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Have you got any interesting experimental data to compare your
calculations and then have a better idea of the U value?<br>
The U value is not only dependent of the element and orbital, but
also on the dimensionality and interactions (covalency) of the
compound. <br>
I would then say that the choice of the U value is always matter of
experience. You can also have a look at the note of G. Madsen and P.
Novak, which is on the WIEN2K website: <br>
<br>
<a class="moz-txt-link-freetext" href="http://www.wien2k.at/reg_user/textbooks/">http://www.wien2k.at/reg_user/textbooks/</a><br>
<br>
- Notes about constraint LDA calculations to determine U (pdf)<br>
<br>
Regards<br>
<br>
Xavier<br>
<br>
<br>
<br>
On 04/18/2012 01:59 PM, Madhav Ghimire wrote:
<blockquote
cite="mid:CAFRWjkPV=Zp1U-m+ftgYQ=oDFs9n9hyG24qfPLaBsjgag6Qstw@mail.gmail.com"
type="cite">Dear Dr. Xavier,<br>
Thank you very much for your prompt reply and for your useful
comments. Now I understand how it works. <br>
For the Platinates or iridates because of their extended orbitals,
the U value is small as explained from literature. So basing on
the literatures, I choose 1-2 eV. to check the nature of band
structure. Do you think it to be very small. If so please let me
know upto how much i can use.<br>
Will there be any influence if U=5 eV for this type of compound is
chosen. Of course, at high U with inclusion of spin-orbit
coupling, it shows interesting behavior in band structure. But I
fail to plot the fat bands in band structure to identify spin up
and spin down together. <br>
By the way, Do you know whether we can fix the value of magnetic
moment of Pt along any specified direction. <br>
If so, I would like to know how and in which file.<br>
Thanks again.<br>
best wishes<br>
M. P. Ghimire <br>
<br>
<div class="gmail_quote">2012/4/18 Rocquefelte <span dir="ltr"><<a
moz-do-not-send="true"
href="mailto:Xavier.Rocquefelte@cnrs-imn.fr">Xavier.Rocquefelte@cnrs-imn.fr</a>></span><br>
<blockquote class="gmail_quote" style="margin: 0pt 0pt 0pt
0.8ex; border-left: 1px solid rgb(204, 204, 204);
padding-left: 1ex;">
<div bgcolor="#ffffff" text="#000000"> I didn't look in detail
your email but here is the solution. <br>
<br>
You should use the following case.indm file:<br>
<br>
--------------------------
<div class="im"><br>
-9. Emin cutoff energy<br>
2 number of atoms for which density
matrix is calculated<br>
4 <span style="color: rgb(255, 0, 0);">1 2</span>
index of 1st atom, number of L's, L1<br>
5 <span style="color: rgb(255, 0, 0);">1 2</span>
dtto for 2nd atom, repeat NATOM times<br>
0 0 r-index, (l,s)index <br>
</div>
--------------------------<br>
<br>
and case.inorb file:<br>
<br>
--------------------------
<div class="im"><br>
1 2 0 nmod, natorb, ipr<br>
PRATT 1.0 BROYD/PRATT, mixing<br>
4 <span style="color: rgb(255, 0, 0);">1 2</span>
iatom nlorb, lorb<br>
5 <span style="color: rgb(255, 0, 0);">1</span> 2
iatom nlorb, lorb<br>
1 nsic 0..AFM, 1..SIC,
2..HFM<br>
0.1471 0.036 U J (Ry) Note: we recommend to
use U_eff = U-J and J=0<br>
0.1471 0.036 U J<br>
</div>
--------------------------<br>
<br>
To explain, let's take an example:
<div class="im"><br>
<br>
4 <span style="color: rgb(255, 0, 0);">1 2</span>
index of 1st atom, number of L's, L1<br>
<br>
</div>
Here you are saying that you will add a Hubbard term for
atom 4, and only for one l-value, which is l = 2. <br>
This treatment will be applied to the valence state of Pt,
i.e. the 5d orbital of Pt. You should not specify the
principal quantum number in this file (n-value). <br>
<br>
I hope this reply will clarify the situation.<br>
<br>
Regards<br>
<br>
Xavier<br>
<br>
P.S.1: It is recommended to use a Ueff value, i.e. Ueff = U
-J. <br>
In you case it will correspond to the following case.inorb
file:<br>
<br>
<br>
--------------------------
<div class="im"><br>
1 2 0 nmod, natorb, ipr<br>
PRATT 1.0 BROYD/PRATT, mixing<br>
4 <span style="color: rgb(255, 0, 0);">1 2</span>
iatom nlorb, lorb<br>
5 <span style="color: rgb(255, 0, 0);">1</span> 2
iatom nlorb, lorb<br>
1 nsic 0..AFM, 1..SIC,
2..HFM<br>
</div>
0.1435 0.000 U J (Ry) Note: we recommend to use
U_eff = U-J and J=0<br>
0.1435 0.000 U J<br>
--------------------------<br>
<br>
P.S.2: The Hubbard term you are using is quite small (2eV).
