9/13/2023 0 Comments Negative vibrations![]() First off, let's note that a transition state (TS) is, loosely speaking, in-between a reactant and a product (state). I will answer from a practical point of view. So, one is looking for either one negative force constant or one imaginary frequency, not a negative frequency. Where $f_$ as usual, and the frequency is imaginary. So, let our PES be given by a function $f(x,y)$. Or, in chemistry (and in math) in order to do this, we construct the Hessian matrix and find its determinant. Note that to confirm we are at a saddle point, in mathematics one could do the second-derivative test. As you say, there are two directions where we go down in energy and two where we go up in energy which are along or perpendicular to the gradient vector. If we are at a transition state, then we are at a saddle point on the potential energy surface (PES). Well, let's just look at the simple case you mention. thinking about a reaction profile curve is too simplistic), or is this okay but i'm missing some fundamental transition state theory? ![]() ![]() This makes less sense to me, as, from a transition state there are (in 2D space) two possible ways to go the wrong way (down in energy) - clearly one is more favourable as it leads to a lower energy intermediate/product, but none the less, the other pathway exists. When looking for a transition state, the common goal is to end up with one (and exactly one) negative frequency, with all others being positive. This makes sense (at least looking at it simplistically) as if we travel in any direction we're always going up (else we're not in a minimum). ![]() When finding a minimum (local or global), the vibrational frequencies should all be positive. The vibrational analysis must be performed at the the optimized geometry. The vibrational frequencies are related to second derivatives a minimum will have only positive frequencies while transition state should have one negative frequency. ![]()
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