Re: [SIESTA-L] Total energy

2022-06-23 Por tôpico RCP 

Hi again,

On 22/06/2022 19:27, Francisco Garcia wrote:

Thank you very much for the reference!Â

I'll use E=1/2(Etot +FreeEng) for my E-V equation of state fit.

I noticed that for semiconductors, Etot = FreeEng; it is in metallic systems 


Sure, because semiconductors do have an energy gap. Smearing (i.e. electronic
temperature) is just a trick used for metals in order to smooth their Fermi
surface so that some BZ integrals become easier to perform.
Take care,
Roberto

that I see differences between Etot and FreeEng (with FreeEng being lower). In 
this regard, using the average of the two energies for T=0 makes sense.


Thanks again!

On Wed, Jun 22, 2022 at 3:01 PM RCP > wrote:


Hello,
May I contribute my 0.01 ?. According to Eq.(21) in,

   Kresse & Furthmüller, Comp. Mat. Sci. 6 (1996) 15-50

a good estimate for the (T=0K) DFT energy is 1/2(Etot +FreeEng), and
this is the recipe I've always used.

Regards,
Roberto

On 22/06/2022 13:11, Francisco Garcia wrote:
 > Thanks  for your response Prof. Artacho!
 >
 > So if I have a situation like below for a large metallic system in which
FreeEng
 > is lower than Etot, should I use Etot? Specifically, if I want to plot
the E-V
 > curve from single point runs to obtain the equation of state I should use
Etot
 > instead of FreeEng? Or if I want to compute the cohesive energy, Etot
should be
 > used instead of FreeEng?
 >
 > siesta: Program's energy decomposition (eV):
 > siesta: Ebs     =      -859.157108
 > siesta: Eions   =      9568.777238
 > siesta: Ena     =       277.651800
 > siesta: Ekin    =      4247.415878
 > siesta: Enl     =     -1698.956164
 > siesta: Eso     =         0.00
 > siesta: Edftu   =         0.00
 > siesta: DEna    =       162.636813
 > siesta: DUscf   =        26.981770
 > siesta: DUext   =         0.00
 > siesta: Exc     =     -6715.533109
 > siesta: eta*DQ  =         0.00
 > siesta: Emadel  =         0.00
 > siesta: Emeta   =         0.00
 > siesta: Emolmec =         0.00
 > siesta: Ekinion =         0.00
 > siesta: Eharris =      -13268.580292
 > siesta: Etot    =        -13268.580250
 > siesta: FreeEng =    -13269.911499
 >
 > On Tue, Jun 21, 2022 at 2:07 PM Emilio Artacho mailto:e.arta...@nanogune.eu>
 > >> wrote:
 >
 >
 >>     On 20 Jun 2022, at 17:16, Francisco Garcia
mailto:garcia.ff@gmail.com>
 >>     >> wrote:
 >>
 >>     Dear users,
 >>
 >>     In metallic systems with a fairly sizable electronic smearing
temperature
 >>     T, is it accurate to claim that
 >>
 >>     (i) in a single point calculation, the free energy is the
representative
 >>     energy of the system (due to the addition of -TS to the total
energy U)
 >
 >     No, the -TS term for the phononic entropy is missing.Â
 >
 >     It is the free energy defining the finite temperature equilibrium
for the
 >     purely electronic
 >     problem for fixed external potential (fixed nuclei), as in
Mermin’s finite
 >     -T DFT.
 >
 >
 >>     (ii) in a variable cell optimization, the enthalpy is the
representative
 >>     energy of the system (due to the addition of the PV term to the
energy).
 >
 >     Yes, it is the free energy of the system (minimum defines
equilibrium) for T=0.
 >
 >     Emilio
 >
 >>
 >>     Thanks!
 >>     Â
 >>
 >>     --
 >>     SIESTA is supported by the Spanish Research Agency (AEI) and by 
the
 >>     European H2020 MaX Centre of Excellence 
(http://www.max-centre.eu/)
 >
 >     --
 >     Emilio Artacho
 >
 >     Theory Group, Nanogune, 20018 San Sebastian, Spain, and
 >     Theory of Condensed Matter, Department of Physics,
 >     Cavendish Laboratory, University of Cambridge, Cambridge CB3
0HE, UK
 >
 >
 >
 >
 >
 >     --
 >     SIESTA is supported by the Spanish Research Agency (AEI) and by
the European
 >     H2020 MaX Centre of Excellence (http://www.max-centre.eu/)
 >
 >
 >

-- 
Dr. Roberto C. Pasianot         Phone: 54 11 4839 6709

Gcia. Materiales, CAC-CNEA      FAX  : 54 11 6772 7362
Avda. Gral. Paz 1499           pasia...@cne

Re: [SIESTA-L] Total energy

2022-06-23 Por tôpico Francisco Garcia 
Thank you very much for the reference!

