Scattering from a crystal
Our next example is a phenomenon in which we have to analyze the interference of probability amplitudes somewhat carefully. We look at the process of the scattering of neutrons from a crystal. Suppose we have a crystal which has a lot of atoms with nuclei at their centers, arranged in a periodic array, and a neutron beam that comes from far away. We can label the various nuclei in the crystal by an index $i$, where $i$ runs over the integers $1$, $2$, $3$, …, $N$, with $N$ equal to the total number of atoms. The problem is to calculate the probability of getting a neutron into a counter with the arrangement shown in Fig. 3–5. For any particular atom $i$, the amplitude that the neutron arrives at the counter $C$ is the amplitude that the neutron gets from the source $S$ to nucleus $i$, multiplied by the amplitude $a$ that it gets scattered there, multiplied by the amplitude that it gets from $i$ to the counter $C$. Let’s write that down:  \begin{equation} \label{Eq:III:3:11} \braket{\text{neutron at $C$}}{\text{neutron from $S$}}_{\text{via $i$}}= \braket{C}{i}\,a\,\braket{i}{S}. \end{equation} 
 \begin{gather} \braket{\text{neutron at $C$}}{\text{neutron from $S$}}_{\text{via $i$}}\notag\\[.5ex] \label{Eq:III:3:11} =\braket{C}{i}\,a\,\braket{i}{S}. \end{gather}  In writing this equation we have assumed that the scattering amplitude $a$ is the same for all atoms. We have here a large number of apparently indistinguishable routes. They are indistinguishable because a low-energy neutron is scattered from a nucleus without knocking the atom out of its place in the crystal—no “record” is left of the scattering. According to the earlier discussion, the total amplitude for a neutron at $C$ involves a sum of Eq. (3.11) over all the atoms:  \begin{equation} \label{Eq:III:3:12} \braket{\text{neutron at $C$}}{\text{neutron from $S$}}= \sum_{i=1}^N\braket{C}{i}\,a\,\braket{i}{S}. \end{equation} 
 \begin{gather} \braket{\text{neutron at $C$}}{\text{neutron from $S$}}\notag\\[.75ex] \label{Eq:III:3:12} =\sum_{i=1}^N\braket{C}{i}\,a\,\braket{i}{S}. \end{gather}  Because we are adding amplitudes of scattering from atoms with different space positions, the amplitudes will have different phases giving the characteristic interference pattern that we have already analyzed in the case of the scattering of light from a grating. The neutron intensity as a function of angle in such an experiment is indeed often found to show tremendous variations, with very sharp interference peaks and almost nothing in between—as shown in Fig. 3–6(a). However, for certain kinds of crystals it does not work this way, and there is—along with the interference peaks discussed above—a general background of scattering in all directions. We must try to understand the apparently mysterious reasons for this. Well, we have not considered one important property of the neutron. It has a spin of one-half, and so there are two conditions in which it can be: either spin “up” (say perpendicular to the page in Fig. 3–5) or spin “down.”  If the nuclei of the crystal have no spin, the neutron spin doesn’t have any effect. But when the nuclei of the crystal also have a spin, say a spin of one-half, you will observe the background of smeared-out scattering described above. The explanation is as follows. If the neutron has one direction of spin and the atomic nucleus has the same spin, then no change of spin can occur in the scattering process. If the neutron and atomic nucleus have opposite spin, then scattering can occur by two processes, one in which the spins are unchanged and another in which the spin directions are exchanged. This rule for no net change of the sum of the spins is analogous to our classical law of conservation of angular momentum. We can begin to understand the phenomenon if we assume that all the scattering nuclei are set up with spins in one direction. A neutron with the same spin will scatter with the expected sharp interference distribution. What about one with opposite spin? If it scatters without spin flip, then nothing is changed from the above; but if the two spins flip over in the scattering, we could, in principle, find out which nucleus had done the scattering, since it would be the only one with spin turned over. Well, if we can tell which atom did the scattering, what have the other atoms got to do with it? Nothing, of course. The scattering is exactly the same as that from a single