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

For angular momentum quantum number \bgroup\color{black}$\ell$\egroup, there are \bgroup\color{black}$(2\ell + 1)$\egroup different m states, while for spin we have 2 states \bgroup\color{black}$\chi_{\pm}$\egroup. Hence the composite system has \bgroup\color{black}$2(2\ell + 1)$\egroup states total.

Max \bgroup\color{black}$j_z = \ell + {1\over 2}$\egroup so we have a state with \bgroup\color{black}$j=\ell+{1\over 2}$\egroup. This makes up \bgroup\color{black}$(2j+1)=(2\ell+2)$\egroup states, leaving

\begin{displaymath}\bgroup\color{black}(4\ell + 2) - (2\ell + 2) = 2\ell = \left(2\left(\ell-{1\over 2}\right) +1\right)\egroup\end{displaymath}

Thus we have a state with \bgroup\color{black}$j=\ell-{1\over 2}$\egroup and that's all.

Jim Branson 2013-04-22