# (How) is a single photon polarized?

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

Related Quantum Physics News on Phys.org
Photons are bosons and have spin $$\pm$$ 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.

dextercioby
Homework Helper
Actually the spin quantum number of a photon is 1 and +/- 1 is the helicity (helicity quantum numbers).

Does the spin quantum number or the helicity quantum number of the photon correspond to the classical concept of electromagnetic wave polarization?

Ok let me just writ it out it's simplest this way
The free electromagnetic potential is given by:

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

Now, $$e^\mu ({\mathbf{p}},\sigma )$$ that represents the direction of the field is called a polarization vector.
$$\sigma$$ the helicity can be thought of as in which direction the polarization vector is rotating clockwise or counterclockwise.

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:

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