# Linear light polarization and photon spin

The following is from Wikipedia: "Circularly polarized electromagnetic waves are composed of photons with only one type of spin, either right- or left-hand. Linearly polarized waves consist of equal numbers of right and left hand spinning photons, with their phase synchronized so they superpose to give oscillation in a plane." Now my question: Can someone elaborate this explanation and explain the linear polarization considering photon spin. I mean by explains the photon spin only, as the explanation based on direction of electric field in the light wave is well known.

## 1 Answer

Firstly, spin in paraxial light beams manifests as helicity. (One can also show that gauge invariance reduces the internal degrees of freedom to just two, but let's not go there now.) So the circular polarization states of light are the helicity eigenstates. This gives us a two-dimensional parameter space in terms of which all states of polarization of paraxial light can be expressed. However, this two dimensional space has the shape (topology) of a spherical surface, which is called the Poincare sphere.

Basically, what it comes down to is that one can represent and state of polarization with a little complex-valued unit vector form as a linear combination of the complex-valued unit vectors for the two circular states of polarization $$\hat{v} = \hat{L} \alpha + \hat{R} \beta ,$$ where $|\alpha|^2+|\beta|^2=1$ and $\hat{L}$ and $\hat{R}$ represent the unit vectors for left- and right circular polarization, respectively. If the light propagates in the $z$-direction. Then $$\hat{L},\hat{R} = \hat{x}\pm i\hat{y} .$$ (The assignment of the signs depends on convention.)

As a little exercise, you can proof to yourself that an appropriate superposition of these two vectors will give you linear polarization.