Dear Chris,
in the end it (of course) doesn't (and shouldn't) matter where you
center what.
You can also center the system at z=0 and the dipole at z=0.5. Just take
care
that the distance between system and dipole is large enough.
Regards
Thomas
On 27.04.2018 12:06, Christoph Wolf wrote:
Dear Thomas,
thank you for your detailed reply!
If I understood this correctly, the ideal situation would be to have
the slab in the center of the cell
Atom xx yy 0.5
(in crystal coordinates)
to center the dipole at 0% (=100% due to PBC) a reasonable choice
would be emaxpos=0.95 and eopreg=0.10. In the dipole example they
located the atoms around z=0 of the cells and put the dipole close to
the center of the cell.
Muchas Gracias/Vielen Dank from Spain,
Christoph
On Fri, Apr 27, 2018 at 10:59 AM, Thomas Brumme
<[email protected] <mailto:[email protected]>>
wrote:
Dear Chris,
both planes of the dipole (the one at emaxpos and the one with the
opposite charge at emaxpos+eopreg) have to be in the vacuum region.
In fact, there should be enough space such that the wavefunctions are
essentially zero at the dipole planes. However, if the dipole is
too large,
charge can spill into the vacuum region as plane waves are not
localized
on the system and the charge could be in a lower energy state at the
dipole. In other words. don't use 50 Angstrom of vacuum as this
will lead
to a very low minimum in the total potential at the dipole.
See also this paper:
https://journals.aps.org/prb/abstract/10.1103/PhysRevB.85.045121
<https://journals.aps.org/prb/abstract/10.1103/PhysRevB.85.045121>
In this paper charged systems are discussed but similar things
apply to
the dipole correction. Thus, if your system is centered at 50% of the
cell, center the dipole at zero and converge things with
increasing the
size along z.
Regards
Thomas
On 26.04.2018 14:00, Christoph Wolf wrote:
Dear all,
After trying for a few days I am still a bit puzzled by the
"proper application" of the dipole correction. To test this I
have made a sheet of graphene added hydrogen below and fluorine
above. I then apply the following corrections:
tefield = .true.
dipfield =.true.
and
eamp = 0.00
edir = 3
emaxpos = 0.80 !(=16 Angstrom)
eopreg = 0.10 ! (=2 Angstrom)
The cell is 20 A in total. As I shift the layer from 0% of the
cell to 50% cell (whilst keeping above emaxpos at 80% and eopreg
at 10% of the cell) the Fermi level shifts slightly (~0.2-0.5 eV
difference) and the electrostatic potential (pp.x plot num 11 and
then planar average using average.x as in the work-function
example) is only "flat" in the vacuum region when the sample is
about 3A from the bottom of the cell (i.e. the z coordinate of
atoms has to be larger than 3 A).
Reading the pw.x input I was under the impression that only
emaxpos has to fall into the vacuum but is there also a "rule of
thumb" for eopreg?
Thanks in advance for your help!
Best,
Chris
PS: I saw the related discussion, but it does not really answer
this I think...
http://qe-forge.org/pipermail/pw_forum/2009-December/089951.html
<http://qe-forge.org/pipermail/pw_forum/2009-December/089951.html>
--
Postdoctoral Researcher
Center for Quantum Nanoscience, Institute for Basic Science
Ewha Womans University, Seoul, South Korea
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Dr. rer. nat. Thomas Brumme
Wilhelm-Ostwald-Institute for Physical and Theoretical Chemistry
Leipzig University
Phillipp-Rosenthal-Strasse 31
04103 Leipzig
Tel: +49 (0)341 97 36456
email:[email protected] <mailto:[email protected]>
--
Postdoctoral Researcher
Center for Quantum Nanoscience, Institute for Basic Science
Ewha Womans University, Seoul, South Korea
--
Dr. rer. nat. Thomas Brumme
Wilhelm-Ostwald-Institute for Physical and Theoretical Chemistry
Leipzig University
Phillipp-Rosenthal-Strasse 31
04103 Leipzig
Tel: +49 (0)341 97 36456
email: [email protected]
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