Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Correlation energy between two electrons

  1. Oct 11, 2012 #1
    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:
    [itex]\left\langle\frac{1}{|\vec{r}_1 - \vec{r}_2|}\right\rangle =[/itex][itex]\int_{-\infty}^{\infty}[/itex][itex]d\vec{r}_1d\vec{r}_2[/itex][itex]e^{-2\alpha(r_1 + r_2)}[/itex][itex]\frac{1}{|\vec{r_1}-\vec{r_2}|}[/itex][itex][/itex]
    [itex]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[/itex]
    I know that the answer is [itex]\frac{5\pi^2}{16^2}[/itex], 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 [itex]\pi^2[/itex], 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. jcsd
  3. Oct 12, 2012 #2
    Hello.
    Try using the law of cosines to expand the denominator and do one integral at a time.
     
  4. Oct 12, 2012 #3
    Thank you for your reply.
    This is one of the things which I've been trying to do. However I end up with a result which is a very complicated integral. Expanding the difference of the length of the two vectors gives:
    [itex]|\vec{r}_1 - \vec{r}_2| = \sqrt{r_1^2 + r_2^2 - 2r_1 r_2 \cos\beta}[/itex] Where [itex]\cos\beta = \cos\theta_1\cos\theta_2 + \sin\theta_1\sin\theta_2 \cos(\phi_1 - \phi_2)[/itex]
    I then need to expand [itex]d\vec{r}_1[/itex] and [itex]d\vec{r}_2[/itex], which can be done in spherical coordinates as: [itex]d\vec{r}_i = r_i^2 \sin\theta_i dr_id\theta_id\phi_i, i = 1,2[/itex] or [itex]d\vec{r}_1d\vec{r}_2 = r_1^2r_2^2dr_1dr_2d\cos\theta_1d\cos\theta_2d\phi_1d\phi_2[/itex]
    This is were I end up with the integral
    [itex]\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}}[/itex]
    I can't seem to find a usefull relation between [itex]d\cos\theta_1d\cos\theta_2d\phi_1d\phi_2[/itex] and [itex]d\beta[/itex] or [itex]\cos\beta[/itex]. Any ideas on how I can proceed?
    Thanks in advance.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook