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More wave integrals

  1. Sep 3, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex] p_x = - \int\int\int \sqrt{\frac{3}{32{\pi}^2}} {sin}^3 \theta cos \phi e^{i\phi} \sqrt{\frac{1}{8a^6}} \frac{2r^4 e^{-3r/2a}}{a\sqrt{3}} dr d\theta d\phi [/tex]

    We are to prove that [tex] p_x = - 2^7 \frac{a}{3^5} [/tex]

    2. Relevant equations
    The constants collapse to
    [tex] \frac{1}{8a^4\pi}[/tex]

    I have already combined the coefficients for spherical coordinates.

    3. The attempt at a solution

    With respect to r,

    [tex] \int_0^{\infty} r^4 e^{-3r/2a} dr = (- \frac{2a}{3})^5 = 6. - 2^7 \frac{a^5}{3^5} [/tex]

    With respect to [tex] \theta [/tex],

    [tex] \int_0^{\pi} {sin}^3 \theta d\theta = \frac{4}{3} [/tex] (trust me)

    With respect to [tex] \phi [/tex]

    [tex] \int_0^{2\pi} cos\phi e^{i\phi} d\phi = \int_0^{2\phi} \frac{1}{2} {cos}^2 \phi + \frac{1}{2} + icos\phi sin\phi d\phi = \pi [/tex]

    This gives us [tex] p_x = 2^7 \frac{a}{3^5} [/tex]

    So where did my minus go?? I Don't think there's a big problem with the integration but not sure if I've made any careless mistakes.
  2. jcsd
  3. Sep 3, 2008 #2


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    If you are claiming the r integral is negative, how can that be? The integrand is positive. If a<0 then the integral is divergent. Where are you getting these horrible integrals?
  4. Sep 4, 2008 #3
    It's from Mathematical Methods for Physics and Engineering. The question actually cites two wavefunctions for the position of a hydrogen electron and then asks you to find the dipole matrix element [tex]p_x[/tex] between the two. Whatever the case, I have combined the two wavefunctions and the volume element in spherical coordinates, which gives you the monstrosity above.

    Well, this integral could be worse, at least the variables are separable.

    I was thinking my r-integral was wrong too, but I'm not sure where.

    a isn't <0 because it's the Bohr radius. Thanks for your help Dick.
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