# D4H symmetry based calculations of square planar tetra hydrogen

Dear All,

I am a Psi-4 newbie and wanted to do an RHF on the square planar H4 molecule

mol = psi4.geometry("""
H 0.45 0.45 0.0
H 0.45 -0.45 0.0
H -0.45 0.45 0.0
H -0.45 -0.45 0.0
“”")
psi4.set_options({‘basis’: ‘sto-3g’,
‘scf_type’: ‘pk’,
‘e_convergence’: 1e-8,
‘d_convergence’: 1e-8})
scf_e, wfn = psi4.energy(‘SCF’, return_wfn=True)

I think Psi-4 detects it to be of D2h symmetry and gives the SCF energy, MO coefficients, and MO energies accordingly. The orbital energies calculated by psi-4 are

-0.844036
-0.116682
0.220582
1.191747

However, I wanted psi-4 to use the full point group D4h symmetry which should give the following orbital
energies (NWChem)

-0.8440
0.0523
0.0523
1.1928

Is it possible to do full point group calculations somehow? I have read the manual which says that psi-4
reverts to the highest abelian group but I have also seen multiple publications that do D4H calculations on H4 square planar molecule using psi4! So I was wondering if there is a way to do it?

No, Psi4 does not support non-Abelian point groups. You are either misreading those publications, or they’re misreporting what they did.

I’m confused by your statement that “Psi-4 detects it to be of D2h symmetry and gives the SCF energy, MO coefficients, and MO energies accordingly.” It’s not the case that the D2h solution is fundamentally different than the D4h solution. Rather, there are multiple solutions. If all of them can be classified by D4h solutions, then they can also all be classified by D2h solutions. The D2h classification is just less fine.

It sounds like NWChem and Psi4 are giving different SCF solutions. The first step is to figure out which solution you want. Do you want the lowest energy? Do you know for sure that you want the NWChem solution? Do you know what the orbital occupation should be in D4h symmetry?

Thanks for your response. I am quite positive those publications used Psi4 to do FCI calculations of the H4 molecule as a function of the angle between the four H atoms in a planar structure. Maybe they had additional codes to do the exact square planar calculations (90-degree angle between four H molecules).

“The first step is to figure out which solution you want” - I wanted the D4H square planar solution which results in degenerate orbitals (HOMO-LUMO) of the H4 molecule which is possible only through NWChem calculations it seems.

“Do you know what the orbital occupation should be in D4h symmetry?” yes, in the D4H symmetry the wavefunction (HF) is a linear combination of the two possible states { (1a_1g)^2 (2e_u)^2 and (1a_1g)^2 (3e_u)^2 } which are degenerate (0.0523 Ea).