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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$\begingroup$ 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. $\endgroup$– KenshinJan 29, 2013 at 2:58
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1$\begingroup$ There is not really anything like the "same" photon. $\endgroup$– GunnishJan 29, 2013 at 11:54
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3 Answers
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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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1$\begingroup$ 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. $\endgroup$ Jan 29, 2013 at 3:31
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3$\begingroup$ Is this your personal opinion or could you give a reference? $\endgroup$ Jan 29, 2013 at 4:24
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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.
answered Jan 29, 2013 at 16:14
dmckee --- ex-moderator kittendmckee --- ex-moderator kitten
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A more detailed explanation of the interaction with QED The photon is absorbed by the electron, the electron is excited, the electron decays, a new photon is emitted. There is some finite, non-zero time between the absorption and emission of the photons. The process is illustrated in the following Feynman diagram from here:
answered Jan 29, 2013 at 3:30
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$\begingroup$ Yes this is what I thought. That absorption and re-emission would occur with a time delay. So what really happens in the scenario given by the asker? Does the photon reflect, or is it absorbed and re-emitted? $\endgroup$– KenshinJan 29, 2013 at 3:50
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$\begingroup$ The photon is absorbed and re-emitted $\endgroup$ Jan 29, 2013 at 4:37
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$\begingroup$ How do you know this is the case. Do you have any references to experimental evidence confirming the time delay that we would expect from the absorption/re-emission case? $\endgroup$– KenshinJan 29, 2013 at 4:45
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7$\begingroup$ You may be taking the Feynman diagrams too seriously. Recall that they stand in for terms in a perturbation series, and look at the labeling of the external line on those two diagrams: though they are drawn confusingly, one corresponds to the out-going photon being emitted before the in-coming photon is absorbed. The electron is not "excited" and it does not "decay", it is "off-shell" on the internal line (experimental evidence of excited lepton states would win the Nobel prize in a heartbeat). $\endgroup$ Jan 29, 2013 at 16:10