Chemistry - On the spin-adaptation of 2nd order Møller-Plesset perturbation theory

$\newcommand{\Ket}[1]{\left|#1\right>}$ $\newcommand{\Bra}[1]{\left<#1\right|}$ $\newcommand{\BraKet}[2] { {\left<#1} \left|#2 \right>}$
It turns out the trick is not a matter of exploiting symmetry or alternate derivations, but rather a simple bookkeeping trick!

$E^{(2)}= \sum_{ijab}^{SF} \frac {\Bra{ij}\left.ab\right>^{2} -\Bra{ij}\left.ab\right>\Bra{ij}\\right> +\Bra{ij}\\right>^{2} } {\Delta _{ab}^{ij}}$ can be broken into three sums:

$E^{(2)}= \sum_{ijab}^{SF} \frac {\Bra{ij}\left.ab\right>^{2}} {\Delta _{ab}^{ij}} -\sum_{ijab}^{SF} \frac{\Bra{ij}\left.ab\right>\Bra{ij}\\right> } {\Delta _{ab}^{ij}} +\sum_{ijab}^{SF} \frac{\Bra{ij}\\right>^{2} } {\Delta _{ab}^{ij}}$

The beauty is that the indices of each summation are independent of the indices of the other summations. As such, I am free to rename a as b and b as a in the third summation:

$\sum_{ijab}^{SF} \frac{\Bra{ij}\\right>^{2} } {\Delta _{ab}^{ij}}= \sum_{ijba}^{SF} \frac{\Bra{ij}\left.ab\right>^{2} } {\Delta _{ba}^{ij}}$

Which, because a and b behave the same in the summation (both over unoccupied MOs) and in the denominator (both represent the energy of an unoccupied MO), can be rearranged as:

$\sum_{ijab}^{SF} \frac{\Bra{ij}\left.ab\right>^{2} } {\Delta _{ab}^{ij}}$

(In pseudo math, $\sum_{ijba}^{SF} = \sum_{ijab}^{SF}$ and ${\Delta _{ab}^{ij}}= {\Delta _{ba}^{ij}}$). Putting it all together, we get

$E^{(2)}= \sum_{ijab}^{SF} \frac {\Bra{ij}\left.ab\right>^{2}} {\Delta _{ab}^{ij}} -\sum_{ijab}^{SF} \frac{\Bra{ij}\left.ab\right>\Bra{ij}\\right> } {\Delta _{ab}^{ij}} +\sum_{ijab}^{SF} \frac{\Bra{ij}\left.ab\right>^{2} } {\Delta _{ab}^{ij}} = 2\sum_{ijab}^{SF} \frac {\Bra{ij}\left.ab\right>^{2}} {\Delta _{ab}^{ij}} -\sum_{ijab}^{SF} \frac{\Bra{ij}\left.ab\right>\Bra{ij}\\right> } {\Delta _{ab}^{ij}} = \sum_{ijab}^{SF} \frac{2\Bra{ij}{ab}\left.\right>^{2}-\Bra{ij}{ba}\left.\right>\Bra{ij}{ab}\left.\right>}{\Delta _{ab}^{ij}}$

Which proves the equality.