That calculation I did quite some time ago. So, I cannot remember my
reasoning behind it. Since I didn't see an explanation for it in
Constraint_U.pdf and didn't know how Madsen and Pavel got the weights, I
likely just assumed the ε3d_dn would be the same as the ε3d_up.
I could be wrong, but
Thank you Dr. Gavin Abo
I agree that the occupation for spin-up (the addition of an electron) is 4
and 5
ε3d_up = (4*0.109629972+5*0.126221709)/9 = 0.1188476037
but when we remove an electron for spin-Dn we have to put 4 and 3 or 4
and5?
ε3d_dn = -(4*-0.292400769+(5or3)*-0.274004694)/ ...
Yes, I believe that is correct that Constraint_U.pdf is performing the
same calculation as Anisimov's paper.
As illustrated nicely on slide 11 in
http://www.fhi-berlin.mpg.de/~xinguo/talks/jiang_cdft-coffeetalk.pdf ,
just as you say, an electron is added for spin-up and an electron is
removed
Dear Gavin Abo;
I agree with you regarding the calculation of Ueff using the method
described in the following file
http://susi.theochem.tuwien.ac.at/reg_user/textbooks/Constraint_U.pdf
But according to Anisimov's paper , J=E3d'up'(n+0.5,n-0.5)- E3d'
dn'(n+0.5,n-0.5)
So we add an electron for the
Sorry, I don't know, but maybe more details about your calculation would
be needed for someone else to answer your question.
For example, are you doing something similar to the Constraint U
exercise [
http://susi.theochem.tuwien.ac.at/reg_user/textbooks/Constraint_U.pdf ],
which uses the F^0_
Dear Prof. Tran and WIEN2k users;
I found in Anisimov's paper the following relation to calculate J
J=E3d'up'(n+0.5,n-0.5)- E3d'dn'(n+0.5,n-0.5)
but I think that the energy of the spin-up electrons is more stable
compared to that of Dn electrons and that's why I found a negative energy
(Do you th
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