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Today’s Outline - February 21, 2017
• Problem 6.21
• Problem 7.4
• Problem 7.14
• Problem 7.15
• Problem 7.16
• Problem 7.19
• Problem 6.39
Midterm Exam #1Thursday, February 23, 2017, Room 036 Rettaliata Engineering
Covers through HW #05, equation sheet provided
Homework Assignment #06:Chapter 8: 1,3,5,6,9,13due Thursday, March 02, 2017
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 1 / 23
Today’s Outline - February 21, 2017
• Problem 6.21
• Problem 7.4
• Problem 7.14
• Problem 7.15
• Problem 7.16
• Problem 7.19
• Problem 6.39
Midterm Exam #1Thursday, February 23, 2017, Room 036 Rettaliata Engineering
Covers through HW #05, equation sheet provided
Homework Assignment #06:Chapter 8: 1,3,5,6,9,13due Thursday, March 02, 2017
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 1 / 23
Today’s Outline - February 21, 2017
• Problem 6.21
• Problem 7.4
• Problem 7.14
• Problem 7.15
• Problem 7.16
• Problem 7.19
• Problem 6.39
Midterm Exam #1Thursday, February 23, 2017, Room 036 Rettaliata Engineering
Covers through HW #05, equation sheet provided
Homework Assignment #06:Chapter 8: 1,3,5,6,9,13due Thursday, March 02, 2017
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 1 / 23
Today’s Outline - February 21, 2017
• Problem 6.21
• Problem 7.4
• Problem 7.14
• Problem 7.15
• Problem 7.16
• Problem 7.19
• Problem 6.39
Midterm Exam #1Thursday, February 23, 2017, Room 036 Rettaliata Engineering
Covers through HW #05, equation sheet provided
Homework Assignment #06:Chapter 8: 1,3,5,6,9,13due Thursday, March 02, 2017
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 1 / 23
Today’s Outline - February 21, 2017
• Problem 6.21
• Problem 7.4
• Problem 7.14
• Problem 7.15
• Problem 7.16
• Problem 7.19
• Problem 6.39
Midterm Exam #1Thursday, February 23, 2017, Room 036 Rettaliata Engineering
Covers through HW #05, equation sheet provided
Homework Assignment #06:Chapter 8: 1,3,5,6,9,13due Thursday, March 02, 2017
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 1 / 23
Today’s Outline - February 21, 2017
• Problem 6.21
• Problem 7.4
• Problem 7.14
• Problem 7.15
• Problem 7.16
• Problem 7.19
• Problem 6.39
Midterm Exam #1Thursday, February 23, 2017, Room 036 Rettaliata Engineering
Covers through HW #05, equation sheet provided
Homework Assignment #06:Chapter 8: 1,3,5,6,9,13due Thursday, March 02, 2017
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 1 / 23
Today’s Outline - February 21, 2017
• Problem 6.21
• Problem 7.4
• Problem 7.14
• Problem 7.15
• Problem 7.16
• Problem 7.19
• Problem 6.39
Midterm Exam #1Thursday, February 23, 2017, Room 036 Rettaliata Engineering
Covers through HW #05, equation sheet provided
Homework Assignment #06:Chapter 8: 1,3,5,6,9,13due Thursday, March 02, 2017
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 1 / 23
Today’s Outline - February 21, 2017
• Problem 6.21
• Problem 7.4
• Problem 7.14
• Problem 7.15
• Problem 7.16
• Problem 7.19
• Problem 6.39
Midterm Exam #1Thursday, February 23, 2017, Room 036 Rettaliata Engineering
Covers through HW #05, equation sheet provided
Homework Assignment #06:Chapter 8: 1,3,5,6,9,13due Thursday, March 02, 2017
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 1 / 23
Today’s Outline - February 21, 2017
• Problem 6.21
• Problem 7.4
• Problem 7.14
• Problem 7.15
• Problem 7.16
• Problem 7.19
• Problem 6.39
Midterm Exam #1Thursday, February 23, 2017, Room 036 Rettaliata Engineering
Covers through HW #05, equation sheet provided
Homework Assignment #06:Chapter 8: 1,3,5,6,9,13due Thursday, March 02, 2017
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 1 / 23
Today’s Outline - February 21, 2017
• Problem 6.21
• Problem 7.4
• Problem 7.14
• Problem 7.15
• Problem 7.16
• Problem 7.19
• Problem 6.39
Midterm Exam #1Thursday, February 23, 2017, Room 036 Rettaliata Engineering
Covers through HW #05, equation sheet provided
Homework Assignment #06:Chapter 8: 1,3,5,6,9,13due Thursday, March 02, 2017
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 1 / 23
Problem 6.21
Consider the eight n = 2 states, |2 l j mj〉. Find the energy of each state,under weak-field Zeeman splitting, and construct a diagram like Figure6.11 to show how the energies evolve as Bext increases. Label each lineclearly, and indicate its slope.
The zero field energy of the 8 states in the |n l j mj〉 basis set is given bythe fine structure correction to the hydrogen energy levels
Enj = −13.6
n2
[1 +
α2
n2
(n
j + 12
− 3
4
)]
in the weak Zeeman regime these energies are perturbed by the magneticfield correction
EZ = µBgJBextmj , gJ =
[1 +
j(j + 1)− l(l + 1) + s(s + 1)
2j(j + 1)
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 2 / 23
Problem 6.21
Consider the eight n = 2 states, |2 l j mj〉. Find the energy of each state,under weak-field Zeeman splitting, and construct a diagram like Figure6.11 to show how the energies evolve as Bext increases. Label each lineclearly, and indicate its slope.
The zero field energy of the 8 states in the |n l j mj〉 basis set is given bythe fine structure correction to the hydrogen energy levels
Enj = −13.6
n2
[1 +
α2
n2
(n
j + 12
− 3
4
)]
in the weak Zeeman regime these energies are perturbed by the magneticfield correction
EZ = µBgJBextmj , gJ =
[1 +
j(j + 1)− l(l + 1) + s(s + 1)
2j(j + 1)
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 2 / 23
Problem 6.21
Consider the eight n = 2 states, |2 l j mj〉. Find the energy of each state,under weak-field Zeeman splitting, and construct a diagram like Figure6.11 to show how the energies evolve as Bext increases. Label each lineclearly, and indicate its slope.
The zero field energy of the 8 states in the |n l j mj〉 basis set is given bythe fine structure correction to the hydrogen energy levels
Enj = −13.6
n2
[1 +
α2
n2
(n
j + 12
− 3
4
)]
in the weak Zeeman regime these energies are perturbed by the magneticfield correction
EZ = µBgJBextmj , gJ =
[1 +
j(j + 1)− l(l + 1) + s(s + 1)
2j(j + 1)
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 2 / 23
Problem 6.21
Consider the eight n = 2 states, |2 l j mj〉. Find the energy of each state,under weak-field Zeeman splitting, and construct a diagram like Figure6.11 to show how the energies evolve as Bext increases. Label each lineclearly, and indicate its slope.
The zero field energy of the 8 states in the |n l j mj〉 basis set is given bythe fine structure correction to the hydrogen energy levels
Enj = −13.6
n2
[1 +
α2
n2
(n
j + 12
− 3
4
)]
in the weak Zeeman regime these energies are perturbed by the magneticfield correction
EZ = µBgJBextmj , gJ =
[1 +
j(j + 1)− l(l + 1) + s(s + 1)
2j(j + 1)
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 2 / 23
Problem 6.21
Consider the eight n = 2 states, |2 l j mj〉. Find the energy of each state,under weak-field Zeeman splitting, and construct a diagram like Figure6.11 to show how the energies evolve as Bext increases. Label each lineclearly, and indicate its slope.
The zero field energy of the 8 states in the |n l j mj〉 basis set is given bythe fine structure correction to the hydrogen energy levels
Enj = −13.6
n2
[1 +
α2
n2
(n
j + 12
− 3
4
)]
in the weak Zeeman regime these energies are perturbed by the magneticfield correction
EZ = µBgJBextmj ,
gJ =
[1 +
j(j + 1)− l(l + 1) + s(s + 1)
2j(j + 1)
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 2 / 23
Problem 6.21
Consider the eight n = 2 states, |2 l j mj〉. Find the energy of each state,under weak-field Zeeman splitting, and construct a diagram like Figure6.11 to show how the energies evolve as Bext increases. Label each lineclearly, and indicate its slope.
The zero field energy of the 8 states in the |n l j mj〉 basis set is given bythe fine structure correction to the hydrogen energy levels
Enj = −13.6
n2
[1 +
α2
n2
(n
j + 12
− 3
4
)]
in the weak Zeeman regime these energies are perturbed by the magneticfield correction
EZ = µBgJBextmj , gJ =
[1 +
j(j + 1)− l(l + 1) + s(s + 1)
2j(j + 1)
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 2 / 23
Problem 7.4
(a) Prove the following corollary to the variational principle: If〈ψ|ψgs〉 = 0, then 〈H〉 ≥ Efe , where Efe is the energy of the firstexcited state.
(b) Find the best bound on the first excited state of the one-dimensionalharmonic oscillator using the trial function
ψ(x) = Axe−bx2
(a) Starting with the definition ofthe arbitrary function ψ
since this function is defined to beorthogonal to the ground state ψ1
ψ =∞∑n=1
cnψn
0 = 〈ψ1|ψ〉 =∞∑n=1
cn〈ψ1|ψn〉 = c1
thus the expectation value ofthe Hamiltonian is 〈H〉 =
∞∑n=2
En|cn|2 ≥ Efe
∞∑n=2
|cn|2 = Efe
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 3 / 23
Problem 7.4
(a) Prove the following corollary to the variational principle: If〈ψ|ψgs〉 = 0, then 〈H〉 ≥ Efe , where Efe is the energy of the firstexcited state.
(b) Find the best bound on the first excited state of the one-dimensionalharmonic oscillator using the trial function
ψ(x) = Axe−bx2
(a) Starting with the definition ofthe arbitrary function ψ
since this function is defined to beorthogonal to the ground state ψ1
ψ =∞∑n=1
cnψn
0 = 〈ψ1|ψ〉 =∞∑n=1
cn〈ψ1|ψn〉 = c1
thus the expectation value ofthe Hamiltonian is 〈H〉 =
∞∑n=2
En|cn|2 ≥ Efe
∞∑n=2
|cn|2 = Efe
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 3 / 23
Problem 7.4
(a) Prove the following corollary to the variational principle: If〈ψ|ψgs〉 = 0, then 〈H〉 ≥ Efe , where Efe is the energy of the firstexcited state.
(b) Find the best bound on the first excited state of the one-dimensionalharmonic oscillator using the trial function
ψ(x) = Axe−bx2
(a) Starting with the definition ofthe arbitrary function ψ
since this function is defined to beorthogonal to the ground state ψ1
ψ =∞∑n=1
cnψn
0 = 〈ψ1|ψ〉 =∞∑n=1
cn〈ψ1|ψn〉 = c1
thus the expectation value ofthe Hamiltonian is 〈H〉 =
∞∑n=2
En|cn|2 ≥ Efe
∞∑n=2
|cn|2 = Efe
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 3 / 23
Problem 7.4
(a) Prove the following corollary to the variational principle: If〈ψ|ψgs〉 = 0, then 〈H〉 ≥ Efe , where Efe is the energy of the firstexcited state.
(b) Find the best bound on the first excited state of the one-dimensionalharmonic oscillator using the trial function
ψ(x) = Axe−bx2
(a) Starting with the definition ofthe arbitrary function ψ
since this function is defined to beorthogonal to the ground state ψ1
ψ =∞∑n=1
cnψn
0 = 〈ψ1|ψ〉 =∞∑n=1
cn〈ψ1|ψn〉 = c1
thus the expectation value ofthe Hamiltonian is 〈H〉 =
∞∑n=2
En|cn|2 ≥ Efe
∞∑n=2
|cn|2 = Efe
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 3 / 23
Problem 7.4
(a) Prove the following corollary to the variational principle: If〈ψ|ψgs〉 = 0, then 〈H〉 ≥ Efe , where Efe is the energy of the firstexcited state.
(b) Find the best bound on the first excited state of the one-dimensionalharmonic oscillator using the trial function
ψ(x) = Axe−bx2
(a) Starting with the definition ofthe arbitrary function ψ
since this function is defined to beorthogonal to the ground state ψ1
ψ =∞∑n=1
cnψn
0 = 〈ψ1|ψ〉
=∞∑n=1
cn〈ψ1|ψn〉 = c1
thus the expectation value ofthe Hamiltonian is 〈H〉 =
∞∑n=2
En|cn|2 ≥ Efe
∞∑n=2
|cn|2 = Efe
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 3 / 23
Problem 7.4
(a) Prove the following corollary to the variational principle: If〈ψ|ψgs〉 = 0, then 〈H〉 ≥ Efe , where Efe is the energy of the firstexcited state.
(b) Find the best bound on the first excited state of the one-dimensionalharmonic oscillator using the trial function
ψ(x) = Axe−bx2
(a) Starting with the definition ofthe arbitrary function ψ
since this function is defined to beorthogonal to the ground state ψ1
ψ =∞∑n=1
cnψn
0 = 〈ψ1|ψ〉 =∞∑n=1
cn〈ψ1|ψn〉
= c1
thus the expectation value ofthe Hamiltonian is 〈H〉 =
∞∑n=2
En|cn|2 ≥ Efe
∞∑n=2
|cn|2 = Efe
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 3 / 23
Problem 7.4
(a) Prove the following corollary to the variational principle: If〈ψ|ψgs〉 = 0, then 〈H〉 ≥ Efe , where Efe is the energy of the firstexcited state.
(b) Find the best bound on the first excited state of the one-dimensionalharmonic oscillator using the trial function
ψ(x) = Axe−bx2
(a) Starting with the definition ofthe arbitrary function ψ
since this function is defined to beorthogonal to the ground state ψ1
ψ =∞∑n=1
cnψn
0 = 〈ψ1|ψ〉 =∞∑n=1
cn〈ψ1|ψn〉 = c1
thus the expectation value ofthe Hamiltonian is 〈H〉 =
∞∑n=2
En|cn|2 ≥ Efe
∞∑n=2
|cn|2 = Efe
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 3 / 23
Problem 7.4
(a) Prove the following corollary to the variational principle: If〈ψ|ψgs〉 = 0, then 〈H〉 ≥ Efe , where Efe is the energy of the firstexcited state.
(b) Find the best bound on the first excited state of the one-dimensionalharmonic oscillator using the trial function
ψ(x) = Axe−bx2
(a) Starting with the definition ofthe arbitrary function ψ
since this function is defined to beorthogonal to the ground state ψ1
ψ =∞∑n=1
cnψn
0 = 〈ψ1|ψ〉 =∞∑n=1
cn〈ψ1|ψn〉 = c1
thus the expectation value ofthe Hamiltonian is
〈H〉 =∞∑n=2
En|cn|2 ≥ Efe
∞∑n=2
|cn|2 = Efe
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 3 / 23
Problem 7.4
(a) Prove the following corollary to the variational principle: If〈ψ|ψgs〉 = 0, then 〈H〉 ≥ Efe , where Efe is the energy of the firstexcited state.
(b) Find the best bound on the first excited state of the one-dimensionalharmonic oscillator using the trial function
ψ(x) = Axe−bx2
(a) Starting with the definition ofthe arbitrary function ψ
since this function is defined to beorthogonal to the ground state ψ1
ψ =∞∑n=1
cnψn
0 = 〈ψ1|ψ〉 =∞∑n=1
cn〈ψ1|ψn〉 = c1
thus the expectation value ofthe Hamiltonian is 〈H〉 =
∞∑n=2
En|cn|2
≥ Efe
∞∑n=2
|cn|2 = Efe
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 3 / 23
Problem 7.4
(a) Prove the following corollary to the variational principle: If〈ψ|ψgs〉 = 0, then 〈H〉 ≥ Efe , where Efe is the energy of the firstexcited state.
(b) Find the best bound on the first excited state of the one-dimensionalharmonic oscillator using the trial function
ψ(x) = Axe−bx2
(a) Starting with the definition ofthe arbitrary function ψ
since this function is defined to beorthogonal to the ground state ψ1
ψ =∞∑n=1
cnψn
0 = 〈ψ1|ψ〉 =∞∑n=1
cn〈ψ1|ψn〉 = c1
thus the expectation value ofthe Hamiltonian is 〈H〉 =
∞∑n=2
En|cn|2 ≥ Efe
∞∑n=2
|cn|2
= Efe
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 3 / 23
Problem 7.4
(a) Prove the following corollary to the variational principle: If〈ψ|ψgs〉 = 0, then 〈H〉 ≥ Efe , where Efe is the energy of the firstexcited state.
