Re: [SIESTA-L] Valence configuration of samarium

2008-04-10 Thread apostnik
 Hi,

 I have made the changes according to u and defined the pseudo of Sm as

Hi -
If u refers to me, mind that I never advised you on the pseudo radii
for Sm...
I only meant that inclusing 4f is technically no error - but it will
probably lead to physically wrong results. Whatever...

Regarding your problems with calculations which sometimes stop
at an error in basis block, sometimes not:
you should carefully read the SIESTA doc file about the syntaxis
of the basis block. Namely:

 but when I define the PAO basis like
 %block PAO.Basis
  Sm  2  # Label, l-shells
   n=6   0   2 P   1 # n, l, Nzeta, Polarization,NzetaPol
 0.000  0.000
 1.000  1.000
   n=6   1   2
 0.000  0.000
 1.000  1.000
   n=5   2   2
 0.000  0.000
 1.000  1.000
   n=4   3   2
 0.000  0.000
 1.000  1.000

- yes, it fails, because you declared 2 l-shells (1st line of the Sm block)
but in the following you pass the info for 4 l-shells. I thought we were
already past this problem...

 Again I mabe the changes in PAO basis as
 %block PAO.Basis
 Sm  2  # Label, l-shells
  n=6   0   2 P   1   # n, l, Nzeta, Polarization, NzetaPol
0.000  0.000
1.000  1.000
  n=4   3   2
0.000  0.000
1.000  1.000
 Te  2# Species label, number of l-shells
  n=5   0   2 P   1   # n, l, Nzeta, Polarization, NzetaPol
0.000  0.000
1.000  1.000
  n=5   1   2 # n, l, Nzeta
0.000  0.000
1.000  1.000
  %endblock PAO.Basis

 then the program run successfully but lattice parameter and band gap is
 far from our expt values.

Now this is formally correct (number of l-shells declared equals
number of l-shells passed), but the basis (at least for Sm) is not
complete enough - you'd need 5d in the basis.
And after you do this, do not expect to get band structure right,
because 4f states will most probably be at a wrong energy...

Good luck

Andrei Postnikov


Re: [SIESTA-L] Valence configuration of samarium

2008-04-09 Thread Eduardo Anglada


Excuse-me, but have u noticed that you cut off for all basis is ZERO?!
It means that all of your electrons are confined to the nucleous...


Sorry but, No. It means: use siesta heuristics (energy shift and split  
norm) in order to determine the rc.


Regards,
Eduardo






all of then, n this way u have no bonds formation and a huge
electronic repulsion I would say ... U need to especiffy your cut off
radius resoanably!

Cheers..

NH
On Wed, Apr 9, 2008 at 1:13 PM, Nidhi Sharma [EMAIL PROTECTED]  
wrote:

Hi,

I have made the changes according to u and defined the pseudo of Sm  
as

pg  Samarium
   tm2  3.0 # PS flavor, logder R
n=Sm c=ca  # Symbol, XC flavor,{ |r|s}

  0.0   0.0   0.0   0.0   0.0   0.0
  114   # norbs_core, norbs_valence
   60  2.00  0.00   # 6s2
   61  0.00  0.00   # 6p0
   52  0.00  0.00   # 5d0
   43  6.00  0.00   # 4f6
 3.11  4.10  3.11  3.11  0.00  0.00
and Te as
pg  Tellurum
   tm2  3.0
n=Te c=ca
  0.0   0.0   0.0   0.0   0.0   0.0
   93

   50 2.00  0.00
   51 4.00  0.00
   52 0.00  0.00
  2.57 2.63 2.57
but when I define the PAO basis like

%block PAO.Basis
Sm 2 # Label, l-shells
n=6 0 2 P 1 # n, l, Nzeta, Polarization, NzetaPol
0.000 0.000
1.000 1.000
n=6 1 2

0.000 0.000
1.000 1.000
n=5 2 2

0.000 0.000
1.000 1.000
n=4 3 2

0.000 0.000
1.000 1.000
Te 2 # Species label, number of l-shells
n=5 0 2 P 1 # n, l, Nzeta, Polarization, NzetaPol
0.000 0.000
1.000 1.000
n=5 1 2 # n, l, Nzeta
0.000 0.000
1.000 1.000
n=5 2 2 # n, l, Nzeta
0.000 0.000
1.000 1.000
%endblock PAO.Basis


