The annihilation of positronium
We would like next to take an example which is very pretty. It is quite interesting and, although somewhat complicated, we hope not too much so. Our example is the system called positronium, which is an “atom” made up of an electron and a positron—a bound state of an e$^+$ and an e$^-$. It is like a hydrogen atom, except that a positron replaces the proton. This object has—like the hydrogen atom—many states. Also like the hydrogen, the ground state is split into a “hyperfine structure” by the interaction of the magnetic moments. The spins of the electron and positron are each one-half, and they can be either parallel or antiparallel to any given axis. (In the ground state there is no other angular momentum due to orbital motion.) So there are four states: three are the substates of a spin-one system, all with the same energy; and one is a state of spin zero with a different energy. The energy splitting is, however, much larger than the $1420$ megacycles of hydrogen because the positron magnetic moment is so much stronger—$1000$ times stronger—than the proton moment.  The most important difference, however, is that positronium cannot last forever. The positron is the antiparticle of the electron; they can annihilate each other. The two particles disappear completely—converting their rest energy into radiation, which appears as $\gamma$-rays (photons). In the disintegration, two particles with a finite rest mass go into two or more objects which have zero rest mass.3
 We begin by analyzing the disintegration of the spin-zero state of the positronium. It disintegrates into two $\gamma$-rays with a lifetime of about $10^{-10}$ second. Initially, we have a positron and an electron close together and with spins antiparallel, making the positronium system. After the disintegration there are two photons going out with equal and opposite momenta (Fig. 18–5). The momenta must be equal and opposite, because the total momentum after the disintegration must be zero, as it was before, if we are taking the case of annihilation at rest. If the positronium is not at rest, we can ride with it, solve the problem, and then transform everything back to the lab system. (See, we can do anything now; we have all the tools.)  First, we note that the angular distribution is not very interesting. Since the initial state has spin zero, it has no special axis—it is symmetric under all rotations. The final state must then also be symmetric under all rotations. That means that all angles for the disintegration are equally likely—the amplitude is the same for a photon to go in any direction. Of course, once we find one of the photons in some direction the other must be opposite.  The only remaining question, which we now want to look at, is about the polarization of the photons. Let’s call the directions of motion of the two photons the plus and minus $z$-axes. We can use any representations we want for the polarization states of the photons; we will choose for our description right and left circular polarization—always with respect to the directions of motion.4 Right away, we can see that if the photon going upward is RHC, then angular momentum will be conserved if the downward going photon is also RHC. Each will carry $+1$ unit of angular momentum with respect to its momentum direction, which means plus and minus one unit about the $z$-axis. The total will be zero, and the angular momentum after the disintegration will be the same as before. See Fig. 18–6.  The same arguments show that if the upward going photon is RHC, the downward cannot be LHC. Then the final state would have two units of angular momentum. This is not permitted if the initial state has spin zero. Note that such a final state is also not possible for the other positronium ground state of spin one, because it can have a maximum of one unit of angular momentum in any direction.  Now we want to show that two-photon annihilation is not possible at all from the spin-one state. You might think that if we took the $j=1$, $m=0$ state—which has zero angular momentum about the $z$-axis—it should be like the spin-zero state, and could disintegrate into two RHC photons. Certainly, the disintegration sketched in Fig. 18–7(a) conserves angular momentum about the $z$-axis. But now look what happens if we rotate this system around the $y$-axis by $180^\circ$; we get the picture shown in Fig. 18–7(b). It is exactly the same as in part (a) of the figure. All we have done is interchange the two photons. Now photons are Bose particles; if we interchange them, the amplitude has the same sign, so the amplitude for the disintegration in part (b) must be the same as in part (a). But we have assumed that the initial object is spin one. And when we rotate a spin-one object in a state with $m=0$ by $180^\circ$ about the $y$-axis, its amplitudes change sign (see Table 17–2 for $\theta=\pi$). So the amplitudes for (a) and (b) in Fig. 18–7 should have opposite signs; the spin-one state cannot disintegrate into two photons.  