The neutral K-meson
We will now describe a two-state system in the world of the strange particles—a system for which quantum mechanics gives a most remarkable prediction. To describe it completely would involve us in a lot of stuff about strange particles, so we will, unfortunately, have to cut some corners. We can only give an outline of how a certain discovery was made—to show you the kind of reasoning that was involved. It begins with the discovery by Gell-Mann and Nishijima of the concept of strangeness and of a new law of conservation of strangeness. It was when Gell-Mann and Pais were analyzing the consequences of these new ideas that they came across the prediction of a most remarkable phenomenon we are going to describe. First, though, we have to tell you a little about “strangeness.”  We must begin with what are called the strong interactions of nuclear particles. These are the interactions which are responsible for the strong nuclear forces—as distinct, for instance, from the relatively weaker electromagnetic interactions. The interactions are “strong” in the sense that if two particles get close enough to interact at all, they interact in a big way and produce other particles very easily. The nuclear particles have also what is called a “weak interaction” by which certain things can happen, such as beta decay, but always very slowly on a nuclear time scale—the weak interactions are many, many orders of magnitude weaker than the strong interactions and even much weaker than electromagnetic interactions.  When the strong interactions were being studied with the big accelerators, people were surprised to find that certain things that “should” happen—that were expected to happen—did not occur. For instance, in some interactions a particle of a certain type did not appear when it was expected. Gell-Mann and Nishijima noticed that many of these peculiar happenings could be explained at once by inventing a new conservation law: the conservation of strangeness. They proposed that there was a new kind of attribute associated with each particle—which they called its “strangeness” number—and that in any strong interaction the “quantity of strangeness” is conserved.  Suppose, for instance, that a high-energy negative K-meson—with, say, an energy of many GeV—collides with a proton. Out of the interaction may come many other particles: $\pi$-mesons, K-mesons, lambda particles, sigma particles—any of the mesons or baryons listed in Table 2–2 of Vol. I. It is observed, however, that only certain combinations appear, and never others. Now certain conservation laws were already known to apply. First, energy and momentum are always conserved. The total energy and momentum after an event must be the same as before the event. Second, there is the conservation of electric charge which says that the total charge of the outgoing particles must be equal to the total charge carried by the original particles. In our example of a K-meson and a proton coming together, the following reactions do occur: \begin{align} &\Kminus+\text{p}\to\text{p}+\Kminus+\pi^++\pi^-+\pi^0\notag \\ \label{Eq:III:11:38} \kern{-3em}\text{or}\\ % ebook remove % ebook insert: \makebox[0em]{\kern{-5em}\text{or}}\\ &\Kminus+\text{p}\to\Sigma^-+\pi^+.\notag \end{align} We would never get:  \begin{equation} \label{Eq:III:11:39} \Kminus+\text{p}\to\text{p}+\Kminus+\pi^+ \quad\text{or}\quad \Kminus+\text{p}\to\Lambda^0+\pi^+, \end{equation} 
 \begin{align} &\Kminus+\text{p}\to\text{p}+\Kminus+\pi^+\notag\\ \label{Eq:III:11:39} \kern{-3em}\text{or}\\ % ebook remove % ebook insert: \makebox[0em]{\kern{-5em}\text{or}}\\ &\Kminus+\text{p}\to\Lambda^0+\pi^+,\notag \end{align}  because of the conservation of charge. It was also known that the number of baryons is conserved. The number of baryons out must be equal to the number of baryons in. For this law, an antiparticle of a baryon is counted as minus one baryon. This means that we can—and do—see \begin{align} &\Kminus+\text{p}\to\Lambda^0+\pi^0\notag \\ \label{Eq:III:11:40} \kern{-3em}\text{or}\\ % ebook remove % ebook insert: \makebox[0em]{\kern{-5em}\text{or}}\\ &\Kminus+\text{p}\to\text{p}+\Kminus+\text{p}+\overline{\text{p}}\notag \end{align} (where $\overline{\text{p}}$ is the antiproton, which carries a negative charge). But we never see \begin{align} &\Kminus+\text{p}\to\Kminus+\pi^++\pi^0\notag \\ \label{Eq:III:11:41} \kern{-3em}\text{or}\\ % ebook remove % ebook insert: \makebox[0em]{\kern{-5em}\text{or}}\\ &\Kminus+\text{p}\to\text{p}+\Kminus+\text{n}\notag \end{align} (even when there is plenty of energy), because baryons would not be conserved.  