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+---
+title: "Fabry-Pérot cavity"
+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 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 $\idv{E_m}{x}$
+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 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.