$\require{mediawiki-texvc}$ Many authors in their research papers involving VASP, report for the k-mesh sampling with a statement like this:

"Gamma-Centered mesh with a resolution of $2\pi \times 0.02 ~~ \pu{\AA}^{-1}$."

Could you please tell me what its Quantum ESPRESSO equivalent is?

  • 1
    $\begingroup$ +1 But your title says 0.03 and the question says 0.02. Also, surely you can choose the mesh size in both programs? $\endgroup$ Jul 29, 2021 at 6:03
  • $\begingroup$ Thank you Sir, I have corrected it out in the title now. $\endgroup$
    – astha
    Jul 29, 2021 at 6:05

1 Answer 1


This can be understood from solid-state physics.

First of all, for the investigated system with lattice constants $\vec{a}_1,\vec{a}_2$ and $\vec{a}_3$ (length), the reciprocal lattice vectors (1/length) can be calculated as: \begin{align} \vec{b}_1 & = 2\pi \dfrac{\vec{a}_2\times\vec{a}_3}{\Omega} \\ \vec{b}_2 & = 2\pi \dfrac{\vec{a}_3\times\vec{a}_1}{\Omega} \\ \vec{b}_3 & = 2\pi \dfrac{\vec{a}_1\times\vec{a}_2}{\Omega} \\ \end{align} in which $\Omega=\vec{a}_1\cdot[\vec{a_2}\times\vec{a}_3]$ is the volume of the investigated system.

Then, back to the Kohn-Sham equation solvers like QE and VASP, they will do many integrals in reciprocal space with numerical techniques, which means that you should sample the reciprocal space along $\vec{b}_1, \vec{b}_2$ and $\vec{b}_3$. For example, if you adopt a $N_1 \times N_2 \times N_3$ k-mesh, you can calculated the resolution as \begin{align} \dfrac{\vec{b}_1}{N_1}, \qquad \dfrac{\vec{b}_2}{N_2}, \qquad \dfrac{\vec{b}_3}{N_3}, \end{align} respectively. Basically, this just tell you how they are sampling the simulated cell in reciprocal space.

Hope it helps.


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