Transforming to a rotated coordinate system
We consider again the “improved” Stern-Gerlach apparatus described in the last chapter. A beam of spin one-half particles, entering at the left, would, in general, be split into two beams, as shown schematically in Fig. 6-1. (There were three beams for spin one.) As before, the beams are put back together again unless one or the other of them is blocked off by a “stop” which intercepts the beam at its half-way point. In the figure we show an arrow which points in the direction of the increase of the magnitude of the field—say toward the magnet pole with the sharp edges. This arrow we take to represent the “up” axis of any particular apparatus. It is fixed relative to the apparatus and will allow us to indicate the relative orientations when we use several apparatuses together. We also assume that the direction of the magnetic field in each magnet is always the same with respect to the arrow.  We will say that those atoms which go in the “upper” beam are in the $(+)$ state with respect to that apparatus and that those in the “lower” beam are in the $(-)$ state. (There is no “zero” state for spin one-half particles.)  Now suppose we put two of our modified Stern-Gerlach apparatuses in sequence, as shown in Fig. 6-2(a). The first one, which we call $S$, can be used to prepare a pure $(+S)$ or a pure $(-S)$ state by blocking one beam or the other. [As shown it prepares a pure $(+S)$ state.] For each condition, there is some amplitude for a particle that comes out of $S$ to be in either the $(+T)$ or the $(-T)$ beam of the second apparatus. There are, in fact, just four amplitudes: the amplitude to go from $(+S)$ to $(+T)$, from $(+S)$ to $(-T)$, from $(-S)$ to $(+T)$, from $(-S)$ to $(-T)$. These amplitudes are just the four coefficients of the transformation matrix $R_{ji}$ to go from the $S$-representation to the $T$-representation. We can consider that the first apparatus “prepares” a particular state in one representation and that the second apparatus “analyzes” that state in terms of the second representation. The kind of question we want to answer, then, is this: If an atom has been prepared in a given condition—say the $(+S)$ state—by blocking one of the beams in the apparatus $S$, what is the chance that it will get through the second apparatus $T$ if this is set for, say, the $(-T)$ state. The result will depend, of course, on the angles between the two systems $S$ and $T$.  We should explain why it is that we could have any hope of finding the coefficients $R_{ji}$ by deduction. You know that it is almost impossible to believe that if a particle has its spin lined up in the $+z$-direction, that there is some chance of finding the same particle with its spin pointing in the $+x$-direction—or in any other direction at all. In fact, it is almost impossible, but not quite. It is so nearly impossible that there is only one way it can be done, and that is the reason we can find out what that unique way is.  The first kind of argument we can make is this. Suppose we have a setup like the one in Fig. 6-2(a), in which we have the two apparatuses $S$ and $T$, with $T$ cocked at the angle $\alpha$ with respect to $S$, and we let only the $(+)$ beam through $S$ and the $(-)$ beam through $T$. We would observe a certain number for the probability that the particles coming out of $S$ get through $T$. Now suppose we make another measurement with the apparatus of Fig. 6-2(b). The relative orientation of $S$ and $T$ is the same, but the whole system sits at a different angle in space. We want to assume that both of these experiments give the same number for the chance that a particle in a pure state with respect to $S$ will get into some particular state with respect to $T$. We are assuming, in other words, that the result of any experiment of this type is the same—that the physics is the same—no matter how the whole apparatus is oriented in space. (You say, “That’s obvious.” But it is an assumption, and it is “right” only if it is actually what happens.)  That means that the coefficients $R_{ji}$ depend only on the relation in space of $S$ and $T$, and not on the absolute situation of $S$ and $T$. To say this in another way, $R_{ji}$ depends only on the rotation which carries $S$ to $T$, for evidently what is the same in Fig. 6-2(a) and Fig. 6-2(b) is the three-dimensional rotation which would carry apparatus $S$ into the orientation of apparatus $T$. When the transformation matrix $R_{ji}$ depends only on a rotation, as it does here, it is called a rotation matrix.  For our next step we will need one more piece of information. Suppose we add a third apparatus which we can call $U$, which follows $T$ at some arbitrary angle, as in Fig. 6-3(a). (It’s beginning to look horrible, but that’s the fun of abstract thinking—you can make the most weird experiments just by drawing lines!) Now what is the $S\to T\to U$ transformation? What we really want to ask for is the amplitude to go from some state with respect to $S$ to some other state with respect to $U$, when we know the transformation from $S$ to $T$ and from $T$ to $U$. We are then asking about an experiment in which both channels of $T$ are open. We can get the answer by applying Eq. (6.5) twice in succession. For going from the $S$-representation to the $T$-representation, we have \begin{equation} \label{Eq:III:6:6} C_j'=\sum_iR_{ji}^{TS}C_i, \end{equation} where we put the superscripts $TS$ on the $R$, so that we can distinguish it from the coefficients $R^{UT}$ we will have for going from $T$ to $U$.  