Vector sum schemes for LS coupling & jj coupling

• A
The difference between light and very heavy atoms reflects itself in these two schemes.
My question is why one scheme for the vector sum is necessarily the right & suitable sum model for one case, and the 2nd scheme suits the 2nd case ?
In other words, why & how the relative magnitude of the respective terms in the Hamiltonian, namely for LS & the central field (residual), results in the order the summing of the angular momenta (orbital & spin) ?
In my eyes, this is a fundamental question regarding the physical picture & the appropriate vector model.
Thanks, Bentzy.

In other words, why does the stronger interaction precede in the order of addition ?

DrClaude
Mentor

You have a base Hamiltonian (central field) and additional terms that act as a perturbation to the base Hamiltonian. But you can also take one of these additional terms and add it to the base Hamiltonian, and treat the other term as a perturbation. You can also see it as a perturbation to a perturbation.

This is what is happening here. If spin-orbit coupling dominates, you can find the states in the presence of LS, and then the residual part will be a perturbation to this. In some rare cases, the residual part dominates, and spin-orbit coupling will be a perturbation to that.

I'm afraid you don't see the main point in my question, and target something else, despite my additional, emphasizing comment. Pertubation theory is known, this isn't the point. Let me rephrase the question by being concrete & specific: e.g., in the case of light atoms the central force governs over thr LS interaction, thus we add first all the orbital angular momenta alone, and the spin angular momenta alone, and only then combine the respective resultants of the to get J. Why is this the only right way to add the various momenta vectors in this case ? How does this, and just this, describe the physical picture mathematically ?
(Similarly, but vice versa, is done for the very heavy atoms.)
Pertubation, pertubation, but why does it require a unique scheme of addion ?
By te way, one of these rare & extreme cases is Mercury (Hg), the emission spectra of which I deal with.

DrClaude
Mentor
Why is this the only right way to add the various momenta vectors in this case ? How does this, and just this, describe the physical picture mathematically ?
The premise is false. There is never a single way to add angular momenta, but there is a preferred way that will give physically valuable information (a nice "picture") of what is going on.

To give a concrete example, you can always describe the electronic states using ##|L M_L S M_S \rangle## as basis states, but in cases where the residual interaction term dominates, these basis states are not constants of motion to any degree of approximation.

Pertubation, pertubation, but why does it require a unique scheme of addion ?
Because of the separation between what is a perturbation and what is not. By separating out a small perturbation, the goal is to have unperturbed eigenstates that are only slightly affected by the perturbation, hence give a good representation of the physics.

Returning to your previous reply, It seems that my phrasing isn't clear enough, rather than a false premise, since there was no premise in the 1st place. I just tried to point out the difference in the order of adding the various momenta to get to the total J, depending on which interaction dominates. That's the meaning of "the only right way" in my question - why you add 1st all the l's separately, & all the s's separately, in the case of light atoms, and only then add up L & S to get J, and not pairs of respective l & s to get j's etc., as we do in the case of very heavy atoms. The opposite question goes for the heavy atoms case.
I know that the difference between the 2 models stems from different suitable basis-states. I think I need more clarification on this.

DrClaude
Mentor
I know that the difference between the 2 models stems from different suitable basis-states.
That's the crux of it. Mathematically, there is no "right" way to add the individual angular momenta. It is simply that one way of doing the summing up will be more useful to describe the physical state.

Consider two electrons and the bases resulting from the two summing approaches, ##|L,S\rangle## and ##|j_1,j_2\rangle##. Neither of these are eigenstates of the full Hamiltonian, but for light atoms one will get, for example, eigenstates ##|\phi\rangle## for which
$$| \langle L',S' | \phi\rangle |^2 \approx 0.99$$
hence ##|L',S'\rangle## is a good approximation to the actual eigenstate. Using ##|j_1,j_2\rangle## states, there is no way to write a good approximation for ##|\phi\rangle## using a single ##|j_1',j_2'\rangle##. One would have to use a superposition of the ##|j_1,j_2\rangle##'s to get a good approximation.

The reason why that is the case is that, starting from the central field Hamiltonian eigenstates ##|l_1,m_{l1},m_{s1},l_2,m_{l2},m_{s2}\rangle##, the residual Hamiltonian couples these states in such a way that the eigenstates are of the type ##|L, M_L, m_{s1},m_{s2}\rangle##, coupling together the orbital angular momenta of the electrons. The spin-orbit interaction then couples these, the the full eigenstates are very close to the ##|L,S\rangle## states.

The situation is (partially) reversed for heavy atoms.