I wonder if the origin of the permanent dipole moment calculated by Gaussian (16) is always the center of mass (0,0,0 if "Nosymm is used)? For the neutral molecule it does not matter, but I am interested in the case when the molecule is not neutral.


1 Answer 1


If my memory serves correctly (this CCL post agrees), the origin used for computing the dipole is the center of nuclear charge:

$$R=\frac{\sum_i Z_ir_i}{\sum_i Z_i}$$

As you noted, the origin doesn't matter if the molecule is neutral and more generally the lowest order non-vanishing multipole will be independent of the origin while any higher order multipole will have some origin dependent terms, see this Physics SE post.

  • 1
    $\begingroup$ +1. Wow, did you know that Physics SE post from earlier? I'll briefly point out that this question was also asked on the CFOUR forum and John Stanton's answer was that in CFOUR the origin is the center of mass rather than the center of charge. It's interesting that Gaussian and CFOUR have made a different choice here :) $\endgroup$ Oct 6, 2023 at 1:24
  • $\begingroup$ Thanks for your response! Well, I also saw that CCL post, but for some reasons I am sure you (and the author of the CCL post) are right. I will try to clarify the issue as much as possible by sending a query to Gaussian and will report the Gaussian's answer here ... $\endgroup$
    – EvGeniy
    Oct 13, 2023 at 11:06
  • $\begingroup$ Well, according to the answer from Gaussian, Tyberius and the author of the CCL post seem to be right :) because in Gaussian the standard origin of the molecule, i.e., 0,0,0, is by default the center of nuclear charge (I actually thought that it was the center of mass) and this is also the origin of the dipole moment. But one can change the default with the option Symm=COM, which makes the origin of the molecule to be the center of mass. $\endgroup$
    – EvGeniy
    Oct 13, 2023 at 14:56

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