Re: [Wien] REDTEF:How do change I my struct file?

[Lyudmila Dobysheva]

28.08.2015 06:43, Marzieh Gh wrote:

You said:  In the positions of your struct file there are some
"rounding errors":

ATOM   1: X=0.37456984 Y=0.77849296 Z=0.34557914
MULT= 4  ISPLIT= 8
ATOM   1: X=0.62543016 Y=0.22150704 Z=0.09557915  <--
ATOM   1: X=0.12543016 Y=0.27849296 Z=0.09557915  <--
ATOM   1: X=0.87456984 Y=0.72150704 Z=0.34557914

obviously, this should be always 14 (or 15), but not mixed.



I don’t know I do things  , to edit struct file  or do some things or... .
How do change I my struct file? What do I do?


You should change manually the struct file:
either
>> ATOM   1: X=0.37456984 Y=0.77849296 Z=0.34557914
>> MULT= 4  ISPLIT= 8
>> ATOM   1: X=0.62543016 Y=0.22150704 Z=0.09557914  <--
>> ATOM   1: X=0.12543016 Y=0.27849296 Z=0.09557914  <--
>> ATOM   1: X=0.87456984 Y=0.72150704 Z=0.34557914
or
>> ATOM   1: X=0.37456984 Y=0.77849296 Z=0.34557915  <--
>> MULT= 4  ISPLIT= 8
>> ATOM   1: X=0.62543016 Y=0.22150704 Z=0.09557915
>> ATOM   1: X=0.12543016 Y=0.27849296 Z=0.09557915
>> ATOM   1: X=0.87456984 Y=0.72150704 Z=0.34557915  <--

> The same occurs for atoms 25 and 26 !

Best wishes
  Lyudmila Dobysheva
--
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426001 Izhevsk, ul.Kirova 132
RUSSIA
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Re: [Wien] REDTEF:How do change I my struct file?

[Saeid Jalali]

Dear Marzieh,Running "x patchsymm" sometime can solve the rounding errors of 
positions. But, maybe your structure file is not correctly created at all. In 
this case, patchsymm cannot fix the problem, as the structure file is basically 
wrong. Creating a correct structure file is a fundamental and private task 
which would not be expected to be done by the others. The comment of Peter were 
valuable and enough, if you got it and try to fix the problem yourself. 
Anyway I suggest before making a supercell, try to see whether you can run your 
converted cif file or not. If not, try to first convert correctly the cif file 
to a correct structure file. To this end, you can apply cif2struct script on 
your desired cif file, 9011274.cif--in this way pay attention to the warning 
message of "unknown element: OT, type in correct name :".
After converting the file, you would also pay attention to RMT errors, and more 
importantly symmetry and sgroup warnings.  Furthermore, for your case you would 
be careful about leakage in lstart due to existing P atoms in your case -- 
maybe -6.0 Ry is not suitable for the P atoms.
Finally, if you could run the structure file, you can try to make a supercell 
from your case. I hope that in this case, your problem will be eventually fixed.
Good luck, 
     Sincerely yours,
S. Jalali
/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/
Saeid Jalali Asadabadi,
Associate Professor of Physics,
Department of Physics, Faculty of Science,
University of Isfahan (UI), Hezar Gerib Avenue,
81744 Isfahan, Iran.
Phones:
Dep. of Phys.   :+98-031-3793 2435
Office               :+98-031-3793 4776
Fax No.            :+98-031-3793 2800
E-mail           
                           :sjal...@sci.ui.ac.ir
                          :sjal...@phys.ui.ac.ir
                          :saeid.jalali.asadab...@gmail.com
                           :s_jalal...@yahoo.com
Homepage           :http://sci.ui.ac.ir/~sjalali
www                    :http://www.ui.ac.ir
/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/_/ 


 On Friday, August 28, 2015 8:14 AM, Marzieh Gh  
wrote:
   

