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

[Paolo Giannozzi]

This is what the header of average.f90 says:

 !      Compute planar and macroscopic averages of a quantity (e.g. charge)
  !      in real space on a 3D FFT mesh. The quantity is read from a file
  !      produced by "pp.x", or from multiple files as follows:
  !          Q(i,j,k) = \sum_n w_n q_n(i,j,k)
  !      where q_n is the quantity for file n, w_n is a user-supplied weight
  !      The planar average is defined as
  !         p(k) = \sum_{i=1}^{N_1} \sum_{j=1}^{N_2} Q(i,j,k) / (N_1 N_2)
  !      along direction 3, and the like for directions 1 and 2;
  !      N_1, N_2, N_3 are the three dimensions of the 3D FFT.
  !      Note that if Q is a charge density whose integral is Z_v:
  !         Z_v = \int p(z) dV = \sum_k p(k) \Omega/N_3
  !      where \Omega is the size of the unit cell (or supercell)
  !      The planar average is then interpolated on the specified number
  !      of points supplied in input and written to file "avg.dat"
  !      The macroscopic average is defined as
  !         m(z) = \int_z^{z+a} p(z) dz
  !      where a is the size of the window (supplied in input)

If it doesn't  correspond to reality, please report

Paolo

On Fri, Oct 26, 2018 at 3:31 AM 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 <dingfu.shao at gmail.com>*
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-- 
Paolo Giannozzi, Dip. Scienze Matematiche Informatiche e Fisiche,
Univ. Udine, via delle Scienze 208, 33100 Udine, Italy
Phone +39-0432-558216, fax +39-0432-558222
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