Assuming the amplitudes to be in the base states of the $U$-representation $C_k''$, we can relate them to the $T$-amplitudes by using Eq. (6.5) once more; we get \begin{equation} \label{Eq:III:6:7} C_k''=\sum_jR_{kj}^{UT}C_j'. \end{equation} Now we can combine Eqs. (6.6) and (6.7) to get the transformation to $U$ directly from $S$. Substituting $C_j'$ from Eq. (6.6) in Eq. (6.7), we have \begin{equation} \label{Eq:III:6:8} C_k''=\sum_jR_{kj}^{UT}\sum_iR_{ji}^{TS}C_i. \end{equation} Or, since $i$ does not appear in $R_{kj}^{UT}$, we can put the $i$-summation also in front, and write \begin{equation} \label{Eq:III:6:9} C_k''=\sum_i\sum_jR_{kj}^{UT}R_{ji}^{TS}C_i. \end{equation} This is the formula for a double transformation.  Notice, however, that so long as all the beams in $T$ are unblocked, the state coming out of $T$ is the same as the one that went in. We could just as well have made a transformation from the $S$-representation directly to the $U$-representation. It should be the same as putting the $U$ apparatus right after $S$, as in Fig. 6-3(b). In that case, we would have written \begin{equation} \label{Eq:III:6:10} C_k''=\sum_iR_{ki}^{US}C_i, \end{equation} with the coefficients $R_{ki}^{US}$ belonging to this transformation. Now, clearly, Eqs. (6.9) and (6.10) should give the same amplitudes $C_k''$, and this should be true no matter what the original state $\phi$ was which gave us the amplitudes $C_i$. So it must be that \begin{equation} \label{Eq:III:6:11} R_{ki}^{US}=\sum_jR_{kj}^{UT}R_{ji}^{TS}. \end{equation} In other words, for any rotation $S\to U$ of a reference base, which is viewed as a compounding of two successive rotations $S\to T$ and $T\to U$, the rotation matrix $R_{ki}^{US}$ can be obtained from the matrices of the two partial rotations by Eq. (6.11). If you wish, you can find Eq. (6.11) directly from Eq. (6.1), for it is only a different notation for $\braket{kU}{iS}=\sum_j\braket{kU}{jT}\braket{jT}{iS}$.   To be thorough, we should add the following parenthetical remarks. They are not terribly important, however, so you can skip to the next section if you want. What we have said is not quite right. We cannot really say that Eq. (6.9) and Eq. (6.10) must give exactly the same amplitudes. Only the physics should be the same; all the amplitudes could be different by some common phase factor like $e^{i\delta}$ without changing the result of any calculation about the real world. So, instead of Eq. (6.11), all we can say, really, is that \begin{equation} \label{Eq:III:6:12} e^{i\delta}R_{ki}^{US}=\sum_jR_{kj}^{UT}R_{ji}^{TS}, \end{equation} where $\delta$ is some real constant. What this extra factor of $e^{i\delta}$ means, of course, is that the amplitudes we get if we use the matrix $R^{US}$ might all differ by the same phase $(e^{-i\delta})$ from the amplitude we would get using the two rotations $R^{UT}$ and $R^{TS}$. We know that it doesn’t matter if all amplitudes are changed by the same phase, so we could just ignore this phase factor if we wanted to. It turns out, however, that if we define all of our rotation matrices in a particular way, this extra phase factor will never appear—the $\delta$ in Eq. (6.12) will always be zero. Although it is not important for the rest of our arguments, we can give a quick proof by using a mathematical theorem about determinants. [If you don’t yet know much about determinants, don’t worry about the proof and just skip to the definition of Eq. (6.15).]   First, we should say that Eq. (6.11) is the mathematical definition of a “product” of two matrices. (It is just convenient to be able to say: “$R^{US}$ is the product of $R^{UT}$ and $R^{TS}$.” )  Second, there is a theorem of mathematics—which you can easily prove for the two-by-two matrices we have here—which says that the determinant of a “product” of two matrices is the product of their determinants. Applying this theorem to Eq. (6.12), we get \begin{equation} \label{Eq:III:6:13} e^{i2\delta}\,(\Det R^{US})=(\Det R^{UT})\cdot(\Det R^{TS}). \end{equation} (We leave off the subscripts, because they don’t tell us anything useful.) Yes, the $2\delta$ is right. Remember that we are dealing with two-by-two matrices; every term in the matrix $R_{ki}^{US}$ is multiplied by $e^{i\delta}$, so each product in the determinant—which has two factors—gets multiplied by $e^{i2\delta}$. Now let’s take the square root of Eq. (6.13) and divide it into Eq. (6.12); we get \begin{equation} \label{Eq:III:6:14} \frac{R_{ki}^{US}}{\sqrt{\Det R^{US}}}=\sum_j \frac{R_{kj}^{UT}}{\sqrt{\Det R^{UT}}}\, \frac{R_{ji}^{TS}}{\sqrt{\Det R^{TS}}}. \end{equation} The extra phase factor has disappeared.   Now it turns out that if we want all of our amplitudes in any given representation to be normalized (which means, you remember, that $\sum_i\braket{\phi}{i}\braket{i}{\phi}=1$), the rotation matrices will all have determinants that are pure imaginary exponentials, like $e^{i\alpha}$. (We won’t prove it; you will see that it always comes out that way.) So we can, if we wish, choose to make all our rotation matrices $R$ have a unique phase by making $\Det R=1$. It is done like this. Suppose we find a rotation matrix $R$ in some arbitrary way. We make it a rule to “convert” it to “standard form” by defining \begin{equation} \label{Eq:III:6:15} R_{\text{standard}}=\frac{R}{\sqrt{\Det R}}. \end{equation} We can do this because we are just multiplying each term of $R$ by the same phase factor, to get the phases we want. In what follows, we will always assume that our matrices have been put in the “standard form” ; then we can use Eq. (6.11) without having any extra phase factors. 