Prove the commutation relation in Equation 9.74. Hint: First show that

$\mathbf{[}{\mathbf{L}}^{\mathbf{2}}\mathbf{,}\mathbf{z}\mathbf{]}\mathbf{=}\mathbf{2}\mathit{i}\mathit{h}\mathbf{(}\mathbf{x}{\mathbf{L}}_{\mathbf{y}}\mathbf{-}\mathbf{y}{\mathbf{L}}_{\mathbf{x}}\mathbf{-}\mathbf{i}\mathbf{h}\mathbf{z}\mathbf{)}$

Use this, and the fact that localid="1657963185161" $\mathbf{r}\mathbf{.}\mathbf{L}\mathbf{=}\mathbf{r}\mathbf{.}\mathbf{(}\mathbf{r}\mathbf{\times }\mathbf{p}\mathbf{)}\mathbf{=}\mathbf{0}$, to demonstrate that

$\mathbf{[}{\mathbf{L}}^{\mathbf{2}}\mathbf{,}\mathbf{[}{\mathbf{L}}^{\mathbf{2}}\mathbf{,}\mathbf{z}\mathbf{]}\mathbf{]}\mathbf{=}\mathbf{2}{\mathit{h}}^{\mathbf{2}}\mathbf{(}\mathbf{z}{\mathbf{L}}^{\mathbf{2}}\mathbf{+}{\mathbf{L}}^{\mathbf{2}}\mathbf{z}\mathbf{)}$

The generalization from z to r is trivial.

Fundamental quantum mechanical relationships that link successive operations on the wave function or state vector of two operators (L1 and L2) in opposing orders, i.e., L1 L2 and L2 L1. The operators' algebra is defined by the commutation relations.

$[L2,z]=[Lx2\text{},z]+[Ly2\text{},z]+[Lz2\text{},z]=Lx\text{}[Lx\text{},z]+[Lx\text{},z]Lx\text{}+Ly\text{}[Ly\text{},z]+[Ly\text{},z]Ly\text{}+Lz\text{}[Lz\text{},z]+[Lz\text{},z]Lz\text{}.$

$\text{But}\{\begin{array}{l}[L_x,z]=[y{p}_z-z{p}_y,z]=[y{p}_z,z]-[z{p}_y,z]=y[p_z,z]=-i\hslash y\\ [L_y,z]=[z{p}_x-x{p}_z,z]=[z{p}_x,z]-[x{p}_z,z]=-x[p_z,z]=i\hslash x\\ [L_z,z]=[x{p}_y-y{p}_x,z]=[x{p}_y,z]-[y{p}_x,z]=0\end{array}$

$\text{So:\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}[L^2,z]=L_x(-i\hslash y)+(-i\hslash y)L_x+L_y\left(i\hslash x\right)+\left(i\hslash x\right)L_y=i\hslash (-L_{x}y-y{L}_x+L_{y}x+x{L}_y).$

$\text{But}\{\begin{array}{l}{L}_{x}y=L_{x}y-y{L}_x+y{L}_x=[L_x,y]+y{L}_x=i\hslash z+y{L}_x\\ L_{y}x=L_{y}x-x{L}_y+x{L}_y=[L_y,x]+x{L}_y=-i\hslash z+x{L}_y\end{array}$

$\begin{array}{l}\text{So:}[L^2,z]=i\hslash (2x{L}_y-i\hslash z-2y{L}_x-i\hslash z)\Rightarrow [L^2,z]=2i\hslash (x{L}_y-y{L}_x-i\hslash z)\\ [L^2,[L^2,z]]=2i\hslash \{[L^2,x{L}_y]-[L^2,y{L}_x]-i\hslash [L^2,z]\}\\ =2i\hslash \{[L^2,x]L_y+x[L^2,L_y]-[L^2,y]L_x-y[L^2,L_x]-i\hslash (L^{2}z-z{L}^2)\}\\ [L^2,L_y]=[L^2,L_x]=0\end{array}$

so,

$[L^2,L_x]=0,[L^2,L_y]=0,[L^2,L_z]=0$(4.102).

$\begin{array}{l}[L^2,[L^2,z]]=2i\hslash \{2i\hslash (y{L}_z-z{L}_y-i\hslash x)L_y-2i\hslash (z{L}_x-x{L}_z-i\hslash y)L_x-i\hslash (L^{2}z-z{L}^2)\}[L2,[L2,z\left]\right]=2i,\\ or[L^2,[L^2,z]]=-2{\hslash }^2(2y{L}_z{L}_y\underset{-2z(L_x^2+L_y^2+L_z^2)+2z{L}_z^2}{\underbrace{-2z{L}_y^2-2z{L}_x^2}}-2i\hslash x{L}_y+2x{L}_z{L}_x+2i\hslash y{L}_x-L^{2}z+z\end{array}$

$\begin{array}{l}=-2{\hslash }^2(2y{L}_z{L}_y-2i\hslash x{L}_y+2x{L}_z{L}_x+2i\hslash y{L}_x+2z{L}_z^2-2z{L}^2-L^{2}z+z{L}^2)\\ =2{\hslash }^2(z{L}^2+L^{2}z)-4{\hslash }^2[\underset{L_{z}y}{\underbrace{(y{L}_z-i\hslash x)}}{L}_y+\underset{L_{z}x}{\underbrace{(x{L}_z+i\hslash y)}}{L}_x+z{L}_z{L}_z]\\ =2{\hslash }^2(z{L}^2+L^{2}z)-4{\hslash }^2\underset{L_z(r\cdot L)=0}{\underbrace{(L_{z}y{L}_y+L_{z}x{L}_x+L_{z}z{L}_z)}}=2{\hslash }^2(z{L}^2+L^{2}z)\end{array}$