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Energy transfer during photon interaction

  1. Nov 8, 2012 #1
    Let's consider a mach-zehnder apparatus

    The picture is below:

    http://en.wikipedia.org/wiki/File:Mach-zender-interferometer.png


    1. When the photon encounters (and emerges from) a half-silvered mirror does it loose any energy? The photon does change phase when it moves through a half silvered mirror, however does it loose energy?

    2. Does a photon loose energy when its reflected of a full silvered mirror?

    3. Does the photon loose any energy if one of the paths is blocked? (and its still detected at detector 1 or 2)
     
  2. jcsd
  3. Nov 8, 2012 #2
    A photon off a silver mirror energy reduction is negligible in this setup. Energy loss results in longer wavelength light. Such processes are referred to as inelastic scattering, Raman scattering, or the overused term "non-linear process".

    The effect of the mirror partial reflection and transmission.
     
  4. Nov 8, 2012 #3
    Thanks Iforgot.

    now to the last question, i.e.:

    if one of the arms/paths of the mach-zehnder is blocked and the photon happens to NOT take that path.....

    is there any change in the energy/wave-length of the photon?
     
  5. Nov 9, 2012 #4
    nope.
     
  6. Nov 9, 2012 #5
    Be careful when you talk about the "phase of a photon" - if you mean the phase of the electrical field, this is not well-defined (the expectation value of the electrical field of a single photon is always zero). If you want a well-defined phase, you need a coherent mixture of states with different photon numbers (see Haroche's Nobel prize).
     
  7. Nov 9, 2012 #6
    is the energy reduction/loss less than a quanta?

    i don't fully understand the concept of quanta.

    Is it that:

    Quanta is the smallest unit of physical entity, however can increments less than a quanta be added?
     
    Last edited: Nov 9, 2012
  8. Nov 9, 2012 #7
    (Take my comments with a grain of salt, as I tend to focus primarily on experiments. )

    The conventional understanding is that the energy reduction or addition cannot be in increments less than a quanta.

    I find it convenient to delude myself by just accepting the wave interpretation, but allowing for discrete wave amplitudes.
     
  9. Nov 11, 2012 #8
    is the energy loss (thought to be) in multiples of "the quanta"?
     
  10. Nov 11, 2012 #9

    Vanadium 50

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    You keep asking that, and you keep being told there is no energy loss.
     
  11. Nov 12, 2012 #10
    San K. I think I see what your problem is. You know momentum is being transferred when a photon reflects off a surface, e.g. 'a sail'. From the imparted 'momentum' to the "sail", one can calculate the 'energy' transferred from the photon to the "sail ". Resulting in lower photon energy. You're asking if there is a limit to how small this energy reduction can be?

    I want to say, 'no', but I'm not sure.

    However, this phenomena is exploited for a pretty clever technique called "doppler radar".
     
  12. Nov 12, 2012 #11
    you are correct Iforgot. Thanks my friend.

    I think QM would say that the energy reduction cannot be smaller than a quantum (whatever a quantum means in this context/system)

    on a separate, but similar, note - for example would the (difference in energy of) photons from the say EM/vibgyor spectrum then be in multiples of quanta?
     
    Last edited: Nov 12, 2012
  13. Nov 12, 2012 #12
    EM/vibgyor spectrum: If your system was in an infinite well, yes, they would be quantized. As the size of the well increases, the difference between energy levels would get smaller and smaller.

    The word quantized is a word we (including me) are guilty of throwing around too casually. Unfortunately, the real answer to your questions lies in tediously slogging through the Dirac-Schrodinger equation and other QED equations I have no clue about.

    How quantization arises from these equations eventually becomes evident. In the process of solving these equations, one gets a feel for when concepts of quantization come into play.
     
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