(b) Find the best bound on the first excited state of the one-dimensionalharmonic oscillator using the trial function
ψ(x) = Axe−bx2
(a) Starting with the definition ofthe arbitrary function ψ
since this function is defined to beorthogonal to the ground state ψ1
ψ =∞∑n=1
cnψn
0 = 〈ψ1|ψ〉 =∞∑n=1
cn〈ψ1|ψn〉 = c1
thus the expectation value ofthe Hamiltonian is 〈H〉 =
∞∑n=2
En|cn|2 ≥ Efe
∞∑n=2
|cn|2 = Efe
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 3 / 23
Problem 7.4 (cont.)
(b) start by computing the normalization constant,
1 = |A|2∫ ∞−∞
x2e−2bx2dx = |A|22
1
8b
√π
2b−→ |A|2 = 4b
√2b
π
the kinetic energy,
〈T 〉 = − ~2
2m|A|2
∫ ∞−∞
xe−bx2 d
dx2
(xe−bx
2)dx
= − ~2
2m4b
√2b
π
∫ ∞−∞
xe−bx2(−2bx − 4bx + 4b2x3)e−bx
2
= − ~2
2m4b
√2b
π
∫ ∞−∞
(−6bx2 + 4b2x4)e−2bx2dx
= −2~2bm
√2b
π2
[−6b
1
8b
√π
2b+ 4b2
3
32b2
√π
2b
]=
3~2b2m
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 4 / 23
Problem 7.4 (cont.)
(b) start by computing the normalization constant,
1 = |A|2∫ ∞−∞
x2e−2bx2dx
= |A|221
8b
√π
2b−→ |A|2 = 4b
√2b
π
the kinetic energy,
〈T 〉 = − ~2
2m|A|2
∫ ∞−∞
xe−bx2 d
dx2
(xe−bx
2)dx
= − ~2
2m4b
√2b
π
∫ ∞−∞
xe−bx2(−2bx − 4bx + 4b2x3)e−bx
2
= − ~2
2m4b
√2b
π
∫ ∞−∞
(−6bx2 + 4b2x4)e−2bx2dx
= −2~2bm
√2b
π2
[−6b
1
8b
√π
2b+ 4b2
3
32b2
√π
2b
]=
3~2b2m
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 4 / 23
Problem 7.4 (cont.)
(b) start by computing the normalization constant,
1 = |A|2∫ ∞−∞
x2e−2bx2dx = |A|22
1
8b
√π
2b
−→ |A|2 = 4b
√2b
π
the kinetic energy,
〈T 〉 = − ~2
2m|A|2
∫ ∞−∞
xe−bx2 d
dx2
(xe−bx
2)dx
= − ~2
2m4b
√2b
π
∫ ∞−∞
xe−bx2(−2bx − 4bx + 4b2x3)e−bx
2
= − ~2
2m4b
√2b
π
∫ ∞−∞
(−6bx2 + 4b2x4)e−2bx2dx
= −2~2bm
√2b
π2
[−6b
1
8b
√π
2b+ 4b2
3
32b2
√π
2b
]=
3~2b2m
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 4 / 23
Problem 7.4 (cont.)
(b) start by computing the normalization constant,
1 = |A|2∫ ∞−∞
x2e−2bx2dx = |A|22
1
8b
√π
2b−→ |A|2 = 4b
√2b
π
the kinetic energy,
〈T 〉 = − ~2
2m|A|2
∫ ∞−∞
xe−bx2 d
dx2
(xe−bx
2)dx
= − ~2
2m4b
√2b
π
∫ ∞−∞
xe−bx2(−2bx − 4bx + 4b2x3)e−bx
2
= − ~2
2m4b
√2b
π
∫ ∞−∞
(−6bx2 + 4b2x4)e−2bx2dx
= −2~2bm
√2b
π2
[−6b
1
8b
√π
2b+ 4b2
3
32b2
√π
2b
]=
3~2b2m
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 4 / 23
Problem 7.4 (cont.)
(b) start by computing the normalization constant,
1 = |A|2∫ ∞−∞
x2e−2bx2dx = |A|22
1
8b
√π
2b−→ |A|2 = 4b
√2b
π
the kinetic energy,
〈T 〉 = − ~2
2m|A|2
∫ ∞−∞
xe−bx2 d
dx2
(xe−bx
2)dx
= − ~2
2m4b
√2b
π
∫ ∞−∞
xe−bx2(−2bx − 4bx + 4b2x3)e−bx
2
= − ~2
2m4b
√2b
π
∫ ∞−∞
(−6bx2 + 4b2x4)e−2bx2dx
= −2~2bm
√2b
π2
[−6b
1
8b
√π
2b+ 4b2
3
32b2
√π
2b
]=
3~2b2m
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 4 / 23
Problem 7.4 (cont.)
(b) start by computing the normalization constant,
1 = |A|2∫ ∞−∞
x2e−2bx2dx = |A|22
1
8b
√π
2b−→ |A|2 = 4b
√2b
π
the kinetic energy,
〈T 〉 = − ~2
2m|A|2
∫ ∞−∞
xe−bx2 d
dx2
(xe−bx
2)dx
= − ~2
2m4b
√2b
π
∫ ∞−∞
xe−bx2(−2bx − 4bx + 4b2x3)e−bx
2
= − ~2
2m4b
√2b
π
∫ ∞−∞
(−6bx2 + 4b2x4)e−2bx2dx
= −2~2bm
√2b
π2
[−6b
1
8b
√π
2b+ 4b2
3
32b2
√π
2b
]=
3~2b2m
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 4 / 23
Problem 7.4 (cont.)
(b) start by computing the normalization constant,
1 = |A|2∫ ∞−∞
x2e−2bx2dx = |A|22
1
8b
√π
2b−→ |A|2 = 4b
√2b
π
the kinetic energy,
〈T 〉 = − ~2
2m|A|2
∫ ∞−∞
xe−bx2 d
dx2
(xe−bx
2)dx
= − ~2
2m4b
√2b
π
∫ ∞−∞
xe−bx2(−2bx − 4bx + 4b2x3)e−bx
2
= − ~2
2m4b
√2b
π
∫ ∞−∞
(−6bx2 + 4b2x4)e−2bx2dx
= −2~2bm
√2b
π2
[−6b
1
8b
√π
2b+ 4b2
3
32b2
√π
2b
]=
3~2b2m
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 4 / 23
Problem 7.4 (cont.)
(b) start by computing the normalization constant,
1 = |A|2∫ ∞−∞
x2e−2bx2dx = |A|22
1
8b
√π
2b−→ |A|2 = 4b
√2b
π
the kinetic energy,
〈T 〉 = − ~2
2m|A|2
∫ ∞−∞
xe−bx2 d
dx2
(xe−bx
2)dx
= − ~2
2m4b
√2b
π
∫ ∞−∞
xe−bx2(−2bx − 4bx + 4b2x3)e−bx
2
= − ~2
2m4b
√2b
π
∫ ∞−∞
(−6bx2 + 4b2x4)e−2bx2dx
= −2~2bm
√2b
π2
[−6b
1
8b
√π
2b+ 4b2
3
32b2
√π
2b
]=
3~2b2m
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 4 / 23
Problem 7.4 (cont.)
(b) start by computing the normalization constant,
1 = |A|2∫ ∞−∞
x2e−2bx2dx = |A|22
1
8b
√π
2b−→ |A|2 = 4b
√2b
π
the kinetic energy,
〈T 〉 = − ~2
2m|A|2
∫ ∞−∞
xe−bx2 d
dx2
(xe−bx
2)dx
= − ~2
2m4b
√2b
π
∫ ∞−∞
xe−bx2(−2bx − 4bx + 4b2x3)e−bx
2
= − ~2
2m4b
√2b
π
∫ ∞−∞
(−6bx2 + 4b2x4)e−2bx2dx
= −2~2bm
√2b
π2
[−6b
1
8b
√π
2b+ 4b2
3
32b2
√π
2b
]
=3~2b2m
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 4 / 23
Problem 7.4 (cont.)
(b) start by computing the normalization constant,
1 = |A|2∫ ∞−∞
x2e−2bx2dx = |A|22
1
8b
√π
2b−→ |A|2 = 4b
√2b
π
the kinetic energy,
〈T 〉 = − ~2
2m|A|2
∫ ∞−∞
xe−bx2 d
dx2
(xe−bx
2)dx
= − ~2
2m4b
√2b
π
∫ ∞−∞
xe−bx2(−2bx − 4bx + 4b2x3)e−bx
2
= − ~2
2m4b
√2b
π
∫ ∞−∞
(−6bx2 + 4b2x4)e−2bx2dx
= −2~2bm
√2b
π2
[−6b
1
8b
√π
2b+ 4b2
3
32b2
√π
2b
]=
3~2b2m
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 4 / 23
Problem 7.4 (cont.)
and the potential energy
〈V 〉 =1
2mω2
∫ ∞−∞
xe−bx2x2(xe−bx
2)dx =
1
2mω2
∫ ∞−∞
x4e−2bx2dx
=1
2mω24b
√2b
π2
3
32b2
√π
2b=
3mω2
8b
combining to get the expectationvalue of the Hamiltonian
taking the derivative to minimizewrt b
〈H〉 =3~2b2m
+3mω2
8b∂〈H〉∂b
=3~2
2m− 3mω2
8b2= 0
b2 =m2ω2
4~2−→ b =
mω
2~
〈H〉min =3~2
2m
mω
2~+
3mω2
8
2~mω
= ~ω(
3
4+
3
4
)=
3
2~
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 5 / 23
Problem 7.4 (cont.)
and the potential energy
〈V 〉 =1
2mω2
∫ ∞−∞
xe−bx2x2(xe−bx
2)dx
=1
2mω2
∫ ∞−∞
x4e−2bx2dx
=1
2mω24b
√2b
π2
3
32b2
√π
2b=
3mω2
8b
combining to get the expectationvalue of the Hamiltonian
taking the derivative to minimizewrt b
〈H〉 =3~2b2m
+3mω2
8b∂〈H〉∂b
=3~2
2m− 3mω2
8b2= 0
b2 =m2ω2
4~2−→ b =
mω
2~
〈H〉min =3~2
2m
mω
2~+
3mω2
8
2~mω
= ~ω(
3
4+
3
4
)=
3
2~
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 5 / 23
Problem 7.4 (cont.)
and the potential energy
〈V 〉 =1
2mω2
∫ ∞−∞
xe−bx2x2(xe−bx
2)dx =
1
2mω2
∫ ∞−∞
x4e−2bx2dx
=1
2mω24b
√2b
π2
3
32b2
√π
2b=
3mω2
8b
combining to get the expectationvalue of the Hamiltonian
taking the derivative to minimizewrt b
〈H〉 =3~2b2m
+3mω2
8b∂〈H〉∂b
=3~2
2m− 3mω2
8b2= 0
b2 =m2ω2
4~2−→ b =
mω
2~
〈H〉min =3~2
2m
mω
2~+
3mω2
8
2~mω
= ~ω(
3
4+
3
4
)=
3
2~
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 5 / 23
Problem 7.4 (cont.)
and the potential energy
〈V 〉 =1
2mω2
∫ ∞−∞
xe−bx2x2(xe−bx
2)dx =
1
2mω2
∫ ∞−∞
x4e−2bx2dx
=1
2mω24b
√2b
π2
3
32b2
√π
2b
=3mω2
8b
combining to get the expectationvalue of the Hamiltonian
taking the derivative to minimizewrt b
〈H〉 =3~2b2m
+3mω2
8b∂〈H〉∂b
=3~2
2m− 3mω2
8b2= 0
b2 =m2ω2
4~2−→ b =
mω
2~
〈H〉min =3~2
2m
mω
2~+
3mω2
8
2~mω
= ~ω(
3
4+
3
4
)=
3
2~
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 5 / 23
Problem 7.4 (cont.)
and the potential energy
〈V 〉 =1
2mω2
∫ ∞−∞
xe−bx2x2(xe−bx
2)dx =
1
2mω2
∫ ∞−∞
x4e−2bx2dx
=1
2mω24b
√2b
π2
3
32b2
√π
2b=
3mω2
8b
combining to get the expectationvalue of the Hamiltonian
taking the derivative to minimizewrt b
〈H〉 =3~2b2m
+3mω2
8b∂〈H〉∂b
=3~2
2m− 3mω2
8b2= 0
b2 =m2ω2
4~2−→ b =
mω
2~
〈H〉min =3~2
2m
mω
2~+
3mω2
8
2~mω
= ~ω(
3
4+
3
4
)=
3
2~
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 5 / 23
Problem 7.4 (cont.)
and the potential energy
〈V 〉 =1
2mω2
∫ ∞−∞
xe−bx2x2(xe−bx
2)dx =
1
2mω2
∫ ∞−∞
x4e−2bx2dx
=1
2mω24b
√2b
π2
3
32b2
√π
2b=
3mω2
8b
combining to get the expectationvalue of the Hamiltonian
taking the derivative to minimizewrt b
〈H〉 =3~2b2m
+3mω2
8b∂〈H〉∂b
=3~2
2m− 3mω2
8b2= 0
b2 =m2ω2
4~2−→ b =
mω
2~
〈H〉min =3~2
2m
mω
2~+
3mω2
8
2~mω
= ~ω(
3
4+
3
4
)=
3
2~
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 5 / 23
Problem 7.4 (cont.)
and the potential energy
〈V 〉 =1
2mω2
∫ ∞−∞
xe−bx2x2(xe−bx
2)dx =
1
2mω2
∫ ∞−∞
x4e−2bx2dx
=1
2mω24b
√2b
π2
3
32b2
√π
2b=
3mω2
8b
combining to get the expectationvalue of the Hamiltonian
taking the derivative to minimizewrt b
〈H〉 =3~2b2m
+3mω2
8b
∂〈H〉∂b
=3~2
2m− 3mω2
8b2= 0
b2 =m2ω2
4~2−→ b =
mω
2~
〈H〉min =3~2
2m
mω
2~+
3mω2
8
2~mω
= ~ω(
3
4+
3
4
)=
3
2~
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 5 / 23
Problem 7.4 (cont.)
and the potential energy
〈V 〉 =1
2mω2
∫ ∞−∞
xe−bx2x2(xe−bx
2)dx =
1
2mω2
∫ ∞−∞
x4e−2bx2dx
=1
2mω24b
√2b
π2
3
32b2
√π
2b=
3mω2
8b
combining to get the expectationvalue of the Hamiltonian
taking the derivative to minimizewrt b
〈H〉 =3~2b2m
+3mω2
8b
∂〈H〉∂b
=3~2
2m− 3mω2
8b2= 0
b2 =m2ω2
4~2−→ b =
mω
2~
〈H〉min =3~2
2m
mω
2~+
3mω2
8
2~mω
= ~ω(
3
4+
3
4
)=
3
2~
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 5 / 23
Problem 7.4 (cont.)
and the potential energy
〈V 〉 =1
2mω2
∫ ∞−∞
xe−bx2x2(xe−bx
2)dx =
1
2mω2
∫ ∞−∞
x4e−2bx2dx
=1
2mω24b
√2b
π2
3
32b2
√π
2b=
3mω2
8b
combining to get the expectationvalue of the Hamiltonian
taking the derivative to minimizewrt b
〈H〉 =3~2b2m
+3mω2
8b∂〈H〉∂b
=3~2
2m− 3mω2
8b2= 0
b2 =m2ω2
4~2−→ b =
mω
2~
〈H〉min =3~2
2m
mω
2~+
3mω2
8
2~mω
= ~ω(
3
4+
3
4
)=
3
2~
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 5 / 23
Problem 7.4 (cont.)
and the potential energy
〈V 〉 =1
2mω2
∫ ∞−∞
xe−bx2x2(xe−bx
2)dx =
1
2mω2
∫ ∞−∞
x4e−2bx2dx
=1
2mω24b
√2b
π2
3
32b2
√π
2b=
3mω2
8b
combining to get the expectationvalue of the Hamiltonian
taking the derivative to minimizewrt b
〈H〉 =3~2b2m
+3mω2
8b∂〈H〉∂b
=3~2
2m− 3mω2
8b2= 0
b2 =m2ω2
4~2
−→ b =mω
2~
〈H〉min =3~2
2m
mω
2~+
3mω2
8
2~mω
= ~ω(
3
4+
3
4
)=
3
2~
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 5 / 23
Problem 7.4 (cont.)
and the potential energy
〈V 〉 =1
2mω2
∫ ∞−∞
xe−bx2x2(xe−bx
2)dx =
1
2mω2
∫ ∞−∞
x4e−2bx2dx
=1
2mω24b
√2b
π2
3
32b2
√π
2b=
3mω2
8b
combining to get the expectationvalue of the Hamiltonian
taking the derivative to minimizewrt b
〈H〉 =3~2b2m
+3mω2
8b∂〈H〉∂b
=3~2
2m− 3mω2
8b2= 0
b2 =m2ω2
4~2−→ b =
mω
2~
〈H〉min =3~2
2m
mω
2~+
3mω2
8
2~mω
= ~ω(
3
4+
3
4
)=
3
2~
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 5 / 23
Problem 7.4 (cont.)
