The spinning electron in a magnetic field
Example number one: We start with a constant field in the $z$-direction. There are just the two stationary states with energies $\mp\mu B_z$. Suppose we add a small field in the $x$-direction. Then the equations look like our old two-state problem. We get the flip-flop business once more, and the energy levels are split a little farther apart. Now let’s let the $x$-component of the field vary with time—say, as $\cos\omega t$. The equations are then the same as we had when we put an oscillating electric field on the ammonia molecule in Chapter 9. You can work out the details in the same way. You will get the result that the oscillating field causes transitions from the $+z$-state to the $-z$-state—and vice versa—when the horizontal field oscillates near the resonant frequency $\omega_0=2\mu B_z/\hbar$. This gives the quantum mechanical theory of the magnetic resonance phenomena we described in Chapter 35 of Volume II. It is also possible to make a maser which uses a spin one-half system. A Stern-Gerlach apparatus is used to produce a beam of particles polarized in, say, the $+z$-direction, which are sent into a cavity in a constant magnetic field. The oscillating fields in the cavity can couple with the magnetic moment and induce transitions which give energy to the cavity. Now let’s look at the following question. Suppose we have a magnetic field $\FLPB$ which points in the direction whose polar angle is $\theta$ and azimuthal angle is $\phi$, as in Fig. 10–10. Suppose, additionally, that there is an electron which has been prepared with its spin pointing along this field. What are the amplitudes $C_1$ and $C_2$ for such an electron? In other words, calling the state of the electron $\ket{\psi}$, we want to write \begin{equation*} \ket{\psi}=\ketsl{\slOne}C_1+\ketsl{\slTwo}C_2, \end{equation*}  where $C_1$ and $C_2$ are \begin{equation*} C_1=\braket{\slOne}{\psi},\quad C_2=\braket{\slTwo}{\psi}, \end{equation*}  where by $\ketsl{\slOne}$ and $\ketsl{\slTwo}$ we mean the same thing we used to call $\ket{+}$ and $\ket{-}$ (referred to our chosen $z$-axis). The answer to this question is also in our general equations for two-state systems. First, we know that since the electron’s spin is parallel to $\FLPB$ it is in a stationary state with energy $E_{\slI}=-\mu B$. Therefore, both $C_1$ and $C_2$ must vary as $e^{-iE_{\slI}t/\hbar}$, as in (9.18); and their coefficients $a_1$ and $a_2$ are given by (10.5), namely, \begin{equation} \label{Eq:III:10:24} \frac{a_1}{a_2}=\frac{H_{12}}{E_{\slI}-H_{11}}. \end{equation}  An additional condition is that $a_1$ and $a_2$ should be normalized so that $\abs{a_1}^2+\abs{a_2}^2=1$. We can take $H_{11}$ and $H_{12}$ from (10.22) using  \begin{equation*} B_z=B\cos\theta,\quad B_x=B\sin\theta\cos\phi,\quad B_y=B\sin\theta\sin\phi. \end{equation*} 
 \begin{align*} B_z&=B\cos\theta,\\[.6ex] B_x&=B\sin\theta\cos\phi,\\[.6ex] B_y&=B\sin\theta\sin\phi. \end{align*}  So we have \begin{equation} \begin{aligned} H_{11}&=-\mu B\cos\theta,\\[1ex] H_{12}&=-\mu B\sin\theta\,(\cos\phi-i\sin\phi). \end{aligned} \label{Eq:III:10:25} \end{equation}  The last factor in the second equation is, incidentally, $e^{-i\phi}$, so it is simpler to write \begin{equation} \label{Eq:III:10:26} H_{12}=-\mu B\sin\theta\,e^{-i\phi}. \end{equation}  Using these matrix elements in Eq. (10.24)—and canceling $-\mu B$ from numerator and denominator—we find \begin{equation} \label{Eq:III:10:27} \frac{a_1}{a_2}=\frac{\sin\theta\,e^{-i\phi}}{1-\cos\theta}. \end{equation}  With this ratio and the normalization condition, we can find both $a_1$ and $a_2$. That’s not hard, but we can make a short cut with a little trick. Notice that $1-\cos\theta=2\sin^2\,(\theta/2)$, and that $\sin\theta=2\sin\,(\theta/2)\cos\,(\theta/2)$. Then Eq. (10.27) is equivalent to \begin{equation} \label{Eq:III:10:28} \frac{a_1}{a_2}=\frac{\cos\dfrac{\theta}{2}\,e^{-i\phi}} {\sin\dfrac{\theta}{2}}. \end{equation}  So one possible answer is \begin{equation} \label{Eq:III:10:29} a_1=\cos\frac{\theta}{2}\,e^{-i\phi},\quad a_2=\sin\frac{\theta}{2}, \end{equation}  since it fits with (10.28) and also makes \begin{equation*} \abs{a_1}^2+\abs{a_2}^2=1. \end{equation*}  As you know, multiplying both $a_1$ and $a_2$ by an arbitrary phase factor doesn’t change anything. People generally prefer to make Eqs. (10.29) more symmetric by multiplying both by $e^{i\phi/2}$. So the form usually used is \begin{equation} \label{Eq:III:10:30} a_1=\cos\frac{\theta}{2}\,e^{-i\phi/2},\quad a_2=\sin\frac{\theta}{2}\,e^{+i\phi/2}, \end{equation}  and this is the answer to our question. The numbers $a_1$ and $a_2$ are the amplitudes to find an electron with its spin up or down along the $z$-axis when we know that its spin is along the axis at $\theta$ and $\phi$. (The amplitudes $C_1$ and $C_2$ are just $a_1$ and $a_2$ times $e^{-iE_{\slI}t/\hbar}$.) Now we notice an interesting thing. The strength $B$ of the magnetic field does not appear anywhere in (10.30). The result is clearly the same in the limit that $B$ goes to zero. This means that we have answered in general the question of how to represent a particle whose spin is along an arbitrary axis. The amplitudes of (10.30) are the projection amplitudes for spin one-half particles corresponding to the projection amplitudes we gave in Chapter 5 [Eqs. (5.38)] for spin-one particles. We can now find the amplitudes for filtered beams of spin one-half particles to go through any particular Stern-Gerlach filter. Let $\ket{+z}$ represent a state with spin up along the $z$-axis, and $\ket{-z}$ represent the spin down state. If $\ket{+z'}$ represents a state with spin up along a $z'$-axis which makes the polar angles $\theta$ and $\phi$ with the $z$-axis, then in the notation of Chapter 5, we have  \begin{equation} \label{Eq:III:10:31} \braket{+z}{+z'}=\cos\frac{\theta}{2}\,e^{-i\phi/2},\quad \braket{-z}{+z'}=\sin\frac{\theta}{2}\,e^{+i\phi/2}. \end{equation} 