I'll use E=1/2(Etot +FreeEng) for my E-V equation of state fit.

I noticed that for semiconductors, Etot = FreeEng; it is in metallic
systems that I see differences between Etot and FreeEng (with FreeEng being
lower). In this regard, using the average of the two energies for T=0 makes
sense.

Thanks again!

On Wed, Jun 22, 2022 at 3:01 PM RCP  wrote:

> Hello,
> May I contribute my 0.01 ?. According to Eq.(21) in,
>
>Kresse & Furthmüller, Comp. Mat. Sci. 6 (1996) 15-50
>
> a good estimate for the (T=0K) DFT energy is 1/2(Etot +FreeEng), and
> this is the recipe I've always used.
>
> Regards,
> Roberto
>
> On 22/06/2022 13:11, Francisco Garcia wrote:
> > Thanks  for your response Prof. Artacho!
> >
> > So if I have a situation like below for a large metallic system in which
> FreeEng
> > is lower than Etot, should I use Etot? Specifically, if I want to plot
> the E-V
> > curve from single point runs to obtain the equation of state I should
> use Etot
> > instead of FreeEng? Or if I want to compute the cohesive energy, Etot
> should be
> > used instead of FreeEng?
> >
> > siesta: Program's energy decomposition (eV):
> > siesta: Ebs     =      -859.157108
> > siesta: Eions   =      9568.777238
> > siesta: Ena     =       277.651800
> > siesta: Ekin    =      4247.415878
> > siesta: Enl     =     -1698.956164
> > siesta: Eso     =         0.00
> > siesta: Edftu   =         0.00
> > siesta: DEna    =       162.636813
> > siesta: DUscf   =        26.981770
> > siesta: DUext   =         0.00
> > siesta: Exc     =     -6715.533109
> > siesta: eta*DQ Â = Â  Â  Â  Â  0.00
> > siesta: Emadel  =         0.00
> > siesta: Emeta   =         0.00
> > siesta: Emolmec = Â  Â  Â  Â  0.00
> > siesta: Ekinion = Â  Â  Â  Â  0.00
> > siesta: Eharris = Â  Â Â  -13268.580292
> > siesta: Etot    =        -13268.580250
> > siesta: FreeEng = Â  Â -13269.911499
> >
> > On Tue, Jun 21, 2022 at 2:07 PM Emilio Artacho  > > wrote:
> >
> >
> >> On 20 Jun 2022, at 17:16, Francisco Garcia  >> > wrote:
> >>
> >> Dear users,
> >>
> >> In metallic systems with a fairly sizable electronic smearing
> temperature
> >> T, is it accurate to claim that
> >>
> >> (i) in a single point calculation, the free energy is the
> representative
> >> energy of the system (due to the addition of -TS to the total
> energy U)
> >
> > No, the -TS term for the phononic entropy is missing.Â
> >
> > It is the free energy defining the finite temperature equilibrium
> for the
> > purely electronic
> > problem for fixed external potential (fixed nuclei), as in
> Mermin’s finite
> > -T DFT.
> >
> >
> >> (ii) in a variable cell optimization, the enthalpy is the
> representative
> >> energy of the system (due to the addition of the PV term to the
> energy).
> >
> > Yes, it is the free energy of the system (minimum defines
> equilibrium) for T=0.
> >
> > Emilio
> >
> >>
> >> Thanks!
> >> Â
> >>
> >> --
> >> SIESTA is supported by the Spanish Research Agency (AEI) and by the
> >> European H2020 MaX Centre of Excellence (http://www.max-centre.eu/)
> >
> > --
> > Emilio Artacho
> >
> > Theory Group, Nanogune, 20018 San Sebastian, Spain, and
> > Theory of Condensed Matter, Department of Physics,
> > Cavendish Laboratory, University of Cambridge, Cambridge CB3 0HE, UK
> >
> >
> >
> >
> >
> > --
> > SIESTA is supported by the Spanish Research Agency (AEI) and by the
> European
> > H2020 MaX Centre of Excellence (http://www.max-centre.eu/)
> >
> >
> >
>
> --
> Dr. Roberto C. Pasianot Phone: 54 11 4839 6709
> Gcia. Materiales, CAC-CNEA  FAX  : 54 11 6772 7362
> Avda. Gral. Paz 1499pasia...@cnea.gov.ar
> 1650 San Martin, Buenos Aires
> ARGENTINA
>

