[molpro-user] hessian always calculated numerically for root=2 ump2, rccsd, rccsd(t) geometry opts
werner at theochem.uni-stuttgart.de
Sat Feb 17 09:19:54 GMT 2007
By default, the hessian is computed in the first step of transition state
optimizations. Minimum searches do not compute the hessian by default but
use a model hessian instead.
If analytical gradients are available (e.g. uhf/rhf/mcscf) the hessian is computed
by finite differences from gradients. If gradients are not available
(ump2, rccsd, and rccsd(t)) the hessian is computed by finite differences
from energies. This needs many more displacements.
In order to avoid the numerical calculation of the rccsd(t) hessian,
you might be able to compute the hessian using, e.g., rhf or mcscf, and then
use this in the rccsd(t) optimization. See manual, READHESS option!
On Fr, 16 Feb 2007, Benj FitzPatrick wrote:
>I am trying to find transition states for a couple of open shell systems with
>2006.1 (patch level 34). When it gets to the first OPT where it calculates the
>hessian I get differing results based on which level of theory I use. If I use
>uhf/rhf/mcscf the hessian is computed using analytic gradients it it makes 9
>calculations. For ump2, rccsd, and rccsd(t) it wants to do 90 calculations to
>get the hessian. To me this looks like it is trying to use finite differences.
> This only happens when I'm looking at a transition state, if I use root=1 the
>behavior of ump2, etc. is the same as uhf. Is this behavior expected? If no,
>why is molpro doing this and is it possible to get around this? I included an
>input file below.
>University of Chicago
> ***,ump2 irc of c3h5o going from the epoxide to o having the radical ***
> H, 1, B1;
> H, 1, B2, 2, A1;
> C, 1, B3, 3, A2, 2, D1;
> H, 4, B4, 1, A3, 3, D2;
Prof. Hans-Joachim Werner
Institute for Theoretical Chemistry
University of Stuttgart
D-70569 Stuttgart, Germany
Tel.: (0049) 711 / 685 64400
Fax.: (0049) 711 / 685 64442
e-mail: werner at theochem.uni-stuttgart.de
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