today’s outline - february 02, 2017csrri.iit.edu/~segre/phys406/17s/lecture_07.pdf · today’s...
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Today’s Outline - February 02, 2017
• Hyperfine interaction
• 21cm line of hydrogen
• Variational method
• Harmonic oscillator
• Delta function
• Helium atom
Homework Assignment #04:Chapter 6:36; 7:1,4,7,13,16due Tuesday, February 14, 2017
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 1 / 18
Today’s Outline - February 02, 2017
• Hyperfine interaction
• 21cm line of hydrogen
• Variational method
• Harmonic oscillator
• Delta function
• Helium atom
Homework Assignment #04:Chapter 6:36; 7:1,4,7,13,16due Tuesday, February 14, 2017
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 1 / 18
Today’s Outline - February 02, 2017
• Hyperfine interaction
• 21cm line of hydrogen
• Variational method
• Harmonic oscillator
• Delta function
• Helium atom
Homework Assignment #04:Chapter 6:36; 7:1,4,7,13,16due Tuesday, February 14, 2017
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 1 / 18
Today’s Outline - February 02, 2017
• Hyperfine interaction
• 21cm line of hydrogen
• Variational method
• Harmonic oscillator
• Delta function
• Helium atom
Homework Assignment #04:Chapter 6:36; 7:1,4,7,13,16due Tuesday, February 14, 2017
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 1 / 18
Today’s Outline - February 02, 2017
• Hyperfine interaction
• 21cm line of hydrogen
• Variational method
• Harmonic oscillator
• Delta function
• Helium atom
Homework Assignment #04:Chapter 6:36; 7:1,4,7,13,16due Tuesday, February 14, 2017
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 1 / 18
Today’s Outline - February 02, 2017
• Hyperfine interaction
• 21cm line of hydrogen
• Variational method
• Harmonic oscillator
• Delta function
• Helium atom
Homework Assignment #04:Chapter 6:36; 7:1,4,7,13,16due Tuesday, February 14, 2017
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 1 / 18
Today’s Outline - February 02, 2017
• Hyperfine interaction
• 21cm line of hydrogen
• Variational method
• Harmonic oscillator
• Delta function
• Helium atom
Homework Assignment #04:Chapter 6:36; 7:1,4,7,13,16due Tuesday, February 14, 2017
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 1 / 18
Today’s Outline - February 02, 2017
• Hyperfine interaction
• 21cm line of hydrogen
• Variational method
• Harmonic oscillator
• Delta function
• Helium atom
Homework Assignment #04:Chapter 6:36; 7:1,4,7,13,16due Tuesday, February 14, 2017
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 1 / 18
Hyperfine interaction review
The proton’s magnetic moment in-teacts with the electron’s spin mag-netic moment to create the hyper-fine splitting
~µp =gpe
2mp
~Sp, ~µe = − gee
2me
~Se
the proton’s magnetic field seen by the electron is
~Bp =µ0
4πr3[3(~µp · r̂)r̂ − ~µp] +
2µ03~µpδ
3(~r)
the hyperfine interaction is simply H ′hf = −~µe · ~Bp which gives an energycorrection of
E(1)hf =
µ0gpe2
8πmpme
⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se
r3
⟩+µ0gpe
2
3mpme〈~Sp · ~Se〉|ψ(0)|2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 2 / 18
Hyperfine interaction review
The proton’s magnetic moment in-teacts with the electron’s spin mag-netic moment to create the hyper-fine splitting
~µp =gpe
2mp
~Sp
, ~µe = − gee
2me
~Se
the proton’s magnetic field seen by the electron is
~Bp =µ0
4πr3[3(~µp · r̂)r̂ − ~µp] +
2µ03~µpδ
3(~r)
the hyperfine interaction is simply H ′hf = −~µe · ~Bp which gives an energycorrection of
E(1)hf =
µ0gpe2
8πmpme
⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se
r3
⟩+µ0gpe
2
3mpme〈~Sp · ~Se〉|ψ(0)|2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 2 / 18
Hyperfine interaction review
The proton’s magnetic moment in-teacts with the electron’s spin mag-netic moment to create the hyper-fine splitting
~µp =gpe
2mp
~Sp, ~µe = − gee
2me
~Se
the proton’s magnetic field seen by the electron is
~Bp =µ0
4πr3[3(~µp · r̂)r̂ − ~µp] +
2µ03~µpδ
3(~r)
the hyperfine interaction is simply H ′hf = −~µe · ~Bp which gives an energycorrection of
E(1)hf =
µ0gpe2
8πmpme
⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se
r3
⟩+µ0gpe
2
3mpme〈~Sp · ~Se〉|ψ(0)|2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 2 / 18
Hyperfine interaction review
The proton’s magnetic moment in-teacts with the electron’s spin mag-netic moment to create the hyper-fine splitting
~µp =gpe
2mp
~Sp, ~µe = − gee
2me
~Se
the proton’s magnetic field seen by the electron is
~Bp =µ0
4πr3[3(~µp · r̂)r̂ − ~µp] +
2µ03~µpδ
3(~r)
the hyperfine interaction is simply H ′hf = −~µe · ~Bp which gives an energycorrection of
E(1)hf =
µ0gpe2
8πmpme
⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se
r3
⟩+µ0gpe
2
3mpme〈~Sp · ~Se〉|ψ(0)|2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 2 / 18
Hyperfine interaction review
The proton’s magnetic moment in-teacts with the electron’s spin mag-netic moment to create the hyper-fine splitting
~µp =gpe
2mp
~Sp, ~µe = − gee
2me
~Se
the proton’s magnetic field seen by the electron is
~Bp =µ0
4πr3[3(~µp · r̂)r̂ − ~µp] +
2µ03~µpδ
3(~r)
the hyperfine interaction is simply H ′hf = −~µe · ~Bp which gives an energycorrection of
E(1)hf =
µ0gpe2
8πmpme
⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se
r3
⟩+µ0gpe
2
3mpme〈~Sp · ~Se〉|ψ(0)|2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 2 / 18
Hyperfine interaction review
The proton’s magnetic moment in-teacts with the electron’s spin mag-netic moment to create the hyper-fine splitting
~µp =gpe
2mp
~Sp, ~µe = − gee
2me
~Se
the proton’s magnetic field seen by the electron is
~Bp =µ0
4πr3[3(~µp · r̂)r̂ − ~µp] +
2µ03~µpδ
3(~r)
the hyperfine interaction is simply H ′hf = −~µe · ~Bp
which gives an energycorrection of
E(1)hf =
µ0gpe2
8πmpme
⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se
r3
⟩+µ0gpe
2
3mpme〈~Sp · ~Se〉|ψ(0)|2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 2 / 18
Hyperfine interaction review
The proton’s magnetic moment in-teacts with the electron’s spin mag-netic moment to create the hyper-fine splitting
~µp =gpe
2mp
~Sp, ~µe = − gee
2me
~Se
the proton’s magnetic field seen by the electron is
~Bp =µ0
4πr3[3(~µp · r̂)r̂ − ~µp] +
2µ03~µpδ
3(~r)
the hyperfine interaction is simply H ′hf = −~µe · ~Bp which gives an energycorrection of
E(1)hf =
µ0gpe2
8πmpme
⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se
r3
⟩+µ0gpe
2
3mpme〈~Sp · ~Se〉|ψ(0)|2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 2 / 18
Hyperfine interaction review
The proton’s magnetic moment in-teacts with the electron’s spin mag-netic moment to create the hyper-fine splitting
~µp =gpe
2mp
~Sp, ~µe = − gee
2me
~Se
the proton’s magnetic field seen by the electron is
~Bp =µ0
4πr3[3(~µp · r̂)r̂ − ~µp] +
2µ03~µpδ
3(~r)
the hyperfine interaction is simply H ′hf = −~µe · ~Bp which gives an energycorrection of
E(1)hf =
µ0gpe2
8πmpme
⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se
r3
⟩+µ0gpe
2
3mpme〈~Sp · ~Se〉|ψ(0)|2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 2 / 18
Spin-spin coupling
E(1)hf =
µ0gpe2
8πmpme
⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se
r3
⟩+µ0gpe
2
3mpme〈~Sp · ~Se〉|ψ(0)|2
for states with l = 0 the first term vanishes and we are left with an energycorrection which depends on a coupling between the spin of the protonand the spin of the electron
for the ground state,
|ψ100(0)|2 =1
πa3
the total spin is now a good opera-tor of the perturbed Hamiltonian
using the usual trick we have
E(1)hf =
µ0gpe2
3πmpmea3〈~Sp · ~Se〉
~S ≡ ~Se + ~Sp
~Se · ~Sp = 12(S2 − S2
e − S2p )
〈~Sp · ~Se〉 =~2
2[s(s + 1)− se(se + 1)− sp(sp + 1)]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 3 / 18
Spin-spin coupling
E(1)hf =
(((((((
(((((((
((((µ0gpe2
8πmpme
⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se
r3
⟩+µ0gpe
2
3mpme〈~Sp · ~Se〉|ψ(0)|2
for states with l = 0 the first term vanishes and we are left with an energycorrection which depends on a coupling between the spin of the protonand the spin of the electron
for the ground state,
|ψ100(0)|2 =1
πa3
the total spin is now a good opera-tor of the perturbed Hamiltonian
using the usual trick we have
E(1)hf =
µ0gpe2
3πmpmea3〈~Sp · ~Se〉
~S ≡ ~Se + ~Sp
~Se · ~Sp = 12(S2 − S2
e − S2p )
〈~Sp · ~Se〉 =~2
2[s(s + 1)− se(se + 1)− sp(sp + 1)]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 3 / 18
Spin-spin coupling
E(1)hf =
(((((((
(((((((
((((µ0gpe2
8πmpme
⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se
r3
⟩+µ0gpe
2
3mpme〈~Sp · ~Se〉|ψ(0)|2
for states with l = 0 the first term vanishes and we are left with an energycorrection which depends on a coupling between the spin of the protonand the spin of the electron
for the ground state,
|ψ100(0)|2 =1
πa3
the total spin is now a good opera-tor of the perturbed Hamiltonian
using the usual trick we have
E(1)hf =
µ0gpe2
3πmpmea3〈~Sp · ~Se〉
~S ≡ ~Se + ~Sp
~Se · ~Sp = 12(S2 − S2
e − S2p )
〈~Sp · ~Se〉 =~2
2[s(s + 1)− se(se + 1)− sp(sp + 1)]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 3 / 18
Spin-spin coupling
E(1)hf =
(((((((
(((((((
((((µ0gpe2
8πmpme
⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se
r3
⟩+µ0gpe
2
3mpme〈~Sp · ~Se〉|ψ(0)|2
for states with l = 0 the first term vanishes and we are left with an energycorrection which depends on a coupling between the spin of the protonand the spin of the electron
for the ground state,
|ψ100(0)|2 =1
πa3
the total spin is now a good opera-tor of the perturbed Hamiltonian
using the usual trick we have
E(1)hf =
µ0gpe2
3πmpmea3〈~Sp · ~Se〉
~S ≡ ~Se + ~Sp
~Se · ~Sp = 12(S2 − S2
e − S2p )
〈~Sp · ~Se〉 =~2
2[s(s + 1)− se(se + 1)− sp(sp + 1)]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 3 / 18
Spin-spin coupling
E(1)hf =
(((((((
(((((((
((((µ0gpe2
8πmpme
⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se
r3
⟩+µ0gpe
2
3mpme〈~Sp · ~Se〉|ψ(0)|2
for states with l = 0 the first term vanishes and we are left with an energycorrection which depends on a coupling between the spin of the protonand the spin of the electron
for the ground state,
|ψ100(0)|2 =1
πa3
the total spin is now a good opera-tor of the perturbed Hamiltonian
using the usual trick we have
E(1)hf =
µ0gpe2
3πmpmea3〈~Sp · ~Se〉
~S ≡ ~Se + ~Sp
~Se · ~Sp = 12(S2 − S2
e − S2p )
〈~Sp · ~Se〉 =~2
2[s(s + 1)− se(se + 1)− sp(sp + 1)]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 3 / 18
Spin-spin coupling
E(1)hf =
(((((((
(((((((
((((µ0gpe2
8πmpme
⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se
r3
⟩+µ0gpe
2
3mpme〈~Sp · ~Se〉|ψ(0)|2
for states with l = 0 the first term vanishes and we are left with an energycorrection which depends on a coupling between the spin of the protonand the spin of the electron
for the ground state,
|ψ100(0)|2 =1
πa3
the total spin is now a good opera-tor of the perturbed Hamiltonian
using the usual trick we have
E(1)hf =
µ0gpe2
3πmpmea3〈~Sp · ~Se〉
~S ≡ ~Se + ~Sp
~Se · ~Sp = 12(S2 − S2
e − S2p )
〈~Sp · ~Se〉 =~2
2[s(s + 1)− se(se + 1)− sp(sp + 1)]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 3 / 18
Spin-spin coupling
E(1)hf =
(((((((
(((((((
((((µ0gpe2
8πmpme
⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se
r3
⟩+µ0gpe
2
3mpme〈~Sp · ~Se〉|ψ(0)|2
for states with l = 0 the first term vanishes and we are left with an energycorrection which depends on a coupling between the spin of the protonand the spin of the electron
for the ground state,
|ψ100(0)|2 =1
πa3
the total spin is now a good opera-tor of the perturbed Hamiltonian
using the usual trick we have
E(1)hf =
µ0gpe2
3πmpmea3〈~Sp · ~Se〉
~S ≡ ~Se + ~Sp
~Se · ~Sp = 12(S2 − S2
e − S2p )
〈~Sp · ~Se〉 =~2
2[s(s + 1)− se(se + 1)− sp(sp + 1)]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 3 / 18
Spin-spin coupling
E(1)hf =
(((((((
(((((((
((((µ0gpe2
8πmpme
⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se
r3
⟩+µ0gpe
2
3mpme〈~Sp · ~Se〉|ψ(0)|2
for states with l = 0 the first term vanishes and we are left with an energycorrection which depends on a coupling between the spin of the protonand the spin of the electron
for the ground state,
|ψ100(0)|2 =1
πa3
the total spin is now a good opera-tor of the perturbed Hamiltonian
using the usual trick we have
E(1)hf =
µ0gpe2
3πmpmea3〈~Sp · ~Se〉
~S ≡ ~Se + ~Sp
~Se · ~Sp = 12(S2 − S2
e − S2p )
〈~Sp · ~Se〉 =~2
2[s(s + 1)− se(se + 1)− sp(sp + 1)]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 3 / 18
Spin-spin coupling
E(1)hf =
(((((((
(((((((
((((µ0gpe2
8πmpme
⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se
r3
⟩+µ0gpe
2
3mpme〈~Sp · ~Se〉|ψ(0)|2
for states with l = 0 the first term vanishes and we are left with an energycorrection which depends on a coupling between the spin of the protonand the spin of the electron
for the ground state,
|ψ100(0)|2 =1
πa3
the total spin is now a good opera-tor of the perturbed Hamiltonian
using the usual trick we have
E(1)hf =
µ0gpe2
3πmpmea3〈~Sp · ~Se〉
~S ≡ ~Se + ~Sp
~Se · ~Sp = 12(S2 − S2
e − S2p )
〈~Sp · ~Se〉 =~2
2[s(s + 1)− se(se + 1)− sp(sp + 1)]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 3 / 18
The 21 cm hydrogen line
E(1)hf =
µ0gpe2
3πmpmea3〈~Sp · ~Se〉
=µ0gpe
2
3πmpmea3~2
2[s(s + 1)− se(se + 1)− sp(sp + 1)]
both the proton and the electronhave spin 1/2
but the total spin can be in eithera singlet or a triplet state
E(1)hf =
4gp~4
3mpmec2a4
{+1
4 , triplet
−34 , singlet
∆E = 5.88× 10−6eV
ν =∆E
h= 1420 MHz→ λ = 21cm
se(se + 1) = sp(sp + 1) = 34
