[molpro-user] Strange results FCI for scandium atom

[Peter Knowles]

Dear Lorenzo,

All that happens with wf,spin= in FCI is that you specify the S_z quantum number of the Slater determinants that form the basis;  it does not use a basis of configuration state functions that are eigenfunctions of S^2 as well. Therefore there is no reason why you must see doublet when you specify spin=1 - you can get quartet as well. Your results in symmetry 6 are 2D, 4F, 4F.

If you change to spin=1,sym=4;state,3 you happen to get 2D, 4F and something else. I'm afraid that the algorithm for choosing which states to track is not so sophisticated for the case that there is non-abelian symmetry, and you can be lucky or unlucky as to whether states of higher symmetry do or do not couple in.

Peter

On 26 Nov 2013, at 19:32, Lorenzo Lodi <l.lodi at ucl.ac.uk> wrote:

> Consider the attached input for three electronic states of the scandium
> atom (runs in ~5 seconds).
> The CASSCF energies come out with the correct degeneracies and with
> quite reasonable values.
> I then tried frozen-core FCI using the CASSCF orbitals and found some
> rather strange results.
> For example, I get for the energies
> !FCI STATE 1.6 Energy               -759.774071038492
> !FCI STATE 2.6 Energy               -759.715754355764
> !FCI STATE 3.6 Energy               -759.715754350740
> !FCI STATE 1.1 Energy               -759.715754355736
> 
> The first three energies look correct fully reasonable (ground 2D term
> and two degenerate component of 2F term); however the fourth energy,
> which should be the 4F term, comes out exactly the same as the 2F state,
> as though even if FCI says `Spin quantum number:          1.5' it
> actually used 0.5.
> Another problem is that, in the first FCI block, if I specify a
> different symmetry (1 or 4) I get different results, which I don't think
> should be the case.
> 
> Any thoughts?
> 
> Lorenzo
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Prof. Peter J. Knowles             
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