and the potential energy
〈V 〉 =1
2mω2
∫ ∞−∞
xe−bx2x2(xe−bx
2)dx =
1
2mω2
∫ ∞−∞
x4e−2bx2dx
=1
2mω24b
√2b
π2
3
32b2
√π
2b=
3mω2
8b
combining to get the expectationvalue of the Hamiltonian
taking the derivative to minimizewrt b
〈H〉 =3~2b2m
+3mω2
8b∂〈H〉∂b
=3~2
2m− 3mω2
8b2= 0
b2 =m2ω2
4~2−→ b =
mω
2~
〈H〉min =3~2
2m
mω
2~+
3mω2
8
2~mω
= ~ω(
3
4+
3
4
)=
3
2~
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 5 / 23
Problem 7.4 (cont.)
and the potential energy
〈V 〉 =1
2mω2
∫ ∞−∞
xe−bx2x2(xe−bx
2)dx =
1
2mω2
∫ ∞−∞
x4e−2bx2dx
=1
2mω24b
√2b
π2
3
32b2
√π
2b=
3mω2
8b
combining to get the expectationvalue of the Hamiltonian
taking the derivative to minimizewrt b
〈H〉 =3~2b2m
+3mω2
8b∂〈H〉∂b
=3~2
2m− 3mω2
8b2= 0
b2 =m2ω2
4~2−→ b =
mω
2~
〈H〉min =3~2
2m
mω
2~+
3mω2
8
2~mω
= ~ω(
3
4+
3
4
)
=3
2~
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 5 / 23
Problem 7.4 (cont.)
and the potential energy
〈V 〉 =1
2mω2
∫ ∞−∞
xe−bx2x2(xe−bx
2)dx =
1
2mω2
∫ ∞−∞
x4e−2bx2dx
=1
2mω24b
√2b
π2
3
32b2
√π
2b=
3mω2
8b
combining to get the expectationvalue of the Hamiltonian
taking the derivative to minimizewrt b
〈H〉 =3~2b2m
+3mω2
8b∂〈H〉∂b
=3~2
2m− 3mω2
8b2= 0
b2 =m2ω2
4~2−→ b =
mω
2~
〈H〉min =3~2
2m
mω
2~+
3mω2
8
2~mω
= ~ω(
3
4+
3
4
)=
3
2~
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 5 / 23
Problem 7.14
If the photon had a nonzero mass (mγ 6= 0), the Coulomb potential wouldbe replaced with the Yukawa potential,
V (~r) = − e2
4πε0
e−µr
r
where µ = mγc/~. Estimate the binding energy of a “hydrogen” atomwith this potential. Assume µa� 1.
First need to choose a trial function and the one which makes most senseto simplify the calculation of 〈H〉 is a hydrogen-like function
ψ =1√πb3
e−r/b
where b is a variable parameter replacing the Bohr radius a
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 6 / 23
Problem 7.14
If the photon had a nonzero mass (mγ 6= 0), the Coulomb potential wouldbe replaced with the Yukawa potential,
V (~r) = − e2
4πε0
e−µr
r
where µ = mγc/~. Estimate the binding energy of a “hydrogen” atomwith this potential. Assume µa� 1.
First need to choose a trial function and the one which makes most senseto simplify the calculation of 〈H〉 is a hydrogen-like function
ψ =1√πb3
e−r/b
where b is a variable parameter replacing the Bohr radius a
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 6 / 23
Problem 7.14
If the photon had a nonzero mass (mγ 6= 0), the Coulomb potential wouldbe replaced with the Yukawa potential,
V (~r) = − e2
4πε0
e−µr
r
where µ = mγc/~. Estimate the binding energy of a “hydrogen” atomwith this potential. Assume µa� 1.
First need to choose a trial function and the one which makes most senseto simplify the calculation of 〈H〉 is a hydrogen-like function
ψ =1√πb3
e−r/b
where b is a variable parameter replacing the Bohr radius a
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 6 / 23
Problem 7.14
If the photon had a nonzero mass (mγ 6= 0), the Coulomb potential wouldbe replaced with the Yukawa potential,
V (~r) = − e2
4πε0
e−µr
r
where µ = mγc/~. Estimate the binding energy of a “hydrogen” atomwith this potential. Assume µa� 1.
First need to choose a trial function and the one which makes most senseto simplify the calculation of 〈H〉 is a hydrogen-like function
ψ =1√πb3
e−r/b
where b is a variable parameter replacing the Bohr radius a
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 6 / 23
Problem 7.14 (cont.)
With this trial function, 〈T 〉 can be writ-ten down from our knowledge of the hy-drogen atom
computing 〈V 〉〈T 〉 =
~2
2mb2
〈V 〉 = − e2
4πε0
4π
πb3
∫ ∞0
e−2r/be−µr
rr2 dr = − e2
4πε0
4
b3
∫ ∞0
e−(µ+2/b)r r dr
= − e2
4πε0
4
b31
(µ+ 2b )2
= − e2
4πε0
1
b(1 + µb2 )2
〈H〉 =~2
2mb2− e2
4πε0
1
b(1 + µb2 )2
∂〈H〉∂b
= − h2
mb3+
e2
4πε0
[1
b2(1 + µb2 )2
+µ
b(1 + µb2 )3
]
= − h2
mb3+
e2
4πε0
(1 + 3µb2 )
b2(1 + µb2 )3
= 0
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 7 / 23
Problem 7.14 (cont.)
With this trial function, 〈T 〉 can be writ-ten down from our knowledge of the hy-drogen atom
computing 〈V 〉
〈T 〉 =~2
2mb2
〈V 〉 = − e2
4πε0
4π
πb3
∫ ∞0
e−2r/be−µr
rr2 dr = − e2
4πε0
4
b3
∫ ∞0
e−(µ+2/b)r r dr
= − e2
4πε0
4
b31
(µ+ 2b )2
= − e2
4πε0
1
b(1 + µb2 )2
〈H〉 =~2
2mb2− e2
4πε0
1
b(1 + µb2 )2
∂〈H〉∂b
= − h2
mb3+
e2
4πε0
[1
b2(1 + µb2 )2
+µ
b(1 + µb2 )3
]
= − h2
mb3+
e2
4πε0
(1 + 3µb2 )
b2(1 + µb2 )3
= 0
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 7 / 23
Problem 7.14 (cont.)
With this trial function, 〈T 〉 can be writ-ten down from our knowledge of the hy-drogen atom
computing 〈V 〉〈T 〉 =
~2
2mb2
〈V 〉 = − e2
4πε0
4π
πb3
∫ ∞0
e−2r/be−µr
rr2 dr = − e2
4πε0
4
b3
∫ ∞0
e−(µ+2/b)r r dr
= − e2
4πε0
4
b31
(µ+ 2b )2
= − e2
4πε0
1
b(1 + µb2 )2
〈H〉 =~2
2mb2− e2
4πε0
1
b(1 + µb2 )2
∂〈H〉∂b
= − h2
mb3+
e2
4πε0
[1
b2(1 + µb2 )2
+µ
b(1 + µb2 )3
]
= − h2
mb3+
e2
4πε0
(1 + 3µb2 )
b2(1 + µb2 )3
= 0
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 7 / 23
Problem 7.14 (cont.)
With this trial function, 〈T 〉 can be writ-ten down from our knowledge of the hy-drogen atom
computing 〈V 〉〈T 〉 =
~2
2mb2
〈V 〉 = − e2
4πε0
4π
πb3
∫ ∞0
e−2r/be−µr
rr2 dr
= − e2
4πε0
4
b3
∫ ∞0
e−(µ+2/b)r r dr
= − e2
4πε0
4
b31
(µ+ 2b )2
= − e2
4πε0
1
b(1 + µb2 )2
〈H〉 =~2
2mb2− e2
4πε0
1
b(1 + µb2 )2
∂〈H〉∂b
= − h2
mb3+
e2
4πε0
[1
b2(1 + µb2 )2
+µ
b(1 + µb2 )3
]
= − h2
mb3+
e2
4πε0
(1 + 3µb2 )
b2(1 + µb2 )3
= 0
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 7 / 23
Problem 7.14 (cont.)
With this trial function, 〈T 〉 can be writ-ten down from our knowledge of the hy-drogen atom
computing 〈V 〉〈T 〉 =
~2
2mb2
〈V 〉 = − e2
4πε0
4π
πb3
∫ ∞0
e−2r/be−µr
rr2 dr = − e2
4πε0
4
b3
∫ ∞0
e−(µ+2/b)r r dr
= − e2
4πε0
4
b31
(µ+ 2b )2
= − e2
4πε0
1
b(1 + µb2 )2
〈H〉 =~2
2mb2− e2
4πε0
1
b(1 + µb2 )2
∂〈H〉∂b
= − h2
mb3+
e2
4πε0
[1
b2(1 + µb2 )2
+µ
b(1 + µb2 )3
]
= − h2
mb3+
e2
4πε0
(1 + 3µb2 )
b2(1 + µb2 )3
= 0
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 7 / 23
Problem 7.14 (cont.)
With this trial function, 〈T 〉 can be writ-ten down from our knowledge of the hy-drogen atom
computing 〈V 〉〈T 〉 =
~2
2mb2
〈V 〉 = − e2
4πε0
4π
πb3
∫ ∞0
e−2r/be−µr
rr2 dr = − e2
4πε0
4
b3
∫ ∞0
e−(µ+2/b)r r dr
= − e2
4πε0
4
b31
(µ+ 2b )2
= − e2
4πε0
1
b(1 + µb2 )2
〈H〉 =~2
2mb2− e2
4πε0
1
b(1 + µb2 )2
∂〈H〉∂b
= − h2
mb3+
e2
4πε0
[1
b2(1 + µb2 )2
+µ
b(1 + µb2 )3
]
= − h2
mb3+
e2
4πε0
(1 + 3µb2 )
b2(1 + µb2 )3
= 0
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 7 / 23
Problem 7.14 (cont.)
With this trial function, 〈T 〉 can be writ-ten down from our knowledge of the hy-drogen atom
computing 〈V 〉〈T 〉 =
~2
2mb2
〈V 〉 = − e2
4πε0
4π
πb3
∫ ∞0
e−2r/be−µr
rr2 dr = − e2
4πε0
4
b3
∫ ∞0
e−(µ+2/b)r r dr
= − e2
4πε0
4
b31
(µ+ 2b )2
= − e2
4πε0
1
b(1 + µb2 )2
〈H〉 =~2
2mb2− e2
4πε0
1
b(1 + µb2 )2
∂〈H〉∂b
= − h2
mb3+
e2
4πε0
[1
b2(1 + µb2 )2
+µ
b(1 + µb2 )3
]
= − h2
mb3+
e2
4πε0
(1 + 3µb2 )
b2(1 + µb2 )3
= 0
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 7 / 23
Problem 7.14 (cont.)
With this trial function, 〈T 〉 can be writ-ten down from our knowledge of the hy-drogen atom
computing 〈V 〉〈T 〉 =
~2
2mb2
〈V 〉 = − e2
4πε0
4π
πb3
∫ ∞0
e−2r/be−µr
rr2 dr = − e2
4πε0
4
b3
∫ ∞0
e−(µ+2/b)r r dr
= − e2
4πε0
4
b31
(µ+ 2b )2
= − e2
4πε0
1
b(1 + µb2 )2
〈H〉 =~2
2mb2− e2
4πε0
1
b(1 + µb2 )2
∂〈H〉∂b
= − h2
mb3+
e2
4πε0
[1
b2(1 + µb2 )2
+µ
b(1 + µb2 )3
]
= − h2
mb3+
e2
4πε0
(1 + 3µb2 )
b2(1 + µb2 )3
= 0
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 7 / 23
Problem 7.14 (cont.)
With this trial function, 〈T 〉 can be writ-ten down from our knowledge of the hy-drogen atom
computing 〈V 〉〈T 〉 =
~2
2mb2
〈V 〉 = − e2
4πε0
4π
πb3
∫ ∞0
e−2r/be−µr
rr2 dr = − e2
4πε0
4
b3
∫ ∞0
e−(µ+2/b)r r dr
= − e2
4πε0
4
b31
(µ+ 2b )2
= − e2
4πε0
1
b(1 + µb2 )2
〈H〉 =~2
2mb2− e2
4πε0
1
b(1 + µb2 )2
∂〈H〉∂b
= − h2
mb3+
e2
4πε0
[1
b2(1 + µb2 )2
+µ
b(1 + µb2 )3
]
= − h2
mb3+
e2
4πε0
(1 + 3µb2 )
b2(1 + µb2 )3
= 0
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 7 / 23
Problem 7.14 (cont.)
With this trial function, 〈T 〉 can be writ-ten down from our knowledge of the hy-drogen atom
computing 〈V 〉〈T 〉 =
~2
2mb2
〈V 〉 = − e2
4πε0
4π
πb3
∫ ∞0
e−2r/be−µr
rr2 dr = − e2
4πε0
4
b3
∫ ∞0
e−(µ+2/b)r r dr
= − e2
4πε0
4
b31
(µ+ 2b )2
= − e2
4πε0
1
b(1 + µb2 )2
〈H〉 =~2
2mb2− e2
4πε0
1
b(1 + µb2 )2
∂〈H〉∂b
= − h2
mb3+
e2
4πε0
[1
b2(1 + µb2 )2
+µ
b(1 + µb2 )3
]
= − h2
mb3+
e2
4πε0
(1 + 3µb2 )
b2(1 + µb2 )3
= 0
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 7 / 23
Problem 7.14 (cont.)
∂〈H〉∂b
= − h2
mb3+
e2
4πε0
(1 + 3µb2 )
b2(1 + µb2 )3
= 0
~2
m
(4πε0e2
)= b
(1 + 3µb2 )
(1 + µb2 )3
= a the left side is simply the Bohr ra-dius, a
this cubic equation can be solved approximately by remembering thatµa� 1
a ≈ b
(1 +
3µb
2
)[1− 3µb
2+ 6
(µb
2
)2]≈ b
[1− 9
4(µb)2 +
6
4(µb)2
]≈ b
[1− 3
4(µb)2
]−→ b ≈ a
1− 34(µb)2
≈ a
[1 +
3
4(µa)2
]
〈H〉min =~2
2ma2
[1 +
3
4(µa)2
]2− e2
4πε0
1
a[1 + 3
4(µa)2] [
1 + 12(µa)
]2
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 8 / 23
Problem 7.14 (cont.)
∂〈H〉∂b
= − h2
mb3+
e2
4πε0
(1 + 3µb2 )
b2(1 + µb2 )3
= 0
~2
m
(4πε0e2
)= b
(1 + 3µb2 )
(1 + µb2 )3
= a the left side is simply the Bohr ra-dius, a
this cubic equation can be solved approximately by remembering thatµa� 1
a ≈ b
(1 +
3µb
2
)[1− 3µb
2+ 6
(µb
2
)2]≈ b
[1− 9
4(µb)2 +
6
4(µb)2
]≈ b
[1− 3
4(µb)2
]−→ b ≈ a
1− 34(µb)2
≈ a
[1 +
3
4(µa)2
]
〈H〉min =~2
2ma2
[1 +
3
4(µa)2
]2− e2
4πε0
1
a[1 + 3
4(µa)2] [
1 + 12(µa)
]2
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 8 / 23
Problem 7.14 (cont.)
∂〈H〉∂b
= − h2
mb3+
e2
4πε0
(1 + 3µb2 )
b2(1 + µb2 )3
= 0
~2
m
(4πε0e2
)= b
(1 + 3µb2 )
(1 + µb2 )3
= a the left side is simply the Bohr ra-dius, a
this cubic equation can be solved approximately by remembering thatµa� 1
a ≈ b
(1 +
3µb
2
)[1− 3µb
2+ 6
(µb
2
)2]≈ b
[1− 9
4(µb)2 +
6
4(µb)2
]≈ b
[1− 3
4(µb)2
]−→ b ≈ a
1− 34(µb)2
≈ a
[1 +
3
4(µa)2
]
〈H〉min =~2
2ma2
[1 +
3
4(µa)2
]2− e2
4πε0
1
a[1 + 3
4(µa)2] [
1 + 12(µa)
]2
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 8 / 23
Problem 7.14 (cont.)
∂〈H〉∂b
= − h2
mb3+
e2
4πε0
(1 + 3µb2 )
b2(1 + µb2 )3
= 0
~2
m
(4πε0e2
)= b
(1 + 3µb2 )
(1 + µb2 )3
= a
the left side is simply the Bohr ra-dius, a
this cubic equation can be solved approximately by remembering thatµa� 1
a ≈ b
(1 +
3µb
2
)[1− 3µb
2+ 6
(µb
2
)2]≈ b
[1− 9
4(µb)2 +
6
4(µb)2
]≈ b
[1− 3
4(µb)2
]−→ b ≈ a
1− 34(µb)2
≈ a
[1 +
3
4(µa)2
]
〈H〉min =~2
2ma2
[1 +
3
4(µa)2
]2− e2
4πε0
1
a[1 + 3
4(µa)2] [
1 + 12(µa)
]2
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 8 / 23
Problem 7.14 (cont.)
