I am currently in the progress of trying to simulate the vertical excited-states of the open-shell molecule CCl3, and stumbled upon a weird problem:
After optimizing the ground state geometry, I obtained a C3v-symmetric ground state. Unfortunately, when I now start a EOM-CCSD calculation to look at the first few excited states, after loading/re-orientating the molecule, it is all of a sudden Cs-symmetric. Even more puzzling is the following: when I add the key-word CC_SYMMETRY = 0 (nothing else changed), it now determines the point group to be C3v, but of course the excited states are now optimized assuming C1 symmetry.
Does anyone have any idea what’s going on? How can I obtain the excited states of CCl3 in the correct symmetry? Relaxing the symmetry threshold from 5 to 3 did not help unfortunately…
Thanks in advance for your help,
Anja
This situation can be reproduced with CHF3. With METHOD=HF:
Standard Nuclear Orientation (Angstroms)
I Atom X Y Z
----------------------------------------------------------------
1 C 0.0000000008 0.0000000013 0.3509718273
2 H 0.0000000013 0.0000000022 1.4457718519
3 F 1.2543992952 -0.0000000000 -0.1315408490
4 F -0.6271996476 1.0863416562 -0.1315408490
5 F -0.6271996476 -1.0863416562 -0.1315408490
----------------------------------------------------------------
Molecular Point Group C3v NOp = 6
Largest Abelian Subgroup Cs NOp = 2
Nuclear Repulsion Energy = 132.95944891 hartrees
but with METHOD=CCSD:
----------------------------------------------------------------
Standard Nuclear Orientation (Angstroms)
I Atom X Y Z
----------------------------------------------------------------
1 C 0.0000000124 0.3509718273 0.0000000000
2 H 0.0000000202 1.4457718519 0.0000000000
3 F 1.2543992952 -0.1315408490 0.0000000000
4 F -0.6271996885 -0.1315408411 -1.0863416377
5 F -0.6271996885 -0.1315408411 1.0863416377
----------------------------------------------------------------
Molecular Point Group Cs NOp = 2
Largest Abelian Subgroup Cs NOp = 2
Nuclear Repulsion Energy = 132.95944891 hartrees
It’s not entirely clear why Q-Chem does not determine that molecular point group symmetry is C3v. However, it is largely irrelevant as Q-Chem always switches to the largest Abelian subgroup in CC/EOM/ADC jobs if the true point group symmetry of the molecule is non-Abelian. Hence the treatment of C3v as Cs in the EOM calculation. Excited state irreps in the Cs point group must be mapped back to the irreps in the C3v point group by the user when interpreting the output.
Using CC_SYMMETRY=FALSE disables symmetry altogether in CC/EOM calculations such that Q-Chem does not even attempt to determine the irreps of the orbitals. The whole thing is treated as if it were C1 symmetry.