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