# Triplet states and Fine Structure

If we have a system, with total spin angular momentum given by $$S$$, then we have spin multiplicity equal to, $$2S+1$$. This spin multiplicity basically tells us the different spin states, this system can have, based on the different values of $$m_s$$.

However, if we have also been given total angular momentum $$L$$, and we have $$L\ge S$$, then we know that $$2S+1$$ also gives us the multiplicity of J if we write the states in the coupled basis.

My question is, in uncoupled basis, the triplet states are defined by the different values of $$m_s$$. For example, $$|L,S,m_s\rangle = |1,1,-1\rangle,|1,1,0\rangle,|1,1,1\rangle$$ is an example of a triplet defined by $$m_s$$.

However, in coupled basis representation, $$|L,S,J\rangle = |1,1,0\rangle,|1,1,1\rangle,|1,1,2\rangle$$ is an example of a triplet defined by $$J$$.

There is no one-to-one correspondence between the states, but they can be related using Clebsch Gordon coefficients I think.

So, my question is, do we define triplets on the basis of $$m_s$$ in uncoupled basis or when we ignore angular momentum, and by using $$j$$ in the coupled basis, in spin-orbit coupling for example ? In spin-orbit coupling, we ignore $$m_s$$ because it's not a good quantum number, and yet we get the triplet state for $$S=1$$, but because of values of J. Are total angular momentum triplets and spin triplets the same?

If I'm wrong, can someone tell me the correct explanation?