# On degeneracies of energy levels in atoms

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.

• For an example of a soluble system in which $l$ affects the energy, see the isotropic harmonic oscillator (i.e., $V \propto r^2$), for which $E = \hbar \omega (2 n_r + l + \frac32)$. Oct 4, 2022 at 18:27

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.