csrri.iit.educsrri.iit.edu/~segre/phys406/16s/lecture_05.pdfproblem 6.5 consider a charged particle...
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Today’s Outline - January 27, 2016
• Problem 6.5
• Problem 6.8
• Problem 6.16
• Zeeman effect
Homework Assignment #03:Chapter 6:9,17,21,24,25,32due Wednesday, February 3, 2016
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 1 / 20
Today’s Outline - January 27, 2016
• Problem 6.5
• Problem 6.8
• Problem 6.16
• Zeeman effect
Homework Assignment #03:Chapter 6:9,17,21,24,25,32due Wednesday, February 3, 2016
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 1 / 20
Today’s Outline - January 27, 2016
• Problem 6.5
• Problem 6.8
• Problem 6.16
• Zeeman effect
Homework Assignment #03:Chapter 6:9,17,21,24,25,32due Wednesday, February 3, 2016
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 1 / 20
Today’s Outline - January 27, 2016
• Problem 6.5
• Problem 6.8
• Problem 6.16
• Zeeman effect
Homework Assignment #03:Chapter 6:9,17,21,24,25,32due Wednesday, February 3, 2016
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 1 / 20
Today’s Outline - January 27, 2016
• Problem 6.5
• Problem 6.8
• Problem 6.16
• Zeeman effect
Homework Assignment #03:Chapter 6:9,17,21,24,25,32due Wednesday, February 3, 2016
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 1 / 20
Today’s Outline - January 27, 2016
• Problem 6.5
• Problem 6.8
• Problem 6.16
• Zeeman effect
Homework Assignment #03:Chapter 6:9,17,21,24,25,32due Wednesday, February 3, 2016
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 1 / 20
Problem 6.5
Consider a charged particle in the one dimensional harmonic oscillatorpotential. Suppose we turn on a weak electric field (E), so that thepotential energy is shifted by an amount H ′ = −qEx .
(a) Show that there is no first order change in the energy levels, andcalculate the second-order correction.
(b) The Schrodinger equation can be solved directly in this case by asubstitution of variables x ′ = x − (qE/mω2). Find the exact energies,and show they are consistent with perturbation theory.
(a) The first order correctionin perturbation theory is simplythe expectation value of the per-turbing Hamiltonian.
but the operator x can berewritten using raising and low-ering operators
E(1)n = 〈ψ(0)
n |H ′|ψ(0)n 〉
= −qE〈ψ(0)n |x |ψ(0)
n 〉 = −qE〈n|x |n〉
x =
√~
2mω(a+ + a−)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 2 / 20
Problem 6.5
Consider a charged particle in the one dimensional harmonic oscillatorpotential. Suppose we turn on a weak electric field (E), so that thepotential energy is shifted by an amount H ′ = −qEx .
(a) Show that there is no first order change in the energy levels, andcalculate the second-order correction.
(b) The Schrodinger equation can be solved directly in this case by asubstitution of variables x ′ = x − (qE/mω2). Find the exact energies,and show they are consistent with perturbation theory.
(a) The first order correctionin perturbation theory is simplythe expectation value of the per-turbing Hamiltonian.
but the operator x can berewritten using raising and low-ering operators
E(1)n = 〈ψ(0)
n |H ′|ψ(0)n 〉
= −qE〈ψ(0)n |x |ψ(0)
n 〉 = −qE〈n|x |n〉
x =
√~
2mω(a+ + a−)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 2 / 20
Problem 6.5
Consider a charged particle in the one dimensional harmonic oscillatorpotential. Suppose we turn on a weak electric field (E), so that thepotential energy is shifted by an amount H ′ = −qEx .
(a) Show that there is no first order change in the energy levels, andcalculate the second-order correction.
(b) The Schrodinger equation can be solved directly in this case by asubstitution of variables x ′ = x − (qE/mω2). Find the exact energies,and show they are consistent with perturbation theory.
(a) The first order correctionin perturbation theory is simplythe expectation value of the per-turbing Hamiltonian.
but the operator x can berewritten using raising and low-ering operators
E(1)n = 〈ψ(0)
n |H ′|ψ(0)n 〉
= −qE〈ψ(0)n |x |ψ(0)
n 〉 = −qE〈n|x |n〉
x =
√~
2mω(a+ + a−)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 2 / 20
Problem 6.5
Consider a charged particle in the one dimensional harmonic oscillatorpotential. Suppose we turn on a weak electric field (E), so that thepotential energy is shifted by an amount H ′ = −qEx .
(a) Show that there is no first order change in the energy levels, andcalculate the second-order correction.
(b) The Schrodinger equation can be solved directly in this case by asubstitution of variables x ′ = x − (qE/mω2). Find the exact energies,and show they are consistent with perturbation theory.
(a) The first order correctionin perturbation theory is simplythe expectation value of the per-turbing Hamiltonian.
but the operator x can berewritten using raising and low-ering operators
E(1)n = 〈ψ(0)
n |H ′|ψ(0)n 〉
= −qE〈ψ(0)n |x |ψ(0)
n 〉
= −qE〈n|x |n〉
x =
√~
2mω(a+ + a−)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 2 / 20
Problem 6.5
Consider a charged particle in the one dimensional harmonic oscillatorpotential. Suppose we turn on a weak electric field (E), so that thepotential energy is shifted by an amount H ′ = −qEx .
(a) Show that there is no first order change in the energy levels, andcalculate the second-order correction.
(b) The Schrodinger equation can be solved directly in this case by asubstitution of variables x ′ = x − (qE/mω2). Find the exact energies,and show they are consistent with perturbation theory.
(a) The first order correctionin perturbation theory is simplythe expectation value of the per-turbing Hamiltonian.
but the operator x can berewritten using raising and low-ering operators
E(1)n = 〈ψ(0)
n |H ′|ψ(0)n 〉
= −qE〈ψ(0)n |x |ψ(0)
n 〉 = −qE〈n|x |n〉
x =
√~
2mω(a+ + a−)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 2 / 20
Problem 6.5
Consider a charged particle in the one dimensional harmonic oscillatorpotential. Suppose we turn on a weak electric field (E), so that thepotential energy is shifted by an amount H ′ = −qEx .
(a) Show that there is no first order change in the energy levels, andcalculate the second-order correction.
(b) The Schrodinger equation can be solved directly in this case by asubstitution of variables x ′ = x − (qE/mω2). Find the exact energies,and show they are consistent with perturbation theory.
(a) The first order correctionin perturbation theory is simplythe expectation value of the per-turbing Hamiltonian.
but the operator x can berewritten using raising and low-ering operators
E(1)n = 〈ψ(0)
n |H ′|ψ(0)n 〉
= −qE〈ψ(0)n |x |ψ(0)
n 〉 = −qE〈n|x |n〉
x =
√~
2mω(a+ + a−)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 2 / 20
Problem 6.5
Consider a charged particle in the one dimensional harmonic oscillatorpotential. Suppose we turn on a weak electric field (E), so that thepotential energy is shifted by an amount H ′ = −qEx .
(a) Show that there is no first order change in the energy levels, andcalculate the second-order correction.
(b) The Schrodinger equation can be solved directly in this case by asubstitution of variables x ′ = x − (qE/mω2). Find the exact energies,and show they are consistent with perturbation theory.
(a) The first order correctionin perturbation theory is simplythe expectation value of the per-turbing Hamiltonian.
but the operator x can berewritten using raising and low-ering operators
E(1)n = 〈ψ(0)
n |H ′|ψ(0)n 〉
= −qE〈ψ(0)n |x |ψ(0)
n 〉 = −qE〈n|x |n〉
x =
√~
2mω(a+ + a−)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 2 / 20
Problem 6.5 (cont.)
The expectation value of x isthus
this can be solved using theproperties of the raising andlowering operators
〈x〉 =
√~
2mω
∫ψ(0)∗n (a+ + a−)ψ
(0)n dx
a+ψ(0)n =
√n + 1ψ
(0)n+1
a−ψ(0)n =
√nψ
(0)n−1
〈x〉 =
√~
2mω
[∫ψ(0)∗n a+ψ
(0)n dx +
∫ψ(0)∗n a−ψ
(0)n dx
]=
√~
2mω
[√n + 1
��������∫ψ(0)∗n ψ
(0)n+1 dx +
√n��������∫ψ(0)∗n ψ
(0)n−1 dx
]≡ 0
thus, the second order perturbation must be calculated
E(2)n =
∑m 6=n
∣∣∣⟨ψ(0)m |H ′|ψ(0)
n
⟩∣∣∣2E(0)n − E
(0)m
= (qE)2∑m 6=n
∣∣∣⟨ψ(0)m |x |ψ(0)
n
⟩∣∣∣2~ω(n −m)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 3 / 20
Problem 6.5 (cont.)
The expectation value of x isthus
this can be solved using theproperties of the raising andlowering operators
〈x〉 =
√~
2mω
∫ψ(0)∗n (a+ + a−)ψ
(0)n dx
a+ψ(0)n =
√n + 1ψ
(0)n+1
a−ψ(0)n =
√nψ
(0)n−1
〈x〉 =
√~
2mω
[∫ψ(0)∗n a+ψ
(0)n dx +
∫ψ(0)∗n a−ψ
(0)n dx
]=
√~
2mω
[√n + 1
��������∫ψ(0)∗n ψ
(0)n+1 dx +
√n��������∫ψ(0)∗n ψ
(0)n−1 dx
]≡ 0
thus, the second order perturbation must be calculated
E(2)n =
∑m 6=n
∣∣∣⟨ψ(0)m |H ′|ψ(0)
n
⟩∣∣∣2E(0)n − E
(0)m
= (qE)2∑m 6=n
∣∣∣⟨ψ(0)m |x |ψ(0)
n
⟩∣∣∣2~ω(n −m)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 3 / 20
Problem 6.5 (cont.)
The expectation value of x isthus
this can be solved using theproperties of the raising andlowering operators
〈x〉 =
√~
2mω
∫ψ(0)∗n (a+ + a−)ψ
(0)n dx
a+ψ(0)n =
√n + 1ψ
(0)n+1
a−ψ(0)n =
√nψ
(0)n−1
〈x〉 =
√~
2mω
[∫ψ(0)∗n a+ψ
(0)n dx +
∫ψ(0)∗n a−ψ
(0)n dx
]=
√~
2mω
[√n + 1
��������∫ψ(0)∗n ψ
(0)n+1 dx +
√n��������∫ψ(0)∗n ψ
(0)n−1 dx
]≡ 0
thus, the second order perturbation must be calculated
E(2)n =
∑m 6=n
∣∣∣⟨ψ(0)m |H ′|ψ(0)
n
⟩∣∣∣2E(0)n − E
(0)m
= (qE)2∑m 6=n
∣∣∣⟨ψ(0)m |x |ψ(0)
n
⟩∣∣∣2~ω(n −m)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 3 / 20
Problem 6.5 (cont.)
The expectation value of x isthus
this can be solved using theproperties of the raising andlowering operators
〈x〉 =
√~
2mω
∫ψ(0)∗n (a+ + a−)ψ
(0)n dx
a+ψ(0)n =
√n + 1ψ
(0)n+1
a−ψ(0)n =
√nψ
(0)n−1
〈x〉 =
√~
2mω
[∫ψ(0)∗n a+ψ
(0)n dx +
∫ψ(0)∗n a−ψ
(0)n dx
]=
√~
2mω
[√n + 1
��������∫ψ(0)∗n ψ
(0)n+1 dx +
√n��������∫ψ(0)∗n ψ
(0)n−1 dx
]≡ 0
thus, the second order perturbation must be calculated
E(2)n =
∑m 6=n
∣∣∣⟨ψ(0)m |H ′|ψ(0)
n
⟩∣∣∣2E(0)n − E
(0)m
= (qE)2∑m 6=n
∣∣∣⟨ψ(0)m |x |ψ(0)
n
⟩∣∣∣2~ω(n −m)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 3 / 20
Problem 6.5 (cont.)
The expectation value of x isthus
this can be solved using theproperties of the raising andlowering operators
〈x〉 =
√~
2mω
∫ψ(0)∗n (a+ + a−)ψ
(0)n dx
a+ψ(0)n =
√n + 1ψ
(0)n+1
a−ψ(0)n =
√nψ
(0)n−1
〈x〉 =
√~
2mω
[∫ψ(0)∗n a+ψ
(0)n dx +
∫ψ(0)∗n a−ψ
(0)n dx
]=
√~
2mω
[√n + 1
��������∫ψ(0)∗n ψ
(0)n+1 dx +
√n��������∫ψ(0)∗n ψ
(0)n−1 dx
]≡ 0
thus, the second order perturbation must be calculated
E(2)n =
∑m 6=n
∣∣∣⟨ψ(0)m |H ′|ψ(0)
n
⟩∣∣∣2E(0)n − E
(0)m
= (qE)2∑m 6=n
∣∣∣⟨ψ(0)m |x |ψ(0)
n
⟩∣∣∣2~ω(n −m)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 3 / 20
Problem 6.5 (cont.)
The expectation value of x isthus
this can be solved using theproperties of the raising andlowering operators
〈x〉 =
√~
2mω
∫ψ(0)∗n (a+ + a−)ψ
(0)n dx
a+ψ(0)n =
√n + 1ψ
(0)n+1
a−ψ(0)n =
√nψ
(0)n−1
〈x〉 =
√~
2mω
[∫ψ(0)∗n a+ψ
(0)n dx +
∫ψ(0)∗n a−ψ
(0)n dx
]
=
√~
2mω
[√n + 1
��������∫ψ(0)∗n ψ
(0)n+1 dx +
√n��������∫ψ(0)∗n ψ
(0)n−1 dx
]≡ 0
thus, the second order perturbation must be calculated
E(2)n =
∑m 6=n
∣∣∣⟨ψ(0)m |H ′|ψ(0)
n
⟩∣∣∣2E(0)n − E
(0)m
= (qE)2∑m 6=n
∣∣∣⟨ψ(0)m |x |ψ(0)
n
⟩∣∣∣2~ω(n −m)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 3 / 20
Problem 6.5 (cont.)
The expectation value of x isthus
this can be solved using theproperties of the raising andlowering operators
〈x〉 =
√~
2mω
∫ψ(0)∗n (a+ + a−)ψ
(0)n dx
a+ψ(0)n =
√n + 1ψ
(0)n+1
a−ψ(0)n =
√nψ
(0)n−1
〈x〉 =
√~
2mω
[∫ψ(0)∗n a+ψ
(0)n dx +
∫ψ(0)∗n a−ψ
(0)n dx
]=
√~
2mω
[√n + 1
∫ψ(0)∗n ψ
(0)n+1 dx +
√n
∫ψ(0)∗n ψ
(0)n−1 dx
]
≡ 0
thus, the second order perturbation must be calculated
E(2)n =
∑m 6=n
∣∣∣⟨ψ(0)m |H ′|ψ(0)
n
⟩∣∣∣2E(0)n − E
(0)m
= (qE)2∑m 6=n
∣∣∣⟨ψ(0)m |x |ψ(0)
n
⟩∣∣∣2~ω(n −m)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 3 / 20
Problem 6.5 (cont.)
The expectation value of x isthus
this can be solved using theproperties of the raising andlowering operators
〈x〉 =
√~
2mω
∫ψ(0)∗n (a+ + a−)ψ
(0)n dx
a+ψ(0)n =
√n + 1ψ
(0)n+1
a−ψ(0)n =
√nψ
(0)n−1
〈x〉 =
√~
2mω
[∫ψ(0)∗n a+ψ
(0)n dx +
∫ψ(0)∗n a−ψ
(0)n dx
]=
√~
2mω
[√n + 1
��������∫ψ(0)∗n ψ
(0)n+1 dx +
√n��������∫ψ(0)∗n ψ
(0)n−1 dx
]≡ 0
thus, the second order perturbation must be calculated
E(2)n =
∑m 6=n
∣∣∣⟨ψ(0)m |H ′|ψ(0)
n
⟩∣∣∣2E(0)n − E
(0)m
= (qE)2∑m 6=n
∣∣∣⟨ψ(0)m |x |ψ(0)
n
⟩∣∣∣2~ω(n −m)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 3 / 20
Problem 6.5 (cont.)
The expectation value of x isthus
this can be solved using theproperties of the raising andlowering operators
〈x〉 =
√~
2mω
∫ψ(0)∗n (a+ + a−)ψ
(0)n dx
a+ψ(0)n =
√n + 1ψ
(0)n+1
a−ψ(0)n =
√nψ
(0)n−1
〈x〉 =
√~
2mω
[∫ψ(0)∗n a+ψ
(0)n dx +
∫ψ(0)∗n a−ψ
(0)n dx
]=
√~
2mω
[√n + 1
��������∫ψ(0)∗n ψ
(0)n+1 dx +
√n��������∫ψ(0)∗n ψ
(0)n−1 dx
]≡ 0
thus, the second order perturbation must be calculated
E(2)n =
∑m 6=n
∣∣∣⟨ψ(0)m |H ′|ψ(0)
n
⟩∣∣∣2E(0)n − E
(0)m
= (qE)2∑m 6=n
∣∣∣⟨ψ(0)m |x |ψ(0)
n
⟩∣∣∣2~ω(n −m)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 3 / 20
Problem 6.5 (cont.)
