today’s outline - february 02, 2017csrri.iit.edu/~segre/phys406/17s/lecture_07.pdf · today’s...

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

• Delta function

• Helium atom

• Delta function

• Helium atom

• Delta function

• Helium atom

• Delta function

• Helium atom

• Delta function

• Helium atom

• Delta function

• Helium atom

• Delta function

• Helium atom

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

~Sp, ~µe = − gee

the proton’s magnetic field seen by the electron is

~Bp =µ0

4πr3[3(~µp · r̂)r̂ − ~µp] +

2µ03~µpδ

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

⟩+µ0gpe

3mpme〈~Sp · ~Se〉|ψ(0)|2

~µp =gpe

, ~µe = − gee

~Bp =µ0

4πr3[3(~µp · r̂)r̂ − ~µp] +

2µ03~µpδ

E(1)hf =

µ0gpe2

8πmpme

⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se

⟩+µ0gpe

3mpme〈~Sp · ~Se〉|ψ(0)|2

~µp =gpe

~Sp, ~µe = − gee

~Bp =µ0

4πr3[3(~µp · r̂)r̂ − ~µp] +

2µ03~µpδ

E(1)hf =

µ0gpe2

8πmpme

⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se

⟩+µ0gpe

3mpme〈~Sp · ~Se〉|ψ(0)|2

~µp =gpe

~Sp, ~µe = − gee

~Bp =µ0

4πr3[3(~µp · r̂)r̂ − ~µp] +

2µ03~µpδ

E(1)hf =

µ0gpe2

8πmpme

⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se

⟩+µ0gpe

3mpme〈~Sp · ~Se〉|ψ(0)|2

~µp =gpe

~Sp, ~µe = − gee

~Bp =µ0

4πr3[3(~µp · r̂)r̂ − ~µp] +

2µ03~µpδ

E(1)hf =

µ0gpe2

8πmpme

⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se

⟩+µ0gpe

3mpme〈~Sp · ~Se〉|ψ(0)|2

~µp =gpe

~Sp, ~µe = − gee

~Bp =µ0

4πr3[3(~µp · r̂)r̂ − ~µp] +

2µ03~µpδ

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

⟩+µ0gpe

3mpme〈~Sp · ~Se〉|ψ(0)|2

~µp =gpe

~Sp, ~µe = − gee

~Bp =µ0

4πr3[3(~µp · r̂)r̂ − ~µp] +

2µ03~µpδ

E(1)hf =

µ0gpe2

8πmpme

⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se

⟩+µ0gpe

3mpme〈~Sp · ~Se〉|ψ(0)|2

~µp =gpe

~Sp, ~µe = − gee

~Bp =µ0

4πr3[3(~µp · r̂)r̂ − ~µp] +

2µ03~µpδ

E(1)hf =

µ0gpe2

8πmpme

⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se

⟩+µ0gpe

3mpme〈~Sp · ~Se〉|ψ(0)|2

Spin-spin coupling

E(1)hf =

µ0gpe2

8πmpme

⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se

⟩+µ0gpe

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

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)]

Spin-spin coupling

E(1)hf =

(((((((

((((µ0gpe2

8πmpme

⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se

⟩+µ0gpe

3mpme〈~Sp · ~Se〉|ψ(0)|2

|ψ100(0)|2 =1

E(1)hf =

µ0gpe2

~S ≡ ~Se + ~Sp

~Se · ~Sp = 12(S2 − S2

e − S2p )

〈~Sp · ~Se〉 =~2

2[s(s + 1)− se(se + 1)− sp(sp + 1)]

Spin-spin coupling

E(1)hf =

(((((((

((((µ0gpe2

8πmpme

⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se

⟩+µ0gpe

3mpme〈~Sp · ~Se〉|ψ(0)|2

|ψ100(0)|2 =1

E(1)hf =

µ0gpe2

~S ≡ ~Se + ~Sp

~Se · ~Sp = 12(S2 − S2

e − S2p )

〈~Sp · ~Se〉 =~2

2[s(s + 1)− se(se + 1)− sp(sp + 1)]

Spin-spin coupling

E(1)hf =

(((((((

((((µ0gpe2

8πmpme

⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se

⟩+µ0gpe

3mpme〈~Sp · ~Se〉|ψ(0)|2

|ψ100(0)|2 =1

E(1)hf =

µ0gpe2

~S ≡ ~Se + ~Sp

~Se · ~Sp = 12(S2 − S2

e − S2p )

〈~Sp · ~Se〉 =~2

2[s(s + 1)− se(se + 1)− sp(sp + 1)]

Spin-spin coupling

E(1)hf =

(((((((

((((µ0gpe2

8πmpme

⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se

⟩+µ0gpe

3mpme〈~Sp · ~Se〉|ψ(0)|2

|ψ100(0)|2 =1

E(1)hf =

µ0gpe2

~S ≡ ~Se + ~Sp

~Se · ~Sp = 12(S2 − S2

e − S2p )

〈~Sp · ~Se〉 =~2

2[s(s + 1)− se(se + 1)− sp(sp + 1)]

Spin-spin coupling

E(1)hf =

(((((((

((((µ0gpe2

8πmpme

⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se

⟩+µ0gpe

3mpme〈~Sp · ~Se〉|ψ(0)|2

|ψ100(0)|2 =1

E(1)hf =

µ0gpe2

~S ≡ ~Se + ~Sp

~Se · ~Sp = 12(S2 − S2

e − S2p )

〈~Sp · ~Se〉 =~2

2[s(s + 1)− se(se + 1)− sp(sp + 1)]

Spin-spin coupling

E(1)hf =

(((((((

((((µ0gpe2

8πmpme

⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se

⟩+µ0gpe

3mpme〈~Sp · ~Se〉|ψ(0)|2

|ψ100(0)|2 =1

E(1)hf =

µ0gpe2

~S ≡ ~Se + ~Sp

~Se · ~Sp = 12(S2 − S2

e − S2p )

〈~Sp · ~Se〉 =~2

2[s(s + 1)− se(se + 1)− sp(sp + 1)]

Spin-spin coupling

E(1)hf =

(((((((

((((µ0gpe2

8πmpme

⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se

⟩+µ0gpe

3mpme〈~Sp · ~Se〉|ψ(0)|2

|ψ100(0)|2 =1

E(1)hf =

µ0gpe2

~S ≡ ~Se + ~Sp

~Se · ~Sp = 12(S2 − S2

e − S2p )

〈~Sp · ~Se〉 =~2

2[s(s + 1)− se(se + 1)− sp(sp + 1)]

Spin-spin coupling

E(1)hf =

(((((((

((((µ0gpe2

8πmpme

⟨3(~Sp · r̂)(~Se · r̂)− ~Sp · ~Se

⟩+µ0gpe

3mpme〈~Sp · ~Se〉|ψ(0)|2

|ψ100(0)|2 =1

E(1)hf =

µ0gpe2

~S ≡ ~Se + ~Sp

~Se · ~Sp = 12(S2 − S2

e − S2p )

〈~Sp · ~Se〉 =~2

2[s(s + 1)− se(se + 1)− sp(sp + 1)]

The 21 cm hydrogen line

E(1)hf =

µ0gpe2

=µ0gpe

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 =

3mpmec2a4

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(1)hf =

µ0gpe2

=µ0gpe

3πmpmea3~2

2[s(s + 1)− se(se + 1)− sp(sp + 1)]

E(1)hf =

3mpmec2a4

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(1)hf =

µ0gpe2

=µ0gpe

3πmpmea3~2

2[s(s + 1)− se(se + 1)− sp(sp + 1)]

E(1)hf =

3mpmec2a4

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(1)hf =

µ0gpe2

=µ0gpe

3πmpmea3~2

2[s(s + 1)− se(se + 1)− sp(sp + 1)]

E(1)hf =

3mpmec2a4

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(1)hf =

µ0gpe2

=µ0gpe

3πmpmea3~2

2[s(s + 1)− se(se + 1)− sp(sp + 1)]

but the total spin can be in eithera singlet

or a triplet state

E(1)hf =

3mpmec2a4

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(1)hf =

µ0gpe2

=µ0gpe

3πmpmea3~2

2[s(s + 1)− se(se + 1)− sp(sp + 1)]

