### 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 two rows in the ex­pan­sion cover the pos­si­bil­ity that 1, with the first one the pos­si­bil­ity that 2 and the sec­ond one the pos­si­bil­ity that 3; note that then there are no choices left for . Sim­i­larly the sec­ond two rows cover the two pos­si­bil­i­ties that 2, and the third that 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.