D.52 Simplification of the Hartree-Fock energy

This note derives the expectation energy for a wave function given by a single Slater determinant.

First note that if you multiply out a Slater determinant

you are going to get terms, or Hartree products if you want, of the form

where the numbers

To make the above more concrete, consider the example of a Slater
determinant of three single-electron functions. It writes out to,
taking

The first two rows in the expansion cover the possibility that

Next, recall that the Hamiltonian consists of single-electron
Hamiltonians

Note that overlines will be used to distinguish the wave function in the right hand side of the inner product from the one in the left hand side. Also note that to take this inner product, you have to integrate over

But multiple integrals, and sums, can be factored into single
integrals, and sums, as long as the integrands and limits only involve
single variables. So you can factor out the inner product as

Now you can start the weeding-out process, because the single-electron
functions are orthonormal. So factors in this product are zero unless
all of the following requirements are met:

Note that

So what is left? Well, with all the

Also, the two signs are equal, because with all the

where for brevity the remaining inner product was called

Next, how many such terms are there for a given electron

You might have guessed that answer from the start. Since the inner product integral is the same for all electrons, the subscripts

The good news is that the reasoning to get the Coulomb and exchange
contributions is pretty much the same. A single electron to electron
repulsion term

You can again split into a product of individual inner products,
except that you cannot split between electrons

For given values of

Next,

The first possibility gives rise to the Coulomb terms, the second to the exchange ones. Note that the former case represents an inner product involving a Hartree product with itself, and the latter case an inner product of a Hartree product with the Hartree product that is the same save for the fact that it has

Consider the Coulomb terms first. For those the two Hartree products
in the inner product are the same, so their signs multiply to one.
Also, their spin states will be the same, so that inner product will
be one too. And as noted there are

with

Again, the

You now need to sum over all pairs of electrons with

The factor

The exchange integrals go exactly the same way; the only differences
are that the Hartree product in the right hand side of the inner
product has the values of

and then the total contribution is

Finally, you can leave the constraint