When positronium is formed you would expect it to end up in the spin-zero state $1/4$ of the time and in the spin-one state (with $m=-1$, $0$, or $+1$) $3/4$ of the time. So $1/4$ of the time you would get two-photon annihilations. The other $3/4$ of the time there can be no two-photon annihilations. There is still an annihilation, but it has to go with three photons. It is harder for it to do that and the lifetime is $1000$ times longer—about $10^{-7}$ second. This is what is observed experimentally. We will not go into any more of the details of the spin-one annihilation.  So far we have that if we only worry about angular momentum, the spin-zero state of the positronium can go into two RHC photons. There is also another possibility: it can go into two LHC photons as shown in Fig. 18–8. The next question is, what is the relation between the amplitudes for these two possible decay modes? We can find out from the conservation of parity.  To do that, however, we need to know the parity of the positronium. Now theoretical physicists have shown in a way that is not easy to explain that the parity of the electron and the positron—its antiparticle—must be opposite, so that the spin-zero ground state of positronium must be odd. We will just assume that it is odd, and since we will get agreement with experiment, we can take that as sufficient proof.  Let’s see then what happens if we make an inversion of the process in Fig. 18–6. When we do that, the two photons reverse directions and polarizations. The inverted picture looks just like Fig. 18–8. Assuming that the parity of the positronium is odd, the amplitudes for the two processes in Figs. 18–6 and 18–8 must have the opposite sign. Let’s let $\ket{R_1R_2}$ stand for the final state of Fig. 18–6 in which both photons are RHC, and let $\ket{L_1L_2}$ stand for the final state of Fig. 18–8, in which both photons are LHC. The true final state—let’s call it $\ket{F}$—must be \begin{equation} \label{Eq:III:18:19} \ket{F}=\ket{R_1R_2}-\ket{L_1L_2}. \end{equation} Then an inversion changes the $R$’s into $L$’s and gives the state \begin{equation} \label{Eq:III:18:20} P\,\ket{F}=\ket{L_1L_2}-\ket{R_1R_2}=-\,\ket{F}, \end{equation} which is the negative of (18.19). So the final state $\ket{F}$ has negative parity, which is the same as the initial spin-zero state of the positronium. This is the only final state that conserves both angular momentum and parity. There is some amplitude that the disintegration into this state will occur, which we don’t need to worry about now, however, since we are only interested in questions about the polarization.  What does the final state of (18.19) mean physically? One thing it means is the following: If we observe the two photons in two detectors which can be set to count separately the RHC or LHC photons, we will always see two RHC photons together, or two LHC photons together. That is, if you stand on one side of the positronium and someone else stands on the opposite side, you can measure the polarization and tell the other guy what polarization he will get. You have a $50$-$50$ chance of catching a RHC photon or a LHC photon; whichever one you get, you can predict that he will get the same.  Since there is a $50$-$50$ chance for RHC or LHC polarization, it sounds as though it might be like linear polarization. Let’s ask what happens if we observe the photon in counters that accept only linearly polarized light. For $\gamma$-rays it is not as easy to measure the polarization as it is for light; there is no polarizer which works well for such short wavelengths. But let’s imagine that there is, to make the discussion easier. Suppose that you have a counter that only accepts light with $x$-polarization, and that there is a guy on the other side that also looks for linear polarized light with, say, $y$-polarization. What is the chance you will pick up the two photons from an annihilation? What we need to ask is the amplitude that $\ket{F}$ will be in the state $\ket{x_1y_2}$. In other words, we want the amplitude \begin{equation*} \braket{x_1y_2}{F}, \end{equation*} which is, of course, just \begin{equation} \label{Eq:III:18:21} \braket{x_1y_2}{R_1R_2}-\braket{x_1y_2}{L_1L_2}. \end{equation}  Now although we are working with two-particle amplitudes for the two photons, we can handle them just as we did the single particle amplitudes, since each particle acts independently of the other. That means that the amplitude $\braket{x_1y_2}{R_1R_2}$ is just the product of the two independent amplitudes $\braket{x_1}{R_1}$ and $\braket{y_2}{R_2}$. Using Table 17–3, these two amplitudes are $1/\sqrt{2}$ and $i/\sqrt{2}$—so  \begin{equation*} \braket{x_1y_2}{R_1R_2}=+\frac{i}{2}. \end{equation*}  Similarly, we find that  \begin{equation*} \braket{x_1y_2}{L_1L_2}=-\frac{i}{2}. \end{equation*}  Subtracting these two amplitudes according to (18.21), we get that \begin{equation} \label{Eq:III:18:22} \braket{x_1y_2}{F}=+i. \end{equation} So there is a unit probability5 that if you get a photon in your $x$-polarized detector, the other guy will get a photon in his $y$-polarized detector.  Now suppose that the other guy sets his counter for $x$-polarization the same as yours. He would never get a count when you got one. If you work it through, you will find that \begin{equation} \label{Eq:III:18:23} \braket{x_1x_2}{F}=0. \end{equation} It will, naturally, also work out that if you set your counter for $y$-polarization he will get coincident counts only if he is set for $x$-polarization.  