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.

  • $\begingroup$ Also take a look at this almost identical question on Physics.SE. $\endgroup$
    – Wildcat
    Nov 16, 2016 at 10:47

3 Answers 3


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.

  • $\begingroup$ Ah. This helps very much to clarify some of the other answers. Just out of curiosity, are there an infinite number of possible virtual states which can be formed as a superposition of two eigenstates? I would assume for a single frequency of incident light the system will enter the same superposition state each time, but if I change the frequency of light, does the system enter a different virtual state or are the virtual states also quantized and thus only possible for certain frequencies of light? $\endgroup$
    – jheindel
    Nov 20, 2016 at 19:34
  • $\begingroup$ Also that article was extremely helpful and more or less answered the question in my comment above. Thanks for the reference. $\endgroup$
    – jheindel
    Nov 20, 2016 at 20:28

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'

  • $\begingroup$ I think this makes a decent deal of sense, especially keeping in mind the fact this is a scattering process when I'm more used to thinking about absorption and emission processes. What, however, does the $E_v$ term mean? I get that it's the energy of the virtual state, but I don't understand where this would come from? That is, is this energy the eigenvalue of any particular operator? The perturbed Hamiltonian I suppose. If that's true, is there not a corresponding eigenstate? That makes the state seem more real than virtual, but simply with a low probability of being occupied. $\endgroup$
    – jheindel
    Nov 18, 2016 at 22:14
  • $\begingroup$ In exact state theory, $\hat{H}\left|v\right> = E_v\left|v\right>$. The Hamiltonian is the unperturbed one, the trick in response theory is to express perturbed states as combinations of unperturbed excited states. $\endgroup$ Nov 20, 2016 at 14:15
  • 1
    $\begingroup$ I could well be totally wrong here, but I think that the virtual state is in fact the tail of a real excited state which has a much higher energy than that of the photon (in non-resonant scattering). The excited state is (a) never sensibly populated and (b) it is not a stationary state so has a finite lifetime. Thus the width of this state (in energy) gives the scattering some very , very small probability of occurring when the photon is non-resonant. $\endgroup$
    – porphyrin
    Nov 21, 2016 at 20:33
  • $\begingroup$ Could one postulate the virtual state to be related to the Heisenberg Uncertainty principle? That as long as the duration of the excitation is short enough, the large energy uncertainty could accommodate a range of virtual states? Not sure if this makes sense though because then scattering could occur for any arbitrary incident frequency. $\endgroup$
    – Blaise
    Sep 1, 2018 at 10:50

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.

  • $\begingroup$ you are not correct when you write that a virtual excitation causes 'absorption' & there is a 'short-lived intermediate' state as this also implies absorption leading to formation of a state. Raman is scattering; the excitation photon is destroyed and a vibrational quantum and a new photon from the vacuum state is produced (not emitted by the molecule). No time is involved. The change in electronic polarisability is the origin of the scattering. $\endgroup$
    – porphyrin
    Nov 17, 2016 at 8:57
  • $\begingroup$ @porphyrin, but I did not say that this virtual excitation actually happens, just that things can be understood in that way. All this is virtual, as I mentioned few times in my answer. Scattering is real, the mathematical machinery involving virtual state to describe it is not. $\endgroup$
    – Wildcat
    Nov 17, 2016 at 13:39
  • $\begingroup$ yes I understand what you are aiming at but I thought that saying ' is thought of a being absorbed' etc. could be misleading. There are physical processes involved via the polarisability. The virtual state is in effect the polarisability of some upper electronic state in the molecule whose energy is far removed from the Raman excitation frequency. $\endgroup$
    – porphyrin
    Nov 17, 2016 at 14:25

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