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.

  • 2
    $\begingroup$ 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 $E = \hbar \omega (2 n_r + l + \frac32)$. $\endgroup$ Commented Oct 4, 2022 at 18:27

1 Answer 1



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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.