[QE-users] Dipole corrections (dipfield) and the position of the slab
thomas.brumme at uni-leipzig.de
Fri Apr 27 12:18:51 CEST 2018
in the end it (of course) doesn't (and shouldn't) matter where you
You can also center the system at z=0 and the dipole at z=0.5. Just take
that the distance between system and dipole is large enough.
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,
> On Fri, Apr 27, 2018 at 10:59 AM, Thomas Brumme
> <thomas.brumme at uni-leipzig.de <mailto:thomas.brumme at uni-leipzig.de>>
> 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
> 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:
> 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.
> 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.
>> 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!
>> PS: I saw the related discussion, but it does not really answer
>> this I think...
>> Postdoctoral Researcher
>> Center for Quantum Nanoscience, Institute for Basic Science
>> Ewha Womans University, Seoul, South Korea
>> users mailing list
>> users at lists.quantum-espresso.org
>> <mailto:users at lists.quantum-espresso.org>
> 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:thomas.brumme at uni-leipzig.de <mailto:thomas.brumme at uni-leipzig.de>
> 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
Tel: +49 (0)341 97 36456
email: thomas.brumme at uni-leipzig.de
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