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(How) is a single photon polarized?

  1. Aug 13, 2007 #1
    What quantum mechanical property of the photon determines its polarization?
    Can a single photon be unpolarized?
    Can a single photon be linear polarized?
    Can a single photon be circular polarized?
    Can a single photon's polarization be changed?

    Can classical and quantum mechanical polarization of an electromagnetic wave/photon coexist?


    Richard
     
  2. jcsd
  3. Aug 13, 2007 #2
    Photons are bosons and have spin [tex]\pm[/tex] 1, corresponding to two different circular polarization modes. One photon has one of these two, and other modes of polarization are achieved by superposition of more photons.
     
  4. Aug 13, 2007 #3

    dextercioby

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    Actually the spin quantum number of a photon is 1 and +/- 1 is the helicity (helicity quantum numbers).
     
  5. Aug 13, 2007 #4
    Does the spin quantum number or the helicity quantum number of the photon correspond to the classical concept of electromagnetic wave polarization?
     
  6. Aug 16, 2007 #5
    The Wikipedia article on Photon Polarization says that the spin corresponds to classical polarization.
     
  7. Aug 17, 2007 #6
    Ok let me just writ it out it's simplest this way
    The free electromagnetic potential is given by:

    [tex]
    a^\mu (x) = (2\pi )^{ - 3/2} \int {\frac{{d^3 p}}
    {{\sqrt {2p^0 } }}} \sum\limits_\sigma {\left[ {e^{ip \cdot x} e^\mu ({\mathbf{p}},\sigma )a({\mathbf{p}},\sigma ) + e^{ - p \cdot x} e^{\mu *} ({\mathbf{p}},\sigma )a^\dag ({\mathbf{p}},\sigma )} \right]}
    [/tex]

    Now, [tex]
    e^\mu ({\mathbf{p}},\sigma )
    [/tex] that represents the direction of the field is called a polarization vector.
    [tex]
    \sigma
    [/tex] the helicity can be thought of as in which direction the polarization vector is rotating clockwise or counterclockwise.
     
  8. Aug 17, 2007 #7
    This very similar to the classical situation.

    Classically for a single monochromatic plane wave, if you pick your coordinates properly, the wave can be written in the general form:

    [tex]
    {\mathbf{A}} = B\cos (kz - \omega t){\mathbf{x}} \pm C\sin (kz - \omega t){\mathbf{y}}
    [/tex] where B and C are the polarizations of the wave, elliptical in general, and the [tex]
    \pm[/tex] symbol represents the helicities which just determine the direction of rotation of the polarized wave.
     
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