= Solution
Use the position-space <momentum operator> $P_j=-i\hbar\partial_j$ and set $F=(x+iy)z$. Direct differentiation gives
$$
L_zF=-i\hbar(x\partial_y-y\partial_x)F=-i\hbar(ix-y)z=\hbar F.
$$
To 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 $L_j=-i\hbar D_j$. Expanding the products, including derivatives of their variable coefficients, gives
$$
D_x^2+D_y^2+D_z^2=r^2\Delta-E^2-E,\qquad E=x\partial_x+y\partial_y+z\partial_z.
$$
For 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 $E^2F=4F$. Therefore
$$
\boxed{\mathbf L^2F=6\hbar^2F,\qquad L_zF=\hbar F}.
$$
This 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>.
Back to article page
