Interfering amplitudes
How can it be that in going from (5.15) to (5.17)—by opening more channels—we let fewer atoms through? This is the old, deep mystery of quantum mechanics—the interference of amplitudes. It’s the same kind of thing we first saw in the two-slit interference experiment with electrons. We saw that we could get fewer electrons at some places with both slits open than we got with one slit open. It works quantitatively this way. We can write the amplitude that an atom will get through $T$ and $S'$ in the apparatus of (5.17) as the sum of three amplitudes, one for each of the three beams in $T$; the sum is equal to zero: \begin{equation} \begin{alignedat}{3} &\braket{\OS}{+T}&\braket{+T}{+S}&\;&+&\\[.2ex] &\braket{\OS}{\OT}&\,\braket{\OT}{+S}&&+&\\[.2ex] &\braket{\OS}{-T}&\braket{-T}{+S}&&=&\;0. \end{alignedat} \label{Eq:III:5:18} \end{equation} None of the three individual amplitudes is zero—for example, the absolute square of the second amplitude is $\gamma\alpha$, see (5.15)—but the sum is zero. We would have also the same answer if $S'$ were set to select the $(-S)$ state. However, in the setup of (5.16), the answer is different. If we call $a$ the amplitude to get through $T$ and $S'$, in this case we have5 \begin{equation} \begin{alignedat}{3} a\;=\;&\braket{+S}{+T}&\braket{+T}{+S}&\;&+&\\[.2ex] &\braket{+S}{\OT}&\,\braket{\OT}{+S}&&+&\\[.2ex] &\braket{+S}{-T}&\braket{-T}{+S}&&=&\;1. \end{alignedat} \label{Eq:III:5:19} \end{equation}  In the experiment (5.16) the beam has been split and recombined. Humpty Dumpty has been put back together again. The information about the original $(+S)$ state is retained—it is just as though the $T$ apparatus were not there at all. This is true whatever is put after the “wide-open” $T$ apparatus. We could follow it with an $R$ filter—a filter at some odd angle—or anything we want. The answer will always be the same as if the atoms were taken directly from the first $S$ filter. So this is the important principle: A $T$ filter—or any filter—with wide-open masks produces no change at all. We should make one additional condition. The wide-open filter must not only transmit all three beams, but it must also not produce unequal disturbances on the three beams. For instance, it should not have a strong electric field near one beam and not the others. The reason is that even if this extra disturbance would still let all the atoms through the filter, it could change the phases of some of the amplitudes. Then the interference would be changed, and the amplitudes in Eqs. (5.18) and (5.19) would be different. We will always assume that there are no such extra disturbances. Let’s rewrite Eqs. (5.18) and (5.19) in an improved notation. We will let $i$ stand for any one of the three states $(+T)$, $(\OT)$, or $(-T)$; then the equations can be written: \begin{equation} \label{Eq:III:5:20} \sum_{\text{all $i$}}\braket{\OS}{i}\braket{i}{+S}=0 \end{equation}  and \begin{equation} \label{Eq:III:5:21} \sum_{\text{all $i$}}\braket{+S}{i}\braket{i}{+S}=1. \end{equation}  Similarly, for an experiment where $S'$ is replaced by a completely arbitrary filter $R$, we have \begin{equation} \label{Eq:III:5:22} \SGP{S} \!\! \SG{T} \!\! \SGP{R} \!\!\!\!\cdotp \end{equation}  The results will always be the same as if the $T$ apparatus were left out and we had only \begin{equation*} \SGP{S} \!\! \SGP{R} \!\!\!\!\cdotp \end{equation*}  Or, expressed mathematically, \begin{equation} \label{Eq:III:5:23} \sum_{\text{all $i$}}\braket{+R}{i}\braket{i}{+S}= \braket{+R}{+S}. \end{equation}  This is our fundamental law, and it is generally true so long as $i$ stands for the three base states of any filter. You will notice that in the experiment (5.22) there is no special relation of $S$ and $R$ to $T$. Furthermore, the arguments would be the same no matter what states they selected. To write the equation in a general way, without having to refer to the specific states selected by $S$ and $R$, let’s call $\phi$ (“phi”) the state prepared by the first filter (in our special example, $+S$) and $\chi$ (“khi”) the state tested by the final filter (in our example, $+R$). Then we can state our fundamental law of Eq. (5.23) in the form \begin{equation} \label{Eq:III:5:24} \braket{\chi}{\phi}= \sum_{\text{all $i$}}\braket{\chi}{i}\braket{i}{\phi}, \end{equation}  where $i$ is to range over the three base states of some particular filter. We want to emphasize again what we mean by base states. They are like the three states which can be selected by