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"I conducted a convergence test for the degauss value in Quantum Espresso input with smearing = 'cold'.

The results, however, exhibited a non-monotonic behavior contrary to the expected trend of decreasing calculation time with increasing degauss value.

The results are summarized in the spreadsheet provided below.

degauss pressure energy, $E$ $\Delta E$ Time (mm:ss)
0.004 12.94 -1986.759246 - 37:56
0.014 12.94 -1986.758622 0.00062393 37:43
0.024 12.88 -1986.756794 0.00182777 39:27
0.034 12.75 -1986.751534 0.00525965 43:14
0.044 12.78 -1986.741627 0.00990765 41:32
0.054 12.89 -1986.728302 0.01332447 38:44

This unexpected outcome prompts me to seek clarification on the underlying factors influencing this behavior. Could someone provide insights into why the calculation time does not follow the anticipated pattern? Additionally, how should I interpret this non-monotonic behavior in the context of my simulations? Any suggestions for further analysis or adjustments to my approach would be greatly appreciated."

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    $\begingroup$ I assume that the difference in simulation times is due to more or fewer scf steps depending on degauss. Maybe you could plot the number of scf steps to reach convergence too? Do you know what the degauss variable does? It certainly affects convergence! $\endgroup$ Commented Apr 2 at 16:10
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    $\begingroup$ @Tyler sterling can you give me a brief insight on degauss variable's effect on convergence $\endgroup$ Commented Apr 3 at 4:36
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    $\begingroup$ For metals, there is a fermi surface where, at 0 K, the occupations discontinuously change from 0 to 1 (or 2 without spin). The density is calculated from occupied states only. When eigenvalues are recalculated at each step, small changes in bands near the Fermi surface result in large changes in the density (since occupations change discontinuously). i.e. the density will never converge. Smearing makes the occupations change continuously from 0 to 1. Changing the smearing width (degauss) affects convergence by changing how quickly occupations go to 0. $\endgroup$ Commented Apr 3 at 14:40
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    $\begingroup$ Please check the number of steps to converge for each degauss and let us know. If we figure it out, I can write an answer to the question. $\endgroup$ Commented Apr 3 at 14:42
  • $\begingroup$ @ Tyler Sterling thank you for your excellent explanation,it is really thought provoking. NO.of steps for convergence for corresponding degauss values 0.004,0.014,0.024,0.034,0.044,0.054 are 16,16,17,19,18 and 16. $\endgroup$ Commented Apr 3 at 17:42

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I summarized our comments above into an answer that you should accept if it answers your question! :)

The difference in simulation times is due to more or fewer SCF steps depending on degauss, as you show in your comment above. The speed of convergence depends on degauss, which is expected.

For metals, there is a Fermi surface where, at 0 K, the occupations discontinuously change from 0 to 1 (or 2 without spin). The density is calculated from occupied states only. When eigenvalues are recalculated at each step, small changes in bands near the Fermi surface result in large changes in the density (since occupations change discontinuously). Since the density jumps discontinuously from one SCF step to the next, there will always be a large residual and the SCF cycle will never converge!

Smearing makes the occupations change smoothly from 0 to 1, so even if the eigenvalues near the Fermi level change, the density will change smoothly and convergence will improve (within reason). Still, pathological choices of degauss=1000 or something are probably not a good idea. You should pick degauss as small as possible for a given k-point grid while still maintaining quick convergence and sensible physical properties.

I hope I have helped! Cheers, Ty

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    $\begingroup$ thank you so much @Tyler Sterling ,it really helped. $\endgroup$ Commented Apr 3 at 19:04

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