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'Mathematical' frequencies and photons

  1. Jan 27, 2007 #1
    It is known that any non-sinusoidal wave can be regarded as a combination of sinusoidal waves of different frecuencies, with the ‘weight’ of the different frecuencies given by the function called Fourier transform. On the other hand, if we have an electromagnetic wave, we know that it is the combined result of many photons of different energies, and it is generally assumed that the interval of the Fourier space where Fourier transform is not zero gives us the energies of the photons in the beam, through the famous relation E=h·f where f is the frecuency and h the Plank constant.

    But if we have a short pulse of radio waves rising and falling abruptly, although the main frecuency is in the radio region, the Fourier transform may have also very high frecuencies, physically impossible to assign as photon energies.

    Or think in a continuous (cuasi)monocromatic wave source, a detector, and a shutter between them, which allow us to switch the wave on and off. For the source, the radiation have only one frecuency, but after the shutter, the fourier transform of the wave has also other frecuencies. However photons should have the same energy!

    So ¿in which cases or under which conditions can we relate the ‘matematical’ frecuencies given by fourier analysis with photons energies?
  2. jcsd
  3. Jan 28, 2007 #2


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    We can always relate the mathematical frequencies in a classical wave to the corresponding free-field photons.
    The case of the rectangular time window is an idealisation, and every physically realisable "shutter" will have a far smoother transition than a perfect heaviside function, and as such, seriously limit the bandwidth.
    Concerning your monochromatic source + shutter, clearly the shutter interacts with the monochromatic beam, and this interaction will be written down as creation and destruction operators, eliminating "monochromatic" photons, and creating others. So the photons after the shutter are not simply a "subsample" of photons before the shutter. Photons aren't bullets, they are states of excitation of the electromagnetic field.
  4. Jan 28, 2007 #3
    Let me know if I understand what you say,

    Suppose some effect that can happen in the detector only above some photon energy, like the production of an electron-hole pair. Supposse that the monochromatic wave before the shutter do not have enough frecuency(energy) to produce this effect. With a fast mechanical shutter we can get a wider bandwidth (it is not necessary a perfect rectangular time window). Do you mean that we could increase the probability of the effect only with the shutter?
    Last edited: Jan 28, 2007
  5. Jan 29, 2007 #4


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    Sure (in principle). This energy will come from the interaction of the EM field with the shutter.

    However, I should add that in practice, a mechanical shutter will give you such a small bandwidth increase, that for all practical purposes this will be entirely neglegible whenever quantum effects (such as pair creation) are to be considered.
    Last edited: Jan 29, 2007
  6. Jan 29, 2007 #5
    Are you sure it couldn't be observed in lab conditions?

    What is the order of magnitude of the maximum bandwidth that we can got technically (by any means) in microwaves or radiowaves?. Or more generally, in which region of the spectrum is it possible to achieve the higher quotient bandwidth/frecuency?

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