Numerical Methods in Quantum Mechanics - Dipartimento di Fisica

course a non-local operator which depends upon the orbitals φ k . Let us look
now for a solution under the form of an expansion on a basis of functions:
φ k (r) = ∑ i c(k) i b i (r). We find the Rothaan-Hartree-Fock equations:
F c (k) = ɛ k Sc (k) (7.22)
where c (k) = (c (k)
1 , c(k) 2 , . . . , c(k) N
) is the vector of the expansion coefficients, S is
the superposition matrix, F is the matrix of the Fock operator on the basis set
functions:
F ij = 〈b i |F|b j 〉, S ij = 〈b i |b j 〉. (7.23)
that after some algebra can be written as
⎛
F ij = f ij + ∑ N/2
∑ ∑
⎝2
l m k=1
c (k)∗
l
⎞
c (k) ⎠
m
(
g iljm − 1 )
2 g ijlm , (7.24)
where, with the notations introduced in this chapter:
∫
f ij = b ∗ i (r 1 )f 1 b j (r 1 )d 3 r 1 , (7.25)
∫
g iljm = b ∗ i (r 1 )b j (r 1 )g 12 b ∗ l (r 2 )b m (r 2 )d 3 r 1 d 3 r 2 . (7.26)
The sum over states between parentheses in Eq.(7.24) is called density matrix.
The two terms in the second parentheses come respectively from the Hartree
and the exchange potentials.
The problem of Eq.(7.22) is more complex than a normal secular problem
solvable by diagonalization, since the Fock matrix, Eq.(7.24), depends upon its
own eigenvectors. It is however possible to reconduct the solution to a selfconsistent
procedure, in which at each step a fixed matrix is diagonalized (or,
for a non-orthonormal basis, a generalized diagonalization is performed at each
step).
Code helium hf gauss.f90 2 (or helium hf gauss.c 3 ) solves Hartree-Fock
equations for the ground state of He atom, using a basis set of S Gaussians. The
basic ingredients are the same as in code hydrogen gauss (for the calculation
of single-electron matrix elements and for matrix diagonalization). Moreover
we need an expression for the g iljm matrix elements introduced in Eq.(7.26).
Using the properties of products of Gaussians, Eq.(5.16), these can be written
in terms of the integral
∫
I =
e −αr2 1 e
−βr 2 2 1
r 12
d 3 r 1 d 3 r 2 . (7.27)
Let us look for a variable change that makes (r 1 −r 2 ) 2 to appear in the exponent
of the gaussians:
[
αr1 2 + βr2 2 = γ (r 1 − r 2 ) 2 + (ar 1 + br 2 ) 2] (7.28)
⎡
⎛ √ ⎞
αβ ⎢
= ⎣(r 1 − r 2 ) 2 + ⎝√ α
α + β
β r β
1 +
α r 2⎠2 ⎤ ⎥ ⎦ . (7.29)
2 http://www.fisica.uniud.it/%7Egiannozz/Corsi/MQ/Software/F90/helium hf gauss.f90
3 http://www.fisica.uniud.it/%7Egiannozz/Corsi/MQ/Software/C/helium hf gauss.c
58