# Helmholtz Equation & Modes

The Helmholtz equation is not unique to to its use in optics; it is a time-independent form of the wave equation, so it can arise in any situation that can be described by the wave equation. It is likely more common that people with any familiarity at all in physics have heard more talk of the time-independent Schrodinger equation. It turns out the Helmholtz equation is a direct analog of the time-independent Schrodinger! Read more in the “Quantum Mechanics” portion of this website if this is interesting to you.

DERIVATION

Starting with the wave equation,

$$\nabla^2E = \mu\epsilon\frac{\partial^2E}{\partial{t}^2}$$

where we have assumed that “E” is simply the amplitude of one of the vector components of $\vec{E}$ i.e. it is a scalar amplitude of x, y, or z. One method to solve the equation that is commonly used in quantum mechanics as well (look at the Ansatz equation and spherical harmonics) is to propose a solution with separable components, i.e. $E = f(r)g(t)$. We will try our hand at that here too! Assuming a spatially dependent electric field amplitude and a time harmonic phase,

$$E(r,t) = E(r)e^{i\omega{t}}$$

we can then substitute this into the wave equation and use our previous definition of the wave vector $k^2 = \omega^2\mu\epsilon$ to find:

$$\nabla^2E(r)e^{i\omega{t}} = -\omega^2\mu\epsilon E(r)e^{i\omega{t}}$$

or

$$(\nabla^2 + k^2)E(r) = 0 \label{helmholtz}$$

where Equation \eqref{helmholtz} is the Helmholtz equation. What this tells us is that a purely harmonic wave, meaning one that oscillates at a single frequency $\omega$, yields a spatial differential equation for $E(r)$ when substituted into the wave equation. If there was a finite bandwidth, then the wave would not oscillate purely harmonically and thus would not be described by the Helmholtz. Note that our previous method of solving the wave equation with plane waves could still be done; the “$r$” dependence would simply be carried in the $E(r)$ term. In fact, if you just hold your horses for one moment, we will be proving this in two sections.

MODAL SOLUTION FORM

If we rearrange the Helmholtz equation, we can obtain the more familiar eigenvalue problem form:

$$\nabla^2E(r) = -k^2E(r)$$

where the Laplacian $\nabla^2$ is an operator and $-k^2$ is a constant, or eigenvalue of the equation. This means that whenever the operator acts on a mode (eigenvector) of the equation, it yield the same mode (eigenvector) multiplied by a constant (the corresponding eigenvalue). Really, all this is, is a refresher for those of you who are familiar with quantum mechanics and a sneak peek of those of you who haven’t yet taken a quantum mechanics course, because all you do in a quantum mechanics course is find solutions to eigenvalue problems :-).

FREE SPACE

In the free space assumption, the most useful form of solutions are plane waves (see, you just have to trust me that they’re useful despite their lost touch with reality because we use them everywhere!). In the purely spatially dependent form, this is given as:

$$E_k(r) = \hat{\epsilon}_ke^{-i\hat{k}\cdot{r}}E_0$$

This is trivial to prove that it is an eigenmode of the Helmholtz because we have already done this in our solution to the free space wave equation. Again, although the plane wave doesn’t describe real waves because the plane wave given here would oscillate in all space with the same phase and frequency, this provides a useful basis for representing real waves. Here, I am using the word “basis” as a mathematical term, because real waves with complicated spatial dependencies can be represented in terms of an expansion of the corresponding basis plane waves.

BOX MODES

An interesting case to consider is the propagation of electromagnetic waves in a box, i.e. free space propagation with boundaries. If we consider the box to be metal, then we would have that the tangential component of the electric field must vanish at the boundaries due to the boundary conditions of metals/dielectrics. We will prove this boundary condition in Section 3, but for now we shall simply use the result of applying this boundary condition with the Helmholtz equation to find that $E(r) = E_x\hat{x} + E_y\hat{y} + E_z\hat{z}$ satisfies the Helmholtz with constituents given as:

$$E_x = E_{x,0}cos(\frac{m_x\pi{x}}{L})sin(\frac{m_y\pi{y}}{L})sin(\frac{m_z\pi{z}}{L})$$

$$E_y = E_{y,0}sin(\frac{m_x\pi{x}}{L})cos(\frac{m_y\pi{y}}{L})sin(\frac{m_z\pi{z}}{L})$$

$$E_z = E_{z,0}sin(\frac{m_x\pi{x}}{L})sin(\frac{m_y\pi{y}}{L})cos(\frac{m_z\pi{z}}{L})$$

where these come directly from the boundary conditions.

NORMALIZATION AND ORTHOGONALITY

These next two conditions for the modes of the Helmholtz come directly from the properties of an orthonormal set of basis vectors, which is what we want in order to construct the solution for any arbitrary spatial dependence.

Normalization:

For any choice of spatially dependent modes $f(r)$, they must satisfy:

$$\int^{L/2}_{-L/2}|f(r)|^2d^3r = 1$$

where here I arbitrarily chose the bounds of the integration for the box modes case, but the bounds would be dictated by the physical boundaries of your problem.

Orthogonality:

Consider the inner product of two plane wave box modes:

$$\int^{L/2}_{-L/2}f_k^*(r)f_{k’}(r)d^3r = \frac{1}{L^3}\int^{L/2}_{-L/2}e^{-i(k’-k)\cdot{r}}d^3r$$

We can see here that if $k \neq k’$, then the integral is zero. If they are equal then the integral is equal to one. Thus, we may write:

$$\frac{1}{L^3}\int^{L/2}_{-L/2}e^{-i(k’-k)\cdot{r}}d^3r = \delta(k-k’)$$

MODE EXPANSIONS

Given the above good news that plane waves form an orthonormal set, we can expand an arbitrary field in terms of plane waves as:

$$E(r,t) = \hat{\epsilon}E_0e^{i\omega{t}}\sum_kU_k(r)a_k$$

where “$\epsilon$” is the polarization unit vector, “$E_0$” is the field amplitude, and “$a_k$” is the mode amplitude. If we assume that “$k$” is in Cartesian coordinates, then the form of the expansion is given as:

$$E(r,t) = \hat{\epsilon}E_0e^{i\omega{t}}(\sum_{m_x}a_{mx}e^{-ik_xx})(\sum_{m_y}a_{my}e^{-ik_yy})(\sum_{m_z}a_{mz}e^{-ik_zz})\frac{1}{L^3} \label{plane_exp}$$

Now here’s where we can find another important aspect of dealing with optics: polarization! By applying Gauss’s Law in a vacuum, $\nabla \cdot E = 0$, to Equation \eqref{plane_exp}, we would see that this implies $k \cdot E = 0$. Since the directional aspect of $E$ is contained in its polarization direction $\hat{\epsilon}$, the equation $k \cdot E = 0$ means that $k \cdot \hat{\epsilon} = 0$. Thus, this gives us the result that orthogonal to the $k$-vector are two independent directions for the electric field. As we will see in later sections and in particular under the “Light Propagation in Crystals” tab, this result has meant the creation of many optical components that take advantage of these orthogonal polarization directions including fibers!

DENSITY OF STATES