E(1)hf =

3mpmec2a4

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(1)hf =

µ0gpe2

=µ0gpe

3πmpmea3~2

2[s(s + 1)− se(se + 1)− sp(sp + 1)]

E(1)hf =

3mpmec2a4

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(1)hf =

µ0gpe2

=µ0gpe

3πmpmea3~2

2[s(s + 1)− se(se + 1)− sp(sp + 1)]

E(1)hf =

3mpmec2a4

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(1)hf =

µ0gpe2

=µ0gpe

3πmpmea3~2

2[s(s + 1)− se(se + 1)− sp(sp + 1)]

E(1)hf =

3mpmec2a4

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(1)hf =

µ0gpe2

=µ0gpe

3πmpmea3~2

2[s(s + 1)− se(se + 1)− sp(sp + 1)]

E(1)hf =

3mpmec2a4

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(1)hf =

µ0gpe2

=µ0gpe

3πmpmea3~2

2[s(s + 1)− se(se + 1)− sp(sp + 1)]

E(1)hf =

3mpmec2a4

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

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

Hψn = Enψn

the normalization condition for ψ requires

1 = 〈ψ|ψ〉 =

⟨∑m

∣∣∣∣∣ ∑n

⟩=∑m

c∗mcn〈ψm|ψn〉 =∑n

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 .

ψ =∑n

Hψn = Enψn

1 = 〈ψ|ψ〉 =

⟨∑m

∣∣∣∣∣ ∑n

⟩=∑m

Variational theorem

ψ =∑n

Hψn = Enψn

1 = 〈ψ|ψ〉 =

⟨∑m

∣∣∣∣∣ ∑n

⟩=∑m

Variational theorem

ψ =∑n

Hψn = Enψn

1 = 〈ψ|ψ〉 =

⟨∑m

∣∣∣∣∣ ∑n

⟩=∑m

Variational theorem

ψ =∑n

Hψn = Enψn

1 = 〈ψ|ψ〉 =

⟨∑m

∣∣∣∣∣ ∑n

⟩=∑m

Variational theorem

ψ =∑n

Hψn = Enψn

1 = 〈ψ|ψ〉 =

⟨∑m

∣∣∣∣∣ ∑n

⟩=∑m

Variational theorem

ψ =∑n

Hψn = Enψn

1 = 〈ψ|ψ〉 =

⟨∑m

∣∣∣∣∣ ∑n

⟩=∑m

Variational theorem

ψ =∑n

Hψn = Enψn

1 = 〈ψ|ψ〉

⟨∑m

∣∣∣∣∣ ∑n

⟩=∑m

Variational theorem

ψ =∑n

Hψn = Enψn

1 = 〈ψ|ψ〉 =

⟨∑m

∣∣∣∣∣ ∑n

Variational theorem

ψ =∑n

Hψn = Enψn

1 = 〈ψ|ψ〉 =

⟨∑m

∣∣∣∣∣ ∑n

⟩=∑m

c∗mcn〈ψm|ψn〉

Variational theorem

ψ =∑n

Hψn = Enψn

1 = 〈ψ|ψ〉 =

⟨∑m

∣∣∣∣∣ ∑n

⟩=∑m

Proof of variational theorem

If we take the expectation value of the Hamiltonian with this arbitrarywave function, ψ we have

〈H〉 =

⟨∑m

∣∣∣∣∣ H∑n

⟩=∑m

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

|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.

〈H〉 =

⟨∑m

∣∣∣∣∣ H∑n

En|cn|2

〈H〉 ≥ Egs

|cn|2 = Egs

〈H〉 =

⟨∑m

∣∣∣∣∣ H∑n

⟩=∑m

c∗mEncn〈ψm|ψn〉

En|cn|2

〈H〉 ≥ Egs

|cn|2 = Egs

〈H〉 =

⟨∑m

∣∣∣∣∣ H∑n

⟩=∑m

En|cn|2

〈H〉 ≥ Egs

|cn|2 = Egs

〈H〉 =

⟨∑m

∣∣∣∣∣ H∑n

⟩=∑m

En|cn|2

〈H〉 ≥ Egs

|cn|2 = Egs

〈H〉 =

⟨∑m

∣∣∣∣∣ H∑n

⟩=∑m

En|cn|2

〈H〉 ≥ Egs

〈H〉 =

⟨∑m

∣∣∣∣∣ H∑n

⟩=∑m

En|cn|2

〈H〉 ≥ Egs

|cn|2 = Egs

〈H〉 =

⟨∑m

∣∣∣∣∣ H∑n

⟩=∑m

En|cn|2

〈H〉 ≥ Egs

|cn|2 = Egs

〈H〉 =

⟨∑m

∣∣∣∣∣ H∑n

⟩=∑m

En|cn|2

〈H〉 ≥ Egs

|cn|2 = Egs

Example 7.1 - Harmonic oscillator

Find the ground state energy for the 1-D harmonic oscillator using thevariational theorem

H = − ~2

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

〈H〉 = 〈T 〉+ 〈V 〉

H = − ~2

2mω2x2

ψ(x) = Ae−bx2

1 = |A|2∫ ∞−∞

e−2bx2dx = |A|2

〈H〉 = 〈T 〉+ 〈V 〉

H = − ~2

2mω2x2

ψ(x) = Ae−bx2

1 = |A|2∫ ∞−∞

e−2bx2dx = |A|2

〈H〉 = 〈T 〉+ 〈V 〉

H = − ~2

2mω2x2

ψ(x) = Ae−bx2

1 = |A|2∫ ∞−∞

e−2bx2dx = |A|2

〈H〉 = 〈T 〉+ 〈V 〉

H = − ~2

2mω2x2

ψ(x) = Ae−bx2

1 = |A|2∫ ∞−∞

e−2bx2dx = |A|2

〈H〉 = 〈T 〉+ 〈V 〉

H = − ~2

2mω2x2

ψ(x) = Ae−bx2

1 = |A|2∫ ∞−∞

e−2bx2dx

= |A|2√

〈H〉 = 〈T 〉+ 〈V 〉

H = − ~2

2mω2x2

ψ(x) = Ae−bx2

1 = |A|2∫ ∞−∞

e−2bx2dx = |A|2

〈H〉 = 〈T 〉+ 〈V 〉

H = − ~2

2mω2x2

ψ(x) = Ae−bx2

1 = |A|2∫ ∞−∞

e−2bx2dx = |A|2

〈H〉 = 〈T 〉+ 〈V 〉

H = − ~2

2mω2x2

ψ(x) = Ae−bx2

1 = |A|2∫ ∞−∞

e−2bx2dx = |A|2

〈H〉 = 〈T 〉+ 〈V 〉

H = − ~2

2mω2x2

ψ(x) = Ae−bx2

1 = |A|2∫ ∞−∞

e−2bx2dx = |A|2

〈H〉 = 〈T 〉+ 〈V 〉

Example 7.1 (cont.)

The kinetic and po-tential energies are in-stances of the Gaus-sian integral

giving

〈H〉 =~2b2m

〈T 〉 = − ~2

2m|A|2

∫ ∞−∞

e−bx2 d2

(e−bx

= − ~2

)1/2∫ ∞−∞

(4b2x2 − 2b

)e−2bx

= − ~2

)1/2 [4b2√

32b3− 2b

〈V 〉 =1

2mω2|A|2

∫ ∞−∞

x2e−2bx2dx =

get the upper bound on energy by minimizing

db〈H〉 =

2m− mω2

8b2→ b =

2~→ 〈H〉min =

Example 7.1 (cont.)

giving

〈H〉 =~2b2m

〈T 〉 = − ~2

2m|A|2

∫ ∞−∞

e−bx2 d2

(e−bx

= − ~2

)1/2∫ ∞−∞

(4b2x2 − 2b

)e−2bx

= − ~2

)1/2 [4b2√

32b3− 2b

〈V 〉 =1

2mω2|A|2

∫ ∞−∞

x2e−2bx2dx =

db〈H〉 =

2m− mω2

8b2→ b =

2~→ 〈H〉min =

Example 7.1 (cont.)