∂〈H〉∂b
= − h2
mb3+
e2
4πε0
(1 + 3µb2 )
b2(1 + µb2 )3
= 0
~2
m
(4πε0e2
)= b
(1 + 3µb2 )
(1 + µb2 )3
= a the left side is simply the Bohr ra-dius, a
this cubic equation can be solved approximately by remembering thatµa� 1
a ≈ b
(1 +
3µb
2
)[1− 3µb
2+ 6
(µb
2
)2]≈ b
[1− 9
4(µb)2 +
6
4(µb)2
]≈ b
[1− 3
4(µb)2
]−→ b ≈ a
1− 34(µb)2
≈ a
[1 +
3
4(µa)2
]
〈H〉min =~2
2ma2
[1 +
3
4(µa)2
]2− e2
4πε0
1
a[1 + 3
4(µa)2] [
1 + 12(µa)
]2
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 8 / 23
Problem 7.14 (cont.)
∂〈H〉∂b
= − h2
mb3+
e2
4πε0
(1 + 3µb2 )
b2(1 + µb2 )3
= 0
~2
m
(4πε0e2
)= b
(1 + 3µb2 )
(1 + µb2 )3
= a the left side is simply the Bohr ra-dius, a
this cubic equation can be solved approximately by remembering thatµa� 1
a ≈ b
(1 +
3µb
2
)[1− 3µb
2+ 6
(µb
2
)2]≈ b
[1− 9
4(µb)2 +
6
4(µb)2
]≈ b
[1− 3
4(µb)2
]−→ b ≈ a
1− 34(µb)2
≈ a
[1 +
3
4(µa)2
]
〈H〉min =~2
2ma2
[1 +
3
4(µa)2
]2− e2
4πε0
1
a[1 + 3
4(µa)2] [
1 + 12(µa)
]2
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 8 / 23
Problem 7.14 (cont.)
∂〈H〉∂b
= − h2
mb3+
e2
4πε0
(1 + 3µb2 )
b2(1 + µb2 )3
= 0
~2
m
(4πε0e2
)= b
(1 + 3µb2 )
(1 + µb2 )3
= a the left side is simply the Bohr ra-dius, a
this cubic equation can be solved approximately by remembering thatµa� 1
a ≈ b
(1 +
3µb
2
)[1− 3µb
2+ 6
(µb
2
)2]
≈ b
[1− 9
4(µb)2 +
6
4(µb)2
]≈ b
[1− 3
4(µb)2
]−→ b ≈ a
1− 34(µb)2
≈ a
[1 +
3
4(µa)2
]
〈H〉min =~2
2ma2
[1 +
3
4(µa)2
]2− e2
4πε0
1
a[1 + 3
4(µa)2] [
1 + 12(µa)
]2
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 8 / 23
Problem 7.14 (cont.)
∂〈H〉∂b
= − h2
mb3+
e2
4πε0
(1 + 3µb2 )
b2(1 + µb2 )3
= 0
~2
m
(4πε0e2
)= b
(1 + 3µb2 )
(1 + µb2 )3
= a the left side is simply the Bohr ra-dius, a
this cubic equation can be solved approximately by remembering thatµa� 1
a ≈ b
(1 +
3µb
2
)[1− 3µb
2+ 6
(µb
2
)2]≈ b
[1− 9
4(µb)2 +
6
4(µb)2
]
≈ b
[1− 3
4(µb)2
]−→ b ≈ a
1− 34(µb)2
≈ a
[1 +
3
4(µa)2
]
〈H〉min =~2
2ma2
[1 +
3
4(µa)2
]2− e2
4πε0
1
a[1 + 3
4(µa)2] [
1 + 12(µa)
]2
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 8 / 23
Problem 7.14 (cont.)
∂〈H〉∂b
= − h2
mb3+
e2
4πε0
(1 + 3µb2 )
b2(1 + µb2 )3
= 0
~2
m
(4πε0e2
)= b
(1 + 3µb2 )
(1 + µb2 )3
= a the left side is simply the Bohr ra-dius, a
this cubic equation can be solved approximately by remembering thatµa� 1
a ≈ b
(1 +
3µb
2
)[1− 3µb
2+ 6
(µb
2
)2]≈ b
[1− 9
4(µb)2 +
6
4(µb)2
]
≈ b
[1− 3
4(µb)2
]−→ b ≈ a
1− 34(µb)2
≈ a
[1 +
3
4(µa)2
]
〈H〉min =~2
2ma2
[1 +
3
4(µa)2
]2− e2
4πε0
1
a[1 + 3
4(µa)2] [
1 + 12(µa)
]2
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 8 / 23
Problem 7.14 (cont.)
∂〈H〉∂b
= − h2
mb3+
e2
4πε0
(1 + 3µb2 )
b2(1 + µb2 )3
= 0
~2
m
(4πε0e2
)= b
(1 + 3µb2 )
(1 + µb2 )3
= a the left side is simply the Bohr ra-dius, a
this cubic equation can be solved approximately by remembering thatµa� 1
a ≈ b
(1 +
3µb
2
)[1− 3µb
2+ 6
(µb
2
)2]≈ b
[1− 9
4(µb)2 +
6
4(µb)2
]≈ b
[1− 3
4(µb)2
]
−→ b ≈ a
1− 34(µb)2
≈ a
[1 +
3
4(µa)2
]
〈H〉min =~2
2ma2
[1 +
3
4(µa)2
]2− e2
4πε0
1
a[1 + 3
4(µa)2] [
1 + 12(µa)
]2
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 8 / 23
Problem 7.14 (cont.)
∂〈H〉∂b
= − h2
mb3+
e2
4πε0
(1 + 3µb2 )
b2(1 + µb2 )3
= 0
~2
m
(4πε0e2
)= b
(1 + 3µb2 )
(1 + µb2 )3
= a the left side is simply the Bohr ra-dius, a
this cubic equation can be solved approximately by remembering thatµa� 1
a ≈ b
(1 +
3µb
2
)[1− 3µb
2+ 6
(µb
2
)2]≈ b
[1− 9
4(µb)2 +
6
4(µb)2
]≈ b
[1− 3
4(µb)2
]−→ b ≈ a
1− 34(µb)2
≈ a
[1 +
3
4(µa)2
]
〈H〉min =~2
2ma2
[1 +
3
4(µa)2
]2− e2
4πε0
1
a[1 + 3
4(µa)2] [
1 + 12(µa)
]2
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 8 / 23
Problem 7.14 (cont.)
∂〈H〉∂b
= − h2
mb3+
e2
4πε0
(1 + 3µb2 )
b2(1 + µb2 )3
= 0
~2
m
(4πε0e2
)= b
(1 + 3µb2 )
(1 + µb2 )3
= a the left side is simply the Bohr ra-dius, a
this cubic equation can be solved approximately by remembering thatµa� 1
a ≈ b
(1 +
3µb
2
)[1− 3µb
2+ 6
(µb
2
)2]≈ b
[1− 9
4(µb)2 +
6
4(µb)2
]≈ b
[1− 3
4(µb)2
]−→ b ≈ a
1− 34(µb)2
≈ a
[1 +
3
4(µa)2
]
〈H〉min =~2
2ma2
[1 +
3
4(µa)2
]2− e2
4πε0
1
a[1 + 3
4(µa)2] [
1 + 12(µa)
]2
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 8 / 23
Problem 7.14 (cont.)
∂〈H〉∂b
= − h2
mb3+
e2
4πε0
(1 + 3µb2 )
b2(1 + µb2 )3
= 0
~2
m
(4πε0e2
)= b
(1 + 3µb2 )
(1 + µb2 )3
= a the left side is simply the Bohr ra-dius, a
this cubic equation can be solved approximately by remembering thatµa� 1
a ≈ b
(1 +
3µb
2
)[1− 3µb
2+ 6
(µb
2
)2]≈ b
[1− 9
4(µb)2 +
6
4(µb)2
]≈ b
[1− 3
4(µb)2
]−→ b ≈ a
1− 34(µb)2
≈ a
[1 +
3
4(µa)2
]
〈H〉min =~2
2ma2
[1 +
3
4(µa)2
]2− e2
4πε0
1
a[1 + 3
4(µa)2] [
1 + 12(µa)
]2
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 8 / 23
Problem 7.14 (cont.)
∂〈H〉∂b
= − h2
mb3+
e2
4πε0
(1 + 3µb2 )
b2(1 + µb2 )3
= 0
~2
m
(4πε0e2
)= b
(1 + 3µb2 )
(1 + µb2 )3
= a the left side is simply the Bohr ra-dius, a
this cubic equation can be solved approximately by remembering thatµa� 1
a ≈ b
(1 +
3µb
2
)[1− 3µb
2+ 6
(µb
2
)2]≈ b
[1− 9
4(µb)2 +
6
4(µb)2
]≈ b
[1− 3
4(µb)2
]−→ b ≈ a
1− 34(µb)2
≈ a
[1 +
3
4(µa)2
]
〈H〉min =~2
2ma2
[1 +
3
4(µa)2
]2− e2
4πε0
1
a[1 + 3
4(µa)2] [
1 + 12(µa)
]2C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 8 / 23
Problem 7.14 (cont.)
〈H〉min =~2
2ma2
[1 +
3
4(µa)2
]2− e2
4πε0
1
a[1 + 3
4(µa)2] [
1 + 12(µa)
]2
≈ ~2
2ma2
[1− 2
3
4(µa)2
]− e2
4πε0a
[1− 3
4(µa)2
][1− 2
µa
2+
3
2
(µa2
)2]≈ −E1
[1− 3
2(µa)2
]+ 2E1
[1− µa +
����3
4(µa)2 −
����3
4(µa)2
]≈ E1
[1− 2(µa) +
3
2(µa)2
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 9 / 23
Problem 7.14 (cont.)
〈H〉min =~2
2ma2
[1 +
3
4(µa)2
]2− e2
4πε0
1
a[1 + 3
4(µa)2] [
1 + 12(µa)
]2≈ ~2
2ma2
[1− 2
3
4(µa)2
]− e2
4πε0a
[1− 3
4(µa)2
][1− 2
µa
2+
3
2
(µa2
)2]
≈ −E1
[1− 3
2(µa)2
]+ 2E1
[1− µa +
����3
4(µa)2 −
����3
4(µa)2
]≈ E1
[1− 2(µa) +
3
2(µa)2
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 9 / 23
Problem 7.14 (cont.)
〈H〉min =~2
2ma2
[1 +
3
4(µa)2
]2− e2
4πε0
1
a[1 + 3
4(µa)2] [
1 + 12(µa)
]2≈ ~2
2ma2
[1− 2
3
4(µa)2
]− e2
4πε0a
[1− 3
4(µa)2
][1− 2
µa
2+
3
2
(µa2
)2]≈ −E1
[1− 3
2(µa)2
]+ 2E1
[1− µa +
3
4(µa)2 − 3
4(µa)2
]
≈ E1
[1− 2(µa) +
3
2(µa)2
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 9 / 23
Problem 7.14 (cont.)
〈H〉min =~2
2ma2
[1 +
3
4(µa)2
]2− e2
4πε0
1
a[1 + 3
4(µa)2] [
1 + 12(µa)
]2≈ ~2
2ma2
[1− 2
3
4(µa)2
]− e2
4πε0a
[1− 3
4(µa)2
][1− 2
µa
2+
3
2
(µa2
)2]≈ −E1
[1− 3
2(µa)2
]+ 2E1
[1− µa +
����3
4(µa)2 −
��
��3
4(µa)2
]
≈ E1
[1− 2(µa) +
3
2(µa)2
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 9 / 23
Problem 7.14 (cont.)
〈H〉min =~2
2ma2
[1 +
3
4(µa)2
]2− e2
4πε0
1
a[1 + 3
4(µa)2] [
1 + 12(µa)
]2≈ ~2
2ma2
[1− 2
3
4(µa)2
]− e2
4πε0a
[1− 3
4(µa)2
][1− 2
µa
2+
3
2
(µa2
)2]≈ −E1
[1− 3
2(µa)2
]+ 2E1
[1− µa +
����3
4(µa)2 −
��
��3
4(µa)2
]≈ E1
[1− 2(µa) +
3
2(µa)2
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 9 / 23
Problem 7.15
Suppose you are given a quantum system whose Hamiltonian H0 admitsjust two eigenstates, ψa (with energy Ea) and ψb (with energy Eb). Theyare orthogonal, normalized and non-degenerate (assume Ea < Eb). Nowturn on a perturbation H ′, with the following matrix elements:⟨
H ′⟩
=
(0 hh 0
), h = constant
a. Find the exact eigenvalues of the perturbing Hamiltonian.
b. Estimate the energies of the perturbed system using second-orderperturbation theory.
c. Estimate the ground state energy of the perturbed system using thevariational principle with a trial wavefuction of the form
ψ = (cosφ)ψa + (sinφ)ψb
where φ is an adjustable parameter.
d. Compare the answers to the sections above.
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 10 / 23
Problem 7.15 (cont.)
a. Start by solving the eigenvaluedeterminant
this is simply a quadratic equation
0 =
∣∣∣∣ Ea − λ hh Eb − λ
∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2
0 = λ2 − (Ea + Eb)λ+ EaEb − h2
E± =1
2
[(Ea + Eb)±
√(Ea + Eb)2 − 4EaEb + 4h2
]=
1
2
[(Ea + Eb)±
√(Eb − Ea)2 + 4h2
]b. The first order perturbation correction is zero as can be seen from thematrix for H ′ which has zero in the diagonal elements. The second orderperturbation correction is
E(2)a =
| 〈ψb|H ′|ψa〉 |2
Ea − Eb= − h2
Eb − Ea; E
(2)b =
h2
Eb − Ea
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 11 / 23
Problem 7.15 (cont.)
a. Start by solving the eigenvaluedeterminant
this is simply a quadratic equation
0 =
∣∣∣∣ Ea − λ hh Eb − λ
∣∣∣∣
0 = (Ea − λ)(Eb − λ)− h2
0 = λ2 − (Ea + Eb)λ+ EaEb − h2
E± =1
2
[(Ea + Eb)±
√(Ea + Eb)2 − 4EaEb + 4h2
]=
1
2
[(Ea + Eb)±
√(Eb − Ea)2 + 4h2
]b. The first order perturbation correction is zero as can be seen from thematrix for H ′ which has zero in the diagonal elements. The second orderperturbation correction is
E(2)a =
| 〈ψb|H ′|ψa〉 |2
Ea − Eb= − h2
Eb − Ea; E
(2)b =
h2
Eb − Ea
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 11 / 23
Problem 7.15 (cont.)
a. Start by solving the eigenvaluedeterminant
this is simply a quadratic equation
0 =
∣∣∣∣ Ea − λ hh Eb − λ
∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2
0 = λ2 − (Ea + Eb)λ+ EaEb − h2
E± =1
2
[(Ea + Eb)±
√(Ea + Eb)2 − 4EaEb + 4h2
]=
1
2
[(Ea + Eb)±
√(Eb − Ea)2 + 4h2
]b. The first order perturbation correction is zero as can be seen from thematrix for H ′ which has zero in the diagonal elements. The second orderperturbation correction is
E(2)a =
| 〈ψb|H ′|ψa〉 |2
Ea − Eb= − h2
Eb − Ea; E
(2)b =
h2
Eb − Ea
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 11 / 23
Problem 7.15 (cont.)
a. Start by solving the eigenvaluedeterminant
this is simply a quadratic equation
0 =
∣∣∣∣ Ea − λ hh Eb − λ
∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2
0 = λ2 − (Ea + Eb)λ+ EaEb − h2
E± =1
2
[(Ea + Eb)±
√(Ea + Eb)2 − 4EaEb + 4h2
]=
1
2
[(Ea + Eb)±
√(Eb − Ea)2 + 4h2
]b. The first order perturbation correction is zero as can be seen from thematrix for H ′ which has zero in the diagonal elements. The second orderperturbation correction is
E(2)a =
| 〈ψb|H ′|ψa〉 |2
Ea − Eb= − h2
Eb − Ea; E
(2)b =
h2
Eb − Ea
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 11 / 23
Problem 7.15 (cont.)
a. Start by solving the eigenvaluedeterminant
this is simply a quadratic equation
0 =
∣∣∣∣ Ea − λ hh Eb − λ
∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2
0 = λ2 − (Ea + Eb)λ+ EaEb − h2
E± =1
2
[(Ea + Eb)±
√(Ea + Eb)2 − 4EaEb + 4h2
]=
1
2
[(Ea + Eb)±
√(Eb − Ea)2 + 4h2
]b. The first order perturbation correction is zero as can be seen from thematrix for H ′ which has zero in the diagonal elements. The second orderperturbation correction is
E(2)a =
| 〈ψb|H ′|ψa〉 |2
Ea − Eb= − h2
Eb − Ea; E
(2)b =
h2
Eb − Ea
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 11 / 23
Problem 7.15 (cont.)
a. Start by solving the eigenvaluedeterminant
this is simply a quadratic equation
0 =
∣∣∣∣ Ea − λ hh Eb − λ
∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2
0 = λ2 − (Ea + Eb)λ+ EaEb − h2
E± =1
2
[(Ea + Eb)±
√(Ea + Eb)2 − 4EaEb + 4h2
]
=1
2
[(Ea + Eb)±
√(Eb − Ea)2 + 4h2
]b. The first order perturbation correction is zero as can be seen from thematrix for H ′ which has zero in the diagonal elements. The second orderperturbation correction is
E(2)a =
| 〈ψb|H ′|ψa〉 |2
Ea − Eb= − h2
Eb − Ea; E
(2)b =
h2
Eb − Ea
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 11 / 23
Problem 7.15 (cont.)
a. Start by solving the eigenvaluedeterminant
this is simply a quadratic equation
0 =
∣∣∣∣ Ea − λ hh Eb − λ
∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2
0 = λ2 − (Ea + Eb)λ+ EaEb − h2
E± =1
2
[(Ea + Eb)±
√(Ea + Eb)2 − 4EaEb + 4h2
]=
1
2
[(Ea + Eb)±
√(Eb − Ea)2 + 4h2
]
b. The first order perturbation correction is zero as can be seen from thematrix for H ′ which has zero in the diagonal elements. The second orderperturbation correction is
E(2)a =
| 〈ψb|H ′|ψa〉 |2
Ea − Eb= − h2
Eb − Ea; E
(2)b =
h2
Eb − Ea
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 11 / 23
Problem 7.15 (cont.)
a. Start by solving the eigenvaluedeterminant
this is simply a quadratic equation
0 =
∣∣∣∣ Ea − λ hh Eb − λ
∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2
0 = λ2 − (Ea + Eb)λ+ EaEb − h2
E± =1
2
[(Ea + Eb)±
√(Ea + Eb)2 − 4EaEb + 4h2
]=
1
2
[(Ea + Eb)±
√(Eb − Ea)2 + 4h2
]b. The first order perturbation correction is zero as can be seen from thematrix for H ′ which has zero in the diagonal elements.