These laws, however, do not explain the strange fact that the following reactions—which do not immediately appear to be especially different from some of those in (11.38) or (11.40)—are also never observed: \begin{align} &\Kminus+\text{p}\to\text{p}+\Kminus+\Kzero\notag \\ \kern{-3em}\text{or}\notag\\ % ebook remove % ebook insert: \makebox[0em]{\kern{-5em}\text{or}}\notag\\ \label{Eq:III:11:42} &\Kminus+\text{p}\to\text{p}+\pi^-\\ \kern{-3em}\text{or}\notag\\ % ebook remove % ebook insert: \makebox[0em]{\kern{-5em}\text{or}}\notag\\ &\Kminus+\text{p}\to\Lambda^0+\Kzero.\notag \end{align} The explanation is the conservation of strangeness. With each particle goes a number—its strangeness $S$—and there is a law that in any strong interaction, the total strangeness out must equal the total strangeness that went in. The proton and antiproton ($\text{p}$, $\overline{\text{p}}$), the neutron and antineutron ($\text{n}$, $\overline{\text{n}}$), and the $\pi$-mesons ($\pi^+$, $\pi^0$, $\pi^-$) all have the strangeness number zero; the $\Kplus$ and $\Kzero$ mesons have strangeness $+1$; the $\Kminus$ and $\Kzerobar$ (the anti-$\Kzero$),3 the $\Lambda^0$ and the $\Sigma$-particles ($+$, $0$, $-$) have strangeness $-1$. There is also a particle with strangeness $-2$—the $\Xi$-particle (capital “ksi”)—and perhaps others as yet unknown. We have made a list of these strangenesses in Table 11–4.  Let’s see how the strangeness conservation works in some of the reactions we have written down. If we start with a $\Kminus$ and a proton, we have a total strangeness of $(-1+0)=-1$. The conservation of strangeness says that the strangeness of products after the reaction must also add up to $-1$. You see that that is so for the reactions of (11.38) and (11.40). But in the reactions of (11.42) the strangeness of the right-hand side is zero in each case. Such reactions do not conserve strangeness, and do not occur. Why? Nobody knows. Nobody knows any more than what we have just told you about this. Nature just works that way.  Now let’s look at the following reaction: a $\pi^-$ hits a proton. You might, for instance, get a $\Lambda^0$ particle plus a neutral K-particle—two neutral particles. Now which neutral K do you get? Since the $\Lambda$-particle has a strangeness $-1$ and the $\pi$ and $\text{p}^+$ have a strangeness zero, and since this is a fast production reaction, the strangeness must not change. The K-particle must have strangeness $+1$—it must therefore be the $\Kzero$. The reaction is \begin{equation*} \pi^-+\text{p}\to\Lambda^0+\Kzero, \end{equation*} with \begin{equation*} S=0+0=-1++1\quad(\text{conserved}). \end{equation*} If the $\Kzerobar$ were there instead of the $\Kzero$, the strangeness on the right would be $-2$—which nature does not permit, since the strangeness on the left side is zero. On the other hand, a $\Kzerobar$ can be produced in other reactions, such as \begin{gather*} \text{n}+\text{n}\to\text{n}+\overline{\text{p}}+\Kzerobar+\Kplus,\\ \\ S=0+0=0+0+-1++1 \end{gather*} or \begin{gather*} \Kminus+\text{p}\to\text{n}+\Kzerobar,\\ \\ S=-1+0=0+-1. \end{gather*}  You may be thinking, “That’s all a lot of stuff, because how do you know whether it is a $\Kzerobar$ or a $\Kzero$?  