Can phase information be extracted from envelope slowdown? That is, if I can track the envelope of a pulse of light as it travels through a medium, but I can't see the phase of that pulse (maybe the frequency of the light is too high such that it's too fast for a detector), is there a way that use the information about the motion of the envelope to figure out information about the phase?

For example, suppose I want to find the difference in the phase of the e-field of light that's traveling through two different mediums. If I send a pulse through medium A, and then see that the pulse moves slower than when it travels through medium B, can I obtain any information about the phase of the pulse? You can assume the light is almost-monochromatic and has an envelope pulse shape.

My thought is that since $v_p = \frac{\omega}{k}$ and $v_g = \frac{\partial \omega}{\partial k}$ then, $\omega = \int dk v_g$ But I'm not sure how I'd integrate the group velocity for a specific example. And if there isn't a direct way to extract phase information from different group velocities, is there any information that can be extracted from paying attention to the envelope?


Here's my best stab at an answer; I'm assuming that you are able to look at some sort of spatial envelope $|\psi(x, t)|^2$ over time and you are interested in the phase as a function of the frequency present, some $\phi(k)$.

So my rough intuition for group velocities always comes from a Gaussian wave-packet, $A\exp(-a(k-k_0)^2).$ Turning that into a sum of plane waves with varying phases $\phi(k)$ gives, $$\psi(x, t) = A \int_{-\infty}^{\infty} dk~\exp\Big(-\frac a2(k-k_0)^2 + i \big[(k - k_0) x - \omega(k)~ t + \phi(k)\big]\Big).$$ Indeed a plane wave with frequency $\omega$ should travel with wave speed $\omega/k$ on this account, but if we expand everything to second order about $k_0$ we actually find:$$\psi(x, t) = A e^{i(\phi(k_0) - \omega(k_0) t)}\int_{\mathbb R} dk~\exp\big(-\frac{\tilde a}2(k-k_0)^2 + i \big[x -\omega'(k_0)~ t + \phi'(k_0)\big] (k-k_0) \big),$$ where $\tilde a = a + i \phi''(k_0) - i\omega''(k_0)t.$ Applying the normal Gaussian Fourier transform formula would give: $$\psi(x, t) = A \sqrt{\frac{2\pi}{\tilde a}} \exp\left(-\frac{(x-\omega_0't+\phi_0')^2}{2\tilde a} + i(\phi_0 - \omega_0 t)\right),$$describing a wave packet which moves forward with time at a speed $\omega_0'$. On the right we see that $\phi_0=\phi(k_0)$ just gets totally absorbed into the phase of the wave at $t=0$, while the wave starts out at position $x_0 = -\phi_0',$ so that's only semi-useful information as well. The first really useful information is $\phi_0'',$ which is only present in this $\tilde a$ term. Still, that is somewhat useful:$$\frac{(x-\omega'_0t+\phi'_0)^2}{\bar a} = \frac{(x-\omega'_0t+\phi'_0)^2}{a^2 + (\omega''_0 t - \phi''_0)^2}(a + i \omega''_0 t - i\phi'').$$So: in a medium with linear dispersion, $\omega_0''=0$ and you can't use the spatial envelope of the function to tell the difference between the second derivative of $\phi$ and the wave-packet's spectral width $a$ without this phase information.

However if you have a nonlinear medium, wave packets that fly through for different amounts of time will have increasing widths, and there is a way to work out $\left.\frac{d^2\phi}{dk^2}\right|_{k=k_0}$ from the way that this width is increasing. For higher order derivatives you'd need to work out corresponding nonlinearities in the above transform, I suppose.


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