Interpreting group velocity of free particle wave packet

Question

I am trying to understand the concept of group velocity of a free particle wave packet: $$\Psi(x,t) = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty} \phi(k)e^{ikx}e^{-\frac{i \hbar k^2 t}{2m}}dk.$$

Where $$\omega(k) = \frac{\hbar k^2}{2m}.$$ Assuming that $\phi(k)$ is narrowly peaked about some particular value $k_0$ we can Taylor expand $\omega(k)$ about $k_0$ to get $$\omega(k) = \frac{\hbar k_0^2}{2m} + \frac{\hbar k_0}{m}(k-k_0)+ \frac{\hbar}{2m}(k-k_0)^2.$$ Then after defining variable $\kappa := k-k_0$ we get $$\Psi(x,t) \approx \frac{1}{\sqrt{2 \pi}}e^{\frac{-i \hbar k_0^2 t}{2m}}e^{\frac{i \hbar k_0^2 t}{m}} \int^{\infty}_{-\infty}\phi(\kappa + k_0)e^{i(\kappa + k_0)(x-\frac{\hbar k_0 t}{m})}dk.$$

This integral is the superposition of waves of the form $$e^{i(\kappa + k_0)(x-\frac{\hbar k_0 t}{m})}.$$ However notice that each of these waves have the same speed $\frac{\hbar k_0}{m}$. Then in terms of the energy $$v_{\text{group}} = \frac{d \omega(k_0)}{dk} = \frac{\hbar k_0}{m} = \sqrt{\frac{2 E_0}{m}}.$$ Hence for stationary state $\Psi_{k_0}$ we have that $$v_{\text{group}} = 2 v_{\text{phase}}.$$

Apparently we can then evaluate the group velocity for each $k$, hence $v_{\text{group}} = \frac{d \omega(k)}{dk}$.

Question: I understand the above calculation. But qualitatively I don't understand how the group velocity can be a function of $k$ as opposed to just a single value (independent of $k$) which describes how fast the envelope propagates? I know that each stationary state has it's own phase velocity but I thought that superimposing these waves of different speeds produces an envelope which travels at a single velocity $v_{\text{group}}$?

Answer 1

A qualitative understanding can be gained from the animation provided by 'TheGhostOfPerdition', each small peak on the dark blue wave (envelope of the wave packet) has an associated $k_{0}$ for which $\phi(k)$ peaks at $k = k_0$. So you can work out the velocity $v_g : = \frac{d \omega}{d k}|_{k = k_0}$ of the dark blue (envelope of wave packet) using the scheme that you outlined, you have thus found the group velocity for that peak. The velocity might differ for another peak. When there is distortion (the component wavelengths move at different speeds), you could get a envelope of the wave packet which changes shape as it moves as different points move at different speeds.

In quantum mechanics, you are often just interested in one of these peaks since you want the wave packet to go to zero outside a small neighborhood for the wave packet to be normalizable (so you have destructive interference outside this small neighborhood).

Answer 2

Question: I understand the above calculation. But qualitatively I don't understand how the group velocity can be a function of k as opposed to just a single value (independent of k) which describes how fast the envelope propagates? I know that each stationary state has it's own phase velocity but I thought that superimposing these waves of different speeds produces an envelope which travels at a single velocity $v_{group}$?

As you note yourself, your wave packet is narrowly peaked about a specific wave vector, $k_0$. Thus, although the packet is composed of many waves with different wave vectors, they are all close to this wave vector. Simplistically, one could say that they lie within a band $[k_0-\Delta k, k_0+\Delta k]$ with $\Delta k \ll k_0$.

For a wave packet centered about a different wave vector, say $q_0$, the group velocity would be different. Thus, the group velocity is a function of the wave vector that the packet is centered about.

In other words, one should distinguish the wave vector in the Fourier transform, $k$, which labels the waves in the packet, and the wave vector about which the packet is centered, $k_0$. The group velocity is a function of the latter, but not the former.

As @Alex has pointed in their answer, a clearer definition of the group velocity is therefore $$ v_g : = \left.\frac{d \omega}{d k}\right|_{k = k_0}$$