Subject: Symmetry: In Reply to Dan Severence

From: "R.G.A. Bone" <[email-masked]>
Date: Wed, 27 Apr 94 10:14:17 -0700

In reply to Dan Severence's comments:

i) Gaussian calculations.

>> "Gaussian" finds symmetry present in the input nuclear coordinates and
>> applies it unless you explicitly switch it off.  But geometry optimizations
>> will typically, irritatingly, quit as soon as a change in point group occurs.
>
>      If you wish to force the symmetry, this can be averted by using the 
>same variable name for the symmetric lengths and angles, thus there is
>no possibility of a change in point group.
>

Yes, Dan is correct, provided that you can do it.  Constraining the 
internal coordinates in the Z-matrix is the way out.  But I had one fiendish
example once (in "C_i" symmetry) where it was not possible to do this: two
related angles could not be specified by the same internal coordinate and
the optimization failed as soon as these angles became slightly inequivalent.

ii) I was very careful in my comment about the gradient.

>> The gradient of the energy w.r.t. nuclear displacements should transform as
>> the totally symmetric 'irrep' in whatever point group you happen to be in.
>>For 'closed-shell' systems (no electronic degeneracies) this means in practice
>> that 'symmetric' structures are usually stationary points.  The only 
>
>      There are actually a number of cases where this is indeed not the
>  case.  Amides, for instance, tend not to be planar but are slightly
>  puckered at the N.  It often doesn't take much energy to force it to
>  be planar, but non-planar is the minimum at any levels of theory that
>  I've seen.  Also, quite bulky molecules often avoid a symmetric 
>  form in favor of spreading out the steric interactions.
>

Note that I said that symmetric structures are usually STATIONARY POINTS,
not 'MINIMA'.  Nothing I said contradicted the example of a non-planar amide.
But, symmetry can not be interpreted to be 'local' either.  I was about to
venture that the configuration in which the bonds about the N-atom are planar
would probably be a 'transition state' but of course such a structure may not
be a stationary point at all if there are other 'out-of-plane' atoms in the
molecule which are not related to one another by reflection in the plane.  
(Remember those non-bonded interactions ...).  If the _whole_ molecule is 
planar or has a plane of symmetry then (if closed-shell, and in the absence of
totally symmetric distortions) it will be a stationary point  but I couldn't
tell you, a priori, what sort.

N.B., Just because symmetric structures are usually stationary, does not mean 
the converse, either, viz:  stationary points do not have to be 'symmetric'.

There are clearly a number of pitfalls and potential confusions in the 
discussion of symmetry.  I will shortly post a useful reference-list to CCL
which not only provides background to my own comments but should hopefully
be a useful almanac to anybody with an interest in this field.

Watch this space .........

Richard Bone

================================================================================

R. G. A. Bone.
Molecular Research Institute,
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