MIXING ANGLE/DDR PROCEDURE

[Jack Klos]

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
}