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

It turns out the trick is not a matter of exploiting symmetry or alternate derivations, but rather a simple bookkeeping trick!

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

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

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:

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

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

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

Which proves the equality.