Backscattering Is Caused by Reciprocity, Not by Rough Sidewalls — Epoche C2
The effect and the usual diagnosis Send light down a silicon wire waveguide and bend it sharply. Some of the light comes back. The standard diagnosis is that the sidewalls are rough, the etch is imperfect, and a better process would fix it. The diagnosis is not wrong about the roughness, but it identifies the wrong cause. Roughness sets the size of the reflection; it does not create the possibility of one. The possibility comes from a symmetry of Maxwell's equations, and no amount of polishing removes a symmetry. This essay follows that causal chain to its end, and then examines the one intervention that acts on the cause rather than the symptom. Units are SI throughout, as in the reviews cited, with the permittivity written as a dimensionless relative tensor $\varepsilon$. The cause: reciprocity supplies a destination Consider a straight waveguide supporting a single guided mode with electric field $\mathbf{E}_f$ travelling forwards at frequency $\omega$. Lorentz reciprocity states that if the permittivity and permeability tensors are symmetric — which they are for any ordinary dielectric, however impure — then the scattering matrix of the structure is symmetric, and a mode running in one direction is always accompanied by a partner at the same frequency running in the other. Call its field $\mathbf{E}_b$. Now introduce any static perturbation $\delta\varepsilon(\mathbf{r})$: a wall bump, a bend, an index step. First-order perturbation theory on Maxwell's equations gives a coupling coefficient between the two modes, $$\kappa_{bf} \;=\; \frac{\omega\varepsilon_0}{4}\int \delta\varepsilon(\mathbf{r})\; \mathbf{E}_b^{*}(\mathbf{r})\cdot\mathbf{E}_f(\mathbf{r})\; d^3r ,$$ which is the optical analogue of the quantum matrix element $\langle b|\delta H|f\rangle$, with $\delta\varepsilon$ in the role of the perturbing potential and the mode overlap in the role of the wavefunction overlap. The structure of this expression is the whole argument. It has two factors: the size of the perturbation, and the overlap of the two modes. Because reciprocity makes $\mathbf{E}_b$ essentially the time-reverse of $\mathbf{E}_f$, the two fields occupy the same region and the overlap integral is generically of the same order as the corresponding forward-scattering term. The backward mode is not a poorly matched destination. It is the best-matched destination available. This is why the effect is generic. Any perturbation at all couples forwards to backwards, because the destination state exists and overlaps well. A perfectly smooth 90° bend still reflects, since the bend itself is a perturbation of the straight-guide problem. Why improving fabrication cannot remove the cause Since $\kappa_{bf}$ is linear in $\delta\varepsilon$, the reflected power scales as the square of the roughness amplitude. Take a typical silicon strip waveguide with root-mean-square sidewall roughness $\sigma = 2$ nm and propagation loss of about 2 dB cm$^{-1}$, of which the backscattered part is a substantial fraction. Suppose one wants to reduce that by two orders of magnitude, to 0.02 dB cm$^{-1}$. Because the loss goes as $\sigma^2$, the roughness must fall by $\sqrt{100} = 10$, from 2 nm to 0.2 nm. The step height on a silicon (001) surface is a quarter of the lattice constant, $5.43/4 = 1.36$ Å $= 0.136$ nm. The present roughness is therefore $2/0.136 = 14.7$ atomic steps, and the target is $0.2/0.136 = 1.5$ atomic steps. Fabrication improvement of this kind runs into the crystal itself, and even a perfect crystal would still reflect at bends. The returns are quadratic and the floor is not zero. The intervention: remove the destination The alternative is to attack the other factor in $\kappa_{bf}$ — not $\delta\varepsilon$, but the existence of $\mathbf{E}_b$. The tool is a photonic band structure with a non-zero topological invariant. For each band $n$ of a two-dimensional periodic structure one defines the Berry connection $\mathbf{A}_n(\mathbf{k}) = i\langle u_{n\mathbf{k}}|\nabla_\mathbf{k}|u_{n\mathbf{k}}\rangle$, where $|u_{n\mathbf{k}}\rangle$ is the periodic part of the Bloch mode at wavevector $\mathbf{k}$, and its curl, the Berry curvature $\Omega_n = \nabla_\mathbf{k}\times\mathbf{A}_n$. The Chern number is its total flux through the Brillouin zone, $$C_n = \frac{1}{2\pi}\int_{BZ}\Omega_n(\mathbf{k})\,d^2k \;\in\; \mathbb{Z} .$$ It is an integer for the same reason a magnetic flux through a closed surface is quantised: the Brillouin zone is a torus, and the phase of the Bloch mode must be single-valued around it. This is the invariant Thouless and colleagues introduced for the quantum Hall effect in 1982, transplanted to photons by Haldane and Raghu in 2008. The consequence that matters is bulk–boundary correspondence: at an interface between regions whose bands below the gap carry different total Chern number, the number of edge modes crossing the gap equals the difference, and they all cross with the same sign of slope. A mode whose dispersion $\omega(k)$ rises monotonically across the gap has a group velocity $d\omega/dk$ of one sign only. So at a frequency inside the gap there is exactly one edge state at that boundary, and it moves forwards. The matrix element $\kappa_{bf}$ is not made small; it is undefined, because there is no state $\mathbf{E}_b$. Wang and colleagues demonstrated this in 2009 in a gyromagnetic photonic crystal at microwave frequencies, showing transmission past an obstacle large enough to block the guide entirely. The price A one-way state cannot be had for free, and the reason is a two-line argument. If time-reversal symmetry holds, then every mode at $(\omega,\mathbf{k})$ has a partner at $(\omega,-\mathbf{k})$, and the Berry curvature is odd, $\Omega_n(-\mathbf{k}) = -\Omega_n(\mathbf{k})$. An odd function integrated over the symmetric Brillouin zone gives zero, so $C_n = 0$. A non-zero Chern number therefore requires broken time-reversal symmetry — most directly through a magneto-optic material