ABINIT Discussion Forums

As you know, the activation energy for ion conductivity can be determined by simulating the oxide ion at a number of intermediate positions along the migration path in the cell. The following code deals with the oxide ion in the interstitial position (oxygen atom is fixed by x in the interstitial position), but it optimizes the lattice so that the total energy is almost the same as in the case where this oxygen atom is in the node position and it never stops to optimize. And never stops to lower the total energy. In the cases where this oxygen atom was in the node position the cell was successfully optimized (in these cases I've decided none of the atoms are fixed).
Here is my code for the case (GGA.fhi pseudopotentials):


Maxim.

Hello,

please start a new thread for a new problem.

You should look at the atomic positions, to see where the atoms are going.

As optcell is 0 you are not optimizing the lattice (I presume you mean optimizing all the other atomic positions). It is possible all the other atoms will move back to put the O in the "node" position again. We need output to see what is actually happening.

Also, your statement that the energy keeps going down is surprising - what exactly do you mean? And does it go below the "node" or bulk total energy?

Matthieu

Ssory,

I mean I wanted to calculate the activation energy by placing oxygen in different positions in the cell and substracting the highest value of the total energy from the lowest one. If I skip optimization and use only crystallographical data on a and c, I had about 4 eV activation energy (too large). So I've decided to optimize positions firstly. Should I perform optimization with optcell=1 also, and what are the corresponding values of ionmov? I guess that the ion in the interstitial position should require a volume relaxation.

Thanks
Maxim

mverstra wrote:It is possible all the other atoms will move back to put the O in the "node" position again.

Maybee I need to fix positions of atoms situated at the boundaries of the cell?

Maxim

Fixing an atom or 2 (at least in the direction of the O path) is a good idea, but you need to find out if it is addressing the problem. So, first, check what the relaxation is doing to the atoms, and why the energy is decreasing (it should decrease a bit).

optcell 0 is probably most physical, as the overall lattice will not expand, only the local atomic configuration.

Matthieu

maxim wrote:As you know, the activation energy for ion conductivity can be determined by simulating the oxide ion at a number of intermediate positions along the migration path in the cell. The following code deals with the oxide ion in the interstitial position (oxygen atom is fixed by x in the interstitial position), but it optimizes the lattice so that the total energy is almost the same as in the case where this oxygen atom is in the node position and it never stops to optimize. And never stops to lower the total energy. In the cases where this oxygen atom was in the node position the cell was successfully optimized (in these cases I've decided none of the atoms are fixed).
Here is my code for the case (GGA.fhi pseudopotentials):

Code: Select all

chkprim 0

#Parameters
nstep  100
ecut 18
ixc 11
iscf 7
toldfe 1.0d-6 
#Definition of occupation numbers
occopt 7
tsmear 0.8 eV

#Optimization of the nuclear positions
optcell 0
ionmov  3
ntime  40
tolmxf 1.0d-5
natfixx   1
iatfixx   5 

natom 47
ntypat 3
typat 2 1 3 3 3 3 2 1 3 3 3 1 1 3 3 3 3 1 1 3 3 3 3 1 1 3 3 3 3 1 1 3 3 3 3 1 1 3 3 3 3 1 1 3 3 3 3
znucl 40 58 8
xred 0.000000000000000 0.000000000000000 0.000000000000000
0.250000000000000 0.250000000000000 0.250000000000000
0.000000000000000 0.250000000000000 0.103150000000000
0.250000000000000 0.000000000000000 0.146850000000000
0.125000000000000 0.375000000000000 0.375000000000000
0.250000000000000 0.000000000000000 0.396850000000000
0.000000000000000 0.500000000000000 0.000000000000000
0.250000000000000 0.750000000000000 0.250000000000000
0.000000000000000 0.750000000000000 0.103150000000000
0.250000000000000 0.500000000000000 0.146850000000000
0.000000000000000 0.750000000000000 0.353150000000000
0.000000000000000 0.000000000000000 0.500000000000000
0.250000000000000 0.250000000000000 0.750000000000000
0.000000000000000 0.250000000000000 0.603150000000000
0.250000000000000 0.000000000000000 0.646850000000000
0.000000000000000 0.250000000000000 0.853150000000000
0.250000000000000 0.000000000000000 0.896850000000000
0.000000000000000 0.500000000000000 0.500000000000000
0.250000000000000 0.750000000000000 0.750000000000000
0.000000000000000 0.750000000000000 0.603150000000000
0.250000000000000 0.500000000000000 0.646850000000000
0.000000000000000 0.750000000000000 0.853150000000000
0.250000000000000 0.500000000000000 0.896850000000000
0.500000000000000 0.000000000000000 0.000000000000000
0.750000000000000 0.250000000000000 0.250000000000000
0.500000000000000 0.250000000000000 0.103150000000000
0.750000000000000 0.000000000000000 0.146850000000000
0.500000000000000 0.250000000000000 0.353150000000000
0.750000000000000 0.000000000000000 0.396850000000000
0.500000000000000 0.500000000000000 0.000000000000000
0.750000000000000 0.750000000000000 0.250000000000000
0.500000000000000 0.750000000000000 0.103150000000000
0.750000000000000 0.500000000000000 0.146850000000000
0.500000000000000 0.750000000000000 0.353150000000000
0.750000000000000 0.500000000000000 0.396850000000000
0.500000000000000 0.000000000000000 0.500000000000000
0.750000000000000 0.250000000000000 0.750000000000000
0.500000000000000 0.250000000000000 0.603150000000000
0.750000000000000 0.000000000000000 0.646850000000000
0.500000000000000 0.250000000000000 0.853150000000000
0.750000000000000 0.000000000000000 0.896850000000000
0.500000000000000 0.500000000000000 0.500000000000000
0.750000000000000 0.750000000000000 0.750000000000000
0.500000000000000 0.750000000000000 0.603150000000000
0.750000000000000 0.500000000000000 0.646850000000000
0.500000000000000 0.750000000000000 0.853150000000000
0.750000000000000 0.500000000000000 0.896850000000000

kptrlatt 3 0 0
         0 3 0
         0 0 3
shiftk 0.5 0.5 0.5
kptrlen 4.1155E+01
nkpt 14

acell 7.2594000000 7.2594000000 10.4434000000 angstrom

diemac 4.0

Maxim.

Thanks you for the post.
Hi guys, Im a newbie. Nice to join this forum.

mverstra wrote:Fixing an atom or 2 (at least in the direction of the O path) is a good idea, but you need to find out if it is addressing the problem. So, first, check what the relaxation is doing to the atoms, and why the energy is decreasing (it should decrease a bit).

optcell 0 is probably most physical, as the overall lattice will not expand, only the local atomic configuration.

Matthieu

Thanks you for the post.
There are some interesting things at http://me.stanford.edu/faculty/publicat ... _FINAL.pdf. They simulated (see page 3) the interstitial position by relaxing all atoms in three directions except for the case of interstitial atom and his some neigbours (see fig. 2): these atoms were fixed at the plane. And they used the internal optimization only as you adviced. Ill try all these things later.