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Today’s Outline - February 13, 2020
• Homework problems
• Hydrogen molecule
• Problem 8.24
Homework Assignment #05:Chapter 7:20,21,24,28,29,37due Tuesday, February 18, 2020
Homework Assignment #06:Chapter 7:38,44,45; 8:1,2,4due Tuesday, February 25, 2020
Midterm Exam #1Thursday, February 27, 2020
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 1 / 21
Today’s Outline - February 13, 2020
• Homework problems
• Hydrogen molecule
• Problem 8.24
Homework Assignment #05:Chapter 7:20,21,24,28,29,37due Tuesday, February 18, 2020
Homework Assignment #06:Chapter 7:38,44,45; 8:1,2,4due Tuesday, February 25, 2020
Midterm Exam #1Thursday, February 27, 2020
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 1 / 21
Today’s Outline - February 13, 2020
• Homework problems
• Hydrogen molecule
• Problem 8.24
Homework Assignment #05:Chapter 7:20,21,24,28,29,37due Tuesday, February 18, 2020
Homework Assignment #06:Chapter 7:38,44,45; 8:1,2,4due Tuesday, February 25, 2020
Midterm Exam #1Thursday, February 27, 2020
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 1 / 21
Today’s Outline - February 13, 2020
• Homework problems
• Hydrogen molecule
• Problem 8.24
Homework Assignment #05:Chapter 7:20,21,24,28,29,37due Tuesday, February 18, 2020
Homework Assignment #06:Chapter 7:38,44,45; 8:1,2,4due Tuesday, February 25, 2020
Midterm Exam #1Thursday, February 27, 2020
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 1 / 21
Today’s Outline - February 13, 2020
• Homework problems
• Hydrogen molecule
• Problem 8.24
Homework Assignment #05:Chapter 7:20,21,24,28,29,37due Tuesday, February 18, 2020
Homework Assignment #06:Chapter 7:38,44,45; 8:1,2,4due Tuesday, February 25, 2020
Midterm Exam #1Thursday, February 27, 2020
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 1 / 21
Today’s Outline - February 13, 2020
• Homework problems
• Hydrogen molecule
• Problem 8.24
Homework Assignment #05:Chapter 7:20,21,24,28,29,37due Tuesday, February 18, 2020
Homework Assignment #06:Chapter 7:38,44,45; 8:1,2,4due Tuesday, February 25, 2020
Midterm Exam #1Thursday, February 27, 2020
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 1 / 21
Today’s Outline - February 13, 2020
• Homework problems
• Hydrogen molecule
• Problem 8.24
Homework Assignment #05:Chapter 7:20,21,24,28,29,37due Tuesday, February 18, 2020
Homework Assignment #06:Chapter 7:38,44,45; 8:1,2,4due Tuesday, February 25, 2020
Midterm Exam #1Thursday, February 27, 2020
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 1 / 21
Problem 7.17
Find the (lowest-order) relativistic correction to the energy levels of theone-dimensional harmonic oscillator
We start with equation (7.54) which defines the form of the relativisticcorrection in terms of the potential and the energy of the harmonicoscillator
E 1r = − 1
2mc2[E 2 − 2E 〈V 〉+ 〈V 2〉
]= − 1
2mc2
[(n + 1
2
)2 ~2ω2 − 2(n + 1
2
)~ω 1
2mω2〈x2〉+ 1
4m2ω4〈x4〉
]recall that the x2 and x4 operators can be written in terms of the raisingand lowering operators
x2 =~
2mω
(a2+ + a+a− + a−a+ + a2−
)x4 =
~4m2ω2
(a2+ + a+a− + a−a+ + a2−
) (a2+ + a+a− + a−a+ + a2−
)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 2 / 21
Problem 7.17
Find the (lowest-order) relativistic correction to the energy levels of theone-dimensional harmonic oscillator
We start with equation (7.54) which defines the form of the relativisticcorrection in terms of the potential and the energy of the harmonicoscillator
E 1r = − 1
2mc2[E 2 − 2E 〈V 〉+ 〈V 2〉
]= − 1
2mc2
[(n + 1
2
)2 ~2ω2 − 2(n + 1
2
)~ω 1
2mω2〈x2〉+ 1
4m2ω4〈x4〉
]recall that the x2 and x4 operators can be written in terms of the raisingand lowering operators
x2 =~
2mω
(a2+ + a+a− + a−a+ + a2−
)x4 =
~4m2ω2
(a2+ + a+a− + a−a+ + a2−
) (a2+ + a+a− + a−a+ + a2−
)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 2 / 21
Problem 7.17
Find the (lowest-order) relativistic correction to the energy levels of theone-dimensional harmonic oscillator
We start with equation (7.54) which defines the form of the relativisticcorrection in terms of the potential and the energy of the harmonicoscillator
E 1r = − 1
2mc2[E 2 − 2E 〈V 〉+ 〈V 2〉
]
= − 1
2mc2
[(n + 1
2
)2 ~2ω2 − 2(n + 1
2
)~ω 1
2mω2〈x2〉+ 1
4m2ω4〈x4〉
]recall that the x2 and x4 operators can be written in terms of the raisingand lowering operators
x2 =~
2mω
(a2+ + a+a− + a−a+ + a2−
)x4 =
~4m2ω2
(a2+ + a+a− + a−a+ + a2−
) (a2+ + a+a− + a−a+ + a2−
)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 2 / 21
Problem 7.17
Find the (lowest-order) relativistic correction to the energy levels of theone-dimensional harmonic oscillator
We start with equation (7.54) which defines the form of the relativisticcorrection in terms of the potential and the energy of the harmonicoscillator
E 1r = − 1
2mc2[E 2 − 2E 〈V 〉+ 〈V 2〉
]= − 1
2mc2
[(n + 1
2
)2 ~2ω2 − 2(n + 1
2
)~ω 1
2mω2〈x2〉+ 1
4m2ω4〈x4〉
]
recall that the x2 and x4 operators can be written in terms of the raisingand lowering operators
x2 =~
2mω
(a2+ + a+a− + a−a+ + a2−
)x4 =
~4m2ω2
(a2+ + a+a− + a−a+ + a2−
) (a2+ + a+a− + a−a+ + a2−
)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 2 / 21
Problem 7.17
Find the (lowest-order) relativistic correction to the energy levels of theone-dimensional harmonic oscillator
We start with equation (7.54) which defines the form of the relativisticcorrection in terms of the potential and the energy of the harmonicoscillator
E 1r = − 1
2mc2[E 2 − 2E 〈V 〉+ 〈V 2〉
]= − 1
2mc2
[(n + 1
2
)2 ~2ω2 − 2(n + 1
2
)~ω 1
2mω2〈x2〉+ 1
4m2ω4〈x4〉
]recall that the x2 and x4 operators can be written in terms of the raisingand lowering operators
x2 =~
2mω
(a2+ + a+a− + a−a+ + a2−
)x4 =
~4m2ω2
(a2+ + a+a− + a−a+ + a2−
) (a2+ + a+a− + a−a+ + a2−
)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 2 / 21
Problem 7.17
Find the (lowest-order) relativistic correction to the energy levels of theone-dimensional harmonic oscillator
We start with equation (7.54) which defines the form of the relativisticcorrection in terms of the potential and the energy of the harmonicoscillator
E 1r = − 1
2mc2[E 2 − 2E 〈V 〉+ 〈V 2〉
]= − 1
2mc2
[(n + 1
2
)2 ~2ω2 − 2(n + 1
2
)~ω 1
2mω2〈x2〉+ 1
4m2ω4〈x4〉
]recall that the x2 and x4 operators can be written in terms of the raisingand lowering operators
x2 =~
2mω
(a2+ + a+a− + a−a+ + a2−
)
x4 =~
4m2ω2
(a2+ + a+a− + a−a+ + a2−
) (a2+ + a+a− + a−a+ + a2−
)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 2 / 21
Problem 7.17
Find the (lowest-order) relativistic correction to the energy levels of theone-dimensional harmonic oscillator
We start with equation (7.54) which defines the form of the relativisticcorrection in terms of the potential and the energy of the harmonicoscillator
E 1r = − 1
2mc2[E 2 − 2E 〈V 〉+ 〈V 2〉
]= − 1
2mc2
[(n + 1
2
)2 ~2ω2 − 2(n + 1
2
)~ω 1
2mω2〈x2〉+ 1
4m2ω4〈x4〉
]recall that the x2 and x4 operators can be written in terms of the raisingand lowering operators
x2 =~
2mω
(a2+ + a+a− + a−a+ + a2−
)x4 =
~4m2ω2
(a2+ + a+a− + a−a+ + a2−
) (a2+ + a+a− + a−a+ + a2−
)C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 2 / 21
Problem 7.17 (cont.)
When taking the expectation values of the x2 and x4 operators, we recallthat the only terms which survive are those with equal number of raisingand lowering operators
〈x2〉 =~
2mω〈n| [a+a− + a−a+] |n〉
=~
2mω[n + (n + 1)] 〈n|n〉 =
~mω
[n + 1
2
]〈x4〉 =
~2
4m2ω2〈n|[a2+a
2− + a+a−a+a− + a+a−a−a+
+a−a+a+a− + a−a+a−a+ + a2−a2+
]|n〉
=~2
4m2ω2
[n(n − 1) + n2 + (n + 1)n + n(n + 1)
+(n + 1)2 + (n + 1)(n + 2)]〈n|n〉
=~2
4m2ω2
[6n2 + 6n + 3
]
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 3 / 21
Problem 7.17 (cont.)
When taking the expectation values of the x2 and x4 operators, we recallthat the only terms which survive are those with equal number of raisingand lowering operators
〈x2〉 =~
2mω〈n| [a+a− + a−a+] |n〉
=~
2mω[n + (n + 1)] 〈n|n〉 =
~mω
[n + 1
2
]〈x4〉 =
~2
4m2ω2〈n|[a2+a
2− + a+a−a+a− + a+a−a−a+
+a−a+a+a− + a−a+a−a+ + a2−a2+
]|n〉
=~2
4m2ω2
[n(n − 1) + n2 + (n + 1)n + n(n + 1)
+(n + 1)2 + (n + 1)(n + 2)]〈n|n〉
=~2
4m2ω2
[6n2 + 6n + 3
]
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 3 / 21
Problem 7.17 (cont.)
When taking the expectation values of the x2 and x4 operators, we recallthat the only terms which survive are those with equal number of raisingand lowering operators
〈x2〉 =~
2mω〈n| [a+a− + a−a+] |n〉
=~
2mω[n + (n + 1)] 〈n|n〉
=~mω
[n + 1
2
]〈x4〉 =
~2
4m2ω2〈n|[a2+a
2− + a+a−a+a− + a+a−a−a+
+a−a+a+a− + a−a+a−a+ + a2−a2+
]|n〉
=~2
4m2ω2
[n(n − 1) + n2 + (n + 1)n + n(n + 1)
+(n + 1)2 + (n + 1)(n + 2)]〈n|n〉
=~2
4m2ω2
[6n2 + 6n + 3
]
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 3 / 21
Problem 7.17 (cont.)
When taking the expectation values of the x2 and x4 operators, we recallthat the only terms which survive are those with equal number of raisingand lowering operators
〈x2〉 =~
2mω〈n| [a+a− + a−a+] |n〉
=~
2mω[n + (n + 1)] 〈n|n〉 =
~mω
[n + 1
2
]
〈x4〉 =~2
4m2ω2〈n|[a2+a
2− + a+a−a+a− + a+a−a−a+
+a−a+a+a− + a−a+a−a+ + a2−a2+
]|n〉
=~2
4m2ω2
[n(n − 1) + n2 + (n + 1)n + n(n + 1)
+(n + 1)2 + (n + 1)(n + 2)]〈n|n〉
=~2
4m2ω2
[6n2 + 6n + 3
]
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 3 / 21
Problem 7.17 (cont.)
When taking the expectation values of the x2 and x4 operators, we recallthat the only terms which survive are those with equal number of raisingand lowering operators
〈x2〉 =~
2mω〈n| [a+a− + a−a+] |n〉
=~
2mω[n + (n + 1)] 〈n|n〉 =
~mω
[n + 1
2
]〈x4〉 =
~2
4m2ω2〈n|[a2+a
2− + a+a−a+a− + a+a−a−a+
+a−a+a+a− + a−a+a−a+ + a2−a2+
]|n〉
=~2
4m2ω2
[n(n − 1) + n2 + (n + 1)n + n(n + 1)
+(n + 1)2 + (n + 1)(n + 2)]〈n|n〉
=~2
4m2ω2
[6n2 + 6n + 3
]
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 3 / 21
Problem 7.17 (cont.)
When taking the expectation values of the x2 and x4 operators, we recallthat the only terms which survive are those with equal number of raisingand lowering operators
〈x2〉 =~
2mω〈n| [a+a− + a−a+] |n〉
=~
2mω[n + (n + 1)] 〈n|n〉 =
~mω
[n + 1
2
]〈x4〉 =
~2
4m2ω2〈n|[a2+a
2− + a+a−a+a− + a+a−a−a+
+a−a+a+a− + a−a+a−a+ + a2−a2+
]|n〉
=~2
4m2ω2
[n(n − 1) + n2 + (n + 1)n + n(n + 1)
+(n + 1)2 + (n + 1)(n + 2)]〈n|n〉
=~2
4m2ω2
[6n2 + 6n + 3
]
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 3 / 21
Problem 7.17 (cont.)
When taking the expectation values of the x2 and x4 operators, we recallthat the only terms which survive are those with equal number of raisingand lowering operators
〈x2〉 =~
2mω〈n| [a+a− + a−a+] |n〉
=~
2mω[n + (n + 1)] 〈n|n〉 =
~mω
[n + 1
2
]〈x4〉 =
~2
4m2ω2〈n|[a2+a
2− + a+a−a+a− + a+a−a−a+
+a−a+a+a− + a−a+a−a+ + a2−a2+
]|n〉
=~2
4m2ω2
[n(n − 1) + n2 + (n + 1)n + n(n + 1)
+(n + 1)2 + (n + 1)(n + 2)]〈n|n〉
=~2
4m2ω2
[6n2 + 6n + 3
]C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 3 / 21
Problem 7.17 ( cont.)
Thus we have
E 1r = − 1
2mc2
[(n + 1
2
)2 ~2ω2 −(n + 1
2
)~mω3〈x2〉+ 1
4m2ω4〈x4〉
]〈x2〉 =
~mω
(n + 1
2
), 〈x4〉 =
~2
4m2ω2
(6n2 + 6n + 3
)
E 1r = − 1
2mc2
[(n + 1
2
)2 ~2ω2 −(n + 1
2
)2 ~2ω2 + 116
(6n2 + 6n + 3
)~2ω2
]= − 3
32
~2ω2
mc2(2n2 − 2n + 1
)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 4 / 21
Problem 7.17 ( cont.)
Thus we have
E 1r = − 1
2mc2
[(n + 1
2
)2 ~2ω2 −(n + 1
2
)~mω3〈x2〉+ 1
4m2ω4〈x4〉
]〈x2〉 =
~mω
(n + 1
2
), 〈x4〉 =
~2
4m2ω2
(6n2 + 6n + 3
)
E 1r = − 1
2mc2
[(n + 1
2
)2 ~2ω2 −(n + 1
2
)2 ~2ω2 + 116
(6n2 + 6n + 3
)~2ω2
]
= − 3
32
~2ω2
mc2(2n2 − 2n + 1
)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 4 / 21
Problem 7.17 ( cont.)
Thus we have
E 1r = − 1
2mc2
[(n + 1
2
)2 ~2ω2 −(n + 1
2
)~mω3〈x2〉+ 1
4m2ω4〈x4〉
]〈x2〉 =
~mω
(n + 1
2
), 〈x4〉 =
~2
4m2ω2
(6n2 + 6n + 3
)
E 1r = − 1
2mc2
[(n + 1
2
)2 ~2ω2 −(n + 1
2
)2 ~2ω2 + 116
(6n2 + 6n + 3
)~2ω2
]= − 3
32
~2ω2
mc2(2n2 − 2n + 1
)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 4 / 21
Problem 7.33
Estimate the correction to the ground state energy of hydrogen due to thefinite size of the nucleus. Treat the proton as a uniformly chargedspherical shell of radius b, so the potential energy of an electron inside theshell is constant: −e2/(4πε0b); this isn’t very realistic, but is the simplestmodel, and it will give us the right order of magnitude. Expand yourresults in powers of the small parameter (b/a), where a is the Bohr radius,and keep only the leading term, so your final answer takes the form
∆E
E= A
(b
a
)nYour business is to determine the constant A and the power n. Finally putin b ≈ 10−15m (roughly the radius of the proton) and work out the actualnumber. How does it compare with fine structure and hyperfine structure?
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 5 / 21
Problem 7.33 (cont.)
The perturbation to the Hamilto-nian is only present in the regionr < b, inside the “proton”
the energy correction to the groundstate is
H ′ = − e2
4πε0
(1
b− 1
r
)∆E = 〈ψ100|H ′|ψ100〉
ψ100 =1√πa3
e−r/a
∆E = − e2
4πε0
1
πa34π
∫ b
0
(1
b− 1
r
)e−2r/ar2 dr
= − e2
πε0a3
(
1
b
∫ b
0r2e−2r/a dr −
∫ b
0re−2r/a dr
)= − e2
πε0a3
{
1
b
[−a
2r2e−2r/a + a
(a2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
−[(a
2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
}
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 6 / 21
Problem 7.33 (cont.)
The perturbation to the Hamilto-nian is only present in the regionr < b, inside the “proton”
the energy correction to the groundstate is
H ′ = − e2
4πε0
(1
b− 1
r
)
∆E = 〈ψ100|H ′|ψ100〉
ψ100 =1√πa3
e−r/a
∆E = − e2
4πε0
1
πa34π
∫ b
0
(1
b− 1
r
)e−2r/ar2 dr
= − e2
πε0a3
(
1
b
∫ b
0r2e−2r/a dr −
∫ b
0re−2r/a dr
)= − e2
πε0a3
{
1
b
[−a
2r2e−2r/a + a
(a2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
−[(a
2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
}
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 6 / 21
Problem 7.33 (cont.)
The perturbation to the Hamilto-nian is only present in the regionr < b, inside the “proton”
the energy correction to the groundstate is
H ′ = − e2
4πε0
(1
b− 1
r
)
∆E = 〈ψ100|H ′|ψ100〉
ψ100 =1√πa3
e−r/a
∆E = − e2
4πε0
1
πa34π
∫ b
0
(1
b− 1
r
)e−2r/ar2 dr
= − e2
πε0a3
(
1
b
∫ b
0r2e−2r/a dr −
∫ b
0re−2r/a dr
)= − e2
πε0a3
{
1
b
[−a
2r2e−2r/a + a
(a2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
−[(a
2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
}
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 6 / 21
Problem 7.33 (cont.)
