[Wien] (no subject)
Madhav Ghimire
ghimire.mpg at gmail.com
Wed Apr 18 13:59:22 CEST 2012
Dear Dr. Xavier,
Thank you very much for your prompt reply and for your useful comments.
Now I understand how it works.
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.
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.
By the way, Do you know whether we can fix the value of magnetic moment of
Pt along any specified direction.
If so, I would like to know how and in which file.
Thanks again.
best wishes
M. P. Ghimire
2012/4/18 Rocquefelte <Xavier.Rocquefelte at cnrs-imn.fr>
> **
> I didn't look in detail your email but here is the solution.
>
> You should use the following case.indm file:
>
> --------------------------
>
> -9. Emin cutoff energy
> 2 number of atoms for which density matrix is
> calculated
> 4 1 2 index of 1st atom, number of L's, L1
> 5 1 2 dtto for 2nd atom, repeat NATOM times
> 0 0 r-index, (l,s)index
> --------------------------
>
> and case.inorb file:
>
> --------------------------
>
> 1 2 0 nmod, natorb, ipr
> PRATT 1.0 BROYD/PRATT, mixing
> 4 1 2 iatom nlorb, lorb
> 5 1 2 iatom nlorb, lorb
> 1 nsic 0..AFM, 1..SIC, 2..HFM
> 0.1471 0.036 U J (Ry) Note: we recommend to use U_eff = U-J
> and J=0
> 0.1471 0.036 U J
> --------------------------
>
> To explain, let's take an example:
>
>
> 4 1 2 index of 1st atom, number of L's, L1
>
> Here you are saying that you will add a Hubbard term for atom 4, and only
> for one l-value, which is l = 2.
> 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).
>
> I hope this reply will clarify the situation.
>
> Regards
>
> Xavier
>
> P.S.1: It is recommended to use a Ueff value, i.e. Ueff = U -J.
> In you case it will correspond to the following case.inorb file:
>
>
> --------------------------
>
> 1 2 0 nmod, natorb, ipr
> PRATT 1.0 BROYD/PRATT, mixing
> 4 1 2 iatom nlorb, lorb
> 5 1 2 iatom nlorb, lorb
> 1 nsic 0..AFM, 1..SIC, 2..HFM
> 0.1435 0.000 U J (Ry) Note: we recommend to use U_eff = U-J
> and J=0
> 0.1435 0.000 U J
> --------------------------
>
> P.S.2: The Hubbard term you are using is quite small (2eV). Is it what you
> really need for your system?
>
>
>
>
>
> On 04/18/2012 01:14 PM, Madhav Ghimire wrote:
>
> Dear Dr. Xavier and wien users,
> 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:
> (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
> -9. Emin cutoff energy
> 2 number of atoms for which density matrix is
> calculated
> 4 2 5 index of 1st atom, number of L's, L1
> 5 2 5 dtto for 2nd atom, repeat NATOM times
> 0 0 r-index, (l,s)index
> 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.
> 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.
> (ii) For case.inorb, below is the input value I tried to use
> 1 2 0 nmod, natorb, ipr
> PRATT 1.0 BROYD/PRATT, mixing
> 1 2 5 iatom nlorb, lorb
> 2 2 5 iatom nlorb, lorb
> 1 nsic 0..AFM, 1..SIC, 2..HFM
> 0.1471 0.036 U J (Ry) Note: we recommend to use U_eff = U-J
> and J=0
> 0.1471 0.036 U J
> As in case.indm, I have similar type of problem. As mentioned in
> userguide,
> nlorb; number of orbital moments for which exact exchange shall be
> calculated
> lorb: orbital numbers .
> Here, I could not differentiate between the number of orbital moments and
> orbital numbers. Does it mean the same or they are different.
> Could you kindly help to correct this two files.
> (iii) Can we fix the orbital moment of Pt.
> (iv) Can we set the minimum energy to more than 400 eV as the ground state
> energy of oxygen is approx. 400 eV.
>
> Thank you in advance
> M. P. Ghimire
> NIMS, Japan
>
> 2012/4/18 Rocquefelte <Xavier.Rocquefelte at cnrs-imn.fr>
>
>> It seems that your message is incomplete.
>> However, I am surprized to see nlorb = 5!
>>
>> If you are interested in "d" orbitals, nlorb should be 2, and if it is
>> "f" orbitals, nlorb should be 3.
>>
>> Best Regards
>>
>> Xavier
>>
>>
>>
>>
>>
>> On 04/18/2012 09:26 AM, Madhav Ghimire wrote:
>>
>>
>> Dear wien2k users,
>> I am facing some problems in putting the value of n nlorb and
>> lorb as provided in userguide.
>> -9. Emin cutoff energy
>> 2 number of atoms for which density matrix is
>> calculated
>> 4 2 5 index of 1st atom, number of L's, L1
>> 5 2 5 dtto for 2nd atom, repeat NATOM times
>> 0 0 r-index, (l,s)index
>> for
>> --
>> M. P. Ghimire
>>
>>
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>>
>
>
> --
> M. P. Ghimire
>
>
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--
M. P. Ghimire
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