Computational Physics - Department of Physics

15.4 Derivation of the Hartree-Fock equations 507

If the state we act on has spin up, we obtain two terms from the Hartree part,

N

∑

μ= 1

Vμd(ri),

and since the interaction does not depend on spin we end up with a total contribution for
helium
N
∑
μ= 1

Vμd(ri)ψλ(ri) =

(

2

∫

φ 100 ∗ (rj)

1

ri j

φ 100 (rj)drj

)

ψλ(ri),

one from spin up and one from spin down. Since the energy for spin up or spin down is the
same we can then write the action of the Hartree term as

N

∑

μ= 1

Vμd(ri)ψλ(ri) =

(

2

∫

φ 100 ∗ (rj)

1

ri jφ^100 (rj)drj

)

ψ 100 ↑(ri).

(the spin inψ 100 ↑is irrelevant)
What we need to code for helium is then

Φ(ri)u 10 = 2 V 10 d(ri)u 10 (ri) = 2

∫∞

0

|u 10 (rj)|^21

r>

drj)u 10 (ri).

withr>=max(ri,rj). What about the exchange or Fock term

Vμex(ri)ψλ(ri) =

(∫

ψ∗μ(rj)

1

ri j

ψλ(rj)drj

)

ψμ(ri)?

We must be careful here with

Vμex(ri)ψλ(ri) =

(∫

ψ∗μ(rj)

1

ri j

ψλ(rj)drj

)

ψμ(ri),

because the spins ofμandλhave to be the same due to the constraint

〈sμmμs|sλmλs〉=δmμs,mλs.

This means that ifmμs=↑thenmλs=↑and ifmμs=↓thenmλs=↓. That is

Vμex(ri)ψλ(ri) =δmμs,mλs

(∫

ψ∗μ(rj)

1

ri j

ψλ(rj)drj

)

ψμ(ri),

The consequence is that for the 1 s↑(and the same for 1 s↓) state we get only one contribu-
tion from the Fock term, namely

N

∑

μ= 1

Vμex(ri)ψ 100 ↑(ri) =δmμs,↑

(∫

ψμ∗(rj)

1

ri jψ^100 ↑(rj)drj

)

ψμ(ri),

resulting in
N
∑
μ= 1

Vμex(ri)ψ 100 ↑(ri) =

(∫

ψ 100 ∗ ↑(rj)

1

ri jψ^100 ↑(rj)drj

)

ψ 100 ↑(ri).

The final Fock term for helium is then

N

∑

μ= 1

Vμex(ri)ψ 100 ↑(ri) =

(∫

ψ 100 ∗ ↑(rj)

1

ri j

ψ 100 ↑(rj)drj

)

ψ 100 ↑(ri),