When I run a linear-response time-dependent DFT calculation in my favourite electronic structure program, for each excitation I'm given an oscillator strength, $f_i$, and transition dipole moment $$\boldsymbol{\mu}_i = \int\Psi_0\hat{\mu}\Psi_i \text{d} \bf{r}$$ What formulae relates the transition dipole moments, the oscillator strength and the observed absorption relative intensity? And more specifically, how can I calculate the relative experimental intensity of TDDFT transitions from the oscillator strengths and transition dipole moments?

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    $\begingroup$ You may consider posting it on matter modelling SE (which is not to say in any way that this is unsuitable for chem SE) $\endgroup$
    – S R Maiti
    May 12, 2021 at 10:29
  • $\begingroup$ This might help researchgate.net/deref/… then "take me there". Pdf. $\endgroup$
    – Alchimista
    May 12, 2021 at 10:54

1 Answer 1


I'll address the last part of your question. I think the first part

What formulae relates the transition dipole moments, the oscillator strength and the observed absorption relative intensity?

of your question could stand on its own.

The oscillator strength is connected to the transition dipole moment vector via $$ f_{if} = \frac{2}{3}\omega_{if}|\vec \mu_{if}|^2 $$ in atomic units,i.e. $m_e=\hbar =1$, $\hbar \omega_{if}=(E_f-E_i)$. The equation in SI units can be found on Wikipedia. $i,f$ stand for the initial and final electronic state involved in the transition.

For spontaneous emission, you have to multiply $f_{if}$ by an additional factor of $\omega_{if}^2$ since spontaneous emission is proportional to $\omega_{if}^3|\mu_{if}|^2$.

The oscillator strength is already a dimensionless relative quantity and the ratio of different oscillator strengths should correspond to the ratio's in an absorption spectrum measurement. Just normalize everything by the highest value if you are going for a plot that is comparable with a relative intensities plot where the peaks are also normalized .

It is important to note that the calculation only describes a purely electronic transition and yields only a single line at the energy difference of the involved adiabatic electronic states. The result of a real measurement is not a single line and includes many effects that are not described by a simple excited electronic states calculation.

But the calculation gives you an estimate where the center of the peaks should be and how intense they should be. A simple TD-DFT calculation does not give you information about the detailed lineshape. The lineshape is typically assumed to have some form and then added by broadening a stick spectrum based on the oscillator strenghts with a spectral lineshape function. The width of the lineshape function is then adjusted to match the experimental measurement manually. Common lineshape functions are here.

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    $\begingroup$ So the adsorption intensity is directly proportional to the oscillator strength, not with standing in/homogenous broadening, finite temperature and other factors giving finite line width? $\endgroup$
    – user213305
    May 12, 2021 at 13:30
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    $\begingroup$ Yes, as long as two different electronic stats are involved in the transition. It would not matter for example in a pure infra red spectrum where we do not excite electrons and the electronic state doesn't change. But as long as you also excite electronic states, you get this as a multiplicative factor for your peak intensity. $\endgroup$
    – Hans Wurst
    May 12, 2021 at 13:42
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    $\begingroup$ Temperature effects, Doppler broadening, Life time broadening, etc are all additional effects that have to be incorporated by other means. They are not part of a simple excited state calculation. That is why you need to broaden the stick spectrum with a suitable lineshape function. This additional broadening with a lineshape function is used to model the effects. It is non trivial to predict the correct lineshape in detail, and typically its adjusted by comparison with the experimental data. $\endgroup$
    – Hans Wurst
    May 12, 2021 at 13:46
  • $\begingroup$ I'm using an extremely simplified adiabatic approach, instantaneous excitation at the FC point, no relaxation, purely electronic etc. $\endgroup$
    – user213305
    May 12, 2021 at 13:53

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