The second orderperturbation correction is
E(2)a =
| 〈ψb|H ′|ψa〉 |2
Ea − Eb= − h2
Eb − Ea; E
(2)b =
h2
Eb − Ea
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 11 / 23
Problem 7.15 (cont.)
a. Start by solving the eigenvaluedeterminant
this is simply a quadratic equation
0 =
∣∣∣∣ Ea − λ hh Eb − λ
∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2
0 = λ2 − (Ea + Eb)λ+ EaEb − h2
E± =1
2
[(Ea + Eb)±
√(Ea + Eb)2 − 4EaEb + 4h2
]=
1
2
[(Ea + Eb)±
√(Eb − Ea)2 + 4h2
]b. The first order perturbation correction is zero as can be seen from thematrix for H ′ which has zero in the diagonal elements. The second orderperturbation correction is
E(2)a =
| 〈ψb|H ′|ψa〉 |2
Ea − Eb= − h2
Eb − Ea; E
(2)b =
h2
Eb − Ea
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 11 / 23
Problem 7.15 (cont.)
a. Start by solving the eigenvaluedeterminant
this is simply a quadratic equation
0 =
∣∣∣∣ Ea − λ hh Eb − λ
∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2
0 = λ2 − (Ea + Eb)λ+ EaEb − h2
E± =1
2
[(Ea + Eb)±
√(Ea + Eb)2 − 4EaEb + 4h2
]=
1
2
[(Ea + Eb)±
√(Eb − Ea)2 + 4h2
]b. The first order perturbation correction is zero as can be seen from thematrix for H ′ which has zero in the diagonal elements. The second orderperturbation correction is
E(2)a =
| 〈ψb|H ′|ψa〉 |2
Ea − Eb
= − h2
Eb − Ea; E
(2)b =
h2
Eb − Ea
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 11 / 23
Problem 7.15 (cont.)
a. Start by solving the eigenvaluedeterminant
this is simply a quadratic equation
0 =
∣∣∣∣ Ea − λ hh Eb − λ
∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2
0 = λ2 − (Ea + Eb)λ+ EaEb − h2
E± =1
2
[(Ea + Eb)±
√(Ea + Eb)2 − 4EaEb + 4h2
]=
1
2
[(Ea + Eb)±
√(Eb − Ea)2 + 4h2
]b. The first order perturbation correction is zero as can be seen from thematrix for H ′ which has zero in the diagonal elements. The second orderperturbation correction is
E(2)a =
| 〈ψb|H ′|ψa〉 |2
Ea − Eb= − h2
Eb − Ea;
E(2)b =
h2
Eb − Ea
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 11 / 23
Problem 7.15 (cont.)
a. Start by solving the eigenvaluedeterminant
this is simply a quadratic equation
0 =
∣∣∣∣ Ea − λ hh Eb − λ
∣∣∣∣0 = (Ea − λ)(Eb − λ)− h2
0 = λ2 − (Ea + Eb)λ+ EaEb − h2
E± =1
2
[(Ea + Eb)±
√(Ea + Eb)2 − 4EaEb + 4h2
]=
1
2
[(Ea + Eb)±
√(Eb − Ea)2 + 4h2
]b. The first order perturbation correction is zero as can be seen from thematrix for H ′ which has zero in the diagonal elements. The second orderperturbation correction is
E(2)a =
| 〈ψb|H ′|ψa〉 |2
Ea − Eb= − h2
Eb − Ea; E
(2)b =
h2
Eb − Ea
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 11 / 23
Problem 7.15 (cont.)
The energies, to second order are thus
E− ≈ Ea −h2
Eb − Ea; E+ ≈ Eb +
h2
Eb − Ea
c. The expectation value of the Hamiltonian using the trial wavefunction is
〈H〉 =⟨cosφψa + sinφψb|H0 + H ′| cosφψa + sinφψb
⟩= cos2 φ 〈ψa|H0|ψa〉+ sin2 φ 〈ψb|H0|ψb〉
+ sinφ cosφ⟨ψa|H ′|ψb
⟩+ sinφ cosφ
⟨ψb|H ′|ψa
⟩= Ea cos2 φ+ Eb sin2 φ+ 2h sinφ cosφ
minimizing
0 =∂ 〈H〉∂φ
= −Ea2 cosφ sinφ+ Eb2 sinφ cosφ+ 2h(cos2 φ− sin2 φ)
= (Eb − Ea) sin 2φ+ 2h cos 2φ −→ tan 2φ = − 2h
Eb − Ea
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 12 / 23
Problem 7.15 (cont.)
The energies, to second order are thus
E− ≈ Ea −h2
Eb − Ea;
E+ ≈ Eb +h2
Eb − Ea
c. The expectation value of the Hamiltonian using the trial wavefunction is
〈H〉 =⟨cosφψa + sinφψb|H0 + H ′| cosφψa + sinφψb
⟩= cos2 φ 〈ψa|H0|ψa〉+ sin2 φ 〈ψb|H0|ψb〉
+ sinφ cosφ⟨ψa|H ′|ψb
⟩+ sinφ cosφ
⟨ψb|H ′|ψa
⟩= Ea cos2 φ+ Eb sin2 φ+ 2h sinφ cosφ
minimizing
0 =∂ 〈H〉∂φ
= −Ea2 cosφ sinφ+ Eb2 sinφ cosφ+ 2h(cos2 φ− sin2 φ)
= (Eb − Ea) sin 2φ+ 2h cos 2φ −→ tan 2φ = − 2h
Eb − Ea
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 12 / 23
Problem 7.15 (cont.)
The energies, to second order are thus
E− ≈ Ea −h2
Eb − Ea; E+ ≈ Eb +
h2
Eb − Ea
c. The expectation value of the Hamiltonian using the trial wavefunction is
〈H〉 =⟨cosφψa + sinφψb|H0 + H ′| cosφψa + sinφψb
⟩= cos2 φ 〈ψa|H0|ψa〉+ sin2 φ 〈ψb|H0|ψb〉
+ sinφ cosφ⟨ψa|H ′|ψb
⟩+ sinφ cosφ
⟨ψb|H ′|ψa
⟩= Ea cos2 φ+ Eb sin2 φ+ 2h sinφ cosφ
minimizing
0 =∂ 〈H〉∂φ
= −Ea2 cosφ sinφ+ Eb2 sinφ cosφ+ 2h(cos2 φ− sin2 φ)
= (Eb − Ea) sin 2φ+ 2h cos 2φ −→ tan 2φ = − 2h
Eb − Ea
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 12 / 23
Problem 7.15 (cont.)
The energies, to second order are thus
E− ≈ Ea −h2
Eb − Ea; E+ ≈ Eb +
h2
Eb − Ea
c. The expectation value of the Hamiltonian using the trial wavefunction is
〈H〉 =⟨cosφψa + sinφψb|H0 + H ′| cosφψa + sinφψb
⟩= cos2 φ 〈ψa|H0|ψa〉+ sin2 φ 〈ψb|H0|ψb〉
+ sinφ cosφ⟨ψa|H ′|ψb
⟩+ sinφ cosφ
⟨ψb|H ′|ψa
⟩= Ea cos2 φ+ Eb sin2 φ+ 2h sinφ cosφ
minimizing
0 =∂ 〈H〉∂φ
= −Ea2 cosφ sinφ+ Eb2 sinφ cosφ+ 2h(cos2 φ− sin2 φ)
= (Eb − Ea) sin 2φ+ 2h cos 2φ −→ tan 2φ = − 2h
Eb − Ea
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 12 / 23
Problem 7.15 (cont.)
The energies, to second order are thus
E− ≈ Ea −h2
Eb − Ea; E+ ≈ Eb +
h2
Eb − Ea
c. The expectation value of the Hamiltonian using the trial wavefunction is
〈H〉 =⟨cosφψa + sinφψb|H0 + H ′| cosφψa + sinφψb
⟩
= cos2 φ 〈ψa|H0|ψa〉+ sin2 φ 〈ψb|H0|ψb〉+ sinφ cosφ
⟨ψa|H ′|ψb
⟩+ sinφ cosφ
⟨ψb|H ′|ψa
⟩= Ea cos2 φ+ Eb sin2 φ+ 2h sinφ cosφ
minimizing
0 =∂ 〈H〉∂φ
= −Ea2 cosφ sinφ+ Eb2 sinφ cosφ+ 2h(cos2 φ− sin2 φ)
= (Eb − Ea) sin 2φ+ 2h cos 2φ −→ tan 2φ = − 2h
Eb − Ea
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 12 / 23
Problem 7.15 (cont.)
The energies, to second order are thus
E− ≈ Ea −h2
Eb − Ea; E+ ≈ Eb +
h2
Eb − Ea
c. The expectation value of the Hamiltonian using the trial wavefunction is
〈H〉 =⟨cosφψa + sinφψb|H0 + H ′| cosφψa + sinφψb
⟩= cos2 φ 〈ψa|H0|ψa〉+ sin2 φ 〈ψb|H0|ψb〉
+ sinφ cosφ⟨ψa|H ′|ψb
⟩+ sinφ cosφ
⟨ψb|H ′|ψa
⟩
= Ea cos2 φ+ Eb sin2 φ+ 2h sinφ cosφ
minimizing
0 =∂ 〈H〉∂φ
= −Ea2 cosφ sinφ+ Eb2 sinφ cosφ+ 2h(cos2 φ− sin2 φ)
= (Eb − Ea) sin 2φ+ 2h cos 2φ −→ tan 2φ = − 2h
Eb − Ea
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 12 / 23
Problem 7.15 (cont.)
The energies, to second order are thus
E− ≈ Ea −h2
Eb − Ea; E+ ≈ Eb +
h2
Eb − Ea
c. The expectation value of the Hamiltonian using the trial wavefunction is
〈H〉 =⟨cosφψa + sinφψb|H0 + H ′| cosφψa + sinφψb
⟩= cos2 φ 〈ψa|H0|ψa〉+ sin2 φ 〈ψb|H0|ψb〉
+ sinφ cosφ⟨ψa|H ′|ψb
⟩+ sinφ cosφ
⟨ψb|H ′|ψa
⟩= Ea cos2 φ+ Eb sin2 φ+ 2h sinφ cosφ
minimizing
0 =∂ 〈H〉∂φ
= −Ea2 cosφ sinφ+ Eb2 sinφ cosφ+ 2h(cos2 φ− sin2 φ)
= (Eb − Ea) sin 2φ+ 2h cos 2φ −→ tan 2φ = − 2h
Eb − Ea
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 12 / 23
Problem 7.15 (cont.)
The energies, to second order are thus
E− ≈ Ea −h2
Eb − Ea; E+ ≈ Eb +
h2
Eb − Ea
c. The expectation value of the Hamiltonian using the trial wavefunction is
〈H〉 =⟨cosφψa + sinφψb|H0 + H ′| cosφψa + sinφψb
⟩= cos2 φ 〈ψa|H0|ψa〉+ sin2 φ 〈ψb|H0|ψb〉
+ sinφ cosφ⟨ψa|H ′|ψb
⟩+ sinφ cosφ
⟨ψb|H ′|ψa
⟩= Ea cos2 φ+ Eb sin2 φ+ 2h sinφ cosφ
minimizing
0 =∂ 〈H〉∂φ
= −Ea2 cosφ sinφ+ Eb2 sinφ cosφ+ 2h(cos2 φ− sin2 φ)
= (Eb − Ea) sin 2φ+ 2h cos 2φ −→ tan 2φ = − 2h
Eb − Ea
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 12 / 23
Problem 7.15 (cont.)
The energies, to second order are thus
E− ≈ Ea −h2
Eb − Ea; E+ ≈ Eb +
h2
Eb − Ea
c. The expectation value of the Hamiltonian using the trial wavefunction is
〈H〉 =⟨cosφψa + sinφψb|H0 + H ′| cosφψa + sinφψb
⟩= cos2 φ 〈ψa|H0|ψa〉+ sin2 φ 〈ψb|H0|ψb〉
+ sinφ cosφ⟨ψa|H ′|ψb
⟩+ sinφ cosφ
⟨ψb|H ′|ψa
⟩= Ea cos2 φ+ Eb sin2 φ+ 2h sinφ cosφ
minimizing
0 =∂ 〈H〉∂φ
= −Ea2 cosφ sinφ+ Eb2 sinφ cosφ+ 2h(cos2 φ− sin2 φ)
= (Eb − Ea) sin 2φ+ 2h cos 2φ −→ tan 2φ = − 2h
Eb − Ea
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 12 / 23
Problem 7.15 (cont.)
The energies, to second order are thus
E− ≈ Ea −h2
Eb − Ea; E+ ≈ Eb +
h2
Eb − Ea
c. The expectation value of the Hamiltonian using the trial wavefunction is
〈H〉 =⟨cosφψa + sinφψb|H0 + H ′| cosφψa + sinφψb
⟩= cos2 φ 〈ψa|H0|ψa〉+ sin2 φ 〈ψb|H0|ψb〉
+ sinφ cosφ⟨ψa|H ′|ψb
⟩+ sinφ cosφ
⟨ψb|H ′|ψa
⟩= Ea cos2 φ+ Eb sin2 φ+ 2h sinφ cosφ
minimizing
0 =∂ 〈H〉∂φ
= −Ea2 cosφ sinφ+ Eb2 sinφ cosφ+ 2h(cos2 φ− sin2 φ)
= (Eb − Ea) sin 2φ+ 2h cos 2φ
−→ tan 2φ = − 2h
Eb − Ea
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 12 / 23
Problem 7.15 (cont.)
