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Wavefunction vs EM wave of a Photon

  1. Feb 4, 2012 #1
    In a single photon at a time double slit experiment. Is it the wave function or electromagnetic wave of a photon that is interfering? If both, what is the contribution of each? Remember that the electromagnetic wave is not the wave function of the photon.

    In a single photon, it has wave function, electromagnetic wave, and quantum field.

    How large is the electromagnetic wave of a single photon and how large is the quantum field? And what is the difference between it? Can anyone point to a site with an illustration or something?

    I want to be conversant with a photon properties first because in QFT, they are said to make a leap of faith that since photon and electron and other particle are fundamental particles. They should be the same, hence Second Quantization is invoked for electron where just like the photons they are field quanta of their respective quantum field (electron quantum field for example). This is due to the electromagnetic field successfully applied with canonical quantization producing the field quanta or photons. So they applied it to all particles even if they don't have any electromagnetic field. The success of QED gave them the confidence to go on in their leap of faith.

    In a single electron at a time double slit experiment, We can't assume only the wave function is present. There must be a corresponding quantum field associated with it. In a photon, it has properties of magnetic field and electric field producing electromagnetic field. Maybe this is why its quantum field (which is simply the electromagnetic field) can be detected? I heard there is an equivalent in the electron field. So what would it take to measure the electron field (or matter field)?
  2. jcsd
  3. Feb 4, 2012 #2


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    There is an excellent set of slides by Arnold Neumaier, who often contributes to PF, at the link below. He discusses how the concept of a wavefunction whose square is equal to the probability of finding the photon, is simply not applicable to a photon. This means that the photon as a particle is inherently nonlocal. In QFT, the photon is typically described by specifying its vector potential A(x,t).

  4. Feb 4, 2012 #3
    I understand you're really interested in this stuff, but I'm getting the impression
    you're trying to go too fast. Quantum physics, specially the relativistic theory, is
    pretty complex, you really need to take your time with it. Otherwise all you'll
    achieve is making yourself more confused.

    Some of the assumptions you're making above are not correct. In particular, there
    is no electromagnetic wave associated to a single photon. The typical electromagnetic
    waves (actually, modes) one studies in classical physics correspond in the quantum
    theory to states with no definite number of photons.

    The electron field itself isn't a hermitian operator, hence why it's not an observable
    (this won't make sense unless you've studied quantum mechanics beyond the
    layman's level). However, you can construct observable (hermitian) operators by
    taking suitable products of the field operators. These are typically fields themselves,
    and are bona fide observables as well.
  5. Feb 4, 2012 #4
    Interesting. So these are kinda like duality. It's like a graviton can't be put in the differential manifold of GR. Likewise, a photon can't be put in the classical electromagnetic wave. So thinking in purely photon way. How is the changing electric field and changing magnetic field encoded in the photon or what is their equivalent in the photon?

    I have ideas of what hermitian and operators are. Can you give example what results if you construct "observable (hermitian) operators by taking suitable products of the field operators", what exactly would come out? Is it position or spin?
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