Symmetry In the DIRAC interface to MRCC

Dear Kallay,
In the DIRAC interface to MRCC, when I use the full Dirac-Coulomb Hamiltonian, the results of mrcc are different between C2v and D2h symmetries. This result agree with Prof. Kallay, who asked you before. But when I use the ECP Hamiltonian, the results of mrcc seem the same between C2v and D2h symmetries. The manual of section 5.3 shows I can only use the Abelian double groups C*2v and C*2h and their subgroups. So, I'd like to know that there's a connection between the Hamiltonian and the symmetry.

The other question is how I Specify the symmetry when calculating excitation energy using the DIRAC interface to MRCC.
For example:
Calculate three states of Au symmetry for Zn atom.

Best regards,
Gaodongdong

Dear Gaodongdong,
If you use ECPs, you have an effective one-component Hamiltonian with normal point group symmetries. So you can also use the D2h point group, which is then Abelian. C2v is a subgroup of D2h, so for a molecule of D2h symmetry or for an atom the energies computed with D2h and C2v are identical.
To set the symmetry of the excited state you should use the symm keyword and give the serial number of the irrep as defined in Dirac.
Note that if you use ECPs, you can also use mrcc in standalone mode, and then you can specify the irrep by its symbol (see Sect. 13 in the manual).

Dear Kallay,
First of all thank you very much for your reply. But I still can't solve the second problem. For example, I Calculate three states of A1 symmetry for Ne atom. When calculating excitation energy using mrcc in standalone mode, I set symm=3 in the MINP file.
But, when using the DIRAC interface to MRCC, I set symm=3 will be an error. I doubt I have missed something useful. But I’m sorry I can’t find other symm key words about the excited states in the manual. I'd like to know whether I can specify the symmetry in this case.
Attachments is DIRAC and MRCC input files.
Best regards,
Gaodongdong

Dear Gaodongdong,
If you want to get three states of A1 symmetry, you should set
nstate=3
symm=1
then you will get the ground state and two low-lying excited states of A1 symmetry, most probably the two lowest excited A1 states.

Dear Kallay,
According to your meaning, the symm represents the order of the irreducible representation. which have been expressed in the 89 pages of the manual. If I want to get three states of B2 symmetry, I should set
nstate=3
symm=3
The program will go wrong.

But If I want to get three states of B2 symmetry, I should set
nstate=3
symm=3
The Program will be carried out smoothly.

Best regards,
Gaodongdong

Dear Gaodongdong,
If you have a molecule of C2v symmetry and set
nstate=3
symm=3
you will get three states of B2 symmetry in the case of a CI calculation, while in a LR-CC calculation you will get the ground state--irrespectively of what symmetry it has--and two excited states of B2 symmetry.