I am unsure how to draw a feynman diagram for a reaction that occurs as follows

$$ K^0 --> l^+\nu_l\pi^- $$

Any tips would be helpful.

  • $\begingroup$ Do you know which quarks are inside kaons and pions? $\endgroup$ – G. Smith Mar 26 at 23:55
  • $\begingroup$ The long-lived neutral kaon is called the KL ("K-long"), decays primarily into three pions, and has a mean lifetime of 5.18×10−8 s. The short-lived neutral kaon is called the KS ("K-short"), decays primarily into two pions, and has a mean lifetime 8.958×10−11 $\endgroup$ – Rick Mar 27 at 0:09
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    $\begingroup$ Hint: the strange antiquark will turn to an up antiquark by the emission of a W +. $\endgroup$ – Cosmas Zachos Mar 27 at 13:33
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    $\begingroup$ @Rick Isn't your comment a deeply irrelevant red herring that confuses the OP? $\endgroup$ – Cosmas Zachos Mar 27 at 13:35
  • $\begingroup$ No it was a way to prompt more thought... there are a lot of Feynman Diagrams of Kaon decays online. hep.ph.ic.ac.uk/~dauncey/will/lecture16.pdf $\endgroup$ – Rick Mar 27 at 17:29

Drawing the most appropriate Feynman diagram can be a little tricky sometimes. I find it best to work backwards with the general rule of thumb being to try to minimize the number of vertices. We know that the lepton pair $l^{+}, \nu_{l}$ must come from a $W^{+}$ boson as that's the only mediator that conserves charge and can violate flavor. Since we are dealing with the $K^{0}$ meson whose quark configuration is $d \bar{s}$ and a daughter meson $\pi^{-}$ whose configuration is $d \bar{u}$, we notice that both the $K^{0}$ and $\pi^{-}$ contain a $d$-quark. This $d$-quark will act as a spectator quark (will remain unchanged in the scattering process). Now all that's left is recognizing that $W$ bosons connect the positively charged quarks to the negative and vice verse with the exception of not connecting particles to antiparticles. So for an example a $q^{+} \longrightarrow q^{-} + W^{+}$ where the positive superscript is merely to denote a positive charge and the negative denotes negative charge.

Hopefully that's sufficient information to guide you to the proper Feynman Diagram.


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