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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?

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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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