[molpro-user] Helium UHF energy

[Peter Knowles]

The article you cite is not at that point discussing UHF, but the G1 method,
which for this system is the same as Spin-Coupled theory, which in this
system is identical to 2-in-2 CASSCF. Try the following!Peter

geometry={he}
basis,v6z(s)
rhf
{mcscf;occ,2}
{mcscf;occ,2;vb}


2009/10/4 Lorenzo Lodi <l.lodi at ucl.ac.uk>

> Gerald Knizia wrote:
> > On Thursday 01 October 2009 19:24, Lorenzo Lodi wrote:
> >> I was playing around with Molpro and I noticed that, for the helium
> >> atom, I obtain the same energy for RHF and UHF (I think UHF should be
> >> lower).
> >
> > For closed-shell systems near equilibrium, RHF and UHF are supposed
> > to give the same energy values. If they don't (i.e., if you're getting
> > symmetry-broken UHF solutions), it usually is an indication that both
> > are bad and that you're dealing with a multi-reference case.
>
> So you're saying that, in this particular case of the helium atom, RHF and
> UHF energies should be the same? This seems to contradict what I've read in
> various sources. For example see pags. 1243-1245 of Hibbert, Rep Prog Phys
> 38, 1217-1338 (1975).
>
> Let me quote an example for this paper.
> If we use as variational function the RHF-type function
> psi(r1,r2) = u(r1)u(r2)*spin_singlet
> with u(r) = exp(-a r) we find as optimal a the value 2-5/16=1.6875 and an
> energy E=-2.84766 hartree.
> This is, of course, a well known textbook example.
>
> On the other hand if we use the UHF-type trial function
> psi(r1,r2) = [u(r1)v(r2)+v(r1)u(r2)]*spin_singlet
> with u(r) and v(r) again simple exponentials u(r)=exp(-a r), v(r)=exp(-b r)
> it is found that the optimal a,b are a=2.1832 and b=1.885 and the energy is
> E=-2.87566 hartree, hence lower.
>
> So the UHF energy should be lower than the RHF too.
> Or am I missing something?
>
>
> Lorenzo
>
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-- 
Prof. Peter J. Knowles
School of Chemistry, Cardiff University, Main Building, Park Place, Cardiff
CF10 3AT, UK
Email KnowlesPJ at Cardiff.ac.uk
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