 Solution

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.