Feynman diagram for semileptonic decay of neutral kaon

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.

• Do you know which quarks are inside kaons and pions? – G. Smith Mar 26 at 23:55
• 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 – Rick Mar 27 at 0:09
• Hint: the strange antiquark will turn to an up antiquark by the emission of a W +. – Cosmas Zachos Mar 27 at 13:33
• @Rick Isn't your comment a deeply irrelevant red herring that confuses the OP? – Cosmas Zachos Mar 27 at 13:35
• 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 – 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.