They look exactly the same. They are antiparticles of each other, so they have exactly the same mass, and both have zero electric charge. How do you distinguish them?” By the reactions they produce. For example, a $\Kzerobar$ can interact with matter to produce a $\Lambda$-particle, like this: \begin{equation*} \Kzerobar+\text{p}\to\Lambda^0+\pi^+, \end{equation*} but a $\Kzero$ cannot. There is no way a $\Kzero$ can produce a $\Lambda$-particle when it interacts with ordinary matter (protons and neutrons).4 So the experimental distinction between the $\Kzero$ and the $\Kzerobar$ would be that one of them will and one of them will not produce $\Lambda$'s.  One of the predictions of the strangeness theory is then this—if, in an experiment with high-energy pions, a $\Lambda$-particle is produced with a neutral K-meson, then that neutral K-meson going into other pieces of matter will never produce a $\Lambda$. The experiment might run something like this. You send a beam of $\pi^-$-mesons into a large hydrogen bubble chamber. A $\pi^-$ track disappears, but somewhere else a pair of tracks appear (a proton and a $\pi^-$) indicating that a $\Lambda$-particle has disintegrated5 —see Fig. 11-5. Then you know that there is a $\Kzero$ somewhere which you cannot see.  You can, however, figure out where it is going by using the conservation of momentum and energy. [It could reveal itself later by disintegrating into two charged particles, as shown in Fig. 11-5(a).]  As the $\Kzero$ goes flying along, it may interact with one of the hydrogen nuclei (protons), producing perhaps some other particles. The prediction of the strangeness theory is that it will never produce a $\Lambda$-particle in a simple reaction like, say, \begin{equation*} \Kzero+\text{p}\to\Lambda^0+\pi^+, \end{equation*} although a $\Kzerobar$ can do just that. That is, in a bubble chamber a $\Kzerobar$ might produce the event sketched in Fig. 11-5(b)—in which the $\Lambda^0$ is seen because it decays—but a $\Kzero$ will not. That’s the first part of our story. That’s the conservation of strangeness.  The conservation of strangeness is, however, not perfect. There are very slow disintegrations of the strange particles—decays taking a long6  time like $10^{-10}$ second in which the strangeness is not conserved. These are called the “weak” decays. For example, the $\Kzero$ disintegrates into a pair of $\pi$-mesons ($+$ and $-$) with a lifetime of $10^{-10}$ second. That was, in fact, the way K-particles were first seen. Notice that the decay reaction \begin{equation*} \Kzero\to\pi^++\pi^- \end{equation*} does not conserve strangeness, so it cannot go “fast” by the strong interaction; it can only go through the weak decay process.  Now the $\Kzerobar$ also disintegrates in the same way—into a $\pi^+$ and a $\pi^-$—and also with the same lifetime \begin{equation*} \Kzerobar\to\pi^-+\pi^+. \end{equation*} Again we have a weak decay because it does not conserve strangeness. There is a principle that for any reaction there is the corresponding reaction with “matter” replaced by “antimatter” and vice versa. Since the $\Kzerobar$ is the antiparticle of the $\Kzero$, it should decay into the antiparticles of the $\pi^+$ and $\pi^-$, but the antiparticle of a $\pi^+$ is the $\pi^-$. (Or, if you prefer, vice versa. It turns out that for the $\pi$-mesons it doesn’t matter which one you call “matter.”) So as a consequence of the weak decays, the $\Kzero$ and $\Kzerobar$ can go into the same final products. When “seen” through their decays—as in a bubble chamber—they look like the same particle. Only their strong interactions are different.  At last we are ready to describe the work of Gell-Mann and Pais. They first noticed that since the $\Kzero$ and the $\Kzerobar$ can both turn into states of two $\pi$-mesons there must be some amplitude that a $\Kzero$ can turn into a $\Kzerobar$, and also that a $\Kzerobar$ can turn into a $\Kzero$. Writing the reactions as one does in chemistry, we would have \begin{equation} \label{Eq:III:11:43} \Kzero\rightleftharpoons\pi^-+\pi^+\rightleftharpoons\Kzerobar. \end{equation} These reactions imply that there is some amplitude per unit time, say $-i/\hbar$ times $\bracket{\Kzerobar}{\text{W}}{\Kzero}$, that a $\Kzero$ will turn into a $\Kzerobar$ through the weak interaction responsible for the decay into two $\pi$-mesons. And