8
$\begingroup$

I am looking for chemical potential from literature:

(1) https://journals.aps.org/prb/abstract/10.1103/PhysRevB.85.115104
(2) https://www.nature.com/articles/npjcompumats201510

I compared the method and data they provided. They use the Fitted Elemental Reference Energy method via GGA+U on VASP employing PAW. U values in these paper are a little different. As far as I understand from the papers, the chemical potential is equal to total energy of an isolated atom in a box (for example, 10x10x10). Am I right? Then I think it should be similar. However, found that the data are somehow inconsistent.For example:

                             Paper (2) (Table 1)                       Paper (1) (Table V: Appendix)
Fe                             2.200                                             -6.15
Mn                             1.987                                             -7.00
Co                             1.987                                             -4.75

Can anyone help me to explain?

$\endgroup$
  • $\begingroup$ Are these papers using different codes / levels of theory? Total energies are often arbitrary and only differences can be looked at. $\endgroup$ – Tristan Maxson Sep 20 at 11:32
  • $\begingroup$ I think chemical potential (reference) is not the isolated atoms in a box but from the elemental ground states, for example, metallic Fe for Fe. Note that the absolute value of chemical potential in DFT is only meaningful with respect to a particular set of pseudopotentials (just like the actual values of the energy themselves). $\endgroup$ – bzbzbz Sep 20 at 11:42
  • $\begingroup$ @TristanMaxson as I understand from these papers, they both used VASP with same method. $\endgroup$ – Binh Thien Sep 20 at 14:47
  • $\begingroup$ @bzbzbz I am still confused about "the elemental ground states". For example, Fe has 2 phases bcc and fcc. When we put several atoms together and they can interact with each others. The total energy now consists of interaction energy between atoms, then I think the obtained result is not typical for element, and it depend on different phases. $\endgroup$ – Binh Thien Sep 20 at 15:04
  • $\begingroup$ +1. Welcome to our community, and thank you for contributing your question here! We hope to see much more of you !!! $\endgroup$ – Nike Dattani Sep 20 at 18:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.