-- 
SIESTA is supported by the Spanish Research Agency (AEI) and by the European 
H2020 MaX Centre of Excellence (http://www.max-centre.eu/)

Re: [SIESTA-L] Total energy

2022-06-23 Por tôpico RCP 

Hello,
May I contribute my 0.01 ?. According to Eq.(21) in,

  Kresse & Furthmüller, Comp. Mat. Sci. 6 (1996) 15-50

a good estimate for the (T=0K) DFT energy is 1/2(Etot +FreeEng), and
this is the recipe I've always used.

Regards,
Roberto

On 22/06/2022 13:11, Francisco Garcia wrote:

Thanks  for your response Prof. Artacho!

So if I have a situation like below for a large metallic system in which FreeEng 
is lower than Etot, should I use Etot? Specifically, if I want to plot the E-V 
curve from single point runs to obtain the equation of state I should use Etot 
instead of FreeEng? Or if I want to compute the cohesive energy, Etot should be 
used instead of FreeEng?


siesta: Program's energy decomposition (eV):
siesta: Ebs     =      -859.157108
siesta: Eions   =      9568.777238
siesta: Ena     =       277.651800
siesta: Ekin    =      4247.415878
siesta: Enl     =     -1698.956164
siesta: Eso     =         0.00
siesta: Edftu   =         0.00
siesta: DEna    =       162.636813
siesta: DUscf   =        26.981770
siesta: DUext   =         0.00
siesta: Exc     =     -6715.533109
siesta: eta*DQ  =         0.00
siesta: Emadel  =         0.00
siesta: Emeta   =         0.00
siesta: Emolmec =         0.00
siesta: Ekinion =         0.00
siesta: Eharris =      -13268.580292
siesta: Etot    =        -13268.580250
siesta: FreeEng =    -13269.911499

On Tue, Jun 21, 2022 at 2:07 PM Emilio Artacho > wrote:




On 20 Jun 2022, at 17:16, Francisco Garcia mailto:garcia.ff@gmail.com>> wrote:

Dear users,

In metallic systems with a fairly sizable electronic smearing temperature
T, is it accurate to claim that

(i) in a single point calculation, the free energy is the representative
energy of the system (due to the addition of -TS to the total energy U)


No, the -TS term for the phononic entropy is missing.Â

It is the free energy defining the finite temperature equilibrium for the
purely electronic
problem for fixed external potential (fixed nuclei), as in Mermin’s finite
-T DFT.



(ii) in a variable cell optimization, the enthalpy is the representative
energy of the system (due to the addition of the PV term to the energy).


Yes, it is the free energy of the system (minimum defines equilibrium) for 
T=0.

Emilio



Thanks!
Â

-- 
SIESTA is supported by the Spanish Research Agency (AEI) and by the

European H2020 MaX Centre of Excellence (http://www.max-centre.eu/)


--
Emilio Artacho

Theory Group, Nanogune, 20018 San Sebastian, Spain, and
Theory of Condensed Matter, Department of Physics,
Cavendish Laboratory, University of Cambridge, Cambridge CB3 0HE, UK





-- 
SIESTA is supported by the Spanish Research Agency (AEI) and by the European

H2020 MaX Centre of Excellence (http://www.max-centre.eu/)





--
Dr. Roberto C. Pasianot Phone: 54 11 4839 6709
Gcia. Materiales, CAC-CNEA  FAX  : 54 11 6772 7362
Avda. Gral. Paz 1499pasia...@cnea.gov.ar
1650 San Martin, Buenos Aires
ARGENTINA

-- 
SIESTA is supported by the Spanish Research Agency (AEI) and by the European 
H2020 MaX Centre of Excellence (http://www.max-centre.eu/)

Re: [SIESTA-L] Total energy

2022-06-22 Por tôpico Francisco Garcia 
Thanks  for your response Prof. Artacho!