giving

〈H〉 =~2b2m

〈T 〉 = − ~2

2m|A|2

∫ ∞−∞

e−bx2 d2

(e−bx

= − ~2

)1/2∫ ∞−∞

(4b2x2 − 2b

)e−2bx

= − ~2

)1/2 [4b2√

32b3− 2b

〈V 〉 =1

2mω2|A|2

∫ ∞−∞

x2e−2bx2dx =

db〈H〉 =

2m− mω2

8b2→ b =

2~→ 〈H〉min =

Example 7.1 (cont.)

giving

〈H〉 =~2b2m

〈T 〉 = − ~2

2m|A|2

∫ ∞−∞

e−bx2 d2

(e−bx

= − ~2

)1/2∫ ∞−∞

(4b2x2 − 2b

)e−2bx

= − ~2

)1/2 [4b2√

32b3− 2b

=~2b2m

〈V 〉 =1

2mω2|A|2

∫ ∞−∞

x2e−2bx2dx =

db〈H〉 =

2m− mω2

8b2→ b =

2~→ 〈H〉min =

Example 7.1 (cont.)

giving

〈H〉 =~2b2m

〈T 〉 = − ~2

2m|A|2

∫ ∞−∞

e−bx2 d2

(e−bx

= − ~2

)1/2∫ ∞−∞

(4b2x2 − 2b

)e−2bx

= − ~2

)1/2 [4b2√

32b3− 2b

〈V 〉 =1

2mω2|A|2

∫ ∞−∞

x2e−2bx2dx =

db〈H〉 =

2m− mω2

8b2→ b =

2~→ 〈H〉min =

Example 7.1 (cont.)

giving

〈H〉 =~2b2m

〈T 〉 = − ~2

2m|A|2

∫ ∞−∞

e−bx2 d2

(e−bx

= − ~2

)1/2∫ ∞−∞

(4b2x2 − 2b

)e−2bx

= − ~2

)1/2 [4b2√

32b3− 2b

〈V 〉 =1

2mω2|A|2

∫ ∞−∞

x2e−2bx2dx

db〈H〉 =

2m− mω2

8b2→ b =

2~→ 〈H〉min =

Example 7.1 (cont.)

giving

〈H〉 =~2b2m

〈T 〉 = − ~2

2m|A|2

∫ ∞−∞

e−bx2 d2

(e−bx

= − ~2

)1/2∫ ∞−∞

(4b2x2 − 2b

)e−2bx

= − ~2

)1/2 [4b2√

32b3− 2b

〈V 〉 =1

2mω2|A|2

∫ ∞−∞

x2e−2bx2dx =

db〈H〉 =

2m− mω2

8b2→ b =

2~→ 〈H〉min =

Example 7.1 (cont.)

giving

〈H〉 =~2b2m

〈T 〉 = − ~2

2m|A|2

∫ ∞−∞

e−bx2 d2

(e−bx

= − ~2

)1/2∫ ∞−∞

(4b2x2 − 2b

)e−2bx

= − ~2

)1/2 [4b2√

32b3− 2b

〈V 〉 =1

2mω2|A|2

∫ ∞−∞

x2e−2bx2dx =

db〈H〉 =

2m− mω2

8b2→ b =

2~→ 〈H〉min =

Example 7.1 (cont.)

giving

〈H〉 =~2b2m

〈T 〉 = − ~2

2m|A|2

∫ ∞−∞

e−bx2 d2

(e−bx

= − ~2

)1/2∫ ∞−∞

(4b2x2 − 2b

)e−2bx

= − ~2

)1/2 [4b2√

32b3− 2b

〈V 〉 =1

2mω2|A|2

∫ ∞−∞

x2e−2bx2dx =

db〈H〉 =

2m− mω2

8b2→ b =

2~→ 〈H〉min =

Example 7.1 (cont.)

giving

〈H〉 =~2b2m

〈T 〉 = − ~2

2m|A|2

∫ ∞−∞

e−bx2 d2

(e−bx

= − ~2

)1/2∫ ∞−∞

(4b2x2 − 2b

)e−2bx

= − ~2

)1/2 [4b2√

32b3− 2b

〈V 〉 =1

2mω2|A|2

∫ ∞−∞

x2e−2bx2dx =

db〈H〉

2m− mω2

8b2→ b =

2~→ 〈H〉min =

Example 7.1 (cont.)

giving

〈H〉 =~2b2m

〈T 〉 = − ~2

2m|A|2

∫ ∞−∞

e−bx2 d2

(e−bx

= − ~2

)1/2∫ ∞−∞

(4b2x2 − 2b

)e−2bx

= − ~2

)1/2 [4b2√

32b3− 2b

〈V 〉 =1

2mω2|A|2

∫ ∞−∞

x2e−2bx2dx =

db〈H〉 =

2m− mω2

→ b =mω

2~→ 〈H〉min =

Example 7.1 (cont.)

giving

〈H〉 =~2b2m

〈T 〉 = − ~2

2m|A|2

∫ ∞−∞

e−bx2 d2

(e−bx

= − ~2

)1/2∫ ∞−∞

(4b2x2 − 2b

)e−2bx

= − ~2

)1/2 [4b2√

32b3− 2b

〈V 〉 =1

2mω2|A|2

∫ ∞−∞

x2e−2bx2dx =

db〈H〉 =

2m− mω2

8b2→ b =

→ 〈H〉min =1

Example 7.1 (cont.)

giving

〈H〉 =~2b2m

〈T 〉 = − ~2

2m|A|2

∫ ∞−∞

e−bx2 d2

(e−bx

= − ~2

)1/2∫ ∞−∞

(4b2x2 − 2b

)e−2bx

= − ~2

)1/2 [4b2√

32b3− 2b

〈V 〉 =1

2mω2|A|2

∫ ∞−∞

x2e−2bx2dx =

db〈H〉 =

2m− mω2

8b2→ b =

2~→ 〈H〉min =

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

dx2− αδ(x)

ψ(x) = Ae−bx2

〈V 〉 = −α|A|2∫ ∞−∞

δ(x)e−2bx2dx

=− α√

〈H〉 =~2b2m− α

db〈H〉 =

2m− α√

b =2m2α2

π~4−→ 〈H〉min = −mα2

π~2≥ −mα2

2~2= Egs

We know that

〈T 〉 =~2b2m

minimizing wrt b

H = − ~2

dx2− αδ(x)

ψ(x) = Ae−bx2

〈V 〉 = −α|A|2∫ ∞−∞

δ(x)e−2bx2dx

=− α√

〈H〉 =~2b2m− α

db〈H〉 =

2m− α√

b =2m2α2

π~4−→ 〈H〉min = −mα2

π~2≥ −mα2

2~2= Egs

We know that

〈T 〉 =~2b2m

minimizing wrt b

H = − ~2

dx2− αδ(x)

ψ(x) = Ae−bx2

〈V 〉 = −α|A|2∫ ∞−∞

δ(x)e−2bx2dx

=− α√

〈H〉 =~2b2m− α

db〈H〉 =

2m− α√

b =2m2α2

π~4−→ 〈H〉min = −mα2

π~2≥ −mα2

2~2= Egs

We know that

〈T 〉 =~2b2m

minimizing wrt b

H = − ~2

dx2− αδ(x)

ψ(x) = Ae−bx2

〈V 〉 = −α|A|2∫ ∞−∞

δ(x)e−2bx2dx

=− α√

〈H〉 =~2b2m− α

db〈H〉 =

2m− α√

b =2m2α2

π~4−→ 〈H〉min = −mα2

π~2≥ −mα2

2~2= Egs

We know that

〈T 〉 =~2b2m

minimizing wrt b

H = − ~2

dx2− αδ(x)

ψ(x) = Ae−bx2

〈V 〉 = −α|A|2∫ ∞−∞

δ(x)e−2bx2dx

=− α√

〈H〉 =~2b2m− α

db〈H〉 =

2m− α√

b =2m2α2

π~4−→ 〈H〉min = −mα2

π~2≥ −mα2

2~2= Egs

We know that

〈T 〉 =~2b2m

minimizing wrt b

H = − ~2

dx2− αδ(x)

ψ(x) = Ae−bx2

〈V 〉 = −α|A|2∫ ∞−∞

δ(x)e−2bx2dx

=− α√

〈H〉 =~2b2m− α

db〈H〉 =

2m− α√

b =2m2α2

π~4−→ 〈H〉min = −mα2

π~2≥ −mα2

2~2= Egs

We know that

〈T 〉 =~2b2m

minimizing wrt b

H = − ~2

dx2− αδ(x)

