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Homework Help: Integration (related to QM).

  1. Dec 18, 2009 #1

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    Questions 3 and 4 in the attachment.


    3. The attempt at a solution
    3. [tex]\int d\omega_1 d\omega_2 /|r1-r2|=(2\pi)^2 \int_{0}^{\pi} d\theta_1 \int_{0}^{\pi} d\theta_2 \frac{1}{\sqrt{r_1^2+r_2^2-2r_1r_2cos(\theta_1+\theta_2)}[/tex]
    don't know how to proceed from here?
    for question 4 I got to the integral:
    [tex]\int_{0}^{\infty}\int_{-1}^{1}dcos(\theta)x^2exp(-(|x-x_A|+|x-x_B|)/a)dx[/tex]
    Now I can assume that x_A is at the origin and x_B=Rx, where R is the seperation between the two atoms, i.e the exponenet becomes: [tex]exp(-(x+\sqrt{x^2+R^2-2Rxcos(\theta))[/tex], but still how do I proceed from here?

    Thanks in advance.
    here's the attachment in case the link doesn't show.
     

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    Last edited: Dec 18, 2009
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  3. Dec 18, 2009 #2

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    I don't see why my first tex code doesn't showup.
     
  4. Dec 18, 2009 #3
    There is a curly bracket missing at the end.

    [tex]\int d\omega_1 d\omega_2 /|r1-r2|=(2\pi)^2 \int_{0}^{\pi} d\theta_1 \int_{0}^{\pi} d\theta_2 \frac{1}{\sqrt{r_1^2+r_2^2-2r_1r_2cos(\theta_1+\theta_2)}}[/tex]

    Remember that

    [tex]d\Omega = sin\theta d\theta d\phi[/tex]

    Also, why do you have a sum of the angles inside cosine? Choose coordinates such taht z-axis is along one of the vectors.
     
  5. Dec 19, 2009 #4

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    So it's should be something like this:
    [tex]\frac{1}{\sqrt{r_1^2+r_2^2-2r_1r_2cos(\theta_1)}}[/tex]
    Now if r1>r2, by expandig this by taylor series, I get that the first term is 1/r1, and the other terms depend on cosines of theta1, which get terminated with the integration.

    Can someone help me with the integral in question 4?

    EDIT:
    Only odd powers of cosines get terminated in integration, I guess that the other terms get cancelled cause they contain powers greater than 1 of 1/r1, though I believe this result should be exact.
     
    Last edited: Dec 19, 2009
  6. Dec 19, 2009 #5
    You are making this too complicated. After integrating over theta2 (which is trivial) your relult is proportional to

    [tex]\int_{0}^{\pi} \frac{sin\theta_1 d\theta_1}{\sqrt{r_1^2+r_2^2-2r_1r_2cos\theta_1}}[/tex]

    Which is easily integrated because the integrand is of the form f'(g)*g'. Do a substitution u=cos(theta1) if that makes it more clear.
     
    Last edited: Dec 19, 2009
  7. Dec 19, 2009 #6

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    OK, thanks I solved this.

    Btw can you help me with my second question?

    Thanks in advance.
     
  8. Dec 19, 2009 #7
    Can't really think of a way right now, so let me know if you solve it/get the the solution.
     
  9. Dec 20, 2009 #8

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    I think I solved it.
    What I argued is that I assume that x vector is in the direction of x-hat, and that x_A is in the origin and x_B is also in the direction of x-hat but with magnitude of R which is the seperation between A and B.
    I got to the next integral to solve:
    [tex]\int_{0}^{\infty}x^2 e^{-(|x-R|+x)/a}dx[/tex]
    which is easy to calculate.
     
  10. Dec 20, 2009 #9

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    Never mind, I found the answer I was looking for in QM-vol 2 of Cohen Tannoudji from pages: 1170-1175.
     
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