The expectation value of x isthus
this can be solved using theproperties of the raising andlowering operators
〈x〉 =
√~
2mω
∫ψ(0)∗n (a+ + a−)ψ
(0)n dx
a+ψ(0)n =
√n + 1ψ
(0)n+1
a−ψ(0)n =
√nψ
(0)n−1
〈x〉 =
√~
2mω
[∫ψ(0)∗n a+ψ
(0)n dx +
∫ψ(0)∗n a−ψ
(0)n dx
]=
√~
2mω
[√n + 1
��������∫ψ(0)∗n ψ
(0)n+1 dx +
√n��������∫ψ(0)∗n ψ
(0)n−1 dx
]≡ 0
thus, the second order perturbation must be calculated
E(2)n =
∑m 6=n
∣∣∣⟨ψ(0)m |H ′|ψ(0)
n
⟩∣∣∣2E(0)n − E
(0)m
= (qE)2∑m 6=n
∣∣∣⟨ψ(0)m |x |ψ(0)
n
⟩∣∣∣2~ω(n −m)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 3 / 20
Problem 6.5 (cont.)
The expectation value of x isthus
this can be solved using theproperties of the raising andlowering operators
〈x〉 =
√~
2mω
∫ψ(0)∗n (a+ + a−)ψ
(0)n dx
a+ψ(0)n =
√n + 1ψ
(0)n+1
a−ψ(0)n =
√nψ
(0)n−1
〈x〉 =
√~
2mω
[∫ψ(0)∗n a+ψ
(0)n dx +
∫ψ(0)∗n a−ψ
(0)n dx
]=
√~
2mω
[√n + 1
��������∫ψ(0)∗n ψ
(0)n+1 dx +
√n��������∫ψ(0)∗n ψ
(0)n−1 dx
]≡ 0
thus, the second order perturbation must be calculated
E(2)n =
∑m 6=n
∣∣∣⟨ψ(0)m |H ′|ψ(0)
n
⟩∣∣∣2E(0)n − E
(0)m
= (qE)2∑m 6=n
∣∣∣⟨ψ(0)m |x |ψ(0)
n
⟩∣∣∣2~ω(n −m)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 3 / 20
Problem 6.5 (cont.)
E(2)n =
(qE)2
~ω~
2mω
∑m 6=n
∣∣∣⟨ψ(0)m |(a+ + a−)|ψ(0)
n
⟩∣∣∣2(n −m)
=(qE)2
2mω2
∑m 6=n
∣∣∣
√n + 1
⟨ψ(0)m
∣∣∣ψ(0)n+1
⟩+√n⟨ψ(0)m
∣∣∣ψ(0)n−1
⟩
∣∣∣2(n −m)
=(qE)2
2mω2
∑m 6=n
∣∣∣
√n + 1δm,n+1 +
√nδm,n−1
∣∣∣2(n −m)
=(qE)2
2mω2
[
n + 1
n − (n + 1)+
n
n − (n − 1)
]=
(qE)2
2mω2[−(n + 1) + n] = − (qE)2
2mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 4 / 20
Problem 6.5 (cont.)
E(2)n =
(qE)2
~ω~
2mω
∑m 6=n
∣∣∣⟨ψ(0)m |(a+ + a−)|ψ(0)
n
⟩∣∣∣2(n −m)
=(qE)2
2mω2
∑m 6=n
∣∣∣
√n + 1
⟨ψ(0)m
∣∣∣ψ(0)n+1
⟩+√n⟨ψ(0)m
∣∣∣ψ(0)n−1
⟩
∣∣∣2(n −m)
=(qE)2
2mω2
∑m 6=n
∣∣∣
√n + 1δm,n+1 +
√nδm,n−1
∣∣∣2(n −m)
=(qE)2
2mω2
[
n + 1
n − (n + 1)+
n
n − (n − 1)
]=
(qE)2
2mω2[−(n + 1) + n] = − (qE)2
2mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 4 / 20
Problem 6.5 (cont.)
E(2)n =
(qE)2
~ω~
2mω
∑m 6=n
∣∣∣⟨ψ(0)m |(a+ + a−)|ψ(0)
n
⟩∣∣∣2(n −m)
=(qE)2
2mω2
∑m 6=n
∣∣∣√n + 1⟨ψ(0)m
∣∣∣ψ(0)n+1
⟩+
√n⟨ψ(0)m
∣∣∣ψ(0)n−1
⟩
∣∣∣2(n −m)
=(qE)2
2mω2
∑m 6=n
∣∣∣
√n + 1δm,n+1 +
√nδm,n−1
∣∣∣2(n −m)
=(qE)2
2mω2
[
n + 1
n − (n + 1)+
n
n − (n − 1)
]=
(qE)2
2mω2[−(n + 1) + n] = − (qE)2
2mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 4 / 20
Problem 6.5 (cont.)
E(2)n =
(qE)2
~ω~
2mω
∑m 6=n
∣∣∣⟨ψ(0)m |(a+ + a−)|ψ(0)
n
⟩∣∣∣2(n −m)
=(qE)2
2mω2
∑m 6=n
∣∣∣√n + 1⟨ψ(0)m
∣∣∣ψ(0)n+1
⟩+√n⟨ψ(0)m
∣∣∣ψ(0)n−1
⟩ ∣∣∣2(n −m)
=(qE)2
2mω2
∑m 6=n
∣∣∣
√n + 1δm,n+1 +
√nδm,n−1
∣∣∣2(n −m)
=(qE)2
2mω2
[
n + 1
n − (n + 1)+
n
n − (n − 1)
]=
(qE)2
2mω2[−(n + 1) + n] = − (qE)2
2mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 4 / 20
Problem 6.5 (cont.)
E(2)n =
(qE)2
~ω~
2mω
∑m 6=n
∣∣∣⟨ψ(0)m |(a+ + a−)|ψ(0)
n
⟩∣∣∣2(n −m)
=(qE)2
2mω2
∑m 6=n
∣∣∣√n + 1⟨ψ(0)m
∣∣∣ψ(0)n+1
⟩+√n⟨ψ(0)m
∣∣∣ψ(0)n−1
⟩ ∣∣∣2(n −m)
=(qE)2
2mω2
∑m 6=n
∣∣∣
√n + 1δm,n+1 +
√nδm,n−1
∣∣∣2(n −m)
=(qE)2
2mω2
[
n + 1
n − (n + 1)+
n
n − (n − 1)
]=
(qE)2
2mω2[−(n + 1) + n] = − (qE)2
2mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 4 / 20
Problem 6.5 (cont.)
E(2)n =
(qE)2
~ω~
2mω
∑m 6=n
∣∣∣⟨ψ(0)m |(a+ + a−)|ψ(0)
n
⟩∣∣∣2(n −m)
=(qE)2
2mω2
∑m 6=n
∣∣∣√n + 1⟨ψ(0)m
∣∣∣ψ(0)n+1
⟩+√n⟨ψ(0)m
∣∣∣ψ(0)n−1
⟩ ∣∣∣2(n −m)
=(qE)2
2mω2
∑m 6=n
∣∣∣√n + 1δm,n+1 +
√nδm,n−1
∣∣∣2(n −m)
=(qE)2
2mω2
[
n + 1
n − (n + 1)+
n
n − (n − 1)
]=
(qE)2
2mω2[−(n + 1) + n] = − (qE)2
2mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 4 / 20
Problem 6.5 (cont.)
E(2)n =
(qE)2
~ω~
2mω
∑m 6=n
∣∣∣⟨ψ(0)m |(a+ + a−)|ψ(0)
n
⟩∣∣∣2(n −m)
=(qE)2
2mω2
∑m 6=n
∣∣∣√n + 1⟨ψ(0)m
∣∣∣ψ(0)n+1
⟩+√n⟨ψ(0)m
∣∣∣ψ(0)n−1
⟩ ∣∣∣2(n −m)
=(qE)2
2mω2
∑m 6=n
∣∣∣√n + 1δm,n+1 +√nδm,n−1
∣∣∣2(n −m)
=(qE)2
2mω2
[
n + 1
n − (n + 1)+
n
n − (n − 1)
]=
(qE)2
2mω2[−(n + 1) + n] = − (qE)2
2mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 4 / 20
Problem 6.5 (cont.)
E(2)n =
(qE)2
~ω~
2mω
∑m 6=n
∣∣∣⟨ψ(0)m |(a+ + a−)|ψ(0)
n
⟩∣∣∣2(n −m)
=(qE)2
2mω2
∑m 6=n
∣∣∣√n + 1⟨ψ(0)m
∣∣∣ψ(0)n+1
⟩+√n⟨ψ(0)m
∣∣∣ψ(0)n−1
⟩ ∣∣∣2(n −m)
=(qE)2
2mω2
∑m 6=n
∣∣∣√n + 1δm,n+1 +√nδm,n−1
∣∣∣2(n −m)
=(qE)2
2mω2
[
n + 1
n − (n + 1)+
n
n − (n − 1)
]
=(qE)2
2mω2[−(n + 1) + n] = − (qE)2
2mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 4 / 20
Problem 6.5 (cont.)
E(2)n =
(qE)2
~ω~
2mω
∑m 6=n
∣∣∣⟨ψ(0)m |(a+ + a−)|ψ(0)
n
⟩∣∣∣2(n −m)
=(qE)2
2mω2
∑m 6=n
∣∣∣√n + 1⟨ψ(0)m
∣∣∣ψ(0)n+1
⟩+√n⟨ψ(0)m
∣∣∣ψ(0)n−1
⟩ ∣∣∣2(n −m)
=(qE)2
2mω2
∑m 6=n
∣∣∣√n + 1δm,n+1 +√nδm,n−1
∣∣∣2(n −m)
=(qE)2
2mω2
[n + 1
n − (n + 1)+
n
n − (n − 1)
]
=(qE)2
2mω2[−(n + 1) + n] = − (qE)2
2mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 4 / 20
Problem 6.5 (cont.)
E(2)n =
(qE)2
~ω~
2mω
∑m 6=n
∣∣∣⟨ψ(0)m |(a+ + a−)|ψ(0)
n
⟩∣∣∣2(n −m)
=(qE)2
2mω2
∑m 6=n
∣∣∣√n + 1⟨ψ(0)m
∣∣∣ψ(0)n+1
⟩+√n⟨ψ(0)m
∣∣∣ψ(0)n−1
⟩ ∣∣∣2(n −m)
=(qE)2
2mω2
∑m 6=n
∣∣∣√n + 1δm,n+1 +√nδm,n−1
∣∣∣2(n −m)
=(qE)2
2mω2
[n + 1
n − (n + 1)+
n
n − (n − 1)
]
=(qE)2
2mω2[−(n + 1) + n] = − (qE)2
2mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 4 / 20
Problem 6.5 (cont.)
E(2)n =
(qE)2
~ω~
2mω
∑m 6=n
∣∣∣⟨ψ(0)m |(a+ + a−)|ψ(0)
n
⟩∣∣∣2(n −m)
=(qE)2
2mω2
∑m 6=n
∣∣∣√n + 1⟨ψ(0)m
∣∣∣ψ(0)n+1
⟩+√n⟨ψ(0)m
∣∣∣ψ(0)n−1
⟩ ∣∣∣2(n −m)
=(qE)2
2mω2
∑m 6=n
∣∣∣√n + 1δm,n+1 +√nδm,n−1
∣∣∣2(n −m)
=(qE)2
2mω2
[n + 1
n − (n + 1)+
n
n − (n − 1)
]=
(qE)2
2mω2[−(n + 1) + n]
= − (qE)2
2mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 4 / 20
Problem 6.5 (cont.)
E(2)n =
(qE)2
~ω~
2mω
∑m 6=n
∣∣∣⟨ψ(0)m |(a+ + a−)|ψ(0)
n
⟩∣∣∣2(n −m)
=(qE)2
2mω2
∑m 6=n
∣∣∣√n + 1⟨ψ(0)m
∣∣∣ψ(0)n+1
⟩+√n⟨ψ(0)m
∣∣∣ψ(0)n−1
⟩ ∣∣∣2(n −m)
=(qE)2
2mω2
∑m 6=n
∣∣∣√n + 1δm,n+1 +√nδm,n−1
∣∣∣2(n −m)
=(qE)2
2mω2
[n + 1
n − (n + 1)+
n
n − (n − 1)
]=
(qE)2
2mω2[−(n + 1) + n] = − (qE)2
2mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 4 / 20
Problem 6.5 (cont.)
(b) The full Schrodingerequation, including theelectric field term is
applying the suggestedsubstitution to the quan-tity in the parenthses
Eψ = − ~2
2m
d2ψ
dx2+(12mω
2x2 − qEx)ψ
x ′ = x −(
qEmω2
)
()= 1
2mω2
[x ′ +
qEmω2
]2− qE
[x ′ +
qEmω2
]= 1
2mω2x ′2
+ 12mω
2x ′2qEmω2
+ 12mω
2 (qE)2
m2ω4− qEx ′ − (qE)2
mω2
= 12mω
2x ′2 +���qEx ′ + 1
2
(qE)2
mω2−�
��qEx ′ − (qE)2
mω2
= 12mω
2x ′2 − 12
(qE)2
mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 5 / 20
Problem 6.5 (cont.)
(b) The full Schrodingerequation, including theelectric field term is
applying the suggestedsubstitution to the quan-tity in the parenthses
Eψ = − ~2
2m
d2ψ
dx2+(12mω
2x2 − qEx)ψ
x ′ = x −(
qEmω2
)
()= 1
2mω2
[x ′ +
qEmω2
]2− qE
[x ′ +
qEmω2
]= 1
2mω2x ′2
+ 12mω
2x ′2qEmω2
+ 12mω
2 (qE)2
m2ω4− qEx ′ − (qE)2
mω2
= 12mω
2x ′2 +���qEx ′ + 1
2
(qE)2
mω2−�
��qEx ′ − (qE)2
mω2
= 12mω
2x ′2 − 12
(qE)2
mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 5 / 20
Problem 6.5 (cont.)
(b) The full Schrodingerequation, including theelectric field term is
applying the suggestedsubstitution to the quan-tity in the parenthses
Eψ = − ~2
2m
d2ψ
dx2+(12mω
2x2 − qEx)ψ
x ′ = x −(
qEmω2
)
()= 1
2mω2
[x ′ +
qEmω2
]2− qE
[x ′ +
qEmω2
]= 1
2mω2x ′2
+ 12mω
2x ′2qEmω2
+ 12mω
2 (qE)2
m2ω4− qEx ′ − (qE)2
mω2
= 12mω
2x ′2 +���qEx ′ + 1
2
(qE)2
mω2−�
��qEx ′ − (qE)2
mω2
= 12mω
2x ′2 − 12
(qE)2
mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 5 / 20
Problem 6.5 (cont.)
(b) The full Schrodingerequation, including theelectric field term is
applying the suggestedsubstitution to the quan-tity in the parenthses
Eψ = − ~2
2m
d2ψ
dx2+(12mω
2x2 − qEx)ψ
x ′ = x −(
qEmω2
)
()= 1
2mω2
[x ′ +
qEmω2
]2− qE
[x ′ +
qEmω2
]= 1
2mω2x ′2
+ 12mω
2x ′2qEmω2
+ 12mω
2 (qE)2
m2ω4− qEx ′ − (qE)2
mω2
= 12mω
2x ′2 +���qEx ′ + 1
2
(qE)2
mω2−�
��qEx ′ − (qE)2
mω2
= 12mω
2x ′2 − 12
(qE)2
mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 5 / 20
Problem 6.5 (cont.)
(b) The full Schrodingerequation, including theelectric field term is
applying the suggestedsubstitution to the quan-tity in the parenthses
Eψ = − ~2
2m
d2ψ
dx2+(12mω
2x2 − qEx)ψ
x ′ = x −(
qEmω2
)
()= 1
2mω2
[x ′ +
qEmω2
]2− qE
[x ′ +
qEmω2
]= 1
2mω2x ′2
+ 12mω
2x ′2qEmω2
+ 12mω
2 (qE)2
m2ω4− qEx ′ − (qE)2
mω2
= 12mω
2x ′2 +���qEx ′ + 1
2
(qE)2
mω2−�
��qEx ′ − (qE)2
mω2
= 12mω
2x ′2 − 12
(qE)2
mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 5 / 20
Problem 6.5 (cont.)