ψ(x) = Ae−bx2

〈V 〉 = −α|A|2∫ ∞−∞

δ(x)e−2bx2dx

=− α√

〈H〉 =~2b2m− α

db〈H〉 =

2m− α√

b =2m2α2

π~4−→ 〈H〉min = −mα2

π~2≥ −mα2

2~2= Egs

We know that

〈T 〉 =~2b2m

minimizing wrt b

H = − ~2

dx2− αδ(x)

ψ(x) = Ae−bx2

〈V 〉 = −α|A|2∫ ∞−∞

δ(x)e−2bx2dx

=− α√

〈H〉 =~2b2m− α

db〈H〉 =

2m− α√

b =2m2α2

π~4−→ 〈H〉min = −mα2

π~2≥ −mα2

2~2= Egs

We know that

〈T 〉 =~2b2m

minimizing wrt b

H = − ~2

dx2− αδ(x)

ψ(x) = Ae−bx2

〈V 〉 = −α|A|2∫ ∞−∞

δ(x)e−2bx2dx

=− α√

〈H〉 =~2b2m− α

db〈H〉 =

2m− α√

b =2m2α2

π~4−→ 〈H〉min = −mα2

π~2≥ −mα2

2~2= Egs

We know that

〈T 〉 =~2b2m

minimizing wrt b

H = − ~2

dx2− αδ(x)

ψ(x) = Ae−bx2

〈V 〉 = −α|A|2∫ ∞−∞

δ(x)e−2bx2dx

=− α√

〈H〉 =~2b2m− α

db〈H〉

2m− α√

b =2m2α2

π~4−→ 〈H〉min = −mα2

π~2≥ −mα2

2~2= Egs

We know that

〈T 〉 =~2b2m

minimizing wrt b

H = − ~2

dx2− αδ(x)

ψ(x) = Ae−bx2

〈V 〉 = −α|A|2∫ ∞−∞

δ(x)e−2bx2dx

=− α√

〈H〉 =~2b2m− α

db〈H〉 =

2m− α√

b =2m2α2

π~4−→ 〈H〉min = −mα2

π~2≥ −mα2

2~2= Egs

We know that

〈T 〉 =~2b2m

minimizing wrt b

H = − ~2

dx2− αδ(x)

ψ(x) = Ae−bx2

〈V 〉 = −α|A|2∫ ∞−∞

δ(x)e−2bx2dx

=− α√

〈H〉 =~2b2m− α

db〈H〉 =

2m− α√

b =2m2α2

−→ 〈H〉min = −mα2

π~2≥ −mα2

2~2= Egs

We know that

〈T 〉 =~2b2m

minimizing wrt b

H = − ~2

dx2− αδ(x)

ψ(x) = Ae−bx2

〈V 〉 = −α|A|2∫ ∞−∞

δ(x)e−2bx2dx

=− α√

〈H〉 =~2b2m− α

db〈H〉 =

2m− α√

b =2m2α2

π~4−→ 〈H〉min = −mα2

≥ −mα2

2~2= Egs

We know that

〈T 〉 =~2b2m

minimizing wrt b

H = − ~2

dx2− αδ(x)

ψ(x) = Ae−bx2

〈V 〉 = −α|A|2∫ ∞−∞

δ(x)e−2bx2dx

=− α√

〈H〉 =~2b2m− α

db〈H〉 =

2m− α√

b =2m2α2

π~4−→ 〈H〉min = −mα2

π~2≥ −mα2

2~2= Egs

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

A(a− x), a2 ≤ x ≤ a

0, otherwise

first determine the normalization constant A

1 = |A|2[∫ a/2

0x2 dx +

a/2(a− x)2 dx

]= |A|2

∣∣∣∣a/20

− (a− x)3

∣∣∣∣aa/2

= |A|2[a3

24− 0− 0 +

]= |A|2 a

12→ A =

Example 7.3

ψ(x) =

Ax , 0 ≤ x ≤ a

A(a− x), a2 ≤ x ≤ a

0, otherwise

1 = |A|2[∫ a/2

0x2 dx +

a/2(a− x)2 dx

]= |A|2

∣∣∣∣a/20

− (a− x)3

∣∣∣∣aa/2

= |A|2[a3

24− 0− 0 +

]= |A|2 a

12→ A =

Example 7.3

ψ(x) =

Ax , 0 ≤ x ≤ a

A(a− x), a2 ≤ x ≤ a

0, otherwise

1 = |A|2[∫ a/2

0x2 dx +

a/2(a− x)2 dx

]= |A|2

∣∣∣∣a/20

− (a− x)3

∣∣∣∣aa/2

= |A|2[a3

24− 0− 0 +

]= |A|2 a

12→ A =

Example 7.3

ψ(x) =

Ax , 0 ≤ x ≤ a

A(a− x), a2 ≤ x ≤ a

0, otherwise

1 = |A|2[∫ a/2

0x2 dx +

a/2(a− x)2 dx

]= |A|2

∣∣∣∣a/20

− (a− x)3

∣∣∣∣aa/2

= |A|2[a3

24− 0− 0 +

]= |A|2 a

12→ A =

Example 7.3

ψ(x) =

Ax , 0 ≤ x ≤ a

A(a− x), a2 ≤ x ≤ a

0, otherwise

1 = |A|2[∫ a/2

0x2 dx +

a/2(a− x)2 dx

= |A|2[x3

∣∣∣∣a/20

− (a− x)3

∣∣∣∣aa/2

= |A|2[a3

24− 0− 0 +

]= |A|2 a

12→ A =

Example 7.3

ψ(x) =

Ax , 0 ≤ x ≤ a

A(a− x), a2 ≤ x ≤ a

0, otherwise

1 = |A|2[∫ a/2

0x2 dx +

a/2(a− x)2 dx

]= |A|2

∣∣∣∣a/20

− (a− x)3

∣∣∣∣aa/2

= |A|2[a3

24− 0− 0 +

]= |A|2 a

12→ A =

Example 7.3

ψ(x) =

Ax , 0 ≤ x ≤ a

A(a− x), a2 ≤ x ≤ a

0, otherwise

1 = |A|2[∫ a/2

0x2 dx +

a/2(a− x)2 dx

]= |A|2

∣∣∣∣a/20

− (a− x)3

∣∣∣∣aa/2

= |A|2[a3

24− 0− 0 +

= |A|2 a3

12→ A =

Example 7.3

ψ(x) =

Ax , 0 ≤ x ≤ a

A(a− x), a2 ≤ x ≤ a

0, otherwise

1 = |A|2[∫ a/2

0x2 dx +

a/2(a− x)2 dx

]= |A|2

∣∣∣∣a/20

− (a− x)3

∣∣∣∣aa/2

= |A|2[a3

24− 0− 0 +

]= |A|2 a

→ A =2

Example 7.3

ψ(x) =

Ax , 0 ≤ x ≤ a

A(a− x), a2 ≤ x ≤ a

0, otherwise

1 = |A|2[∫ a/2

0x2 dx +

a/2(a− x)2 dx

]= |A|2

∣∣∣∣a/20

− (a− x)3

∣∣∣∣aa/2

= |A|2[a3

24− 0− 0 +

]= |A|2 a

12→ A =

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

+A, 0 ≤ x ≤ a

−A, a2 ≤ x ≤ a

0, otherwise

and the second derivative is threedelta functions, one at each edgeand one at the center of the well

dx2= Aδ(x)− 2Aδ(x − a

+ Aδ(x − a)

Ax , 0 ≤ x ≤ a

A(a− x), a2 ≤ x ≤ a

0, otherwise

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

+A, 0 ≤ x ≤ a

−A, a2 ≤ x ≤ a

0, otherwise

+ Aδ(x − a)

Ax , 0 ≤ x ≤ a

A(a− x), a2 ≤ x ≤ a

0, otherwise

Example 7.3 (cont.)

+A, 0 ≤ x ≤ a

−A, a2 ≤ x ≤ a

0, otherwise

+ Aδ(x − a)

Ax , 0 ≤ x ≤ a

A(a− x), a2 ≤ x ≤ a

0, otherwise

Example 7.3 (cont.)

