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During a convergence test of (MoS2 2x2x1) monolayer using Quantum ESPRESSO, I tried several K-point Monkhorst-Pack grids. According to my understanding, the calculation time and computational cost should decrease with decreasing K-points density.

However as I tried 2x2x1 K-point grid, the calculation was found to be not converging (it stopped after electron_maxstep 80). Upon changing K-points to 3x3x1, I found that the calculation is able to converge within few iterations (~17 iterations). Similarly, It converged for (4x4x1), (5x5x1), and (6x6x1) grids.

My question: what is the physical reason behind this non-convergence at this particular K-point?

For your reference, the following are my basic input and output files:

Input file input.in

output file for 2x2x1 sampling output.out

output file for 3x3x1 sampling output.out

output file for 4x4x1 sampling output.out

output file for 5x5x1 sampling output.out

K-point convergence data summarized conv.ods

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  • $\begingroup$ Please see this: mattermodeling.meta.stackexchange.com/q/393/5, please also fix the grammar. $\endgroup$ Mar 31 at 10:51
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    $\begingroup$ Are you aware that the letter "i" needs to be capitalized when used on its own for the word that you used in your question? $\endgroup$ Mar 31 at 11:44
  • $\begingroup$ I do not know the true cause as of yet, but when you have such reality-destroying situations, you should simplify the problem as much as you can. For example, you should consider only the primitive unit cell, and let QE fill in the cell parameters as much as you can. i.e. insert only the lattice parameter, not define the cell. You should also set one of the Mo to be the origin and the S to be positive and negative z, i.e. maximise symmetry. Your atomic positions should be machine precision 1/3 and 1/6 and so forth. $\endgroup$ Mar 31 at 12:17
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    $\begingroup$ And your system is showing that there is no magnetisation. You should switch that off to save on computation time and resources. $\endgroup$ Mar 31 at 12:22
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    $\begingroup$ Every k point that is in (221) is in (441) and in (661), so that it is more likely to just be a fluke in the computation. I'm trying to get you to get (221) to work. $\endgroup$ Mar 31 at 12:44

1 Answer 1

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Yes, if you have fewer k points, there is less to calculate, so one iteration will be faster.

However, there is also another effect: if your discretization is poor, the minimization of the energy becomes ill-behaved and you end up having to spend many more iterations to get the self-consistent field problem to converge.

It seems to me that the issue here is that you don't have a sufficiently good description of the k-point dependence with the $2\times2\times1$ grid.

In a periodic calculation, you have to do integrals over the first Brillouin zone. If you have a too small k-point grid, this integral will be badly approximated. To extract any sort of observables from your calculation, you need to converge the k-point grid such that the total energy is reasonably converged to the thermodynamic limit. Since you've only attached one converged calculation, it is hard to judge how far you are from the thermodynamic limit.

Addendum: as pointed out by Tyler Sterling in the comment below, a converged energy doesn't mean that the property you want to extract is also converged, so you may want to also converge the pressure and forces, too, for example.

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  • $\begingroup$ your answer makes sense .Thank you. I have attached other converged calculations too. $\endgroup$ Apr 1 at 13:41
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    $\begingroup$ +1 But I've also noticed that for a k-grid that converges total energy to my satisfaction, forces and pressure are sometimes still undergonverged. I usually also converge k-points wrt. to pressure and forces too! $\endgroup$ Apr 1 at 20:08
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    $\begingroup$ @TylerSterling thanks for the addendum; indeed, different properties have different sensitivity to the wave function, and it is always important to converge the calculation for the property you actually want at the end! $\endgroup$ Apr 2 at 7:51

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