(b) The full Schrodingerequation, including theelectric field term is
applying the suggestedsubstitution to the quan-tity in the parenthses
Eψ = − ~2
2m
d2ψ
dx2+(12mω
2x2 − qEx)ψ
x ′ = x −(
qEmω2
)
()= 1
2mω2
[x ′ +
qEmω2
]2− qE
[x ′ +
qEmω2
]
= 12mω
2x ′2
+ 12mω
2x ′2qEmω2
+ 12mω
2 (qE)2
m2ω4− qEx ′ − (qE)2
mω2
= 12mω
2x ′2 +���qEx ′ + 1
2
(qE)2
mω2−�
��qEx ′ − (qE)2
mω2
= 12mω
2x ′2 − 12
(qE)2
mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 5 / 20
Problem 6.5 (cont.)
(b) The full Schrodingerequation, including theelectric field term is
applying the suggestedsubstitution to the quan-tity in the parenthses
Eψ = − ~2
2m
d2ψ
dx2+(12mω
2x2 − qEx)ψ
x ′ = x −(
qEmω2
)
()= 1
2mω2
[x ′ +
qEmω2
]2− qE
[x ′ +
qEmω2
]= 1
2mω2x ′2
+ 12mω
2x ′2qEmω2
+ 12mω
2 (qE)2
m2ω4− qEx ′ − (qE)2
mω2
= 12mω
2x ′2 +���qEx ′ + 1
2
(qE)2
mω2−�
��qEx ′ − (qE)2
mω2
= 12mω
2x ′2 − 12
(qE)2
mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 5 / 20
Problem 6.5 (cont.)
(b) The full Schrodingerequation, including theelectric field term is
applying the suggestedsubstitution to the quan-tity in the parenthses
Eψ = − ~2
2m
d2ψ
dx2+(12mω
2x2 − qEx)ψ
x ′ = x −(
qEmω2
)
()= 1
2mω2
[x ′ +
qEmω2
]2− qE
[x ′ +
qEmω2
]= 1
2mω2x ′2 + 1
2mω2x ′
2qEmω2
+ 12mω
2 (qE)2
m2ω4− qEx ′ − (qE)2
mω2
= 12mω
2x ′2 +���qEx ′ + 1
2
(qE)2
mω2−�
��qEx ′ − (qE)2
mω2
= 12mω
2x ′2 − 12
(qE)2
mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 5 / 20
Problem 6.5 (cont.)
(b) The full Schrodingerequation, including theelectric field term is
applying the suggestedsubstitution to the quan-tity in the parenthses
Eψ = − ~2
2m
d2ψ
dx2+(12mω
2x2 − qEx)ψ
x ′ = x −(
qEmω2
)
()= 1
2mω2
[x ′ +
qEmω2
]2− qE
[x ′ +
qEmω2
]= 1
2mω2x ′2 + 1
2mω2x ′
2qEmω2
+ 12mω
2 (qE)2
m2ω4
− qEx ′ − (qE)2
mω2
= 12mω
2x ′2 +���qEx ′ + 1
2
(qE)2
mω2−�
��qEx ′ − (qE)2
mω2
= 12mω
2x ′2 − 12
(qE)2
mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 5 / 20
Problem 6.5 (cont.)
(b) The full Schrodingerequation, including theelectric field term is
applying the suggestedsubstitution to the quan-tity in the parenthses
Eψ = − ~2
2m
d2ψ
dx2+(12mω
2x2 − qEx)ψ
x ′ = x −(
qEmω2
)
()= 1
2mω2
[x ′ +
qEmω2
]2− qE
[x ′ +
qEmω2
]= 1
2mω2x ′2 + 1
2mω2x ′
2qEmω2
+ 12mω
2 (qE)2
m2ω4− qEx ′
− (qE)2
mω2
= 12mω
2x ′2 +���qEx ′ + 1
2
(qE)2
mω2−�
��qEx ′ − (qE)2
mω2
= 12mω
2x ′2 − 12
(qE)2
mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 5 / 20
Problem 6.5 (cont.)
(b) The full Schrodingerequation, including theelectric field term is
applying the suggestedsubstitution to the quan-tity in the parenthses
Eψ = − ~2
2m
d2ψ
dx2+(12mω
2x2 − qEx)ψ
x ′ = x −(
qEmω2
)
()= 1
2mω2
[x ′ +
qEmω2
]2− qE
[x ′ +
qEmω2
]= 1
2mω2x ′2 + 1
2mω2x ′
2qEmω2
+ 12mω
2 (qE)2
m2ω4− qEx ′ − (qE)2
mω2
= 12mω
2x ′2 +���qEx ′ + 1
2
(qE)2
mω2−�
��qEx ′ − (qE)2
mω2
= 12mω
2x ′2 − 12
(qE)2
mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 5 / 20
Problem 6.5 (cont.)
(b) The full Schrodingerequation, including theelectric field term is
applying the suggestedsubstitution to the quan-tity in the parenthses
Eψ = − ~2
2m
d2ψ
dx2+(12mω
2x2 − qEx)ψ
x ′ = x −(
qEmω2
)
()= 1
2mω2
[x ′ +
qEmω2
]2− qE
[x ′ +
qEmω2
]= 1
2mω2x ′2 + 1
2mω2x ′
2qEmω2
+ 12mω
2 (qE)2
m2ω4− qEx ′ − (qE)2
mω2
= 12mω
2x ′2 + qEx ′ + 12
(qE)2
mω2− qEx ′ − (qE)2
mω2
= 12mω
2x ′2 − 12
(qE)2
mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 5 / 20
Problem 6.5 (cont.)
(b) The full Schrodingerequation, including theelectric field term is
applying the suggestedsubstitution to the quan-tity in the parenthses
Eψ = − ~2
2m
d2ψ
dx2+(12mω
2x2 − qEx)ψ
x ′ = x −(
qEmω2
)
()= 1
2mω2
[x ′ +
qEmω2
]2− qE
[x ′ +
qEmω2
]= 1
2mω2x ′2 + 1
2mω2x ′
2qEmω2
+ 12mω
2 (qE)2
m2ω4− qEx ′ − (qE)2
mω2
= 12mω
2x ′2 +���qEx ′ + 1
2
(qE)2
mω2−�
��qEx ′ − (qE)2
mω2
= 12mω
2x ′2 − 12
(qE)2
mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 5 / 20
Problem 6.5 (cont.)
(b) The full Schrodingerequation, including theelectric field term is
applying the suggestedsubstitution to the quan-tity in the parenthses
Eψ = − ~2
2m
d2ψ
dx2+(12mω
2x2 − qEx)ψ
x ′ = x −(
qEmω2
)
()= 1
2mω2
[x ′ +
qEmω2
]2− qE
[x ′ +
qEmω2
]= 1
2mω2x ′2 + 1
2mω2x ′
2qEmω2
+ 12mω
2 (qE)2
m2ω4− qEx ′ − (qE)2
mω2
= 12mω
2x ′2 +���qEx ′ + 1
2
(qE)2
mω2−�
��qEx ′ − (qE)2
mω2
= 12mω
2x ′2 − 12
(qE)2
mω2
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 5 / 20
Problem 6.5 (cont.)
Noting that dx ′ ≡ dx , the Schrodinger equation for the transformedcoordinate is now
− ~2
2m
d2ψ
dx ′2+
(12mω
2x ′2 − (qE)2
2mω2
)ψ = Eψ
− ~2
2m
d2ψ
dx ′2+ 1
2mω2x ′2 =
[E +
(qE)2
2mω2
]ψ ≡ E ′ψ
this is simply the eigenvalueequation of a harmonic oscilla-tor with a shifted energy
the corrected energy is therefore
E ′ =
[En +
(qE)2
2mω2
]=(n + 1
2
)~ω
En =(n + 1
2
)~ω − (qE)2
2mω2
the second term is exactly the second order perturbation correctionobtained in part (a)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 6 / 20
Problem 6.5 (cont.)
Noting that dx ′ ≡ dx , the Schrodinger equation for the transformedcoordinate is now
− ~2
2m
d2ψ
dx ′2+
(12mω
2x ′2 − (qE)2
2mω2
)ψ = Eψ
− ~2
2m
d2ψ
dx ′2+ 1
2mω2x ′2 =
[E +
(qE)2
2mω2
]ψ ≡ E ′ψ
this is simply the eigenvalueequation of a harmonic oscilla-tor with a shifted energy
the corrected energy is therefore
E ′ =
[En +
(qE)2
2mω2
]=(n + 1
2
)~ω
En =(n + 1
2
)~ω − (qE)2
2mω2
the second term is exactly the second order perturbation correctionobtained in part (a)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 6 / 20
Problem 6.5 (cont.)
Noting that dx ′ ≡ dx , the Schrodinger equation for the transformedcoordinate is now
− ~2
2m
d2ψ
dx ′2+
(12mω
2x ′2 − (qE)2
2mω2
)ψ = Eψ
− ~2
2m
d2ψ
dx ′2+ 1
2mω2x ′2 =
[E +
(qE)2
2mω2
]ψ
≡ E ′ψ
this is simply the eigenvalueequation of a harmonic oscilla-tor with a shifted energy
the corrected energy is therefore
E ′ =
[En +
(qE)2
2mω2
]=(n + 1
2
)~ω
En =(n + 1
2
)~ω − (qE)2
2mω2
the second term is exactly the second order perturbation correctionobtained in part (a)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 6 / 20
Problem 6.5 (cont.)
Noting that dx ′ ≡ dx , the Schrodinger equation for the transformedcoordinate is now
− ~2
2m
d2ψ
dx ′2+
(12mω
2x ′2 − (qE)2
2mω2
)ψ = Eψ
− ~2
2m
d2ψ
dx ′2+ 1
2mω2x ′2 =
[E +
(qE)2
2mω2
]ψ
≡ E ′ψ
this is simply the eigenvalueequation of a harmonic oscilla-tor with a shifted energy
the corrected energy is therefore
E ′ =
[En +
(qE)2
2mω2
]=(n + 1
2
)~ω
En =(n + 1
2
)~ω − (qE)2
2mω2
the second term is exactly the second order perturbation correctionobtained in part (a)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 6 / 20
Problem 6.5 (cont.)
Noting that dx ′ ≡ dx , the Schrodinger equation for the transformedcoordinate is now
− ~2
2m
d2ψ
dx ′2+
(12mω
2x ′2 − (qE)2
2mω2
)ψ = Eψ
− ~2
2m
d2ψ
dx ′2+ 1
2mω2x ′2 =
[E +
(qE)2
2mω2
]ψ ≡ E ′ψ
this is simply the eigenvalueequation of a harmonic oscilla-tor with a shifted energy
the corrected energy is therefore
E ′ =
[En +
(qE)2
2mω2
]
=(n + 1
2
)~ω
En =(n + 1
2
)~ω − (qE)2
2mω2
the second term is exactly the second order perturbation correctionobtained in part (a)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 6 / 20
Problem 6.5 (cont.)
Noting that dx ′ ≡ dx , the Schrodinger equation for the transformedcoordinate is now
− ~2
2m
d2ψ
dx ′2+
(12mω
2x ′2 − (qE)2
2mω2
)ψ = Eψ
− ~2
2m
d2ψ
dx ′2+ 1
2mω2x ′2 =
[E +
(qE)2
2mω2
]ψ ≡ E ′ψ
this is simply the eigenvalueequation of a harmonic oscilla-tor with a shifted energy
the corrected energy is therefore
E ′ =
[En +
(qE)2
2mω2
]=(n + 1
2
)~ω
En =(n + 1
2
)~ω − (qE)2
2mω2
the second term is exactly the second order perturbation correctionobtained in part (a)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 6 / 20
Problem 6.5 (cont.)
Noting that dx ′ ≡ dx , the Schrodinger equation for the transformedcoordinate is now
− ~2
2m
d2ψ
dx ′2+
(12mω
2x ′2 − (qE)2
2mω2
)ψ = Eψ
− ~2
2m
d2ψ
dx ′2+ 1
2mω2x ′2 =
[E +
(qE)2
2mω2
]ψ ≡ E ′ψ
this is simply the eigenvalueequation of a harmonic oscilla-tor with a shifted energy
the corrected energy is therefore
E ′ =
[En +
(qE)2
2mω2
]=(n + 1
2
)~ω
En =(n + 1
2
)~ω − (qE)2
2mω2
the second term is exactly the second order perturbation correctionobtained in part (a)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 6 / 20
Problem 6.5 (cont.)
Noting that dx ′ ≡ dx , the Schrodinger equation for the transformedcoordinate is now
− ~2
2m
d2ψ
dx ′2+
(12mω
2x ′2 − (qE)2
2mω2
)ψ = Eψ
− ~2
2m
d2ψ
dx ′2+ 1
2mω2x ′2 =
[E +
(qE)2
2mω2
]ψ ≡ E ′ψ
this is simply the eigenvalueequation of a harmonic oscilla-tor with a shifted energy
the corrected energy is therefore
E ′ =
[En +
(qE)2
2mω2
]=(n + 1
2
)~ω
En =(n + 1
2
)~ω − (qE)2
2mω2
the second term is exactly the second order perturbation correctionobtained in part (a)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 6 / 20
Problem 6.5 (cont.)
Noting that dx ′ ≡ dx , the Schrodinger equation for the transformedcoordinate is now
− ~2
2m
d2ψ
dx ′2+
(12mω
2x ′2 − (qE)2
2mω2
)ψ = Eψ
− ~2
2m
d2ψ
dx ′2+ 1
2mω2x ′2 =
[E +
(qE)2
2mω2
]ψ ≡ E ′ψ
this is simply the eigenvalueequation of a harmonic oscilla-tor with a shifted energy
the corrected energy is therefore
E ′ =
[En +
(qE)2
2mω2
]=(n + 1
2
)~ω
En =(n + 1
2
)~ω − (qE)2
2mω2
the second term is exactly the second order perturbation correctionobtained in part (a)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 6 / 20
Problem 6.8
Suppose we perturb the infinite cubical well
V (x , y , z) =
{0, 0 < x , y , z < a
∞ otherwise
by putting a delta function “bump” at the point ( a4 ,a2 ,
3a4 ):
H ′ = a3V0δ(x − a4)δ(y − a
2)δ(z − 3a4 )
Find the first-order corrections to the energy of the ground state and the(triply degenerate) first excited states.
The ground state ψ111 is non-degenerate but the first excited state is triplydegenerate: ψa = ψ112, ψb = ψ121, and ψc = ψ211.
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 7 / 20
Problem 6.8
Suppose we perturb the infinite cubical well
V (x , y , z) =
{0, 0 < x , y , z < a
∞ otherwise
by putting a delta function “bump” at the point ( a4 ,a2 ,
3a4 ):
H ′ = a3V0δ(x − a4)δ(y − a
2)δ(z − 3a4 )
Find the first-order corrections to the energy of the ground state and the(triply degenerate) first excited states.
The ground state ψ111 is non-degenerate but the first excited state is triplydegenerate: ψa = ψ112, ψb = ψ121, and ψc = ψ211.
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 7 / 20
Problem 6.8 (cont.)
First compute the first order correction to the ground state
E(1)111 =
(2
a
)3a3V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(π2
)sin2
(3π4
)= 8V0 · 12 · 1 ·
12 = 2V0
for the first excited state, we need to calculate the elements of the “W”matrix and then diagonalize it
Waa = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(2πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(π2
)sin2
(6π4
)= 8V0 · 12 · 1 · 1 = 4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 8 / 20
Problem 6.8 (cont.)
First compute the first order correction to the ground state
E(1)111 =
(2
a
)3a3V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(π2
)sin2
(3π4
)= 8V0 · 12 · 1 ·
12 = 2V0
for the first excited state, we need to calculate the elements of the “W”matrix and then diagonalize it
Waa = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(2πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(π2
)sin2
(6π4
)= 8V0 · 12 · 1 · 1 = 4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 8 / 20
Problem 6.8 (cont.)
First compute the first order correction to the ground state
E(1)111 =
(2
a
)3a3V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(π2
)sin2
(3π4
)
= 8V0 · 12 · 1 ·12 = 2V0
for the first excited state, we need to calculate the elements of the “W”matrix and then diagonalize it
Waa = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(2πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(π2
)sin2
(6π4
)= 8V0 · 12 · 1 · 1 = 4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 8 / 20
Problem 6.8 (cont.)
First compute the first order correction to the ground state
E(1)111 =
(2
a
)3a3V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(π2
)sin2
(3π4
)= 8V0 · 12 · 1 ·
12
= 2V0
for the first excited state, we need to calculate the elements of the “W”matrix and then diagonalize it
Waa = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(2πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(π2
)sin2
(6π4
)= 8V0 · 12 · 1 · 1 = 4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 8 / 20
Problem 6.8 (cont.)
First compute the first order correction to the ground state
E(1)111 =
(2
a
)3a3V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(π2
)sin2
(3π4
)= 8V0 · 12 · 1 ·
12 = 2V0
for the first excited state, we need to calculate the elements of the “W”matrix and then diagonalize it
Waa = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(2πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(π2
)sin2
(6π4
)= 8V0 · 12 · 1 · 1 = 4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 8 / 20
Problem 6.8 (cont.)
