Rotations about the $\boldsymbol{z}$-axis
We are now ready to find the transformation matrix $R_{ji}$ between two different representations. With our rule for compounding rotations and our assumption that space has no preferred direction, we have the keys we need for finding the matrix of any arbitrary rotation. There is only one solution. We begin with the transformation which corresponds to a rotation about the $z$-axis. Suppose we have two apparatuses $S$ and $T$ placed in series along a straight line with their axes parallel and pointing out of the page, as shown in Fig. 6-4(a). We take our “$z$-axis” in this direction. Surely, if the beam goes “up” (toward $+z$) in the $S$ apparatus, it will do the same in the $T$ apparatus. Similarly, if it goes down in $S$, it will go down in $T$. Suppose, however, that the $T$ apparatus were placed at some other angle, but still with its axis parallel to the axis of $S$, as in Fig. 6-4(b). Intuitively, you would say that a $(+)$ beam in $S$ would still go with a $(+)$ beam in $T$, because the fields and field gradients are still in the same physical direction. And that would be quite right. Also, a $(-)$ beam in $S$ would still go into a $(-)$ beam in $T$. The same result would apply for any orientation of $T$ in the $xy$-plane of $S$. What does this tell us about the relation between $C_+'=\braket{+T}{\psi}$, $C_-'=\braket{-T}{\psi}$ and $C_+=\braket{+S}{\psi}$, $C_-=\braket{-S}{\psi}$? You might conclude that any rotation about the $z$-axis of the “frame of reference” for base states leaves the amplitudes to be “up” and “down” the same as before.  We could write $C_+'=C_+$ and $C_-'=C_-$—but that is wrong. All we can conclude is that for such rotations the probabilities to be in the “up” beam are the same for the $S$ and $T$ apparatuses. That is, \begin{equation*} \abs{C_+'}=\abs{C_+}\quad \text{and}\quad \abs{C_-'}=\abs{C_-}. \end{equation*} We cannot say that the phases of the amplitudes referred to the $T$ apparatus may not be different for the two different orientations in (a) and (b) of Fig. 6-4.  The two apparatuses in (a) and (b) of Fig. 6-4 are, in fact, different, as we can see in the following way. Suppose that we put an apparatus in front of $S$ which produces a pure $(+x)$ state. (The $x$-axis points toward the bottom of the figure.) Such particles would be split into $(+z)$ and $(-z)$ beams in $S$, but the two beams would be recombined to give a $(+x)$ state again at $P_1$—the exit of $S$. The same thing happens again in $T$. If we follow $T$ by a third apparatus $U$, whose axis is in the $(+x)$ direction and, as shown in Fig. 6-5(a), all the particles would go into the $(+)$ beam of $U$. Now imagine what happens if $T$ and $U$ are swung around together by $90^\circ$ to the positions shown in Fig. 6-5(b). Again, the $T$ apparatus puts out just what it takes in, so the particles that enter $U$ are in a $(+x)$ state with respect to $S$. But $U$ now analyzes for the $(+y)$ state with respect to $S$, which is different. (By symmetry, we would now expect only one-half of the particles to get through.)  What could have changed? The apparatuses $T$ and $U$ are still in the same physical relationship to each other. Can the physics be changed just because $T$ and $U$ are in a different orientation? Our original assumption is that it should not. It must be that the amplitudes with respect to $T$ are different in the two cases shown in Fig. 6-5—and, therefore, also in Fig. 6-4. There must be some way for a particle to know that it has turned the corner at $P_1$. How could it tell? Well, all we have decided is that the magnitudes of $C_+'$ and $C_+$ are the same in the two cases, but they could—in fact, must—have different phases. We conclude that $C_+'$ and $C_+$ must be related by \begin{equation*} C_+'=e^{i\lambda}C_+, \end{equation*} and that $C_-'$ and $C_-$ must be related by \begin{equation*} C_-'=e^{i\mu}C_-, \end{equation*} where $\lambda$ and $\mu$ are real numbers which must be related in some way to the angle between $S$ and $T$.  The only thing we can say at the moment about $\lambda$ and $\mu$ is that they must not be equal [except for the special case shown in Fig. 6-5(a), when $T$ is in the same orientation as $S$]. We have seen that equal phase changes in all amplitudes have no physical consequence. For the same reason, we can always add the same arbitrary amount to both $\lambda$ and $\mu$ without changing anything. So we are permitted to choose to make $\lambda$ and $\mu$ equal to plus and minus the same number. That is, we can always take \begin{equation*} \lambda'=\lambda-\frac{(\lambda+\mu)}{2},\quad \mu'=\mu-\frac{(\lambda+\mu)}{2}. \end{equation*} Then \begin{equation*} \lambda'=\frac{\lambda}{2}-\frac{\mu}{2}=-\mu'. \end{equation*} So we adopt the convention2 that $\mu=-\lambda$. We have then the general rule that for a rotation of the reference apparatus by some angle about the $z$-axis, the transformation is \begin{equation} \label{Eq:III:6:16} C_+'=e^{+i\lambda}C_+,\quad C_-'=e^{-i\lambda}C_-. \end{equation} The absolute values are the same, only the phases are different. These phase factors are responsible for the different results in the two experiments of Fig. 6-5.  