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I am trying to calculate the dipole moment of an excited state, using "Gaussian" software. My input commands were:

opt freq td=(singlets,nstates=5,root=1) cam-b3lyp/6-31g(d) geom=connectivity

as I wanted to optimize the geometry of $S_1$ state (starting from optimized $S_0$ structure) and calculate the $S_1$ dipole moment of the optimized structure. I am wondering, is this is the correct way to do the excited state dipole moment calculation? Moreover, is the Dipole moment in an output after the last "Population analysis using the SCF Density" the correct dipole moment of the excited state? I have looked for the answers over the Internet, however, information regarding this issue is very limited.

Thank you very much in advance for your help!

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2 Answers 2

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This looks like it has been addressed on the CCL previously.

To summarize and add details for your case, Gaussian can directly calculate the excited state dipole, but you need to add the keyword Density=current to get it to use the density from the excited rather than the ground state. The excited dipole should then be under a header like this:

**********************************************************************
Population analysis using the CI density.
**********************************************************************

This will print twice during an optimization: once for the initial geometry and once for the final geometry.

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  • $\begingroup$ Thank you very much! That was exactly what I was looking for. I have seen that correspondence with "Gaussian" developers, however, I did not think the density keyword was necessary, as by default "Gaussian" should take "Density=Current". I have also consulted on this in "Multiwfn" forum and there was also an answer that "Density" keyword must be included. Thank you! $\endgroup$
    – aerospace
    Nov 18, 2021 at 9:13
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There's actually more than one way to calculate the dipole moment, and you can check whether or not the numbers you're obtaining from td=(singlets,nstates=5,root=1) are making sense by comparing your result to what you'd get with a more "manual" approach in which you can see exactly what's going on.

The simplest formula is simply the expectation value of the dipole moment operator:

$$\tag{1} \langle \psi | \mathbf{\hat{\mu} }| \psi\rangle. $$

However, it can be much harder to get wavefunctions than energies, and a first-order error in the wavefunction can lead to a second-order error in the energy, so it would be convenient if we could calculate the dipole moment via only energy calculations (also, some electronic structure methods or implementations in software may be easily able to give energies but not wavefunctions). In fact we can calculate the dipole moment with only simple energy calculations.

It can be shown that for an electric field $\mathcal{E}$ we have (see section 4 of this PDF, for example):

$$\langle \psi | \mathbf{\hat{\mu} }| \psi\rangle = -\frac{\textrm{d}E}{\textrm{d}\mathcal{E}}\tag{2}.$$

This is related to perturbation theory and/or linear response theory, and it allows us to calculate Eq. 1 without ever needing the wavefunction, simply by calculating the energy multiple times and doing finite differences to approximate the derivative (if you want $n$ components of the dipole moment, you'll need to do $n+1$ extra energy calculations in addition to the zero-field energy calculation, but even for 3-components of the dipole moment, I often wouldn't mind doing 5 energy calculations if it could help me avoid requiring use of the wavefunction).

Whenever I hear "finite differences" I cringe a bit since I almost always prefer to do things analytically if possible, but this paper by Ernzerhof (the 'E' in PBE) et al. actually explains that using finite differences can be better than trying to calculate the dipole moment more directly. The same is said in section 3.1 of this paper by Lodi & Tennyson, along with more references that recommend the finite differences approach over the more "direct" approach. You can see the differences between numerical values obtained via finite differences versus DFPT here and the difference between finite differences and using the linear response here.

So if you're not sure if the direct approach is giving you sensible answers, you can just do a simple energy calculation $2 + n$ times if you want $n$ components of the dipole moment, and it might not be a bad idea to do this second method anyway, to see whether or not the more "direct" approach is giving sensible results!

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  • $\begingroup$ Why $n+2$? I would think you could do the zero field calculation and then $n$ additional calculations with E-field in each direction. $\endgroup$
    – Tyberius
    Nov 17, 2021 at 21:28
  • $\begingroup$ The Lodi and Tennyson reference I gave says: "The major inconvenience of the ED approach is that if analytical derivatives with respect to the perturbation strength are not implemented the derivatives have to be computed as finite differences and this involves at least n+1 separate energy computations if n components are to be determined, see e.g. section 4.1 of [278] for a related discussion" and I recall Eamon Conway telling me at a conference in 2018 that he was doing 5 energy calculations for each point on the dipole moment surface, but it seems it should only have been 4. $\endgroup$ Nov 17, 2021 at 21:39
  • $\begingroup$ I suppose he could have been doing a more accurate finite difference, possibly for a material where the one direction of the dipole was known to be zero or had to be by symmetry. If they did a central finite difference to get 2 dipole components, that would lead to 5 calculations assuming they still needed the energy at the field (the dipole components themselves would only require a calculation with a forward and reverse field in that direction). $\endgroup$
    – Tyberius
    Nov 17, 2021 at 22:31
  • $\begingroup$ @Tyberius This might have been what happened, since it was the water molecule (symmetric), but Tennyson (the author of the paper that said n+1 separate calculations) was his co-author and I'm sure he was using the same method that Tennyson was describing, rather than a more accurate finite difference approximation. I'll ask him on Messenger to confirm! $\endgroup$ Nov 17, 2021 at 23:29

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