0
$\begingroup$

I tried to parameterize a particular x-y-z-t dihedral angle of novel molecule using a Quantum Mechanical scan, but I got the strange result below. Molecule PES with dihedral

The energy changes sharply around 130° and $E_\text{dihedral}(0°) \neq E_{dihedral}(360°)$. I performed PES scan in gaussian09 and psi4 and I got the same results in different softwares.

I'm uncertain what approach to take to fix the scan. I saw some parameterization articles which didn't perform PES scan 0-360 degree. Other authors performed scan in the range of 0-60 degree or 0-120 degree and I didn't predict I will face a result like this. What is going wrong with my scan and how can I fix the issue? Would performing single point energy for each angle instead of PES scan help?

Improve this question
$\endgroup$
4
  • $\begingroup$ What are the units on the figure? $\endgroup$ – Buck Thorn Apr 11 '20 at 19:11
  • $\begingroup$ I'm voting to close this question as off-topic because I'd bounce it back too if I was the referee. The plots should be the same at 0/360 and the discontinuity adds doubt to the methodology. $\endgroup$ – Todd Minehardt Apr 11 '20 at 22:33
  • $\begingroup$ of course you can vote to close this question... but this is not off-topic. your strange comment is off-topic. Nobody asked you "hey if you were referee what would you do". why are you saying this? Are you trying to get attention? this is a computational chemistry problem (an important one) and we found a solution. TO say "should be equal" is not a solution. You should be more respectful to your olders, teenager. $\endgroup$ – Barış Kurt Apr 12 '20 at 4:05
  • $\begingroup$ Karsten Theis, Mathew Mahindaratne, Todd Minehardt, Mithoron, Jan,Let me make a criticism, please: It is really funny to mark an article about to be published in an academic journal with 3.5 impact as a "zero-effort homework". lol :) But I respect that. I hope someday you will get nobel prize. $\endgroup$ – Barış Kurt Apr 19 '20 at 13:07
3
$\begingroup$

Even without seeing your molecule, I can say the referee is very likely correct. If you think of doing a scan of the dihedral of hydrogen peroxide, by symmetry the path from $\theta=180\to360$ should just be the reverse of $\theta=0\to180$. In your case, I don't know the symmetry around the dihedral, but at a bare minimum the energy should be the same after a $360$ degree rotation. The fact that it isn't suggests that some scan step (probably that point around $130$) irreparably distorted the molecule geometry so that additional steps didn't bring it back towards the initial geometry.

Generating the scan point by point may help, but you still have to ensure the rotations you are making are physical and aren't distorting the molecule too much. Sharp changes in the energy between steps are a sign that something is going wrong.

You don't mention this detail in your question, but its also important to consider whether what you want is a rigid scan (where you just change this angle) or a relaxed scan (where you change this angle and then allow the rest of the structure to relax).

Improve this answer
$\endgroup$
4
  • $\begingroup$ I found the solution !. moderation can close this topic. but the solution has nothing to do with rigid or relax. $\endgroup$ – Barış Kurt Apr 11 '20 at 20:19
  • $\begingroup$ You could add an answer, it may help other users with a similar problem. $\endgroup$ – Tyberius Apr 11 '20 at 20:20
  • $\begingroup$ In regards to your previous comment, you did also ask whether the referees statement was reasonable. As I said, generating the scan steps point by point could help, but you still have to make sure you aren't unduly distorting the rest of the molecule to the point where most of the change in energy is due to the distortion rather than the changing dihedral. Making sure that ~130 geometry doesn't, for example, cause atoms to nearly overlap would help. You can also do two scans from 0 to -180 and 0 to 180 so that distortions along the way are less likely to propagate to later steps. $\endgroup$ – Tyberius Apr 11 '20 at 20:36
  • $\begingroup$ Thanks Tyberius . I will also try two-step method. (0-180 and 0 180) $\endgroup$ – Barış Kurt Apr 12 '20 at 4:07

Not the answer you're looking for? Browse other questions tagged or ask your own question.