Question about the E0 energies for the dataset

[naveenmohandas]

Hi,
I was looking to fine-tune the foundation model. I needed the E0 energy for a system, to be consistent with the original dataset I wanted to know how it was calculated for the MPTRJ dataset. I saw in the documentation that we can use isolated atom energies if available or using linear regression if it is not available.
So, were the E0 energies used for the MPTRJ dataset determined based on linear regression or were the isolated atom energies calculated for each element?

[ilyes319]

Hi,
MACE-MP-0 uses E0s estimated with linear regression. We have released another model that has E0s computed with DFT, and I recommend you using for finetuning: https://github.com/ACEsuit/mace-mp/releases/tag/mace_mp_0b.
Note that the E0s should be consistent with YOUR DFT settings.

[alvaroprat97]

If the objective is to compute formation energy though, would the model not learn that offset? Even if we use slightly inaccurate E0s during training, it is but a (small) scalar correction that is implicit in the E(Atomization(model)) right?

[alvaroprat97]

Assuming we use the exact same E0s in finetuning and in testing of course

[ilyes319]

When we use exact E0s, we enforce that MACE has no bias by construction. The only way to get the exact isolated atom energy is to fix any bias in the model to be exactly reproducing that. Not using the exact isolated atom energy will not prevent the model from being accurate in most places, but we observed that it leads to much worse reactivity. Look at Fig 8 here for example: https://arxiv.org/pdf/2205.06643.

[gabor1]

This is a design choice. I like fixing the E0 for the isolated atom because then it is correct when evaluated on the isolated atom with the model. There is no other construction when that statement is true. If you don't do this, then even if you include the isolated atom in the training set, it's only a small contribution to the loss, so will not be very accurate. As you train and vary the model, the formation energy will fluctuate because your learned shifts will move around. There is no opportunity to learn and adjust for the isolated atom because there are no neighbors.

[gabor1]

Another way to think about it that is mathematically identical is to feed atomuzation energies to the training and analytically construct the model to give 0.0 for all isolated atoms. It is formally identical to what we do.