there is the corresponding amplitude $\bracket{\Kzero}{\text{W}}{\Kzerobar}$ for the reverse process. Because matter and antimatter behave in exactly the same way, these two amplitudes are numerically equal; we’ll call them both $A$: \begin{equation} \label{Eq:III:11:44} \bracket{\Kzerobar}{\text{W}}{\Kzero}= \bracket{\Kzero}{\text{W}}{\Kzerobar}=A. \end{equation}  Now—said Gell-Mann and Pais—here is an interesting situation. What people have been calling two distinct states of the world—the $\Kzero$ and the $\Kzerobar$—should really be considered as one two-state system, because there is an amplitude to go from one state to the other. For a complete treatment, one would, of course, have to deal with more than two states, because there are also the states of $2\pi$'s, and so on; but since they were mainly interested in the relation of $\Kzero$ and $\Kzerobar$, they did not have to complicate things and could make the approximation of a two-state system. The other states were taken into account to the extent that their effects appeared implicitly in the amplitudes of Eq. (11.44).  Accordingly, Gell-Mann and Pais analyzed the neutral particle as a two-state system. They began by choosing as their two base states the states $\ket{\Kzero}$ and $\ket{\Kzerobar}$. (From here on, the story goes very much as it did for the ammonia molecule.) Any state $\ket{\psi}$ of the neutral K-particle could then be described by giving the amplitudes that it was in either base state. We’ll call these amplitudes \begin{equation} \label{Eq:III:11:45} C_+=\braket{\Kzero}{\psi},\quad C_-=\braket{\Kzerobar}{\psi}. \end{equation}  The next step was to write the Hamiltonian equations for this two-state system. If there were no coupling between the $\Kzero$ and the $\Kzerobar$, the equations would be simply \begin{equation} \begin{aligned} i\hbar\,\ddt{C_+}{t}&=E_0C_+,\\[2ex] i\hbar\,\ddt{C_-}{t}&=E_0C_-. \end{aligned} \label{Eq:III:11:46} \end{equation} But since there is the amplitude $\bracket{\Kzero}{\text{W}}{\Kzerobar}$ for the $\Kzerobar$ to turn into a $\Kzero$ there should be the additional term \begin{equation*} \bracket{\Kzero}{\text{W}}{\Kzerobar}C_-=AC_- \end{equation*} added to the right-hand side of the first equation. And similarly, the term $AC_+$ should be inserted in the equation for the rate of change of $C_-$.  But that’s not all. When the two-pion effect is taken into account there is an additional amplitude for the $\Kzero$ to turn into itself through the process \begin{equation*} \Kzero\to\pi^-+\pi^+\to\Kzero. \end{equation*} The additional amplitude, which we would write $\bracket{\Kzero}{\text{W}}{\Kzero}$, is just equal to the amplitude $\bracket{\Kzerobar}{\text{W}}{\Kzero}$, since the amplitudes to go to and from a pair of $\pi$-mesons are identical for the $\Kzero$ and the $\Kzerobar$. If you wish, the argument can be written out in detail like this. First write7 \begin{equation*} \bracket{\Kzerobar}{\text{W}}{\Kzero}= \bracket{\Kzerobar}{\text{W}}{2\pi} \bracket{2\pi}{\text{W}}{\Kzero} \end{equation*} and \begin{equation*} \bracket{\Kzero}{\text{W}}{\Kzero}= \bracket{\Kzero}{\text{W}}{2\pi} \bracket{2\pi}{\text{W}}{\Kzero}. \end{equation*} Because of the symmetry of matter and antimatter \begin{equation*} \bracket{2\pi}{\text{W}}{\Kzero}= \bracket{2\pi}{\text{W}}{\Kzerobar}, \end{equation*} and also \begin{equation*} \bracket{\Kzero}{\text{W}}{2\pi}= \bracket{\Kzerobar}{\text{W}}{2\pi}. \end{equation*} It then follows that $\bracket{\Kzero}{\text{W}}{\Kzero}= \bracket{\Kzerobar}{\text{W}}{\Kzero}$, and also that $\bracket{\Kzerobar}{\text{W}}{\Kzero}= \bracket{\Kzero}{\text{W}}{\Kzerobar}$, as we said earlier. Anyway, there are the two additional amplitudes $\bracket{\Kzero}{\text{W}}{\Kzero}$ and $\bracket{\Kzerobar}{\text{W}}{\Kzerobar}$, both equal to $A$, which should be included in the Hamiltonian equations. The first gives a term $AC_+$ on the right-hand side of the equation for $dC_+/dt$, and the second gives a new term $AC_-$ in the equation for $dC_-/dt$. Reasoning this way, Gell-Mann and Pais