**Table of Contents:**

- General Description
- Orthogonality of Modes

**GENERAL DESCRIPTION**

As mentioned in the last section, the phase velocity of light in an anisotropic medium depends on the state of polarization as well as the direction of propagation within a medium. This is because of how the light interacts with the optic axis, which is an axis in the crystal around which there is symmetry in the medium. Actually, it is more a direction in the medium versus an actual axis, but you will see the “optic axis” direction in other literature labeled as a line. Apologies on behalf of all the people in optics who came before me, as they had questionable intent when they made that so confusing! Anyhow, you may imagine then that if there is symmetric along the optic axis, then light travelling in the direction of the optic axis i.e. parallel to it would not be disturbed. This is in fact correct! The interesting things, and the reason for this whole section, happen when light travels perpendicularly or obliquely (at an arbitrary angle) to the optic axis. To begin our analysis, then, let’s start with the assumption of monochromatic $E$ and $H$ plane waves propagating in the anisotropic media with equations given as:

\begin{equation} \vec{E} = Ee^{i(\omega{t}-k\cdot{r})} \end{equation}

\begin{equation} \vec{H} = He^{i(\omega{t}-k\cdot{r})} \end{equation}

where we can define $k$ as the wave vector $k = (\omega/c)ns$, with “$s$” being a unit vector in the direction of propagation. Then, substituting these equations into Faraday’s Law and Ampere’s law, we can get:

\begin{equation} k \times E = \omega\mu{H} \end{equation}

\begin{equation} k \times H = -\omega\epsilon{E} \end{equation}

We can now choose to eliminate $H$ (as opposed to E), such that we are left with the scalar wave equation:

\begin{equation} k \times (k \times E) + \omega^2\mu\epsilon{E} = 0 \label{wave_k}\end{equation}

which we can see is indeed in the form of a wave equation because of the double cross product and the velocity coefficient of the second term. Note that we generalized the first term in the above equation to a double spatial derivative in our original derivation of the wave equation for free space in “Classical Optics” since there we were primarily only concerned with propagation in one direction; here, unfortunately, we cannot make that same approximation. Now, since we will always be rotating to a space such that we can have a principle coordinate system, we will have a diagonal dielectric tensor given as:

\begin{equation} \epsilon = \begin{pmatrix} \epsilon_x & 0 & 0 \\ 0 & \epsilon_y & 0 \\ 0 & 0 & \epsilon_x \end{pmatrix} \end{equation}

Thus, when we do the cross-multiplication math for the $k \times (k \times E)$ term, we can rewrite Equation wave_k in matrix form as:

\begin{equation} \begin{pmatrix} \omega^2\mu\epsilon_x – k_y^2 – k_z^2 & k_xk_y & k_xk_z \\ k_yk_x & \omega^2\mu\epsilon_y – k_x^2 – k_z^2 & k_yk_z \\ k_zk_x & k_zk_y & \omega^2\mu\epsilon_z – k_x^2 – k_y^2 \end{pmatrix} \begin{pmatrix} E_x \\ E_y \\ E_z \end{pmatrix} = 0 \label{new_wave}\end{equation}

Note that this is exactly equivalent to Equation \eqref{wave_k}, just written as the product of matrices! As some of you may know from linear algebra, in order for the solution to a matrix to not yield trivial solutions (i.e. $k_x = k_y = k_z = 0$), the determinant of the matrix has to equal zero. We won’t go into the reasons for that here, but most introductory linear algebra texts will suffice to explains this; or, for you lazy people, there is a short description here. Anywho, the reason we care about this is because we can then find a relationship between $\omega$ and $\vec{k}$ that is solvable! The determinant is given as:

\begin{equation} det\begin{bmatrix} \omega^2\mu\epsilon_x – k_y^2 – k_z^2 & k_xk_y & k_xk_z \\ k_yk_x & \omega^2\mu\epsilon_y – k_x^2 – k_z^2 & k_yk_z \\ k_zk_x & k_zk_y & \omega^2\mu\epsilon_z – k_x^2 – k_y^2 \end{bmatrix} = 0 \end{equation}

I’m not going to write out this proof because I am also lazy BUT if you don’t believe me, hopefully you believe Matlab! Here is a snippet of the determinant of Equation \eqref{new_wave} solved:

Now note here that the answer is in the form of a sphere. Once plotted, this is called the sphere of wave normals! This is shown below:

As we can see from the figure above, there are two directions labeled “Optic axis”, where there is only one intersection. What this means is that a light propagating in that direction will only see one refractive index and thus any wave propagating along this will not accrue differently polarized elements. For any other direction of propagation, though, we can see that a k-vector would intercept two different points in the inner and outer surfaces. These points correspond to the allowed magnitudes of the k-vectors for that propagation direction, which move with a velocity $\omega/k$.

Now that we have gotten this result, we can also see that we can solve Equation \eqref{new_wave} for the electric field $E_x$, $E_y$, and $E_z$ directions. The result of this is given as:

\begin{equation} \begin{pmatrix} \frac{k_x}{k^2-\omega^2\mu\epsilon_x} \\ \frac{k_y}{k^2-\omega^2\mu\epsilon_y} \\ \frac{k_z}{k^2-\omega^2\mu\epsilon_z} \end{pmatrix} \end{equation}

https://www.microscopyu.com/techniques/polarized-light/principles-of-birefringence