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If I understand it correctly the anomalously large total magnetic moment of the proton inside the nucleus of the atom is because it is a combinatoric particle and there are contributions by the quarks' spin and orbital angular momenta and also the possible orbital angular momentum of the proton around the neutron or other proton etc...

However, I am interested purely in the Dirac spin magnetic dipole moment value of the proton generated purely by its spin angular momentum and not any orbital angular momentum or other components.

This WP article does not give a clear answer to this although it maybe implied that it is equal with one nuclear magneton $μ_{Ν}$:

$$ \mu_{\mathrm{N}}=\frac{e \hbar}{2 m_{\mathrm{p}}} $$ $$ \mu_{N}=5.050783699(31) \times 10^{-27} \mathrm{~J} / \mathrm{T} $$

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This is a question that doesn't really have an answer, because you can't really separate the proton's angular momentum into "spin" and "orbital" parts.

For electronic systems, you have pretty good quantum numbers for both spin and orbital angular momentum. But in nuclear systems, the spin-orbit interaction is large, and states with well-defined energies don't correspond to states with well-defined $S$ or $L$. This is true for states within nuclei and for excitations of the nucleons themselves.

A structureless particle whose angular momentum comes entirely from spin would have a g-factor of $g=2$, with the first virtual-particle correction $a = \frac{g-2}{2} = \frac{\alpha}{2\pi} \approx 10^{-3}$ . The nucleons have $g_\text{neutron} = -3.8$ and $g_\text{proton} = +5.6$.

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  • $\begingroup$ Well, very good explanation, however you could effectively assign the proton as a Dirac particle and extract its effective spin magnetic moment given by the equation shown in the question. It is not by coincidence that both equations for the Bohr magneton and Nuclear magneton for g-factor=2 are identical and differentiate only in the mass. IMO, this is the whole meaning of the Nuclear Magneton concept and value. Thus the effective value of the proton's spin magnetic moment if it is considered as a Dirac particle. $\endgroup$
    – Markoul11
    Apr 11, 2022 at 16:59
  • $\begingroup$ Keep in mind that the magnetic moment of the zero-charge neutron is all anomalous. $\endgroup$
    – rob ♦
    Apr 11, 2022 at 17:23
  • $\begingroup$ More perplexing news about the proton-neutron spin in this recent Nature publication: arxiv.org/abs/2103.03333 The authors collaboration (more than 100 authors) report a newly experimental found large discrepancy on the precession. $\endgroup$
    – Markoul11
    Apr 12, 2022 at 7:24
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    $\begingroup$ Nature Physics, not Nature. We don't have a good theory of how the nucleon behaves in terms of its quarks. Chiral perturbation theory is promising, but apparently mispredicts that experiment. The neutron's $g$-factor was not measured in that paper. $\endgroup$
    – rob ♦
    Apr 12, 2022 at 15:12

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