# Difference between spin-orbit coupling and $LS$ coupling (Russell-Saunders)

I'm having some trouble understand what the difference is between these two. It seems as though there are kind of the same, but that spin-orbit coupling reduces to $$LS$$ coupling under certain circumstances.

But, I can't seem to make sense of it. So I was hoping anyone could explain the difference briefly, and maybe explain when you use one of them, instead of the other.

Basically, in atomic physics, you would have two electrons, each with an angular momentum $l_1$ and $l_2$ and spin $s_1$ and $s_2$, and you want to couple all those to get the best approximation for the resulting spectrum.

So you have two options:

1- You couple $l_1$ and $l_2$ to $L$, and $s_1$ and $s_2$ to $S$, and then you couple $L$ with $S$ to get $LS$.

2- You couple $l_1$ and $s_1$ to $j_1$, and $l_2$ and $s_2$ to $j_2$, and then couple $j_1$ and $j_2$ to to get $JJ$.

So you have two ways to couple those, and the choice depends on how far the electrons are from each other where the specific angular momentum coupling is more pronounced. So if the electrons are close to each other, then you use LS coupling. While if you have them far apart, you use JJ coupling.

I hope this helps.

• A nice succinct answer that presents the main point. Notice that in '$LS$ coupling' the interaction between $L$ and $S$ is completely ignored in the first instance. The name has the sense 'coupling which leads to $L$ and $S$' not 'coupling between $L$ and $S$' (but there is such a coupling at the second stage where the spin-orbit interaction is allowed for). Jul 10, 2021 at 14:15

The answer by "The Quantum Physicist" is correct. LS-coupling and jj-coupling are two extreme ways of applying perturbation theory to an electronic configuration in the central-field approximation. In LS-coupling one first ignores the spin-orbit coupling and perturb the level with the residual angular part of the electrostatic repulsion between valence electrons. This gives the "LS-coupling atomic term". It is called "LS-coupling" because the quantum numbers L and S provide a diagonal representation to the Hamiltonian. When the atomic term has been established, one can include the spin-orbit interaction as a second layer of perturbation. That will then give the "fine-structure levels" specified by J. One way to put it is that the "LS-coupling approximation" the INTER-electronic interactions swamp the INTRA-electronic spin-orbit ones.

In jj-coupling, it is the other way around. The strong spin-orbit interaction will then has as an effect that L and S no longer are good quantum numbers. The proper representation is then the ensemble of individual j for the valence electrons (for a 3-electron system, a more logical name would be jjj-coupling). This gives "jj-coupling terms" and the angular part of the inter-electronic interaction will then be the second layer of perturbation.

Note that both schemes are approximations. LS-coupling is typically quite good for light atoms, and quite far in the periodic system. This is because of the Z^4 scaling for spin-orbit coupling. Really good jj-coupling is rare for neutral atoms in ground states. One good exemple of jj-coupling is Pb. Several intermediate coupling schemes exist.

An additional comment that might be in place that "LS-coupling" is sometimes confused as being the same thing as "spin-orbit coupling". That is absolutely incorrect.

For light atomic systems with small value of nuclear charge, the electrostatic hamiltonian term is much stronger than the spin orbit coupling that increases with increasing Z, therefore in this case we use LS coupling. This means that we must first combine the orbital angular momenta of all the particles quantum mechanically to get L and do a similar thing for the spin angular momenta to obtain S of the whole system. You then combine L and S to obtain total angular momentum J. For high Z systems (say greater than Fe), the spin-orbital term dominates over electrostatic term which is now considered as perturbation. Under these conditions we first form j1 which is coupling of l1 and s1 for the first particle, proceed like this for all the particles then finally add all terms like j1,j2,j3.. and obtain j. Note that capital letters are used for LSJ coupling and lower case letters are used for jj coupling [(j1,j2,j3...)j].

L-S coupling is the appropriate way of description of small small electronic configuration ($Z\leq10$). In case of atoms having a large number of electrons, J-J coupling works.

• I have not show here how LS COUPLING and JJ COUPLING WORKS. I think you have knowledge about it.I just gave a basic difference. Oct 5, 2015 at 19:42
• If you think you should show it, then show it and make your answer better. Oct 5, 2015 at 19:45
• Please do not use all capitals, as that is the internet-equivalent of shouting at someone. Oct 5, 2015 at 20:00

In my opinion there is logical difference between spin-orbit coupling and LS coupling. One particular electron possesses angular momentum l and spin s.when a coupling takes place between l and s of one particular electron then it is known as spin-orbit coupling which gives resultant j. But in case of an atom having two balance electron if angular momentum l1 and l2 and spin s1 and s2 couplings are preferred then they form resultant angular momentum L and spin S. When these two quantities combine they form resultant LS coupling generally known as LS coupling. This will be true in case of three or more valance electrons also.

I hope it will be helpful.

*The strong spin orbit interaction gives rise to singlet and triplet terms with relatively large energy difference in L.S. coupling.

• The strong spin orbit interaction gives rise to widely separated (j1,j2 ) terms when weak j1,j2 interaction is taken into account each term is splitted into two closely spaced components having different J values.
• did you mean to say residual electrostatic interaction n the first sentence? Jul 10, 2021 at 14:06