For a low level field, the energy splittings are given by $$\Delta E(L, S, J, m_J) = \mu_B m_j g_L B_z$$ where $g_L$ is a g-factor.

This predicts a separation into $(2J+1)$ levels, so completely lifting the degeneracy of the $J$ states.

This is what I've always been taught: applying an external magnetic field lifts any degeneracy.

However, now we turn to the case of a strong field, i.e. assuming the perturbing magnetic field is much greater than the effects of spin-orbit coupling.

Using perturbation theory, my lecture notes derive that the energy splittings are now $$\Delta E(L, S, m_l, m_s) = \mu_B B_z (m_l + 2m_s)$$

However, the notes then note that now we still have some accidental degenerate levels. Notice that when $m_l + 2m_s = 0$, we don't have any splittings. So for a $^2P$ state $m_l = +1, m_s = -\frac{1}{2}$ and $m_l = -1, m_s = \frac{1}{2}$ do not split and have the same energy.

My question is how is the possible? For a low field, the levels completely split and then when you crank up the field a bit more so you end up in the higher field region, some of the levels become degenerate again or am I misunderstanding something? Why is this happening?

Edit - attached an exam question about this to question the point about being 'approximately' degenerate. This question seems to suggest there is a specific definitive degeneracy.

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1 Answer 1


Exact symmetries are used to explain exact degeneracies. The system you're talking about here is not actually exactly degenerate, because the energy splittings your lecture notes presented are just the first term in an infinite series, so we don't need an exact symmetry to explain it.

However, it is true that in the limit $B \to \infty$ one gets closer and closer to degeneracy, so we expect we should have an arbitrarily good symmetry in this limit. In this limit, the electric field from the nucleus doesn't matter, so all we have is a constant magnetic field. That means the symmetry is rotational symmetry about the $z$ axis. States with the same $m_l + 2 m_s$ are degenerate because they have the same angular momentum about the $z$ axis.

  • $\begingroup$ I'm not sure what you mean by the system is not exactly degenerate. We had an exam question saying 'which levels are remain degenerate under the application of the field?'. So would such a question technically be wrong because it's not really degenerate? $\endgroup$ Apr 18, 2019 at 15:27
  • $\begingroup$ Secondly, am I right in saying that what we have is the following: low field -> complete splitting of energy levels (non degeneracy) then increasing field -> getting closer to a degeneracy again? This seems to imply that there exists a value for the field such that we have 'maximum non-degeneracy'? $\endgroup$ Apr 18, 2019 at 15:28
  • $\begingroup$ @PhysicsMathsLove That exam question is a bit unclear, but it presumably means to ask which remain approximately degenerate. $\endgroup$
    – knzhou
    Apr 18, 2019 at 15:29
  • $\begingroup$ I've added the exam question to the main thread... it seems to suggest there is a definitive degeneracy? $\endgroup$ Apr 18, 2019 at 15:32
  • $\begingroup$ @PhysicsMathsLove Not really. For example, it starts by talking about "previously" degenerate energy levels, even without a magnetic field. That's not exactly true either, there's fine structure, hyperfine splitting... $\endgroup$
    – knzhou
    Apr 18, 2019 at 17:53

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