Tuesday, 16 June 2020

particle physics - Understanding Kaon Mixing


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. |KL=12(|K0+|¯K0) |KS=12(|K0|¯K0) Here, |KL goes to two pions while|KS decays to three pion system. The thing I don't understand is that in nature what exists in |K0 and |¯K0; how do they sometimes combine to give a CP violating state and sometimes CP preserving?



Answer



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



  • K0 and ¯K0 are the kaons produced by strong interaction. They have a definite isospin and strangeness quantum numbers made respectively of dˉs and ˉds. However they can oscillate meaning that they transform spontaneously in each other: K0¯K0 through a box diagram involving the weak interaction. Clearly, that means that K0 or ¯K0 cannot have a definite lifetime (since they transform continuously).

  • KL and KS are the kaon having a definite lifetime, the ones propagating and observed in the detectors.

  • K1 and K2 are the linear combinations of K0 and ¯K0 that are CP eigentates: K1=12(K0¯K0) and K2=12(K0+¯K0). Kaons decay into 2 pions states or 3 pions states. It turns out that the CP quantum numbers are such that: CP(2π)=CP(K1) and CP(3π)=CP(K2).



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 KL, the only surviving kaons having a lifetime long enough to survive. If CP was conserved, you would expect to observe only: KL=K23π The equality KL=K2 results from phase space considerations: the phase space in the 2π states is much larger than in 3π, so we expect K1 to have a much smaller lifetime than K2, thus the identification KL=K2. However, in 0.2% of the case we do observe KL2π. So what can you conclude?




  • either, there is a direct CP violation: KL is really a K2 CP eigenstate but CP is violated in the decay process itself allowing K22π,




  • or there is an indirect CP violation: KL is not a pure K2 but a mixture of K1 and K2: KL=11+ϵ2(K2+ϵK1) ϵ 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.




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