## Thursday, 3 January 2008

### Slater determinants

I am having some time these days to catch up on the basics [1]. This could therefore be another dry mathematical post. On the other hand: you may have been wondering all your life what exchange interaction was. Then this is your chance, in case you are open minded when it comes to looking at a few summation signs and angle brackets.

Wave functions are typically built on spin molecular orbitals, uj(i), i=(xi, yi, zi, msi) (i.e. "i" is an abbreviation for the 4 coordinates from last post). MOs are functions that involve one electron. To get a multielectron function we could put each electron into an MO and multiply the functions:

$u_1(1)u_2(2)...u_n(n)=\prod_{i=1}^n u_i(i)$

In these functions electrons are statistically independent, therefore also uncorrelated (which is a weaker statement).

The product would violate the Pauli principle. In order to comply with it, we introduce a Slater determinant.

$\mid u_1 u_2 ... u_n \mid := \frac{1}{\sqrt{(n!)}}\sum_{\pi \in S_n} sgn(\pi) \prod_{i=1}^n u_{\pi(i)}(i)$

This is the explicit form of a determinant. You form the sum of all possible products with electrons in different orbitals with an appropriate sign (-1 or 1). (Sn is the symmetric group of all permutations; definition of sgn).

In a Slater determinant, electrons are not stastically independent anymore. But "correlation" usually means going beyond the formation of a Slater determinant.

Let's evaluate a matrix element between two Slater determinants.

$D= \mid u_1 ... u_n \mid,
E= \mid v_1 ... v_n \mid$

All the orbitals are taken out of an orthonormal system. We will notice that the determinant vanishes unless most of the orbitals are equals. Hence we arrange the orbital in a way that equal orbitals are moved up front. Let p orbitals be equal.

$p \leq n
u_i = v_i;~ i = 1, ..., p$

The operator A shall be a sum of similar 1- and 2-electron operators:

$A=\sum_{i=1}^n f_i+ \sum_{i=1}^{n-1} \sum_{j=1+1}^n g_{ij}$

Now we can write down the equation:

$n!=
<\sum_{\pi \in S_n} sgn(\pi) \prod_{i=1}^n u_{\pi(i)}(i)|\sum_{j=1}^n f_j+
+ \sum_{k=1}^{n-1} \sum_{l=k+1}^n g_{kl}|\sum_{\sigma \in S_n} sgn(\sigma) \prod_{i=1}^n v_{\sigma(i)}(i)>=$

Because of linearity, we can take out the summations

$=\sum_{\pi \in S_n} sgn(\pi) \sum_{\sigma \in S_n} sgn(\sigma) ( \sum_{i=1}^n<\prod_{i=1}^n u_{\pi(i)}(i)| f_i|\prod_{i=1}^n v_{\sigma(i)}(i)>+
+ \sum_{k=1}^{n-1} \sum_{l=k+1}^n<\prod_{i=1}^n u_{\pi(i)}(i)| g_{kl}|\prod_{i=1}^n v_{\sigma(i)}(i)>)=$

Because of the antisymmetry, the integrals for all the electrons are the same

$=\sum_{\pi \in S_n} sgn(\pi) \sum_{\sigma \in S_n} sgn(\sigma) (n<\prod_{i=1}^n u_{\pi(i)}(i)| f_1|\prod_{i=1}^n v_{\sigma(i)}(i)>+
+ \frac{n(n-1)}{2}<\prod_{i=1}^n u_{\pi(i)}(i)| g_{12}|\prod_{i=1}^n v_{\sigma(i)}(i)>)=$

Now if you look at it as a multiple integrals, you will notice that you can take out all the functions that are not affected by the operator. (<...> means the integral over the respective configurational space)

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

You notice that most of the orbitals have to be the same, in order for those products not to vanish. How to go on and where the spin comes in, next time.

[1] It kind of helped that I broke my hand on my first snowboarding day this year ... There's still going to be february.