# Zeeman effect - accidental degeneracy for high fields?

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.

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.