## Details in Dipole Moment Calculations

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dinpajooh
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### Details in Dipole Moment Calculations

Hi,

I am calculating the full dipole moment matrix for a given cytochrome C, whose charge is -2 and I need to know more details about dipole moment calculations in (LS)DALTON:

Q1- Are the Cartesian components of the dipole moment shown in the output based on a center of mass, center of charge, or geometric centre for the origin?

Q2- In another software program, if one uses the keyword "symmetry", the molecule is rotated to the principal axis such that the center of mass is located in the origin of the Cartesian coordinate system. However, the keyword "Nosymmetry" suppress any orientation changes. After checking some test cases in (LS)DALTON, I am wondering any rotations are involved in (LS)DALTON calculations or is the original input orientation the reference description for all information regarding the wavefunction, dipole moments, etc. ?
If any rotations are involved during (LS)DALTON calculations I need to the corresponding details.

Q3- In fact, I did the same preliminary calculations for HBr molecule in that software program and LSDALTON. The output file for LSDALTON is attached.

Thanks,

############################################################
LSDALTON RESULTS:
Permanent dipole moment
-----------------------
au Debye 10**-30 C m
0.418333 1.06330 3.54678

Dipole moment components
------------------------
Debye x y z
-1.06330 -0.444564E-12 0.206660E-14

DipoleMomentMatrix: Transition moments between the groundstate and the excited states

STATE Freq(eV) X Y Z
1 6.342258 -0.0000000 -0.0000045 0.1686733
2 6.342258 0.0000000 0.1686738 0.0000063
3 11.555971 -1.5671080 -0.0000000 -0.0000000
4 11.602212 0.0000000 0.0001120 -0.6030157
5 11.602212 0.0000000 -0.6030134 -0.0001129
6 14.465625 0.0000000 0.0002274 -0.6054355

############################################################
Attachments
LSDALTON.OUT
LSDALTON output file for HBr molecule.
Last edited by dinpajooh on 21 Sep 2015, 17:32, edited 2 times in total.

kennethruud
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### Re: Details in Dipole Moment Calculations

I am not an expert on LSDalton, but let me anyway try to answer your questions:

1) Dalton will calculate the dipole moment with respect to the center of charge, which coincides with the center of mass unless you change the center of charge. My assumption would be that LSDalton calculates the dipole moment relative to the Cartesian origin.

2) I would not expect there to be any rotations are translations of the molecule in LSDalton (but you see this very easily by comparing the geometry in the output file with that of your input file).

3) You also report here the dipole transition moments, so I am uncertain what you are actually asking here, but the wave function is only converged to a predefined threshold, e.g. 1.0D-5 for Dalton, and for the transition dipole moments, we also have to determine the perturbed densities, which is normally 1.0D-4 for Dalton. This means that any digits beyond 1.0D-5 (dipole moment) and 1.0D-4 (transition dipole moment) has no physical meaning. As for the states that are degenerate, any rotation among orbitals are allowed, and thus the distribution of the transition dipole moment among the two degenerate states is not an observable. You will not that the sum of the squares of the transition dipole moments are the same in both calculations.

Best regards,

Kenneth

taylor
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### Re: Details in Dipole Moment Calculations

Kenneth's reply arrived while I was composing this so I have modified it accordingly. But I have to point out that your posting implies a lack of familiarity with the treatment of symmetry in Dalton/LSDalton that is covered at great length in both the manual and in (many) other postings on this forum --- a few minutes reading the manual and various forum postings (for example, the topmost item in the "Running Dalton" topic!) might have answered several aspects of your questions.

First, LSDalton does not use symmetry at all. It prints the coordinates it uses (or a subset of them for large molecules) so you can see immediately where the coordinate origin is, and the properties are computed with respect to this origin. As Kenneth points out, irrespective of whether symmetry is used explicitly, the presence of degenerate irreducible representations (as is the case for a diatomic such as HBr) means that the results are not unique to within a unitary transformation of the components of the degenerate representations.

Second, Dalton (not LSDalton) will by default (no options specified) try to determine the molecular symmetry to within D2h or a subgroup. This will usually involve translation to the centre of mass and rotation (see below) to a coordinate system determined by the principal moments of inertia. For most systems (but not always for systems possessing higher symmetry) this will provide a unique orientation that other codes might also choose, although they may differ within labelling of the axes. I stressed "rotation" above because the transformation to the principal axis system must preserve the handedness of the molecule if it is dissymmetric: the transformation must have a positive determinant (that is, be a proper rotation).

Third, users and developers of Dalton have the luxury of being able to say anything about our code and to compare it at will with other software that we may have licences for. This is not true of some software, and in particular it is not true of Gaussian, which you seem to reference specifically in your posting. Have you read the Gaussian licence agreement that you or your supervisor/department/institution signed with Gaussian Inc? Are you aware that any comparison of Gaussian with other programs that may illustrate performance issues or errors with Gaussian is expressly forbidden under the terms of the licence? I repeat, we at Dalton/LSDalton simply don't care. Nor do the developers of other free software such as CFOUR, ORCA, GAMESS, etc., nor even some non-free software such as Molpro or Molcas. But Gaussian..., well, I suggest you read the licence and think carefully before posting! But before you read the Gaussian licence, you would help yourself by reading about symmetry on this forum, and in the Dalton manual.

Best regards
Pete

taylor
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### Re: Details in Dipole Moment Calculations

In particular, so you know, I would think that your explicitly posting a Gaussian output to this forum for comparing with an LSDalton output is a breach of the Gaussian licence conditions. To repeat again, we don't care, but Gaussian Inc might...

Best regards
Pete

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