First compute the first order correction to the ground state
E(1)111 =
(2
a
)3a3V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(π2
)sin2
(3π4
)= 8V0 · 12 · 1 ·
12 = 2V0
for the first excited state, we need to calculate the elements of the “W”matrix and then diagonalize it
Waa = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(2πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(π2
)sin2
(6π4
)= 8V0 · 12 · 1 · 1 = 4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 8 / 20
Problem 6.8 (cont.)
First compute the first order correction to the ground state
E(1)111 =
(2
a
)3a3V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(π2
)sin2
(3π4
)= 8V0 · 12 · 1 ·
12 = 2V0
for the first excited state, we need to calculate the elements of the “W”matrix and then diagonalize it
Waa = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(2πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(π2
)sin2
(6π4
)= 8V0 · 12 · 1 · 1 = 4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 8 / 20
Problem 6.8 (cont.)
First compute the first order correction to the ground state
E(1)111 =
(2
a
)3a3V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(π2
)sin2
(3π4
)= 8V0 · 12 · 1 ·
12 = 2V0
for the first excited state, we need to calculate the elements of the “W”matrix and then diagonalize it
Waa = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(2πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(π2
)sin2
(6π4
)
= 8V0 · 12 · 1 · 1 = 4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 8 / 20
Problem 6.8 (cont.)
First compute the first order correction to the ground state
E(1)111 =
(2
a
)3a3V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(π2
)sin2
(3π4
)= 8V0 · 12 · 1 ·
12 = 2V0
for the first excited state, we need to calculate the elements of the “W”matrix and then diagonalize it
Waa = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(2πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(π2
)sin2
(6π4
)= 8V0 · 12 · 1 · 1
= 4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 8 / 20
Problem 6.8 (cont.)
First compute the first order correction to the ground state
E(1)111 =
(2
a
)3a3V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(π2
)sin2
(3π4
)= 8V0 · 12 · 1 ·
12 = 2V0
for the first excited state, we need to calculate the elements of the “W”matrix and then diagonalize it
Waa = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(2πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(π2
)sin2
(6π4
)= 8V0 · 12 · 1 · 1 = 4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 8 / 20
Problem 6.8 (cont.)
Wbb = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(2πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(2π2
)sin2
(3π4
)= 8V0 · 12 · 0 · 1 = 0
Wcc = 8V0
∫ a
0sin2
(2πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(2π4
)sin2
(π2
)sin2
(3π4
)= 8V0 · 1 · 1 · 12 = 4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 9 / 20
Problem 6.8 (cont.)
Wbb = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(2πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(2π2
)sin2
(3π4
)
= 8V0 · 12 · 0 · 1 = 0
Wcc = 8V0
∫ a
0sin2
(2πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(2π4
)sin2
(π2
)sin2
(3π4
)= 8V0 · 1 · 1 · 12 = 4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 9 / 20
Problem 6.8 (cont.)
Wbb = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(2πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(2π2
)sin2
(3π4
)= 8V0 · 12 · 0 · 1
= 0
Wcc = 8V0
∫ a
0sin2
(2πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(2π4
)sin2
(π2
)sin2
(3π4
)= 8V0 · 1 · 1 · 12 = 4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 9 / 20
Problem 6.8 (cont.)
Wbb = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(2πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(2π2
)sin2
(3π4
)= 8V0 · 12 · 0 · 1 = 0
Wcc = 8V0
∫ a
0sin2
(2πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(2π4
)sin2
(π2
)sin2
(3π4
)= 8V0 · 1 · 1 · 12 = 4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 9 / 20
Problem 6.8 (cont.)
Wbb = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(2πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(2π2
)sin2
(3π4
)= 8V0 · 12 · 0 · 1 = 0
Wcc = 8V0
∫ a
0sin2
(2πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(2π4
)sin2
(π2
)sin2
(3π4
)= 8V0 · 1 · 1 · 12 = 4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 9 / 20
Problem 6.8 (cont.)
Wbb = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(2πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(2π2
)sin2
(3π4
)= 8V0 · 12 · 0 · 1 = 0
Wcc = 8V0
∫ a
0sin2
(2πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(2π4
)sin2
(π2
)sin2
(3π4
)
= 8V0 · 1 · 1 · 12 = 4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 9 / 20
Problem 6.8 (cont.)
Wbb = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(2πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(2π2
)sin2
(3π4
)= 8V0 · 12 · 0 · 1 = 0
Wcc = 8V0
∫ a
0sin2
(2πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(2π4
)sin2
(π2
)sin2
(3π4
)= 8V0 · 1 · 1 · 12
= 4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 9 / 20
Problem 6.8 (cont.)
Wbb = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin2
(2πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin2
(2π2
)sin2
(3π4
)= 8V0 · 12 · 0 · 1 = 0
Wcc = 8V0
∫ a
0sin2
(2πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin2
(πa z)δ(x − 3a
4 )dz
= 8V0 sin2(2π4
)sin2
(π2
)sin2
(3π4
)= 8V0 · 1 · 1 · 12 = 4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 9 / 20
Problem 6.8 (cont.)
Wab = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin(2πa y)
sin(πa y)δ(x − a
2)dy
∫ a
0sin(πa z)
sin(2πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin(2π2
)sin(π2
)sin(3π4
)sin(6π4
)= 0 = Wbc
Wac = 8V0
∫ a
0sin(2πa x)
sin(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin(πa z)sin(2πa z)δ(x − 3a
4 )dz
= 8V0 sin(2π4
)sin(π4
)sin2
(π2
)sin(3π4
)sin(6π4
)= 8V0 · 1 · 1√
2· 1 · 1√
2· (−1) = −4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 10 / 20
Problem 6.8 (cont.)
Wab = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin(2πa y)
sin(πa y)δ(x − a
2)dy
∫ a
0sin(πa z)
sin(2πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin(2π2
)sin(π2
)sin(3π4
)sin(6π4
)
= 0 = Wbc
Wac = 8V0
∫ a
0sin(2πa x)
sin(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin(πa z)sin(2πa z)δ(x − 3a
4 )dz
= 8V0 sin(2π4
)sin(π4
)sin2
(π2
)sin(3π4
)sin(6π4
)= 8V0 · 1 · 1√
2· 1 · 1√
2· (−1) = −4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 10 / 20
Problem 6.8 (cont.)
Wab = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin(2πa y)
sin(πa y)δ(x − a
2)dy
∫ a
0sin(πa z)
sin(2πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin(2π2
)sin(π2
)sin(3π4
)sin(6π4
)= 0
= Wbc
Wac = 8V0
∫ a
0sin(2πa x)
sin(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin(πa z)sin(2πa z)δ(x − 3a
4 )dz
= 8V0 sin(2π4
)sin(π4
)sin2
(π2
)sin(3π4
)sin(6π4
)= 8V0 · 1 · 1√
2· 1 · 1√
2· (−1) = −4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 10 / 20
Problem 6.8 (cont.)
Wab = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin(2πa y)
sin(πa y)δ(x − a
2)dy
∫ a
0sin(πa z)
sin(2πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin(2π2
)sin(π2
)sin(3π4
)sin(6π4
)= 0 = Wbc
Wac = 8V0
∫ a
0sin(2πa x)
sin(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin(πa z)sin(2πa z)δ(x − 3a
4 )dz
= 8V0 sin(2π4
)sin(π4
)sin2
(π2
)sin(3π4
)sin(6π4
)= 8V0 · 1 · 1√
2· 1 · 1√
2· (−1) = −4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 10 / 20
Problem 6.8 (cont.)
Wab = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin(2πa y)
sin(πa y)δ(x − a
2)dy
∫ a
0sin(πa z)
sin(2πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin(2π2
)sin(π2
)sin(3π4
)sin(6π4
)= 0 = Wbc
Wac = 8V0
∫ a
0sin(2πa x)
sin(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin(πa z)sin(2πa z)δ(x − 3a
4 )dz
= 8V0 sin(2π4
)sin(π4
)sin2
(π2
)sin(3π4
)sin(6π4
)= 8V0 · 1 · 1√
2· 1 · 1√
2· (−1) = −4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 10 / 20
Problem 6.8 (cont.)
Wab = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin(2πa y)
sin(πa y)δ(x − a
2)dy
∫ a
0sin(πa z)
sin(2πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin(2π2
)sin(π2
)sin(3π4
)sin(6π4
)= 0 = Wbc
Wac = 8V0
∫ a
0sin(2πa x)
sin(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin(πa z)sin(2πa z)δ(x − 3a
4 )dz
= 8V0 sin(2π4
)sin(π4
)sin2
(π2
)sin(3π4
)sin(6π4
)
= 8V0 · 1 · 1√2· 1 · 1√
2· (−1) = −4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 10 / 20
Problem 6.8 (cont.)
Wab = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin(2πa y)
sin(πa y)δ(x − a
2)dy
∫ a
0sin(πa z)
sin(2πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin(2π2
)sin(π2
)sin(3π4
)sin(6π4
)= 0 = Wbc
Wac = 8V0
∫ a
0sin(2πa x)
sin(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin(πa z)sin(2πa z)δ(x − 3a
4 )dz
= 8V0 sin(2π4
)sin(π4
)sin2
(π2
)sin(3π4
)sin(6π4
)= 8V0 · 1 · 1√
2· 1 · 1√
2· (−1)
= −4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 10 / 20
Problem 6.8 (cont.)
Wab = 8V0
∫ a
0sin2
(πa x)δ(x − a
4)dx
×∫ a
0sin(2πa y)
sin(πa y)δ(x − a
2)dy
∫ a
0sin(πa z)
sin(2πa z)δ(x − 3a
4 )dz
= 8V0 sin2(π4
)sin(2π2
)sin(π2
)sin(3π4
)sin(6π4
)= 0 = Wbc
Wac = 8V0
∫ a
0sin(2πa x)
sin(πa x)δ(x − a
4)dx
×∫ a
0sin2
(πa y)δ(x − a
2)dy
∫ a
0sin(πa z)sin(2πa z)δ(x − 3a
4 )dz
= 8V0 sin(2π4
)sin(π4
)sin2
(π2
)sin(3π4
)sin(6π4
)= 8V0 · 1 · 1√
2· 1 · 1√
2· (−1) = −4V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 10 / 20
Problem 6.8 (cont.)
The “W”matrix for these three de-generate states is
it can be diagonalized by comput-ing a determinant
W = 4V0
1 0 −10 0 0−1 0 1
= 4V0D
0 = det(D − λI ) =
∣∣∣∣∣∣(1− λ) 0 −1
0 −λ 0−1 0 (1− λ)
∣∣∣∣∣∣ = −λ(1− λ)2 + λ
0 = λ[1− (1− λ)2] −→ λ = 0, (1− λ)2 = 1
±1 = 1− λ −→ λ = 1∓ 1 = 0, 2
The first order energy corrections to the energies of the first excited stateare 0, 0, and 8V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 11 / 20
Problem 6.8 (cont.)
The “W”matrix for these three de-generate states is
it can be diagonalized by comput-ing a determinant
W = 4V0
1 0 −10 0 0−1 0 1
= 4V0D
0 = det(D − λI ) =
∣∣∣∣∣∣(1− λ) 0 −1
0 −λ 0−1 0 (1− λ)
∣∣∣∣∣∣ = −λ(1− λ)2 + λ
0 = λ[1− (1− λ)2] −→ λ = 0, (1− λ)2 = 1
±1 = 1− λ −→ λ = 1∓ 1 = 0, 2
The first order energy corrections to the energies of the first excited stateare 0, 0, and 8V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 11 / 20
Problem 6.8 (cont.)
The “W”matrix for these three de-generate states is
it can be diagonalized by comput-ing a determinant
W = 4V0
1 0 −10 0 0−1 0 1
= 4V0D
0 = det(D − λI ) =
∣∣∣∣∣∣(1− λ) 0 −1
0 −λ 0−1 0 (1− λ)
∣∣∣∣∣∣ = −λ(1− λ)2 + λ
0 = λ[1− (1− λ)2] −→ λ = 0, (1− λ)2 = 1
±1 = 1− λ −→ λ = 1∓ 1 = 0, 2
The first order energy corrections to the energies of the first excited stateare 0, 0, and 8V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 11 / 20
Problem 6.8 (cont.)
The “W”matrix for these three de-generate states is
it can be diagonalized by comput-ing a determinant
W = 4V0
1 0 −10 0 0−1 0 1
= 4V0D
0 = det(D − λI ) =
∣∣∣∣∣∣(1− λ) 0 −1
0 −λ 0−1 0 (1− λ)
∣∣∣∣∣∣ = −λ(1− λ)2 + λ
0 = λ[1− (1− λ)2] −→ λ = 0, (1− λ)2 = 1
±1 = 1− λ −→ λ = 1∓ 1 = 0, 2
The first order energy corrections to the energies of the first excited stateare 0, 0, and 8V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 11 / 20
Problem 6.8 (cont.)
The “W”matrix for these three de-generate states is
it can be diagonalized by comput-ing a determinant
W = 4V0
1 0 −10 0 0−1 0 1
= 4V0D
0 = det(D − λI ) =
∣∣∣∣∣∣(1− λ) 0 −1
0 −λ 0−1 0 (1− λ)
∣∣∣∣∣∣
= −λ(1− λ)2 + λ
0 = λ[1− (1− λ)2] −→ λ = 0, (1− λ)2 = 1
±1 = 1− λ −→ λ = 1∓ 1 = 0, 2
The first order energy corrections to the energies of the first excited stateare 0, 0, and 8V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 11 / 20
Problem 6.8 (cont.)
The “W”matrix for these three de-generate states is
it can be diagonalized by comput-ing a determinant
W = 4V0
1 0 −10 0 0−1 0 1
= 4V0D
0 = det(D − λI ) =
∣∣∣∣∣∣(1− λ) 0 −1
0 −λ 0−1 0 (1− λ)
∣∣∣∣∣∣ = −λ(1− λ)2 + λ
0 = λ[1− (1− λ)2] −→ λ = 0, (1− λ)2 = 1
±1 = 1− λ −→ λ = 1∓ 1 = 0, 2
The first order energy corrections to the energies of the first excited stateare 0, 0, and 8V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 11 / 20
Problem 6.8 (cont.)
The “W”matrix for these three de-generate states is
it can be diagonalized by comput-ing a determinant
W = 4V0
1 0 −10 0 0−1 0 1
= 4V0D
0 = det(D − λI ) =
∣∣∣∣∣∣(1− λ) 0 −1
0 −λ 0−1 0 (1− λ)
∣∣∣∣∣∣ = −λ(1− λ)2 + λ
0 = λ[1− (1− λ)2]
−→ λ = 0, (1− λ)2 = 1
±1 = 1− λ −→ λ = 1∓ 1 = 0, 2
The first order energy corrections to the energies of the first excited stateare 0, 0, and 8V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 11 / 20
Problem 6.8 (cont.)
The “W”matrix for these three de-generate states is
it can be diagonalized by comput-ing a determinant
W = 4V0
1 0 −10 0 0−1 0 1
= 4V0D
0 = det(D − λI ) =
∣∣∣∣∣∣(1− λ) 0 −1
0 −λ 0−1 0 (1− λ)
∣∣∣∣∣∣ = −λ(1− λ)2 + λ
0 = λ[1− (1− λ)2] −→ λ = 0, (1− λ)2 = 1
±1 = 1− λ −→ λ = 1∓ 1 = 0, 2
The first order energy corrections to the energies of the first excited stateare 0, 0, and 8V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 11 / 20
Problem 6.8 (cont.)
The “W”matrix for these three de-generate states is
it can be diagonalized by comput-ing a determinant
W = 4V0
1 0 −10 0 0−1 0 1
= 4V0D
0 = det(D − λI ) =
∣∣∣∣∣∣(1− λ) 0 −1
0 −λ 0−1 0 (1− λ)
∣∣∣∣∣∣ = −λ(1− λ)2 + λ
0 = λ[1− (1− λ)2] −→ λ = 0, (1− λ)2 = 1
±1 = 1− λ
−→ λ = 1∓ 1 = 0, 2
The first order energy corrections to the energies of the first excited stateare 0, 0, and 8V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 11 / 20
Problem 6.8 (cont.)
The “W”matrix for these three de-generate states is
it can be diagonalized by comput-ing a determinant
W = 4V0
1 0 −10 0 0−1 0 1
= 4V0D
0 = det(D − λI ) =
∣∣∣∣∣∣(1− λ) 0 −1
0 −λ 0−1 0 (1− λ)
∣∣∣∣∣∣ = −λ(1− λ)2 + λ
0 = λ[1− (1− λ)2] −→ λ = 0, (1− λ)2 = 1
±1 = 1− λ −→ λ = 1∓ 1
= 0, 2
The first order energy corrections to the energies of the first excited stateare 0, 0, and 8V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 11 / 20
Problem 6.8 (cont.)
