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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?

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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
1  
@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

The multiplicity $2S+1$ actually tells you how many degenerate spin states there are, each labelled with the total spin projection quantum number $M_S$ (this is from the total spin projection operator $\hat{S_z}$(conventionally taken to be in the z-direction) whose eigenvalues are $\hbar M_S$). The possible values of $M_s$ are $-S\le M_S\le S$ in integer step. As $S=1$ or $0$ for any two fermions, $M_S=1, 0, -1$. As you can see, there are 3 values of $M_S$ i.e., 3 eigenvalues for the same energy spin eigenstate, so the multiplicity is 3. If you calculate $2S+1=2\times 1+1=3$ you get 3 as expected.

This principle is the same for the total angular momentum states labelled with the total angular momentum projection number $M_J$. Again, note that the multiplicity $2J+1$ tells you the number of different values of $M_J$ for a particular value of $J$, not the number of different energy levels (each of which has a different value of $J$)

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