s = 0 → s(s + 1) = 0
s = 1 → s(s + 1) = 2
∆E
s=1
s=0
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 4 / 18
The 21 cm hydrogen line
E(1)hf =
µ0gpe2
3πmpmea3〈~Sp · ~Se〉
=µ0gpe
2
3πmpmea3~2
2[s(s + 1)− se(se + 1)− sp(sp + 1)]
both the proton and the electronhave spin 1/2
but the total spin can be in eithera singlet or a triplet state
E(1)hf =
4gp~4
3mpmec2a4
{+1
4 , triplet
−34 , singlet
∆E = 5.88× 10−6eV
ν =∆E
h= 1420 MHz→ λ = 21cm
se(se + 1) = sp(sp + 1) = 34
s = 0 → s(s + 1) = 0
s = 1 → s(s + 1) = 2
∆E
s=1
s=0
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 4 / 18
The 21 cm hydrogen line
E(1)hf =
µ0gpe2
3πmpmea3〈~Sp · ~Se〉
=µ0gpe
2
3πmpmea3~2
2[s(s + 1)− se(se + 1)− sp(sp + 1)]
both the proton and the electronhave spin 1/2
but the total spin can be in eithera singlet or a triplet state
E(1)hf =
4gp~4
3mpmec2a4
{+1
4 , triplet
−34 , singlet
∆E = 5.88× 10−6eV
ν =∆E
h= 1420 MHz→ λ = 21cm
se(se + 1) = sp(sp + 1) = 34
s = 0 → s(s + 1) = 0
s = 1 → s(s + 1) = 2
∆E
s=1
s=0
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 4 / 18
The 21 cm hydrogen line
E(1)hf =
µ0gpe2
3πmpmea3〈~Sp · ~Se〉
=µ0gpe
2
3πmpmea3~2
2[s(s + 1)− se(se + 1)− sp(sp + 1)]
both the proton and the electronhave spin 1/2
but the total spin can be in eithera singlet or a triplet state
E(1)hf =
4gp~4
3mpmec2a4
{+1
4 , triplet
−34 , singlet
∆E = 5.88× 10−6eV
ν =∆E
h= 1420 MHz→ λ = 21cm
se(se + 1) = sp(sp + 1) = 34
s = 0 → s(s + 1) = 0
s = 1 → s(s + 1) = 2
∆E
s=1
s=0
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 4 / 18
The 21 cm hydrogen line
E(1)hf =
µ0gpe2
3πmpmea3〈~Sp · ~Se〉
=µ0gpe
2
3πmpmea3~2
2[s(s + 1)− se(se + 1)− sp(sp + 1)]
both the proton and the electronhave spin 1/2
but the total spin can be in eithera singlet
or a triplet state
E(1)hf =
4gp~4
3mpmec2a4
{+1
4 , triplet
−34 , singlet
∆E = 5.88× 10−6eV
ν =∆E
h= 1420 MHz→ λ = 21cm
se(se + 1) = sp(sp + 1) = 34
s = 0 → s(s + 1) = 0
s = 1 → s(s + 1) = 2
∆E
s=1
s=0
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 4 / 18
The 21 cm hydrogen line
E(1)hf =
µ0gpe2
3πmpmea3〈~Sp · ~Se〉
=µ0gpe
2
3πmpmea3~2
2[s(s + 1)− se(se + 1)− sp(sp + 1)]
both the proton and the electronhave spin 1/2
but the total spin can be in eithera singlet or a triplet state
E(1)hf =
4gp~4
3mpmec2a4
{+1
4 , triplet
−34 , singlet
∆E = 5.88× 10−6eV
ν =∆E
h= 1420 MHz→ λ = 21cm
se(se + 1) = sp(sp + 1) = 34
s = 0 → s(s + 1) = 0
s = 1 → s(s + 1) = 2
∆E
s=1
s=0
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 4 / 18
The 21 cm hydrogen line
E(1)hf =
µ0gpe2
3πmpmea3〈~Sp · ~Se〉
=µ0gpe
2
3πmpmea3~2
2[s(s + 1)− se(se + 1)− sp(sp + 1)]
both the proton and the electronhave spin 1/2
but the total spin can be in eithera singlet or a triplet state
E(1)hf =
4gp~4
3mpmec2a4
{+1
4 , triplet
−34 , singlet
∆E = 5.88× 10−6eV
ν =∆E
h= 1420 MHz→ λ = 21cm
se(se + 1) = sp(sp + 1) = 34
s = 0 → s(s + 1) = 0
s = 1 → s(s + 1) = 2
∆E
s=1
s=0
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 4 / 18
The 21 cm hydrogen line
E(1)hf =
µ0gpe2
3πmpmea3〈~Sp · ~Se〉
=µ0gpe
2
3πmpmea3~2
2[s(s + 1)− se(se + 1)− sp(sp + 1)]
both the proton and the electronhave spin 1/2
but the total spin can be in eithera singlet or a triplet state
E(1)hf =
4gp~4
3mpmec2a4
{+1
4 , triplet
−34 , singlet
∆E = 5.88× 10−6eV
ν =∆E
h= 1420 MHz→ λ = 21cm
se(se + 1) = sp(sp + 1) = 34
s = 0 → s(s + 1) = 0
s = 1 → s(s + 1) = 2
∆E
s=1
s=0
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 4 / 18
The 21 cm hydrogen line
E(1)hf =
µ0gpe2
3πmpmea3〈~Sp · ~Se〉
=µ0gpe
2
3πmpmea3~2
2[s(s + 1)− se(se + 1)− sp(sp + 1)]
both the proton and the electronhave spin 1/2
but the total spin can be in eithera singlet or a triplet state
E(1)hf =
4gp~4
3mpmec2a4
{+1
4 , triplet
−34 , singlet
∆E = 5.88× 10−6eV
ν =∆E
h= 1420 MHz→ λ = 21cm
se(se + 1) = sp(sp + 1) = 34
s = 0 → s(s + 1) = 0
s = 1 → s(s + 1) = 2
∆E
s=1
s=0
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 4 / 18
The 21 cm hydrogen line
E(1)hf =
µ0gpe2
3πmpmea3〈~Sp · ~Se〉
=µ0gpe
2
3πmpmea3~2
2[s(s + 1)− se(se + 1)− sp(sp + 1)]
both the proton and the electronhave spin 1/2
but the total spin can be in eithera singlet or a triplet state
E(1)hf =
4gp~4
3mpmec2a4
{+1
4 , triplet
−34 , singlet
∆E = 5.88× 10−6eV
ν =∆E
h= 1420 MHz
→ λ = 21cm
se(se + 1) = sp(sp + 1) = 34
s = 0 → s(s + 1) = 0
s = 1 → s(s + 1) = 2
∆E
s=1
s=0
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 4 / 18
The 21 cm hydrogen line
E(1)hf =
µ0gpe2
3πmpmea3〈~Sp · ~Se〉
=µ0gpe
2
3πmpmea3~2
2[s(s + 1)− se(se + 1)− sp(sp + 1)]
both the proton and the electronhave spin 1/2
but the total spin can be in eithera singlet or a triplet state
E(1)hf =
4gp~4
3mpmec2a4
{+1
4 , triplet
−34 , singlet
∆E = 5.88× 10−6eV
ν =∆E
h= 1420 MHz→ λ = 21cm
se(se + 1) = sp(sp + 1) = 34
s = 0 → s(s + 1) = 0
s = 1 → s(s + 1) = 2
∆E
s=1
s=0
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 4 / 18
Variational theorem
Suppose that we wish to calculate the ground state energy, Egs , of asystem with Hamiltonian H which cannot be solved exactly (a verycommon occurrence!).
Using the variational principle it is possible toobtain an upper bound on Egs .
If ψ is an arbitrary normalized wavefunction, we can write
where ψn are the (unknown) eigen-functions of H, which form a com-plete set
ψ =∑n
cnψn
Hψn = Enψn
the normalization condition for ψ requires
1 = 〈ψ|ψ〉 =
⟨∑m
cmψm
∣∣∣∣∣ ∑n
cnψn
⟩=∑m
∑n
c∗mcn〈ψm|ψn〉 =∑n
|cn|2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 5 / 18
Variational theorem
Suppose that we wish to calculate the ground state energy, Egs , of asystem with Hamiltonian H which cannot be solved exactly (a verycommon occurrence!). Using the variational principle it is possible toobtain an upper bound on Egs .
If ψ is an arbitrary normalized wavefunction, we can write
where ψn are the (unknown) eigen-functions of H, which form a com-plete set
ψ =∑n
cnψn
Hψn = Enψn
the normalization condition for ψ requires
1 = 〈ψ|ψ〉 =
⟨∑m
cmψm
∣∣∣∣∣ ∑n
cnψn
⟩=∑m
∑n
c∗mcn〈ψm|ψn〉 =∑n
|cn|2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 5 / 18
Variational theorem
Suppose that we wish to calculate the ground state energy, Egs , of asystem with Hamiltonian H which cannot be solved exactly (a verycommon occurrence!). Using the variational principle it is possible toobtain an upper bound on Egs .
If ψ is an arbitrary normalized wavefunction, we can write
where ψn are the (unknown) eigen-functions of H, which form a com-plete set
ψ =∑n
cnψn
Hψn = Enψn
the normalization condition for ψ requires
1 = 〈ψ|ψ〉 =
⟨∑m
cmψm
∣∣∣∣∣ ∑n
cnψn
⟩=∑m
∑n
c∗mcn〈ψm|ψn〉 =∑n
|cn|2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 5 / 18
Variational theorem
Suppose that we wish to calculate the ground state energy, Egs , of asystem with Hamiltonian H which cannot be solved exactly (a verycommon occurrence!). Using the variational principle it is possible toobtain an upper bound on Egs .
If ψ is an arbitrary normalized wavefunction, we can write
where ψn are the (unknown) eigen-functions of H, which form a com-plete set
ψ =∑n
cnψn
Hψn = Enψn
the normalization condition for ψ requires
1 = 〈ψ|ψ〉 =
⟨∑m
cmψm
∣∣∣∣∣ ∑n
cnψn
⟩=∑m
∑n
c∗mcn〈ψm|ψn〉 =∑n
|cn|2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 5 / 18
Variational theorem
Suppose that we wish to calculate the ground state energy, Egs , of asystem with Hamiltonian H which cannot be solved exactly (a verycommon occurrence!). Using the variational principle it is possible toobtain an upper bound on Egs .
If ψ is an arbitrary normalized wavefunction, we can write
where ψn are the (unknown) eigen-functions of H, which form a com-plete set
ψ =∑n
cnψn
Hψn = Enψn
the normalization condition for ψ requires
1 = 〈ψ|ψ〉 =
⟨∑m
cmψm
∣∣∣∣∣ ∑n
cnψn
⟩=∑m
∑n
c∗mcn〈ψm|ψn〉 =∑n
|cn|2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 5 / 18
Variational theorem
Suppose that we wish to calculate the ground state energy, Egs , of asystem with Hamiltonian H which cannot be solved exactly (a verycommon occurrence!). Using the variational principle it is possible toobtain an upper bound on Egs .
If ψ is an arbitrary normalized wavefunction, we can write
where ψn are the (unknown) eigen-functions of H, which form a com-plete set
ψ =∑n
cnψn
Hψn = Enψn
the normalization condition for ψ requires
1 = 〈ψ|ψ〉 =
⟨∑m
cmψm
∣∣∣∣∣ ∑n
cnψn
⟩=∑m
∑n
c∗mcn〈ψm|ψn〉 =∑n
|cn|2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 5 / 18
Variational theorem
Suppose that we wish to calculate the ground state energy, Egs , of asystem with Hamiltonian H which cannot be solved exactly (a verycommon occurrence!). Using the variational principle it is possible toobtain an upper bound on Egs .
If ψ is an arbitrary normalized wavefunction, we can write
where ψn are the (unknown) eigen-functions of H, which form a com-plete set
ψ =∑n
cnψn
Hψn = Enψn
the normalization condition for ψ requires
1 = 〈ψ|ψ〉 =
⟨∑m
cmψm
∣∣∣∣∣ ∑n
cnψn
⟩=∑m
∑n
c∗mcn〈ψm|ψn〉 =∑n
|cn|2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 5 / 18
Variational theorem
Suppose that we wish to calculate the ground state energy, Egs , of asystem with Hamiltonian H which cannot be solved exactly (a verycommon occurrence!). Using the variational principle it is possible toobtain an upper bound on Egs .
If ψ is an arbitrary normalized wavefunction, we can write
where ψn are the (unknown) eigen-functions of H, which form a com-plete set
ψ =∑n
cnψn
Hψn = Enψn
the normalization condition for ψ requires
1 = 〈ψ|ψ〉
=
⟨∑m
cmψm
∣∣∣∣∣ ∑n
cnψn
⟩=∑m
∑n
c∗mcn〈ψm|ψn〉 =∑n
|cn|2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 5 / 18
Variational theorem
Suppose that we wish to calculate the ground state energy, Egs , of asystem with Hamiltonian H which cannot be solved exactly (a verycommon occurrence!). Using the variational principle it is possible toobtain an upper bound on Egs .
If ψ is an arbitrary normalized wavefunction, we can write
where ψn are the (unknown) eigen-functions of H, which form a com-plete set
ψ =∑n
cnψn
Hψn = Enψn
the normalization condition for ψ requires
1 = 〈ψ|ψ〉 =
⟨∑m
cmψm
∣∣∣∣∣ ∑n
cnψn
⟩
=∑m
∑n
c∗mcn〈ψm|ψn〉 =∑n
|cn|2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 5 / 18
Variational theorem
Suppose that we wish to calculate the ground state energy, Egs , of asystem with Hamiltonian H which cannot be solved exactly (a verycommon occurrence!). Using the variational principle it is possible toobtain an upper bound on Egs .
If ψ is an arbitrary normalized wavefunction, we can write
where ψn are the (unknown) eigen-functions of H, which form a com-plete set
ψ =∑n
cnψn
Hψn = Enψn
the normalization condition for ψ requires
1 = 〈ψ|ψ〉 =
⟨∑m
cmψm
∣∣∣∣∣ ∑n
cnψn
⟩=∑m
∑n
c∗mcn〈ψm|ψn〉
=∑n
|cn|2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 5 / 18
Variational theorem
Suppose that we wish to calculate the ground state energy, Egs , of asystem with Hamiltonian H which cannot be solved exactly (a verycommon occurrence!). Using the variational principle it is possible toobtain an upper bound on Egs .
If ψ is an arbitrary normalized wavefunction, we can write
where ψn are the (unknown) eigen-functions of H, which form a com-plete set
ψ =∑n
cnψn
Hψn = Enψn
the normalization condition for ψ requires
1 = 〈ψ|ψ〉 =
⟨∑m
cmψm
∣∣∣∣∣ ∑n
cnψn
⟩=∑m
∑n
c∗mcn〈ψm|ψn〉 =∑n
|cn|2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 5 / 18
Proof of variational theorem
If we take the expectation value of the Hamiltonian with this arbitrarywave function, ψ we have
〈H〉 =
⟨∑m
cmψm
∣∣∣∣∣ H∑n
cnψn
⟩=∑m
∑n
c∗mEncn〈ψm|ψn〉 =∑n
En|cn|2
since the ground state energy mustbe the smallest eigenvalue of theHamiltonian, then Egs ≤ En andwe can write
〈H〉 ≥ Egs
∑n
|cn|2 = Egs
This is the so-called variational principle which allows us to compute anupper bound on the ground state energy and, if we make a judiciouschoice of arbitrary wave function, ψ (ie. not arbitrary at all!) we can getvery close to the actual ground state energy.
In practice this means we minimize the value of the Hamiltonianexpectation value to achieve this upper bound.
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 6 / 18
Proof of variational theorem
If we take the expectation value of the Hamiltonian with this arbitrarywave function, ψ we have
〈H〉 =
⟨∑m
cmψm
∣∣∣∣∣ H∑n
cnψn
⟩
=∑m
∑n
c∗mEncn〈ψm|ψn〉 =∑n
En|cn|2
since the ground state energy mustbe the smallest eigenvalue of theHamiltonian, then Egs ≤ En andwe can write
〈H〉 ≥ Egs
∑n
|cn|2 = Egs
This is the so-called variational principle which allows us to compute anupper bound on the ground state energy and, if we make a judiciouschoice of arbitrary wave function, ψ (ie. not arbitrary at all!) we can getvery close to the actual ground state energy.
In practice this means we minimize the value of the Hamiltonianexpectation value to achieve this upper bound.
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 6 / 18
Proof of variational theorem
If we take the expectation value of the Hamiltonian with this arbitrarywave function, ψ we have
〈H〉 =
⟨∑m
cmψm
∣∣∣∣∣ H∑n
cnψn
⟩=∑m
∑n
c∗mEncn〈ψm|ψn〉
=∑n
En|cn|2
since the ground state energy mustbe the smallest eigenvalue of theHamiltonian, then Egs ≤ En andwe can write
〈H〉 ≥ Egs
∑n
|cn|2 = Egs
This is the so-called variational principle which allows us to compute anupper bound on the ground state energy and, if we make a judiciouschoice of arbitrary wave function, ψ (ie. not arbitrary at all!) we can getvery close to the actual ground state energy.