The energies, to second order are thus
E− ≈ Ea −h2
Eb − Ea; E+ ≈ Eb +
h2
Eb − Ea
c. The expectation value of the Hamiltonian using the trial wavefunction is
〈H〉 =⟨cosφψa + sinφψb|H0 + H ′| cosφψa + sinφψb
⟩= cos2 φ 〈ψa|H0|ψa〉+ sin2 φ 〈ψb|H0|ψb〉
+ sinφ cosφ⟨ψa|H ′|ψb
⟩+ sinφ cosφ
⟨ψb|H ′|ψa
⟩= Ea cos2 φ+ Eb sin2 φ+ 2h sinφ cosφ
minimizing
0 =∂ 〈H〉∂φ
= −Ea2 cosφ sinφ+ Eb2 sinφ cosφ+ 2h(cos2 φ− sin2 φ)
= (Eb − Ea) sin 2φ+ 2h cos 2φ −→ tan 2φ = − 2h
Eb − Ea
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 12 / 23
Problem 7.15 (cont.)
defining
ε ≡ 2h
Eb − Ea
rearranging to solve forsin 2φ
similarly, we can solve forcos 2φ
− ε = tan 2φ =sin 2φ
cos 2φ=
sin 2φ√1− sin2 2φ
sin2 2φ = ε2(1− sin2 2φ)
sin2 2φ (1 + ε2) = ε2
sin 2φ =±ε√
1 + ε2
cos2 2φ = 1− ε2
1 + ε2=
1
1 + ε2
cos 2φ =∓1√1 + ε2
thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy
cos2 φ =1
2(1 + cos 2φ), sin2 φ =
1
2(1− cos 2φ)
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 13 / 23
Problem 7.15 (cont.)
defining
ε ≡ 2h
Eb − Ea
rearranging to solve forsin 2φ
similarly, we can solve forcos 2φ
− ε = tan 2φ
=sin 2φ
cos 2φ=
sin 2φ√1− sin2 2φ
sin2 2φ = ε2(1− sin2 2φ)
sin2 2φ (1 + ε2) = ε2
sin 2φ =±ε√
1 + ε2
cos2 2φ = 1− ε2
1 + ε2=
1
1 + ε2
cos 2φ =∓1√1 + ε2
thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy
cos2 φ =1
2(1 + cos 2φ), sin2 φ =
1
2(1− cos 2φ)
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 13 / 23
Problem 7.15 (cont.)
defining
ε ≡ 2h
Eb − Ea
rearranging to solve forsin 2φ
similarly, we can solve forcos 2φ
− ε = tan 2φ =sin 2φ
cos 2φ
=sin 2φ√
1− sin2 2φ
sin2 2φ = ε2(1− sin2 2φ)
sin2 2φ (1 + ε2) = ε2
sin 2φ =±ε√
1 + ε2
cos2 2φ = 1− ε2
1 + ε2=
1
1 + ε2
cos 2φ =∓1√1 + ε2
thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy
cos2 φ =1
2(1 + cos 2φ), sin2 φ =
1
2(1− cos 2φ)
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 13 / 23
Problem 7.15 (cont.)
defining
ε ≡ 2h
Eb − Ea
rearranging to solve forsin 2φ
similarly, we can solve forcos 2φ
− ε = tan 2φ =sin 2φ
cos 2φ=
sin 2φ√1− sin2 2φ
sin2 2φ = ε2(1− sin2 2φ)
sin2 2φ (1 + ε2) = ε2
sin 2φ =±ε√
1 + ε2
cos2 2φ = 1− ε2
1 + ε2=
1
1 + ε2
cos 2φ =∓1√1 + ε2
thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy
cos2 φ =1
2(1 + cos 2φ), sin2 φ =
1
2(1− cos 2φ)
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 13 / 23
Problem 7.15 (cont.)
defining
ε ≡ 2h
Eb − Ea
rearranging to solve forsin 2φ
similarly, we can solve forcos 2φ
− ε = tan 2φ =sin 2φ
cos 2φ=
sin 2φ√1− sin2 2φ
sin2 2φ = ε2(1− sin2 2φ)
sin2 2φ (1 + ε2) = ε2
sin 2φ =±ε√
1 + ε2
cos2 2φ = 1− ε2
1 + ε2=
1
1 + ε2
cos 2φ =∓1√1 + ε2
thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy
cos2 φ =1
2(1 + cos 2φ), sin2 φ =
1
2(1− cos 2φ)
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 13 / 23
Problem 7.15 (cont.)
defining
ε ≡ 2h
Eb − Ea
rearranging to solve forsin 2φ
similarly, we can solve forcos 2φ
− ε = tan 2φ =sin 2φ
cos 2φ=
sin 2φ√1− sin2 2φ
sin2 2φ = ε2(1− sin2 2φ)
sin2 2φ (1 + ε2) = ε2
sin 2φ =±ε√
1 + ε2
cos2 2φ = 1− ε2
1 + ε2=
1
1 + ε2
cos 2φ =∓1√1 + ε2
thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy
cos2 φ =1
2(1 + cos 2φ), sin2 φ =
1
2(1− cos 2φ)
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 13 / 23
Problem 7.15 (cont.)
defining
ε ≡ 2h
Eb − Ea
rearranging to solve forsin 2φ
similarly, we can solve forcos 2φ
− ε = tan 2φ =sin 2φ
cos 2φ=
sin 2φ√1− sin2 2φ
sin2 2φ = ε2(1− sin2 2φ)
sin2 2φ (1 + ε2) = ε2
sin 2φ =±ε√
1 + ε2
cos2 2φ = 1− ε2
1 + ε2=
1
1 + ε2
cos 2φ =∓1√1 + ε2
thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy
cos2 φ =1
2(1 + cos 2φ), sin2 φ =
1
2(1− cos 2φ)
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 13 / 23
Problem 7.15 (cont.)
defining
ε ≡ 2h
Eb − Ea
rearranging to solve forsin 2φ
similarly, we can solve forcos 2φ
− ε = tan 2φ =sin 2φ
cos 2φ=
sin 2φ√1− sin2 2φ
sin2 2φ = ε2(1− sin2 2φ)
sin2 2φ (1 + ε2) = ε2
sin 2φ =±ε√
1 + ε2
cos2 2φ = 1− ε2
1 + ε2=
1
1 + ε2
cos 2φ =∓1√1 + ε2
thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy
cos2 φ =1
2(1 + cos 2φ), sin2 φ =
1
2(1− cos 2φ)
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 13 / 23
Problem 7.15 (cont.)
defining
ε ≡ 2h
Eb − Ea
rearranging to solve forsin 2φ
similarly, we can solve forcos 2φ
− ε = tan 2φ =sin 2φ
cos 2φ=
sin 2φ√1− sin2 2φ
sin2 2φ = ε2(1− sin2 2φ)
sin2 2φ (1 + ε2) = ε2
sin 2φ =±ε√
1 + ε2
cos2 2φ = 1− ε2
1 + ε2=
1
1 + ε2
cos 2φ =∓1√1 + ε2
thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy
cos2 φ =1
2(1 + cos 2φ), sin2 φ =
1
2(1− cos 2φ)
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 13 / 23
Problem 7.15 (cont.)
defining
ε ≡ 2h
Eb − Ea
rearranging to solve forsin 2φ
similarly, we can solve forcos 2φ
− ε = tan 2φ =sin 2φ
cos 2φ=
sin 2φ√1− sin2 2φ
sin2 2φ = ε2(1− sin2 2φ)
sin2 2φ (1 + ε2) = ε2
sin 2φ =±ε√
1 + ε2
cos2 2φ = 1− ε2
1 + ε2
=1
1 + ε2
cos 2φ =∓1√1 + ε2
thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy
cos2 φ =1
2(1 + cos 2φ), sin2 φ =
1
2(1− cos 2φ)
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 13 / 23
Problem 7.15 (cont.)
defining
ε ≡ 2h
Eb − Ea
rearranging to solve forsin 2φ
similarly, we can solve forcos 2φ
− ε = tan 2φ =sin 2φ
cos 2φ=
sin 2φ√1− sin2 2φ
sin2 2φ = ε2(1− sin2 2φ)
sin2 2φ (1 + ε2) = ε2
sin 2φ =±ε√
1 + ε2
cos2 2φ = 1− ε2
1 + ε2=
1
1 + ε2
cos 2φ =∓1√1 + ε2
thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy
cos2 φ =1
2(1 + cos 2φ), sin2 φ =
1
2(1− cos 2φ)
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 13 / 23
Problem 7.15 (cont.)
defining
ε ≡ 2h
Eb − Ea
rearranging to solve forsin 2φ
similarly, we can solve forcos 2φ
− ε = tan 2φ =sin 2φ
cos 2φ=
sin 2φ√1− sin2 2φ
sin2 2φ = ε2(1− sin2 2φ)
sin2 2φ (1 + ε2) = ε2
sin 2φ =±ε√
1 + ε2
cos2 2φ = 1− ε2
1 + ε2=
1
1 + ε2
cos 2φ =∓1√1 + ε2
thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy
cos2 φ =1
2(1 + cos 2φ), sin2 φ =
1
2(1− cos 2φ)
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 13 / 23
Problem 7.15 (cont.)
defining
ε ≡ 2h
Eb − Ea
rearranging to solve forsin 2φ
similarly, we can solve forcos 2φ
− ε = tan 2φ =sin 2φ
cos 2φ=
sin 2φ√1− sin2 2φ
sin2 2φ = ε2(1− sin2 2φ)
sin2 2φ (1 + ε2) = ε2
sin 2φ =±ε√
1 + ε2
cos2 2φ = 1− ε2
1 + ε2=
1
1 + ε2
cos 2φ =∓1√1 + ε2
thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy
cos2 φ =1
2(1 + cos 2φ), sin2 φ =
1
2(1− cos 2φ)
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 13 / 23
Problem 7.15 (cont.)
defining
ε ≡ 2h
Eb − Ea
rearranging to solve forsin 2φ
similarly, we can solve forcos 2φ
− ε = tan 2φ =sin 2φ
cos 2φ=
sin 2φ√1− sin2 2φ
sin2 2φ = ε2(1− sin2 2φ)
sin2 2φ (1 + ε2) = ε2
sin 2φ =±ε√
1 + ε2
cos2 2φ = 1− ε2
1 + ε2=
1
1 + ε2
cos 2φ =∓1√1 + ε2
thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy
cos2 φ =1
2(1 + cos 2φ),
sin2 φ =1
2(1− cos 2φ)
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 13 / 23
Problem 7.15 (cont.)
defining
ε ≡ 2h
Eb − Ea
rearranging to solve forsin 2φ
similarly, we can solve forcos 2φ
− ε = tan 2φ =sin 2φ
cos 2φ=
sin 2φ√1− sin2 2φ
sin2 2φ = ε2(1− sin2 2φ)
sin2 2φ (1 + ε2) = ε2
sin 2φ =±ε√
1 + ε2
cos2 2φ = 1− ε2
1 + ε2=
1
1 + ε2
cos 2φ =∓1√1 + ε2
thus, using the half-angle relations, we get expressions for sinφ and cosφneeded to compute the upper bound on the energy
cos2 φ =1
2(1 + cos 2φ), sin2 φ =
1
2(1− cos 2φ)
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 13 / 23
Problem 7.15 (cont.)
cos2 φ =1
2
(1∓ 1√
1 + ε2
),
sin2 φ =1
2
(1± 1√
1 + ε2
), sin 2φ =
±ε√1 + ε2
substituting into the expectation value of the Hamiltonian
〈H〉 = Ea cos2 φ+ Eb sin2 φ+ h sin 2φ
〈H〉min =1
2Ea
(1∓ 1√
1 + ε2
)+
1
2Eb
(1± 1√
1 + ε2
)± h
ε√1 + ε2
=1
2
[Ea + Eb ±
Eb − Ea + 2hε√1 + ε2
]
=1
2
Ea + Eb ±Eb − Ea + 2h 2h
Eb−Ea√1 + 4h2
(Eb−Ea)2
=
1
2
[Ea + Eb ±
√(Eb − Ea)2 + 4h2
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 14 / 23
Problem 7.15 (cont.)
cos2 φ =1
2
(1∓ 1√
1 + ε2
), sin2 φ =
1
2
(1± 1√
1 + ε2
),
sin 2φ =±ε√
1 + ε2
substituting into the expectation value of the Hamiltonian
〈H〉 = Ea cos2 φ+ Eb sin2 φ+ h sin 2φ
〈H〉min =1
2Ea
(1∓ 1√
1 + ε2
)+
1
2Eb
(1± 1√
1 + ε2
)± h
ε√1 + ε2
=1
2
[Ea + Eb ±
Eb − Ea + 2hε√1 + ε2
]
=1
2
Ea + Eb ±Eb − Ea + 2h 2h
Eb−Ea√1 + 4h2
(Eb−Ea)2
=
1
2
[Ea + Eb ±
√(Eb − Ea)2 + 4h2
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 14 / 23
Problem 7.15 (cont.)
cos2 φ =1
2
(1∓ 1√
1 + ε2
), sin2 φ =
1
2
(1± 1√
1 + ε2
), sin 2φ =
±ε√1 + ε2
substituting into the expectation value of the Hamiltonian
〈H〉 = Ea cos2 φ+ Eb sin2 φ+ h sin 2φ
〈H〉min =1
2Ea
(1∓ 1√
1 + ε2
)+
1
2Eb
(1± 1√
1 + ε2
)± h
ε√1 + ε2
=1
2
[Ea + Eb ±
Eb − Ea + 2hε√1 + ε2
]
=1
2
Ea + Eb ±Eb − Ea + 2h 2h
Eb−Ea√1 + 4h2
(Eb−Ea)2
=
1
2
[Ea + Eb ±
√(Eb − Ea)2 + 4h2
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 14 / 23
Problem 7.15 (cont.)
cos2 φ =1
2
(1∓ 1√
1 + ε2
), sin2 φ =
1
2
(1± 1√
1 + ε2
), sin 2φ =
±ε√1 + ε2
substituting into the expectation value of the Hamiltonian
〈H〉 = Ea cos2 φ+ Eb sin2 φ+ 2h sinφ cosφ
〈H〉min =1
2Ea
(1∓ 1√
1 + ε2
)+
1
2Eb
(1± 1√
1 + ε2
)± h
ε√1 + ε2
=1
2
[Ea + Eb ±
Eb − Ea + 2hε√1 + ε2
]
=1
2
Ea + Eb ±Eb − Ea + 2h 2h
Eb−Ea√1 + 4h2
(Eb−Ea)2
=
1
2
[Ea + Eb ±
√(Eb − Ea)2 + 4h2
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 14 / 23
Problem 7.15 (cont.)
cos2 φ =1
2
(1∓ 1√
1 + ε2
), sin2 φ =
1
2
(1± 1√
1 + ε2
), sin 2φ =
±ε√1 + ε2
substituting into the expectation value of the Hamiltonian
〈H〉 = Ea cos2 φ+ Eb sin2 φ+ h sin 2φ
〈H〉min =1
2Ea
(1∓ 1√
1 + ε2
)+
1
2Eb
(1± 1√
1 + ε2
)± h
ε√1 + ε2
=1
2
[Ea + Eb ±
Eb − Ea + 2hε√1 + ε2
]
=1
2
Ea + Eb ±Eb − Ea + 2h 2h
Eb−Ea√1 + 4h2
(Eb−Ea)2
=
1
2
[Ea + Eb ±
√(Eb − Ea)2 + 4h2
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 14 / 23
Problem 7.15 (cont.)
cos2 φ =1
2
(1∓ 1√
1 + ε2
), sin2 φ =
1
2
(1± 1√
1 + ε2
), sin 2φ =
±ε√1 + ε2
substituting into the expectation value of the Hamiltonian
〈H〉 = Ea cos2 φ+ Eb sin2 φ+ h sin 2φ
〈H〉min =1
2Ea
(1∓ 1√
1 + ε2
)+
1
2Eb
(1± 1√
1 + ε2
)± h
ε√1 + ε2
=1
2
[Ea + Eb ±
Eb − Ea + 2hε√1 + ε2
]
=1
2
Ea + Eb ±Eb − Ea + 2h 2h
Eb−Ea√1 + 4h2
(Eb−Ea)2
=
1
2
[Ea + Eb ±
√(Eb − Ea)2 + 4h2
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 14 / 23
Problem 7.15 (cont.)
cos2 φ =1
2
(1∓ 1√
1 + ε2
), sin2 φ =
1
2
(1± 1√
1 + ε2
), sin 2φ =
±ε√1 + ε2
substituting into the expectation value of the Hamiltonian
〈H〉 = Ea cos2 φ+ Eb sin2 φ+ h sin 2φ
〈H〉min =1
2Ea
(1∓ 1√
1 + ε2
)+
1
2Eb
(1± 1√
1 + ε2
)± h
ε√1 + ε2
=1
2
[Ea + Eb ±
Eb − Ea + 2hε√1 + ε2
]
=1
2
Ea + Eb ±Eb − Ea + 2h 2h
Eb−Ea√1 + 4h2
(Eb−Ea)2
=
1
2
[Ea + Eb ±
√(Eb − Ea)2 + 4h2
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 14 / 23
Problem 7.15 (cont.)
cos2 φ =1
2
(1∓ 1√
1 + ε2
), sin2 φ =
1
2
(1± 1√
1 + ε2
), sin 2φ =
±ε√1 + ε2
substituting into the expectation value of the Hamiltonian
〈H〉 = Ea cos2 φ+ Eb sin2 φ+ h sin 2φ
〈H〉min =1
2Ea
(1∓ 1√
1 + ε2
)+
1
2Eb
(1± 1√
1 + ε2
)± h
ε√1 + ε2
=1
2
[Ea + Eb ±
Eb − Ea + 2hε√1 + ε2
]
=1
2
Ea + Eb ±Eb − Ea + 2h 2h
Eb−Ea√1 + 4h2
(Eb−Ea)2
=1
2
[Ea + Eb ±
√(Eb − Ea)2 + 4h2
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 14 / 23
Problem 7.15 (cont.)
cos2 φ =1
2
(1∓ 1√
1 + ε2
), sin2 φ =
1
2
(1± 1√
1 + ε2
), sin 2φ =
±ε√1 + ε2
substituting into the expectation value of the Hamiltonian
〈H〉 = Ea cos2 φ+ Eb sin2 φ+ h sin 2φ
〈H〉min =1
2Ea
(1∓ 1√
1 + ε2
)+
1
2Eb
(1± 1√
1 + ε2
)± h
ε√1 + ε2
=1
2
[Ea + Eb ±
Eb − Ea + 2hε√1 + ε2
]
=1
2
Ea + Eb ±Eb − Ea + 2h 2h
Eb−Ea√1 + 4h2
(Eb−Ea)2
=
1
2
[Ea + Eb ±
√(Eb − Ea)2 + 4h2
]C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 14 / 23
Problem 7.15 (cont.)
