The “precession” of a spin one-half particle
Notice that we have not assumed anything special about the potential energy—it is just that energy whose derivative gives a force. For instance, in the Stern-Gerlach experiment we had the energy $U=-\FLPmu\cdot\FLPB$, which gives a force if $\FLPB$ has a spatial variation. If we wanted to give a quantum mechanical description, we would have said that the particles in one beam had an energy that varied one way and that those in the other beam had an opposite energy variation. (We could put the magnetic energy $U$ into the potential energy $V$ or into the “internal” energy $W$; it doesn’t matter.) Because of the energy variation, the waves are refracted, and the beams are bent up or down. (We see now that quantum mechanics would give us the same bending as we would compute from the classical mechanics.) From the dependence of the amplitude on potential energy we would also expect that if a particle sits in a uniform magnetic field along the $z$-direction, its probability amplitude must be changing with time according to \begin{equation*} e^{-(i/\hbar)(-\mu_zB)t}. \end{equation*}  (We can consider that this is, in effect, a definition of $\mu_z$.) In other words, if we place a particle in a uniform field $B$ for a time $\tau$, its probability amplitude will be multiplied by \begin{equation*} e^{-(i/\hbar)(-\mu_zB)\tau} \end{equation*}  over what it would be in no field. Since for a spin one-half particle, $\mu_z$ can be either plus or minus some number, say $\mu$, the two possible states in a uniform field would have their phases changing at the same rate but in opposite directions. The two amplitudes get multiplied by \begin{equation} \label{Eq:III:7:34} e^{\pm(i/\hbar)\mu B\tau}. \end{equation}  This result has some interesting consequences. Suppose we have a spin one-half particle in some state that is not purely spin up or spin down. We can describe its condition in terms of the amplitudes to be in the pure up and pure down states. But in a magnetic field, these two states will have phases changing at a different rate. So if we ask some question about the amplitudes, the answer will depend on how long it has been in the field. As an example, we consider the disintegration of the muon in a magnetic field. When muons are produced as disintegration products of $\pi$-mesons, they are polarized (in other words, they have a preferred spin direction). The muons, in turn, disintegrate—in about $2.2$ microseconds on the average—emitting an electron and two neutrinos: \begin{equation*} \mu\to e+\nu+\bar{\nu}. \end{equation*}  In this disintegration it turns out that (for at least the highest energies) the electrons are emitted preferentially in the direction opposite to the spin direction of the muon. Suppose then that we consider the experimental arrangement shown in Fig. 7–9. If polarized muons enter from the left and are brought to rest in a block of material at $A$, they will, a little while later, disintegrate. The electrons emitted will, in general, go off in all possible directions. Suppose, however, that the muons all enter the stopping block at $A$ with their spins in the $x$-direction. Without a magnetic field there would be some angular distribution of decay directions; we would like to know how this distribution is changed by the presence of the magnetic field. We expect that it may vary in some way with time. We can find out what happens by asking, for any moment, what the amplitude is that the muon will be found in the $(+x)$ state. We can state the problem in the following way: A muon is known to have its spin in the $+x$-direction at $t=0$; what is the amplitude that it will be in the same state at the time $\tau$? Now we do not have any rule for the behavior of a spin one-half particle in a magnetic field at right angles to the spin, but we do know what happens to the spin up and spin down states with respect to the field—their amplitudes get multiplied by the factor (7.34). Our procedure then is to choose the representation in which the base states are spin up and spin down with respect to the $z$-direction (the field direction). Any question can then be expressed with reference