---
title: "Fabry-Pérot cavity"
sort_title: "Fabry-Perot cavity" # sic
date: 2021-09-18
categories:
- Physics
- Optics
- Laser theory
layout: "concept"
---

In its simplest form, a **Fabry-Pérot cavity**
is a region of light-transmitting medium surrounded by two parallel mirrors,
which may let some of the light escape.
Such a setup can be used as e.g. a laser cavity or an interferometer.
Below, we treat this simple system as an exercise
for calculating *quasinormal modes* in 1D,
i.e. modes with complex resonances.

We divide the $$x$$-axis into three domains: left $$L$$, center $$C$$, and right $$R$$.
The cavity $$C$$ has length $$\ell$$ and is centered on $$x = 0$$.
Let $$n_L$$, $$n_C$$ and $$n_R$$ be the respective domains' refractive indices:

{% include image.html file="sketch-full.png" width="70%"
    alt="Cavity structure" %}



## Microscopic cavity

In its simplest "microscopic" form, the reflection at the boundaries
is simply caused by the index differences there.
Consider this ansatz for the [electric field](/know/concept/electric-field/) $$E_m(x)$$,
where $$m$$ is the mode:

$$\begin{aligned}
    E_m(x)
    = \begin{cases}
        A_1 e^{- i k_m n_L x} & \mathrm{for}\; x < -\ell/2 \\
        A_2 e^{- i k_m n_C x} + A_3 e^{i k_m n_C x} & \mathrm{for}\; \!-\!\ell/2 < x < \ell/2 \\
        A_4 e^{i k_m n_R x} & \mathrm{for}\; x > \ell/2
    \end{cases}
\end{aligned}$$

The goal is to find the modes' wavenumbers $$k_m$$.
First, we demand that $$E_m$$ and its derivative $$\idv{E_m}{x}$$
are continuous at the boundaries $$x = \pm \ell/2$$:

$$\begin{aligned}
    A_1 e^{i k_m n_L \ell/2}
    &= A_2 e^{i k_m n_C \ell/2} + A_3 e^{- i k_m n_C \ell/2}
    \\
    A_4 e^{i k_m n_R \ell/2}
    &= A_2 e^{- i k_m n_C \ell/2} + A_3 e^{i k_m n_C \ell/2}
\end{aligned}$$

$$\begin{aligned}
    - i k_m n_L A_1 e^{i k_m n_L \ell/2}
    &= - i k_m n_C A_2 e^{i k_m n_C \ell/2} + i k_m n_C A_3 e^{- i k_m n_C \ell/2}
    \\
    i k_m n_R A_4 e^{i k_m n_R \ell/2}
    &= - i k_m n_C A_2 e^{- i k_m n_C \ell/2} + i k_m n_C A_3 e^{i k_m n_C \ell/2}
\end{aligned}$$

Rearranging the four equations above yields the following linear system:

$$\begin{aligned}
    0
    &= A_1 - A_2 e^{i k_m (n_C - n_L) \ell/2} - A_3 e^{- i k_m (n_C + n_L) \ell/2}
    \\
    0
    &= A_2 e^{- i k_m (n_C + n_R) \ell/2} + A_3 e^{i k_m (n_C - n_R) \ell/2} - A_4
    \\
    0
    &= n_L A_1 + n_C \big( A_3 e^{- i k_m (n_C + n_L) \ell/2} - A_2 e^{i k_m (n_C - n_L) \ell/2} \big)
    \\
    0
    &= n_C \big( A_3 e^{i k_m (n_C - n_R) \ell/2} - A_2 e^{- i k_m (n_C + n_R) \ell/2} \big) - n_R A_4
\end{aligned}$$

Which can be rewritten in matrix form as follows, with the system matrix on the left:

$$\begin{aligned}
    \begin{bmatrix}
        1 & -e^{i k_m (n_C - n_L) \ell/2} & -e^{- i k_m (n_C + n_L) \ell/2} & 0 \\
        0 & e^{- i k_m (n_C + n_R) \ell/2} & e^{i k_m (n_C - n_R) \ell/2} & -1 \\
        n_L & -n_C e^{i k_m (n_C - n_L) \ell/2} & n_C e^{- i k_m (n_C + n_L) \ell/2} & 0 \\
        0 & -n_C e^{- i k_m (n_C + n_R) \ell/2} & n_C e^{i k_m (n_C - n_R) \ell/2} & -n_R
    \end{bmatrix}
    \cdot
    \begin{bmatrix}
        A_1 \\ A_2 \\ A_3 \\ A_4
    \end{bmatrix}
    =
    \begin{bmatrix}
        0 \\ 0 \\ 0 \\ 0
    \end{bmatrix}
\end{aligned}$$

We do not want to satisfy this equation
by simply setting $$A_1$$, $$A_2$$, $$A_3$$ and $$A_4$$,
so we demand that the system matrix is not invertible,
i.e. its determinant is zero:

$$\begin{aligned}
    0 =
    &- n_C (n_L + n_R) \big( e^{i k_m (2 n_C - n_L - n_R) \ell/2} + e^{- i k_m (2 n_C + n_L + n_R) \ell/2} \big)
    \\
    &+ (n_C^2 + n_L n_R) \big( e^{i k_m (2 n_C - n_L - n_R) \ell/2} - e^{- i k_m (2 n_C + n_L + n_R) \ell/2} \big)
\end{aligned}$$