In practice this means we minimize the value of the Hamiltonianexpectation value to achieve this upper bound.
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 6 / 18
Proof of variational theorem
If we take the expectation value of the Hamiltonian with this arbitrarywave function, ψ we have
〈H〉 =
⟨∑m
cmψm
∣∣∣∣∣ H∑n
cnψn
⟩=∑m
∑n
c∗mEncn〈ψm|ψn〉 =∑n
En|cn|2
since the ground state energy mustbe the smallest eigenvalue of theHamiltonian, then Egs ≤ En andwe can write
〈H〉 ≥ Egs
∑n
|cn|2 = Egs
This is the so-called variational principle which allows us to compute anupper bound on the ground state energy and, if we make a judiciouschoice of arbitrary wave function, ψ (ie. not arbitrary at all!) we can getvery close to the actual ground state energy.
In practice this means we minimize the value of the Hamiltonianexpectation value to achieve this upper bound.
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 6 / 18
Proof of variational theorem
If we take the expectation value of the Hamiltonian with this arbitrarywave function, ψ we have
〈H〉 =
⟨∑m
cmψm
∣∣∣∣∣ H∑n
cnψn
⟩=∑m
∑n
c∗mEncn〈ψm|ψn〉 =∑n
En|cn|2
since the ground state energy mustbe the smallest eigenvalue of theHamiltonian, then Egs ≤ En andwe can write
〈H〉 ≥ Egs
∑n
|cn|2 = Egs
This is the so-called variational principle which allows us to compute anupper bound on the ground state energy and, if we make a judiciouschoice of arbitrary wave function, ψ (ie. not arbitrary at all!) we can getvery close to the actual ground state energy.
In practice this means we minimize the value of the Hamiltonianexpectation value to achieve this upper bound.
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 6 / 18
Proof of variational theorem
If we take the expectation value of the Hamiltonian with this arbitrarywave function, ψ we have
〈H〉 =
⟨∑m
cmψm
∣∣∣∣∣ H∑n
cnψn
⟩=∑m
∑n
c∗mEncn〈ψm|ψn〉 =∑n
En|cn|2
since the ground state energy mustbe the smallest eigenvalue of theHamiltonian, then Egs ≤ En andwe can write
〈H〉 ≥ Egs
∑n
|cn|2
= Egs
This is the so-called variational principle which allows us to compute anupper bound on the ground state energy and, if we make a judiciouschoice of arbitrary wave function, ψ (ie. not arbitrary at all!) we can getvery close to the actual ground state energy.
In practice this means we minimize the value of the Hamiltonianexpectation value to achieve this upper bound.
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 6 / 18
Proof of variational theorem
If we take the expectation value of the Hamiltonian with this arbitrarywave function, ψ we have
〈H〉 =
⟨∑m
cmψm
∣∣∣∣∣ H∑n
cnψn
⟩=∑m
∑n
c∗mEncn〈ψm|ψn〉 =∑n
En|cn|2
since the ground state energy mustbe the smallest eigenvalue of theHamiltonian, then Egs ≤ En andwe can write
〈H〉 ≥ Egs
∑n
|cn|2 = Egs
This is the so-called variational principle which allows us to compute anupper bound on the ground state energy and, if we make a judiciouschoice of arbitrary wave function, ψ (ie. not arbitrary at all!) we can getvery close to the actual ground state energy.
In practice this means we minimize the value of the Hamiltonianexpectation value to achieve this upper bound.
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 6 / 18
Proof of variational theorem
If we take the expectation value of the Hamiltonian with this arbitrarywave function, ψ we have
〈H〉 =
⟨∑m
cmψm
∣∣∣∣∣ H∑n
cnψn
⟩=∑m
∑n
c∗mEncn〈ψm|ψn〉 =∑n
En|cn|2
since the ground state energy mustbe the smallest eigenvalue of theHamiltonian, then Egs ≤ En andwe can write
〈H〉 ≥ Egs
∑n
|cn|2 = Egs
This is the so-called variational principle which allows us to compute anupper bound on the ground state energy and, if we make a judiciouschoice of arbitrary wave function, ψ (ie. not arbitrary at all!) we can getvery close to the actual ground state energy.
In practice this means we minimize the value of the Hamiltonianexpectation value to achieve this upper bound.
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 6 / 18
Proof of variational theorem
If we take the expectation value of the Hamiltonian with this arbitrarywave function, ψ we have
〈H〉 =
⟨∑m
cmψm
∣∣∣∣∣ H∑n
cnψn
⟩=∑m
∑n
c∗mEncn〈ψm|ψn〉 =∑n
En|cn|2
since the ground state energy mustbe the smallest eigenvalue of theHamiltonian, then Egs ≤ En andwe can write
〈H〉 ≥ Egs
∑n
|cn|2 = Egs
This is the so-called variational principle which allows us to compute anupper bound on the ground state energy and, if we make a judiciouschoice of arbitrary wave function, ψ (ie. not arbitrary at all!) we can getvery close to the actual ground state energy.
In practice this means we minimize the value of the Hamiltonianexpectation value to achieve this upper bound.
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 6 / 18
Example 7.1 - Harmonic oscillator
Find the ground state energy for the 1-D harmonic oscillator using thevariational theorem
H = − ~2
2m
d2
dx2+
1
2mω2x2
Assume a solution of theform
where b is the variational pa-rameter and A is the normal-ization parameter
Now compute the expecta-tion value of the Hamilto-nian, which must be an up-per bound on the groundstate energy
ψ(x) = Ae−bx2
1 = |A|2∫ ∞−∞
e−2bx2dx = |A|2
√π
2b
A =
(2b
π
)1/4
〈H〉 = 〈T 〉+ 〈V 〉
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 7 / 18
Example 7.1 - Harmonic oscillator
Find the ground state energy for the 1-D harmonic oscillator using thevariational theorem
H = − ~2
2m
d2
dx2+
1
2mω2x2
Assume a solution of theform
where b is the variational pa-rameter and A is the normal-ization parameter
Now compute the expecta-tion value of the Hamilto-nian, which must be an up-per bound on the groundstate energy
ψ(x) = Ae−bx2
1 = |A|2∫ ∞−∞
e−2bx2dx = |A|2
√π
2b
A =
(2b
π
)1/4
〈H〉 = 〈T 〉+ 〈V 〉
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 7 / 18
Example 7.1 - Harmonic oscillator
Find the ground state energy for the 1-D harmonic oscillator using thevariational theorem
H = − ~2
2m
d2
dx2+
1
2mω2x2
Assume a solution of theform
where b is the variational pa-rameter and A is the normal-ization parameter
Now compute the expecta-tion value of the Hamilto-nian, which must be an up-per bound on the groundstate energy
ψ(x) = Ae−bx2
1 = |A|2∫ ∞−∞
e−2bx2dx = |A|2
√π
2b
A =
(2b
π
)1/4
〈H〉 = 〈T 〉+ 〈V 〉
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 7 / 18
Example 7.1 - Harmonic oscillator
Find the ground state energy for the 1-D harmonic oscillator using thevariational theorem
H = − ~2
2m
d2
dx2+
1
2mω2x2
Assume a solution of theform
where b is the variational pa-rameter and A is the normal-ization parameter
Now compute the expecta-tion value of the Hamilto-nian, which must be an up-per bound on the groundstate energy
ψ(x) = Ae−bx2
1 = |A|2∫ ∞−∞
e−2bx2dx = |A|2
√π
2b
A =
(2b
π
)1/4
〈H〉 = 〈T 〉+ 〈V 〉
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 7 / 18
Example 7.1 - Harmonic oscillator
Find the ground state energy for the 1-D harmonic oscillator using thevariational theorem
H = − ~2
2m
d2
dx2+
1
2mω2x2
Assume a solution of theform
where b is the variational pa-rameter and A is the normal-ization parameter
Now compute the expecta-tion value of the Hamilto-nian, which must be an up-per bound on the groundstate energy
ψ(x) = Ae−bx2
1 = |A|2∫ ∞−∞
e−2bx2dx = |A|2
√π
2b
A =
(2b
π
)1/4
〈H〉 = 〈T 〉+ 〈V 〉
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 7 / 18
Example 7.1 - Harmonic oscillator
Find the ground state energy for the 1-D harmonic oscillator using thevariational theorem
H = − ~2
2m
d2
dx2+
1
2mω2x2
Assume a solution of theform
where b is the variational pa-rameter and A is the normal-ization parameter
Now compute the expecta-tion value of the Hamilto-nian, which must be an up-per bound on the groundstate energy
ψ(x) = Ae−bx2
1 = |A|2∫ ∞−∞
e−2bx2dx
= |A|2√
π
2b
A =
(2b
π
)1/4
〈H〉 = 〈T 〉+ 〈V 〉
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 7 / 18
Example 7.1 - Harmonic oscillator
Find the ground state energy for the 1-D harmonic oscillator using thevariational theorem
H = − ~2
2m
d2
dx2+
1
2mω2x2
Assume a solution of theform
where b is the variational pa-rameter and A is the normal-ization parameter
Now compute the expecta-tion value of the Hamilto-nian, which must be an up-per bound on the groundstate energy
ψ(x) = Ae−bx2
1 = |A|2∫ ∞−∞
e−2bx2dx = |A|2
√π
2b
A =
(2b
π
)1/4
〈H〉 = 〈T 〉+ 〈V 〉
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 7 / 18
Example 7.1 - Harmonic oscillator
Find the ground state energy for the 1-D harmonic oscillator using thevariational theorem
H = − ~2
2m
d2
dx2+
1
2mω2x2
Assume a solution of theform
where b is the variational pa-rameter and A is the normal-ization parameter
Now compute the expecta-tion value of the Hamilto-nian, which must be an up-per bound on the groundstate energy
ψ(x) = Ae−bx2
1 = |A|2∫ ∞−∞
e−2bx2dx = |A|2
√π
2b
A =
(2b
π
)1/4
〈H〉 = 〈T 〉+ 〈V 〉
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 7 / 18
Example 7.1 - Harmonic oscillator
Find the ground state energy for the 1-D harmonic oscillator using thevariational theorem
H = − ~2
2m
d2
dx2+
1
2mω2x2
Assume a solution of theform
where b is the variational pa-rameter and A is the normal-ization parameter
Now compute the expecta-tion value of the Hamilto-nian, which must be an up-per bound on the groundstate energy
ψ(x) = Ae−bx2
1 = |A|2∫ ∞−∞
e−2bx2dx = |A|2
√π
2b
A =
(2b
π
)1/4
〈H〉 = 〈T 〉+ 〈V 〉
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 7 / 18
Example 7.1 - Harmonic oscillator
Find the ground state energy for the 1-D harmonic oscillator using thevariational theorem
H = − ~2
2m
d2
dx2+
1
2mω2x2
Assume a solution of theform
where b is the variational pa-rameter and A is the normal-ization parameter
Now compute the expecta-tion value of the Hamilto-nian, which must be an up-per bound on the groundstate energy
ψ(x) = Ae−bx2
1 = |A|2∫ ∞−∞
e−2bx2dx = |A|2
√π
2b
A =
(2b
π
)1/4
〈H〉 = 〈T 〉+ 〈V 〉
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 7 / 18
Example 7.1 (cont.)
The kinetic and po-tential energies are in-stances of the Gaus-sian integral
giving
〈H〉 =~2b2m
+mω2
8b
〈T 〉 = − ~2
2m|A|2
∫ ∞−∞
e−bx2 d2
dx2
(e−bx
2)dx
= − ~2
2m
(2b
π
)1/2∫ ∞−∞
(4b2x2 − 2b
)e−2bx
2dx
= − ~2
2m
(2b
π
)1/2 [4b2√
π
32b3− 2b
√π
2b
]=
~2b2m
〈V 〉 =1
2mω2|A|2
∫ ∞−∞
x2e−2bx2dx =
mω2
8b
get the upper bound on energy by minimizing
0 =d
db〈H〉 =
~2
2m− mω2
8b2→ b =
mω
2~→ 〈H〉min =
1
2~ω
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 8 / 18
Example 7.1 (cont.)
The kinetic and po-tential energies are in-stances of the Gaus-sian integral
giving
〈H〉 =~2b2m
+mω2
8b
〈T 〉 = − ~2
2m|A|2
∫ ∞−∞
e−bx2 d2
dx2
(e−bx
2)dx
= − ~2
2m
(2b
π
)1/2∫ ∞−∞
(4b2x2 − 2b
)e−2bx
2dx
= − ~2
2m
(2b
π
)1/2 [4b2√
π
32b3− 2b
√π
2b
]=
~2b2m
〈V 〉 =1
2mω2|A|2
∫ ∞−∞
x2e−2bx2dx =
mω2
8b
get the upper bound on energy by minimizing
0 =d
db〈H〉 =
~2
2m− mω2
8b2→ b =
mω
2~→ 〈H〉min =
1
2~ω
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 8 / 18
Example 7.1 (cont.)
The kinetic and po-tential energies are in-stances of the Gaus-sian integral
giving
〈H〉 =~2b2m
+mω2
8b
〈T 〉 = − ~2
2m|A|2
∫ ∞−∞
e−bx2 d2
dx2
(e−bx
2)dx
= − ~2
2m
(2b
π
)1/2∫ ∞−∞
(4b2x2 − 2b
)e−2bx
2dx
= − ~2
2m
(2b
π
)1/2 [4b2√
π
32b3− 2b
√π
2b
]=
~2b2m
〈V 〉 =1
2mω2|A|2
∫ ∞−∞
x2e−2bx2dx =
mω2
8b
get the upper bound on energy by minimizing
0 =d
db〈H〉 =
~2
2m− mω2
8b2→ b =
mω
2~→ 〈H〉min =
1
2~ω
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 8 / 18
Example 7.1 (cont.)
The kinetic and po-tential energies are in-stances of the Gaus-sian integral
giving
〈H〉 =~2b2m
+mω2
8b
〈T 〉 = − ~2
2m|A|2
∫ ∞−∞
e−bx2 d2
dx2
(e−bx
2)dx
= − ~2
2m
(2b
π
)1/2∫ ∞−∞
(4b2x2 − 2b
)e−2bx
2dx
= − ~2
2m
(2b
π
)1/2 [4b2√
π
32b3− 2b
√π
2b
]
=~2b2m
〈V 〉 =1
2mω2|A|2
∫ ∞−∞
x2e−2bx2dx =
mω2
8b
get the upper bound on energy by minimizing
0 =d
db〈H〉 =
~2
2m− mω2
8b2→ b =
mω
2~→ 〈H〉min =
1
2~ω
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 8 / 18
Example 7.1 (cont.)
The kinetic and po-tential energies are in-stances of the Gaus-sian integral
giving
〈H〉 =~2b2m
+mω2
8b
〈T 〉 = − ~2
2m|A|2
∫ ∞−∞
e−bx2 d2
dx2
(e−bx
2)dx
= − ~2
2m
(2b
π
)1/2∫ ∞−∞
(4b2x2 − 2b
)e−2bx
2dx
= − ~2
2m
(2b
π
)1/2 [4b2√
π
32b3− 2b
√π
2b
]=
~2b2m
〈V 〉 =1
2mω2|A|2
∫ ∞−∞
x2e−2bx2dx =
mω2
8b
get the upper bound on energy by minimizing
0 =d
db〈H〉 =
~2
2m− mω2
8b2→ b =
mω
2~→ 〈H〉min =
1
2~ω
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 8 / 18
Example 7.1 (cont.)
The kinetic and po-tential energies are in-stances of the Gaus-sian integral
giving
〈H〉 =~2b2m
+mω2
8b
〈T 〉 = − ~2
2m|A|2
∫ ∞−∞
e−bx2 d2
dx2
(e−bx
2)dx
= − ~2
2m
(2b
π
)1/2∫ ∞−∞
(4b2x2 − 2b
)e−2bx
2dx
= − ~2
2m
(2b
π
)1/2 [4b2√
π
32b3− 2b
√π
2b
]=
~2b2m
〈V 〉 =1
2mω2|A|2
∫ ∞−∞
x2e−2bx2dx
=mω2
8b
get the upper bound on energy by minimizing
0 =d
db〈H〉 =
~2
2m− mω2
8b2→ b =
mω
2~→ 〈H〉min =
1
2~ω
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 8 / 18
Example 7.1 (cont.)
