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I was reading about Delta+ baryon decay modes and found a EM decay mode

Delta+ (uud) -> Proton (uud) + photon

However when I attempt at making Feynman diagram of this, I got this

enter image description here

And a similar one with photon out of d quark.

My doubt is that, viewing from the quark COM reference frame, a photon emission violates conservation of momentum. How is that possible?

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  • $\begingroup$ replace the on-shell photon with an off-shell gluon. $\endgroup$ – JEB Mar 11 at 22:38
  • $\begingroup$ What prevents the affected quark from being virtual before or after the photon emission? $\endgroup$ – Cosmas Zachos Mar 11 at 22:43
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The quark radiating the photon is an internal line in the diagram, i.e. virtual. This means that one should look for energy conservation in the total process where the four vectors are real, i.e. on mass shell. The mass difference of the Delta to the proton leaves enough energy for the gamma to be a real four vector.

See even more complicated Feynman diagrams for delta decay.

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Well momentum is conserved because a delta+ baryon has a 3/2 spin as the resulting photon(+1 spin) and proton(+1/2 spin) add up to 3/2 spin. That photon caries the extra energy from the delta+ baryon that did not become part of the proton.

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