Are virtual states eigenstates of physical observables?

Question

In quantum mechanics, physical observables are represented by self-adjoint operators. A "virtual state" is supposedly a state which is unmeasurable, due to being too "short lived". To keep things simple, let's not even think about virtual particles or quantum field theory. My question applies there as well, but also to ordinary single-particle quantum mechanics--for example, to quantum tunneling or Raman scattering.

In quantum tunneling, there is a thin energy barrier which would confine a classical particle to within one of two separate allowed regions. It's "forbidden" to be there since the kinetic energy of a particle within the zone is negative. (In more abstract terms, its state there is a solution to the Wick-rotated time independent Schrödinger equation but not the ordinary time independent Schrödinger equation). Nevertheless, a quantum particle within one of the allowed regions can still tunnel through the barrier to the other region as long as it's thin enough, even if the expected value of its energy in both initial and final states is less than the barrier height. This happens because the wavefunctions for the initial and final states overlap in the forbidden zone, so there is a non-zero amplitude for the transition. I understand that part, but what always confuses me a bit is thinking about the tail of the wavefunction that extends into the forbidden zone. According to the Born Rule, this can be interpreted as a non-zero probability for the particle to be found in the forbidden zone if its position is measured.

However, as I understand it, in practice this is never supposed to happen. I've seen it explained as "virtual states are not eigenstates of any physical observable". Is that really a correct statement? If so, how do we know that, is there a proof? Or is it more of a pragmatic statement like "the tunneling happens so fast, you don't have time to measure it while it's happening"? or "the barrier is so thin, we don't have instruments accurate enough to measure it there?"

Same thing with Raman scattering. a photon gets absorbed by some material, which puts it into a superposition of many things happening, one of which is an electron momentarily getting excited into a "virtual state" whose energy is greater than the sum of the photon's energy and the electron's original energy... after which it drops back down into a different state, emitting a photon of different energy than the incoming photon. Presumably, you can never observe the intermediate state because it doesn't satisfy the relationships it's supposed to classically. But is that because it's impossible to measure it in that state? Or just improbable?

Possibly what I'm asking might boil down to: does the mathematical existence of a self adjoint operator (which has some particular state as one of its eigenstates) guarantee that there must be some way to physically measure that state, given enough technology? Or is the set of physically observable states actually somewhat smaller than the set of all self adjoint operators in the Hilbert space? And if it is a 1-to-1 mapping, at least in principle, does that mean there is a proof somewhere that virtual states are not eigenstates of any self-adjoint operator? It would seem to me that, if nothing else, you could just construct such an operator from any state by creating a projector from the outer product of the state with its conjugate $|\psi><\psi|$. And if it isn't a 1-to-1 mapping, how do you know which ones correspond to actual physical observables?

Update: I notice the Wikipedia page on virtual state makes an even stronger claim than the one I've typically seen elsewhere, saying "virtual states are not eigenfunctions of any operator", citing a textbook on Nonlinear Optical Microscopy as the source. If true, that would exclude even operators with non-real eigenvalues. (That seems even more implausible to me... although when I look at the original source, it seems like this might be a misinterpretation of the intended meaning.)

Answer 1

The wikipedia page called "virtual state" is too brief to be useful. The problem is that it does not ever define what it means by a virtual state.

I think the term "virtual state" is often used to refer to a part of an expansion of the propagator in quantum mechanics. In this situation you might for example have $$ | \psi(t) \rangle = U | \psi(0) \rangle $$ where $U$ might include terms like $$ \sum_i \hat{V}_2 | \psi_i \rangle \langle \psi_i | \hat{V}_1 $$ so in this case the system might be said to be going from $| \psi(0) \rangle$ to $| \psi(t) \rangle$ via $| \psi_i \rangle$, so each such $| \psi_i \rangle$ might be called a virtual state.

Anything which can be expressed as a ket is an eigenstate of a hermitian operator. For we can always write down the operator $$ \hat{Q} = | \psi_i \rangle \langle \psi_i |. $$ This has an eigenstate $| \psi_i \rangle$ with eigenvalue 1. (Such a state will not usually be an energy eigenstate.)

As for detecting a particle in a region of space where it is tunneling: sure, why not? For example, set up some grids so as to arrange for an electron to tunnel from one place to another, and then shine some light into the region where the electron's energy is less than its potential energy. The electron will scatter the light.