So if I have a situation like below for a large metallic system in which
FreeEng is lower than Etot, should I use Etot? Specifically, if I want to
plot the E-V curve from single point runs to obtain the equation of state I
should use Etot instead of FreeEng? Or if I want to compute the cohesive
energy, Etot should be used instead of FreeEng?

siesta: Program's energy decomposition (eV):
siesta: Ebs =  -859.157108
siesta: Eions   =  9568.777238
siesta: Ena =   277.651800
siesta: Ekin=  4247.415878
siesta: Enl = -1698.956164
siesta: Eso = 0.00
siesta: Edftu   = 0.00
siesta: DEna=   162.636813
siesta: DUscf   =26.981770
siesta: DUext   = 0.00
siesta: Exc = -6715.533109
siesta: eta*DQ  = 0.00
siesta: Emadel  = 0.00
siesta: Emeta   = 0.00
siesta: Emolmec = 0.00
siesta: Ekinion = 0.00
siesta: Eharris =  -13268.580292
siesta: Etot=-13268.580250
siesta: FreeEng =-13269.911499

On Tue, Jun 21, 2022 at 2:07 PM Emilio Artacho 
wrote:

>
> On 20 Jun 2022, at 17:16, Francisco Garcia 
> wrote:
>
> Dear users,
>
> In metallic systems with a fairly sizable electronic smearing temperature
> T, is it accurate to claim that
>
> (i) in a single point calculation, the free energy is the representative
> energy of the system (due to the addition of -TS to the total energy U)
>
>
> No, the -TS term for the phononic entropy is missing.
>
> It is the free energy defining the finite temperature equilibrium for the
> purely electronic
> problem for fixed external potential (fixed nuclei), as in Mermin’s finite
> -T DFT.
>
>
> (ii) in a variable cell optimization, the enthalpy is the representative
> energy of the system (due to the addition of the PV term to the energy).
>
>
> Yes, it is the free energy of the system (minimum defines equilibrium) for
> T=0.
>
> Emilio
>
>
> Thanks!
>
>
> --
> SIESTA is supported by the Spanish Research Agency (AEI) and by the
> European H2020 MaX Centre of Excellence (http://www.max-centre.eu/)
>
>
> --
> Emilio Artacho
>
> Theory Group, Nanogune, 20018 San Sebastian, Spain, and
> Theory of Condensed Matter, Department of Physics,
> Cavendish Laboratory, University of Cambridge, Cambridge CB3 0HE, UK
>
>
>
>
>
> --
> SIESTA is supported by the Spanish Research Agency (AEI) and by the
> European H2020 MaX Centre of Excellence (http://www.max-centre.eu/)
>

-- 
SIESTA is supported by the Spanish Research Agency (AEI) and by the European 
H2020 MaX Centre of Excellence (http://www.max-centre.eu/)

Re: [SIESTA-L] Total energy

2022-06-21 Por tôpico Emilio Artacho 

On 20 Jun 2022, at 17:16, Francisco Garcia 
mailto:garcia.ff@gmail.com>> wrote:

Dear users,

In metallic systems with a fairly sizable electronic smearing temperature T, is 
it accurate to claim that

(i) in a single point calculation, the free energy is the representative energy 
of the system (due to the addition of -TS to the total energy U)

No, the -TS term for the phononic entropy is missing.

It is the free energy defining the finite temperature equilibrium for the 
purely electronic
problem for fixed external potential (fixed nuclei), as in Mermin’s finite -T 
DFT.


(ii) in a variable cell optimization, the enthalpy is the representative energy 
of the system (due to the addition of the PV term to the energy).

Yes, it is the free energy of the system (minimum defines equilibrium) for T=0.

Emilio


Thanks!


--
SIESTA is supported by the Spanish Research Agency (AEI) and by the European 
H2020 MaX Centre of Excellence (http://www.max-centre.eu/)

--
Emilio Artacho

Theory Group, Nanogune, 20018 San Sebastian, Spain, and
Theory of Condensed Matter, Department of Physics,
Cavendish Laboratory, University of Cambridge, Cambridge CB3 0HE, UK





-- 
SIESTA is supported by the Spanish Research Agency (AEI) and by the European 
H2020 MaX Centre of Excellence (http://www.max-centre.eu/)

[SIESTA-L] Total energy

2022-06-20 Por tôpico Francisco Garcia 
Dear users,

In metallic systems with a fairly sizable electronic smearing temperature
T, is it accurate to claim that

(i) in a single point calculation, the free energy is the representative
energy of the system (due to the addition of -TS to the total energy U)

(ii) in a variable cell optimization, the enthalpy is the representative
energy of the system (due to the addition of the PV term to the energy).

Thanks!

-- 
SIESTA is supported by the Spanish Research Agency (AEI) and by the European 
H2020 MaX Centre of Excellence (http://www.max-centre.eu/)

Re: [SIESTA-L] Total energy of my supercell was calculated to be lower than that of the optimized unit cell.

