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In publications, I usually see the statement "A dense k-mesh of 40,000 k-points was used in the Brillouin zone". An example of this is in this paper. But in the paper they have used a k-point grid of 8x8x8 and 9x9x9 for the SCF calculation.

How do they arrive at the number "40000"?

Thanks :)

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  • $\begingroup$ Can you please include a legitimate link to the paper? Your earlier link is not allowed. Please avoid this in the future. Btw, you don't need to upload the full paper, just include a link. Most of us have access to scientific articles. $\endgroup$
    – Xivi76
    Mar 24, 2021 at 20:06
  • $\begingroup$ +1. I've switched the link from SciHub tot he original journal's version at @Xivi76's request. Hydrogen: are you saying that several papers use 40,000 and you don't know why they all use the same number, or are you saying that in this specific case where 40000 was used, you don't know why they chose 40000? $\endgroup$
    – Nike Dattani
    Mar 24, 2021 at 20:15

1 Answer 1

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The computational content of the cited paper includes two parts:

  • DFT calculations (such as band structures) with QE;
  • Transport calculations (such as Seebeck coefficients) with BoltzTraP based on DFT calculations.

The $8 \times 8 \times 8$ and $9 \times 9 \times 9$ are chosen for DFT calculations. However, the dense k-mesh with $40000$ points in the first Brillouin zone is for transport calculation. Why transport calculation needs so dense k-mesh is related to the semiclassical method that BoltzTraP adopted.

You may take a look at the computational method of this paper also calculated with BoltzTraP:

Hope it helps.

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