Polarization of Monochromatic Plane Waves

Table of Contents:

  1. General Elliptic Properties
  2. Linear and Circular Polarization


First, the most important thing to know when we are talking about this business of polarization is specifically what we are describing is the path traced by the tip of the electric field vector. Now that we know that, let’s consider how we have arrived at this definition. First, hopefully by now (if you started with the “Classical Optics” section, as you perhaps should) you have picked up that the propagation of light is, by convention, primarily described by its electric field (it would be completely analogous to use the magnetic field; we are just creatures of habit here in the optics community). If we describe the propagation of the light in the $z$-direction, then the electric field would propagate in the $xy$-plane. We know this because $k$, $E$, and $B$, form an orthogonal triad in free space. It turns out the $x$ and $y$ components of the electric field vector can oscillate independently at a given frequency. Thus, to consider the propagation of this field in space, we must consider the vector addition of these two orthogonal components, which would give a vector that lies between them. As the electric field oscillates in space, this will begin to trace out a shape. To describe this, let’s use math!

To continue with our wave propagating in the $z$ direction, we have:

\begin{equation} E(z,t) = Re[Ae^{i(\omega{t}-kz)}] \end{equation}

where $A$ is our complex vector in the $xy$ plane. We will effectively be analyzing this vector when talking about polarization, with the inclusion of the frequency and direction of travel to give the electric field. Now, what uniquely determines the vector in the $xy$ plane is both its amplitude and phase, right? That’s all we’ve ever needed to describe the electric field, really (we’re a simple bunch that don’t ask for much, us optics nerds). So with that in mind, we can describe just the vector in the $xy$ plane as $A = \hat{x}A_xe^{i\delta_x} + \hat{y}A_ye^{i\delta_y}$. However, in order to make this a field, we have to add the direction that the field is propagating. Thus, we can describe the electric field in the $xy$-plane via the two separate equations:

\begin{equation} E_x = A_xcos(\omega{t}-kz + \delta_x) \end{equation}

\begin{equation} E_y = A_ycos(\omega{t}-kz + \delta_y) \end{equation}

Now that we have the electric field relation, though, we can eliminate the time $t$ and propagation direction dependence $z$ between these equations to obtain:

\begin{equation} (\frac{E_x}{A_x})^2 + (\frac{E_y}{A_y})^2-2\frac{cos\delta}{A_xA_y}E_xE_y = sin^2\delta \label{conic}\end{equation}

where the phase difference between the two directions of the electric field are encapsulated in $\delta$ as:

\begin{equation} \delta = \delta_y – \delta_x \end{equation}


We can look at the Equation \eqref{conic} to see that if the difference between the $x$ and $y$ phases satisfies:

\begin{equation} \delta = \delta_y – \delta_x = m\pi \end{equation}

then the Equation \eqref{conic} describes the path of a straight line. We call this linear polarization!

The other special case occurs when we have the condition:

\begin{equation} \delta = \delta_y – \delta_x = \pm \frac{1}{2}\pi \end{equation}

and that the amplitudes in the $x$ and $y$ direction are equal, i.e.:

\begin{equation} A_y = A_x \end{equation}

Finally, we can define the ellipticity of a polarization ellipse as:

\begin{equation} e = \pm\frac{b}{a} \end{equation}

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