Base states
These results illustrate one of the basic principles of quantum mechanics: Any atomic system can be separated by a filtering process into a certain set of what we will call base states, and the future behavior of the atoms in any single given base state depends only on the nature of the base state—it is independent of any previous history.3 The base states depend, of course, on the filter used; for instance, the three states $(+T)$, $(\OT)$, and $(-T)$ are one set of base states; the three states $(+S)$, $(\OS)$, and $(-S)$ are another. There are any number of possibilities each as good as any other. We should be careful to say that we are considering good filters which do indeed produce “pure” beams. If, for instance, our Stern-Gerlach apparatus didn’t produce a good separation of the three beams so that we could not separate them cleanly by our masks, then we could not make a complete separation into base states. We can tell if we have pure base states by seeing whether or not the beams can be split again in another filter of the same kind. If we have a pure $(+T)$ state, for instance, all the atoms will go through  \begin{equation*} \SGP{T}\!\!\!\!, \end{equation*}  and none will go through  \begin{equation*} \SGZ{T}\!\!\!\!, \end{equation*}  or through  \begin{equation*} \SGM{T}\!\!\!\!\cdotp \end{equation*}  Our statement about base states means that it is possible to filter to some pure state, so that no further filtering by an identical apparatus is possible. We must also point out that what we are saying is exactly true only in rather idealized situations. In any real Stern-Gerlach apparatus, we would have to worry about diffraction by the slits that could cause some atoms to go into states corresponding to different angles, or about whether the beams might contain atoms with different excitations of their internal states, and so on. We have idealized the situation so that we are talking only about the states that are split in a magnetic field; we are ignoring things having to do with position, momentum, internal excitations, and the like. In general, one would need to consider also base states which are sorted out with respect to such things also. But to keep the concepts simple, we are considering only our set of three states, which is sufficient for the exact treatment of the idealized situation in which the atoms don’t get torn up in going through the apparatus, or otherwise badly treated, and come to rest when they leave the apparatus. You will note that we always begin our thought experiments by taking a filter with only one channel open, so that we start with some definite base state. We do this because atoms come out of a furnace in various states determined at random by the accidental happenings inside the furnace. (It gives what is called an “unpolarized” beam.) This randomness involves probabilities of the “classical” kind—as in coin tossing—which are different from the quantum mechanical probabilities we are worrying about now. Dealing with an unpolarized beam would get us into additional complications that are better to avoid until after we understand the behavior of polarized beams. So don’t try to consider at this point what happens if the first apparatus lets more than one beam through. (We will tell you how you can handle such cases at the end of the chapter.) Let’s now go back and see what happens when we go from a base state for one filter to a base state for a different filter. Suppose we start again with \begin{equation*} \SGP{S} \!\! \SGZ{T}\!\!\!\!\cdotp \end{equation*}  The atoms which come out of $T$ are in the base state $(\OT)$ and have no memory that they were once in the state $(+S)$. Some people would say that in the filtering by $T$ we have “lost the information” about the previous state $(+S)$ because we have “disturbed” the atoms when we separated them into three beams in the apparatus $T$. But that is not true. The past information is not lost by the separation into three beams, but by the blocking masks that are put in—as we can see by the following set of experiments. We start with a $+S$ filter and will call $N$ the number of atoms that come through it. If we follow this by a $\OT$ filter, the number of atoms that come out is some fraction of the original number, say $\alpha N$. If we then put another $+S$ filter, only some fraction $\beta$ of these atoms will get to the far end. We can indicate this in the following way:  \begin{equation} \label{Eq:III:5:14} \SGP{S}   \kern{-12pt} \raise{7pt} {\overset{N}{\longrightarrow}}\! \SGZ{T}   \kern{-12pt} \raise{7pt} {\overset{\alpha N}{\longrightarrow}}\! \SGP{S'}  \kern{-12pt} \raise{7pt} {\overset{\beta \alpha N}{\longrightarrow .}} \end{equation}  If our third apparatus $S'$ selected a different state, say the $(\OS)$ state, a different fraction, say $\gamma$, would get through.4 We would have \begin{equation} \label{Eq:III:5:15} \SGP{S}   \kern{-12pt} \raise{7pt} {\overset{N}{\longrightarrow}}\! \SGZ{T}   \kern{-12pt} \raise{7pt} {\overset{\alpha N}{\longrightarrow}}\! \SGZ{S'}  \kern{-12pt} \raise{7pt} {\overset{\gamma \alpha N}{\longrightarrow .}} \end{equation}  Now suppose we repeat these two experiments but remove all the masks from $T$. We would then find the remarkable results as follows: \begin{equation} \label{Eq:III:5:16} \SGP{S}   \kern{-12pt} \raise{7pt} {\overset{N}{\longrightarrow}}\! \SG{T}   \kern{-12pt} \raise{7pt} {\overset{N}{\longrightarrow}}\! \SGP{S'}  \kern{-12pt} \raise{7pt} {\overset{N}{\longrightarrow ,}} \end{equation}  \begin{equation} \label{Eq:III:5:17} \SGP{S}   \kern{-12pt} \raise{7pt} {\overset{N}{\longrightarrow}}\! \SG{T}   \kern{-12pt} \raise{7pt} {\overset{N}{\longrightarrow}}\! \SGZ{S'}  \kern{-12pt} \raise{7pt} {\overset{0}{\longrightarrow .}} \end{equation}  All the atoms get through $S'$ in the first case, but none in the second case! This is one of the great laws of quantum mechanics. That nature works this way is not self-evident, but the results we have given correspond for our idealized situation to the quantum mechanical behavior observed in innumerable experiments.