# Group velocity of a wavepacket vs its mean phase velocity

1. Mar 6, 2015

### jfizzix

The mean velocity of a wavepacket given by the general wavefunction:
$\Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int dk A(k)e^{i(k x - \omega(k) t)}$,
can be expressed in two ways.

First, we have that it's the time derivative of the mean position (i.e., its mean group velocity):
$\frac{d \langle x\rangle}{dt}=\int dk |A(k)|^{2} \frac{d\omega(k)}{d k}\approx \frac{d\omega}{dk}$ at center frequency.

Second, we have that it is the averaged velocity of all the plane wave components of the wavepacket (i.e., the mean phase velocity):
$\langle \frac{\omega(k)}{k}\rangle=\int dk |A(k)|^{2} \frac{\omega(k)}{k}.$

My questions are these:
When are these two formulations of "the mean velocity" equivalent?
Which (if either) best corresponds to the speed of information transfer, or energy flow?

2. Mar 7, 2015

### Staff: Mentor

They have the same speed if you do not have dispersion, which means $\frac{d\omega}{dk}$ is constant (then $\frac{d\omega}{dk}=\frac{\omega}{k}$).
There might be some exotic special case where the phase velocity matches group velocity even with some dispersion in some frequency range, not sure.
Energy flow is related to group velocity. The speed of information transfer is more complicated (front velocity), but for usual data transmission group velocity matters.

3. Mar 7, 2015

### jfizzix

So what does the mean phase velocity tell us, exactly? If each plane-wave component is moving at a given phase velocity, and we average those over all plane wave components, what does that number tell us?

4. Mar 7, 2015

### Staff: Mentor

It tells you how fast the phase and therefore peaks are advancing, which is important for refraction, for example. If you get significant dispersion within your wave packet, then you cannot average over the phase velocities any more and you have to consider the different frequencies separately.