Experiments with filtered atoms
Now comes the big question: What happens if the second apparatus is tipped to a different angle, so that its field axis is no longer parallel to the first? It could be not only tipped, but also pointed in a different direction—for instance, it could take the beam off at $90^\circ$ with respect to the original direction. To take it easy at first, let’s first think about an arrangement in which the second Stern-Gerlach experiment is tilted by some angle $\alpha$ about the $y$-axis, as shown in Fig. 5–6. We’ll call the second apparatus $T$. Suppose that we now set up the following experiment: \begin{equation*} \SGP{S} \!\! \SGP{T} \!\!\!\!, \end{equation*}  or the experiment:  \begin{equation*} \SGP{S} \!\! \SGZ{T} \!\!\!\!\cdotp \end{equation*}  What comes out at the far end in these cases? The answer is this: If the atoms are in a definite state with respect to $S$, they are not in the same state with respect to $T$—a $(+S)$ state is not also a $(+T)$ state. There is, however, a certain amplitude to find the atom in a $(+T)$ state—or a $(\OT)$ state or a $(-T)$ state. In other words, as careful as we have been to make sure that we have the atoms in a definite condition, the fact of the matter is that if it goes through an apparatus which is tilted at a different angle it has, so to speak, to “reorient” itself—which it does, don’t forget, by luck. We can put only one particle through at a time, and then we can only ask the question: What is the probability that it gets through? Some of the atoms that have gone through $S$ will end in a $(+T)$ state, some of them will end in a $(\OT)$, and some in a $(-T)$ state—all with different odds. These odds can be calculated by the absolute squares of complex amplitudes; what we want is some mathematical method, or quantum mechanical description, for these amplitudes. What we need to know are various quantities like \begin{equation*} \braket{-T}{+S}, \end{equation*}  by which we mean the amplitude that an atom initially in the $(+S)$ state can get into the $(-T)$ condition (which is not zero unless $T$ and $S$ are lined up parallel to each other). There are other amplitudes like \begin{equation*} \braket{+T}{\OS},\quad \text{or}\quad \braket{\OT}{-S},\quad \text{etc}. \end{equation*}  There are, in fact, nine such amplitudes—another matrix—that a theory of particles should tell us how to calculate. Just as $F=ma$ tells us how to calculate what happens to a classical particle in any circumstance, the laws of quantum mechanics permit us to determine the amplitude that a particle will get through a particular apparatus. The central problem, then, is to be able to calculate—for any given tilt angle $\alpha$, or in fact for any orientation whatever—the nine amplitudes: \begin{equation} \begin{alignedat}{3} \braket{+T}{+S}&,&\quad \braket{+T}{\OS}&,&\quad \braket{+T}{-S}&,\\ \braket{\OT}{+S}&,&\quad \braket{\OT}{\OS}&,&\quad \braket{\OT}{-S}&,\\ \braket{-T}{+S}&,&\quad \braket{-T}{\OS}&,&\quad \braket{-T}{-S}&. \end{alignedat} \label{Eq:III:5:9} \end{equation}  We can already figure out some relations among these amplitudes. First, according to our definitions, the absolute square \begin{equation*} \abs{\braket{+T}{+S}}^2 \end{equation*}  is the probability that an atom in a $(+S)$ state will enter a $(+T)$ state. We will often find it more convenient to write such squares in the equivalent form \begin{equation*} \braket{+T}{+S}\braket{+T}{+S}\cconj. \end{equation*}  In the same notation the number \begin{equation*} \braket{\OT}{+S}\braket{\OT}{+S}\cconj \end{equation*}  is the probability that a particle in the $(+S)$ state will enter the $(\OT)$ state, and \begin{equation*} \braket{-T}{+S}\braket{-T}{+S}\cconj \end{equation*}  is the probability that it will enter the $(-T)$ state. But the way our apparatuses are made, every atom which enters the $T$ apparatus must be found in some one of the three states of the $T$ apparatus—there’s nowhere else for a given kind of atom to go. So the sum of the three probabilities we’ve just written must be equal to $100$ percent. We have the relation \begin{equation} \begin{alignedat}{3} &\braket{+T}{+S}&\braket{+T}{+S}&\cconj\;&+&\\[.2ex] &\braket{\OT}{+S}&\,\braket{\OT}{+S}&\cconj&+&\\[.2ex] &\braket{-T}{+S}&\braket{-T}{+S}&\cconj&=&\;1. \end{alignedat} \label{Eq:III:5:10} \end{equation}  There are, of course, two other such equations that we get if we start with a $(\OS)$ or a $(-S)$ state. But they are all we can easily get, so we’ll go on to some other general questions.