one of our Stern-Gerlach apparatuses. One condition is that if you have a base state, then the future is independent of the past. Another condition is that if you have a complete set of base states, Eq. (5.24) is true for any set of beginning and ending states $\phi$ and $\chi$. There is, however, no unique set of base states. We began by considering base states with respect to a particular apparatus $T$. We could equally well consider a different set of base states with respect to an apparatus $S$, or with respect to $R$, etc.6 We usually speak of the base states “in a certain representation.” Another condition on a set of base states in any particular representation is that they are all completely different. By that we mean that if we have a $(+T)$ state, there is no amplitude for it to go into a $(\OT)$ or a $(-T)$ state. If we let $i$ and $j$ stand for any two base states of a particular set, the general rules discussed in connection with (5.8) are that \begin{equation*} \braket{j}{i}=0 \end{equation*}  for all $i$ and $j$ that are not equal. Of course, we know that \begin{equation*} \braket{i}{i}=1. \end{equation*}  These two equations are usually written as \begin{equation} \label{Eq:III:5:25} \braket{j}{i}=\delta_{ji}, \end{equation}  where $\delta_{ji}$ (the “Kronecker delta”) is a symbol that is defined to be zero for $i\neq j$, and to be one for $i=j$. Equation (5.25) is not independent of the other laws we have mentioned. It happens that we are not particularly interested in the mathematical problem of finding the minimum set of independent axioms that will give all the laws as consequences.7 We are satisfied if we have a set that is complete and not apparently inconsistent. We can, however, show that Eqs. (5.25) and (5.24) are not independent. Suppose we let $\phi$ in Eq. (5.24) represent one of the base states of the same set as $i$, say the $j$th state; then we have \begin{equation*} \braket{\chi}{j}= \sum_i\braket{\chi}{i}\braket{i}{j}. \end{equation*}  But Eq. (5.25) says that $\braket{i}{j}$ is zero unless $i=j$, so the sum becomes just $\braket{\chi}{j}$ and we have an identity, which shows that the two laws are not independent. We can see that there must be another relation among the amplitudes if both Eqs. (5.10) and (5.24) are true. Equation (5.10) is \begin{alignat*}{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{alignat*}  If we write Eq. (5.24), letting both $\phi$ and $\chi$ be the state $(+S)$, the left-hand side is $\braket{+S}{+S}$, which is clearly${}=1$; so we get once more Eq. (5.19), \begin{alignat*}{3} &\braket{+S}{+T}&\braket{+T}{+S}&\;&+&\\[.2ex] &\braket{+S}{\OT}&\,\braket{\OT}{+S}&&+&\\[.2ex] &\braket{+S}{-T}&\braket{-T}{+S}&&=&\;1. \end{alignat*}    These two equations are consistent (for all relative orientations of the $T$ and $S$ apparatuses) only if \begin{align*} \braket{+S}{+T}&=\braket{+T}{+S}\cconj,\\[1ex] \braket{+S}{\OT}&=\braket{\OT}{+S}\cconj,\\[1ex] \braket{+S}{-T}&=\braket{-T}{+S}\cconj. \end{align*}  And it follows that for any states $\phi$ and $\chi$, \begin{equation} \label{Eq:III:5:26} \braket{\phi}{\chi}=\braket{\chi}{\phi}\cconj. \end{equation}  If this were not true, probability wouldn’t be “conserved,” and particles would get “lost.” Before going on, we want to summarize the three important general laws about amplitudes. They are Eqs. (5.24), (5.25), and (5.26): \begin{equation} \begin{alignedat}{2} &\text{$\phantom{II}$I}&\quad \braket{j}{i}&=\delta_{ji},\\[1ex] &\text{$\phantom{I}$II}&\quad \braket{\chi}{\phi}&= \sum_{\text{all $i$}} \braket{\chi}{i}\braket{i}{\phi},\\[1ex] &\text{III}&\quad \braket{\phi}{\chi}&= \braket{\chi}{\phi}\cconj. \end{alignedat} \label{Eq:III:5:27} \end{equation}  In these equations the $i$ and $j$ refer to all the base states of some one representation, while $\phi$ and $\chi$ represent any possible states of the atom. It is important to note that II is valid only if the sum is carried out over all the base states of the system (in our case, three: $+T$, $\OT$, $-T$). These laws say nothing about what we should choose for a base for our set of base states. We began by using a $T$ apparatus, which is a Stern-Gerlach experiment with some arbitrary orientation; but any other orientation, say $W$, would be just as good. We would have a different set of states to use for $i$ and $j$, but all the laws would still be good—there is no unique set. One of the great games of quantum mechanics is to make use of the fact that things can be calculated in more than one way.