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