Computational Physics

Exercises 85

The matrix to be diagonalised is given by

Fpq=hpq+

∑

rs

CrCsQprqs

with

Qprqs=〈pr|g|qs〉=

∫

d^3 r 1 d^3 r 2 φp(r 1 )φr(r 2 )

1

r 12

φq(r 1 )φs(r 2 ).

As we are dealing with two electrons only, we do not have to sum over the different

orbitalsk.

In Problem 4.8 you have already programmed the expressions for the matrix

elementshpq, so these pose no problems. The two-electron matrix elementsQprqsare

rather complicated; they are given in Section 4.8. The resulting expressions can be

written in the form:

〈χpχr|g|χqχs〉= 2

√

AB

π(A+B)

SpqSsrF 0 (t).

Here,tis defined as

t=

(αp+αq)(αr+αs)

(αp+αq+αr+αs)

|RA−RB|^2 ;

RA=

αpRp+αqRq

αp+αq

RB=

αrRr+αsRs

αr+αs

,

andAandBas
A=αp+αq
B=αr+αs.
Spqis the overlap matrix.
You can now use these matrix elements in a program which has the same structure
as that of the helium atom.
Check: for a distance 1 a.u. between the atoms, one finds for the ground state
vectorC:
(0.092 561 548 6, 0.165 180 118, 0.120 122 665, 0.021 154 565 7,
0.092 561 548 6, 0.165 180 118, 0.120 122 665, 0.021 154 565 7)
and an energy−1.078 547 61 (nuclear repulsion+1 included!).
4.10 In a restricted Hartree–Fock (RHF) calculation using a finite basisχp(r), the kinetic
and Coulomb integrals must be calculated. We can gain speed by using the
symmetry of these matrices, for example
〈χp|∇^2 |χq〉=〈χq|∇^2 |χp〉.
We do not assume other symmetries to be present in the system.