Why the angle of nuclear magnetic moment of the proton with an external magnetic field is 54,7°?Why it is not completely parallel to the ext field?

Question

When trying to understand the basics of Nuclear Magnetic Resonance I always found that the angle between the nuclear magnetic moment of the proton and the allied external magnetic field is 54,7° but I cannot find the explanation nor the math to get to that number. And moreover I cannot understand why the nuclear magnetic moment does not completely align to the external magnetic field direction since that should be the lower energy configuration.

Answer 1

The spin angular momentum for a particle with spin $s$ is given by:

$$ S = \hbar\sqrt{s(s+1)} $$

so for a proton with spin $\tfrac12$ the angular momentum is $\hbar\sqrt{3}/2$.

However when you put a spin $\tfrac12$ particle in an external field the component of the angular momentum along the field can only have values $S_z = \pm\hbar/2$. So the angle of the spin to the field has to be given by:

$$ \hbar\sqrt{3}/2 \cos\theta = \hbar/2 $$

Hence:

$$ \theta = \cos^{-1}(1/\sqrt{3}) \approx 54.7^\circ $$

The reason why the angular momentum is not exactly aligned with the external magnetic field is a little involved, but a simple way of looking at it is using the uncertainty principle. When applied to spin the uncertainty principle tells us that only one of the three components $s_x$, $s_y$ and $s_z$ can be known precisely. For our proton we know $s_z = \pm\hbar/2$ precisely, but we don't know $s_x$ or $s_y$ at all so that's fine. However if the spin were aligned along the $z$ axis not only would we know $s_z = \hbar\sqrt{s(s+1)}$ but we would also know $s_x = s_y = 0$. That means we would know all three components precisely and this contradicts the uncertainty principle.