Dear sychen,
If the scan worked you should get a section like:
***********************************************
Scan completed.
Summary of the potential surface scan:
N D3 SCF
---- --------- -----------
1 180.0000 -78.78894
2 160.0000 -78.78749
3 140.0000 -78.78459
4 120.0000 -78.78313
5 100.0000 -78.78459
6 80.0000 -78.78749
7 60.0000 -78.78894
8 40.0000 -78.78749
9 20.0000 -78.78459
10 0.0000 -78.78313
11 -20.0000 -78.78459
12 -40.0000 -78.78749
13 -60.0000 -78.78894
14 -80.0000 -78.78749
15 -100.0000 -78.78459
16 -120.0000 -78.78313
17 -140.0000 -78.78459
18 -160.0000 -78.78749
19 -180.0000 -78.78894
---- --------- -----------
*********************************************
almost at the very end of the output. The above output is from a simple
example for the ethane torsion that I just did using the following
input:
**********************************************
# rhf/3-21g scan nosym
example scan, ethane
0 1
C
C 1 B1
H 1 B2 2 A1
H 1 B3 2 A2 3 D1
H 1 B4 2 A3 3 D2
H 2 B5 1 A4 3 D3
H 2 B6 1 A5 6 D4
H 2 B7 1 A6 6 D5
B1 1.500250
B2 1.117137
B3 1.117137
B4 1.117137
B5 1.117137
B6 1.117137
B7 1.117137
A1 110.724835
A2 110.724835
A3 110.724835
A4 110.724835
A5 110.724835
A6 110.724835
D1 120.0
D2 -120.0
D3 180.0 s 18 -20.0
D4 -120.0
D5 120.0
*******************************************
There are a few things to note.
This changes the value of the dihedral angle "D3" from 180 degrees by
-20 degrees over 18 steps.
I prefer to use a z-matrix in a scan as I can better control what's
going on than with "redundant internal coordinates".
It is usually a good idea to turn symmetry off (nosym) as often the
scanning coordinate will break the symmetry (as it does in this case).
The above scan is a "rigid scan". This means that the structures are not
optimized at each step, the dihedral is simply changed and the energy
calculated.
If you replaced "scan" with "opt=z-matrix" you would optimize all other
parameters (except D3) at each step. If the system is not too big then
this is probably the way to go.
I guess you are already aware but the energy units are Hartrees and 1
Hartree is equivalent to 627.5 kcal/mol.
I think that's about all. If you have any more questions, please don't
hesitate to ask.
Good Luck.
David.
On Fri, 2003-08-08 at 12:25, yuann wrote:
> Dear All,
>
> Is there any reference or document to understand the
> followings,
> 'The relative energies of three stationary
> points on the energy surfaces were selected to derive
> the corresponding torsional parameters.',
> which is described in Junmei et al. 1999.
>
> It seems to be that the method is described in Appendix C,
> 'Parameter Development'. If so, how to derive the
> specific torsional energy from Gaussian98's output?
> because I could just get the whole energy of the
> molecule after the potential energy surface scans.
> I'm sorry if this is an easy question, because I'm
> not very familiar to Gaussian.
>
> Thank you for your help in advance.
>
> Best Regards,
> sychen.
>
>
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--
---------------------------------------
Dr. David Smith
Department of Chemistry
Ludwig Maximilians University
Butenandt-Str. 5-13, D-81377 Munich
Germany
Tel.: +49 (0)89 2180 77740
Fax.: +49 (0)89 2180 77738
e-mail: David.Smith.cup.uni-muenchen.de
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Received on Fri Aug 08 2003 - 14:53:00 PDT
