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, 2013 at 20:59

1 Answer 1


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, 2013 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, 2013 at 12:23
  • $\begingroup$ you're not the only one... $\endgroup$
    – TMOTTM
    Mar 5, 2014 at 13:19

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