√

⟨3212 |Hfs |32

⟩−√

⟨1212 |Hfs |12

⟩+⟨101

212 |HZ |12 −

1211⟩

= −γ√

29 [4− 3] + γ

√29 [8− 3] = 4

3 γ =⟨ψ6|H ′|ψ5

⟩⟨ψ7|H ′|ψ8

⟩=√

⟨32 −

12 |Hfs |32 −

⟩−√

⟨12 −

12 |Hfs |12 −

⟩+⟨101

2 −12 |HZ |12

121−1

⟩= −γ

√29 [4− 3] + γ

√29 [8− 3] = 4

3 γ =⟨ψ8|H ′|ψ7

The W-matrix

Defining the fine structure and Zeeman terms with

γ ≡ (α/8)2 13.6 eV, β ≡ µBBext

W =−

5γ−β 0 0 0 0 0 0 00 5γ+β 0 0 0 0 0 00 0 γ−2β 0 0 0 0 00 0 0 γ+2β 0 0 0 0

0 0 0 0 73γ−β −4

3 γ 0 0

0 0 0 0 −4√2

3 γ 113 γ 0 0

0 0 0 0 0 0 73γ+β −4

0 0 0 0 0 0 −4√2

3 γ 113 γ

ψ1, ψ2, ψ3, and ψ4 are clearly already eigenfunctions of the fullperturbation as we saw in the original calculation. The other 4eigenfunctions can be solved by solving two 2× 2 blocks and thennegating the solutions

