Counting states for Arbitrary Plus spin

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