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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 $μ_{Ν}$:

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

rob
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Markoul11
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1 Answers1

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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 $g_\text{proton} = +5.6$.

rob
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