The W-matrix

γ ≡ (α/8)2 13.6 eV, β ≡ µBBext

W =−

5γ−β 0 0 0 0 0 0 00 5γ+β 0 0 0 0 0 00 0 γ−2β 0 0 0 0 00 0 0 γ+2β 0 0 0 0

0 0 0 0 73γ−β −4

3 γ 0 0

0 0 0 0 −4√2

3 γ 113 γ 0 0

0 0 0 0 0 0 73γ+β −4

0 0 0 0 0 0 −4√2

3 γ 113 γ

The W-matrix

γ ≡ (α/8)2 13.6 eV, β ≡ µBBext

W =−

5γ−β 0 0 0 0 0 0 00 5γ+β 0 0 0 0 0 00 0 γ−2β 0 0 0 0 00 0 0 γ+2β 0 0 0 0

0 0 0 0 73γ−β −4

3 γ 0 0

0 0 0 0 −4√2

3 γ 113 γ 0 0

0 0 0 0 0 0 73γ+β −4

0 0 0 0 0 0 −4√2

3 γ 113 γ

The W-matrix

γ ≡ (α/8)2 13.6 eV, β ≡ µBBext

W =−

5γ−β

0 0 0 0 0 0 00 5γ+β 0 0 0 0 0 00 0 γ−2β 0 0 0 0 00 0 0 γ+2β 0 0 0 0

0 0 0 0 73γ−β −4

3 γ 0 0

0 0 0 0 −4√2

3 γ 113 γ 0 0

0 0 0 0 0 0 73γ+β −4

0 0 0 0 0 0 −4√2

3 γ 113 γ

The W-matrix

γ ≡ (α/8)2 13.6 eV, β ≡ µBBext

W =−

5γ−β

0 0 0 0 0 0 00

5γ+β

0 0 0 0 0 00 0 γ−2β 0 0 0 0 00 0 0 γ+2β 0 0 0 0

0 0 0 0 73γ−β −4

3 γ 0 0

0 0 0 0 −4√2

3 γ 113 γ 0 0

0 0 0 0 0 0 73γ+β −4

0 0 0 0 0 0 −4√2

3 γ 113 γ

The W-matrix

γ ≡ (α/8)2 13.6 eV, β ≡ µBBext

W =−

5γ−β

0 0 0 0 0 0 00

5γ+β

0 0 0 0 0 00 0

γ−2β

0 0 0 0 00 0 0 γ+2β 0 0 0 0

0 0 0 0 73γ−β −4

3 γ 0 0

0 0 0 0 −4√2

3 γ 113 γ 0 0

0 0 0 0 0 0 73γ+β −4

0 0 0 0 0 0 −4√2

3 γ 113 γ

The W-matrix

γ ≡ (α/8)2 13.6 eV, β ≡ µBBext

W =−

5γ−β

0 0 0 0 0 0 00

5γ+β

0 0 0 0 0 00 0

γ−2β

0 0 0 0 00 0 0

γ+2β

0 0 0 0

0 0 0 0 73γ−β −4

3 γ 0 0

0 0 0 0 −4√2

3 γ 113 γ 0 0

0 0 0 0 0 0 73γ+β −4

0 0 0 0 0 0 −4√2

3 γ 113 γ

The W-matrix

γ ≡ (α/8)2 13.6 eV, β ≡ µBBext

W =−

5γ−β

0 0 0 0 0 0 00

5γ+β

0 0 0 0 0 00 0

γ−2β

0 0 0 0 00 0 0

γ+2β

0 0 0 0

73γ−β

−4√2

3 γ 0 0

0 0 0 0 −4√2

3 γ 113 γ 0 0

0 0 0 0 0 0 73γ+β −4

0 0 0 0 0 0 −4√2

3 γ 113 γ

The W-matrix

γ ≡ (α/8)2 13.6 eV, β ≡ µBBext

W =−

5γ−β

0 0 0 0 0 0 00

5γ+β

0 0 0 0 0 00 0

γ−2β

0 0 0 0 00 0 0

γ+2β

0 0 0 0

73γ−β

−4√2

3 γ 0 0

0 0 0 0 −4√2

113 γ

0 0 0 0 0 0 73γ+β −4

0 0 0 0 0 0 −4√2

3 γ 113 γ

The W-matrix

γ ≡ (α/8)2 13.6 eV, β ≡ µBBext

W =−

5γ−β

0 0 0 0 0 0 00

5γ+β

0 0 0 0 0 00 0

γ−2β

0 0 0 0 00 0 0

γ+2β

0 0 0 0

73γ−β

−4√2

3 γ 0 0

0 0 0 0 −4√2

113 γ

0 0 0 0 0 0

73γ+β

−4√2

0 0 0 0 0 0 −4√2

3 γ 113 γ

The W-matrix

γ ≡ (α/8)2 13.6 eV, β ≡ µBBext

W =−

5γ−β

0 0 0 0 0 0 00

5γ+β

0 0 0 0 0 00 0

γ−2β

0 0 0 0 00 0 0

γ+2β

0 0 0 0

73γ−β

−4√2

3 γ 0 0

0 0 0 0 −4√2

113 γ

0 0 0 0 0 0

73γ+β

−4√2

0 0 0 0 0 0 −4√2

113 γ

The W-matrix

γ ≡ (α/8)2 13.6 eV, β ≡ µBBext

W =−

5γ−β

0 0 0 0 0 0 00

5γ+β

0 0 0 0 0 00 0

γ−2β

0 0 0 0 00 0 0

γ+2β

0 0 0 0

73γ−β − 4

0 0 0 0

− 4√2

3 γ 113 γ

0 0 0 0 0 0

73γ+β

−4√2

0 0 0 0 0 0 −4√2

113 γ

The W-matrix

γ ≡ (α/8)2 13.6 eV, β ≡ µBBext

W =−

5γ−β

0 0 0 0 0 0 00

5γ+β

0 0 0 0 0 00 0

γ−2β

0 0 0 0 00 0 0

γ+2β

0 0 0 0

73γ−β − 4

0 0 0 0

− 4√2

3 γ 113 γ

0 0 0 0 0 0

73γ+β − 4

0 0 0 0 0 0

− 4√2

3 γ 113 γ

The W-matrix

γ ≡ (α/8)2 13.6 eV, β ≡ µBBext

W =−

5γ−β 0 0 0 0 0 0 00 5γ+β 0 0 0 0 0 00 0 γ−2β 0 0 0 0 00 0 0 γ+2β 0 0 0 0

0 0 0 0 73γ−β − 4

3 γ 0 0

0 0 0 0 − 4√2

3 γ 113 γ 0 0

0 0 0 0 0 0 73γ+β − 4

0 0 0 0 0 0 − 4√2

3 γ 113 γ

The W-matrix

γ ≡ (α/8)2 13.6 eV, β ≡ µBBext

W =−

5γ−β 0 0 0 0 0 0 00 5γ+β 0 0 0 0 0 00 0 γ−2β 0 0 0 0 00 0 0 γ+2β 0 0 0 0

0 0 0 0 73γ−β − 4

3 γ 0 0

0 0 0 0 − 4√2

3 γ 113 γ 0 0

0 0 0 0 0 0 73γ+β − 4

0 0 0 0 0 0 − 4√2

3 γ 113 γ

ψ1, ψ2, ψ3, and ψ4 are clearly already eigenfunctions of the fullperturbation as we saw in the original calculation.

The other 4eigenfunctions can be solved by solving two 2× 2 blocks and thennegating the solutions

