# Chemistry - What is a virtual state?

## Solution 1:

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.

## Solution 2:

Ordinary absorption and emission processes depend on the matrix element of the electric dipole moment operator. In Raman transitions it is the matrix elements of the molecular polarisability between the initial and final states that are important. This polarisability difference can be considered to be the 'virtual state'. The Raman scattering probability involve terms from a second order perturbation expansion of the interaction between the molecule and the radiation field which have the form $$\sum_v\frac{\langle f|\mu\cdot \epsilon|v \rangle \langle v|\mu\cdot \epsilon|i \rangle f}{h\nu_{ex}-(E_v-E_f)}$$

where $\epsilon$ is the electric field of the lightwave and $v$ a 'virtual' state. However, transitions to and from these do not occur but are the means by which the quantum formalism takes place. This is somewhat unsatisfactory because to evaluate these integrals something must be known about the wavefunctions $\nu$ which is that the wavefunction will be proportional to the symmetry species of the vibration involved in the point group of the molecules. In a Raman process only the initial and final states have any population.

There are several different types of Raman processes; spontaneous, stimulated, coherent, resonant, non-resonant and methods such as CARS & CSRS. see Mukamel 'Nonlinear Optical Spectroscopy' ; Yariv 'Quantum Electronics'; Marcuse 'Engineering Quantum Electrodynamics'; Herzberg 'Infrared & Raman Spectra'

## Solution 3:

It looks simply as mind games: in principle, we are free to explain any observed phenomenon any way we like unless our explanation do not contradict well-established foundations of physics. For instance, in the case of Raman spectroscopy we observe changes in rovibrational states of molecules, but the problem is that the incident radiation is not resonant with any real state, so the corresponding direct transitions are not possible.

To sidestep the problem, we could explain Raman scattering as follows:

• Virtual excitation: first, a photon is thought as of being absorbed by a molecule exciting it from a rovibrational state of some electronic state to a shortly-lived intermediate state called the virtual state.
• Virtual de-excitation: then, a photon is thought as of being emitted by a molecule causing de-excitation from the virtual state back to a rovibrational state of the same electronic state.
• Both these processes are not real physical processes, it is all just a mental construct. And if the wavelength of the emitted photon is different from the incident one (Raman scattering), then, in order to conserve the energy, some change in rovibrational state of the molecule is though of as taking place in the course of these virtual processes.

The above described mental scheme relies on the so-called energy-time uncertainty principle, due to which a photon can be (though of as being) absorbed without conservation of energy as long as it is (though of as being) emitted a short time after.

Now, the only problem is where do we get these virtual states mathematically to do some predictions? And this is where perturbation theory come into the picture: virtual states are constructed by perturbation theory treating the incident electromagnetic wave as time-dependent perturbation.