How does the "speed" of the wavefunction of the electron relate to the speed of the electron particle? (first year undergrad)
Is the speed of the electron the same as the speed of its wavefunction?
I've got the equation E=ħω but I don't really get what ω is, am I right in thinking the energy E is the sum of the potential and the kinetic energies of the particle? I know c=ω/k, but is the c the speed of the electron of just something to do with the shape of its probability curve?
(This isn't very well worded sorry. Basically, I'm confused.
 A: To elaborate the point of WillO, note that velocity is a property of the state of a particle, not of the particle itself. Just like a particle can be in a state that is a linear superposition of states of different momenta, it can also be in a state that is a linear superposition of states of different velocities, which of course doesn't make any sense in classical theory.
But as you will study in scattering theory, while trying to get the form of differential cross section(see Weinberg, Lectures on Quantum Mechanics chapter 8.2 or Quantum Theory of Fields chapter 3.4), you need to consider the velocity of a particle in the framework of quantum mechanics. That velocity, in free theories, is defined to be the ratio $\frac{\vec p}{m}$ and it gives exactly the speed of a free particle of definite momentum. In interacting theories or theories with external potential, the same form $\frac{\vec p}{m}$ may be considered as the velocity of a particle, but here $\vec p$ is the momentum of the corresponding asymptotic "in" or "out" states which are guaranteed to be free states.