The W-matrix

γ ≡ (α/8)2 13.6 eV, β ≡ µBBext

W =−

5γ−β 0 0 0 0 0 0 00 5γ+β 0 0 0 0 0 00 0 γ−2β 0 0 0 0 00 0 0 γ+2β 0 0 0 0

0 0 0 0 73γ−β −4

3 γ 0 0

0 0 0 0 −4√2

3 γ 113 γ 0 0

0 0 0 0 0 0 73γ+β −4

0 0 0 0 0 0 −4√2

3 γ 113 γ

Block 1 solution

The first block is solved bydiagonalizing the 2 × 2 sub-matrix

0 = det

∣∣∣∣∣ 73γ−β − λ −4

−4√2

3 γ 113 γ− λ

∣∣∣∣∣0 = (

3γ−β − λ)(

3γ− λ)− 32

= λ2 +77

9γ2 − 32

9γ2 − 11

3βγ − 18

3γλ+ βλ

= λ2 − λ(6γ − β) +

9γ2 − 11

)using the quadratic equation and recalling that the solution must benegated

− λ± = 3γ − β

√36γ2 − 12βγ + β2 − 20γ2 +

λ±= −3γ +β

2±√

4γ2 +2

3γβ +

Block 1 solution

0 = det

∣∣∣∣∣ 73γ−β − λ −4

−4√2

3 γ 113 γ− λ

∣∣∣∣∣

0 = (7

3γ−β − λ)(

3γ− λ)− 32

= λ2 +77

9γ2 − 32

9γ2 − 11

3βγ − 18

3γλ+ βλ

= λ2 − λ(6γ − β) +

9γ2 − 11

− λ± = 3γ − β

√36γ2 − 12βγ + β2 − 20γ2 +

λ±= −3γ +β

2±√

4γ2 +2

3γβ +

Block 1 solution

0 = det

∣∣∣∣∣ 73γ−β − λ −4

−4√2

3 γ 113 γ− λ

∣∣∣∣∣0 = (

3γ−β − λ)(

3γ− λ)− 32

= λ2 +77

9γ2 − 32

9γ2 − 11

3βγ − 18

3γλ+ βλ

= λ2 − λ(6γ − β) +

9γ2 − 11

− λ± = 3γ − β

√36γ2 − 12βγ + β2 − 20γ2 +

λ±= −3γ +β

2±√

4γ2 +2

3γβ +

Block 1 solution

0 = det

∣∣∣∣∣ 73γ−β − λ −4

−4√2

3 γ 113 γ− λ

∣∣∣∣∣0 = (

3γ−β − λ)(

3γ− λ)− 32

= λ2 +77

9γ2 − 32

9γ2 − 11

3βγ − 18

3γλ+ βλ

= λ2 − λ(6γ − β) +

9γ2 − 11

− λ± = 3γ − β

√36γ2 − 12βγ + β2 − 20γ2 +

λ±= −3γ +β

2±√

4γ2 +2

3γβ +

Block 1 solution

0 = det

∣∣∣∣∣ 73γ−β − λ −4

−4√2

3 γ 113 γ− λ

∣∣∣∣∣0 = (

3γ−β − λ)(

3γ− λ)− 32

= λ2 +77

9γ2 − 32

9γ2 − 11

3βγ − 18

3γλ+ βλ

= λ2 − λ(6γ − β) +

9γ2 − 11

− λ± = 3γ − β

√36γ2 − 12βγ + β2 − 20γ2 +

λ±= −3γ +β

2±√

4γ2 +2

3γβ +

Block 1 solution

0 = det

∣∣∣∣∣ 73γ−β − λ −4

−4√2

3 γ 113 γ− λ

∣∣∣∣∣0 = (

3γ−β − λ)(

3γ− λ)− 32

= λ2 +77

9γ2 − 32

9γ2 − 11

3βγ − 18

3γλ+ βλ

= λ2 − λ(6γ − β) +

9γ2 − 11

using the quadratic equation and recalling that the solution must benegated

− λ± = 3γ − β

√36γ2 − 12βγ + β2 − 20γ2 +

λ±= −3γ +β

2±√

4γ2 +2

3γβ +

Block 1 solution

0 = det

∣∣∣∣∣ 73γ−β − λ −4

−4√2

3 γ 113 γ− λ

∣∣∣∣∣0 = (

3γ−β − λ)(

3γ− λ)− 32

= λ2 +77

9γ2 − 32

9γ2 − 11

3βγ − 18

3γλ+ βλ

= λ2 − λ(6γ − β) +

9γ2 − 11

− λ± = 3γ − β

√36γ2 − 12βγ + β2 − 20γ2 +

λ±= −3γ +β

2±√

4γ2 +2

3γβ +

Block 1 solution

0 = det

∣∣∣∣∣ 73γ−β − λ −4

−4√2

3 γ 113 γ− λ

∣∣∣∣∣0 = (

3γ−β − λ)(

3γ− λ)− 32

= λ2 +77

9γ2 − 32

9γ2 − 11

3βγ − 18

3γλ+ βλ

= λ2 − λ(6γ − β) +

9γ2 − 11

− λ± = 3γ − β

√36γ2 − 12βγ + β2 − 20γ2 +

λ±= −3γ +β

2±√

4γ2 +2

3γβ +

Block 1 solution

0 = det

∣∣∣∣∣ 73γ−β − λ −4

−4√2

3 γ 113 γ− λ

∣∣∣∣∣0 = (

3γ−β − λ)(

3γ− λ)− 32

= λ2 +77

9γ2 − 32

9γ2 − 11

3βγ − 18

3γλ+ βλ

= λ2 − λ(6γ − β) +

9γ2 − 11

− λ± = 3γ − β

√36γ2 − 12βγ + β2 − 20γ2 +

λ±= −3γ +β

2±√

4γ2 +2

3γβ +

Block 1 solution

0 = det

∣∣∣∣∣ 73γ−β − λ −4

−4√2

3 γ 113 γ− λ

∣∣∣∣∣0 = (

3γ−β − λ)(

3γ− λ)− 32

= λ2 +77

9γ2 − 32

9γ2 − 11

3βγ − 18

3γλ+ βλ

= λ2 − λ(6γ − β) +

9γ2 − 11

− λ± = 3γ − β

√36γ2 − 12βγ + β2 − 20γ2 +

λ±= −3γ +β

2±√

4γ2 +2

3γβ +

Block 2 solution

The second block is solved inthe same way

0 = det

∣∣∣∣∣ 73γ+β − λ −4

−4√2

3 γ 113 γ− λ

∣∣∣∣∣0 = (

3γ+β − λ)(

3γ− λ)− 32

= λ2 +77

9γ2 − 32

9γ2 +

3βγ − 18

3γλ− βλ

= λ2 − λ(6γ + β) +

9γ2 +

)using the quadratic equation and negating the solution

− λ± = 3γ +β

√36γ2 + 12βγ + β2 − 20γ2 − 44

λ±= −3γ − β

2±√

4γ2 − 2

3γβ +

Block 2 solution

The second block is solved inthe same way 0 = det

∣∣∣∣∣ 73γ+β − λ −4

−4√2

3 γ 113 γ− λ

∣∣∣∣∣

0 = (7

3γ+β − λ)(

3γ− λ)− 32

= λ2 +77

9γ2 − 32

9γ2 +

3βγ − 18

3γλ− βλ

= λ2 − λ(6γ + β) +

9γ2 +

− λ± = 3γ +β

√36γ2 + 12βγ + β2 − 20γ2 − 44

λ±= −3γ − β

2±√

4γ2 − 2

3γβ +

Block 2 solution

∣∣∣∣∣ 73γ+β − λ −4

−4√2

3 γ 113 γ− λ

∣∣∣∣∣0 = (

3γ+β − λ)(

3γ− λ)− 32

= λ2 +77

9γ2 − 32

9γ2 +

3βγ − 18

3γλ− βλ

= λ2 − λ(6γ + β) +

9γ2 +

− λ± = 3γ +β

√36γ2 + 12βγ + β2 − 20γ2 − 44

λ±= −3γ − β

2±√

4γ2 − 2

3γβ +

Block 2 solution

∣∣∣∣∣ 73γ+β − λ −4

−4√2

3 γ 113 γ− λ

∣∣∣∣∣0 = (

3γ+β − λ)(

3γ− λ)− 32

= λ2 +77

9γ2 − 32

9γ2 +

3βγ − 18

3γλ− βλ

= λ2 − λ(6γ + β) +

9γ2 +

− λ± = 3γ +β

√36γ2 + 12βγ + β2 − 20γ2 − 44

λ±= −3γ − β

2±√

4γ2 − 2

3γβ +

Block 2 solution

∣∣∣∣∣ 73γ+β − λ −4

−4√2

3 γ 113 γ− λ

∣∣∣∣∣0 = (

3γ+β − λ)(

3γ− λ)− 32

= λ2 +77

9γ2 − 32

9γ2 +

3βγ − 18

3γλ− βλ

= λ2 − λ(6γ + β) +

9γ2 +

− λ± = 3γ +β

√36γ2 + 12βγ + β2 − 20γ2 − 44

λ±= −3γ − β

2±√

4γ2 − 2

3γβ +

Block 2 solution

∣∣∣∣∣ 73γ+β − λ −4

−4√2

3 γ 113 γ− λ

∣∣∣∣∣0 = (

3γ+β − λ)(

3γ− λ)− 32

= λ2 +77

9γ2 − 32

9γ2 +

3βγ − 18

3γλ− βλ

= λ2 − λ(6γ + β) +

9γ2 +

using the quadratic equation and negating the solution

− λ± = 3γ +β

√36γ2 + 12βγ + β2 − 20γ2 − 44

λ±= −3γ − β

2±√

4γ2 − 2

3γβ +

Block 2 solution

∣∣∣∣∣ 73γ+β − λ −4

−4√2

3 γ 113 γ− λ

∣∣∣∣∣0 = (

3γ+β − λ)(

3γ− λ)− 32

= λ2 +77

9γ2 − 32

9γ2 +

3βγ − 18

3γλ− βλ

= λ2 − λ(6γ + β) +

9γ2 +

− λ± = 3γ +β

√36γ2 + 12βγ + β2 − 20γ2 − 44

λ±= −3γ − β

2±√

4γ2 − 2

3γβ +

Block 2 solution

∣∣∣∣∣ 73γ+β − λ −4

−4√2

3 γ 113 γ− λ

∣∣∣∣∣0 = (

3γ+β − λ)(

3γ− λ)− 32

= λ2 +77

9γ2 − 32

9γ2 +

3βγ − 18

3γλ− βλ

= λ2 − λ(6γ + β) +

9γ2 +

− λ± = 3γ +β

√36γ2 + 12βγ + β2 − 20γ2 − 44

λ±= −3γ − β

2±√

4γ2 − 2

3γβ +

Block 2 solution

∣∣∣∣∣ 73γ+β − λ −4

−4√2

3 γ 113 γ− λ

∣∣∣∣∣0 = (

3γ+β − λ)(

3γ− λ)− 32

= λ2 +77

9γ2 − 32

9γ2 +

3βγ − 18

3γλ− βλ

= λ2 − λ(6γ + β) +

9γ2 +

− λ± = 3γ +β

√36γ2 + 12βγ + β2 − 20γ2 − 44

λ±= −3γ − β

2±√

4γ2 − 2

3γβ +

Block 2 solution

∣∣∣∣∣ 73γ+β − λ −4

−4√2

3 γ 113 γ− λ

∣∣∣∣∣0 = (

3γ+β − λ)(

3γ− λ)− 32

= λ2 +77

9γ2 − 32

9γ2 +

3βγ − 18

3γλ− βλ

= λ2 − λ(6γ + β) +

9γ2 +

− λ± = 3γ +β

√36γ2 + 12βγ + β2 − 20γ2 − 44

λ±= −3γ − β

2±√

4γ2 − 2

3γβ +

Problem 7.17

Although the Schrodinger equation for helium itself cannot be solvedexactly, there exist “helium-like” systems that do admit exact solutions. Asimple example is “rubber-band helium,” in which the Coulomb forces arereplaced by Hooke’s law forces:

H = − ~2

1 +∇22

(r21 + r22

)− λ

4mω2|~r1 −~r2|2

(a) Show that by the change of variables below, this Hamiltonian can besolved analytically

~u ≡ 1√2

(~r1 +~r2) ~v ≡ 1√2

(~r1 −~r2)

(b) What is the exact ground state energy for this system?

