The HF-LCAO Model - Molecular Modelling - page 259

16.3.1 The HF-LCAO Equations

For the record, since I will need to refer to the HF Hamiltonian many times, here they are
(for the closed shell system as shown in Figure 16.1, and assuming real basis functions):

χ i ( r 1 ) h (1) ( r 1 ) χ j ( r 1 ) dτ

h ij =

P kl χ i ( r 1 )χ j ( r 1 )

n n

+ g ( r 1 , r 2 ) χ k ( r 2 ) χ l ( r 2 ) dτ 1 dτ 2 ˆ

(16.19)

k = 1 l = 1

P kl χ i ( r 1 )χ k ( r 1 )

n n

1
2

− g ( r 1 , r 2 ) χ j ( r 2 ) χ l ( r 2 ) dτ 1 dτ 2 ˆ

k = 1 l = 1

Over the years, many workers have devised iterative procedures for solving the problem.
The simplest procedure is as follows.

• Choose a molecular geometry.
• Choose a suitable basis set.
• Calculate all integrals over the basis functions and store them.
• Make an educated guess at the HF-LCAO coefficients U occ .
• Calculate P and h F (the time-consuming step).
• Solve the matrix eigenvalue problem to give the new U occ .
• Check for convergence (test ε el and/or P ).
• Exit or go back three steps.

There are other procedures. I will give you a numerical example later, and explain some
of the methods that people use in order to speed up their calculation. Naturally, having
progressed this far in the text, you will know that the HF-LCAO calculation simply gives
one point on a molecular potential energy surface, as defined by the Born-Oppenheimer
approximation. If your interest in life is molecular geometry optimization then you will
have to follow the same kind of procedures as with MM in Chapter 5; there is a very big
difference in that MM energies can be calculated in almost no time at all, whilst HF-LCAO
energies consume considerable resource.

16.4 Electronic Energy

As noted in Equation (16.4), the HF-LCAO electronic energy is given by

1
2 Tr ( PG )

ε el = Tr ( Ph 1 ) +

Tr Ph F −

1
2 Tr ( PG )

=

The HF-LCAOmatrix eigenvalue equation is h F c ε Sc , and the lowest energy m solutions

determine the electronic ground state of a closed shell molecule. The sum of orbital energies
ε orb is therefore

=

ε orb = 2 (ε A + ε B +···+ ε M )