quantum chemistry - What is a virtual state?

In talking about Raman spectroscopy, one finds the Stokes line is simply the difference between the energy of an incoming photon and an emitted photon. This energy corresponds to a vibrational transition in terms of energy. Yet, the light sent at the molecule is usually much higher energy than that vibrational transition (or else we would just be doing IR).

The explanation I've heard is that the state which holds all that energy in between the ground vibrational state and decaying back to the first vibrationally excited state is a virtual state.

From doing some reading, I have gathered that a virtual state is called virtual not because it's imaginary, but because it's not an eigenfunction of the hamiltonian of the system.

Where does this state come from? I've heard that it drops out of some perturbation theory. Because Raman is intricately connected with polarizabilities, I imagine this might be perturbations to the time-dependent Schroedinger equation.

So, to put some actual questions out there:

What is meant by "virtual state"? In what sense is it virtual?

Can anyone show me the perturbation theory where this comes from? Even if it is just a reference, that would be greatly appreciated.

I have also seen this is somehow connected with Feshbach Resonance which I know nothing about, so an answer which addresses that point would be a good one.

Answer

In most Raman experiments, the incident radiation is not near (or at) an absorbing wavelength, and so you will never access a real, honest-to-God excited state (stationary state). (If it was close to an absorbing wavelength, we would be taking about a resonance Raman experiment.)

Now obviously something is going on between the molecule and the incident light. What you are really doing is preparing a superposition state. A superposition of real excited states, or stationary states, of your Hamiltonian. This is what we call a "virtual state," and it is not an eigenstate of your Hamiltonian, so its energy is undefined. (You are free to compute an expectation value, though.) This virtual state doesn't last forever, and, when it decays, scatters radiation away in some direction (it doesn't matter where, unless you are doing an angle-resolved Raman experiment).

(By the way, if you look closely at the equation porphyrin gave, the sum over $v$ virtual states is actually a sum over real $v$ excited stationary states, such that $E_v$ is defined. Otherwise that expression is nonsense. This equation essentially is the sum of overlaps of the initial state into the states that make up your virtual state, and then multiplied by the sum of overlaps of the excited states that make up the virtual state into the final state. This is the Kramers-Heisenberg-Dirac expression for Raman amplitudes.)

If the superposition collapses back to the ground state, the scattered radiation will have the same wavelength as the incident light, and we have Rayleigh scattering. If the superposition collapses to an excited vibrational states, we have Stokes Raman. If your initial state was vibrationally excited and you collapse to a lower vibrational state, you get anti-Stokes Raman. In the actual experiment you will observe all types and can see a Rayleigh line in the spectra. Most spectrometers will filter this wavelength out, however, leaving just the Raman peaks. Usually the virtual state is dominated by the ground state and most of the scattered light is Rayleigh. This is why Raman spectroscopy was too weak to be useful until we had better optical components.

Obviously this isn't the only way to think about Raman. You can use path integrals and/or QED if you want, which is much closer to Wildcat's answer. In fact, I think you must use QED if you want to work out virtual state lifetimes. Here is a good reference if you want to dig a bit deeper:

Williams, Stewart O., and Dan G. Imre. "Raman spectroscopy: time-dependent pictures." The Journal of Physical Chemistry 92, no. 12 (1988): 3363-3374.