[molpro-user] ECP for Ar+He

[Hans-Joachim Werner]

Dear Andrey,
I don’t see that problem when running your input with the current distribution (2015.1). Which program version did you use? There must be something wrong, since certainly DK should not be used with ECPs. Please send the failing output.
Best regards
Joachim

---
Prof. Dr. Hans-Joachim Werner
Institut für Theoretische Chemie
Universität Stuttgart
Pfaffenwaldring 55
70569 Stuttgart, Germany
e-mail: werner at theochem.uni-stuttgart.de




> Am 25.01.2018 um 15:03 schrieb Andrey Pershin <anchizh93 at gmail.com>:
> 
> 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
> 
> 	Без вирусов. www.avast.ru
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