The perturbation to the Hamilto-nian is only present in the regionr < b, inside the “proton”
the energy correction to the groundstate is
H ′ = − e2
4πε0
(1
b− 1
r
)∆E = 〈ψ100|H ′|ψ100〉
ψ100 =1√πa3
e−r/a
∆E = − e2
4πε0
1
πa34π
∫ b
0
(1
b− 1
r
)e−2r/ar2 dr
= − e2
πε0a3
(
1
b
∫ b
0r2e−2r/a dr −
∫ b
0re−2r/a dr
)= − e2
πε0a3
{
1
b
[−a
2r2e−2r/a + a
(a2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
−[(a
2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
}
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 6 / 21
Problem 7.33 (cont.)
The perturbation to the Hamilto-nian is only present in the regionr < b, inside the “proton”
the energy correction to the groundstate is
H ′ = − e2
4πε0
(1
b− 1
r
)∆E = 〈ψ100|H ′|ψ100〉
ψ100 =1√πa3
e−r/a
∆E = − e2
4πε0
1
πa34π
∫ b
0
(1
b− 1
r
)e−2r/ar2 dr
= − e2
πε0a3
(
1
b
∫ b
0r2e−2r/a dr −
∫ b
0re−2r/a dr
)= − e2
πε0a3
{
1
b
[−a
2r2e−2r/a + a
(a2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
−[(a
2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
}
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 6 / 21
Problem 7.33 (cont.)
The perturbation to the Hamilto-nian is only present in the regionr < b, inside the “proton”
the energy correction to the groundstate is
H ′ = − e2
4πε0
(1
b− 1
r
)∆E = 〈ψ100|H ′|ψ100〉
ψ100 =1√πa3
e−r/a
∆E = − e2
4πε0
1
πa34π
∫ b
0
(1
b− 1
r
)e−2r/ar2 dr
= − e2
πε0a3
(
1
b
∫ b
0r2e−2r/a dr −
∫ b
0re−2r/a dr
)= − e2
πε0a3
{
1
b
[−a
2r2e−2r/a + a
(a2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
−[(a
2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
}
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 6 / 21
Problem 7.33 (cont.)
The perturbation to the Hamilto-nian is only present in the regionr < b, inside the “proton”
the energy correction to the groundstate is
H ′ = − e2
4πε0
(1
b− 1
r
)∆E = 〈ψ100|H ′|ψ100〉
ψ100 =1√πa3
e−r/a
∆E = − e2
4πε0
1
πa34π
∫ b
0
(1
b− 1
r
)e−2r/ar2 dr
= − e2
πε0a3
(
1
b
∫ b
0r2e−2r/a dr −
∫ b
0re−2r/a dr
)
= − e2
πε0a3
{
1
b
[−a
2r2e−2r/a + a
(a2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
−[(a
2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
}
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 6 / 21
Problem 7.33 (cont.)
The perturbation to the Hamilto-nian is only present in the regionr < b, inside the “proton”
the energy correction to the groundstate is
H ′ = − e2
4πε0
(1
b− 1
r
)∆E = 〈ψ100|H ′|ψ100〉
ψ100 =1√πa3
e−r/a
∆E = − e2
4πε0
1
πa34π
∫ b
0
(1
b− 1
r
)e−2r/ar2 dr
= − e2
πε0a3
(1
b
∫ b
0r2e−2r/a dr
−∫ b
0re−2r/a dr
)
= − e2
πε0a3
{
1
b
[−a
2r2e−2r/a + a
(a2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
−[(a
2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
}
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 6 / 21
Problem 7.33 (cont.)
The perturbation to the Hamilto-nian is only present in the regionr < b, inside the “proton”
the energy correction to the groundstate is
H ′ = − e2
4πε0
(1
b− 1
r
)∆E = 〈ψ100|H ′|ψ100〉
ψ100 =1√πa3
e−r/a
∆E = − e2
4πε0
1
πa34π
∫ b
0
(1
b− 1
r
)e−2r/ar2 dr
= − e2
πε0a3
(1
b
∫ b
0r2e−2r/a dr −
∫ b
0re−2r/a dr
)
= − e2
πε0a3
{
1
b
[−a
2r2e−2r/a + a
(a2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
−[(a
2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
}
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 6 / 21
Problem 7.33 (cont.)
The perturbation to the Hamilto-nian is only present in the regionr < b, inside the “proton”
the energy correction to the groundstate is
H ′ = − e2
4πε0
(1
b− 1
r
)∆E = 〈ψ100|H ′|ψ100〉
ψ100 =1√πa3
e−r/a
∆E = − e2
4πε0
1
πa34π
∫ b
0
(1
b− 1
r
)e−2r/ar2 dr
= − e2
πε0a3
(1
b
∫ b
0r2e−2r/a dr −
∫ b
0re−2r/a dr
)= − e2
πε0a3
{
1
b
[−a
2r2e−2r/a + a
(a2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
−[(a
2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
}C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 6 / 21
Problem 7.33 (cont.)
The perturbation to the Hamilto-nian is only present in the regionr < b, inside the “proton”
the energy correction to the groundstate is
H ′ = − e2
4πε0
(1
b− 1
r
)∆E = 〈ψ100|H ′|ψ100〉
ψ100 =1√πa3
e−r/a
∆E = − e2
4πε0
1
πa34π
∫ b
0
(1
b− 1
r
)e−2r/ar2 dr
= − e2
πε0a3
(1
b
∫ b
0r2e−2r/a dr −
∫ b
0re−2r/a dr
)= − e2
πε0a3
{1
b
[−a
2r2e−2r/a + a
(a2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
−[(a
2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
}C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 6 / 21
Problem 7.33 (cont.)
The perturbation to the Hamilto-nian is only present in the regionr < b, inside the “proton”
the energy correction to the groundstate is
H ′ = − e2
4πε0
(1
b− 1
r
)∆E = 〈ψ100|H ′|ψ100〉
ψ100 =1√πa3
e−r/a
∆E = − e2
4πε0
1
πa34π
∫ b
0
(1
b− 1
r
)e−2r/ar2 dr
= − e2
πε0a3
(1
b
∫ b
0r2e−2r/a dr −
∫ b
0re−2r/a dr
)= − e2
πε0a3
{1
b
[−a
2r2e−2r/a + a
(a2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
−[(a
2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
}C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 6 / 21
Problem 7.33 (cont.)
∆E = − e2
πε0a3
{1
b
[−a
2r2e−2r/a + a
(a2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
−[(a
2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
}
= − e2
πε0a3
[
− a
2bb2e−2b/a +
a3
4be−2b/a
(−2b
a− 1
)
− a2
4e−2b/a
(−2b
a− 1
)+
a3
4b− a2
4
]= − e2
πε0a3
[
e−2b/a(−���ab
2− a2
2− a3
4b+
���ab
2+
a2
4
)+
a2
4
(ab− 1)
]= − e2
πε0a3
[e−2b/a
(−a2
4
)(ab
+ 1)
+a2
4
(ab− 1)]
=e2
4πε0a
[(1− a
b
)+(
1 +a
b
)e−2b/a
]
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 7 / 21
Problem 7.33 (cont.)
∆E = − e2
πε0a3
{1
b
[−a
2r2e−2r/a + a
(a2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
−[(a
2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
}
= − e2
πε0a3
[
− a
2bb2e−2b/a +
a3
4be−2b/a
(−2b
a− 1
)− a2
4e−2b/a
(−2b
a− 1
)+
a3
4b− a2
4
]
= − e2
πε0a3
[
e−2b/a(−���ab
2− a2
2− a3
4b+
���ab
2+
a2
4
)+
a2
4
(ab− 1)
]= − e2
πε0a3
[e−2b/a
(−a2
4
)(ab
+ 1)
+a2
4
(ab− 1)]
=e2
4πε0a
[(1− a
b
)+(
1 +a
b
)e−2b/a
]
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 7 / 21
Problem 7.33 (cont.)
∆E = − e2
πε0a3
{1
b
[−a
2r2e−2r/a + a
(a2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
−[(a
2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
}
= − e2
πε0a3
[− a
2bb2e−2b/a
+a3
4be−2b/a
(−2b
a− 1
)− a2
4e−2b/a
(−2b
a− 1
)+
a3
4b− a2
4
]
= − e2
πε0a3
[
e−2b/a(−���ab
2− a2
2− a3
4b+
���ab
2+
a2
4
)+
a2
4
(ab− 1)
]= − e2
πε0a3
[e−2b/a
(−a2
4
)(ab
+ 1)
+a2
4
(ab− 1)]
=e2
4πε0a
[(1− a
b
)+(
1 +a
b
)e−2b/a
]
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 7 / 21
Problem 7.33 (cont.)
∆E = − e2
πε0a3
{1
b
[−a
2r2e−2r/a + a
(a2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
−[(a
2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
}
= − e2
πε0a3
[− a
2bb2e−2b/a +
a3
4be−2b/a
(−2b
a− 1
)
− a2
4e−2b/a
(−2b
a− 1
)
+a3
4b
− a2
4
]
= − e2
πε0a3
[
e−2b/a(−���ab
2− a2
2− a3
4b+
���ab
2+
a2
4
)+
a2
4
(ab− 1)
]= − e2
πε0a3
[e−2b/a
(−a2
4
)(ab
+ 1)
+a2
4
(ab− 1)]
=e2
4πε0a
[(1− a
b
)+(
1 +a
b
)e−2b/a
]
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 7 / 21
Problem 7.33 (cont.)
∆E = − e2
πε0a3
{1
b
[−a
2r2e−2r/a + a
(a2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
−[(a
2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
}
= − e2
πε0a3
[− a
2bb2e−2b/a +
a3
4be−2b/a
(−2b
a− 1
)− a2
4e−2b/a
(−2b
a− 1
)+
a3
4b− a2
4
]
= − e2
πε0a3
[
e−2b/a(−���ab
2− a2
2− a3
4b+
���ab
2+
a2
4
)+
a2
4
(ab− 1)
]= − e2
πε0a3
[e−2b/a
(−a2
4
)(ab
+ 1)
+a2
4
(ab− 1)]
=e2
4πε0a
[(1− a
b
)+(
1 +a
b
)e−2b/a
]
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 7 / 21
Problem 7.33 (cont.)
∆E = − e2
πε0a3
{1
b
[−a
2r2e−2r/a + a
(a2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
−[(a
2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
}
= − e2
πε0a3
[− a
2bb2e−2b/a +
a3
4be−2b/a
(−2b
a− 1
)− a2
4e−2b/a
(−2b
a− 1
)+
a3
4b− a2
4
]= − e2
πε0a3
[
e−2b/a(−���ab
2− a2
2− a3
4b+
���ab
2+
a2
4
)+
a2
4
(ab− 1)
]
= − e2
πε0a3
[e−2b/a
(−a2
4
)(ab
+ 1)
+a2
4
(ab− 1)]
=e2
4πε0a
[(1− a
b
)+(
1 +a
b
)e−2b/a
]
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 7 / 21
Problem 7.33 (cont.)
∆E = − e2
πε0a3
{1
b
[−a
2r2e−2r/a + a
(a2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
−[(a
2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
}
= − e2
πε0a3
[− a
2bb2e−2b/a +
a3
4be−2b/a
(−2b
a− 1
)− a2
4e−2b/a
(−2b
a− 1
)+
a3
4b− a2
4
]= − e2
πε0a3
[e−2b/a
(−ab
2− a2
2− a3
4b+
ab
2+
a2
4
)
+a2
4
(ab− 1)
]
= − e2
πε0a3
[e−2b/a
(−a2
4
)(ab
+ 1)
+a2
4
(ab− 1)]
=e2
4πε0a
[(1− a
b
)+(
1 +a
b
)e−2b/a
]
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 7 / 21
Problem 7.33 (cont.)
∆E = − e2
πε0a3
{1
b
[−a
2r2e−2r/a + a
(a2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
−[(a
2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
}
= − e2
πε0a3
[− a
2bb2e−2b/a +
a3
4be−2b/a
(−2b
a− 1
)− a2
4e−2b/a
(−2b
a− 1
)+
a3
4b− a2
4
]= − e2
πε0a3
[e−2b/a
(−ab
2− a2
2− a3
4b+
ab
2+
a2
4
)+
a2
4
(ab− 1)]
= − e2
πε0a3
[e−2b/a
(−a2
4
)(ab
+ 1)
+a2
4
(ab− 1)]
=e2
4πε0a
[(1− a
b
)+(
1 +a
b
)e−2b/a
]
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 7 / 21
Problem 7.33 (cont.)
∆E = − e2
πε0a3
{1
b
[−a
2r2e−2r/a + a
(a2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
−[(a
2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
}
= − e2
πε0a3
[− a
2bb2e−2b/a +
a3
4be−2b/a
(−2b
a− 1
)− a2
4e−2b/a
(−2b
a− 1
)+
a3
4b− a2
4
]= − e2
πε0a3
[e−2b/a
(−���ab
2− a2
2− a3
4b+
���ab
2+
a2
4
)+
a2
4
(ab− 1)]
= − e2
πε0a3
[e−2b/a
(−a2
4
)(ab
+ 1)
+a2
4
(ab− 1)]
=e2
4πε0a
[(1− a
b
)+(
1 +a
b
)e−2b/a
]
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 7 / 21
Problem 7.33 (cont.)
∆E = − e2
πε0a3
{1
b
[−a
2r2e−2r/a + a
(a2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
−[(a
2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
}
= − e2
πε0a3
[− a
2bb2e−2b/a +
a3
4be−2b/a
(−2b
a− 1
)− a2
4e−2b/a
(−2b
a− 1
)+
a3
4b− a2
4
]= − e2
πε0a3
[e−2b/a
(−���ab
2− a2
2− a3
4b+
���ab
2+
a2
4
)+
a2
4
(ab− 1)]
= − e2
πε0a3
[e−2b/a
(−a2
4
)(ab
+ 1)
+a2
4
(ab− 1)]
=e2
4πε0a
[(1− a
b
)+(
1 +a
b
)e−2b/a
]
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 7 / 21
Problem 7.33 (cont.)
∆E = − e2
πε0a3
{1
b
[−a
2r2e−2r/a + a
(a2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
−[(a
2
)2e−2r/a
(−2r
a− 1
)∣∣∣∣b0
}
= − e2
πε0a3
[− a
2bb2e−2b/a +
a3
4be−2b/a
(−2b
a− 1
)− a2
4e−2b/a
(−2b
a− 1
)+
a3
4b− a2
4
]= − e2
πε0a3
[e−2b/a
(−���ab
2− a2
2− a3
4b+
���ab
2+
a2
4
)+
a2
4
(ab− 1)]
= − e2
πε0a3
[e−2b/a
(−a2
4
)(ab
+ 1)
+a2
4
(ab− 1)]
=e2
4πε0a
[(1− a
b
)+(
1 +a
b
)e−2b/a
]C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 7 / 21
Problem 7.33 (cont.)
∆E =e2
4πε0a
[(1− a
b
)+(
1 +a
b
)e−2b/a
]
if we take λ = 2b/a� 1, we can expand the exponential
∆E =e2
4πε0a
[(1− 2
λ
)+
(1 +
2
λ
)(1− λ+
λ2
2− λ3
6+ · · ·
)]=
e2
4πε0a
[�1−
���2
λ+ �1 +
���2
λ−�λ− �2 +
λ2
2+�λ−
λ3
6− λ2
3+ O(λ3) + · · ·
]=
e2
4πε0a
[λ2
6+ O(λ3) + O(λ4) + · · ·
]≈ e2
4πε0a
4b2
6a2=
e2b2
6πε0a3
the energy of the ground state isbut since
a =4πε0~2
me2E = − me4
2(4πε0)2~2= − e2
8πε0a
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 8 / 21
Problem 7.33 (cont.)
∆E =e2
4πε0a
[(1− a
b
)+(
1 +a
b
)e−2b/a
]if we take λ = 2b/a� 1, we can expand the exponential
∆E =e2
4πε0a
[(1− 2
λ
)+
(1 +
2
λ
)(1− λ+
λ2
2− λ3
6+ · · ·
)]=
e2
4πε0a
[�1−
���2
λ+ �1 +
���2
λ−�λ− �2 +
λ2
2+�λ−
λ3
6− λ2
3+ O(λ3) + · · ·
]=
e2
4πε0a
[λ2
6+ O(λ3) + O(λ4) + · · ·
]≈ e2
4πε0a
4b2
6a2=
e2b2
6πε0a3
the energy of the ground state isbut since
a =4πε0~2
me2E = − me4
2(4πε0)2~2= − e2
8πε0a
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 8 / 21
Problem 7.33 (cont.)
∆E =e2
4πε0a
[(1− a
b
)+(
1 +a
b
)e−2b/a
]if we take λ = 2b/a� 1, we can expand the exponential
∆E =e2
4πε0a
[(1− 2
λ
)+
(1 +
2
λ
)(1− λ+
λ2
2− λ3
6+ · · ·
)]
=e2
4πε0a
[�1−
���2
λ+ �1 +
���2
λ−�λ− �2 +
λ2
2+�λ−
λ3
6− λ2
3+ O(λ3) + · · ·
]=
e2
4πε0a
[λ2
6+ O(λ3) + O(λ4) + · · ·
]≈ e2
4πε0a
4b2
6a2=
e2b2
6πε0a3
the energy of the ground state isbut since
a =4πε0~2
me2E = − me4
2(4πε0)2~2= − e2
8πε0a
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 8 / 21
Problem 7.33 (cont.)
∆E =e2
4πε0a
[(1− a
b
)+(
1 +a
b
)e−2b/a
]if we take λ = 2b/a� 1, we can expand the exponential
∆E =e2
4πε0a
[(1− 2
λ
)+
(1 +
2
λ
)(1− λ+
λ2
2− λ3
6+ · · ·
)]=
e2
4πε0a
[1− 2
λ+ 1 +
2
λ− λ− 2 +
λ2
2+ λ− λ3
6− λ2
3+ O(λ3) + · · ·
]
=e2
4πε0a
[λ2
6+ O(λ3) + O(λ4) + · · ·
]≈ e2
4πε0a
4b2
6a2=
e2b2
6πε0a3
the energy of the ground state isbut since
a =4πε0~2
me2E = − me4
2(4πε0)2~2= − e2
8πε0a
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 8 / 21
Problem 7.33 (cont.)
∆E =e2
4πε0a
[(1− a
b
)+(
1 +a
b
)e−2b/a
]if we take λ = 2b/a� 1, we can expand the exponential
∆E =e2
4πε0a
[(1− 2
λ
)+
(1 +
2
λ
)(1− λ+
λ2
2− λ3
6+ · · ·
)]=
e2
4πε0a
[�1−
2
λ+ �1 +
2
λ− λ− �2 +
λ2
2+ λ− λ3
6− λ2
3+ O(λ3) + · · ·
]
=e2
4πε0a
[λ2
6+ O(λ3) + O(λ4) + · · ·
]≈ e2
4πε0a
4b2
6a2=
e2b2
6πε0a3
the energy of the ground state isbut since
a =4πε0~2
me2E = − me4
2(4πε0)2~2= − e2
8πε0a
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 8 / 21
Problem 7.33 (cont.)
