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

$$\Psi= {\left\vert{\rm det}\;\pe1//b//,\pe2//b//,\pe3//b//,\ldots\right\rangle}$$

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

$$\frac{\pm}{\sqrt{I!}}\; \pe{n_1}/{\skew0\vec r}_1/b/z1/\; ... ...kew0\vec r}_2/b/z2/\; \pe{n_3}/{\skew0\vec r}_3/b/z3/\; \ldots$$

where the num­bers $n_1,n_2,n_3,\ldots$ of the sin­gle-elec­tron states can have val­ues from 1 to $I$, but they must be all dif­fer­ent. So there are $I!$ such terms: there are $I$ pos­si­bil­i­ties among $1,2,3,\ldots,I$ for the num­ber $n_1$ of the sin­gle-elec­tron state for elec­tron 1, which leaves $I-1$ re­main­ing pos­si­bil­i­ties for the num­ber $n_2$ of the sin­gle-elec­tron state for elec­tron 2, $I-2$ re­main­ing pos­si­bil­i­ties for $n_3$, etcetera. That means a to­tal of $I(I-1)(I-2)\ldots2 1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $I!$ 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 $n$ val­ues, it changes the sign of the term. It has to be, be­cause ex­chang­ing $n$ 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 $\sqrt{I!}$ to the other side for con­ve­nience,

$$\begin{eqnarray*} \lefteqn{{\left\vert{\rm det}\;\pe1//b//,\pe2//b//,\pe3//b//\... ..._1/b/z1/ \pe2/{\skew0\vec r}_2/b/z2/ \pe1/{\skew0\vec r}_3/b/z3/ \end{eqnarray*}$$

The first two rows in the ex­pan­sion cover the pos­si­bil­ity that $n_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, with the first one the pos­si­bil­ity that $n_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 and the sec­ond one the pos­si­bil­ity that $n_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3; note that then there are no choices left for $n_3$. Sim­i­larly the sec­ond two rows cover the two pos­si­bil­i­ties that $n_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2, and the third that $n_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3. You see that there are 3! = 6 Hartree prod­uct terms to­tal.

Next, re­call that the Hamil­ton­ian con­sists of sin­gle-​elec­tron Hamil­to­ni­ans $h^{\rm e}_i$ and elec­tron-pair re­pul­sion po­ten­tials $v^{\rm ee}_{i{\underline i}}$. The ex­pec­ta­tion value of a sin­gle elec­tron Hamil­ton­ian $h^{\rm e}_i$ will be done first. In form­ing the in­ner prod­uct $\langle\Psi\vert h^{\rm e}_i\vert\Psi\rangle$, and tak­ing $\Psi$ 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

$$\begin{array}{@{}l@{}} \displaystyle \Big\langle \frac{\... ...\overline n_I}/{\skew0\vec r}_I/b/zI/ \Big\rangle \end{array}$$

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 $3I$ scalar po­si­tion co­or­di­nates, and sum over $I$ 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

$$\begin{array}{l} \displaystyle \frac{\pm}{\strut\sqrt{I!}... ...e{\overline n_I}/{\skew0\vec r}_I/b/zI/\Big\rangle \end{array}$$

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:

$$n_1=\overline n_1,\; n_2=\overline n_2,\; \ldots,\; n_{i-... ...,\; n_{i+1}=\overline n_{i+1},\; \ldots,\; n_I=\overline n_I$$

Note that $\langle\pe{n_i}/{\skew0\vec r}_i/b/zi/\vert h^{\rm e}_i\vert\pe{\overline{n}_i}/{\skew0\vec r}_i/b/zi/\rangle$ does not re­quire $n_i$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\overline{n}_i$ 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 $n_1,n_2,n_3,\ldots$ in an in­di­vid­ual term are all dif­fer­ent. So the num­bers $n_1,n_2,\ldots,n_{i-1},n_{i+1},\ldots$ in­clude all the num­bers that are not equal to $n_i$. Then so do $\overline{n}_1$, $\overline{n}_2$, ..., $\overline{n}_{i-1}$, $\overline{n}_{i+1}$ ,..., be­cause they are the same. And since $\overline{n}_i$ must be dif­fer­ent from all of those, it can only be equal to $n_i$ any­way.

