Computational Physics

74 The Hartree–Fock method

The kinetic integral:This is given by

〈1s,α,A|−∇^2 |1s,β,B〉=

∫

d^3 r∇e−α(r−RA)

2

∇e−β(r−RB)

2

(4.101)

where we have used Green’s theorem. Working out the gradients and using the
Gaussian product theorem, we arrive at

〈1s,α,A|−∇^2 |1s,β,B〉= 4 αβ

∫

d^3 r(r−RA)(r−RB)Ke−γ(r−RP)

2

(4.102)

withγ,KandRPgiven above. Substitutingu=r−RPand using the fact that
integrals antisymmetric inuvanish, we arrive at:

〈1s,α,A|−∇^2 |1s,β,B〉

= 16 αβKπ

[∫∞

0

duu^4 e−γu+(RP−RA)·(RP−RB)

∫∞

0

duu^2 e−γu

]

=

αβ

α+β

[

6 − 4

αβ

α+β

|RA−RB|^2

]

×

(

π

α+β

) 3 / 2

exp[−αβ/(α+β)|RA−RB|^2 ]. (4.103)

The nuclear attraction integral:This integral is more difficult than the previous
ones, because after applying the Gaussian product theorem we are still left with the
1 /rCoulomb term of the nucleus (whose position does in general not coincide with
the centre of the orbital). To reduce the integral to a simpler form, we use Fourier
transforms:
ˆf(k)=

∫

d^3 rf(r)e−ik·r. (4.104)

The inverse transformation is given by

f(r)=( 2 π)−^3

∫

d^3 kfˆ(k)eik·r. (4.105)

The Dirac delta-function can be written as

δ(r)=( 2 π)−^3

∫

d^3 keik·r. (4.106)

The Coulomb integral is given by

〈1s,α,A|−Z/rC|1s,β,B〉=−Z

∫

d^3 rKe−γ|r−RP|

2

|r−RC|−^1. (4.107)

The Fourier transform of 1/ris 4π/k^2 , as can be seen for example by Fourier
transforming the Poisson equation

−∇^2

1

r

= 4 πδ(r). (4.108)