The precession of atomic magnets
One of the consequences of having the magnetic moment proportional to the angular momentum is that an atomic magnet placed in a magnetic field will precess. First we will argue classically. Suppose that we have the magnetic moment $\FLPmu$ suspended freely in a uniform magnetic field. It will feel a torque $\FLPtau$, equal to $\FLPmu\times\FLPB$, which tries to bring it in line with the field direction. But the atomic magnet is a gyroscope—it has the angular momentum $\FLPJ$. Therefore the torque due to the magnetic field will not cause the magnet to line up. Instead, the magnet will precess, as we saw when we analyzed a gyroscope in Chapter 20 of Volume I. The angular momentum—and with it the magnetic moment—precesses about an axis parallel to the magnetic field. We can find the rate of precession by the same method we used in Chapter 20 of the first volume.  Suppose that in a small time $\Delta t$ the angular momentum changes from $\FLPJ$ to $\FLPJ'$, as drawn in Fig. 34-3, staying always at the same angle $\theta$ with respect to the direction of the magnetic field $\FLPB$. Let’s call $\omega_p$ the angular velocity of the precession, so that in the time $\Delta t$ the angle of precession is $\omega_p\,\Delta t$. From the geometry of the figure, we see that the change of angular momentum in the time $\Delta t$ is \begin{equation*} \Delta J=(J\sin\theta)(\omega_p\,\Delta t). \end{equation*} So the rate of change of the angular momentum is \begin{equation} \label{Eq:II:34:8} \ddt{J}{t}=\omega_pJ\sin\theta, \end{equation} which must be equal to the torque: \begin{equation} \label{Eq:II:34:9} \tau=\mu B\sin\theta. \end{equation} The angular velocity of precession is then \begin{equation} \label{Eq:II:34:10} \omega_p=\frac{\mu}{J}\,B. \end{equation}  Substituting $\mu/J$ from Eq. (34.6), we see that for an atomic system \begin{equation} \label{Eq:II:34:11} \omega_p=g\,\frac{q_eB}{2m}; \end{equation} the precession frequency is proportional to $B$. It is handy to remember that for an atom (or electron) \begin{equation} \label{Eq:II:34:12} f_p=\frac{\omega_p}{2\pi}=(\text{$1.4$ megacycles/gauss})gB, \end{equation} and that for a nucleus \begin{equation} \label{Eq:II:34:13} f_p=\frac{\omega_p}{2\pi}=(\text{$0.76$ kilocycles/gauss})gB. \end{equation} (The formulas for atoms and nuclei are different only because of the different conventions for $g$ for the two cases.)  According to the classical theory, then, the electron orbits—and spins—in an atom should precess in a magnetic field. Is it also true quantum-mechanically? It is essentially true, but the meaning of the “precession” is different. In quantum mechanics one cannot talk about the direction of the angular momentum in the same sense as one does classically; nevertheless, there is a very close analogy—so close that we continue to call it “precession.” We will discuss it later when we talk about the quantum-mechanical point of view. 