Dear users,
I have been comparing the output from a NEB calculations and one from a CI-NEB calculation for the
dissociation of a NO molecule on a Pt surface.
The result from the neb calculation, as obtained using the nebbarrier.pl script is:
0 0.000000 0.000000 0.000000 0
1 0.904029 0.502353 -1.203445 1
2 1.809576 2.242689 -4.132918 2
3 2.716485 2.012534 0.927999 3
4 3.624264 1.392145 1.313531 4
5 4.530816 0.790637 0.156333 5
6 5.437633 0.346388 0.000000 6
While the result from a ci-neb calculation (starting with the final images from the neb calculation) is:
0 0.000000 0.000000 0.000000 0
1 1.807779 0.347128 -0.671574 1
2 3.619005 2.084444 -0.002807 2
3 5.031200 1.407191 0.035186 3
4 6.443959 0.897233 0.370033 4
5 7.857079 0.859869 0.492205 5
6 9.270229 0.346388 0.000000 6
I find this strange, since path is now much longer with a CI-NEB calculation and the CI-NEB now gives a smaller barrier. The INCAR file is (for the neb calculation):
SPRING=-5
IMAGES=5
ISTART = 0
ICHARG = 2
ENMAX = 400
ALGO = V
EDIFF=1E-4
EDIFFG=-0.030
LREAL=A
ROPT=3*0.0005
NSW=600
POTIM = 0.1
IBRION = 3
ISMEAR=2 ; SIGMA=0.2
SMASS=1.0
LWAVE=.FALSE.
LVTOT=.FALSE.
IDIPOL=3
LDIPOL=.TRUE.
LCLIMB=.FALSE.
LCHARG=.FALSE.
The only thing that I can think of is that the linear interpolation between the initial and the final state is not very accurate, resulting in a bad initial
guess of a transition state. As a consequence the neb calculation does not give a very accurate estimation of the minimum energy path. Is this correct or
is there something I am not considering correctly?
Thank you in advance for your answers.
CI-NEB calculation vs. NEB calculation
Moderator: moderators
Re: CI-NEB calculation vs. NEB calculation
This looks like an advertisement for the climbing image!
In the regular neb calculation, the barrier is between images 2 and 3. The high force along the band (column 4) shows that neither image is particularly close to the saddle. An interpolation would give a barrier which overestimates the true barrier by several tenths of an eV.
In the climbing image calculation, image 2 is right at the saddle and you get a good estimation for the energy barrier.
You would have to look at the geometries to see why one path is so much longer than the other. If the true saddle is well-off the linear path, the climbing image would extend the length of the band to get there. Also, as you say, it is possible that there are two different pathways for the reaction. One last thing to make sure is that there is no net translation of your system (you should have frozen Pt atoms) or of the top Pt layer.
The nice thing about the climbing image is that you have the saddle and the barrier. If you are still suspicious of the mechanism, you can minimize from either side of the saddle and make sure that you reach your specified endpoints.
In the regular neb calculation, the barrier is between images 2 and 3. The high force along the band (column 4) shows that neither image is particularly close to the saddle. An interpolation would give a barrier which overestimates the true barrier by several tenths of an eV.
In the climbing image calculation, image 2 is right at the saddle and you get a good estimation for the energy barrier.
You would have to look at the geometries to see why one path is so much longer than the other. If the true saddle is well-off the linear path, the climbing image would extend the length of the band to get there. Also, as you say, it is possible that there are two different pathways for the reaction. One last thing to make sure is that there is no net translation of your system (you should have frozen Pt atoms) or of the top Pt layer.
The nice thing about the climbing image is that you have the saddle and the barrier. If you are still suspicious of the mechanism, you can minimize from either side of the saddle and make sure that you reach your specified endpoints.
Re: CI-NEB calculation vs. NEB calculation
Thank you for the reply.
I will do the last check that you suggested.
As movement of the Pt atoms, only the top two Pt layers are allowed to relax (the rest of the Pt atoms are kept frozen). Thus, there cannot be a no net translation of the system if I understand everything correctly.
I will do the last check that you suggested.
As movement of the Pt atoms, only the top two Pt layers are allowed to relax (the rest of the Pt atoms are kept frozen). Thus, there cannot be a no net translation of the system if I understand everything correctly.