[molpro-user] Why SCF results do not give the lowest energy?

[Ricardo Mata]

Due to recent changes in the SCF program, the problem only occurs in
2006.1 . For 2006.1 users, {rhf,shifta=-0.2; ...} solves the problem
at the cost of a slower convergence. For the patched 2008.1, aside the
better defaults, the option NITSH might prove to be helpful.

On Tue, Nov 11, 2008 at 11:58 AM,  <cjrosa at ualg.pt> wrote:
>
> I want to calculate the SCF energy of H2O triplet in geometry close to
> the C2V symmetry. For this  geometry that I study the state of lower
> energy is the 3B1 followed by 3B2 and 3A2 respectively. I have no
> problem in calculate each one of these energies. For this calculation
> I use the follow input:
>
> ***,h2o
>
> BASIS=aug-cc-pvqz
>
> R=7.0
> RHH=1.402
> RH=(RHH/2.0)*toang
> ROHH=R*toang
>
> geomtyp=xyz
> geometry={
> 3
> Cartesian coordinates of each atom
> O1,0.0000000000, ROHH*cos(90.0)     ,  ROHH*sin(90.0)
> H1,0.0000000000,-RH                 , 0.000000000
> H2,0.0000000000, RH                 , 0.000000000
> }
>
> {RHF;
> WF,10,?,2;                           !? can be 2 for B1, 3 for B2 or 4 for A2
> }
> ---
>
> The problem occurs when I use lower symmetry to study the system.
> If I use Cs symmetry and I ask for the lower A?? energy I get the 3A2
> energy instead of 3B1 energy. (That is, I get the second energy of
> state A??!!!) . The unique difference in the input is:
>
> ...
> geometry={
> X;
> 3
> Cartesian coordinates of each atom
> O1,0.0000000000, ROHH*cos(90.0)    , ROHH*sin(90.0)
> H1,0.0000000000,-RH                ,0.0000000000
> H2,0.0000000000, RH                , 0.000000000
> }
> ...
> {RHF;
> WF,10,2,2                              !or WF,10,1,2 for the A? state
> }
> ---
>
> In the same way, if I use C1 symmetry I get the 3A2 energy (the third
> energy!!!). The unique difference in the input is:
> ...
> geometry={
> NOSYM;
> 3
> Cartesian coordinates of each atom
> O1,0.0000000000, ROHH*cos(90.0)    , ROHH*sin(90.0)
> H1,0.0000000000,-RH                ,0.0000000000
> H2,0.0000000000, RH                , 0.000000000
> }
> ...
> {RHF;
> WF,10,1,2
> }
> ---
> That is, if I use lower symmetry I never get the 3B1 energy
> Could anyone explain me why it happens. Thank you.
>
>
>
> ----------------------------------------------------------------
> This message was sent using IMP, the Internet Messaging Program.
>
> _______________________________________________
> Molpro-user mailing list
> Molpro-user at molpro.net
> http://www.molpro.net/mailman/listinfo/molpro-user
>