Eigenvectors of $L_x$

We will do it as if we don't already know that the eigenvalues are $m\hbar$.

$$\begin{eqnarray*} &L_x\psi=a\psi \\ & \left(\matrix{0&1&0\cr 1&0&1\cr 0&1&0}\... ... & \left\vert\matrix{-b&1&0\cr 1&-b&1\cr 0&1&-b}\right\vert=0 \end{eqnarray*}$$

where $a={\hbar\over\sqrt{2}}b$.

$$\begin{eqnarray*} & -b(b^2-1)-1(-b-0)=0 \\ & b(b^2-2)=0 \end{eqnarray*}$$

There are three solutions to this equation: $b=0$, $b=+\sqrt{2}$, and $b=-\sqrt{2}$ or $a=0$, $a=+\hbar$, and $a=-\hbar$. These are the eigenvalues we expected for $\ell=1$. For each of these three eigenvalues, we should go back and find the corresponding eigenvector by using the matrix equation.

$$\bgroup\color{black}\left(\matrix{0&1&0\cr 1&0&1\cr 0&1&0}\... ...right)=b\left(\matrix{\psi_1\cr \psi_2\cr \psi_3}\right)\egroup$$

$$\bgroup\color{black}\left(\matrix{\psi_2\cr \psi_1+\psi_3\cr ... ...right)=b\left(\matrix{\psi_1\cr \psi_2\cr \psi_3}\right)\egroup$$

Up to a normalization constant, the solutions are:

$$\begin{eqnarray*} \psi_{+\hbar}=c\left(\matrix{{1\over\sqrt{2}}\cr 1\cr {1\over\... ...ft(\matrix{{-1\over\sqrt{2}}\cr 1\cr {-1\over\sqrt{2}}}\right). \end{eqnarray*}$$

We should normalize these eigenvectors to represent one particle. For example:

$$\begin{eqnarray*} \langle\psi_{+\hbar}\vert\psi_{+\hbar}\rangle&=&1 \\ \vert c\... ...r\sqrt{2}}}\right)&=&2\vert c\vert^2=1 \\ c&=&{1\over\sqrt{2}}. \end{eqnarray*}$$

Try calculating the eigenvectors of $L_y$.
You already know what the eigenvalues are.