The program automatically added hydrogens when using symmetry

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Just Got Here
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When I was running a TDDFT calculation for NH3 molecule with the following input geometry:

geometry
N 0.0000 0.0000 0.0000
H 0.0000 -0.9377 -0.3816
H 0.8121 0.4689 -0.3816
H -0.8121 0.4689 -0.3816
end

The job failed with the error information "tddft_init: non-Abelian symmetry not permitted".

The point group of NH3 is C3v which is indeed not Abelian. The corresponding Abelian subgroup is Cs. So I added this symmetry in the input file like this:

geometry
N 0.0000 0.0000 0.0000
H 0.0000 -0.9377 -0.3816
H 0.8121 0.4689 -0.3816
H -0.8121 0.4689 -0.3816
symmetry cs
end

However, in the output file the program added three hydrogen atoms and it became a strange geometry:

                            Geometry "geometry" -> ""
-------------------------

Output coordinates in angstroms (scale by  1.889725989 to convert to a.u.)

 No.       Tag          Charge          X              Y              Z
---- ---------------- ---------- -------------- -------------- --------------
1 N 7.0000 0.00000000 -0.00001538 0.00000000
2 H 1.0000 0.00000000 -0.93771538 -0.38160000
3 H 1.0000 0.00000000 -0.93771538 0.38160000
4 H 1.0000 0.81210000 0.46888462 -0.38160000
5 H 1.0000 0.81210000 0.46888462 0.38160000
6 H 1.0000 -0.81210000 0.46888462 -0.38160000
7 H 1.0000 -0.81210000 0.46888462 0.38160000

This is definitely not correct. Only the atom 1, 2, 4 and 6 are in the input file. I have the same problem for the CH4 molecule (in which 4 hydrogen atoms were added).

I also tried the Z-matrix using the variables that make it completely symmetric. But there is still the same error.

Now the only way that I can run this job for this molecule is to completely turn off the symmetry. But this makes it very difficult to analyze the result since it is impossible to identify the states based on symmetry.

I don't understand why the program automatically added several hydrogen atoms. Is there any solutions for this kind of problem?

Thanks very much.

Forum Vet
Threads 7
Posts 1355
When using the Cs point group, NWChem assume the mirror plane being perpendicular to the z-axis (i.e. xy plane).
Therefore, you need to rotate the molecule to satisfy this constraint as shown here

geometry
 N   0.0000   0.0000  0.0000
 H  -0.9377  -0.3816  0.0000
 H   0.4689  -0.3816  0.8121
 H   0.4689  -0.3816 -0.8121
 symmetry cs
end

Just Got Here
Threads 1
Posts 2
Quote:Edoapra May 21st 6:34 pm
When using the Cs point group, NWChem assume the mirror plane being perpendicular to the z-axis (i.e. xy plane).
Therefore, you need to rotate the molecule to satisfy this constraint as shown here

geometry
 N   0.0000   0.0000  0.0000
 H  -0.9377  -0.3816  0.0000
 H   0.4689  -0.3816  0.8121
 H   0.4689  -0.3816 -0.8121
 symmetry cs
end


Thanks so much! This works now. However, I still have trouble for the methane molecule. The full point group is Td but it is also not Abelian. The Abelian subgroup is C2v. From the manual, in the C2v point group "the z axis is the C2 axis and the ?v may be either the xz or the yz planes". This is the input:

geometry
symmetry group c2v
C 0.0000 0.0000 0.0000
H 0.6276 0.6276 0.6276
H 0.6276 -0.6276 -0.6276
H -0.6276 0.6276 -0.6276
H -0.6276 -0.6276 0.6276
end

I checked in a graphics program. The z axis is the C2 axis and both xy and yz planes are the ?v. But the program still added 4 hydrogen atoms:

 No.       Tag          Charge          X              Y              Z
---- ---------------- ---------- -------------- -------------- --------------
1 C 6.0000 0.00000000 0.00000000 0.00000000
2 H 1.0000 0.62760000 0.62760000 0.62760000
3 H 1.0000 -0.62760000 -0.62760000 0.62760000
4 H 1.0000 0.62760000 -0.62760000 0.62760000
5 H 1.0000 -0.62760000 0.62760000 0.62760000
6 H 1.0000 0.62760000 -0.62760000 -0.62760000
7 H 1.0000 -0.62760000 0.62760000 -0.62760000
8 H 1.0000 0.62760000 0.62760000 -0.62760000
9 H 1.0000 -0.62760000 -0.62760000 -0.62760000

So what's wrong here?

Thanks very much.

Clicked A Few Times
Threads 10
Posts 24
If you enumerate and locate all the atoms in a compound, then you can't use symmetry point groups, only C1 symmetry.

In Principle, you may specify methane using Td symmetry and need only state the coordinates of
Carbon at (0,0,0) and
Hydrogen at (0.6276,0.6276,0.6276)

The justification is that an optimization using symmetry converges faster. You can describe benzene as 12 atoms in a C1 group or two atoms in a D6h group. The latter is significantly quicker to converge ... if you can describe it correctly.

In Practice, I use C1 symmetry by default according to the Principle of Maximum Laziness, a 1st Law corollary, and try to use Z-matrix form for molecules having more than 3 atoms.

Kill Specify them all and let God NWChem sort it out.
Edited On 10:21:46 PM PDT - Thu, May 21st 2015 by Drhaney

Clicked A Few Times
Threads 10
Posts 24
Good symmetry examples in the Documentation section:

http://www.nwchem-sw.org/index.php/Release65:Geometry_examples

Forum Vet
Threads 7
Posts 1355
Here we go

H positions at 1.087*sin(109.5/2) 0. 1.087*cos(109.5/2)


geometry
symmetry group c2v
 C                     0.000000000    0.000000000     0.000000000
 H                     0.887531783    0.000000000     0.627579743
 H                    -0.887531783    0.00000000      0.627579743
 H                     0.000000000    0.887531783    -0.627579743
 H                     0.000000000   -0.887531783    -0.627579743
end
Edited On 12:52:27 PM PDT - Tue, May 26th 2015 by Edoapra


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