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According to my knowledge, each computational chemistry code optimizes a molecule keeping the input orientation of the molecule. The population and spectral analysis is performed on the optimized geometry.

My Question is, if a molecule is slightly different from a perfect point group representation in Cartesian coordinates(i.e tilted from perfect Cartesian planes), will there be any consequence?

If so, how much difference can we expect in the molecular properties like ml-resolved partial occupation, spectral data, molecular orbital coefficients, Molecular orbital plots, electron density and it's plots, spin moments?

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    $\begingroup$ +1 and welcome to the site! But I do think this needs to be more specific. Using "etc" makes this open ended. For example I can tell you that 0.01 Angstroms away from equilibrium, the energy of the N2 molecule is almost 1 milli-Hartree different, but I cannot answer about what happens to "partial occupations, spectra, etc." especially since "etc" can mean anything. $\endgroup$ May 28, 2020 at 14:29
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    $\begingroup$ Related, internal energy and other molecular properties are independent of translation and rotation in the absence of any external field. However, because of the way the density functional integration grids are set up there can be some differences depending on orientation. This caused a big ruckous about 1 year ago. Would that answer be relevant here? $\endgroup$
    – Cody Aldaz
    May 28, 2020 at 18:31
  • $\begingroup$ when programs/algorithms employ symmetry they reorient the molecule for example see here. Would this answer your question? $\endgroup$
    – Cody Aldaz
    May 28, 2020 at 20:47

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Actually, quantum chemistry codes tend to reorient molecules to make calculations more reproducible, since as was already mentioned in a comment above, the use of quadrature makes density functional calculations somewhat dependent on the absolute orientation of the molecule, see e.g. the paper by Gill, Johnson, and Pople.

As to the second part of your question, if a molecule is slightly different from a perfect point group representation, this will affect its symmetries that may have repercussions onto observables due to e.g. the Jahn-Teller effect. Quantum chemistry codes usually employ the symmetry of the molecule to speed up calculations, but if you choose a too high level of symmetry, you might not get the correct solution!

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