 Dear Prof.Blaha
I am running supercell of KTiOPO4 (direction z) using 9011274.cif file
but I got Error in DSTART:
Error in DSTART
'ROTDEF' - no symmetry operation found.
'ROTDEF' - for jatom, index 1 2
'ROTDEF' - atomposition of jatom 0.3745698  0.7784930  0.3455791
'ROTDEF' - atomposition of index 0.6254302  0.2215070  0.0955792

You said:  In the positions of your struct file there are some
"rounding errors":
You should be able to find this from the error messages:
> 'ROTDEF' - atomposition of jatom 0.3745698  0.7784930  0.3455791
> 'ROTDEF' - atomposition of index 0.6254302  0.2215070  0.0955792
So look for these positions:
> ATOM  1: X=0.37456984 Y=0.77849296 Z=0.34557914
>            MULT= 4          ISPLIT= 8
> ATOM  1: X=0.62543016 Y=0.22150704 Z=0.09557915  <--
> ATOM  1: X=0.12543016 Y=0.27849296 Z=0.09557915  <--
> ATOM  1: X=0.87456984 Y=0.72150704 Z=0.34557914
obviously, this should be always 14 (or 15), but not mixed.
The same occurs for atoms 25 and 26 !

I don’t know I do things  , to edit struct file  or do some things or... .
How do change I my struct file? What do I do?
please help me
Best Regards,


-- 
Marzieh Ghoohestani
PhD Student of Computational Nano Physics
Nano Research Center, Department of Physics
University of Technology, Isfahan, Iran
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[Wien] question again :ROTDEF: How do change I my struct file?

[Marzieh Gh]

Dear Prof. Blaha  & Lyudmila Dobysheva


I I have changed manually the struct file (supercell of KTiOPO4) according
to your guidance, But I got Error in DSTART again:

Error in DSTART
 'ROTDEF' - no symmetry operation found.
 'ROTDEF' - for jatom, index   1   2

 'ROTDEF' - atomposition of jatom   0.3745698   0.7784930   0.3455791

 'ROTDEF' - atomposition of index   0.6254302   0.2215070   0.0955791

Please help me

Best Regards,


-- 
Marzieh Ghoohestani
PhD Student of Computational Nano Physics
Nano Research Center, Department of Physics
University of Technology, Isfahan, Iran
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Re: [Wien] question again :ROTDEF: How do change I my struct file?

[Laurence Marks]

In your struct file you have (had?) a multiplicity of 4 for many atoms. You
only had either none or 2 symmetry operations. The multiplicity of the
sites cannot be larger than the number of symmetry operations.

Either:

a) Edit by hand the case.struct file to have the symmetry operations you
believe are right. Then do "x patchsymm", read the output, look at the file
it produces. Use the computer above your eyes, not just the one at your
keyboard. If it is right copy it to case.struct. Rerun the initialization
from the beginning.

b) Do the full initialization again. If the code says " do you want to use
the new struct file" (or similar) answer "Y". Don't ignore what it says.

---
Professor Laurence Marks
Department of Materials Science and Engineering
Northwestern University
http://www.numis.northwestern.edu
Corrosion in 4D http://MURI4D.numis.northwestern.edu
Co-Editor, Acta Cryst A
"Research is to see what everybody else has seen, and to think what nobody
else has thought"
Albert Szent-Gyorgi
On Aug 28, 2015 07:08, "Marzieh Gh"  wrote:

> Dear Prof. Blaha  & Lyudmila Dobysheva
> 
>
> I I have changed manually the struct file (supercell of KTiOPO4) according
> to your guidance, But I got Error in DSTART again:
>
> Error in DSTART
>  'ROTDEF' - no symmetry operation found.
>  'ROTDEF' - for jatom, index   1   2
>
>  'ROTDEF' - atomposition of jatom   0.3745698   0.7784930   0.3455791
>
>  'ROTDEF' - atomposition of index   0.6254302   0.2215070   0.0955791
>
> Please help me
>
> Best Regards,
>
>
> --
> Marzieh Ghoohestani
> PhD Student of Computational Nano Physics
> Nano Research Center, Department of Physics
> University of Technology, Isfahan, Iran
>
>
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[Wien] regarding structure change

