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How to derive the selection rules $\Delta L= \pm 1$, $\Delta S=0$ for electron spectroscopy?

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up vote 4 down vote accepted

these are selection rules for the electric dipole radiation. The transition from the excited state to a lower state of an atom is governed by the following matrix element: $$C \cdot \langle f | \hat{p} | i\rangle$$ You could also write the sum of positions of electrons $\Sigma\hat{r}$ instead of their total momentum $\hat{p}$ in the middle. This simple form of the operator in the middle is because at low enough frequencies, i.e. long enough wavelength, the atom simply finds itself in a uniform field, anf the electric potential $\phi$ for a uniform field is linear in $\hat{r}$. Equivalently, the matrix element may be converted from $\hat{r}$ to $\hat{p}$ and vice versa by realizing that the commutator of the Hamiltonian with $\hat{x}$ is proportional to $\hat{p}$.

At any rate, the operator in the middle is a 3-dimensional vector that only acts on the positions or momenta of the electrons, not their spins. So it has to commute with the total spin operator $\hat{S}$ of the electrons, and $\Delta S=0$ as a consequence.

On the other hand, it is a vector, i.e. a $j=1$ object as far as the SO(3) transformations go, and by combining its $j=1$ angular momentum with the orbital angular momentum $L_i$ of the initial state, you may get the final $L_f$ being $L_i\pm 1$ or $L_i$ according to the basic rules of the addition of angular momentum. You may imagine that you're just adding two vectors of specified lengths, $L_i$ and $1$, and depending on their relative angle, the length of the sum may go from $L_i-1$ to $L_i+1$.

Your list didn't allow $L_f=L_i$ and I think you are right that it is typically forbidden as well because it violates parity. The angular momentum to parity goes like $(-1)^L$, I guess, but because the vector operator is parity-odd, the initial and final states have to differ by their parity as well, which means that only $\Delta L=\pm 1$ is allowed. Everyone, please correct me if I am saying something incorrectly.

Best wishes Lubos

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essentially at low frequencies is permissible to make a mean field approximation and treat it as a perturbation. Is that right? – Andrea Amoretti Jan 18 '11 at 14:02
Yes, absolutely. Of course, the mean field approximation won't see that one has emitted a photon but it doesn't matter - the changed state of the atom knows about this thing. The low-frequency is related to the fact that the dipole is enough - the $\cos(x)$ wave may be expanded and only the linear terms matter at the distance scales of the atom. – Luboš Motl Jan 19 '11 at 13:48
What I can't figure out is why I'm the only person who's upvoted this since it was written a month and a half ago. Am I the only person who goes around upvoting things? Hmmmmmmm.. Turns out that I am: and the only one who upvotes competing answers: – Carl Brannen Mar 7 '11 at 4:51
@CarlBrannen: I totally agree with you that this answer should receive more upvotes. I did my part ;) – Philipp Feb 20 '12 at 23:28
What is going to happen if spin orbit interaction is considered where spin and orbit momentum are neither good quantum number? – Laurent Mar 27 '14 at 23:20

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