Units of the wavefunction and velocity
Hi, I was wondering what units would the wavefunction obtained from Kwant have? I was thinking they would have the units 1/sqrt(nm.eV) (since my energies are in eV and lengths are in nm) as the modes are normalized according to [image: Inline image 1] How would I calculate the velocity of the mode, if the modes are normalized to carry unit current? Thanks, Harshad
Dear Hardshad, I might be missing some factors here, but the unit of current is elementary charge / unit of time. The unit of time is hbar / unit of energy that you used in defining your tightbinding model. Best, Anton On Mon, Jan 23, 2017 at 3:49 PM, Harshad Sahasrabudhe <hsahasra@purdue.edu> wrote:
Hi,
I was wondering what units would the wavefunction obtained from Kwant have? I was thinking they would have the units 1/sqrt(nm.eV) (since my energies are in eV and lengths are in nm) as the modes are normalized according to
[image: Inline image 1]
How would I calculate the velocity of the mode, if the modes are normalized to carry unit current?
Thanks, Harshad
Hi Prof. Akhmerov, 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? Thanks, Harshad On Mon, Jan 23, 2017 at 9:53 AM, Anton Akhmerov <anton.akhmerov+kd@gmail.com
wrote:
Dear Hardshad,
I might be missing some factors here, but the unit of current is elementary charge / unit of time. The unit of time is hbar / unit of energy that you used in defining your tightbinding model.
Best, Anton
On Mon, Jan 23, 2017 at 3:49 PM, Harshad Sahasrabudhe <hsahasra@purdue.edu
wrote:
Hi,
I was wondering what units would the wavefunction obtained from Kwant have? I was thinking they would have the units 1/sqrt(nm.eV) (since my energies are in eV and lengths are in nm) as the modes are normalized according to
[image: Inline image 1]
How would I calculate the velocity of the mode, if the modes are normalized to carry unit current?
Thanks, Harshad
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 𝜓2 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.
participants (2)

Anton Akhmerov

Harshad Sahasrabudhe