23 Jan
2017
23 Jan
'17
7:11 p.m.
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.