2018-12-05 Por tôpico RCP 

Hi Yang Mino,

Try running the perfect lattice case also using 39 unit cells. Check
the parameters (e.g. k-point mesh) are the same as in the stacking
fault run. Thus you'd have two calculatins fully comparable.
Hopefully, the deffect energy will be positive.
Good luck,

Roberto

On 04/12/18 06:54, 밀러 wrote:

Dear users.

I'm trying to calculate the energy of stacking fault energy of wurtzite
GaN.

I firstly have optimized the unit cell of GaN. I changed the volumes of
a unit cell from 98 to 110% using an experimental value and applied
siesta calculation with lattice parameters of different a/c ratio with
respect to the changed volumes.

The lowest energy was found at the lattice constant of 1.626 c/a ration
at 108.35% volume. The cutting Energy was tested and determined to be
450eV.
The energy of the single unit cell is 4.9121eV.

Then I made stacking fault supercell which contained 39 unit-cell. The
energy of the supercell was calculated -162667.255043eV.

The defect energy for a supercell is -1.683143eV. (wall energy is
55.6meV/Ang^2) The minus energy does not seem to be normal. It should be
plus as far as I know.
So, I had checked again the volume optimization process but found no error.

What I suspect is that there could be small errors in the optimized
volume. But it seems like that the slight difference in volumes is not
well distinguished. Actually, the energy of 108.35% volume and 108.47%
were the same in my calculations. Does not siesta distinguish the
energies difference of the cells under very slight lattice parameters
difference?

I don't know where I have mistakes.

Advise me, please.



양민호 (Yang Mino, Ph.D.)

Senior researcher|Korea Basic Science Institute

[SIESTA-L] Total energy of my supercell was calculated to be lower than that of the optimized unit cell.

2018-12-04 Por tôpico 밀러 







Dear users.I'm trying to calculate the energy of stacking fault energy of wurtzite GaN. I firstly have optimized the unit cell of GaN. I changed the volumes of a unit cell from 98 to 110% using an experimental value and applied siesta calculation with lattice parameters of different a/c ratio with respect to the changed volumes. The lowest energy was found at the lattice constant of 1.626 c/a ration at 108.35% volume. The cutting Energy was tested and determined to be 450eV. The energy of the single unit cell is 4.9121eV. Then I made stacking fault supercell which contained 39 unit-cell. The energy of the supercell was calculated -162667.255043eV. The defect energy for a supercell is -1.683143eV. (wall energy is 55.6meV/Ang^2) The minus energy does not seem to be normal. It should be plus as far as I know. So, I had checked again the volume optimization process but found no error.  What I suspect is that there could be small errors in the optimized volume. But it seems like that the slight difference in volumes is not well distinguished. Actually, the energy of 108.35% volume and 108.47% were the same in my calculations. Does not siesta distinguish the energies difference of the cells under very slight lattice parameters difference?I don't know where I have mistakes. Advise me, please.
양민호 (Yang Mino, Ph.D.)Senior researcher|Korea Basic Science Institute









SystemName  GaN 
SystemLabel caratio_1626 
NumberOfAtoms   4 
NumberOfSpecies 2

%block ChemicalSpeciesLabel
 1  7  N  # Species index, atomic number, species label
 2  31  Ga  # Species index, atomic number, species label
%endblock ChemicalSpeciesLabel

%block kgrid_Monkhorst_Pack
  15   0   00.0
  0   15   00.0
  0   0  150.0
%endblock kgrid_Monkhorst_Pack

MeshCutoff  480. Ry

LatticeConstant  1.0 Ang

%block LatticeParameters
  3.2771 3.2771 5.3286 90. 90. 120.0
%endblock LatticeParameters

AtomicCoordinatesFormat Fractional
%block AtomicCoordinatesAndAtomicSpecies
  0.3270  0.6730  0.3809885049989710  1
  0.6730  0.3270  0.8809885049989710  1   
  0.3270  0.6730  0.0040114855010280  2   
  0.6730  0.3270  0.5040114855010280  2
%endblock AtomicCoordinatesAndAtomicSpecies

SpinPolarized false
xc.functional LDA
xc.authorsCA

SolutionMethoddiagon

MaxSCFIterations  200
DM.MixingWeight   0.02
DM.Tolerance  1.d-5
DM.NumberPulay 4
ElectronicTemperature  300 K

DM.UseSaveDM  true
UseSaveData  true
WriteMullikenPop   1

%block kgrid_Monkhorst_Pack
  9   0   00.1
  0   9   00.1
  0   0   60.1
%endblock kgrid_Monkhorst_Pack

WriteCoorXmol   T

%block PAO.Basis
Ga  3
n=4   0   2
   0.000  0.000
   1.000  1.000
n=4   1   2  P  1
   0.000  0.000
   1.000  1.000
n=3   2  1
   0.000
   1.000
N   2
 n=2   0   2
   0.000  0.000
   1.000  1.000
 n=2   1   2 P   1
   0.000  0.000
   1.000  1.000
%endblock PAO.Basis

[SIESTA-L] Total energy of my supercell was calculated to be lower than that of the optimized unit cell.

