In Szabo and Ostlund's Modern Quantum Chemistry, the procedure of single determinant energy minimization is presented. Omitting the whole procedure, I have a question about functional variation during the derivation of the Hartree-Fock equations.

Given the single determinant $| \Psi_0 \rangle = | \chi_1 \chi_2 \ldots \chi_a \chi_b \ldots \chi_N \rangle$, the energy $E_0 = \langle \Psi_0 | \mathscr{H} | \Psi_0 \rangle$ is a functional of the spin orbitals $\{ \chi_a \}$. $E_0$ is the expectation value of the single determinant $\vert \Psi_0 \rangle$,

$$ E_0[\{ \chi_a \} ] = \sum\limits_{a=1}^N [a\vert h \vert a] + \frac{1}{2} \sum\limits_{a=1}^N \sum\limits_{b=1}^N [aa \vert bb] - [ab \vert ba]. $$

This equation can be varied:

$$ \begin{align*} \delta E_0 &= \sum \limits_{a=1}^N [\delta \chi_a | h | \chi_a] + [\chi_a | h | \delta \chi_a] \\ &+ \frac{1}{2} \sum \limits_{a=1}^N \sum \limits_{b=1}^N [\delta \chi_a \chi_a | \chi_b \chi_b] + [ \chi_a \delta \chi_a | \chi_b \chi_b] + [ \chi_a \chi_a | \delta \chi_b \chi_b] + [\chi_a \chi_a | \chi_b \delta \chi_b] \\ &-\frac{1}{2} \sum \limits_{a=1}^N \sum \limits_{b=1}^N [\delta \chi_a \chi_b | \chi_b \chi_a] + [ \chi_a \delta \chi_b | \chi_b \chi_a] + [ \chi_a \chi_b | \delta \chi_b \chi_a] + [\chi_a \chi_b | \chi_b \delta \chi_a] \label{1}\tag{1} \end{align*} $$

Authors suggest to the reader as an exercise to manipulate this equation for $\delta E_0$ to show that

$$ \delta E_0 = \sum \limits_{a=1}^N [\delta \chi_a | h |\chi_a] + \sum \limits_{a=1}^N \sum \limits_{b=1}^N [\delta \chi_a \chi_a | \chi_b \chi_b] - [\delta \chi_a \chi_b | \chi_b \chi_a] + \text{c.c.} \label{2}\tag{2} $$

It is clear that the first sum in $\eqref{1}$ can be easily converted to that in $\eqref{2}$ because

$$ [\delta \chi_a | h | \chi_a]^* = [\chi_a | h | \delta \chi_a] $$

Analogously, for the second sum in $\eqref{1}$ one can show that

$$ [\delta \chi_a \chi_a | \chi_b \chi_b]^* = [\chi_a \delta \chi_a | \chi_b \chi_b] $$


$$ [\chi_a \chi_a | \delta \chi_b \chi_b]^* = [\chi_a \chi_a | \chi_b \delta\chi_b]. $$

I did the same manipulations with the third sum in $\eqref{1}$ and obtained

$$ \begin{align*} \delta E_0 &= \sum \limits_{a=1}^N [\delta \chi_a | h | \chi_a] \\ &+ \frac{1}{2} \sum \limits_{a=1}^N \sum \limits_{b=1}^N [\delta \chi_a \chi_a | \chi_b \chi_b] + [ \chi_a \chi_a | \delta \chi_b \chi_b] \\ &- \frac{1}{2} \sum \limits_{a=1}^N \sum \limits_{b=1}^N [\delta \chi_a \chi_b | \chi_b \chi_a] + [ \chi_a \delta \chi_b | \chi_b \chi_a] + \text{c.c.} \label{3}\tag{3} \end{align*} $$

So, comparing $\eqref{3}$ and $\eqref{2}$ I suppose that one should show that both terms in each sum are equal, i.e.

$$ [\delta \chi_a \chi_a | \chi_b \chi_b] = [ \chi_a \chi_a | \delta \chi_b \chi_b] $$


$$ [\delta \chi_a \chi_b | \chi_b \chi_a] = [ \chi_a \delta \chi_b | \chi_b \chi_a]. $$

So, how to do that?

  • $\begingroup$ I'd try integration-by-parts. $\endgroup$ Nov 29 '13 at 20:59

Is it not just because of the fact that $[aa|bb]$ is in chemists notation, meaning that $[aa|bb] = [a(1)a(1)|b(2)b(2)]$ and so you can just switch them around?

  • $\begingroup$ Thanks. I've already solved the problem. The idea is to change indices in a double sum. ∑a,b[δχaχa|χbχb]=∑a,b[δχbχb|χaχa]=∑a,b[χaχa|δχbχb]. I asked this question at physics.stackexchange.com physics.stackexchange.com/questions/80718/… $\endgroup$ Dec 5 '13 at 11:22
  • $\begingroup$ Just once I'd like to see a full description of HF that kept consistent, fully explicit notation all the way through. $\endgroup$
    – Aesin
    Dec 23 '13 at 12:23
  • $\begingroup$ you're not the only one... $\endgroup$
    – TMOTTM
    Mar 5 '14 at 13:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.