I asked this question at chemistry.stackexchange.com, but the attendance of that source is a little bit lower than here.

I would like to ask a question about mathematical derivation of the HF equations. Some moments related to the functional variation are not clear for me. I read a book of Szabo and Ostlund "Modern quantum chemistry". There the procedure of the Single determinant energy minimization is presented. Omitting the whole procedure i present the most obscure moment. 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 $\vert \Psi_0 \rangle$,

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

Here $[ij|kl]$ is the chemist's notation for a two-electron integral involving spin orbitals:

$[ij|kl] = [\chi_i \chi_j | \chi_k \chi_l] = \int d\mathbf{x_1} d\mathbf{x_2} \chi_i^{*}(\mathbf{x_1}) \chi_j(\mathbf{x_1}) r_{12}^{-1} \chi_k^{*}(\mathbf{x_2}) \chi_l(\mathbf{x_2}) $

And one-electron integral is

$[i|h|j] = \int d \mathbf{x_1} \chi_i^{*}(\mathbf{x_1}) h(\mathbf{r_1}) \chi_j(\mathbf{x_1})$

This equation can be varied

(1) $\delta E_0 = \sum \limits_{a=1}^N [\delta \chi_a | h | \chi_a] + [\chi_a | h | \delta \chi_a] $

$+ \frac{1}{2} \sum \limits_{a=1}^N \sum \limits_{b=1}^N [\delta \chi_a \chi_a | \chi_b \chi_b] + [ \chi_a \delta \chi_a | \chi_b \chi_b] + [ \chi_a \chi_a | \delta \chi_b \chi_b] + [\chi_a \chi_a | \chi_b \delta \chi_b]$

$-\frac{1}{2} \sum \limits_{a=1}^N \sum \limits_{b=1}^N [\delta \chi_a \chi_b | \chi_b \chi_a] + [ \chi_a \delta \chi_b | \chi_b \chi_a] + [ \chi_a \chi_b | \delta \chi_b \chi_a] + [\chi_a \chi_b | \chi_b \delta \chi_a]$

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

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

It is clear that the first sum in (1) can be easily converted to that in (2) because

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

Analogously for the second sum in (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 (1) and obtained

(3) $\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] $

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

So, comparing (3) and (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?

  • 1
    $\begingroup$ I think that you have to show that the sum over $a$ and $b$ of these quantities is equal. For instance, if $a=b$, the change of variable $x_1\to x_2$ and $x_2\to x_1$ does the job. I think that the sum of $a=1$, $b=2$ plus $a=2$, $b=1$ gives two time $a=1$, $b=2$, etc. $\endgroup$ – Adam Oct 14 '13 at 13:44
  • $\begingroup$ Crossposted from: chemistry.stackexchange.com/q/6475 $\endgroup$ – Qmechanic Oct 14 '13 at 14:02
  • $\begingroup$ @Adam, change of variables only gives $[\delta \chi_a \chi_a | \chi_b \chi_b] = [ \chi_b \chi_b | \delta \chi_a \chi_a ]$. So the transition from $\delta \chi_a$ to $\delta \chi_b$ is not clear. $\endgroup$ – jacksonslsmg4 Oct 14 '13 at 14:12
  • 1
    $\begingroup$ Yes, but $\sum_{a,b}[\delta \chi_a\chi_a|\chi_b\chi_b]=\sum_{a,b}[\delta \chi_b\chi_b|\chi_a\chi_a]=\sum_{a,b}[ \chi_a\chi_a|\delta\chi_b\chi_b]$, i.e. you have to change the variables in the integral and in the sum. $\endgroup$ – Adam Oct 14 '13 at 14:26
  • $\begingroup$ Wow. How simple. Thank you very much! I didn't think about the sum indices. $\endgroup$ – jacksonslsmg4 Oct 14 '13 at 17:01

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