, 5 min, 975 words
Tags: physics particle-physics
As I've previously mentioned, the K mesons have a glorious history in messing with particle physicists. The discovery of the K mesons and $\Lambda$ baryons was a puzzle for physicists. Here were particles that were produced in strong interactions but decayed with lifetimes more than a thousand times those typical of strong decays. Instead, the decays were by weak interactions. This, of course, begs the question: why can't the K and $\Lambda$ decay strongly? To explain the phenomenon, physicists invented the concept of strangeness, and stated that strong and electromagnetic interactions had to conserve it. So once a particle was produced that had nonzero strangeness, the only way it could decay in the absence of other strange particles to interact with was by the weak force, which explained the extraordinarily long lifetimes (on the order of a nanosecond instead of a femtosecond) of the strange particles.
Strangeness is an additive quantum number, so a combined state including one particle with -1 strangeness and another with +1 strangeness (like the $K^+ + \Lambda^0$) has a total strangeness of zero. Therefore under charge conjugation, a particle's strangeness switches sign. In other words, a particle and its antiparticle have opposite strangeness.
Nowadays, with our knowledge of quarks, we say that a strange quark $s$ has strangeness -1, and an anti-strange quark $\overline{s}$ has strangeness +1. Previously, these were arbitrary designations in which the $K^+$ (quark content $u\overline{s}$) had strangeness +1 and other particles' strangeness values were determined by examination of strange production processes. Anyhow, enough about strangeness - let's get to the interesting bit!
Neutral K mesons (or just kaons) are produced in strong and electromagnetic interactions, usually by pair production of strange quarks, which produces a meson along with a strange hadron. Strangeness is conserved in this interaction, so the two possible neutral K mesons to be produced have quark contents of $d\overline{s}$ ($K^0$) and $\overline{d}s$ ($\overline{K}^0$). This is in sharp contrast to the neutral pion ($\pi^0$), which is a superposition of two states $u\overline{u}$ and $d\overline{d}$. Since these states are indistinguishable by any quantum number, the neutral pion is its own antiparticle. The distinguishing feature here is that the $K^0$ and $\overline{K}^0$ have different values for strangeness.
The real weirdness kicks in when the particle decays. It decays in a weak interaction, so strangeness is not conserved. This means that the particle's strangeness eigenstate ($K^0$ or $\overline{K}^0$) is irrelevant to the decay mode. Instead, what matters is the particle's CP eigenstate, since CP is (mostly) conserved in weak interactions. But of course the strangeness eigenstates are not also CP eigenstates. What can we do? Examine the two superpositions of strangeness eigenstates that are CP eigenstates! We know that the charge conjugate of the $K^0$ is the $\overline{K}^0$, and vice versa. It turns out that both strangeness eigenstates have odd parity, so we find that the eigenstates of CP are superpositions of $K^0$ and $\overline{K}^0$ with either a plus or minus sign between them. The even CP eigenstate is called $K_1$ (creative name, right?), and the CP-odd eigenstate is called (wait for it...) $K_2$.
Any state involving neutral kaons can be expressed as a superposition of the strangeness eigenstates or the CP eigenstates. In particular, if a bunch of $K^0$ mesons are produced, we can view them as half $K_1$'s and half $K_2$'s. Or a bunch of $K_2$'s can be viewed as half $K^0$'s and half $\overline{K}^0$'s. It's a matter of which basis you choose to express the state of the system.
In weak decays, the CP eigenstate is (mostly) conserved, so $K_1$ has to decay to a CP-even final state. In particular, it decays into two pions. By similar logic, $K_2$ can decay into three pions. As a result of something called phase space that determines the probabilities of various decays, the $2\pi$ decay is about a hundred times faster than the $3\pi$ decay. So the two CP eigenstates have distinct lifetimes. That means that if we start with a pure $K^0$ beam, the $K_1$ component decays away very quickly in a flurry of $2\pi$ decays, so a short while later, only $K_2$'s are left.
There are some philosophical implications to the distinction between CP and strangeness eigenstates. In particular, what is a particle? Is it the thing produced in a strong interaction, or the thing that decays away in a weak interaction? Is it the thing with a definite lifetime, or the thing with a definite production cross-section? In the end, though, these are primarily questions of philosophy rather than physics, and it makes little difference which states you define as particles.
Oh, and another thing. Not content with introducing strangeness and violating parity conservation, the kaons were also the first proof of CP violation. If CP were conserved, the strangeness and CP eigenstates of the neutral kaon would be all we needed to analyze the (somewhat bizarre) system. In that case, after all the $K_1$'s had decayed away, we would expect to see only $3\pi$ decays of $K_2$'s. Instead, Cronin and Fitch observed that a beam of pure $K^0$, after traveling over fifty feet (plenty of time for all the $K_1$ components to decay away), around one in five hundred of the decays were in fact $2\pi$ decays. That means that the long-lived neutral kaon (which gets yet another name, $K_L$, for K-long) is actually a mixture of both $K_2$ and $K_1$, with substantially more $K_2$. And that means that like parity, CP is not a perfect symmetry in weak interactions. Another score for kaons in that fascinating game of Confuse the Physicists.