We’ve written the electric field of light waves both as vectors and as scalars, but in reality the electric field is always a vector field; when we write it as a scalar, it should be understood that the vector direction of the field is constant. The direction that the vector electric field points is the direction of polarization, not to be confused with the direction of propagation – for plane waves, those two directions are perpendicular to one another! For the phasor electromagnetic plane wave

$latex \displaystyle \vec{E} = \hat{a}_x A e^{i k z}$

the direction of polarization is $latex \hat{a}_x$ and the direction of propagation is $latex \hat{a}_z$. This plane wave is linearly polarized – if the plane wave is viewed head-on, the electric field vector would appear to be oscillating up and down in a straight line along the $latex x$-axis (if it could be seen somehow).

Linear polarization is not the only possibility. Another polarization type is circular or elliptical polarization, in which there are two linear polarization components which are $latex 90^o$ out of phase with each other. An example of a circularly polarized wave is

$latex \displaystyle \vec{E} = (\hat{a}_x + i \hat{a}_y) A e^{i k z}$

This is a phasor field. To get a better idea of how it behaves, let’s bring it back to the time domain:

$latex \displaystyle \vec{E}(z,t) = Re[ (\hat{a}_x + i \hat{a}_y) A e^{i k z} e^{-i \omega t} ] $

Assuming $latex A$ is real, this becomes

$latex \displaystyle \vec{E}(z,t) = \hat{a}_x A \cos{(kz – \omega t)} -\hat{a}_y A \sin{(kz – \omega t)} $

Let’s consider this wave at location $latex z = 0$. How does the polarization evolve as time advances? At time $latex t = 0$, the electric field points entirely in the $latex \hat{a}_x$ direction. If the period of oscillation is $latex T$, at time $latex t = T/4$ the electric field points entirely in the $latex \hat{a}_y$ direction. At time $latex t = T / 2$, the electric field points entirely in the $latex -\hat{a}_x$ direction. As you can see, the electric field direction is rotating as time increases. In fact, if we put the thumb of our right hand in the direction of propagation ($latex z$), our fingers curl in the direction that the polarization rotates. Thus, this is called ‘right hand circular polarization’. To achieve left hand circular polarization, we only need to change the sign on the $latex \hat{a}_y$ component of the electric field.

Wikipedia has some fantastic visualizations of circularly polarized electromagnetic waves, I strongly encourage you to check them out: https://en.wikipedia.org/wiki/Circular_polarization

In the circularly polarized wave written above, the amplitudes of the two vector components are equal. In practice, of course, they may be unequal, in which case the rotating polarization describes an ellipse, and the wave is said to be elliptically polarized.

The key to a circularly or elliptically polarized wave is the electric field having two vector components that are $latex 90^o$ **out of phase** with each other. An electric field with two vector components in-phase with each other is still linearly polarized, but the total polarization vector is at an angle with respect to the axes.

How is linearly or circularly polarized light produced? Laser light is often linearly polarized by design. We can force light to have linear polarization by passing it through a polarizer. A simple linear polarizer may consist of a set of parallel thin metallic wires laid upon a dielectric platform. The wires will strongly reflect waves whose polarization aligns with the wires, as these waves will induce currents along the wires. Waves whose polarization is perpendicular to the wires will be transmitted much more strongly (albeit not perfectly). The intensity transmission for a typical linear polarizer is given by

$latex \displaystyle I_{trans} = I_0 \cos^2{\theta} $

where $latex \theta$ is the angle between the transmission axis of the polarizer (the direction perpendicular to the wires) and the polarization direction of the incoming wave. $latex I_0$ is the incident intensity.

Another implementation of a linear polarizer makes use of the Brewster angle. At the interface between two materials of differing refractive index, there is an angle of incidence for which one linear polarization (the transverse magnetic or TM polarization) has zero reflectivity. Shining a light beam on an interface at this angle will produce reflected light which is purely linearly polarized transverse electric (TE).

In order to produce circular or elliptical polarization from linearly polarized light, it’s necessary to ‘delay’ one component of the polarization, so that the two components are out of phase. Some materials are birefringent, which means that they exhibit different refractive indecies depending on the direction of the electric field polarization. So, for example, $latex \hat{a}_y$ polarized light might experience a refractive index $latex n_{slow}$, while $latex \hat{a}_x$ polarized light might experience a different refractive index $latex n_{fast}$. Since the phase velocity and the wavenumber $latex k$ depend on refractive index, a birefringent material can produce the necessary phase delay. As we’ve seen, the amount of phase picked up by a wave propagating distance $latex d$ is $latex k d$. Consider linearly polarized light with both $latex \hat{a}_x$ and $latex \hat{a}_y$ polarization components:

$latex \displaystyle \vec{E} = (\hat{a}_x + \hat{a}_y) A e^{i k z}$

Suppose this wave propagates a distance $latex d$ through the birefringent material we described above. The phase difference between the two polarization components will be

$latex \displaystyle \Delta \phi = (k_{slow} – k_{fast}) d = \frac{2 \pi d}{\lambda_0} (n_{slow} – n_{fast})$

So the wave will become

$latex \displaystyle \vec{E} = (\hat{a}_x + e^{i \Delta \phi} \hat{a}_y) A e^{i k_{fast} z}$

We can therefore produce circularly polarized light by choosing $latex d$ so that $latex \Delta \phi = \pi / 2$ . This is called a ‘quarter-wave plate’ as it delays one polarization component by a quarter wavelength.