Hello, I am considering the case of the total spin when adding three spin 1/2s. The combined system has dimension 2x2x2=8. The possible values for the total spin quantum number are: [tex] \left([1/2] \otimes [1/2]) \otimes [1/2] = ( \oplus ) \otimes [1/2] =[3/2] \oplus [1/2] \oplus [1/2] [/tex] so either s=3/2 or s=1/2. s=3/2 is 4 fold degenerate and s=1/2 is 2 fold degenerate. But this sums up to only 6 eigenvectors. Where are the missing two eigenvectors? It seems to me that S^2 and S_z no longer form a C.S.C.O in this case, since for s=1/2 both m=1/2 and m=-1/2 still have to be 2-fold degenerate. Is this correct? And with which operator is this degeneracy usually resolved?