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Need help regarding transition dipole moment

  1. Aug 11, 2014 #1
    Hi, I have been trying to get the expression for the transition dipole moment of hydrogen but I am not able to get the expression. Hope someone can help me with that.

    I want to evaluate [itex]\vec{d}(v)=<v|\hat{r}|0>[/itex] where v is the free state and 0 is the 1s wave function for hydrogen.

    [itex]\vec{d}(v)=\int dĪ„\frac{\alpha^{3/4}}{\pi^{1/2}}exp(-\sqrt{\alpha}r)(-ih\frac{d}{dp})exp(-ipr)[/itex]

    After integration by parts i got [itex]\frac{\alpha^{3/4}}{\pi^{1/2}}\left(\frac{1}{\sqrt{\alpha}+ip}\right)^{2}[/itex]

    However, it should be [itex]\vec{d}(v)=i\left(\frac{2^{7/2}}{\pi}\alpha^{5/4}\right)\frac{p}{\left( p^2+\alpha\right)^3}[/itex].

    Can someone point out where I went wrong?
     
  2. jcsd
  3. Aug 12, 2014 #2

    Meir Achuz

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    exp(-ipr)should be [itex]exp(-i{\bf p\cdot r})[/itex],
    and [itex]-i\hbar\partial_p[/itex] can just be [itex]{\bf r}[/itex].
    Then you need angular integration.
     
    Last edited: Aug 12, 2014
  4. Aug 12, 2014 #3
    Actually I am having trouble doing the integration due to the plane wave part since there is a dot product over there. I have been looking up the meaning of dipole moment and transition dipole moment but can't find much information. Is there any recommended books to read up on this? Thanks!
     
  5. Aug 13, 2014 #4

    Meir Achuz

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    [itex]{\bf p\cdot r}=pr\cos(\theta)[/itex]. Then the angular integration is easy.
     
    Last edited: Aug 13, 2014
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