[QE-users] apply_U in Hubbard U implementation

[Aditya Putatunda]

Dear Prof Giannozzi,

I am writing to clarify part of the output being printed by pw.x. For the
case of EuS (with Eu.pbe-spdfn-rrkjus_psl.1.0.0.UPF pseudopotential using
QE7.5), I have activated Hubbard U using:

HUBBARD atomic
 U Eu-4f 7.0
 U Eu-4d 1.0

Which results into this snippet from the output:

     Exchange-correlation= SLA  PW   PBX  PBC
                           (   1   4   3   4   0   0   0)
     Hubbard projectors: atomic
     Hubbard parameters of DFT+U (Dudarev formulation) in eV:
     U(Eu-4f) =  7.0000
     U(Eu-4d) =  1.0000

     Internal variables: lda_plus_u = T, lda_plus_u_kind = 0

     Internal variables: orbital_resolved = F, hub_pot_fix = F, apply_U =  F

Could you kindly please clarify what is the meaning of "apply_U=F"? Can
this be safely ignored?

Thank you.
Regards.
Aditya Putatunda
CNR-SPIN & UniMiB
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