The problem of the pressure output by i-pi with vasp.6.3.0

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wuwvij utxoxr

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Mar 25, 2024, 9:31:10 PMMar 25
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Hello everyone! When I use i-PI with VASP 6.3.0, I find that the pressure output from i-PI is always a multiple of beads, while there is no such issue when I use i-PI with Quantum Espresso (QE). What could be the reason for this? I have attached my input and output files.

i-pi_vasp.6.3.0.tar.gz

wuwvij utxoxr

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Mar 26, 2024, 12:24:48 AMMar 26
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when using the NVT ensemble, i-PI with Quantum Espresso (QE) also has this problem.

Michele Ceriotti

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Mar 26, 2024, 9:21:41 AMMar 26
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Actually I wouldn't be sure this is an issue. Pressure will be different depending on the number of beads (in NVT for sure, and in NPT until it equilibrates). Have you tried running this example with a cheap-to-run empirical forcefield to get an idea of how this converges with number of beads, and what is the sampling time? You can even use a LJ potential with half-reasonable parameters to get an order-of-magnitude estimate of what's reasonable. In principle, I see no reason why there should be a difference between QE and VASP.
Michele

Mariana Rossi

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Mar 26, 2024, 9:35:05 AMMar 26
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Could you elaborate a bit on what you are seeing? If I read the msg I understand that the temperature is nbeads*T and pressure is nbeads*P only if you use VASP (?). That would be very odd. What is the problem exactly?

wuwvij utxoxr

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Mar 26, 2024, 11:25:50 AMMar 26
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Based on further testing, I found that both QE and VASP have the same issue. When in an NVT ensemble, the pressure P with beads n is n times the pressure P with beads 1, while the temperature is the same. Is the actual pressure of the system equals to P/n?

Michele Ceriotti

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Mar 26, 2024, 12:44:35 PMMar 26
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OK now I see better where the confusion comes from. When you are doing a PI simulation, pressure_md, as well as kinetic_md and a bunch of other properties are computed for the "extended system" that comprises the combination of all P beads. These quantities have NO PHYSICAL MEANING: for instance the kinetic energy will be _on average_ PNkbT3/2, which is not the kinetic energy of the quantum system. In some cases we do scale these properties by P, typically for the potential because the potential is actually a well-defined quantum estimator, or for "temperature" which is useful to check equilibration. We could discuss about scaling them but then there would be even more people making the mistake of taking them seriously. 
So, instead, use pressure_cv which is the right quantum estimator.

wuwvij utxoxr

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Mar 27, 2024, 4:48:24 AMMar 27
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I understand now, thank you for your prompt response.
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