Does Compton effect take place because at energies higher than required for photoelectric effect, the electron would be destabilized after absorbing a photon of such a large amount of energy and therefore dissipates some reproducing another photon and using some kinetic energy itself to be scattered?
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Both the electron and the photon are elementary particles, and the interactions of elementary particles need quantum mechanics. Feynman diagrams are an iconal representation of the calculations necessary to get measurable predictions for the interactions of elementary particles. In this case Compton scattering:
Elementary particles interact at a point, they are point particles. Here there are two diagrams contributing to lowest order, a real photon hitting a real electron as input, a real photon and a real electron as output. In between there exists a virtual electron, within an integral. It is called "virtual" because it is off mass shell.
There is no "destabilization", unless one means the off mass shell propagating "electron", with the quantum numbers of the electron but not its mass. Calculating the diagrams will give the probability distributions for the incoming photon to transmit energy to the exiting electron and by momentum and energy conservation the outgoing photon will have a lower energy. ( There exists inverse Compton effect , where a low energy photon gains energy but it is a state relevant to astrophysics)
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$\begingroup$ Just to be sure in the left diagram if we flip the outgoing electron and photon is still the same process right? $\endgroup$– DabedJun 8, 2021 at 18:21
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$\begingroup$ @Dabed yes, the functions entering the calculation are the same for a given vertex set up $\endgroup$– anna vJun 8, 2021 at 18:50
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$\begingroup$ thanks, another doubt came to me now you wrote that "In between there exists a virtual electron, within an integral" where could I look an example of that? (so if I'm getting it correctly in the left diagram we can flip the electron and photon because in any vertex both are going in/out but in the right diagram we can't flip the electron and photon because in any vertex one is going in while the other is going out) $\endgroup$– DabedJun 8, 2021 at 19:54
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1$\begingroup$ @Dabed yes, it is a mater of convention, in the first diagram, but it is changing the interaction in the second, because of the assumed time axis. see arxiv.org/pdf/1602.04182.pdf page 17 $\endgroup$– anna vJun 9, 2021 at 4:43