[molpro-user] ECP for Ar+He

Fri Jan 26 07:59:55 CET 2018

Dear Andrey,

I don’t know why the program has activated the DK integral code, but in any event it was going to fail anyway since the number of electrons and occupations are all wrong for the case of a 10 electron ECP on Ar : the rhf and below seem to be written for an all-electron calculation.

What version of Molpro are you using?

regards,

-Kirk

On Jan 25, 2018, at 6:03 AM, Andrey Pershin <anchizh93 at gmail.com<mailto:anchizh93 at gmail.com>> wrote:

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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