Units of current operator
Dear Kwant developers,
What are the units of the output of the current density operator? The units of the density operator is per area per energy, so the units of current are (e/h)* per area * energy?
Thanks a lot.
Bests,
Marc
Dear Marc,
Please have a look at the corresponding documentation: https://kwantproject.org/doc/1/reference/generated/kwant.operator.Current#k... https://kwantproject.org/doc/1/reference/generated/kwant.operator.Current#kwant.operator.Current
The density n_i = psi_i^* psi_i has unit 1/energy.
The current between two sites i and j j_ij = i (psi_i^* H_ij psi_j  cc ) therefore is dimensionless if I am not mistaken.
The question about adding e/h or area is not related to Kwant but to how you actually discretized your initial continuous problem.
Best regards,
Xavier
Le 13 févr. 2020 à 21:38, Marc Vila marc.vila@icn2.cat a écrit :
Dear Kwant developers,
What are the units of the output of the current density operator? The units of the density operator is per area per energy, so the units of current are (e/h)* per area * energy?
Thanks a lot.
Bests,
Marc
Dear Xavier,
Thanks for your quick answer. The reason behind this question is that I'm trying to fit the spin density extracted from the Kwant spin density operator with a formula derived from the spin diffusion equations (to extract e.g. the spin diffusion length or spin Hall angle). In this reference (https://arxiv.org/abs/1908.11354) they seem to complement the information in the Kwant documentation, and effectively, the units of the (spin) density are 1/energy. However, the solution of the spin density in the diffusion equations is 1/length^2 (in 2 dimensions). So a priori we did not know how to relate them, so we thought of dividing the output of the spin density operator with the output of the current density operator to get rid of the energy terms in the Kwant output. But if the current is unitless then I guess this is not useful... Therefore, let me ask you another question, how can I relate the spin density extracted from kwant with the one derived from the diffusion equations?
Thanks a lot again.
Best regards,
Marc
 Marc Vila Tusell La Caixa  Severo Ochoa PhD in the Theoretical and Computational Nanoscience Group Catalan Institute of Nanoscience and Nanotechnology (ICN2) Barcelona Institute of Science and Technology (BIST)
Additional information:
https://scholar.google.es/citations?user=h2V4iNIAAAAJ&hl=es
http://icn2.cat/en/theoreticalandcomputationalnanosciencegroup
https://www.researchgate.net/profile/Marc_Vila_Tusell
https://www.becarioslacaixa.net/marcvilatusellBI00042?nav=true
https://orcid.org/000000019118421X
________________________________ From: Xavier Waintal xavier.waintal@cea.fr Sent: Friday, February 14, 2020 7:59 AM To: Marc Vila Cc: kwantdiscuss@kwantproject.org Subject: Re: [Kwant] Units of current operator
Dear Marc,
Please have a look at the corresponding documentation: https://kwantproject.org/doc/1/reference/generated/kwant.operator.Current#k...
The density n_i = psi_i^* psi_i has unit 1/energy.
The current between two sites i and j j_ij = i (psi_i^* H_ij psi_j  cc ) therefore is dimensionless if I am not mistaken.
The question about adding e/h or area is not related to Kwant but to how you actually discretized your initial continuous problem.
Best regards,
Xavier
Le 13 f?vr. 2020 ? 21:38, Marc Vila <marc.vila@icn2.catmailto:marc.vila@icn2.cat> a ?crit :
Dear Kwant developers,
What are the units of the output of the current density operator? The units of the density operator is per area per energy, so the units of current are (e/h)* per area * energy?
Thanks a lot.
Bests,
Marc
participants (2)

Marc Vila

Xavier Waintal