For a simple Fabry-Perot cavity formed by a dielectric slab, the quasi-normal mode frequencies (i.e. $\omega=i \frac{c}{2nL}\ln{r^2}$, where $nL$ is the optical path difference between ends of the cavity, $c$ is the speed of light, and $r$ is the reflection coefficient off of one edge), when projected onto the real axis, correspond precisely to points of zero reflection (i.e., something like $\omega=\frac{m\pi c}{nL},\forall m\in\mathbb{Z}$). For some reason, I had the intuition that this generalised to the resonances of more general structures (specifically structures formed by varying a real susceptibility) in 1D, I guess maybe informed by the insertion of defects to allow transmission through photonic crystals, but I realise that I actually have little/no proof of this being the case.

I'm fairly confident that this result doesn't hold in higher dimensions, but am uncertain whether or not this is a known result in 1D wave equations, at least.

Is there some sort of general result I can use to this effect, or is hoping for some such result just wishful thinking, and I should just knuckle down and compute the real parts of my QNMs properly.

And, as a follow-up, my intuition for resonance in 1D was built around the idea that, at least in the 1D lossless case (I'm assuming that measuring 'straight through' transmission in 3D acts similarly to lossy systems in 1D - with channels coupling the light out in other directions just acting like any other channel for loss, which would peak when $\omega$ is near the real part of a QNM frequency), QMNs contribute perfect transmission peaks with a width in frequency space given by the decay rate. Presumably they do still contribute features with width scaling like their decay rate, but if my idea above is wrong, what feature-structure, if any, would you expect from such a resonance?


1 Answer 1


This statement is in general not true.

Let us first look at how this comes about in this specific geometry. The reflection coefficient of a dielectric slab of length $L$ and refractive index $n$ surrounded by vacuum is $$r(k) = r_{01} \frac{1-e^{2ikLn}}{1-r_{01}^2e^{2ikLn}}\,,$$ where $r_{01}=\frac{1-n}{1+n}$ is the Fresnel reflection coefficient of the interface and $k=\omega/c$ is the wave number at which we measure the reflectivity.

From this, we can derive the locations $k_\mathrm{zero}$ at which $r(k_\mathrm{zero})=0$ and the quasimode frequencies $k_\mathrm{pole}$, which correspond to the poles where $r(k_\mathrm{zero})=\infty$ (i.e. where the denominator is zero). This gives the following solutions $$k_{m,\mathrm{zero}}Ln = \pi m\,,$$ $$k_{m,\mathrm{pole}}Ln = \pi m + i\ln{r_{01}}\,.$$ We see that the real parts of the two are indeed identical, which is the coincidence observed by the OP.

However, we also see already in this simple geometry that this statement is only true if $\ln{r_{01}}$ is real. If we add some absorption, this can easily be circumvented and is not a strict locking.

Note, however, that the zeros then also move away from the real axis. In 1D geometries with multiple layers, one can break the coincidence even without absorption and keeping the zero on the real axis (see e.g. https://arxiv.org/abs/2107.11775).

The way to think about this is that the quasimodes are the resonances of the cavity. These will correspond to dips in the reflection spectrum, where the minima indeed approximately correspond the poles for good cavities/well separated resonances. This can be understood from the behavior of the real and imaginary parts of an isolated Lorentzian $\frac{1}{k-k_0 + i\gamma}$. However, if the cavity becomes bad/lossy, the overlap between the resonances can break this correspondence. With this argument in mind, it is therefore not so surprising that $\mathrm{Re}[k_\mathrm{zero}]$ and $\mathrm{Re}[k_\mathrm{pole}]$ can differ.

The surprising thing is rather that in the specific geometry above, the correspondence is exact for real $n$. This is really a special feature of the single layer geometry. If we look at the solutions above, the real part of $k_{m,\mathrm{zero}}Ln$ and $k_{m,\mathrm{pole}}Ln$ is purely determined by the multiplicity of the logarithm. This is because a round trip inside this cavity is trivial, since it does not involve any interfaces.

Note that there are various fascinating phenomena which result from the coalescence between poles and reflection zeros, see e.g. https://arxiv.org/abs/1909.04017 and https://arxiv.org/abs/1603.02325.

  • 1
    $\begingroup$ I feel like this works as a great argument for the strength of the field inside the cavity, and fits my experience for the scattering, but don't see as clearly why it is the case. I know in general the link between QNMs and scattering coefficients is non-trivial (journals.aps.org/prx/pdf/10.1103/PhysRevX.7.021035), but any further intuition would be appreciated. $\endgroup$ Commented Jul 22, 2022 at 13:50
  • 1
    $\begingroup$ @DoublyNegative No worries, I think it is a very well phrased and interesting question! I think I get also your follow-up question. A few notes on this: 1. The main point of my answer is that the correspondence between poles and minima/maxima is only true for good cavity and also not a strict correspondence there, just a rough rule. $\endgroup$ Commented Jul 22, 2022 at 14:14
  • 1
    $\begingroup$ 2. The reason for the maximum/minimum - transmission/reflection correspondence is that for a good cavity in 1D, you will hit a good mirror at some point. Therefore, off-resonance everything comes back. Again, in general, you can get almost any behavior here, from the reverse to Fano resonances. $\endgroup$ Commented Jul 22, 2022 at 14:15
  • 1
    $\begingroup$ 3. Constructing a quasi-mode expansion of the scattering matrix is indeed non-trivial. That paper is a good resource, but there has been a lot more discussion on this (beyond the scope of this comment, I would suggest a separate question for that). Let me note though that the scattering matrix and the field inside the cavity (or the Green's function for that matter, which has a well-known quasi-mode expansion) have the same poles in the complex plane. That correspondence is exact and independent of the minima/maxima, which are not strictly locked. $\endgroup$ Commented Jul 22, 2022 at 14:19
  • 1
    $\begingroup$ 4. The Lorentzian in my argument was just to add intuition. Indeed, the correspondence can break in two ways: if you have complex residues or if other poles contribute. Both are possible, see the first linked paper in my answer. $\endgroup$ Commented Jul 22, 2022 at 14:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.