[mandeep hooda]

 Thanks Sir for help. I am writing this mail in reference to my previous
mail SUBJECT " regarding symmetry change." In w2web i have not selected
xcrysden. I have  clicked on "createZrTe3.klist_band  where it is already
showing simple cubic and there is no option of monoclinic structure.
Initially when I solved some other structures I got as I expected. But now
it shows simple cubic for tetragonal as well as monoclinic structure. I
have attached html  picture of ZrTe3. I have followed the steps shown in
this , nothing more. I  hope you will get the point where I have made
mistake. I am really thankful  for your help.




  Thanks and Regards
Mandeep Kumar
Title: ZrTe3@localhost




		
			
			

		




w2web
Sorry, but for w2web you need a frame-enabled web browser.

   
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Re: [Wien] regarding structure change

[Gavin Abo]

The html file that you attached contains no picture.

As you have said, there is no option for a monoclinic structure (no 
monoclinic klist_band template).


So you have to create your own monoclinic klist_band file, and that is 
what the "from xcrysden" option is for.


1. You have to use XCrySDen ("Generate k-mesh using XCrysden" button) to 
create your xcrysden.klist or hand type your own mesh into xcrysden.klist.


Section "3.11.4 Bandstructure" in the WIEN2k usersguide 
[http://www.wien2k.at/reg_user/textbooks/usersguide.pdf]:


For a few crystal structures template files are supplied in the 
SRC-directory, you can also *use**
**XCRYSDEN (save it as xcrysden.klist) to generate a k-mesh or type in 
your own mesh*.


2. Select "from xcrysden" and click the "create case.klist_band" button



On 8/28/2015 7:14 AM, mandeep hooda wrote:
Thanks Sir for help. I am writing this mail in reference to my 
previous mail SUBJECT " regarding symmetry change." In w2web i have 
not selected xcrysden. I have  clicked on "createZrTe3.klist_band  
where it is already showing simple cubic and there is no option of 
monoclinic structure. Initially when I solved some other structures I 
got as I expected. But now it shows simple cubic for tetragonal as 
well as monoclinic structure. I have attached html  picture of ZrTe3. 
I have followed the steps shown in this , nothing more. I  hope you 
will get the point where I have made mistake. I am really thankful  
for your help.





  Thanks and Regards
Mandeep Kumar
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[Wien] Photon energies in XES ...

[Vladimir Timoshevskii]

Dear Wien2k users and developers,

I am working with experimentalists and try to simulate the XES measured by
soft x-ray detector, coupled with electron microscope. So, the ionization
source in this setup is the electronic gun of the TEM. The test compound is
hexagonal layered BN, which was quite well studied before, including the
similar setup. I would greatly appreciate if you could share your opinion
on the following 2 issues, which I am facing now:

i) Is it possible, in principle, to obtain correct photon energies instead
of shifting spectrum by hand to the EF position? I understand, that the
position of the core level (B-1s) is sensitive to the form of the potential
well, and the closer my potential is to the real one, the better is the
position of the core level. I tried different XC-functionals, and found
that actually the atomic-like Hartree-Fock gives the best results: the
whole spectrum (B K-edge) is shifted to higher energies, closer to
experiment, and the spectrum shape is also much better. However, there is
still ~10eV shift, relative to the experimental spectrum. So, the
XC-functional alone does not solve this problem ...

