Re: [Pw_forum] Cohesive energy of bulk Iridium

2015-10-05 Thread Venkataramana Imandi
 Dear sir,

I have done calculations according your suggestions.
I got cohesive energy at 0 K is as follows.
934.7 kJ/mol
749.6 kJ/mol (with dispersion correction)
731.7 kJ/mol (with 2x2x1 k-point)
670.0 kJ/mol (Experiment at 0 K)

I increased plane wave cutoff to 35 to 40 Ryd., there is no change in the
values.

I got surface energy as follows.
11.7 kJ/mol/Angstrom^2
28.4 kJ/mol/Angstrom^2 (with dispersion correction)
28.4 kJ/mol/Angstrom^2 (with 2x2x1 k-point)
14.9 kJ/mol/Angstrom^2 (Experiment at 0 K)

How to refine my results in such way that i will get close to experimental
results by tuning input parameters. Please suggest and your information is
valuable to me.
In single Iridium case, the input file contains additional nspin=2 and
starting_magnetization(1)=1 parameters and I didn't include those
parameters in the bulk Iridium case.
I am attaching one input file with dispersion correction as shown below.


 Structure: Ir fcc(111) ; (3x2x2) SUPER-CELL

 
 6 Layers slab


&control
calculation='relax',
prefix='ir_ch',
nstep=5000,
   etot_conv_thr=1.0D-5,
   forc_conv_thr=1.0D-4,
   pseudo_dir = '/home/venkat/ORR1/PPS1'
 /
&system
   ibrav=0,
nat=72,
ntyp=1,
  ecutwfc = 35.D0,
  ecutrho=350.D0,
 nosym=.true.,
occupations='smearing',
smearing='m-p',
 degauss=0.07D0,
 vdw_corr='grimme-d2'
/
&electrons
   electron_maxstep=2000,
  diagonalization='david',
   mixing_beta = 0.7D0,
conv_thr =  1.0D-8,
   scf_must_converge=.true.
  mixing_mode = 'local-TF' ,
  startingpot = 'atomic' ,
   startingwfc = 'atomic' ,
/
&ions
ion_dynamics='bfgs'
/
ATOMIC_SPECIES
Ir   192.217   Ir.pw91-n-rrkjus_psl.0.2.3.UPF

CELL_PARAMETERS angstrom
  8.14374880.0.
  0.0009.403591120.
  0.0000.   13.29868610

ATOMIC_POSITIONS angstrom
Ir   6.78645733  5.48542817  6.64934305
Ir   5.42916587  7.83632593  6.64934305
Ir   5.42916587  4.70179556  4.43289536
Ir   6.78645733  7.05269336  4.43289536
Ir   5.42916587  6.26906075  2.21644768
Ir   6.78645733  8.61995854  2.21644768
Ir   4.07187440  5.48542817  6.64934305
Ir   2.71458293  7.83632593  6.64934305
Ir   2.71458293  4.70179556  4.43289536
Ir   4.07187440  7.05269336  4.43289536
Ir   2.71458293  6.26906075  2.21644768
Ir   4.07187440  8.61995854  2.21644768
Ir   1.35729146  5.48542817  6.64934305
Ir   0.  7.83632593  6.64934305
Ir   0.  4.70179556  4.43289536
Ir   1.35729146  7.05269336  4.43289536
Ir   0.  6.26906075  2.21644768
Ir   1.35729146  8.61995854  2.21644768
Ir   6.78645733  0.78363261  6.64934305
Ir   5.42916587  3.13453037  6.64934305
Ir   5.42916587  0.  4.43289536
Ir   6.78645733  2.35089780  4.43289536
Ir   5.42916587  1.56726519  2.21644768
Ir   6.78645733  3.91816298  2.21644768
Ir   4.07187440  0.78363261  6.64934305
Ir   2.71458293  3.13453037  6.64934305
Ir   2.71458293  0.  4.43289536
Ir   4.07187440  2.35089780  4.43289536
Ir   2.71458293  1.56726519  2.21644768
Ir   4.07187440  3.91816298  2.21644768
Ir   1.35729146  0.78363261  6.64934305
Ir   0.  3.13453037  6.64934305
Ir   0.  0.  4.43289536
Ir   1.35729146  2.35089780  4.43289536
Ir   0.  1.56726519  2.21644768
Ir   1.35729146  3.91816298  2.21644768
Ir   6.78645733  5.48542817  0.
Ir   5.42916587  7.83632593  0.
Ir   5.42916587  4.70179556 -2.21644769
Ir   6.78645733  7.05269336 -2.21644769
Ir   5.42916587  6.26906075 -4.43289537
Ir   6.78645733  8.61995854 -4.43289537
Ir   4.07187440  5.48542817  0.
Ir   2.71458293  7.83632593  0.

