86 The Hartree–Fock method
(a) Suppose the basis containsMbasis functions, at least how many of these matrix
elements must be calculated?
(b) How many two-electron matrix elements
〈pr|g|qs〉
must be calculated?
Now suppose that the molecule for which we are performing the calculation
consists of three identical atoms, located on an equilateral triangle:
On every atom, we haveMbasis functions which we denote byχpA(r)etc.,
p=1,...,M. The basis functions on the different atoms have the same form:
χpA(r)=χp(r−RA),
χpB(r)=χp(r−RB)
and similarly forC.(c) How many different kinetic and Coulomb matrix elements must be calculated in
this case?
(d) How many different two-electron matrix elements must be calculated?
(e) Suppose now thatMis very large. What is then the gain in speed when using all
symmetries of the matrix elements instead of using no symmetry at all?4.11 Suppose a Hartree–Fock calculation is carried out for a linear chain ofKidentical
atoms andNelectrons, whereKis a large number. The distance between two
successive atoms isa. For each atom, the same set ofMbasis functions is used. We
assume that the overlap between two wave functions centred around two atoms at a
distance larger thanpa(pis some integer) vanishes.
(a) How many elements of the Hamilton matrix are nonzero (for largeK)?
(b) How many nonzero two-electron matrix elements〈pr|g|qs〉do we have (for
largeK)?
4.12 [C] In this problem, we modify the hydrogen calculation as carried out in
Problem 4.9 to a Hartree–Fock calculation – remember that in the previous version
we solved the Hartree equation and not the Hartree–Fock equation. Furthermore, we
consider exploiting the symmetry in order to speed up the calculation. As we have