ii) This is a more fundamental question, and is actually related to the
first one. I guess, the main reason for the photon energy underestimation
is the presence of the core hole, which shifts the ionized core level to
lower energies. I did several test calculations of B-K spectrum using
supercells of diffferent sizes with a core hole in B 1s. Indeed, by playing
with fractional B 1s occupation (trying to catch the "transition state"),
it seems to be possible to shift the whole spectrum to experimental
position. But in this case, what about the "rule of the final state"?
According to this rule, the hole must be created in the valence band (and
screened out), and the core lavel must be filled. This is what we normally
assume ... Does that mean that the XES calculations with hole in the core
are unphysical, in spite of giving better photon energies? May be, the
situation here, especially when we use TEM electronic gun for core
ionization,  is more complicated? In my opinion, the valence-core
transitions are happening in the potential, already distorted by the
presence of the core hole. Am I right? Then, how it agrees with the "rule
of the final state"? Any thoughts on that would be highly appreciated!

Thanks a lot in advance!

Vladimir Timoshevskii
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[Wien] Photon energies in XES ...

[Vladimir Timoshevskii]

Dear Wien2k users and developers,

I am working with experimentalists and try to simulate the XES measured by
soft x-ray detector, coupled with electron microscope. So, the ionization
source in this setup is the electronic gun of the TEM. The test compound is
hexagonal layered BN, which was quite well studied before, including the
similar setup. I would greatly appreciate if you could share your opinion
on the following 2 issues, which I am facing now:

i) Is it possible, in principle, to obtain correct photon energies instead
of shifting spectrum by hand to the EF position? I understand, that the
position of the core level (B-1s) is sensitive to the form of the potential
well, and the closer my potential is to the real one, the better is the
position of the core level. I tried different XC-functionals, and found
that actually the atomic-like Hartree-Fock gives the best results: the
whole spectrum (B K-edge) is shifted to higher energies, closer to
experiment, and the spectrum shape is also much better. However, there is
still ~10eV shift, relative to the experimental spectrum. So, the
XC-functional alone does not solve this problem ...

ii) This is a more fundamental question, and is actually related to the
first one. I guess, the main reason for the photon energy underestimation
is the presence of the core hole, which shifts the ionized core level to
lower energies. I did several test calculations of B-K spectrum using
supercells of diffferent sizes with a core hole in B 1s. Indeed, by playing
with fractional B 1s occupation (trying to catch the "transition state"),
it seems to be possible to shift the whole spectrum to experimental
position. But in this case, what about the "rule of the final state"?
According to this rule, the hole must be created in the valence band (and
screened out), and the core lavel must be filled. This is what we normally
assume ... Does that mean that the XES calculations with hole in the core
are unphysical, in spite of giving better photon energies? May be, the
situation here, especially when we use TEM electronic gun for core
ionization,  is more complicated? In my opinion, the valence-core
transitions are happening in the potential, already distorted by the
presence of the core hole. Am I right? Then, how it agrees with the "rule
of the final state"? Any thoughts on that would be highly appreciated!

Thanks a lot in advance!

Vladimir Timoshevskii
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Re: [Wien] Photon energies in XES ...

[Laurence Marks]

There is a lot of literature on this. Have you looked at (for instance) the
work of Les Allen, as well as the work of the group who wrote the TELNES
script?

N.B., "electron gun", not "electronic gun". Electronic microscope is an all
too common typo.

On Fri, Aug 28, 2015 at 10:19 AM, Vladimir Timoshevskii <
vladi...@physics.mcgill.ca> wrote:

