Computational Physics

4.6 Basis functions 63

of the orbital parts only, read

J ̃(r)φ(r)= 2

∑

l

∫

d^3 r′φ∗l(r′)φl(r′)

1

|r′−r|

φ(r) and

K ̃(r)φ(r)=

∑

l

∫

d^3 r′φl∗(r′)φ(r′)

1

|r′−r|

φl(r).

(4.61)

In contrast withEq. (4.60), the sums overlrun over half the number of electrons
because the spin degrees of freedom have been summed over. The Fock operator
now becomes

F ̃(r)=h(r)+ 2 J ̃(r)−K ̃(r). (4.62)

From now on, we shall only use this spatial form of the Fock operator and drop the
tilde from the operators in (4.61) and (4.62). The corresponding expression for the
energy is found analogously and is given by

Eg= 2

∑

k

〈φk|h|φk〉+

∑

k

( 2 〈φk|J|φk〉−〈φk|K|φk〉). (4.63)

It is possible to solve the Fock equation using a finite basis set, in the same spirit
as the helium calculation of Section 4.3.2. The spin part of the basis functions is
simplyα(s)orβ(s)(spin-up and -down respectively) and the orbital partχp(r)
needs to be specified – this will be done in the next section. For a given basisχp(r),
we obtain the following matrix equation, which is known as theRoothaan equation:

FCk=SCk, (4.64)

similar to(4.51), but nowSis the overlap matrix for theorbitalbasisχp(r)and the
matrixFis given by

Fpq=hpq+

∑

k

∑

rs

Crk∗Csk( 2 〈pr|g|qs〉−〈pr|g|sq〉) (4.65)

where

hpq=〈p|h|q〉=

∫

d^3 rχp∗(r)

[

−

1

2

∇^2 −

∑

n

Zn

|Rn−r|

]

χq(r), (4.66)

and

〈pr|g|qs〉=

∫

d^3 r 1 d^3 r 2 χp∗(r 1 )χr∗(r 2 )

1

|r 1 −r 2 |

χq(r 1 )χs(r 2 ). (4.67)

klabels the orbitalsφkandp,q,randslabel the basis functions. Generally, sums
over labelskandlrun over the occupied orbitals, and sums overp,q,r,srun over
the functions in the basis set.