∆E =e2
4πε0a
[(1− a
b
)+(
1 +a
b
)e−2b/a
]if we take λ = 2b/a� 1, we can expand the exponential
∆E =e2
4πε0a
[(1− 2
λ
)+
(1 +
2
λ
)(1− λ+
λ2
2− λ3
6+ · · ·
)]=
e2
4πε0a
[�1−
���2
λ+ �1 +
���2
λ− λ− �2 +
λ2
2+ λ− λ3
6− λ2
3+ O(λ3) + · · ·
]
=e2
4πε0a
[λ2
6+ O(λ3) + O(λ4) + · · ·
]≈ e2
4πε0a
4b2
6a2=
e2b2
6πε0a3
the energy of the ground state isbut since
a =4πε0~2
me2E = − me4
2(4πε0)2~2= − e2
8πε0a
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 8 / 21
Problem 7.33 (cont.)
∆E =e2
4πε0a
[(1− a
b
)+(
1 +a
b
)e−2b/a
]if we take λ = 2b/a� 1, we can expand the exponential
∆E =e2
4πε0a
[(1− 2
λ
)+
(1 +
2
λ
)(1− λ+
λ2
2− λ3
6+ · · ·
)]=
e2
4πε0a
[�1−
���2
λ+ �1 +
���2
λ−�λ− �2 +
λ2
2+�λ−
λ3
6− λ2
3+ O(λ3) + · · ·
]
=e2
4πε0a
[λ2
6+ O(λ3) + O(λ4) + · · ·
]≈ e2
4πε0a
4b2
6a2=
e2b2
6πε0a3
the energy of the ground state isbut since
a =4πε0~2
me2E = − me4
2(4πε0)2~2= − e2
8πε0a
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 8 / 21
Problem 7.33 (cont.)
∆E =e2
4πε0a
[(1− a
b
)+(
1 +a
b
)e−2b/a
]if we take λ = 2b/a� 1, we can expand the exponential
∆E =e2
4πε0a
[(1− 2
λ
)+
(1 +
2
λ
)(1− λ+
λ2
2− λ3
6+ · · ·
)]=
e2
4πε0a
[�1−
���2
λ+ �1 +
���2
λ−�λ− �2 +
λ2
2+�λ−
λ3
6− λ2
3+ O(λ3) + · · ·
]=
e2
4πε0a
[λ2
6+ O(λ3) + O(λ4) + · · ·
]
≈ e2
4πε0a
4b2
6a2=
e2b2
6πε0a3
the energy of the ground state isbut since
a =4πε0~2
me2E = − me4
2(4πε0)2~2= − e2
8πε0a
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 8 / 21
Problem 7.33 (cont.)
∆E =e2
4πε0a
[(1− a
b
)+(
1 +a
b
)e−2b/a
]if we take λ = 2b/a� 1, we can expand the exponential
∆E =e2
4πε0a
[(1− 2
λ
)+
(1 +
2
λ
)(1− λ+
λ2
2− λ3
6+ · · ·
)]=
e2
4πε0a
[�1−
���2
λ+ �1 +
���2
λ−�λ− �2 +
λ2
2+�λ−
λ3
6− λ2
3+ O(λ3) + · · ·
]=
e2
4πε0a
[λ2
6+ O(λ3) + O(λ4) + · · ·
]≈ e2
4πε0a
4b2
6a2
=e2b2
6πε0a3
the energy of the ground state isbut since
a =4πε0~2
me2E = − me4
2(4πε0)2~2= − e2
8πε0a
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 8 / 21
Problem 7.33 (cont.)
∆E =e2
4πε0a
[(1− a
b
)+(
1 +a
b
)e−2b/a
]if we take λ = 2b/a� 1, we can expand the exponential
∆E =e2
4πε0a
[(1− 2
λ
)+
(1 +
2
λ
)(1− λ+
λ2
2− λ3
6+ · · ·
)]=
e2
4πε0a
[�1−
���2
λ+ �1 +
���2
λ−�λ− �2 +
λ2
2+�λ−
λ3
6− λ2
3+ O(λ3) + · · ·
]=
e2
4πε0a
[λ2
6+ O(λ3) + O(λ4) + · · ·
]≈ e2
4πε0a
4b2
6a2=
e2b2
6πε0a3
the energy of the ground state isbut since
a =4πε0~2
me2E = − me4
2(4πε0)2~2= − e2
8πε0a
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 8 / 21
Problem 7.33 (cont.)
∆E =e2
4πε0a
[(1− a
b
)+(
1 +a
b
)e−2b/a
]if we take λ = 2b/a� 1, we can expand the exponential
∆E =e2
4πε0a
[(1− 2
λ
)+
(1 +
2
λ
)(1− λ+
λ2
2− λ3
6+ · · ·
)]=
e2
4πε0a
[�1−
���2
λ+ �1 +
���2
λ−�λ− �2 +
λ2
2+�λ−
λ3
6− λ2
3+ O(λ3) + · · ·
]=
e2
4πε0a
[λ2
6+ O(λ3) + O(λ4) + · · ·
]≈ e2
4πε0a
4b2
6a2=
e2b2
6πε0a3
the energy of the ground state is
but since
a =4πε0~2
me2
E = − me4
2(4πε0)2~2
= − e2
8πε0a
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 8 / 21
Problem 7.33 (cont.)
∆E =e2
4πε0a
[(1− a
b
)+(
1 +a
b
)e−2b/a
]if we take λ = 2b/a� 1, we can expand the exponential
∆E =e2
4πε0a
[(1− 2
λ
)+
(1 +
2
λ
)(1− λ+
λ2
2− λ3
6+ · · ·
)]=
e2
4πε0a
[�1−
���2
λ+ �1 +
���2
λ−�λ− �2 +
λ2
2+�λ−
λ3
6− λ2
3+ O(λ3) + · · ·
]=
e2
4πε0a
[λ2
6+ O(λ3) + O(λ4) + · · ·
]≈ e2
4πε0a
4b2
6a2=
e2b2
6πε0a3
the energy of the ground state isbut since
a =4πε0~2
me2E = − me4
2(4πε0)2~2
= − e2
8πε0a
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 8 / 21
Problem 7.33 (cont.)
∆E =e2
4πε0a
[(1− a
b
)+(
1 +a
b
)e−2b/a
]if we take λ = 2b/a� 1, we can expand the exponential
∆E =e2
4πε0a
[(1− 2
λ
)+
(1 +
2
λ
)(1− λ+
λ2
2− λ3
6+ · · ·
)]=
e2
4πε0a
[�1−
���2
λ+ �1 +
���2
λ−�λ− �2 +
λ2
2+�λ−
λ3
6− λ2
3+ O(λ3) + · · ·
]=
e2
4πε0a
[λ2
6+ O(λ3) + O(λ4) + · · ·
]≈ e2
4πε0a
4b2
6a2=
e2b2
6πε0a3
the energy of the ground state isbut since
a =4πε0~2
me2E = − me4
2(4πε0)2~2= − e2
8πε0a
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 8 / 21
Problem 7.33 (cont.)
The ratio of ∆E to E isthus
taking a ≈ 5× 10−11 and
b ≈ 10−15
fine structure
hyperfine structure
∆E
E=
(e2b2
6πε0a3
)(−8πε0a
e2
)= −4
3
(b
a
)2≈ −4
3
(10−15
5× 10−11
)2≈ −5× 10−10
∆E
E≈ α2 = 5× 10−5
∆E
E≈ me
mpα2 = 3× 10−8
this correction is 100 times smaller than the characteristic energy of thehyperfine structure
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 9 / 21
Problem 7.33 (cont.)
The ratio of ∆E to E isthus
taking a ≈ 5× 10−11 and
b ≈ 10−15
fine structure
hyperfine structure
∆E
E=
(e2b2
6πε0a3
)(−8πε0a
e2
)
= −4
3
(b
a
)2≈ −4
3
(10−15
5× 10−11
)2≈ −5× 10−10
∆E
E≈ α2 = 5× 10−5
∆E
E≈ me
mpα2 = 3× 10−8
this correction is 100 times smaller than the characteristic energy of thehyperfine structure
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 9 / 21
Problem 7.33 (cont.)
The ratio of ∆E to E isthus
taking a ≈ 5× 10−11 and
b ≈ 10−15
fine structure
hyperfine structure
∆E
E=
(e2b2
6πε0a3
)(−8πε0a
e2
)= −4
3
(b
a
)2
≈ −4
3
(10−15
5× 10−11
)2≈ −5× 10−10
∆E
E≈ α2 = 5× 10−5
∆E
E≈ me
mpα2 = 3× 10−8
this correction is 100 times smaller than the characteristic energy of thehyperfine structure
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 9 / 21
Problem 7.33 (cont.)
The ratio of ∆E to E isthus
taking a ≈ 5× 10−11 and
b ≈ 10−15
fine structure
hyperfine structure
∆E
E=
(e2b2
6πε0a3
)(−8πε0a
e2
)= −4
3
(b
a
)2
≈ −4
3
(10−15
5× 10−11
)2≈ −5× 10−10
∆E
E≈ α2 = 5× 10−5
∆E
E≈ me
mpα2 = 3× 10−8
this correction is 100 times smaller than the characteristic energy of thehyperfine structure
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 9 / 21
Problem 7.33 (cont.)
The ratio of ∆E to E isthus
taking a ≈ 5× 10−11 and
b ≈ 10−15
fine structure
hyperfine structure
∆E
E=
(e2b2
6πε0a3
)(−8πε0a
e2
)= −4
3
(b
a
)2≈ −4
3
(10−15
5× 10−11
)2
≈ −5× 10−10
∆E
E≈ α2 = 5× 10−5
∆E
E≈ me
mpα2 = 3× 10−8
this correction is 100 times smaller than the characteristic energy of thehyperfine structure
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 9 / 21
Problem 7.33 (cont.)
The ratio of ∆E to E isthus
taking a ≈ 5× 10−11 and
b ≈ 10−15
fine structure
hyperfine structure
∆E
E=
(e2b2
6πε0a3
)(−8πε0a
e2
)= −4
3
(b
a
)2≈ −4
3
(10−15
5× 10−11
)2≈ −5× 10−10
∆E
E≈ α2 = 5× 10−5
∆E
E≈ me
mpα2 = 3× 10−8
this correction is 100 times smaller than the characteristic energy of thehyperfine structure
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 9 / 21
Problem 7.33 (cont.)
The ratio of ∆E to E isthus
taking a ≈ 5× 10−11 and
b ≈ 10−15
fine structure
hyperfine structure
∆E
E=
(e2b2
6πε0a3
)(−8πε0a
e2
)= −4
3
(b
a
)2≈ −4
3
(10−15
5× 10−11
)2≈ −5× 10−10
∆E
E≈ α2 = 5× 10−5
∆E
E≈ me
mpα2 = 3× 10−8
this correction is 100 times smaller than the characteristic energy of thehyperfine structure
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 9 / 21
Problem 7.33 (cont.)
The ratio of ∆E to E isthus
taking a ≈ 5× 10−11 and
b ≈ 10−15
fine structure
hyperfine structure
∆E
E=
(e2b2
6πε0a3
)(−8πε0a
e2
)= −4
3
(b
a
)2≈ −4
3
(10−15
5× 10−11
)2≈ −5× 10−10
∆E
E≈ α2 = 5× 10−5
∆E
E≈ me
mpα2 = 3× 10−8
this correction is 100 times smaller than the characteristic energy of thehyperfine structure
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 9 / 21
Problem 7.33 (cont.)
The ratio of ∆E to E isthus
taking a ≈ 5× 10−11 and
b ≈ 10−15
fine structure
hyperfine structure
∆E
E=
(e2b2
6πε0a3
)(−8πε0a
e2
)= −4
3
(b
a
)2≈ −4
3
(10−15
5× 10−11
)2≈ −5× 10−10
∆E
E≈ α2 = 5× 10−5
∆E
E≈ me
mpα2 = 3× 10−8
this correction is 100 times smaller than the characteristic energy of thehyperfine structure
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 9 / 21
Problem 7.36
For the isotropic 3D harmonic oscillator, discuss the effect (in first order)of the perturbation
H ′ = λx2yz
(for some constant λ) on (a) the ground state, and (b) the (triplydegenerate) first excited state.
(a) The unperturbed ground state is simply a product state and the firstorder correction to the energy is
|0 0 0〉 = ψ0(x)ψ0(y)ψ0(z)
E 10 = 〈0 0 0|H ′|0 0 0〉 = λ〈0|x2|0〉x〈0|y |0〉y 〈0|z |0〉z = 0
(b) The triply degenerate excited states are
|0 0 1〉 = ψ0(x)ψ0(y)ψ1(z)
|0 1 0〉 = ψ0(x)ψ1(y)ψ0(z), |1 0 0〉 = ψ1(x)ψ0(y)ψ0(z)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 10 / 21
Problem 7.36
For the isotropic 3D harmonic oscillator, discuss the effect (in first order)of the perturbation
H ′ = λx2yz
(for some constant λ) on (a) the ground state, and (b) the (triplydegenerate) first excited state.
(a) The unperturbed ground state is simply a product state
and the firstorder correction to the energy is
|0 0 0〉 = ψ0(x)ψ0(y)ψ0(z)
E 10 = 〈0 0 0|H ′|0 0 0〉 = λ〈0|x2|0〉x〈0|y |0〉y 〈0|z |0〉z = 0
(b) The triply degenerate excited states are
|0 0 1〉 = ψ0(x)ψ0(y)ψ1(z)
|0 1 0〉 = ψ0(x)ψ1(y)ψ0(z), |1 0 0〉 = ψ1(x)ψ0(y)ψ0(z)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 10 / 21
Problem 7.36
For the isotropic 3D harmonic oscillator, discuss the effect (in first order)of the perturbation
H ′ = λx2yz
(for some constant λ) on (a) the ground state, and (b) the (triplydegenerate) first excited state.
(a) The unperturbed ground state is simply a product state
and the firstorder correction to the energy is
|0 0 0〉 = ψ0(x)ψ0(y)ψ0(z)
E 10 = 〈0 0 0|H ′|0 0 0〉 = λ〈0|x2|0〉x〈0|y |0〉y 〈0|z |0〉z = 0
(b) The triply degenerate excited states are
|0 0 1〉 = ψ0(x)ψ0(y)ψ1(z)
|0 1 0〉 = ψ0(x)ψ1(y)ψ0(z), |1 0 0〉 = ψ1(x)ψ0(y)ψ0(z)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 10 / 21
Problem 7.36
For the isotropic 3D harmonic oscillator, discuss the effect (in first order)of the perturbation
H ′ = λx2yz
(for some constant λ) on (a) the ground state, and (b) the (triplydegenerate) first excited state.
(a) The unperturbed ground state is simply a product state and the firstorder correction to the energy is
|0 0 0〉 = ψ0(x)ψ0(y)ψ0(z)
E 10 = 〈0 0 0|H ′|0 0 0〉 = λ〈0|x2|0〉x〈0|y |0〉y 〈0|z |0〉z = 0
(b) The triply degenerate excited states are
|0 0 1〉 = ψ0(x)ψ0(y)ψ1(z)
|0 1 0〉 = ψ0(x)ψ1(y)ψ0(z), |1 0 0〉 = ψ1(x)ψ0(y)ψ0(z)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 10 / 21
Problem 7.36
For the isotropic 3D harmonic oscillator, discuss the effect (in first order)of the perturbation
H ′ = λx2yz
(for some constant λ) on (a) the ground state, and (b) the (triplydegenerate) first excited state.
(a) The unperturbed ground state is simply a product state and the firstorder correction to the energy is
|0 0 0〉 = ψ0(x)ψ0(y)ψ0(z)
E 10 = 〈0 0 0|H ′|0 0 0〉
= λ〈0|x2|0〉x〈0|y |0〉y 〈0|z |0〉z = 0
(b) The triply degenerate excited states are
|0 0 1〉 = ψ0(x)ψ0(y)ψ1(z)
|0 1 0〉 = ψ0(x)ψ1(y)ψ0(z), |1 0 0〉 = ψ1(x)ψ0(y)ψ0(z)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 10 / 21
Problem 7.36
For the isotropic 3D harmonic oscillator, discuss the effect (in first order)of the perturbation
H ′ = λx2yz
(for some constant λ) on (a) the ground state, and (b) the (triplydegenerate) first excited state.
(a) The unperturbed ground state is simply a product state and the firstorder correction to the energy is
|0 0 0〉 = ψ0(x)ψ0(y)ψ0(z)
E 10 = 〈0 0 0|H ′|0 0 0〉 = λ〈0|x2|0〉x〈0|y |0〉y 〈0|z |0〉z
= 0
(b) The triply degenerate excited states are
|0 0 1〉 = ψ0(x)ψ0(y)ψ1(z)
|0 1 0〉 = ψ0(x)ψ1(y)ψ0(z), |1 0 0〉 = ψ1(x)ψ0(y)ψ0(z)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 10 / 21
Problem 7.36
For the isotropic 3D harmonic oscillator, discuss the effect (in first order)of the perturbation
H ′ = λx2yz
(for some constant λ) on (a) the ground state, and (b) the (triplydegenerate) first excited state.
(a) The unperturbed ground state is simply a product state and the firstorder correction to the energy is
|0 0 0〉 = ψ0(x)ψ0(y)ψ0(z)
E 10 = 〈0 0 0|H ′|0 0 0〉 = λ〈0|x2|0〉x〈0|y |0〉y 〈0|z |0〉z = 0
(b) The triply degenerate excited states are
|0 0 1〉 = ψ0(x)ψ0(y)ψ1(z)
|0 1 0〉 = ψ0(x)ψ1(y)ψ0(z), |1 0 0〉 = ψ1(x)ψ0(y)ψ0(z)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 10 / 21
Problem 7.36
For the isotropic 3D harmonic oscillator, discuss the effect (in first order)of the perturbation
H ′ = λx2yz
(for some constant λ) on (a) the ground state, and (b) the (triplydegenerate) first excited state.
(a) The unperturbed ground state is simply a product state and the firstorder correction to the energy is
|0 0 0〉 = ψ0(x)ψ0(y)ψ0(z)
E 10 = 〈0 0 0|H ′|0 0 0〉 = λ〈0|x2|0〉x〈0|y |0〉y 〈0|z |0〉z = 0
(b) The triply degenerate excited states are
|0 0 1〉 = ψ0(x)ψ0(y)ψ1(z)
|0 1 0〉 = ψ0(x)ψ1(y)ψ0(z), |1 0 0〉 = ψ1(x)ψ0(y)ψ0(z)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 10 / 21
Problem 7.36
For the isotropic 3D harmonic oscillator, discuss the effect (in first order)of the perturbation
H ′ = λx2yz
(for some constant λ) on (a) the ground state, and (b) the (triplydegenerate) first excited state.
