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We all have studied in introduction to quantum mechanics about Compton Effect. In all the books I have read, it says that classical theory can't explain the shift in wavelength because the incident EM wave will oscillate the electron at the frequency of light, and the oscillating electron will emit radiation of the same frequency. This is all well and good, but while reading Quantum Physics[Berkeley Series],the author mentioned something among the lines that ( not exact sentence) , the em waves oscillate electrons and that in turn produces the light of same frequency, also some loosely bound electrons are ejected from the atom which radiate the light of the slightly different frequency.

Is this explanation correct, if so what was the problem in mathematically describing this theory? Also , if this is the case, how is the ejected electron different from the ones in photoelectric effect.

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Compton scattering, unlike the photoelectric effect, can occur for a free electron. For a free electron, the classical theory can't explain the shift in wavelength. A theory of Compton scattering has to explain all observations, not just some of them, so it needs to explain the case where the electron is free.

Of course there is no such thing as a perfectly free electron, since there is no such thing as an exactly vanishing electric field. However, the electrons in ordinary matter are an excellent approximation to free electrons, because the atomic binding energies, on the order of eV, are negligible compared to the MeV energies of the photons.

There is also a problem with classical explanations of the photoelectric effect and Compton scattering, because they can't explain the entanglement between electrons in different atoms. Without this entanglement, each atom's probability of being ionized is independent, and then conservation of energy and momentum hold only at the statistical level. This lack of correlation was a prediction of the BKS theory, which was disproved by the Nobel prize-winning 1925 Bothe-Geiger experiment.

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  • $\begingroup$ Do you know of any experiment, where the Compton formula was observed as valid for actual free electron, like an electron in a vacuum tube emitted from a cathode? The electrons in a piece of graphite Compton used are, as you say, correlated with each other (since they interact with each other), so it is not clear that a single electron or a rarified cloud of electrons that interact only weakly would scatter the radiation the same way. $\endgroup$
    – Ján Lalinský
    Commented Oct 9, 2018 at 10:07
  • $\begingroup$ @JánLalinský: I'm not sure how you would define an "actual" free electron. If the energy scale for the electron's interaction with the ambient electric field is $E$, then to me, "free" means $E\ll1\ \text{MeV}$, which is already true for electrons in atoms. If you want even smaller values of $E$, then I don't know, but maybe this is observed in astrophysical contexts. $\endgroup$
    – user4552
    Commented Oct 11, 2018 at 15:33
  • $\begingroup$ What I meant is a situation where the electron interacts with the radiation and any interactions with other electrons are not relevant (to spectrum of radiation near the primary frequency). What I mean is more like a much lower electron density, say >1000x less electrons per unit volume than in solid graphite. So something like very rare gas of electrons in a vacuum tube. The interaction between the electrons should be much lower then. $\endgroup$
    – Ján Lalinský
    Commented Oct 11, 2018 at 17:15
  • $\begingroup$ The reason I am asking is that when primary wave interacts with a piece of graphite, the secondary radiation from the vast number of electrons there easily matches in intensity the primary radiation. This should not happen for a situation which I would call "free electron" interacting with the primary radiation. $\endgroup$
    – Ján Lalinský
    Commented Oct 11, 2018 at 17:43

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