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Tricky Integral, H-Atom Dipole Transition Matrix Elements

  1. Apr 27, 2008 #1
    1. Problem

    Evaluate

    [tex]\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} z^{2} e^{-A \sqrt{x^{2}+y^{2}+z^{2}}} dxdydz[/tex]

    2. Useful Formulae

    none

    3. Attempt at Solution

    Well, this is part of a much larger problem. I am trying to compute the dipole moment matrix elements for a Hydrogen 2P -> 1S transition, and it involves this integral. In particular, this is the "z-component" of the |2,1,0> -> |1,0,0> transition - I've already found the x- and y-components to be zero for this transition.

    I've spent nearly an hour looking through online integral tables, and I've found nothing useful. Can anyone help? Thanks!
     
  2. jcsd
  3. Apr 27, 2008 #2
    switch to spherical coordinates and it should be doable.
    It will probably take an integration by parts once or twice. (don't forget your jacobian)
     
  4. Apr 27, 2008 #3
    I started in spherical, and had a tough time with it. But maybe I was turning to the definite integral tables too early. Looks like if I do integration by parts two (or maybe three?) times, it should take a familiar form.

    Alright, I'll give it a shot.
     
  5. Apr 27, 2008 #4
    Great, I got it. Thanks.

    Quick question, though. What is:

    [tex]\int_{0}^{2 \pi} \sin \phi d \phi[/tex]

    Isn't it zero? But then all of my matrix elements would go to zero when evaluated in spherical coordinates (where [itex]\phi[/itex] is evaluate from [itex]0 \rightarrow 2 \pi[/itex]). It seems like cheating to evaluate it half-way and multiply by 2, giving 4 - but is that the trick?
     
  6. Apr 27, 2008 #5
    that integral over 2pi is zero; but you integrate theta over 2 pi, and phi over pi. Does that make sense?
    phi is the angle coming from the z axis, and theta is the angle in the xy plane. only one of those 2 angles needs to go from 0 to 2pi, and the other from 0 to pi; by convention this is phi and not theta --> and that is tied to the jacobian r^2 * sinphi dr dphi dtheta
     
  7. Apr 27, 2008 #6
    Ah, I must've had my convention backwards. That fixes everything - thanks!
     
  8. Apr 27, 2008 #7
    No prob!
     
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