Kaon oscillations

Question

Neutral Kaons CP eigenstates can be built as following:

$$|K_1\rangle=\frac{|K^0\rangle-\bar{|K^0\rangle}}{\sqrt{2}}$$ $$|K_2\rangle=\frac{|K^0\rangle+\bar{|K^0\rangle}}{\sqrt{2}}$$

So that we obtain a CP-even and a CP-odd state:

$$CP|K_1\rangle=|K_1\rangle$$ $$CP|K_2\rangle=-|K_2\rangle$$

It is experimentally observed that $|K_1\rangle$ decay rapidly in two pions while $|K_2\rangle$ has a longer lifetime and decays into three pions. Therefore, $|K_1\rangle$ is commonly called $|K_S\rangle$ (S for short) and $|K_2\rangle$ is called $|K_L\rangle$ (L for long).

My question is very simple: is it possible that $|K_S\rangle$ and $|K_L\rangle$ oscillate, as $|K^0\rangle$ and $\bar{|K^0\rangle}$ do?

Answer 1

You say that $K_0$ and $\bar{K}_0$ oscillate because while you create a state with definite strangeness (either $K_0$ or $\bar{K}_0$ from a strong decay) what you observe decaying are mass eigenstates ($K_S$ and $K_L$) that are a linear combination of $K_0$ and $\bar{K}_0$.

What this means is that you start with a state of definite strangeness, you write it as a combination of $K_S$ and $K_L$, you make those states evolve with time and you see that the probability of finding a state with a certain strangeness is either $\cos^2(\frac{\Delta m}{2} t)$ or $\sin^2(\frac{\Delta m}{2} t)$ depending on which strangeness you started with.

On the other hand, $K_S$ and $K_L$ are mass eigenstates, so they evolve with a pure phase. This means that if you start with one state and you "leave him alone" it will not change (until it decays).

You can still see the fact you can also write $K_S$ and $K_L$ as combination of $K_0$ and $\bar{K}_0$ in neutral Kaon regeneration experiments.

Answer 2

In the basis for $K^0$ there is an oscillation between $K_L$ and $K_S$. However, this is not the basis we make measurements in. We measure the decay products, which are the pions, and what is found is there is a probability for the production of $2$ or $3$ pions that match the lifetime of $K_S$ and $K_L$ respectively.