[molpro-user] Orbital energies in HF
Lorenzo Lodi
l.lodi at ucl.ac.uk
Mon Oct 26 17:51:20 GMT 2009
Omololu wrote:
> Hi,
>
> Pls how do I obtain the Hartree-Fock orbital energies from
> a MOLPRO calculation. I tried using the orbprint directive
> but I do not understand what the columns mean:
> Orb Occ Energy Couls-En Coefficients
>
you are correct in that the ORBPRINT directive specifies to print
information on the molecular orbitals.
This is my understanding of how it works:
ORBPRINT,0 => will print info only on the occupied orbitals
ORBPRINT,n => will print info on all the occupied and first "n" virtual
orbitals for each symmetry.
The meaning of the columns is as follows:
"Orb" is a label in the format "n.k" where "k" represent the symmetry of
the orbital and "n" is a progressive number counting the orbitals with
symmetry "k".
"Occ" is the occupation number for the orbital, "0" being unoccupied
(virtual), "2" doubly occupied and "+" or "-" singly occupied by
respectively an alpha- or beta-spin electron.
"Energy" is the canonical orbital energy, i.e. the eigenvalue or the
Fock matrix
"Couls-En" is the Coulson-Neilson [C.A. Coulson, A.H. Neilson, Discuss.
Faraday Soc., 1963, 35, 71] orbital energy, which is
a special partitioning of the total Hartree-Fock energy.
I'll quote what Peter Knowles said in this forum on 15 May 2009:
"'Couls-En' is the sum of the Fock eigenvalue and the diagonal element
of the 1-electron hamiltonian matrix. For a closed-shell system, the
sum of these gives the total energy, not including the nuclear-nuclear
Coulomb contribution."
"Coefficients" are the exapansion coefficients of the orbital in the
symmetry-adapted basis set which is being used.
> Also, when I tried the simple case of the H atom, I could
> only get out the groundstate energy (-0.5 Eh). How would
> I get the other -0.5/n^2 energy levels and their
> corresponding orbitals?
You won't get the excited-state energies as energies of the virtual
orbitals. I think for this you'll need to perform a separate HF
calculation specifying a different symmetry or a MCSCF calculation...
I hope it helps,
Lorenzo
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