Hello,

I am trying to calculate bader charge for a perfect graphene sheet with lattice parameter a=8.52168, b=9.84000, c=10.0000. The supercell is a cubic one containing 32 carbon atoms.

Since this is a perfect one in which one carbon atom binds with 3 adjacent atoms and one unused delocalized electron, I expect the bader result to be the charge of 4 for one carbon atom.

Based on my little knowledge about this area including NGX,Y,ZF effect on the accuracy (I 'm not pretty sure though) and with the INCAR settings included below, with increasing the value of them(NGX,Y,ZF),

I tried several times for getting the expected results.

For the first time, I used NGXF= 96, NGYF= 108 and NGZF=112, and then *2 for second trial. The first result gave me a periodical values of charge with difference between atoms was from about 0.12 to about 0.3.

Though the second one was much better than the previous one in my opinion, since the periodicity decreased from 8 to 2 with the charge difference was only 0.0154, that of the 'third' trial with the NGX,Y,ZF value of 3 times compared to the first one was not better at all.

The reason why I thought this way was the periodicity increased once again to 8 and the difference was increased though smaller than those of first trial.

Simply, I need to know some about this bader calculation.

1. I want to figure out what are the criteria that determine the accuracy of this calculation (parameter or flag)

2. I want to know how to set the NGX,Y,ZF and/or NGX,Y,Z for my objective.

3. I want to find whether there would be an error or irrelevant flag in my INCAR settings.

Here is my INCAR details.

Startparameter for this Run:

NWRITE = 2; LPETIM=F write-flag & timer

ISTART = 1 job : 0-new 1-cont 2-samecut

ISPIN = 2; NPAR = 2; LAECHG = T; SYMPREC = 1E-8

Electronic Relaxation 1

ENCUT = 400.00 eV; NELM = 90; NELMIN = 4; EDIFFG = -5E-02; EDIFF = 1E-04 stopping-criterion for ELM

prec = high; RWIGS = 0.863

Ionic Relaxation

NSW = 0 number of steps for IOM; IBRION = -1 ionic relax: 0-MD 1-quasi-New 2-CG

DOS related values:

ISMEAR = 1; SIGMA = 0.05

# ISMEAR = -5

Electronic Relaxation 2

IALGO = 48 algorithm

LDIAG = T sub-space diagonalisation

LREAL = A ; LWAVE = T

# LELF = T; # LPARD = T; # IBAND = 97

# NGX = 100; NGY =100; NGZ =100

NGXF = 198; NGYF = 216; NGZF = 224

Any comments on any part of my question will be very grateful for me.

Thank you very much.

Best,

Denny

## Questions for bader calculation on perfect graphene sheet

**Moderator:** moderators

### Re: Questions for bader calculation on perfect graphene shee

The convergence of charges should converge linearly with the number of grid points. When you have periodic structures with symmetry, the convergence may not be monotonic because of how the FFT grid is aligned with your atomic structure.

We are currently implementing a method which will improve the accuracy without having to go to very fine grids, but for now, the only solution is to look for convergence with respect to the grid density.

We are currently implementing a method which will improve the accuracy without having to go to very fine grids, but for now, the only solution is to look for convergence with respect to the grid density.

### Re: Questions for bader calculation on perfect graphene shee

Thank you very much for the reply. But can you just point out why increasing NGX,Y,ZF more than twice doesn't give me better value than twice of the initial value even though there are more points for FFT grid? (NGXF = 98, NGYF= 108, NGZF= 112) Is it related to NGX,Y,Z values?

### Re: Questions for bader calculation on perfect graphene shee

If the grid errors were random they would drop by a factor of 2 with a doubling of the grid. If your molecule has a symmetry which matches the grid, however, there can be a systematic error so that this factor of 2 reduction will only happen on average.

### Re: Questions for bader calculation on perfect graphene shee

Thank you once again for the answer.

Actually, I'd like to ask about two more things related to NGX,Y,ZF values.

1. I, recently, found that the NGXF values in the INCAR files are different from that of OUTCAR files. When I set it as 198 which is about twice of the initial value (98), it changed to 200 in the OUTCAR file. Similarly, when I set it as 202 and 205, the values in OUTCAR was same at 210. It seems hard to accept that there are random difference between them. Since the change of NGXF in OUTCAR makes difference in the bader charge output (ACF.dat), I would like to know how can I find the reason of this phenomena.

2. Unfortunately, due to the lack of my knowledge, it was hard to understand that my molecule(?) might have a symmetry matching with the grid (set by NGX,Y,ZF).

Can the symmetry make error or unreliable results for finer grid? (NGXF = 298, NGYF = 324, NGZF = 336 was strangely gave me result which was worse compared to the case of NGXF = 198 (actually 200 in OUTCAR as stated in question 1.), NGYF = 216 and NGZF = 224)

Sorry for the long questions and thank you very much in advance.

Best,

Denny

Actually, I'd like to ask about two more things related to NGX,Y,ZF values.

1. I, recently, found that the NGXF values in the INCAR files are different from that of OUTCAR files. When I set it as 198 which is about twice of the initial value (98), it changed to 200 in the OUTCAR file. Similarly, when I set it as 202 and 205, the values in OUTCAR was same at 210. It seems hard to accept that there are random difference between them. Since the change of NGXF in OUTCAR makes difference in the bader charge output (ACF.dat), I would like to know how can I find the reason of this phenomena.

2. Unfortunately, due to the lack of my knowledge, it was hard to understand that my molecule(?) might have a symmetry matching with the grid (set by NGX,Y,ZF).

Can the symmetry make error or unreliable results for finer grid? (NGXF = 298, NGYF = 324, NGZF = 336 was strangely gave me result which was worse compared to the case of NGXF = 198 (actually 200 in OUTCAR as stated in question 1.), NGYF = 216 and NGZF = 224)

Sorry for the long questions and thank you very much in advance.

Best,

Denny