Re: [Pw_forum] Cohesive energy of bulk Iridium

2015-10-04 Thread Venkataramana Imandi
Dear prof. Stefano,
Thank you very much for kind your help.
I will check results.

On Sat, Oct 3, 2015 at 2:59 PM, Venkataramana Imandi <
venkataramana.ima...@gmail.com> wrote:

> Dear prof. *STEFANO DE GIRONCOLI*
>
> Many thanks for spontaneous reply. On the basis of your answer, If I
> understood correctly, I can use nspin=2 for atomic Iridium and bulk Iridium
> (since Iridium is paramagnetic from literature data). However, I have to
> specify starting_magnetization in the input file in the both atomic Iridium
> and bulk Iridium
> input files. The keywords list information says that values range between
> -1 (all spins down for the valence electrons of atom type 'i') to 1 (all
> spins up). Iridium has three unpaired electrons in the spin up.
> 1. It indicates can I use starting_magnetization(1)=1 along with nspin=2.
> 2. If not that value, what value I have to use, I don't know, please
> suggest me.
> 3. I am not using full relativistic pseudopotential for Iridium, so, I can
> skip nspin=4, am I correct ?.
>
>  Please verify my assumptions.
>  The reply of previous thread of clean stop of running job, now I got
> clean stop during running job.
>
>
> On Sat, Oct 3, 2015 at 12:52 PM, Venkataramana Imandi <
> venkataramana.ima...@gmail.com> wrote:
>
>>
>> Dear all
>>
>> I want to calculate ground state total energy of single Iridium neutral
>> gaseous atom.
>> The electronic configuration of Iridium atom is [Xe].4f^14.5d^7.6s^2 and
>> in the 5d orbital, three unpaired electrons are there.
>> So, the resultant spin multiplicity is 4. Hence, in keywords list,
>> nspin=4 or noncolin=.true. is essential.
>> I am asking that whether nspin means spin multiplicity or not ?.
>> Am i correct for determining total energy of Iridium with nspin=4 or
>> noncolin=.true. ?
>> In case of bulk Iridium (total atoms=72), then can I skip nspin=4 or
>> noncolin=.true. ?
>> In case of bulk Iridium electrons can get paired or not ?
>>
>> Finally I want to calculate cohesive energy of bulk Iridium.
>>
>> I am extremely say sorry, if questions are fundamental and trivial.
>>
>> Any suggestions are appreciated and thanks in anticipated.
>>
>>
>> venkataramana
>> PhD student
>> IIT Bombay
>> Mumbai
>>
>
>
>
> --
> venkataramana
>



-- 
venkataramana
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Re: [Pw_forum] Cohesive energy of bulk Iridium

2015-10-03 Thread stefano de gironcoli

dear Venkataramana Imandi
 nspin=2 + starting_magnetization/=0 should be ok for your case i think.

 the value of the starting magnetization should not be very important. 
Its role is to break the up/down symmetry in the first iteration and 
then the code should reach self consistency.
 Unless the system has many competing solutions with different 
magnetizations the final scf value of the magnetization should be the same.
 The initial value may affect the number of iterations needed to reach 
self-consistency but should not affect the final self consistent 
results. You may want to explore this dependence if you plan to make 
many calculations with different supercells to study defect or other 
properties.
 Starting magnetization can be used to explore different magnetic 
configurations (ferromagnetic, AFM, more complex configurations if many 
inequivalent atoms/sites are present).

In bulk Iridium i think a simple FM solution should be enough.

best

stefano

On 03/10/2015 11:29, Venkataramana Imandi wrote:

Dear prof. *STEFANO DE GIRONCOLI*

Many thanks for spontaneous reply. On the basis of your answer, If I 
understood correctly, I can use nspin=2 for atomic Iridium and bulk 
Iridium
(since Iridium is paramagnetic from literature data). However, I have 
to specify starting_magnetization in the input file in the both atomic 
Iridium and bulk Iridium
input files. The keywords list information says that values range 
between -1 (all spins down for the valence electrons of atom type 'i') 
to 1 (all spins up). Iridium has three unpaired electrons in the spin up.

1. It indicates can I use starting_magnetization(1)=1 along with nspin=2.
2. If not that value, what value I have to use, I don't know, please 
suggest me.
3. I am not using full relativistic pseudopotential for Iridium, so, I 
can skip nspin=4, am I correct ?.


 Please verify my assumptions.
 The reply of previous thread of clean stop of running job, now I got 
clean stop during running job.


