## Saturday, 5 January 2008

### Coulomb and exchange

Let's keep going with what we did last time. It will get even more exciting than it already was.

Well I think it's nice to see what's behind some of those interactions and if I write it down here I'll find it again. But I admit that I usually read blogs with pictures rather than lots of weird signs.

The first expression was

$n \sum_{\pi \in S_n} \sum_{\sigma \in S_n} sgn(\pi) sgn(\sigma) \prod_{i=2}^n $

Let us assume that 2 or more orbitals are different between the Slater determinants (p$\leq$n-2). Then one of the terms in every product has to be 0 and the expression vanishes.

If p=n-1, then we have to claim the following:

$\pi(1)=\sigma(1)=n
\forall i\in \{2,...,n\}:\pi(i)=\sigma(i)$

This of course means that π=σ. There are now (n-1)! possible permutations that work with this. And the expression reduces to: (the n! cancels against the n! we started out with)

$n!$

If p=n, we also need π=σ to have all the factors unequal to 0 (and equal to 1). The expression reduces to

$n \sum_{\pi \in S_n} sgn(\pi)^2 $

The squared sign is of course 1. For every i there are (n-1)! peruations π with π(1)=i and ui=vi. Then we have
$n! \sum_{i=1}^n $

The second expression was
$\frac{n(n-1)}{2} \sum_{\pi \in S_n} sgn(\pi) \sum_{\sigma \in S_n} sgn(\sigma) \prod_{i=3}^n
)$

This time it turns out that one of the overlap integrals (and hence the whole product) equals zero if p$\leq$n-3.

I'll only talk about the case p=n-2. If you want to have no overlap integral equal to 0, you first need

$\forall i\in \{3,...,n\}:\pi(i)=\sigma(i)$

Aside from that you have to make sure that the mutually different orbitals are in the last integral. 1 and 2 have to be mapped onto n-1 and n.

$\pi(\{1,2\})=\{n-1,n\}
\sigma(\{1,2\})=\{n-1,n\}$

As a next step you can acknowledge that π and σ have to either be the same or different by one transposition.

$\sigma=\pi
or
\sigma=\pi(12)=:\pi'$

In the second case their signs are opposite. We get the following expression.

$\frac{n(n-1)}{2} \sum_{\pi \in S_n} sgn(\pi) (sgn(\pi) +
+sgn(\pi') )$

There are appropriately (n-2)!*2 permutations π that fulfill the requirements above and since the signs are opposite, this reduces to

$n! (- )$

The derivations seem to be similar for p=n-1 and p=n. As Levine tells me (who was in turn told by Parr) we get for p=n-1

$n! \sum_{j=1}^n (- )$

and for p=n

$n! \sum_{i=1}^{n-1} \sum_{j=i+1}^n (- )$

(the n! is only there because I started with an extra n!)

The first integral is the Coulomb integral (with g12=1/r12). It corresponds to the Coulomb repulsion of two blurred electron clouds corresponding to the MOs. The second integral, called exchange integral, comes in because of the Slater determinant and is a consequence of the Pauli principle.

The next question is using the spin functions from two posts ago and see if some of the integrals vanish. And compare this to what you expected.