[QE-users] [EXT] Total energy of charged slabs

[Andreussi, Oliviero]

Hi Zhenyao,

I am writing to give some remarks on simulating negatively charged slabs, as I am afraid that the correct answer requires to know what the package “sxdefectalign2d” does exactly (which I don’t).

A charged slab that is infinitely periodic along x and y must have an infinite energy. It is not just a matter of interactions with periodic replicas, even if you remove the periodicity along z, this is not enough to make your energy finite. If you correct for PBC on the perpendicular direction, the energy should still be a function of the cell size, as a uniformly charged slab generates a constant field in space, so the electrostatic energy of the system is proportional to the cell volume. If the charged defect is only in the unit cell, but the periodic cells along x and y have no charged defects, than the energy is finite, but I am not sure if this is what “sxdefectalign2d” does, as it seems to me kind of tricky to implement.

If what you wanted to study is an infinitely periodic charged slab (say because it is connected to a potential difference), in order for the energy to be finite and well defined you need to have some compensating charge somewhere. If you are modeling the interface of a material in an electrochemical cell, one possibility is to include a model for ionic diffuse layer in the space above/below your slab. This can be done with Environ (www.quantum-environ.org<http://www.quantum-environ.org>), which is a library for modeling continuum embeddings coupled with QE.

Standard DFT (e.g. PBE) suffers from self-interaction errors, among other approximations. As a result, in negative systems you may have that the extra electron in an anion is not bound, i.e. it’s orbital energy is greater than zero. Thus, the stable configuration for the extra electron would be delocalized over the entire cell. If your simulation cell is small, the free electron states may be slightly higher in energy than your band energy and the extra electron could still converge, with a positive band energy. This is what happens in quantum chemistry codes when you use DFT to simulate anions: having localized basis sets force the extra electron to stay on the atoms, instead of wandering around as a free electron.

I don’t know if this helps, but I would suggest to be particularly careful with charged slabs.

Best,

Oliviero Andreussi
--
Assistant Professor
Department of Physics
University of North Texas
Email: oliviero.andreussi at unt.edu<mailto:oliviero.andreussi at unt.edu>
Phone: +1-(940)-369-5316
Skype: olivieroandreussi
Web: https://www.materialab.org

On Aug 16, 2021, at 4:30 PM, Zhenyao Fang <zhenyaof at sas.upenn.edu<mailto:zhenyaof at sas.upenn.edu>> wrote:

Dear pwscf users,

I am performing some calculations on negatively charged slabs (with one additional electron), and I’m mainly focusing on the energy difference between charged system and neutral system. With ions fixed, I found that the charged system has a higher energy than the neutral system, instead of being lower.

The energy difference dE = E(q=-1) - E(q=0) is around 3.944 eV. The LUMO position of the neutral slab is 3.462 eV, and the vacuum level for the neutral slab is 7.654 eV. Besides, by inspecting the charge difference plots, the additional charge is localized inside the slab. I understand that there could be problems about jellium background, so I used the package “sxdefectalign2d” to correct the interactions due to periodic images, and the total energy of the charged slab remains the same as the size of vacuum increases.

I’m quite confused by the fact that the total energy would increase with one more electron. Since the additional charge is mainly inside the slab, it should lower the energy and be stabilised inside the slab. Therefore, I was wondering how I can reconcile these controversies. Besides, another related question is how we define the zero energy in QE? In other words, does LUMO larger or smaller than zero possibly imply the stability of the system?

I am looking forward to your replies. Thanks in advance.

Best,
Zhenyao Fang





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