Bests,

Marc

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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)

Sent: Monday, October 22, 2018 10:40 AM
To: Marc Vila
Cc: kwant-discuss
Subject: Re: [Kwant] Units of density

Dear Marc,

There is no problem in a density of probability being larger than one. In fact, this doesn’t prevent it from being normalizable. Example: the probability of the transmission T in random cavities P(T)=1/(2 sqrt(T)) for T in [0, 1]. This function is normalizable despite the fact that it diverges near T=0.

Now, for the wavefunction, you should remember that for infinite systems (open systems) the wavefucntion is not normalizable in the usual way.
In fact, we say that it is normalizable in the sens of Dirac-distributions. (delta(k-k'))

The unit of your density is therefore 1/a      with 'a' the latice constant. The hopping 't' and the parameter 'a' are related, that is way, changing the hopping gives you a different result.
I hope this helps.

On Mon, Oct 22, 2018 at 9:40 AM Marc Vila <marc.vila@icn2.cat> wrote:

Dear Kwant developers,

I've found in other threads in the mailing list that the units of current is for example (unit of charge)/(hbar/unit of energy) (https://www.mail-archive.com/kwant-discuss@kwant-project.org/msg01100.html). Also, the local density of states has units of energy/volume (https://www.mail-archive.com/kwant-discuss@kwant-project.org/msg00169.html​).

My question is, what is the units of the output of the density operator? Is it energy/volume as well? I ask this because intuitively I view it as the square of the wavefunction, but it gives me values larger than 1 for each site when there is only 1 mode involved (see attached picture) ​so it is not just the probability of findinge the electron at that site because this should be maximum 1. I have also noticed that the values I get in the colorbar depend on the value of my hopping (e.g. case of graphene), but overall I'm not so sure of the units.

Thank you again for your help.

Kind regards,

Marc

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