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[Michel Cote <Michel.Cote@umontreal.ca>]

Hello Hua,

Le 07-07-11 à 12:25, Hua Bao a écrit :

Dear Konstantin Rushchanskii and all,

Thank you for your suggestions.

I have two more questions here.

1.After I changed the ecut to 30, the band structure became unreasonable. I didn't see any band gap for the anatase TiO2. And it is supposed to have  a gap of 3.2eV even with LDA functional. Then what is the possible the problems here?

2. I checked the force result and find some forces are exactly zero. See below. It seems not correct here. What may lead to this result?

   1      0.00000000000000     0.00000000000000     0.00000000000000

   2      0.00000000000000     0.00000000000000     0.00748690006424

   3      0.00000000000000     0.00000000000000    -0.00748690006424

   4      0.00000000000000     0.00000000000000     0.00000000000000

   5      0.00000000000000     0.00000000000000    -0.00748690006424

   6      0.00000000000000     0.00000000000000     0.00748690006424

The forces are zero because of symmetry, an atom pulled equally in all directions will result in a zero force on this atom. The other forces are small so it seems that your structure is well relaxed.

I am not sure why you find no gap, I am not familiar with this structure. Are many fill bands does the structure have? Are you looking at the right place on the band structure?

Michel

Thanks,

Hua

Input file,

ndtset 2

xangst        .......

ntypat 2                   znucl 22 8

   #Definition of the unit cell

acell  3.7842 3.7842 9.5146  angstrom  

#Definition of the atoms

natom 12           typat  1 2 2 1 2 2 1 2 1 2 2 2        

#Definition of the planewave basis set

ecut 30.0         # Maximal kinetic energy cut-off, in Hartree

chkprim 0

#Dataset 1

kptopt1 1

ngkpt1 4 4 4

prtden1 1

toldfe1 1.0d-6

#Dataset 2

#Definition of the k-point grid

iscf2 -2

getden2 -1

kptopt2 -7                  # taken by default to be 0.0 0.0 0.0

nband2 64

ndivk2 21 21 21 21 21 21 21

kptbounds2  0.0 0.0 0.0 #gamma

          0.5 0.0 0.0 #X

          0.5 0.0 0.5 #R

          0.0 0.0 0.5 #Z

          0.0 0.0 0.0 #gamma

          0.5 0.5 0.0 #M

          0.5 0.5 0.5 #A

          0.0 0.0 0.5 #5

tolwfr2 1.0d-12

enunit2 1

prteig2 1

#Definition of the SCF procedure

nstep 100          # Maximal number of SCF cycles

Konstantin Rushchanskii wrote:

Dear Hua Bao,

ecut 15.0         # Maximal kinetic energy cut-off, in Hartree

ecut 15 Ha is very low for Ti and for O.

If you use TM's pspnc pseudopotentials from abinit.org, the simplest way to find the minimal cutoff is to look for the maximal value of e99.9 parameter in the header of pseudopotential files.

In your case it is 28.075 (the cut-off energy for p-orbital of oxygen).

Regards,

KR

_____________________________________

Michel Cote

Departement de physique

Universite de Montreal

emai: Michel.Cote@umontreal.ca

web: http://www.phys.umontreal.ca/~michel_cote