The “W”matrix for these three de-generate states is
it can be diagonalized by comput-ing a determinant
W = 4V0
1 0 −10 0 0−1 0 1
= 4V0D
0 = det(D − λI ) =
∣∣∣∣∣∣(1− λ) 0 −1
0 −λ 0−1 0 (1− λ)
∣∣∣∣∣∣ = −λ(1− λ)2 + λ
0 = λ[1− (1− λ)2] −→ λ = 0, (1− λ)2 = 1
±1 = 1− λ −→ λ = 1∓ 1 = 0, 2
The first order energy corrections to the energies of the first excited stateare 0, 0, and 8V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 11 / 20
Problem 6.8 (cont.)
The “W”matrix for these three de-generate states is
it can be diagonalized by comput-ing a determinant
W = 4V0
1 0 −10 0 0−1 0 1
= 4V0D
0 = det(D − λI ) =
∣∣∣∣∣∣(1− λ) 0 −1
0 −λ 0−1 0 (1− λ)
∣∣∣∣∣∣ = −λ(1− λ)2 + λ
0 = λ[1− (1− λ)2] −→ λ = 0, (1− λ)2 = 1
±1 = 1− λ −→ λ = 1∓ 1 = 0, 2
The first order energy corrections to the energies of the first excited stateare 0, 0, and 8V0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 11 / 20
Problem 6.8 (cont.)
In order to find the eigen-functions of the “D” matrix,solve the eigenvalue equation
this leads to two equationsonly since ψb is not coupledto either of the other twostates
first choose λ = 0, thenchoose λ = 2
1 0 −10 0 0−1 0 1
αβγ
= λ
αβγ
α− γ = λα, γ − α = λγ
λ = 0 −→ α = γ
λ = 0 −→ α = −γ
The new eigenfuctions are thus
1√2
(ψ112 + ψ211), ψ121,1√2
(ψ112 − ψ211)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 12 / 20
Problem 6.8 (cont.)
In order to find the eigen-functions of the “D” matrix,solve the eigenvalue equation
this leads to two equationsonly since ψb is not coupledto either of the other twostates
first choose λ = 0, thenchoose λ = 2
1 0 −10 0 0−1 0 1
αβγ
= λ
αβγ
α− γ = λα, γ − α = λγ
λ = 0 −→ α = γ
λ = 0 −→ α = −γ
The new eigenfuctions are thus
1√2
(ψ112 + ψ211), ψ121,1√2
(ψ112 − ψ211)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 12 / 20
Problem 6.8 (cont.)
In order to find the eigen-functions of the “D” matrix,solve the eigenvalue equation
this leads to two equationsonly since ψb is not coupledto either of the other twostates
first choose λ = 0, thenchoose λ = 2
1 0 −10 0 0−1 0 1
αβγ
= λ
αβγ
α− γ = λα, γ − α = λγ
λ = 0 −→ α = γ
λ = 0 −→ α = −γ
The new eigenfuctions are thus
1√2
(ψ112 + ψ211), ψ121,1√2
(ψ112 − ψ211)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 12 / 20
Problem 6.8 (cont.)
In order to find the eigen-functions of the “D” matrix,solve the eigenvalue equation
this leads to two equationsonly since ψb is not coupledto either of the other twostates
first choose λ = 0, thenchoose λ = 2
1 0 −10 0 0−1 0 1
αβγ
= λ
αβγ
α− γ = λα,
γ − α = λγ
λ = 0 −→ α = γ
λ = 0 −→ α = −γ
The new eigenfuctions are thus
1√2
(ψ112 + ψ211), ψ121,1√2
(ψ112 − ψ211)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 12 / 20
Problem 6.8 (cont.)
In order to find the eigen-functions of the “D” matrix,solve the eigenvalue equation
this leads to two equationsonly since ψb is not coupledto either of the other twostates
first choose λ = 0, thenchoose λ = 2
1 0 −10 0 0−1 0 1
αβγ
= λ
αβγ
α− γ = λα, γ − α = λγ
λ = 0 −→ α = γ
λ = 0 −→ α = −γ
The new eigenfuctions are thus
1√2
(ψ112 + ψ211), ψ121,1√2
(ψ112 − ψ211)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 12 / 20
Problem 6.8 (cont.)
In order to find the eigen-functions of the “D” matrix,solve the eigenvalue equation
this leads to two equationsonly since ψb is not coupledto either of the other twostates
first choose λ = 0,
thenchoose λ = 2
1 0 −10 0 0−1 0 1
αβγ
= λ
αβγ
α− γ = λα, γ − α = λγ
λ = 0 −→ α = γ
λ = 0 −→ α = −γ
The new eigenfuctions are thus
1√2
(ψ112 + ψ211), ψ121,1√2
(ψ112 − ψ211)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 12 / 20
Problem 6.8 (cont.)
In order to find the eigen-functions of the “D” matrix,solve the eigenvalue equation
this leads to two equationsonly since ψb is not coupledto either of the other twostates
first choose λ = 0,
thenchoose λ = 2
1 0 −10 0 0−1 0 1
αβγ
= λ
αβγ
α− γ = λα, γ − α = λγ
λ = 0 −→ α = γ
λ = 0 −→ α = −γ
The new eigenfuctions are thus
1√2
(ψ112 + ψ211), ψ121,1√2
(ψ112 − ψ211)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 12 / 20
Problem 6.8 (cont.)
In order to find the eigen-functions of the “D” matrix,solve the eigenvalue equation
this leads to two equationsonly since ψb is not coupledto either of the other twostates
first choose λ = 0, thenchoose λ = 2
1 0 −10 0 0−1 0 1
αβγ
= λ
αβγ
α− γ = λα, γ − α = λγ
λ = 0 −→ α = γ
λ = 0 −→ α = −γ
The new eigenfuctions are thus
1√2
(ψ112 + ψ211), ψ121,1√2
(ψ112 − ψ211)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 12 / 20
Problem 6.8 (cont.)
In order to find the eigen-functions of the “D” matrix,solve the eigenvalue equation
this leads to two equationsonly since ψb is not coupledto either of the other twostates
first choose λ = 0, thenchoose λ = 2
1 0 −10 0 0−1 0 1
αβγ
= λ
αβγ
α− γ = λα, γ − α = λγ
λ = 0 −→ α = γ
λ = 0 −→ α = −γ
The new eigenfuctions are thus
1√2
(ψ112 + ψ211), ψ121,1√2
(ψ112 − ψ211)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 12 / 20
Problem 6.8 (cont.)
In order to find the eigen-functions of the “D” matrix,solve the eigenvalue equation
this leads to two equationsonly since ψb is not coupledto either of the other twostates
first choose λ = 0, thenchoose λ = 2
1 0 −10 0 0−1 0 1
αβγ
= λ
αβγ
α− γ = λα, γ − α = λγ
λ = 0 −→ α = γ
λ = 0 −→ α = −γ
The new eigenfuctions are thus
1√2
(ψ112 + ψ211), ψ121,1√2
(ψ112 − ψ211)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 12 / 20
Problem 6.8 (cont.)
In order to find the eigen-functions of the “D” matrix,solve the eigenvalue equation
this leads to two equationsonly since ψb is not coupledto either of the other twostates
first choose λ = 0, thenchoose λ = 2
1 0 −10 0 0−1 0 1
αβγ
= λ
αβγ
α− γ = λα, γ − α = λγ
λ = 0 −→ α = γ
λ = 0 −→ α = −γ
The new eigenfuctions are thus
1√2
(ψ112 + ψ211), ψ121,1√2
(ψ112 − ψ211)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 12 / 20
Problem 6.8 (cont.)
In order to find the eigen-functions of the “D” matrix,solve the eigenvalue equation
this leads to two equationsonly since ψb is not coupledto either of the other twostates
first choose λ = 0, thenchoose λ = 2
1 0 −10 0 0−1 0 1
αβγ
= λ
αβγ
α− γ = λα, γ − α = λγ
λ = 0 −→ α = γ
λ = 0 −→ α = −γ
The new eigenfuctions are thus
1√2
(ψ112 + ψ211), ψ121,1√2
(ψ112 − ψ211)
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 12 / 20
Problem 6.16
Evaluate the following commutators:
[~L · ~S ,~L], [~L · ~S , ~S ], [~L · ~S ,~J], [~L · ~S , L2], [~L · ~S ,S2], [~L · ~S , J2]
For [~L · ~S ,~L], start by breaking into components
[~L · ~S , Lx ] = [(LxSx + LySy + LzSz), Lx ]
= Sx [Lx , Lx ] + Sy [Ly , Lx ] + Sz [Lz , Lx ]
= Sx(0) + Sy (−i~Lz) + Sz(i~Ly ) = i~(LySz − LzSy )
= i~(~L× ~S)x
[~L · ~S , Ly ] = i~(~L× ~S)y
[~L · ~S , Lz ] = i~(~L× ~S)z −→ [~L · ~S ,~L] = i~(~L× ~S)
[~L · ~S , ~S ] = i~(~S × ~L)
[~L · ~S ,~J] = [~L · ~S ,~L] + [~L · ~S , ~S ] = i~(~L× ~S + ~S × ~L) = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 13 / 20
Problem 6.16
Evaluate the following commutators:
[~L · ~S ,~L], [~L · ~S , ~S ], [~L · ~S ,~J], [~L · ~S , L2], [~L · ~S ,S2], [~L · ~S , J2]
For [~L · ~S ,~L], start by breaking into components
[~L · ~S , Lx ] = [(LxSx + LySy + LzSz), Lx ]
= Sx [Lx , Lx ] + Sy [Ly , Lx ] + Sz [Lz , Lx ]
= Sx(0) + Sy (−i~Lz) + Sz(i~Ly ) = i~(LySz − LzSy )
= i~(~L× ~S)x
[~L · ~S , Ly ] = i~(~L× ~S)y
[~L · ~S , Lz ] = i~(~L× ~S)z −→ [~L · ~S ,~L] = i~(~L× ~S)
[~L · ~S , ~S ] = i~(~S × ~L)
[~L · ~S ,~J] = [~L · ~S ,~L] + [~L · ~S , ~S ] = i~(~L× ~S + ~S × ~L) = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 13 / 20
Problem 6.16
Evaluate the following commutators:
[~L · ~S ,~L], [~L · ~S , ~S ], [~L · ~S ,~J], [~L · ~S , L2], [~L · ~S ,S2], [~L · ~S , J2]
For [~L · ~S ,~L], start by breaking into components
[~L · ~S , Lx ]
= [(LxSx + LySy + LzSz), Lx ]
= Sx [Lx , Lx ] + Sy [Ly , Lx ] + Sz [Lz , Lx ]
= Sx(0) + Sy (−i~Lz) + Sz(i~Ly ) = i~(LySz − LzSy )
= i~(~L× ~S)x
[~L · ~S , Ly ] = i~(~L× ~S)y
[~L · ~S , Lz ] = i~(~L× ~S)z −→ [~L · ~S ,~L] = i~(~L× ~S)
[~L · ~S , ~S ] = i~(~S × ~L)
[~L · ~S ,~J] = [~L · ~S ,~L] + [~L · ~S , ~S ] = i~(~L× ~S + ~S × ~L) = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 13 / 20
Problem 6.16
Evaluate the following commutators:
[~L · ~S ,~L], [~L · ~S , ~S ], [~L · ~S ,~J], [~L · ~S , L2], [~L · ~S ,S2], [~L · ~S , J2]
For [~L · ~S ,~L], start by breaking into components
[~L · ~S , Lx ] = [(LxSx + LySy + LzSz), Lx ]
= Sx [Lx , Lx ] + Sy [Ly , Lx ] + Sz [Lz , Lx ]
= Sx(0) + Sy (−i~Lz) + Sz(i~Ly ) = i~(LySz − LzSy )
= i~(~L× ~S)x
[~L · ~S , Ly ] = i~(~L× ~S)y
[~L · ~S , Lz ] = i~(~L× ~S)z −→ [~L · ~S ,~L] = i~(~L× ~S)
[~L · ~S , ~S ] = i~(~S × ~L)
[~L · ~S ,~J] = [~L · ~S ,~L] + [~L · ~S , ~S ] = i~(~L× ~S + ~S × ~L) = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 13 / 20
Problem 6.16
Evaluate the following commutators:
[~L · ~S ,~L], [~L · ~S , ~S ], [~L · ~S ,~J], [~L · ~S , L2], [~L · ~S ,S2], [~L · ~S , J2]
For [~L · ~S ,~L], start by breaking into components
[~L · ~S , Lx ] = [(LxSx + LySy + LzSz), Lx ]
= Sx [Lx , Lx ] + Sy [Ly , Lx ] + Sz [Lz , Lx ]
= Sx(0) + Sy (−i~Lz) + Sz(i~Ly ) = i~(LySz − LzSy )
= i~(~L× ~S)x
[~L · ~S , Ly ] = i~(~L× ~S)y
[~L · ~S , Lz ] = i~(~L× ~S)z −→ [~L · ~S ,~L] = i~(~L× ~S)
[~L · ~S , ~S ] = i~(~S × ~L)
[~L · ~S ,~J] = [~L · ~S ,~L] + [~L · ~S , ~S ] = i~(~L× ~S + ~S × ~L) = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 13 / 20
Problem 6.16
Evaluate the following commutators:
[~L · ~S ,~L], [~L · ~S , ~S ], [~L · ~S ,~J], [~L · ~S , L2], [~L · ~S ,S2], [~L · ~S , J2]
For [~L · ~S ,~L], start by breaking into components
[~L · ~S , Lx ] = [(LxSx + LySy + LzSz), Lx ]
= Sx [Lx , Lx ] + Sy [Ly , Lx ] + Sz [Lz , Lx ]
= Sx(0) + Sy (−i~Lz) + Sz(i~Ly )
= i~(LySz − LzSy )
= i~(~L× ~S)x
[~L · ~S , Ly ] = i~(~L× ~S)y
[~L · ~S , Lz ] = i~(~L× ~S)z −→ [~L · ~S ,~L] = i~(~L× ~S)
[~L · ~S , ~S ] = i~(~S × ~L)
[~L · ~S ,~J] = [~L · ~S ,~L] + [~L · ~S , ~S ] = i~(~L× ~S + ~S × ~L) = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 13 / 20
Problem 6.16
Evaluate the following commutators:
[~L · ~S ,~L], [~L · ~S , ~S ], [~L · ~S ,~J], [~L · ~S , L2], [~L · ~S ,S2], [~L · ~S , J2]
For [~L · ~S ,~L], start by breaking into components
[~L · ~S , Lx ] = [(LxSx + LySy + LzSz), Lx ]
= Sx [Lx , Lx ] + Sy [Ly , Lx ] + Sz [Lz , Lx ]
= Sx(0) + Sy (−i~Lz) + Sz(i~Ly ) = i~(LySz − LzSy )
= i~(~L× ~S)x
[~L · ~S , Ly ] = i~(~L× ~S)y
[~L · ~S , Lz ] = i~(~L× ~S)z −→ [~L · ~S ,~L] = i~(~L× ~S)
[~L · ~S , ~S ] = i~(~S × ~L)
[~L · ~S ,~J] = [~L · ~S ,~L] + [~L · ~S , ~S ] = i~(~L× ~S + ~S × ~L) = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 13 / 20
Problem 6.16
Evaluate the following commutators:
[~L · ~S ,~L], [~L · ~S , ~S ], [~L · ~S ,~J], [~L · ~S , L2], [~L · ~S ,S2], [~L · ~S , J2]
For [~L · ~S ,~L], start by breaking into components
[~L · ~S , Lx ] = [(LxSx + LySy + LzSz), Lx ]
= Sx [Lx , Lx ] + Sy [Ly , Lx ] + Sz [Lz , Lx ]
= Sx(0) + Sy (−i~Lz) + Sz(i~Ly ) = i~(LySz − LzSy )
= i~(~L× ~S)x
[~L · ~S , Ly ] = i~(~L× ~S)y
[~L · ~S , Lz ] = i~(~L× ~S)z −→ [~L · ~S ,~L] = i~(~L× ~S)
[~L · ~S , ~S ] = i~(~S × ~L)
[~L · ~S ,~J] = [~L · ~S ,~L] + [~L · ~S , ~S ] = i~(~L× ~S + ~S × ~L) = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 13 / 20
Problem 6.16
Evaluate the following commutators:
[~L · ~S ,~L], [~L · ~S , ~S ], [~L · ~S ,~J], [~L · ~S , L2], [~L · ~S ,S2], [~L · ~S , J2]
For [~L · ~S ,~L], start by breaking into components
[~L · ~S , Lx ] = [(LxSx + LySy + LzSz), Lx ]
= Sx [Lx , Lx ] + Sy [Ly , Lx ] + Sz [Lz , Lx ]
= Sx(0) + Sy (−i~Lz) + Sz(i~Ly ) = i~(LySz − LzSy )
= i~(~L× ~S)x
[~L · ~S , Ly ] = i~(~L× ~S)y
[~L · ~S , Lz ] = i~(~L× ~S)z −→ [~L · ~S ,~L] = i~(~L× ~S)
[~L · ~S , ~S ] = i~(~S × ~L)
[~L · ~S ,~J] = [~L · ~S ,~L] + [~L · ~S , ~S ] = i~(~L× ~S + ~S × ~L) = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 13 / 20
Problem 6.16
Evaluate the following commutators:
[~L · ~S ,~L], [~L · ~S , ~S ], [~L · ~S ,~J], [~L · ~S , L2], [~L · ~S ,S2], [~L · ~S , J2]
For [~L · ~S ,~L], start by breaking into components
[~L · ~S , Lx ] = [(LxSx + LySy + LzSz), Lx ]
= Sx [Lx , Lx ] + Sy [Ly , Lx ] + Sz [Lz , Lx ]
= Sx(0) + Sy (−i~Lz) + Sz(i~Ly ) = i~(LySz − LzSy )
= i~(~L× ~S)x
[~L · ~S , Ly ] = i~(~L× ~S)y
[~L · ~S , Lz ] = i~(~L× ~S)z
−→ [~L · ~S ,~L] = i~(~L× ~S)
[~L · ~S , ~S ] = i~(~S × ~L)
[~L · ~S ,~J] = [~L · ~S ,~L] + [~L · ~S , ~S ] = i~(~L× ~S + ~S × ~L) = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 13 / 20
