Procedure to compare DFT vs MLFF

[exk301]

Hello!

This question could sounds very dumb, but I wanted to make sure if I understood correctly how the MLFF works in vasp recently?
I trained my force field and I tried to verify if the force field has been trained well against DFT. Assuming I have ML_FF file available, I ran geometry optimization on my random test structure and compared the final energy from trained force field (In my case, surface energy matters)

This is the INCAR files I have used for DFT and, INCAR to read MLFF for energy comparison:

DFT:

ENCUT = 400
PRECT = Normal
ALGO = Fast
LREAL = A
ISMEAR = 0
SIGMA = 0.1
EDIFF = 1E-05
EDIFFG = -0.05
NELM = 200
LCHARG = .FALSE.
LWAVE = .FALSE.
POTIM = 0.5 # 3 femtoseconds
IBRION = 2
ISIF = 3
NSW = 6000


MLFF:

IBRION = 2
NSW = 1000
ISIF = 3

ML_LMLFF = T
ML_ISTART = 2

I'm not sure if My INCAR tags are correct. If this correct, i'm not sure if I need to or how to remove the kinetic energy term from the energy coming from the 2nd INCAR as I trained my MLFF at temperature 700K in AIMD.

Thank you very much for your feedback in advance!!

[andreas.singraber]

Hello!

Indeed comparing the results of geometry optimization can be a valuable test to verify the quality of the generated MLFF. I cannot comment on the quality of the ab-initio settings but as far as I can see you set correctly the tags to compare the ab-initio with the prediction-only MLFF run.

If you compare the potential energies of both methods you do not need to subtract any kinetic energy that was applied during MLFF training. The MLFF is trained on the potential energy, the forces and stress coming from the ab-initio data. Upon prediction, it only returns potential energies, no kinetic energy contribution is included. Simply put, the thermostats applied during on-the-fly training are used to drive the atoms out of their equilibrium positions, so that more diverse training data can be collected. Temperature is not introduced to train the kinetic energy itself but rather the potential energies at positions away from the optimal geometries.

Usually, training at higher temperatures "includes" also correct predictions close to optimal geometries. However, it may be beneficial to train also at some lower temperatures but whether this is really necessary depends a lot on your actual system.

I hope this helps, does this answer your question?

All the best,
Andreas Singraber

[exk301]

Hello andreas.singraber!!

Yes! That answers my questions !!! I appreciate for your response!!!!