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Electric Dipole Selection Rules

  • #1

Homework Statement



For [tex]\Delta[/tex]l = 0 the transition rate can be obtained by evaluating the electric dipole matrix elements
given by

[tex]\vec{I}[/tex] = [tex]\int[/tex] [tex]\Psi^{*}_{1,0,0}[/tex] (e [tex]\vec{r}[/tex]) [tex]\Psi_{2,0,0}[/tex] d[tex]\tau[/tex]

Homework Equations





The Attempt at a Solution



I've got the two wave functions, neither of which have a theta or phi dependance, so when multiplied by the r vector, I should just get their r components. Evaluating this integral is simple, but I'm not sure if I understand what the answer means.
The selection rule for l is [tex]\Delta[/tex]l =[tex]\pm[/tex]1, so doesn't that mean that this case, where [tex]\Delta[/tex]l = 0 shouldn't be allowed? I might be completely off track, but I thought that the integral would give me 0, proving this, but that's not the value I'm getting. The actual calculation here isn't difficult, but I think I'm missing something conceptually.
 

Answers and Replies

  • #2
malawi_glenn
Science Advisor
Homework Helper
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the operator has odd parity, angular wave functions has parity [tex](-1)^l [/tex].

So [itex]\Psi_{2,0,0}[/itex] means [itex] n=2, l = 0, m = 0 [/itex] right ?

If that is the case, then you see that the total integrand has odd parity, and integration over whole space will give you zero.
 
Last edited:
  • #3
Makes sense.. I actually talked to the prof about the question, and it turns out we had to split r into components, and evaluate all three integrals explicitly.. it was a bit annoying, but I got it sorted out. Thanks for the help!
 
  • #4
malawi_glenn
Science Advisor
Homework Helper
4,786
22
yeah, if you have explicit wave functions, then you just work it out. I was trying to explain the general idea behind the selection rules :)
 

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