(c) Solve for an upper bound to the ground state using the variationalprinciple

H = − ~2

1 +∇22

(r21 + r22

)− λ

4mω2|~r1 −~r2|2

~u ≡ 1√2

(~r1 +~r2) ~v ≡ 1√2

(~r1 −~r2)

First apply the change of variables to the r21 + r22 term of the potentialenergy

~r1 =1√2

(~u + ~v) ~r2 =1√2

(~u − ~v)

r21 + r22 =1

(u2 + 2~u · ~v + v2

+ u2 − 2~u · ~v + v2

)= u2 + v2

the same can be done with the derivative terms, ∇21 +∇2

2 by expandingeach coordinate’s derivative in turn

∂x1=

∂ux∂x1

∂vx∂x1

=1√2

∂ux+

∂x2=

∂ux− ∂

H = − ~2

1 +∇22

(r21 + r22

)− λ

4mω2|~r1 −~r2|2

~u ≡ 1√2

(~r1 +~r2) ~v ≡ 1√2

(~r1 −~r2)

~r1 =1√2

(~u + ~v) ~r2 =1√2

(~u − ~v)

r21 + r22 =1

(u2 + 2~u · ~v + v2

+ u2 − 2~u · ~v + v2

)= u2 + v2

∂x1=

∂ux∂x1

∂vx∂x1

=1√2

∂ux+

∂x2=

∂ux− ∂

H = − ~2

1 +∇22

(r21 + r22

)− λ

4mω2|~r1 −~r2|2

~u ≡ 1√2

(~r1 +~r2) ~v ≡ 1√2

(~r1 −~r2)

~r1 =1√2

(~u + ~v)

~r2 =1√2

(~u − ~v)

r21 + r22 =1

(u2 + 2~u · ~v + v2

+ u2 − 2~u · ~v + v2

)= u2 + v2

∂x1=

∂ux∂x1

∂vx∂x1

=1√2

∂ux+

∂x2=

∂ux− ∂

H = − ~2

1 +∇22

(r21 + r22

)− λ

4mω2|~r1 −~r2|2

~u ≡ 1√2

(~r1 +~r2) ~v ≡ 1√2

(~r1 −~r2)

~r1 =1√2

(~u + ~v) ~r2 =1√2

(~u − ~v)

r21 + r22 =1

(u2 + 2~u · ~v + v2

+ u2 − 2~u · ~v + v2

)= u2 + v2

∂x1=

∂ux∂x1

∂vx∂x1

=1√2

∂ux+

∂x2=

∂ux− ∂

H = − ~2

1 +∇22

(r21 + r22

)− λ

4mω2|~r1 −~r2|2

~u ≡ 1√2

(~r1 +~r2) ~v ≡ 1√2

(~r1 −~r2)

~r1 =1√2

(~u + ~v) ~r2 =1√2

(~u − ~v)

r21 + r22 =1

(u2 + 2~u · ~v + v2

+ u2 − 2~u · ~v + v2

= u2 + v2

∂x1=

∂ux∂x1

∂vx∂x1

=1√2

∂ux+

∂x2=

∂ux− ∂

H = − ~2

1 +∇22

(r21 + r22

)− λ

4mω2|~r1 −~r2|2

~u ≡ 1√2

(~r1 +~r2) ~v ≡ 1√2

(~r1 −~r2)

~r1 =1√2

(~u + ~v) ~r2 =1√2

(~u − ~v)

r21 + r22 =1

(u2 + 2~u · ~v + v2 + u2 − 2~u · ~v + v2

= u2 + v2

∂x1=

∂ux∂x1

∂vx∂x1

=1√2

∂ux+

∂x2=

∂ux− ∂

H = − ~2

1 +∇22

(r21 + r22

)− λ

4mω2|~r1 −~r2|2

~u ≡ 1√2

(~r1 +~r2) ~v ≡ 1√2

(~r1 −~r2)

~r1 =1√2

(~u + ~v) ~r2 =1√2

(~u − ~v)

r21 + r22 =1

(u2 + 2~u · ~v + v2 + u2 − 2~u · ~v + v2

)= u2 + v2

∂x1=

∂ux∂x1

∂vx∂x1

=1√2

∂ux+

∂x2=

∂ux− ∂

H = − ~2

1 +∇22

(r21 + r22

)− λ

4mω2|~r1 −~r2|2

~u ≡ 1√2

(~r1 +~r2) ~v ≡ 1√2

(~r1 −~r2)

~r1 =1√2

(~u + ~v) ~r2 =1√2

(~u − ~v)

r21 + r22 =1

(u2 + 2~u · ~v + v2 + u2 − 2~u · ~v + v2

)= u2 + v2

∂x1=

∂ux∂x1

∂vx∂x1

=1√2

∂ux+

∂x2=

∂ux− ∂

H = − ~2

1 +∇22

(r21 + r22

)− λ

4mω2|~r1 −~r2|2

~u ≡ 1√2

(~r1 +~r2) ~v ≡ 1√2

(~r1 −~r2)

~r1 =1√2

(~u + ~v) ~r2 =1√2

(~u − ~v)

r21 + r22 =1

(u2 + 2~u · ~v + v2 + u2 − 2~u · ~v + v2

)= u2 + v2

∂x1=

∂ux∂x1

∂vx∂x1

=1√2

∂ux+

∂x2=

∂ux− ∂

H = − ~2

1 +∇22

(r21 + r22

)− λ

4mω2|~r1 −~r2|2

~u ≡ 1√2

(~r1 +~r2) ~v ≡ 1√2

(~r1 −~r2)

~r1 =1√2

(~u + ~v) ~r2 =1√2

(~u − ~v)

r21 + r22 =1

(u2 + 2~u · ~v + v2 + u2 − 2~u · ~v + v2

)= u2 + v2

∂x1=

∂ux∂x1

∂vx∂x1

=1√2

∂ux+

∂x2=

∂ux− ∂

H = − ~2

1 +∇22

(r21 + r22

)− λ

4mω2|~r1 −~r2|2

~u ≡ 1√2

(~r1 +~r2) ~v ≡ 1√2

(~r1 −~r2)

~r1 =1√2

(~u + ~v) ~r2 =1√2

(~u − ~v)

r21 + r22 =1

(u2 + 2~u · ~v + v2 + u2 − 2~u · ~v + v2

)= u2 + v2

∂x1=

∂ux∂x1

∂vx∂x1

=1√2

∂ux+

∂x2=

∂ux− ∂

H = − ~2

1 +∇22

(r21 + r22

)− λ

4mω2|~r1 −~r2|2

~u ≡ 1√2

(~r1 +~r2) ~v ≡ 1√2

(~r1 −~r2)

