I have just started studying MRI physics and was reading F.Bloch’s paper on Nuclear Induction.


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?

  • $\begingroup$ That's odd phrasing, since it implies that it's "magnetic momentum" rather than magnetic moment $\endgroup$ – Nihar Karve Nov 19 '20 at 2:19
  • $\begingroup$ @NiharKarve Yes, it is indeed odd. This sentence describes the variation with time of the vector M (resultant nuclear magnetic moment per unit volume). I have also edited the text to give the next paragraph which implies that he is talking about the magnetic moment. $\endgroup$ – Julian Nov 19 '20 at 7:13
  • 5
    $\begingroup$ Does this answer your question? Relation between magnetic moment and angular momentum -- classic theory $\endgroup$ – Nihar Karve Nov 19 '20 at 10:02
  • $\begingroup$ @NiharKarve I had seen this before and it says that the direction of the magnetic moment and the angular momentum is perpendicular to the plane of the loop and hence they are parallel but this a classical approach to it. I want to know if this is always the case since the magnetic moment of any particle is due to its intrinsic angular momentum. Also, if it is only true for only electrons or for all particles like protons $\endgroup$ – Julian Nov 19 '20 at 13:37
  • $\begingroup$ Well, it is true that the spin magnetic moment is always (anti)-parallel to the spin angular moment - that's the basis for the definition of the g-factor. I believe you can invoke the Dirac equation to see this. $\endgroup$ – Nihar Karve Nov 19 '20 at 13:48

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.

  • $\begingroup$ Thank you. This equation makes more sense. $\endgroup$ – Julian Nov 19 '20 at 14:38

Not the answer you're looking for? Browse other questions tagged or ask your own question.