12
$\begingroup$

This is from the entry for LORBIT in the VASP Wiki.

For LORBIT >= 11 and ISYM = 2 the partial charge densities are not correctly symmetrized and can result in different charges for symmetrically equivalent partial charge densities.

For example, in the calculations of $\ce{Ni_3Al}$ with spin-orbit coupling (ISPIN = 2), I got different charges and magnetic moments for each of the three $\ce{Ni}$ atoms when I used symmetry (ISYM = 1) whereas with symmetry off (ISYM = 0) the partial charges and magnetic moments on each $\ce{Ni}$ atom were equal. In both cases though, the total charge and magnetic moment remained the same.

That brings me to the electronic Density of States (DoS), specifically partial DoS. I read on the VASP forum that ISYM = 0 is required for a proper projection of the DoS on each atom. That exact page is inaccessible at the moment, but you can get a sense of the idea from the discussion on this page as well.

A general workflow to calculate electronic DoS using VASP:

  1. Converge and relax the system as usual.
  2. Increase the # of k-points, and turn ICHARG = 11 so that the charge density from step 1 (written in the CHGCAR file) is used. Also, no relaxation in this step with IBRION = -1 and NSW = 0.

The doubt: Should I take ISYM = 0 in step 2 or during the last relaxation in step 1? The latter gets too expensive. I also have a doubt if there is an intermediate step involved (let's call it step 1.5):

1.5. After proper convergence and relaxation of the system, increase the # of $k$-points, and do one final relaxation.

$2^{nd}$ doubt: If step 1.5 is required then do we turn ISYM = 0 for this one or not? An ISIF = 3 type of complete relaxation with a higher # of $k$-points and ISYM = 0 is extremely expensive.

Any guidance/discussion is welcome.

Improve this question
$\endgroup$
1

2 Answers 2

Reset to default
7
$\begingroup$

Based on the comments I shared above, it seems that the quality of the charge density is fine. It is just the projection onto atoms that may not be done correctly. If this is indeed the case, then you don't need to re-relax the structure (Step #1). You should be able to read in the charge density and set ISYM=0 (symmetry disabled) and be okay. Disclaimer: Only one way to find out -- give it a try! If it's expensive, drop the $k$-points just for a simple test.

As for your proposed Step #1.5, that's a separate question altogether. The answer to that question is whether increasing the $k$-points will notably alter the geometry. If increasing the $k$-points keeps essentially the same geometry, then you can likely increase it to get a more accurate density of states without worrying that the structure is not a local minimum in the potential energy surface. This will entirely depend on the difference in $k$-points and the system of interest.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Just wondering if going beyond the converged # of k-points would have any effect on the geometry? By converged # of k-points, I mean converged based on the internal energy outputted by a DFT code. $\endgroup$
    – Hitanshu Sachania
    Commented Aug 24, 2020 at 11:48
  • 1
    $\begingroup$ It is always possible that altering the number of $k$-points can alter the geometry. However, if the number of $k$-points is large enough that the energy is essentially converged, the forces are likely not to differ such that the structure is largely (or entirely) unchanged. $\endgroup$
    – Andrew Rosen
    Commented Aug 24, 2020 at 14:49
  • $\begingroup$ Yup, it is in situations like these that questions like mattermodeling.stackexchange.com/questions/2015/… come to mind. $\endgroup$
    – Hitanshu Sachania
    Commented Aug 24, 2020 at 17:07
  • 1
    $\begingroup$ A better question to ask yourself is does it really matter if your structure is not exactly relaxed? For a DOS, if you are extremely close -- but not at -- the local minimum in the potential energy surface, it very well could be just fine. $\endgroup$
    – Andrew Rosen
    Commented Aug 24, 2020 at 17:51
  • $\begingroup$ Interesting. Because my knowledge in these areas seems quite limited now, are structures around a minimum in the PES similar in symmetry? $\endgroup$
    – Hitanshu Sachania
    Commented Aug 24, 2020 at 18:18
8
$\begingroup$

The typical calculation flow for the density of states is:

  • Geometric relaxation to obtain the lowest-energy structure (CONTCAR) [1relax] (For your system, you should do the spin-polarized calculation by setting ISPIN=2).
  • Using the relaxed structure to perform the electronic self-consistent calculation to obtain converged charge density [2scf].
  • Using the converged charge density to calculate the density of states [3dos].

At the final step, you can increase the k sampling to obtain a smooth curve.

For the geometric relaxation and self-consistent calculation, you can use the default value of ISYM.

As for the effect of ISYM on the partial density of states, you can test different ISYM with LORBIT=1 by plotting the DODCAR. Usually, you can just use the default value of ISYM, which will give you reasonable results.

For your initial observation:

In the spin-polarized coupled (ISPIN=2) calculations with Ni3Al, I got different charges and magnetic moments on each of the three Ni atoms when I kept symmetrization option ISYM on i.e. ISYM = 1. On turning it off (ISYM = 0), partial charges and magnetic moments on each ๐‘๐‘– atom became equal. In both cases though, total charge and magnetic moment remained the same.

Only the total charge and magnetic moment are meaningful. The comparison for the charges and magnetic moments on each Ni atom for different ISYM is meaningless. You can compare the charges and magnetic moments for different atoms with the same ISYM value.

Improve this answer
$\endgroup$
2
  • $\begingroup$ What is the point of the third step? Why not take the DOS from the second step? You get the DOS from the self-consistent static calculation anyway, right? $\endgroup$
    – Andrew Rosen
    Commented Aug 22, 2020 at 22:00
  • 3
    $\begingroup$ Short answer: for saving time. Step three will use the much denser k sampling to plot smooth DOS based on the converged charge density from step two with appropriate k sampling. $\endgroup$
    – Jack
    Commented Aug 23, 2020 at 0:00

You must log in to answer this question.

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