[QE-users] Dipole corrections (dipfield) and the position of the slab

Christoph Wolf wolf.christoph at qns.science
Fri Apr 27 12:06:16 CEST 2018

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> 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
> 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
> --
> Postdoctoral Researcher
> Center for Quantum Nanoscience, Institute for Basic Science
> Ewha Womans University, Seoul, South Korea
> _______________________________________________
> users mailing listusers at lists.quantum-espresso.orghttps://lists.quantum-espresso.org/mailman/listinfo/users
> --
> 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

Postdoctoral Researcher
Center for Quantum Nanoscience, Institute for Basic Science
Ewha Womans University, Seoul, South Korea
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