Symmetry elements Z, XY, cnnot be honoured

[Mohamed Abdelrasoul]

Dear Molpro users,

I am trying to generate a Potential Energy Surface (PES) of a few lowest excited states of Benzene along its normal modes, its symmetry at equilibrium geometry is defined automatically by the molpro package to be D2h ( it is D6h in literature), however, when the molecule is displaced along a certain normal mode to calculate the energy of same excited states, the symmetry is changed to be D2, so I have tried to fix the symmetry to be D2h using the following line: SYMMETRY, XY,Z but it failed as you can see below.

I would be grateful if you could help me to fix this issue?

"Using spherical harmonics

 The request for symmetry elements Z,XY, cannot be honoured

 Prototype coordinates
              1             2             3             4             5             6             7             8
    1    -4.6414692    -2.6132456    -1.3066228    -2.3207346     1.3066228     2.3207346     2.6132456     4.6414692
    2     0.0000000     0.0000000     2.2631371     4.0196302     2.2631371     4.0196302     0.0000000     0.0000000
    3     0.0000000     0.0000000     0.0335522     0.0713630    -0.0335522    -0.0713630     0.0000000     0.0000000
              9            10            11            12
    1     1.3066228     2.3207346    -1.3066228    -2.3207346
    2    -2.2631371    -4.0196302    -2.2631371    -4.0196302
    3     0.0335522     0.0713630    -0.0335522    -0.0713630
 ? Error
 ? Symmetry problem
 ? The problem occurs in zmatrix.f:zmat_orient

 GLOBAL ERROR fehler on processor   0     "                                   

Best regards,

Mohamed

[Piotr Gniewek]

Dear Mohamed, 

I have several pieces of advice/information for you, regarding your question about benzene calculations in Molpro. 

1. Molpro can use only those symmetry point groups that have elements of order at most 2, that is: D2h, C2v, C2h, D2, Cs, C2, Ci. This means that you cannot use the full symmetry group of the equilibrium structure of benzene molecule, which is D6h. In Molpro calculations for equilibrium geometry of benzene you can use D2h, which is the largest available subgroup of D6h.

You can read more about symmetry in Molpro calculations in the documentation: https://www.molpro.net/manual/doku.php?id=general_program_structure#symmetry

2. The equilibrium geometry of benzene molecule has D6h symmetry, but geometries displaced along normal modes will in general have lower symmetry! Only geometries obtained as snapshots of fully symmetric, that is a_{1g} vibrationla modes, will have D6h symmetry. For snapshots of vibrational modes of type different than a_{1g} you will have symmetry lower than D6h.
 
This means that you have to investigate each vibrational mode carefully. You have to carefully determine which electronic states of displaced geometries correspond to which electtronic states of the equilibrium geometry. (Careful investigation of orbital coefficients, occupations and energies might help with that.)

For learning about vibrational modes of benzene I've found useful the following: Figure 1.14 on page 71 of "Vibrational Spectra of Benzene Derivatives" by  G. Varsanyi.

3. The SYMMETRY keyword in Molpro cannot give your system more symmetry than you have!

For example: if you have a benzene molecule displaced along an e_{2u} vibrational mode you cannot have D2h symmetry because this system doesn't have the corresponding symmetry elements (you don't have three prependicular planes of reflection, you don't have a center of inversion). You can see for example in panel 16a of Figure 1.14 in the book mentioned above.

In fact you can only use SYMMETRY to request calculations in the full symmetry of your arrangement of atoms, or lower symmetry. You cannot "increase" symmetry with this keyword.

4. You need to be careful when specifying symmetry generators in Molpro. The commas matter a lot!

SYMMETRY,XY,Z; means you want the symmetry group given by two symmetry generators:
XY = change sign of x and y coordinates = C_2 rotation along z axis
Z = change sign of z coordinate = reflection in xy plane

On the other hand
SYMMETRY,X,Y,Z; means you want the symmetry group given by three symmetry generators:
X = change sign of x coordinate = reflection in yz plane
Y = change sign of y coordinate = reflection in xz plane
Z = change sign of z coordinate = reflection in xy plane

To request D2h group you would use:
SYMMETRY,X,Y,Z;

The instruction that was used in your email:
SYMMETRY,XY,Z;
would request not D2h calculation, but C2h one! (Those are different symmetry groups, D2h has eight elements, C2h has four elements).

So commas matter a lot here!

Read more in the documentation: https://www.molpro.net/manual/doku.php?id=general_program_structure#symmetry

But let me stress again that you cannot artificially increase symmetry using SYMMETRY keyword. Molpro can use only the symmetry that your arrangement of atoms has in the given calculation. It is impossible to artificially "increase" symmetry using a keyword.

With best regards,
Piotr Gniewek

[Abdeljabbar Jaddi]

Hi,

I have the same problem, i did the same stages. But still the same problem. can anyone help me to resolve the issue?

**input

  ***,tr
       gprint,orbitals,civector
       angstrom;
        geometry={C1
                  C2,C1,6.0
                  O3,C1,r2,C2,a1
                  H4,C1,r3,C2,a2,O3,d1
                  H5,C2,r4,C1,a3,O3,d2
                  H6,C2,r5,C1,a4,H5,d3
                  H7,C2,r6,C1,a5,H5,-d3}
       r1=6.0 ang, r2=1.21326512 ang, r3=1.09970580 ang, r4=1.07962307 ang, r5=1.10626409 ang, r6=1.10626409 ang, a1=124.38656784, a2=114.97088682, a3=111.11894725, a4=109.08515994, a5=109.08515994, d1=180, d2=180, d3=120
       basis=avtz
       int;
       {rhf-scf;wf,24,1,0;}
       {multi;
       occ,11,3;
       closed,5,0;
      wf,24,1,0;state,1
      wf,24,2,0;state,1
       canonical,ci;}
      {rs2,MIX=1,INIT
       wf,24,1,0;state,1,1;}
      optg

**** out

ITER.   ENERGY(OLD)    ENERGY(NEW)      DE          GRADMAX     GRADNORM    GRADRMS     STEPMAX     STEPLEN     STEPRMS   CPU-time
   1  -153.40319166  -153.41608971    -0.01289805  0.04293766  0.07263483  0.02014528  0.24241190  0.41567069  0.03197467    200.81
 no convergence in dis1, ss=   1.95942755247553
 no convergence in dis1, ss=   1.88364810300258
 The request for symmetry elements X, cannot be honoured

 ? Error
 ? Symmetry problem
 ? The problem occurs in zmatrix.f:zmat_orient

 GLOBAL ERROR fehler on processor   0