Group Velocity vs Phase Velocity

When we left off, I told you I had come up against a paradox when applying the principle of least action to the trajectory of an electron. I started by doing it for baseballs. But I didn't do what Feynman does, where he minimizes the action between the starting and ending point. I assumed an initial trajectory, and claimed that the subsequent trajectory had to remain perpendicular to surfaces of equal action. By comparing trajectories of corresponding action at different elevations, I showed the only way to get this to work out was to have the trajectories curve. And I actually got the correct curvature for a baseball.

Something nasty happens when we try to apply this to electrons. Instead of a baseball under the influence of gravity, we send an electron beam between the plates of a charged capacitor. The "surfaces of equal action" naturally become the surfaces of constant phase for the electron wave function. They look like this:

So the upper part of the wave is going faster, making the whole thing bend. But wait...in quantum mechanics, wave number corresponds to momentum...so according to wavelength, it's the lower part of the wave that's going faster. Which one is right?

The paradox is resolved when we examine the group velocity of the wave. Electron waves are not like light waves, where all frequencies travel at the same speed. For light waves, you can easily calculate the velocity as the frequency times the wavelength, and this looks like it should work for electrons too. But you may have heard people talk about group velocity and phase velocity. I'm not about to explain why it works this way. That's a long story. But whether or not you understand the math of it, you may have heard that for electron waves you need to look at the group velocity to see how fast the electron is actually moving. 

The formula for group velocity is a little different from the obvious formula for phase velocity. This is how it goes:

Let's see if we can calculate the group velocities for the electron wave along the two trajectories shown here, and see if the faster wave isn't actually going slower!

We can start with the equation for the wave along the lower trajectory. Using the complex exponential, it is just:

To get the group velocity, we need to figure out what is w (omega) as a function of k. Well, omega is the frequency, which lines up with the energy, and k is the wave number, which lines up with the momentum. We can arbitrarily declare the potential energy along the flat trajectory to be zero, which makes the energy simply proportional to the square of the momentum.  We can arbitrarily choose our units so  k=1, and then the proportionality becomes exact:

So the group velocity it 2, which is twice the phase velocity.

How about the upper path? Well, the total energy of the upper electron is the same, because the two paths started out as a simple wave in free space. But when they enter the capacitor, the one closer to the negative plate (the upper path) loses kinetic energy and gains potential energy. Since we show (in our diagram) that the wavelength is one percent greater, we interpret this as a one percent decrease in momentum. Remembering that total energy is given by w and kinetic energy by k-squared, we must have:

But remember that k for the upper path is numerically 0.99  which means the P.E. along that path must be 0.02 to keep w constant=1. You can see this is correct because for k=0.99 we then get w=1, which are the correct wave parameters for the function as sketched.

Again, if we differentiate w with respect to k, the P.E. term disappears (since it's a constant) and we get once again dw/dk = 2k.

But on the upper path, k = 0.99, so the group velocity is one percent less than on the lower path. Even though the phase velocity was one percent greater. So now we can understand why the electron wave bends away from the direction of slower electrons, while a light wave in a refractive medium bends towards the direction of slower "photons".