~r1 =1√2

(~u + ~v) ~r2 =1√2

(~u − ~v)

r21 + r22 =1

(u2 + 2~u · ~v + v2 + u2 − 2~u · ~v + v2

)= u2 + v2

∂x1=

∂ux∂x1

∂vx∂x1

=1√2

∂ux+

∂x2=

∂ux− ∂

H = − ~2

1 +∇22

(r21 + r22

)− λ

4mω2|~r1 −~r2|2

~u ≡ 1√2

(~r1 +~r2) ~v ≡ 1√2

(~r1 −~r2)

~r1 =1√2

(~u + ~v) ~r2 =1√2

(~u − ~v)

r21 + r22 =1

(u2 + 2~u · ~v + v2 + u2 − 2~u · ~v + v2

)= u2 + v2

∂x1=

∂ux∂x1

∂vx∂x1

=1√2

∂ux+

∂x2=

∂ux− ∂

)C. Segre (IIT) PHYS 406 - Spring 2018 February 01, 2018 14 / 18

∂x1=

∂ux+

∂x2=

∂ux− ∂

∂x21=

∂ux+

∂u2x

+ 2∂2

∂ux∂vx+

∂v2x

∂x22=

∂ux− ∂

∂u2x

− 2∂2

∂ux∂vx+

∂v2x

)(∂2

∂x21+

∂x22

∂u2x+

∂v2x

)−→ ∇2

1 +∇22 = ∇2

u +∇2v

∂x1=

∂ux+

∂x2=

∂ux− ∂

∂x21=

∂ux+

∂u2x

+ 2∂2

∂ux∂vx+

∂v2x

∂x22=

∂ux− ∂

∂u2x

− 2∂2

∂ux∂vx+

∂v2x

)(∂2

∂x21+

∂x22

∂u2x+

∂v2x

)−→ ∇2

1 +∇22 = ∇2

u +∇2v

∂x1=

∂ux+

∂x2=

∂ux− ∂

∂x21=

∂ux+

∂u2x

+ 2∂2

∂ux∂vx+

∂v2x

∂x22=

∂ux− ∂

∂u2x

− 2∂2

∂ux∂vx+

∂v2x

)(∂2

∂x21+

∂x22

∂u2x+

∂v2x

)−→ ∇2

1 +∇22 = ∇2

u +∇2v

∂x1=

∂ux+

∂x2=

∂ux− ∂

∂x21=

∂ux+

∂u2x

+ 2∂2

∂ux∂vx+

∂v2x

∂x22=

∂ux− ∂

∂u2x

− 2∂2

∂ux∂vx+

∂v2x

)(∂2

∂x21+

∂x22

∂u2x+

∂v2x

)−→ ∇2

1 +∇22 = ∇2

u +∇2v

∂x1=

∂ux+

∂x2=

∂ux− ∂

∂x21=

∂ux+

∂u2x+ 2

∂ux∂vx

∂v2x

∂x22=

∂ux− ∂

∂u2x

− 2∂2

∂ux∂vx+

∂v2x

)(∂2

∂x21+

∂x22

∂u2x+

∂v2x

)−→ ∇2

1 +∇22 = ∇2

u +∇2v

∂x1=

∂ux+

∂x2=

∂ux− ∂

∂x21=

∂ux+

∂u2x+ 2

∂ux∂vx+

∂v2x

∂x22=

∂ux− ∂

∂u2x

− 2∂2

∂ux∂vx+

∂v2x

)(∂2

∂x21+

∂x22

∂u2x+

∂v2x

)−→ ∇2

1 +∇22 = ∇2

u +∇2v

∂x1=

∂ux+

∂x2=

∂ux− ∂

∂x21=

∂ux+

∂u2x+ 2

∂ux∂vx+

∂v2x

∂x22=

∂ux− ∂

∂u2x

− 2∂2

∂ux∂vx+

∂v2x

)(∂2

∂x21+

∂x22

∂u2x+

∂v2x

)−→ ∇2

1 +∇22 = ∇2

u +∇2v

∂x1=

∂ux+

∂x2=

∂ux− ∂

∂x21=

∂ux+

∂u2x+ 2

∂ux∂vx+

∂v2x

∂x22=

∂ux− ∂

∂u2x

− 2∂2

∂ux∂vx+

∂v2x

)(∂2

∂x21+

∂x22

∂u2x+

∂v2x

)−→ ∇2

1 +∇22 = ∇2

u +∇2v

∂x1=

∂ux+

∂x2=

∂ux− ∂

∂x21=

∂ux+

∂u2x+ 2

∂ux∂vx+

∂v2x

∂x22=

∂ux− ∂

∂u2x

− 2∂2

∂ux∂vx+

∂v2x

∂x21+

∂x22

∂u2x+

∂v2x

)−→ ∇2

1 +∇22 = ∇2

u +∇2v

∂x1=

∂ux+

∂x2=

∂ux− ∂

∂x21=

∂ux+

∂u2x+ 2

∂ux∂vx+

∂v2x

∂x22=

∂ux− ∂

∂u2x− 2

∂ux∂vx

∂v2x

∂x21+

∂x22

∂u2x+

∂v2x

)−→ ∇2

1 +∇22 = ∇2

u +∇2v

∂x1=

∂ux+

∂x2=

∂ux− ∂

∂x21=

∂ux+

∂u2x+ 2

∂ux∂vx+

∂v2x

∂x22=

∂ux− ∂

∂u2x− 2

∂ux∂vx+

∂v2x

∂x21+

∂x22

∂u2x+

∂v2x

)−→ ∇2

1 +∇22 = ∇2

u +∇2v

∂x1=

∂ux+

∂x2=

∂ux− ∂

∂x21=

∂ux+

∂u2x+ 2

∂ux∂vx+

∂v2x

∂x22=

∂ux− ∂

∂u2x− 2

∂ux∂vx+

∂v2x

)(∂2

∂x21+

∂x22

∂u2x+

∂v2x

)−→ ∇2

1 +∇22 = ∇2

u +∇2v

∂x1=

∂ux+

∂x2=

∂ux− ∂

∂x21=

∂ux+

∂u2x+ 2

∂ux∂vx+

∂v2x

∂x22=

∂ux− ∂

∂u2x− 2

∂ux∂vx+

∂v2x

)(∂2

∂x21+

∂x22

∂u2x+

∂v2x

−→ ∇21 +∇2

2 = ∇2u +∇2

∂x1=

∂ux+

∂x2=

∂ux− ∂

∂x21=

∂ux+

∂u2x+ 2

∂ux∂vx+

∂v2x

∂x22=

∂ux− ∂

∂u2x− 2

∂ux∂vx+

∂v2x

)(∂2

∂x21+

∂x22

∂u2x+

∂v2x

)−→ ∇2

1 +∇22 = ∇2

u +∇2v

Putting it all together . . .

