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?

  • $\begingroup$ I would like to add a question to this. How do we relate masses of $\left|K^{0}\right>$ and $\left|\bar{K^{0}}\right>$ to that of $\left|K_{L}\right>$and$\left|K_{S}\right>$? $\endgroup$ Commented Oct 5, 2015 at 9:33

1 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.

  • $\begingroup$ Thank you for your answer but this literature in their in books. I guess I didn't mention my problem clearly. Can you explain how $ K_{L}$and $K_{S}$ exist in nature. Is their any other way to detect one of these without looking at decay products? $\endgroup$ Commented Oct 5, 2015 at 9:14
  • $\begingroup$ yes, the lifetime of both kaons is long enough to reach the hadronic calorimeters. The $K_{s,L}$ then interact with the nucleon creating an hadronic shower. But you can't determine if this was a $K_S$ or a $K_L$. In addition, since they react via strong interaction it means that the $K_{S,L}$ are projected on $K^0$ or $\overline{K^0}$ by the detection... $\endgroup$
    – Paganini
    Commented Oct 5, 2015 at 9:43
  • $\begingroup$ Now, this is confusing. Answer a binary question. Which two are the physical states out of $\left|K^{0,\bar{0}}\right>$ and $\left|K_{L,S}\right>$? $\endgroup$ Commented Oct 5, 2015 at 10:09
  • $\begingroup$ both are physical states… If by physical state you mean state that propagate (having a definite lifetime), it is $K_{S,L}$. If by physical state, you mean the one that can be physically detected (or created) $K^0$ is the answer. As usual with QM, both exists, just a matter of the way they are observed. $\endgroup$
    – Paganini
    Commented Oct 5, 2015 at 10:20
  • $\begingroup$ Thank you. I will look into some papers regarding how these two sets of states are detected. $\endgroup$ Commented Oct 5, 2015 at 10:54

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