Is the photon truly not absorbed in Raman scattering?

Question

In reading about Raman Scattering, I was thinking while reading it "okay, incident photo absorbed by molecule, molecule goes to higher energy vibrational state, molecule re-emits photon with either higher or lower energy than incident...how is this different than light interacting with any sort of possible transition?"

They attempt to answer that question lower on the Wikipedia page, here, saying:

The Raman effect differs from the process of fluorescence. For the latter, the incident light is completely absorbed and the system is transferred to an excited state from which it can go to various lower states only after a certain resonance lifetime. The result of both processes is in essence the same: A photon with a frequency different from that of the incident photon is produced and the molecule is brought to a higher or lower energy level. But the major difference is that the Raman effect can take place for any frequency of incident light. In contrast to the fluorescence effect, the Raman effect is therefore not a resonant effect. In practice, this means that a fluorescence peak is anchored at a specific frequency, whereas a Raman peak maintains a constant separation from the excitation frequency.

I've read this thread as well, but it still doesn't clear up much. It also seems like there was disagreement between the different answers and the wikipedia page. For example, the chosen answer there says

In contrast to fluorescence, there is no excited state in Raman scattering

Whereas the Wikipedia article clearly uses the concept of excited vibrational states:

The energy difference between the absorbed and emitted photon corresponds to the energy difference between two resonant states of the material and is independent of the absolute energy of the photon.

It seems like many people are making a distinction between a photon being just scattered as opposed to absorbed and emitted, whereas John Rennie's answer and comments seem to be saying that the photon is always absorbed, though the order of events may differ from fluorescence.

So can someone give me a more clear picture of this? Is the photon really not absorbed in Raman scattering? Why isn't it resonant if there are different energy levels?

Answer 1

I think the difference between luminescence and Raman scattering lies in whether or not the mixed photon-molecule state maintains coherence with the exciting radiation. In Raman scattering, we imagine that coherence is maintained. In luminescence we imagine that coherence is disturbed by any one of a variety of interactions: for example, collisions with other entities in the case of molecules, and interactions with phonons in the case of solids. Disturbing interactions are either missing in Raman scattering, or the process occurs too quickly for the disturbance to mess up the coherence. In such cases, the properties of the scattered light depend essentially on the properties of the incident light. If coherence is upset, the molecule has no "memory" of where the energy came from, so the emission is like absorption and re-emission. Note that in some cases of disturbance, the energy of the photon can be distributed to phonons, or other molecules via collisions, and the energy never reappears as light: absorption has occurred.

As for resonant fluorescence occurring only at resonance: that's simply a result of the fact that the mixed state has a longer lifetime when resonance, giving more time for disturbances to upset coherence. fluorescence still occurs off-resonance, but it will be weaker than when on resonance.

I also think that in isolated molecules, such as can be produced in supersonic expansion, it's not possible to upset coherence. Any photon that couples to a molecule must eventually re-radiate the energy coherently. Neither absorption nor "fluorescence" can occur. Perhaps a better way to say it is that there is no distinction between Raman scattering and luminescence for isolated molecules.

Answer 2

In Raman scattering, the molecule absorbs the photon into a virtual state, which doesn't actually exist. Unlike an excited state, the molecule can't stay in that state for longer than a time $\Delta t$ where $\Delta t \Delta E \leq \hbar/2$ - the Heisenberg uncertainty relation.

A virtual state can have any energy level, though, and that's the reason for the statement you quote about Raman being a nonresonant effect. The photon coming in doesn't have to have a specific energy that will boost the molecule exactly to an excited state.

The vibrational state, on the other hand, is a kind of excited state; but it's the end state of the Raman process, not the intermediate state. (See also the answer I linked to above.)