UT theoretical chemistry code forum

Hello all.

I meet a problem in CI-NEB calculation to find the diffusion path of an atom on some surface. At the beginning I used 4 images with CI turned on, and got a weird geometric path of diffusion with energies like that:
0 0.00488800 -632.69314600 0.00000000
1 0.03114200 -631.77079900 0.92234700
2 0.02935700 -630.97151600 1.72163000
3 0.04859100 -631.81289900 0.88024700
4 0.04957100 -631.83806000 0.85508600
5 0.00497900 -631.84525600 0.84789000

Then I turned down the CI and got a more 'regular' diffusion path with energies:
0 0.00488800 -632.69314600 0.00000000
1 0.04730100 -631.81015600 0.88299000
2 0.03371100 -631.52114400 1.17200200
3 0.04334600 -631.75091200 0.94223400
4 0.04878200 -631.82580100 0.86734500
5 0.00497900 -631.84525600 0.84789000

To get the exact transition state of this diffusion, I interpolated the images on the base of the result of the second calculation, turned on the CI, and the weird result appeared again:
0 0.00488800 -632.69314600 0.00000000
1 0.11204800 -631.80954300 0.88360300
2 0.18528600 -631.02241100 1.67073500
3 0.24075100 -630.89844500 1.79470100
4 0.19671000 -631.74894800 0.94419800
5 0.25517000 -631.77662900 0.91651700
6 0.35579300 -631.80840500 0.88474100
7 0.03660800 -631.83039100 0.86275500
8 0.10738200 -631.83967500 0.85347100
9 0.00497900 -631.84525600 0.84789000
The most strange thing is that the geometric configurations of image 3 and 4 are very similar but the energy differ in 0.85eV. Further more, comparing the informations of the OSZICARs:
F= -.63089817E+03 E0= -.63089817E+03 d E =0.195702E-03 mag= 1.0000
for image 03
and
F= -.63174701E+03 E0= -.63174701E+03 d E =-.710089E-02 mag= 3.0000
for image 04,
it can be seen that the 'mag' values are totally different, again these two images are very similar in geometric configurations.
So I took single point calculations of these two image and got similar energies of about 0.95eV, which indicates that the energy estimated in NEB calculation of image 3 is largely different from which in single point calculation.


So I took this 8 image calculation without CI again and converged in a reasonable path:
0 0.00488800 -632.69314600 0.00000000
1 0.03475300 -632.62576200 0.06738400
2 0.04909100 -632.05208000 0.64106600
3 0.03966500 -631.33535300 1.35779300
4 0.03654200 -631.56127500 1.13187100
5 0.04271800 -631.73241200 0.96073400
6 0.04747600 -631.79912700 0.89401900
7 0.01875600 -631.82794100 0.86520500
8 0.02513200 -631.84020100 0.85294500
9 0.00497900 -631.84525600 0.84789000

The CI calculation on the base of result of the 4th calculation and the path went bad again:
0 0.00488800 -632.69314600 0.00000000
1 0.12846200 -631.98405200 0.70909400
2 0.17357500 -631.06113100 1.63201500
3 0.20635800 -630.98363800 1.70950800
4 0.20410400 -631.82819200 0.86495400
5 0.10395300 -631.82879100 0.86435500
6 0.07415500 -631.83491500 0.85823100
7 0.02683900 -631.83953100 0.85361500
8 0.02959700 -631.84305300 0.85009300
9 0.00497900 -631.84525600 0.84789000

So this calculation can only get reasonable result when CI is turned off, even in calculation with 8 image from the result of no-CI calculation.

There are 3 other calculations with the same problem. The similarity of these task is that breaks up of bonds happen in these paths and the 'mag' value changes from 1 to 3 in this breaking. The CI calculation move some images from bond-remained configuration to bond-broken one, but the mag value remains in 1, which seems to be queer.

That's the problem I met, a bit long story, hoping some one have an idea:)

It's a little hard to know what's going on without the details of the calculation. But it sounds like the magnetic moment could be the answer. If some images are converging to a higher electronic state (and particularly near the saddle for the climbing image) they can have higher energies and forces which are different from the other images on the ground state.

To check this, you might turn off spin polarization and see if the problems go away. You can also set the magnetic moment on each atom, in an attempt to get each image into the ground state. If you can get a stable path with these approaches, you can then turn on spin and see if there are any local relaxations due to spin polarization.

My guess, however, is that the differences observed with the climbing image are indirect, that is, a climbing image can drive an image on an excited state to higher energy. The root of the problem, though, is likely with the electronic optimization.

Thanks a lot for your suggestions, I will try~