H = − ~2

1 +∇22

(r21 + r22

)− λ

4mω2|~r1 −~r2|2

= − ~2

u +∇2v

(u2 + v2

)− λ

4mω2 1

2|(~u + ~v)− (~u − ~v)|2

= − ~2

u +∇2v

(u2 + v2

)− λ

4mω22v2

[− ~2

2m∇2

2mω2u2

[− ~2

2m∇2

2mω2v2− 1

2λmω2v2

]this is now two separated harmonicoscillators, the second with an addednegative potential energy term

the ground state energies of this three-dimensional oscillator can simply bewritten down

2~ω√

1− λ

Egs =3

2~ω(1 +

√1− λ)

H = − ~2

1 +∇22

(r21 + r22

)− λ

4mω2|~r1 −~r2|2

= − ~2

u +∇2v

(u2 + v2

)− λ

4mω2 1

2|(~u + ~v)− (~u − ~v)|2

= − ~2

u +∇2v

(u2 + v2

)− λ

4mω22v2

[− ~2

2m∇2

2mω2u2

[− ~2

2m∇2

2mω2v2− 1

2λmω2v2

2~ω√

1− λ

Egs =3

2~ω(1 +

√1− λ)

H = − ~2

1 +∇22

(r21 + r22

)− λ

4mω2|~r1 −~r2|2

= − ~2

u +∇2v

(u2 + v2

)− λ

4mω2 1

2|(~u + ~v)− (~u − ~v)|2

= − ~2

u +∇2v

(u2 + v2

)− λ

4mω22v2

[− ~2

2m∇2

2mω2u2

[− ~2

2m∇2

2mω2v2− 1

2λmω2v2

2~ω√

1− λ

Egs =3

2~ω(1 +

√1− λ)

H = − ~2

1 +∇22

(r21 + r22

)− λ

4mω2|~r1 −~r2|2

= − ~2

u +∇2v

(u2 + v2

)− λ

4mω2 1

2|(~u + ~v)− (~u − ~v)|2

= − ~2

u +∇2v

(u2 + v2

)− λ

4mω22v2

[− ~2

2m∇2

2mω2u2

[− ~2

2m∇2

2mω2v2− 1

2λmω2v2

2~ω√

1− λ

Egs =3

2~ω(1 +

√1− λ)

H = − ~2

1 +∇22

(r21 + r22

)− λ

4mω2|~r1 −~r2|2

= − ~2

u +∇2v

(u2 + v2

)− λ

4mω2 1

2|(~u + ~v)− (~u − ~v)|2

= − ~2

u +∇2v

(u2 + v2

)− λ

4mω22v2

[− ~2

2m∇2

2mω2u2

[− ~2

2m∇2

2mω2v2− 1

2λmω2v2

2~ω√

1− λ

Egs =3

2~ω(1 +

√1− λ)

H = − ~2

1 +∇22

(r21 + r22

)− λ

4mω2|~r1 −~r2|2

= − ~2

u +∇2v

(u2 + v2

)− λ

4mω2 1

2|(~u + ~v)− (~u − ~v)|2

= − ~2

u +∇2v

(u2 + v2

)− λ

4mω22v2

[− ~2

2m∇2

2mω2u2

[− ~2

2m∇2

2mω2v2 − 1

2λmω2v2

this is now two separated harmonicoscillators, the second with an addednegative potential energy term

2~ω√

1− λ

Egs =3

2~ω(1 +

√1− λ)

H = − ~2

1 +∇22

(r21 + r22

)− λ

4mω2|~r1 −~r2|2

= − ~2

u +∇2v

(u2 + v2

)− λ

4mω2 1

2|(~u + ~v)− (~u − ~v)|2

= − ~2

u +∇2v

(u2 + v2

)− λ

4mω22v2

[− ~2

2m∇2

2mω2u2

[− ~2

2m∇2

2mω2v2− 1

2λmω2v2

2~ω√

1− λ

Egs =3

2~ω(1 +

√1− λ)

H = − ~2

1 +∇22

(r21 + r22

)− λ

4mω2|~r1 −~r2|2

= − ~2

u +∇2v

(u2 + v2

)− λ

4mω2 1

2|(~u + ~v)− (~u − ~v)|2

= − ~2

u +∇2v

(u2 + v2

)− λ

4mω22v2

[− ~2

2m∇2

2mω2u2

[− ~2

2m∇2

2mω2v2− 1

2λmω2v2

2~ω√

1− λ

Egs =3

2~ω(1 +

√1− λ)

H = − ~2

1 +∇22

(r21 + r22

)− λ

4mω2|~r1 −~r2|2

= − ~2

u +∇2v

(u2 + v2

)− λ

4mω2 1

2|(~u + ~v)− (~u − ~v)|2

= − ~2

u +∇2v

(u2 + v2

)− λ

4mω22v2

[− ~2

2m∇2

2mω2u2

[− ~2

2m∇2

2mω2v2− 1

2λmω2v2

2~ω√

1− λ

Egs =3

2~ω(1 +

√1− λ)

H = − ~2

1 +∇22

(r21 + r22

)− λ

4mω2|~r1 −~r2|2

= − ~2

u +∇2v

(u2 + v2

)− λ

4mω2 1

2|(~u + ~v)− (~u − ~v)|2

= − ~2

u +∇2v

(u2 + v2

)− λ

4mω22v2

[− ~2

2m∇2

2mω2u2

[− ~2

2m∇2

2mω2v2− 1

2λmω2v2

2~ω√

1− λ

Egs =3

2~ω(1 +

√1− λ)

H = − ~2

1 +∇22

(r21 + r22

)− λ

4mω2|~r1 −~r2|2

= − ~2

u +∇2v

(u2 + v2

)− λ

4mω2 1

2|(~u + ~v)− (~u − ~v)|2

= − ~2

u +∇2v

(u2 + v2

)− λ

4mω22v2

[− ~2

2m∇2

2mω2u2

[− ~2

2m∇2

2mω2v2− 1

2λmω2v2

2~ω√

1− λ

Egs =3

2~ω(1 +

√1− λ)

