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Photons and electrons

  1. Jan 23, 2006 #1


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    Assuming that process of electrons releasing photons isn't instaneous, what happens during the transition?

    I'm also curious why lasers work. Photons passing through high energy molecules cause the eletrons to release photons with the same phase and direction, and I'm wondering how this release happens relative to the orbital path of the electron at the during the release period. The photons direction is always the same, but what is the electron's path and speed during the period of release?
  2. jcsd
  3. Jan 24, 2006 #2
    I would say that, in some cases, the photons direction lies in a well defied plane, but this assertion of photon's direction assumes you have a well defined wave behavior which you call photon, which I guess it is not true.

    Best Regards

  4. Jan 24, 2006 #3
    Photon absorption and release from electrons would be interesting even if it were instantaneous...and it may be. How does a "particle" with mass, energy and momentum absorb or release a "particle" that has energy but no mass nor momentum?
    I guess string theory provides the latest hypothetical insights where it is conjectured all particles are vibrating strings. String theory proposes one string (an electron) either aborsbs another vibrating string or splits like an amobea into two strings with different vibrational characteristics, one beingthe photon. But why electrons only absorb and emit photons in quanta matching hypothetical orbit levels/energies is not fully understood.
    Having rules and math for interactions to describe what is observed doesn't necessarily mean the underlying detail is fully understood. In ten or twenty years there may well be another theory to explain it!!
  5. Jan 24, 2006 #4


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    In the specific case you mention, it is difficult to really work out what I'm going to sketch in detail - I even wonder if it can be done within the actual theoretical framework of QED, where we only have a working machinery for asymptotic states (the S-matrix).
    But we could imagine the following: the coupling between the atomic system (the electron in a potential, if it were) and the free photon state (the state of the unperturbed EM field) results in a unitary time evolution operator U(t) which acts upon the initial state |excited atom>|n photons> and results into something which takes on the form:
    U(t) |excited atom>|n photons> = a(t) |excited atom>|n photons> + b(t) |ground state atom>|n+1 photons>

    where a(t) is a falling amplitude with time and b(t) is a rising amplitude with time, such that after some time T, a(T) is essentially 0, and b(T) is essentially 1.

    So the quantum state transits smoothly from the |excited atom>|n photons> state at t=0 to the |ground state atom>|n+1 photons> at time T.
    But of course a measurement of this state will make us decide, probabilistically, between the first or the second state. The longer we wait, the more chance we have to perceive the photon emission. The probability to observe an emission is given by |b(t)|^2.
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