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Tricky Integral, H-Atom Dipole Transition Matrix Elements

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!
 

Answers and Replies

275
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)
 
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.
 
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?
 
275
2
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?
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
 
Ah, I must've had my convention backwards. That fixes everything - thanks!
 
275
2
No prob!
 

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