Question

Determine the commutator of ${\mathbf{S}}^{\mathbf{2}}$with${\mathbf{S}}_{\mathbf{Z}}^{\left(1\right)}$(where$\mathbf{S}\mathbf{\equiv }{\mathbf{S}}^{\left(1\right)}\mathbf{+}{\mathbf{S}}^{\left(2\right)}$) Generalize your result to show that

$\left[S^2,S^{\left(1\right)}\right]\mathbf{=}\mathbf{2}\mathbf{I}\sout{\mathbf{h}}\left(S^{\left(1\right)}\times S^{\left(2\right)}\right)$

Comment: Because ${\mathbf{S}}_{\mathbf{z}}^{\left(1\right)}$does not commute with ${\mathbf{S}}^{\mathbf{2}}$, we cannot hope to find states that are simultaneous eigenvectors of both. In order to form eigenstates of${\mathbf{S}}^{\mathbf{2}}$weneed linear combinations of eigenstates of${\mathbf{S}}_{\mathbf{z}}^{\left(1\right)}$. This is precisely what the Clebsch-Gordan coefficients (in Equation 4.185) do for us, On the other hand, it follows by obvious inference from Equation 4.187that the sumrole="math" localid="1655980965321" ${\mathit{S}}^{\left(1\right)}\mathbf{+}{\mathit{S}}^{\left(2\right)}$does commute withdata-custom-editor="chemistry" ${\mathit{S}}^{\mathbf{2}}$, which is a special case of something we already knew (see Equation 4.103).

Short answer

$$\begin{gathered} S^2,S_z^{\left(1\right)}=2i\sout{h}{S}_x^{\left(1\right)}{S}_y^{\left(2\right)}-S_y^{\left(1\right)}{S}_x^{\left(2\right)} \\ Onemayapplythefollowinggeneralizationtotheresult: \\ {S}^2,S^{\left(1\right)}=2i\sout{h}{S}^{\left(1\right)}\times S^{\left(2\right)} \end{gathered}$$

Step-by-step solution

Define Eigenstate

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).

Determine the commutator of S2 with Sz(1)

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

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

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

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