> Dear Wien2k users and developers,
>
> I am working with experimentalists and try to simulate the XES measured by
> soft x-ray detector, coupled with electron microscope. So, the ionization
> source in this setup is the electronic gun of the TEM. The test compound is
> hexagonal layered BN, which was quite well studied before, including the
> similar setup. I would greatly appreciate if you could share your opinion
> on the following 2 issues, which I am facing now:
>
> i) Is it possible, in principle, to obtain correct photon energies instead
> of shifting spectrum by hand to the EF position? I understand, that the
> position of the core level (B-1s) is sensitive to the form of the potential
> well, and the closer my potential is to the real one, the better is the
> position of the core level. I tried different XC-functionals, and found
> that actually the atomic-like Hartree-Fock gives the best results: the
> whole spectrum (B K-edge) is shifted to higher energies, closer to
> experiment, and the spectrum shape is also much better. However, there is
> still ~10eV shift, relative to the experimental spectrum. So, the
> XC-functional alone does not solve this problem ...
>
> ii) This is a more fundamental question, and is actually related to the
> first one. I guess, the main reason for the photon energy underestimation
> is the presence of the core hole, which shifts the ionized core level to
> lower energies. I did several test calculations of B-K spectrum using
> supercells of diffferent sizes with a core hole in B 1s. Indeed, by playing
> with fractional B 1s occupation (trying to catch the "transition state"),
> it seems to be possible to shift the whole spectrum to experimental
> position. But in this case, what about the "rule of the final state"?
> According to this rule, the hole must be created in the valence band (and
> screened out), and the core lavel must be filled. This is what we normally
> assume ... Does that mean that the XES calculations with hole in the core
> are unphysical, in spite of giving better photon energies? May be, the
> situation here, especially when we use TEM electronic gun for core
> ionization,  is more complicated? In my opinion, the valence-core
> transitions are happening in the potential, already distorted by the
> presence of the core hole. Am I right? Then, how it agrees with the "rule
> of the final state"? Any thoughts on that would be highly appreciated!
>
> Thanks a lot in advance!
>
> Vladimir Timoshevskii
>
>
>



-- 
Professor Laurence Marks
Department of Materials Science and Engineering
Northwestern University
www.numis.northwestern.edu
Corrosion in 4D: MURI4D.numis.northwestern.edu
Co-Editor, Acta Cryst A
"Research is to see what everybody else has seen, and to think what nobody
else has thought"
Albert Szent-Gyorgi
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[Wien] Eorb =0 in LDA+U case.outputfiles

[Paresh Chandra Rout]

Dear all,
I am getting Eorb=0 in  case.outorbup and case.outorndn files. I had
applied U values to 9th and 10th atom of my system. The complex and real
part of potential for spin up and dn are also providing zero for all m
values . Does that make any sense for the LDA+U calculation ? Would anybody
kindly explain a little bit  about the out put files whether it is correct
or I have reproduced  it incorrectly ? Here I have attached the output
files.  Any help would be highly appreciated .



Kind Regards
Paresh Chandra Rout
Research Scholar
Indian Institute of Science Education and Research


BiFeReO6.outputorbdn
Description: Binary data


BiFeReO6.outputorbup
Description: Binary data


BiFeReO6.outputorb
Description: Binary data
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Re: [Wien] Eorb =0 in LDA+U case.outputfiles

[Gavin Abo]

I don't see any lines in your BiFeReO6.outputorb[up/dn] like:

Atom 9   L=  2 U=  0.300 J=  0.000 Ry
Atom 10 L=  2 U=  0.300 J=  0.000 Ry

So there might still be a problem with your BiFeReO6.inorb file.

Did you adjust BiFeReO6.inorb and do the scf calculation again?

If you only want to apply U values to atoms 9 and 10, then remove atoms 
11 and 12 from the inorb and indm files.


On 8/28/2015 10:00 AM, Paresh Chandra Rout wrote:

Dear all,
I am getting Eorb=0 in  case.outorbup and case.outorndn files. I had 
applied U values to 9th and 10th atom of my system. The complex and 
real part of potential for spin up and dn are also providing zero for 
all m values . Does that make any sense for the LDA+U calculation ? 
Would anybody kindly explain a little bit  about the out put files 
whether it is correct or I have reproduced  it incorrectly ? Here I 
have attached the output files.  Any help would be highly appreciated .




