Total energy of model DMFT

[guuuj]

Hi developers,
I am trying to use solid_dmft to compute the total energy E_tot for my model Hamiltonian. This can be done because all solid_dmft need from DFT part is the wannier90_hr.dat (right?), and if I can convert my model Hamiltonian into the wannier90_hr.dat format, then the solid_dmft can be used to do the DMFT calculation for my model Hamiltonian.
In real DFT+DMFT calculation, the total energy from solid_dmft should be added to the DFT total energy to get the real total energy for the system, this means that the solid_dmft treat the DFT total energy as reference, and gives the DMFT correction for the DFT total energy (right?).
My question is, in my model DMFT calculation, is there any "reference energy" that should be added to the solid_dmft result? Or the solid_dmft already gives the exact total energy for my model Hamiltonian?
Best regard.

[alberto-carta]

Hi guuuj,

I answer briefly but let me know if you want a more in depth answer.

The short answer is that you add the energy output of solid_dmft on top of the energy you obtain by solving the tight binding model without the interaction part (i.e. the sum over all occupied energies of the eigenstates of the tight binding model)

As far as I remember we decided that in the case of non charge-self consistent DFT+DMFT (i.e. only post processing a DFT result and doing DMFT on top) we only show the DMFT part of the energy. This is because without charge self-consistency the whole DFT+DMFT process does not correspond to a stationary point of the Gibbs energy/Luttinger-Ward functional and taking energy differences might be misleading.

Alberto

[guuuj]

Thank you. In fact, I found that when I set U and J=0, the total energy from solid_dmft is 0, this means that what solid_dmft gives is the correction. And I found that the total energy of the tight binding model without the interaction part is already given in the output of solid_dmft, which is the "Kinetic energy contribution...", this is the sum over all occupied energies of the eigenstates of the tight binding model, right?

[alberto-carta]

Yes precisely