1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Selection rule for electric dipole transitions in x direction

  1. Dec 4, 2013 #1
    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]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]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


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    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


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    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


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    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]






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


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    Sounds good.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Threads - Selection rule electric Date
Integral to show selection rules Mar 19, 2016
Emission lines of calcium Nov 3, 2014
Selection Rule confusion Aug 3, 2013
Electric Dipole Selection Rules Nov 16, 2007
Selection Rules (Electric Dipole) Nov 10, 2005