Assume that the numbers of k-points in each direction are sufficient.

Example - Lengths of lattice vectors: $2\: 2\: 4$

Required k-mesh: $9\times 9\times 5$

Would a $9\times 9\times 9$ k-mesh work? Will it affect the calculation in any way?

Intuitively, I would think it wouldn't affect the calculation because $9$ in the 3rd lattice vector direction is still higher than the required $5$.

  • $\begingroup$ How do you know that $9\times9\times5$ is the required k-mesh? Normally, you do a convergence test in order to determine the k-values. $\endgroup$ – Camps Aug 8 '20 at 19:11
  • $\begingroup$ Hi @Camps, the question was hypothetical. I assumed since the last vector is twice as long as the first two, it would require half the # of k-points for convergence. $\endgroup$ – Hitanshu Sachania Aug 8 '20 at 19:16

It will obviously affect the calculation time, which would be the only reason I would discourage you based on my knowledge. It should not harm anything to continue with a higher symmetry k-mesh, but I believe there are problems with a lower symmetry k-mesh.

Lattice of (2, 2, 4) can use a k-mesh of (9, 9, 5)

Lattice of (4, 4, 4) should probably not use a mesh of (9, 9, 5)

I am by no means an expert in this topic though. I am just aware of warnings I have seen while running different codes.

  • $\begingroup$ Time is definitely an effect, but nothing else should go wrong I suppose. Going by the example in the question a lattice vector of length 4 gives converged results for a mesh size of 5 in its direction, so wouldn't 9x9x5 still work for a cell with lattice = 4 4 4 (ideally, it needs only 5x5x5)? $\endgroup$ – Hitanshu Sachania Aug 8 '20 at 19:26
  • $\begingroup$ +1 for such a rapid response! It could easily become a Hot Network Question: chat.stackexchange.com/rooms/109935/hot-network-questions- $\endgroup$ – Nike Dattani Aug 8 '20 at 19:28

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