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.)

  • $\begingroup$ Your question seems to be the same as this: Can a particle be physically observed inside a quantum barrier? $\endgroup$
    – Ruslan
    Nov 29, 2021 at 9:47
  • $\begingroup$ It is a special case of the more general question I'm asking. But even for that question, I don't see any satisfactory answers there. The top answer doesn't attempt to prove, explain, or theoretically justify the claim it makes, instead it just states without any supporting evidence that evanescent waves are an example of quantum tunneling. If that's true, I can't immediately see how or why, as there isn't any barrier involved at all. And the 2nd most upvoted answer confuses the opposite side of a barrier with the forbidden zone, dodging the question as well! $\endgroup$ Nov 29, 2021 at 19:50
  • $\begingroup$ Also, all 3 of the top answers there appear to contradict the Wikipedia page on virtual state, which claims: 1. "virtual states are not eigenfunctions of any operator" 2. "No measurement of a system will show one to be occupied" and 3. "While each virtual state has an associated energy, no direct measurement of its energy is possible" en.wikipedia.org/wiki/Virtual_state I would like someone who can explain whether Wikipedia is wrong or those answers are all wrong... or if there is something I'm missing which would allow them not to be contradictory. $\endgroup$ Nov 29, 2021 at 20:03
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    $\begingroup$ This Q&A over at Chemistry.SE might be worth reading. $\endgroup$ Nov 29, 2021 at 21:08
  • $\begingroup$ @MichaelSeifert Thanks! The top answer there was very helpful, I think it clears up the main confusion I had around this: the reason why no measurement of the system will show one to be occupied has nothing to do with the state itself being unmeasurable or unobservable, it's just that the state is never actually occupied; if it were, then it wouldn't be called virtual! $\endgroup$ Nov 30, 2021 at 8:34

1 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.

  • $\begingroup$ I think your first point is important: virtual states almost always seem to show up as part of a sum over intermediate states. In that context, they are never technically occupied during the process, they just contribute to the sum. In the case of tunneling then, I guess the sum is over position eigenstates... but while tunneling it never actually goes into an eigenstate of position. Assuming $\Delta x$ is greater than the barrier width, that would mean the operator $V(x)$ is also not well-defined... so maybe it's meaningless to try to compute $E - V(x)$ while tunneling? $\endgroup$ Nov 30, 2021 at 9:08
  • $\begingroup$ But then, what happens to $E$ if you shine light on the forbidden region and detect it there? I guess it must then change from the energy eigenstate with $E = E_{\rm initial}$ into a state with undefined energy where $\langle H\rangle = E_{\rm initial}?$ But now $V(x)$ is well-defined and greater than $\langle E \rangle$, so... wouldn't that mean the average kinetic energy $\langle T\rangle = \langle\frac{p^2}{2m}\rangle$ is negative? What does it mean to get back negative kinetic energy from a physical measurement? $\endgroup$ Nov 30, 2021 at 9:24
  • $\begingroup$ @reductionista A very precise position measurement would give a state of high momentum uncertainty, and thus also kinetic energy. In order to infer that the particle is localized in the classically forbidden region the $x$-precision has to be high enough to give a $\langle p_x^2 \rangle$ which makes the k.e. come out positive (I think: you can check). $\endgroup$ Nov 30, 2021 at 10:24
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    $\begingroup$ @reductionista $\Delta p^2 = \langle p^2\rangle - \langle p \rangle^2$ hence $\langle p^2 \rangle \ge \Delta p^2$. $\endgroup$ Dec 1, 2021 at 10:06
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    $\begingroup$ @reductionista If a measurement locates the particle in the barrier then the wavefunction has changed such that it has more k.e. The energy came from the interaction with the measuring device. $\endgroup$ Dec 1, 2021 at 10:09

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