concluded that the Hamiltonian equations for the $\Kzero\,\Kzerobar$ system should be \begin{equation} \begin{aligned} i\hbar\,\ddt{C_+}{t}&=E_0C_++AC_-+AC_+,\\[2ex] i\hbar\,\ddt{C_-}{t}&=E_0C_-+AC_++AC_-. \end{aligned} \label{Eq:III:11:47} \end{equation}  We must now correct something we have said in earlier chapters: that two amplitudes like $\bracket{\Kzero}{\text{W}}{\Kzerobar}$ and $\bracket{\Kzerobar}{\text{W}}{\Kzero}$ which are the reverse of each other, are always complex conjugates. That was true when we were talking about particles that did not decay. But if particles can decay—and can, therefore, become “lost”—the two amplitudes are not necessarily complex conjugates. So the equality of (11.44) does not mean that the amplitudes are real numbers; they are in fact complex numbers. The coefficient $A$ is, therefore, complex; and we can’t just incorporate it into the energy $E_0$.  Having played often with electron spins and such, our heroes knew that the Hamiltonian equations of (11.47) meant that there was another pair of base states which could also be used to represent the K-particle system and which would have especially simple behaviors. They said, “Let’s take the sum and difference of these two equations. Also, let’s measure all our energies from $E_0$, and use units for energy and time that make $\hbar=1$.” (That’s what modern theoretical physicists always do. It doesn’t change the physics but makes the equations take on a simple form.) Their result:  \begin{equation} \label{Eq:III:11:48} i\,\ddt{}{t}\,(C_++C_-)=2A(C_++C_-),\quad i\,\ddt{}{t}\,(C_+-C_-)=0. \end{equation} 
  \begin{equation} \begin{aligned} i\,\ddt{}{t}\,(C_++C_-)&=2A(C_++C_-),\\[2ex] i\,\ddt{}{t}\,(C_+-C_-)&=0. \end{aligned} \label{Eq:III:11:48} \end{equation}  
 It is apparent that the combinations of amplitudes $(C_++C_-)$ and $(C_+-C_-)$ act independently from each other (corresponding, of course, to the stationary states we have been studying earlier). So they concluded that it would be more convenient to use a different representation for the K-particle. They defined the two states  \begin{equation} \label{Eq:III:11:49} \ket{\text{K}_1}=\frac{1}{\sqrt{2}}\, (\ket{\Kzero}+\ket{\Kzerobar}),\quad \ket{\text{K}_2}=\frac{1}{\sqrt{2}}\, (\ket{\Kzero}-\ket{\Kzerobar}). \end{equation} 
  \begin{equation} \begin{aligned} \ket{\text{K}_1}&=\frac{1}{\sqrt{2}}\, (\ket{\Kzero}+\ket{\Kzerobar}),\\[2ex] \ket{\text{K}_2}&=\frac{1}{\sqrt{2}}\, (\ket{\Kzero}-\ket{\Kzerobar}). \end{aligned} \label{Eq:III:11:49} \end{equation}   They said that instead of thinking of the $\Kzero$ and $\Kzerobar$ mesons, we can equally well think in terms of the two “particles” (that is, “states”) K$_1$ and K$_2$. (These correspond, of course, to the states we have usually called $\ketsl{\slI}$ and $\ketsl{\slII}$. We are not using our old notation because we want now to follow the notation of the original authors—and the one you will see in physics seminars.)  Now Gell-Mann and Pais didn’t do all this just to get different names for the particles—there is also some strange new physics in it. Suppose that $C_1$ and $C_2$ are the amplitudes that some state $\ket{\psi}$ will be either a K$_1$ or a K$_2$ meson: \begin{equation*} C_1=\braket{\text{K}_1}{\psi},\quad C_2=\braket{\text{K}_2}{\psi}. \end{equation*} From the equations of (11.49),  \begin{equation} \label{Eq:III:11:50} C_1=\frac{1}{\sqrt{2}}\,(C_++C_-),\quad C_2=\frac{1}{\sqrt{2}}\,(C_+-C_-). \end{equation} 
 \begin{equation} \begin{aligned} C_1&=\frac{1}{\sqrt{2}}\,(C_++C_-),\\[2ex] C_2&=\frac{1}{\sqrt{2}}\,(C_+-C_-). \end{aligned} \label{Eq:III:11:50} \end{equation}  Then the Eqs. (11.48) become \begin{equation} \label{Eq:III:11:51} i\,\ddt{C_1}{t}=2AC_1,\quad i\,\ddt{C_2}{t}=0. \end{equation} The solutions are \begin{equation} \label{Eq:III:11:52} C_1(t)=C_1(0)e^{-i2At},\quad C_2(t)=C_2(0), \end{equation} where, of course, $C_1(0)$ and $C_2(0)$ are the amplitudes at $t=0$.  