Re: [SIESTA-L] net charge calculation

2007-04-19 Thread Yurko Natanzon

ok, there some papers which use Mulliken population with SIESTA. For
example, this one:
http://arxiv.org/pdf/cond-mat/0008340
The authors say that differences (of net charges and BOP-s) are less
sensitive to the choice of basis sets, so they can be meaningful. Can
anybody confirm this?

On 19/04/07, Yurko Natanzon [EMAIL PROTECTED] wrote:

Dear Adam Gali, Andrei Postnikov,
thank you for explanation. Actually, I'm not interested in particular
numbers, but want to observe changes in ionic charges of atoms,
neighboring to dopant and the dependence of such changes on a dopant
concentration (comparabale to undoped system). Is it possible to do it
with Mulliken analysis? And how does the charge depend on the basis?
For example, I use standard DZP basis automatically generated by
SIESTA, but for SiO2 I get negative charge of SI, and for GeO2 I get
positive one. What can I do, if I want to use Ge and Si as dopants and
compare their influence on net charges of nieghboring atoms?

On 19/04/07, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:

   Dear Yurko Natanzon,

 take care! Mulliken-charges could be meaningless by using diffusing
 orbitals. Draw the net charge density around the atoms (you can do it by
 SIESTA and utility programs provided with) and you can see that O is
 negatively polarized while Si is positively polarized opposite what
 Mulliken-charges would indicate. I suggest to implement and use
 Bader-charges for analyses which DOES NOT depend on the basis set.

 Yours,
  Adam Gali

  Species: Si
  Atom  Qatom  Qorb
3s  3s  3py 3pz 3px 3py 3pz 3px
3Pdxy   3Pdyz   3Pdz2   3Pdxz   3Pdx2-y2
1  4.026   0.404   0.474   0.308   0.152   0.308   0.254   0.394   0.254
   0.338   0.330   0.259   0.330   0.219
4  4.026   0.404   0.474   0.308   0.152   0.308   0.254   0.394   0.254
   0.338   0.330   0.259   0.330   0.219
 
  Species: O
  Atom  Qatom  Qorb
2s  2s  2py 2pz 2px 2py 2pz 2px
2Pdxy   2Pdyz   2Pdz2   2Pdxz   2Pdx2-y2
2  5.987   1.136   0.383   1.391   1.164   1.391   0.139   0.226   0.139
   0.003   0.005   0.003   0.005   0.003
3  5.987   1.136   0.383   1.391   1.164   1.391   0.139   0.226   0.139
   0.003   0.005   0.003   0.005   0.003
5  5.987   1.136   0.383   1.391   1.164   1.391   0.139   0.226   0.139
   0.003   0.005   0.003   0.005   0.003
6  5.987   1.136   0.383   1.391   1.164   1.391   0.139   0.226   0.139
   0.003   0.005   0.003   0.005   0.003
 
  mulliken: Qtot =   32.000
 
  then oxygen has positive ionic charge (6-5.987), and silicon has
  negative (4-4.026). Does it make any sense??
 
  On 19/04/07, Vasilii Artyukhov [EMAIL PROTECTED] wrote:
   Just subtract the ionic charge from the Mulliken population, you will get
   -.514 for O and +1.028 for Ti.
  
   2007/4/19, Yurko Natanzon [EMAIL PROTECTED] :
Dear SIESTers,
I wonder how to calculate net ionic charge with SIESTA. For example, I
have TiO2 anatase supercell with 12 atoms. Pseudopotential for Ti has
valence electrons in 3s2 3p6 3d2 4s2 (12 electrons), O has 2s2 2p4 (6
electrons). Then i set WriteMullikenPop 2 and obtain
10.972 for each Ti
6.514 fo O
Total charge is 10.972*4+6.514*8 = 96 which equals the total number of
valence electrons. Then how to obtain net electric charges of Ti and O
atoms? If it is the difference between number of valence electrons and
a charge obtained by SIESTA, than it is much lower, than is reported
in other calculations. Net charge should be around +2 for Ti and -1
for O.
   