The kinetic and po-tential energies are in-stances of the Gaus-sian integral
giving
〈H〉 =~2b2m
+mω2
8b
〈T 〉 = − ~2
2m|A|2
∫ ∞−∞
e−bx2 d2
dx2
(e−bx
2)dx
= − ~2
2m
(2b
π
)1/2∫ ∞−∞
(4b2x2 − 2b
)e−2bx
2dx
= − ~2
2m
(2b
π
)1/2 [4b2√
π
32b3− 2b
√π
2b
]=
~2b2m
〈V 〉 =1
2mω2|A|2
∫ ∞−∞
x2e−2bx2dx =
mω2
8b
get the upper bound on energy by minimizing
0 =d
db〈H〉 =
~2
2m− mω2
8b2→ b =
mω
2~→ 〈H〉min =
1
2~ω
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 8 / 18
Example 7.1 (cont.)
The kinetic and po-tential energies are in-stances of the Gaus-sian integral
giving
〈H〉 =~2b2m
+mω2
8b
〈T 〉 = − ~2
2m|A|2
∫ ∞−∞
e−bx2 d2
dx2
(e−bx
2)dx
= − ~2
2m
(2b
π
)1/2∫ ∞−∞
(4b2x2 − 2b
)e−2bx
2dx
= − ~2
2m
(2b
π
)1/2 [4b2√
π
32b3− 2b
√π
2b
]=
~2b2m
〈V 〉 =1
2mω2|A|2
∫ ∞−∞
x2e−2bx2dx =
mω2
8b
get the upper bound on energy by minimizing
0 =d
db〈H〉 =
~2
2m− mω2
8b2→ b =
mω
2~→ 〈H〉min =
1
2~ω
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 8 / 18
Example 7.1 (cont.)
The kinetic and po-tential energies are in-stances of the Gaus-sian integral
giving
〈H〉 =~2b2m
+mω2
8b
〈T 〉 = − ~2
2m|A|2
∫ ∞−∞
e−bx2 d2
dx2
(e−bx
2)dx
= − ~2
2m
(2b
π
)1/2∫ ∞−∞
(4b2x2 − 2b
)e−2bx
2dx
= − ~2
2m
(2b
π
)1/2 [4b2√
π
32b3− 2b
√π
2b
]=
~2b2m
〈V 〉 =1
2mω2|A|2
∫ ∞−∞
x2e−2bx2dx =
mω2
8b
get the upper bound on energy by minimizing
0 =d
db〈H〉 =
~2
2m− mω2
8b2→ b =
mω
2~→ 〈H〉min =
1
2~ω
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 8 / 18
Example 7.1 (cont.)
The kinetic and po-tential energies are in-stances of the Gaus-sian integral
giving
〈H〉 =~2b2m
+mω2
8b
〈T 〉 = − ~2
2m|A|2
∫ ∞−∞
e−bx2 d2
dx2
(e−bx
2)dx
= − ~2
2m
(2b
π
)1/2∫ ∞−∞
(4b2x2 − 2b
)e−2bx
2dx
= − ~2
2m
(2b
π
)1/2 [4b2√
π
32b3− 2b
√π
2b
]=
~2b2m
〈V 〉 =1
2mω2|A|2
∫ ∞−∞
x2e−2bx2dx =
mω2
8b
get the upper bound on energy by minimizing
0 =d
db〈H〉
=~2
2m− mω2
8b2→ b =
mω
2~→ 〈H〉min =
1
2~ω
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 8 / 18
Example 7.1 (cont.)
The kinetic and po-tential energies are in-stances of the Gaus-sian integral
giving
〈H〉 =~2b2m
+mω2
8b
〈T 〉 = − ~2
2m|A|2
∫ ∞−∞
e−bx2 d2
dx2
(e−bx
2)dx
= − ~2
2m
(2b
π
)1/2∫ ∞−∞
(4b2x2 − 2b
)e−2bx
2dx
= − ~2
2m
(2b
π
)1/2 [4b2√
π
32b3− 2b
√π
2b
]=
~2b2m
〈V 〉 =1
2mω2|A|2
∫ ∞−∞
x2e−2bx2dx =
mω2
8b
get the upper bound on energy by minimizing
0 =d
db〈H〉 =
~2
2m− mω2
8b2
→ b =mω
2~→ 〈H〉min =
1
2~ω
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 8 / 18
Example 7.1 (cont.)
The kinetic and po-tential energies are in-stances of the Gaus-sian integral
giving
〈H〉 =~2b2m
+mω2
8b
〈T 〉 = − ~2
2m|A|2
∫ ∞−∞
e−bx2 d2
dx2
(e−bx
2)dx
= − ~2
2m
(2b
π
)1/2∫ ∞−∞
(4b2x2 − 2b
)e−2bx
2dx
= − ~2
2m
(2b
π
)1/2 [4b2√
π
32b3− 2b
√π
2b
]=
~2b2m
〈V 〉 =1
2mω2|A|2
∫ ∞−∞
x2e−2bx2dx =
mω2
8b
get the upper bound on energy by minimizing
0 =d
db〈H〉 =
~2
2m− mω2
8b2→ b =
mω
2~
→ 〈H〉min =1
2~ω
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 8 / 18
Example 7.1 (cont.)
The kinetic and po-tential energies are in-stances of the Gaus-sian integral
giving
〈H〉 =~2b2m
+mω2
8b
〈T 〉 = − ~2
2m|A|2
∫ ∞−∞
e−bx2 d2
dx2
(e−bx
2)dx
= − ~2
2m
(2b
π
)1/2∫ ∞−∞
(4b2x2 − 2b
)e−2bx
2dx
= − ~2
2m
(2b
π
)1/2 [4b2√
π
32b3− 2b
√π
2b
]=
~2b2m
〈V 〉 =1
2mω2|A|2
∫ ∞−∞
x2e−2bx2dx =
mω2
8b
get the upper bound on energy by minimizing
0 =d
db〈H〉 =
~2
2m− mω2
8b2→ b =
mω
2~→ 〈H〉min =
1
2~ω
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 8 / 18
Example 7.2 - Delta function
Find the ground state energy of adelta function potential
Start with the same Gaussian trialfunction as for the harmonic oscil-lator
We know that
〈T 〉 =~2b2m
we need to compute 〈V 〉
the upper bound for the energy isthus
minimizing wrt b
H = − ~2
2m
d2
dx2− αδ(x)
ψ(x) = Ae−bx2
〈V 〉 = −α|A|2∫ ∞−∞
δ(x)e−2bx2dx
=− α√
2b
π
〈H〉 =~2b2m− α
√2b
π
0 =d
db〈H〉 =
~2
2m− α√
2πb
b =2m2α2
π~4−→ 〈H〉min = −mα2
π~2≥ −mα2
2~2= Egs
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 9 / 18
Example 7.2 - Delta function
Find the ground state energy of adelta function potential
Start with the same Gaussian trialfunction as for the harmonic oscil-lator
We know that
〈T 〉 =~2b2m
we need to compute 〈V 〉
the upper bound for the energy isthus
minimizing wrt b
H = − ~2
2m
d2
dx2− αδ(x)
ψ(x) = Ae−bx2
〈V 〉 = −α|A|2∫ ∞−∞
δ(x)e−2bx2dx
=− α√
2b
π
〈H〉 =~2b2m− α
√2b
π
0 =d
db〈H〉 =
~2
2m− α√
2πb
b =2m2α2
π~4−→ 〈H〉min = −mα2
π~2≥ −mα2
2~2= Egs
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 9 / 18
Example 7.2 - Delta function
Find the ground state energy of adelta function potential
Start with the same Gaussian trialfunction as for the harmonic oscil-lator
We know that
〈T 〉 =~2b2m
we need to compute 〈V 〉
the upper bound for the energy isthus
minimizing wrt b
H = − ~2
2m
d2
dx2− αδ(x)
ψ(x) = Ae−bx2
〈V 〉 = −α|A|2∫ ∞−∞
δ(x)e−2bx2dx
=− α√
2b
π
〈H〉 =~2b2m− α
√2b
π
0 =d
db〈H〉 =
~2
2m− α√
2πb
b =2m2α2
π~4−→ 〈H〉min = −mα2
π~2≥ −mα2
2~2= Egs
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 9 / 18
Example 7.2 - Delta function
Find the ground state energy of adelta function potential
Start with the same Gaussian trialfunction as for the harmonic oscil-lator
We know that
〈T 〉 =~2b2m
we need to compute 〈V 〉
the upper bound for the energy isthus
minimizing wrt b
H = − ~2
2m
d2
dx2− αδ(x)
ψ(x) = Ae−bx2
〈V 〉 = −α|A|2∫ ∞−∞
δ(x)e−2bx2dx
=− α√
2b
π
〈H〉 =~2b2m− α
√2b
π
0 =d
db〈H〉 =
~2
2m− α√
2πb
b =2m2α2
π~4−→ 〈H〉min = −mα2
π~2≥ −mα2
2~2= Egs
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 9 / 18
Example 7.2 - Delta function
Find the ground state energy of adelta function potential
Start with the same Gaussian trialfunction as for the harmonic oscil-lator
We know that
〈T 〉 =~2b2m
we need to compute 〈V 〉
the upper bound for the energy isthus
minimizing wrt b
H = − ~2
2m
d2
dx2− αδ(x)
ψ(x) = Ae−bx2
〈V 〉 = −α|A|2∫ ∞−∞
δ(x)e−2bx2dx
=− α√
2b
π
〈H〉 =~2b2m− α
√2b
π
0 =d
db〈H〉 =
~2
2m− α√
2πb
b =2m2α2
π~4−→ 〈H〉min = −mα2
π~2≥ −mα2
2~2= Egs
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 9 / 18
Example 7.2 - Delta function
Find the ground state energy of adelta function potential
Start with the same Gaussian trialfunction as for the harmonic oscil-lator
We know that
〈T 〉 =~2b2m
we need to compute 〈V 〉
the upper bound for the energy isthus
minimizing wrt b
H = − ~2
2m
d2
dx2− αδ(x)
ψ(x) = Ae−bx2
〈V 〉 = −α|A|2∫ ∞−∞
δ(x)e−2bx2dx
=− α√
2b
π
〈H〉 =~2b2m− α
√2b
π
0 =d
db〈H〉 =
~2
2m− α√
2πb
b =2m2α2
π~4−→ 〈H〉min = −mα2
π~2≥ −mα2
2~2= Egs
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 9 / 18
Example 7.2 - Delta function
Find the ground state energy of adelta function potential
Start with the same Gaussian trialfunction as for the harmonic oscil-lator
We know that
〈T 〉 =~2b2m
we need to compute 〈V 〉
the upper bound for the energy isthus
minimizing wrt b
H = − ~2
2m
d2
dx2− αδ(x)
ψ(x) = Ae−bx2
〈V 〉 = −α|A|2∫ ∞−∞
δ(x)e−2bx2dx
=− α√
2b
π
〈H〉 =~2b2m− α
√2b
π
0 =d
db〈H〉 =
~2
2m− α√
2πb
b =2m2α2
π~4−→ 〈H〉min = −mα2
π~2≥ −mα2
2~2= Egs
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 9 / 18
Example 7.2 - Delta function
Find the ground state energy of adelta function potential
Start with the same Gaussian trialfunction as for the harmonic oscil-lator
We know that
〈T 〉 =~2b2m
we need to compute 〈V 〉
the upper bound for the energy isthus
minimizing wrt b
H = − ~2
2m
d2
dx2− αδ(x)
ψ(x) = Ae−bx2
〈V 〉 = −α|A|2∫ ∞−∞
δ(x)e−2bx2dx
=− α√
2b
π
〈H〉 =~2b2m− α
√2b
π
0 =d
db〈H〉 =
~2
2m− α√
2πb
b =2m2α2
π~4−→ 〈H〉min = −mα2
π~2≥ −mα2
2~2= Egs
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 9 / 18
Example 7.2 - Delta function
Find the ground state energy of adelta function potential
Start with the same Gaussian trialfunction as for the harmonic oscil-lator
We know that
〈T 〉 =~2b2m
we need to compute 〈V 〉
the upper bound for the energy isthus
minimizing wrt b
H = − ~2
2m
d2
dx2− αδ(x)
ψ(x) = Ae−bx2
〈V 〉 = −α|A|2∫ ∞−∞
δ(x)e−2bx2dx
=− α√
2b
π
〈H〉 =~2b2m− α
√2b
π
0 =d
db〈H〉 =
~2
2m− α√
2πb
b =2m2α2
π~4−→ 〈H〉min = −mα2
π~2≥ −mα2
2~2= Egs
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 9 / 18
Example 7.2 - Delta function
Find the ground state energy of adelta function potential
Start with the same Gaussian trialfunction as for the harmonic oscil-lator
We know that
〈T 〉 =~2b2m
we need to compute 〈V 〉
the upper bound for the energy isthus
minimizing wrt b
H = − ~2
2m
d2
dx2− αδ(x)
ψ(x) = Ae−bx2
〈V 〉 = −α|A|2∫ ∞−∞
δ(x)e−2bx2dx
=− α√
2b
π
〈H〉 =~2b2m− α
√2b
π
0 =d
db〈H〉
=~2
2m− α√
2πb
b =2m2α2
π~4−→ 〈H〉min = −mα2
π~2≥ −mα2
2~2= Egs
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 9 / 18
Example 7.2 - Delta function
Find the ground state energy of adelta function potential
Start with the same Gaussian trialfunction as for the harmonic oscil-lator
We know that
〈T 〉 =~2b2m
we need to compute 〈V 〉
the upper bound for the energy isthus
minimizing wrt b
H = − ~2
2m
d2
dx2− αδ(x)
ψ(x) = Ae−bx2
〈V 〉 = −α|A|2∫ ∞−∞
δ(x)e−2bx2dx
=− α√
2b
π
〈H〉 =~2b2m− α
√2b
π
0 =d
db〈H〉 =
~2
2m− α√
2πb
b =2m2α2
π~4−→ 〈H〉min = −mα2
π~2≥ −mα2
2~2= Egs
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 9 / 18
Example 7.2 - Delta function
Find the ground state energy of adelta function potential
Start with the same Gaussian trialfunction as for the harmonic oscil-lator
We know that
〈T 〉 =~2b2m
we need to compute 〈V 〉
the upper bound for the energy isthus
minimizing wrt b
H = − ~2
2m
d2
dx2− αδ(x)
ψ(x) = Ae−bx2
〈V 〉 = −α|A|2∫ ∞−∞
δ(x)e−2bx2dx
=− α√
2b
π
〈H〉 =~2b2m− α
√2b
π
0 =d
db〈H〉 =
~2
2m− α√
2πb
b =2m2α2
π~4
−→ 〈H〉min = −mα2
π~2≥ −mα2
2~2= Egs
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 9 / 18
Example 7.2 - Delta function
Find the ground state energy of adelta function potential
Start with the same Gaussian trialfunction as for the harmonic oscil-lator
We know that
〈T 〉 =~2b2m
we need to compute 〈V 〉
the upper bound for the energy isthus
minimizing wrt b
H = − ~2
2m
d2
dx2− αδ(x)
ψ(x) = Ae−bx2
〈V 〉 = −α|A|2∫ ∞−∞
δ(x)e−2bx2dx
=− α√
2b
π
〈H〉 =~2b2m− α
√2b
π
0 =d
db〈H〉 =
~2
2m− α√
2πb
b =2m2α2
π~4−→ 〈H〉min = −mα2
π~2
≥ −mα2
2~2= Egs
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 9 / 18
Example 7.2 - Delta function
Find the ground state energy of adelta function potential
Start with the same Gaussian trialfunction as for the harmonic oscil-lator
We know that
〈T 〉 =~2b2m
we need to compute 〈V 〉
the upper bound for the energy isthus
minimizing wrt b
H = − ~2
2m
d2
dx2− αδ(x)
ψ(x) = Ae−bx2
〈V 〉 = −α|A|2∫ ∞−∞
δ(x)e−2bx2dx
=− α√
2b
π
〈H〉 =~2b2m− α
√2b
π
0 =d
db〈H〉 =
~2
2m− α√
2πb
b =2m2α2
π~4−→ 〈H〉min = −mα2
π~2≥ −mα2
2~2= Egs
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 9 / 18
Example 7.3
It is even possible to work with wave functions with discontinuousderivatives
Find an upper bound on the groundstate energy of the infinite squarewell using a triangular wave func-tion
ψ(x) =
Ax , 0 ≤ x ≤ a
2
A(a− x), a2 ≤ x ≤ a
0, otherwise
first determine the normalization constant A
1 = |A|2[∫ a/2
0x2 dx +
∫ a
a/2(a− x)2 dx
]= |A|2
[x3
3
∣∣∣∣a/20
− (a− x)3
3
∣∣∣∣aa/2
]
= |A|2[a3
24− 0− 0 +
a3
24
]= |A|2 a
3
12→ A =
2
a
√3
a
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 10 / 18
