Molcas Forum

tumitumi123

Member

Registered: 2017-06-13

Posts: 16

something wrong with casscf calculation in C2v symmetry.

Dear all
I use MOLCAS@UU to calculate PES of O2+ in spin multiplicity 4.
First I optimized the geometry, and use the optimized orbital file jobiph as an initial guess to calculate other points.
Without symmetry, i can get all the PESs, but they mixed together, which is not easy to seperate them and to do diabatization(since in molcas@UU there is no DQVD in RASSI).
Therefore I tried to use C2v to get PEC according to symmetry. But i can not find the sigma ground state.
The orbitals at equilibrium of O2+ is: sigma_g, sigma_u, sigma_g, sigma_u, pi_ux, pi_uy, sigma_g, pi_gx, pi_gy, sigma_u
with first two orbital inactive, the sigma ground state configuration should be 2222uuu0, which mean one electron escape from sigma_g.
with symmetry the configuration should be:
conf/sym   1111 22 44     Coeff  Weight
                 22u2 2u 2u   
but I didn't see these configuration in the result. I even didn't see 2u in the 2, 4 irrep, there only configurations like 20, 02, uu, du, ud in 2 and 4 irrep.
I wonder is this a bug in molcas@UU. anything can i do to fix this?
Here i attached my input and part of output.

&gateway
coord
2
bohr
O  0 0 -$RR1
O  0 0 $RR2
basis=cc-pvtz
group=xy y
&SEWARD
R02O02
amfi
>> COPY $Home/$Project.JobIph $WorkDir/JOBOLD
jobiph
&RASSCF
spin=4
nactel=11 0 0
inactive=2 0 0 0
ras2=4 2 0 2
ciroot=22 22 1
symmetry=1
linear

      ************************************************************************************************************************
                                                      Wave function printout:
                       occupation of active orbitals, and spin coupling of open shells (u,d: Spin up or down)
      ************************************************************************************************************************

      Note: transformation to natural orbitals
      has been made, which may change the order of the CSFs.

      printout of CI-coefficients larger than  0.05 for root  1
      energy=    -149.058247
      conf/sym  1111 22 44     Coeff  Weight
            33  22u0 22 uu   0.69657 0.48521
            34  22u0 uu 22   0.69657 0.48521
            41  2udu 22 uu   0.07077 0.00501
            42  2udu uu 22   0.07077 0.00501
            51  2uud 22 uu   0.05032 0.00253
            52  2uud uu 22   0.05032 0.00253

      printout of CI-coefficients larger than  0.05 for root  2
      energy=    -149.054602
      conf/sym  1111 22 44     Coeff  Weight
            33  22u0 22 uu  -0.69626 0.48478
            34  22u0 uu 22   0.69626 0.48478
            41  2udu 22 uu  -0.07063 0.00499
            42  2udu uu 22   0.07063 0.00499
            51  2uud 22 uu  -0.05038 0.00254
            52  2uud uu 22   0.05038 0.00254

      printout of CI-coefficients larger than  0.05 for root  3
      energy=    -149.034457
      conf/sym  1111 22 44     Coeff  Weight
             1  222u uu 20  -0.65805 0.43302
             2  222u 20 uu   0.65805 0.43302
             5  222u 02 uu  -0.23189 0.05377
             8  222u uu 02   0.23189 0.05377
            12  22u2 ud uu  -0.05462 0.00298
            15  22u2 uu ud  -0.05270 0.00278

Ignacio

Administrator

From: Uppsala

Registered: 2015-11-03

Posts: 1,085

Re: something wrong with casscf calculation in C2v symmetry.

with symmetry the configuration should be:
conf/sym   1111 22 44     Coeff  Weight
                 22u2 2u 2u

22u0 2u 2u, I presume.

symmetry=1

Try with symmetry=3

I hope someone will correct me if I'm wrong (I'm not an expert with symmetry), but take a look at the "Symmetry information" in the output:

                             E   C2(z) s(xz) s(yz)
                    a1       1     1     1     1  z
                    b1       1    -1     1    -1  x, xz, Ry
                    a2       1     1    -1    -1  xy, Rz, I
                    b2       1    -1    -1     1  y, yz, Rx

Doubly-occupied orbitals don't count, so you want your electrons in "a1", "b1" and "b2". You multiply the signatures and get 1*1*1=1, 1*-1*-1=1, 1*1*-1=-1, 1*-1*1=-1. Now "1, 1, -1, -1" is the signature of "a2", so your "22u0 2u 2u" configuration has symmetry "a2" (number 3).