Counting states for \bgroup\color{black}$\ell=3$\egroup Plus spin \bgroup\color{black}${1\over 2}$\egroup

For \bgroup\color{black}$\ell=3$\egroup there are \bgroup\color{black}$2\ell+1=7$\egroup different eigenstates of \bgroup\color{black}$L_z$\egroup. There are two different eigenstates of \bgroup\color{black}$S_z$\egroup for spin \bgroup\color{black}${1\over 2}$\egroup. We can have any combination of these states, implying \bgroup\color{black}$2\times 7=14$\egroup possible product states like \bgroup\color{black}$Y_{31}\chi_+$\egroup.

We will argue based on adding vectors... that there will be two total angular momentum states that can be made up from the 14 product states, \bgroup\color{black}$j=\ell\pm{1\over 2}$\egroup, in this case \bgroup\color{black}$j={5\over 2}$\egroup and \bgroup\color{black}$j={7\over 2}$\egroup. Each of these has \bgroup\color{black}$2j+1$\egroup states, that is 6 and 8 states respectively. Since \bgroup\color{black}$6+8=14$\egroup this gives us the right number of states.

Jim Branson 2013-04-22