Drawing a Feynman Diagram for K^{*+}→K^0 + \pi^+

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SUMMARY

The discussion focuses on drawing a Feynman diagram for the reaction K^{*+} → K^0 + π^+. The K^{*+} meson consists of an u quark and an anti-s quark, while the K^0 meson comprises a d quark and an anti-s quark, and the π^+ meson is made up of an u quark and an anti-d quark. The participant shared a diagram created using a demo program and sought feedback, noting the representation of a virtual gluon that facilitates the quark-antiquark pair production. The participant expressed uncertainty about whether the process is a weak decay, emphasizing the presence of virtual gluons in the K^{*+} bound state.

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  • Familiarity with particle interactions, specifically gluon exchange
  • Basic concepts of quantum chromodynamics (QCD)
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broegger
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Hi

I have to draw a Feynman diagram for the following reaction:

K^{*+} \rightarrow K^0 + \pi^+

The K^{*+}-meson is composed of an u and an anti-s quark, the K^0-meson is composed of a d and an anti-s quark, and the \pi^+-meson is composed of an u and an anti-d quark. I have drawn something like the following:

(EDIT: I have attached a picture of my diagram drawn in some demo-program instead, since I couldn't do it properly in text mode. The curly line represents a gluon creating a quark-antiquark pair. I have no experience drawing these things, so any feedback is very welcome :smile:)
 

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Last edited:
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Here's something i did in "Paint". It looks very much like the diagrams you find in Griffiths' text on Elementary Particles.

I hope it's not a weak decay. I assumed the gluon was virtual (K^{*+} is a bound state, so there should be plenty of virtual gluons), so that's how two real quarks came out (one being the antiparticle of the other).

Daniel.

P.S. The arrow on the outgoing antiquark should be reversed and time "flies" from left to right.
 

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