The Angular Momentum Matrices

We can easily derive the matrices representing the angular momentum operators for $\ell =1$.

$$\bgroup\color{black}L_x={\hbar\over\sqrt{2}}\left(\matrix{0&1... ... L_z=\hbar\left(\matrix{1&0&0\cr 0&0&0\cr 0&0&-1}\right)\egroup$$

The matrices must satisfy the same commutation relations as the differential operators.

$$\bgroup\color{black}[L_x,L_y]=i\hbar L_z\egroup$$

We verify this with an explicit computation of the commutator.

Since these matrices represent physical variables, we expect them to be Hermitian. That is, they are equal to their conjugate transpose. Note that they are also traceless.

Let's compute an expectation value of $L_x$ in the matrix representation for the general state $\psi$.

$$\begin{eqnarray*} \langle\psi\vert L_x\vert\psi\rangle & = & \left(\matrix{\psi... ...}(\psi^*_1\psi_2 + \psi^*_2(\psi_1+\psi_3) + \psi^*_3\psi_2) \\ \end{eqnarray*}$$