Example 7.3
It is even possible to work with wave functions with discontinuousderivatives
Find an upper bound on the groundstate energy of the infinite squarewell using a triangular wave func-tion
ψ(x) =
Ax , 0 ≤ x ≤ a
2
A(a− x), a2 ≤ x ≤ a
0, otherwise
first determine the normalization constant A
1 = |A|2[∫ a/2
0x2 dx +
∫ a
a/2(a− x)2 dx
]= |A|2
[x3
3
∣∣∣∣a/20
− (a− x)3
3
∣∣∣∣aa/2
]
= |A|2[a3
24− 0− 0 +
a3
24
]= |A|2 a
3
12→ A =
2
a
√3
a
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 10 / 18
Example 7.3
It is even possible to work with wave functions with discontinuousderivatives
Find an upper bound on the groundstate energy of the infinite squarewell using a triangular wave func-tion
ψ(x) =
Ax , 0 ≤ x ≤ a
2
A(a− x), a2 ≤ x ≤ a
0, otherwise
first determine the normalization constant A
1 = |A|2[∫ a/2
0x2 dx +
∫ a
a/2(a− x)2 dx
]= |A|2
[x3
3
∣∣∣∣a/20
− (a− x)3
3
∣∣∣∣aa/2
]
= |A|2[a3
24− 0− 0 +
a3
24
]= |A|2 a
3
12→ A =
2
a
√3
a
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 10 / 18
Example 7.3
It is even possible to work with wave functions with discontinuousderivatives
Find an upper bound on the groundstate energy of the infinite squarewell using a triangular wave func-tion
ψ(x) =
Ax , 0 ≤ x ≤ a
2
A(a− x), a2 ≤ x ≤ a
0, otherwise
first determine the normalization constant A
1 = |A|2[∫ a/2
0x2 dx +
∫ a
a/2(a− x)2 dx
]= |A|2
[x3
3
∣∣∣∣a/20
− (a− x)3
3
∣∣∣∣aa/2
]
= |A|2[a3
24− 0− 0 +
a3
24
]= |A|2 a
3
12→ A =
2
a
√3
a
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 10 / 18
Example 7.3
It is even possible to work with wave functions with discontinuousderivatives
Find an upper bound on the groundstate energy of the infinite squarewell using a triangular wave func-tion
ψ(x) =
Ax , 0 ≤ x ≤ a
2
A(a− x), a2 ≤ x ≤ a
0, otherwise
first determine the normalization constant A
1 = |A|2[∫ a/2
0x2 dx +
∫ a
a/2(a− x)2 dx
]
= |A|2[x3
3
∣∣∣∣a/20
− (a− x)3
3
∣∣∣∣aa/2
]
= |A|2[a3
24− 0− 0 +
a3
24
]= |A|2 a
3
12→ A =
2
a
√3
a
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 10 / 18
Example 7.3
It is even possible to work with wave functions with discontinuousderivatives
Find an upper bound on the groundstate energy of the infinite squarewell using a triangular wave func-tion
ψ(x) =
Ax , 0 ≤ x ≤ a
2
A(a− x), a2 ≤ x ≤ a
0, otherwise
first determine the normalization constant A
1 = |A|2[∫ a/2
0x2 dx +
∫ a
a/2(a− x)2 dx
]= |A|2
[x3
3
∣∣∣∣a/20
− (a− x)3
3
∣∣∣∣aa/2
]
= |A|2[a3
24− 0− 0 +
a3
24
]= |A|2 a
3
12→ A =
2
a
√3
a
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 10 / 18
Example 7.3
It is even possible to work with wave functions with discontinuousderivatives
Find an upper bound on the groundstate energy of the infinite squarewell using a triangular wave func-tion
ψ(x) =
Ax , 0 ≤ x ≤ a
2
A(a− x), a2 ≤ x ≤ a
0, otherwise
first determine the normalization constant A
1 = |A|2[∫ a/2
0x2 dx +
∫ a
a/2(a− x)2 dx
]= |A|2
[x3
3
∣∣∣∣a/20
− (a− x)3
3
∣∣∣∣aa/2
]
= |A|2[a3
24− 0− 0 +
a3
24
]
= |A|2 a3
12→ A =
2
a
√3
a
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 10 / 18
Example 7.3
It is even possible to work with wave functions with discontinuousderivatives
Find an upper bound on the groundstate energy of the infinite squarewell using a triangular wave func-tion
ψ(x) =
Ax , 0 ≤ x ≤ a
2
A(a− x), a2 ≤ x ≤ a
0, otherwise
first determine the normalization constant A
1 = |A|2[∫ a/2
0x2 dx +
∫ a
a/2(a− x)2 dx
]= |A|2
[x3
3
∣∣∣∣a/20
− (a− x)3
3
∣∣∣∣aa/2
]
= |A|2[a3
24− 0− 0 +
a3
24
]= |A|2 a
3
12
→ A =2
a
√3
a
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 10 / 18
Example 7.3
It is even possible to work with wave functions with discontinuousderivatives
Find an upper bound on the groundstate energy of the infinite squarewell using a triangular wave func-tion
ψ(x) =
Ax , 0 ≤ x ≤ a
2
A(a− x), a2 ≤ x ≤ a
0, otherwise
first determine the normalization constant A
1 = |A|2[∫ a/2
0x2 dx +
∫ a
a/2(a− x)2 dx
]= |A|2
[x3
3
∣∣∣∣a/20
− (a− x)3
3
∣∣∣∣aa/2
]
= |A|2[a3
24− 0− 0 +
a3
24
]= |A|2 a
3
12→ A =
2
a
√3
a
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 10 / 18
Example 7.3 (cont.)
In order to get the expectationvalue of H, we need the secondderivative of the wave function.
The first derivative is simply a stepfunction
dψ
dx=
+A, 0 ≤ x ≤ a
2
−A, a2 ≤ x ≤ a
0, otherwise
and the second derivative is threedelta functions, one at each edgeand one at the center of the well
d2ψ
dx2= Aδ(x)− 2Aδ(x − a
2)
+ Aδ(x − a)
ψ =
Ax , 0 ≤ x ≤ a
2
A(a− x), a2 ≤ x ≤ a
0, otherwise
ψ
xa0
Aa
2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 11 / 18
Example 7.3 (cont.)
In order to get the expectationvalue of H, we need the secondderivative of the wave function.The first derivative is simply a stepfunction
dψ
dx=
+A, 0 ≤ x ≤ a
2
−A, a2 ≤ x ≤ a
0, otherwise
and the second derivative is threedelta functions, one at each edgeand one at the center of the well
d2ψ
dx2= Aδ(x)− 2Aδ(x − a
2)
+ Aδ(x − a)
ψ =
Ax , 0 ≤ x ≤ a
2
A(a− x), a2 ≤ x ≤ a
0, otherwise
ψ
xa0
Aa
2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 11 / 18
Example 7.3 (cont.)
In order to get the expectationvalue of H, we need the secondderivative of the wave function.The first derivative is simply a stepfunction
dψ
dx=
+A, 0 ≤ x ≤ a
2
−A, a2 ≤ x ≤ a
0, otherwise
and the second derivative is threedelta functions, one at each edgeand one at the center of the well
d2ψ
dx2= Aδ(x)− 2Aδ(x − a
2)
+ Aδ(x − a)
ψ =
Ax , 0 ≤ x ≤ a
2
A(a− x), a2 ≤ x ≤ a
0, otherwise
dψdx
A
x
a
A
x
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 11 / 18
Example 7.3 (cont.)
In order to get the expectationvalue of H, we need the secondderivative of the wave function.The first derivative is simply a stepfunction
dψ
dx=
+A, 0 ≤ x ≤ a
2
−A, a2 ≤ x ≤ a
0, otherwise
and the second derivative is threedelta functions, one at each edgeand one at the center of the well
d2ψ
dx2= Aδ(x)− 2Aδ(x − a
2)
+ Aδ(x − a)
ψ =
Ax , 0 ≤ x ≤ a
2
A(a− x), a2 ≤ x ≤ a
0, otherwise
dψdx
A
x
a
A
x
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 11 / 18
Example 7.3 (cont.)
In order to get the expectationvalue of H, we need the secondderivative of the wave function.The first derivative is simply a stepfunction
dψ
dx=
+A, 0 ≤ x ≤ a
2
−A, a2 ≤ x ≤ a
0, otherwise
and the second derivative is threedelta functions, one at each edgeand one at the center of the well
d2ψ
dx2= Aδ(x)− 2Aδ(x − a
2)
+ Aδ(x − a)
ψ =
Ax , 0 ≤ x ≤ a
2
A(a− x), a2 ≤ x ≤ a
0, otherwise
d2ψdx2
x
a/2
a
A
x
-2A
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 11 / 18
Example 7.3 (cont.)
The expectation value of the Hamiltonian is then
〈H〉 = −~2A2m
∫ψ∗(x)[δ(x)− 2δ(x − a
2)δ(x − a)]ψ(x) dx
= −~2A2m
[����|ψ(0)|2 − 2|ψ( a2)|2 +���
�|ψ(a)|2] =~2A2a
2m=
12~2
2ma2
comparing to the exact ground state energy
Egs =π2~2
2ma2<
12~2
2ma2= 〈H〉
this demonstrates the variational principle even if there is no freeparameter with which to minimize the energy
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 12 / 18
Example 7.3 (cont.)
The expectation value of the Hamiltonian is then
〈H〉 = −~2A2m
∫ψ∗(x)[δ(x)− 2δ(x − a
2)δ(x − a)]ψ(x) dx
= −~2A2m
[����|ψ(0)|2 − 2|ψ( a2)|2 +���
�|ψ(a)|2] =~2A2a
2m=
12~2
2ma2
comparing to the exact ground state energy
Egs =π2~2
2ma2<
12~2
2ma2= 〈H〉
this demonstrates the variational principle even if there is no freeparameter with which to minimize the energy
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 12 / 18
Example 7.3 (cont.)
The expectation value of the Hamiltonian is then
〈H〉 = −~2A2m
∫ψ∗(x)[δ(x)− 2δ(x − a
2)δ(x − a)]ψ(x) dx
= −~2A2m
[|ψ(0)|2 − 2|ψ( a2)|2 + |ψ(a)|2]
=~2A2a
2m=
12~2
2ma2
comparing to the exact ground state energy
Egs =π2~2
2ma2<
12~2
2ma2= 〈H〉
this demonstrates the variational principle even if there is no freeparameter with which to minimize the energy
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 12 / 18
Example 7.3 (cont.)
The expectation value of the Hamiltonian is then
〈H〉 = −~2A2m
∫ψ∗(x)[δ(x)− 2δ(x − a
2)δ(x − a)]ψ(x) dx
= −~2A2m
[����|ψ(0)|2 − 2|ψ( a2)|2 +���
�|ψ(a)|2] =~2A2a
2m
=12~2
2ma2
comparing to the exact ground state energy
Egs =π2~2
2ma2<
12~2
2ma2= 〈H〉
this demonstrates the variational principle even if there is no freeparameter with which to minimize the energy
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 12 / 18
Example 7.3 (cont.)
The expectation value of the Hamiltonian is then
〈H〉 = −~2A2m
∫ψ∗(x)[δ(x)− 2δ(x − a
2)δ(x − a)]ψ(x) dx
= −~2A2m
[����|ψ(0)|2 − 2|ψ( a2)|2 +���
�|ψ(a)|2] =~2A2a
2m=
12~2
2ma2
comparing to the exact ground state energy
Egs =π2~2
2ma2<
12~2
2ma2= 〈H〉
this demonstrates the variational principle even if there is no freeparameter with which to minimize the energy
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 12 / 18
Example 7.3 (cont.)
The expectation value of the Hamiltonian is then
〈H〉 = −~2A2m
∫ψ∗(x)[δ(x)− 2δ(x − a
2)δ(x − a)]ψ(x) dx
= −~2A2m
[����|ψ(0)|2 − 2|ψ( a2)|2 +���
�|ψ(a)|2] =~2A2a
2m=
12~2
2ma2
comparing to the exact ground state energy
Egs =π2~2
2ma2<
12~2
2ma2= 〈H〉
this demonstrates the variational principle even if there is no freeparameter with which to minimize the energy
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 12 / 18
Example 7.3 (cont.)
The expectation value of the Hamiltonian is then
〈H〉 = −~2A2m
∫ψ∗(x)[δ(x)− 2δ(x − a
2)δ(x − a)]ψ(x) dx
= −~2A2m
[����|ψ(0)|2 − 2|ψ( a2)|2 +���
�|ψ(a)|2] =~2A2a
2m=
12~2
2ma2
comparing to the exact ground state energy
Egs =π2~2
2ma2
<12~2
2ma2= 〈H〉
this demonstrates the variational principle even if there is no freeparameter with which to minimize the energy
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 12 / 18
Example 7.3 (cont.)
The expectation value of the Hamiltonian is then
〈H〉 = −~2A2m
∫ψ∗(x)[δ(x)− 2δ(x − a
2)δ(x − a)]ψ(x) dx
= −~2A2m
[����|ψ(0)|2 − 2|ψ( a2)|2 +���
�|ψ(a)|2] =~2A2a
2m=
12~2
2ma2
comparing to the exact ground state energy
Egs =π2~2
2ma2
<
12~2
2ma2= 〈H〉
this demonstrates the variational principle even if there is no freeparameter with which to minimize the energy
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 12 / 18
Example 7.3 (cont.)
The expectation value of the Hamiltonian is then
〈H〉 = −~2A2m
∫ψ∗(x)[δ(x)− 2δ(x − a
2)δ(x − a)]ψ(x) dx
= −~2A2m
[����|ψ(0)|2 − 2|ψ( a2)|2 +���
�|ψ(a)|2] =~2A2a
2m=
12~2
2ma2
comparing to the exact ground state energy
Egs =π2~2
2ma2
<
12~2
2ma2= 〈H〉
this demonstrates the variational principle even if there is no freeparameter with which to minimize the energy
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 12 / 18
Example 7.3 (cont.)
The expectation value of the Hamiltonian is then
〈H〉 = −~2A2m
∫ψ∗(x)[δ(x)− 2δ(x − a
2)δ(x − a)]ψ(x) dx
= −~2A2m
[����|ψ(0)|2 − 2|ψ( a2)|2 +���
�|ψ(a)|2] =~2A2a
2m=
12~2
2ma2
comparing to the exact ground state energy
Egs =π2~2
2ma2<
12~2
2ma2= 〈H〉
this demonstrates the variational principle even if there is no freeparameter with which to minimize the energy
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 12 / 18
Example 7.3 (cont.)
