I have a simple question: is it reasonable to compare the total energy of different magnetic solutions (all the same parameters except for the spin configuration) obtained with LDA+U?
My doubt arises from the following consideration: the energy stemming from the U term represents a correction to the DFT results and incorporates a series of approximations (the evaluation of orbital occupancy for example), so, in one sentence, am I partially measuring how keen is the system to the LDA+U correction?
I've seen this type of comparisons in literature but I do not find a clear justification to this approach.
Thanks,
Pietro
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a comparison of DFT+U total energies is compulsaroy, because spin-polarized DFT (which you apply +U to) has several local minima. Remember that we solve a self-consistent equation -- the Kohn-Sham equation + "possible corrections" -- in an iterative manner, and thus do not arrive at the global energy minimum. For a fixed set of parameters you arrive at different solution upon varying the initial conditions and the mixing. So once you accept DFT+U, you have to compare total energies of different configurations in order to make a statement about the "DFT+U" ground state.
The approximation in the +U implementation are controlled in contrast to the approximation you have made in the exchange and correlation potential in comparison to the compounds you use it for.
What do you mean by "keenness"?
All the best,
Marc
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Within the DFT+U ansatz, localized states φi largely retain their atomic nature and, therefore, can be expanded in term of an atomic-orbital basis set. (quoted from here ;)
So my idea of "keenness" is: how much the d (or f) states of a system retain the same localized character across various spin configurations?
If they are different I may decide that the best solution is the one with the most localized states? Or (similarly) the one that allows to have the smallest fractional occupations?
I think that if the energy difference between various solutions is larger than the Hubbard contribution there is no problem: in this case different occupation of the d orbitals leads to different hybridizations so the physics is dominated by the (in principle exact) KS Hamiltonian.
However, for example in my specific case, I want to compare a ferromagnetic with an antiferromagnetic solution. The energy difference between the two is of the order of tens of meV while the Hubbard contribution is, of course, much larger (see attachment for the values). Here my choice is dominated by the Hubbard contribution (which is different for the two solutions).
Can I safely state that the one with the lowest energy is the ground state?
Is there a (small) fraction of the Hubbard contribution to the total energy that stems from a different projection onto localized states? How small is it?
Thanks,
Pietro
Side note: this question came to my mind while working with plane-waves where the projection on different basis (for example atomic orbitals or wannier functions) gives different values for the Hubbard contribution.
As Marc stated DFT+U is a variational approach, if the double counting is kept constant and if you do not change your definition of correlated orbitals. Hence, if you keep the definition of correlated orbitals and do not change the strength of "U", the solution with lowest energy is the ground state in your calculation.
Hence, you can safely state that the lowest energy solution is the ground state in your variational calculation, when you have considered all possible symmetries.
The example you provide that seems to contradict this is an example where you drastically alter the "correlated electron states". For e.g. FeO: in a local description the correlated electrons would most often be all Fe-d, but in a Wannier description it has to be a mixture of Fe-d and O-p states as they strongly hybridize, irrespectively how well you localize the Wannier function. Then the "U" will naturally depend on whether you consider only the interaction between local Fe-d states or also include O-p states ...
Best,
Lars
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Dear Elk developers and users,
I have a simple question: is it reasonable to compare the total energy of different magnetic solutions (all the same parameters except for the spin configuration) obtained with LDA+U?
My doubt arises from the following consideration: the energy stemming from the U term represents a correction to the DFT results and incorporates a series of approximations (the evaluation of orbital occupancy for example), so, in one sentence, am I partially measuring how keen is the system to the LDA+U correction?
I've seen this type of comparisons in literature but I do not find a clear justification to this approach.
Thanks,
Pietro
Dear Pietro,
a comparison of DFT+U total energies is compulsaroy, because spin-polarized DFT (which you apply +U to) has several local minima. Remember that we solve a self-consistent equation -- the Kohn-Sham equation + "possible corrections" -- in an iterative manner, and thus do not arrive at the global energy minimum. For a fixed set of parameters you arrive at different solution upon varying the initial conditions and the mixing. So once you accept DFT+U, you have to compare total energies of different configurations in order to make a statement about the "DFT+U" ground state.
The approximation in the +U implementation are controlled in contrast to the approximation you have made in the exchange and correlation potential in comparison to the compounds you use it for.
What do you mean by "keenness"?
All the best,
Marc
Dear Marc,
first of all, thanks for your answer.
I try to clarify on my (probably wrong) idea.
Within the DFT+U ansatz, localized states φi largely retain their atomic nature and, therefore, can be expanded in term of an atomic-orbital basis set. (quoted from here ;)
So my idea of "keenness" is: how much the d (or f) states of a system retain the same localized character across various spin configurations?
If they are different I may decide that the best solution is the one with the most localized states? Or (similarly) the one that allows to have the smallest fractional occupations?
I think that if the energy difference between various solutions is larger than the Hubbard contribution there is no problem: in this case different occupation of the d orbitals leads to different hybridizations so the physics is dominated by the (in principle exact) KS Hamiltonian.
However, for example in my specific case, I want to compare a ferromagnetic with an antiferromagnetic solution. The energy difference between the two is of the order of tens of meV while the Hubbard contribution is, of course, much larger (see attachment for the values). Here my choice is dominated by the Hubbard contribution (which is different for the two solutions).
Thanks,
Pietro
Side note: this question came to my mind while working with plane-waves where the projection on different basis (for example atomic orbitals or wannier functions) gives different values for the Hubbard contribution.
Hi Pietro!
As Marc stated DFT+U is a variational approach, if the double counting is kept constant and if you do not change your definition of correlated orbitals. Hence, if you keep the definition of correlated orbitals and do not change the strength of "U", the solution with lowest energy is the ground state in your calculation.
Hence, you can safely state that the lowest energy solution is the ground state in your variational calculation, when you have considered all possible symmetries.
The example you provide that seems to contradict this is an example where you drastically alter the "correlated electron states". For e.g. FeO: in a local description the correlated electrons would most often be all Fe-d, but in a Wannier description it has to be a mixture of Fe-d and O-p states as they strongly hybridize, irrespectively how well you localize the Wannier function. Then the "U" will naturally depend on whether you consider only the interaction between local Fe-d states or also include O-p states ...
Best,
Lars
Now it's clear, thank you both!
Kind regards,
Pietro