Build a variational trial functionfrom the single oscillator solutions

by symmetry the scaler product in-tegrates to zero

ψ(~r1,~r2) =(mωπ~

)3/2e−mω(r

21+r22 )/2~

〈H〉 =3

2~ω +

2~ω + 〈Vee〉

= 3~ω + 〈Vee〉〈Vee〉 = −λ

(mωπ~

)3 ∫e−mω(r

21+r22 )/~(~r1 −~r2)2 d3~r1d

= −λ4mω2

(mωπ~

)3 ∫e−mω(r

21+r22 )/~(r21 −��2~r1 ·~r2 + r22 ) d3~r1d

= −λ4mω2

(mωπ~

∫e−mω(r

21+r22 )/~r21 d

3~r1d3~r2

= −λ2mω2

(mωπ~

)3(4π)2

∫ ∞0

e−mωr22 /~r22 dr2

∫ ∞0

= −λ8m4ω5

√π~mω

]= −3

4λ~ω

ψ(~r1,~r2) =(mωπ~

)3/2e−mω(r

21+r22 )/2~

〈H〉 =3

2~ω +

2~ω + 〈Vee〉

= 3~ω + 〈Vee〉〈Vee〉 = −λ

(mωπ~

)3 ∫e−mω(r

21+r22 )/~(~r1 −~r2)2 d3~r1d

= −λ4mω2

(mωπ~

)3 ∫e−mω(r

21+r22 )/~(r21 −��2~r1 ·~r2 + r22 ) d3~r1d

= −λ4mω2

(mωπ~

∫e−mω(r

21+r22 )/~r21 d

3~r1d3~r2

= −λ2mω2

(mωπ~

)3(4π)2

∫ ∞0

= −λ8m4ω5

√π~mω

]= −3

4λ~ω

ψ(~r1,~r2) =(mωπ~

)3/2e−mω(r

21+r22 )/2~

〈H〉 =3

2~ω +

2~ω + 〈Vee〉

= 3~ω + 〈Vee〉〈Vee〉 = −λ

(mωπ~

)3 ∫e−mω(r

21+r22 )/~(~r1 −~r2)2 d3~r1d

= −λ4mω2

(mωπ~

)3 ∫e−mω(r

21+r22 )/~(r21 −��2~r1 ·~r2 + r22 ) d3~r1d

= −λ4mω2

(mωπ~

∫e−mω(r

21+r22 )/~r21 d

3~r1d3~r2

= −λ2mω2

(mωπ~

)3(4π)2

∫ ∞0

= −λ8m4ω5

√π~mω

]= −3

4λ~ω

ψ(~r1,~r2) =(mωπ~

)3/2e−mω(r

21+r22 )/2~

〈H〉 =3

2~ω +

2~ω + 〈Vee〉

= 3~ω + 〈Vee〉

〈Vee〉 = −λ4mω2

(mωπ~

)3 ∫e−mω(r

21+r22 )/~(~r1 −~r2)2 d3~r1d

= −λ4mω2

(mωπ~

)3 ∫e−mω(r

21+r22 )/~(r21 −��2~r1 ·~r2 + r22 ) d3~r1d

= −λ4mω2

(mωπ~

∫e−mω(r

21+r22 )/~r21 d

3~r1d3~r2

= −λ2mω2

(mωπ~

)3(4π)2

∫ ∞0

= −λ8m4ω5

√π~mω

]= −3

4λ~ω

ψ(~r1,~r2) =(mωπ~

)3/2e−mω(r

21+r22 )/2~

〈H〉 =3

2~ω +

2~ω + 〈Vee〉

= 3~ω + 〈Vee〉〈Vee〉 = −λ

(mωπ~

)3 ∫e−mω(r

21+r22 )/~(~r1 −~r2)2 d3~r1d

= −λ4mω2

(mωπ~

)3 ∫e−mω(r

21+r22 )/~(r21 −��2~r1 ·~r2 + r22 ) d3~r1d

= −λ4mω2

(mωπ~

∫e−mω(r

21+r22 )/~r21 d

3~r1d3~r2

= −λ2mω2

(mωπ~

)3(4π)2

∫ ∞0

= −λ8m4ω5

√π~mω

]= −3

4λ~ω

ψ(~r1,~r2) =(mωπ~

)3/2e−mω(r

21+r22 )/2~

〈H〉 =3

2~ω +

2~ω + 〈Vee〉

= 3~ω + 〈Vee〉〈Vee〉 = −λ

(mωπ~

)3 ∫e−mω(r

21+r22 )/~(~r1 −~r2)2 d3~r1d

= −λ4mω2

(mωπ~

)3 ∫e−mω(r

21+r22 )/~(r21 −��2~r1 ·~r2 + r22 ) d3~r1d

= −λ4mω2

(mωπ~

∫e−mω(r

21+r22 )/~r21 d

3~r1d3~r2

= −λ2mω2

(mωπ~

)3(4π)2

∫ ∞0

= −λ8m4ω5

√π~mω

]= −3

4λ~ω

ψ(~r1,~r2) =(mωπ~

)3/2e−mω(r

21+r22 )/2~

〈H〉 =3

2~ω +

2~ω + 〈Vee〉

= 3~ω + 〈Vee〉〈Vee〉 = −λ

(mωπ~

)3 ∫e−mω(r

21+r22 )/~(~r1 −~r2)2 d3~r1d

= −λ4mω2

(mωπ~

)3 ∫e−mω(r

21+r22 )/~(r21 − 2~r1 ·~r2 + r22 ) d3~r1d

= −λ4mω2

(mωπ~

∫e−mω(r

21+r22 )/~r21 d

3~r1d3~r2

= −λ2mω2

(mωπ~

)3(4π)2

∫ ∞0

= −λ8m4ω5

√π~mω

]= −3

4λ~ω

ψ(~r1,~r2) =(mωπ~

)3/2e−mω(r

21+r22 )/2~

〈H〉 =3

2~ω +

2~ω + 〈Vee〉

= 3~ω + 〈Vee〉〈Vee〉 = −λ

(mωπ~

)3 ∫e−mω(r

21+r22 )/~(~r1 −~r2)2 d3~r1d

= −λ4mω2

(mωπ~

)3 ∫e−mω(r

21+r22 )/~(r21 −��2~r1 ·~r2 + r22 ) d3~r1d

= −λ4mω2

(mωπ~

∫e−mω(r

21+r22 )/~r21 d

3~r1d3~r2

= −λ2mω2

(mωπ~

)3(4π)2

∫ ∞0

= −λ8m4ω5

√π~mω

]= −3

4λ~ω

ψ(~r1,~r2) =(mωπ~

)3/2e−mω(r

21+r22 )/2~

〈H〉 =3

2~ω +

2~ω + 〈Vee〉

= 3~ω + 〈Vee〉〈Vee〉 = −λ

(mωπ~

)3 ∫e−mω(r

21+r22 )/~(~r1 −~r2)2 d3~r1d

= −λ4mω2

(mωπ~

)3 ∫e−mω(r

21+r22 )/~(r21 −��2~r1 ·~r2 + r22 ) d3~r1d

= −λ4mω2

(mωπ~

∫e−mω(r

21+r22 )/~r21 d

3~r1d3~r2

= −λ2mω2

(mωπ~

)3(4π)2

∫ ∞0

= −λ8m4ω5

√π~mω

]= −3

4λ~ω

ψ(~r1,~r2) =(mωπ~

)3/2e−mω(r

21+r22 )/2~

〈H〉 =3

2~ω +

2~ω + 〈Vee〉

= 3~ω + 〈Vee〉〈Vee〉 = −λ

(mωπ~

)3 ∫e−mω(r

