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I saw this video which provides a clear, though simple, introduction to the argument. I think the main thing missing is a proof for:

  • $\Delta \ell = \pm 1 $
  • $\Delta m_\ell = 0, \pm 1$
  • $\Delta m_s = 0$ (which is not even mentioned)

which is presented as sort of a given. I saw a bunch of sites. Some say it's due to Cleibsch-Gordan coefficients, others begin with the Quadratic Stark Shift... maybe it's all connected, but I'm still confused.

ric.san
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1 Answers1

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There is a mathematical and a physical way to look at it. From the physical point of view, photons carry angular momentum $\pm 1$. Thus, due to the conservation of the total angular momentum, the absorption/emission of a photon is possible only between the states whose total angular momentum differs by $1$. Since only one electron participate sin an absorption/emission event, flipping its spin could account for this, but this does not necessarily happen - hence the rule for the orbital angular momentum.

Mathematically, this follows from calculating the matrix element of the atom-photon interaction between the relevant atomic orbitals (i.e., the matrix element that enters the Fermi golden rule). In the dipole approximation this interaction is proprotional to $\mathbf{r}$, i.e. we calculate $$ \int d\mathbf{r}\psi_{final}^*(\mathbf{r})\mathbf{r}\psi_{initial}(\mathbf{r}) $$ The transitions are possible only when this matrix element is not zero. This is where Cleibsch-Gordan coefficients or some other math may enter - depending on the specific situation one is dealing with.

Remark: here is my short derivation of the dipole approximation (in somewhat different context)

Roger V.
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  • "The transitions are possible only when this matrix element is zero"... Uhm, why? – ric.san Mar 22 '21 at 09:36
  • Because this is the matrix element in the Fermi golden rule - it it is zero, then the transition rate is zero, i.e., no transitions. – Roger V. Mar 22 '21 at 09:37
  • Excuse me, you said before "transitions are possible" and now "no transitions". – ric.san Mar 22 '21 at 09:39
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    Sorry, that was a typo. It should be non-zero. – Roger V. Mar 22 '21 at 09:40
  • Ok, one final question... if $\psi_{final}$ and $\psi_{initial}$ are two eigenstates relative to different eigenvalues, they should be orthogonal, so why isn't the matrix element always zero? (After all, I'm sort of projecting a vector onto two orthogonal components ... that should give me zero no matter what) – ric.san Mar 22 '21 at 09:44
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    Because you are computing perturbations, the dipole term shifts the eigenvalues and eigenfunctions and creates overlaps which are computed from the "free" or unperturbed part – ohneVal Mar 22 '21 at 09:45
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    @ric.san Note that there is also $\mathbf{r}$ in the integral - it is not the same integral that is zero according to the orthogonality relation. – Roger V. Mar 22 '21 at 09:52