[molpro-user] Helium UHF energy

Lorenzo Lodi l.lodi at ucl.ac.uk
Sun Oct 4 22:24:51 BST 2009

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?


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