21+r22 )/~(~r1 −~r2)2 d3~r1d

= −λ4mω2

(mωπ~

)3 ∫e−mω(r

21+r22 )/~(r21 −��2~r1 ·~r2 + r22 ) d3~r1d

= −λ4mω2

(mωπ~

∫e−mω(r

21+r22 )/~r21 d

3~r1d3~r2

= −λ2mω2

(mωπ~

)3(4π)2

∫ ∞0

= −λ8m4ω5

√π~mω

]= −3

4λ~ω

ψ(~r1,~r2) =(mωπ~

)3/2e−mω(r

21+r22 )/2~

〈H〉 =3

2~ω +

2~ω + 〈Vee〉

= 3~ω + 〈Vee〉〈Vee〉 = −λ

(mωπ~

)3 ∫e−mω(r

21+r22 )/~(~r1 −~r2)2 d3~r1d

= −λ4mω2

(mωπ~

)3 ∫e−mω(r

21+r22 )/~(r21 −��2~r1 ·~r2 + r22 ) d3~r1d

= −λ4mω2

(mωπ~

∫e−mω(r

21+r22 )/~r21 d

3~r1d3~r2

= −λ2mω2

(mωπ~

)3(4π)2

∫ ∞0

= −λ8m4ω5

√π~mω

= −3

4λ~ω

ψ(~r1,~r2) =(mωπ~

)3/2e−mω(r

21+r22 )/2~

〈H〉 =3

2~ω +

2~ω + 〈Vee〉

= 3~ω + 〈Vee〉〈Vee〉 = −λ

(mωπ~

)3 ∫e−mω(r

21+r22 )/~(~r1 −~r2)2 d3~r1d

= −λ4mω2

(mωπ~

)3 ∫e−mω(r

21+r22 )/~(r21 −��2~r1 ·~r2 + r22 ) d3~r1d

= −λ4mω2

(mωπ~

∫e−mω(r

21+r22 )/~r21 d

3~r1d3~r2

= −λ2mω2

(mωπ~

)3(4π)2

∫ ∞0

= −λ8m4ω5

√π~mω

]= −3

4λ~ω

The expectation value of theHamiltonian is thus

comparing to the exact answer

〈H〉 = 3~ω − 3

4λ~ω

= 3~ω(

1− λ

〈H〉 = 32~ω

(2− λ

)2− λ

1− λ2

1− λ+ λ2

divide by the constants

subtract 1

square and apply bino-mial expansion

Egs = 32~ω(1 +

√1− λ)

1 +√

1− λ√

1− λ1− λ

clearly 〈H〉 > Egs

〈H〉 = 3~ω − 3

4λ~ω

= 3~ω(

1− λ

〈H〉 = 32~ω

(2− λ

)2− λ

1− λ2

1− λ+ λ2

subtract 1

Egs = 32~ω(1 +

√1− λ)

1 +√

1− λ√

1− λ1− λ

〈H〉 = 3~ω − 3

4λ~ω

= 3~ω(

1− λ

〈H〉 = 32~ω

(2− λ

)2− λ

1− λ2

1− λ+ λ2

subtract 1

Egs = 32~ω(1 +

√1− λ)

1 +√

1− λ√

1− λ1− λ

〈H〉 = 3~ω − 3

4λ~ω

= 3~ω(

1− λ

〈H〉 = 32~ω

(2− λ

)2− λ

1− λ2

1− λ+ λ2

subtract 1

Egs = 32~ω(1 +

√1− λ)

1 +√

1− λ√

1− λ1− λ

〈H〉 = 3~ω − 3

4λ~ω

= 3~ω(

1− λ

〈H〉 = 32~ω

(2− λ

2− λ2

1− λ2

1− λ+ λ2

subtract 1

Egs = 32~ω(1 +

√1− λ)

1 +√

1− λ√

1− λ1− λ

〈H〉 = 3~ω − 3

4λ~ω

= 3~ω(

1− λ

〈H〉 = 32~ω

(2− λ

2− λ2

1− λ2

1− λ+ λ2

subtract 1

Egs = 32~ω(1 +

√1− λ)

1 +√

1− λ√

1− λ1− λ

〈H〉 = 3~ω − 3

4λ~ω

= 3~ω(

1− λ

〈H〉 = 32~ω

(2− λ

2− λ2

1− λ2

1− λ+ λ2

subtract 1

Egs = 32~ω(1 +

√1− λ)

1 +√

1− λ√

1− λ1− λ

〈H〉 = 3~ω − 3

4λ~ω

= 3~ω(

1− λ

〈H〉 = 32~ω

(2− λ

)2− λ

1− λ2

1− λ+ λ2

subtract 1

Egs = 32~ω(1 +

√1− λ)

1 +√

1− λ√

1− λ1− λ

〈H〉 = 3~ω − 3

4λ~ω

= 3~ω(

1− λ

〈H〉 = 32~ω

(2− λ

)2− λ

1− λ2

1− λ+ λ2

subtract 1

Egs = 32~ω(1 +

√1− λ)

1 +√

1− λ

√1− λ

1− λ

〈H〉 = 3~ω − 3

4λ~ω

= 3~ω(

1− λ

〈H〉 = 32~ω

(2− λ

)2− λ

1− λ2

1− λ+ λ2

subtract 1

Egs = 32~ω(1 +

√1− λ)

1 +√

1− λ

√1− λ

1− λ

〈H〉 = 3~ω − 3

4λ~ω

= 3~ω(

1− λ

〈H〉 = 32~ω

(2− λ

)2− λ

1− λ2

1− λ+ λ2

subtract 1

Egs = 32~ω(1 +

√1− λ)

1 +√

1− λ

√1− λ

1− λ

〈H〉 = 3~ω − 3

4λ~ω

= 3~ω(

1− λ

〈H〉 = 32~ω

(2− λ

)2− λ

1− λ2

1− λ+ λ2

subtract 1

Egs = 32~ω(1 +

√1− λ)

1 +√

1− λ√

1− λ

〈H〉 = 3~ω − 3

4λ~ω

= 3~ω(

1− λ

〈H〉 = 32~ω

(2− λ

)2− λ

1− λ2

1− λ+ λ2

subtract 1

Egs = 32~ω(1 +

√1− λ)

1 +√

1− λ√

1− λ

〈H〉 = 3~ω − 3

4λ~ω

= 3~ω(

1− λ

〈H〉 = 32~ω

(2− λ

)2− λ

1− λ2

1− λ+ λ2

subtract 1

Egs = 32~ω(1 +

√1− λ)

1 +√

1− λ√

1− λ

〈H〉 = 3~ω − 3

4λ~ω

= 3~ω(

1− λ

〈H〉 = 32~ω

(2− λ

)2− λ

1− λ2

1− λ+ λ2

subtract 1

Egs = 32~ω(1 +

√1− λ)

1 +√

1− λ√

1− λ1− λ

〈H〉 = 3~ω − 3

4λ~ω

= 3~ω(

1− λ

〈H〉 = 32~ω

(2− λ

)2− λ

1− λ2

1− λ+ λ2

subtract 1

Egs = 32~ω(1 +

√1− λ)

1 +√

1− λ√

1− λ1− λ

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