(a) The unperturbed ground state is simply a product state and the firstorder correction to the energy is
|0 0 0〉 = ψ0(x)ψ0(y)ψ0(z)
E 10 = 〈0 0 0|H ′|0 0 0〉 = λ〈0|x2|0〉x〈0|y |0〉y 〈0|z |0〉z = 0
(b) The triply degenerate excited states are
|0 0 1〉 = ψ0(x)ψ0(y)ψ1(z)
|0 1 0〉 = ψ0(x)ψ1(y)ψ0(z), |1 0 0〉 = ψ1(x)ψ0(y)ψ0(z)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 10 / 21
Problem 7.36 (cont.)
The matrix elements of the perturbing Hamiltonian are thus
〈0 0 1|H ′|0 0 1〉 = λ〈0|x2|0〉x〈0|y |0〉y 〈1|z |1〉z = 0
〈0 1 0|H ′|0 1 0〉 = λ〈0|x2|0〉x〈1|y |1〉y 〈0|z |0〉z = 0
〈1 0 0|H ′|1 0 0〉 = λ〈1|x2|1〉x〈0|y |0〉y 〈0|z |0〉z = 0
〈0 0 1|H ′|0 1 0〉 = λ〈0|x2|0〉x〈0|y |1〉y 〈1|z |0〉z = λ
(~
2mω
)2〈0 0 1|H ′|1 0 0〉 = λ〈0|x2|1〉x〈0|y |0〉y 〈1|z |0〉z = 0
〈0 1 0|H ′|1 0 0〉 = λ〈0|x2|1〉x〈1|y |0〉y 〈0|z |0〉z = 0
since the only two matrix elements that are non-zero mix the |0 0 1〉 and|0 1 0〉 states, the eigenvalues of the H ′ matrix are
E = 0,±λ(
~2mω
)2
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 11 / 21
Problem 7.36 (cont.)
The matrix elements of the perturbing Hamiltonian are thus
〈0 0 1|H ′|0 0 1〉 = λ〈0|x2|0〉x〈0|y |0〉y 〈1|z |1〉z = 0
〈0 1 0|H ′|0 1 0〉 = λ〈0|x2|0〉x〈1|y |1〉y 〈0|z |0〉z = 0
〈1 0 0|H ′|1 0 0〉 = λ〈1|x2|1〉x〈0|y |0〉y 〈0|z |0〉z = 0
〈0 0 1|H ′|0 1 0〉 = λ〈0|x2|0〉x〈0|y |1〉y 〈1|z |0〉z = λ
(~
2mω
)2〈0 0 1|H ′|1 0 0〉 = λ〈0|x2|1〉x〈0|y |0〉y 〈1|z |0〉z = 0
〈0 1 0|H ′|1 0 0〉 = λ〈0|x2|1〉x〈1|y |0〉y 〈0|z |0〉z = 0
since the only two matrix elements that are non-zero mix the |0 0 1〉 and|0 1 0〉 states, the eigenvalues of the H ′ matrix are
E = 0,±λ(
~2mω
)2
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 11 / 21
Problem 7.36 (cont.)
The matrix elements of the perturbing Hamiltonian are thus
〈0 0 1|H ′|0 0 1〉 = λ〈0|x2|0〉x〈0|y |0〉y 〈1|z |1〉z = 0
〈0 1 0|H ′|0 1 0〉 = λ〈0|x2|0〉x〈1|y |1〉y 〈0|z |0〉z = 0
〈1 0 0|H ′|1 0 0〉 = λ〈1|x2|1〉x〈0|y |0〉y 〈0|z |0〉z = 0
〈0 0 1|H ′|0 1 0〉 = λ〈0|x2|0〉x〈0|y |1〉y 〈1|z |0〉z = λ
(~
2mω
)2〈0 0 1|H ′|1 0 0〉 = λ〈0|x2|1〉x〈0|y |0〉y 〈1|z |0〉z = 0
〈0 1 0|H ′|1 0 0〉 = λ〈0|x2|1〉x〈1|y |0〉y 〈0|z |0〉z = 0
since the only two matrix elements that are non-zero mix the |0 0 1〉 and|0 1 0〉 states, the eigenvalues of the H ′ matrix are
E = 0,±λ(
~2mω
)2
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 11 / 21
Problem 7.36 (cont.)
The matrix elements of the perturbing Hamiltonian are thus
〈0 0 1|H ′|0 0 1〉 = λ〈0|x2|0〉x〈0|y |0〉y 〈1|z |1〉z = 0
〈0 1 0|H ′|0 1 0〉 = λ〈0|x2|0〉x〈1|y |1〉y 〈0|z |0〉z = 0
〈1 0 0|H ′|1 0 0〉 = λ〈1|x2|1〉x〈0|y |0〉y 〈0|z |0〉z = 0
〈0 0 1|H ′|0 1 0〉 = λ〈0|x2|0〉x〈0|y |1〉y 〈1|z |0〉z = λ
(~
2mω
)2〈0 0 1|H ′|1 0 0〉 = λ〈0|x2|1〉x〈0|y |0〉y 〈1|z |0〉z = 0
〈0 1 0|H ′|1 0 0〉 = λ〈0|x2|1〉x〈1|y |0〉y 〈0|z |0〉z = 0
since the only two matrix elements that are non-zero mix the |0 0 1〉 and|0 1 0〉 states, the eigenvalues of the H ′ matrix are
E = 0,±λ(
~2mω
)2
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 11 / 21
Problem 7.36 (cont.)
The matrix elements of the perturbing Hamiltonian are thus
〈0 0 1|H ′|0 0 1〉 = λ〈0|x2|0〉x〈0|y |0〉y 〈1|z |1〉z = 0
〈0 1 0|H ′|0 1 0〉 = λ〈0|x2|0〉x〈1|y |1〉y 〈0|z |0〉z = 0
〈1 0 0|H ′|1 0 0〉 = λ〈1|x2|1〉x〈0|y |0〉y 〈0|z |0〉z = 0
〈0 0 1|H ′|0 1 0〉 = λ〈0|x2|0〉x〈0|y |1〉y 〈1|z |0〉z
= λ
(~
2mω
)2〈0 0 1|H ′|1 0 0〉 = λ〈0|x2|1〉x〈0|y |0〉y 〈1|z |0〉z = 0
〈0 1 0|H ′|1 0 0〉 = λ〈0|x2|1〉x〈1|y |0〉y 〈0|z |0〉z = 0
since the only two matrix elements that are non-zero mix the |0 0 1〉 and|0 1 0〉 states, the eigenvalues of the H ′ matrix are
E = 0,±λ(
~2mω
)2
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 11 / 21
Problem 7.36 (cont.)
The matrix elements of the perturbing Hamiltonian are thus
〈0 0 1|H ′|0 0 1〉 = λ〈0|x2|0〉x〈0|y |0〉y 〈1|z |1〉z = 0
〈0 1 0|H ′|0 1 0〉 = λ〈0|x2|0〉x〈1|y |1〉y 〈0|z |0〉z = 0
〈1 0 0|H ′|1 0 0〉 = λ〈1|x2|1〉x〈0|y |0〉y 〈0|z |0〉z = 0
〈0 0 1|H ′|0 1 0〉 = λ〈0|x2|0〉x〈0|y |1〉y 〈1|z |0〉z = λ
(~
2mω
)2
〈0 0 1|H ′|1 0 0〉 = λ〈0|x2|1〉x〈0|y |0〉y 〈1|z |0〉z = 0
〈0 1 0|H ′|1 0 0〉 = λ〈0|x2|1〉x〈1|y |0〉y 〈0|z |0〉z = 0
since the only two matrix elements that are non-zero mix the |0 0 1〉 and|0 1 0〉 states, the eigenvalues of the H ′ matrix are
E = 0,±λ(
~2mω
)2
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 11 / 21
Problem 7.36 (cont.)
The matrix elements of the perturbing Hamiltonian are thus
〈0 0 1|H ′|0 0 1〉 = λ〈0|x2|0〉x〈0|y |0〉y 〈1|z |1〉z = 0
〈0 1 0|H ′|0 1 0〉 = λ〈0|x2|0〉x〈1|y |1〉y 〈0|z |0〉z = 0
〈1 0 0|H ′|1 0 0〉 = λ〈1|x2|1〉x〈0|y |0〉y 〈0|z |0〉z = 0
〈0 0 1|H ′|0 1 0〉 = λ〈0|x2|0〉x〈0|y |1〉y 〈1|z |0〉z = λ
(~
2mω
)2〈0 0 1|H ′|1 0 0〉 = λ〈0|x2|1〉x〈0|y |0〉y 〈1|z |0〉z = 0
〈0 1 0|H ′|1 0 0〉 = λ〈0|x2|1〉x〈1|y |0〉y 〈0|z |0〉z = 0
since the only two matrix elements that are non-zero mix the |0 0 1〉 and|0 1 0〉 states, the eigenvalues of the H ′ matrix are
E = 0,±λ(
~2mω
)2
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 11 / 21
Problem 7.36 (cont.)
The matrix elements of the perturbing Hamiltonian are thus
〈0 0 1|H ′|0 0 1〉 = λ〈0|x2|0〉x〈0|y |0〉y 〈1|z |1〉z = 0
〈0 1 0|H ′|0 1 0〉 = λ〈0|x2|0〉x〈1|y |1〉y 〈0|z |0〉z = 0
〈1 0 0|H ′|1 0 0〉 = λ〈1|x2|1〉x〈0|y |0〉y 〈0|z |0〉z = 0
〈0 0 1|H ′|0 1 0〉 = λ〈0|x2|0〉x〈0|y |1〉y 〈1|z |0〉z = λ
(~
2mω
)2〈0 0 1|H ′|1 0 0〉 = λ〈0|x2|1〉x〈0|y |0〉y 〈1|z |0〉z = 0
〈0 1 0|H ′|1 0 0〉 = λ〈0|x2|1〉x〈1|y |0〉y 〈0|z |0〉z = 0
since the only two matrix elements that are non-zero mix the |0 0 1〉 and|0 1 0〉 states, the eigenvalues of the H ′ matrix are
E = 0,±λ(
~2mω
)2
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 11 / 21
Problem 7.36 (cont.)
The matrix elements of the perturbing Hamiltonian are thus
〈0 0 1|H ′|0 0 1〉 = λ〈0|x2|0〉x〈0|y |0〉y 〈1|z |1〉z = 0
〈0 1 0|H ′|0 1 0〉 = λ〈0|x2|0〉x〈1|y |1〉y 〈0|z |0〉z = 0
〈1 0 0|H ′|1 0 0〉 = λ〈1|x2|1〉x〈0|y |0〉y 〈0|z |0〉z = 0
〈0 0 1|H ′|0 1 0〉 = λ〈0|x2|0〉x〈0|y |1〉y 〈1|z |0〉z = λ
(~
2mω
)2〈0 0 1|H ′|1 0 0〉 = λ〈0|x2|1〉x〈0|y |0〉y 〈1|z |0〉z = 0
〈0 1 0|H ′|1 0 0〉 = λ〈0|x2|1〉x〈1|y |0〉y 〈0|z |0〉z = 0
since the only two matrix elements that are non-zero mix the |0 0 1〉 and|0 1 0〉 states, the eigenvalues of the H ′ matrix are
E = 0,±λ(
~2mω
)2
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 11 / 21
Problem 7.36 (cont.)
The matrix elements of the perturbing Hamiltonian are thus
〈0 0 1|H ′|0 0 1〉 = λ〈0|x2|0〉x〈0|y |0〉y 〈1|z |1〉z = 0
〈0 1 0|H ′|0 1 0〉 = λ〈0|x2|0〉x〈1|y |1〉y 〈0|z |0〉z = 0
〈1 0 0|H ′|1 0 0〉 = λ〈1|x2|1〉x〈0|y |0〉y 〈0|z |0〉z = 0
〈0 0 1|H ′|0 1 0〉 = λ〈0|x2|0〉x〈0|y |1〉y 〈1|z |0〉z = λ
(~
2mω
)2〈0 0 1|H ′|1 0 0〉 = λ〈0|x2|1〉x〈0|y |0〉y 〈1|z |0〉z = 0
〈0 1 0|H ′|1 0 0〉 = λ〈0|x2|1〉x〈1|y |0〉y 〈0|z |0〉z = 0
since the only two matrix elements that are non-zero mix the |0 0 1〉 and|0 1 0〉 states, the eigenvalues of the H ′ matrix are
E = 0,±λ(
~2mω
)2C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 11 / 21
Hydrogen molecule
We can make an approximation to the energy of the full hydrogenmolecule by adding the second electron to the Hamiltonian
H = − ~2
2m
(∇2
1 +∇22
)+
e2
4πε0
(1
r12+
1
R− 1
r1− 1
r ′1− 1
r2− 1
r ′2
)
R
r1r2
r12
p
pr1
r2
For the variational wave function use the symmetricand antisymmetric combinations of two hydrogenatoms
ψ±(~r1,~r2) = A±[ψ0(r1)ψ0(r ′2)± ψ0(r ′1)ψ0(r2)
]what about spin?ψ+ is symmetric and requires χsinglet
ψ− is antisymmetric and requires χtriplet
it is not clear which will have the lowest energy so they both must becomputed
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 12 / 21
Hydrogen molecule
We can make an approximation to the energy of the full hydrogenmolecule by adding the second electron to the Hamiltonian
H = − ~2
2m
(∇2
1 +∇22
)+
e2
4πε0
(1
r12+
1
R− 1
r1− 1
r ′1− 1
r2− 1
r ′2
)
R
r1r2
r12
p
pr1
r2
For the variational wave function use the symmetricand antisymmetric combinations of two hydrogenatoms
ψ±(~r1,~r2) = A±[ψ0(r1)ψ0(r ′2)± ψ0(r ′1)ψ0(r2)
]what about spin?ψ+ is symmetric and requires χsinglet
ψ− is antisymmetric and requires χtriplet
it is not clear which will have the lowest energy so they both must becomputed
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 12 / 21
Hydrogen molecule
We can make an approximation to the energy of the full hydrogenmolecule by adding the second electron to the Hamiltonian
H = − ~2
2m
(∇2
1 +∇22
)+
e2
4πε0
(1
r12+
1
R− 1
r1− 1
r ′1− 1
r2− 1
r ′2
)
R
r1r2
r12
p
pr1
r2
For the variational wave function use the symmetricand antisymmetric combinations of two hydrogenatoms
ψ±(~r1,~r2) = A±[ψ0(r1)ψ0(r ′2)± ψ0(r ′1)ψ0(r2)
]what about spin?ψ+ is symmetric and requires χsinglet
ψ− is antisymmetric and requires χtriplet
it is not clear which will have the lowest energy so they both must becomputed
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 12 / 21
Hydrogen molecule
We can make an approximation to the energy of the full hydrogenmolecule by adding the second electron to the Hamiltonian
H = − ~2
2m
(∇2
1 +∇22
)+
e2
4πε0
(1
r12+
1
R− 1
r1− 1
r ′1− 1
r2− 1
r ′2
)
R
r1r2
r12
p
pr1
r2
For the variational wave function use the symmetricand antisymmetric combinations of two hydrogenatoms
ψ±(~r1,~r2) = A±[ψ0(r1)ψ0(r ′2)± ψ0(r ′1)ψ0(r2)
]what about spin?
