20
$\begingroup$

Plane-wave basis sets are useful in the calculation of periodic systems and can be used in combination with pseudopotentials. However, when dealing with surfaces the system is no longer periodic in, at least, one dimension. Moreover, in modeling adsorption processes, it is necessary to include a surface and the molecule attached with a suitable basis set.

As far as I know, a plane-wave basis set should be expanded over the whole simulation box and the periodicity in three dimensions is required. However, some studies on adsorption were made using this type of basis (for instance, J. Phys. Chem. 2009, 113, 9256-9274), although the periodicity is broken. At first glance, it seems like a localized basis set is better for surfaces with molecules attached.

  1. Are plane-waves basis sets really useful for surfaces or adsorption processes?
  2. What advantages (if there are) have plane-waves over localized basis sets in modeling surfaces?
$\endgroup$
7
  • 1
    $\begingroup$ The periodicity is not broken in these cases. Rather, there is artificial vacuum space added in the z dimension so that the surface does not interact with itself. There are still 3D periodic boundary conditions -- but one of the dimensions is largely vacuum space. $\endgroup$ Apr 30, 2020 at 19:43
  • $\begingroup$ @AndrewRosen I agree, can you please post that as an answer? $\endgroup$
    – Cody Aldaz
    Apr 30, 2020 at 23:22
  • $\begingroup$ @CodyAldaz that doesn't consider the part of the molecule, being a localized inhomogeneous system. $\endgroup$
    – Verktaj
    Apr 30, 2020 at 23:25
  • $\begingroup$ @Verktaj I'm not sure what this means. Plane-wave basis sets are still being used. Your molecule will still be modeled in a box with periodic boundary conditions in 3 dimensions, at least in the example you highlighted and indeed in most DFT codes. Care to elaborate further? $\endgroup$ Apr 30, 2020 at 23:26
  • $\begingroup$ @CodyAldaz I understand that a plane wave basis set is uniform across the whole simulation box. In case of a surface alone, plane waves covers the vacuum space also. If we add a molecule, doesn't its inhomogeneity require a larger number of plane waves and thus reduce the computational efficiency? $\endgroup$
    – Verktaj
    Apr 30, 2020 at 23:38

1 Answer 1

13
$\begingroup$

In the example you highlighted and indeed in most plane-wave DFT codes, there is periodicity in all three dimensions including for surface slab calculations. In the case of a surface slab, vacuum space is commonly added in the $z$ dimension. The vacuum space is there so that an adsorbate can bind of course, but it's also there because of the boundary conditions. A vacuum space ensures that the adsorbate-slab complex does not interact with itself over the periodic boundary, provided the vacuum space is large enough. In this way, you're modeling what is effectively a 2D system while still having 3D periodic boundary conditions as part of the DFT calculation. If an interactive example would be helpful, I recommend John Kitchin's DFT ebook, specifically Section 5.

The answer to both your questions is actually, for the most part, one in the same. The reason plane-wave basis sets are so useful is how well they lend themselves to periodic DFT calculations. In this case, you can represent a crystalline or bulk system with much fewer atoms than you would be able to do if you relied on a Gaussian basis set. A given metal might have a primitive unit cell of only a few atoms, whereas modeling the same metal with the same degree of accuracy without finite boundary conditions might require several hundred atoms to prevent edge effects and ensure the system is large enough. A localized, Gaussian basis set is not inherently better than a plane-wave basis set. The latter, however, is naturally better suited for adsorption problems in general.

As a side-note, because it can be helpful for to visualize orbitals in a Gaussian basis set, there are several algorithms that project plane-wave bands to Gaussian type orbitals, such as the periodic NBO code of Dunnington and Schmidt. This is mostly for gaining insight into the electronic structure of the chemical process, rather than a question of accuracy.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .