# Compton Scattering

Compton Scattering essentially states that when a photon of a given wavelength hits an electron the energy level of the electron changes and the photon has its wavelength changed. This seems to be implying that it is the same photon that is reflected outwards. Do we know it is the same photon of a changed wavelength, or is it possible that the original photon merely pushed a 2nd photon out of the electron? (This would mean the original photon is not changing wavelength, that is meerly absorbed).

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I don't think the electron "absorbs" the full photon. I'm not certain, but I think the excess energy is "reflected" from the photon rather than "absorbed" and "re-emitted" by the electron. –  Mew Jan 29 '13 at 2:58
There is not really anything like the "same" photon. –  Gunnish Jan 29 '13 at 11:54

It is my understanding that we cannot tell which of the two scenarios takes place, moreover, it does not matter, as photons having the same characteristics (such as momentum, energy, and polarization) are indistinguishable.

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This is not correct. There is a finite amount of time between absorption and re-emission of the photon---allowing the two possibilities to be distinguished. See my answer. –  zhermes Jan 29 '13 at 3:31
Is this your personal opinion or could you give a reference? –  akhmeteli Jan 29 '13 at 4:24

It is conventional to describe the out-going photon as a "different" particle. Basically because photons have only one independent property: their wavenumber ($\vec{k}$) and that is how we label them. Further as they experience no time there is not opportunity for them to change it.

This is consistent with quantum field theory where a process like this will be written with a destruction operator on the in-coming wavenumber and a creation operator on the out-going wavenumber.

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