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I am currently working on a machine learning with spectroscopy. I had learnt about the Franck-Condon principle and the vibronic coupling in photochemistry courses in my undergraduate lectures. However, in computational chemistry when electronic spectra (absorption) are calculated with TD-DFT for example, I find the usage of the terms "vertical excitation" and "0-0 transition".

I am struggling to understand what they mean and what their link with the peak in experimental absoroption spectrum is. For example, here is a UV-visible absorption spectrum of BODIPY dye:

BODIPY UV-visible absorption spectrum with peak at 498 nm

Now I used to think that the absorption peak here (498 nm) corresponds closest to results of the "vertical excitation", since the light absorption in experiment is faster than nuclear movement. But recently I saw a paper where they were using "0-0 transitions" to compare against experiment. I had the impression from the photochemistry lectures that the "0-0 transition" is usually not the peak in experimental spectrum due to Franck-Condon effects:

Frank-Condon excitation

Could someone please help me understand these terms and what their link to the experimental spectrum is?

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2 Answers 2

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Yes, the absorption maximum is usually close to (but not necessarily exactly equal to) the vertical excitation from the ground state, due to the Franck-Condon principle. However, this does not mean that the 0-0 excitation energy is not experimentally measurable. The point is that, when one approximates the ground and excited state potential energy surfaces as quadratic surfaces with the same quadrature, one can show that the 0-0 energy is exactly the average of the Franck-Condon absorption energy and the Franck-Condon fluorescence energy. Therefore, one can derive an experimental estimate of the 0-0 energy from an average of the experimental absorption and fluorescence energies. Some similar but slightly different alternatives are (1) averaging the absorption and fluorescence wavelengths instead (which is not equivalent to averaging the excitation energies due to the non-linear relationship between wavelengths and excitation energies), and (2) plotting both the absorption and fluorescence spectra, find the intersection point of the two spectra, and read off the corresponding excitation energy. I'm not sure which of the three strategies is the best, but mathematically they should give quite similar results for systems whose Stokes shift is not too large; when the Stokes shift is very large, all methods should become unreliable since the quadratic approximation of the PESs break down (EDIT: this holds for most, if not all, realistic molecular PESs, but does not necessarily hold for artificial PESs, such as those that are by construction globally quadratic. Thanks Hans Wurst for pointing this out).

Moreover, some experiments may probe the 0-0 energy directly. One obvious example is the high-precision experiments that exhaustively enumerate all low-lying vibronic levels, fit potential energy surfaces based on the data and read off the 0-0 energy from the PESs. Another example is to extract the 0-0 energy from ground state and excited state reactions. Say you have measured the equilibrium constants of the reactions $A + B \to A^+ + B^-$ and $A^* + C \to A^+ + C^-$, where $A^*$ is one of the excited states of $A$, and you know the redox potentials of the $B/B^-$ and $C/C^-$ couples. Then you can calculate the 0-0 Gibbs free energy of $A$; extrapolation to zero temperature yields the 0-0 energy of $A$.

Finally, there is a trivial case for gas phase measurements of very small molecules. In this case the vibronic levels are usually highly resolved, and chances are that the 0-0 excitation appears as an isolated peak that does not overlap excessively with other vibronic peaks. In this case you can directly obtain the 0-0 excitation energy from experiment, provided that you can correctly assign the 0-0 peak.

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  • $\begingroup$ +1 for a thorough, timely, and valuable answer! I added another answer which serves more to "add" to yours rather than to contradict it. $\endgroup$ Jul 7 at 18:30
  • $\begingroup$ +1 Thanks a lot! Could you please explain this part: absorption maximum is usually close to (but **not necessarily exactly equal to**) the vertical excitation, because that's the main point I am getting confused on! Why are they not necessarily equal? $\endgroup$
    – S R Maiti
    Jul 8 at 21:02
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    $\begingroup$ @SRMaiti Suppose the absorption spectrum is fully vibrationally resolved, and we gradually increase the difference of the equilibrium structures of the ground and excited states. The vertical excitation energy will increase continuously, but the absorption maximum will increase in a stepwise fashion, because it only changes when the strength of the $0\to\nu'+1$ excitation exceeds that of the $0\to\nu'$ excitation. When the spectrum is not vibrationally resolved, the latter change is also continuous, but small discrepancies between vertical and Franck-Condon energies still remain. $\endgroup$
    – wzkchem5
    Jul 9 at 7:41
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    $\begingroup$ @HansWurst Thanks for pointing out this. My statement only holds for "realistic" PESs, which always have some anharmonicity. It certainly does not hold for globally quadratic PESs, but anyway these PESs do not occur in real molecules. I have modified my answer accordingly. $\endgroup$
    – wzkchem5
    Jul 9 at 16:07
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    $\begingroup$ @SRMaiti Yes, exactly $\endgroup$
    – wzkchem5
    Jul 9 at 16:11
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"I find the usage of the terms "vertical excitation" and "0-0 transition". I am struggling to understand what they mean"

