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Dear Wien2k users,<br><br>
Recently I have read the lecture note by Dr. Madsen and Dr. Novak
"calculate the effective U in APW methods". <br><br>
There are something is unclear for me a bit. We really need your
help.<br><br>
The question is about how to set up NiO.inst for (4.5 up
spin, 3 down spin) calculation (everthing is clear for calculation with
(4.5,4) ). <br><br>
In (4.5,3) case, Ni has <b>7.5</b> 3d electrons meanwhile the
total number of Ni valance electrons is 10, Ni- 3d8_ 4s2. Is it correct
that we have to set up the last 2.5 electron for Ni 4s? If it is so the
following file is set up<br><br>
Ni (Ni 1, is considered as impurity)<br>
Ar 3 5<br>
3, 2,2.0 P<br>
3, 2,2.0 P<br>
3,-3,2.5 P<br>
3,-3,1.0 P ===> total 7.5 electron for
Ni_ 3d <br>
4,-1,1.25 N =====> max is 1?<br>
4,-1,1.25 N =====> max is 1?<br><br>
But I'm so afraid that I'm wrong. Or I understand wrongly about
case.inst? The User guide page 72-73, table 6.6 show that the max
occupation for <font color="#0000FF"><b>kappa = -1</b></font> is
<font color="#0000FF"><b>2</b></font>. But here we split them into spin
up and spin down therefore the max occupation is 1 for each spin up/down.
Am I correct? <br><br>
- If not, please advice me on this. <br><br>
- If it is true, could you help me to set up the NiO.inst file for the
calculation of (4.5 spin up ,3 spin down). <br><br>
The same question is for Fe2O3 (the article Europhys. Letters. 69, 777
(2005)). Fe3+ has five 3d electrons. We don't know how to set up the
case.inst for both calculations: (3 spin up, 2.5 spin down) and (3 spin
up,1.5 spin down).<br><br>
The Wien2k suggested the electronic configuration for case.inst
with 3d6_4s2 (total 8 valance electrons). If we set up 5.5 or 4.5
electrons for Fe_3d and how about the last 2.5 and 3.5 electron?<br><br>
Please help me to clarify this puzzle. I deeply thank for your
help.<br><br>
Best regards,<br>
Khuong<br>
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