(How) is a single photon polarized?

  • Thread starter Richard J
  • Start date
  • #1
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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
 

Answers and Replies

  • #2
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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.
 
  • #3
dextercioby
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Actually the spin quantum number of a photon is 1 and +/- 1 is the helicity (helicity quantum numbers).
 
  • #4
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Does the spin quantum number or the helicity quantum number of the photon correspond to the classical concept of electromagnetic wave polarization?
 
  • #6
192
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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.
 
  • #7
192
6
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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