Electronics transitions of a molecule I have some doubts on electronics transitions of molecules in Franck-Condon principle:
1. The principle states that if a molecule absorbs a photon of the appropriate energy, we could have both electronic and vibrational transitions and that is more likely to have a vibrational transition between states that have two "similar" wavefunctions. The first question is:
It is ok to me that it is more likely to have a vibrational transition between states that have two "similar" wavefunctions but of course if we see a transition between $\nu''=0$ state and $\nu'=6$ state means that the absorbed photon had at least the energy separating $\nu''=0$ and $\nu'=6$, right? With reference to the image, the "blue arrow transition" is more lilely to happen but we need a photon with the right energy, more or equal to the "lenght" of the arrow, right?
2. If the energy of the incoming photon is enough for that transition, is it mandatory for the molecule to also have a vibrational transition or it can jump just from electronic G.S. to electronic first excited state remaining on the same vibrational level?
3. Again with reference to the image, the potential drawn higher refers to an excited electronic state, if we are talking about molecules, what do we precisely mean by that? If just one electron of the electronic clouds is excited by a photon, the new arrangement of the molecule need to be drawn higher than the ground state in which all electrons were in their ground states? And, if two photons are excited, the new potential must be drawn as a different one much higher? Same if three electrons are excited or two electrons are excited to second excited state or third excited state?

 A: The vibronic states of your molecule are a product of electronic and vibrational states. Let the electronic eigenfunctions be $\varphi_i(r,R)$ with electronic quantum number $i$ and the vibrational states be $\chi_{ij}(R)$, where $i$ gives the adiabatic electronic state and $j$ the vibrational state. Note that vibrational states of molecules are always associated with an electronic state since they are defined as eigenfunctions of a particular electronic adiabatic Hamiltonian. The index specifying the electronic state is often omitted when there is only a single electronic state of interest, but generally, you should  specify the adiabatic electronic state to which the vibrational eigenstate belongs. In your image ticks are used to indicate the electronic state $'=1$, $''=0$ but I find it more reasonable to simply use the quantum number of the electronic state via second index.
The probability for a transition within the dipole approximation between two vibronic states $$\phi_{ij}\rightarrow \phi_{fk}$$ with  $$\phi_{ij}(r,R) \equiv \varphi_i(r,R)\chi_{ij}(R)  $$ is then proportional to
$$
|\langle \phi_{ij}|\hat \mu |\phi_{fk} \rangle|^2  = \left | \int dr dR \ (\varphi_i(r,R)\chi_{ij}(R))^* \hat \mu (\varphi_f(r,R)\chi_{fk}(R))\right |^2
$$
We can integrate first over der electronic coordinates $r$, which yields
$$
\langle \phi_{ij}|\hat \mu |\phi_{fk} \rangle  =  \int dR \ \chi_{ij}^*(R) \mu_{if}(R) \chi_{fk}(R)
$$
where I have defined the electronic transition dipole moment $$
\mu_{if}(R) = \int dr \varphi_i^*(r,R)\hat \mu \varphi_f(r,R)$$
The transition dipole moment is then approximated by its zero order term of a Taylor expansion around the equilibrium geometry
$$
\mu_{if}(R)\approx \mu_{if}({R_0})
$$
This approximation turns the transition dipole moment into a constant that can be pulled outside of the integral.
$$
\langle \phi_{ij}|\hat \mu |\phi_{fk} \rangle  \approx  \mu_{if}(R_0) \int dR \  \chi_{ij}^*(R)  \chi_{fk}(R)
$$
This is commonly known as Franck-Condon approximation and the basis for the Franck-Condon principle. We see that the probability for the transition is depending on the overlap integral of the vibrational states
$$
\langle \chi_{ij} |\chi_{fk}\rangle = \int dR \  \chi^*_{ij}(R) \chi_{fk}(R)
$$
$$
|\langle \phi_{ij}|\hat \mu |\phi_{fk} \rangle |^2 \approx |\mu_{if}(R_0)|^2 |\langle \chi_{ij} |\chi_{fk}\rangle|^2
$$
Adressing 1:
Energy conservation requires that the photon mediating the transition has an energy corresponding to the difference of the involved states. The vibronic states are eigenfunctions of the molecular Hamiltonian such that
$$
\hat H|\phi_{ij}\rangle = E_{ij}|\phi_{ij}\rangle
$$
The photon thus needs to have an energy of $$E_{ph} = |E_{ij} - E_{fk}|$$
The energy of the photon may not deviate from this difference, i.e. you can not use an higher energy photon to induce this particular transition. It has to match the exact difference if you want to induce a transition between these two states. In reality there are some effects that weaken this condition and the probability is not exactly zero for small energy deviations, but in principle the energy should match. It can not be arbitrarily larger. The probability for the transition goes down quickly for both deviations, it doesn't matter if the deviation is smaller or larger than the exact match.
Adressing 2:
You cannot separate the transition as in your description. You have to specify both the electronic and vibrational quantum numbers of all involved states to specify a specific vibronic transition. You can have transitions of the form $ij=00$ and $fk=10$ such that $j=k$, i.e. with the same vibrational quantum number. This would be the transition from the vibrational groundstate of one electronic state to the vibrational groundstate of the other electronic state. For this transition to happen, we need a photon with the proper energy i.e. $E_{ph}=E_{00}-E_{10}$ and the probabilty has to be unequal zero,
$$
|\mu_{01}(R_0)|^2 |\langle \chi_{00}|\chi_{10}\rangle|^2\neq 0
$$
Adressing 3:
An excited state is not necessarily equal to exciting a single electron. This is a crude and over simplified picture that is only working for some examples/models. An electronic state of a molecule describes all electrons in their entirety and an excited state must not correspond to a single excitation of one electron hopping to an higher lying orbital. I will not address this in detail, as this goes way beyond the scope of this question. Another point is that we usually talk only about single-photon processes when talk about basic absorption or emission. My equations are also  for this case. I also assumed that we work within the framework of time dependent perturbation theory up to first order, which is the common method to model basic absorption and emission spectra. But again, addressing all of this goes beyond the scope. Multi-photon processes are not required for an understanding of basic absorption spectra of molecules.
