= Solution
Write
$$
\Lambda_\ell=\ell(\ell+1).
$$
Taking the stated double-commutator identity between the two eigenstates gives, if the matrix element is nonzero,
$$
\left(\Lambda_{\ell'}-\Lambda_\ell\right)^2
=2\left(\Lambda_{\ell'}+\Lambda_\ell\right).
$$
For nonnegative integers $\ell,\ell'$, this equation has precisely the possibilities
$$
|\ell'-\ell|=1
\qquad\text{or}\qquad
\ell'=\ell=0.
$$
Indeed, $\ell'=\ell$ forces $\ell=0$, while for unequal values the difference
$(\ell'-\ell)(\ell'+\ell+1)$ satisfies the equation only when its first factor has magnitude one.
Under parity, $z\mapsto-z$, while a hydrogen state of orbital angular momentum $\ell$ has parity $(-1)^\ell$. Therefore a nonzero matrix element requires
$$
(-1)^{\ell'+\ell+1}=1,
$$
so $\ell'$ and $\ell$ must have opposite parity. This excludes every case $\ell'=\ell$, including $\ell'=\ell=0$. Combining the results gives
$$
\boxed{\Delta\ell=\pm1,\qquad\Delta m=0}.
$$
Back to article page
