From:	m10!frisch %-% at %-% uunet.UU.NET (Michael Frisch)
Date:	Thu, 15 Oct 92 21:28:03 EDT
Subject:	Re: Spin density calculations for benzene (fwd)

    On Oct 14  3:32pm, Christopher Cramer wrote:
    
    > expense of the empty orbital. Mike Frisch may weigh in to tell us how the
    > resultant wavefunction from G92 may be other than D6h if that symmetry is
    > imposed from the start, but this is one of those cases where symmetry would
    > clearly produce an unstable wavefunction in any case.
    
    and Dave Bernholdt wrote:

    I'm not Mike Frisch, but I probably weigh more :-)
    
    Gaussian, and most other quantum chemistry codes are limited to
    abelian point groups for practical reasons.  The largest abelian
    subgroup of D6h is D2h, so that's what the calculation would have been
    done in.
    
    The HF wavefunction is forced to maintain the symmetry imposed by the
    _calculation_, but in this case it is not the symmetry of the nuclear
    framework.  Therefore, the wavefunction (and hence the spin densities)
    will satisfy D2h symmetry, but not necessarily D6h.
    
    Some codes (I don't remember of Gaussian does or not) will analyze the
    symmetry of the wavefunction after an SCF calculation and compare it
    to the symmetry of the nuclear framework.  This can be a very useful
    thing to check if you're in the habit of studying "challenging"
    systems.
    
    When studying Jahn-Teller systems, be careful about which potential
    surface you're on too.
    
I'll try not to be too ponderous (just verbose) :-), but packages vary in
how they handle symmetry, and I'd like to clear up what Gaussian does and
what one can conclude about some of these cases.

Gaussian does use abelian symmetry to speed up the calculation, but it does
not apply any symmetry constraints on the orbitals.  They normally come out
symmetric because that represents the lowest energy solution.  There are two
ways of using symmetry to reduce the work of forming the Fock matrix (and
computing the integrals, for direct SCF).  In the method used by default,
the Fock matrix is formed from the symmetry-unique integrals and then
symmetrized.  This is only valid if the density matrix has the molecular
symmetry, and this is checked for by the program.  The other method
(using the operators of the point group to replicate the unique integrals
while forming the Fock matrix) is available as an option, and does not
impose any symmetry constraints on the wavefunction.

Since Gaussian does not impose any symmetry on the solution, if the
wavefunction/density comes out D2h for nuclear symmetry D6h, that IS telling
you something about the physics of the molecule.

Sometimes the wavefunction and density have the full molecular symmetry
but there is a lower (broken) symmetry solution.  In this case, the
high symmetry wavefunction is a saddle point in orbital (Fock) space.
This can be tested for via a stability calculation -- the stability
matrix will have a negative eigenvalue corresponding to mixing occupied
and virtual orbitals to break the symmetry.

Gaussian does analyze the symmetries of the orbitals, which as Dave points
out can be useful in understanding what's happening in a difficult system.
That shows up the case of a highly symmetric system whose wavefunction has
broken symmetry.  However, in a careful study of these systems one also
needs to test stability in order to recognize the opposite case -- the
wavefunction stayed symmetric during the SCF cycles, but there is really a
lower solution.

In the case of a Jahn-Teller distortion, a set of degenerate orbitals is
only partially occupied.  The result in a Hartree-Fock calculation is that
the degeneracy is split, because HF occupied orbitals are optimized for the
neutral molecule, but the virtual orbitals "see" n+1 rather than n
electrons.  Consequently, the density is not symmetric, leading to
non-symmetric populations and also forces which break symmetry, leading to a
distorted structure if an unconstrained optimization is performed.

There can't be a Jann-Teller distortion in a molecule with a non-degenerate
point group such as C2v, but a similar effect can result from using an
excited-state wavefunction.  If the SCF proceedure happens to converge to a
state of a different symmetry than the ground state, then a frequency
calculation will show an imaginary frequency, which normally implies a
saddle point on the nuclear potential energy surface.  In the case of an
excited state SCF solution, however, what is happening is that the CPHF
procedure (which finds the derivatives of the density with respect to the
nuclear coordinates) determines that if a distortion is made which breaks
symmetry, then the lower energy (ground state) SCF solution can be mixed in,
lowering the energy.  This case can be distinguished from the usual (nuclear
saddle point) case by a stability calculation.  I think this is par of what
Dave meant about being careful about what potential surface you're on, and
that's good advice all the time, not just for Jahn-Teller systems.

Finally, someone else suggested that for delocalized systems one can get
spurious localized solutions using Hartree-Fock, and thus one needs MCSCF.
This is true with RESTRICTED open-shell Hartree-Fock.  However, UNRESTRICTED
Hartree-Fock can and does usually give properly delocalized orbitals.  The
simplest example is allyl radical, for which ROHF gives a localized solution
with one single and one double bond, and hence optimization with ROHF will
give a non-symmetric system.  UHF on the same system gives the proper C2v
symmetry.  There are certainly many systems which are not well described by
a single determinant, but one can't make blanket generalizations (at least,
not for UHF) based on whether the system has high symmetry.

Mike Frisch
-------


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