ψ+ is symmetric and requires χsinglet
ψ− is antisymmetric and requires χtriplet
it is not clear which will have the lowest energy so they both must becomputed
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 12 / 21
Hydrogen molecule
We can make an approximation to the energy of the full hydrogenmolecule by adding the second electron to the Hamiltonian
H = − ~2
2m
(∇2
1 +∇22
)+
e2
4πε0
(1
r12+
1
R− 1
r1− 1
r ′1− 1
r2− 1
r ′2
)
R
r1r2
r12
p
pr1
r2
For the variational wave function use the symmetricand antisymmetric combinations of two hydrogenatoms
ψ±(~r1,~r2) = A±[ψ0(r1)ψ0(r ′2)± ψ0(r ′1)ψ0(r2)
]what about spin?ψ+ is symmetric and requires χsinglet
ψ− is antisymmetric and requires χtriplet
it is not clear which will have the lowest energy so they both must becomputed
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 12 / 21
Hydrogen molecule
We can make an approximation to the energy of the full hydrogenmolecule by adding the second electron to the Hamiltonian
H = − ~2
2m
(∇2
1 +∇22
)+
e2
4πε0
(1
r12+
1
R− 1
r1− 1
r ′1− 1
r2− 1
r ′2
)
R
r1r2
r12
p
pr1
r2
For the variational wave function use the symmetricand antisymmetric combinations of two hydrogenatoms
ψ±(~r1,~r2) = A±[ψ0(r1)ψ0(r ′2)± ψ0(r ′1)ψ0(r2)
]what about spin?ψ+ is symmetric and requires χsinglet
ψ− is antisymmetric and requires χtriplet
it is not clear which will have the lowest energy so they both must becomputed
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 12 / 21
Normalizing the trial function
ψ±(~r1,~r2) = A±[ψ0(r1)ψ0(r ′2)± ψ0(r ′1)ψ0(r2)
]First normalize both trial functions
1 =
∫ ∫|ψ±(~r1,~r2)|2 d3~r1 d
3~r2
= A2±
[∫ψ0(r1)2 d3~r1
∫ψ0(r ′2)2 d3~r2 +
∫ψ0(r ′1)2 d3~r1
∫ψ0(r2)2 d3~r2
±2
∫ψ0(r1)ψ0(r ′1) d3~r1
∫ψ0(r ′2)ψ0(r2) d3~r2
]
since the original wavefuctions werenormalized and we just computedthe overlap integral, I for the hy-drogen molecule ion
A± =1√
2(1± I 2)
I = e−R/a
[1 +
R
a+
1
3
(R
a
)2]
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 13 / 21
Normalizing the trial function
ψ±(~r1,~r2) = A±[ψ0(r1)ψ0(r ′2)± ψ0(r ′1)ψ0(r2)
]First normalize both trial functions
1 =
∫ ∫|ψ±(~r1,~r2)|2 d3~r1 d
3~r2
= A2±
[∫ψ0(r1)2 d3~r1
∫ψ0(r ′2)2 d3~r2 +
∫ψ0(r ′1)2 d3~r1
∫ψ0(r2)2 d3~r2
±2
∫ψ0(r1)ψ0(r ′1) d3~r1
∫ψ0(r ′2)ψ0(r2) d3~r2
]
since the original wavefuctions werenormalized and we just computedthe overlap integral, I for the hy-drogen molecule ion
A± =1√
2(1± I 2)
I = e−R/a
[1 +
R
a+
1
3
(R
a
)2]
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 13 / 21
Normalizing the trial function
ψ±(~r1,~r2) = A±[ψ0(r1)ψ0(r ′2)± ψ0(r ′1)ψ0(r2)
]First normalize both trial functions
1 =
∫ ∫|ψ±(~r1,~r2)|2 d3~r1 d
3~r2
= A2±
[∫ψ0(r1)2 d3~r1
∫ψ0(r ′2)2 d3~r2
+
∫ψ0(r ′1)2 d3~r1
∫ψ0(r2)2 d3~r2
±2
∫ψ0(r1)ψ0(r ′1) d3~r1
∫ψ0(r ′2)ψ0(r2) d3~r2
]
since the original wavefuctions werenormalized and we just computedthe overlap integral, I for the hy-drogen molecule ion
A± =1√
2(1± I 2)
I = e−R/a
[1 +
R
a+
1
3
(R
a
)2]
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 13 / 21
Normalizing the trial function
ψ±(~r1,~r2) = A±[ψ0(r1)ψ0(r ′2)± ψ0(r ′1)ψ0(r2)
]First normalize both trial functions
1 =
∫ ∫|ψ±(~r1,~r2)|2 d3~r1 d
3~r2
= A2±
[∫ψ0(r1)2 d3~r1
∫ψ0(r ′2)2 d3~r2 +
∫ψ0(r ′1)2 d3~r1
∫ψ0(r2)2 d3~r2
±2
∫ψ0(r1)ψ0(r ′1) d3~r1
∫ψ0(r ′2)ψ0(r2) d3~r2
]
since the original wavefuctions werenormalized and we just computedthe overlap integral, I for the hy-drogen molecule ion
A± =1√
2(1± I 2)
I = e−R/a
[1 +
R
a+
1
3
(R
a
)2]
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 13 / 21
Normalizing the trial function
ψ±(~r1,~r2) = A±[ψ0(r1)ψ0(r ′2)± ψ0(r ′1)ψ0(r2)
]First normalize both trial functions
1 =
∫ ∫|ψ±(~r1,~r2)|2 d3~r1 d
3~r2
= A2±
[∫ψ0(r1)2 d3~r1
∫ψ0(r ′2)2 d3~r2 +
∫ψ0(r ′1)2 d3~r1
∫ψ0(r2)2 d3~r2
±2
∫ψ0(r1)ψ0(r ′1) d3~r1
∫ψ0(r ′2)ψ0(r2) d3~r2
]
since the original wavefuctions werenormalized and we just computedthe overlap integral, I for the hy-drogen molecule ion
A± =1√
2(1± I 2)
I = e−R/a
[1 +
R
a+
1
3
(R
a
)2]
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 13 / 21
Normalizing the trial function
ψ±(~r1,~r2) = A±[ψ0(r1)ψ0(r ′2)± ψ0(r ′1)ψ0(r2)
]First normalize both trial functions
1 =
∫ ∫|ψ±(~r1,~r2)|2 d3~r1 d
3~r2
= A2±
[∫ψ0(r1)2 d3~r1
∫ψ0(r ′2)2 d3~r2 +
∫ψ0(r ′1)2 d3~r1
∫ψ0(r2)2 d3~r2
±2
∫ψ0(r1)ψ0(r ′1) d3~r1
∫ψ0(r ′2)ψ0(r2) d3~r2
]
since the original wavefuctions werenormalized and we just computedthe overlap integral, I for the hy-drogen molecule ion
A± =1√
2(1± I 2)
I = e−R/a
[1 +
R
a+
1
3
(R
a
)2]
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 13 / 21
Normalizing the trial function
ψ±(~r1,~r2) = A±[ψ0(r1)ψ0(r ′2)± ψ0(r ′1)ψ0(r2)
]First normalize both trial functions
1 =
∫ ∫|ψ±(~r1,~r2)|2 d3~r1 d
3~r2
= A2±
[∫ψ0(r1)2 d3~r1
∫ψ0(r ′2)2 d3~r2 +
∫ψ0(r ′1)2 d3~r1
∫ψ0(r2)2 d3~r2
±2
∫ψ0(r1)ψ0(r ′1) d3~r1
∫ψ0(r ′2)ψ0(r2) d3~r2
]
since the original wavefuctions werenormalized and we just computedthe overlap integral, I for the hy-drogen molecule ion
A± =1√
2(1± I 2)
I = e−R/a
[1 +
R
a+
1
3
(R
a
)2]
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 13 / 21
Normalizing the trial function
ψ±(~r1,~r2) = A±[ψ0(r1)ψ0(r ′2)± ψ0(r ′1)ψ0(r2)
]First normalize both trial functions
1 =
∫ ∫|ψ±(~r1,~r2)|2 d3~r1 d
3~r2
= A2±
[∫ψ0(r1)2 d3~r1
∫ψ0(r ′2)2 d3~r2 +
∫ψ0(r ′1)2 d3~r1
∫ψ0(r2)2 d3~r2
±2
∫ψ0(r1)ψ0(r ′1) d3~r1
∫ψ0(r ′2)ψ0(r2) d3~r2
]
since the original wavefuctions werenormalized and we just computedthe overlap integral, I for the hy-drogen molecule ion
A± =1√
2(1± I 2)
I = e−R/a
[1 +
R
a+
1
3
(R
a
)2]C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 13 / 21
Computing the energies
Using the same methods as for the hydrogen molecule ion, the kineticenergy
and the electron potential (all 4 terms) are⟨− ~2
2m∇2
1
⟩= E1 +
(e2
4πε0a
)1± IX
1± I 2=
⟨− ~2
2m∇2
2
⟩4
⟨− e2
4πε0r1
⟩= −4
2
(e2
4πε0a
)1 + D ± 2IX
1± I 2
the electron-electron energy becomes
〈Vee〉 =
(e2
4πε0a
)D2 ± X2
1± I 2
D2 =
∫ ∫|ψ0(r1)|2 a
r12|ψ0(r ′2)|2 d3~r1 d
3~r2
X2 =
∫ ∫ψ0(r1)ψ0(r ′1)
a
r12ψ0(r2)ψ0(r ′2) d3~r1 d
3~r2
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 14 / 21
Computing the energies
Using the same methods as for the hydrogen molecule ion, the kineticenergy
and the electron potential (all 4 terms) are
⟨− ~2
2m∇2
1
⟩= E1 +
(e2
4πε0a
)1± IX
1± I 2
=
⟨− ~2
2m∇2
2
⟩4
⟨− e2
4πε0r1
⟩= −4
2
(e2
4πε0a
)1 + D ± 2IX
1± I 2
the electron-electron energy becomes
〈Vee〉 =
(e2
4πε0a
)D2 ± X2
1± I 2
D2 =
∫ ∫|ψ0(r1)|2 a
r12|ψ0(r ′2)|2 d3~r1 d
3~r2
X2 =
∫ ∫ψ0(r1)ψ0(r ′1)
a
r12ψ0(r2)ψ0(r ′2) d3~r1 d
3~r2
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 14 / 21
Computing the energies
Using the same methods as for the hydrogen molecule ion, the kineticenergy
and the electron potential (all 4 terms) are
⟨− ~2
2m∇2
1
⟩= E1 +
(e2
4πε0a
)1± IX
1± I 2=
⟨− ~2
2m∇2
2
⟩
4
⟨− e2
4πε0r1
⟩= −4
2
(e2
4πε0a
)1 + D ± 2IX
1± I 2
the electron-electron energy becomes
〈Vee〉 =
(e2
4πε0a
)D2 ± X2
1± I 2
D2 =
∫ ∫|ψ0(r1)|2 a
r12|ψ0(r ′2)|2 d3~r1 d
3~r2
X2 =
∫ ∫ψ0(r1)ψ0(r ′1)
a
r12ψ0(r2)ψ0(r ′2) d3~r1 d
3~r2
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 14 / 21
Computing the energies
Using the same methods as for the hydrogen molecule ion, the kineticenergy and the electron potential (all 4 terms) are⟨
− ~2
2m∇2
1
⟩= E1 +
(e2
4πε0a
)1± IX
1± I 2=
⟨− ~2
2m∇2
2
⟩
4
⟨− e2
4πε0r1
⟩= −4
2
(e2
4πε0a
)1 + D ± 2IX
1± I 2
the electron-electron energy becomes
〈Vee〉 =
(e2
4πε0a
)D2 ± X2
1± I 2
D2 =
∫ ∫|ψ0(r1)|2 a
r12|ψ0(r ′2)|2 d3~r1 d
3~r2
X2 =
∫ ∫ψ0(r1)ψ0(r ′1)
a
r12ψ0(r2)ψ0(r ′2) d3~r1 d
3~r2
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 14 / 21
Computing the energies
Using the same methods as for the hydrogen molecule ion, the kineticenergy and the electron potential (all 4 terms) are⟨
− ~2
2m∇2
1
⟩= E1 +
(e2
4πε0a
)1± IX
1± I 2=
⟨− ~2
2m∇2
2
⟩4
⟨− e2
4πε0r1
⟩= −4
2
(e2
4πε0a
)1 + D ± 2IX
1± I 2
the electron-electron energy becomes
〈Vee〉 =
(e2
4πε0a
)D2 ± X2
1± I 2
D2 =
∫ ∫|ψ0(r1)|2 a
r12|ψ0(r ′2)|2 d3~r1 d
3~r2
X2 =
∫ ∫ψ0(r1)ψ0(r ′1)
a
r12ψ0(r2)ψ0(r ′2) d3~r1 d
3~r2
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 14 / 21
Computing the energies
Using the same methods as for the hydrogen molecule ion, the kineticenergy and the electron potential (all 4 terms) are⟨
− ~2
2m∇2
1
⟩= E1 +
(e2
4πε0a
)1± IX
1± I 2=
⟨− ~2
2m∇2
2
⟩4
⟨− e2
4πε0r1
⟩= −4
2
(e2
4πε0a
)1 + D ± 2IX
1± I 2
the electron-electron energy becomes
〈Vee〉 =
(e2
4πε0a
)D2 ± X2
1± I 2
D2 =
∫ ∫|ψ0(r1)|2 a
r12|ψ0(r ′2)|2 d3~r1 d
3~r2
X2 =
∫ ∫ψ0(r1)ψ0(r ′1)
a
r12ψ0(r2)ψ0(r ′2) d3~r1 d
3~r2
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 14 / 21
Computing the energies
Using the same methods as for the hydrogen molecule ion, the kineticenergy and the electron potential (all 4 terms) are⟨
− ~2
2m∇2
1
⟩= E1 +
(e2
4πε0a
)1± IX
1± I 2=
⟨− ~2
2m∇2
2
⟩4
⟨− e2
4πε0r1
⟩= −4
2
(e2
4πε0a
)1 + D ± 2IX
1± I 2
the electron-electron energy becomes
〈Vee〉 =
(e2
4πε0a
)D2 ± X2
1± I 2
D2 =
∫ ∫|ψ0(r1)|2 a
r12|ψ0(r ′2)|2 d3~r1 d
3~r2
X2 =
∫ ∫ψ0(r1)ψ0(r ′1)
a
r12ψ0(r2)ψ0(r ′2) d3~r1 d
3~r2
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 14 / 21
Computing the energies
Using the same methods as for the hydrogen molecule ion, the kineticenergy and the electron potential (all 4 terms) are⟨
− ~2
2m∇2
1
⟩= E1 +
(e2
4πε0a
)1± IX
1± I 2=
⟨− ~2
2m∇2
2
⟩4
⟨− e2
4πε0r1
⟩= −4
2
(e2
4πε0a
)1 + D ± 2IX
1± I 2
the electron-electron energy becomes
〈Vee〉 =
(e2
4πε0a
)D2 ± X2
1± I 2
D2 =
∫ ∫|ψ0(r1)|2 a
r12|ψ0(r ′2)|2 d3~r1 d
3~r2
X2 =
∫ ∫ψ0(r1)ψ0(r ′1)
a
r12ψ0(r2)ψ0(r ′2) d3~r1 d
3~r2
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 14 / 21
Computing the energies
Using the same methods as for the hydrogen molecule ion, the kineticenergy and the electron potential (all 4 terms) are⟨
− ~2
2m∇2
1
⟩= E1 +
(e2
4πε0a
)1± IX
1± I 2=
⟨− ~2
2m∇2
2
⟩4
⟨− e2
4πε0r1
⟩= −4
2
(e2
4πε0a
)1 + D ± 2IX
1± I 2
the electron-electron energy becomes
〈Vee〉 =
(e2
4πε0a
)D2 ± X2
1± I 2
D2 =
∫ ∫|ψ0(r1)|2 a
r12|ψ0(r ′2)|2 d3~r1 d
3~r2
X2 =
∫ ∫ψ0(r1)ψ0(r ′1)
a
r12ψ0(r2)ψ0(r ′2) d3~r1 d
3~r2
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 14 / 21
Hydrogen molecule energy
The total energy of the hydrogen molecule for these two trial functions isthus
〈H±〉 = 2E1
[1− a
R+
2D − D2 ± (2IX − X2)
1± I 2
]
654321
R/a
E/E1
-2.0
-1.5
-1.0
-0.5
0
-2.5
the singlet is about10 eV lower than thetriplet
bond length1.64a −→ 1.40a
binding energy3.15 eV −→ 4.75 eV
spin-spin interaction isnegligible
this is called the Heitler-London approximation
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 15 / 21
Hydrogen molecule energy
The total energy of the hydrogen molecule for these two trial functions isthus
〈H±〉 = 2E1
[1− a
R+
2D − D2 ± (2IX − X2)
1± I 2
]
654321
R/a
E/E1
-2.0
-1.5
-1.0
-0.5
0
-2.5
the singlet is about10 eV lower than thetriplet
bond length1.64a −→ 1.40a
binding energy3.15 eV −→ 4.75 eV
spin-spin interaction isnegligible
this is called the Heitler-London approximation
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 15 / 21
Hydrogen molecule energy
The total energy of the hydrogen molecule for these two trial functions isthus
〈H±〉 = 2E1
[1− a
R+
2D − D2 ± (2IX − X2)
1± I 2
]
654321
R/a
E/E1
-2.0
-1.5
-1.0
-0.5
0
-2.5
the singlet is about10 eV lower than thetriplet
bond length1.64a −→ 1.40a
binding energy3.15 eV −→ 4.75 eV
spin-spin interaction isnegligible
this is called the Heitler-London approximation
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 15 / 21
Hydrogen molecule energy
The total energy of the hydrogen molecule for these two trial functions isthus
〈H±〉 = 2E1
[1− a
R+
2D − D2 ± (2IX − X2)
1± I 2
]
654321
R/a
E/E1
-2.0
-1.5
-1.0
-0.5
0
-2.5
the singlet is about10 eV lower than thetriplet
bond length1.64a −→ 1.40a
binding energy3.15 eV −→ 4.75 eV
spin-spin interaction isnegligible
this is called the Heitler-London approximation
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 15 / 21
Hydrogen molecule energy
The total energy of the hydrogen molecule for these two trial functions isthus
〈H±〉 = 2E1
[1− a
R+
2D − D2 ± (2IX − X2)
1± I 2
]
654321
R/a
E/E1
-2.0
-1.5
-1.0
-0.5
0
-2.5
the singlet is about10 eV lower than thetriplet
bond length1.64a −→ 1.40a
binding energy3.15 eV −→ 4.75 eV
spin-spin interaction isnegligible
this is called the Heitler-London approximation
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 15 / 21
Hydrogen molecule energy
The total energy of the hydrogen molecule for these two trial functions isthus
〈H±〉 = 2E1
[1− a
R+
2D − D2 ± (2IX − X2)
1± I 2
]
654321
R/a
E/E1
-2.0
-1.5
-1.0
-0.5
0
-2.5
the singlet is about10 eV lower than thetriplet
bond length1.64a −→ 1.40a
binding energy3.15 eV −→ 4.75 eV
spin-spin interaction isnegligible
this is called the Heitler-London approximation
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 15 / 21
Hydrogen molecule energy
The total energy of the hydrogen molecule for these two trial functions isthus
〈H±〉 = 2E1
[1− a
R+
2D − D2 ± (2IX − X2)
1± I 2
]
654321
R/a
E/E1
-2.0
-1.5
-1.0
-0.5
0
-2.5
the singlet is about10 eV lower than thetriplet
bond length1.64a −→ 1.40a
binding energy3.15 eV −→ 4.75 eV
spin-spin interaction isnegligible
this is called the Heitler-London approximation
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 15 / 21
Hydrogen molecule energy
The total energy of the hydrogen molecule for these two trial functions isthus
〈H±〉 = 2E1
[1− a
R+
2D − D2 ± (2IX − X2)
1± I 2
]
654321
R/a
E/E1
-2.0
-1.5
-1.0
-0.5
0
-2.5
the singlet is about10 eV lower than thetriplet
bond length1.64a −→ 1.40a
binding energy3.15 eV −→ 4.75 eV
spin-spin interaction isnegligible
this is called the Heitler-London approximationC. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 15 / 21
Problem 8.24
Although the Schrodinger equation for helium itself cannot be solvedexactly, there exist “helium-like” systems that do admit exact solutions. Asimple example is “rubber-band helium,” in which the Coulomb forces arereplaced by Hooke’s law forces:
H = − ~2
2m
(∇2
1 +∇22
)+
1
2mω2
(r21 + r22
)− λ
4mω2|~r1 −~r2|2
(a) Show that by the change of variables below, this Hamiltonian can besolved analytically
~u ≡ 1√2
(~r1 +~r2) ~v ≡ 1√2
(~r1 −~r2)
(b) What is the exact ground state energy for this system?
(c) Solve for an upper bound to the ground state using the variationalprinciple
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 16 / 21
Problem 8.24 (cont.)
H = − ~2
2m
(∇2
1 +∇22
)+
1
2mω2
(r21 + r22
)− λ
4mω2|~r1 −~r2|2
~u ≡ 1√2
(~r1 +~r2) ~v ≡ 1√2
(~r1 −~r2)
First apply the change of variables to the r21 + r22 term of the potentialenergy
~r1 =1√2
(~u + ~v) ~r2 =1√2
(~u − ~v)
r21 + r22 =1
2
(u2 + 2~u · ~v + v2
+ u2 − 2~u · ~v + v2
)= u2 + v2
the same can be done with the derivative terms, ∇21 +∇2
2 by expandingeach coordinate’s derivative in turn
∂
∂x1=
∂
∂ux
∂ux∂x1
+∂
∂vx
∂vx∂x1
=1√2
(∂
∂ux+
∂
∂vx
),∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 17 / 21
Problem 8.24 (cont.)
