Subject: Summary on symmetry, part two

From: Patrick Bultinck <[email-masked]>
Date: Fri, 20 May 1994 15:41:02 +0200 (MET DST)

Netters,

A few weeks ago I started a discussion on the use of symmetry for 
optimizations. I got quite a lot of answers... and at the time I sent the 
CCl a summary I clearly stated that if any more answers were to reach me, 
I would send an appendix to the net.

Patrick, University of Ghent, Belgium

So here it is :

Dear Netters,

We have seen lots of symmetry related messages.  It seems to me there are
some confusions regarding using symmetry in calcualtions.  IMHO, to use
symmetry or not is totally intuitively-based decision.  There is no a priori
reason to prove that symmetric or asymmetric geometry should be minimum or
not.  But chemical sense (VSEPR rules based) always favors to start from
symmetric geom.  Take TiCl4 (or simply CH4) as an example, it makes little
sense to start from C1 symmetry, though you'll get to Td geom eventually and
costly.  As molecules are bigger and more complicated, using symmetry is
always a good starting point.  Though high symmetry geom might not be
minimum, it also gives us certain info anout the structure. The example is
MoH6 which is not Oh structure!

For some multi-well potential surfaces, even you start from C1 symmetry,
there is no guarrantee you'll reach a global minimum.  Therefore using lower
symmetry is not a good starting point, unless your intuition tells you to do
so (how from the beginning?  I don't know :-)

There are some confusions about minimum (saddle point) and stationary point.
We all refer these terms to OPTIMIZED structure, not randomly drawn
structure.  Using any symmetry constraint, you can always get to a
stationary point (counter-examples? I'd like to hear one :-), meaning you
can always optimize it to a point where all gradients vanish (within a given
threshold).  But the stationary point may not be a minimum.  At times you
have to break or lower symmetry to reach a minimum (after you are sure
>from a freq. calcn.)

All in all, computational chemistry is still chemistry.  Do whatever makes
chemical sense. 

Happy computing...

--------                
   _         Tang, Huang
 _| ~-       Chem. Dept, U. of Houston
 \,  _}      Houston, Texas 77204-5641
   \(        e-mail:  [email-masked]
             phone:   (O)  (713) 743-3269  

Apropos the recent discussion on symmetry and stationary
points:

With the exception of Jahn-Teller systems, if the symmetry
of the wavefunction is the same as that of the nuclear
coordinates, then, for an appropriate set of coordinates,
there will always exist a stationary point.  That is easy
to prove from symmetry theory.

Jahn-Teller systems are special - for such systems there
cannot be a stationary point, and a distorting force will
exist which will lower the symmetry.  If a center of symmetry
exists, that will be preserved under the distortion.

Of more interest are systems for which the wavefunction does
NOT have the symmetry of the nuclear coordinates, particularly
in low-symmetry systems (to avoid confusion due to Jahn-Teller
considerations). 

An example of such a system is fluoroethylene, forced into a 
twisted Cs symmetry (the plane of the CH2 being perpendicular 
to that of the CHF).  If the symmetry of the wavefunction
is forced to be Cs, then the energy of the system will be much
higher than if the wavefunction is unconstrained.  Conventionally,
by a self-consistent field, we mean that the electronic energy
is an irreducible minimum.  Given this, it follows that the
forces due to the electronic structure at self-consistency
will not reflect Cs symmetry (Obviously they will be C1).
In that case, there cannot exist a stationary point in the
Cs symmetry.

This does not invalidate the Jahn-Teller effect.  It is only
a logical extension.  To see how, consider a Jahn-Teller system
such as singlet D2d ethylene.  Under the Jahn-Teller theorem,
a distortion of the wave-function from D2d symmetry lower the energy.  
The extension consists of simply making the minimal disturbance 
to the system so as to destroy the high symmetry, but not 
enough to affect the energetics too much.  In such a case 
(C2H3F being an example), the distorting force will still exist.

Jimmy Stewart

Dear Dr. Stewart: It is somewhat strange to read your message about stationary  points and the Jahn-Teller effect, as if the problem has not been studied for many years by many authors and several books were not published. You may not be aware of all thes
e publications and, apparently you are not alone; this is why I  refer you to the book of R.Englman and to my book with V.Polinger (Springer, N.Y., 1989) where this and many other related problems are considered in detail.   
Regards,  Isaac-- 
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Isaac B. Bersuker		 | E-mail: 
Dept. of Chemistry	         | [email-masked]
Univeristy of Texas at Austin    | Vox: (512) 471-4671
Austin, TX 78712 	         | Fax: (512) 471-8696
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