These equations say that if a neutral K-particle starts out in the state $\ket{\text{K}_1}$ at $t=0$ [then $C_1(0)=1$ and $C_2(0)=0$], the amplitudes at the time $t$ are \begin{equation*} C_1(t)=e^{-i2At},\quad C_2(t)=0. \end{equation*}  Remembering that $A$ is a complex number, it is convenient to take $2A=\alpha-i\beta$. (Since the imaginary part of $2A$ turns out to be negative, we write it as minus $i\beta$.) With this substitution, $C_1(t)$ reads \begin{equation} \label{Eq:III:11:53} C_1(t)=C_1(0)e^{-\beta t}e^{-i\alpha t}. \end{equation} The probability of finding a K$_1$ particle at $t$ is the absolute square of this amplitude, which is $e^{-2\beta t}$. And, from Eqs. (11.52), the probability of finding the K$_2$ state at any time is zero. That means that if you make a K-particle in the state $\ket{\text{K}_1}$, the probability of finding it in the same state decreases exponentially with time—but you will never find it in state $\ket{\text{K}_2}$. Where does it go?  It disintegrates into two $\pi$-mesons with the mean life $\tau=1/2\beta$ which is, experimentally, $10^{-10}$ sec. We made provisions for that when we said that $A$ was complex.  On the other hand, Eq. (11.52) says that if we make a K-particle completely in the K$_2$ state, it stays that way forever. Well, that’s not really true. It is observed experimentally to disintegrate into three $\pi$-mesons, but $600$ times slower than the two-pion decay we have described. So there are some other small terms we have left out in our approximation. But so long as we are considering only the two-pion decay, the K$_2$ lasts “forever.”  Now to finish the story of Gell-Mann and Pais. They went on to consider what happens when a K-particle is produced with a $\Lambda^0$ particle in a strong interaction. Since it must then have a strangeness of $+1$, it must be produced in the $\Kzero$ state. So at $t=0$ it is neither a K$_1$ nor a K$_2$ but a mixture. The initial conditions are \begin{equation*} C_+(0)=1,\quad C_-(0)=0. \end{equation*} But that means—from Eq. (11.50)—that \begin{equation*} C_1(0)=\frac{1}{\sqrt{2}},\quad C_2(0)=\frac{1}{\sqrt{2}}, \end{equation*} and—from Eqs. (11.52) and (11.53)—that \begin{equation} \label{Eq:III:11:54} C_1(t)=\frac{1}{\sqrt{2}}\, e^{-\beta t}e^{-i\alpha t},\quad C_2(t)=\frac{1}{\sqrt{2}}. \end{equation} Now remember that $\Kzero$ and $\Kzerobar$ are each linear combinations of K$_1$ and K$_2$. In Eqs. (11.54) the amplitudes have been chosen so that at $t=0$ the $\Kzerobar$ parts cancel each other out by interference, leaving only a $\Kzero$ state. But the $\ket{\text{K}_1}$ state changes with time, and the $\ket{\text{K}_2}$ state does not. After $t=0$ the interference of $C_1$ and $C_2$ will give finite amplitudes for both $\Kzero$ and $\Kzerobar$.  What does all this mean? Let’s go back and think of the experiment we sketched in Fig. 11-5. A $\pi^-$ meson has produced a $\Lambda^0$ particle and a $\Kzero$ meson which is tooting along through the hydrogen in the chamber. As it goes along, there is some small but uniform chance that it will collide with a hydrogen nucleus. At first, we thought that strangeness conservation would prevent the K-particle from making a $\Lambda^0$ in such an interaction. Now, however, we see that that is not right. For although our K-particle starts out as a $\Kzero$—which cannot make a $\Lambda^0$—it does not stay this way. After a while, there is some amplitude that it will have flipped to the $\Kzerobar$ state. We can, therefore, sometimes expect to see a $\Lambda^0$ produced along the K-particle track. The chance of this happening is given by the amplitude $C_-$, which we can [by using Eq. (11.50) backwards] relate to $C_1$ and $C_2$. The relation is \begin{equation} \label{Eq:III:11:55} C_-=\frac{1}{\sqrt{2}}\,(C_1-C_2)= \tfrac{1}{2}(e^{-\beta t}e^{-i\alpha t}-1). \end{equation} As our K-particle goes along, the probability that it will “act like” a $\Kzerobar$ is equal to $\abs{C_-}^2$, which is \begin{equation} \label{Eq:III:11:56} \abs{C_-}^2=\tfrac{1}{4} (1+e^{-2\beta t}-2e^{-\beta t}\cos\alpha t). \end{equation} A complicated and strange result!  