= Solution
The states $|n,\ell,m\rangle$ diagonalize $L^2$:
$$
L^2|n,\ell,m\rangle
=\hbar^2\ell(\ell+1)|n,\ell,m\rangle.
$$
The perturbation therefore introduces no mixing. The <orbital-angular-momentum-squared perturbation of hydrogen> leaves
$$
|2,0,0\rangle
$$
at energy
$$
\boxed{E_2},
$$
while
$$
|2,1,-1\rangle,\quad |2,1,0\rangle,\quad |2,1,1\rangle
$$
remain triply degenerate at
$$
\boxed{E_2+2g\hbar^2}.
$$
These statements are exact because the unperturbed Hamiltonian commutes with $L^2$.
Back to article page
