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Is there a frequency eigenstate for light?

  1. Dec 7, 2013 #1
    I thought it was the coherent state, but since that is an eigenstate of the annihilation operator, and the annihilation operator is not hermitian, then it has no corresponding observable, and I'm assuming that one can observe frequency.

    Thanks.
     
  2. jcsd
  3. Dec 7, 2013 #2

    Simon Bridge

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    Considering that ##E=h\nu##, isn't that the same as asking if there is an energy eigenstate.
     
  4. Dec 8, 2013 #3
    Well, inside a cube of light, the total energy depends on both the number of photons and the energy (frequency) of each photon. Since the number of photons is by itself an operator, I thought there would have to be a separate one for frequency.

    On the other hand, if frequency is a parameter instead of an operator, then maybe not.

    I was really wondering about the mutual compatibility of certain kinds of knowledge of light. Is it possible, for example, to know that I have exactly, say, three photons inside of a cube, and to also know the exact frequency of each one?

    (And if so, would this simply mean that I would have no knowledge of their phase as well? I do remember reading in an old Dirac paper that energy and phase were conjugate variables. And what about polarity? Is that simply another parameter and therefore not a problem? And so on . . . )
     
  5. Dec 8, 2013 #4

    tom.stoer

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    I think this is what the Fock space provides. There is a Hilbert space vector

    ##|n_1,n_2,\ldots\rangle##

    where we can in principle know all occupation numbers for all states. So for each state we know the frequency and the energy (as parameters determined by the size of the box) and we know the occupation number.
     
  6. Dec 8, 2013 #5
    Doesn't the uncertainty principle demand that if the spacetime dimensions of the box are finite, then it is impossible to know exactly the energy/momentum content inside, and thus both the number of photons and the energy/momentum/frequency of each one simultaneously?
     
  7. Dec 8, 2013 #6

    tom.stoer

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    Why do you think so?

    Suppose you have a state

    ##|0,\ldots,0,1,0,\ldots\rangle##

    with "1" at the i-the position. Then the occupation number is

    ##N_i = 1##

    the frequency is

    ##\omega = \omega_i##

    and the energy is

    ##E = \hbar\omega_i##
     
  8. Dec 8, 2013 #7
    I'm thinking about the wave/particle dual nature of light. To have light energy contained in a finite region of spacetime, does it not have to be like a pulse, that is, a mix of many wavelengths instead of just one?
     
  9. Dec 8, 2013 #8

    tom.stoer

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    You are right, for a wave packet instead of a plane wave energy and frequency are subject to uncertainty.
     
  10. Dec 8, 2013 #9
    So then, for a plane wave with an exactly defined wavelength, if we take a box of finite dimensions inside of it, can it still contain an exactly known number of photons? Does measuring the photon number change this wavelength certainty?
     
  11. Dec 8, 2013 #10

    tom.stoer

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    Theoretically if the box is perfectly mirrored it allowes for exact plane waves and exactly defined frequencies
     
  12. Dec 8, 2013 #11
    Oh I didn't mean a closed box with reflecting sides, I just meant, "inside this particular cube of spacetime," and in particular I was imagining a (non physical) cube with edges much smaller than the wavelength of the light itself.
     
  13. Dec 8, 2013 #12
    Actually this reminds me, I asked a question similar to this one almost three years ago, and it was there that I first learned about coherent light.

    https://www.physicsforums.com/showthread.php?t=470413

    But if light with a well-defined wavelength is coherent, and coherent light is light with an indefinite photon number, then . . well, you can see why I would think that wavelength and photon number are non-commuting observables.
     
  14. Dec 8, 2013 #13

    Simon Bridge

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    You want to look at the relationship between the wave and photon descriptions of light?
    The number of photons is proportional to the intensity of the light.

    But I think you are mixing models here.
     
  15. Dec 9, 2013 #14

    Cthugha

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    You are mixing up different meanings of coherence here. First-order coherence is what you measure in a double slit or Michelson interferometer and is related to how well defined the phase of your field is. This is indeed related to the spectral purity of your light field. The power spectral density and the first-order field coherence form a Fourier transform pair. Or simply speaking: A spectrally narrow field will have a long coherence time and a well defined phase over long times and a spectrally broad field will have a short coherence time and the phase will randomize over short time scales. All of these quantities are still classical and on the field level. On this level coherence is just a matter of long or short coherence times, but there is no completely incoherent light.

    Now, second-order coherence is the meaning of coherence in the sense of quantum optics and the coherence found in lasers. Here, a coherent state means that a state is an eigenstate of the photon annihilation operator and it also means that joint detection rates factorize. Basically, that means that you do not get any additional information by detecting a photon. For such states the photon number uncertainty is small, but present (it is the uncertainty given by a Poissonian distribution). However, this is not related to wavelength. The uncertainty in the width of the spectral distribution and the uncertainty in photon number are two different and not necessarily related sources for uncertainty of the total energy. For example you could have spectrally broad single photon states (no photon number uncertainty, large spectral uncertainty) or spectrally filtered thermal light (huge photon number uncertainty, but small spectral uncertainty).

    The pairs of noncommuting quantities you get are time-energy and photon number-phase. Both are "weak" uncertainty relations because there is neither a strict time, nor a strict phase operator for light.
     
  16. Dec 9, 2013 #15
    Thanks Cthugha, much to chew on here.
     
  17. Dec 10, 2013 #16
    Just a thought, I don't know if it means anything: The product of time and energy, as well as that of position and momentum, have the dimensions of Planck's constant, while neither photon number nor phase have dimensions at all.
     
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