+A, 0 ≤ x ≤ a

−A, a2 ≤ x ≤ a

0, otherwise

+ Aδ(x − a)

Ax , 0 ≤ x ≤ a

A(a− x), a2 ≤ x ≤ a

0, otherwise

Example 7.3 (cont.)

+A, 0 ≤ x ≤ a

−A, a2 ≤ x ≤ a

0, otherwise

+ Aδ(x − a)

Ax , 0 ≤ x ≤ a

A(a− x), a2 ≤ x ≤ a

0, otherwise

d2ψdx2

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

comparing to the exact ground state energy

Egs =π2~2

2ma2= 〈H〉

this demonstrates the variational principle even if there is no freeparameter with which to minimize the energy

Example 7.3 (cont.)

〈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

Egs =π2~2

2ma2= 〈H〉

Example 7.3 (cont.)

〈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

Egs =π2~2

2ma2= 〈H〉

Example 7.3 (cont.)

〈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

Egs =π2~2

2ma2= 〈H〉

Example 7.3 (cont.)

〈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

Egs =π2~2

2ma2= 〈H〉

Example 7.3 (cont.)

〈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

Egs =π2~2

2ma2= 〈H〉

Example 7.3 (cont.)

〈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

Egs =π2~2

2ma2= 〈H〉

Example 7.3 (cont.)

〈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

Egs =π2~2

2ma2= 〈H〉

Example 7.3 (cont.)

〈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

Egs =π2~2

2ma2= 〈H〉

Example 7.3 (cont.)

〈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

Egs =π2~2

2ma2= 〈H〉

Example 7.3 (cont.)

〈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

Egs =π2~2

2ma2= 〈H〉

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

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

|~r1 − ~r2|

which is not exactly soluble

Egs = −78.975 eV

-e|r1-r2|

|r1||r2|

Helium atom

H = − ~2

2m(∇2

1 +∇22)− e2

4πε0

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

Vee=e2

4πε0

|~r1 − ~r2|

Egs = −78.975 eV

-e|r1-r2|

|r1||r2|

Helium atom

H = − ~2

2m(∇2

1 +∇22)− e2

4πε0

r2− 1

|~r1 − ~r2|

Vee=e2

4πε0

|~r1 − ~r2|

Egs = −78.975 eV

-e|r1-r2|

|r1||r2|

Helium atom

H = − ~2

2m(∇2

1 +∇22)− e2

4πε0

r2− 1

|~r1 − ~r2|

Vee=e2

4πε0

|~r1 − ~r2|

Egs = −78.975 eV

-e|r1-r2|

|r1||r2|

Helium atom

H = − ~2

2m(∇2

1 +∇22)− e2

4πε0

r2− 1

|~r1 − ~r2|

Vee=e2

4πε0

|~r1 − ~r2|

Egs = −78.975 eV

-e|r1-r2|

|r1||r2|

Helium atom

H = − ~2

2m(∇2

1 +∇22)− e2

4πε0

r2− 1

|~r1 − ~r2|

Vee=e2

4πε0

|~r1 − ~r2|

Egs = −78.975 eV

-e|r1-r2|

|r1||r2|

Helium atom

H = − ~2

2m(∇2

1 +∇22)− e2

4πε0

r2− 1

|~r1 − ~r2|

Vee=e2

4πε0

|~r1 − ~r2|

Egs = −78.975 eV

-e|r1-r2|

|r1||r2|

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〉

4πε0

)2 ∫ e−4r1/a e−4r2/a

|~r1 − ~r2|d3~r1 d

ψ(~r1, ~r2)

≡ ψ100(~r1)ψ100(~r2) =8

Hψ0 = (8E1 + Vee)ψ0

〈H〉 = (8E1 + 〈Vee〉),

Vee =e2/4πε0|~r1 − ~r2|

〈Vee〉 =

4πε0

∫e−4r1/a e−4r2/a

|~r1 − ~r2|d3~r1 d

ψ(~r1, ~r2) ≡ ψ100(~r1)ψ100(~r2)