Then I got this kind of error on the command line

reinit:
---
reinit: System Name: SmTe
reinit:
---
reinit: System Label: SmTe
reinit:
---



initatom: Reading input for the pseudopotentials and atomic orbitals
--
Species number: 1 Label: Sm Atomic number: 62
Species number: 2 Label: Te Atomic number: 52
Ground state valence configuration: 6s02 4f06
Reading pseudopotential information in formatted form from Sm.psf
Ground state valence configuration: 5s02 5p04
Reading pseudopotential information in formatted form from Te.psf
WRONG species symbol in PAO.Basis: n

Stopping Program from Node: 0
Again I mabe the changes in PAO basis as
%block PAO.Basis
Sm  2  # Label, l-shells
n=6   0   2 P   1   # n, l, Nzeta, Polarization,  
NzetaPol

  0.000  0.000
  1.000  1.000
n=4   3   2

  0.000  0.000
  1.000  1.000
Te  2# Species label, number of l-shells
n=5   0   2 P   1   # n, l, Nzeta, Polarization,  
NzetaPol

  0.000  0.000
  1.000  1.000
n=5   1   2 # n, l, Nzeta
  0.000  0.000
  1.000  1.000
%endblock PAO.Basis

then the program run successfully but lattice parameter and band  
gap is far
from our expt values.  May you please check my .fdf file and guide  
me where
should make the appropriate changes.  I m enclosing .fdf file, .out  
file,

psf file and band structure (looks like metallic) that we obtained.

Many thanks in advance.


[EMAIL PROTECTED] wrote:

You mix up several things; I doubt it will help to resolve your  
problems

but let us address them one by one.

Sm valence configuration. 4f states are quite localized and
probably (in reality, not in DFT the calculation) are not any near
to the band gap. If you include them in valence states and in the  
basis,

your trouble will be not performing the calculation as such,
but their wrong calculated positioning (at the Fermi level).
If you attribute them to core... I don't know there is an easy way
to do this, because the 4f shell is not fully occupied.
Search for previous calculations (any method, with DFT and beyond)
on RE chalcogenides, and on any RE calculations using  
pseudopotentials.


Dear users, as we know the valence configuration of Sm is 4f6,6s2.  
In
order to combine it with chalcogenides it is necessary to make the  
net
ionic charge of Sm to 2, means we have to consider the 4f6 in the  
core.


Why necessary? The net ionic charge, whatever its definition,  
will come
out of your calculation somehow. To begin with, you start from  
neutral

atoms, and they remain neutral, whether you attribute 4f to the core
or to the valence states...


When we define the PAO basis set as

%block PAO.Basis
Sm 2 # Label, l-shells
n=6 0 2 P 1 # n, l, Nzeta, Polarization, NzetaPol
0.000 0.000
1.000 1.000


Don't forget to include 5d in the basis; they are IMPORTANT.


Te 2 # Species label, number of l-shells
n=5 0 2 P 1 # n, l, Nzeta, Polarization, NzetaPol
0.000 0.000
1.000 

Re: [SIESTA-L] Valence configuration of samarium

2008-04-08 Thread Marcos Verissimo Alves
Nidhi,

If you need 4f in the valence as semicore, then your line should have
something like

n=4  3   2

4f, double-zeta. Of course the number of zetas can vary, as well as the
number of polarization functions. What is the electronic configuration of
your Sm pseudo?

Marcos

Vous avez écrit / You have written / Lei ha scritto / Você escreveu...
Nidhi Sharma
 Hi to all,

 Dear users, as we know the valence configuration of Sm is 4f6,6s2. In
 order to combine it with chalcogenides it is necessary to make the net
 ionic charge of Sm to 2, means we have to consider the 4f6 in the core.
 When we define the PAO basis set as

 %block PAO.Basis
 Sm  2  # Label, l-shells
  n=6   0   2 P   1   # n, l, Nzeta, Polarization, NzetaPol
0.000  0.000
1.000  1.000
   Te  2# Species label, number of l-shells
  n=5   0   2 P   1   # n, l, Nzeta, Polarization, NzetaPol
0.000  0.000
1.000  1.000
  n=5   1   2 # n, l, Nzeta
0.000  0.000
1.000  1.000
  %endblock PAO.Basis

 it will display the following message
 reinit:
 ---
 reinit: System Name: SmTe
 reinit:
 ---
 reinit: System Label: SmTe
 reinit:
 ---

 initatom: Reading input for the pseudopotentials and atomic orbitals
 --
  Species number:1  Label: Sm Atomic number:  62
  Species number:2  Label: Te Atomic number:  52
 Ground state valence configuration:   6s02  4f06
 Reading pseudopotential information in formatted form from Sm.psf
 Ground state valence configuration:   5s02  5p04
 Reading pseudopotential information in formatted form from Te.psf
 Bad format of (n), l, nzeta line in PAO.Basis
 Stopping Program from Node:0

 If I include the 4f6 in basis set it will make the net charge 8 and behave
 as a semi core.

