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I recently came across this Feynman diagram: enter image description here

For a more simplistic diagram, I suppose even this would be adequate: enter image description here

As you can see in these diagrams, they radiate these virtual photons. The virtual photons have no charge or mass, and there is no apparent visible change in the energy of the electron after it radiates this photon. I'm aware that the probability level of the incident occurring drops by 1% for every extra vertex in the diagram, so low as the probability can get, it can never touch 0. What stops the electron from radiating multiple photons that can turn into other particles? What change occurs in the electron itself after the emission of the photon?

In the case of quarks and gluons, as seen in the diagram below, the quarks are losing the color charge of the gluon. Therefore, some change must occur in the electron after it radiates the photon. enter image description here

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    $\begingroup$ "The virtual photons have no charge or mass" - so-called virtual photons are off mass shell, i.e., virtual photons are not massless: As a consequence, a real photon is massless and thus has only two polarization states, whereas a virtual one, being effectively massive, has three polarization states. $\endgroup$ Dec 30 '20 at 4:31
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    $\begingroup$ there is clearly no visible change in the energy of the electron after it radiates this photon How is that clear from the diagram? $\endgroup$
    – G. Smith
    Dec 30 '20 at 4:39
  • $\begingroup$ the quarks are losing the color charge of the gluon. Therefore, some change must occur in the electron after it radiates the photon The electric charge of an electron does not change when it radiates a photon. $\endgroup$
    – G. Smith
    Dec 30 '20 at 4:41
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    $\begingroup$ @AlfredCentauri That's interesting, Thanks for the information! $\endgroup$ Dec 30 '20 at 4:47
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    $\begingroup$ My question was to do with what change occurs. The electron’s energy, momentum, and, if I remember correctly, spin state can change. $\endgroup$
    – G. Smith
    Dec 30 '20 at 4:49
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My understanding is that this is due to momentum conservation.

A gamma ray cannot create an e-/e+ pair without interacting with a nucleus in order to conserve momentum. This is why the virtual gamma ray here can only create a virtual (off-shell) e-/e+ pair.

The change that occurs in the electron after the photon emission is just a change in energy/momentum.

Additionally, your comment about an electron creating multiple gammas - the only vertex allowed is two fermions with a photon, so emitting multiple gammas must be done in the fashion shown in the first diagram (in multiple steps)

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