Commutator of \bgroup\color{black}$L_x$\egroup and \bgroup\color{black}$L_y$\egroup

Angular momentum is defined by

\begin{displaymath}\bgroup\color{black} \vec{L}=\vec{r}\times\vec{p} .\egroup\end{displaymath}

So the components of angular momentum are

\begin{displaymath}\bgroup\color{black} L_z=xp_y-yp_x \egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black} L_x=yp_z-zp_y \egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black} L_y=zp_x-xp_z .\egroup\end{displaymath}

We wish to compute \bgroup\color{black}$[L_x,L_y]$\egroup which has all the coordinates and momenta in it.

The only operators that do not commute are the coordinates and their conjugate momenta.

\begin{displaymath}\bgroup\color{black} [x,y]=0 \egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black} [p_x,p_y]=0 \egroup\end{displaymath}


\begin{displaymath}\bgroup\color{black} [p_i,r_j]={\hbar\over i}\delta_{ij} \egroup\end{displaymath}

So now we just need to compute.

\begin{eqnarray*}[L_x,L_y]&=&[yp_z-zp_y,zp_x-xp_z]\\
&=&[yp_z,zp_x]-[yp_z,xp_z...
...-0-0+x[z,p_z]p_y\\
&=&{\hbar\over i}(yp_x-xp_y)=i\hbar L_z \\
\end{eqnarray*}


It is not necessary (or wise) to use the differential operators and a wave function crutch to compute commutators like this one. Use the known basic commutators when you can.

Jim Branson 2013-04-22