H = − ~2
2m
(∇2
1 +∇22
)+
1
2mω2
(r21 + r22
)− λ
4mω2|~r1 −~r2|2
~u ≡ 1√2
(~r1 +~r2) ~v ≡ 1√2
(~r1 −~r2)
First apply the change of variables to the r21 + r22 term of the potentialenergy
~r1 =1√2
(~u + ~v) ~r2 =1√2
(~u − ~v)
r21 + r22 =1
2
(u2 + 2~u · ~v + v2
+ u2 − 2~u · ~v + v2
)= u2 + v2
the same can be done with the derivative terms, ∇21 +∇2
2 by expandingeach coordinate’s derivative in turn
∂
∂x1=
∂
∂ux
∂ux∂x1
+∂
∂vx
∂vx∂x1
=1√2
(∂
∂ux+
∂
∂vx
),∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 17 / 21
Problem 8.24 (cont.)
H = − ~2
2m
(∇2
1 +∇22
)+
1
2mω2
(r21 + r22
)− λ
4mω2|~r1 −~r2|2
~u ≡ 1√2
(~r1 +~r2) ~v ≡ 1√2
(~r1 −~r2)
First apply the change of variables to the r21 + r22 term of the potentialenergy
~r1 =1√2
(~u + ~v)
~r2 =1√2
(~u − ~v)
r21 + r22 =1
2
(u2 + 2~u · ~v + v2
+ u2 − 2~u · ~v + v2
)= u2 + v2
the same can be done with the derivative terms, ∇21 +∇2
2 by expandingeach coordinate’s derivative in turn
∂
∂x1=
∂
∂ux
∂ux∂x1
+∂
∂vx
∂vx∂x1
=1√2
(∂
∂ux+
∂
∂vx
),∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 17 / 21
Problem 8.24 (cont.)
H = − ~2
2m
(∇2
1 +∇22
)+
1
2mω2
(r21 + r22
)− λ
4mω2|~r1 −~r2|2
~u ≡ 1√2
(~r1 +~r2) ~v ≡ 1√2
(~r1 −~r2)
First apply the change of variables to the r21 + r22 term of the potentialenergy
~r1 =1√2
(~u + ~v) ~r2 =1√2
(~u − ~v)
r21 + r22 =1
2
(u2 + 2~u · ~v + v2
+ u2 − 2~u · ~v + v2
)= u2 + v2
the same can be done with the derivative terms, ∇21 +∇2
2 by expandingeach coordinate’s derivative in turn
∂
∂x1=
∂
∂ux
∂ux∂x1
+∂
∂vx
∂vx∂x1
=1√2
(∂
∂ux+
∂
∂vx
),∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 17 / 21
Problem 8.24 (cont.)
H = − ~2
2m
(∇2
1 +∇22
)+
1
2mω2
(r21 + r22
)− λ
4mω2|~r1 −~r2|2
~u ≡ 1√2
(~r1 +~r2) ~v ≡ 1√2
(~r1 −~r2)
First apply the change of variables to the r21 + r22 term of the potentialenergy
~r1 =1√2
(~u + ~v) ~r2 =1√2
(~u − ~v)
r21 + r22 =1
2
(u2 + 2~u · ~v + v2
+ u2 − 2~u · ~v + v2
)
= u2 + v2
the same can be done with the derivative terms, ∇21 +∇2
2 by expandingeach coordinate’s derivative in turn
∂
∂x1=
∂
∂ux
∂ux∂x1
+∂
∂vx
∂vx∂x1
=1√2
(∂
∂ux+
∂
∂vx
),∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 17 / 21
Problem 8.24 (cont.)
H = − ~2
2m
(∇2
1 +∇22
)+
1
2mω2
(r21 + r22
)− λ
4mω2|~r1 −~r2|2
~u ≡ 1√2
(~r1 +~r2) ~v ≡ 1√2
(~r1 −~r2)
First apply the change of variables to the r21 + r22 term of the potentialenergy
~r1 =1√2
(~u + ~v) ~r2 =1√2
(~u − ~v)
r21 + r22 =1
2
(u2 + 2~u · ~v + v2 + u2 − 2~u · ~v + v2
)
= u2 + v2
the same can be done with the derivative terms, ∇21 +∇2
2 by expandingeach coordinate’s derivative in turn
∂
∂x1=
∂
∂ux
∂ux∂x1
+∂
∂vx
∂vx∂x1
=1√2
(∂
∂ux+
∂
∂vx
),∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 17 / 21
Problem 8.24 (cont.)
H = − ~2
2m
(∇2
1 +∇22
)+
1
2mω2
(r21 + r22
)− λ
4mω2|~r1 −~r2|2
~u ≡ 1√2
(~r1 +~r2) ~v ≡ 1√2
(~r1 −~r2)
First apply the change of variables to the r21 + r22 term of the potentialenergy
~r1 =1√2
(~u + ~v) ~r2 =1√2
(~u − ~v)
r21 + r22 =1
2
(u2 + 2~u · ~v + v2 + u2 − 2~u · ~v + v2
)= u2 + v2
the same can be done with the derivative terms, ∇21 +∇2
2 by expandingeach coordinate’s derivative in turn
∂
∂x1=
∂
∂ux
∂ux∂x1
+∂
∂vx
∂vx∂x1
=1√2
(∂
∂ux+
∂
∂vx
),∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 17 / 21
Problem 8.24 (cont.)
H = − ~2
2m
(∇2
1 +∇22
)+
1
2mω2
(r21 + r22
)− λ
4mω2|~r1 −~r2|2
~u ≡ 1√2
(~r1 +~r2) ~v ≡ 1√2
(~r1 −~r2)
First apply the change of variables to the r21 + r22 term of the potentialenergy
~r1 =1√2
(~u + ~v) ~r2 =1√2
(~u − ~v)
r21 + r22 =1
2
(u2 + 2~u · ~v + v2 + u2 − 2~u · ~v + v2
)= u2 + v2
the same can be done with the derivative terms, ∇21 +∇2
2 by expandingeach coordinate’s derivative in turn
∂
∂x1=
∂
∂ux
∂ux∂x1
+∂
∂vx
∂vx∂x1
=1√2
(∂
∂ux+
∂
∂vx
),∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 17 / 21
Problem 8.24 (cont.)
H = − ~2
2m
(∇2
1 +∇22
)+
1
2mω2
(r21 + r22
)− λ
4mω2|~r1 −~r2|2
~u ≡ 1√2
(~r1 +~r2) ~v ≡ 1√2
(~r1 −~r2)
First apply the change of variables to the r21 + r22 term of the potentialenergy
~r1 =1√2
(~u + ~v) ~r2 =1√2
(~u − ~v)
r21 + r22 =1
2
(u2 + 2~u · ~v + v2 + u2 − 2~u · ~v + v2
)= u2 + v2
the same can be done with the derivative terms, ∇21 +∇2
2 by expandingeach coordinate’s derivative in turn
∂
∂x1=
∂
∂ux
∂ux∂x1
+∂
∂vx
∂vx∂x1
=1√2
(∂
∂ux+
∂
∂vx
),∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 17 / 21
Problem 8.24 (cont.)
H = − ~2
2m
(∇2
1 +∇22
)+
1
2mω2
(r21 + r22
)− λ
4mω2|~r1 −~r2|2
~u ≡ 1√2
(~r1 +~r2) ~v ≡ 1√2
(~r1 −~r2)
First apply the change of variables to the r21 + r22 term of the potentialenergy
~r1 =1√2
(~u + ~v) ~r2 =1√2
(~u − ~v)
r21 + r22 =1
2
(u2 + 2~u · ~v + v2 + u2 − 2~u · ~v + v2
)= u2 + v2
the same can be done with the derivative terms, ∇21 +∇2
2 by expandingeach coordinate’s derivative in turn
∂
∂x1=
∂
∂ux
∂ux∂x1
+∂
∂vx
∂vx∂x1
=1√2
(∂
∂ux+
∂
∂vx
),∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 17 / 21
Problem 8.24 (cont.)
H = − ~2
2m
(∇2
1 +∇22
)+
1
2mω2
(r21 + r22
)− λ
4mω2|~r1 −~r2|2
~u ≡ 1√2
(~r1 +~r2) ~v ≡ 1√2
(~r1 −~r2)
First apply the change of variables to the r21 + r22 term of the potentialenergy
~r1 =1√2
(~u + ~v) ~r2 =1√2
(~u − ~v)
r21 + r22 =1
2
(u2 + 2~u · ~v + v2 + u2 − 2~u · ~v + v2
)= u2 + v2
the same can be done with the derivative terms, ∇21 +∇2
2 by expandingeach coordinate’s derivative in turn
∂
∂x1=
∂
∂ux
∂ux∂x1
+∂
∂vx
∂vx∂x1
=1√2
(∂
∂ux+
∂
∂vx
),∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 17 / 21
Problem 8.24 (cont.)
H = − ~2
2m
(∇2
1 +∇22
)+
1
2mω2
(r21 + r22
)− λ
4mω2|~r1 −~r2|2
~u ≡ 1√2
(~r1 +~r2) ~v ≡ 1√2
(~r1 −~r2)
First apply the change of variables to the r21 + r22 term of the potentialenergy
~r1 =1√2
(~u + ~v) ~r2 =1√2
(~u − ~v)
r21 + r22 =1
2
(u2 + 2~u · ~v + v2 + u2 − 2~u · ~v + v2
)= u2 + v2
the same can be done with the derivative terms, ∇21 +∇2
2 by expandingeach coordinate’s derivative in turn
∂
∂x1=
∂
∂ux
∂ux∂x1
+∂
∂vx
∂vx∂x1
=1√2
(∂
∂ux+
∂
∂vx
),
∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 17 / 21
Problem 8.24 (cont.)
H = − ~2
2m
(∇2
1 +∇22
)+
1
2mω2
(r21 + r22
)− λ
4mω2|~r1 −~r2|2
~u ≡ 1√2
(~r1 +~r2) ~v ≡ 1√2
(~r1 −~r2)
First apply the change of variables to the r21 + r22 term of the potentialenergy
~r1 =1√2
(~u + ~v) ~r2 =1√2
(~u − ~v)
r21 + r22 =1
2
(u2 + 2~u · ~v + v2 + u2 − 2~u · ~v + v2
)= u2 + v2
the same can be done with the derivative terms, ∇21 +∇2
2 by expandingeach coordinate’s derivative in turn
∂
∂x1=
∂
∂ux
∂ux∂x1
+∂
∂vx
∂vx∂x1
=1√2
(∂
∂ux+
∂
∂vx
),∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 17 / 21
Problem 8.24 (cont.)
∂
∂x1=
1√2
(∂
∂ux+
∂
∂vx
)∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)
∂2
∂x21=
1√2
∂
∂x1
(∂
∂ux+
∂
∂vx
)=
1
2
(∂
∂ux+
∂
∂vx
)(∂
∂ux+
∂
∂vx
)=
1
2
(∂2
∂u2x
+ 2∂2
∂ux∂vx+
∂2
∂v2x
)∂2
∂x22=
1√2
∂
∂x2
(∂
∂ux− ∂
∂vx
)=
1
2
(∂
∂ux− ∂
∂vx
)(∂
∂ux− ∂
∂vx
)=
1
2
(∂2
∂u2x
− 2∂2
∂ux∂vx+
∂2
∂v2x
)(∂2
∂x21+
∂2
∂x22
)=
(∂2
∂u2x+
∂2
∂v2x
)−→ ∇2
1 +∇22 = ∇2
u +∇2v
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 18 / 21
Problem 8.24 (cont.)
∂
∂x1=
1√2
(∂
∂ux+
∂
∂vx
)∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)∂2
∂x21=
1√2
∂
∂x1
(∂
∂ux+
∂
∂vx
)
=1
2
(∂
∂ux+
∂
∂vx
)(∂
∂ux+
∂
∂vx
)=
1
2
(∂2
∂u2x
+ 2∂2
∂ux∂vx+
∂2
∂v2x
)∂2
∂x22=
1√2
∂
∂x2
(∂
∂ux− ∂
∂vx
)=
1
2
(∂
∂ux− ∂
∂vx
)(∂
∂ux− ∂
∂vx
)=
1
2
(∂2
∂u2x
− 2∂2
∂ux∂vx+
∂2
∂v2x
)(∂2
∂x21+
∂2
∂x22
)=
(∂2
∂u2x+
∂2
∂v2x
)−→ ∇2
1 +∇22 = ∇2
u +∇2v
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 18 / 21
Problem 8.24 (cont.)
∂
∂x1=
1√2
(∂
∂ux+
∂
∂vx
)∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)∂2
∂x21=
1√2
∂
∂x1
(∂
∂ux+
∂
∂vx
)=
1
2
(∂
∂ux+
∂
∂vx
)(∂
∂ux+
∂
∂vx
)
=1
2
(∂2
∂u2x
+ 2∂2
∂ux∂vx+
∂2
∂v2x
)∂2
∂x22=
1√2
∂
∂x2
(∂
∂ux− ∂
∂vx
)=
1
2
(∂
∂ux− ∂
∂vx
)(∂
∂ux− ∂
∂vx
)=
1
2
(∂2
∂u2x
− 2∂2
∂ux∂vx+
∂2
∂v2x
)(∂2
∂x21+
∂2
∂x22
)=
(∂2
∂u2x+
∂2
∂v2x
)−→ ∇2
1 +∇22 = ∇2
u +∇2v
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 18 / 21
Problem 8.24 (cont.)
∂
∂x1=
1√2
(∂
∂ux+
∂
∂vx
)∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)∂2
∂x21=
1√2
∂
∂x1
(∂
∂ux+
∂
∂vx
)=
1
2
(∂
∂ux+
∂
∂vx
)(∂
∂ux+
∂
∂vx
)=
1
2
(∂2
∂u2x
+ 2∂2
∂ux∂vx+
∂2
∂v2x
)
∂2
∂x22=
1√2
∂
∂x2
(∂
∂ux− ∂
∂vx
)=
1
2
(∂
∂ux− ∂
∂vx
)(∂
∂ux− ∂
∂vx
)=
1
2
(∂2
∂u2x
− 2∂2
∂ux∂vx+
∂2
∂v2x
)(∂2
∂x21+
∂2
∂x22
)=
(∂2
∂u2x+
∂2
∂v2x
)−→ ∇2
1 +∇22 = ∇2
u +∇2v
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 18 / 21
Problem 8.24 (cont.)
∂
∂x1=
1√2
(∂
∂ux+
∂
∂vx
)∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)∂2
∂x21=
1√2
∂
∂x1
(∂
∂ux+
∂
∂vx
)=
1
2
(∂
∂ux+
∂
∂vx
)(∂
∂ux+
∂
∂vx
)=
1
2
(∂2
∂u2x+ 2
∂2
∂ux∂vx
+∂2
∂v2x
)
∂2
∂x22=
1√2
∂
∂x2
(∂
∂ux− ∂
∂vx
)=
1
2
(∂
∂ux− ∂
∂vx
)(∂
∂ux− ∂
∂vx
)=
1
2
(∂2
∂u2x
− 2∂2
∂ux∂vx+
∂2
∂v2x
)(∂2
∂x21+
∂2
∂x22
)=
(∂2
∂u2x+
∂2
∂v2x
)−→ ∇2
1 +∇22 = ∇2
u +∇2v
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 18 / 21
Problem 8.24 (cont.)
∂
∂x1=
1√2
(∂
∂ux+
∂
∂vx
)∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)∂2
∂x21=
1√2
∂
∂x1
(∂
∂ux+
∂
∂vx
)=
1
2
(∂
∂ux+
∂
∂vx
)(∂
∂ux+
∂
∂vx
)=
1
2
(∂2
∂u2x+ 2
∂2
∂ux∂vx+
∂2
∂v2x
)
∂2
∂x22=
1√2
∂
∂x2
(∂
∂ux− ∂
∂vx
)=
1
2
(∂
∂ux− ∂
∂vx
)(∂
∂ux− ∂
∂vx
)=
1
2
(∂2
∂u2x
− 2∂2
∂ux∂vx+
∂2
∂v2x
)(∂2
∂x21+
∂2
∂x22
)=
(∂2
∂u2x+
∂2
∂v2x
)−→ ∇2
1 +∇22 = ∇2
u +∇2v
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 18 / 21
Problem 8.24 (cont.)
∂
∂x1=
1√2
(∂
∂ux+
∂
∂vx
)∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)∂2
∂x21=
1√2
∂
∂x1
(∂
∂ux+
∂
∂vx
)=
1
2
(∂
∂ux+
∂
∂vx
)(∂
∂ux+
∂
∂vx
)=
1
2
(∂2
∂u2x+ 2
∂2
∂ux∂vx+
∂2
∂v2x
)∂2
∂x22=
1√2
∂
∂x2
(∂
∂ux− ∂
∂vx
)
=1
2
(∂
∂ux− ∂
∂vx
)(∂
∂ux− ∂
∂vx
)=
1
2
(∂2
∂u2x
− 2∂2
∂ux∂vx+
∂2
∂v2x
)(∂2
∂x21+
∂2
∂x22
)=
(∂2
∂u2x+
∂2
∂v2x
)−→ ∇2
1 +∇22 = ∇2
u +∇2v
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 18 / 21
Problem 8.24 (cont.)
∂
∂x1=
1√2
(∂
∂ux+
∂
∂vx
)∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)∂2
∂x21=
1√2
∂
∂x1
(∂
∂ux+
∂
∂vx
)=
1
2
(∂
∂ux+
∂
∂vx
)(∂
∂ux+
∂
∂vx
)=
1
2
(∂2
∂u2x+ 2
∂2
∂ux∂vx+
∂2
∂v2x
)∂2
∂x22=
1√2
∂
∂x2
(∂
∂ux− ∂
∂vx
)=
1
2
(∂
∂ux− ∂
∂vx
)(∂
∂ux− ∂
∂vx
)
=1
2
(∂2
∂u2x
− 2∂2
∂ux∂vx+
∂2
∂v2x
)(∂2
∂x21+
∂2
∂x22
)=
(∂2
∂u2x+
∂2
∂v2x
)−→ ∇2
1 +∇22 = ∇2
u +∇2v
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 18 / 21
Problem 8.24 (cont.)
∂
∂x1=
1√2
(∂
∂ux+
∂
∂vx
)∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)∂2
∂x21=
1√2
∂
∂x1
(∂
∂ux+
∂
∂vx
)=
1
2
(∂
∂ux+
∂
∂vx
)(∂
∂ux+
∂
∂vx
)=
1
2
(∂2
∂u2x+ 2
∂2
∂ux∂vx+
∂2
∂v2x
)∂2
∂x22=
1√2
∂
∂x2
(∂
∂ux− ∂
∂vx
)=
1
2
(∂
∂ux− ∂
∂vx
)(∂
∂ux− ∂
∂vx
)=
1
2
(∂2
∂u2x
− 2∂2
∂ux∂vx+
∂2
∂v2x
)
(∂2
∂x21+
∂2
∂x22
)=
(∂2
∂u2x+
∂2
∂v2x
)−→ ∇2
1 +∇22 = ∇2
u +∇2v
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 18 / 21
Problem 8.24 (cont.)
