### 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 of the single-electron states can have values from 1 to , but they must be all different. So there are such terms: there are possibilities among for the number of the single-electron state for electron 1, which leaves remaining possibilities for the number of the single-electron state for electron 2, remaining possibilities for , etcetera. That means a total of terms. As far as the sign of the term is concerned, just don't worry about it. The only thing to remember is that whenever you exchange two values, it changes the sign of the term. It has to be, because exchanging values is equivalent to exchanging electrons, and the complete wave function must change sign under that.

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

The first row in the expansion covers the possibility that 1, with the first term the possibility that 2 and the second term the possibility that 3; note that there are then no choices left for . The second row covers the possibilities in which 2, and the third 3. You see that there are 3! = 6 Hartree product terms total.

Next, recall that the Hamiltonian consists of single-​electron Hamiltonians and electron-pair repulsion potentials . The expectation value of a single electron Hamiltonian will be done first. In forming the inner product , and taking apart into its Hartree product terms as above, you are going to end up with a large number of individual terms that all look like

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 scalar position coordinates, and sum over spin values.

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 does not require for a nonzero value, since the single-electron functions are most definitely not eigenfunctions of the single-electron Hamiltonians, (you would wish things were that easy!) But now remember that the numbers in an individual term are all different. So the numbers include all the numbers that are not equal to . Then so do , , ..., , ,..., because they are the same. And since must be different from all of those, it can only be equal to anyway.

So what is left? Well, with all the values equal to the corresponding values, all the plain inner products are one on account of orthonormality, and the only thing left is:

Also, the two signs are equal, because with all the values equal to the corresponding values, the wave function term in the right side of the inner product is the exact same one as in the left side. So the signs multiply to 1, and you can further factor out the spin inner product, which is one since the spin states are normalized:

where for brevity the remaining inner product was called . Normally you would call it , but an inner product integral does not care what the integration variable is called, so the thing has the same value regardless what the electron is. Only the value of the single-electron function number makes a difference.

Next, how many such terms are there for a given electron and single-electron function number ? Well, for a given value for electron , there are possible values left among for the value of the first of the other electrons, then left for the second of the other electrons, etcetera. So there are a total of such terms. Since 1/, if you sum them all together you get a total contribution from terms in which electron is in state equal to . Summing over the electrons kills off the factor 1 and so you finally get the total energy due to the single-electron Hamiltonians as

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

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 between an electron numbered and another numbered makes a contribution to the expectation energy equal to , and if you multiply out , you get terms of the general form:

You can again split into a product of individual inner products, except that you cannot split between electrons and since involves both electrons in a nontrivial way. Still, you get again that all the other values must be the same as the corresponding values, eliminating those inner products from the expression:

For given values of and , there are equivalent terms, since that is the number of possibilities left for the -values of the other electrons.

Next, and must together be the same pair of numbers as and , since they must be the two numbers left by the set of numbers not equal to and . But that still leaves two possibilities, they can be in the same order or in reversed order:

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 and reversed, or equivalently, electrons and exchanged.

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 equivalent terms for given and , so for each pair of electrons and , and each pair of states and , you get one term

with

Again, the are the same regardless of what and are; they depend only on what and are. So the subscripts and were left out, after setting and .

You now need to sum over all pairs of electrons with and pairs of single-electron function numbers . Since there are a total of electron pairs, it takes out the factor 1/, and you get a contribution to the energy

The factor was added since for every electron pair, you are summing both and , and that counts the same energy twice.

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 reversed, producing a change of sign, and that the inner product of the spins is not trivial. Define

and then the total contribution is

Finally, you can leave the constraint on the sums away since , so they cancel each other.