Computational Physics

50 The Hartree–Fock method

overr 1 leads to
∑

pq

(

hpq+

∑

rs

CrCsQprqs

)

Cq=E′

∑

pq

SpqCq (4.14)

with

hpq=

〈

χp

∣∣

∣

∣−

1

2

∇^2 −

2

r

∣∣

∣

∣χq

〉

; (4.15a)

Qprqs=

∫

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 ) and (4.15b)

Spq=〈χp|χq〉. (4.15c)

Unfortunately,(4.14)is not a generalised eigenvalue equation because of the pres-
ence of the variablesCrandCsbetween the brackets on the left hand side. However,
if we carry out the self-consistency iteration process as indicated in the previous
section, theCrandCsare keptfixed, and finding theCqin(4.14)reduces to solving
a generalised eigenvalue equation. We then replaceCr,Csby the solution found
and start the same procedure again.
The matrix elements from (4.15) remain to be found. We shall use Gaussian
l=0 basis functions (s-functions), just as in the case of the hydrogen atom (see
Section 3.2.2). Of course, the optimal exponentsαpoccurring in the Gaussian
s-basis functionsχp,

χp(r)=e−αpr

2

, (4.16)

are different from those of the hydrogen atom. Again, rather than solve the non-
linear variational problem, which involves not only the prefactorsCpbut also the
exponentsαpas parameters of the wave function, we shall take the optimal values
calculated from a different program which we do not go into here. They are

α 1 =0.298 073

α 2 =1.242 567

α 2 =5.782 948,

α 4 =38.474 970.

The matrix elements of the kinetic and the Coulomb energy are similar to those
calculated for the hydrogen atom (see Eq. (3.29)), except for an extra factor of 2 in
the nuclear attraction (due to the nuclear charge). In Section 4.8, the matrix element
Qprqswill be calculated; the result is given by

Qprqs=

2 π^5 /^2

(αp+αq)(αr+αs)

√

αp+αq+αr+αs

. (4.17)