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How to choose the value of nbnd in quantum espresso and what is the difference between occupations = "fixed" and occupations = "smearing"

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nbnd is the number of valence orbitals/bands your system has. As for the occupations smearing is a mathematical tool which 'smears' out the band energy diagram as in changes the occupation number to do convergence of energy. For fixed no such things happen and occupations is set to fixed.

I'd suggest read this to see the effect of different type of smearing. And also ref this for more info on smearing

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I am only answering the nbnd part. If I can I will later edit this and add the smearing part too. Kindly note that it's usually encouraged to ask one question per post. You might want to consider asking the smearing question in a separate post.

nbnd specifies the number of bands to be calculated. So, the band structure with nbnd=20 would be more dense (there will be more band lines) than the one with nbnd=10. It should be selected depending on the type of your material, and depending on the type of your calculation.

For spin-unpolarized (i.e. regular) calculations: if your material is an insulator, nbnd should at least be equal to the number of valence bands (i.e., the number of electrons / 2). If your material is a metal, then 20% more than the number of valence bands should be chosen (minimum 4 more) [source].

For spin-polarized calculation, a useful rule of thumb is to double the number of bands. Since we consider both spin‐up and spin‐down of an electron, the number of Kohn‐Sham states, i.e., nbnd should be selected more than twice the number of electrons [1].

[1] Hung, Nguyen Tuan; Nugraha, Ahmad R. T.; Saito, Riichiro, Quantum ESPRESSO course for solid-state physics, ZBL07655576. Page 133.

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    $\begingroup$ Just to add to @AbdulMuhaymin's answer. If your material is an insulator, the better choice would be to set nbnd to (number of valence electrons/2) + 4 since it helps with numerical stability. $\endgroup$ Jun 16, 2023 at 5:54

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