8
$\begingroup$

In quantum mechanics / Raman spectroscopy, what is a virtual state? What is the difference between a virtual state and a superposition of states? Can you simply think of the virtual state as a superposition of eigenstates?

If you have the complete set of eigenstates for a system, then you can represent any configuration of the system as a linear combination of the eigenstates, so it seems that you should be able to represent the virtual state that way. Is there something more to the virtual state than that?

$\endgroup$

1 Answer 1

7
$\begingroup$

There are differences in the spin that the term "virtual state" gets depending on context. In the context of Raman scattering, one sometimes refers to transitions to virtual states. Other terminology used is "virtual transition" to a real state. The latter terminology is a little easier to understand in my way of thinking. And note that the "virtual state" can be constructed as a superposition of real states, as you point out. So whatever we say about virtual transitions to real states immediately applies to the virtual state picture. But the virtual state picture is more complicated.

In either point of view, the incident radiation in Raman scattering is not resonant with any real state, so real transitions are not possible. We imagine that the system makes a transition to a real state, and a quantum of EM excitation is destroyed. Because the real state is not resonant with the radiation, energy is not conserved. This is possible as long as the lifetime of the system in that state is short. Heisenberg's uncertainty principle allows violations of conservation of energy for short time intervals: $\Delta E\Delta t \leq \hbar/2$.

Here's a semi-classical take on Raman scattering.

Consider a diatomic molecule in its ground state. Upon irradiation, it makes a virtual transition to an excited state, and stays there for a short period of time. But the characteristics of the chemical bond in the excited state are different from what they were in the ground state. In particular, the equilibrium length of the bond will be different. The molecule is promoted to the excited state, but it's initial bond length up there is not the equilibrium length for the excited state. So the atoms feel a force and move towards the new equilibrium length. But in a short period of time the system returns back to the ground electronic state. But now the bond length is no longer the ground state equilibrium length; the atoms feel a force and move towards the original bond length. Now there's nothing to stop them: the atoms are oscillate back and forth past the original equilibrium length: it's vibrating.

$\endgroup$
3
  • 1
    $\begingroup$ Thank you. So if we think of our state space as being electronic and vibrational degrees of freedom, then the incident photon creates a superposition of states, the lifetime of this superposition is short, but when it re-radiates there is now occupation of one of the excited vibrational states. Is that what you're saying? $\endgroup$
    – dllahr
    Sep 12, 2014 at 17:34
  • $\begingroup$ I think that's what I'm saying. :) $\endgroup$
    – garyp
    Sep 12, 2014 at 17:44
  • 1
    $\begingroup$ "Heisenberg's uncertainty principle allows violations of conservation of energy for short time intervals" that is not true, it's just a misinterpretation of the uncertainty principle. It only means that if a state is not an eigenstate oh the Hamiltonian (and thus has a non-zero variance in energy), it has a lifetime smaller or equal than ℏ/2ΔE. However, there is no violation in the conservation of energy whatsoever. $\endgroup$
    – Ajayu
    Sep 22, 2017 at 23:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.