atom. To include this effect, the mathematical formulation of Eq. (3.12) must be modified since we haven’t described the states completely in that analysis. Let’s start with all neutrons from the source having spin up and all the nuclei of the crystal having spin down. First, we would like the amplitude that at the counter the spin of the neutron is up and all spins of the crystal are still down. This is not different from our previous discussion. We will let $a$ be the amplitude to scatter with no spin flip. The amplitude for scattering from the $i$th atom is, of course,  \begin{equation*} \braket{\text{$C_{\text{up}}$, crystal all down}} {\text{$S_{\text{up}}$, crystal all down}}= \braket{C}{i}\,a\,\braket{i}{S}. \end{equation*} 
 \begin{gather*} \braket{\text{$C_{\text{up}}$, crystal all down}} {\text{$S_{\text{up}}$, crystal all down}}\\[.5ex] =\braket{C}{i}\,a\,\braket{i}{S}. \end{gather*}  Since all the atomic spins are still down, the various alternatives (different values of $i$) cannot be distinguished. There is clearly no way to tell which atom did the scattering. For this process, all the amplitudes interfere. We have another case, however, where the spin of the detected neutron is down although it started from $S$ with spin up. In the crystal, one of the spins must be changed to the up direction—let’s say that of the $k$th atom. We will assume that there is the same scattering amplitude with spin flip for every atom, namely $b$. (In a real crystal there is the disagreeable possibility that the reversed spin moves to some other atom, but let’s take the case of a crystal for which this probability is very low.) The scattering amplitude is then  \begin{equation} \label{Eq:III:3:13} \braket{\text{$C_{\text{down}}$, nucleus $k$ up}} {\text{$S_{\text{up}}$, crystal all down}}= \braket{C}{k}\,b\,\braket{k}{S}. \end{equation} 
 \begin{gather} \braket{\text{$C_{\text{down}}$, nucleus $k$ up}} {\text{$S_{\text{up}}$, crystal all down}}\notag\\[.5ex] \label{Eq:III:3:13} =\braket{C}{k}\,b\,\braket{k}{S}. \end{gather}  If we ask for the probability of finding the neutron spin down and the $k$th nucleus spin up, it is equal to the absolute square of this amplitude, which is simply $\abs{b}^2$ times $\abs{\braket{C}{k}\braket{k}{S}}^2$. The second factor is almost independent of location in the crystal, and all phases have disappeared in taking the absolute square. The probability of scattering from any nucleus in the crystal with spin flip is now \begin{equation*} \abs{b}^2\sum_{k=1}^N\abs{\braket{C}{k}\braket{k}{S}}^2, \end{equation*}  which will show a smooth distribution as in Fig. 3–6(b). You may argue, “I don’t care which atom is up.” Perhaps you don’t, but nature knows; and the probability is, in fact, what we gave above—there isn’t any interference. On the other hand, if we ask for the probability that the spin is up at the detector and all the atoms still have spin down, then we must take the absolute square of \begin{equation*} \sum_{i=1}^N\braket{C}{i}\,a\,\braket{i}{S}. \end{equation*}  Since the terms in this sum have phases, they do interfere, and we get a sharp interference pattern. If we do an experiment in which we don’t observe the spin of the detected neutron, then both kinds of events can occur; and the separate probabilities add. The total probability (or counting rate) as a function of angle then looks like the graph in Fig. 3–6(c). Let’s review the physics of this experiment. If you could, in principle, distinguish the alternative final states (even though you do not bother to do so), the total, final probability is obtained by calculating the probability for each state (not the amplitude) and then adding them together. If you cannot distinguish the final states even in principle, then the probability amplitudes must be summed before taking the absolute square to find the actual probability. The thing you should notice particularly is that if you were to try to represent the neutron by a wave alone, you would get the same kind of distribution for the scattering of a down-spinning neutron as for an up-spinning neutron. You would have to say that the “wave” would come from all the different atoms and interfere just as for the up-spinning one with the same wavelength. But we know that is not the way it works. So as we stated earlier, we must be careful not to attribute too much reality to the waves in space. They are useful for certain problems but not for all.