Kind Regards
Paresh Chandra Rout
Research Scholar
Indian Institute of Science Education and Research
-12.  Emin cutoff energy
 2   number of atoms for which density matrix is calculated
 9  1  2  index of 1st atom, number of L's, L1
 10 1  2  dtto for 2nd atom, repeat NATOM times
 0 0   r-index, (l,s)index  1  2  0 nmod, natorb, ipr
PRATT  1.0BROYD/PRATT, mixing
  9 1 2  iatom nlorb, lorb
  10 1 2  iatom nlorb, lorb
  1  nsic 0..AMF, 1..SIC, 2..HFM
   0.30 0.00U J (Ry)   Note: we recommend to use U_eff = U-J and J=0
   0.30 0.00U J___
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Re: [Wien] Photon energies in XES ...

[Peter Blaha]

You are mixing 2 concepts:

a) Yes, the final state rule applies and thus for XES calculations you 
should NOT use a core hole, but use the ground state DOS.


b) Strictly speaking, only E-tot is a valid quantity in DFT. In 
particular, the eigenvalues are in principle NOT excitation energies.
However, for delocalized states (eg. valence electrons) experience has 
shown that the eigenvalues (bandstructure,DOS) can be used to interprete 
experimental spectra.
This is, however, no true for localized states (core states), because 
these eigenvalues are not ionization energies but are defined as the 
derivative of the total energy with respect to occupation of the 
corresponding state.

 e_i =d E / d n_i
So one can use "Slaters transition state" approximation and make a 
supercell calculation, where on one atom HALF an electron should be 
removed. After scf, check the eigenvalue (with respect to EF) and you 
should obtain a significantly better binding energy of this core state 
(and the corresponding absolute XES transition energy.
(Typically the error of a 1s core state is reduced from eg. 10 eV to 
about 1 eV.
(about XPS energies and Slaters transition state see my lecture notes of 
our workshops on our web site).


Am 28.08.2015 um 17:19 schrieb Vladimir Timoshevskii:

Dear Wien2k users and developers,

I am working with experimentalists and try to simulate the XES measured
by soft x-ray detector, coupled with electron microscope. So, the
ionization source in this setup is the electronic gun of the TEM. The
test compound is hexagonal layered BN, which was quite well studied
before, including the similar setup. I would greatly appreciate if you
could share your opinion on the following 2 issues, which I am facing now:

i) Is it possible, in principle, to obtain correct photon energies
instead of shifting spectrum by hand to the EF position? I understand,
that the position of the core level (B-1s) is sensitive to the form of
the potential well, and the closer my potential is to the real one, the
better is the position of the core level. I tried different
XC-functionals, and found that actually the atomic-like Hartree-Fock
gives the best results: the whole spectrum (B K-edge) is shifted to
higher energies, closer to experiment, and the spectrum shape is also
much better. However, there is still ~10eV shift, relative to the
experimental spectrum. So, the XC-functional alone does not solve this
problem ...

ii) This is a more fundamental question, and is actually related to the
first one. I guess, the main reason for the photon energy
underestimation is the presence of the core hole, which shifts the
ionized core level to lower energies. I did several test calculations of
B-K spectrum using supercells of diffferent sizes with a core hole in B
1s. Indeed, by playing with fractional B 1s occupation (trying to catch
the "transition state"), it seems to be possible to shift the whole
spectrum to experimental position. But in this case, what about the
"rule of the final state"? According to this rule, the hole must be
created in the valence band (and screened out), and the core lavel must
be filled. This is what we normally assume ... Does that mean that the
XES calculations with hole in the core are unphysical, in spite of
giving better photon energies? May be, the situation here, especially
when we use TEM electronic gun for core ionization,  is more
complicated? In my opinion, the valence-core transitions are happening
in the potential, already distorted by the presence of the core hole. Am
I right? Then, how it agrees with the "rule of the final state"? Any
thoughts on that would be highly appreciated!

Thanks a lot in advance!

