[molpro-user] Radical Cation Wavefunction

Hans-Joachim Werner werner at theochem.uni-stuttgart.de
Sat Apr 16 12:27:42 CEST 2016

the Hartree-Fock program at least tries to find the lowest state and to determine the state symmetry automatically. But this is based on orbital energies and there
is no guarantee that the correct state is always found. The meaning of the state symmetries is explained in the manual, see also the quickstart manual for examples. For example, in a linear molecule oriented along the z-axis, a Sigma^+, state has symmetry 1, Pi_x, Pi_y (symmetries 2,3), Sigma^- symmetry 4, Delta_{x^2-y^2} symmetry 1, Delta_{xy} symmetry 4 (but note that some of these states require more than 1 Slater determinant, and then mcscf muust be used for a proper description).
Best regards
Joachim Werner  
Prof. Dr. Hans-Joachim Werner
Institute for Theoretical Chemistry
University of Stuttgart
Pfaffenwaldring 55
D-70569 Stuttgart
Tel: +49 711 / 685 64400
Fax: +49 711 / 685 64442
email: werner at theochem.uni-stuttgart.de

> Am 15.04.2016 um 14:56 schrieb MD Simulation <mdsimulationgroup at gmail.com>:
> Hello Molpro Users,
> I'm very new Molpro but have used NWChem and Gaussian quite a bit.  I'm attempting to use CCSD(T) on a radical cation organic compound.  My question is, how necessary is it to set the wavefunction symmetry and occupied orbitals (I'm not certain how to determine either of those) in the WF card and OCC card?  In NWChem or Gaussian I would simply set the charge (equal to 1) and multiplicity (equal to 2), so in this case will I get the correct results by simply using:
> """
> wf,charge=1,spin=1
> """
> Thanks for the help.  And as a side note, Molpro is really fast!
> Cheers,
> Alan
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