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