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 tight-binding model.
Best,
Anton
On Mon, Jan 23, 2017 at 3:49 PM, Harshad Sahasrabudhe
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 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 tight-binding model. Best,
Anton On Mon, Jan 23, 2017 at 3:49 PM, Harshad Sahasrabudhe 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 |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.
participants (2)
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Anton Akhmerov
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Harshad Sahasrabudhe