Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Selection Rules for EM Dipole Radiation

  1. Jul 4, 2010 #1
    hey,
    I was asking myself a few questions about the selection rules for EM dipole radiation which occurs if electrons "jump" into lower bound states according to the selection rules.

    now I know that the full explanation about matrix elements of the dipole operator comes from fermi's golden rule... which I will learn next semester... but apart from that...we have 2 considerations:

    1) Parity... parity has to change... ok i got this
    2) Conservation of angular momentum:
    we have
    ji=j[tex]\gamma[/tex] + jf
    and we assume that
    j[tex]\gamma[/tex]=1

    WHY is this angular momentum 1??
    I guess it has something to do that the cartesian components of the Dipole operator are proportional to linear compinations of the spherical harmonics Y1,m but I don't get this point at all....

    I'd highly appreciate your help!
    thanks in advance
     
  2. jcsd
  3. Jul 4, 2010 #2
    Ok Photons have Spin 1... therefore they have j=1

    but i guess you can also explain it through expanding the dipole radiation into sperical harmonics and that only those with l=1 can express a dipole or something... could that be it?
     
  4. Jul 4, 2010 #3

    alxm

    User Avatar
    Science Advisor

    Yes, that's correct. I'll skip the derivation and just state that the matrix element for the dipole moment transition is:
    [tex]\mathbf{r}_{n',n} = \int u_{n'}^*\sum_i\mathbf{r}_i u_n d\tau[/tex]
    I.e. zero unless you have a mean change in location along some axis. (which is intuitive, right?)

    Using the rationale from Bethe and Salpeter, the single-electron atom wave-function is:
    [tex]u_{nlm} = \frac{1}{\sqrt{2\pi}}R_{nl}(r)P_{lm}(\vartheta)e^{im\phi}[/tex]

    So using [tex]z = r\cos \vartheta[/tex], the matrix element for the transition on the z axis is:
    [tex]z_{nlm}^{n'l'm'} = \int u^*_{n'l'm'}z u_{nlm} d\tau = \int_0^\infty r^2 R_{n'l'}(r)R_{nl}(r) dr * \int_0^\pi P_{l'm'}(\vartheta)P_{lm}(\vartheta)\cos (\vartheta)\sin (\vartheta) d\vartheta * \int_0^{2\pi}\frac{1}{2\pi}e^{i(m-m')\phi} d\phi[/tex]

    The last integral will be zero unless [tex]m = m'[/tex], in which case it's exactly 1, so [tex]\Delta m = 0[/tex], and the second integral will vanish unless [tex]\Delta l = \pm 1[/tex] (a bit trickier, turn the legendre functions into the related spherical harmonics, one of which becomes [tex]l \rightarrow l \pm 1[/tex]. Then you have from the othogonality relations of spherical harmonics that the integral will be zero unless [tex]l = l' \pm 1[/tex])

    And you can of course do the same for any other coordinate and arrive at the same rules.
     
    Last edited: Jul 4, 2010
  5. Jul 4, 2010 #4

    Dr Transport

    User Avatar
    Science Advisor
    Gold Member

    The [tex] \theta [/tex] equation can be integrated using the following recursion relation

    [tex] (2l+1)cos(\theta)P_{l,m}(cos(\theta)) = (l-m+1)P_{l+1,m}(cos(\theta))+(l+m)P_{l-1,m}(cos(\theta)) [/tex]

    which shows that [tex] \Delta l = \pm 1 [/tex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Selection Rules for EM Dipole Radiation
Loading...