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pkg://aces3_3.0.6.orig.tar.gz:9616554/ACESIII/a2util/symcor/  info  downloads



I was asked many occasions how symcor works, especially
whether it is doing it's supposed job most effeciently.
One of the questions that is frequently asked is whether
symcor is actually taking advantage of the symmetry. For
example, in a C2V molecule such as H2O, are we duplicating
the Force Constant matrix calculations for symmetry
equivalent Hydrogens. Another question that was raised
recently has to deal with the mode of point selction;
single sided or double sided differentiation. In order to
answer these questions,
one has to inner working of the symcor member
executable (ME).

The function of symcor is to set up a table of geometry
points which will be used to generate energies which
will eventually be used to calculate numerical gradients
or Hessians. The symcor ME is used during finite
difference frequency calculations with analytical gradients
and numerical Hessians,
finite difference frequency calculations with numerical
gradients and Hessians and geometry optimizations with numerical
gradients. The symcor ME is not needed for
analytical gradient geometry optimizations. We will address
each mode of operation separately. Let's look at
some of the preliminaries;

 The first step is to make symmetry assignments
 to molecular vibrations (See Group Theory Cotton).
 The basis vectors that are used in here are
 (x,y,z) coordinates of each atoms. The actual
 subroutine that handle this operation is vibinf.f
 This is word by word computer implementation of
vibrational analysis descirbed in Cotton Pg. 309-315.
 Next step is to symmetry adapted vibrational coordiantes.
 This is done in symadq.f. It is somewhat difficult algorithm
 to understand. Once again the basis vectors are (x,y,z)
 coordinates of each nuclei. We take basis vectors for
 atom (eg. x coordinate on atom) and apply symmetry operations
 of the group and multiply by the character of irrep and
 sum over all the operations. It should be clear to people
 for some basis vectros this sum can be zero. That means
 that particular motion doest not constitue a vibration,
 rotation or translation. For eg. H2O, C2V (molecule in
 YZ plane) the x coordinates on H atoms does not contribute
 any of the motions due to symmetry. Only reason I am
 going to such details here is to answer the question
 whether we are doing unnecessary duplicate
 energy or energy gradient calculations.
 The answer is NO!!!! All the displacements are symmetry
 adapted. After appropriate normalizations, we will
 end up with a set of symmerty adapted Cartesian displacements
 (respect to an arbitrary origin). Now one has to transform
 them to the center-of-mass and principle axis (diagonalize
 the moment of inertia matrix). Here it is achived in traprj.f
 and rotprj.f. In my opinion rotational and translational
 projections are unfortunate names: they tend to imply that this
 is a some esoteric operation (it took me a while to figure
 out what the hell is going on those two routines).

 At this point we are ready to generate points for various
 options for numerical derivatives; Note that we always
 run a energy calculation at the reference geometry (E_0).

 NUMERICAL (both first and second derv.) FREQUANCY CALCULATIONS:

  The relevent routines are setpts.f and doener.f. At this point
  we know how many vibrations in each symmetry block. Assume that
  we have N vib. modes in a given symmetry block (q_1...q_n).
  The derivatives that we are trying to calculate are DE/Dq_iDq_j
  for all i and j.
  To get a numerical Hessian for a given mode, we need at least
  3 dispalced points. The reference (E_0), a positive
  displacement and a negative displacement. If particular
  vibration happens to have a plane or center of inversion,
  negative displacement is
  identical except for a oposite sign. In other words
  no explicit energy calculation is not required.

  Maximum # of points = N*(N+1), (or N*(N+1)/2 for A1 modes)
                      + Reference


  The most of the above discussion apply. However, the total
  displacement has to preserve the symmetry of the molecule.
  That means we only do consider totaly symmetric irreps.
  Also, we only need the derivatives DE/Dq_iDq_j (i=j).

 Maximum # of points = 2*N + Reference


  The relevent routines are setpts.f and dograd.f. At this point
  we know how many vibrations in each symmetry block. Assume that
  you have N modes in a symmetry block (q_1...q_n)
  To calculate Hessian, we need a gradient at a two displaced points.
  What we trying to calculate is DG/Dq_i for all i. For vibrations
  that have plane or center od inversion, we only need positive dispalcements
  since the negative displacement is identical to positive
  one but with an opposite sign.

 Maximum # of Points = 2*N or (N for A1 modes).


 In light of this investigations we can make the
 following conclusions.

  1. There is no repeat of energy calculations for symmetry
     redundent centers. The program take full advantage of
     Abelian symmetry.

  2. The numerical derivatives are calculated in double
     sided fashion. However, for totally symmetric
     irreps single or double sided makes no difference
     since we actualy do not do the energy calculation
     for the negative dispalcement (It is identical
     to the positive displaced value). In the case
     of numerical gradients and Hessians, the routine
     of interest is ener2fcm.f and for the case
     of anlytical gradients and numrical Hessian
     it is grad2fcm.f

  Logic in ener2fcm:

     For totally symmetric irreps:

     Digonal of the Force constant matrix,
                             DE/Dq_iDq_i = 2*(E_pi-E_0)/STPSZ**2

     where E_pi is the positive dispalced energy of ith mode and

STPSZ is the stepsize.

     Off diagonal elements of the FC matrix
                             DE/Dq_iDq_j = (2*(E_pipj-E_0) -
                                               E_pi - E_0) -
                                               E_pj - E_0))/2*STPSZ**2

     E_pipj is the positive dispalced energy of the ith mode
     while jth mode is held constant or vice versa.

     For other irreps:

     Digonal of the Force constant matrix,
                             DE/Dq_iDq_i = (E_pi+E_mi-2*E_0)/STPSZ**2

     where E_mi is the negative dispalced energy of ith mode and
     STPSZ is the stepsize.

     Off diagonal elements of the FC matrix
                             DE/Dq_iDq_j = (E_pipj + E_mimj - 2*E_0) -
                                               E_Pi +E_mi - 2*E_0) -
                                               E_Pj +E_mj - 2*E_0))

     E_mimj is the negative dispalced energy of the ith mode
     while jth mode is held constant or vice versa.

  Logic in grad2fcm is straightforward.

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