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I learnt that Compton scattering means an incoming photon hits an electron at rest and electron goes in a deviated angle $\phi$ as well as a "giving off" a photon at an angle $\theta$. I suppose that all photons have the same properties, so is the resultant photon the same photon as the original one (like reflection but with an increase in wavelength)? Or does the electron absorb the photon and then release a new photon?

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  • $\begingroup$ Perhaps that’s a matter of semantics? $\endgroup$
    – Ben51
    Feb 9, 2021 at 2:47

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It is not only semantic, but more a question of physical point view.

In the usual (relativistic) derivation of the Compton shift in wavelenght it is depicted as a colision betwen two "particles", so the answer would be "the same photon".

However, there is no mandatory need of photons to interpret the Compton effect. If you quantize the matter (electron) and not the EM field (photon), the "collision" is a regular Bragg reflexion or diffraction of the EM wave on the "gratting" formed by the electronic wave fonction (superposition of incoming and outgoing electron). In this point of view, the frequency of the EM wave relains unchanged in the frame where this matter standing wave is stationnary, and the frequency shift in the lab frame simply result from (relativistic) Doppler effect. The answer would be "no photon at all".

Furthermore, in the frame of full quantized theory, you could try to draw the Feynmann diagram associated to this effect, which (at lowest order) would imply a vertex with two photon lines. Such a vertex is allowed in the frame of Coulomb gauge quanization, but not in the, thought physicaly equivalent, (at least at moderate energy) Lorentz gauge quantization, which will describe it as a two step process : absorbtion plus emission. Hence you will answer "the same photon" in the former point of view, and "an other photon" in the latter.

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