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

[Giovanni Cantele]

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
            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 <mailto:dingfu.shao at gmail.com>_______________________________________________
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