[QE-users] Unit for the output of average.x

[Dingfu Shao]

Dear Giovanni and Paolo,

Thanks very much for your suggestions.

I reconsidered the definition of the local density of states (say,
LD(x,y,z,E)). Since it is "local", the unit of it should be
states/eV/bohr^3 or electrons/eV/bohr^3. Therefore, the integration of it
within an energy window should lead to the charge density in this energy
window: electrons/bohr^3 . Therefore, if we choose the energy window from
the lowest energy to the Fermi energy, we should get exactly the total
charge density.

So I did some tests following Giovanni's suggestion, using a simple case of
momolayer MoS2, which has "number of electrons  =   26.00". The area of its
xy plane is S. For those tests, the previous scf and nscf calculations are
the same, with LDA USPP,  the occupation= 'fixed', and k-points of 40*40*1
for scf and 100*100*1 for nscf.

1. I  calculated the total charge density (rho(x,y,z)) using pp.x with
plot_num = 0, and then calculate the planar average of it (rho_avg(z)).
Then I integrated it by \int (S* rho_avg(z)) dz and I got 25.89. It is
close to 26.00, maybe a more accurate value can be obtained by a
calculation with denser k points.

2. I calculated the integrated local density of states  ( ILD(x,y,z)) from
-65 eV (this energy is below the lower band) to Fermi energy using pp.x
with plot_num =10, and and then calculate the planar average of it (
ILD_avg(z) ). When I integrated it by \int (S* ILD_avg(z)) dz, I got 29.25,
which is much larger than the total number of electrons of 26.

 So it seems the the test 2 doesn't correctly reflect the  reality. I am
not sure it is due to something happened with plot_num = 10 in pp.x, or
just I understand this incorrectly.

Any suggestions? Thank you very much!

Best,

Ding-Fu






From: Giovanni Cantele <giovanni.cantele at spin.cnr.it>
> To: Quantum Espresso users Forum <users at lists.quantum-espresso.org>
> Cc:
> Bcc:
> Date: Fri, 26 Oct 2018 09:51:40 +0200
> Subject: Re: [QE-users] Unit for the output of average.x
> Dear Ding-Fu,
> as far as I remember there is a surface factor that you need to adjust
> units.
> For sure on the ascissa axis the coordinate is in bohr.
> The planar average give you back a quantity with the same units as the
> averaged quantity
> (e.g. if you star from charge density in electrons/bohr^3 you get an
> averaged electron density in electrons/bohr^3),
> being defined as (let us suppose that you average in the plane defined by
> a1 and a2 vectors):
> rho_avg(z) = ( 1 / S ) * integral( dx dy rho(x,y,z) )
> That means that if you perform
>  integral( dz rho_avg(z) )
> you get
> number of electrons / S
> If you need number of electrons than just multiply by S with
> S = cross_product( a1, a2 )
> (in bohr^2)
> Just try, better if you do it with the total charge density, to check if
> the integral returns you
> the number of electrons.
> I’m sorry but I cannot check directly if I remember correctly at the
> moment, but
> this should work.
> Giovanni
> --
> Giovanni Cantele, PhD
> CNR-SPIN
> c/o Dipartimento di Fisica
> Universita' di Napoli "Federico II"
> Complesso Universitario M. S. Angelo - Ed. 6
> Via Cintia, I-80126, Napoli, Italy
> e-mail: giovanni.cantele at spin.cnr.it
> <giovanni.cantele at spin.cnr.it>            gcantele at gmail.com
> Phone: +39 081 676910
> Skype contact: giocan74
> Web page: https://sites.google.com/view/giovanni-cantele
> On 26 Oct 2018, at 03:31, Dingfu Shao <dingfu.shao at gmail.com> wrote:
> Dear QE developers and users:
> I am wondering what should be the unit of the planar average data got from
> the average.x
>  I am calculating the planar average of charge density within a energy
> window. What I did is firstly using pp to get the integrated local density
> of states (ILDOS) of that energy window with plot_num=10, then using
> average.x to get the planar average.
> In this case, what is the unit of the second column (say, rho(z)) of  the
> output file? I thought since the DOS has a unit of states/eV, the
> integration of DOS within a energy window should get some states or
> electrons. Then the unit of rho(z) should be electron/bohr. But seems it
> is not. In my case the energy window I concerned contains one electron,
> However, if I directly integrate  rho(z), I can only get a very small
> value. If I assume the unit is electron/(bohr^3), the integretion of  rho(z)*A
> is also smaller than one (here A is the area of xy plane).
> Can you help me about it? Thank you very much!
> Best,
> Ding-Fu
>
>
> *Ding-Fu Shao, Ph. D.*
> *Department of Physics and Astronomy, University of Nebraska-Lincoln*
> *Lincoln, NE **68588-0299*
> *Email: dingfu.shao at gmail.com <dingfu.shao at gmail.com>*
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