Hψ0 = (8E1 + Vee)ψ0

〈H〉 = (8E1 + 〈Vee〉),

Vee =e2/4πε0|~r1 − ~r2|

〈Vee〉 =

4πε0

)2 ∫ e−4r1/a e−4r2/a

|~r1 − ~r2|d3~r1 d

ψ(~r1, ~r2) ≡ ψ100(~r1)ψ100(~r2) =8

Hψ0 = (8E1 + Vee)ψ0

〈H〉 = (8E1 + 〈Vee〉),

Vee =e2/4πε0|~r1 − ~r2|

〈Vee〉 =

4πε0

)2 ∫ e−4r1/a e−4r2/a

|~r1 − ~r2|d3~r1 d

ψ(~r1, ~r2) ≡ ψ100(~r1)ψ100(~r2) =8

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〉 =

4πε0

)2 ∫ e−4r1/a e−4r2/a

|~r1 − ~r2|d3~r1 d

ψ(~r1, ~r2) ≡ ψ100(~r1)ψ100(~r2) =8

Hψ0 = (8E1 + Vee)ψ0

〈H〉 = (8E1 + 〈Vee〉),

Vee =e2/4πε0|~r1 − ~r2|

〈Vee〉 =

4πε0

)2 ∫ e−4r1/a e−4r2/a

|~r1 − ~r2|d3~r1 d

ψ(~r1, ~r2) ≡ ψ100(~r1)ψ100(~r2) =8

Hψ0 = (8E1 + Vee)ψ0

〈H〉 = (8E1 + 〈Vee〉),

Vee =e2/4πε0|~r1 − ~r2|

〈Vee〉 =

4πε0

)2 ∫ e−4r1/a e−4r2/a

|~r1 − ~r2|d3~r1 d

ψ(~r1, ~r2) ≡ ψ100(~r1)ψ100(~r2) =8

Hψ0 = (8E1 + Vee)ψ0

〈H〉 = (8E1 + 〈Vee〉),

Vee =e2/4πε0|~r1 − ~r2|

〈Vee〉 =

4πε0

)2 ∫ e−4r1/a e−4r2/a

|~r1 − ~r2|d3~r1 d

ψ(~r1, ~r2) ≡ ψ100(~r1)ψ100(~r2) =8

Hψ0 = (8E1 + Vee)ψ0

〈H〉 = (8E1 + 〈Vee〉), Vee =e2/4πε0|~r1 − ~r2|

〈Vee〉 =

4πε0

)2 ∫ e−4r1/a e−4r2/a

|~r1 − ~r2|d3~r1 d

ψ(~r1, ~r2) ≡ ψ100(~r1)ψ100(~r2) =8

Hψ0 = (8E1 + Vee)ψ0

〈H〉 = (8E1 + 〈Vee〉), Vee =e2/4πε0|~r1 − ~r2|

〈Vee〉 =

4πε0

)2 ∫ e−4r1/a e−4r2/a

|~r1 − ~r2|d3~r1 d

Setting up the integrals

〈Vee〉 =

4πε0

)2 ∫ e−4r1/a e−4r2/a

|~r1 − ~r2|d3~r1 d

|r -r |1 2

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

〈Vee〉 =

4πε0

)2 ∫ e−4r1/a e−4r2/a

|~r1 − ~r2|d3~r1 d

|r -r |1 2

|~r1 − ~r2|

r21 + r22 − 2r1r2 cos θ2

〈Vee〉 =

4πε0

)2 ∫ e−4r1/a e−4r2/a

|~r1 − ~r2|d3~r1 d

|r -r |1 2

|~r1 − ~r2|

r21 + r22 − 2r1r2 cos θ2

〈Vee〉 =

4πε0

)2 ∫ e−4r1/a e−4r2/a

|~r1 − ~r2|d3~r1 d

|r -r |1 2

|~r1 − ~r2|

r21 + r22 − 2r1r2 cos θ2

〈Vee〉 =

4πε0

)2 ∫ e−4r1/a e−4r2/a

|~r1 − ~r2|d3~r1 d

|r -r |1 2

|~r1 − ~r2| =√

r21 + r22 − 2r1r2 cos θ2

The r2 integral

〈Vee〉 =

4πε0

)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

∫ π

sin θ2√r21 + r22 − 2r1r2 cos θ2

√r21 + r22 − 2r1r2 cos θ2

∣∣∣∣∣∣π

[(r1 + r2)− |r1 − r2|

2/r1, r2 < r1

2/r2, r2 > r1

The r2 integral

〈Vee〉 =

4πε0

)2 ∫ e−4r1/a e−4r2/a√r21 + r22 − 2r1r2 cos θ2

d3~r1 d3~r2

I2 ≡

r21 + r22 − 2r1r2 cos θ2

dr2 dθ2 dφ2

∫ π

sin θ2√r21 + r22 − 2r1r2 cos θ2

√r21 + r22 − 2r1r2 cos θ2

∣∣∣∣∣∣π

[(r1 + r2)− |r1 − r2|

2/r1, r2 < r1

2/r2, r2 > r1

The r2 integral

〈Vee〉 =

4πε0

)2 ∫ e−4r1/a e−4r2/a√r21 + r22 − 2r1r2 cos θ2

d3~r1 d3~r2

I2 ≡

r21 + r22 − 2r1r2 cos θ2

dr2 dθ2 dφ2

∫ π

sin θ2√r21 + r22 − 2r1r2 cos θ2

√r21 + r22 − 2r1r2 cos θ2

∣∣∣∣∣∣π

[(r1 + r2)− |r1 − r2|

2/r1, r2 < r1

2/r2, r2 > r1

The r2 integral

〈Vee〉 =

4πε0

)2 ∫ e−4r1/a e−4r2/a√r21 + r22 − 2r1r2 cos θ2

d3~r1 d3~r2

I2 ≡∫

e−4r2/a r22 sin θ2√r21 + r22 − 2r1r2 cos θ2

dr2 dθ2 dφ2

∫ π

sin θ2√r21 + r22 − 2r1r2 cos θ2

√r21 + r22 − 2r1r2 cos θ2

∣∣∣∣∣∣π

[(r1 + r2)− |r1 − r2|

2/r1, r2 < r1

2/r2, r2 > r1

The r2 integral

〈Vee〉 =

4πε0

)2 ∫ e−4r1/a e−4r2/a√r21 + r22 − 2r1r2 cos θ2

d3~r1 d3~r2

I2 ≡∫

dr2 dθ2 dφ2

∫ π

sin θ2√r21 + r22 − 2r1r2 cos θ2

√r21 + r22 − 2r1r2 cos θ2

∣∣∣∣∣∣π

[(r1 + r2)− |r1 − r2|

2/r1, r2 < r1

2/r2, r2 > r1

The r2 integral

〈Vee〉 =

4πε0

)2 ∫ e−4r1/a e−4r2/a√r21 + r22 − 2r1r2 cos θ2

d3~r1 d3~r2

I2 ≡∫

dr2 dθ2 dφ2

∫ π

sin θ2√r21 + r22 − 2r1r2 cos θ2

√r21 + r22 − 2r1r2 cos θ2

∣∣∣∣∣∣π

[(r1 + r2)− |r1 − r2|

2/r1, r2 < r1

2/r2, r2 > r1

The r2 integral

〈Vee〉 =

4πε0

)2 ∫ e−4r1/a e−4r2/a√r21 + r22 − 2r1r2 cos θ2

d3~r1 d3~r2

I2 ≡∫

dr2 dθ2 dφ2

∫ π

sin θ2√r21 + r22 − 2r1r2 cos θ2

dθ2 =

√r21 + r22 − 2r1r2 cos θ2

∣∣∣∣∣∣π

[(r1 + r2)− |r1 − r2|

2/r1, r2 < r1

2/r2, r2 > r1

The r2 integral

〈Vee〉 =

4πε0

)2 ∫ e−4r1/a e−4r2/a√r21 + r22 − 2r1r2 cos θ2

d3~r1 d3~r2

I2 ≡∫

dr2 dθ2 dφ2

∫ π

sin θ2√r21 + r22 − 2r1r2 cos θ2

dθ2 =

√r21 + r22 − 2r1r2 cos θ2

∣∣∣∣∣∣π

[√r21 + r22 + 2r1 r2 −

√r21 + r22 − 2r1 r2

2/r1, r2 < r1

2/r2, r2 > r1

The r2 integral

〈Vee〉 =

4πε0

)2 ∫ e−4r1/a e−4r2/a√r21 + r22 − 2r1r2 cos θ2

d3~r1 d3~r2

I2 ≡∫

dr2 dθ2 dφ2

∫ π

sin θ2√r21 + r22 − 2r1r2 cos θ2

dθ2 =

√r21 + r22 − 2r1r2 cos θ2

∣∣∣∣∣∣π

[(r1 + r2)− |r1 − r2|

2/r1, r2 < r1

2/r2, r2 > r1

The r2 integral

〈Vee〉 =

4πε0

)2 ∫ e−4r1/a e−4r2/a√r21 + r22 − 2r1r2 cos θ2

d3~r1 d3~r2

I2 ≡∫

dr2 dθ2 dφ2

∫ π

sin θ2√r21 + r22 − 2r1r2 cos θ2

dθ2 =

√r21 + r22 − 2r1r2 cos θ2

∣∣∣∣∣∣π

[(r1 + r2)− |r1 − r2|

2/r1, r2 < r1

2/r2, r2 > r1

The r2 integral

〈Vee〉 =

4πε0

)2 ∫ e−4r1/a e−4r2/a√r21 + r22 − 2r1r2 cos θ2

d3~r1 d3~r2

I2 ≡∫

dr2 dθ2 dφ2

∫ π

sin θ2√r21 + r22 − 2r1r2 cos θ2

dθ2 =

√r21 + r22 − 2r1r2 cos θ2

∣∣∣∣∣∣π

