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Selection rule for electric dipole transitions in x direction

  1. Dec 4, 2013 #1
    Hello,
    I have this matrix element

    [itex]\langle n', l', m'| x |n, l, m \rangle[/itex]

    where [itex]x=rsin\theta cos\phi[/itex] in spherical polars

    Which I write explicitly as:

    [itex]=\int \int \int R_{n', l'} R_{n, l} A_{l', m'} A_{l, m} P^{m'}_{l'} P^{m}_{l} e^{im\phi} e^{-im\phi} rsin\theta cos\phi r^{2} sin \theta d \theta d \phi dr[/itex]


    [itex]=A_{l', m'} A_{l, m} \int dr R_{n', l'} r^{3} R_{n, l} \bullet \int d \theta sin^{2} \theta P^{m'}_{l'} P^{m}_{l} \bullet \int^{2\pi}_{0} e^{i\phi(m-m')} cos \phi d \phi[/itex]

    The question just wants me to do the integral for the [itex]d \phi[/itex] bit then show that it is only non zero for [itex]m=m' \pm 1[/itex] hence [itex](m-m')= \pm 1[/itex]

    So I let [itex]i(m-m')=a[/itex]

    [itex]I=\int^{2\pi}_{0} e^{a\phi} cos \phi[/itex]

    [itex]u=e^{a\phi}[/itex]
    [itex]du=a e^{a\phi}[/itex]
    [itex]dv=cos \phi[/itex]
    [itex]v=sin \phi[/itex]

    [itex]I=sin \phi e^{a\phi}-a \int^{2\pi}_{0} sin \phi e^{a\phi}[/itex]

    for the integral here I do the same again

    [itex]u=e^{a\phi}[/itex]
    [itex]du=a e^{a\phi}[/itex]
    [itex]dv=sin \phi[/itex]
    [itex]v=-cos \phi[/itex]

    [itex]I_{2}=-cos \phi e^{a\phi} +a \int^{2\pi}_{0} cos \phi e^{a\phi}[/itex]

    So in total

    [itex]\int^{2\pi}_{0} e^{a\phi} cos \phi=sin \phi e^{a\phi}+acos \phi e^{a\phi}-a^{2}\int^{2\pi}_{0} e^{a\phi} cos\phi[/itex]

    Then I take the integral on the RHS over to the left, factor out the integral and divide through by [itex]1+a^{2}[/itex] giving:

    [itex]\int^{2\pi}_{0} e^{a\phi}cos \phi=\frac{sin\phi e^{a\phi}+acos\phi e^{a\phi}}{1+a^{2}}[/itex]

    [itex]= [\frac{sin\phi e^{i(m-m')\phi}+i(m-m')cos\phi e^{i(m-m')\phi}}{1-(m-m')^{2}}]^{2\pi}_{0}[/itex]



    Is this integration ok? I kind of made it up as I went along as I wasn't sure how to integrate this function.
    Assuming it is could someone explain to me how it follows that it is only non zero if [itex]m=m' \pm 1[/itex]
     
    Last edited: Dec 4, 2013
  2. jcsd
  3. Dec 4, 2013 #2

    TSny

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    OK, except when you substituted for ##a## in the denominator in the last step.
     
  4. Dec 4, 2013 #3
    Thanks, I've fixed that in the original post
     
  5. Dec 4, 2013 #4

    TSny

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    So, what do you get when m ≠ m' ±1 and what do you get when m = m' ±1?
     
  6. Dec 4, 2013 #5
    I'm having trouble getting passed the fact that for [itex]m=m'\pm 1[/itex] the denominator will be zero
    e.g for [itex]m=m'+1[/itex]

    [itex]= [\frac{sin\phi e^{i\phi}+icos\phi e^{i\phi}}{1-1}]^{2\pi}_{0}[/itex]
     
  7. Dec 4, 2013 #6

    TSny

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    Right. For this case you will need to go back and substitute [itex]m=m'\pm 1[/itex] before evaluating the integral.
     
  8. Dec 4, 2013 #7
    OK, I think I get this now, so for example for [itex](m-m')=+1[/itex]

    [itex]\int^{2\pi}_{0} e^{i(m-m')\phi}cos\phi=\int^{2\pi}_{0} e^{i \phi}cos\phi[/itex]

    [itex]=\int^{2\pi}_{0} e^{i \phi} \frac{e^{i\phi}+e^{-i\phi}}{2}[/itex]

    [itex]=\frac{1}{2}\int^{2\pi}_{0}e^{2i\phi}+1[/itex]

    [itex]=\frac{1}{2}[\frac{e^{2i\phi}}{2i}+\phi]^{2\pi}_{0}[/itex]

    [itex]=\frac{1}{2}[\frac{cos(2\phi)+isin(2\phi)}{2i}+\phi]^{2\pi}_{0}[/itex]

    [itex]=\frac{1}{2}[\frac{1}{2i}+2\pi-\frac{1}{2i}-0][/itex]

    [itex]=\pi[/itex]

    Similarly for the case when [itex](m-m')=-1[/itex]

    [itex]\int^{2\pi}_{0} e^{i(m-m')\phi}cos\phi=\int^{2\pi}_{0} e^{-i \phi}cos\phi[/itex]

    [itex]=\int^{2\pi}_{0} e^{-i \phi} \frac{e^{i\phi}+e^{-i\phi}}{2}[/itex]

    [itex]=\frac{1}{2}\int^{2\pi}_{0}e^{-2i\phi}+1[/itex]

    [itex]=\frac{1}{2}[-\frac{e^{-2i\phi}}{2i}+\phi]^{2\pi}_{0}[/itex]

    [itex]=\frac{1}{2}[\phi -\frac{cos(2\phi)-isin(2\phi)}{2i}]^{2\pi}_{0}[/itex]

    [itex]=\frac{1}{2}[2\pi-\frac{1}{2i}+\frac{1}{2i}][/itex]

    [itex]=\pi[/itex]

    When [itex]m=m'\pm 1[/itex] one of the exponential terms multiplies with another to make 1. This means after integration there is a non trigonometric term which will become zero when zero is substituted in to it. However when [itex]m\neq m'\pm 1[/itex] both terms have an exponential in after the integral is evaluated so the two terms you get after subbing in the [itex]2\pi[/itex] will equal the two terms you get after subbing in the zero. Does that sound right? Thanks for the help.
     
  9. Dec 4, 2013 #8

    TSny

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    Sounds good.
     
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