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Use the position-space momentum operator $P_j=-i\hslash {\partial }_j$ and set $F=(x+iy)z$. Direct differentiation givesTo compute the squared orbital angular momentum, write $D_x=y{\partial }_z-z{\partial }_y$, $D_y=z{\partial }_x-x{\partial }_z$, $D_z=x{\partial }_y-y{\partial }_x$, so $L_j=-i\hslash D_j$. Expanding the products, including derivatives of their variable coefficients, givesFor example $D_x^2=y^2{\partial }_z^2+z^2{\partial }_y^2-2yz{\partial }_y{\partial }_z-y{\partial }_y-z{\partial }_z$; summing the three cyclic expressions proves the identity.

Here $\Delta F=0$ and $EF=2F$, so $E^{2}F=4F$. ThereforeThis is the angular momentum of a homogeneous harmonic polynomial, with angular quantum numbers $\ell =2,m=1$. The polynomial is a nonzero angular eigenfunction; by itself it is not a normalizable full-space wavefunction.

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 $l=2$, $m=1$.