to the amplitudes for these states. Let’s say that $\psi(t)$ represents the muon state. When it enters the block $A$, its state is $\psi(0)$, and we want to know $\psi(\tau)$ at the later time $\tau$. If we represent the two base states by $(+z)$ and $(-z)$ we know the two amplitudes $\braket{+z}{\psi(0)}$ and $\braket{-z}{\psi(0)}$—we know these amplitudes because we know that $\psi(0)$ represents a state with the spin in the $(+x)$ state. From the results of the last chapter, these amplitudes are2 \begin{equation} \begin{aligned} \braket{+z}{+x}&=C_+=\frac{1}{\sqrt{2}},\\[1ex] \braket{-z}{+x}&=C_-=\frac{1}{\sqrt{2}}. \end{aligned} \label{Eq:III:7:35} \end{equation}  They happen to be equal. Since these amplitudes refer to the condition at $t=0$, let’s call them $C_+(0)$ and $C_-(0)$. Now we know what happens to these two amplitudes with time. Using (7.34), we have \begin{equation} \begin{aligned} C_+(t)&=C_+(0)e^{-(i/\hbar)\mu Bt},\\[1ex] C_-(t)&=C_-(0)e^{+(i/\hbar)\mu Bt}. \end{aligned} \label{Eq:III:7:36} \end{equation}  But if we know $C_+(t)$ and $C_-(t)$, we have all there is to know about the condition at $t$. The only trouble is that what we want to know is the probability that at $t$ the spin will be in the $+x$-direction. Our general rules can, however, take care of this problem. We write that the amplitude to be in the $(+x)$ state at time $t$, which we may call $A_+(t)$, is  \begin{equation*} A_+(t)=\braket{+x}{\psi(t)}= \braket{+x}{+z}\braket{+z}{\psi(t)}+ \braket{+x}{-z}\braket{-z}{\psi(t)} \end{equation*} 
  \begin{align*} A_+(t)&=\braket{+x}{\psi(t)}\\[.5ex] &=\braket{+x}{+z}\braket{+z}{\psi(t)}+ \braket{+x}{-z}\braket{-z}{\psi(t)}\notag \end{align*}  or \begin{equation} \label{Eq:III:7:37} A_+(t)=\braket{+x}{+z}C_+(t)+\braket{+x}{-z}C_-(t). \end{equation}  Again using the results of the last chapter—or better the equality $\braket{\phi}{\chi}=\braket{\chi}{\phi}\cconj$ from Chapter 5—we know that \begin{equation*} \braket{+x}{+z}=\frac{1}{\sqrt{2}},\quad \braket{+x}{-z}=\frac{1}{\sqrt{2}}. \end{equation*}  So we know all the quantities in Eq. (7.37). We get \begin{equation*} A_+(t)=\tfrac{1}{2}e^{(i/\hbar)\mu Bt}+ \tfrac{1}{2}e^{-(i/\hbar)\mu Bt}, \end{equation*}  or \begin{equation*} A_+(t)=\cos\frac{\mu B}{\hbar}\,t. \end{equation*}  A particularly simple result! Notice that the answer agrees with what we expect for $t=0$. We get $A_+(0)=1$, which is right, because we assumed that the muon was in the $(+x)$ state at $t=0$. The probability $P_+$ that the muon will be found in the $(+x)$ state at $t$ is $(A_+)^2$ or \begin{equation*} P_+=\cos^2\frac{\mu Bt}{\hbar}. \end{equation*}  The probability oscillates between zero and one, as shown in Fig. 7–10. Note that the probability returns to one for $\mu Bt/\hbar=\pi$ (not $2\pi$). Because we have squared the cosine function, the probability repeats itself with the frequency $2\mu B/\hbar$. Thus, we find that the chance of catching the decay electron in the electron counter of Fig. 7–9 varies periodically with the length of time the muon has been sitting in the magnetic field. The frequency depends on the magnetic moment $\mu$. The magnetic moment of the muon has, in fact, been measured in just this way.  We can, of course, use the same method to answer any other questions about the muon decay. For example, how does the chance of detecting a decay electron in the $y$-direction at $90^\circ$ to the $x$-direction but still at right angles to the field depend on $t$? If you work it out, the probability to be in the $(+y)$ state varies as $\cos^2\{(\mu Bt/\hbar)-\pi/4\}$, which oscillates with the same period but reaches its maximum one-quarter cycle later, when $\mu Bt/\hbar=\pi/4$. In fact, what is happening is that as time goes on, the muon goes through a succession of states which correspond to complete polarization in a direction that is continually rotating about the $z$-axis. We can describe this by saying that the spin is precessing at the frequency \begin{equation} \label{Eq:III:7:38} \omega_p=\frac{2\mu B}{\hbar}. \end{equation}  You can begin to see the form that our quantum mechanical description will take when we are describing how things behave in time.