Yes, that is correct. The equation that you quoted is indeed the probability current

Thank you. Then to calculate the velocity, should I just divide the probability current by the integral of |𝜓|2 over the unit cell?


On Mon, Jan 23, 2017 at 10:00 AM, Anton Akhmerov <anton.akhmerov+kd@gmail.com> wrote:
> Sorry I made a mistake in the units of 'I'. Isn't 'I' here the probability current and not the charge current? We would just have to multiply 'I' by e to get the charge current, right?

Yes, that is correct. The equation that you quoted is indeed the
probability current.