 Solution

Applying the angular momentum operator directly givesFor an eigenvalue $\lambda$, comparison with $\lambda (f(x)+iy)z$ gives $\hslash x=\lambda f(x)$ and $\hslash f^{\prime }(x)=\lambda$. The first identity excludes $\lambda =0$; differentiating it and using the second gives ${\lambda }^2={\hslash }^2$. No additive constant in $f$ is permitted. The possibilities are $f(x)=x$, $\lambda =\hslash$, and $f(x)=-x$, $\lambda =-\hslash$.

For $f=x$, the polynomial $\psi =(x+iy)z$ is homogeneous of degree two and satisfies $\Delta \psi =0$. Expanding the three squared orbital angular momentum operators gives, with $D=x{\partial }_x+y{\partial }_y+z{\partial }_z$,Here $D\psi =2\psi$ and $D^2\psi =4\psi$. Therefore ${\hat{L}}^2\psi =6{\hslash }^2\psi$. This is the degree-two case of angular momentum of a homogeneous harmonic polynomial; its angular labels are $l=2$, $m=1$.