On Sat, Oct 3, 2015 at 12:52 PM, Venkataramana Imandi 
> wrote:



Dear all

I want to calculate ground state total energy of single Iridium
neutral gaseous atom.
The electronic configuration of Iridium atom is
[Xe].4f^14.5d^7.6s^2 and in the 5d orbital, three unpaired
electrons are there.
So, the resultant spin multiplicity is 4. Hence, in keywords list,
nspin=4 or noncolin=.true. is essential.
I am asking that whether nspin means spin multiplicity or not ?.
Am i correct for determining total energy of Iridium with nspin=4
or noncolin=.true. ?
In case of bulk Iridium (total atoms=72), then can I skip nspin=4
or noncolin=.true. ?
In case of bulk Iridium electrons can get paired or not ?

Finally I want to calculate cohesive energy of bulk Iridium.

I am extremely say sorry, if questions are fundamental and trivial.

Any suggestions are appreciated and thanks in anticipated.


venkataramana
PhD student
IIT Bombay
Mumbai




--
venkataramana


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Re: [Pw_forum] Cohesive energy of bulk Iridium

2015-10-03 Thread Venkataramana Imandi
Dear prof. *STEFANO DE GIRONCOLI*

Many thanks for spontaneous reply. On the basis of your answer, If I
understood correctly, I can use nspin=2 for atomic Iridium and bulk Iridium
(since Iridium is paramagnetic from literature data). However, I have to
specify starting_magnetization in the input file in the both atomic Iridium
and bulk Iridium
input files. The keywords list information says that values range between
-1 (all spins down for the valence electrons of atom type 'i') to 1 (all
spins up). Iridium has three unpaired electrons in the spin up.
1. It indicates can I use starting_magnetization(1)=1 along with nspin=2.
2. If not that value, what value I have to use, I don't know, please
suggest me.
3. I am not using full relativistic pseudopotential for Iridium, so, I can
skip nspin=4, am I correct ?.

 Please verify my assumptions.
 The reply of previous thread of clean stop of running job, now I got clean
stop during running job.


On Sat, Oct 3, 2015 at 12:52 PM, Venkataramana Imandi <
venkataramana.ima...@gmail.com> wrote:

>
> Dear all
>
> I want to calculate ground state total energy of single Iridium neutral
> gaseous atom.
> The electronic configuration of Iridium atom is [Xe].4f^14.5d^7.6s^2 and
> in the 5d orbital, three unpaired electrons are there.
> So, the resultant spin multiplicity is 4. Hence, in keywords list, nspin=4
> or noncolin=.true. is essential.
> I am asking that whether nspin means spin multiplicity or not ?.
> Am i correct for determining total energy of Iridium with nspin=4 or
> noncolin=.true. ?
> In case of bulk Iridium (total atoms=72), then can I skip nspin=4 or
> noncolin=.true. ?
> In case of bulk Iridium electrons can get paired or not ?
>
> Finally I want to calculate cohesive energy of bulk Iridium.
>
> I am extremely say sorry, if questions are fundamental and trivial.
>
> Any suggestions are appreciated and thanks in anticipated.
>
>
> venkataramana
> PhD student
> IIT Bombay
> Mumbai
>



-- 
venkataramana
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Re: [Pw_forum] Cohesive energy of bulk Iridium

2015-10-03 Thread stefano de gironcoli

nspin is used to define whether you are performing a
non magnetic (nspin=1,default), collinear magnetic (nspin=2) or 
non-collinear/fully relativistic (nspin=4) calculation.


In the atomic case if you want to fix the total spin (the up/dw density 
unbalance) you can use
the total_magnetization variable (there use to be a redundant 
multiplicity variable but i think it has been suppressed).


In most cases you just run nspin=2 calculations specifying a non-zero 
starting_magnetization to break the up/dw symmetry


stefano

On 03/10/2015 09:22, Venkataramana Imandi wrote:


Dear all

I want to calculate ground state total energy of single Iridium 
neutral gaseous atom.
The electronic configuration of Iridium atom is [Xe].4f^14.5d^7.6s^2 
and in the 5d orbital, three unpaired electrons are there.
So, the resultant spin multiplicity is 4. Hence, in keywords list, 
nspin=4 or noncolin=.true. is essential.

I am asking that whether nspin means spin multiplicity or not ?.
Am i correct for determining total energy of Iridium with nspin=4 or 
noncolin=.true. ?
In case of bulk Iridium (total atoms=72), then can I skip nspin=4 or 
noncolin=.true. ?

In case of bulk Iridium electrons can get paired or not ?

Finally I want to calculate cohesive energy of bulk Iridium.

I am extremely say sorry, if questions are fundamental and trivial.

Any suggestions are appreciated and thanks in anticipated.


venkataramana
PhD student
IIT Bombay
Mumbai


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