If the fluorescence is the re-emitting of a photon with a larger wavelenght due to the transition from a higher energy state to a lower energy state the case of resonance Raman (where there aren't any virtual states involved) seems be equal to the fluorescence. Which differences are there?
okay, old question, but:
recall that the virtual state can always be decomposed as a sum over all real states, and the contribution of every real state to the superposition is determined by the matrix element and the difference in energy from the real state to the photon energy; usual perturbation-theory sort of equation where you divide by energy; by the way this is how you get to the usual polarizability equations for non-res raman from the virtual state sort of equation;
in resonance raman, if you tune sufficiently close to a discrete absorption band, then that one state will dominate the sum, and you no longer consider things to be from a virtual state
if you've done that, resonance raman is equivalent to resonance fluorescence
recall resonant fluorescence is basically when you've made your gas dilute enough that collisional cooling is slow compared to the radiative lifetime, so you emit from whatever level you excited to, instead of cooling to v=0 upstairs before emitting
so there are two key things needed to make resonance raman the same as fluorescence: you have to tune so close to a resonance that a only a single excited state is contributing, and you have to be comparing to the fluorescence you get in the limit of no excited state relaxation occurring before the emission
The main difference between Raman scattering and fluorescence is the excited state lifetime. Fluorescence excited states are longer-lived than the 'virtual' states associated with Raman scattering.
In fluorescence, absorption of light excites an electron to a higher energy state. The lifetimes of these excited states are long enough where the geometry of the molecule can relax to accommodate this new electron configuration. However, this new geometry is not equivalent to the lowest energy structure, so when the molecule returns to the ground state, the energy of the system will be slightly higher than where it started. This means that the fluorescence energy is always lower than the excitation energy (this is true for molecules, not atoms).
For Raman scattering, light absorption promotes the molecule to a very short-lived 'virtual' excited state. When this state relaxes, it can return to a different vibrational or rotational sublevel of the ground state than where the molecule started. This leads to slightly different energies than the elastically scattered light, which is at the same energy as the incoming radiation source.
Resonance Raman is a special case where the excitation energy is tuned to a particular electronic transition of a molecule. This has the advantage that vibrational transitions associated with the electronic change are substantially enhanced compared to other Raman transitions, which simplifies the Raman spectrum. Because you are exciting a specific electronic transition, though, it is possible to get a substantial fluorescence background as well.