--
Yurko Natanzon
PhD Student
Henryk Niewodniczański Institute of Nuclear Physics
Polish Academy of Sciences
ul. Radzikowskiego 152,
31-342 Kraków, Poland
Email: [EMAIL PROTECTED], [EMAIL PROTECTED]
   
  
  
 
 
  --
  Yurko Natanzon
  PhD Student
  Henryk Niewodniczański Institute of Nuclear Physics
  Polish Academy of Sciences
  ul. Radzikowskiego 152,
  31-342 Kraków, Poland
  Email: [EMAIL PROTECTED], [EMAIL PROTECTED]
 
 

 
---
 Dr. Gali ÁdámAdam Gali, PhD

 Budapesti Műszaki és Department of Atomic Physics,
 Gazdaságtudományi Egyetem,   Budapest University of Technology and
 Atomfizika Tanszék   Economics
 Budapest, Budafoki út 8.,    Budafoki út 8., H-, Budapest,
  Hungary

 telefon: 463-1580telephone: [36]-(1)-463-1580
 fax: 463-4357fax:  [36]-(1)-463-4357

 e-mail: [EMAIL PROTECTED]
http://www.fat.bme.hu/homepages/galia/index.en.html
 
---


--

Re: [SIESTA-L] net charge calculation

2007-04-19 Thread Adam Gali

  Dear Yurko Natanzon,

take care! Mulliken-charges could be meaningless by using diffusing 
orbitals. Draw the net charge density around the atoms (you can do it by 
SIESTA and utility programs provided with) and you can see that O is 
negatively polarized while Si is positively polarized opposite what 
Mulliken-charges would indicate. I suggest to implement and use 
Bader-charges for analyses which DOES NOT depend on the basis set.

Yours,
 Adam Gali

 Species: Si
 Atom  Qatom  Qorb
   3s  3s  3py 3pz 3px 3py 3pz 3px
   3Pdxy   3Pdyz   3Pdz2   3Pdxz   3Pdx2-y2
   1  4.026   0.404   0.474   0.308   0.152   0.308   0.254   0.394   0.254
  0.338   0.330   0.259   0.330   0.219
   4  4.026   0.404   0.474   0.308   0.152   0.308   0.254   0.394   0.254
  0.338   0.330   0.259   0.330   0.219
 
 Species: O
 Atom  Qatom  Qorb
   2s  2s  2py 2pz 2px 2py 2pz 2px
   2Pdxy   2Pdyz   2Pdz2   2Pdxz   2Pdx2-y2
   2  5.987   1.136   0.383   1.391   1.164   1.391   0.139   0.226   0.139
  0.003   0.005   0.003   0.005   0.003
   3  5.987   1.136   0.383   1.391   1.164   1.391   0.139   0.226   0.139
  0.003   0.005   0.003   0.005   0.003
   5  5.987   1.136   0.383   1.391   1.164   1.391   0.139   0.226   0.139
  0.003   0.005   0.003   0.005   0.003
   6  5.987   1.136   0.383   1.391   1.164   1.391   0.139   0.226   0.139
  0.003   0.005   0.003   0.005   0.003
 
 mulliken: Qtot =   32.000
 
 then oxygen has positive ionic charge (6-5.987), and silicon has
 negative (4-4.026). Does it make any sense??
 
 On 19/04/07, Vasilii Artyukhov [EMAIL PROTECTED] wrote:
  Just subtract the ionic charge from the Mulliken population, you will get
  -.514 for O and +1.028 for Ti.
 
  2007/4/19, Yurko Natanzon [EMAIL PROTECTED] :
   Dear SIESTers,
   I wonder how to calculate net ionic charge with SIESTA. For example, I
   have TiO2 anatase supercell with 12 atoms. Pseudopotential for Ti has
   valence electrons in 3s2 3p6 3d2 4s2 (12 electrons), O has 2s2 2p4 (6
   electrons). Then i set WriteMullikenPop 2 and obtain
   10.972 for each Ti
   6.514 fo O
   Total charge is 10.972*4+6.514*8 = 96 which equals the total number of
   valence electrons. Then how to obtain net electric charges of Ti and O
   atoms? If it is the difference between number of valence electrons and
   a charge obtained by SIESTA, than it is much lower, than is reported
   in other calculations. Net charge should be around +2 for Ti and -1
   for O.
  
   --
   Yurko Natanzon
   PhD Student
   Henryk Niewodniczański Institute of Nuclear Physics
   Polish Academy of Sciences
   ul. Radzikowskiego 152,
   31-342 Kraków, Poland
   Email: [EMAIL PROTECTED], [EMAIL PROTECTED]
  
 
 
 
 
 -- 
 Yurko Natanzon
 PhD Student
 Henryk Niewodniczański Institute of Nuclear Physics
 Polish Academy of Sciences
 ul. Radzikowskiego 152,
 31-342 Kraków, Poland
 Email: [EMAIL PROTECTED], [EMAIL PROTECTED]
 
 

---
Dr. Gali ÁdámAdam Gali, PhD

Budapesti Műszaki és Department of Atomic Physics,
Gazdaságtudományi Egyetem,   Budapest University of Technology and
Atomfizika Tanszék   Economics
Budapest, Budafoki út 8.,    Budafoki út 8., H-, Budapest,
 Hungary

telefon: 463-1580telephone: [36]-(1)-463-1580
fax: 463-4357fax:  [36]-(1)-463-4357

e-mail: [EMAIL PROTECTED]
   http://www.fat.bme.hu/homepages/galia/index.en.html
---

Re: [SIESTA-L] net charge calculation

2007-04-19 Thread Andrei Postnikov
On Thu, 19 Apr 2007, Yurko Natanzon wrote:

| Well, this works for TiO2, but when I tried to do it with SiO2, I've got:
| 
| then oxygen has positive ionic charge (6-5.987), and silicon has
| negative (4-4.026). Does it make any sense??