The expectation value of the Hamiltonian is then
〈H〉 = −~2A2m
∫ψ∗(x)[δ(x)− 2δ(x − a
2)δ(x − a)]ψ(x) dx
= −~2A2m
[����|ψ(0)|2 − 2|ψ( a2)|2 +���
�|ψ(a)|2] =~2A2a
2m=
12~2
2ma2
comparing to the exact ground state energy
Egs =π2~2
2ma2<
12~2
2ma2= 〈H〉
this demonstrates the variational principle even if there is no freeparameter with which to minimize the energy
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 12 / 18
Helium atom
The helium atom is an ideal application of the variational principle. Thefull Hamiltonian includes an electron-electron interaction which we ignoredwhen we first discussed it
H = − ~2
2m(∇2
1 +∇22)− e2
4πε0
(2
r1+
2
r2− 1
|~r1 − ~r2|
)We wish to compute the groundstate energy, which is the total en-ergy required to remove both elec-trons. This has been measured tobe
the problem is the electron-electroninteraction
Vee=e2
4πε0
1
|~r1 − ~r2|
which is not exactly soluble
Egs = −78.975 eV
+2e
-e
-e|r1-r2|
|r1||r2|
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 13 / 18
Helium atom
The helium atom is an ideal application of the variational principle. Thefull Hamiltonian includes an electron-electron interaction which we ignoredwhen we first discussed it
H = − ~2
2m(∇2
1 +∇22)− e2
4πε0
(2
r1+
2
r2− 1
|~r1 − ~r2|
)
We wish to compute the groundstate energy, which is the total en-ergy required to remove both elec-trons. This has been measured tobe
the problem is the electron-electroninteraction
Vee=e2
4πε0
1
|~r1 − ~r2|
which is not exactly soluble
Egs = −78.975 eV
+2e
-e
-e|r1-r2|
|r1||r2|
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 13 / 18
Helium atom
The helium atom is an ideal application of the variational principle. Thefull Hamiltonian includes an electron-electron interaction which we ignoredwhen we first discussed it
H = − ~2
2m(∇2
1 +∇22)− e2
4πε0
(2
r1+
2
r2− 1
|~r1 − ~r2|
)We wish to compute the groundstate energy, which is the total en-ergy required to remove both elec-trons. This has been measured tobe
the problem is the electron-electroninteraction
Vee=e2
4πε0
1
|~r1 − ~r2|
which is not exactly soluble
Egs = −78.975 eV
+2e
-e
-e|r1-r2|
|r1||r2|
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 13 / 18
Helium atom
The helium atom is an ideal application of the variational principle. Thefull Hamiltonian includes an electron-electron interaction which we ignoredwhen we first discussed it
H = − ~2
2m(∇2
1 +∇22)− e2
4πε0
(2
r1+
2
r2− 1
|~r1 − ~r2|
)We wish to compute the groundstate energy, which is the total en-ergy required to remove both elec-trons. This has been measured tobe
the problem is the electron-electroninteraction
Vee=e2
4πε0
1
|~r1 − ~r2|
which is not exactly soluble
Egs = −78.975 eV
+2e
-e
-e|r1-r2|
|r1||r2|
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 13 / 18
Helium atom
The helium atom is an ideal application of the variational principle. Thefull Hamiltonian includes an electron-electron interaction which we ignoredwhen we first discussed it
H = − ~2
2m(∇2
1 +∇22)− e2
4πε0
(2
r1+
2
r2− 1
|~r1 − ~r2|
)We wish to compute the groundstate energy, which is the total en-ergy required to remove both elec-trons. This has been measured tobe
the problem is the electron-electroninteraction
Vee=e2
4πε0
1
|~r1 − ~r2|
which is not exactly soluble
Egs = −78.975 eV
+2e
-e
-e|r1-r2|
|r1||r2|
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 13 / 18
Helium atom
The helium atom is an ideal application of the variational principle. Thefull Hamiltonian includes an electron-electron interaction which we ignoredwhen we first discussed it
H = − ~2
2m(∇2
1 +∇22)− e2
4πε0
(2
r1+
2
r2− 1
|~r1 − ~r2|
)We wish to compute the groundstate energy, which is the total en-ergy required to remove both elec-trons. This has been measured tobe
the problem is the electron-electroninteraction
Vee=e2
4πε0
1
|~r1 − ~r2|
which is not exactly soluble
Egs = −78.975 eV
+2e
-e
-e|r1-r2|
|r1||r2|
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 13 / 18
Helium atom
The helium atom is an ideal application of the variational principle. Thefull Hamiltonian includes an electron-electron interaction which we ignoredwhen we first discussed it
H = − ~2
2m(∇2
1 +∇22)− e2
4πε0
(2
r1+
2
r2− 1
|~r1 − ~r2|
)We wish to compute the groundstate energy, which is the total en-ergy required to remove both elec-trons. This has been measured tobe
the problem is the electron-electroninteraction
Vee=e2
4πε0
1
|~r1 − ~r2|
which is not exactly soluble
Egs = −78.975 eV
+2e
-e
-e|r1-r2|
|r1||r2|
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 13 / 18
Trial function for helium
If the electron-electron term is ignored, the exact solution of the heliumatom is a combination of the two hydrogenic atom wave functions
ψ(~r1, ~r2)
≡ ψ100(~r1)ψ100(~r2) =8
πa3e−2r1/ae−2r2/a
the energy of this solution is 8E1 = −109 eV, so we can get a betterapproximation by applying the variational principle with this trial function
Hψ0 = (8E1 + Vee)ψ0
〈H〉 = (8E1 + 〈Vee〉),
Vee =e2/4πε0|~r1 − ~r2|
〈Vee〉
=
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a
|~r1 − ~r2|d3~r1 d
3~r2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 14 / 18
Trial function for helium
If the electron-electron term is ignored, the exact solution of the heliumatom is a combination of the two hydrogenic atom wave functions
ψ(~r1, ~r2)
≡ ψ100(~r1)ψ100(~r2) =8
πa3e−2r1/ae−2r2/a
the energy of this solution is 8E1 = −109 eV, so we can get a betterapproximation by applying the variational principle with this trial function
Hψ0 = (8E1 + Vee)ψ0
〈H〉 = (8E1 + 〈Vee〉),
Vee =e2/4πε0|~r1 − ~r2|
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2
∫e−4r1/a e−4r2/a
|~r1 − ~r2|d3~r1 d
3~r2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 14 / 18
Trial function for helium
If the electron-electron term is ignored, the exact solution of the heliumatom is a combination of the two hydrogenic atom wave functions
ψ(~r1, ~r2) ≡ ψ100(~r1)ψ100(~r2)
=8
πa3e−2r1/ae−2r2/a
the energy of this solution is 8E1 = −109 eV, so we can get a betterapproximation by applying the variational principle with this trial function
Hψ0 = (8E1 + Vee)ψ0
〈H〉 = (8E1 + 〈Vee〉),
Vee =e2/4πε0|~r1 − ~r2|
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a
|~r1 − ~r2|d3~r1 d
3~r2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 14 / 18
Trial function for helium
If the electron-electron term is ignored, the exact solution of the heliumatom is a combination of the two hydrogenic atom wave functions
ψ(~r1, ~r2) ≡ ψ100(~r1)ψ100(~r2) =8
πa3e−2r1/ae−2r2/a
the energy of this solution is 8E1 = −109 eV, so we can get a betterapproximation by applying the variational principle with this trial function
Hψ0 = (8E1 + Vee)ψ0
〈H〉 = (8E1 + 〈Vee〉),
Vee =e2/4πε0|~r1 − ~r2|
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a
|~r1 − ~r2|d3~r1 d
3~r2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 14 / 18
Trial function for helium
If the electron-electron term is ignored, the exact solution of the heliumatom is a combination of the two hydrogenic atom wave functions
ψ(~r1, ~r2) ≡ ψ100(~r1)ψ100(~r2) =8
πa3e−2r1/ae−2r2/a
the energy of this solution is 8E1 = −109 eV,
so we can get a betterapproximation by applying the variational principle with this trial function
Hψ0 = (8E1 + Vee)ψ0
〈H〉 = (8E1 + 〈Vee〉),
Vee =e2/4πε0|~r1 − ~r2|
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a
|~r1 − ~r2|d3~r1 d
3~r2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 14 / 18
Trial function for helium
If the electron-electron term is ignored, the exact solution of the heliumatom is a combination of the two hydrogenic atom wave functions
ψ(~r1, ~r2) ≡ ψ100(~r1)ψ100(~r2) =8
πa3e−2r1/ae−2r2/a
the energy of this solution is 8E1 = −109 eV, so we can get a betterapproximation by applying the variational principle with this trial function
Hψ0 = (8E1 + Vee)ψ0
〈H〉 = (8E1 + 〈Vee〉),
Vee =e2/4πε0|~r1 − ~r2|
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a
|~r1 − ~r2|d3~r1 d
3~r2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 14 / 18
Trial function for helium
If the electron-electron term is ignored, the exact solution of the heliumatom is a combination of the two hydrogenic atom wave functions
ψ(~r1, ~r2) ≡ ψ100(~r1)ψ100(~r2) =8
πa3e−2r1/ae−2r2/a
the energy of this solution is 8E1 = −109 eV, so we can get a betterapproximation by applying the variational principle with this trial function
Hψ0 = (8E1 + Vee)ψ0
〈H〉 = (8E1 + 〈Vee〉),
Vee =e2/4πε0|~r1 − ~r2|
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a
|~r1 − ~r2|d3~r1 d
3~r2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 14 / 18
Trial function for helium
If the electron-electron term is ignored, the exact solution of the heliumatom is a combination of the two hydrogenic atom wave functions
ψ(~r1, ~r2) ≡ ψ100(~r1)ψ100(~r2) =8
πa3e−2r1/ae−2r2/a
the energy of this solution is 8E1 = −109 eV, so we can get a betterapproximation by applying the variational principle with this trial function
Hψ0 = (8E1 + Vee)ψ0
〈H〉 = (8E1 + 〈Vee〉),
Vee =e2/4πε0|~r1 − ~r2|
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a
|~r1 − ~r2|d3~r1 d
3~r2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 14 / 18
Trial function for helium
If the electron-electron term is ignored, the exact solution of the heliumatom is a combination of the two hydrogenic atom wave functions
ψ(~r1, ~r2) ≡ ψ100(~r1)ψ100(~r2) =8
πa3e−2r1/ae−2r2/a
the energy of this solution is 8E1 = −109 eV, so we can get a betterapproximation by applying the variational principle with this trial function
Hψ0 = (8E1 + Vee)ψ0
〈H〉 = (8E1 + 〈Vee〉), Vee =e2/4πε0|~r1 − ~r2|
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a
|~r1 − ~r2|d3~r1 d
3~r2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 14 / 18
Trial function for helium
If the electron-electron term is ignored, the exact solution of the heliumatom is a combination of the two hydrogenic atom wave functions
ψ(~r1, ~r2) ≡ ψ100(~r1)ψ100(~r2) =8
πa3e−2r1/ae−2r2/a
the energy of this solution is 8E1 = −109 eV, so we can get a betterapproximation by applying the variational principle with this trial function
Hψ0 = (8E1 + Vee)ψ0
〈H〉 = (8E1 + 〈Vee〉), Vee =e2/4πε0|~r1 − ~r2|
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a
|~r1 − ~r2|d3~r1 d
3~r2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 14 / 18
Setting up the integrals
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a
|~r1 − ~r2|d3~r1 d
3~r2
r1r2
θ2
φ2
|r -r |1 2
x2
y2
z2
the integral mixes the coordinates but canbe simplified by careful selection of thetwo coordinate systems
taking ~r1 along the z-axis of the ~r2 coor-dinate system, gives
|~r1 − ~r2|
=√
r21 + r22 − 2r1r2 cos θ2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 15 / 18
Setting up the integrals
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a
|~r1 − ~r2|d3~r1 d
3~r2
r1r2
θ2
φ2
|r -r |1 2
x2
y2
z2
the integral mixes the coordinates but canbe simplified by careful selection of thetwo coordinate systems
taking ~r1 along the z-axis of the ~r2 coor-dinate system, gives
|~r1 − ~r2|
=√
r21 + r22 − 2r1r2 cos θ2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 15 / 18
Setting up the integrals
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a
|~r1 − ~r2|d3~r1 d
3~r2
r1r2
θ2
φ2
|r -r |1 2
x2
y2
z2
the integral mixes the coordinates but canbe simplified by careful selection of thetwo coordinate systems