Problem 6.16
Evaluate the following commutators:
[~L · ~S ,~L], [~L · ~S , ~S ], [~L · ~S ,~J], [~L · ~S , L2], [~L · ~S ,S2], [~L · ~S , J2]
For [~L · ~S ,~L], start by breaking into components
[~L · ~S , Lx ] = [(LxSx + LySy + LzSz), Lx ]
= Sx [Lx , Lx ] + Sy [Ly , Lx ] + Sz [Lz , Lx ]
= Sx(0) + Sy (−i~Lz) + Sz(i~Ly ) = i~(LySz − LzSy )
= i~(~L× ~S)x
[~L · ~S , Ly ] = i~(~L× ~S)y
[~L · ~S , Lz ] = i~(~L× ~S)z −→ [~L · ~S ,~L] = i~(~L× ~S)
[~L · ~S , ~S ] = i~(~S × ~L)
[~L · ~S ,~J] = [~L · ~S ,~L] + [~L · ~S , ~S ] = i~(~L× ~S + ~S × ~L) = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 13 / 20
Problem 6.16
Evaluate the following commutators:
[~L · ~S ,~L], [~L · ~S , ~S ], [~L · ~S ,~J], [~L · ~S , L2], [~L · ~S ,S2], [~L · ~S , J2]
For [~L · ~S ,~L], start by breaking into components
[~L · ~S , Lx ] = [(LxSx + LySy + LzSz), Lx ]
= Sx [Lx , Lx ] + Sy [Ly , Lx ] + Sz [Lz , Lx ]
= Sx(0) + Sy (−i~Lz) + Sz(i~Ly ) = i~(LySz − LzSy )
= i~(~L× ~S)x
[~L · ~S , Ly ] = i~(~L× ~S)y
[~L · ~S , Lz ] = i~(~L× ~S)z −→ [~L · ~S ,~L] = i~(~L× ~S)
[~L · ~S , ~S ] = i~(~S × ~L)
[~L · ~S ,~J] = [~L · ~S ,~L] + [~L · ~S , ~S ] = i~(~L× ~S + ~S × ~L) = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 13 / 20
Problem 6.16
Evaluate the following commutators:
[~L · ~S ,~L], [~L · ~S , ~S ], [~L · ~S ,~J], [~L · ~S , L2], [~L · ~S ,S2], [~L · ~S , J2]
For [~L · ~S ,~L], start by breaking into components
[~L · ~S , Lx ] = [(LxSx + LySy + LzSz), Lx ]
= Sx [Lx , Lx ] + Sy [Ly , Lx ] + Sz [Lz , Lx ]
= Sx(0) + Sy (−i~Lz) + Sz(i~Ly ) = i~(LySz − LzSy )
= i~(~L× ~S)x
[~L · ~S , Ly ] = i~(~L× ~S)y
[~L · ~S , Lz ] = i~(~L× ~S)z −→ [~L · ~S ,~L] = i~(~L× ~S)
[~L · ~S , ~S ] = i~(~S × ~L)
[~L · ~S ,~J]
= [~L · ~S ,~L] + [~L · ~S , ~S ] = i~(~L× ~S + ~S × ~L) = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 13 / 20
Problem 6.16
Evaluate the following commutators:
[~L · ~S ,~L], [~L · ~S , ~S ], [~L · ~S ,~J], [~L · ~S , L2], [~L · ~S ,S2], [~L · ~S , J2]
For [~L · ~S ,~L], start by breaking into components
[~L · ~S , Lx ] = [(LxSx + LySy + LzSz), Lx ]
= Sx [Lx , Lx ] + Sy [Ly , Lx ] + Sz [Lz , Lx ]
= Sx(0) + Sy (−i~Lz) + Sz(i~Ly ) = i~(LySz − LzSy )
= i~(~L× ~S)x
[~L · ~S , Ly ] = i~(~L× ~S)y
[~L · ~S , Lz ] = i~(~L× ~S)z −→ [~L · ~S ,~L] = i~(~L× ~S)
[~L · ~S , ~S ] = i~(~S × ~L)
[~L · ~S ,~J] = [~L · ~S ,~L] + [~L · ~S , ~S ]
= i~(~L× ~S + ~S × ~L) = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 13 / 20
Problem 6.16
Evaluate the following commutators:
[~L · ~S ,~L], [~L · ~S , ~S ], [~L · ~S ,~J], [~L · ~S , L2], [~L · ~S ,S2], [~L · ~S , J2]
For [~L · ~S ,~L], start by breaking into components
[~L · ~S , Lx ] = [(LxSx + LySy + LzSz), Lx ]
= Sx [Lx , Lx ] + Sy [Ly , Lx ] + Sz [Lz , Lx ]
= Sx(0) + Sy (−i~Lz) + Sz(i~Ly ) = i~(LySz − LzSy )
= i~(~L× ~S)x
[~L · ~S , Ly ] = i~(~L× ~S)y
[~L · ~S , Lz ] = i~(~L× ~S)z −→ [~L · ~S ,~L] = i~(~L× ~S)
[~L · ~S , ~S ] = i~(~S × ~L)
[~L · ~S ,~J] = [~L · ~S ,~L] + [~L · ~S , ~S ] = i~(~L× ~S + ~S × ~L)
= 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 13 / 20
Problem 6.16
Evaluate the following commutators:
[~L · ~S ,~L], [~L · ~S , ~S ], [~L · ~S ,~J], [~L · ~S , L2], [~L · ~S ,S2], [~L · ~S , J2]
For [~L · ~S ,~L], start by breaking into components
[~L · ~S , Lx ] = [(LxSx + LySy + LzSz), Lx ]
= Sx [Lx , Lx ] + Sy [Ly , Lx ] + Sz [Lz , Lx ]
= Sx(0) + Sy (−i~Lz) + Sz(i~Ly ) = i~(LySz − LzSy )
= i~(~L× ~S)x
[~L · ~S , Ly ] = i~(~L× ~S)y
[~L · ~S , Lz ] = i~(~L× ~S)z −→ [~L · ~S ,~L] = i~(~L× ~S)
[~L · ~S , ~S ] = i~(~S × ~L)
[~L · ~S ,~J] = [~L · ~S ,~L] + [~L · ~S , ~S ] = i~(~L× ~S + ~S × ~L) = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 13 / 20
Problem 6.16 (cont.)
[~L · ~S , L2]
= [(LxSx + LySy + LzSz), L2] = 0
[~L · ~S , S2] = [(LxSx + LySy + LzSz),S2] = 0
[~L · ~S , J2] = [~L · ~S , L2] + [~L · ~S ,S2] + 2[~L · ~S ,~L · ~S ] = 0 + 0 + 0 = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 14 / 20
Problem 6.16 (cont.)
[~L · ~S , L2] = [(LxSx + LySy + LzSz), L2]
= 0
[~L · ~S , S2] = [(LxSx + LySy + LzSz),S2] = 0
[~L · ~S , J2] = [~L · ~S , L2] + [~L · ~S ,S2] + 2[~L · ~S ,~L · ~S ] = 0 + 0 + 0 = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 14 / 20
Problem 6.16 (cont.)
[~L · ~S , L2] = [(LxSx + LySy + LzSz), L2] = 0
[~L · ~S , S2] = [(LxSx + LySy + LzSz),S2] = 0
[~L · ~S , J2] = [~L · ~S , L2] + [~L · ~S ,S2] + 2[~L · ~S ,~L · ~S ] = 0 + 0 + 0 = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 14 / 20
Problem 6.16 (cont.)
[~L · ~S , L2] = [(LxSx + LySy + LzSz), L2] = 0
[~L · ~S , S2]
= [(LxSx + LySy + LzSz),S2] = 0
[~L · ~S , J2] = [~L · ~S , L2] + [~L · ~S ,S2] + 2[~L · ~S ,~L · ~S ] = 0 + 0 + 0 = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 14 / 20
Problem 6.16 (cont.)
[~L · ~S , L2] = [(LxSx + LySy + LzSz), L2] = 0
[~L · ~S , S2] = [(LxSx + LySy + LzSz),S2]
= 0
[~L · ~S , J2] = [~L · ~S , L2] + [~L · ~S ,S2] + 2[~L · ~S ,~L · ~S ] = 0 + 0 + 0 = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 14 / 20
Problem 6.16 (cont.)
[~L · ~S , L2] = [(LxSx + LySy + LzSz), L2] = 0
[~L · ~S , S2] = [(LxSx + LySy + LzSz),S2] = 0
[~L · ~S , J2] = [~L · ~S , L2] + [~L · ~S ,S2] + 2[~L · ~S ,~L · ~S ] = 0 + 0 + 0 = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 14 / 20
Problem 6.16 (cont.)
[~L · ~S , L2] = [(LxSx + LySy + LzSz), L2] = 0
[~L · ~S , S2] = [(LxSx + LySy + LzSz),S2] = 0
[~L · ~S , J2]
= [~L · ~S , L2] + [~L · ~S ,S2] + 2[~L · ~S ,~L · ~S ] = 0 + 0 + 0 = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 14 / 20
Problem 6.16 (cont.)
[~L · ~S , L2] = [(LxSx + LySy + LzSz), L2] = 0
[~L · ~S , S2] = [(LxSx + LySy + LzSz),S2] = 0
[~L · ~S , J2] = [~L · ~S , L2]
+ [~L · ~S ,S2] + 2[~L · ~S ,~L · ~S ] = 0 + 0 + 0 = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 14 / 20
Problem 6.16 (cont.)
[~L · ~S , L2] = [(LxSx + LySy + LzSz), L2] = 0
[~L · ~S , S2] = [(LxSx + LySy + LzSz),S2] = 0
[~L · ~S , J2] = [~L · ~S , L2] + [~L · ~S ,S2]
+ 2[~L · ~S ,~L · ~S ] = 0 + 0 + 0 = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 14 / 20
Problem 6.16 (cont.)
[~L · ~S , L2] = [(LxSx + LySy + LzSz), L2] = 0
[~L · ~S , S2] = [(LxSx + LySy + LzSz),S2] = 0
[~L · ~S , J2] = [~L · ~S , L2] + [~L · ~S ,S2] + 2[~L · ~S ,~L · ~S ]
= 0 + 0 + 0 = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 14 / 20
Problem 6.16 (cont.)
[~L · ~S , L2] = [(LxSx + LySy + LzSz), L2] = 0
[~L · ~S , S2] = [(LxSx + LySy + LzSz),S2] = 0
[~L · ~S , J2] = [~L · ~S , L2] + [~L · ~S ,S2] + 2[~L · ~S ,~L · ~S ] = 0 + 0 + 0 = 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 14 / 20
Zeeman effect
When an atom is in a uniform mag-netic field ~Bext , the energy levelsare shifted by the Zeeman effect
~µs = − e
m~S ~µl = − e
2m~L
H ′Z = −(~µl + ~µs) · ~Bext
H ′Z =e
2m(~L + 2~S) · ~Bext
The nature of the Zeeman effect is dependent on the relative strengths ofthe external and internal (spin-orbit) magnetic fields
Bext � Bint weak-fieldBext ≈ Bint intermediate-fieldBext � Bint strong-field
depending on the regime, we can use different kinds of perturbation theory
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 15 / 20
Zeeman effect
When an atom is in a uniform mag-netic field ~Bext , the energy levelsare shifted by the Zeeman effect
~µs = − e
m~S ~µl = − e
2m~L
H ′Z = −(~µl + ~µs) · ~Bext
H ′Z =e
2m(~L + 2~S) · ~Bext
The nature of the Zeeman effect is dependent on the relative strengths ofthe external and internal (spin-orbit) magnetic fields
Bext � Bint weak-fieldBext ≈ Bint intermediate-fieldBext � Bint strong-field
depending on the regime, we can use different kinds of perturbation theory
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 15 / 20
Zeeman effect
When an atom is in a uniform mag-netic field ~Bext , the energy levelsare shifted by the Zeeman effect
~µs = − e
m~S
~µl = − e
2m~L
H ′Z = −(~µl + ~µs) · ~Bext
H ′Z =e
2m(~L + 2~S) · ~Bext
The nature of the Zeeman effect is dependent on the relative strengths ofthe external and internal (spin-orbit) magnetic fields
Bext � Bint weak-fieldBext ≈ Bint intermediate-fieldBext � Bint strong-field
depending on the regime, we can use different kinds of perturbation theory
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 15 / 20
Zeeman effect
When an atom is in a uniform mag-netic field ~Bext , the energy levelsare shifted by the Zeeman effect
~µs = − e
m~S ~µl = − e
2m~L
H ′Z = −(~µl + ~µs) · ~Bext
H ′Z =e
2m(~L + 2~S) · ~Bext
The nature of the Zeeman effect is dependent on the relative strengths ofthe external and internal (spin-orbit) magnetic fields
Bext � Bint weak-fieldBext ≈ Bint intermediate-fieldBext � Bint strong-field
depending on the regime, we can use different kinds of perturbation theory
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 15 / 20
Zeeman effect
When an atom is in a uniform mag-netic field ~Bext , the energy levelsare shifted by the Zeeman effect
~µs = − e
m~S ~µl = − e
2m~L
H ′Z = −(~µl + ~µs) · ~Bext
H ′Z =e
2m(~L + 2~S) · ~Bext
The nature of the Zeeman effect is dependent on the relative strengths ofthe external and internal (spin-orbit) magnetic fields
Bext � Bint weak-fieldBext ≈ Bint intermediate-fieldBext � Bint strong-field
depending on the regime, we can use different kinds of perturbation theory
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 15 / 20
Zeeman effect
When an atom is in a uniform mag-netic field ~Bext , the energy levelsare shifted by the Zeeman effect
~µs = − e
m~S ~µl = − e
2m~L
H ′Z = −(~µl + ~µs) · ~Bext
H ′Z =e
2m(~L + 2~S) · ~Bext
The nature of the Zeeman effect is dependent on the relative strengths ofthe external and internal (spin-orbit) magnetic fields
Bext � Bint weak-fieldBext ≈ Bint intermediate-fieldBext � Bint strong-field
depending on the regime, we can use different kinds of perturbation theory
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 15 / 20
Zeeman effect
When an atom is in a uniform mag-netic field ~Bext , the energy levelsare shifted by the Zeeman effect
~µs = − e
m~S ~µl = − e
2m~L
H ′Z = −(~µl + ~µs) · ~Bext
H ′Z =e
2m(~L + 2~S) · ~Bext
The nature of the Zeeman effect is dependent on the relative strengths ofthe external and internal (spin-orbit) magnetic fields
Bext � Bint weak-field
Bext ≈ Bint intermediate-fieldBext � Bint strong-field
depending on the regime, we can use different kinds of perturbation theory
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 15 / 20
Zeeman effect
When an atom is in a uniform mag-netic field ~Bext , the energy levelsare shifted by the Zeeman effect
~µs = − e
m~S ~µl = − e
2m~L
H ′Z = −(~µl + ~µs) · ~Bext
H ′Z =e
2m(~L + 2~S) · ~Bext
The nature of the Zeeman effect is dependent on the relative strengths ofthe external and internal (spin-orbit) magnetic fields
Bext � Bint weak-fieldBext ≈ Bint intermediate-field
Bext � Bint strong-field
depending on the regime, we can use different kinds of perturbation theory
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 15 / 20
Zeeman effect
When an atom is in a uniform mag-netic field ~Bext , the energy levelsare shifted by the Zeeman effect
~µs = − e
m~S ~µl = − e
2m~L
H ′Z = −(~µl + ~µs) · ~Bext
H ′Z =e
2m(~L + 2~S) · ~Bext
The nature of the Zeeman effect is dependent on the relative strengths ofthe external and internal (spin-orbit) magnetic fields
Bext � Bint weak-fieldBext ≈ Bint intermediate-fieldBext � Bint strong-field
depending on the regime, we can use different kinds of perturbation theory
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 15 / 20
Zeeman effect
When an atom is in a uniform mag-netic field ~Bext , the energy levelsare shifted by the Zeeman effect
~µs = − e
m~S ~µl = − e
2m~L
H ′Z = −(~µl + ~µs) · ~Bext
H ′Z =e
2m(~L + 2~S) · ~Bext
The nature of the Zeeman effect is dependent on the relative strengths ofthe external and internal (spin-orbit) magnetic fields
Bext � Bint weak-fieldBext ≈ Bint intermediate-fieldBext � Bint strong-field
depending on the regime, we can use different kinds of perturbation theory
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 15 / 20
Problem 6.20
Consider the following relationship between angular momentum andmagnetic field:
~B =1
4πε0
e
mc2r3~L
use it to estimate the internal field in hydrogen, and characterizequantitatively a “strong” and “weak” Zeeman field.
