# What is the process transforming $K_L$ mesons into $K_S$ mesons?

First of all, I could not find any good book/webpage where I can read about K mesons, can somebody point me to a good reading? The second point is about $K_L$ and $K_S$ mesons. As I can read, they are eigenstates of $K^0$ with opposite strangeness. In principle they should decay into pions, but I was wondering if a $K_L$ can spontaneously change into a $K_S$ instead of decay into pions or there would be an interaction with protons/nuclei needed for this transformation to occur.

• Your link already explains regeneration: The freely flying $K_L$ and $K_S$ of course do not mix--that's the point of them being free hamiltonian eigenstates! But interaction of $K_L$ with a nuclear target regenerates $K_S$. What are you asking? Related. Have you tried any introductory book on particle physics? May 9, 2018 at 14:19
• my question is what is the process taking place in the generation of K_S, could you explain it using a Feynmann diagram? May 10, 2018 at 4:52
• Again, there is no regeneration in vacuum, as indicated in the WP article you cite, which has all the Feynman diagrams you'd ever need. The book by Perkins explains how, in matter, coherent strong interactions---think zillions of pions, gluons, etc---interact differently with the notional (fake, strong eigen-)states $K^0$ (elastic & charge-exchange scattering) and $\bar{K}^0$ (absorption) so treat these two differently and "tilt" $K_L$ back to $K_S$ by a bit. But if you wish to further understand that, you must remove the crank stuff from your question and say so. May 10, 2018 at 13:54

• $K_L$s cannot spontaneously transform to $K_S$s in free space. They are eigenstates of the free hamiltonian and maintain their identity. Their mass difference is $\Delta m \sim 10^{-14} m$. To effect a $K_L\to K_S$ transformation ("regeneration") they must go through matter and interact coherently and strongly with its nuclei: absolutely nothing spontaneous about it. I would strongly recommend changing your title, as the present one is odd.

Let's ignore CP violation to avoid mission creep and muddying the waters in the process, so just call $$|K_{L}\rangle \sim \dfrac{1}{\sqrt2}\left( |K^{0}\rangle - |\bar{K^{0}}\rangle \right) \\ |K_{S}\rangle \sim\dfrac{1}{\sqrt2}\left(|K^{0}\rangle + |\bar{K^{0}}\rangle \right).$$

The $|K^{0}\rangle , |\bar{K^{0}}\rangle$ states are production states, that is they have well-defined strangeness, but they do not propagate or decay. The $|K^{0}\rangle\sim \bar{s} d$ is the isopartner of $K^+$, with S=1; while the $|\bar{K^{0}}\rangle\sim \bar{d}s$ is the isopartner of $K^-$, with S=-1.

They differ in strangeness by two, so only a doubly weak interaction mixes them, the celebrated box diagram. When this is taken into account, the free eigenstates are $K_L, K_S$ with very different lifetimes, so a beam of neutral kaons will quickly purify to a beam of surviving $K_L$, which are longer lived. In vacuum, no conversion to $K_S$s.

As this beam enters a nuclear target, the two "strongly distinct" states (since strangeness is preserved in the strong interactions) are treated very very differently. The $K^0$ scatters its quarks by the exchange of gluons, pions, whatever, and exchanges quarks with the nucleons and back: elastic & charge-exchange scattering. You may write 137000 Feynman diagrams. Coherently, and then goes out, slightly diminished.

The $\bar{K^{0}}$ is different. It may do all of the above, but it also has additional channels: Its s quark may replace a nucleon quark to form a hyperon, Λ or Σ, so it is absorbed and gone. Whatever the hyperon does (decay), it does not return to the original state, so the absorption crudely degrades the $\bar{K^{0}}$ beam more, and increases strangeness, if you must think of it that way. (People would frown, with justice).

The crucial thing is that, since the two components of the $K_L$ wave function were absorbed and phase-shifted differently, many times (coherent scattering), the outgoing wave function is not $\dfrac{1}{\sqrt2}\left( |K^{0}\rangle - |\bar{K^{0}}\rangle \right)$ anymore, but a slightly distorted version thereof. When resolved to free propagation eigenstates, the distortion amounts to a component of $K_S$ now present: $K_S$ has been regenerated after strong interaction with matter.

The standard textbook by D. Perkins should answer any questions you might have.

• Is the 137000 number related to the fine structure constant? :p May 10, 2018 at 17:20
• Indeed; the secret handshake of physicists underscoring good-faith communication channels. May 10, 2018 at 18:26