MIXING ANGLE/DDR PROCEDURE

Prof. H.-J. Werner werner at theochem.uni-stuttgart.de
Wed Feb 9 11:03:15 GMT 2000


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> MIXING ANGLE/DDR PROCEDURE
>
>   ------------------------------------------------------------------------
>
>    * To: MOLPRO-USERS <molpro-user at tcpc3.bham.ac.uk>
>    * Subject: MIXING ANGLE/DDR PROCEDURE
>    * From: Jack Klos <jklos at ouchem.chem.oakland.edu>
>    * Date: Thu, 27 Jan 2000 11:46:06 -0400
>    * Sender: owner-molprouser at tc.bham.ac.uk
>
>   ------------------------------------------------------------------------
>
> Dear Sir,
>
> I have a problem with values of mixing angle between states calculated
> by DDR procedure. Let me explain this on the example which I took
> from Molpro2000 examples:h2s_diab1.com.
> Firs run is done for original example and a reference geometry is taken
> for r1=2.5 and loop is from r=2.5 to 2.6 as showed below:
>
>  ***,h2s Diabatization
>  memory,3,m
>
>  gprint,orbitals,civector
>
>  geometry{x;noorient
>           s;
>           h1,s,r1;
>           h2,s,r2,h1,theta}
>
>  basis=avdz
>
>  r1=2.5
> !Reference geometry
>  theta=[92]
>
>  r=[2.50,2.55,2.60]
> !Displaced geometries
> .....
> The results from this example are as follows:
> ++++++++++++++++++++++++++++++++++++
>  Diabatic energies for H2S, obtained from CI-vectors
>
>         R       E1                         E2
> H11CI                   H22CI         H21CI          MIXCI
>       2.50 -398.64296319 -398.63384782 -398.64296319 -398.63384782
> 0.00000000        0.00
>       2.55 -398.64572746 -398.63666636 -398.64509901 -398.63729481
> -0.00230207       15.27
>       2.60 -398.64911752 -398.63771802 -398.64662578 -398.64020976
> -0.00471125       27.87
>
>  Diabatic energies for H2S, obtained from CI-vectors and orbital
> correction
>
>         R       E1                             E2
> H11                H22                   H21           MIXTOT
>       2.50 -398.64296319 -398.63384782 -398.64296319 -398.63384782
> 0.00000000        0.00
>       2.55 -398.64572746 -398.63666636 -398.64509941 -398.63729441
> -0.00230139       15.26
>       2.60 -398.64911752 -398.63771802 -398.64662526 -398.64021027
> -0.00471160       27.88
> +++++++++++++++++++++++++++++++++++++
>
> When I will set the reference geometry to be at r1=2.6 and I will do
> reversed loop (r=[2.6,2.55,2.5])
> as in this fragment of input:
> .................................
> ***,h2s Diabatization
>  memory,3,m
>
>  gprint,orbitals,civector
>
>  geometry{x;noorient
>           s;
>           h1,s,r1;
>           h2,s,r2,h1,theta}
>
>  basis=avdz
>
>  r1=2.6
> !Reference geometry
>  theta=[92]
>
>  r=[2.60,2.55,2.50]
> !Displaced geometries
> .........................
> (The rest of input is the same as in the example h2s_diab.com)
>
> I got quite different results in comparison to the previous original
> example
> (both adiabatic and diabatic energies and mixing angle):
> ++++++++++++++++++++++++++++++++++++++++++++++++++++++++
> Diabatic energies for H2S, obtained from CI-vectors
>
>         R       E1            E2            H11CI         H22CI
> H21CI          MIXCI
>       2.50 -398.64911752 -398.63771803 -398.64665123 -398.64018432
> 0.00469381      -27.72
>       2.55 -398.64979586 -398.64251959 -398.64888720 -398.64342826
> 0.00240542      -20.69
>       2.60 -398.65053196 -398.64613332 -398.65053196 -398.64613332
> 0.00000000        0.00
>
>  Diabatic energies for H2S, obtained from CI-vectors and orbital
> correction
>
>         R       E1            E2            H11           H22
> H21           MIXTOT
>       2.50 -398.64911752 -398.63771803 -398.64665042 -398.64018514
> 0.00469437      -27.72
>       2.55 -398.64979586 -398.64251959 -398.64888635 -398.64342910
> 0.00240637      -20.70
>       2.60 -398.65053196 -398.64613332 -398.65053196 -398.64613332
> 0.00000000        0.00
> +++++++++++++++++++++++++++++++++++++++++++++++++++++++++
>
> I have the same problem with my system, so I would like to ask you for
> an
> advice what can be the reason of this differencies when you go from
> shorter
> distancies to larger or in the reversed manner.
> Version of Molpro:2000.1
> Thanks in advance
> Best regards
> Jacek Klos
>
> {
> Jacek Klos
> Department of Chemistry
> University of Warsaw
> Pasteura 1,02-093 Warsaw POLAND
> jakl at tiger.chem.uw.edu.pl
>
> Present Address:
> Department of Chemistry
> Oakland University
> Rochester, MI 48309 U.S.A
> Phone: (248)-370-2092
> jklos at ouchem.chem.oakland.edu
>
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> To add an item, send mail to molpro-user at tc.bham.ac.uk
>
>
> ------------------------------------------------------------------------Th

The problem is that you don't perform calculations at the same geometries;
in the first case, you have for (r1,r2) (2.5,2.5), (2.5,2.55), (2.5,2.6), in
the
second case (2.6,2.5) (identical), (2.6,2.55), (2.6,2.50). Of course, the
latter
two yield different results!





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