Today let’s have a look to a single particle, going around without any potential. So we get the same time-independent Schrödinger equation as we got inside the infinite square well:

$$ -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi $$

A single solution for this is the same as for the infinite well, but this time we will use the exponential form, as that will help us later:

$$ \psi(x) = Ae^{ikx} + Be^{-ikx} \hspace{0.5cm}\text{where }k=\frac{\sqrt{2mE}}{\hbar} $$

Now adding the time dependent bits (the “wiggle factor”):

$$ \psi(x) = Ae^{ik\left(x - \frac{\hbar k}{2m}t\right)} + Be^{-ik\left(x-\frac{\hbar k}{2m}t\right)} $$

Here we recognize this as two waves, one moving right and one moving left.

But, this is a function that can’t be normalized, this means that there’s no such thing as a free particle with a specific energy.

But lucky us, we can normalize an infinite sum of them, so we can use exactly that :)

This is what we call a wave packet.

Then, using an integral instead (and keeping only one of the waves, as the other is just a ):

$$ \psi(x, t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\phi(k)e^{i\left(kx-\frac{\hbar k^2}{2m}t\right)} $$

And, using the Plancherel’s theorem (and the Fourier transform), we end up with the solution for $\phi(k)$:

$$ \phi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\psi(x, 0)e^{-ikx}dx $$

And now we only depend on $\psi(x, 0)$ to have the whole system defined, win!

Given all this, let’s assume that we have a wave packet that has a peak at a $k_0$ value (if there’s no distinguishable shape for the packet, the notion of “packet velocity” has no meaning). Given that, we can then approximate $\omega$ by Taylor expansion and we can keep only the first two terms:

$$ \omega(k) \approx \omega_0 + \omega’_0(k - k_0) $$

Now centering the integral ($r = k - k_0$) at $k_0$, and doing a bit of mangling, we get:

$$ \psi(x, t) \approx \frac{1}{\sqrt{2\pi}}e^{i(k_0x-\omega_0t)}\int_{-\infty}^{+\infty}\phi(k_0 + r)e^{ir(x-\omega’_0r)t}dr $$

$$ \omega = \frac{\hbar k^2}{2m} $$

And this can be recognized as a wave (the left exponential) traveling at speed $\frac{\omega_0}{k}$, bundled in an envelop wave (the right exponential) that travels at the speed $\omega’_0$. So we have (at $k=k_0$):

$$ v_\text{phase} = \frac{\omega}{k} = \frac{\hbar k}{2m} $$

$$ v_\text{group} = \omega’(k) = \frac{\hbar k}{m} = 2 v_\text{phase} $$

You can find an animation on the jupyter notebook here.

Note that you might have to run locally for the interactive plot to show.