Categories: Laser theory, Optics, Physics.

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 xx-axis into three domains: left LL, center CC, and right RR. The cavity CC has length \ell and is centered on x=0x = 0. Let nLn_L, nCn_C and nRn_R be the respective domains’ refractive indices:

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 Em(x)E_m(x), where mm is the mode:

Em(x)={A1eikmnLxfor  x</2A2eikmnCx+A3eikmnCxfor   ⁣ ⁣/2<x</2A4eikmnRxfor  x>/2\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 kmk_m. First, we demand that EmE_m and its derivative dEm/dx\idv{E_m}{x} are continuous at the boundaries x=±/2x = \pm \ell/2:

A1eikmnL/2=A2eikmnC/2+A3eikmnC/2A4eikmnR/2=A2eikmnC/2+A3eikmnC/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} ikmnLA1eikmnL/2=ikmnCA2eikmnC/2+ikmnCA3eikmnC/2ikmnRA4eikmnR/2=ikmnCA2eikmnC/2+ikmnCA3eikmnC/2\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:

0=A1A2eikm(nCnL)/2A3eikm(nC+nL)/20=A2eikm(nC+nR)/2+A3eikm(nCnR)/2A40=nLA1+nC(A3eikm(nC+nL)/2A2eikm(nCnL)/2)0=nC(A3eikm(nCnR)/2A2eikm(nC+nR)/2)nRA4\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:

[1eikm(nCnL)/2eikm(nC+nL)/200eikm(nC+nR)/2eikm(nCnR)/21nLnCeikm(nCnL)/2nCeikm(nC+nL)/200nCeikm(nC+nR)/2nCeikm(nCnR)/2nR][A1A2A3A4]=[0000]\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 simply satisfy this equation by setting A1A_1, A2A_2, A3A_3 and A4A_4, so we demand that the system matrix is not invertible, i.e. its determinant is zero:

0=nC(nL+nR)(eikm(2nCnLnR)/2+eikm(2nC+nL+nR)/2)+(nC2+nLnR)(eikm(2nCnLnR)/2eikm(2nC+nL+nR)/2)\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 eikm(nL+nR)/2e^{i k_m (n_L + n_R) \ell / 2} and decompose the exponentials into sines and cosines:

0=i2(nC2+nLnR)sin(kmnC)2nC(nL+nR)cos(kmnC)\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:

0=tan(kmnC)+inC(nL+nR)nC2+nLnR\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 A1A_1, A2A_2, A3A_3 or A4A_4 freely, and then the others are determined by kmk_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 rLr_L and rRr_R. We use the same ansatz:

Em(x)={A1eikmnLxfor  x</2A2eikmnCx+A3eikmnCxfor   ⁣ ⁣/2<x</2A4eikmnRxfor  /2<x\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, A3A_3 is the reflection of A2A_2, and on the right, A2A_2 is the reflection of A3A_3, where the reflected amplitudes are determined by the coefficients rLr_L and rRr_R, respectively:

A3eikmnC/2=rLA2eikmnC/2A2eikmnC/2=rRA3eikmnC/2\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:

[1rReikmnCrLeikmnC1][A2A3]=[00]\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:

0=1rLrRei2kmnC\begin{aligned} 0 &= 1 - r_L r_R e^{i 2 k_m n_C \ell} \end{aligned}

Isolating this for kmk_m yields the following modes, where mm is an arbitrary integer:

km=ln(rLrR)+i2πmi2nC\begin{aligned} \boxed{ k_m = - \frac{\ln(r_L r_R) + i 2 \pi m}{i 2 n_C \ell} } \end{aligned}

These kmk_m satisfy the matrix equation above. Thanks to linearity, we can choose one of A2A_2 or A3A_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 ⁣ ⁣rL1 \!-\! r_L and 1 ⁣ ⁣rR1 \!-\! r_R:

A1eikmnL/2=(1rL)A2eikmnC/2A4eikmnR/2=(1rR)A3eikmnC/2\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 A1A_1 and A4A_4 respectively, yielding the following amplitudes:

A1=(1rL)A2eikm(nCnL)/2A4=(1rR)A3eikm(nCnR)/2\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 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, 2020, Optical Society of America.