Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose there are two electrons in an atom with $s_1 = \frac{1}{2}$, $l_1 = 1$ and $s_2 = \frac{1}{2}$, $l_2 = 1$. Hence the total $S$ (of the atom) may be +1 or 0. And total $L$ is either $+2$, $+1$ or $0$.

Now If we consider

$$\begin{align} S=1,L=2 &\to 2S+1=3; J=3,2,1\\ S=0,L=2 &\to 2S+1=1; J=2\\ S=1,L=1 &\to 2S+1=3; J=2,1,0\\ S=0,L=1 &\to 2S+1=1; J=1\\ S=1,L=0 &\to 2S+1=3; J=1\end{align}$$

But the last one does not show that $2S+1$ is multiplicity as it has only one $J$ value. Where am I making a mistake?

share|cite|improve this question
The formula $2S+1$ for total multiplicity has nothing to do with spin-orbit coupling, so you should forget about $L$. If you know that a system has spin $S$, then it can have $S_z = -S,-S+1,\ldots,S-1,S$, so you have $2S+1$ possibilities. – Vibert Feb 5 '13 at 20:08
Hi user22180 - I removed the extra question at the end of your post, as we like to limit each post to contain only one question. You can always post it separately. – David Z Feb 5 '13 at 20:54
@Vibert that could be the basis of an answer. – David Z Feb 5 '13 at 20:54
@Vibert , Then why is it called by multiplicity?And how is it different from degeneracy in the ABSENCE of magnetic field? – user22180 Feb 7 '13 at 21:54
@user22180: it's not different at all. The following statement is always true: a system of spin/angular momentum $J$ has multiplicity $2J+1.$ (It is really a maths statement: the dimension of the $SU(2)$ representation $J$ is $2J+1$.) In practice, you need to think which spin you use. For a single electron, it is clearly the electron spin $S$, in a system coupled with orbital angular momentum $L$ you can choose a basis labelled by the total spin $J = L+S.$ – Vibert Feb 8 '13 at 11:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.