The Feynman diagram for the decay $\phi \to K^+K^-$ usually depicts two gluons. (This can be seen e.g. on Wikipedia).

feynman diagram

Why do we need two gluons, instead of just one?

  • $\begingroup$ I've removed some comments that answered the question, and replies to them. Please post answers as answers, and use comments to refine or improve the posts they are attached to. $\endgroup$ – rob May 17 '20 at 4:45

When one draws Feynman diagrams, one has to use correctly all the conservation laws. In the decay diagrams you show they are there: energy, momentum, strangeness, charge, they are all conserved by the input and output particles.

In this imagined by you case


the up- antiup pair appear out of nothing, i.e. there is no energy and momentum conservation in their creation. It will become an allowed diagram if the gluon exchanges a gluon with the pair, one of the large number of exchanges allowed by the strong interactions ( as discussed in the comments that disappeared).

If you are thinking of the figure in your question, (which is missing the antiparticle signs) to eliminate one of the gluons and take (for example) the anti-up to match with the s quark , it is again energy/momentum but in terms of probabilities, the unti-up would have to have the correct energy momentum vector to match the mass of the $K^-$ with the s quark, very improbable.

  • $\begingroup$ Thank you. So if I understand correctly, the Feynman diagram in my question where one gluon is removed, is allowed by all symmetries, however due to energy-momentum conservation very improbable? $\endgroup$ – ersbygre1 May 17 '20 at 5:14
  • $\begingroup$ Yes, think in the center of mass, for both the generated particles of the pair to have the energy momentum four vectors that would match the masses of the Ks .is very improbable in available phase space. $\endgroup$ – anna v May 17 '20 at 5:18
  • $\begingroup$ Great, thank you for the explanation. $\endgroup$ – ersbygre1 May 17 '20 at 5:30
  • $\begingroup$ Stephan I rejected your edit because it no longer was my answer, but the way you would have answered. You can write an answer to your question , and even accept it as the best, no problem. $\endgroup$ – anna v May 17 '20 at 6:47

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