[QE-users] DFT+U calculation and Hubbard_J0 parameter

[Yuvam Bhateja]

Hello experts,

I have been using Quantum ESPRESSO for my project and now I wanted to
include DFT+U in my calculation.

I was using lda_plus_u_kind = 0 (which only uses Hubbard_U parameter) along
with Hubbard_J0.
For example-

lda_plus_u_kind=0
Hubbard_U(1)=5.0
Hubbard_J0(1)=0.48

Using the above configuration my system converged perfectly.

I wanted to know if I am doing something wrong here? Will the value of
Hubbard_J0 be even considered if I am using lda_plus_u_kind=0?

I wished to replicate the results in other softwares like SIESTA which only
uses rotationally invariant scheme (i.e., equivalent to lda_plus_u=1).
Can I use values of U=5.0 and J=0.48 as mentioned above in other DFT codes?

Any suggestions would be very helpful.

Regards
Yuvam Bhateja
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