Colby College

Variation Method- Helium If you guess a solution to the Schrödinger Equation, how do you find out how good a guess you have made? H

^ Ψi = Ei Ψi

can’t be solved exactly:

– h-2

2m ( )∇21 + ∇2

2 Ψi + 1

4πεo

–

Ze2

r1 –

Ze2

r2 +

e2

r12 Ψi = EiΨi

but you have : φ1 guess φ2 better guess φ3 better yet Eφ ≥ Ei

The Variation Theorem guarantees that the trial energy is always greater than the exact energy.

H

^ Ψi = Ei Ψi

∫ Ψ*i H

^ Ψi dτ = ∫ Ψ*

i Ei Ψi dτ

solve for Ei: Ei = ∫ Ψ*

i H ^ Ψi dτ

∫ Ψ*i Ψi dτ

is exact

Eφ = ∫ φ*H

^ φ dτ

∫ φ*φ dτ

Eφ ≥ Ei by the variation theorem

φgs = φ1s(r1) φ1s(r2) = 1π

Zeff

ao

3 e–Zeff r1/ao e–Zeff r2/ao Zeff : variation parameter

Eφ = ∫ φ*

gs H ^ φgs dτ

∫ φ*gs φgs dτ

= e2

4πεo ao

Z2

eff – 278 Zeff see Karplus and Porter 4.1.5

minimize by changing Zeff: dEφ

dZeff = 0 =

e2

4πεo ao (2 Zeff – 27/8)

Zeff = 27/16 = 1.6875

Egs = – 13.6 eV(Z2eff /n21 + Z2

eff /n22) = -77.5 eV (exp. -79.0 eV)

φ1s(r1) and φ1s(r2) are normalized eigenfunctions of one-electron Hamiltonians with charge Zeff :

– h-2

2m∇21 –

Zeff e2

4πεor1 φ1s(r1) = E1 φ1s(r1)

– h-2

2m∇22 –

Zeff e2

4πεor2 φ1s(r2) = E2 φ1s(r2)

normalization: ∫ φ*1s(r1) φ1s(r1) dτ1 = 1 ∫ φ*

1s(r2) φ1s(r2) dτ2 = 1

and eigenvalues: Ε1 = –½

e2

4πεo ao Z2

eff = -13.6 eV Z2eff Ε2 = –½

e2

4πεo ao Z2

eff = -13.6 eV Z2eff

E

Ei – exact

Eφ1

Eφ2

Eφ3

Colby College

The exact Hamiltonian uses the full nuclear charge Z = 2. We can split the Coulomb portion of the exact Hamiltonian using Z = Zeff + (Z – Zeff)

– h-2

2m∇21 –

Ze2

4πεor1 φ1s(r1) =

– h-2

2m∇21 –

Zeff e2

4πεor1 φ1s(r1) –

(Z – Zeff) e2

4πεor1 φ1s(r1)

– h-2

2m∇22 –

Ze2

4πεor2 φ1s(r2) =

– h-2

2m∇22 –

Zeff e2

4πεor2 φ1s(r2) –

(Z – Zeff) e2

4πεor2 φ1s(r2)

Both electrons are in the same 1s-orbital so these two previous equations give the same energies. The electrons are identical, except for spin. To calculate the expectation value of the exact Hamiltonian,

∫ φ*1s(r1) φ*

1s(r2) H ^ φ1s(r1) φ1s(r2) dτ, we find H

^ φ1s(r1) φ1s(r2) using the last two equations:

H

^ φgs = φ1s(r2)

– h-2

2m∇21 –

Ze2

4πεor1 φ1s(r1) + φ1s(r1)

– h-2

2m∇22 –

Ze2

4πεor2 φ1s(r2) +

e2

4πεor12 φ1s(r1)φ1s(r2)

= –

e2

4πεo ao Z2

eff φ1s(r1) φ1s(r2) – 2 (Z – Zeff) e

2

4πεor1 φ1s(r1) φ1s(r2) +

e2

4πεor12 φ1s(r1)φ1s(r2)

The variational energy for our guessed wave function is then:

Eφ = ∫ φ*1s(r1) φ*

1s(r2) H ^ φ1s(r1) φ1s(r2) dτ = –

e2

4πεo ao Z2

eff – 2⟨ (Z – Zeff) e

2

4πεor1 ⟩ + ⟨

e2

4πεor12 ⟩

The first expectation value is the difference in the Coulomb attraction of an electron with a nucleus of charge Z and a nucleus of charge Zeff:

–⟨ (Z – Zeff) e

2

4πεor1 ⟩ = – (Z – Zeff) ∫ φ*

1s(r1) φ*1s(r2)

e2

4πεo r1 φ1s(r1) φ1s(r2) dτ1dτ2

= – (Z – Zeff) ∫ φ*1s(r2) φ1s(r2) dτ2 ∫ φ*

1s(r1) e2

4πεo r1 φ1s(r1) dτ1

normalization ⟨V⟩

The first integral is the normalization for electron-2. The second integral is the expectation value of the potential energy for a one-electron atom with nuclear charge Zeff, ⟨V⟩, see Problem 25.5.

–⟨ (Z – Zeff) e

2

4πεor1 ⟩ = – (Z – Zeff)

Zeff e

2

4πεo ao

We evaluated the expectation value of the electron-electron repulsion for our perturbation treatment of the helium atom. We only need to substitute Zeff for Z:

⟨ e2

4πεor12 ⟩ = ∫ φ*

1s(r1) φ*1s(r2)

e2

4πεor12 φ1s(r1) φ1s(r2) dτ1dτ2 =

5Zeff

8

e2

4πεo ao

Setting Z = 2 gives the final result:

Eφ =

e2

4πεo ao( – Z2

eff – 2 ZZeff + 2 Z2eff +

5Zeff

8 )

Eφ = e2

4πεo ao

Z2

eff – 278 Zeff = 27.211 eV

Z2

eff – 278 Zeff