Re: [Pw_forum] How to read the output of pp.x?

[Jing Wang]

Thanks for the very detailed explanation of the output file. I read the
format explanation in the link you gave again and again and I still can't
figure out what the five columns are. For example in the main part of the
data file (shown as below), if the first three columns are the coordinates
in the real r space with the unit in alat, then what are the left two
columns recording? Which column is the potential? How does the datafile
represent the Vscf(r) function in real space?
.
.
3.248268856E+00  3.159870688E+00  2.515974539E+00  6.186557012E-01
-2.154636042E+00
 -3.866568628E+00 -3.274420891E+00 -2.365272143E+00 -2.086628582E+00
-1.945051309E+00
 -1.918992011E+00 -1.927145835E+00 -1.943940931E+00 -1.943940931E+00
-1.927145835E+00
.
.

Thank you.

Jing Wang
Dept. of Physics,
Georgia Tech, GA

2014-12-19 11:42 GMT-05:00 Paolo Giannozzi :
>
> On Wed, 2014-12-17 at 10:59 -0500, Jing Wang wrote:
>
> > I'm really confused about how can I read the local potential out
> > from the scf calculation. I used pp.x, plot_num=1 to extract the scf
> > potential of the the system, however I can't really understand the
> > output?
>
> the "output" is not meant to be read by humans but by other codes.
>
> > Should it show the potential in cofficients of G-vectors?
>
> no, it is in real space, in the format explained here:
>
> http://www.quantum-espresso.org/wp-content/uploads/Doc//developer_man/developer_man.html#SECTION0008
>
> > Head of the output is like this:
>
> >   25  25 100  25  25 100   2   1
>
> FFT maximum dimensions and grids (nr1x nr2x nr3x nr1 nr2 nr3),
> number of atoms, number of species
>
> >  44.6490  0.  4.
> >  0.  0.  0.
>
> ibrav, celldm(1-3) as given in scf input
>
> >   153.29125753344.00   70.00 1
>
> Gmax in (2pi/a)^2 units, dual (ecutrho=dual*ecutwfc), cutoff (Ry)
>
> >1   C 4.00
> >1   0.00.00.01
> >2   0.5   -0.2886751350.01
>
> atomic species and atomic positions. The plotted quantity in real space
> (array length nr1x*nr2x*nr3) follows
>
> Paolo
>
> >   3.248268856E+00  3.159870688E+00  2.515974539E+00  6.186557012E-01
> > -2.154636042E+00
> >  -3.866568628E+00 -3.274420891E+00 -2.365272143E+00 -2.086628582E+00
> > -1.945051309E+00
> >  -1.918992011E+00 -1.927145835E+00 -1.943940931E+00 -1.943940931E+00
> > -1.927145835E+00
> >  -1.918992010E+00 -1.945051309E+00 -2.086628581E+00 -2.365272143E+00
> > -3.274420890E+00
> >  -3.866568627E+00 -2.154636041E+00  6.186557014E-01  2.515974539E+00
> > 3.159870688E+00
> >   3.159870688E+00  3.159870688E+00  2.802339131E+00  1.452305376E+00
> > -1.033031298E+00
> >  -3.409639826E+00 -3.654745566E+00 -2.575759384E+00 -2.021392120E+00
> > -1.762140769E+00
> >  -1.637123367E+00 -1.599819700E+00 -1.588050229E+00 -1.591678304E+00
> > -1.588050228E+00
> >  -1.599819700E+00 -1.637123366E+00 -1.762140768E+00 -2.021392119E+00
> > -2.575759383E+00
> >  -3.654745566E+00 -3.409639826E+00 -1.033031298E+00  1.452305376E+00
> > 2.802339131E+00
> >   2.515974539E+00  2.794920980E+00  2.515974539E+00  1.452305376E+00
> > -6.088327143E-01
> >  -2.976954364E+00 -3.787496708E+00 -2.827784418E+00 -2.048224903E+00
> > -1.714196738E+00
> >  -1.488710208E+00 -1.380390580E+00 -1.313398615E+00 -1.296957599E+00
> > -1.296957599E+00
> >  -1.313398615E+00 -1.380390580E+00 -1.488710208E+00 -1.714196738E+00
> > -2.048224903E+00
> >  -2.827784418E+00 -3.787496708E+00 -2.976954363E+00 -6.088327142E-01
> > 1.452305376E+00
> >   6.186557014E-01  1.426203806E+00  1.426203806E+00  6.186557014E-01
> > -1.033031298E+00
> >  -2.976954363E+00 -3.792558359E+00 -2.995197487E+00 -2.094921250E+00
> > -1.698073457
> >
> >
> >
> > Thanks.
> >
> >
> > Jing Wang
> >
> > Dept. of Physics
> >
> > Georgia Tech, GA, U.S.A
> >
> > ___
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> > http://pwscf.org/mailman/listinfo/pw_forum
>
> --
>  Paolo Giannozzi, Dept. Chemistry,
>  Univ. Udine, via delle Scienze 208, 33100 Udine, Italy
>  Phone +39-0432-558216, fax +39-0432-558222
>
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>
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[Pw_forum] Computing Vacuum potential

[Elliot Menkah]

Hello Everyone,
I'm trying to calculate the work function of a pure slab as such I'm
computing for the vacuum potential.
When I run my scf calculation and compute the vacuum potential, I get
very small values for the vacuum potential. I get a vacuum potential of
about 0.8 eV and a fermi energy of -0.234. which seems not to make sense.