This, then, is the remarkable prediction of Gell-Mann and Pais: when a $\Kzero$ is produced, the chance that it will turn into a $\Kzerobar$—as it can demonstrate by being able to produce a $\Lambda^0$—varies with time according to Eq. (11.56). This prediction came from using only sheer logic and the basic principles of the quantum mechanics—with no knowledge at all of the inner workings of the K-particle. Since nobody knows anything about the inner machinery, that is as far as Gell-Mann and Pais could go. They could not give any theoretical values for $\alpha$ and $\beta$. And nobody has been able to do so to this date. They were able to give a value of $\beta$ obtained from the experimentally observed rate of decay into two $\pi$'s ($2\beta=10^{10}$ sec$^{-1}$), but they could say nothing about $\alpha$.  We have plotted the function of Eq. (11.56) for two values of $\alpha$ in Fig. 11-6. You can see that the form depends very much on the ratio of $\alpha$ to $\beta$. There is no $\Kzerobar$ probability at first; then it builds up. If $\alpha$ is large, the probability would have large oscillations. If $\alpha$ is small, there will be little or no oscillation—the probability will just rise smoothly to $1/4$.  Now, typically, the K-particle will be travelling at a constant speed near the speed of light. The curves of Fig. 11-6 then also represent the probability along the track of observing a $\Kzerobar$—with typical distances of several centimeters. You can see why this prediction is so remarkably peculiar. You produce a single particle and instead of just disintegrating, it does something else. Sometimes it disintegrates, and other times it turns into a different kind of a particle. Its characteristic probability of producing an effect varies in a strange way as it goes along. There is nothing else quite like it in nature. And this most remarkable prediction was made solely by arguments about the interference of amplitudes.  If there is any place where we have a chance to test the main principles of quantum mechanics in the purest way—does the superposition of amplitudes work or doesn’t it?—this is it. In spite of the fact that this effect has been predicted now for several years, there is no experimental determination that is very clear. There are some rough results which indicate that the $\alpha$ is not zero, and that the effect really occurs—they indicate that $\alpha$ is between $2\beta$ and $4\beta$. That’s all there is, experimentally. It would be very beautiful to check out the curve exactly to see if the principle of superposition really still works in such a mysterious world as that of the strange particles—with unknown reasons for the decays, and unknown reasons for the strangeness.  The analysis we have just described is very characteristic of the way quantum mechanics is being used today in the search for an understanding of the strange particles. All the complicated theories that you may hear about are no more and no less than this kind of elementary hocus-pocus using the principles of superposition and other principles of quantum mechanics of that level. Some people claim that they have theories by which it is possible to calculate the $\beta$ and $\alpha$, or at least the $\alpha$ given the $\beta$, but these theories are completely useless. For instance, the theory that predicts the value of $\alpha$, given the $\beta$, tells us that the value of $\alpha$ should be infinite. The set of equations with which they originally start involves two $\pi$-mesons and then goes from the two $\pi$’s back to a $\Kzero$, and so on. When it’s all worked out, it does indeed produce a pair of equations like the ones we have here; but because there are an infinite number of states of two $\pi$'s, depending on their momenta, integrating over all the possibilities gives an $\alpha$ which is infinite. But nature’s $\alpha$ is not infinite. So the dynamical theories are wrong. It is really quite remarkable that the phenomena which can be predicted at all in the world of the strange particles come from the principles of quantum mechanics at the level at which you are learning them now. 