I have a question regarding the Zeeman effect and perturbation theory in the hydrogen atom.

We have hamiltonian of the hydrogen atom is given by $H_0$, that of spin-orbit coupling given by $H_{\text{so}}$ and that of Zeeman given by $H_z$.

In standard textbook treatments, the Zeeman effect is divided into three regimes:

  • Weak-field Zeeman effect: here, spin-orbit coupling dominates, we take $H_z$ as a perturbation on top of $H_0+H_{\text{so}}$ (which is already a perturbation to $H_0$).

  • Strong-field Zeeman effect: here we take $H_{\text{so}}$ as a perturbation on top of $H_0+H_z$

  • Intermediate-field: here we take $H_{\text{so}}+H_z$ as a pertubation to $H_0$.

it seems to me that the “Strong-field Zeeman effect” is the most accurate treatment and I think it can work for both intermediate and weak fields. Here’s why:

When we take $H_0+H_z$ as our unperturbed hamiltonian, we know the exact solution to this ( it’s the states $ \vert n,l,m_l, m_s\rangle$ ) with the corresponding exact energies for all field strength. Here we apply perturbation theory/approximation only once.

On the other hand, in the “weak” field approach, we get energy corrections by applying perturbation theory to $H_z$ and $H_{\text{so}}$ which are two successive approximations/perturbations.

The same holds for intermediate field approach, even worse, we're forced to resort to degenerate perturbation theory.

So why we don’t use the “strong” field approach? it seems to be the best:

  • It is general, and applicable to all cases regardless of field strength.

  • It is more accurate, since it only involves one approximate result (energy correction due to $H_{\text{so}}$ by perturbation theory) as opposed to the other two approaches which involves two approximate energy corrections.


We can't solve equation by keeping all three terms in the Hamiltonian. Thats why we neglect spin orbit interaction in the very strong field domain, and we take that as main part in the weak field. General approach should work for all range of magnetic fields. But it is not possible and strong field results cannot be applied for weak field. Refer 'Physics of atoms and molecules' by Bransden. I tried to explain this here : https://youtu.be/tVoUTFageok


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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