Please finds attached the input file for the self consistent field(scf) 
calculation.

Could there be anything wrong with the system please.
Do I have to include or account for dipole correction? And how please.
All contributions are welcome.
Thank you.


Warm Regards,

Elliot

-- 
Elliot S. Menkah
Research Student - Computational Chemistry/ Computational Material Science
Theoretical and Computational Chemistry
Dept. of Chemistry
Kwame Nkrumah UNiversity of Sci. and Tech.
Kumasi
Ghana

Tel: +233 243-055-717

Alt Email: elliotsmen...@gmail.com
   elliotsmen...@hotmail.com


   title = 'Ni-1x1-100-2L-METADISE-surf-full-rlx' ,
   calculation = 'scf' ,
   restart_mode = 'from_scratch' ,
   outdir = './tmp' ,
   pseudo_dir = '/home/mmc14/pseudo' ,
   prefix = 'ni-100-1x1-2L-metaD-full-rlx',
   tstress = .true. ,
   tprnfor = .true. ,
/
 
   ibrav = 0,
   celldm(1) = 1.88972599,
   nat = 4,
   ntyp = 1,
   ecutwfc = 60,
   ecutrho = 650,
   occupations='smearing',
   smearing='mp',
   degauss=0.005,
   nbnd=50,
   nspin=2,
   starting_magnetization(1)=0.5,
/
 
conv_thr = 1.0D-6,
mixing_mode = 'local-TF'
mixing_beta = 0.2,
/
 
/
ATOMIC_SPECIES
Ni 58.6934   Ni.pbe-nd-rrkjus.UPF
ATOMIC_POSITIONS (alat)
Ni   3.522092393   3.522092393   0.059971412
Ni   1.761046196   1.761046196   0.059979212
Ni   3.522092393   1.761046196   1.701070884
Ni   1.761046196   3.522092393   1.701070884
K_POINTS automatic
  5 5 1   0 0 0
CELL_PARAMETERS
 +3.5220923930  +0.00  +0.00
 +0.00  +3.5220923930  +0.00
 -0.00  -0.00 +15.5220923930
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Re: [Pw_forum] How to read the output of pp.x?

[Paolo Giannozzi]

On Wed, 2014-12-17 at 10:59 -0500, Jing Wang wrote:

> I'm really confused about how can I read the local potential out 
> from the scf calculation. I used pp.x, plot_num=1 to extract the scf
> potential of the the system, however I can't really understand the
> output? 

the "output" is not meant to be read by humans but by other codes.

> Should it show the potential in cofficients of G-vectors?

no, it is in real space, in the format explained here:
http://www.quantum-espresso.org/wp-content/uploads/Doc//developer_man/developer_man.html#SECTION0008

> Head of the output is like this:

>   25  25 100  25  25 100   2   1

FFT maximum dimensions and grids (nr1x nr2x nr3x nr1 nr2 nr3),
number of atoms, number of species

>  44.6490  0.  4. 
>  0.  0.  0.

ibrav, celldm(1-3) as given in scf input

>   153.29125753344.00   70.00 1

Gmax in (2pi/a)^2 units, dual (ecutrho=dual*ecutwfc), cutoff (Ry)

>1   C 4.00
>1   0.00.00.01
>2   0.5   -0.2886751350.01

atomic species and atomic positions. The plotted quantity in real space
(array length nr1x*nr2x*nr3) follows

