[molpro-user] Double minima problem of CaF ground state with MRCI

[Peterson, Kirk]

Dear Shilin,

I assume it is a problem in your active space, i.e., it probably changes discontinuously with geometry.  I would suggest to closely inspect your orbitals.  Actually I don’t quite understand your choice of  occ, 14,5,5,0.  The default full valence space is 9,3,3,0   which includes the Ca 4s and F 2s and 2p orbitals.  If you want to try to include the Ca 4p and 3d orbitals, the 4p would add  1,1,1,0 and the 3d would add  2,1,1,1, which would be  12,5,5,1 .  Perhaps the extra orbitals you have in symmetry 1 are causing the problem since they are not strongly occupied. I might also expect problems just from trying to include the 4p and 3d since you are correlating very few Ca electrons (since you freeze everything below the 4s).  Just some thoughts.

best regards,

-Kirk


On Jan 11, 2017, at 4:39 AM, 侯世林 <slhou at ouc.edu.cn<mailto:slhou at ouc.edu.cn>> wrote:

Dear Molpro users,
I am calculating the CaF ground state with MRCI/cc-pvxz (x=T,Q,5), and find there are two minima on its potential curve: one at Re(~2.0 angstrom), another at R=(2.5~2.6 angstrom). Active spaces are set from the default to {8,3,3} for nine active electrons.
I also try the ACPF approach; the results are similar to those of MRCI. Attached are the results and input file of ACPF/cc-pcvqz. Note with the ACPF/cc-pcv5z the second minima can be removed. But I think MRCI can give more physical data for this ground state.
Is there something wrong with my input file? How to remove it with MRCI approach?
Any comments and help are appreciated.
Thank you in advance,

Shilin Hou

Results(ACPF/cc-pcvqz):
      R     E1(HF)       DIP1        E2(CASSCF)       DIP2        E3(ACPF)          DIP3
    1.6   -776.2493090   0.17282959   -776.4200188   0.13818992   -776.5204776   0.23656468
    1.8   -776.3148194   0.65163519   -776.4839416   0.60620805   -776.5804423   0.68439505
    1.9   -776.3251340   0.88388333   -776.4937886   0.83099076   -776.5888267   0.90010015
    2.0   -776.3269510   1.11353675   -776.4954508   1.05297386   -776.5893276   1.11336495
    2.1   -776.3233233   1.34279372   -776.4920427   1.27395004   -776.5850192   1.32566006
    2.2   -776.3163261   1.57322444   -776.4858342   1.49535523   -776.5781622   1.53696369
    2.3   -776.3073476   1.80432337   -776.4788920   1.71816131   -776.5709855   1.74820085
    2.4   -776.2973052   2.03726534   -776.4746510   1.94288730   -776.5681313   1.95872143
    2.5   -776.2868008   2.27188611   -776.4804660   2.15765732   -776.5782443   2.16482483
    2.6   -776.2762266   2.50854076   -776.4823416   2.37699001   -776.5861416   2.37097908
    2.7   -776.2658324   2.74714894   -776.4760499   2.60049036   -776.5819935   2.58517957
    2.8   -776.2557688   2.98754186   -776.4673357   2.82661062   -776.5738194   2.80337286


***, CaF ground state
 r=[ 1.6 1.8 1.9  2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 ] ang
 do l=1,#r
 geometry={F;Ca,F,$r(l)}

  basis=cc-pcvqz
 !basis=cc-pcv5z

 { hf
     occ,9,3,3
      closed, 8,3,3
       wf,29,1,1 }

  e(1) = energy
  method(1) = program
  dip(1) = dmz
  bond(1) = $r(l)
  e1(l)= energy
  dip1(l)= dmz

      {casscf
     occ,14,5,5
      closed, 6,2,2
      frozen,6,2,2
      maxiter,40
       wf,29,1,1
      orbital,2140.2      }

e(2) = energy
  method(2) = program
  dip(2) = dmz
  bond(2) = $r(l)
  e2(l)= energy
  dip2(l)= dmz

  ! {MRCI
      {acpf
     occ,14,5,5
     closed, 6,2,2
     core,6,2,2
   maxiter,200,200
       wf,29,1,1
      orbital,2140.2   }

  e(3) = energy
  method(3) = program
  dip(3) = dmz
  bond(3) = $r(l)
  e3(l)= energy
  dip3(l)= dmz

 table,method,bond,e,dip
 enddo
 table, r,e1,dip1, e2,dip2,e3,dip3
---
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