Now we would like to know the law that relates $\lambda$ to the angle between $S$ and $T$. We already know the answer for one case. If the angle is zero, $\lambda$ is zero. Now we will assume that the phase shift $\lambda$ is a continuous function of angle $\phi$ between $S$ and $T$ (see Fig. 6-4) as $\phi$ goes to zero—as only seems reasonable. In other words, if we rotate $T$ from the straight line through $S$ by the small angle $\epsilon$, the $\lambda$ is also a small quantity, say $m\epsilon$, where $m$ is some number. We write it this way because we can show that $\lambda$ must be proportional to $\epsilon$. Suppose we were to put after $T$ another apparatus $T'$ which makes the angle $\epsilon$ with $T$, and, therefore, the angle $2\epsilon$ with $S$. Then, with respect to $T$, we have \begin{equation*} C_+'=e^{i\lambda}C_+, \end{equation*} and with respect to $T'$, we have \begin{equation*} C_+''=e^{i\lambda}C_+'=e^{i2\lambda}C_+. \end{equation*} But we know that we should get the same result if we put $T'$ right after $S$. Thus, when the angle is doubled, the phase is doubled. We can evidently extend the argument and build up any rotation at all by a sequence of infinitesimal rotations. We conclude that for any angle $\phi$, $\lambda$ is proportional to the angle. We can, therefore, write $\lambda=m\phi$.  The general result we get, then, is that for $T$ rotated about the $z$-axis by the angle $\phi$ with respect to $S$ \begin{equation} \label{Eq:III:6:17} C_+'=e^{im\phi}C_+,\quad C_-'=e^{-im\phi}C_-. \end{equation} For the angle $\phi$, and for all rotations we speak of in the future, we adopt the standard convention that a positive rotation is a right-handed rotation about the positive direction of the reference axis. A positive $\phi$ has the sense of rotation of a right-handed screw advancing in the positive $z$-direction.  Now we have to find what $m$ must be. First, we might try this argument: Suppose $T$ is rotated by $360^\circ$; then, clearly, it is right back at zero degrees, and we should have $C_+'=C_+$ and $C_-'=C_-$, or, what is the same thing, $e^{im2\pi}=1$. We get $m=1$. This argument is wrong! To see that it is, consider that $T$ is rotated by $180^\circ$. If $m$ were equal to $1$, we would have $C_+'=$ $e^{i\pi}C_+=$ $-C_+$ and $C_-'=$ $e^{-i\pi}C_-=$ $-C_-$. However, this is just the original state all over again. Both amplitudes are just multiplied by $-1$ which gives back the original physical system. (It is again a case of a common phase change.) This means that if the angle between $T$ and $S$ in Fig. 6-5(b) is increased to $180^\circ$, the system (with respect to $T$) would be indistinguishable from the zero-degree situation, and the particles would again go through the $(+)$ state of the $U$ apparatus. At $180^\circ$, though, the $(+)$ state of the $U$ apparatus is the $(-x)$ state of the original $S$ apparatus. So a $(+x)$ state would become a $(-x)$ state. But we have done nothing to change the original state; the answer is wrong. We cannot have $m=1$.  We must have the situation that a rotation by $360^\circ$ and no smaller angle reproduces the same physical state. This will happen if $m=\tfrac{1}{2}$. Then, and only then, will the first angle that reproduces the same physical state be $\phi=360^\circ$.3 It gives \begin{equation} \left. \begin{aligned} C_+'&=-C_+\\ \\ C_-'&=-C_- \end{aligned} \right\} \text{$360^\circ$ about $z$-axis}. \label{Eq:III:6:18} \end{equation} It is very curious to say that if you turn the apparatus $360^\circ$ you get new amplitudes. They aren’t really new, though, because the common change of sign doesn’t give any different physics. If someone else had decided to change all the signs of the amplitudes because he thought he had turned $360^\circ$, that’s all right; he gets the same physics.4 So our final answer is that if we know the amplitudes $C_+$ and $C_-$ for spin one-half particles with respect to a reference frame $S$, and we then use a base system referred to $T$ which is obtained from $S$ by a rotation of $\phi$ around the $z$-axis, the new amplitudes are given in terms of the old by \begin{equation} \left. \begin{aligned} C_+'=e^{i\phi/2}C_+&\\ \\ C_-'=e^{-i\phi/2}C_-& \end{aligned} \right\} \text{$\phi$ about $z$}. \label{Eq:III:6:19} \end{equation} 