Bose particles and Fermi particles
In the last chapter we began to consider the special rules for the interference that occurs in processes with two identical particles. By identical particles we mean things like electrons which can in no way be distinguished one from another. If a process involves two particles that are identical, reversing which one arrives at a counter is an alternative which cannot be distinguished and—like all cases of alternatives which cannot be distinguished—interferes with the original, unexchanged case. The amplitude for an event is then the sum of the two interfering amplitudes; but, interestingly enough, the interference is in some cases with the same phase and, in others, with the opposite phase. Suppose we have a collision of two particles $a$ and $b$ in which particle $a$ scatters in the direction $1$ and particle $b$ scatters in the direction $2$, as sketched in Fig. 4–1(a). Let’s call $f(\theta)$ the amplitude for this process; then the probability $P_1$ of observing such an event is proportional to $\abs{f(\theta)}^2$. Of course, it could also happen that particle $b$ scattered into counter $1$ and particle $a$ went into counter $2$, as shown in Fig. 4–1(b). Assuming that there are no special directions defined by spins or such, the probability $P_2$ for this process is just $\abs{f(\pi-\theta)}^2$, because it is just equivalent to the first process with counter $1$ moved over to the angle $\pi-\theta$. You might also think that the amplitude for the second process is just $f(\pi-\theta)$. But that is not necessarily so, because there could be an arbitrary phase factor. That is, the amplitude could be \begin{equation*} e^{i\delta}f(\pi-\theta). \end{equation*}  Such an amplitude still gives a probability $P_2$ equal to $\abs{f(\pi-\theta)}^2$. Now let’s see what happens if $a$ and $b$ are identical particles. Then the two different processes shown in the two diagrams of Fig. 4–1 cannot be distinguished. There is an amplitude that either $a$ or $b$ goes into counter $1$, while the other goes into counter $2$. This amplitude is the sum of the amplitudes for the two processes shown in Fig. 4–1. If we call the first one $f(\theta)$, then the second one is $e^{i\delta}f(\pi-\theta)$, where now the phase factor is very important because we are going to be adding two amplitudes. Suppose we have to multiply the amplitude by a certain phase factor when we exchange the roles of the two particles. If we exchange them again we should get the same factor again. But we are then back to the first process. The phase factor taken twice must bring us back where we started—its square must be equal to $1$. There are only two possibilities: $e^{i\delta}$ is equal to $+1$, or is equal to $-1$. Either the exchanged case contributes with the same sign, or it contributes with the opposite sign. Both cases exist in nature, each for a different class of particles. Particles which interfere with a positive sign are called Bose particles and those which interfere with a negative sign are called Fermi particles. The Bose particles are the photon, the mesons, and the graviton. The Fermi particles are the electron, the muon, the neutrinos, the nucleons, and the baryons. We have, then, that the amplitude for the scattering of identical particles is:
  Bose particles: \begin{equation} \label{Eq:III:4:1} (\text{Amplitude direct})+(\text{Amplitude exchanged}). \end{equation}   Fermi particles: \begin{equation} \label{Eq:III:4:2} (\text{Amplitude direct})-(\text{Amplitude exchanged}). \end{equation}  For particles with spin—like electrons—there is an additional complication. We must specify not only the location of the particles but the direction of their spins. It is only for identical particles with identical spin states that the amplitudes interfere when the particles are exchanged. If you think of the scattering of unpolarized beams—which are a mixture of different spin states—there is some extra arithmetic. Now an interesting problem arises when there are two or more particles bound tightly together. For example, an $\alpha$-particle has four particles in it—two neutrons and two protons. When two $\alpha$-particles scatter, there are several possibilities. It may be that during the scattering there is a certain amplitude that one of the neutrons will leap across from one $\alpha$-particle to the other, while a neutron from the other $\alpha$-particle leaps the other way so that the two alphas which come out of the scattering are not the original ones—there has been an exchange of a pair of neutrons. See Fig. 4–2. The amplitude for scattering with an exchange of a pair of neutrons will interfere with the amplitude for scattering with no such exchange, and the interference must be with a minus sign because there has been an exchange of one pair of Fermi particles. On the other hand, if the relative energy of the two $\alpha$-particles is so low that they stay fairly far apart—say, due to the Coulomb repulsion—and there is never any appreciable probability of exchanging any of the internal particles, we can consider the $\alpha$-particle as a simple object, and we do not need to worry about its internal details. In such circumstances, there are only two contributions to the scattering amplitude. Either there is no exchange, or all four of the nucleons are exchanged in the scattering. Since the protons and the neutrons in the $\alpha$-particle are all Fermi particles, an exchange of any pair reverses the sign of the scattering amplitude. So long as there are no internal changes in the $\alpha$-particles, interchanging the two $\alpha$-particles is the same as interchanging four pairs of Fermi particles. There is a change in sign for each pair, so the net result is that the amplitudes combine with a positive sign. The $\alpha$-particle behaves like a Bose particle. So the rule is that composite objects, in circumstances in which the composite object can be considered as a single object, behave like Fermi particles or Bose particles, depending on whether they contain an odd number or an even number of Fermi particles. All the elementary Fermi particles we have mentioned—such as the electron, the proton, the neutron, and so on—have a spin $j=1/2$. If several such Fermi particles are put together to form a composite object, the resulting spin may be either integral or half-integral. For example, the common isotope of helium, He$^4$, which has two neutrons and two protons, has a spin of zero, whereas Li$^7$, which has three protons and four neutrons, has a spin of $3/2$. We will learn later the rules for compounding angular momentum, and will just mention now that every composite object which has a half-integral spin imitates a Fermi particle, whereas every composite object with an integral spin imitates a Bose particle. This brings up an interesting question: Why is it that particles with half-integral spin are Fermi particles whose amplitudes add with the minus sign, whereas particles with integral spin are Bose particles whose amplitudes add with the positive sign? We apologize for the fact that we cannot give you an elementary explanation. An explanation has been worked out by Pauli from complicated arguments of quantum field theory and relativity. He has shown that the two must necessarily go together, but we have not been able to find a way of reproducing his arguments on an elementary level. It appears to be one of the few places in physics where there is a rule which can be stated very simply, but for which no one has found a simple and easy explanation. The explanation is deep down in relativistic quantum mechanics. This probably means that we do not have a complete understanding of the fundamental principle involved. For the moment, you will just have to take it as one of the rules of the world.