[(r1 + r2)− |r1 − r2|

{2/r1, r2 < r1

2/r2, r2 > r1

The r2 integral

〈Vee〉 =

4πε0

)2 ∫ e−4r1/a e−4r2/a√r21 + r22 − 2r1r2 cos θ2

d3~r1 d3~r2

I2 ≡∫

dr2 dθ2 dφ2

∫ π

sin θ2√r21 + r22 − 2r1r2 cos θ2

dθ2 =

√r21 + r22 − 2r1r2 cos θ2

∣∣∣∣∣∣π

[(r1 + r2)− |r1 − r2|

{2/r1, r2 < r1

2/r2, r2 > r1

The r2 integral

The r2 integral is thus split into two parts

I2 = 4π

∫ 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 =

)r22 e−4r2/a

∣∣∣r10

)∫ r1

0r2e−4r2/a dr2

= −(a

)r21 e−4r1/a

−2(a

)2r2e−4r2/a

∣∣∣∣r10

)2 ∫ r1

0e−4r2/a dr2

= −(a

)r21 e−4r1/a − 2

)2r1e−4r1/a

− 2(a

)3e−4r1/a + 2

Iblue =

)r2e−4r2/a

∣∣∣∞r1

)∫ ∞r1

e−4r2/a dr2

)r1e−4r1/a

)2e−4r1/a

I2 =πa3

)e−4r1/a

The r2 integral

I2 = 4π

∫ 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 =

)r22 e−4r2/a

∣∣∣r10

)∫ r1

0r2e−4r2/a dr2

= −(a

)r21 e−4r1/a

−2(a

)2r2e−4r2/a

∣∣∣∣r10

)2 ∫ r1

0e−4r2/a dr2

= −(a

)r21 e−4r1/a − 2

)2r1e−4r1/a

− 2(a

)3e−4r1/a + 2

Iblue =

)r2e−4r2/a

∣∣∣∞r1

)∫ ∞r1

e−4r2/a dr2

)r1e−4r1/a

)2e−4r1/a

I2 =πa3

)e−4r1/a

The r2 integral

I2 = 4π

∫ r1

0r22 e

−4r2/a dr2 +

∫ ∞r1

r2 e−4r2/a dr2

Ired =

)r22 e−4r2/a

∣∣∣r10

)∫ r1

0r2e−4r2/a dr2

= −(a

)r21 e−4r1/a

−2(a

)2r2e−4r2/a

∣∣∣∣r10

)2 ∫ r1

0e−4r2/a dr2

= −(a

)r21 e−4r1/a − 2

)2r1e−4r1/a

− 2(a

)3e−4r1/a + 2

Iblue =

)r2e−4r2/a

∣∣∣∞r1

)∫ ∞r1

e−4r2/a dr2

)r1e−4r1/a

)2e−4r1/a

I2 =πa3

)e−4r1/a

The r2 integral

I2 = 4π

∫ r1

0r22 e

−4r2/a dr2 +

∫ ∞r1

r2 e−4r2/a dr2

Ired =

)r22 e−4r2/a

∣∣∣r10

)∫ r1

0r2e−4r2/a dr2

= −(a

)r21 e−4r1/a

−2(a

)2r2e−4r2/a

∣∣∣∣r10

)2 ∫ r1

0e−4r2/a dr2

= −(a

)r21 e−4r1/a − 2

)2r1e−4r1/a

− 2(a

)3e−4r1/a + 2

Iblue =

)r2e−4r2/a

∣∣∣∞r1

)∫ ∞r1

e−4r2/a dr2

)r1e−4r1/a

)2e−4r1/a

I2 =πa3

)e−4r1/a

The r2 integral

I2 = 4π

∫ r1

0r22 e

−4r2/a dr2 +

∫ ∞r1

r2 e−4r2/a dr2

Ired =

)r22 e−4r2/a

∣∣∣r10

)∫ r1

0r2e−4r2/a dr2

= −(a

)r21 e−4r1/a

−2(a

)2r2e−4r2/a

∣∣∣∣r10

)2 ∫ r1

0e−4r2/a dr2

= −(a

)r21 e−4r1/a − 2

)2r1e−4r1/a

− 2(a

)3e−4r1/a + 2

Iblue =

)r2e−4r2/a

∣∣∣∞r1

)∫ ∞r1

e−4r2/a dr2

)r1e−4r1/a

)2e−4r1/a

I2 =πa3

)e−4r1/a

The r2 integral

I2 = 4π

∫ r1

0r22 e

−4r2/a dr2 +

∫ ∞r1

r2 e−4r2/a dr2

Ired =

)r22 e−4r2/a

∣∣∣r10

)∫ r1

0r2e−4r2/a dr2

= −(a

)r21 e−4r1/a

−2(a

)2r2e−4r2/a

∣∣∣∣r10

)2 ∫ r1

0e−4r2/a dr2

= −(a

)r21 e−4r1/a − 2

)2r1e−4r1/a

− 2(a

)3e−4r1/a + 2

Iblue =

)r2e−4r2/a

∣∣∣∞r1

)∫ ∞r1

e−4r2/a dr2

)r1e−4r1/a

)2e−4r1/a

I2 =πa3

)e−4r1/a

The r2 integral

I2 = 4π

∫ r1

0r22 e

−4r2/a dr2 +

∫ ∞r1

r2 e−4r2/a dr2

Ired = −(a

)r22 e−4r2/a

∣∣∣r10

)∫ r1

0r2e−4r2/a dr2

= −(a

)r21 e−4r1/a

−2(a

)2r2e−4r2/a

∣∣∣∣r10

)2 ∫ r1

0e−4r2/a dr2

= −(a

)r21 e−4r1/a − 2

)2r1e−4r1/a

− 2(a

)3e−4r1/a + 2

Iblue =

)r2e−4r2/a

∣∣∣∞r1

)∫ ∞r1

e−4r2/a dr2

)r1e−4r1/a

)2e−4r1/a

I2 =πa3

)e−4r1/a

The r2 integral

I2 = 4π

∫ r1

0r22 e

−4r2/a dr2 +

∫ ∞r1

r2 e−4r2/a dr2

Ired = −(a

)r22 e−4r2/a

∣∣∣r10

)∫ r1

0r2e−4r2/a dr2

= −(a

)r21 e−4r1/a

−2(a

)2r2e−4r2/a

∣∣∣∣r10

)2 ∫ r1

0e−4r2/a dr2

= −(a

)r21 e−4r1/a − 2

)2r1e−4r1/a

− 2(a

)3e−4r1/a + 2

Iblue =

)r2e−4r2/a

∣∣∣∞r1

)∫ ∞r1

e−4r2/a dr2

)r1e−4r1/a

)2e−4r1/a

I2 =πa3

)e−4r1/a

The r2 integral

I2 = 4π

∫ r1

0r22 e

−4r2/a dr2 +

∫ ∞r1

r2 e−4r2/a dr2

Ired = −(a

)r22 e−4r2/a

∣∣∣r10

)∫ r1

0r2e−4r2/a dr2

= −(a

)r21 e−4r1/a

−2(a

)2r2e−4r2/a

∣∣∣∣r10

)2 ∫ r1

0e−4r2/a dr2

= −(a

)r21 e−4r1/a − 2

)2r1e−4r1/a

− 2(a

)3e−4r1/a + 2

Iblue =

)r2e−4r2/a

∣∣∣∞r1

)∫ ∞r1

e−4r2/a dr2

)r1e−4r1/a

)2e−4r1/a

I2 =πa3

)e−4r1/a

The r2 integral

I2 = 4π

∫ r1

0r22 e

−4r2/a dr2 +

∫ ∞r1

r2 e−4r2/a dr2

Ired = −(a

)r22 e−4r2/a

∣∣∣r10

)∫ r1

0r2e−4r2/a dr2

= −(a

)r21 e−4r1/a −2

)2r2e−4r2/a

∣∣∣∣r10

)2 ∫ r1

0e−4r2/a dr2

= −(a

)r21 e−4r1/a − 2

)2r1e−4r1/a

− 2(a

)3e−4r1/a + 2

Iblue =

)r2e−4r2/a

∣∣∣∞r1

)∫ ∞r1

e−4r2/a dr2

)r1e−4r1/a

)2e−4r1/a

I2 =πa3

)e−4r1/a

The r2 integral

I2 = 4π

∫ r1

0r22 e

−4r2/a dr2 +

∫ ∞r1

r2 e−4r2/a dr2

Ired = −(a

)r22 e−4r2/a

∣∣∣r10

)∫ r1

0r2e−4r2/a dr2

= −(a

)r21 e−4r1/a −2

)2r2e−4r2/a

∣∣∣∣r10

)2 ∫ r1

0e−4r2/a dr2

= −(a

)r21 e−4r1/a − 2

)2r1e−4r1/a

− 2(a

)3e−4r1/a + 2

Iblue =

)r2e−4r2/a

∣∣∣∞r1

)∫ ∞r1

e−4r2/a dr2

)r1e−4r1/a

)2e−4r1/a

I2 =πa3

)e−4r1/a

The r2 integral

I2 = 4π

∫ r1

0r22 e

−4r2/a dr2 +

∫ ∞r1

r2 e−4r2/a dr2

Ired = −(a

)r22 e−4r2/a

∣∣∣r10

)∫ r1

0r2e−4r2/a dr2

= −(a

)r21 e−4r1/a −2

)2r2e−4r2/a

∣∣∣∣r10

)2 ∫ r1

0e−4r2/a dr2

= −(a

)r21 e−4r1/a − 2

)2r1e−4r1/a

− 2(a

)3e−4r1/a + 2

)3Iblue =

)r2e−4r2/a

∣∣∣∞r1