 If I use a already generated pseudo file of Te which include 5s2, 5p4, 4d0
 and 4f0  But how can 4d0 is possible although it contains 10 electrons.
 When I use this file then we get results but band gap in B1 phase is ~10eV
 which is quite far from the expt 0.67eV.

 Please help me how can i resolve the problem of valence charge .

 Thanks in advance.

 Nidhi


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Re: [SIESTA-L] Valence configuration of samarium

2008-04-08 Thread apostnik
You mix up several things; I doubt it will help to resolve your problems
but let us address them one by one.

Sm valence configuration. 4f states are quite localized and
probably (in reality, not in DFT the calculation) are not any near
to the band gap. If you include them in valence states and in the basis,
your trouble will be not performing the calculation as such,
but their wrong calculated positioning (at the Fermi level).
If you attribute them to core... I don't know there is an easy way
to do this, because the 4f shell is not fully occupied.
Search for previous calculations (any method, with DFT and beyond)
on RE chalcogenides, and on any RE calculations using pseudopotentials.

 Dear users, as we know the valence configuration of Sm is 4f6,6s2. In
 order to combine it with chalcogenides it is necessary to make the net
 ionic charge of Sm to 2, means we have to consider the 4f6 in the core.

Why necessary? The net ionic charge, whatever its definition, will come
out of your calculation somehow. To begin with, you start from neutral
atoms, and they remain neutral, whether you attribute 4f to the core
or to the valence states...

 When we define the PAO basis set as

 %block PAO.Basis
 Sm  2  # Label, l-shells
  n=6   0   2 P   1   # n, l, Nzeta, Polarization, NzetaPol
0.000  0.000
1.000  1.000

Don't forget to include 5d in the basis; they are IMPORTANT.

   Te  2# Species label, number of l-shells
  n=5   0   2 P   1   # n, l, Nzeta, Polarization, NzetaPol
0.000  0.000
1.000  1.000
  n=5   1   2 # n, l, Nzeta
0.000  0.000
1.000  1.000
  %endblock PAO.Basis

Your Te might be OK (or not); at least no obvious faults.


 it will display the following message
 reinit:
 reinit: System Label: SmTe
 ---
 initatom: Reading input for the pseudopotentials and atomic orbitals
 --
  Species number:1  Label: Sm Atomic number:  62
  Species number:2  Label: Te Atomic number:  52
 Ground state valence configuration:   6s02  4f06
 Reading pseudopotential information in formatted form from Sm.psf
 Ground state valence configuration:   5s02  5p04
 Reading pseudopotential information in formatted form from Te.psf
 Bad format of (n), l, nzeta line in PAO.Basis
 Stopping Program from Node:0

This is probably because you promised 2 functions in the basis block for Sm
but passed only one (6s). Make it consistent.

 If I include the 4f6 in basis set it will make the net charge 8 and behave
 as a semi core.

This net charge is not exactly your worry. It simply gives you the number
of electrons provided by the atom in question to the valence band,
in does not yet make from Sm a 8+ ion. Similarly, you can choose
the Te configuration either as 5s2 5p4 5d0 (6 valence electrons)
or 5s2 5p4 4d10 (16 valence electrons), it is still the same atom.
Only, you'll have different number of bands. I repeate, the decision
to put Sm 4f in the core or in the valence is only your - difficult,
but free - choice.

Now we come to Te.

 If I use a already generated pseudo file of Te which include 5s2, 5p4, 4d0
 and 4f0  But how can 4d0 is possible although it contains 10 electrons.

This is a misprint in the head line of the Te pseudo. It was generated
with 4d10 in the core and 5d as valence states. (Ask Eduardo Anglada).

 When I use this file then we get results but band gap in B1 phase
 is ~10eV which is quite far from the expt 0.67eV.

This can be due to anything. (In fact an absence of Sm5d in the basis
is a good candidate). Try to look not only at the band gap value
(which will be wrong anyway) but at the density of states,
positioning of different groups of valence bands. The band structure
of RE chalcogenides is well known.

 Please help me how can i resolve the problem of valence charge .

I don't see there is a problem, in fact. The (technical) problem is -
if you want to remove 4f from the valence - how to declare them as core,
even as this shell is not fully occupied. But by default, you can
go ahead with 4f as valence states (in BOTH basis and pseudopotential).
Then you'll see that the positioning of the 4f is wrong, and start
to think how bad this is for the problem your have to solve,
and what to do about it. This is not a SIESTA problem, but one
which appeared before in other DFT calculations.

Good luck,

Andrei Postnikov