I want to estimate the lifetime of an electronic excited state. Lets assume there is only radiative coupling via spontaneous emission from this state to the ground state. I can optimize the excited state, for example the S1 state and calculate the transition dipole moment to the ground state at this S1 equilibrium geometry. Can I calculate an estimate for the lifetime of this state from these values?

  • $\begingroup$ +1 and welcome to our new community! This is a nice first question, so thank you for contributing your question here! We hope to see much more of you in the future. I commented out the part in your question about widths, because we have a policy to only ask one question per post, and while the two questions had some relation to each other, the question about whether or not widths can be calculated without any experimental input, is different from the question in the title and currently in bold (also none of the answers so far talked about widths, so no worries there!). $\endgroup$ Commented Mar 27, 2021 at 4:16

3 Answers 3


tldr; it's related to the excitation energy

The lifetime of an excited molecule assuming there are no non-radiative pathways, is related to the Einstein A coefficient:

$\left(\frac{dn_2}{dt}\right)_\mathrm{spontaneous} = -A_{21}n_2$

where $n_2$ is the population of the excited-state.

I used this reference to get the following equations: J. Braz. Chem. Soc., Vol. 23, No. 12, 2237-2247, 2012. But there probably exists a much more formal derivation elsewhere.

In short it's possible to calculate the radiative lifetime by:

$A_{21} = \frac{1}{\tau_\mathrm{rad}} = 2.142995\times10^{10}|\langle\Psi_1|\mu|\Psi_2\rangle|^2\times E^3_{ve}$

in which $A_{21}$ and $\tau_\mathrm{rad}$ have units, $\mathrm{s}^{-1}$ and s, respectively, $\mu$ represents the transition moment operator, and the vertical energy $E_{ve}$ is in atomic units, $\Psi_1$ and $\Psi_2$ are the ground and excited-state.

  • 2
    $\begingroup$ This looks somewhat different. Any idea why this rate is proportional to the square of the transition frequency instead of the third power ? Id really be interested in a source for this equation. The paper sadly doesn't give a citation for it. There is a small typo, the matrix element is missing a square $\langle \Psi_1|\mu|\Psi_2 \rangle ^2$ btw. $\endgroup$
    – Hans Wurst
    Commented Mar 26, 2021 at 22:42
  • $\begingroup$ Hi, good eye. I hope the answer was helpful. I'm not sure why it's squared, but if I had to guess I'd say it looks like a probability which are usually wave functions squared. I'm also not sure why that paper doesn't use adiabatic excitation energies and dipole moments. But it's the only paper I have in my personal library that has this type of calculation. I will say that it is a well known result so you can definitely find a good derivation of this $\endgroup$
    – Cody Aldaz
    Commented Mar 27, 2021 at 1:01
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    $\begingroup$ The rate is proportional to the cube of the energy (or frequency). See Bernath, Spectra of Atoms and Molecules, 2nd Ed. equation 1.53. The numerical factor that I get for these units is a (tiny) bit different but close enough (2.1420008e+10). $\endgroup$ Commented Mar 30, 2021 at 20:38
  • $\begingroup$ @AntoniodeOliveira-Filho Thank you for the correction. I am not a fan of equations with evaluated constants so i looked up your reference and i assume you mean this equation, $A_{i\rightarrow f}=\frac{16\pi^3\nu^3}{3\varepsilon_0hc^3}\mu^2_{if}=\frac{4\omega^3}{3\varepsilon_0 \pi\hbar c^3}\mu^2_{if}=\frac{4E^3}{3\varepsilon_0 \pi\hbar^4 c^3}\mu^2_{if}$ Do you know why it differs by a factor of 4 from the commonly given rate as here for example, en.wikipedia.org/wiki/… $\endgroup$
    – Hans Wurst
    Commented Apr 1, 2021 at 22:02
  • $\begingroup$ I would really love to know a book that handles the approximations and conditions which are made until one arrives at the spectroscopy equations rigorously. Most texts that i know are very "handwavy" and don't discuss in which cases the equations are applicable and or not. $\endgroup$
    – Hans Wurst
    Commented Apr 1, 2021 at 22:06

The Einstein A coefficient (as mentioned in Cody's answer) tells you the probability per unit time of spontaneous radiative decay from an excited state to a lower state.

The Fermi golden rule (as mentioned in sleepy's answer) also gives a probability per unit time to transition from a specific state, but the destination is usually a group of states in a continuum (not just one discrete state lower than the initial state). It's possible to use Fermi's golden rule for the case where the final state is also just one specific state (not a group of states in a continuum), but you will see this discussed less often than the first case.