∂
∂x1=
1√2
(∂
∂ux+
∂
∂vx
)∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)∂2
∂x21=
1√2
∂
∂x1
(∂
∂ux+
∂
∂vx
)=
1
2
(∂
∂ux+
∂
∂vx
)(∂
∂ux+
∂
∂vx
)=
1
2
(∂2
∂u2x+ 2
∂2
∂ux∂vx+
∂2
∂v2x
)∂2
∂x22=
1√2
∂
∂x2
(∂
∂ux− ∂
∂vx
)=
1
2
(∂
∂ux− ∂
∂vx
)(∂
∂ux− ∂
∂vx
)=
1
2
(∂2
∂u2x− 2
∂2
∂ux∂vx
+∂2
∂v2x
)
(∂2
∂x21+
∂2
∂x22
)=
(∂2
∂u2x+
∂2
∂v2x
)−→ ∇2
1 +∇22 = ∇2
u +∇2v
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 18 / 21
Problem 8.24 (cont.)
∂
∂x1=
1√2
(∂
∂ux+
∂
∂vx
)∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)∂2
∂x21=
1√2
∂
∂x1
(∂
∂ux+
∂
∂vx
)=
1
2
(∂
∂ux+
∂
∂vx
)(∂
∂ux+
∂
∂vx
)=
1
2
(∂2
∂u2x+ 2
∂2
∂ux∂vx+
∂2
∂v2x
)∂2
∂x22=
1√2
∂
∂x2
(∂
∂ux− ∂
∂vx
)=
1
2
(∂
∂ux− ∂
∂vx
)(∂
∂ux− ∂
∂vx
)=
1
2
(∂2
∂u2x− 2
∂2
∂ux∂vx+
∂2
∂v2x
)
(∂2
∂x21+
∂2
∂x22
)=
(∂2
∂u2x+
∂2
∂v2x
)−→ ∇2
1 +∇22 = ∇2
u +∇2v
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 18 / 21
Problem 8.24 (cont.)
∂
∂x1=
1√2
(∂
∂ux+
∂
∂vx
)∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)∂2
∂x21=
1√2
∂
∂x1
(∂
∂ux+
∂
∂vx
)=
1
2
(∂
∂ux+
∂
∂vx
)(∂
∂ux+
∂
∂vx
)=
1
2
(∂2
∂u2x+ 2
∂2
∂ux∂vx+
∂2
∂v2x
)∂2
∂x22=
1√2
∂
∂x2
(∂
∂ux− ∂
∂vx
)=
1
2
(∂
∂ux− ∂
∂vx
)(∂
∂ux− ∂
∂vx
)=
1
2
(∂2
∂u2x− 2
∂2
∂ux∂vx+
∂2
∂v2x
)(∂2
∂x21+
∂2
∂x22
)=
(∂2
∂u2x+
∂2
∂v2x
)−→ ∇2
1 +∇22 = ∇2
u +∇2v
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 18 / 21
Problem 8.24 (cont.)
∂
∂x1=
1√2
(∂
∂ux+
∂
∂vx
)∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)∂2
∂x21=
1√2
∂
∂x1
(∂
∂ux+
∂
∂vx
)=
1
2
(∂
∂ux+
∂
∂vx
)(∂
∂ux+
∂
∂vx
)=
1
2
(∂2
∂u2x+ 2
∂2
∂ux∂vx+
∂2
∂v2x
)∂2
∂x22=
1√2
∂
∂x2
(∂
∂ux− ∂
∂vx
)=
1
2
(∂
∂ux− ∂
∂vx
)(∂
∂ux− ∂
∂vx
)=
1
2
(∂2
∂u2x− 2
∂2
∂ux∂vx+
∂2
∂v2x
)(∂2
∂x21+
∂2
∂x22
)=
(∂2
∂u2x+
∂2
∂v2x
)
−→ ∇21 +∇2
2 = ∇2u +∇2
v
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 18 / 21
Problem 8.24 (cont.)
∂
∂x1=
1√2
(∂
∂ux+
∂
∂vx
)∂
∂x2=
1√2
(∂
∂ux− ∂
∂vx
)∂2
∂x21=
1√2
∂
∂x1
(∂
∂ux+
∂
∂vx
)=
1
2
(∂
∂ux+
∂
∂vx
)(∂
∂ux+
∂
∂vx
)=
1
2
(∂2
∂u2x+ 2
∂2
∂ux∂vx+
∂2
∂v2x
)∂2
∂x22=
1√2
∂
∂x2
(∂
∂ux− ∂
∂vx
)=
1
2
(∂
∂ux− ∂
∂vx
)(∂
∂ux− ∂
∂vx
)=
1
2
(∂2
∂u2x− 2
∂2
∂ux∂vx+
∂2
∂v2x
)(∂2
∂x21+
∂2
∂x22
)=
(∂2
∂u2x+
∂2
∂v2x
)−→ ∇2
1 +∇22 = ∇2
u +∇2v
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 18 / 21
Problem 8.24 (cont.)
Putting it all together . . .
H = − ~2
2m
(∇2
1 +∇22
)+
1
2mω2
(r21 + r22
)− λ
4mω2|~r1 −~r2|2
= − ~2
2m
(∇2
u +∇2v
)+
1
2mω2
(u2 + v2
)− λ
4mω2 1
2|(~u + ~v)− (~u − ~v)|2
= − ~2
2m
(∇2
u +∇2v
)+
1
2mω2
(u2 + v2
)− λ
4mω22v2
=
[− ~2
2m∇2
u +1
2mω2u2
]+
[− ~2
2m∇2
v +1
2mω2v2− 1
2λmω2v2
]this is now two separated harmonicoscillators, the second with an addednegative potential energy term
the ground state energies of this three-dimensional oscillator can simply bewritten down
Eu =3
2~ω
Ev =3
2~ω√
1− λ
Egs =3
2~ω(1 +
√1− λ)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 19 / 21
Problem 8.24 (cont.)
Putting it all together . . .
H = − ~2
2m
(∇2
1 +∇22
)+
1
2mω2
(r21 + r22
)− λ
4mω2|~r1 −~r2|2
= − ~2
2m
(∇2
u +∇2v
)+
1
2mω2
(u2 + v2
)− λ
4mω2 1
2|(~u + ~v)− (~u − ~v)|2
= − ~2
2m
(∇2
u +∇2v
)+
1
2mω2
(u2 + v2
)− λ
4mω22v2
=
[− ~2
2m∇2
u +1
2mω2u2
]+
[− ~2
2m∇2
v +1
2mω2v2− 1
2λmω2v2
]this is now two separated harmonicoscillators, the second with an addednegative potential energy term
the ground state energies of this three-dimensional oscillator can simply bewritten down
Eu =3
2~ω
Ev =3
2~ω√
1− λ
Egs =3
2~ω(1 +
√1− λ)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 19 / 21
Problem 8.24 (cont.)
Putting it all together . . .
H = − ~2
2m
(∇2
1 +∇22
)+
1
2mω2
(r21 + r22
)− λ
4mω2|~r1 −~r2|2
= − ~2
2m
(∇2
u +∇2v
)+
1
2mω2
(u2 + v2
)− λ
4mω2 1
2|(~u + ~v)− (~u − ~v)|2
= − ~2
2m
(∇2
u +∇2v
)+
1
2mω2
(u2 + v2
)− λ
4mω22v2
=
[− ~2
2m∇2
u +1
2mω2u2
]+
[− ~2
2m∇2
v +1
2mω2v2− 1
2λmω2v2
]this is now two separated harmonicoscillators, the second with an addednegative potential energy term
the ground state energies of this three-dimensional oscillator can simply bewritten down
Eu =3
2~ω
Ev =3
2~ω√
1− λ
Egs =3
2~ω(1 +
√1− λ)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 19 / 21
Problem 8.24 (cont.)
Putting it all together . . .
H = − ~2
2m
(∇2
1 +∇22
)+
1
2mω2
(r21 + r22
)− λ
4mω2|~r1 −~r2|2
= − ~2
2m
(∇2
u +∇2v
)+
1
2mω2
(u2 + v2
)− λ
4mω2 1
2|(~u + ~v)− (~u − ~v)|2
= − ~2
2m
(∇2
u +∇2v
)+
1
2mω2
(u2 + v2
)− λ
4mω22v2
=
[− ~2
2m∇2
u +1
2mω2u2
]+
[− ~2
2m∇2
v +1
2mω2v2− 1
2λmω2v2
]this is now two separated harmonicoscillators, the second with an addednegative potential energy term
the ground state energies of this three-dimensional oscillator can simply bewritten down
Eu =3
2~ω
Ev =3
2~ω√
1− λ
Egs =3
2~ω(1 +
√1− λ)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 19 / 21
Problem 8.24 (cont.)
Putting it all together . . .
H = − ~2
2m
(∇2
1 +∇22
)+
1
2mω2
(r21 + r22
)− λ
4mω2|~r1 −~r2|2
= − ~2
2m
(∇2
u +∇2v
)+
1
2mω2
(u2 + v2
)− λ
4mω2 1
2|(~u + ~v)− (~u − ~v)|2
= − ~2
2m
(∇2
u +∇2v
)+
1
2mω2
(u2 + v2
)− λ
4mω22v2
=
[− ~2
2m∇2
u +1
2mω2u2
]
+
[− ~2
2m∇2
v +1
2mω2v2− 1
2λmω2v2
]this is now two separated harmonicoscillators, the second with an addednegative potential energy term
the ground state energies of this three-dimensional oscillator can simply bewritten down
Eu =3
2~ω
Ev =3
2~ω√
1− λ
Egs =3
2~ω(1 +
√1− λ)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 19 / 21
Problem 8.24 (cont.)
Putting it all together . . .
H = − ~2
2m
(∇2
1 +∇22
)+
1
2mω2
(r21 + r22
)− λ
4mω2|~r1 −~r2|2
= − ~2
2m
(∇2
u +∇2v
)+
1
2mω2
(u2 + v2
)− λ
4mω2 1
2|(~u + ~v)− (~u − ~v)|2
= − ~2
2m
(∇2
u +∇2v
)+
1
2mω2
(u2 + v2
)− λ
4mω22v2
=
[− ~2
2m∇2
u +1
2mω2u2
]+
[− ~2
2m∇2
v +1
2mω2v2 − 1
2λmω2v2
]
this is now two separated harmonicoscillators, the second with an addednegative potential energy term
the ground state energies of this three-dimensional oscillator can simply bewritten down
Eu =3
2~ω
Ev =3
2~ω√
1− λ
Egs =3
2~ω(1 +
√1− λ)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 19 / 21
Problem 8.24 (cont.)
Putting it all together . . .
H = − ~2
2m
(∇2
1 +∇22
)+
1
2mω2
(r21 + r22
)− λ
4mω2|~r1 −~r2|2
= − ~2
2m
(∇2
u +∇2v
)+
1
2mω2
(u2 + v2
)− λ
4mω2 1
2|(~u + ~v)− (~u − ~v)|2
= − ~2
2m
(∇2
u +∇2v
)+
1
2mω2
(u2 + v2
)− λ
4mω22v2
=
[− ~2
2m∇2
u +1
2mω2u2
]+
[− ~2
2m∇2
v +1
2mω2v2− 1
2λmω2v2
]this is now two separated harmonicoscillators, the second with an addednegative potential energy term
the ground state energies of this three-dimensional oscillator can simply bewritten down
Eu =3
2~ω
Ev =3
2~ω√
1− λ
Egs =3
2~ω(1 +
√1− λ)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 19 / 21
Problem 8.24 (cont.)
Putting it all together . . .
H = − ~2
2m
(∇2
1 +∇22
)+
1
2mω2
(r21 + r22
)− λ
4mω2|~r1 −~r2|2
= − ~2
2m
(∇2
u +∇2v
)+
1
2mω2
(u2 + v2
)− λ
4mω2 1
2|(~u + ~v)− (~u − ~v)|2
= − ~2
2m
(∇2
u +∇2v
)+
1
2mω2
(u2 + v2
)− λ
4mω22v2
=
[− ~2
2m∇2
u +1
2mω2u2
]+
[− ~2
2m∇2
v +1
2mω2v2− 1
2λmω2v2
]this is now two separated harmonicoscillators, the second with an addednegative potential energy term
the ground state energies of this three-dimensional oscillator can simply bewritten down
Eu =3
2~ω
Ev =3
2~ω√
1− λ
Egs =3
2~ω(1 +
√1− λ)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 19 / 21
Problem 8.24 (cont.)
Putting it all together . . .
H = − ~2
2m
(∇2
1 +∇22
)+
1
2mω2
(r21 + r22
)− λ
4mω2|~r1 −~r2|2
= − ~2
2m
(∇2
u +∇2v
)+
1
2mω2
(u2 + v2
)− λ
4mω2 1
2|(~u + ~v)− (~u − ~v)|2
= − ~2
2m
(∇2
u +∇2v
)+
1
2mω2
(u2 + v2
)− λ
4mω22v2
=
[− ~2
2m∇2
u +1
2mω2u2
]+
[− ~2
2m∇2
v +1
2mω2v2− 1
2λmω2v2
]this is now two separated harmonicoscillators, the second with an addednegative potential energy term
the ground state energies of this three-dimensional oscillator can simply bewritten down
Eu =3
2~ω
Ev =3
2~ω√
1− λ
Egs =3
2~ω(1 +
√1− λ)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 19 / 21
Problem 8.24 (cont.)
Putting it all together . . .
H = − ~2
2m
(∇2
1 +∇22
)+
1
2mω2
(r21 + r22
)− λ
4mω2|~r1 −~r2|2
= − ~2
2m
(∇2
u +∇2v
)+
1
2mω2
(u2 + v2
)− λ
4mω2 1
2|(~u + ~v)− (~u − ~v)|2
= − ~2
2m
(∇2
u +∇2v
)+
1
2mω2
(u2 + v2
)− λ
4mω22v2
=
[− ~2
2m∇2
u +1
2mω2u2
]+
[− ~2
2m∇2
v +1
2mω2v2− 1
2λmω2v2
]this is now two separated harmonicoscillators, the second with an addednegative potential energy term
the ground state energies of this three-dimensional oscillator can simply bewritten down
Eu =3
2~ω
Ev =3
2~ω√
1− λ
Egs =3
2~ω(1 +
√1− λ)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 19 / 21
Problem 8.24 (cont.)
Putting it all together . . .
H = − ~2
2m
(∇2
1 +∇22
)+
1
2mω2
(r21 + r22
)− λ
4mω2|~r1 −~r2|2
= − ~2
2m
(∇2
u +∇2v
)+
1
2mω2
(u2 + v2
)− λ
4mω2 1
2|(~u + ~v)− (~u − ~v)|2
= − ~2
2m
(∇2
u +∇2v
)+
1
2mω2
(u2 + v2
)− λ
4mω22v2
=
[− ~2
2m∇2
u +1
2mω2u2
]+
[− ~2
2m∇2
v +1
2mω2v2− 1
2λmω2v2
]this is now two separated harmonicoscillators, the second with an addednegative potential energy term
the ground state energies of this three-dimensional oscillator can simply bewritten down
Eu =3
2~ω
Ev =3
2~ω√
1− λ
Egs =3
2~ω(1 +
√1− λ)
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 19 / 21
Problem 8.24 (cont.)
Build a variational trial functionfrom the single oscillator solutions
by symmetry the scalar product in-tegrates to zero
ψ(~r1,~r2) =(mωπ~
)3/2e−mω(r
21+r22 )/2~
〈H〉 =3
2~ω +
3
2~ω + 〈Vee〉
= 3~ω + 〈Vee〉〈Vee〉 = −λ
4mω2
(mωπ~
)3 ∫e−mω(r
21+r22 )/~(~r1 −~r2)2 d3~r1d
3~r2
= −λ4mω2
(mωπ~
)3 ∫e−mω(r
21+r22 )/~(r21 −����2~r1 ·~r2 + r22 ) d3~r1d
3~r2
= −λ4mω2
(mωπ~
)32
∫e−mω(r
21+r22 )/~r21 d
3~r1d3~r2
= −λ2mω2
(mωπ~
)3(4π)2
∫ ∞0
e−mωr22 /~r22 dr2
∫ ∞0
e−mωr21 /~r41 dr1
= −λ8m4ω5
π~3
[1
4
~mω
√π~mω
][3
8
~2
m2ω2
√π~mω
]= −3
4λ~ω
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 20 / 21
Problem 8.24 (cont.)
Build a variational trial functionfrom the single oscillator solutions
by symmetry the scalar product in-tegrates to zero
ψ(~r1,~r2) =(mωπ~
)3/2e−mω(r
21+r22 )/2~
〈H〉 =3
2~ω +
3
2~ω + 〈Vee〉
= 3~ω + 〈Vee〉〈Vee〉 = −λ
4mω2
(mωπ~
)3 ∫e−mω(r
21+r22 )/~(~r1 −~r2)2 d3~r1d
3~r2
= −λ4mω2
(mωπ~
)3 ∫e−mω(r
21+r22 )/~(r21 −����2~r1 ·~r2 + r22 ) d3~r1d
3~r2
= −λ4mω2
(mωπ~
)32
∫e−mω(r
21+r22 )/~r21 d
3~r1d3~r2
= −λ2mω2
(mωπ~
)3(4π)2
∫ ∞0
e−mωr22 /~r22 dr2
∫ ∞0
e−mωr21 /~r41 dr1
= −λ8m4ω5
π~3
[1
4
~mω
√π~mω
][3
8
~2
m2ω2
√π~mω
]= −3
4λ~ω
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 20 / 21
Problem 8.24 (cont.)
Build a variational trial functionfrom the single oscillator solutions
by symmetry the scalar product in-tegrates to zero
ψ(~r1,~r2) =(mωπ~
)3/2e−mω(r
21+r22 )/2~
〈H〉 =3
2~ω +
3
2~ω + 〈Vee〉
= 3~ω + 〈Vee〉〈Vee〉 = −λ
4mω2
(mωπ~
)3 ∫e−mω(r
21+r22 )/~(~r1 −~r2)2 d3~r1d
3~r2
= −λ4mω2
(mωπ~
)3 ∫e−mω(r
21+r22 )/~(r21 −����2~r1 ·~r2 + r22 ) d3~r1d
3~r2
= −λ4mω2
(mωπ~
)32
∫e−mω(r
21+r22 )/~r21 d
3~r1d3~r2
= −λ2mω2
(mωπ~
)3(4π)2
∫ ∞0
e−mωr22 /~r22 dr2
∫ ∞0
e−mωr21 /~r41 dr1
= −λ8m4ω5
π~3
[1
4
~mω
√π~mω
][3
8
~2
m2ω2
√π~mω
]= −3
4λ~ω
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 20 / 21
Problem 8.24 (cont.)
