Stern-Gerlach filters in series
Here is an interesting question: Suppose we had atoms filtered into the $(+S)$ state, then we put them through a second filter, say into a $(\OT)$ state, and then through another $+S$ filter. (We’ll call the last filter $S'$ just so we can distinguish it from the first $S$-filter.) Do the atoms remember that they were once in a $(+S)$ state? In other words, we have the following experiment: \begin{equation} \label{Eq:III:5:11} \SGP{S} \!\! \SGZ{T} \!\! \SGP{S'}\!\!\!\!\cdotp \end{equation}  We want to know whether all those that get through $T$ also get through $S'$. They do not. Once they have been filtered by $T$, they do not remember in any way that they were in a $(+S)$ state when they entered $T$. Note that the second $S$ apparatus in (5.11) is oriented exactly the same as the first, so it is still an $S$-type filter. The states filtered by $S'$ are, of course, still $(+S)$, $(\OS)$, and $(-S)$. The important point is this: If the $T$ filter passes only one beam, the fraction that gets through the second $S$ filter depends only on the setup of the $T$ filter, and is completely independent of what precedes it. The fact that the same atoms were once sorted by an $S$ filter has no influence whatever on what they will do once they have been sorted again into a pure beam by a $T$ apparatus. From then on, the probability for getting into different states is the same no matter what happened before they got into the $T$ apparatus.  As an example, let’s compare the experiment of (5.11) with the following experiment: \begin{equation} \label{Eq:III:5:12} \SGZ{S} \!\! \SGZ{T} \!\! \SGP{S'} \end{equation}  in which only the first $S$ is changed. Let’s say that the angle $\alpha$ (between $S$ and $T$) is such that in experiment (5.11) one-third of the atoms that get through $T$ also get through $S'$. In experiment (5.12), although there will, in general, be a different number of atoms coming through $T$, the same fraction of these—one-third—will also get through $S'$. We can, in fact, show from what you have learned earlier that the fraction of the atoms that come out of $T$ and get through any particular $S'$ depends only on $T$ and $S'$, not on anything that happened earlier. Let’s compare experiment (5.12) with \begin{equation} \label{Eq:III:5:13} \SGZ{S} \!\! \SGZ{T} \!\! \SGZ{S'}\!\!\!\!. \end{equation}  The amplitude that an atom that comes out of $S$ will also get through both $T$ and $S'$ is, for the experiments of (5.12), \begin{equation*} \braket{+S}{\OT}\braket{\OT}{\OS}. \end{equation*}  The corresponding probability is \begin{equation*} \abs{\braket{+S}{\OT}\braket{\OT}{\OS}}^2 =\abs{\braket{+S}{\OT}}^2\abs{\braket{\OT}{\OS}}^2. \end{equation*}  The probability for experiment (5.13) is \begin{equation*} \abs{\braket{\OS}{\OT}\braket{\OT}{\OS}}^2 =\abs{\braket{\OS}{\OT}}^2\abs{\braket{\OT}{\OS}}^2. \end{equation*}  The ratio is \begin{equation*} \frac{\abs{\braket{\OS}{\OT}}^2} {\abs{\braket{+S}{\OT}}^2}. \end{equation*}  and depends only on $T$ and $S'$, and not at all on which beam $(+S)$, $(\OS)$, or $(-S)$ is selected by $S$. (The absolute numbers may go up and down together depending on how much gets through $T$.) We would, of course, find the same result if we compared the probabilities that the atoms would go into the plus or the minus states with respect to $S'$, or the ratio of the probabilities to go into the zero or minus states. In fact, since these ratios depend only on which beam is allowed to pass through $T$, and not on the selection made by the first $S$ filter, it is clear that we would get the same result even if the last apparatus were not an $S$ filter. If we use for the third apparatus—which we will now call $R$—one rotated by some arbitrary angle with respect to $T$, we would find that a ratio such as $\abs{\braket{\OR}{\OT}}^2/\abs{\braket{+R}{\OT}}^2$ was independent of which beam was passed by the first filter $S$.