2018-11-08 Por tôpico 밀러 






Dear users.I'm trying to calculate the energy of stacking fault energy of wurtzite GaN. I firstly have optimized the unit cell of GaN. I changed the volumes of a unit cell from 98 to 110% using an experimental value and applied siesta calculation with lattice parameters of different a/c ratio with respect to the changed volumes. The lowest energy was found at the lattice constant of 1.626 c/a ration at 108.35% volume. The cutting Energy was tested and determined to be 450eV. The energy of the single unit cell is 4.9121eV. Then I made stacking fault supercell which contained 39 unit-cell. The energy of the supercell was calculated -162667.255043eV. The defect energy for a supercell is -1.683143eV. (wall energy is 55.6meV/Ang^2) The minus energy does not seem to be normal. It should be plus as far as I know. So, I had checked again the volume optimization process but found no error.  What I suspect is that there could be small errors in the optimized volume. But it seems like that the slight difference in volumes is not well distinguished. Actually, the energy of 108.35% volume and 108.47% were the same in my calculations. Does not siesta distinguish the energies difference of the cells under very slight lattice parameters difference?I don't know where I have mistakes. Advise me, please.
양민호 (Yang Mino, Ph.D.)Senior researcher|Korea Basic Science Institute






SystemName  GaN 
SystemLabel caratio_1626 
NumberOfAtoms   4 
NumberOfSpecies 2

%block ChemicalSpeciesLabel
 1  7  N  # Species index, atomic number, species label
 2  31  Ga  # Species index, atomic number, species label
%endblock ChemicalSpeciesLabel

%block kgrid_Monkhorst_Pack
  15   0   00.0
  0   15   00.0
  0   0  150.0
%endblock kgrid_Monkhorst_Pack

MeshCutoff  480. Ry

LatticeConstant  1.0 Ang

%block LatticeParameters
  3.2771 3.2771 5.3286 90. 90. 120.0
%endblock LatticeParameters

AtomicCoordinatesFormat Fractional
%block AtomicCoordinatesAndAtomicSpecies
  0.3270  0.6730  0.3809885049989710  1
  0.6730  0.3270  0.8809885049989710  1   
  0.3270  0.6730  0.0040114855010280  2   
  0.6730  0.3270  0.5040114855010280  2
%endblock AtomicCoordinatesAndAtomicSpecies

SpinPolarized false
xc.functional LDA
xc.authorsCA

SolutionMethoddiagon

MaxSCFIterations  200
DM.MixingWeight   0.02
DM.Tolerance  1.d-5
DM.NumberPulay 4
ElectronicTemperature  300 K

DM.UseSaveDM  true
UseSaveData  true
WriteMullikenPop   1

%block kgrid_Monkhorst_Pack
  9   0   00.1
  0   9   00.1
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Ga  3
n=4   0   2
   0.000  0.000
   1.000  1.000
n=4   1   2  P  1
   0.000  0.000
   1.000  1.000
n=3   2  1
   0.000
   1.000
N   2
 n=2   0   2
   0.000  0.000
   1.000  1.000
 n=2   1   2 P   1
   0.000  0.000
   1.000  1.000
%endblock PAO.Basis

Re: [SIESTA-L] Total Energy calculations with a H+ ion

2015-06-03 Por tôpico zohreh Siahpoush 
hiI want to relax this nanoribbon , but that is not be relaxedplease help methe 
file and figure were jointed
 


 On Wednesday, June 3, 2015 6:04 PM, Herbert Fruchtl 
 wrote:
   

 H+ has no electrons, so the electronic energy is 0. I don't know to what 
extent 
this is "the right zero" if you have a pseudopotential rather than a 
point-charge nucleus.

Even if this was not the case (i.e. if you have a larger ion), one has to be 
careful with charged unit cells. SIESTA (and any other periodic code I know) 
will add a background charge, because otherwise you have an infinitely charged 
system. I think the background charge changes the Hamiltonian sufficiently so 
that the intuitive calculation of the adsorption energy does not work (you need 
to take differences between energies of charged and uncharged cells).