Build a variational trial functionfrom the single oscillator solutions
by symmetry the scalar product in-tegrates to zero
ψ(~r1,~r2) =(mωπ~
)3/2e−mω(r
21+r22 )/2~
〈H〉 =3
2~ω +
3
2~ω + 〈Vee〉
= 3~ω + 〈Vee〉
〈Vee〉 = −λ4mω2
(mωπ~
)3 ∫e−mω(r
21+r22 )/~(~r1 −~r2)2 d3~r1d
3~r2
= −λ4mω2
(mωπ~
)3 ∫e−mω(r
21+r22 )/~(r21 −����2~r1 ·~r2 + r22 ) d3~r1d
3~r2
= −λ4mω2
(mωπ~
)32
∫e−mω(r
21+r22 )/~r21 d
3~r1d3~r2
= −λ2mω2
(mωπ~
)3(4π)2
∫ ∞0
e−mωr22 /~r22 dr2
∫ ∞0
e−mωr21 /~r41 dr1
= −λ8m4ω5
π~3
[1
4
~mω
√π~mω
][3
8
~2
m2ω2
√π~mω
]= −3
4λ~ω
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 20 / 21
Problem 8.24 (cont.)
Build a variational trial functionfrom the single oscillator solutions
by symmetry the scalar product in-tegrates to zero
ψ(~r1,~r2) =(mωπ~
)3/2e−mω(r
21+r22 )/2~
〈H〉 =3
2~ω +
3
2~ω + 〈Vee〉
= 3~ω + 〈Vee〉〈Vee〉 = −λ
4mω2
(mωπ~
)3 ∫e−mω(r
21+r22 )/~(~r1 −~r2)2 d3~r1d
3~r2
= −λ4mω2
(mωπ~
)3 ∫e−mω(r
21+r22 )/~(r21 −����2~r1 ·~r2 + r22 ) d3~r1d
3~r2
= −λ4mω2
(mωπ~
)32
∫e−mω(r
21+r22 )/~r21 d
3~r1d3~r2
= −λ2mω2
(mωπ~
)3(4π)2
∫ ∞0
e−mωr22 /~r22 dr2
∫ ∞0
e−mωr21 /~r41 dr1
= −λ8m4ω5
π~3
[1
4
~mω
√π~mω
][3
8
~2
m2ω2
√π~mω
]= −3
4λ~ω
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 20 / 21
Problem 8.24 (cont.)
Build a variational trial functionfrom the single oscillator solutions
by symmetry the scalar product in-tegrates to zero
ψ(~r1,~r2) =(mωπ~
)3/2e−mω(r
21+r22 )/2~
〈H〉 =3
2~ω +
3
2~ω + 〈Vee〉
= 3~ω + 〈Vee〉〈Vee〉 = −λ
4mω2
(mωπ~
)3 ∫e−mω(r
21+r22 )/~(~r1 −~r2)2 d3~r1d
3~r2
= −λ4mω2
(mωπ~
)3 ∫e−mω(r
21+r22 )/~(r21 − 2~r1 ·~r2 + r22 ) d3~r1d
3~r2
= −λ4mω2
(mωπ~
)32
∫e−mω(r
21+r22 )/~r21 d
3~r1d3~r2
= −λ2mω2
(mωπ~
)3(4π)2
∫ ∞0
e−mωr22 /~r22 dr2
∫ ∞0
e−mωr21 /~r41 dr1
= −λ8m4ω5
π~3
[1
4
~mω
√π~mω
][3
8
~2
m2ω2
√π~mω
]= −3
4λ~ω
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 20 / 21
Problem 8.24 (cont.)
Build a variational trial functionfrom the single oscillator solutions
by symmetry the scalar product in-tegrates to zero
ψ(~r1,~r2) =(mωπ~
)3/2e−mω(r
21+r22 )/2~
〈H〉 =3
2~ω +
3
2~ω + 〈Vee〉
= 3~ω + 〈Vee〉〈Vee〉 = −λ
4mω2
(mωπ~
)3 ∫e−mω(r
21+r22 )/~(~r1 −~r2)2 d3~r1d
3~r2
= −λ4mω2
(mωπ~
)3 ∫e−mω(r
21+r22 )/~(r21 −����2~r1 ·~r2 + r22 ) d3~r1d
3~r2
= −λ4mω2
(mωπ~
)32
∫e−mω(r
21+r22 )/~r21 d
3~r1d3~r2
= −λ2mω2
(mωπ~
)3(4π)2
∫ ∞0
e−mωr22 /~r22 dr2
∫ ∞0
e−mωr21 /~r41 dr1
= −λ8m4ω5
π~3
[1
4
~mω
√π~mω
][3
8
~2
m2ω2
√π~mω
]= −3
4λ~ω
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 20 / 21
Problem 8.24 (cont.)
Build a variational trial functionfrom the single oscillator solutions
by symmetry the scalar product in-tegrates to zero
ψ(~r1,~r2) =(mωπ~
)3/2e−mω(r
21+r22 )/2~
〈H〉 =3
2~ω +
3
2~ω + 〈Vee〉
= 3~ω + 〈Vee〉〈Vee〉 = −λ
4mω2
(mωπ~
)3 ∫e−mω(r
21+r22 )/~(~r1 −~r2)2 d3~r1d
3~r2
= −λ4mω2
(mωπ~
)3 ∫e−mω(r
21+r22 )/~(r21 −����2~r1 ·~r2 + r22 ) d3~r1d
3~r2
= −λ4mω2
(mωπ~
)32
∫e−mω(r
21+r22 )/~r21 d
3~r1d3~r2
= −λ2mω2
(mωπ~
)3(4π)2
∫ ∞0
e−mωr22 /~r22 dr2
∫ ∞0
e−mωr21 /~r41 dr1
= −λ8m4ω5
π~3
[1
4
~mω
√π~mω
][3
8
~2
m2ω2
√π~mω
]= −3
4λ~ω
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 20 / 21
Problem 8.24 (cont.)
Build a variational trial functionfrom the single oscillator solutions
by symmetry the scalar product in-tegrates to zero
ψ(~r1,~r2) =(mωπ~
)3/2e−mω(r
21+r22 )/2~
〈H〉 =3
2~ω +
3
2~ω + 〈Vee〉
= 3~ω + 〈Vee〉〈Vee〉 = −λ
4mω2
(mωπ~
)3 ∫e−mω(r
21+r22 )/~(~r1 −~r2)2 d3~r1d
3~r2
= −λ4mω2
(mωπ~
)3 ∫e−mω(r
21+r22 )/~(r21 −����2~r1 ·~r2 + r22 ) d3~r1d
3~r2
= −λ4mω2
(mωπ~
)32
∫e−mω(r
21+r22 )/~r21 d
3~r1d3~r2
= −λ2mω2
(mωπ~
)3(4π)2
∫ ∞0
e−mωr22 /~r22 dr2
∫ ∞0
e−mωr21 /~r41 dr1
= −λ8m4ω5
π~3
[1
4
~mω
√π~mω
][3
8
~2
m2ω2
√π~mω
]= −3
4λ~ω
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 20 / 21
Problem 8.24 (cont.)
Build a variational trial functionfrom the single oscillator solutions
by symmetry the scalar product in-tegrates to zero
ψ(~r1,~r2) =(mωπ~
)3/2e−mω(r
21+r22 )/2~
〈H〉 =3
2~ω +
3
2~ω + 〈Vee〉
= 3~ω + 〈Vee〉〈Vee〉 = −λ
4mω2
(mωπ~
)3 ∫e−mω(r
21+r22 )/~(~r1 −~r2)2 d3~r1d
3~r2
= −λ4mω2
(mωπ~
)3 ∫e−mω(r
21+r22 )/~(r21 −����2~r1 ·~r2 + r22 ) d3~r1d
3~r2
= −λ4mω2
(mωπ~
)32
∫e−mω(r
21+r22 )/~r21 d
3~r1d3~r2
= −λ2mω2
(mωπ~
)3(4π)2
∫ ∞0
e−mωr22 /~r22 dr2
∫ ∞0
e−mωr21 /~r41 dr1
= −λ8m4ω5
π~3
[1
4
~mω
√π~mω
][3
8
~2
m2ω2
√π~mω
]= −3
4λ~ω
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 20 / 21
Problem 8.24 (cont.)
Build a variational trial functionfrom the single oscillator solutions
by symmetry the scalar product in-tegrates to zero
ψ(~r1,~r2) =(mωπ~
)3/2e−mω(r
21+r22 )/2~
〈H〉 =3
2~ω +
3
2~ω + 〈Vee〉
= 3~ω + 〈Vee〉〈Vee〉 = −λ
4mω2
(mωπ~
)3 ∫e−mω(r
21+r22 )/~(~r1 −~r2)2 d3~r1d
3~r2
= −λ4mω2
(mωπ~
)3 ∫e−mω(r
21+r22 )/~(r21 −����2~r1 ·~r2 + r22 ) d3~r1d
3~r2
= −λ4mω2
(mωπ~
)32
∫e−mω(r
21+r22 )/~r21 d
3~r1d3~r2
= −λ2mω2
(mωπ~
)3(4π)2
∫ ∞0
e−mωr22 /~r22 dr2
∫ ∞0
e−mωr21 /~r41 dr1
= −λ8m4ω5
π~3
[1
4
~mω
√π~mω
][3
8
~2
m2ω2
√π~mω
]
= −3
4λ~ω
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 20 / 21
Problem 8.24 (cont.)
Build a variational trial functionfrom the single oscillator solutions
by symmetry the scalar product in-tegrates to zero
ψ(~r1,~r2) =(mωπ~
)3/2e−mω(r
21+r22 )/2~
〈H〉 =3
2~ω +
3
2~ω + 〈Vee〉
= 3~ω + 〈Vee〉〈Vee〉 = −λ
4mω2
(mωπ~
)3 ∫e−mω(r
21+r22 )/~(~r1 −~r2)2 d3~r1d
3~r2
= −λ4mω2
(mωπ~
)3 ∫e−mω(r
21+r22 )/~(r21 −����2~r1 ·~r2 + r22 ) d3~r1d
3~r2
= −λ4mω2
(mωπ~
)32
∫e−mω(r
21+r22 )/~r21 d
3~r1d3~r2
= −λ2mω2
(mωπ~
)3(4π)2
∫ ∞0
e−mωr22 /~r22 dr2
∫ ∞0
e−mωr21 /~r41 dr1
= −λ8m4ω5
π~3
[1
4
~mω
√π~mω
][3
8
~2
m2ω2
√π~mω
]= −3
4λ~ω
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 20 / 21
Problem 8.24 (cont.)
The expectation value of theHamiltonian is thus
comparing to the exact answer
〈H〉 = 3~ω − 3
4λ~ω
= 3~ω(
1− λ
4
)
〈H〉 = 32~ω
(2− λ
2
)2− λ
2
1− λ2
1− λ+ λ2
4
divide by the constants
subtract 1
square and apply bino-mial expansion
Egs = 32~ω(1 +
√1− λ)
1 +√
1− λ√
1− λ1− λ
clearly 〈H〉 > Egs
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 21 / 21
Problem 8.24 (cont.)
The expectation value of theHamiltonian is thus
comparing to the exact answer
〈H〉 = 3~ω − 3
4λ~ω
= 3~ω(
1− λ
4
)
〈H〉 = 32~ω
(2− λ
2
)2− λ
2
1− λ2
1− λ+ λ2
4
divide by the constants
subtract 1
square and apply bino-mial expansion
Egs = 32~ω(1 +
√1− λ)
1 +√
1− λ√
1− λ1− λ
clearly 〈H〉 > Egs
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 21 / 21
Problem 8.24 (cont.)
The expectation value of theHamiltonian is thus
comparing to the exact answer
〈H〉 = 3~ω − 3
4λ~ω
= 3~ω(
1− λ
4
)
〈H〉 = 32~ω
(2− λ
2
)2− λ
2
1− λ2
1− λ+ λ2
4
divide by the constants
subtract 1
square and apply bino-mial expansion
Egs = 32~ω(1 +
√1− λ)
1 +√
1− λ√
1− λ1− λ
clearly 〈H〉 > Egs
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 21 / 21
Problem 8.24 (cont.)
The expectation value of theHamiltonian is thus
comparing to the exact answer
〈H〉 = 3~ω − 3
4λ~ω
= 3~ω(
1− λ
4
)
〈H〉 = 32~ω
(2− λ
2
)2− λ
2
1− λ2
1− λ+ λ2
4
divide by the constants
subtract 1
square and apply bino-mial expansion
Egs = 32~ω(1 +
√1− λ)
1 +√
1− λ√
1− λ1− λ
clearly 〈H〉 > Egs
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 21 / 21
Problem 8.24 (cont.)
The expectation value of theHamiltonian is thus
comparing to the exact answer
〈H〉 = 3~ω − 3
4λ~ω
= 3~ω(
1− λ
4
)
〈H〉 = 32~ω
(2− λ
2
)
2− λ2
1− λ2
1− λ+ λ2
4
divide by the constants
subtract 1
square and apply bino-mial expansion
Egs = 32~ω(1 +
√1− λ)
1 +√
1− λ√
1− λ1− λ
clearly 〈H〉 > Egs
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 21 / 21
Problem 8.24 (cont.)
The expectation value of theHamiltonian is thus
comparing to the exact answer
〈H〉 = 3~ω − 3
4λ~ω
= 3~ω(
1− λ
4
)
〈H〉 = 32~ω
(2− λ
2
)
2− λ2
1− λ2
1− λ+ λ2
4
divide by the constants
subtract 1
square and apply bino-mial expansion
Egs = 32~ω(1 +
√1− λ)
1 +√
1− λ√
1− λ1− λ
clearly 〈H〉 > Egs
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 21 / 21
Problem 8.24 (cont.)
The expectation value of theHamiltonian is thus
comparing to the exact answer
〈H〉 = 3~ω − 3
4λ~ω
= 3~ω(
1− λ
4
)
〈H〉 = 32~ω
(2− λ
2
)
2− λ2
1− λ2
1− λ+ λ2
4
divide by the constants
subtract 1
square and apply bino-mial expansion
Egs = 32~ω(1 +
√1− λ)
1 +√
1− λ√
1− λ1− λ
clearly 〈H〉 > Egs
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 21 / 21
Problem 8.24 (cont.)
The expectation value of theHamiltonian is thus
comparing to the exact answer
〈H〉 = 3~ω − 3
4λ~ω
= 3~ω(
1− λ
4
)
〈H〉 = 32~ω
(2− λ
2
)2− λ
2
1− λ2
1− λ+ λ2
4
divide by the constants
subtract 1
square and apply bino-mial expansion
Egs = 32~ω(1 +
√1− λ)
1 +√
1− λ√
1− λ1− λ
clearly 〈H〉 > Egs
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 21 / 21
Problem 8.24 (cont.)
The expectation value of theHamiltonian is thus
comparing to the exact answer
〈H〉 = 3~ω − 3
4λ~ω
= 3~ω(
1− λ
4
)
〈H〉 = 32~ω
(2− λ
2
)2− λ
2
1− λ2
1− λ+ λ2
4
divide by the constants
subtract 1
square and apply bino-mial expansion
Egs = 32~ω(1 +
√1− λ)
1 +√
1− λ
√1− λ
1− λ
clearly 〈H〉 > Egs
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 21 / 21
Problem 8.24 (cont.)
The expectation value of theHamiltonian is thus
comparing to the exact answer
〈H〉 = 3~ω − 3
4λ~ω
= 3~ω(
1− λ
4
)
〈H〉 = 32~ω
(2− λ
2
)2− λ
2
1− λ2
1− λ+ λ2
4
divide by the constants
subtract 1
square and apply bino-mial expansion
Egs = 32~ω(1 +
√1− λ)
1 +√
1− λ
√1− λ
1− λ
clearly 〈H〉 > Egs
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 21 / 21
Problem 8.24 (cont.)
The expectation value of theHamiltonian is thus
comparing to the exact answer
〈H〉 = 3~ω − 3
4λ~ω
= 3~ω(
1− λ
4
)
〈H〉 = 32~ω
(2− λ
2
)2− λ
2
1− λ2
1− λ+ λ2
4
divide by the constants
subtract 1
square and apply bino-mial expansion
Egs = 32~ω(1 +
√1− λ)
1 +√
1− λ
√1− λ
1− λ
clearly 〈H〉 > Egs
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 21 / 21
Problem 8.24 (cont.)
The expectation value of theHamiltonian is thus
comparing to the exact answer
〈H〉 = 3~ω − 3
4λ~ω
= 3~ω(
1− λ
4
)
〈H〉 = 32~ω
(2− λ
2
)2− λ
2
1− λ2
1− λ+ λ2
4
divide by the constants
subtract 1
square and apply bino-mial expansion
Egs = 32~ω(1 +
√1− λ)
1 +√
1− λ√
1− λ
1− λ
clearly 〈H〉 > Egs
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 21 / 21
Problem 8.24 (cont.)
The expectation value of theHamiltonian is thus
comparing to the exact answer
〈H〉 = 3~ω − 3
4λ~ω
= 3~ω(
1− λ
4
)
〈H〉 = 32~ω
(2− λ
2
)2− λ
2
1− λ2
1− λ+ λ2
4
divide by the constants
subtract 1
square and apply bino-mial expansion
Egs = 32~ω(1 +
√1− λ)
1 +√
1− λ√
1− λ
1− λ
clearly 〈H〉 > Egs
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 21 / 21
Problem 8.24 (cont.)
The expectation value of theHamiltonian is thus
comparing to the exact answer
〈H〉 = 3~ω − 3
4λ~ω
= 3~ω(
1− λ
4
)
〈H〉 = 32~ω
(2− λ
2
)2− λ
2
1− λ2
1− λ+ λ2
4
divide by the constants
subtract 1
square and apply bino-mial expansion
Egs = 32~ω(1 +
√1− λ)
1 +√
1− λ√
1− λ
1− λ
clearly 〈H〉 > Egs
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 21 / 21
Problem 8.24 (cont.)
The expectation value of theHamiltonian is thus
comparing to the exact answer
〈H〉 = 3~ω − 3
4λ~ω
= 3~ω(
1− λ
4
)
〈H〉 = 32~ω
(2− λ
2
)2− λ
2
1− λ2
1− λ+ λ2
4
divide by the constants
subtract 1
square and apply bino-mial expansion
Egs = 32~ω(1 +
√1− λ)
1 +√
1− λ√
1− λ1− λ
clearly 〈H〉 > Egs
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 21 / 21
Problem 8.24 (cont.)
The expectation value of theHamiltonian is thus
comparing to the exact answer
〈H〉 = 3~ω − 3
4λ~ω
= 3~ω(
1− λ
4
)
〈H〉 = 32~ω
(2− λ
2
)2− λ
2
1− λ2
1− λ+ λ2
4
divide by the constants
subtract 1
square and apply bino-mial expansion
Egs = 32~ω(1 +
√1− λ)
1 +√
1− λ√
1− λ1− λ
clearly 〈H〉 > Egs
C. Segre (IIT) PHYS 406 - Spring 2020 February 13, 2020 21 / 21
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