$K^{0}$ meson consists of a $d$ quark and an $\bar{s}$ antiquark. Its antiparticle $\bar{K^{0}}$ consists of an $s$ quark and a $d$ antidown quark.

$$|K^{0}\rangle=|d\bar{s}\rangle$$ $$|\bar{K}^{0}\rangle=|\bar{d}s\rangle$$

There are two versions of $K^{0}$ meson. They have the same mass but different decay lifetimes.

K-zero-short: $$K_{S}^{0}=\frac{1}{\sqrt{2}}\left[|K^{0}\rangle+|\bar{K}^{0}\rangle\right]$$ with the observed lifetime $\tau_{S}=90$ ps.

K-zero-long: $$K_{L}^{0}=\frac{1}{\sqrt{2}}\left[|K^{0}\rangle-|\bar{K}^{0}\rangle\right]$$ with the observed lifetime $\tau_{L}=50$ ns.

What is the physical interpretation of the minus sign? I know that $\pm 1$ are the eigenvalues of $CP$ operator and $K_{S}^{0}$ and $K_{L}^{0}$ are its eigenstates. But what is the physical difference between $K_{S}^{0}$ and $K_{L}^{0}$ besides that they decay the way they do.

  • $\begingroup$ Related . The lifetimes and the associated glaring difference in the decay modes are not quite physical? $\endgroup$ – Cosmas Zachos Nov 6 '17 at 16:06
  • $\begingroup$ @CosmasZachos, They are indeed, but I am trying to apply a more phenomenological approach. There must be a reason why those differences appear. It's not their composition. Is it their interaction with some particular quantum fields? Is it the way they couple? $\endgroup$ – Andrei Geanta Nov 7 '17 at 10:12
  • 1
    $\begingroup$ Yes, the way they couple weakly, almost preserving CP, restricts their decay modes, and thence the rest. $\endgroup$ – Cosmas Zachos Nov 7 '17 at 11:34

Strictly speaking it is not true, as you say, that the mass eigenstates are the CP eigenstates in the neutral kaon system. This is exactly where CP violation was observed in first place by Cronin and Fitch.

But lets assume CP symmetry holds (which is a good assumptions since the CP violation is tiny in the neutral kaon system). What is the difference between the two Kaons? It is not the particle content. We define 'particles' by their well defined lifetime and mass. It may look weird that these states are not the same but on the other hand why should they? In the lagrangian to your qft you have terms like $\psi^\dagger M \psi$ where $M$ could be a non-diagonal Matrix. Diagonalizing it creates mixed flavour states but diagonal mass eigenstates. Every oscillation of particles means that the flavour eigenstates are not matching with the mass eigenstates, like in B Meson or neutrino oscillations.

If you now observe two Kaons being produced in the same process but with different lifetimes you will conclude that they have to be different particles. Your minus sign then determines which Kaon will be the long-living and which the short-living one, since CP eigenstates can only decay into final states with the same eigenvalue (assuming CP holds). Neutral Kaons can decay into 2 or 3 pions. Now you could calculate what the CP eigenvalues of a 2 and a 3 pion system and conclude which Kaon will decay into which pion system. This also allows you to see why one of the Kaons will decay faster by looking at the phase space dimension.

I hope this helps a bit.


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