NEB using symmetric slabs

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NEB using symmetric slabs

Post by l.shu »


I am trying to calculate the dissociation if simple molecules on an oxide surface.

I want to use a symmetric slab and I want to put the adsorbates on both sides of the slab to cancel any dipole moment. I begin by finding the initial and final geometries.

For example, I have O2 on opposite sides of the slab as my initial starting point and then I have (O+O) sitting on the opposite sides as the final geoemtry. If I run an NEB calculation does the barrier correspond to the dissociation of 2 O2 molecules and hence I have to divide the barrier I obtain by 2. Or the value I get is simply the barrier of O2->2O?

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Re: NEB using symmetric slabs

Post by graeme »

This is a bit of a tricky question. If no symmetry is applied in the DFT calculation, a reaction on both sides of the slab will be a second order saddle. A good relaxation (without symmetry) should converge to a reaction on one side of the slab and then the other, with barrier roughy half of that for both reactions occurring together. That said, if symmetry is imposed, symmetric reactions on both sides of the slab will save a single negative mode, and as you say, each reaction should have a barrier half of what you calculate.

I suggest checking what you are doing by calculating a reaction barrier on one side of the slab, using a dipole correction.
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