I think that if there are two unstable directions at the saddle point this 
means there is another saddle point with a lower Activation energy nearby. Just 
slide the neb downhill stefano 

Sent from my Samsung Galaxy smartphone.
-------- Original message --------From: Pascal Boulet 
<[email protected]> Date: 13/11/2017  19:45  (GMT+01:00) To: PWSCF 
Forum <[email protected]> Subject: *spam*Re: [Pw_forum] Physical meaning of 
two imaginary phonon frequencies at NEB saddle point and Transition State 
Theory 
Hello,
My guess: you have to remove one of the imaginary frequencies (most probably 
the smallest one) by following the corresponding eigenvector (=moving the atoms 
accordingly). Then, you will have to make sure that the transition state 
connects the reactants and products you are interested in.
HTH,Pascal

Pascal Boulet—Professor in computational chemistry - DEPARTEMENT OF 
CHEMISTRYAix-Marseille University - ST JEROME - Avenue Escadrille Normandie 
Niemen - F-13013 Marseille - FRANCETél: +33(0)4 13 55 18 10 - Fax : +33(0)4 13 
55 18 50Site : http://madirel.univ-amu.fr/pages_web_BOULET_PASCAL/infos - Email 
: [email protected]





Le 13 nov. 2017 à 12:44, JAY Antoine <[email protected]> a écrit :



Dear all,
I have been using NEB to obtain the Delta E and phonon DOS at 
begining/end/saddle point to properly evaluate the vibrationnal entropy 
contribution in the transition state theory.
Commonly, one frequency in "negative" at the saddle point, the one that is 
"responssible" of the transition and the following of its eigen vector from the 
initial state give the minimum energy path.
However, when two phonon frequencies are "negatives" at the saddle point, that 
means that at least 3  energy minima are link together to the same saddle point.
The transition state theory cannot be applied there as the preexponential term 
is the ratio of the 3N-3 phonon frequencies of initial state over the 3N-4 
frequencies of the saddle point, giving the dimension of a time (state life 
time)
Using 3N-5 frequencies without the two negatives ones would not have a physical 
meaning for the initial partition function.

Any idea on how to treat such a problem to evaluate the transition rate would 
be appreciated.

Best regards.

Antoine Jay
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