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Reference: Density Functional Theory: A Practical Introduction, David Sholl, Janice A. Steckel, Chapter 3, Page No. 59

A useful rule of thumb is that calculations that have similar densities of k points in reciprocal space will have similar levels of convergence. This rule of thumb suggests that using 8x8x2k points would give reasonable convergence, where the three numbers refer to the number of k points along the reciprocal lattice vectors $b_1$ , $b_2$ , and $b_3$ , respectively. You should check from the definition of the reciprocal lattice vectors that this choice for the k points will define a set of k points with equal density in each direction in reciprocal space.

I didn't understand what the author wants to convey. Prior to this paragraph, he talks about how many k-points one should take (Monkhorst – Pack sampling) during calculation. Gives an example of bulk Cu which contains 16 atoms in its supercell with lattice vectors given by,

$$ \vec{a}_1 = a(1, 0, 0), \quad \vec{a}_2 = a(0, 1, 0), \quad \vec{a}_3 = a(0, 0, 4) $$

How does the author figure the optimal number of k-points and how does he justify his findings?

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The author does this by saying that we need 8 k-points along each lattice direction with unit length. And since one of the lattice vectors is 4 times longer it needs $1/4$ of the k-points. Remember that in reciprocal space the length of the $b_i$ vectors are (in orthogonal lattices) $b_i=1/a_i$. And to sample with reciprocal vectors with the same density you can infer the number of k-points.

In this case $|a_1|=|a_2|=|a_3|/4$, and in reciprocal space we have $|b_1| =|b_2|=4|b_3|$.

As an example, consider a 1D chain with lattice constant $a$. You sample this with $N_k$ k-points. Now take the same 1D chain and double it to $a'=2a$. To sample exactly the same k-points in reciprocal space you need only $N_k/2$ k-points in the $a'$ system.

So say the lattice is made twice as big along the $a_1$, then you would halve the number of k-points

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  • $\begingroup$ It simply means you are preserving the the density of the points. Am I right? $\endgroup$
    – 147875
    Sep 28 at 8:58
  • $\begingroup$ Isn't the equation should be $|a_1|=|a_2|=4 |a_3| \implies 4|b_1| =4|b_2|=|b_3|$? $\endgroup$
    – 147875
    Sep 28 at 9:50
  • $\begingroup$ Sorry for the typo, see my correction. And yes, you are preserving the density of points. $\endgroup$
    – zeroth
    Sep 28 at 20:32

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