2
$\begingroup$

In Griffiths, the hyperfine structure is described as follows:

enter image description here

enter image description here

So the hyperfine structure is a result of a mechanism called spin-spin coupling, which is the interaction of the spin of the nucleus (proton in this case) with the spin of the electron. The spin of the nucleus has a magnetic moment which creates a magnetic field and this magnetic field creates a torque on the spin/magnetic moment of the electron. So this lifts the degeneracy in the triplet state and the singlet state.

However, when I look up another image, I get the following: enter image description here

where $F = J + S$, where $J$ is the total angular momentum of the electron (spin and orbital angular momentum) and $S$ is the spin of the nucleus/proton. This lifts the degeneracy in the $m_j$ levels (if that is even a correct statement).

So my questions are:

  • Are these two things consistent with each other? (degeneracy in spin states vs. degeneracy in mj levels).

  • Is it even correct to state that hyperfine structure lifts the degeneracy in mj levels? (This is however true for the Zeeman effect), as there is a clear dependence on mj)

  • Is the hyperfine structure a result of the interaction between the spin of the nucleus and the spin of the electron (spin-spin coupling)? Or is it a result of the interaction between the spin of the nucleus and the total angular momentum of the electron (so spin of electron and orbital angular momentum of electron)?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Note that the calculation done in Griffiths is only for the ground state $l=0$ case in hydrogen. Therefore, orbital effects are irrelevant. The more general statement is that the total nuclear spin interacts with the total angular momentum of the electron. See the wikipedia page on hyperfine structure for some more detail. (And, if I remember correctly, Griffiths starts with a somewhat more general case but argues that for the $l=0$ state, most of the other stuff averages out.) $\endgroup$
    – march
    Commented May 20, 2023 at 17:43
  • $\begingroup$ Oh I see, so spin-spin coupling is only relevant for the hyperfine splitting for the ground state and not generally speaking? $\endgroup$
    – Stallmp
    Commented May 20, 2023 at 17:54

2 Answers 2

Reset to default
3
+50
$\begingroup$

Griffiths only calculates the hyperfine splitting for the ground state. However, as he states, his result can be easily generalized for any state with l=0. So, in the ground state, $J = \frac{1}{2}$ since l=0.
With that in mind, here are the answers to your questions:

  • Yes, they are. In the ground state (l=0) you only have one possible m (m=0) F=0/1 both only come from the spin degeneracy. Hence, your second image contains what Griffiths calculated (but showed a little more).
  • In general, it should since there is a term ~$ \vec L$. But keep in mind that in many cases already the fine structure does that.
  • As mentioned above, in case of $l \neq 0$ there is a term in the hyperfine splitting that contains $\vec L$. So it is not only a result of the interaction between the spin of the nucleus and the spin of the electron.
Improve this answer
$\endgroup$
2
$\begingroup$

Yes, they are consistent. Griffiths is describing only the hyperfine structure of the ground state, with zero orbital angular momentum.

enter image description here

For the two $^2P_J$ states, the proton spin may split the $J=1/2$ state into a singlet or a triplet of hyperfine states. But the proton can’t “cancel” the $J=3/2$ state into a singlet; its options are the $F=1$ triplet or the $F=2$ quintuplet.

Improve this answer
$\endgroup$

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.