taking ~r1 along the z-axis of the ~r2 coor-dinate system, gives
|~r1 − ~r2|
=√
r21 + r22 − 2r1r2 cos θ2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 15 / 18
Setting up the integrals
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a
|~r1 − ~r2|d3~r1 d
3~r2
r1r2
θ2
φ2
|r -r |1 2
x2
y2
z2
the integral mixes the coordinates but canbe simplified by careful selection of thetwo coordinate systems
taking ~r1 along the z-axis of the ~r2 coor-dinate system, gives
|~r1 − ~r2|
=√
r21 + r22 − 2r1r2 cos θ2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 15 / 18
Setting up the integrals
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a
|~r1 − ~r2|d3~r1 d
3~r2
r1r2
θ2
φ2
|r -r |1 2
x2
y2
z2
the integral mixes the coordinates but canbe simplified by careful selection of thetwo coordinate systems
taking ~r1 along the z-axis of the ~r2 coor-dinate system, gives
|~r1 − ~r2| =√
r21 + r22 − 2r1r2 cos θ2
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 15 / 18
The r2 integral
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a√r21 + r22 − 2r1r2 cos θ2
d3~r1 d3~r2
the ~r2 integral then can be evaluated
I2 ≡
∫e−4r2/a r22 sin θ2√
r21 + r22 − 2r1r2 cos θ2
dr2 dθ2 dφ2
the φ2 integral gives 2π and the θ2 integrand is a perfect derivative
Iθ =
∫ π
0
sin θ2√r21 + r22 − 2r1r2 cos θ2
dθ2
=
√r21 + r22 − 2r1r2 cos θ2
r1 r2
∣∣∣∣∣∣π
0
=1
r1 r2
[(r1 + r2)− |r1 − r2|
]
=
{
2/r1, r2 < r1
2/r2, r2 > r1
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 16 / 18
The r2 integral
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a√r21 + r22 − 2r1r2 cos θ2
d3~r1 d3~r2
the ~r2 integral then can be evaluated
I2 ≡
∫e−4r2/a r22 sin θ2√
r21 + r22 − 2r1r2 cos θ2
dr2 dθ2 dφ2
the φ2 integral gives 2π and the θ2 integrand is a perfect derivative
Iθ =
∫ π
0
sin θ2√r21 + r22 − 2r1r2 cos θ2
dθ2
=
√r21 + r22 − 2r1r2 cos θ2
r1 r2
∣∣∣∣∣∣π
0
=1
r1 r2
[(r1 + r2)− |r1 − r2|
]
=
{
2/r1, r2 < r1
2/r2, r2 > r1
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 16 / 18
The r2 integral
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a√r21 + r22 − 2r1r2 cos θ2
d3~r1 d3~r2
the ~r2 integral then can be evaluated
I2 ≡
∫e−4r2/a r22 sin θ2√
r21 + r22 − 2r1r2 cos θ2
dr2 dθ2 dφ2
the φ2 integral gives 2π and the θ2 integrand is a perfect derivative
Iθ =
∫ π
0
sin θ2√r21 + r22 − 2r1r2 cos θ2
dθ2
=
√r21 + r22 − 2r1r2 cos θ2
r1 r2
∣∣∣∣∣∣π
0
=1
r1 r2
[(r1 + r2)− |r1 − r2|
]
=
{
2/r1, r2 < r1
2/r2, r2 > r1
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 16 / 18
The r2 integral
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a√r21 + r22 − 2r1r2 cos θ2
d3~r1 d3~r2
the ~r2 integral then can be evaluated
I2 ≡∫
e−4r2/a r22 sin θ2√r21 + r22 − 2r1r2 cos θ2
dr2 dθ2 dφ2
the φ2 integral gives 2π and the θ2 integrand is a perfect derivative
Iθ =
∫ π
0
sin θ2√r21 + r22 − 2r1r2 cos θ2
dθ2
=
√r21 + r22 − 2r1r2 cos θ2
r1 r2
∣∣∣∣∣∣π
0
=1
r1 r2
[(r1 + r2)− |r1 − r2|
]
=
{
2/r1, r2 < r1
2/r2, r2 > r1
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 16 / 18
The r2 integral
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a√r21 + r22 − 2r1r2 cos θ2
d3~r1 d3~r2
the ~r2 integral then can be evaluated
I2 ≡∫
e−4r2/a r22 sin θ2√r21 + r22 − 2r1r2 cos θ2
dr2 dθ2 dφ2
the φ2 integral gives 2π and the θ2 integrand is a perfect derivative
Iθ =
∫ π
0
sin θ2√r21 + r22 − 2r1r2 cos θ2
dθ2
=
√r21 + r22 − 2r1r2 cos θ2
r1 r2
∣∣∣∣∣∣π
0
=1
r1 r2
[(r1 + r2)− |r1 − r2|
]
=
{
2/r1, r2 < r1
2/r2, r2 > r1
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 16 / 18
The r2 integral
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a√r21 + r22 − 2r1r2 cos θ2
d3~r1 d3~r2
the ~r2 integral then can be evaluated
I2 ≡∫
e−4r2/a r22 sin θ2√r21 + r22 − 2r1r2 cos θ2
dr2 dθ2 dφ2
the φ2 integral gives 2π and the θ2 integrand is a perfect derivative
Iθ =
∫ π
0
sin θ2√r21 + r22 − 2r1r2 cos θ2
dθ2
=
√r21 + r22 − 2r1r2 cos θ2
r1 r2
∣∣∣∣∣∣π
0
=1
r1 r2
[(r1 + r2)− |r1 − r2|
]
=
{
2/r1, r2 < r1
2/r2, r2 > r1
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 16 / 18
The r2 integral
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a√r21 + r22 − 2r1r2 cos θ2
d3~r1 d3~r2
the ~r2 integral then can be evaluated
I2 ≡∫
e−4r2/a r22 sin θ2√r21 + r22 − 2r1r2 cos θ2
dr2 dθ2 dφ2
the φ2 integral gives 2π and the θ2 integrand is a perfect derivative
Iθ =
∫ π
0
sin θ2√r21 + r22 − 2r1r2 cos θ2
dθ2 =
√r21 + r22 − 2r1r2 cos θ2
r1 r2
∣∣∣∣∣∣π
0
=1
r1 r2
[(r1 + r2)− |r1 − r2|
]
=
{
2/r1, r2 < r1
2/r2, r2 > r1
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 16 / 18
The r2 integral
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a√r21 + r22 − 2r1r2 cos θ2
d3~r1 d3~r2
the ~r2 integral then can be evaluated
I2 ≡∫
e−4r2/a r22 sin θ2√r21 + r22 − 2r1r2 cos θ2
dr2 dθ2 dφ2
the φ2 integral gives 2π and the θ2 integrand is a perfect derivative
Iθ =
∫ π
0
sin θ2√r21 + r22 − 2r1r2 cos θ2
dθ2 =
√r21 + r22 − 2r1r2 cos θ2
r1 r2
∣∣∣∣∣∣π
0
=1
r1 r2
[√r21 + r22 + 2r1 r2 −
√r21 + r22 − 2r1 r2
]
=
{
2/r1, r2 < r1
2/r2, r2 > r1
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 16 / 18
The r2 integral
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a√r21 + r22 − 2r1r2 cos θ2
d3~r1 d3~r2
the ~r2 integral then can be evaluated
I2 ≡∫
e−4r2/a r22 sin θ2√r21 + r22 − 2r1r2 cos θ2
dr2 dθ2 dφ2
the φ2 integral gives 2π and the θ2 integrand is a perfect derivative
Iθ =
∫ π
0
sin θ2√r21 + r22 − 2r1r2 cos θ2
dθ2 =
√r21 + r22 − 2r1r2 cos θ2
r1 r2
∣∣∣∣∣∣π
0
=1
r1 r2
[(r1 + r2)− |r1 − r2|
]
=
{
2/r1, r2 < r1
2/r2, r2 > r1
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 16 / 18
The r2 integral
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a√r21 + r22 − 2r1r2 cos θ2
d3~r1 d3~r2
the ~r2 integral then can be evaluated
I2 ≡∫
e−4r2/a r22 sin θ2√r21 + r22 − 2r1r2 cos θ2
dr2 dθ2 dφ2
the φ2 integral gives 2π and the θ2 integrand is a perfect derivative
Iθ =
∫ π
0
sin θ2√r21 + r22 − 2r1r2 cos θ2
dθ2 =
√r21 + r22 − 2r1r2 cos θ2
r1 r2
∣∣∣∣∣∣π
0
=1
r1 r2
[(r1 + r2)− |r1 − r2|
]=
{
2/r1, r2 < r1
2/r2, r2 > r1
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 16 / 18
The r2 integral
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a√r21 + r22 − 2r1r2 cos θ2
d3~r1 d3~r2
the ~r2 integral then can be evaluated
I2 ≡∫
e−4r2/a r22 sin θ2√r21 + r22 − 2r1r2 cos θ2
dr2 dθ2 dφ2
the φ2 integral gives 2π and the θ2 integrand is a perfect derivative
Iθ =
∫ π
0
sin θ2√r21 + r22 − 2r1r2 cos θ2
dθ2 =
√r21 + r22 − 2r1r2 cos θ2
r1 r2
∣∣∣∣∣∣π
0
=1
r1 r2
[(r1 + r2)− |r1 − r2|
]=
{2/r1, r2 < r1
2/r2, r2 > r1
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 16 / 18
The r2 integral
〈Vee〉 =
(e2
4πε0
)(8
πa3
)2 ∫ e−4r1/a e−4r2/a√r21 + r22 − 2r1r2 cos θ2
d3~r1 d3~r2
the ~r2 integral then can be evaluated
I2 ≡∫
e−4r2/a r22 sin θ2√r21 + r22 − 2r1r2 cos θ2
dr2 dθ2 dφ2
the φ2 integral gives 2π and the θ2 integrand is a perfect derivative
Iθ =
∫ π
0
sin θ2√r21 + r22 − 2r1r2 cos θ2
dθ2 =
√r21 + r22 − 2r1r2 cos θ2
r1 r2
∣∣∣∣∣∣π
0
=1
r1 r2
[(r1 + r2)− |r1 − r2|
]=
{2/r1, r2 < r1
2/r2, r2 > r1
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 16 / 18
The r2 integral
The r2 integral is thus split into two parts
I2 = 4π
(
1
r1
∫ r1
0r22 e
−4r2/a dr2
+
∫ ∞r1
r2 e−4r2/a dr2
)these can be solved using integration by parts with dv = e−4r2/a dr2
Ired =
−(a
4
)r22 e−4r2/a
∣∣∣r10
+ 2(a
4
)∫ r1
0r2e−4r2/a dr2
= −(a
4
)r21 e−4r1/a
−2(a
4
)2r2e−4r2/a
∣∣∣∣r10
+ 2(a
4
)2 ∫ r1
0e−4r2/a dr2
= −(a
4
)r21 e−4r1/a − 2
(a4
)2r1e−4r1/a
− 2(a
4
)3e−4r1/a + 2
(a4
)3
Iblue =
−(a
4
)r2e−4r2/a
∣∣∣∞r1
+(a
4
)∫ ∞r1
e−4r2/a dr2
= +(a
4
)r1e−4r1/a
+(a
4
)2e−4r1/a
I2 =πa3
8r1
[1−
(1 +
2r1a
)e−4r1/a
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 17 / 18
The r2 integral
The r2 integral is thus split into two parts
I2 = 4π
(
1
r1
∫ r1
0r22 e
−4r2/a dr2
+
∫ ∞r1
r2 e−4r2/a dr2
)
these can be solved using integration by parts with dv = e−4r2/a dr2
Ired =
−(a
4
)r22 e−4r2/a
∣∣∣r10
+ 2(a
4
)∫ r1
0r2e−4r2/a dr2
= −(a
4
)r21 e−4r1/a
−2(a
4
)2r2e−4r2/a
∣∣∣∣r10
+ 2(a
4
)2 ∫ r1
0e−4r2/a dr2
= −(a
4
)r21 e−4r1/a − 2
(a4
)2r1e−4r1/a
− 2(a
4
)3e−4r1/a + 2
(a4
)3
Iblue =
−(a
4
)r2e−4r2/a
∣∣∣∞r1
+(a
4
)∫ ∞r1
e−4r2/a dr2
= +(a
4
)r1e−4r1/a
+(a
4
)2e−4r1/a
I2 =πa3
8r1
[1−
(1 +
2r1a
)e−4r1/a
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 17 / 18
The r2 integral
The r2 integral is thus split into two parts
I2 = 4π
(1
r1
∫ r1
0r22 e
−4r2/a dr2 +
∫ ∞r1
r2 e−4r2/a dr2
)
these can be solved using integration by parts with dv = e−4r2/a dr2
Ired =
−(a
4
)r22 e−4r2/a
∣∣∣r10
+ 2(a
4
)∫ r1
0r2e−4r2/a dr2
= −(a
4
)r21 e−4r1/a
−2(a
4
)2r2e−4r2/a
∣∣∣∣r10
+ 2(a
4
)2 ∫ r1
0e−4r2/a dr2
= −(a
4
)r21 e−4r1/a − 2
(a4
)2r1e−4r1/a
− 2(a
4
)3e−4r1/a + 2
(a4
)3
Iblue =
−(a
4
)r2e−4r2/a
∣∣∣∞r1
+(a
4
)∫ ∞r1
e−4r2/a dr2
= +(a
4
)r1e−4r1/a
+(a
4
)2e−4r1/a
I2 =πa3
8r1
[1−
(1 +
2r1a
)e−4r1/a
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 17 / 18
The r2 integral
The r2 integral is thus split into two parts
I2 = 4π
(1
r1
∫ r1
0r22 e
−4r2/a dr2 +
∫ ∞r1
r2 e−4r2/a dr2
)
these can be solved using integration by parts with dv = e−4r2/a dr2
Ired =
−(a
4
)r22 e−4r2/a
∣∣∣r10
+ 2(a
4
)∫ r1
0r2e−4r2/a dr2
= −(a
4
)r21 e−4r1/a
−2(a
4
)2r2e−4r2/a
∣∣∣∣r10
+ 2(a
4
)2 ∫ r1
0e−4r2/a dr2
= −(a
4
)r21 e−4r1/a − 2
(a4
)2r1e−4r1/a
− 2(a
4
)3e−4r1/a + 2
(a4
)3
Iblue =
−(a
4
)r2e−4r2/a
∣∣∣∞r1
+(a
4
)∫ ∞r1
e−4r2/a dr2
= +(a
4
)r1e−4r1/a
+(a
4
)2e−4r1/a
I2 =πa3
8r1
[1−
(1 +
2r1a
)e−4r1/a
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 17 / 18
The r2 integral
The r2 integral is thus split into two parts
I2 = 4π
(1
r1
∫ r1
0r22 e
−4r2/a dr2 +
∫ ∞r1
r2 e−4r2/a dr2
)these can be solved using integration by parts with dv = e−4r2/a dr2
Ired =
−(a
4
)r22 e−4r2/a
∣∣∣r10
+ 2(a
4
)∫ r1
0r2e−4r2/a dr2
= −(a
4
)r21 e−4r1/a
−2(a
4
)2r2e−4r2/a
∣∣∣∣r10
+ 2(a
4
)2 ∫ r1
0e−4r2/a dr2
= −(a
4
)r21 e−4r1/a − 2
(a4
)2r1e−4r1/a
− 2(a
4
)3e−4r1/a + 2
(a4
)3
Iblue =
−(a
4
)r2e−4r2/a
∣∣∣∞r1
+(a
4
)∫ ∞r1
e−4r2/a dr2
= +(a
4
)r1e−4r1/a
+(a
4
)2e−4r1/a
I2 =πa3
8r1
[1−
(1 +
2r1a
)e−4r1/a
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 17 / 18
The r2 integral
The r2 integral is thus split into two parts
I2 = 4π
(1
r1
∫ r1
0r22 e
−4r2/a dr2 +
∫ ∞r1
r2 e−4r2/a dr2
)these can be solved using integration by parts with dv = e−4r2/a dr2
Ired =
−(a
4
)r22 e−4r2/a
∣∣∣r10
+ 2(a
4
)∫ r1
0r2e−4r2/a dr2
= −(a
4
)r21 e−4r1/a
−2(a
4
)2r2e−4r2/a
∣∣∣∣r10
+ 2(a
4
)2 ∫ r1
0e−4r2/a dr2
= −(a
4
)r21 e−4r1/a − 2
(a4
)2r1e−4r1/a
− 2(a
4
)3e−4r1/a + 2
(a4
)3
Iblue =
−(a
4
)r2e−4r2/a
∣∣∣∞r1
+(a
4
)∫ ∞r1
e−4r2/a dr2
= +(a
4
)r1e−4r1/a
+(a
4
)2e−4r1/a
I2 =πa3
8r1
[1−
(1 +
2r1a
)e−4r1/a
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 17 / 18
The r2 integral
The r2 integral is thus split into two parts
I2 = 4π
(1
r1
∫ r1
0r22 e
−4r2/a dr2 +
∫ ∞r1
r2 e−4r2/a dr2
)these can be solved using integration by parts with dv = e−4r2/a dr2
Ired = −(a
4
)r22 e−4r2/a
∣∣∣r10
+ 2(a
4
)∫ r1
0r2e−4r2/a dr2
= −(a
4
)r21 e−4r1/a
−2(a
4
)2r2e−4r2/a
∣∣∣∣r10
+ 2(a
4
)2 ∫ r1
0e−4r2/a dr2
= −(a
4
)r21 e−4r1/a − 2
(a4
)2r1e−4r1/a
− 2(a
4
)3e−4r1/a + 2
(a4
)3
Iblue =
−(a
4
)r2e−4r2/a
∣∣∣∞r1
+(a
4
)∫ ∞r1
e−4r2/a dr2
= +(a
4
)r1e−4r1/a
+(a
4
)2e−4r1/a
I2 =πa3
8r1
[1−
(1 +
2r1a
)e−4r1/a
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 17 / 18
The r2 integral
The r2 integral is thus split into two parts
I2 = 4π
(1
r1
∫ r1
0r22 e
−4r2/a dr2 +
∫ ∞r1
r2 e−4r2/a dr2