Vladimir Timoshevskii



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Re: [Wien] Eorb =0 in LDA+U case.outputfiles

[Paresh Chandra Rout]

Thanks again for the kind reply. Now it is printing like the two atoms you
mentioned in the outputorb[up/dn] file.
Here I am attaching the output files. I have one more query . Can you tell
me how and where can I get occupations of atoms for which I applied U
values ?

On Fri, Aug 28, 2015 at 9:56 PM, Gavin Abo  wrote:

> I don't see any lines in your BiFeReO6.outputorb[up/dn] like:
>
> Atom 9   L=  2 U=  0.300 J=  0.000 Ry
> Atom 10 L=  2 U=  0.300 J=  0.000 Ry
>
> So there might still be a problem with your BiFeReO6.inorb file.
>
> Did you adjust BiFeReO6.inorb and do the scf calculation again?
>
> If you only want to apply U values to atoms 9 and 10, then remove atoms 11
> and 12 from the inorb and indm files.
>
>
> On 8/28/2015 10:00 AM, Paresh Chandra Rout wrote:
>
>> Dear all,
>> I am getting Eorb=0 in  case.outorbup and case.outorndn files. I had
>> applied U values to 9th and 10th atom of my system. The complex and real
>> part of potential for spin up and dn are also providing zero for all m
>> values . Does that make any sense for the LDA+U calculation ? Would anybody
>> kindly explain a little bit  about the out put files whether it is correct
>> or I have reproduced  it incorrectly ? Here I have attached the output
>> files.  Any help would be highly appreciated .
>>
>>
>>
>> Kind Regards
>> Paresh Chandra Rout
>> Research Scholar
>> Indian Institute of Science Education and Research
>>
>
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>


BiFeReO6.outputorb
Description: Binary data


BiFeReO6.outputorbup
Description: Binary data


BiFeReO6.outputorbdn
Description: Binary data
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Re: [Wien] Eorb =0 in LDA+U case.outputfiles

[Gavin Abo]

The occupations can be found in the corresponding :QTLxxx line in the 
scf file [ 
http://www.mail-archive.com/wien%40zeus.theochem.tuwien.ac.at/msg03288.html 
, 
http://www.mail-archive.com/wien%40zeus.theochem.tuwien.ac.at/msg12761.html 
].


On 8/28/2015 12:16 PM, Paresh Chandra Rout wrote:
Thanks again for the kind reply. Now it is printing like the two atoms 
you mentioned in the outputorb[up/dn] file.
Here I am attaching the output files. I have one more query . Can you 
tell me how and where can I get occupations of atoms for which I 
applied U values ?


On Fri, Aug 28, 2015 at 9:56 PM, Gavin Abo > wrote:


I don't see any lines in your BiFeReO6.outputorb[up/dn] like:

Atom 9   L=  2 U=  0.300 J=  0.000 Ry
Atom 10 L=  2 U=  0.300 J=  0.000 Ry

So there might still be a problem with your BiFeReO6.inorb file.

Did you adjust BiFeReO6.inorb and do the scf calculation again?

If you only want to apply U values to atoms 9 and 10, then remove
atoms 11 and 12 from the inorb and indm files.


On 8/28/2015 10:00 AM, Paresh Chandra Rout wrote:

Dear all,
I am getting Eorb=0 in  case.outorbup and case.outorndn files.
I had applied U values to 9th and 10th atom of my system. The
complex and real part of potential for spin up and dn are also
providing zero for all m values . Does that make any sense for
the LDA+U calculation ? Would anybody kindly explain a little
bit  about the out put files whether it is correct or I have
reproduced  it incorrectly ? Here I have attached the output
files.  Any help would be highly appreciated .



Kind Regards
Paresh Chandra Rout
Research Scholar
Indian Institute of Science Education and Research

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[Wien] No mailing list this weekend

[tran]

Dear WIEN2k users,

The server of the WIEN2k mailing list will be down during the weekend.

Regards,

F. Tran
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