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Compton scattering: electron absorbs then emits a photon?

  1. Jul 29, 2007 #1

    In Compton scattering, does the electron absorbs a photon and then emit another photon with another energy??

    I couldn't understand how would the electron absorb a FRACTION of the photon's energy which is forbidden in QM.

  2. jcsd
  3. Jul 29, 2007 #2
    Like everywhere else in quantum mechanics, Compton scattering can be described in terms of a wavefunction evolving in time. In a fair approximation, this can be a two-particle wavefunction in this case. Individual energies of the electron and the photon are usually fixed in the remote past. However, these energies are no longer definite at the time of collision and in the remote future (when scattering cross-sections are measured). They are described by the wavefunction which spreads over a wide range of particle energies. So, your description "the electron absorbs a photon and then emits another photon with another energy" is too simplistic.

  4. Jul 29, 2007 #3
    The derivation of wavelength shift in Compton scattering uses only the entrance and exit energies and momenta. It does not require a detailed understanding of the interaction. But the Feynman diagrams show an absorption and a re-emission.
  5. Jul 29, 2007 #4
    This is true, because the S-matrix formalism of QFT (which uses Feynman diagrams) is a simplified description of reality. In this formalism we care only about entrance and exit states, and don't ask about what happens in the middle. This simplification is a good match for scattering experiments in high energy physics. However, we shouldn't forget that between entrance and exit states the system undergoes some non-trivial time evolution. This time evolution is not accessible by modern experimental techniques, but it may be accessible in the future.

  6. Jul 30, 2007 #5
    Aren't you cracking eggs with sledgehammers here? I haven't treated compton scattering using the wavefunction yet but to my knowledge what happens is an electron in an atom absorbs a photon and rises a few energy levels, and then releases a photon of a different energy by moving down one or more energy levels but not to the ground state (which it eventually reaches by emitting more photons).

    Is this wrong?
  7. Jul 30, 2007 #6


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    Staff: Mentor

    The electron is ejected from the atom, and the photon is scattered (or re-emitted) with a lower energy (longer wavelength). Other electrons then fall into lower energy levels and they emit photons. The ejected electron is usually absorbed by another atom where it happens to stop after slowing down.
  8. Jul 30, 2007 #7
    As far as I know, the traditional definition of Compton scattering involves two free particles - the photon and the electron. It is not about the photon interacting with an electron bound in an atom. The latter interaction should be treated by very different methods, of course.

  9. Jul 30, 2007 #8


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    traditionally, yeah. but a lot of people call many types of electron-photon scattering "compton." For example, some people I know who do NRIX (non-resonant inelastic x-ray scattering) call the scattering off of valence electrons "valence compton." Etc.

    Anyways, to get back to what the original post was regarding. The kinematics are pretty easy, but it's the dynamics that are interesting: In a non-relativistic description of the electron there are actually two types of interaction. An interaction whose diagram looks like the standard QED vertex (but is different) where two electron lines and one photon line meet, and an interaction whose diagram looks like two electron lines and two photon lines all meeting at the same place. I believe that it is the latter diagram which gives rise to free electron compton. See, for example:

    Eisenberger and Platzman, "compton scattering of x-rays from bound electrons", Phys. Rev. A, vol. 2, p. 415.
  10. Jul 30, 2007 #9


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    The original experimental observations were based on bound electrons in carbon atoms. The binding energies of those electrons is in the eV range as opposed to keV energies of photons. Therefore one could treat the electrons as essentially free. Compton was apparently looking at X-ray diffraction among other things.

    Of course, if an X-ray or gamma-ray scatters off a K or L electron in a much heavier element, then the binding energy is significant.

    See also - http://nobelprize.org/nobel_prizes/physics/laureates/1927/index.html

    or http://nobelprize.org/nobel_prizes/physics/laureates/1927/compton-lecture.pdf (the discussion of the Compton effect begins on page 10 of the pdf file)
    Last edited: Jul 30, 2007
  11. Jul 30, 2007 #10
    You probably wanted to say: as opposed to keV energies of photons.

    Last edited: Jul 30, 2007
  12. Jul 30, 2007 #11


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    Staff: Mentor

    Corrected it. Thanks! :redface:
  13. Jul 31, 2007 #12
    You could also see things in this (simplistic) way: because of Doppler shift, the frequency of the scattered photon is reduced because it starts from a moving frame (the pushed electron).
    A photon's energy infact depends on the ref. frame, that is: the relative speed of source and observer.
    Last edited: Jul 31, 2007
  14. Mar 7, 2011 #13
    Well actually, all of the photon is absorbed by the electron. Some of the energy from the photon is used to remove the electron from the atom and its kinetic energy. The rest is released by the electron. Since E = hc/wavelength, lower energy in the released photon changes its wavelength.
  15. Mar 8, 2011 #14
    Then what [tex]
    \Delta \lambda
    [/tex] stands for in the Compton's formula?
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