Thus the upper bound on the ground state energy is
〈H〉min =1
2
[Ea + Eb −
√(Eb − Ea)2 + 4h2
]d. The variational result is equivalent to the exact solution in this casebecause the unperturbed system only has two eigenstates and because ofthe particular choice of trial function.
The perturbation result, on the other hand, is simply the approximatevalue of the exact solution when h is small
E± =1
2
[(Ea + Eb)± (Eb − Ea)
√1 +
4h2
(Eb − Ea)2
]
≈ 1
2
[(Ea + Eb)± (Eb − Ea)
(1 +
2h2
(Eb − Ea)2
)]
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 15 / 23
Problem 7.15 (cont.)
Thus the upper bound on the ground state energy is
〈H〉min =1
2
[Ea + Eb −
√(Eb − Ea)2 + 4h2
]
d. The variational result is equivalent to the exact solution in this casebecause the unperturbed system only has two eigenstates and because ofthe particular choice of trial function.
The perturbation result, on the other hand, is simply the approximatevalue of the exact solution when h is small
E± =1
2
[(Ea + Eb)± (Eb − Ea)
√1 +
4h2
(Eb − Ea)2
]
≈ 1
2
[(Ea + Eb)± (Eb − Ea)
(1 +
2h2
(Eb − Ea)2
)]
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 15 / 23
Problem 7.15 (cont.)
Thus the upper bound on the ground state energy is
〈H〉min =1
2
[Ea + Eb −
√(Eb − Ea)2 + 4h2
]d. The variational result is equivalent to the exact solution in this casebecause the unperturbed system only has two eigenstates and because ofthe particular choice of trial function.
The perturbation result, on the other hand, is simply the approximatevalue of the exact solution when h is small
E± =1
2
[(Ea + Eb)± (Eb − Ea)
√1 +
4h2
(Eb − Ea)2
]
≈ 1
2
[(Ea + Eb)± (Eb − Ea)
(1 +
2h2
(Eb − Ea)2
)]
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 15 / 23
Problem 7.15 (cont.)
Thus the upper bound on the ground state energy is
〈H〉min =1
2
[Ea + Eb −
√(Eb − Ea)2 + 4h2
]d. The variational result is equivalent to the exact solution in this casebecause the unperturbed system only has two eigenstates and because ofthe particular choice of trial function.
The perturbation result, on the other hand, is simply the approximatevalue of the exact solution when h is small
E± =1
2
[(Ea + Eb)± (Eb − Ea)
√1 +
4h2
(Eb − Ea)2
]
≈ 1
2
[(Ea + Eb)± (Eb − Ea)
(1 +
2h2
(Eb − Ea)2
)]
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 15 / 23
Problem 7.15 (cont.)
Thus the upper bound on the ground state energy is
〈H〉min =1
2
[Ea + Eb −
√(Eb − Ea)2 + 4h2
]d. The variational result is equivalent to the exact solution in this casebecause the unperturbed system only has two eigenstates and because ofthe particular choice of trial function.
The perturbation result, on the other hand, is simply the approximatevalue of the exact solution when h is small
E± =1
2
[(Ea + Eb)± (Eb − Ea)
√1 +
4h2
(Eb − Ea)2
]
≈ 1
2
[(Ea + Eb)± (Eb − Ea)
(1 +
2h2
(Eb − Ea)2
)]
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 15 / 23
Problem 7.15 (cont.)
Thus the upper bound on the ground state energy is
〈H〉min =1
2
[Ea + Eb −
√(Eb − Ea)2 + 4h2
]d. The variational result is equivalent to the exact solution in this casebecause the unperturbed system only has two eigenstates and because ofthe particular choice of trial function.
The perturbation result, on the other hand, is simply the approximatevalue of the exact solution when h is small
E± =1
2
[(Ea + Eb)± (Eb − Ea)
√1 +
4h2
(Eb − Ea)2
]
≈ 1
2
[(Ea + Eb)± (Eb − Ea)
(1 +
2h2
(Eb − Ea)2
)]
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 15 / 23
Problem 7.15 (cont.)
E± ≈1
2
[(Ea + Eb)± (Eb − Ea)± 2h2
Eb − Ea
]
E− ≈1
2
[2Ea −
2h2
Eb − Ea
]≈ Ea −
h2
Eb − Ea
E+ ≈1
2
[2Eb +
2h2
Eb − Ea
]≈ Eb +
h2
Eb − Ea
These are simply the second order perturbation theory results.
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 16 / 23
Problem 7.15 (cont.)
E± ≈1
2
[(Ea + Eb)± (Eb − Ea)± 2h2
Eb − Ea
]
E− ≈1
2
[2Ea −
2h2
Eb − Ea
]
≈ Ea −h2
Eb − Ea
E+ ≈1
2
[2Eb +
2h2
Eb − Ea
]
≈ Eb +h2
Eb − Ea
These are simply the second order perturbation theory results.
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 16 / 23
Problem 7.15 (cont.)
E± ≈1
2
[(Ea + Eb)± (Eb − Ea)± 2h2
Eb − Ea
]
E− ≈1
2
[2Ea −
2h2
Eb − Ea
]≈ Ea −
h2
Eb − Ea
E+ ≈1
2
[2Eb +
2h2
Eb − Ea
]≈ Eb +
h2
Eb − Ea
These are simply the second order perturbation theory results.
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 16 / 23
Problem 7.15 (cont.)
E± ≈1
2
[(Ea + Eb)± (Eb − Ea)± 2h2
Eb − Ea
]
E− ≈1
2
[2Ea −
2h2
Eb − Ea
]≈ Ea −
h2
Eb − Ea
E+ ≈1
2
[2Eb +
2h2
Eb − Ea
]≈ Eb +
h2
Eb − Ea
These are simply the second order perturbation theory results.
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 16 / 23
Problem 7.16
Consider an electron at rest in a uniform magnetic field ~B = Bz k̂ , forwhich the Hamiltonian is
H0 =eBz
mSz
The eigenspinors, and corresponding energies are:{χ+, E+ = −(γBz~)/2
χ−, E− = +(γBz~)/2
Now we turn on a perturbation, in the form of a uniform field in the xdirection
H ′ =eBx
mSx
(a) Find the matrix elements of H ′.
(b) Using the result of problem 7.15(b) find the new ground state energy,in second order perturbation theory.
(c) Using the result of problem 7.15(c), find the variational principlebound on the ground state energy.
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 17 / 23
Problem 7.16 (cont.)
For an electron, γ = −e/m andthus the energies of the two eigen-spinors are
choosing the higher energy to bethe Eb as was in problem 7.15, wehave
E± = ±eBz~2m
Ea ≡ E−, and Eb ≡ E+
χb = χ+ =
(01
)χa = χ− =
(10
)(a) The matrix elements in the original basis set become
〈χa|H ′|χa〉 =eBx
m
(0 1)~
2
(0 11 0
)(01
)=
eBx~2m
(0 1)( 1
0
)= 0
〈χb|H ′|χb〉 =eBx
m
(1 0)~
2
(0 11 0
)(10
)=
eBx~2m
(1 0)( 0
1
)= 0
〈χa|H ′|χb〉 =eBx
m
(0 1)~
2
(0 11 0
)(10
)=
eBx~2m
(0 1)( 0
1
)=
eBx~2m
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 18 / 23
Problem 7.16 (cont.)
For an electron, γ = −e/m andthus the energies of the two eigen-spinors are
choosing the higher energy to bethe Eb as was in problem 7.15, wehave
E± = ±eBz~2m
Ea ≡ E−, and Eb ≡ E+
χb = χ+ =
(01
)χa = χ− =
(10
)(a) The matrix elements in the original basis set become
〈χa|H ′|χa〉 =eBx
m
(0 1)~
2
(0 11 0
)(01
)=
eBx~2m
(0 1)( 1
0
)= 0
〈χb|H ′|χb〉 =eBx
m
(1 0)~
2
(0 11 0
)(10
)=
eBx~2m
(1 0)( 0
1
)= 0
〈χa|H ′|χb〉 =eBx
m
(0 1)~
2
(0 11 0
)(10
)=
eBx~2m
(0 1)( 0
1
)=
eBx~2m
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 18 / 23
Problem 7.16 (cont.)
For an electron, γ = −e/m andthus the energies of the two eigen-spinors are
choosing the higher energy to bethe Eb as was in problem 7.15, wehave
E± = ±eBz~2m
Ea ≡ E−, and Eb ≡ E+
χb = χ+ =
(01
)χa = χ− =
(10
)(a) The matrix elements in the original basis set become
〈χa|H ′|χa〉 =eBx
m
(0 1)~
2
(0 11 0
)(01
)=
eBx~2m
(0 1)( 1
0
)= 0
〈χb|H ′|χb〉 =eBx
m
(1 0)~
2
(0 11 0
)(10
)=
eBx~2m
(1 0)( 0
1
)= 0
〈χa|H ′|χb〉 =eBx
m
(0 1)~
2
(0 11 0
)(10
)=
eBx~2m
(0 1)( 0
1
)=
eBx~2m
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 18 / 23
Problem 7.16 (cont.)
For an electron, γ = −e/m andthus the energies of the two eigen-spinors are
choosing the higher energy to bethe Eb as was in problem 7.15, wehave
E± = ±eBz~2m
Ea ≡ E−, and Eb ≡ E+
χb = χ+ =
(01
)χa = χ− =
(10
)(a) The matrix elements in the original basis set become
〈χa|H ′|χa〉 =eBx
m
(0 1)~
2
(0 11 0
)(01
)=
eBx~2m
(0 1)( 1
0
)= 0
〈χb|H ′|χb〉 =eBx
m
(1 0)~
2
(0 11 0
)(10
)=
eBx~2m
(1 0)( 0
1
)= 0
〈χa|H ′|χb〉 =eBx
m
(0 1)~
2
(0 11 0
)(10
)=
eBx~2m
(0 1)( 0
1
)=
eBx~2m
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 18 / 23
Problem 7.16 (cont.)
For an electron, γ = −e/m andthus the energies of the two eigen-spinors are
choosing the higher energy to bethe Eb as was in problem 7.15, wehave
E± = ±eBz~2m
Ea ≡ E−, and Eb ≡ E+
χb = χ+ =
(01
)
χa = χ− =
(10
)(a) The matrix elements in the original basis set become
〈χa|H ′|χa〉 =eBx
m
(0 1)~
2
(0 11 0
)(01
)=
eBx~2m
(0 1)( 1
0
)= 0
〈χb|H ′|χb〉 =eBx
m
(1 0)~
2
(0 11 0
)(10
)=
eBx~2m
(1 0)( 0
1
)= 0
〈χa|H ′|χb〉 =eBx
m
(0 1)~
2
(0 11 0
)(10
)=
eBx~2m
(0 1)( 0
1
)=
eBx~2m
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 18 / 23
Problem 7.16 (cont.)
For an electron, γ = −e/m andthus the energies of the two eigen-spinors are
choosing the higher energy to bethe Eb as was in problem 7.15, wehave
E± = ±eBz~2m
Ea ≡ E−, and Eb ≡ E+
χb = χ+ =
(01
)χa = χ− =
(10
)
(a) The matrix elements in the original basis set become
〈χa|H ′|χa〉 =eBx
m
(0 1)~
2
(0 11 0
)(01
)=
eBx~2m
(0 1)( 1
0
)= 0
〈χb|H ′|χb〉 =eBx
m
(1 0)~
2
(0 11 0
)(10
)=
eBx~2m
(1 0)( 0
1
)= 0
〈χa|H ′|χb〉 =eBx
m
(0 1)~
2
(0 11 0
)(10
)=
eBx~2m
(0 1)( 0
1
)=
eBx~2m
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 18 / 23
Problem 7.16 (cont.)
For an electron, γ = −e/m andthus the energies of the two eigen-spinors are
choosing the higher energy to bethe Eb as was in problem 7.15, wehave
E± = ±eBz~2m
Ea ≡ E−, and Eb ≡ E+
χb = χ+ =
(01
)χa = χ− =
(10
)(a) The matrix elements in the original basis set become
〈χa|H ′|χa〉 =eBx
m
(0 1)~
2
(0 11 0
)(01
)=
eBx~2m
(0 1)( 1
0
)= 0
〈χb|H ′|χb〉 =eBx
m
(1 0)~
2
(0 11 0
)(10
)=
eBx~2m
(1 0)( 0
1
)= 0
〈χa|H ′|χb〉 =eBx
m
(0 1)~
2
(0 11 0
)(10
)=
eBx~2m
(0 1)( 0
1
)=
eBx~2m
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 18 / 23
Problem 7.16 (cont.)
For an electron, γ = −e/m andthus the energies of the two eigen-spinors are
choosing the higher energy to bethe Eb as was in problem 7.15, wehave
E± = ±eBz~2m
Ea ≡ E−, and Eb ≡ E+
χb = χ+ =
(01
)χa = χ− =
(10
)(a) The matrix elements in the original basis set become
〈χa|H ′|χa〉 =eBx
m
(0 1)~
2
(0 11 0
)(01
)
=eBx~2m
(0 1)( 1
0
)= 0
〈χb|H ′|χb〉 =eBx
m
(1 0)~
2
(0 11 0
)(10
)=
eBx~2m
(1 0)( 0
1
)= 0
〈χa|H ′|χb〉 =eBx
m
(0 1)~
2
(0 11 0
)(10
)=
eBx~2m
(0 1)( 0
1
)=
eBx~2m
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 18 / 23
Problem 7.16 (cont.)
For an electron, γ = −e/m andthus the energies of the two eigen-spinors are
choosing the higher energy to bethe Eb as was in problem 7.15, wehave
E± = ±eBz~2m
Ea ≡ E−, and Eb ≡ E+
χb = χ+ =
(01
)χa = χ− =
(10
)(a) The matrix elements in the original basis set become
〈χa|H ′|χa〉 =eBx
m
(0 1)~
2
(0 11 0
)(01
)=
eBx~2m
(0 1)( 1
0
)= 0
〈χb|H ′|χb〉 =eBx
m
(1 0)~
2
(0 11 0
)(10
)=
eBx~2m
(1 0)( 0
1
)= 0
〈χa|H ′|χb〉 =eBx
m
(0 1)~
2
(0 11 0
)(10
)=
eBx~2m
(0 1)( 0
1
)=
eBx~2m
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 18 / 23
Problem 7.16 (cont.)
For an electron, γ = −e/m andthus the energies of the two eigen-spinors are
choosing the higher energy to bethe Eb as was in problem 7.15, wehave
E± = ±eBz~2m
Ea ≡ E−, and Eb ≡ E+
χb = χ+ =
(01
)χa = χ− =
(10
)(a) The matrix elements in the original basis set become
〈χa|H ′|χa〉 =eBx
m
(0 1)~
2
(0 11 0
)(01
)=
eBx~2m
(0 1)( 1
0
)= 0
〈χb|H ′|χb〉 =eBx
m
(1 0)~
2
(0 11 0
)(10
)
=eBx~2m
(1 0)( 0
1
)= 0
〈χa|H ′|χb〉 =eBx
m
(0 1)~
2
(0 11 0
)(10
)=
eBx~2m
(0 1)( 0
1
)=
eBx~2m
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 18 / 23
Problem 7.16 (cont.)
For an electron, γ = −e/m andthus the energies of the two eigen-spinors are
choosing the higher energy to bethe Eb as was in problem 7.15, wehave
E± = ±eBz~2m
Ea ≡ E−, and Eb ≡ E+
χb = χ+ =
(01
)χa = χ− =
(10
)(a) The matrix elements in the original basis set become
〈χa|H ′|χa〉 =eBx
m
(0 1)~
2
(0 11 0
)(01
)=
eBx~2m
(0 1)( 1
0
)= 0
〈χb|H ′|χb〉 =eBx
m
(1 0)~
2
(0 11 0
)(10
)=
eBx~2m
(1 0)( 0
1
)= 0
〈χa|H ′|χb〉 =eBx
m
(0 1)~
2
(0 11 0
)(10
)=
eBx~2m
(0 1)( 0
1
)=
eBx~2m
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 18 / 23
Problem 7.16 (cont.)
