Angular momentum and magnetic moment I have just started studying MRI physics and was reading F.Bloch’s paper on Nuclear Induction.
https://doi.org/10.1103/PhysRev.70.460
In page 463, it is mentioned,

To obtain this variation does not require the solution of the Schroedinger equation. It is enough to remember the general fact that the quantum-mechanical expectation value of any quantity follows in its time dependence exactly the classical equations of motion and that the magnetic and angular momenta of each nucleus are parallel to each other.


The parallelity between the magnetic moment $\mu$ and the angular momentum a for each nucleus implies $\mu = \gamma a$

Are the magnetic and angular momenta of the proton always parallel to each other?
Why is this so?
 A: The Pauli equation in the weak magnetic field approximation is
$$
\left[\frac{1}{2m}(p^2-q(\vec{L}+2\vec{S})\cdot \vec{B})\right] |\psi\rangle = i\hbar\frac{\partial}{\partial t}|\psi\rangle
$$
which is itself obtained from the non-relativistic limit of the Dirac equation. The $\frac{q}{2m}\vec{L}\cdot \vec{B}$ and $\frac{2q}{2m}\vec{S}\cdot \vec{B}$ terms are exactly the perturbation to the Hamiltonian of the form $-\vec{\mu}\cdot\vec{B}$ in, for example, the Zeeman effect, so we can identify the orbital magnetic moment $\vec\mu_B$ with $\frac{q}{2m}$ and the spin magnetic moment $\vec\mu_S$ with $\frac{q}m\vec S$ - so the magnetic moment is aligned with the spin angular momentum.
A curious observation is that the spin magnetic moment for the electron is twice the classical result (the orbital magnetic momentum) - this factor of two$^\dagger$ is called the g factor, and typically varies for different subatomic particles (analogous results hold)
$\dagger$ Actually, loop diagrams in QED lead to a g-factor slightly greater than two: $2 + \frac{\alpha}{\pi} + \ldots$ as a perturbation series.
