Rotations of 180$^{\boldsymbol{\circ}}$ and
90$^{\boldsymbol{\circ}}$ about $\boldsymbol{y}$
Next, we will try to guess the transformation for a rotation of $T$ with respect to $S$ of $180^\circ$ around an axis perpendicular to the $z$-axis—say, about the $y$-axis. (We have defined the coordinate axes in Fig. 6-1.)  In other words, we start with two identical Stern-Gerlach equipments, with the second one, $T$, turned “upside down” with respect to the first one, $S$, as in Fig. 6-6. Now if we think of our particles as little magnetic dipoles, a particle that is in the $(+S)$ state—so that it goes on the “upper” path in the first apparatus—will also take the “upper” path in the second, so that it will be in the minus state with respect to $T$. (In the inverted $T$ apparatus, both the gradients and the field direction are reversed; for a particle with its magnetic moment in a given direction, the force is unchanged.)  Anyway, what is “up” with respect to $S$ will be “down” with respect to $T$. For these relative positions of $S$ and $T$, then, we know that the transformation must give \begin{equation*} \abs{C_+'}=\abs{C_-},\quad \abs{C_-'}=\abs{C_+}. \end{equation*} As before, we cannot rule out some additional phase factors; we could have (for $180^\circ$ about the $y$-axis) \begin{equation} \label{Eq:III:6:20} C_+'=e^{i\beta}C_-\quad \text{and}\quad C_-'=e^{i\gamma}C_+, \end{equation} where $\beta$ and $\gamma$ are still to be determined.  What about a rotation of $360^\circ$ about the $y$-axis? Well, we already know the answer for a rotation of $360^\circ$ about the $z$-axis—the amplitude to be in any state changes sign. A rotation of $360^\circ$ around any axis always brings us back to the original position. It must be that for any $360^\circ$ rotation, the result is the same as a $360^\circ$ rotation about the $z$-axis—all amplitudes simply change sign. Now suppose we imagine two successive rotations of $180^\circ$ about $y$—using Eq. (6.20)—we should get the result of Eq. (6.18). In other words, \begin{equation*} C_+''=e^{i\beta}C_-'=e^{i\beta}e^{i\gamma}C_+=-C_+ \end{equation*} and \begin{equation} \label{Eq:III:6:21} C_-''=e^{i\gamma}C_+'=e^{i\gamma}e^{i\beta}C_-=-C_-. \end{equation} This means that \begin{equation*} e^{i\beta}e^{i\gamma}=-1\quad \text{or}\quad e^{i\gamma}=-e^{-i\beta}. \end{equation*} So the transformation for a rotation of $180^\circ$ about the $y$-axis can be written \begin{equation} \label{Eq:III:6:22} C_+'=e^{i\beta}C_-,\quad C_-'=-e^{-i\beta}C_+. \end{equation}  The arguments we have just used would apply equally well to a rotation of $180^\circ$ about any axis in the $xy$-plane, although different axes can, of course, give different numbers for $\beta$. However, that is the only way they can differ. Now there is a certain amount of arbitrariness in the number $\beta$, but once it is specified for one axis of rotation in the $xy$-plane it is determined for any other axis. It is conventional to choose to set $\beta=0$ for a $180^\circ$ rotation about the $y$-axis.  To show that we have this choice, suppose we imagine that $\beta$ was not equal to zero for a rotation about the $y$-axis; then we can show that there is some other axis in the $xy$-plane, for which the corresponding phase factor will be zero. Let’s find the phase factor $\beta_A$ for an axis $A$ that makes the angle $\alpha$ with the $y$-axis, as shown in Fig. 6-7(a). (For clarity, the figure is drawn with $\alpha$ equal to a negative number, but that doesn’t matter.) Now if we take a $T$ apparatus which is initially lined up with the $S$ apparatus and is then rotated $180^\circ$ about the axis $A$, its axes—which we will call $x''$, $y''$,  and $z''$—will be as shown in Fig. 6-7(a). The amplitudes with respect to $T$ will then be \begin{equation} \label{Eq:III:6:23} C_+''=e^{i\beta_A}C_-,\quad C_-''=-e^{-i\beta_A}C_+. \end{equation}  We can now think of getting to the same orientation by the two successive rotations shown in (b) and (c) of the figure. First, we imagine an apparatus $U$ which is rotated with respect to $S$ by $180^\circ$ about the $y$-axis. The axes $x'$, $y'$, and $z'$ of $U$ will be as shown in Fig. 6-7(b), and the amplitudes with respect to $U$ are given by (6.22).  