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

[Giovanni Cantele]

Are you sure about you Fermi level? How did you get it? 

GIOVANNI 

Inviato da iPhone

> Il giorno 26 ott 2018, alle ore 19:33, Dingfu Shao <dingfu.shao at gmail.com> ha scritto:
> 
> 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
>>             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
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