The Commutators of the Angular Momentum Operators

$$\bgroup\color{black} [L_x,L_y]\neq 0 ,\egroup$$

however, the square of the angular momentum vector commutes with all the components.

$$\bgroup\color{black}[L^2,L_z]=0\egroup$$

This will give us the operators we need to label states in 3D central potentials.

Lets just compute the commutator.

$$\begin{eqnarray*}[L_x,L_y]&=&[yp_z-zp_y,zp_x-xp_z]=y[p_z,z]p_x+[z,p_z]p_yx \\ &=&{\hbar\over i}[yp_x-xp_y]=i\hbar L_z \\ \end{eqnarray*}$$

Since there is no difference between $x$, $y$ and $z$, we can generalize this to

$$\bgroup\color{black} [L_i,L_j]=i\hbar\epsilon_{ijk}L_k \egroup$$

where $\epsilon_{ijk}$ is the completely antisymmetric tensor and we assume a sum over repeated indices.

$$\bgroup\color{black} \epsilon_{ijk}=-\epsilon_{jik}=-\epsilon_{ikj}=-\epsilon_{kji} \egroup$$

The tensor is equal to 1 for cyclic permutations of 123, equal to -1 for anti-cyclic permutations, and equal to zero if any index is repeated. It is commonly used for a cross product. For example, if

$$\bgroup\color{black} \vec{L}=\vec{r}\times\vec{p} \egroup$$

then

$$\bgroup\color{black} L_i=r_jp_k\epsilon_{ijk} \egroup$$

where we again assume a sum over repeated indices.

Now lets compute commutators of the $L^2$ operator.

$$\begin{eqnarray*} L^2&=&L_x^2+L_y^2+L_z^2 \\ {[L_z,L^2]}&=&[L_z,L_x^2]+[L_z,L... ..._y[L_z,L_y] \\ &=&i\hbar (L_yL_x+L_xL_y-L_xL_y-L_yL_x)=0 \\ \end{eqnarray*}$$

We can generalize this to

$$\bgroup\color{black} [L_i,L^2]=0 .\egroup$$

$L^2$ commutes with every component of $\vec{L}$.