On degeneracies of energy levels in atoms

Question

For hydrogenic atoms the energy levels are (in cgs units): $$ E_n = -\frac{e^2Z^2}{2n^2a_0} $$ This formula shows there's no dependence on quantum numbers $l$ and $m_l$. These so-called degeneracies arise from two distinct aspects:

1. In the Schrödinger equation, the potential $V(\mathbf{r})$ of every hydrogenic atom is radial, namely, it has spherical symmetry: $\displaystyle V(\mathbf{r})\sim r^\alpha, \quad \alpha\in\mathbb{Z}$. This implies an absence of $m_l$.
2. The potential is Coulombic ($\alpha = -1$). This cancels the dependence on $l$.

My question is: can there be a dependence on $l$ but not on $m_l$? If yes, how can it be? The presence of $l$ means that orbitals $s$, $p$, $d$, $f$, etc. have different energies, but since orbital $s$ is the only one with spherical symmetry, the presence of any other kind of orbital would break the spherical symmetry of the potential (point 1.), so I should expect dependence on $m_l$ as well.

Answer 1

Yes.

The fine structure and hyperfine structure, for instance, have energy splittings that go as $\propto J$ and $\propto F$, which depend on the quantum numbers $j$ and $f$ and hence essentially $\ell$ (since $s$, the spin, is fixed).

To break the $m_j$ degeneracies you actually have to break another symmetry. E.g. when you apply a uniform electric field you cause a Stark shift that breaks rotational symmetry which distinguishes among the $|m_j|$, and when you apply a uniform magnetic field you cause a Zeeman shift that breaks time reversal symmetry which distinguishes between the $\pm m_j$.

You have to remember that the spherical symmetry of the potential does not mean every single eigenstate has the same spherical symmetry. Not depending on $m_j$ means you are in a superposition, which will likely have a spherical symmetry. So you need to break that further by picking an axis.