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

$$\Psi= \frac{1}{\sqrt{I!}} \Big\vert{\rm det}(\pe1//b//,\pe2//b//,\pe3//b//,\ldots)\Big\rangle$$

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

$$\frac{\pm}{\sqrt{I!}}\; \pe{n_1}/{\skew0\vec r}_1/b/z1/\... ...ew0\vec r}_2/b/z2/\; \pe{n_3}/{\skew0\vec r}_3/b/z3/\; \ldots$$

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,

$$\begin{eqnarray*} \lefteqn{\sqrt{3!}\;\Big\vert{\rm det}(\pe1//b//,\pe2//b//,\... ...1/b/z1/ \pe2/{\skew0\vec r}_2/b/z2/ \pe1/{\skew0\vec r}_3/b/z3/ \end{eqnarray*}$$

The first row in the expansion covers the possi­bility that $n_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, with the first term the possi­bility that $n_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 and the second term the possi­bility that $n_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3; note that there are then no choices left for $n_3$. The second row covers the possi­bilities in which $n_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2, and the third $n_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ 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

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

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

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

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:

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

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:

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

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:

$$\frac{1}{I!} \Big\langle\pe{n_i}/{\skew0\vec r}_i///\Big... ...w0\vec r}_i///\Big\rangle \equiv \frac{1}{I!} E^{\rm e}_n$$

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

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

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:

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

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:

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

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:

$$\overline n_i=n_i,\; \overline n_{\underline i}=n_{\underl... ...rline n_i=n_{\underline i},\; \overline n_{\underline i}=n_i.$$

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

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

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

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

and then the total contribution is

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

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.