After finding a reaction "Transitional State" by GAMESS, and finding Hessian Eigenvalues, my project instructor said:

seems your imaginary frequency (negative eigenvalue) is not that big (it is around -13.01), which means your reaction is barrier-less.

As I am a beginner in Quantum Chemistry, I could not figure out what is the relation between a small imaginary frequency and being barrier-less.

As I remember from the Arrhenius equation, being barrier-less has something to do with higher temperature leading to a slower reaction, but I could not understand its relation to imaginary frequencies.

  • 1
    $\begingroup$ See for example this related question $\endgroup$ Feb 8, 2015 at 19:51
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    $\begingroup$ Small magnitudes in imaginary vibrational modes could be due to numerical noise if using DFT approaches. You might want to tighten up your convergence or try another method to verify. That being said, I don't agree that this would be barrier-less at all. Of course you could try to take into account zero-point vibrational energy correction to the TS energy (though how does one ZPVE-correct a TS is beyond me) which would raise the energy even more. Just compare the electronic energies of the TS and corresponding minima to see how big the barrier is. $\endgroup$ Feb 9, 2015 at 15:36
  • $\begingroup$ @GeoffHutchison This Marcus theory was awesome ! Thank you $\endgroup$
    – Aug
    Feb 12, 2015 at 3:13
  • $\begingroup$ @LordStryker it was a great suggestion to try another method. Is there any cutoff for saying this is a barrier-less reaction ? for example can we say if the imaginary eigenvalue is larger than a specific number ( or a range of numbers ), it is assumed barrier-less ? $\endgroup$
    – Aug
    Feb 12, 2015 at 3:16
  • $\begingroup$ @Aug I have little experience with barrierless pathways. I only know more orthodox mechanisms where you characterize a minimum energy pathway for a reaction. I've read into this a bit and it looks interesting, however. You may want to read this (math.uni-leipzig.de/~quapp/JKQCPaper2.pdf) as it may be useful. Also, I wouldn't correlate an imaginary mode of vibration with a barrierless reaction. It sounds very wrong indeed to do that. $\endgroup$ Feb 12, 2015 at 13:02

1 Answer 1


Usually, in the reactions with a barrier, the saddle point with one imaginary frequency is taken as TS. However, many unimolecular dissociation reactions are barrierless: the reaction path on potential surface, connecting (minimum of) reactant with (minima of) products, is such that the energy along the path does not have a local maximum. Kinetic treatments of such reactions are more difficult: it is still possible to assign a TS, but it no longer a saddle point. In order to find it, one can use variational TST, but it is somewhat expensive.

A general recommendation after location of a saddle point: run reaction path calculation to see that this saddle point indeed connects the products and reactants that you expect.

About this small eigenvalue: this eigenvalue can easily pop up instead of zero eigenvalues of rotations and translations. Even if your calculation explicitly projects them out, numerical noise can be this order of magnitude (as pointed out above).


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