[Smeagol-discuss] Understanding of some basic concepts

[lan haiping]

Though smeagol calculation include  leads' self-energy accounting for
charges' transport through scattering region,
you should keep in your mind that the situation NEGF can handle now is about
stable cases, which means the scattering-in charges
are equal to the scattering-out part.
This is my basic consideration of this problem.
Any comments are welcome

On Tue, Apr 1, 2008 at 11:59 AM, XIAOHONG ZHENG <exhzheng at gmail.com> wrote:

> Hello, Dear Haiping,
>
> Thank you for your reply. I am happy to discuss with you.
>
> I do not think the neutrality of the scattering region is due to
> calculation by periodic cell. In fact, it is an imposed condition and we
> have to normalize the charge to a fixed value after every SCF step. The
> periodic condition is used only when the Poisson equation is solved by
> FFT. It depends on the electron density. However, the electron density
> is determined by the Green function. If we do not have the self-energy
> in the calculation of the Green function, namely, if we do not take into
> account the effects of the leads,  then the electron charge will be
> conserved. However, in our calculation, the self-energy from the leads
> is considered in the Green function. So charge will not be conserved.
>
> This is my opinion. Any comments are welcome.  Thanks.
>
> Sincerely,
> Xiaohong
>
>
> lan haiping wrote:
> > Dear XIAOHONG:
> >
> > A basic sense, the scattering region calculation is performed under
> > periodic cell, which of course prevents any  extra charge situations.
> >
> > On Tue, Apr 1, 2008 at 5:59 AM, XIAOHONG ZHENG <exhzheng at gmail.com
> > <mailto:exhzheng at gmail.com>> wrote:
> >
> >     Dear Ivan,
> >
> >     Thank you very much for your help. I still have some doubts about
> your
> >     answers.
> >
> >     1. Although the scattering region is supposed to be neutral, the
> >     Poisson
> >     equation is solved  for the electrons, not for the whole charge (
> the
> >     sum of the ionic charge and electronic charge). So the k=0 term
> >     will not
> >     be 0, but proportional to Q/V (namely, the average electron
> >     density). Q
> >     is the total electrons, not zero.
> >
> >     2. In my understanding, the scattering region is not necessarily
> >     neutral
> >     since it is an open system. The scattering region may be charged or
> >     discharged. So, if there is some extra charge in the scattering
> >     region,
> >     I think it might be reasonable. In fact, in a previous post (by
> Prof.
> >     Sanvito?) I find that, due to the self-energy from the leads, the
> >     Hamiltonian becomes non-hermitian now and the charge can not be
> >     exactly
> >     conserved.
> >
> >     Sincerely,
> >     Xiaohong
> >
> >
> >     >> (4) In the solution of Poisson equation by FFT, to dispose the
> >     k = 0
> >     >> term does not affect  the physics, of course, because this term
> >     is just
> >     >> a constant. But what is the advantage of disposing this term,
> >     please?
> >     >> Just for saving computation time? If we keep this term, then the
> >     >> potential matching might not be a problem, especially for cases
> >     where we
> >     >> use different leads. Because now, we do not need to get the
> >     average in
> >     >> the lead for the potential matching in the scattering region.
> >     Of course,
> >     >> we still have to do some matching, but we should do the matching
> >     >> according to the Fermi levels in the three parts. We should
> >     align their
> >     >> Fermi levels to the same energy point and thus shift the hartree
> >     >> potentials of the three parts accordingly. How do you think
> >     about this,
> >     >> please?
> >     >>
> >     >>
> >     > One of the main assumptions in smeagol is that the scattering
> >     region is
> >     > neutral, so that the k=0 term is 0 (small fluctuations of the
> excess
> >     > charge around 0 are OK in practice). If the system has a finite
> >     extra
> >     > charge then this term is not 0 and we have an incorrect
> >     potential with
> >     > the FFT. But there should always be enough leads slices inside the
> >     > scattering region to screen the potential before joining the
> >     leads, this
> >     > always results in a neutral scattering region. This is the case
> >     also for
> >     > different leads.
> >     >
> >     >
> >     > Cheers,
> >     >
> >     >  Ivan
> >     >
> >     _______________________________________________
> >     Smeagol-discuss mailing list
> >     Smeagol-discuss at lists.tchpc.tcd.ie
> >     <mailto:Smeagol-discuss at lists.tchpc.tcd.ie>
> >     http://lists.tchpc.tcd.ie/listinfo/smeagol-discuss
> >
> >
> >
> >
> > --
> > Hai-Ping Lan
> > Department of Electronics ,
> > Peking University , Bejing, 100871
> > lanhaiping at gmail.com <mailto:lanhaiping at gmail.com>, hplan at pku.edu.cn
> > <mailto:hplan at pku.edu.cn>
>
>


-- 
Hai-Ping Lan
Department of Electronics ,
Peking University , Bejing, 100871
lanhaiping at gmail.com, hplan at pku.edu.cn
-------------- next part --------------
An HTML attachment was scrubbed...
URL: http://lists.tchpc.tcd.ie/pipermail/smeagol-discuss/attachments/20080401/6dbe99aa/attachment-0001.html