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I am searching for calculating Bader charge using Quantum ESPRESSO. I am using pseudopotential based DFT calculation. And I have charge density files (cube and XSF formats).

My objective is to determine interstitial charge (charge in the void spaces, or between two atoms).

It would be great to know the step by step procedure to get the Bader charge through Bader analysis. (I have installed Henkelman Bader in my computer. I can install other as well, if suggested)

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1 Answer 1

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As far as you have correctly obtained the charge density files, the procedure is straight forward.

I will assume that you have the all-electron density and the valence electron density. If not, then This can be achieved by pp.x (post processing) utility provided with the Quantum Espresso package. As an example for the valence density:

&inputpp
    prefix  = 'yoursystem'
    outdir = './'
    filplot = 'yoursystem_valence'
    plot_num= 0
/
&plot
nfile = 1
iflag = 3
output_format = 6
fileout = 'yoursystem.cube'
/

and the all-electron density:

&inputpp
    prefix  = 'yoursystem'
    outdir = './'
    filplot = 'yoursystem_allelec'
    plot_num= 21
/
&plot
nfile = 1
iflag = 3
output_format = 6
fileout = 'yoursystem_allelec.cube'
/

The most important difference between the two input files is the plot_num line which specifies the physical quantities (all-electron or valence densities) that are written to the fileout file and then

$ pp.x < yoursystem_val.pp.in > yoursystem_val.pp.out
$ pp.x < yoursystem_all.pp.in > yoursystem_all.pp.out

Now suppose you have Bader program installed you can perform Bader partitioning on the .cube files created by pp.x:

bader yoursystem_valence.cube -ref yoursystem_allelec.cube

as a result, you will get The Bader charges on each atom contained in the ACF.dat file which will look like the following:

#         X           Y           Z       CHARGE      MIN DIST   ATOMIC VOL
 --------------------------------------------------------------------------------
    1  -10.660000   10.660000   10.660000    8.140837     1.940149    64.992061
    2   -5.330000    5.330000    5.330000    7.859147     2.931500   237.846813
 --------------------------------------------------------------------------------
    VACUUM CHARGE:               0.0000
    VACUUM VOLUME:               0.0000
    NUMBER OF ELECTRONS:        16.0000

This example is specifically for NaCl. It shows that out of 16 valence electrons, 8.14 is assigned to Na and 7.86 to Cl. To convert this to ionic charge, you need to check the number of electrons that are fixed in the core. This can be achieved by looking at 'valence configuration' lines in the pseudopotential files for the constituant atoms. for instance, in the case of Na:

Valence configuration:
    nl pn  l   occ       Rcut    Rcut US       E pseu
    2S  1  0  2.00      1.000      1.250    -4.158089
    3S  2  0  1.00      1.000      1.250    -0.198813
    2P  2  1  6.00      0.900      1.300    -2.106125

and for Cl:

Valence configuration:
    nl pn  l   occ       Rcut    Rcut US       E pseu
    3S  1  0  2.00      1.200      1.600    -1.515413
    3P  2  1  5.00      1.300      1.600    -0.629258

The Na pseudopotential indicates that two 1s electrons are fixed to the core, while the Cl pseudopotential indicates that two 1s, two 2s, and six 2p electrons are fixed to the core. From this, we can calculate the total electron number on Na as (2 + 8.14) = 10.14 and on Cl as (2 + 2 + 6 + 7.86) = 17.86. By subtracting these values from the atomic numbers for Na (11) and Cl (17), we arrive at Bader charge-based ionic charge states of Na_^(0.86+) and Cl_^(0.86-)

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