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If a photon's emission is detected is it real or virtual?

  1. Feb 26, 2014 #1
    I understand that one can measure a single photon being absorbed using a photomultiplier tube or CCD.

    Can one measure a single photon being emitted by monitoring the current through an LED or the recoil of an emitting ion?

    Is it therefore possible to detect the same photon both being emitted and later being absorbed?

    After the photon's emission has been detected, but before its absorption has been registered, is it real or virtual?

    I assume a QED calculation would proceed on the assumption that the photon is virtual but if its emission has been measured then surely it should be treated as real?

    Is there a paradox here?
     
  2. jcsd
  3. Feb 26, 2014 #2

    Cthugha

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    In principle you can. In practice you pretty much never do it. While there is some recoil in the photon emission process, the momentum transferred is tiny. You will only be able to measure it, if it sends the emitter into a quantum state that does not have much overlap with the initial state. Otherwise uncertainty "wins" over the momentum transfer. So for incredibly lightweight emitters or other especially prepared emitters, you might be able to witness the emission process. For pretty much any typical light source you encounter in real life, you will not be able to do that.

    That also has some influence on the light field you get. If you can know the amount of momentum transfer, you will also know exactly into which direction the photon will be going. Otherwise, you just get a superposition state and the photon could be going in any direction. You will only find out at the time you detect the photon.
     
  4. Feb 26, 2014 #3

    f95toli

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    This can -at least in principle- certainly be done in cavity-QED type experiments. It is also -again in principle- possible when working with some types of non-linear cavities where the number of photons in one mode can be detected by montoring another mode.

    By "in principle" I mean that this turns out to be tricky in real life, but this is AFAIK mainly due to technological problems (getting a good enough signal-to-noise ratio) and we can be pretty sure this will be possible in a few years (it might even be possible now in other types of experiments, I am not sure).
    This is very much a "hot topic" since making good single photon emitters is important for a number of applications and part of that involves being able to monitor/verify what is going on and make sure that only one photon has been emitted.
     
    Last edited: Feb 26, 2014
  5. Feb 26, 2014 #4

    jtbell

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    Why?
     
  6. Feb 26, 2014 #5
    Well if we assume that the photon emission and absorption are measured then the photon must be virtual as it can not have had a definite momentum and energy like a real photon i.e. it could not have been represented by a plane wave.
     
  7. Feb 26, 2014 #6

    ZapperZ

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    Plane wave? You are talking about single photons, here!

    The photon is real. That is why we have single-photo sources. Virtual photons do not "live" for very long times or very long distances. One also does not detect virtual photons with such detectors or else all our light detectors would be swamped with overwhelming level of noise!

    Zz.
     
  8. Feb 26, 2014 #7

    WannabeNewton

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    "Real" photons are not represented by plane waves. Rather the photon field is represented by a superposition of plane waves (Fourier modes) with creation and annihilation operator coefficients. "Real" photons are really mode excitations of the field i.e. the definite number (energy) eigenstates, but not all states of the field have to be definite energy states.

    As Zz stated, you cannot detect virtual photons. We draw internal lines on Feynman diagrams to represent propagators, we don't actually detect the particles labeling the internal lines.
     
    Last edited: Feb 26, 2014
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