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I am a little confused about what should be the correct number of kpoints that I wanna use. I am currently working on a small cluster so picking a high value(even around 10) gives me a memory fault. My first question is how does choosing the k points vary for nanostructures and for bulk.

For our geometry optimization we use a k-points

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And for calcualtion of DOS we have been using 3 3 3 which I think is less. I am currently exploring how the DOS vary for various values of K Points. And the values vary a lot when I take 4 4 4 instead of 3 3 3 . Why is this happening!

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  • $\begingroup$ Related: mattermodeling.stackexchange.com/q/2347/5 and mattermodeling.stackexchange.com/q/4589/5 $\endgroup$
    – Nike Dattani
    Mar 14, 2022 at 13:44
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    $\begingroup$ A nanocluster is not a periodic system, so the correct way of simulating it with periodic boundary conditions is to use a very large supercell (how large is a convergence parameter) such that you suppress the interactions between the nanocluster and its periodic images. When things are properly converged, you should only use a single $\mathbf{k}$-point as the "bands" will be flat. $\endgroup$
    – ProfM
    Mar 15, 2022 at 8:14
  • $\begingroup$ @ProfM thanks for your reply and in the past day I have understood what you mean by using only a single k point. I have also gone through some literature and I am seeing people using only the 1 1 1 for ground state minimization but I am still a little unsure about using 1 1 1 for different properties. Mainly for DOS and Linear optical properties. I have also read that increasing the number of KPoints actually gives me wrong results. Is that true ?? $\endgroup$
    – Parmeet Singh EP 066
    Mar 15, 2022 at 8:24
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    $\begingroup$ The same argument applies for DOS or optical properties. In a true nanocluster there are no bands, just isolated energy levels. When we simulate it using periodic boundary conditions, a well-converged calculation should lead to completely flat "bands", which means that you only need one $\mathbf{k}$-point, and this could be any $\mathbf{k}$-point as they should all be equivalent. In this context, it is best to think of a nanocluster as a "big molecule" rather than as a bulk material. $\endgroup$
    – ProfM
    Mar 15, 2022 at 10:42
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    $\begingroup$ @ParmeetSinghEP066 why don't you write a self-answer? Future users might find it helpful! $\endgroup$
    – Nike Dattani
    Sep 24, 2022 at 15:38

1 Answer 1

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For Bulk we start with a small number of K-points and then increase it step by step until we get convergence with respect to the target property. This is generally done with respect to the SCF energy that is obtained at the end. The procedure to be used here is quite similar to convergence of cutoff energy as has been shown here.

The key difference between nano structures and Bulk systems is that in nano structures electrons are confined in a few directions(for Quantum Dots electrons are confined in all 3 directions), whereas in bulk they are allowed to move in all directions. This leads to flat bands in the case of a nano structure, due to which a single k-point should be enough.

Another way to look at this is when performing calculations on such systems with plane wave codes we need to introduce vacuum around our system, such that one particular quantum dot does not interact with itself image. In order to know the amount of vacuum to be inserted we need to again perform convergence testing. The addition of vacuum generally leads to big cells. As you must know bigger the real cell smaller the k space. So in order to traverse this small k-space a single k point is enough.

Now only one question stands if taking more k points leads to some problems for optical properties and DOS. I am not sure about this but I think that the answers obtained should be correct, it would just be much more inefficient.

Key takeaway is that if you are performing calculations with Bulk run convergence tests with K point and if calculations are being run with Quantum Dots choose a single k point.

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