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 <[email protected]> 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 <[email protected]> >> To: Quantum Espresso users Forum <[email protected]> >> 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: [email protected] >> [email protected] >> 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 <[email protected]> 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: [email protected] >> _______________________________________________ >> users mailing list >> [email protected] >> https://lists.quantum-espresso.org/mailman/listinfo/users >> > _______________________________________________ > users mailing list > [email protected] > https://lists.quantum-espresso.org/mailman/listinfo/users
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