The 0-0 transition is from $v^{\prime\prime}=0$ to $v^\prime = 0$, whereas "vertical" means that it's vertical on a plot of the energy vs the nuclear geometry: the blue $v^{\prime\prime}=0$ to $v^\prime = 2$ line in your plot.

"Now I used to think that the absorption peak here (498 nm) corresponds closest to results of the "vertical excitation", since the light absorption in experiment is faster than nuclear movement. But recently I saw a paper where they were using "0-0 transitions" to compare against experiment."

It depends on the type of experiment. Without seeing the original paper containing the absorption peak at 498 nm, or the original paper that uses 0-0 transitions to compare to experiment, it's hard to give a general answer.

If the experiment uses a laser to excite the molecule, and the laser's peak wavelength exactly matches the energy gap between $v^{\prime\prime}=0$ and $v^\prime = 0$, it won't be a surprise if the peak in the absorption corresponds to a "0-0 transition". In two-color phooassociation experiments, a molecule is created in an arbitrary $v^\prime$ state, then de-excited to an arbitrary $v^{\prime\prime}$ depending on the two "colors" of the laser.

If you shine light on a molecule with a wide spectrum of equally distribted frequencies, then the peak absorption would likely happen where the Franck-Condon overlap is greatest. This can be calculated by obtaining the wavefunctions of the ground and excited states and doing an integral.

In the case of the BODIPY figure in your question, perhaps you can calculate the $v=0$ energy of the ground state and the excited state, and see if 498nm is much greater than the difference. If it's much greater, then perhaps we're looking at a 0-$n$ transition where $n>0$ (and perhaps it's vertical, but I won't say that it's guaranteed!).

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  • $\begingroup$ +1 Thanks! You wrote that the peak absorption would happen where the Franck Condon overlap would be greatest... Does this mean overlap of vibration wavefunctions? Do they not happen in vertical excitation? Because I though that the overlap would be greatest at the vertical excitation point? $\endgroup$
    – S R Maiti
    Jul 8 at 21:03
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    $\begingroup$ From the Franck-Condon principle Wikipedia page it says "The quantum mechanical formulation of this principle is that the intensity of a vibronic transition is proportional to the square of the overlap integral between the vibrational wavefunctions of the two states that are involved in the transition" indicated to use vibrational wavefunctions. Ro-vibrational wavefunctions should also be okay. However, I think that statement assumes a Born-Oppenheimer separation of vibrational and electronic wavefunctions, which wouldn't be valid near a conical-intersection, for example. $\endgroup$ Jul 8 at 23:32
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    $\begingroup$ Your next question was about whether or not the overlap would be greatest at the vertical excitation point, but which one? The blue arrow in your Franck-Condon diagram intersects several vibrational wavefunctions. $\endgroup$ Jul 8 at 23:33
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    $\begingroup$ How can the vertical excitation energy be the difference between an energy level and something that is not an energy level? Hence, the blue arrow doesn't end where the vertical line intersects the curve (also remember that a PES is an "artificial" thing, which only exists in the BO approximation), but it ends on an actual vibrational energy level. Which vibrational energy level? The one with the optimal overlap, which is calculated as an integral involving wavefunctions. Also when you say "intersects the PES", you're making an assumption about where the line starts (ok for v=0 but not for v>0) $\endgroup$ Jul 10 at 2:22
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    $\begingroup$ When doing DFT calculations with most existing software, you choose a geometry and get the electronic potential energy at that geometry. You can then map out the full PES by repeating this for many geometries. If you ask for the first 5 excited states, you get 5 electronic energies for each geometry, so you would get 5 PESs. You would then have to solve the vibrational Schroedinger equation using each of those PESs, to get vibrational energies. Some electronic structure softwares make approximations that give you E(v=0) so 0-0, but some people might just do E_i(r_min) for all states i. $\endgroup$ Jul 10 at 13:50

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