Selection Rules for EM Dipole Radiation

  • Thread starter Dumbleboar
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  • #1

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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
 

Answers and Replies

  • #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?
 
  • #3
alxm
Science Advisor
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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?
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:
  • #4
Dr Transport
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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].
 

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