# Is it okay to take a cubic Monkhorst-Pack k-mesh for a non-cubic supercell in a DFT calculation?

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 $$5$$.

• 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. – Camps Aug 8 '20 at 19:11
• 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. – Hitanshu Sachania Aug 8 '20 at 19:16