The internal field that gives rise tothe spin-orbit interactions is givenby
if we assume reasonable values forfor L and r the crossover point forthe Zeeman effect can be estimated
~Bint =1
4πε0
e
mc2r3~L
L ≈ ~r ≈ a
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 16 / 20
Problem 6.20
Consider the following relationship between angular momentum andmagnetic field:
~B =1
4πε0
e
mc2r3~L
use it to estimate the internal field in hydrogen, and characterizequantitatively a “strong” and “weak” Zeeman field.
The internal field that gives rise tothe spin-orbit interactions is givenby
if we assume reasonable values forfor L and r the crossover point forthe Zeeman effect can be estimated
~Bint =1
4πε0
e
mc2r3~L
L ≈ ~r ≈ a
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 16 / 20
Problem 6.20
Consider the following relationship between angular momentum andmagnetic field:
~B =1
4πε0
e
mc2r3~L
use it to estimate the internal field in hydrogen, and characterizequantitatively a “strong” and “weak” Zeeman field.
The internal field that gives rise tothe spin-orbit interactions is givenby
if we assume reasonable values forfor L and r the crossover point forthe Zeeman effect can be estimated
~Bint =1
4πε0
e
mc2r3~L
L ≈ ~r ≈ a
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 16 / 20
Problem 6.20
Consider the following relationship between angular momentum andmagnetic field:
~B =1
4πε0
e
mc2r3~L
use it to estimate the internal field in hydrogen, and characterizequantitatively a “strong” and “weak” Zeeman field.
The internal field that gives rise tothe spin-orbit interactions is givenby
if we assume reasonable values forfor L and r
the crossover point forthe Zeeman effect can be estimated
~Bint =1
4πε0
e
mc2r3~L
L ≈ ~r ≈ a
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 16 / 20
Problem 6.20
Consider the following relationship between angular momentum andmagnetic field:
~B =1
4πε0
e
mc2r3~L
use it to estimate the internal field in hydrogen, and characterizequantitatively a “strong” and “weak” Zeeman field.
The internal field that gives rise tothe spin-orbit interactions is givenby
if we assume reasonable values forfor L and r
the crossover point forthe Zeeman effect can be estimated
~Bint =1
4πε0
e
mc2r3~L
L ≈ ~r ≈ a
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 16 / 20
Problem 6.20
Consider the following relationship between angular momentum andmagnetic field:
~B =1
4πε0
e
mc2r3~L
use it to estimate the internal field in hydrogen, and characterizequantitatively a “strong” and “weak” Zeeman field.
The internal field that gives rise tothe spin-orbit interactions is givenby
if we assume reasonable values forfor L and r the crossover point forthe Zeeman effect can be estimated
~Bint =1
4πε0
e
mc2r3~L
L ≈ ~r ≈ a
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 16 / 20
Problem 6.20 (cont.)
Bint ≈1
4πε0
e
mc2a3=
12T
(1.60× 10−19C)(1.05× 10−34J · s)
4π(8.9× 10−12C2/N ·m2)(9.1× 10−31kg)(3× 108m/s2)2(0.53× 10−10m)3
a “strong” Zeeman field is Bext � 10T
a “weak” Zeeman field is Bext � 10T
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 17 / 20
Problem 6.20 (cont.)
Bint ≈1
4πε0
e
mc2a3=
12T
(1.60× 10−19C)(1.05× 10−34J · s)
4π(8.9× 10−12C2/N ·m2)(9.1× 10−31kg)(3× 108m/s2)2(0.53× 10−10m)3
a “strong” Zeeman field is Bext � 10T
a “weak” Zeeman field is Bext � 10T
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 17 / 20
Problem 6.20 (cont.)
Bint ≈1
4πε0
e
mc2a3= 12T
(1.60× 10−19C)(1.05× 10−34J · s)
4π(8.9× 10−12C2/N ·m2)(9.1× 10−31kg)(3× 108m/s2)2(0.53× 10−10m)3
a “strong” Zeeman field is Bext � 10T
a “weak” Zeeman field is Bext � 10T
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 17 / 20
Problem 6.20 (cont.)
Bint ≈1
4πε0
e
mc2a3= 12T
(1.60× 10−19C)(1.05× 10−34J · s)
4π(8.9× 10−12C2/N ·m2)(9.1× 10−31kg)(3× 108m/s2)2(0.53× 10−10m)3
a “strong” Zeeman field is Bext � 10T
a “weak” Zeeman field is Bext � 10T
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 17 / 20
Problem 6.20 (cont.)
Bint ≈1
4πε0
e
mc2a3= 12T
(1.60× 10−19C)(1.05× 10−34J · s)
4π(8.9× 10−12C2/N ·m2)(9.1× 10−31kg)(3× 108m/s2)2(0.53× 10−10m)3
a “strong” Zeeman field is Bext � 10T
a “weak” Zeeman field is Bext � 10T
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 17 / 20
Weak-field Zeeman effect
When Bext � Bint fine structure dominates and the good quantumnumbers are n, l , j , mj
apply first order perturbation the-ory to get
we can rewrite this using ~J = ~L+~S
this can be evaluated by realizingthat ~J is constant and that the ti-ime average of ~S is
E(1)Z =
⟨nljmj
∣∣H ′Z ∣∣ nljmj
⟩=
e
2m~Bext ·
⟨~L + 2~S
⟩=
e
2m~Bext ·
⟨~J + ~S
⟩
~Save =~S · ~JJ2
~J
~L = ~J − ~S → L2 = J2 + S2 − 2~J · ~S
~S · ~J =1
2(J2 + S2 − L2) =
~2
2[j(j + 1) + s(s + 1)− l(l + 1)]
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 18 / 20
Weak-field Zeeman effect
When Bext � Bint fine structure dominates and the good quantumnumbers are n, l , j , mj
apply first order perturbation the-ory to get
we can rewrite this using ~J = ~L+~S
this can be evaluated by realizingthat ~J is constant and that the ti-ime average of ~S is
E(1)Z =
⟨nljmj
∣∣H ′Z ∣∣ nljmj
⟩=
e
2m~Bext ·
⟨~L + 2~S
⟩=
e
2m~Bext ·
⟨~J + ~S
⟩
~Save =~S · ~JJ2
~J
~L = ~J − ~S → L2 = J2 + S2 − 2~J · ~S
~S · ~J =1
2(J2 + S2 − L2) =
~2
2[j(j + 1) + s(s + 1)− l(l + 1)]
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 18 / 20
Weak-field Zeeman effect
When Bext � Bint fine structure dominates and the good quantumnumbers are n, l , j , mj
apply first order perturbation the-ory to get
we can rewrite this using ~J = ~L+~S
this can be evaluated by realizingthat ~J is constant and that the ti-ime average of ~S is
E(1)Z =
⟨nljmj
∣∣H ′Z ∣∣ nljmj
⟩
=e
2m~Bext ·
⟨~L + 2~S
⟩=
e
2m~Bext ·
⟨~J + ~S
⟩
~Save =~S · ~JJ2
~J
~L = ~J − ~S → L2 = J2 + S2 − 2~J · ~S
~S · ~J =1
2(J2 + S2 − L2) =
~2
2[j(j + 1) + s(s + 1)− l(l + 1)]
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 18 / 20
Weak-field Zeeman effect
When Bext � Bint fine structure dominates and the good quantumnumbers are n, l , j , mj
apply first order perturbation the-ory to get
we can rewrite this using ~J = ~L+~S
this can be evaluated by realizingthat ~J is constant and that the ti-ime average of ~S is
E(1)Z =
⟨nljmj
∣∣H ′Z ∣∣ nljmj
⟩=
e
2m~Bext ·
⟨~L + 2~S
⟩
=e
2m~Bext ·
⟨~J + ~S
⟩
~Save =~S · ~JJ2
~J
~L = ~J − ~S → L2 = J2 + S2 − 2~J · ~S
~S · ~J =1
2(J2 + S2 − L2) =
~2
2[j(j + 1) + s(s + 1)− l(l + 1)]
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 18 / 20
Weak-field Zeeman effect
When Bext � Bint fine structure dominates and the good quantumnumbers are n, l , j , mj
apply first order perturbation the-ory to get
we can rewrite this using ~J = ~L+~S
this can be evaluated by realizingthat ~J is constant and that the ti-ime average of ~S is
E(1)Z =
⟨nljmj
∣∣H ′Z ∣∣ nljmj
⟩=
e
2m~Bext ·
⟨~L + 2~S
⟩
=e
2m~Bext ·
⟨~J + ~S
⟩
~Save =~S · ~JJ2
~J
~L = ~J − ~S → L2 = J2 + S2 − 2~J · ~S
~S · ~J =1
2(J2 + S2 − L2) =
~2
2[j(j + 1) + s(s + 1)− l(l + 1)]
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 18 / 20
Weak-field Zeeman effect
When Bext � Bint fine structure dominates and the good quantumnumbers are n, l , j , mj
apply first order perturbation the-ory to get
we can rewrite this using ~J = ~L+~S
this can be evaluated by realizingthat ~J is constant and that the ti-ime average of ~S is
E(1)Z =
⟨nljmj
∣∣H ′Z ∣∣ nljmj
⟩=
e
2m~Bext ·
⟨~L + 2~S
⟩=
e
2m~Bext ·
⟨~J + ~S
⟩
~Save =~S · ~JJ2
~J
~L = ~J − ~S → L2 = J2 + S2 − 2~J · ~S
~S · ~J =1
2(J2 + S2 − L2) =
~2
2[j(j + 1) + s(s + 1)− l(l + 1)]
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 18 / 20
Weak-field Zeeman effect
When Bext � Bint fine structure dominates and the good quantumnumbers are n, l , j , mj
apply first order perturbation the-ory to get
we can rewrite this using ~J = ~L+~S
this can be evaluated by realizingthat ~J is constant and that the ti-ime average of ~S is
E(1)Z =
⟨nljmj
∣∣H ′Z ∣∣ nljmj
⟩=
e
2m~Bext ·
⟨~L + 2~S
⟩=
e
2m~Bext ·
⟨~J + ~S
⟩
~Save =~S · ~JJ2
~J
~L = ~J − ~S → L2 = J2 + S2 − 2~J · ~S
~S · ~J =1
2(J2 + S2 − L2) =
~2
2[j(j + 1) + s(s + 1)− l(l + 1)]
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 18 / 20
Weak-field Zeeman effect
When Bext � Bint fine structure dominates and the good quantumnumbers are n, l , j , mj
apply first order perturbation the-ory to get
we can rewrite this using ~J = ~L+~S
this can be evaluated by realizingthat ~J is constant and that the ti-ime average of ~S is
E(1)Z =
⟨nljmj
∣∣H ′Z ∣∣ nljmj
⟩=
e
2m~Bext ·
⟨~L + 2~S
⟩=
e
2m~Bext ·
⟨~J + ~S
⟩
~Save =~S · ~JJ2
~J
~L = ~J − ~S → L2 = J2 + S2 − 2~J · ~S
~S · ~J =1
2(J2 + S2 − L2) =
~2
2[j(j + 1) + s(s + 1)− l(l + 1)]
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 18 / 20
Weak-field Zeeman effect
When Bext � Bint fine structure dominates and the good quantumnumbers are n, l , j , mj
apply first order perturbation the-ory to get
we can rewrite this using ~J = ~L+~S
this can be evaluated by realizingthat ~J is constant and that the ti-ime average of ~S is
E(1)Z =
⟨nljmj
∣∣H ′Z ∣∣ nljmj
⟩=
e
2m~Bext ·
⟨~L + 2~S
⟩=
e
2m~Bext ·
⟨~J + ~S
⟩
~Save =~S · ~JJ2
~J
~L = ~J − ~S
→ L2 = J2 + S2 − 2~J · ~S
~S · ~J =1
2(J2 + S2 − L2) =
~2
2[j(j + 1) + s(s + 1)− l(l + 1)]
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 18 / 20
Weak-field Zeeman effect
When Bext � Bint fine structure dominates and the good quantumnumbers are n, l , j , mj
apply first order perturbation the-ory to get
we can rewrite this using ~J = ~L+~S
this can be evaluated by realizingthat ~J is constant and that the ti-ime average of ~S is
E(1)Z =
⟨nljmj
∣∣H ′Z ∣∣ nljmj
⟩=
e
2m~Bext ·
⟨~L + 2~S
⟩=
e
2m~Bext ·
⟨~J + ~S
⟩
~Save =~S · ~JJ2
~J
~L = ~J − ~S → L2 = J2 + S2 − 2~J · ~S
~S · ~J =1
2(J2 + S2 − L2) =
~2
2[j(j + 1) + s(s + 1)− l(l + 1)]
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 18 / 20
Weak-field Zeeman effect
When Bext � Bint fine structure dominates and the good quantumnumbers are n, l , j , mj
apply first order perturbation the-ory to get
we can rewrite this using ~J = ~L+~S
this can be evaluated by realizingthat ~J is constant and that the ti-ime average of ~S is
E(1)Z =
⟨nljmj
∣∣H ′Z ∣∣ nljmj
⟩=
e
2m~Bext ·
⟨~L + 2~S
⟩=
e
2m~Bext ·
⟨~J + ~S
⟩
~Save =~S · ~JJ2
~J
~L = ~J − ~S → L2 = J2 + S2 − 2~J · ~S
~S · ~J =1
2(J2 + S2 − L2)
=~2
2[j(j + 1) + s(s + 1)− l(l + 1)]
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 18 / 20
Weak-field Zeeman effect
When Bext � Bint fine structure dominates and the good quantumnumbers are n, l , j , mj
apply first order perturbation the-ory to get
we can rewrite this using ~J = ~L+~S
this can be evaluated by realizingthat ~J is constant and that the ti-ime average of ~S is
E(1)Z =
⟨nljmj
∣∣H ′Z ∣∣ nljmj
⟩=
e
2m~Bext ·
⟨~L + 2~S
⟩=
e
2m~Bext ·
⟨~J + ~S
⟩
~Save =~S · ~JJ2
~J
~L = ~J − ~S → L2 = J2 + S2 − 2~J · ~S
~S · ~J =1
2(J2 + S2 − L2) =
~2
2[j(j + 1) + s(s + 1)− l(l + 1)]
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 18 / 20
Lande g-factor
Thus the expectation value in the Zeeman energy correction becomes
⟨~L + 2~S
⟩=⟨~J + ~S
⟩=
⟨(1 +
~S · ~JJ2
)~J
⟩
=
[1 +
j(j + 1)− l(l + 1) + s(s + 1)
2j(j + 1)
]〈~J〉 ≡ gJ〈~J〉
and the full energy correction becomes
E(1)Z =
e
2mBextgJ~mj
= µBgJBextmj
µB ≡e~2m
= 5.788× 10−5eV/T
the total energy includes both the spin-orbit and Zeeman corrections andthe 2j + 1 states then have unique energies
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 19 / 20
Lande g-factor
Thus the expectation value in the Zeeman energy correction becomes
⟨~L + 2~S
⟩=⟨~J + ~S
⟩
=
⟨(1 +
~S · ~JJ2
)~J
⟩
=
[1 +
j(j + 1)− l(l + 1) + s(s + 1)
2j(j + 1)
]〈~J〉 ≡ gJ〈~J〉
and the full energy correction becomes
E(1)Z =
e
2mBextgJ~mj
= µBgJBextmj
µB ≡e~2m
= 5.788× 10−5eV/T
the total energy includes both the spin-orbit and Zeeman corrections andthe 2j + 1 states then have unique energies
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 19 / 20
Lande g-factor
Thus the expectation value in the Zeeman energy correction becomes
⟨~L + 2~S
⟩=⟨~J + ~S
⟩=
⟨(1 +
~S · ~JJ2
)~J
⟩
=
[1 +
j(j + 1)− l(l + 1) + s(s + 1)
2j(j + 1)
]〈~J〉 ≡ gJ〈~J〉
and the full energy correction becomes
