# Homework Help: Correlation energy between two electrons

1. Oct 11, 2012

### eonsv

Hello. I need to find the expectational value of the correlation energy between two electrons which repel eachother via the classical Coulomb interaction. This is given by the 6-dimensional integral:
$\left\langle\frac{1}{|\vec{r}_1 - \vec{r}_2|}\right\rangle =$$\int_{-\infty}^{\infty}$$d\vec{r}_1d\vec{r}_2$$e^{-2\alpha(r_1 + r_2)}$$\frac{1}{|\vec{r_1}-\vec{r_2}|}$
$r_i = |\vec{r}_i|, \quad \vec{r}_i = x_i \hat{i} + y_i\hat{j} + z_i\hat{k}, \quad i = 1,2$
I know that the answer is $\frac{5\pi^2}{16^2}$, however I can't seem to find a method for solving this type of integrals in any of my books. I have tried to use spherical coordinates, which seemed logical due to the answer having a factor of $\pi^2$, but with no luck. And it is a couple of years since I've been solving integrals like this one, so a nudge in the right direction is greatly appriciated.

(It is a bonus question in one of the projects in the course: computational physics)

2. Oct 12, 2012

### PhysicsGente

Hello.
Try using the law of cosines to expand the denominator and do one integral at a time.

3. Oct 12, 2012

### eonsv

$|\vec{r}_1 - \vec{r}_2| = \sqrt{r_1^2 + r_2^2 - 2r_1 r_2 \cos\beta}$ Where $\cos\beta = \cos\theta_1\cos\theta_2 + \sin\theta_1\sin\theta_2 \cos(\phi_1 - \phi_2)$
I then need to expand $d\vec{r}_1$ and $d\vec{r}_2$, which can be done in spherical coordinates as: $d\vec{r}_i = r_i^2 \sin\theta_i dr_id\theta_id\phi_i, i = 1,2$ or $d\vec{r}_1d\vec{r}_2 = r_1^2r_2^2dr_1dr_2d\cos\theta_1d\cos\theta_2d\phi_1d\phi_2$
$\left\langle\frac{1}{|\vec{r}_1 - \vec{r}_2|}\right\rangle = \int_{-\infty}^{\infty}e^{-2\alpha(r_1 + r_2)} \frac{r_1^2r_2^2dr_1dr_2d\cos\theta_1d\cos\theta_2d\phi_1d\phi_2}{\sqrt{r_1^2 + r_2^2 - 2r_1 r_2\cos\beta}}$
I can't seem to find a usefull relation between $d\cos\theta_1d\cos\theta_2d\phi_1d\phi_2$ and $d\beta$ or $\cos\beta$. Any ideas on how I can proceed?