We multiply by $$e^{i k_m (n_L + n_R) \ell / 2}$$ and
decompose the exponentials into sines and cosines:

$$\begin{aligned}
    0
    = i 2 (n_C^2 + n_L n_R) \sin(k_m n_C \ell)
    - 2 n_C (n_L + n_R) \cos(k_m n_C \ell)
\end{aligned}$$

Finally, some further rearranging gives a convenient transcendental equation,
keeping in mind that $$k_m$$ and the indices $$n_L$$, $$n_C$$ and $$n_R$$
are generally complex numbers:

$$\begin{aligned}
    \boxed{
        0
        = \tan(k_m n_C \ell) + i \frac{n_C (n_L + n_R)}{n_C^2 + n_L n_R}
    }
\end{aligned}$$

Thanks to linearity, we can choose one of the amplitudes
$$A_1$$, $$A_2$$, $$A_3$$ or $$A_4$$ freely,
and then the others are determined by $$k_m$$ and the field's continuity.



## Macroscopic cavity

Next, consider a "macroscopic" Fabry-Pérot cavity
with complex mirror structures at boundaries, e.g. Bragg reflectors.
If the cavity is large enough, we can neglect the mirrors' thicknesses,
and just use their reflection coefficients $$r_L$$ and $$r_R$$.
We use the same ansatz:

$$\begin{aligned}
    E_m(x)
    =
    \begin{cases}
        A_1 e^{-i k_m n_L x} & \mathrm{for}\; x < -\ell/2 \\
        A_2 e^{-i k_m n_C x} + A_3 e^{i k_m n_C x} & \mathrm{for}\; \!-\!\ell/2 < x < \ell/2 \\
        A_4 e^{i k_m n_R x} & \mathrm{for}\; \ell/2 < x
    \end{cases}
\end{aligned}$$

On the left, $$A_3$$ is the reflection of $$A_2$$,
and on the right, $$A_2$$ is the reflection of $$A_3$$,
where the reflected amplitudes are determined
by the coefficients $$r_L$$ and $$r_R$$, respectively:

$$\begin{aligned}
    A_3 e^{- i k_m n_C \ell/2}
    &= r_L A_2 e^{i k_m n_C \ell/2}
    \\
    A_2 e^{-i k_m n_C \ell/2}
    &= r_R A_3 e^{i k_m n_C \ell/2}
\end{aligned}$$

These equations might seem to contradict each other.
We recast them into matrix form:

$$\begin{aligned}
    \begin{bmatrix}
        1 & - r_R e^{i k_m n_C \ell} \\
        - r_L e^{i k_m n_C \ell} & 1
    \end{bmatrix}
    \cdot
    \begin{bmatrix}
        A_2 \\ A_3
    \end{bmatrix}
    =
    \begin{bmatrix}
        0 \\ 0
    \end{bmatrix}
\end{aligned}$$

Again, we demand that the determinant is zero in order to get non-trivial solutions:

$$\begin{aligned}
    0
    &= 1 - r_L r_R e^{i 2 k_m n_C \ell}
\end{aligned}$$

Isolating this for $$k_m$$ yields the following modes,
where $$m$$ is an arbitrary integer:

$$\begin{aligned}
    \boxed{
        k_m
        = - \frac{\ln(r_L r_R) + i 2 \pi m}{i 2 n_C \ell}
    }
\end{aligned}$$

These $$k_m$$ satisfy the matrix equation above.
Thanks to linearity, we can choose one of $$A_2$$ or $$A_3$$,
and then the other is determined by the corresponding reflection equation.

Finally, we look at the light transmitted through the mirrors,
according to $$1 \!-\! r_L$$ and $$1 \!-\! r_R$$:

$$\begin{aligned}
    A_1 e^{i k_m n_L \ell/2}
    &= (1 - r_L) A_2 e^{i k_m n_C \ell/2}
    \\
    A_4 e^{i k_m n_R \ell/2}
    &= (1 - r_R) A_3 e^{i k_m n_C \ell/2}
\end{aligned}$$

We simply isolate for $$A_1$$ and $$A_4$$ respectively,
yielding the following amplitudes:

$$\begin{aligned}
    A_1
    &= (1 - r_L) A_2 e^{i k_m (n_C - n_L) \ell/2}
    \\
    A_4
    &= (1 - r_R) A_3 e^{i k_m (n_C - n_R) \ell/2}
\end{aligned}$$

Note that we have not demanded continuity of the electric field,
because the mirrors are infinitely thin "magic" planes in this case.
If we had instead included the full microscopic mirror structure,
then we would have demanded continuity as before.



## References
1.  P.T. Kristensen, K. Herrmann, F. Intravaia, K. Busch,
    [Modeling electromagnetic resonators using quasinormal modes](https://doi.org/10.1364/AOP.377940),
    2020, Optical Society of America.