Dear Yurko:
they have no more sense than ionic charge in general; 
such numbers are totally dependent on their definition 
(in the case given - on your choice of basis functions),
even more so as you deal with covalent bond.

Similarly funny numbers may occur e.g. in LMTO-ASA
(charges come from integration over overlapping atomic spheres).
It doesn't mean that your calculation is wrong, though.
Just, better not to show these numbers to chemists...

Best regards

Andrei Postnikov

+-- Dr. Andrei Postnikov  Tel. +33-387315873 - mobile +33-666784053 ---+
| Paul Verlaine University - Institute de Physique Electronique et Chimie, |
| Laboratoire de Physique des Milieux Denses, 1 Bd Arago, F-57078 Metz, France |
+-- [EMAIL PROTECTED] -- http://www.home.uni-osnabrueck.de/apostnik/ --+



Re: [SIESTA-L] net charge calculation

2007-04-19 Thread Yurko Natanzon

Well, this works for TiO2, but when I tried to do it with SiO2, I've got:
mulliken: Atomic and Orbital Populations:

Species: Si
Atom  Qatom  Qorb
  3s  3s  3py 3pz 3px 3py 3pz 3px
  3Pdxy   3Pdyz   3Pdz2   3Pdxz   3Pdx2-y2
  1  4.026   0.404   0.474   0.308   0.152   0.308   0.254   0.394   0.254
 0.338   0.330   0.259   0.330   0.219
  4  4.026   0.404   0.474   0.308   0.152   0.308   0.254   0.394   0.254
 0.338   0.330   0.259   0.330   0.219

Species: O
Atom  Qatom  Qorb
  2s  2s  2py 2pz 2px 2py 2pz 2px
  2Pdxy   2Pdyz   2Pdz2   2Pdxz   2Pdx2-y2
  2  5.987   1.136   0.383   1.391   1.164   1.391   0.139   0.226   0.139
 0.003   0.005   0.003   0.005   0.003
  3  5.987   1.136   0.383   1.391   1.164   1.391   0.139   0.226   0.139
 0.003   0.005   0.003   0.005   0.003
  5  5.987   1.136   0.383   1.391   1.164   1.391   0.139   0.226   0.139
 0.003   0.005   0.003   0.005   0.003
  6  5.987   1.136   0.383   1.391   1.164   1.391   0.139   0.226   0.139
 0.003   0.005   0.003   0.005   0.003

mulliken: Qtot =   32.000

then oxygen has positive ionic charge (6-5.987), and silicon has
negative (4-4.026). Does it make any sense??

On 19/04/07, Vasilii Artyukhov [EMAIL PROTECTED] wrote:

Just subtract the ionic charge from the Mulliken population, you will get
-.514 for O and +1.028 for Ti.

2007/4/19, Yurko Natanzon [EMAIL PROTECTED] :
 Dear SIESTers,
 I wonder how to calculate net ionic charge with SIESTA. For example, I
 have TiO2 anatase supercell with 12 atoms. Pseudopotential for Ti has
 valence electrons in 3s2 3p6 3d2 4s2 (12 electrons), O has 2s2 2p4 (6
 electrons). Then i set WriteMullikenPop 2 and obtain
 10.972 for each Ti
 6.514 fo O
 Total charge is 10.972*4+6.514*8 = 96 which equals the total number of
 valence electrons. Then how to obtain net electric charges of Ti and O
 atoms? If it is the difference between number of valence electrons and
 a charge obtained by SIESTA, than it is much lower, than is reported
 in other calculations. Net charge should be around +2 for Ti and -1
 for O.

 --
 Yurko Natanzon
 PhD Student
 Henryk Niewodniczański Institute of Nuclear Physics
 Polish Academy of Sciences
 ul. Radzikowskiego 152,
 31-342 Kraków, Poland
 Email: [EMAIL PROTECTED], [EMAIL PROTECTED]






--
Yurko Natanzon
PhD Student
Henryk Niewodniczański Institute of Nuclear Physics
Polish Academy of Sciences
ul. Radzikowskiego 152,
31-342 Kraków, Poland
Email: [EMAIL PROTECTED], [EMAIL PROTECTED]