)these can be solved using integration by parts with dv = e−4r2/a dr2
Ired = −(a
4
)r22 e−4r2/a
∣∣∣r10
+ 2(a
4
)∫ r1
0r2e−4r2/a dr2
= −(a
4
)r21 e−4r1/a
−2(a
4
)2r2e−4r2/a
∣∣∣∣r10
+ 2(a
4
)2 ∫ r1
0e−4r2/a dr2
= −(a
4
)r21 e−4r1/a − 2
(a4
)2r1e−4r1/a
− 2(a
4
)3e−4r1/a + 2
(a4
)3
Iblue =
−(a
4
)r2e−4r2/a
∣∣∣∞r1
+(a
4
)∫ ∞r1
e−4r2/a dr2
= +(a
4
)r1e−4r1/a
+(a
4
)2e−4r1/a
I2 =πa3
8r1
[1−
(1 +
2r1a
)e−4r1/a
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 17 / 18
The r2 integral
The r2 integral is thus split into two parts
I2 = 4π
(1
r1
∫ r1
0r22 e
−4r2/a dr2 +
∫ ∞r1
r2 e−4r2/a dr2
)these can be solved using integration by parts with dv = e−4r2/a dr2
Ired = −(a
4
)r22 e−4r2/a
∣∣∣r10
+ 2(a
4
)∫ r1
0r2e−4r2/a dr2
= −(a
4
)r21 e−4r1/a
−2(a
4
)2r2e−4r2/a
∣∣∣∣r10
+ 2(a
4
)2 ∫ r1
0e−4r2/a dr2
= −(a
4
)r21 e−4r1/a − 2
(a4
)2r1e−4r1/a
− 2(a
4
)3e−4r1/a + 2
(a4
)3
Iblue =
−(a
4
)r2e−4r2/a
∣∣∣∞r1
+(a
4
)∫ ∞r1
e−4r2/a dr2
= +(a
4
)r1e−4r1/a
+(a
4
)2e−4r1/a
I2 =πa3
8r1
[1−
(1 +
2r1a
)e−4r1/a
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 17 / 18
The r2 integral
The r2 integral is thus split into two parts
I2 = 4π
(1
r1
∫ r1
0r22 e
−4r2/a dr2 +
∫ ∞r1
r2 e−4r2/a dr2
)these can be solved using integration by parts with dv = e−4r2/a dr2
Ired = −(a
4
)r22 e−4r2/a
∣∣∣r10
+ 2(a
4
)∫ r1
0r2e−4r2/a dr2
= −(a
4
)r21 e−4r1/a −2
(a4
)2r2e−4r2/a
∣∣∣∣r10
+ 2(a
4
)2 ∫ r1
0e−4r2/a dr2
= −(a
4
)r21 e−4r1/a − 2
(a4
)2r1e−4r1/a
− 2(a
4
)3e−4r1/a + 2
(a4
)3
Iblue =
−(a
4
)r2e−4r2/a
∣∣∣∞r1
+(a
4
)∫ ∞r1
e−4r2/a dr2
= +(a
4
)r1e−4r1/a
+(a
4
)2e−4r1/a
I2 =πa3
8r1
[1−
(1 +
2r1a
)e−4r1/a
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 17 / 18
The r2 integral
The r2 integral is thus split into two parts
I2 = 4π
(1
r1
∫ r1
0r22 e
−4r2/a dr2 +
∫ ∞r1
r2 e−4r2/a dr2
)these can be solved using integration by parts with dv = e−4r2/a dr2
Ired = −(a
4
)r22 e−4r2/a
∣∣∣r10
+ 2(a
4
)∫ r1
0r2e−4r2/a dr2
= −(a
4
)r21 e−4r1/a −2
(a4
)2r2e−4r2/a
∣∣∣∣r10
+ 2(a
4
)2 ∫ r1
0e−4r2/a dr2
= −(a
4
)r21 e−4r1/a − 2
(a4
)2r1e−4r1/a
− 2(a
4
)3e−4r1/a + 2
(a4
)3
Iblue =
−(a
4
)r2e−4r2/a
∣∣∣∞r1
+(a
4
)∫ ∞r1
e−4r2/a dr2
= +(a
4
)r1e−4r1/a
+(a
4
)2e−4r1/a
I2 =πa3
8r1
[1−
(1 +
2r1a
)e−4r1/a
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 17 / 18
The r2 integral
The r2 integral is thus split into two parts
I2 = 4π
(1
r1
∫ r1
0r22 e
−4r2/a dr2 +
∫ ∞r1
r2 e−4r2/a dr2
)these can be solved using integration by parts with dv = e−4r2/a dr2
Ired = −(a
4
)r22 e−4r2/a
∣∣∣r10
+ 2(a
4
)∫ r1
0r2e−4r2/a dr2
= −(a
4
)r21 e−4r1/a −2
(a4
)2r2e−4r2/a
∣∣∣∣r10
+ 2(a
4
)2 ∫ r1
0e−4r2/a dr2
= −(a
4
)r21 e−4r1/a − 2
(a4
)2r1e−4r1/a
− 2(a
4
)3e−4r1/a + 2
(a4
)3Iblue =
−(a
4
)r2e−4r2/a
∣∣∣∞r1
+(a
4
)∫ ∞r1
e−4r2/a dr2
= +(a
4
)r1e−4r1/a
+(a
4
)2e−4r1/a
I2 =πa3
8r1
[1−
(1 +
2r1a
)e−4r1/a
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 17 / 18
The r2 integral
The r2 integral is thus split into two parts
I2 = 4π
(1
r1
∫ r1
0r22 e
−4r2/a dr2 +
∫ ∞r1
r2 e−4r2/a dr2
)these can be solved using integration by parts with dv = e−4r2/a dr2
Ired = −(a
4
)r22 e−4r2/a
∣∣∣r10
+ 2(a
4
)∫ r1
0r2e−4r2/a dr2
= −(a
4
)r21 e−4r1/a −2
(a4
)2r2e−4r2/a
∣∣∣∣r10
+ 2(a
4
)2 ∫ r1
0e−4r2/a dr2
= −(a
4
)r21 e−4r1/a − 2
(a4
)2r1e−4r1/a − 2
(a4
)3e−4r1/a + 2
(a4
)3
Iblue =
−(a
4
)r2e−4r2/a
∣∣∣∞r1
+(a
4
)∫ ∞r1
e−4r2/a dr2
= +(a
4
)r1e−4r1/a
+(a
4
)2e−4r1/a
I2 =πa3
8r1
[1−
(1 +
2r1a
)e−4r1/a
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 17 / 18
The r2 integral
The r2 integral is thus split into two parts
I2 = 4π
(1
r1
∫ r1
0r22 e
−4r2/a dr2 +
∫ ∞r1
r2 e−4r2/a dr2
)these can be solved using integration by parts with dv = e−4r2/a dr2
Ired = −(a
4
)r22 e−4r2/a
∣∣∣r10
+ 2(a
4
)∫ r1
0r2e−4r2/a dr2
= −(a
4
)r21 e−4r1/a −2
(a4
)2r2e−4r2/a
∣∣∣∣r10
+ 2(a
4
)2 ∫ r1
0e−4r2/a dr2
= −(a
4
)r21 e−4r1/a − 2
(a4
)2r1e−4r1/a − 2
(a4
)3e−4r1/a + 2
(a4
)3Iblue =
−(a
4
)r2e−4r2/a
∣∣∣∞r1
+(a
4
)∫ ∞r1
e−4r2/a dr2
= +(a
4
)r1e−4r1/a
+(a
4
)2e−4r1/a
I2 =πa3
8r1
[1−
(1 +
2r1a
)e−4r1/a
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 17 / 18
The r2 integral
The r2 integral is thus split into two parts
I2 = 4π
(1
r1
∫ r1
0r22 e
−4r2/a dr2 +
∫ ∞r1
r2 e−4r2/a dr2
)these can be solved using integration by parts with dv = e−4r2/a dr2
Ired = −(a
4
)r22 e−4r2/a
∣∣∣r10
+ 2(a
4
)∫ r1
0r2e−4r2/a dr2
= −(a
4
)r21 e−4r1/a −2
(a4
)2r2e−4r2/a
∣∣∣∣r10
+ 2(a
4
)2 ∫ r1
0e−4r2/a dr2
= −(a
4
)r21 e−4r1/a − 2
(a4
)2r1e−4r1/a − 2
(a4
)3e−4r1/a + 2
(a4
)3Iblue = −
(a4
)r2e−4r2/a
∣∣∣∞r1
+(a
4
)∫ ∞r1
e−4r2/a dr2
= +(a
4
)r1e−4r1/a
+(a
4
)2e−4r1/a
I2 =πa3
8r1
[1−
(1 +
2r1a
)e−4r1/a
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 17 / 18
The r2 integral
The r2 integral is thus split into two parts
I2 = 4π
(1
r1
∫ r1
0r22 e
−4r2/a dr2 +
∫ ∞r1
r2 e−4r2/a dr2
)these can be solved using integration by parts with dv = e−4r2/a dr2
Ired = −(a
4
)r22 e−4r2/a
∣∣∣r10
+ 2(a
4
)∫ r1
0r2e−4r2/a dr2
= −(a
4
)r21 e−4r1/a −2
(a4
)2r2e−4r2/a
∣∣∣∣r10
+ 2(a
4
)2 ∫ r1
0e−4r2/a dr2
= −(a
4
)r21 e−4r1/a − 2
(a4
)2r1e−4r1/a − 2
(a4
)3e−4r1/a + 2
(a4
)3Iblue = −
(a4
)r2e−4r2/a
∣∣∣∞r1
+(a
4
)∫ ∞r1
e−4r2/a dr2
= +(a
4
)r1e−4r1/a
+(a
4
)2e−4r1/a
I2 =πa3
8r1
[1−
(1 +
2r1a
)e−4r1/a
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 17 / 18
The r2 integral
The r2 integral is thus split into two parts
I2 = 4π
(1
r1
∫ r1
0r22 e
−4r2/a dr2 +
∫ ∞r1
r2 e−4r2/a dr2
)these can be solved using integration by parts with dv = e−4r2/a dr2
Ired = −(a
4
)r22 e−4r2/a
∣∣∣r10
+ 2(a
4
)∫ r1
0r2e−4r2/a dr2
= −(a
4
)r21 e−4r1/a −2
(a4
)2r2e−4r2/a
∣∣∣∣r10
+ 2(a
4
)2 ∫ r1
0e−4r2/a dr2
= −(a
4
)r21 e−4r1/a − 2
(a4
)2r1e−4r1/a − 2
(a4
)3e−4r1/a + 2
(a4
)3Iblue = −
(a4
)r2e−4r2/a
∣∣∣∞r1
+(a
4
)∫ ∞r1
e−4r2/a dr2
= +(a
4
)r1e−4r1/a
+(a
4
)2e−4r1/a
I2 =πa3
8r1
[1−
(1 +
2r1a
)e−4r1/a
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 17 / 18
The r2 integral
The r2 integral is thus split into two parts
I2 = 4π
(1
r1
∫ r1
0r22 e
−4r2/a dr2 +
∫ ∞r1
r2 e−4r2/a dr2
)these can be solved using integration by parts with dv = e−4r2/a dr2
Ired = −(a
4
)r22 e−4r2/a
∣∣∣r10
+ 2(a
4
)∫ r1
0r2e−4r2/a dr2
= −(a
4
)r21 e−4r1/a −2
(a4
)2r2e−4r2/a
∣∣∣∣r10
+ 2(a
4
)2 ∫ r1
0e−4r2/a dr2
= −(a
4
)r21 e−4r1/a − 2
(a4
)2r1e−4r1/a − 2
(a4
)3e−4r1/a + 2
(a4
)3Iblue = −
(a4
)r2e−4r2/a
∣∣∣∞r1
+(a
4
)∫ ∞r1
e−4r2/a dr2
= +(a
4
)r1e−4r1/a +
(a4
)2e−4r1/a
I2 =πa3
8r1
[1−
(1 +
2r1a
)e−4r1/a
]
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 17 / 18
The r2 integral
The r2 integral is thus split into two parts
I2 = 4π
(1
r1
∫ r1
0r22 e
−4r2/a dr2 +
∫ ∞r1
r2 e−4r2/a dr2
)these can be solved using integration by parts with dv = e−4r2/a dr2
Ired = −(a
4
)r22 e−4r2/a
∣∣∣r10
+ 2(a
4
)∫ r1
0r2e−4r2/a dr2
= −(a
4
)r21 e−4r1/a −2
(a4
)2r2e−4r2/a
∣∣∣∣r10
+ 2(a
4
)2 ∫ r1
0e−4r2/a dr2
= −(a
4
)r21 e−4r1/a − 2
(a4
)2r1e−4r1/a − 2
(a4
)3e−4r1/a + 2
(a4
)3Iblue = −
(a4
)r2e−4r2/a
∣∣∣∞r1
+(a
4
)∫ ∞r1
e−4r2/a dr2
= +(a
4
)r1e−4r1/a +
(a4
)2e−4r1/a
I2 =πa3
8r1
[1−
(1 +
2r1a
)e−4r1/a
]C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 17 / 18
The r1 integral
The expression for 〈Vee〉 becomes
〈Vee〉 =
(e2
4πε0
)(8
πa3
)∫ [1−(
1+2r1a
)e−4r1/a
]e−4r1/ar1 sin θ1 dr1dθ1dφ1
the angular integrals simply give 4π and dropping the subscript on theradial coordinate
〈Vee〉 =
(8e2
πε0a3
)∫ ∞0
[re−4r/a −
(r +
2r2
a
)e−8r/a
]dr
=
(8e2
πε0a3
)[(a4
)2−(a
8
)2− 4
a
(a8
)3]=
(8e2
πε0a3
)[a2
16− a2
64− a2
128
]=
(8e2
πε0a3
)a2[
8− 2− 1
128
]=
5
2
(e2
8πε0a
)= −5
2E1 = +34eV
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 18 / 18
The r1 integral
The expression for 〈Vee〉 becomes
〈Vee〉 =
(e2
4πε0
)(8
πa3
)∫ [1−(
1+2r1a
)e−4r1/a
]e−4r1/ar1 sin θ1 dr1dθ1dφ1
the angular integrals simply give 4π and dropping the subscript on theradial coordinate
〈Vee〉 =
(8e2
πε0a3
)∫ ∞0
[re−4r/a −
(r +
2r2
a
)e−8r/a
]dr
=
(8e2
πε0a3
)[(a4
)2−(a
8
)2− 4
a
(a8
)3]=
(8e2
πε0a3
)[a2
16− a2
64− a2
128
]=
(8e2
πε0a3
)a2[
8− 2− 1
128
]=
5
2
(e2
8πε0a
)= −5
2E1 = +34eV
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 18 / 18
The r1 integral
The expression for 〈Vee〉 becomes
〈Vee〉 =
(e2
4πε0
)(8
πa3
)∫ [1−(
1+2r1a
)e−4r1/a
]e−4r1/ar1 sin θ1 dr1dθ1dφ1
the angular integrals simply give 4π and dropping the subscript on theradial coordinate
〈Vee〉 =
(8e2
πε0a3
)∫ ∞0
[re−4r/a −
(r +
2r2
a
)e−8r/a
]dr
=
(8e2
πε0a3
)[(a4
)2−(a
8
)2− 4
a
(a8
)3]=
(8e2
πε0a3
)[a2
16− a2
64− a2
128
]=
(8e2
πε0a3
)a2[
8− 2− 1
128
]=
5
2
(e2
8πε0a
)= −5
2E1 = +34eV
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 18 / 18
The r1 integral
The expression for 〈Vee〉 becomes
〈Vee〉 =
(e2
4πε0
)(8
πa3
)∫ [1−(
1+2r1a
)e−4r1/a
]e−4r1/ar1 sin θ1 dr1dθ1dφ1
the angular integrals simply give 4π and dropping the subscript on theradial coordinate
〈Vee〉 =
(8e2
πε0a3
)∫ ∞0
[re−4r/a −
(r +
2r2
a
)e−8r/a
]dr
=
(8e2
πε0a3
)[(a4
)2−(a
8
)2− 4
a
(a8
)3]=
(8e2
πε0a3
)[a2
16− a2
64− a2
128
]=
(8e2
πε0a3
)a2[
8− 2− 1
128
]=
5
2
(e2
8πε0a
)= −5
2E1 = +34eV
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 18 / 18
The r1 integral
The expression for 〈Vee〉 becomes
〈Vee〉 =
(e2
4πε0
)(8
πa3
)∫ [1−(
1+2r1a
)e−4r1/a
]e−4r1/ar1 sin θ1 dr1dθ1dφ1
the angular integrals simply give 4π and dropping the subscript on theradial coordinate
〈Vee〉 =
(8e2
πε0a3
)∫ ∞0
[re−4r/a −
(r +
2r2
a
)e−8r/a
]dr
=
(8e2
πε0a3
)[(a4
)2−(a
8
)2− 4
a
(a8
)3]
=
(8e2
πε0a3
)[a2
16− a2
64− a2
128
]=
(8e2
πε0a3
)a2[
8− 2− 1
128
]=
5
2
(e2
8πε0a
)= −5
2E1 = +34eV
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 18 / 18
The r1 integral
The expression for 〈Vee〉 becomes
〈Vee〉 =
(e2
4πε0
)(8
πa3
)∫ [1−(
1+2r1a
)e−4r1/a
]e−4r1/ar1 sin θ1 dr1dθ1dφ1
the angular integrals simply give 4π and dropping the subscript on theradial coordinate
〈Vee〉 =
(8e2
πε0a3
)∫ ∞0
[re−4r/a −
(r +
2r2
a
)e−8r/a
]dr
=
(8e2
πε0a3
)[(a4
)2−(a
8
)2− 4
a
(a8
)3]=
(8e2
πε0a3
)[a2
16− a2
64− a2
128
]
=
(8e2
πε0a3
)a2[
8− 2− 1
128
]=
5
2
(e2
8πε0a
)= −5
2E1 = +34eV
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 18 / 18
The r1 integral
The expression for 〈Vee〉 becomes
〈Vee〉 =
(e2
4πε0
)(8
πa3
)∫ [1−(
1+2r1a
)e−4r1/a
]e−4r1/ar1 sin θ1 dr1dθ1dφ1
the angular integrals simply give 4π and dropping the subscript on theradial coordinate
〈Vee〉 =
(8e2
πε0a3
)∫ ∞0
[re−4r/a −
(r +
2r2
a
)e−8r/a
]dr
=
(8e2
πε0a3
)[(a4
)2−(a
8
)2− 4
a
(a8
)3]=
(8e2
πε0a3
)[a2
16− a2
64− a2
128
]=
(8e2
πε0a3
)a2[
8− 2− 1
128
]
=5
2
(e2
8πε0a
)= −5
2E1 = +34eV
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 18 / 18
The r1 integral
The expression for 〈Vee〉 becomes
〈Vee〉 =
(e2
4πε0
)(8
πa3
)∫ [1−(
1+2r1a
)e−4r1/a
]e−4r1/ar1 sin θ1 dr1dθ1dφ1
the angular integrals simply give 4π and dropping the subscript on theradial coordinate
〈Vee〉 =
(8e2
πε0a3
)∫ ∞0
[re−4r/a −
(r +
2r2
a
)e−8r/a
]dr
=
(8e2
πε0a3
)[(a4
)2−(a
8
)2− 4
a
(a8
)3]=
(8e2
πε0a3
)[a2
16− a2
64− a2
128
]=
(8e2
πε0a3
)a2[
8− 2− 1
128
]=
5
2
(e2
8πε0a
)
= −5
2E1 = +34eV
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 18 / 18
The r1 integral
The expression for 〈Vee〉 becomes
〈Vee〉 =
(e2
4πε0
)(8
πa3
)∫ [1−(
1+2r1a
)e−4r1/a
]e−4r1/ar1 sin θ1 dr1dθ1dφ1
the angular integrals simply give 4π and dropping the subscript on theradial coordinate
〈Vee〉 =
(8e2
πε0a3
)∫ ∞0
[re−4r/a −
(r +
2r2
a
)e−8r/a
]dr
=
(8e2
πε0a3
)[(a4
)2−(a
8
)2− 4
a
(a8
)3]=
(8e2
πε0a3
)[a2
16− a2
64− a2
128
]=
(8e2
πε0a3
)a2[
8− 2− 1
128
]=
5
2
(e2
8πε0a
)= −5
2E1 = +34eV
C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 18 / 18