)∫ ∞r1

e−4r2/a dr2

)r1e−4r1/a

)2e−4r1/a

I2 =πa3

)e−4r1/a

The r2 integral

I2 = 4π

∫ r1

0r22 e

−4r2/a dr2 +

∫ ∞r1

r2 e−4r2/a dr2

Ired = −(a

)r22 e−4r2/a

∣∣∣r10

)∫ r1

0r2e−4r2/a dr2

= −(a

)r21 e−4r1/a −2

)2r2e−4r2/a

∣∣∣∣r10

)2 ∫ r1

0e−4r2/a dr2

= −(a

)r21 e−4r1/a − 2

)2r1e−4r1/a − 2

)3e−4r1/a + 2

Iblue =

)r2e−4r2/a

∣∣∣∞r1

)∫ ∞r1

e−4r2/a dr2

)r1e−4r1/a

)2e−4r1/a

I2 =πa3

)e−4r1/a

The r2 integral

I2 = 4π

∫ r1

0r22 e

−4r2/a dr2 +

∫ ∞r1

r2 e−4r2/a dr2

Ired = −(a

)r22 e−4r2/a

∣∣∣r10

)∫ r1

0r2e−4r2/a dr2

= −(a

)r21 e−4r1/a −2

)2r2e−4r2/a

∣∣∣∣r10

)2 ∫ r1

0e−4r2/a dr2

= −(a

)r21 e−4r1/a − 2

)2r1e−4r1/a − 2

)3e−4r1/a + 2

)3Iblue =

)r2e−4r2/a

∣∣∣∞r1

)∫ ∞r1

e−4r2/a dr2

)r1e−4r1/a

)2e−4r1/a

I2 =πa3

)e−4r1/a

The r2 integral

I2 = 4π

∫ r1

0r22 e

−4r2/a dr2 +

∫ ∞r1

r2 e−4r2/a dr2

Ired = −(a

)r22 e−4r2/a

∣∣∣r10

)∫ r1

0r2e−4r2/a dr2

= −(a

)r21 e−4r1/a −2

)2r2e−4r2/a

∣∣∣∣r10

)2 ∫ r1

0e−4r2/a dr2

= −(a

)r21 e−4r1/a − 2

)2r1e−4r1/a − 2

)3e−4r1/a + 2

)3Iblue = −

)r2e−4r2/a

∣∣∣∞r1

)∫ ∞r1

e−4r2/a dr2

)r1e−4r1/a

)2e−4r1/a

I2 =πa3

)e−4r1/a

The r2 integral

I2 = 4π

∫ r1

0r22 e

−4r2/a dr2 +

∫ ∞r1

r2 e−4r2/a dr2

Ired = −(a

)r22 e−4r2/a

∣∣∣r10

)∫ r1

0r2e−4r2/a dr2

= −(a

)r21 e−4r1/a −2

)2r2e−4r2/a

∣∣∣∣r10

)2 ∫ r1

0e−4r2/a dr2

= −(a

)r21 e−4r1/a − 2

)2r1e−4r1/a − 2

)3e−4r1/a + 2

)3Iblue = −

)r2e−4r2/a

∣∣∣∞r1

)∫ ∞r1

e−4r2/a dr2

)r1e−4r1/a

)2e−4r1/a

I2 =πa3

)e−4r1/a

The r2 integral

I2 = 4π

∫ r1

0r22 e

−4r2/a dr2 +

∫ ∞r1

r2 e−4r2/a dr2

Ired = −(a

)r22 e−4r2/a

∣∣∣r10

)∫ r1

0r2e−4r2/a dr2

= −(a

)r21 e−4r1/a −2

)2r2e−4r2/a

∣∣∣∣r10

)2 ∫ r1

0e−4r2/a dr2

= −(a

)r21 e−4r1/a − 2

)2r1e−4r1/a − 2

)3e−4r1/a + 2

)3Iblue = −

)r2e−4r2/a

∣∣∣∞r1

)∫ ∞r1

e−4r2/a dr2

)r1e−4r1/a

)2e−4r1/a

I2 =πa3

)e−4r1/a

The r2 integral

I2 = 4π

∫ r1

0r22 e

−4r2/a dr2 +

∫ ∞r1

r2 e−4r2/a dr2

Ired = −(a

)r22 e−4r2/a

∣∣∣r10

)∫ r1

0r2e−4r2/a dr2

= −(a

)r21 e−4r1/a −2

)2r2e−4r2/a

∣∣∣∣r10

)2 ∫ r1

0e−4r2/a dr2

= −(a

)r21 e−4r1/a − 2

)2r1e−4r1/a − 2

)3e−4r1/a + 2

)3Iblue = −

)r2e−4r2/a

∣∣∣∞r1

)∫ ∞r1

e−4r2/a dr2

)r1e−4r1/a +

)2e−4r1/a

I2 =πa3

)e−4r1/a

The r2 integral

I2 = 4π

∫ r1

0r22 e

−4r2/a dr2 +

∫ ∞r1

r2 e−4r2/a dr2

Ired = −(a

)r22 e−4r2/a

∣∣∣r10

)∫ r1

0r2e−4r2/a dr2

= −(a

)r21 e−4r1/a −2

)2r2e−4r2/a

∣∣∣∣r10

)2 ∫ r1

0e−4r2/a dr2

= −(a

)r21 e−4r1/a − 2

)2r1e−4r1/a − 2

)3e−4r1/a + 2

)3Iblue = −

)r2e−4r2/a

∣∣∣∞r1

)∫ ∞r1

e−4r2/a dr2

)r1e−4r1/a +

)2e−4r1/a

I2 =πa3

)e−4r1/a

]C. Segre (IIT) PHYS 406 - Spring 2017 February 02, 2017 17 / 18

The r1 integral

The expression for 〈Vee〉 becomes

〈Vee〉 =

4πε0

)∫ [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〉 =

πε0a3

)∫ ∞0

[re−4r/a −

)e−8r/a

πε0a3

)2−(a

)2− 4

πε0a3

16− a2

64− a2

πε0a3

8− 2− 1

8πε0a

)= −5

2E1 = +34eV

The r1 integral

〈Vee〉 =

4πε0

)∫ [1−(

1+2r1a

)e−4r1/a

〈Vee〉 =

πε0a3

)∫ ∞0

[re−4r/a −

)e−8r/a

πε0a3

)2−(a

)2− 4

πε0a3

16− a2

64− a2

πε0a3

8− 2− 1

8πε0a

)= −5

2E1 = +34eV

The r1 integral

〈Vee〉 =

4πε0

)∫ [1−(

1+2r1a

)e−4r1/a

〈Vee〉 =

πε0a3

)∫ ∞0

[re−4r/a −

)e−8r/a

πε0a3

)2−(a

)2− 4

πε0a3

16− a2

64− a2

πε0a3

8− 2− 1

8πε0a

)= −5

2E1 = +34eV

The r1 integral

〈Vee〉 =

4πε0

)∫ [1−(

1+2r1a

)e−4r1/a

〈Vee〉 =

πε0a3

)∫ ∞0

[re−4r/a −

)e−8r/a

πε0a3

)2−(a

)2− 4

πε0a3

16− a2

64− a2

πε0a3

8− 2− 1

8πε0a

)= −5

2E1 = +34eV

The r1 integral

〈Vee〉 =

4πε0

)∫ [1−(

1+2r1a

)e−4r1/a

〈Vee〉 =

πε0a3

)∫ ∞0

[re−4r/a −

)e−8r/a

πε0a3

)2−(a

)2− 4

πε0a3

16− a2

64− a2

πε0a3

8− 2− 1

8πε0a

)= −5

2E1 = +34eV

The r1 integral

〈Vee〉 =

4πε0

)∫ [1−(

1+2r1a

)e−4r1/a

〈Vee〉 =

πε0a3

)∫ ∞0

[re−4r/a −

)e−8r/a

πε0a3

)2−(a

)2− 4

πε0a3

16− a2

64− a2

πε0a3

8− 2− 1

8πε0a

)= −5

2E1 = +34eV

The r1 integral

〈Vee〉 =

4πε0

)∫ [1−(

1+2r1a

)e−4r1/a

〈Vee〉 =

πε0a3

)∫ ∞0

[re−4r/a −

)e−8r/a

πε0a3

)2−(a

)2− 4

πε0a3

16− a2

64− a2

πε0a3

8− 2− 1

8πε0a

)= −5

2E1 = +34eV

The r1 integral

〈Vee〉 =

4πε0

)∫ [1−(

1+2r1a

)e−4r1/a

〈Vee〉 =

πε0a3

)∫ ∞0

[re−4r/a −

)e−8r/a

πε0a3

)2−(a

)2− 4

πε0a3

16− a2

64− a2

πε0a3

8− 2− 1

8πε0a

= −5

2E1 = +34eV

The r1 integral

〈Vee〉 =

4πε0

)∫ [1−(

1+2r1a

)e−4r1/a

〈Vee〉 =

πε0a3

)∫ ∞0

[re−4r/a −

)e−8r/a

πε0a3

)2−(a

)2− 4

πε0a3

16− a2

64− a2

πε0a3

8− 2− 1

8πε0a

)= −5

2E1 = +34eV

today’s outline - february 02, 2017csrri.iit.edu/~segre/phys406/17s/lecture_07.pdf · today’s...

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