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

$$\frac{\pm}{\sqrt{I!}} \frac{\pm}{\sqrt{I!}} \Big\langle\p... ...h^{\rm e}_i\Big\vert\pe{n_i}/{\skew0\vec r}_i/b/zi/\Big\rangle$$

Also, the two signs are equal, be­cause with all the $\overline{n}$ val­ues equal to the cor­re­spond­ing $n$ 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:

$$\frac{1}{I!} \Big\langle\pe{n_i}/{\skew0\vec r}_i///\Big\v... ...skew0\vec r}_i///\Big\rangle \equiv \frac{1}{I!} E^{\rm e}_n$$

where for brevity the re­main­ing in­ner prod­uct was called $E^{\rm e}_n$. Nor­mally you would call it $E^{\rm e}_{n_ii}$, 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 $i$ is. Only the value of the sin­gle-elec­tron func­tion num­ber $n_i$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n$ makes a dif­fer­ence.

Next, how many such terms are there for a given elec­tron $i$ and sin­gle-elec­tron func­tion num­ber $n$? Well, for a given $n$ value for elec­tron $i$, there are $I-1$ pos­si­ble val­ues left among $1,2,3,\ldots$ for the $n$ value of the first of the other elec­trons, then $I-2$ left for the sec­ond of the other elec­trons, etcetera. So there are a to­tal of $(I{-}1)(I{-}2)\ldots1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(I{-}1)!$ such terms. Since $(I{-}1)!$$\raisebox{.5pt}{$/$}$​$I!$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1/$I$, if you sum them all to­gether you get a to­tal con­tri­bu­tion from terms in which elec­tron $i$ is in state $n$ equal to $E^{\rm e}_n$$\raisebox{.5pt}{$/$}$​$I$. Sum­ming over the $I$ elec­trons kills off the fac­tor 1$\raisebox{.5pt}{$/$}$​$I$ and so you fi­nally get the to­tal en­ergy due to the sin­gle-elec­tron Hamil­to­ni­ans as

$$\sum_{n=1}^I E^{\rm e}_n \qquad E^{\rm e}_n = \Big\langl... ...Big\vert h^{\rm e}\Big\vert\pe{n}/{\skew0\vec r}///\Big\rangle$$

You might have guessed that an­swer from the start. Since the in­ner prod­uct in­te­gral is the same for all elec­trons, the sub­scripts $i$ 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 $v^{\rm ee}_{i{\underline i}}$ be­tween an elec­tron num­bered $i$ and an­other num­bered ${\underline i}$ makes a con­tri­bu­tion to the ex­pec­ta­tion en­ergy equal to $\langle\Psi\vert v^{\rm ee}_{i{\underline i}}\vert\Psi\rangle$, and if you mul­ti­ply out $\Psi$, you get terms of the gen­eral form:

$$\begin{array}{@{}l@{}} \displaystyle \frac{1}{I!\strut} ... ...derline i}/b/z{\underline i}/ \ldots \Big\rangle \end{array}$$

You can again split into a prod­uct of in­di­vid­ual in­ner prod­ucts, ex­cept that you can­not split be­tween elec­trons $i$ and ${\underline i}$ since $v^{\rm ee}_{i{\underline i}}$ in­volves both elec­trons in a non­triv­ial way. Still, you get again that all the other $n$ val­ues must be the same as the cor­re­spond­ing $\overline{n}$ val­ues, elim­i­nat­ing those in­ner prod­ucts from the ex­pres­sion:

$$\frac{1}{I!} \Big\langle \pe{n_i}/{\skew0\vec r}_i/b/zi/ ... .../{\skew0\vec r}_{\underline i}/b/z{\underline i}/ \Big\rangle$$

For given val­ues of $n_i$ and $n_{\underline i}$, there are $(I-2)!$ equiv­a­lent terms, since that is the num­ber of pos­si­bil­i­ties left for the $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\overline{n}$-val­ues of the other $I-2$ elec­trons.

