Past exam of the quantum physics course of the University of Oxford / 2016 / 1 / Solution

Insert hydrogen energy eigenstates on either side of the commutator. Because $L_z$ is Hermitian and has eigenvalue $\hslash m$,ThusThis is a necessary electric-dipole selection rule, not a sufficient condition for a nonzero matrix element. For example, $⟨200|z|200⟩=0$ by parity even though $m=m^{\prime }=0$. The angular factor $z=r\cos \theta$ additionally requires $l^{\prime }=l\pm 1$: multiplication of a spherical harmonic by $\cos \theta$ gives only harmonics with the same $m$ and adjacent $l$. The commutator alone establishes only the magnetic quantum number restriction; radial integrals and the remaining electric-dipole selection rules for hydrogen must still be considered.