[molpro-user] ECP for Ar+He

[Andrey Pershin]

Dear colleagues,

I try to calculate potential curves for Ar+He with ECP for Ar. I used this
code:
***,ArHe
   memory,100,M
   gprint,civector,orbital=2;
   geometry={angstrom
   Ar
   He,Ar,R(i)}
   R=[20.,3.3,3.1,3.0,2.9,2.6,2.4,2.]

 basis={
ECP, ar, 10, 4 ;
1; !  g-ul potential
2,1.000000000,0.000000000;
2; !  s-ul potential
2,10.261721000,68.667788010;
2,3.952725000,24.042766290;
2; !  p-ul potential
2,5.392714000,27.730763310;
2,2.699967000,4.045459040;
2; !  d-ul potential
2,8.086235000,-8.137476960;
2,4.016632000,-1.664528080;
1; !  f-ul potential
2,5.208459000,-3.400098450;
spdfgh,1,aug-cc-pCV5Z;C;
SP,1,EVEN,NPRIM=3,RATIO=2.5;
spdfg,2,aug-cc-pV5Z;C;
 }
do i=1,#R
{RHF
wf,20,1,2
}
 {casscf
closed,3,1,1,0
occ,8,3,3,0;
DYNW,2
 wf,20,1,2;state,4
 wf,20,2,2;state,3
 wf,20,3,2;state,3
 wf,20,4,2;state,2
 wf,20,1,0;state,5
 wf,20,2,0;state,3
 wf,20,3,0;state,3
 wf,20,4,0;state,2
}

{ci;closed,3,1,1,0;occ,8,3,3,0;orbital,ignore_error;wf,sym=1,SPIN=2;state,4;save,5101.2}
 hlsdiag(1)=energd(1)
 hlsdiag(2)=energd(2)
 hlsdiag(3)=energd(3)
 hlsdiag(4)=energd(4)
  {ci;closed,3,1,1,0;occ,8,3,3,0;wf,sym=2,SPIN=2;state,3;save,5102.2}
 hlsdiag(5)=energd(1)
 hlsdiag(6)=energd(2)
 hlsdiag(7)=energd(3)
  {ci;closed,3,1,1,0;occ,8,3,3,0;wf,sym=3,SPIN=2;state,3;save,5103.2}
 hlsdiag(8)=energd(1)
 hlsdiag(9)=energd(2)
 hlsdiag(10)=energd(3)
  {ci; closed,3,1,1,0;occ,8,3,3,0;wf,sym=4,SPIN=2;state,2;save,5104.2}
 hlsdiag(11)=energd(1)
 hlsdiag(12)=energd(2)
  {ci; closed,3,1,1,0;occ,8,3,3,0;wf,sym=1,SPIN=0;State,5;save,5105.2}
 hlsdiag(13)=energd(1)
 hlsdiag(14)=energd(2)
 hlsdiag(15)=energd(3)
 hlsdiag(16)=energd(4)
 hlsdiag(17)=energd(5)
  {ci; closed,3,1,1,0;occ,8,3,3,0;wf,sym=2,SPIN=0;state,3;save,5106.2}
 hlsdiag(18)=energd(1)
 hlsdiag(19)=energd(2)
 hlsdiag(20)=energd(3)
  {ci; closed,3,1,1,0;occ,8,3,3,0;wf,sym=3,SPIN=0;state,3;save,5107.2}
 hlsdiag(21)=energd(1)
 hlsdiag(22)=energd(2)
 hlsdiag(23)=energd(3)
  {ci; closed,3,1,1,0;occ,8,3,3,0;wf,sym=4,SPIN=0;state,2;save,5108.2}
 hlsdiag(24)=energd(1)
 hlsdiag(25)=energd(2)

  {ci;hlsmat,ls,5101.2,5102.2,5103.2,5104.2,5105.2,5106.2,5107.2,5108.2}

enddo

On the RHF calculation Molpro wrote this text:

...
ATOMIC COORDINATES

 NR  ATOM    CHARGE       X              Y              Z

   1  AR      8.00    0.000000000    0.000000000   -3.441963392
   2  HE      2.00    0.000000000    0.000000000   34.352559236

 NUCLEAR CHARGE:                   10
 NUMBER OF PRIMITIVE AOS:         442
 NUMBER OF SYMMETRY AOS:          327
 NUMBER OF CONTRACTIONS:          309   ( 116A1  +  75B1  +  75B2  +  43A2
)
 NUMBER OF CORE ORBITALS:           0   (   0A1  +   0B1  +   0B2  +   0A2
)
 NUMBER OF VALENCE ORBITALS:        5   (   3A1  +   1B1  +   1B2  +   0A2
)


 NUCLEAR REPULSION ENERGY    0.42334177

 One-electron integrals computed with SEWARD

 2nd-order Douglas-Kroll-Hess method activated. Optimal DKH parametrization
is used.

 GLOBAL ERROR fehler on processor   0

How to check this problem?

Best Regards,
Andrey

-- 

Postgraduate Student Andrey Pershin
Samara University, Samara, Russia

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