$$K^0 \rightarrow \mu^+ + \mu^-$$ Just like a neutral kaon decays into a pair of pions, can it also decay into a pair of muons? If not, why?

  • $\begingroup$ How will quarks decay purely into leptons in the Standard Model? $\endgroup$ – SRS Jan 2 '20 at 5:58
  • $\begingroup$ @SRS But baryon number is conserved since neutral kaon has baryon number zero. But I do get the logic that quarks cannot decay purely into leptons. I am confused, hence the question. $\endgroup$ – Dirac93 Jan 2 '20 at 7:23
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    $\begingroup$ Kaons can decay to leptons. The $K^+$ can decay to $\mu^+ + \nu_\mu$. So the answer to this question seems far from obvious to me. The downvotes seem unwarranted. $\endgroup$ – John Rennie Jan 2 '20 at 12:44
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    $\begingroup$ This paper suggests the decay to a muon pair is possible. I won't post this as an answer since it's outside my area of expertise. $\endgroup$ – John Rennie Jan 2 '20 at 12:46
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    $\begingroup$ Note: the Particle Data Group lists nonzero branching ratios for $K^0_L\to \mu\mu\gamma$ and $K^0_L\to\mu\mu\gamma\gamma$, and lists $K^0_S\to\mu\mu$ as a $CP$-violating, strangeness-changing decay mode with a branching ratio below $10^{-9}$. $\endgroup$ – rob Jan 3 '20 at 15:13

Yes, this decay is possible, and measured to occur, via diagrams similar to:

enter image description here

involving two W bosons. Note that since it involves one more W boson than the $\rm K^0$ to pion decays, the decay is suppressed pretty strongly.

There are some subtleties involved here regarding CP violation, depending on whether the original particle is a $\rm K^0_L$ or $\rm K^0_S$. But overall this decay can happen.


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