Understanding Kaon Mixing

Question

While studying Neutral-Kaon mixing, I learnt that the way we observe them is through their decays to Pions. But the states that decay are not the particles/antiparticles themselves but their mixture. $$ \left|K_{L}\right>=\dfrac{1}{\sqrt2}(\left|K^{0}\right>+\left|\bar{K^{0}}\right>) $$ $$ \left|K_{S}\right>=\dfrac{1}{\sqrt2}(\left|K^{0}\right>-\left|\bar{K^{0}}\right>)$$ Here, $\left|K_{L}\right>$ goes to two pions while$\left|K_{S}\right>$ decays to three pion system. The thing I don't understand is that in nature what exists in $\left|K^{0}\right>$ and $\left|\bar{K_{0}}\right>$; how do they sometimes combine to give a CP violating state and sometimes CP preserving?

Answer 1

Depending on the way you look at them, there are several kind of kaons:

• $K^0$ and $\overline{K^0}$ are the kaons produced by strong interaction. They have a definite isospin and strangeness quantum numbers made respectively of $d\bar{s}$ and $\bar{d}s$. However they can oscillate meaning that they transform spontaneously in each other: $K^0 \leftrightarrow \overline{K^0}$ through a box diagram involving the weak interaction. Clearly, that means that $K^0$ or $\overline{K^0}$ cannot have a definite lifetime (since they transform continuously).
• $K_L$ and $K_S$ are the kaon having a definite lifetime, the ones propagating and observed in the detectors.
• $K_1$ and $K_2$ are the linear combinations of $K^0$ and $\overline{K^0}$ that are CP eigentates: $K_1 = \frac{1}{\sqrt{2}}(K^0 - \overline{K^0})$ and $K_2 = \frac{1}{\sqrt{2}}(K^0 + \overline{K^0})$. Kaons decay into 2 pions states or 3 pions states. It turns out that the CP quantum numbers are such that: $CP(2\pi) = CP(K_1)$ and $CP(3\pi)=CP(K_2)$.

So far so good. Now, experimentally when we create kaons via the strong interaction and detect them long time after their creation we know that what is detected are actually $K_L$, the only surviving kaons having a lifetime long enough to survive. If CP was conserved, you would expect to observe only: $$ K_L = K_2 \to 3\pi$$ The equality $K_L=K_2$ results from phase space considerations: the phase space in the $2\pi$ states is much larger than in $3\pi$, so we expect $K_1$ to have a much smaller lifetime than $K_2$, thus the identification $K_L=K_2$. However, in 0.2% of the case we do observe $K_L \to 2\pi$. So what can you conclude?

• either, there is a direct CP violation: $K_L$ is really a $K_2$ CP eigenstate but CP is violated in the decay process itself allowing $K_2 \to 2\pi$,

• or there is an indirect CP violation: $K_L$ is not a pure $K_2$ but a mixture of $K_1$ and $K_2$: $$ K_L = \frac{1}{\sqrt{1+\epsilon^2}}(K_2+\epsilon K_1)$$ $\epsilon$ characterizing the amount of indirect CP violation. It turns out that both violations exist! In the kaon system the latter dominated (about 600 times larger than the former). But in other systems, like neutral B meson, direct CP violation is pretty common.