Is it what you really need for your system?
<div>
<div class="h5"><br>
<br>
<br>
<br>
<br>
On 04/18/2012 01:14 PM, Madhav Ghimire wrote:
<blockquote type="cite">
<div>Dear Dr. Xavier and wien users,<br>
Thank you for your kind concern to my problem.
Sorry for the incomplete information. Let me
complete my few queries and cases over which I am
concerned to: <br>
(i) I am having one oxide compound with the presence
of Pt atoms having 5d states. Because of this, I
want to implement U. For the said case, we require
case.indm and case.inorb as suggested in userguide.
My edited case.indm file is as shown below <br>
-9. Emin cutoff energy<br>
2 number of atoms for which
density matrix is calculated<br>
4 <span style="color: rgb(255, 0, 0);">2 5</span>
index of 1st atom, number of L's, L1<br>
5 <span style="color: rgb(255, 0, 0);">2 5</span>
dtto for 2nd atom, repeat NATOM times<br>
0 0 r-index, (l,s)index <br>
In this indm file (marked with red color), I could
not understand how L's=1 and , L1=2 is taken in
userguide. From the the userguide, L's corresponds
to number of l-values for which the density matrix
should be calculated and L1 is the l-values for
which the density matrix should be calculated. <br>
For the case of Pt which is 5d atom it has l=2 and
n=5 with d=5 sublevels. Hence I substituted as shown
above. But, I did not understand whether L's must be
2 or 5. Similar is for the case of L1. And, what
about the r-index.<br>
(ii) For case.inorb, below is the input value I
tried to use<br>
1 2 0 nmod, natorb, ipr<br>
PRATT 1.0 BROYD/PRATT, mixing<br>
1 <span style="color: rgb(255, 0, 0);">2 5</span>
iatom nlorb, lorb<br>
2 <span style="color: rgb(255, 0, 0);">2 5</span>
iatom nlorb, lorb<br>
1 nsic 0..AFM,
1..SIC, 2..HFM<br>
0.1471 0.036 U J (Ry) Note: we recommend
to use U_eff = U-J and J=0<br>
0.1471 0.036 U J<br>
As in case.indm, I have similar type of problem. As
mentioned in userguide, <br>
nlorb; number of orbital moments for which exact
exchange shall be calculated<br>
lorb: orbital numbers .<br>
Here, I could not differentiate between the number
of orbital moments and orbital numbers. Does it mean
the same or they are different.<br>
Could you kindly help to correct this two files. <br>
(iii) Can we fix the orbital moment of Pt. <br>
(iv) Can we set the minimum energy to more than 400
eV as the ground state energy of oxygen is approx.
400 eV.<br>
<br>
Thank you in advance <br>
M. P. Ghimire<br>
NIMS, Japan<br>
<br>
2012/4/18 Rocquefelte <span dir="ltr"><<a
moz-do-not-send="true"
href="mailto:Xavier.Rocquefelte@cnrs-imn.fr"
target="_blank">Xavier.Rocquefelte@cnrs-imn.fr</a>></span><br>
<div class="gmail_quote">
<blockquote class="gmail_quote" style="margin: 0pt
0pt 0pt 0.8ex; border-left: 1px solid rgb(204,
204, 204); padding-left: 1ex;">
<div bgcolor="#ffffff" text="#000000"> It seems
that your message is incomplete. <br>
However, I am surprized to see nlorb = 5!<br>
<br>
If you are interested in "d" orbitals, nlorb
should be 2, and if it is "f" orbitals, nlorb
should be 3. <br>
<br>
Best Regards<br>
<br>
Xavier
<div>
<div><br>
<br>
<br>
<br>
<br>
On 04/18/2012 09:26 AM, Madhav Ghimire
wrote: </div>
</div>
<blockquote type="cite">
<div>
<div><br clear="all">
Dear wien2k users, <br>
I am facing some problems in putting
the value of n nlorb and lorb as
provided in userguide. <br>
-9. Emin cutoff
energy<br>
2 number of atoms
for which density matrix is calculated<br>
4 2 5 index of 1st atom, number
of L's, L1<br>
5 2 5 dtto for 2nd atom, repeat
NATOM times<br>
0 0 r-index, (l,s)index<br>
for <br>
-- <br>
M. P. Ghimire<br>
<br>
</div>
</div>
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