Now notice that we can go from $U$ to $T$ by a rotation about the “$z$-axis” of $U$, namely about $z'$, as shown in Fig. 6-7(c). From the figure you can see that the angle required is two times the angle $\alpha$ but in the opposite direction (with respect to $z'$). Using the transformation of (6.19) with $\phi=-2\alpha$, we get \begin{equation} \label{Eq:III:6:24} C_+''=e^{-i\alpha}C_+',\quad C_-''=e^{+i\alpha}C_-'. \end{equation} Combining Eqs. (6.24) and (6.22), we get that \begin{equation} \label{Eq:III:6:25} C_+''=e^{i(\beta-\alpha)}C_-,\quad C_-''=-e^{-i(\beta-\alpha)}C_+. \end{equation} These amplitudes must, of course, be the same as we got in (6.23). So $\beta_A$ must be related to $\alpha$ and $\beta$ by \begin{equation} \label{Eq:III:6:26} \beta_A=\beta-\alpha. \end{equation} This means that if the angle $\alpha$ between the $A$-axis and the $y$-axis (of $S$) is equal to $\beta$, the transformation for a rotation of $180^\circ$ about $A$ will have $\beta_A=0$.  Now so long as some axis perpendicular to the $z$-axis is going to have $\beta=0$, we may as well take it to be the $y$-axis. It is purely a matter of convention, and we adopt the one in general use. Our result: For a rotation of $180^\circ$ about the $y$-axis, we have \begin{equation} \left. \begin{aligned} C_+'&=C_-\\ \\ C_-'&=-C_+ \end{aligned} \right\} \text{$180^\circ$ about $y$}. \label{Eq:III:6:27} \end{equation}   While we are thinking about the $y$-axis, let’s next ask for the transformation matrix for a rotation of $90^\circ$ about $y$. We can find it because we know that two successive $90^\circ$ rotations about the same axis must equal one $180^\circ$ rotation. We start by writing the transformation for $90^\circ$ in the most general form: \begin{equation} \label{Eq:III:6:28} C_+'=aC_++bC_-,\quad C_-'=cC_++dC_-. \end{equation} A second rotation of $90^\circ$ about the same axis would have the same coefficients: \begin{equation} \label{Eq:III:6:29} C_+''=aC_+'+bC_-',\quad C_-''=cC_+'+dC_-'. \end{equation} Combining Eqs. (6.28) and (6.29), we have \begin{equation} \begin{alignedat}{2} C_+''&=a(aC_++bC_-)&+\,b(cC_++dC_-),\\[1ex] C_-''&=c(aC_++bC_-)&+\,d(cC_++dC_-). \end{alignedat} \label{Eq:III:6:30} \end{equation} However, from (6.27) we know that \begin{equation*} C_+''=C_-,\quad C_-''=-C_+, \end{equation*} so that we must have that \begin{equation} \begin{aligned} ab+bd&=1,\\ a^2+bc&=0,\\ ac+cd&=-1,\\ bc+d^2&=0. \end{aligned} \label{Eq:III:6:31} \end{equation} These four equations are enough to determine all our unknowns: $a$, $b$, $c$, and $d$. It is not hard to do. Look at the second and fourth equations. Deduce that $a^2=d^2$, which means that $a=d$ or else that $a=-d$. But $a=-d$ is out, because then the first equation wouldn’t be right. So $d=a$. Using this, we have immediately that $b=1/2a$ and that $c=-1/2a$. Now we have everything in terms of $a$. Putting, say, the second equation all in terms of $a$, we have \begin{equation*} a^2-\frac{1}{4a^2}=0\quad \text{or}\quad a^4=\frac{1}{4}. \end{equation*} This equation has four different solutions, but only two of them give the standard value for the determinant. We might as well take $a=1/\sqrt{2}$; then5 \begin{alignat*}{2} a&=1/\sqrt{2},&\quad b&=1/\sqrt{2},\\\\ c&=-1/\sqrt{2},&\quad d&=1/\sqrt{2}. \end{alignat*}  In other words, for two apparatuses $S$ and $T$, with $T$ rotated with respect to $S$ by $90^\circ$ about the $y$-axis, the transformation is \begin{equation} \left. \begin{aligned} C_+'&=\frac{1}{\sqrt{2}}\,(C_++C_-)\\ \\ C_-'&=\frac{1}{\sqrt{2}}\,(-C_++C_-) \end{aligned} \right\} \text{$90^\circ$ about $y$}. \label{Eq:III:6:32} \end{equation}  We can, of course, solve these equations for $C_+$ and $C_-$, which will give us the transformation for a rotation of minus $90^\circ$ about $y$. Changing the primes around, we would conclude that \begin{equation} \left. \begin{aligned} C_+'&=\frac{1}{\sqrt{2}}\,(C_+-C_-)\\ \\ C_-'&=\frac{1}{\sqrt{2}}\,(C_++C_-) \end{aligned} \right\} \text{$-90^\circ$ about $y$}. \label{Eq:III:6:33} \end{equation} 