 \begin{equation} \begin{aligned} \braket{+z}{+z'}&=\cos\frac{\theta}{2}\,e^{-i\phi/2},\\[1.5ex] \braket{-z}{+z'}&=\sin\frac{\theta}{2}\,e^{+i\phi/2}. \end{aligned} \label{Eq:III:10:31} \end{equation}  These results are equivalent to what we found in Chapter 6, Eq. (6.36), by purely geometrical arguments. (So if you decided to skip Chapter 6, you now have the essential results anyway.) As our final example lets look again at one which we’ve already mentioned a number of times. Suppose that we consider the following problem. We start with an electron whose spin is in some given direction, then turn on a magnetic field in the $z$-direction for $25$ minutes, and then turn it off. What is the final state? Again let’s represent the state by the linear combination $\ket{\psi}=\ketsl{\slOne}C_1+\ketsl{\slTwo}C_2$. For this problem, however, the states of definite energy are also our base states $\ketsl{\slOne}$ and $\ketsl{\slTwo}$. So $C_1$ and $C_2$ only vary in phase. We know that \begin{equation*} C_1(t)=C_1(0)e^{-iE_{\slI}t/\hbar}=C_1(0)e^{+i\mu Bt/\hbar}, \end{equation*}  and \begin{equation*} C_2(t)=C_2(0)e^{-iE_{\slII}t/\hbar}=C_2(0)e^{-i\mu Bt/\hbar}. \end{equation*}  Now initially we said the electron spin was set in a given direction. That means that initially $C_1$ and $C_2$ are two numbers given by Eqs. (10.30). After we wait for a period of time $T$, the new $C_1$ and $C_2$ are the same two numbers multiplied respectively by $e^{i\mu B_zT/\hbar}$ and $e^{-i\mu B_zT/\hbar}$. What state is that? That’s easy. It’s exactly the same as if the angle $\phi$ had been changed by the subtraction of $2\mu B_zT/\hbar$ and the angle $\theta$ had been left unchanged. That means that at the end of the time $T$, the state $\ket{\psi}$ represents an electron lined up in a direction which differs from the original direction only by a rotation about the $z$-axis through the angle $\Delta\phi=2\mu B_zT/\hbar$. Since this angle is proportional to $T$, we can also say the direction of the spin precesses at the angular velocity $2\mu B_z/\hbar$ around the $z$-axis. This result we discussed several times previously in a less complete and rigorous manner. Now we have obtained a complete and accurate quantum mechanical description of the precession of atomic magnets. It is interesting that the mathematical ideas we have just gone over for the spinning electron in a magnetic field can be applied to any two-state system. That means that by making a mathematical analogy to the spinning electron, any problem about two-state systems can be solved by pure geometry. It works like this. First you shift the zero of energy so that $(H_{11}+H_{22})$ is equal to zero so that $H_{11}=-H_{22}$. Then any two-state problem is formally the same as the electron in a magnetic field. All you have to do is identify $-\mu B_z$ with $H_{11}$ and $-\mu(B_x-iB_y)$ with $H_{12}$. No matter what the physics is originally—an ammonia molecule, or whatever—you can translate it into a corresponding electron problem. So if we can solve the electron problem in general, we have solved all two-state problems. And we have the general solution for the electron! Suppose you have some state to start with that has spin “up” in some direction, and you have a magnetic field $\FLPB$ that points in some other direction. You just rotate the spin direction around the axis of $\FLPB$ with the vector angular velocity $\FLPomega(t)$ equal to a constant times the vector $\FLPB$ (namely $\FLPomega=2\mu\FLPB/\hbar$). As $\FLPB$ varies with time, you keep moving the axis of the rotation to keep it parallel with $\FLPB$, and keep changing the speed of rotation so that it is always proportional to the strength of $\FLPB$. See Fig. 10–11. If you keep doing this, you will end up with a certain final orientation of the spin axis, and the amplitudes $C_1$ and $C_2$ are just given by the projections—using (10.30)—into your coordinate frame. You see, it’s just a geometric problem to keep track of where you end up after all the rotating. Although it’s easy to see what’s involved, this geometric problem (of finding the net result of a rotation with a varying angular velocity vector) is not easy to solve explicitly in the general case. Anyway, we see, in principle, the general solution to any two-state problem. In the next chapter we will look some more into the mathematical techniques for handling the important case of a spin one-half particle—and, therefore, for handling two-state systems in general.