Paolo

>   3.248268856E+00  3.159870688E+00  2.515974539E+00  6.186557012E-01
> -2.154636042E+00
>  -3.866568628E+00 -3.274420891E+00 -2.365272143E+00 -2.086628582E+00
> -1.945051309E+00
>  -1.918992011E+00 -1.927145835E+00 -1.943940931E+00 -1.943940931E+00
> -1.927145835E+00
>  -1.918992010E+00 -1.945051309E+00 -2.086628581E+00 -2.365272143E+00
> -3.274420890E+00
>  -3.866568627E+00 -2.154636041E+00  6.186557014E-01  2.515974539E+00
> 3.159870688E+00
>   3.159870688E+00  3.159870688E+00  2.802339131E+00  1.452305376E+00
> -1.033031298E+00
>  -3.409639826E+00 -3.654745566E+00 -2.575759384E+00 -2.021392120E+00
> -1.762140769E+00
>  -1.637123367E+00 -1.599819700E+00 -1.588050229E+00 -1.591678304E+00
> -1.588050228E+00
>  -1.599819700E+00 -1.637123366E+00 -1.762140768E+00 -2.021392119E+00
> -2.575759383E+00
>  -3.654745566E+00 -3.409639826E+00 -1.033031298E+00  1.452305376E+00
> 2.802339131E+00
>   2.515974539E+00  2.794920980E+00  2.515974539E+00  1.452305376E+00
> -6.088327143E-01
>  -2.976954364E+00 -3.787496708E+00 -2.827784418E+00 -2.048224903E+00
> -1.714196738E+00
>  -1.488710208E+00 -1.380390580E+00 -1.313398615E+00 -1.296957599E+00
> -1.296957599E+00
>  -1.313398615E+00 -1.380390580E+00 -1.488710208E+00 -1.714196738E+00
> -2.048224903E+00
>  -2.827784418E+00 -3.787496708E+00 -2.976954363E+00 -6.088327142E-01
> 1.452305376E+00
>   6.186557014E-01  1.426203806E+00  1.426203806E+00  6.186557014E-01
> -1.033031298E+00
>  -2.976954363E+00 -3.792558359E+00 -2.995197487E+00 -2.094921250E+00
> -1.698073457
> 
> 
> 
> Thanks.
> 
> 
> Jing Wang
> 
> Dept. of Physics
> 
> Georgia Tech, GA, U.S.A
> 
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-- 
 Paolo Giannozzi, Dept. Chemistry, 
 Univ. Udine, via delle Scienze 208, 33100 Udine, Italy
 Phone +39-0432-558216, fax +39-0432-558222 

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Re: [Pw_forum] about C2/c point group symmetry of a monoclinic lattice.

[Andrea Dal Corso]

On Thu, 2014-12-18 at 19:29 +, Mutlu COLAKOGULLARI wrote:
> Dear Carlo, Andrea and Paolo;
>   The taking responds and advices from people who has experienced makes 
> me happy.
>   As Paolo and Carlo also said, cif2qe.sh is not working properly, (at 
> least for this case). Therefore, I used the other codes for double cross 
> check: pymatgen and cif2cell. The conventional cell has 64 atoms whereas its 
> primitive has 32 atoms. 
>   Pymatgen is very proper for POSCAR output because it takes symmetry 
> analysis depend on Curtarolo-Setyawan's paper 
> [http://arxiv.org/abs/1004.2974v1] (MCLC5 - it gives also gamma angle, for 
> this case)...so, it is not matching with QE's ibrav settings depend on cell 
> vectors. Cif2Cell code gives the exactly same CELL_PARAMETERS card with 
> Quantum-ESPRESSO's ibrav=-13...that is the reason why I chose Cif2Cell code.
>   Andrea, it was my fault. I had been complicated among space groups(2/m 
> for sg:12 and 2/C for sg:15) and Herman-Mauguin notations(2/m for space 
> groups 12-15). You said that ”Using ibrav=-13 it is not necessary to add 
> these atoms in the list of atoms inside the unit cell." Andrea, you mean that 
> could I decrease the number of atoms in primitive cell if they have 
> symmetrized by last 4 symmetry operations in this case? By this way, the cell 
> lost half of atoms inside which is good news for computational resources. 
>   
I think so. As far as I understand there are two possibilities:
1) You apply all the 8 symmetry operations and use ibrav=-12 (simple
monoclinic) obtaining 64 atoms.
2) You apply only the first four symmetry operations and use ibrav=-13
(one base centered monoclinic). Note however that in this case you have
to refer the positions to the base centered monoclinic axes, while the
positions that you obtain applying the symmetry are still referred to
the simple monoclinic cell. The base centered monoclinic cell has one
half the number of atoms of the simple monoclinic cell.

In no cases however QE will find 8 symmetry operations. In the first
case because fractional translations will be disabled (the cell  is
actually a supercell). In the second case because there are only 4.
It is not a question of notations, the point group for the space group
C2/c is C_2h (that has 4 operations), there is no way to obtain a larger
point group.

HTH,

Andrea



> With my best wishes,
>Mutlu.
> 
> --Dr. Mutlu COLAKOGULLARITrakya 
> Universitesi Fen FakultesiFizik Bolumu22030 Merkez-EDİRNE
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Andrea Dal CorsoTel. 0039-040-3787428
SISSA, Via Bonomea 265  Fax. 0039-040-3787249
I-34136 Trieste (Italy) e-mail: dalco...@sissa.it


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