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Photon Amplitude

  1. Apr 9, 2013 #1
    Does it make sense to talk about the amplitude of a photon?

    In my mind, the amplitude of the photon is the maximum strength that the electric and magnetic field gets as they oscillate. If you were to change the amplitude of a photon (say increase), then the maximum strength of the e&m fields increase which translates to a photon of higher energy, thus changing the classification of the photon (say the suggested increase would move the photon's energy from UV to x-ray).

    Now, I understand that many of the things I've just said may not exactly be right (and please correct me if it is). My main concern is understanding how amplitudes of EM waves fit in (if at all).

    I'm ultimately asking about this because I had a physics teacher who told us that increasing the amplitude of visible light will result in light with appears to be brighter.

    I think this is fundamentally wrong because brightness of light is related to the amount of photons that are emitted, not the amplitude of the individual photons.

    What's right and what's wrong with all of this?
    Don't beat me up too much with what I've said, but please do correct me.

    Thanks a bunch.
     
  2. jcsd
  3. Apr 9, 2013 #2

    Cthugha

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    Ok, let us start.

    This is not correct. To change energy, you need to have a different wavelength. Amplitude is uncorrelated with photon energy.

    The square of the amplitude of a light field IS the intensity. For a single photon, the amplitude becomes a probability amplitude and the square will become the probability density to detect a photon somewhere. So yes, increasing the amplitude of the light field somewhere will increase brightness there.
     
  4. Apr 9, 2013 #3

    jtbell

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    Staff: Mentor

    And if you have a bazillion photons, which is the case for "normal" light sources, the amplitude of the light field is proportional to the average number of photons.
     
  5. Apr 9, 2013 #4
    So, does this mean that more photons given off from a source will "constructively interfere" to give a greater amplitude and thus a greater brightness? I used the quotes because i'm not sure that's the correct idea here.

    Thanks for the responses so far.
     
  6. Jan 1, 2014 #5
    Does that also mean that all photons have the same amplitude? Also meaning only the photon energy is only associated with the wavelength?
     
  7. Jan 1, 2014 #6
    The AMOUNT of photons relates to the intensity. The amplitude gets normalized so that the probability of finding it somewhere is 1. So the amplitude itself means nothing but its phase and distribution does.
     
  8. Jan 1, 2014 #7

    vanhees71

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    Sigh :-(. I don't know, why everybody uses bad ideas on photons instead of good old classical Maxwell theory to introduce people to classical optics.

    The first thing is that relativistic quantum theory, and particularly photons that are massless spin-1 particles which make them as relativistic and as complicated as anything can get when it comes to relativistic quanta, make only sense in the context of quantum field theory. There is no such thing as a photon's wave function or amplitude that makes a strict sense.

    In terms of quantum electrodynamics, the thing coming closest to a classical electromagnetic wave, which in the optical regime are nowadays nicely realized by lasers or by antenna for radio waves, are coherent states which are coherent superpositions of multi-photon states, i.e., it's an excitation of the quantized electromagnetic field that has not a definite photon number but a pretty definite phase. The intensity of such a coherent state, i.e., the expectation value of its energy density, is indeed proportional to the average photon number, which is a parameter that can be freely chosen to define a coherent state of the electromagnetic field.

    "Natural light" is reprsented by a mixture, i.e., incoherent additions of such classical waves.

    As I said, it's much better to first learn classical Maxwellian electromagnetics and its application to optics. You come very far with that theory already! Then you should learn non-relativistic quantum theory first and only then you can attack relativistic quantum theory, which only makes sense as a quantum field theory. "Wave functions" make only sense in an approximation, where the particles are not too relativistic, e.g., you can treat as a first approximation the relativistic hydrogen atom with a Dirac wave function for an electron moving in a Coulomb field, because there the average speed of the electron is much smaller than the speed of light. This is a clear approximation scheme that is derivable from QED (see Weinberg, Quantum Theory of Fields Vol. 1).
     
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