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I was revisiting the Franck-Condon-principle and was wondering about one thing.

In the Franck-Condon principle, it is stated that if an electronic transition happens, the vibrational wave functions have to be compatible, because the internuclear distance remains relatively the same. The transition is a vertical one.

enter image description here

However, I was wondering the following: Do the vibrational wave functions taking part in the transitions only need to be similar exactly at the internuclear distance R(eq), or have to be overall similar? In the diagram for example, the v”=0 wave function has a maximum at R(eq), and at this equilibrium distance, v’=4 and v’=6 have a maximum roughly at this same position. However, v’=4 has less nodes and seems “overall” (over all R) more similar to v”=0 than v’=6 is similar to v”=0.

So my question is: Does the excited state vibrational wave function only be similar to the ground state wave function at exactly the R(eq) where the transition happens (have a maximum at this position), or be similar “overall” around the R(eq) and R around it?

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  • $\begingroup$ All that is needed is that the wavefunctions overlap. If they only do a little then FC factor is small and vice versa. Its the area under the curve of the product that matters, $| \int \psi_i\psi_fdx |^2$ for initial $i$ and final $f$ states. At the time of absorption the molecule can be at any position determined by the probability given by the wavefunction. As you can see the displacement between electronic states is important as to which levels have biggest FC but this is accounted for in the wavefunctions anyway. $\endgroup$
    – porphyrin
    Commented Jun 12 at 18:03

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All that is needed is that the wavefunctions overlap. If they only do a little then FC factor is small and vice versa. Its the area under the curve of the product that matters, $| \int \psi_i(x)\psi_f(x-q_0)dx |^2$ for initial $i$ and final $f$ states. At the time of absorption the molecule can be at any position determined by the probability given by the wavefunction. As you can see the displacement between electronic states is important as to which levels have biggest FC but this is accounted for in the wavefunctions anyway. When the displacement is slight only low quantum numbers v' have large FC, when displacement is large, larger v' are present with greater FC's than those with smaller v'.

I found an old picture showing this

FC factors

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  • $\begingroup$ So that would mean that the phase of the wavefunctions does not matter, as long as extrema of wavefunctions are at the same distance? And the wavefunctions in general have to overlap well, not only at the internuclear distance of the transition $\endgroup$
    – Mäßige
    Commented Jun 13 at 19:26
  • $\begingroup$ the complex conjugate takes care of the phase. As you can see its the whole wavefunction overlap that matters. $\endgroup$
    – porphyrin
    Commented Jun 14 at 10:21
  • $\begingroup$ Okay true that makes sense. But if only the absolute product matters, why would it be small for 0-20? I mean there are many maxima (but also zero points) so why would the overlap then be small, sorry if I’m asking too much $\endgroup$
    – Mäßige
    Commented Jun 14 at 12:03
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    $\begingroup$ the product is taken then summed (i.e. integrated) then squared. $\endgroup$
    – porphyrin
    Commented Jun 15 at 16:00
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    $\begingroup$ There is no time dependence here. The phase as you call it is removed as the function is $\int \psi^*\psi dx$ where the * indicates the complex conjugate so the product is always real number $\endgroup$
    – porphyrin
    Commented Jun 16 at 15:05

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