Now this all leads to an interesting situation. Suppose you were to set up something like a piece of calcite which separated the photons into $x$-polarized and $y$-polarized beams, and put a counter in each beam. Let’s call one the $x$-counter and the other the $y$-counter. If the guy on the other side does the same thing, you can always tell him which beam his photon is going to go into. Whenever you and he get simultaneous counts, you can see which of your detectors caught the photon and then tell him which of his counters had a photon. Let’s say that in a certain disintegration you find that a photon went into your $x$-counter; you can tell him that he must have had a count in his $y$-counter.  Now many people who learn quantum mechanics in the usual (old-fashioned) way find this disturbing. They would like to think that once the photons are emitted it goes along as a wave with a definite character. They would think that since “any given photon” has some “amplitude” to be $x$-polarized or to be $y$-polarized, there should be some chance of picking it up in either the $x$- or $y$-counter and that this chance shouldn’t depend on what some other person finds out about a completely different photon. They argue that “someone else making a measurement shouldn’t be able to change the probability that I will find something.” Our quantum mechanics says, however, that by making a measurement on photon number one, you can predict precisely what the polarization of photon number two is going to be when it is detected. This point was never accepted by Einstein, and he worried about it a great deal—it became known as the “Einstein-Podolsky-Rosen paradox.” But when the situation is described as we have done it here, there doesn’t seem to be any paradox at all; it comes out quite naturally that what is measured in one place is correlated with what is measured somewhere else. The argument that the result is paradoxical runs something like this: 
If you have a counter which tells you whether your photon is RHC or LHC, you can predict exactly what kind of a photon (RHC or LHC) he will find. 
The photons he receives must, therefore, each be purely RHC or purely LHC, some of one kind and some of the other. 
Surely you cannot alter the physical nature of his photons by changing the kind of observation you make on your photons. No matter what measurements you make on yours, his must still be either RHC or LHC. 
Now suppose he changes his apparatus to split his photons into two linearly polarized beams with a piece of calcite so that all of his photons go either into an $x$-polarized beam or into a $y$-polarized beam. There is absolutely no way, according to quantum mechanics, to tell into which beam any particular RHC photon will go. There is a 50% probability it will go into the $x$-beam and a 50% probability it will go into the $y$-beam. And the same goes for a LHC photon. 
Since each photon is RHC or LHC—according to (2) and (3)—each one must have a 50-50 chance of going into the $x$-beam or the $y$-beam and there is no way to predict which way it will go. 
Yet the theory predicts that if you see your photon go through an $x$-polarizer you can predict with certainty that his photon will go into his $y$-polarized beam. This is in contradiction to (5) so there is a paradox. 

 Nature apparently doesn’t see the “paradox,” however, because experiment shows that the prediction in (6) is, in fact, true. We have already discussed the key to this “paradox” in our very first lecture on quantum mechanical behavior in Chapter 37, Vol. I.6 In the argument above, steps (1), (2), (4), and (6) are all correct, but (3), and its consequence (5), are wrong; they are not a true description of nature. Argument (3) says that by your measurement (seeing a RHC or a LHC photon) you can determine which of two alternative events occurs for him (seeing a RHC or a LHC photon), and that even if you do not make your measurement you can still say that his event will occur either by one alternative or the other. But it was precisely the point of Chapter 37, Vol. I, to point out right at the beginning that this is not so in Nature. Her way requires a description in terms of interfering amplitudes, one amplitude for each alternative. A measurement of which alternative actually occurs destroys the interference, but if a measurement is not made you cannot still say that “one alternative or the other is still occurring.”  If you could determine for each one of your photons whether it was RHC or LHC, and also whether it was $x$-polarized (all for the same photon) there would indeed be a paradox. But you cannot do that—it is an example of the uncertainty principle.  Do you still think there is a “paradox”? Make sure that it is, in fact, a paradox about the behavior of Nature, by setting up an imaginary experiment for which the theory of quantum mechanics would predict inconsistent results via two different arguments. Otherwise the “paradox” is only a conflict between reality and your feeling of what reality “ought to be.”  Do you think that it is not a “paradox,” but that it is still very peculiar? On that we can all agree. It is what makes physics fascinating. 