Experimentally, you never have free protons. Depending on where your H+ comes 
from, you could look at the reaction

G + H3O+ -> GH + H2O

If you place your H3O+ or H2O far above the surface, you can do each side of 
the 
reaction in one calculation, and you always have the same charge.

There must also be a counter ion somewhere...

HTH,

  Herbert

On 03/06/15 13:44, James Lawlor wrote:
> Hi,
>
> I'm trying to do calculate the binding energy of H+ with graphene, so this
> involves finding the total energies of 3 systems - H+ isolated, graphene
> isolated, and the combined system.
>
>  My current method is to use "NetCharge 1.0" in the input files of the 
>isolated
> H+ and the combined systems, which should in theory remove an electron from 
> the
> system. The problem is this returns errors for the H+ as the system is
> essentially a proton and I think this is causing SIESTA to get confused.
>
> Could anyone suggest a possible solution, or perhaps a different method?
>
> Cheers,
>
> James
>
> --
> James Lawlor
> Theory & Modelling Group
> School of Physics
> Trinity College Dublin, Ireland

-- 
Herbert Fruchtl
Senior Scientific Computing Officer
School of Chemistry, School of Mathematics and Statistics
University of St Andrews
--
The University of St Andrews is a charity registered in Scotland:
No SC013532


  

gyne.fdf
Description: application/vnd.fdf

Re: [SIESTA-L] Total Energy calculations with a H+ ion

2015-06-03 Por tôpico Herbert Fruchtl 
H+ has no electrons, so the electronic energy is 0. I don't know to what extent 
this is "the right zero" if you have a pseudopotential rather than a 
point-charge nucleus.


Even if this was not the case (i.e. if you have a larger ion), one has to be 
careful with charged unit cells. SIESTA (and any other periodic code I know) 
will add a background charge, because otherwise you have an infinitely charged 
system. I think the background charge changes the Hamiltonian sufficiently so 
that the intuitive calculation of the adsorption energy does not work (you need 
to take differences between energies of charged and uncharged cells).


Experimentally, you never have free protons. Depending on where your H+ comes 
from, you could look at the reaction


G + H3O+ -> GH + H2O

If you place your H3O+ or H2O far above the surface, you can do each side of the 
reaction in one calculation, and you always have the same charge.


There must also be a counter ion somewhere...

HTH,

  Herbert

On 03/06/15 13:44, James Lawlor wrote:

Hi,

I'm trying to do calculate the binding energy of H+ with graphene, so this
involves finding the total energies of 3 systems - H+ isolated, graphene
isolated, and the combined system.

  My current method is to use "NetCharge 1.0" in the input files of the isolated
H+ and the combined systems, which should in theory remove an electron from the
system. The problem is this returns errors for the H+ as the system is
essentially a proton and I think this is causing SIESTA to get confused.

Could anyone suggest a possible solution, or perhaps a different method?

Cheers,

James

--
James Lawlor
Theory & Modelling Group
School of Physics
Trinity College Dublin, Ireland


--
Herbert Fruchtl
Senior Scientific Computing Officer
School of Chemistry, School of Mathematics and Statistics
University of St Andrews
--
The University of St Andrews is a charity registered in Scotland:
No SC013532

Re: [SIESTA-L] Total Energy calculations with a H+ ion

2015-06-03 Por tôpico I. Camps 
Hello,

Could it be a solution to generate a pseudo for the charged H?

Regards,

Camps

On Wed, Jun 3, 2015 at 10:13 AM, Nicolas Leconte 
wrote:

