@ACuriousMind If you have a chance please see my post.

@Alex No, there is no "physical interpretation" in the sense you are probably demanding.

It's simply an experimental fact (Stern-Gerlach, Einstein-de Haas) that spin behaves like ordinary angular momentum for the pruposes of coupling to the magnetic field, but that doesn't mean that "something is spinning".

@ACuriousMind Do you maybe know why the eigenvectors for the z component of spin are given in the form $(0,1)$ and $(1, 0)$ and not say some other two linearly independent vectors which span two dimensional vectors space. Is it because you could simply rotate and scale any other two such vectors until you get $(0,1)$ and $(1,0)$?

@Alex We usually choose the eigenvectors of the z-spin as the basis for spin space. It's just a convention, you might as well choose the x-spin eigenvectors or the y-spin eigenvectors.

@ACuriousMind Okay thanks. Do you maybe know why it makes sense to then write the magnetic dipole as $$\mu = \gamma S$$ if the components of $S$ are never well defined since we can at most know one of them. Does this only make sense in the classical case?

@Alex Well, it also makes sense in the quantum case, but it's an operator there, not a value.

@ACuriousMind Oh okay, so it is just $\hat{S} = (\hat{S_x}, \hat{S_y}, \hat{S_z})$?

Yep

@ACuriousMind One more thing, am I right in stating that a solid rotating sphere of electric charge has a magnetic dipole moment in the direction of the rotation axis but if we put this object in a uniform magnetic field, it will acquire a Larmor precession of it's dipole moment.

@Alex Sure. The Larmor precession is still happening in QM, though: The expectation value of $\mu$ is not time-independent for a state that doesn't start in an eigenstate of the spin component parallel to the axis of the magnetic field.

@ACuriousMind So we don't have this type of rotation unless we start in an eigenstate parallel to the external magnetic field?

@Alex No, exactly the other way around - you have this rotation only if you start not parallel to the external field. A dipole parallel to the field won't precess classically either.

The Larmor force is $\mu\times B$, which vanishes if both are parallel

@ACuriousMind Yeah of course, that makes sense, thanks.

does anyone understand this new answer? :|