# What is the Quantum ESPRESSO equivalent for VASP's k-mesh sampling resolution?

$$\require{mediawiki-texvc}$$ 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 But your title says 0.03 and the question says 0.02. Also, surely you can choose the mesh size in both programs? Jul 29 at 6:03
• Thank you Sir, I have corrected it out in the title now. Jul 29 at 6:05

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.