> James, you are right about the reason SIESTA is complaining. I had the
> same problem long time ago, and I did not find any other solution but to
> replace H+ with H. Probably not that big a problem because Netcharge is not
> doing anything local anyhow, cfr Nick's comment. And if your graphene
> system is large enough, I doubt the impact to be large.
>
> You might do a test with any other ion and see if the binding energy is
> very different using Netcharge 1 vs Netcharge 0.
>
> Good luck!
> Nicolas
>
> PS : Don't forget to include BSSE corrections... (search the mailing list
> if you don't know how to do that).
>
> On Wed, Jun 3, 2015 at 9:50 PM, Nick Papior Andersen  > wrote:
>
>> NetCharge 1. does not yield an H+ plus graphene (I would be surprised),
>> rather it would more likely be graphene+1 plus H.
>> Be careful here...
>>
>> 2015-06-03 14:44 GMT+02:00 James Lawlor :
>>
>>> Hi,
>>>
>>> I'm trying to do calculate the binding energy of H+ with graphene, so
>>> this involves finding the total energies of 3 systems - H+ isolated,
>>> graphene isolated, and the combined system.
>>>
>>>  My current method is to use "NetCharge 1.0" in the input files of the
>>> isolated H+ and the combined systems, which should in theory remove an
>>> electron from the system. The problem is this returns errors for the H+ as
>>> the system is essentially a proton and I think this is causing SIESTA to
>>> get confused.
>>>
>>> Could anyone suggest a possible solution, or perhaps a different method?
>>>
>>> Cheers,
>>>
>>> James
>>>
>>> --
>>> James Lawlor
>>> Theory & Modelling Group
>>> School of Physics
>>> Trinity College Dublin, Ireland
>>>
>>
>>
>>
>> --
>> Kind regards Nick
>>
>
>

Re: [SIESTA-L] Total Energy calculations with a H+ ion

2015-06-03 Por tôpico Nicolas Leconte 
James, you are right about the reason SIESTA is complaining. I had the same
problem long time ago, and I did not find any other solution but to replace
H+ with H. Probably not that big a problem because Netcharge is not doing
anything local anyhow, cfr Nick's comment. And if your graphene system is
large enough, I doubt the impact to be large.

You might do a test with any other ion and see if the binding energy is
very different using Netcharge 1 vs Netcharge 0.

Good luck!
Nicolas

PS : Don't forget to include BSSE corrections... (search the mailing list
if you don't know how to do that).

On Wed, Jun 3, 2015 at 9:50 PM, Nick Papior Andersen 
wrote:

> NetCharge 1. does not yield an H+ plus graphene (I would be surprised),
> rather it would more likely be graphene+1 plus H.
> Be careful here...
>
> 2015-06-03 14:44 GMT+02:00 James Lawlor :
>
>> Hi,
>>
>> I'm trying to do calculate the binding energy of H+ with graphene, so
>> this involves finding the total energies of 3 systems - H+ isolated,
>> graphene isolated, and the combined system.
>>
>>  My current method is to use "NetCharge 1.0" in the input files of the
>> isolated H+ and the combined systems, which should in theory remove an
>> electron from the system. The problem is this returns errors for the H+ as
>> the system is essentially a proton and I think this is causing SIESTA to
>> get confused.
>>
>> Could anyone suggest a possible solution, or perhaps a different method?
>>
>> Cheers,
>>
>> James
>>
>> --
>> James Lawlor
>> Theory & Modelling Group
>> School of Physics
>> Trinity College Dublin, Ireland
>>
>
>
>
> --
> Kind regards Nick
>

Re: [SIESTA-L] Total Energy calculations with a H+ ion

2015-06-03 Por tôpico Nick Papior Andersen 
NetCharge 1. does not yield an H+ plus graphene (I would be surprised),
rather it would more likely be graphene+1 plus H.
Be careful here...

2015-06-03 14:44 GMT+02:00 James Lawlor :

> Hi,
>
> I'm trying to do calculate the binding energy of H+ with graphene, so this
> involves finding the total energies of 3 systems - H+ isolated, graphene
> isolated, and the combined system.
>
>  My current method is to use "NetCharge 1.0" in the input files of the
> isolated H+ and the combined systems, which should in theory remove an
> electron from the system. The problem is this returns errors for the H+ as
> the system is essentially a proton and I think this is causing SIESTA to
> get confused.
>
> Could anyone suggest a possible solution, or perhaps a different method?
>
> Cheers,
>
> James
>
> --
> James Lawlor
> Theory & Modelling Group
> School of Physics
> Trinity College Dublin, Ireland
>



-- 
Kind regards Nick

[SIESTA-L] Total Energy calculations with a H+ ion

2015-06-03 Por tôpico James Lawlor 
Hi,

I'm trying to do calculate the binding energy of H+ with graphene, so this
involves finding the total energies of 3 systems - H+ isolated, graphene
isolated, and the combined system.

 My current method is to use "NetCharge 1.0" in the input files of the
isolated H+ and the combined systems, which should in theory remove an
electron from the system. The problem is this returns errors for the H+ as
the system is essentially a proton and I think this is causing SIESTA to
get confused.

Could anyone suggest a possible solution, or perhaps a different method?

Cheers,

James

-- 
James Lawlor
Theory & Modelling Group
School of Physics
Trinity College Dublin, Ireland