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Circularly Polarized Photons

  1. May 2, 2010 #1
    I've got three questions basically.

    1. Do photons actually have only an integer spin +/- 1, or do people really only mean the sign of its chirality? The reason I ask is that I am interested in whether the photon spin is related to its frequency. This leads to the next question.
    2. Is the basis for Left and Right circularly polarized photons the 'natural' basis for photons? Or is it the linear polarization basis? Photons have spin and so I am betting that the 'natural' basis is the circularly polarized one. But, if it turns out that in (question 1) they all have the same spin frequency, then it presents a problem - the variable frequencies needed to make circular polarization of different colors of light wouldn't match up then because all the different colors would need different frequencies of 'rotation'.
    3. This last one probably hinges on the first two. Is there a spacial dependence on the probability of finding a circularly polarized photon in a linear polarization state? In other words, will there be periodic points in space where it is more probable to measure a vertical polarizations versus a horizontal polarization - (the periodicity would be based on the frequency of rotation of the circularly polarized photon)?

    Ok, depending on what comes out to be true, the 3rd question might be an analogy for neutrino oscillations - ya know, how there is a periodic dependence on space in what you measure for the probability of a certain state.
  2. jcsd
  3. May 2, 2010 #2
    It is not related to the frequency.

    Any basis is as good as any other. Physicists use circular left and right handed, but it's just a convention.

    Any frequency mod can have it's own polarization and they don't need to match.

    There is no such periodicity.

    Even if it was - one measurement of polarization destroys original photon state, so any subsequent measurements in other points will not give you any more data about the photon.

    You can always express circular polarized photon as a combination of two linearly polarized ones. You can do it at any point and for any two orthogonal polarizations. You seem to think that the amplitudes of linear components change with time, but they don't. It's only their phase that changes, but it doesn't manifest as change of probability.
  4. May 2, 2010 #3
    I am not talking about subsequent measurements of the same photon, but measurements on identical photons emitted coherently from the same source but then measured at different distances.

    So coherent circularly polarized photons don't have changing probabilities of linear polarizations? I find that hard to believe but only from an intuitive standpoint. I am having trouble finding state vectors defined in a nice way to demonstrate these intricacies so maybe you could write down what they are like. I'll guess at what they are below and maybe you can tell me where I am wrong, then show the correct ones.

    What is the state vector for a vertically polarized photon?
    For a horizontally polarized photon?
    For an arbitrary linear polarized photon?
    [tex]\Psi_{\theta}=|\theta\rangle=\sin{\theta}|v\rangle + \cos{\theta}|h\rangle[/tex]

    For a LHCP photon?
    [tex]\Psi_{L}=|L\rangle=\sin{\omega t}|v\rangle + \cos{\omega t}|h\rangle[/tex]
    A RHCP photon?
    [tex]\Psi_{R}=|R\rangle=\sin{(-\omega t)}|v\rangle + \cos{(-\omega t)}|h\rangle[/tex]
    An elliptically polarized photon?
    [tex]\Psi_{Ell}=|Ell\rangle=\alpha\sin{\omega t}|v\rangle + \beta\cos{\omega t}|h\rangle[/tex] where [tex]|\alpha|^{2}+|\beta|^{2}=1[/tex]

    Etc. These are merely guesses at what they'd be because I haven't managed to find a source for this info yet.
  5. May 2, 2010 #4
    No no. It's not like that. The weights should be complex, not real.
    It's more like:
    [tex]|L\rangle=(\exp{(i \omega t)}|v\rangle + \exp{(i \omega t + \pi / 2)}|h\rangle) / \sqrt{2}[/tex]
    Each weight is a complex number that changes with time, but its module remains constant. (Square of module is always [tex]\sqrt{2} / 2[/tex].)

    I'm not sure if I'm exactly correct, but it should go like that.
  6. May 2, 2010 #5
    Ok, but even in that setup there seems to be a spacial dependence to me. Since they move at the speed of light, and [tex]\frac{\omega}{k}=c[/tex], I think there will be a spacial dependence (as long as that state is correct).
  7. May 2, 2010 #6
    Amplitudes of the linear components don't change. Only their phase changes. So at any point you will have the same probability of detecting linearly polarized photon ([tex]\sqrt{2}/2[/tex] of original photon).
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