Next, $\overline{n}_i$ and $\overline{n}_{\underline i}$ must to­gether be the same pair of num­bers as $n_i$ and $n_{\underline i}$, since they must be the two num­bers left by the set of num­bers not equal to $n_i$ and $n_{\underline i}$. But that still leaves two pos­si­bil­i­ties, they can be in the same or­der or in re­versed or­der:

$$\overline n_i=n_i,\; \overline n_{\underline i}=n_{\underli... ...erline n_i=n_{\underline i},\; \overline n_{\underline i}=n_i.$$

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 $n_i$ and $n_{\underline i}$ re­versed, or equiv­a­lently, elec­trons $i$ and ${\underline i}$ 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 $(I-2)!$ equiv­a­lent terms for given $n_i$ and $n_{\underline i}$, so for each pair of elec­trons $i$ and ${\underline i}$ $\raisebox{.2pt}{$\ne$}$ $i$, and each pair of states $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_i$ and ${\underline n}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_{\underline i}$, you get one term

$$\frac{1}{I(I-1)} J_{n{\underline n}}$$

with

$$J_{n{\underline n}} \equiv \Big\langle \pe n/{\skew0\vec ... ... \pe{\underline n}/{\underline{\skew0\vec r}}/// \Big\rangle.$$

Again, the $J_{n{\underline n}}$ are the same re­gard­less of what $i$ and ${\underline i}$ are; they de­pend only on what $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_i$ and ${\underline n}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n_{\underline i}$ are. So the sub­scripts $i$ and ${\underline i}$ were left out, af­ter set­ting ${\skew0\vec r}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\skew0\vec r}_i$ and ${\underline{\skew0\vec r}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\skew0\vec r}_{\underline i}$.

You now need to sum over all pairs of elec­trons with $i$ $\raisebox{.2pt}{$\ne$}$ ${\underline i}$ and pairs of sin­gle-elec­tron func­tion num­bers $n$ $\raisebox{.2pt}{$\ne$}$ ${\underline n}$. Since there are a to­tal of $I(I-1)$ elec­tron pairs, it takes out the fac­tor 1/$I(I-1)$, and you get a con­tri­bu­tion to the en­ergy

$${\textstyle\frac{1}{2}} \sum_{n=1}^I\sum_{\textstyle{{\underline n}=1\atop{\underline n}\ne n}}^I J_{n{\underline n}}$$

The fac­tor $\frac12$ was added since for every elec­tron pair, you are sum­ming both $v^{\rm ee}_{i{\underline i}}$ and $v^{\rm ee}_{{\underline i}{i}}$, 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 $\overline{n}_i$ and $\overline{n}_{\underline i}$ re­versed, pro­duc­ing a change of sign, and that the in­ner prod­uct of the spins is not triv­ial. De­fine

$$K_{n{\underline n}} \equiv \Big\langle \pe n/{\skew0\vec ... ...w0\vec r}/// \pe n/{\underline{\skew0\vec r}}/// \Big\rangle.$$

and then the to­tal con­tri­bu­tion is

$$-{\textstyle\frac{1}{2}} \sum_{n=1}^I\sum_{\textstyle{{\und... ...gle{\updownarrow}_n\vert{\updownarrow}_{\underline n}\rangle^2$$

Fi­nally, you can leave the con­straint ${\underline n}$ $\raisebox{.2pt}{$\ne$}$ $n$ on the sums away since $K_{nn}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $J_{nn}$, so they can­cel each other.