91Ó°ÊÓ

A quantum state whose wave function is an eigenfunction of the linear operator that corresponds to an observable is called an eigenstate. When you measure that observable, the eigenvalue of that wave function is the quantity you see (the eigenvalue could be a vector quantity).

$$\begin{gathered} Firstofall,letusexpand{S}^2. \\ {S}^2=S^{\left(1\right)}+S^{\left(2\right)}.S^{\left(1\right)}+S^{\left(2\right)}=S^{\left(1\right)}+S^{\left(2\right)}+2S^{\left(1\right)}.S^{\left(2\right)} \\ Then, \\ {S}^2,S_z^{\left(1\right)}=S^{\left(1\right)}+S^{\left(2\right)}+{25}^{\left(1\right)}.S^{\left(2\right)},S_Z^{\left(1\right)} \\ Here,basedonthecommutator\text{'}sdistributivity,write: \\ {S}^2,S_z^{\left(1\right)}=S^{\left(1\right)2}{S}_z^{\left(1\right)}+S^{\left(2\right)},z^{\left(1\right)}+2S^{\left(1\right)}.S^{\left(2\right)},S_z^{\left(1\right)} \\ Forthefirsttime,locatethecommutotorasfollows: \\ {S}^{\left(1\right)2},S_z^{\left(1\right)}=S_x^{\left(1\right)2}+S_y^{\left(1\right)2}+S_z^{\left(1\right)2}{S}_z^{\left(1\right)} \\ =S_x^{\left(1\right)2},S_z^{\left(1\right)}+S_y^{\left(1\right)2},S_z^{\left(1\right)}+S_z^{\left(1\right)2},S_z^{\left(1\right)} \end{gathered}$$

$$\begin{gathered} \mathrm{Use}\mathrm{the}\mathrm{identity}\left[\mathrm{AB},\mathrm{C}\right]=\mathrm{A}\left[\mathrm{B},\mathrm{C}\right]+\left[\mathrm{A},\mathrm{C}\right]\mathrm{B}\mathrm{to}\mathrm{simplify}\mathrm{the}\mathrm{problem}.\mathrm{of}\mathrm{course},{\mathrm{S}}_{\mathrm{z}}^{\left(1\right)} \\ \mathrm{is}\mathrm{commute}\mathrm{with}\mathrm{itself}\mathrm{like}\mathrm{any}\mathrm{other}\mathrm{operator},\mathrm{so}{\mathrm{S}}_{\mathrm{z}}^{\left(1\right)2},{\mathrm{S}}_{\mathrm{z}}^{\left(1\right)}=0,\mathrm{thus}, \end{gathered}$$

$$\begin{gathered} {\text{S}}^{\left(1\right)2},{\mathrm{S}}_{\mathrm{z}}^{\left(1\right)}={\mathrm{S}}_{\mathrm{x}}^{\left(1\right)}{\mathrm{S}}_{\mathrm{x}}^{\left(1\right)},{\mathrm{S}}_{\mathrm{z}}^{\left(1\right)}+{\mathrm{S}}_{\mathrm{x}}^{\left(1\right)},{\mathrm{S}}_{\mathrm{z}}^{\left(1\right)}{\mathrm{S}}_{\mathrm{x}}^{\left(1\right)}+{\mathrm{S}}_{\mathrm{y}}^{\left(1\right)}{\mathrm{S}}_{\mathrm{y}}^{\left(1\right)},{\mathrm{S}}_{\mathrm{z}}^{\left(1\right)}{\mathrm{S}}_{\mathrm{y}}^{\left(1\right)} \\ ={\mathrm{S}}_{\mathrm{x}}^{\left(1\right)}-\mathrm{i}\sout{\mathrm{h}}{\mathrm{S}}_{\mathrm{y}}^{\left(1\right)}+-\mathrm{i}\sout{\mathrm{h}}{\mathrm{S}}_{\mathrm{y}}^{\left(1\right)}{\mathrm{S}}_{\mathrm{x}}^{\left(1\right)}+{\mathrm{S}}_{\mathrm{y}}^{\left(1\right)}\mathrm{i}\sout{\mathrm{h}}{\mathrm{S}}_{\mathrm{x}}^{\left(1\right)}+\mathrm{i}\sout{\mathrm{h}}{\mathrm{S}}_{\mathrm{x}}^{\left(1\right)}{\mathrm{S}}_{\mathrm{y}}^{\left(1\right)} \\ =0 \end{gathered}$$

$$\begin{gathered} Thus, \\ {S}^2,S_z^{\left(1\right)}=2i\sout{h}{S}_x^{\left(1\right)}{S}_y^{\left(2\right)}-S_y^{\left(1\right)}{S}_x^{\left(2\right)} \\ Ifdothesameprocedure{S}^2,S_x^{\left(1\right)}and{S}^2,S_y^{\left(1\right)},findthemtobe2i\sout{h}{S}_y^{\left(1\right)}{S}_z^{\left(2\right)}-S_z^{\left(1\right)}{S}_y^{\left(2\right)}and \\ -2i\sout{h}{S}_x^{\left(1\right)}{S}_z^{\left(2\right)}-S_z^{\left(1\right)}{S}_x^{\left(2\right)}respectively,somakeageneralization \\ {S}^2,S^{\left(1\right)}=2i\sout{h}{S}^{\left(1\right)}×S^{\left(2\right)} \end{gathered}$$