Phy5646/AddAngularMomentumProb

Based on exercise 15.1.1. from Principles of Quantum Mechanics, 2nd ed. by Shankar:

Express $\ S^{2}$ as a matrix for two spin-1/2 particles in the direct product basis.

1.) First express $\ S^{2}$ in terms of $\ S_{1}^{2}$, $\ S_{2}^{2}$, $\ S_{1z}$, $\ S_{2z}$, $\ S_{1\pm }$ and $\ S_{2\pm }$: $\ S^{2}=({\vec {S_{1}}}+{\vec {S_{2}}})^{2}=S_{1}^{2}+S_{2}^{2}+2{\vec {S_{1}}}\cdot {\vec {S_{2}}}=S_{1}^{2}+S_{2}^{2}+2(S_{1x}S_{2x}+S_{1y}S_{2y}+S_{1z}S_{2z})=S_{1}^{2}+S_{2}^{2}+2S_{1z}S_{2z}+S_{1+}S_{2-}+S_{1-}S_{2+}$

2.) Then act with this on direct product state $\ |m_{1}m_{2}\rangle$: $\ S^{2}|m_{1}m_{2}\rangle =(S_{1}^{2}+S_{2}^{2}+2S_{1z}S_{2z}+S_{1+}S_{2-}+S_{1-}S_{2+})|m_{1}m_{2}\rangle$ $\ =S_{1}^{2}|m_{1}m_{2}\rangle +S_{2}^{2}|m_{1}m_{2}\rangle +2S_{1z}S_{2z}|m_{1}m_{2}\rangle +S_{1+}S_{2-}|m_{1}m_{2}\rangle +S_{1-}S_{2+}|m_{1}m_{2}\rangle$

$\ =\hbar ^{2}{\frac {1}{2}}\left({\frac {1}{2}}+1\right)|m_{1}m_{2}\rangle +\hbar ^{2}{\frac {1}{2}}\left({\frac {1}{2}}+1\right)|m_{1}m_{2}\rangle +\hbar ^{2}{\sqrt {{\frac {1}{2}}\left({\frac {1}{2}}+1\right)-m_{1}\left(m_{1}+1\right)}}{\sqrt {{\frac {1}{2}}\left({\frac {1}{2}}+1\right)-m_{2}\left(m_{1}-1\right)}}|m_{1}+1;m_{2}-1\rangle$ $\ +\hbar ^{2}{\sqrt {{\frac {1}{2}}\left({\frac {1}{2}}+1\right)-m_{1}\left(m_{1}-1\right)}}{\sqrt {{\frac {1}{2}}\left({\frac {1}{2}}+1\right)-m_{2}\left(m_{1}+1\right)}}|m_{1}-1;m_{2}+1\rangle$ $\ \Rightarrow S^{2}|m_{1}m_{2}\rangle =\hbar ^{2}\left({\frac {3}{2}}|m_{1}m_{2}\rangle +{\sqrt {\left({\frac {3}{2}}-m_{1}(m_{1}+1)\right)\left({\frac {3}{2}}-m_{2}(m_{2}-1)\right)}}|m_{1}+1;m_{2}-1\rangle +{\sqrt {\left({\frac {3}{2}}-m_{1}(m_{1}-1)\right)\left({\frac {3}{2}}-m_{2}(m_{2}+1)\right)}}|m_{1}-1;m_{2}+1\rangle \right)$

${\hbar }^{2}{\begin{pmatrix}2&0&0&0\\0&1&1&0\\0&1&1&0\\0&0&0&2\\\end{pmatrix}}$