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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)?

Stallmp
  • 879

2 Answers2

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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.
Tarik
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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.

rob
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