The Russell-Saunders effect is the same thing as 'spin-orbit interaction, correct?

The reason I am asking is because I was reviewing the Wikipedia page on 'spin-orbit interaction' and it does not mention Russell-Saunders at all...


1 Answer 1


I'm not aware of the Russell–Saunders effect, but the Russell–Saunders coupling scheme is definitely a thing. As you noted, the Wikipedia page on "spin-orbit interaction" doesn't talk about it, but a different Wikipedia page does, and basically tells you the same thing as I will.

The answer is... yes and no.

The word "coupling" refers to the coupling of several sources of angular momentum, namely the spin component and orbital component. Now, the issue is that in an atom you typically have many electrons, and every electron has its own orbital angular momentum $\vec{l}$ and spin angular momentum $\vec{s}$,* so you have many, many sources of angular momentum.

The challenge is to bring all of these together in a way that allows us to describe the electronic state of an atom using its angular momentum properties (for example, this is what a term symbol does). There are two approaches towards coupling all of these angular momenta together:

  1. (Russell–Saunders or LS-coupling) Couple all the individual $\vec{l}$'s together to form one gigantic orbital angular momentum $\vec{L}$, and couple all the individual $\vec{s}$'s together to form one gigantic spin angular momentum $\vec{S}$. Then couple these two together to form the total angular momentum $\vec{J}$.

    If you have studied term symbols before it is probably using the Russell–Saunders scheme, where you calculate $L$, $S$, and $J$, then write the term symbol $^{2S+1}L_J$.

  2. (jj-coupling) For each individual electron, couple $\vec{l}$ and $\vec{s}$ together to form the total angular momentum $\vec{j}$ for that one particular electron. Then bring all the electrons' total angular momentum together to form $\vec{J}$.

    Note that we haven't ever mentioned $L$ and $S$ here, so the term symbols under this coupling scheme are different. Instead, you'd label the term symbols with the individual values of $j$ for each electron. For an example see e.g. Atkins Molecular Quantum Mechanics.

Now, which you use depends on whether electron-electron repulsions or the spin-orbit coupling is a "bigger" term. If spin-orbit coupling is very significant, then it means that the spin and orbital angular momenta on their own (i.e. $\vec{L}$ and $\vec{S}$) are not very useful quantities,† since the interaction between them is large. In this scenario, jj-coupling is a more appropriate way of describing the effects of spin-orbit coupling on the electronic state.

On the other hand, if the spin-orbit coupling is relatively small, then $\vec{L}$ and $\vec{S}$ are useful quantities which are still applicable to the atom's electronic state, so the Russell–Saunders scheme is appropriate.

And of course, sometimes we get stuck in the middle ground where neither scheme is fully appropriate.

TL;DR The Russell–Saunders and jj schemes are both methods which can be used to describe the effects of spin–orbit coupling, but they are not the same thing as spin–orbit coupling.‡

* Well, sort of, anyway: the electrons are indistinguishable, so it's more accurate to say that the $n$ electrons in the atom have $n$ orbital angular momenta $\{\vec{l}_1, \vec{l}_2, \cdots, \vec{l}_n\}$ and $n$ spin angular momenta $\{\vec{s}_1, \vec{s}_2, \cdots, \vec{s}_n\}$.

† To be precise, $L$ and $S$ are not "good quantum numbers" because the operators $\hat{L}$ and $\hat{S}$ don't (approximately) commute with the total Hamiltonian $\hat{H}_0 + \hat{H}_\text{so}$, where $\hat{H}_\text{so}$ is the spin-orbit coupling Hamiltonian and $\hat{H}_0$ is the rest of the Hamiltonian (which does commute with $\hat{L}$ and $\hat{S}$).

‡ If you read the previous footnote, then the spin–orbit coupling itself is represented by the Hamiltonian $\hat{H}_\text{so}$. The two coupling schemes can be thought of as ways of dealing with this term as a perturbation to $\hat{H}_0$. In Russell–Saunders the perturbation is small, and consequently, the "good" quantum numbers are similar to those of $\hat{H}_0$. In jj the perturbation is large.


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