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The problem Lasers do all sorts of cool things in research and in applications, and there are many good reasons for it, including their coherence, frequency stability, and controllability, but for so …

This is chirp induced by dispersion, which is the acoustic version of the same phenomenon for light. … For each individual sample a full analysis is necessary to understand which bit of the circumstances caused the dispersion (so it's hard-to-impossible to tell you any specifics about the individual recordings …

The dynamical origins of the two are extremely different. Surface waves in water are gravity waves, which means that the restoring force trying to bring peaks and troughs back to the mean height is …

In the specific case you posit, you want to consider a pulse with a bandwidth $1/\sigma$ that is narrow enough that the dispersion only has enough space to show a linear variation: $$ \omega(k)=\omega( … This is another natural consequence of dispersion, and it can be a bummer if what you're after is a really short pulse. …

If you phrase the equation of motion as explicitly real valued, $$ \ddot x+\omega_S\dot x+\omega_0^2 x=F(t)=\frac{qE_{0}}{m}\cos(\omega_F t), $$ then indeed you can write down the solution as $$ x(t)= …

In terms of Fourier analysis, then, if you write your solution as $$ f(x,t)=\int^\infty_{-\infty}C(k)e^{i(kx-\omega t)}\mathrm dk \tag{1'} $$ then the dispersion relation requires you to set $\omega^2 …

So, let's consider your example, where we have a free particle in one dimension, with dispersion relation $$ \omega(k)=\frac{\hbar k^2}{2m}, $$ and where we have two states, $|\psi_1\rangle$ and $|\psi … Here's the important part: the dispersion relation doesn't change, and neither does $v_g(k)$ as a function of wavevector, but you do change the inputs to both of them. …

The $\lambda^{-2}$ term works reasonably well to account for the decrease in the refractive index for materials with normal dispersion away from any resonances, and the $\lambda^{-4}$ term can fudge things …

For the dispersion relation to be useful (as opposed to just being true) then yes. … We find the dispersion relation of a wave by putting in a trial solution of the form $e^{i(kx-\omega t)}$ and then working out the conditions on $\omega = \omega(k)$ for the trial plane-wave waveform to …

frac{\epsilon}{c}(\lambda -v)\omega^2 =0 $$ which fails to simplify in ways that make suspect you've made a sign error somewhere, so I'll leave it to you to double-check your workings and pull out the dispersion …

Your second result is a constant times the identity operator, which in matrix representation corresponds to the identity matrix. It is a common notation practice in quantum mechanics to denote both t …

You've got some things right, and some things wrong. I have learned that all light in a vacuum, all electromagnetic waves, travel at the same speed in a vacuum. Yes, that is exactly right. …