Computational Physics - Department of Physics

50815 Many-body approaches to studies of electronic systems: Hartree-Fock theory and Density Functional Theory

which is exactly the same as the Hartree term except for a factor of 2. Else the integral is the
same. We can then write the differential equation
(
−

1

2

d^2

dr^2 +

l(l+ 1 )

2 r^2 −

2

r+Φnl(r)−Fnl(r)

)

unl(r) =enlunl(r).

as (

−

1

2

d^2

dr^2

+

l(l+ 1 )

2 r^2

−

2

r

+ 2 V 10 d(r)

)

u 10 (r)−V 10 ex(r) =e 10 u 10 (r),

or (

−^1

2

d^2

dr^2

−^2

r

+V 10 d(r)

)

u 10 (r) =e 10 u 10 (r),

sincel= 0. The shorthandV 10 ex(r)contains the 1 swave function.
The expression we have obtained are independent of the spin projections and we have
skipped them in the equations.
For beryllium the Slater determinant takes the form

Φ(r 1 ,r 2 ,,r 3 ,r 4 ,α,β,γ,δ) =

√^1

4!

∣∣

∣∣

∣∣

∣∣

ψ 100 ↑(r 1 )ψ 100 ↑(r 2 )ψ 100 ↑(r 3 )ψ 100 ↑(r 4 )

ψ 100 ↓(r 1 )ψ 100 ↓(r 2 )ψ 100 ↓(r 3 )ψ 100 ↓(r 4 )

ψ 200 ↑(r 1 )ψ 200 ↑(r 2 )ψ 200 ↑(r 3 )ψ 200 ↑(r 4 )

ψ 200 ↓(r 1 )ψ 200 ↓(r 2 )ψ 200 ↓(r 3 )ψ 200 ↓(r 4 )

∣∣

∣∣

∣∣

∣∣

,

When we now spell out the Hartree-Fock equations we get two coupled differential equa-
tions, one foru 10 and one foru 20.
The 1 swave function has the same Hartree-Fock contribution as in helium for the 1 sstate,
but the 2 sstate gives two times the Hartree term and one time the Fock term. We get

N

∑

μ= 1

Vμd(ri)ψ 100 ↑(ri) = 2

∫∞

0

drj

(

φ 100 ∗ (rj)

1

ri j

φ 100 (rj)+φ 200 ∗ (rj)

1

ri j

φ 200 (rj)

)

ψ 100 ↑(ri)

= ( 2 V 10 d(ri)+ 2 V 20 d(ri))ψ 100 ↑(ri)

for the Hartree part.
For the Fock term we get (we fix the spin)
N
∑
μ= 1

Vμex(ri)ψ 100 ↑(ri) =

∫∞

0

drjφ 100 ∗ (rj)

1

ri j

φ 100 (rj)ψ 100 ↑(ri)+

∫∞

0

drjφ 200 ∗ (rj)

1

ri j

φ 100 (rj)ψ 200 ↑(ri) =V 10 ex(ri)+V 20 ex(ri).

The first term is the same as we have for the Hartree term with 1 sexcept the factor of two.
The final differential equation is
(
−

1

2

d^2

dr^2

−

4

r

+V 10 d(r)+ 2 V 20 d(r)

)

u 10 (r)−V 20 ex(r) =e 10 u 10 (r).

Note again that theV 20 ex(r)contains the 1 sfunction in the integral, that is

V 20 ex(r) =

∫∞

0

drjφ 200 ∗ (rj)^1

r−rj

φ 100 (rj)ψ 200 ↑(r).

The 2 swave function obtains the following Hartree term (recall that the interaction has no
spin dependence)