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[Anglade Pierre-Matthieu <anglade@gmail.com>]

1.  
- Supposing you are right in saying "I did it but the result was not satisfactory" : you have computed the correct cohesive energy has given by DFT + LDA or GGA. If you are not satisfied you may try to compute the same thing with a CAS or DFT + B3LYP or ...
Now supposing that you made a mistake : do you really think that it is very easy to guess your mistake without ANY actual information ? Would you mind having a look at http://www.abinit.org/community/?text=netiquette ? This would be very helpful for yourself and for people trying to help you. Following the advices there it is not unlikely that your question becomes clearer and the answers more accurate. Since I can't know what you did, I guess all the following may just be junk, but any way let me try :

- nkpt 1 is OK for any large real space cell since the reciprocal space cell gets quite small. However, as usual you must check : is there enough kpt to get convergence (with respect to kpt) ? If no, use more kpt.
- I guess you have not copied every single parameter used to calculate the isolated hydrogen atom ? Do you ? The most simple way to proceed would be to just keep the calculation parameters you have used so far. 
In the case of t23.in it happens that the isolated hydrogen atom is much better described taking into account the spin (because it has a single electron).  That's the reason for using input variable like spinat, nsppol ...

2. Acell has decreased ? 
You mean that for your computer 9*1.2**n < 9*1.2**(n-1) with n>0,  right ?
You'd better install a working fortran compiler, recompile abinit and check that the self tests passe before going any further. 

Endly, let me recast my prayer : try to be as precise as possible in your questions. For instance "( I did it but the result was not satisfactory, maybe I should know more details.)" is plain wrong.  For instance, don't you think that a formulation similar to the following for your questions would have helped a little bit :
Trying to follow the examples of tutorial ... I did :
A bulk  BaTiO3 calculation with input file batio3.in (joined) I get etotal = xxx
Atomisation energies calculations for Ba, Ti and O with input files ... (joined) I get 
the total energy Eba = ..., ETi = ..., EO = ... . 
According to those results, I find the cohesive energy of BaTiO3 to be xxx. This seems incorrect since the experimental/theoretical data from \cite{paperXXXX} and \cite{paper2YYYY} says it should be yyy.
Did I made any mistake ? 

regards

PMA

On Tue, Feb 17, 2009 at 7:53 AM, hiva m <hiva.ms@gmail.com> wrote:

Thanks again for replying me.

1. in lession t23, for calculating the isolated atom energy we have:

#Second dataset : get the total energy of the isolated atom of hydrogen
 natom?2 1            
 nsppol?2 2           
 occopt?2 2          
 nband?2 1 1 
 occ?2 1.0  0.0    
 toldfe?2 1.0d-6                              
 xcart?2 0.0 0.0 0.0  
 spinat?2 0.0 0.0 1.0 
#rprim 1 0 0  0 1 0  0 0 1
nkpt 1 (is it always true for isolated atoms?)
                         
If I take the same trend, calculating the isolated atom energy and subtract it from the total energy of the crystal I have I can have the cohesive energy. ( I did it but the result was not satisfactory, maybe I should know more details.)

2. in what you have kindly written:

ndtset 15
acell: 3*9
acell* 3*1.2

the acell has decreased, why?

On Mon, Feb 16, 2009 at 3:53 PM, Anglade Pierre-Matthieu <anglade@gmail.com> wrote:

Hi,
I'm not sure to understand your concerns ?
For instance suppose that for your bulk system you have
acell 3*9
just add the following into your input (provided you have specified atoms coordinates with xred and not xcart or xangst) :
ndtset 15
acell: 3*9
acell* 3*1.2
prtwf 1
getwfk -1

If you happen to find that the derivative of total energy with respect to volume at any point is close enough to zero you can tell that this is where the cohesive energy is "zero" and then use this as a reference to compute the cohesive energy of your cristal.
Obviously for large volume cells you may want to reduce the number of k-points.

regards

PMA

On Mon, Feb 16, 2009 at 5:06 AM, hiva m <hiva.ms@gmail.com> wrote:

Hi,

Hello,

Thanks for the reply, but I still didn't get it.

You can Just proceed as usual with other codes :

I want to proceed with abinit.
 

increase your cell parameter to the point you can consider the cohesive energy to be zero.

zero cohesive energy?  what does it mean?!                                                                           

Alternatively you may want to do this calculation go reach isolated atom, Ba, Ti, and O.

 

that was a part of my question, calculating the single atom energy has no other regulation?
I know that the functional for the single atom is also the same as the one we use in the condensed system, and that the atomic configuration for which we are going to test the pseudopotential corresponds to the ground state, polarized in spin.

Please let me know if there's any detail to consider for the atomic configuration in  calculating the single atom energy.

Regards,
Hiva

--
Pierre-Matthieu Anglade

-- 
Pierre-Matthieu Anglade