A third and much less well-known way, was used in my paper on "The radiative lifetime of $\ce{Li}(2p)$". We didn't know this when we published the paper, but 2 years later a paper was published by a completely different group saying that we had determined the radiatime lifetime more accurately than any previously determined empirical value, by an order of magnitude (we had no idea that we'd done that until we saw this paper come up 2 years later, otherwise maybe we would have submitted to a different journal!). Our success was due in part to making use of fairly high-accuracy experimental data, but no experimental data is necessary for this procedure. The method is as follows, and I'll explain it for the radiative lifetime of an atom, but it can also be used for molecules:

The above paper's analysis was a bit more complicated because spin-orbit (SO) coupling was treated, but the main procedure is the same whether or not the fine-structure and other added complexities associated with SO coupling are included. Consider the ground state of $\ce{Li2}$ which dissociates into two ground-state (S-state for $\ce{Li}$) atoms: $\ce{Li}(S)+\ce{Li}(S)$, and an excited state of $\ce{Li2}$ which dissociates into one ground-satte atom and one excited-state atom: $\ce{Li}(S)+\ce{Li}(P)$. The long-range potential energy function between the two pieces is given by $C_3/r^3$ rather than the typical Van der Waals or London dispersion form of $C_6/r^6$. The radiative lifetime of the $\ce{Li}(P)$ atoms is given by (see Eq. 37 of the paper, and $\lambda$ is the excitation energy between $\ce{Li}(S)$ and $\ce{Li}(P)$, which is similar to what we saw in Cody's answer):

$$\tag{1} \tau = \frac{3}{4\pi c}\frac{\left(\lambda / 2\pi\right)^3}{C_3}. $$

$C_3$ can be calculated entirely ab initio as in Eq. 47 of this paper, or you can do an ab initio calculation of the potential energy curve of an excited state $\ce{Li2}$ molecule and fit it from there. In this way, the chemistry of a molecule is used to obtain a fundamental atomic property.

So by taking an excited system (the $P$-state of $\ce{Li}$ atom in the above exmaple, but it could also be the S1 molecule in your question), and seeing how it interacts with a ground-state version of that system a long distance away, you can get the radiative lifetime of the excited system.

  • $\begingroup$ I must say that it looks quite non trivial to me to transfer that approach to molecules with more than 2 atoms, but its interesting none the less. As i understand it you would need very good information about the full potential energy surface which is pretty hard to get for large molecules, especially for excited states to obtain the parameters. I also couldn't find any derivation for equation (1), since all citations lead me to G. W. King and J. H. Van Vleck, Phys. Rev. 55, 1165 ~1939, which seems not to be concerned with life times at all. Although its possible that i missed it. $\endgroup$
    – Hans Wurst
    Commented Mar 27, 2021 at 9:55
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    $\begingroup$ This method is interesting! For larger molecules, the oscillator strength is the quantity that is easier to measure (from absorption spectroscopy), from which one can calculate the radiative lifetime. Then by using Eq. (1) in the reversed way, one can get the asymptotic behavior of the PES of the excited dimer of the molecule, which can be useful in the study of weak intermolecular interactions. One may even obtain the ground state C6 coefficients between two molecules from their experimental electronic spectra, using the Casimir-Polder formula. $\endgroup$
    – wzkchem5
    Commented Apr 15, 2023 at 14:15

Fermi golden rule should give you the transition probability (per time), from which you should be able to estimate the lifetime. You should know of be able to estimate the density of states in the final (ground) state, of course.

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    $\begingroup$ I am talking about a discrete transition from S1 to S0. I am only consindering this single transition to a single discrete state. I have heard that about Fermi's Golden Rule and read that before but never seen it done. Can you point me to an example or work it out to some degree ? $\endgroup$
    – Hans Wurst
    Commented Mar 26, 2021 at 18:17
  • $\begingroup$ actually, I have not done anything with it myself in the last 15 years or so, so for the sake of safety you might want to check with a good book on quantum mechanics, but if I understand your problem correctly, you are essentially interested in the rate of spontaneous emission. For that, the density-of-states part describes photons in the free space and it would be something like (8pi/hc^3)*w^2 (w is the frequency, please check the numbers because my memory is terrible). $\endgroup$
    – sleepy
    Commented Mar 26, 2021 at 19:41
  • $\begingroup$ I guess you are thinking about the equation as given here: en.wikipedia.org/wiki/… So all i would have to do is evaluate the equation of $\Gamma_{rad}$ with the transition dipole moment and the transition frequency to obtain the rate. And the life time is then the inverse of this rate if i understand it correctly. $\endgroup$
    – Hans Wurst
    Commented Mar 26, 2021 at 19:52
  • $\begingroup$ yeah, this one. Lifetime is indeed inverse of the rate. $\endgroup$
    – sleepy
    Commented Mar 26, 2021 at 19:55
  • $\begingroup$ I did know that equation but i thought that for molecules the whole procedure might be a little bit more complicated or not applicable at all. $\endgroup$
    – Hans Wurst
    Commented Mar 26, 2021 at 19:58

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