$LS$-coupling vector model precession? In $LS$-coupling it is said (e.g. here, link to Google Books1) that in the $LS$-coupling scheme the individual $\vec l_i$ precess around $\vec L$ and the individual $\vec s_i$ precess around $\vec S$. Every where I have looked, this is simply stated without proof. Hence my question is: How is it possible to show that such precessions do occur?
1Fundamentals of Spectroscopy by B.Narayan
 A: The idea that vectors representing angular momenta are precessing is "semi-classical", i.e., supposed to be an aid in understanding what's going on. In reality such pictures of the "vector model" sometimes serve only to add confusion.
As we know, the rules for angular momenta arise from corresponding operators and their group theory. The truth about $L-S$ coupling or $jj$ coupling is that in lighter atoms the Hamiltonian ($H$) is such that the individual electrons' total angular momenta $j_i$ are not good* quantum numbers, and it is better to use instead the total orbital angular momentum ($L^2$) and the total spin ($S^2$) to describe the atom. In the semi-classical model we would say that individual $\vec l_i$ "precess" around $\vec L$ (and similarly for $\vec s_i$ and $\vec S$). [Note that $S$ enters the picture indirectly due to anti-symmetry of the overall atomic state under electron exchange: the total spin $S$-state must be combined with the (conserved) total $L$-state to be antisymmetric under electron exchange.]
For heavy atoms the spin-orbit or $jj$ coupling dominates, i.e., the total angular momenta $j_i$ of individual electrons are good quantum numbers due to the $\vec l_i\cdot\vec s_i$ terms. Thus the analogy here is that the individual $\vec j_i$ "precess" around $\vec J$.
*Good quantum numbers are eigenvalues of operators that commute with the Hamiltonian so that the corresponding physical quantities are constant in time and thus these quantum numbers can be used to label states.
P.S. It is easy to mistake the term $L-S$ coupling for spin-orbit coupling, but it is really $jj$ coupling that is spin-orbit coupling.
A: From the Hartree theory, (i dont know if you have studied it or not) it is seen that the coulumb interaction of the optically active electrons results in a tendency of them to couple in such a way that the magnitude of the total orbital angular momentum $L'= (L_1 + L_2 + ....$) is constant. This happens simply because in most quantum states the charge distributions of the electrons are not spherically symmetrical (Why, you ask? Because only fully filled shells have spherically symmetric charge distribution; one extra valence electron, or many valence electrons can ruin the symmetric distribution), and so they exert torques on each other. Since the space orientation of the charge distribution of an electron is related to the space orientation of its orbital angular momentum vector, there are torques acting between the angular momentum vectors. The torques do not tend to change the magnitude of the individual orbital angular momentum vectors, but only tend to make them precess about the total orbital angular momentum vector in such a way that its magnitude L' remains constant. The coupling of $L'$ is such that it results in the largest possible value. This is confirmed from the spectral analysis of atoms with several optically active electrons. (Imagine 2 optically active electrons rotating about the nucleus; lowest energy is such that they are at the diametrically opposite ends of the nucleus because that is where the coulumb repulsion is the least between them...hence $L_1$ and $L_2$ are parallel) 
This goes for $S'(= S_1 + S_2 + ....$) as well. In fact, the spins tend to couple in such a way that the total spin is as large as possible because such a configuration leads to the lowest energy according to Hund's first rule.

Individual $Ls = (L_1 , L_2 .....)$ and individual $Ss = (S_1 , S_2 ......)$ couple strongly(so that their resultant is maximum) and precess rapidly about their resultant. But the resultant $L=L'$ and the $S=S'$ couple weakly with each other and precess slowly about their resultant $J=J'$. This is consistent with Hund's rules which says lowest $J$ has the lowest energy (for not greater than half filled orbitals) 
