Question

Question: Evaluate the following commutators :

a)$\left[L·S,L\right]$

b)$\mathbf{[}\mathbf{L}\mathbf{·}\mathbf{S}\mathbf{,}\mathbf{S}\mathbf{]}$

c)role="math" localid="1658226147021" $\mathbf{}\mathbf{[}\mathbf{L}\mathbf{·}\mathbf{S}\mathbf{,}\mathbf{J}\mathbf{]}$

d)$\mathbf{[}\mathbf{L}\mathbf{·}\mathbf{S}\mathbf{,}{\mathbf{L}}^{\mathbf{2}}\mathbf{]}$

e)$\mathbf{[}\mathbf{L}\mathbf{·}\mathbf{S}\mathbf{,}{\mathbf{S}}^{\mathbf{2}}\mathbf{]}$

f)$\mathbf{[}\mathbf{L}\mathbf{·}\mathbf{S}\mathbf{,}{\mathbf{J}}^{\mathbf{2}}\mathbf{]}$

Hint: L and S satisfy the fundamental commutation relations for angular momentum (Equations 4.99 and 4.134 ), but they commute with each other.

$$\begin{gathered} \left[L_X,L_Y\right]\mathbf{=}{\mathbf{ihL}}_{\mathbf{z}}\mathbf{;}\mathbf{[}{\mathbf{L}}_{\mathbf{y}}\mathbf{,}{\mathbf{L}}_{\mathbf{z}}\mathbf{]}\mathbf{=}{\mathbf{ihL}}_{\mathbf{x}}\mathbf{;}\mathbf{[}{\mathbf{L}}_{\mathbf{z}}\mathbf{,}{\mathbf{L}}_{\mathbf{x}}\mathbf{]}\mathbf{=}{\mathbf{ihL}}_{\mathbf{y}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{4}\mathbf{.}\mathbf{99} \\ \mathbf{[}{\mathbf{S}}_{\mathbf{X}}\mathbf{,}{\mathbf{S}}_{\mathbf{Y}}\mathbf{]}\mathbf{=}{\mathbf{ihS}}_{\mathbf{z}}\mathbf{;}\mathbf{[}{\mathbf{S}}_{\mathbf{y}}\mathbf{,}{\mathbf{S}}_{\mathbf{z}}\mathbf{]}\mathbf{=}{\mathbf{ihS}}_{\mathbf{x}}\mathbf{;}\mathbf{[}{\mathbf{S}}_{\mathbf{z}}\mathbf{,}{\mathbf{S}}_{\mathbf{x}}\mathbf{]}\mathbf{=}{\mathbf{ihS}}_{\mathbf{y}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{4}\mathbf{.}\mathbf{134} \end{gathered}$$

Short answer

a)$\mathrm{i}\sout{\mathrm{h}}\left(\mathrm{L}\times \mathrm{S}\right)$

b)$\mathrm{i}\sout{\mathrm{h}}\left(S\times \mathrm{L}\right)$

c) 0

d) 0

e) 0

f) 0

Step-by-step solution

Step 1:(a)

$$\begin{gathered} \left[L·S,L_x\right]=\left[L_x{S}_x+L_y{S}_y+L_z{S}_z,L_X\right] \\ =S_x\left[L_x,L_x\right]+S_y\left[L_y,L_x\right]+S_z\left[L_z,L_x\right] \\ =S_x\left(0\right)+S_y(-i\sout{h}{L}_z)+S_z(-i\sout{h}{L}_y) \\ =i\sout{h}(L_y{S}_z-L_z{S}_y) \\ =i\sout{h}{(L\times S)}_x \end{gathered}$$

Same goes for the other two components, so

$\left[L·S,L\right]=i\sout{h}(L\times S)$

Step 2:(b)

$\left[L·S,S\right]$is identical, only with$L\leftrightarrow S$ :

$\left[L·S,S\right]=i\sout{h}(S\times L)$

:(c)

$$\begin{gathered} \left[L·S,J\right]=\left[L·S,L\right]+\left[L·S,S\right] \\ =i\sout{h}(L\times S+S\times L) \\ =0 \end{gathered}$$

Step 4:(d)

$L^2$Commutes with all components of L (and S),

So$\left[L.S,L^2\right]=0$

Step 5:(e)

Likewise$\left[L.S,S^2\right]=0$

Step 6:(f)

$$\begin{gathered} \left[L.S,{\mathrm{J}}^2\right]=\left[L.S,L^2\right]+\left[L.S,S^2\right]+2\left[L.S,L.S\right] \\ =0+0+0 \\ =\left[L.S,{\mathrm{J}}^2\right]=0 \end{gathered}$$