E(1)Z =
e
2mBextgJ~mj
= µBgJBextmj
µB ≡e~2m
= 5.788× 10−5eV/T
the total energy includes both the spin-orbit and Zeeman corrections andthe 2j + 1 states then have unique energies
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 19 / 20
Lande g-factor
Thus the expectation value in the Zeeman energy correction becomes
⟨~L + 2~S
⟩=⟨~J + ~S
⟩=
⟨(1 +
~S · ~JJ2
)~J
⟩
=
[1 +
j(j + 1)− l(l + 1) + s(s + 1)
2j(j + 1)
]〈~J〉
≡ gJ〈~J〉
and the full energy correction becomes
E(1)Z =
e
2mBextgJ~mj
= µBgJBextmj
µB ≡e~2m
= 5.788× 10−5eV/T
the total energy includes both the spin-orbit and Zeeman corrections andthe 2j + 1 states then have unique energies
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 19 / 20
Lande g-factor
Thus the expectation value in the Zeeman energy correction becomes
⟨~L + 2~S
⟩=⟨~J + ~S
⟩=
⟨(1 +
~S · ~JJ2
)~J
⟩
=
[1 +
j(j + 1)− l(l + 1) + s(s + 1)
2j(j + 1)
]〈~J〉 ≡ gJ〈~J〉
and the full energy correction becomes
E(1)Z =
e
2mBextgJ~mj
= µBgJBextmj
µB ≡e~2m
= 5.788× 10−5eV/T
the total energy includes both the spin-orbit and Zeeman corrections andthe 2j + 1 states then have unique energies
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 19 / 20
Lande g-factor
Thus the expectation value in the Zeeman energy correction becomes
⟨~L + 2~S
⟩=⟨~J + ~S
⟩=
⟨(1 +
~S · ~JJ2
)~J
⟩
=
[1 +
j(j + 1)− l(l + 1) + s(s + 1)
2j(j + 1)
]〈~J〉 ≡ gJ〈~J〉
and the full energy correction becomes
E(1)Z =
e
2mBextgJ~mj
= µBgJBextmj
µB ≡e~2m
= 5.788× 10−5eV/T
the total energy includes both the spin-orbit and Zeeman corrections andthe 2j + 1 states then have unique energies
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 19 / 20
Lande g-factor
Thus the expectation value in the Zeeman energy correction becomes
⟨~L + 2~S
⟩=⟨~J + ~S
⟩=
⟨(1 +
~S · ~JJ2
)~J
⟩
=
[1 +
j(j + 1)− l(l + 1) + s(s + 1)
2j(j + 1)
]〈~J〉 ≡ gJ〈~J〉
and the full energy correction becomes
E(1)Z =
e
2mBextgJ~mj
= µBgJBextmj µB ≡e~2m
= 5.788× 10−5eV/T
the total energy includes both the spin-orbit and Zeeman corrections andthe 2j + 1 states then have unique energies
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 19 / 20
Lande g-factor
Thus the expectation value in the Zeeman energy correction becomes
⟨~L + 2~S
⟩=⟨~J + ~S
⟩=
⟨(1 +
~S · ~JJ2
)~J
⟩
=
[1 +
j(j + 1)− l(l + 1) + s(s + 1)
2j(j + 1)
]〈~J〉 ≡ gJ〈~J〉
and the full energy correction becomes
E(1)Z =
e
2mBextgJ~mj
= µBgJBextmj
µB ≡e~2m
= 5.788× 10−5eV/T
the total energy includes both the spin-orbit and Zeeman corrections andthe 2j + 1 states then have unique energies
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 19 / 20
Lande g-factor
Thus the expectation value in the Zeeman energy correction becomes
⟨~L + 2~S
⟩=⟨~J + ~S
⟩=
⟨(1 +
~S · ~JJ2
)~J
⟩
=
[1 +
j(j + 1)− l(l + 1) + s(s + 1)
2j(j + 1)
]〈~J〉 ≡ gJ〈~J〉
and the full energy correction becomes
E(1)Z =
e
2mBextgJ~mj
= µBgJBextmj
µB ≡e~2m
= 5.788× 10−5eV/T
the total energy includes both the spin-orbit and Zeeman corrections andthe 2j + 1 states then have unique energies
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 19 / 20
Lande g-factor
Thus the expectation value in the Zeeman energy correction becomes
⟨~L + 2~S
⟩=⟨~J + ~S
⟩=
⟨(1 +
~S · ~JJ2
)~J
⟩
=
[1 +
j(j + 1)− l(l + 1) + s(s + 1)
2j(j + 1)
]〈~J〉 ≡ gJ〈~J〉
and the full energy correction becomes
E(1)Z =
e
2mBextgJ~mj = µBgJBextmj µB ≡
e~2m
= 5.788× 10−5eV/T
the total energy includes both the spin-orbit and Zeeman corrections andthe 2j + 1 states then have unique energies
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 19 / 20
Lande g-factor
Thus the expectation value in the Zeeman energy correction becomes
⟨~L + 2~S
⟩=⟨~J + ~S
⟩=
⟨(1 +
~S · ~JJ2
)~J
⟩
=
[1 +
j(j + 1)− l(l + 1) + s(s + 1)
2j(j + 1)
]〈~J〉 ≡ gJ〈~J〉
and the full energy correction becomes
E(1)Z =
e
2mBextgJ~mj = µBgJBextmj µB ≡
e~2m
= 5.788× 10−5eV/T
the total energy includes both the spin-orbit and Zeeman corrections andthe 2j + 1 states then have unique energies
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 19 / 20
Strong-field Zeeman effect
When Bext � Bint , the spin-orbit coupling must be treated as theperturbation and the solutions must be eigenfunctions of the unperturbedwave functions with good quantum numbers.
If Bext is in the z direction,the Zeeman Hamiltonian is
and the energies (withoutfine structure), are
Applying perturbation theoryto the fine structure Hamilto-nian
H ′Z =e
2mBext(Lz + 2Sz)
Enmlms = −13.6eV
n2+ µBBext(ml + 2ms)
E 1fs =
⟨nlmlms
∣∣(H ′r + H ′so)∣∣ nlmlms
⟩〈~S · ~L〉 =
�����〈Sx〉 〈Lx〉+�����〈Sy 〉 〈Ly 〉+ 〈Sz〉 〈Lz〉 = ~2mlms
E 1fs =
13.6eV
n3α2
{3
4n−[l(l + 1)−mlms
l(l + 1/2)(l + 1)
]}, l > 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 20 / 20
Strong-field Zeeman effect
When Bext � Bint , the spin-orbit coupling must be treated as theperturbation and the solutions must be eigenfunctions of the unperturbedwave functions with good quantum numbers.
If Bext is in the z direction,the Zeeman Hamiltonian is
and the energies (withoutfine structure), are
Applying perturbation theoryto the fine structure Hamilto-nian
H ′Z =e
2mBext(Lz + 2Sz)
Enmlms = −13.6eV
n2+ µBBext(ml + 2ms)
E 1fs =
⟨nlmlms
∣∣(H ′r + H ′so)∣∣ nlmlms
⟩〈~S · ~L〉 =
�����〈Sx〉 〈Lx〉+�����〈Sy 〉 〈Ly 〉+ 〈Sz〉 〈Lz〉 = ~2mlms
E 1fs =
13.6eV
n3α2
{3
4n−[l(l + 1)−mlms
l(l + 1/2)(l + 1)
]}, l > 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 20 / 20
Strong-field Zeeman effect
When Bext � Bint , the spin-orbit coupling must be treated as theperturbation and the solutions must be eigenfunctions of the unperturbedwave functions with good quantum numbers.
If Bext is in the z direction,the Zeeman Hamiltonian is
and the energies (withoutfine structure), are
Applying perturbation theoryto the fine structure Hamilto-nian
H ′Z =e
2mBext(Lz + 2Sz)
Enmlms = −13.6eV
n2+ µBBext(ml + 2ms)
E 1fs =
⟨nlmlms
∣∣(H ′r + H ′so)∣∣ nlmlms
⟩〈~S · ~L〉 =
�����〈Sx〉 〈Lx〉+�����〈Sy 〉 〈Ly 〉+ 〈Sz〉 〈Lz〉 = ~2mlms
E 1fs =
13.6eV
n3α2
{3
4n−[l(l + 1)−mlms
l(l + 1/2)(l + 1)
]}, l > 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 20 / 20
Strong-field Zeeman effect
When Bext � Bint , the spin-orbit coupling must be treated as theperturbation and the solutions must be eigenfunctions of the unperturbedwave functions with good quantum numbers.
If Bext is in the z direction,the Zeeman Hamiltonian is
and the energies (withoutfine structure), are
Applying perturbation theoryto the fine structure Hamilto-nian
H ′Z =e
2mBext(Lz + 2Sz)
Enmlms = −13.6eV
n2+ µBBext(ml + 2ms)
E 1fs =
⟨nlmlms
∣∣(H ′r + H ′so)∣∣ nlmlms
⟩〈~S · ~L〉 =
�����〈Sx〉 〈Lx〉+�����〈Sy 〉 〈Ly 〉+ 〈Sz〉 〈Lz〉 = ~2mlms
E 1fs =
13.6eV
n3α2
{3
4n−[l(l + 1)−mlms
l(l + 1/2)(l + 1)
]}, l > 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 20 / 20
Strong-field Zeeman effect
When Bext � Bint , the spin-orbit coupling must be treated as theperturbation and the solutions must be eigenfunctions of the unperturbedwave functions with good quantum numbers.
If Bext is in the z direction,the Zeeman Hamiltonian is
and the energies (withoutfine structure), are
Applying perturbation theoryto the fine structure Hamilto-nian
H ′Z =e
2mBext(Lz + 2Sz)
Enmlms = −13.6eV
n2+ µBBext(ml + 2ms)
E 1fs =
⟨nlmlms
∣∣(H ′r + H ′so)∣∣ nlmlms
⟩〈~S · ~L〉 =
�����〈Sx〉 〈Lx〉+�����〈Sy 〉 〈Ly 〉+ 〈Sz〉 〈Lz〉 = ~2mlms
E 1fs =
13.6eV
n3α2
{3
4n−[l(l + 1)−mlms
l(l + 1/2)(l + 1)
]}, l > 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 20 / 20
Strong-field Zeeman effect
When Bext � Bint , the spin-orbit coupling must be treated as theperturbation and the solutions must be eigenfunctions of the unperturbedwave functions with good quantum numbers.
If Bext is in the z direction,the Zeeman Hamiltonian is
and the energies (withoutfine structure), are
Applying perturbation theoryto the fine structure Hamilto-nian
H ′Z =e
2mBext(Lz + 2Sz)
Enmlms = −13.6eV
n2+ µBBext(ml + 2ms)
E 1fs =
⟨nlmlms
∣∣(H ′r + H ′so)∣∣ nlmlms
⟩〈~S · ~L〉 =
�����〈Sx〉 〈Lx〉+�����〈Sy 〉 〈Ly 〉+ 〈Sz〉 〈Lz〉 = ~2mlms
E 1fs =
13.6eV
n3α2
{3
4n−[l(l + 1)−mlms
l(l + 1/2)(l + 1)
]}, l > 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 20 / 20
Strong-field Zeeman effect
When Bext � Bint , the spin-orbit coupling must be treated as theperturbation and the solutions must be eigenfunctions of the unperturbedwave functions with good quantum numbers.
If Bext is in the z direction,the Zeeman Hamiltonian is
and the energies (withoutfine structure), are
Applying perturbation theoryto the fine structure Hamilto-nian
H ′Z =e
2mBext(Lz + 2Sz)
Enmlms = −13.6eV
n2+ µBBext(ml + 2ms)
E 1fs =
⟨nlmlms
∣∣(H ′r + H ′so)∣∣ nlmlms
⟩
〈~S · ~L〉 =
�����〈Sx〉 〈Lx〉+�����〈Sy 〉 〈Ly 〉+ 〈Sz〉 〈Lz〉 = ~2mlms
E 1fs =
13.6eV
n3α2
{3
4n−[l(l + 1)−mlms
l(l + 1/2)(l + 1)
]}, l > 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 20 / 20
Strong-field Zeeman effect
When Bext � Bint , the spin-orbit coupling must be treated as theperturbation and the solutions must be eigenfunctions of the unperturbedwave functions with good quantum numbers.
If Bext is in the z direction,the Zeeman Hamiltonian is
and the energies (withoutfine structure), are
Applying perturbation theoryto the fine structure Hamilto-nian
H ′Z =e
2mBext(Lz + 2Sz)
Enmlms = −13.6eV
n2+ µBBext(ml + 2ms)
E 1fs =
⟨nlmlms
∣∣(H ′r + H ′so)∣∣ nlmlms
⟩〈~S · ~L〉 =
�����〈Sx〉 〈Lx〉+�����〈Sy 〉 〈Ly 〉+ 〈Sz〉 〈Lz〉 = ~2mlms
E 1fs =
13.6eV
n3α2
{3
4n−[l(l + 1)−mlms
l(l + 1/2)(l + 1)
]}, l > 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 20 / 20
Strong-field Zeeman effect
When Bext � Bint , the spin-orbit coupling must be treated as theperturbation and the solutions must be eigenfunctions of the unperturbedwave functions with good quantum numbers.
If Bext is in the z direction,the Zeeman Hamiltonian is
and the energies (withoutfine structure), are
Applying perturbation theoryto the fine structure Hamilto-nian
H ′Z =e
2mBext(Lz + 2Sz)
Enmlms = −13.6eV
n2+ µBBext(ml + 2ms)
E 1fs =
⟨nlmlms
∣∣(H ′r + H ′so)∣∣ nlmlms
⟩〈~S · ~L〉 = 〈Sx〉 〈Lx〉+ 〈Sy 〉 〈Ly 〉+ 〈Sz〉 〈Lz〉
= ~2mlms
E 1fs =
13.6eV
n3α2
{3
4n−[l(l + 1)−mlms
l(l + 1/2)(l + 1)
]}, l > 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 20 / 20
Strong-field Zeeman effect
When Bext � Bint , the spin-orbit coupling must be treated as theperturbation and the solutions must be eigenfunctions of the unperturbedwave functions with good quantum numbers.
If Bext is in the z direction,the Zeeman Hamiltonian is
and the energies (withoutfine structure), are
Applying perturbation theoryto the fine structure Hamilto-nian
H ′Z =e
2mBext(Lz + 2Sz)
Enmlms = −13.6eV
n2+ µBBext(ml + 2ms)
E 1fs =
⟨nlmlms
∣∣(H ′r + H ′so)∣∣ nlmlms
⟩〈~S · ~L〉 = �����〈Sx〉 〈Lx〉+�����〈Sy 〉 〈Ly 〉+ 〈Sz〉 〈Lz〉
= ~2mlms
E 1fs =
13.6eV
n3α2
{3
4n−[l(l + 1)−mlms
l(l + 1/2)(l + 1)
]}, l > 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 20 / 20
Strong-field Zeeman effect
When Bext � Bint , the spin-orbit coupling must be treated as theperturbation and the solutions must be eigenfunctions of the unperturbedwave functions with good quantum numbers.
If Bext is in the z direction,the Zeeman Hamiltonian is
and the energies (withoutfine structure), are
Applying perturbation theoryto the fine structure Hamilto-nian
H ′Z =e
2mBext(Lz + 2Sz)
Enmlms = −13.6eV
n2+ µBBext(ml + 2ms)
E 1fs =
⟨nlmlms
∣∣(H ′r + H ′so)∣∣ nlmlms
⟩〈~S · ~L〉 = �����〈Sx〉 〈Lx〉+�����〈Sy 〉 〈Ly 〉+ 〈Sz〉 〈Lz〉 = ~2mlms
E 1fs =
13.6eV
n3α2
{3
4n−[l(l + 1)−mlms
l(l + 1/2)(l + 1)
]}, l > 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 20 / 20
Strong-field Zeeman effect
When Bext � Bint , the spin-orbit coupling must be treated as theperturbation and the solutions must be eigenfunctions of the unperturbedwave functions with good quantum numbers.
If Bext is in the z direction,the Zeeman Hamiltonian is
and the energies (withoutfine structure), are
Applying perturbation theoryto the fine structure Hamilto-nian
H ′Z =e
2mBext(Lz + 2Sz)
Enmlms = −13.6eV
n2+ µBBext(ml + 2ms)
E 1fs =
⟨nlmlms
∣∣(H ′r + H ′so)∣∣ nlmlms
⟩〈~S · ~L〉 = �����〈Sx〉 〈Lx〉+�����〈Sy 〉 〈Ly 〉+ 〈Sz〉 〈Lz〉 = ~2mlms
E 1fs =
13.6eV
n3α2
{3
4n−[l(l + 1)−mlms
l(l + 1/2)(l + 1)
]}, l > 0
C. Segre (IIT) PHYS 406 - Spring 2016 January 27, 2016 20 / 20