# Number of unique 2-electron integrals

Consider 2-electron integrals over real basis functions of the form $$(\mu\nu|\lambda\sigma) = \int d\vec{r}_{1}d\vec{r}_{2} \phi_{\mu}(\vec{r}_{1}) \phi_{\nu}(\vec{r}_{1}) r_{12}^{-1} \phi_{\lambda}(\vec{r}_{2}) \phi_{\sigma}(\vec{r}_{2})$$ I am told that for a basis set of size K=100, there are 12,753,775 unique 2-electron integrals of this form.

Symmetry considerations mean that we have less than $K^{4}$ unique integrals, since we can exchange electrons and also exchange the basis functions for each electron without changing the value of the integral.

How could one work out the number of unique integrals?

My method is:

Find the number of unique integrals of the forms $(\mu\nu|\lambda\sigma)$, $(\mu\mu|\lambda\sigma)$, $(\mu\mu|\nu\nu)$ and $(\mu\mu|\mu\mu)$ (where in these integrals, each index is unique unless repeated) and sum these together.

My working gives the wrong answer, though:

$$\frac{4!}{8}{100 \choose 4}+\frac{3!}{4}{100 \choose 3}+\frac{2!}{2}{100 \choose 2}+1!{100 \choose 1} = 12,011,275$$

My rationale is this: for the integral form $(\mu\nu|\lambda\sigma)$, there are ${100 \choose 4}$ unique unordered combinations of basis functions. There are $4!$ ways of arranging these unique basis functions. We can exchange the electrons, basis functions on electron 1 and basis functions on electron 2 without changing the value of the integral, thus halving the number of unique integrals 3 times ($\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}$). Therefore the number of unique integrals of form $(\mu\nu|\lambda\sigma)$ is $\frac{4!}{8}{100 \choose 4}$.

Where am I going wrong?

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The right formula is very similar to yours, $$\frac{4!}{8}{100 \choose 4}+3!{100 \choose 3}+2\times 2!{100 \choose 2}+1!{100 \choose 1} = 12,753,775$$ I think that by comparing the coefficients in front of the (correct) binomial numbers, you may determine how you need to fix your calculation.
Divide the integrals to groups according to how many values of the indices $\mu\nu,\lambda\sigma$ coincide or not. The first term counts all the integrals where all the 4 indices are different and you gave the justification. Then add the integrals for which two of the indices are equal. Then add the 3+1 and 2+2 splitting, and finally, add the 100 integrals where all the 4 indices are equal. –  Luboš Motl Sep 25 '12 at 19:07
I didn't understand very well how your calculate the number of 2-electron integrals. Please, can anyone explain me with a little detail? Also, I found that the number of 2-electron integrals is iqual to $$\frac{1}{8}n(n+1)(n^2+n+2),\hspace{3mm}for\hspace{4mm} n=100 \hspace{4mm}\Longrightarrow \hspace{4mm}12,753,775\hspace{4mm} integrals$$ but I don't know how explain it. Best Regards –  partcula94 Nov 22 '13 at 5:21