path: root/content/know/concept/fabry-perot-cavity/index.pdc

---
title: "Fabry-Pérot cavity"
firstLetter: "F"
publishDate: 2021-09-18
categories:
- Physics
- Optics
- Laser theory

date: 2021-09-18T00:42:59+02:00
draft: false
markup: pandoc
---

# Fabry-Pérot cavity

In its simplest form, a **Fabry-Pérot cavity**
is a region of light-transmitting medium surrounded by two mirrors,
which may transmit some of the incoming light.
Such a setup can be used as e.g. an interferometer or a laser cavity.

Below, we calculate its quasinormal modes in 1D.
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:

<a href="cavity.png">
<img src="cavity.png" style="width:70%">
</a>


## 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 $\dv*{E_m}{x}$
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 want non-trivial solutions, where we
cannot simply satisfy the system by setting $A_1$, $A_2$, $A_3$ and
$A_4$; this constraint will give us an equation for $k_m$. Therefore, we
demand that the system matrix is singular, 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:

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

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.
This is because the mirrors are infinitely thin "magic" planes;
had we instead used the full mirror structure,
then we would have demanded continuity, as you maybe expected.



## 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.