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[hiva m <hiva.ms@gmail.com>]

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