Consider light moving along one dimension at the classical level. I am interested in the situation where a wave front impacts a material with some generic index of refraction $n(\omega)$, and propagates through. My calculations seems to suggest that the very front of the wave travels at exactly $c$, conditional only on $n(\omega)\to 1$ as $|\omega|\to \infty$ (which I believe must be the case). This seems to conflict with common intuition that light is slowed by media, but perhaps it is the case that the front amplitude is strongly suppressed relative to the main amplitude of the wave? Could someone clarify what is going on?
My calculation is quite simple. Suppose the light wave is moving along the positive $x$ direction, and the material is defined as starting at $x=0$. Let the vector potential be $\mathbf A(t,x)=\hat{\mathbf A}\psi(t,x) $ here be given as $\psi(t,0)=\Theta(t)u(t)$ for some smooth function $u(t)$. For simplicity let us take it to be $e^{-0^+t}$ for convergence purposes. Then:
$$\psi(\omega,0)=\int_0^\infty\mathrm{d} t\,\mathrm{e}^{i(\omega+i0^+) t}=\frac{i}{\omega+i0^+} \ .$$
Hence the wave front at any point in the material may be calculated to be: $$\psi(t,x)=\int_\mathbb{R}\frac{\mathrm{d}\omega}{2\pi}\frac{i\mathrm{e}^{-i\omega(t-n(\omega)x/c)}}{\omega+i 0^+} \ . $$
As long as $n(\omega)\to 1$ as $|\omega|\to \infty$, for $x>ct$ we can add to the integral a semicircle contour, such that we enclose the upper complex $\omega$ plane. By the residue theorem we therefore have $\psi(x>ct)=0$ as expected. However, no matter the dispersion, when $x<ct$ we can add the semicircle contour in the lower half plane. Now in general, $n(\omega)$ will have poles in the lower half plane (but none in the upper), so we cannot perform the integral, but unless a miraculous cancellation occurs, surely we have $\psi(x=ct-0^+)\neq0$, independent of the material.