For an electron, γ = −e/m andthus the energies of the two eigen-spinors are
choosing the higher energy to bethe Eb as was in problem 7.15, wehave
E± = ±eBz~2m
Ea ≡ E−, and Eb ≡ E+
χb = χ+ =
(01
)χa = χ− =
(10
)(a) The matrix elements in the original basis set become
〈χa|H ′|χa〉 =eBx
m
(0 1)~
2
(0 11 0
)(01
)=
eBx~2m
(0 1)( 1
0
)= 0
〈χb|H ′|χb〉 =eBx
m
(1 0)~
2
(0 11 0
)(10
)=
eBx~2m
(1 0)( 0
1
)= 0
〈χa|H ′|χb〉 =eBx
m
(0 1)~
2
(0 11 0
)(10
)
=eBx~2m
(0 1)( 0
1
)=
eBx~2m
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 18 / 23
Problem 7.16 (cont.)
For an electron, γ = −e/m andthus the energies of the two eigen-spinors are
choosing the higher energy to bethe Eb as was in problem 7.15, wehave
E± = ±eBz~2m
Ea ≡ E−, and Eb ≡ E+
χb = χ+ =
(01
)χa = χ− =
(10
)(a) The matrix elements in the original basis set become
〈χa|H ′|χa〉 =eBx
m
(0 1)~
2
(0 11 0
)(01
)=
eBx~2m
(0 1)( 1
0
)= 0
〈χb|H ′|χb〉 =eBx
m
(1 0)~
2
(0 11 0
)(10
)=
eBx~2m
(1 0)( 0
1
)= 0
〈χa|H ′|χb〉 =eBx
m
(0 1)~
2
(0 11 0
)(10
)=
eBx~2m
(0 1)( 0
1
)
=eBx~2m
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 18 / 23
Problem 7.16 (cont.)
For an electron, γ = −e/m andthus the energies of the two eigen-spinors are
choosing the higher energy to bethe Eb as was in problem 7.15, wehave
E± = ±eBz~2m
Ea ≡ E−, and Eb ≡ E+
χb = χ+ =
(01
)χa = χ− =
(10
)(a) The matrix elements in the original basis set become
〈χa|H ′|χa〉 =eBx
m
(0 1)~
2
(0 11 0
)(01
)=
eBx~2m
(0 1)( 1
0
)= 0
〈χb|H ′|χb〉 =eBx
m
(1 0)~
2
(0 11 0
)(10
)=
eBx~2m
(1 0)( 0
1
)= 0
〈χa|H ′|χb〉 =eBx
m
(0 1)~
2
(0 11 0
)(10
)=
eBx~2m
(0 1)( 0
1
)=
eBx~2m
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 18 / 23
Problem 7.16 (cont.)
(b) Applying the results of second order perturbation theory (problem7.15b) with h = eBx~/2m
Egs ≈ Ea −h2
(Eb − Ea)= −eBz~
2m− (eBx~/2m)2
(eBz~/m)= − e~
2m
(Bz +
B2x
2Bz
)(c) Taking the variational method results (problem 7.15c), we have
Egs =1
2
[Ea + Eb −
√(Eb − Ea)2 + 4h2
]= −1
2
√(eBz~m
)2
+ 4
(eBx~2m
)2
= − e~2m
√B2z + B2
x
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 19 / 23
Problem 7.16 (cont.)
(b) Applying the results of second order perturbation theory (problem7.15b) with h = eBx~/2m
Egs ≈ Ea −h2
(Eb − Ea)
= −eBz~2m
− (eBx~/2m)2
(eBz~/m)= − e~
2m
(Bz +
B2x
2Bz
)(c) Taking the variational method results (problem 7.15c), we have
Egs =1
2
[Ea + Eb −
√(Eb − Ea)2 + 4h2
]= −1
2
√(eBz~m
)2
+ 4
(eBx~2m
)2
= − e~2m
√B2z + B2
x
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 19 / 23
Problem 7.16 (cont.)
(b) Applying the results of second order perturbation theory (problem7.15b) with h = eBx~/2m
Egs ≈ Ea −h2
(Eb − Ea)= −eBz~
2m− (eBx~/2m)2
(eBz~/m)
= − e~2m
(Bz +
B2x
2Bz
)(c) Taking the variational method results (problem 7.15c), we have
Egs =1
2
[Ea + Eb −
√(Eb − Ea)2 + 4h2
]= −1
2
√(eBz~m
)2
+ 4
(eBx~2m
)2
= − e~2m
√B2z + B2
x
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 19 / 23
Problem 7.16 (cont.)
(b) Applying the results of second order perturbation theory (problem7.15b) with h = eBx~/2m
Egs ≈ Ea −h2
(Eb − Ea)= −eBz~
2m− (eBx~/2m)2
(eBz~/m)= − e~
2m
(Bz +
B2x
2Bz
)
(c) Taking the variational method results (problem 7.15c), we have
Egs =1
2
[Ea + Eb −
√(Eb − Ea)2 + 4h2
]= −1
2
√(eBz~m
)2
+ 4
(eBx~2m
)2
= − e~2m
√B2z + B2
x
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 19 / 23
Problem 7.16 (cont.)
(b) Applying the results of second order perturbation theory (problem7.15b) with h = eBx~/2m
Egs ≈ Ea −h2
(Eb − Ea)= −eBz~
2m− (eBx~/2m)2
(eBz~/m)= − e~
2m
(Bz +
B2x
2Bz
)(c) Taking the variational method results (problem 7.15c), we have
Egs =1
2
[Ea + Eb −
√(Eb − Ea)2 + 4h2
]= −1
2
√(eBz~m
)2
+ 4
(eBx~2m
)2
= − e~2m
√B2z + B2
x
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 19 / 23
Problem 7.16 (cont.)
(b) Applying the results of second order perturbation theory (problem7.15b) with h = eBx~/2m
Egs ≈ Ea −h2
(Eb − Ea)= −eBz~
2m− (eBx~/2m)2
(eBz~/m)= − e~
2m
(Bz +
B2x
2Bz
)(c) Taking the variational method results (problem 7.15c), we have
Egs =1
2
[Ea + Eb −
√(Eb − Ea)2 + 4h2
]
= −1
2
√(eBz~m
)2
+ 4
(eBx~2m
)2
= − e~2m
√B2z + B2
x
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 19 / 23
Problem 7.16 (cont.)
(b) Applying the results of second order perturbation theory (problem7.15b) with h = eBx~/2m
Egs ≈ Ea −h2
(Eb − Ea)= −eBz~
2m− (eBx~/2m)2
(eBz~/m)= − e~
2m
(Bz +
B2x
2Bz
)(c) Taking the variational method results (problem 7.15c), we have
Egs =1
2
[Ea + Eb −
√(Eb − Ea)2 + 4h2
]= −1
2
√(eBz~m
)2
+ 4
(eBx~2m
)2
= − e~2m
√B2z + B2
x
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 19 / 23
Problem 7.16 (cont.)
(b) Applying the results of second order perturbation theory (problem7.15b) with h = eBx~/2m
Egs ≈ Ea −h2
(Eb − Ea)= −eBz~
2m− (eBx~/2m)2
(eBz~/m)= − e~
2m
(Bz +
B2x
2Bz
)(c) Taking the variational method results (problem 7.15c), we have
Egs =1
2
[Ea + Eb −
√(Eb − Ea)2 + 4h2
]= −1
2
√(eBz~m
)2
+ 4
(eBx~2m
)2
= − e~2m
√B2z + B2
x
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 19 / 23
Problem 7.19
The fundamental problem in harnessing nuclear fusion is getting the twoparticles (say, two deutrons) close enough to each other for the attractive(but short range) nuclear force to overcome the Coulomb repulsion. The“bulldozer” method is to heat the particles up to fantastic temperatures,and allow the random collision to bring them together. A more exoticproposal is muon catalysis, in which we construct a “hydrogen moleculeion,” only with deutrons in place of protons and a muon in place of theelectron. Predict the equilibrium separation distance between the deutronsin such a structure, and explain why muons are superior to electrons forthis purpose.
The result we seek is the same as we calculated before but with theelectron mass replaced by the reduced muon mass (becuase the muonmass is a significant fraction of the proton mass)
µµ =mµmd
mµ + md=
mµ2mp
mµ + 2mp
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 20 / 23
Problem 7.19
The fundamental problem in harnessing nuclear fusion is getting the twoparticles (say, two deutrons) close enough to each other for the attractive(but short range) nuclear force to overcome the Coulomb repulsion. The“bulldozer” method is to heat the particles up to fantastic temperatures,and allow the random collision to bring them together. A more exoticproposal is muon catalysis, in which we construct a “hydrogen moleculeion,” only with deutrons in place of protons and a muon in place of theelectron. Predict the equilibrium separation distance between the deutronsin such a structure, and explain why muons are superior to electrons forthis purpose.
The result we seek is the same as we calculated before but with theelectron mass replaced by the reduced muon mass (becuase the muonmass is a significant fraction of the proton mass)
µµ =mµmd
mµ + md=
mµ2mp
mµ + 2mp
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 20 / 23
Problem 7.19
The fundamental problem in harnessing nuclear fusion is getting the twoparticles (say, two deutrons) close enough to each other for the attractive(but short range) nuclear force to overcome the Coulomb repulsion. The“bulldozer” method is to heat the particles up to fantastic temperatures,and allow the random collision to bring them together. A more exoticproposal is muon catalysis, in which we construct a “hydrogen moleculeion,” only with deutrons in place of protons and a muon in place of theelectron. Predict the equilibrium separation distance between the deutronsin such a structure, and explain why muons are superior to electrons forthis purpose.
The result we seek is the same as we calculated before but with theelectron mass replaced by the reduced muon mass (becuase the muonmass is a significant fraction of the proton mass)
µµ =mµmd
mµ + md
=mµ2mp
mµ + 2mp
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 20 / 23
Problem 7.19
The fundamental problem in harnessing nuclear fusion is getting the twoparticles (say, two deutrons) close enough to each other for the attractive(but short range) nuclear force to overcome the Coulomb repulsion. The“bulldozer” method is to heat the particles up to fantastic temperatures,and allow the random collision to bring them together. A more exoticproposal is muon catalysis, in which we construct a “hydrogen moleculeion,” only with deutrons in place of protons and a muon in place of theelectron. Predict the equilibrium separation distance between the deutronsin such a structure, and explain why muons are superior to electrons forthis purpose.
The result we seek is the same as we calculated before but with theelectron mass replaced by the reduced muon mass (becuase the muonmass is a significant fraction of the proton mass)
µµ =mµmd
mµ + md=
mµ2mp
mµ + 2mp
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 20 / 23
Problem 7.19
The mass of the muon is
mµ = 207me
and the proton mass is
mp ≈ 1836me
µµ =2mµmp
mµ + 2mp
µµ ≈2 · 207 · 1836m2
e
(207 + 2 · 1836)me
≈ 7.60104× 105me
3.879× 103≈ 196me
so by replacing the electron with a muon and the protons with deutrons,the new “Bohr radius” is reduced by a factor of almost 200 as is theequilibrium distance of the molecule
Req =2.49a
196= 6.73× 10−13m
the muon brings the deutrons much closer and thus rasies the probabilityof fusion
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 21 / 23
Problem 7.19
The mass of the muon is
mµ = 207me
and the proton mass is
mp ≈ 1836me
µµ =2mµmp
mµ + 2mp
µµ ≈2 · 207 · 1836m2
e
(207 + 2 · 1836)me
≈ 7.60104× 105me
3.879× 103≈ 196me
so by replacing the electron with a muon and the protons with deutrons,the new “Bohr radius” is reduced by a factor of almost 200 as is theequilibrium distance of the molecule
Req =2.49a
196= 6.73× 10−13m
the muon brings the deutrons much closer and thus rasies the probabilityof fusion
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 21 / 23
Problem 7.19
The mass of the muon is
mµ = 207me
and the proton mass is
mp ≈ 1836me
µµ =2mµmp
mµ + 2mp
µµ ≈2 · 207 · 1836m2
e
(207 + 2 · 1836)me
≈ 7.60104× 105me
3.879× 103≈ 196me
so by replacing the electron with a muon and the protons with deutrons,the new “Bohr radius” is reduced by a factor of almost 200 as is theequilibrium distance of the molecule
Req =2.49a
196= 6.73× 10−13m
the muon brings the deutrons much closer and thus rasies the probabilityof fusion
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 21 / 23
Problem 7.19
The mass of the muon is
mµ = 207me
and the proton mass is
mp ≈ 1836me
µµ =2mµmp
mµ + 2mp
µµ ≈2 · 207 · 1836m2
e
(207 + 2 · 1836)me
≈ 7.60104× 105me
3.879× 103
≈ 196me
so by replacing the electron with a muon and the protons with deutrons,the new “Bohr radius” is reduced by a factor of almost 200 as is theequilibrium distance of the molecule
Req =2.49a
196= 6.73× 10−13m
the muon brings the deutrons much closer and thus rasies the probabilityof fusion
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 21 / 23
Problem 7.19
The mass of the muon is
mµ = 207me
and the proton mass is
mp ≈ 1836me
µµ =2mµmp
mµ + 2mp
µµ ≈2 · 207 · 1836m2
e
(207 + 2 · 1836)me
≈ 7.60104× 105me
3.879× 103≈ 196me
so by replacing the electron with a muon and the protons with deutrons,the new “Bohr radius” is reduced by a factor of almost 200 as is theequilibrium distance of the molecule
Req =2.49a
196= 6.73× 10−13m
the muon brings the deutrons much closer and thus rasies the probabilityof fusion
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 21 / 23
Problem 7.19
The mass of the muon is
mµ = 207me
and the proton mass is
mp ≈ 1836me
µµ =2mµmp
mµ + 2mp
µµ ≈2 · 207 · 1836m2
e
(207 + 2 · 1836)me
≈ 7.60104× 105me
3.879× 103≈ 196me
so by replacing the electron with a muon and the protons with deutrons,the new “Bohr radius” is reduced by a factor of almost 200 as is theequilibrium distance of the molecule
Req =2.49a
196= 6.73× 10−13m
the muon brings the deutrons much closer and thus rasies the probabilityof fusion
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 21 / 23
Problem 7.19
The mass of the muon is
mµ = 207me
and the proton mass is
mp ≈ 1836me
µµ =2mµmp
mµ + 2mp
µµ ≈2 · 207 · 1836m2
e
(207 + 2 · 1836)me
≈ 7.60104× 105me
3.879× 103≈ 196me
so by replacing the electron with a muon and the protons with deutrons,the new “Bohr radius” is reduced by a factor of almost 200 as is theequilibrium distance of the molecule
Req =2.49a
196
= 6.73× 10−13m
the muon brings the deutrons much closer and thus rasies the probabilityof fusion
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 21 / 23
Problem 7.19
The mass of the muon is
mµ = 207me
and the proton mass is
mp ≈ 1836me
µµ =2mµmp
mµ + 2mp
µµ ≈2 · 207 · 1836m2
e
(207 + 2 · 1836)me
≈ 7.60104× 105me
3.879× 103≈ 196me
so by replacing the electron with a muon and the protons with deutrons,the new “Bohr radius” is reduced by a factor of almost 200 as is theequilibrium distance of the molecule
Req =2.49a
196= 6.73× 10−13m
the muon brings the deutrons much closer and thus rasies the probabilityof fusion
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 21 / 23
Problem 7.19
The mass of the muon is
mµ = 207me
and the proton mass is
mp ≈ 1836me
µµ =2mµmp
mµ + 2mp
µµ ≈2 · 207 · 1836m2
e
(207 + 2 · 1836)me
≈ 7.60104× 105me
3.879× 103≈ 196me
so by replacing the electron with a muon and the protons with deutrons,the new “Bohr radius” is reduced by a factor of almost 200 as is theequilibrium distance of the molecule
Req =2.49a
196= 6.73× 10−13m
the muon brings the deutrons much closer and thus rasies the probabilityof fusion
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 21 / 23
Problem 6.39
In a crystal, the electric field of neighboring ions perturbs the energy levelsof an atom. As a crude model, imagine that a hydrogen atom issurrounded by three pairs of point charges, as shown in Figure 6.15.
(a) Assuming that r � d1, r � d2, and r � d3, show that
H ′ = V0 + 3(β1x2 + β2y
2 + β3z2)− (β1 + β2 + β3)r2
where
βi ≡ −e
4πε0
qid2i
, and V0 = 2(β1d21 + β2d2
2 + β3d23)
(b) Find the lowest order correction to the ground state energy
(c) Calculate the first-order corrections to the energy of the first excitedstates (n = 2). Into how many levels does the four-fold degeneratesystem split, in the cases of (i) cubic, (ii) tetragonal, and (iii)orthorhombic symmetries?
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 22 / 23
Problem 6.39 (cont.)
y
z
x
d1
d2
d3
d1d3
d2
q2q2
q1
q1
q3
q3
C. Segre (IIT) PHYS 406 - Spring 2017 February 21, 2017 23 / 23