What are virtual bound states or virtual levels of a finite square well? The lecture note here says that

... for a short-range or abrupt-sided potential there exist quasi-bound or virtual single-particle states which have positive energy. A long-range potential like the Coulomb potential has no such states.

See the discrete blue lines in the figure below on the top of the well.

For a finite square well:
$$V(x)=\left\{\begin{array}{ll}
-V_0, & -a \leq x \leq a, \\
0, & |x|>a
\end{array}\right.
$$
the solution to the time-independent Schrodinger equation tells that the energy levels for $-V_0<E<0$ are discrete but for $E>0$ are continuous. If so, I am not able to understand how can we have discrete (virtual) states as shown in the figure above?
 A: An educated guess:
For an infinite square well, the bound states are those which satisfy $\psi(x)=0$ at the edges of the well, so that the wavefunction goes continuously to zero probability of detecting the particle outside of the well.  Such bound states satisfy the condition that the well size is a half-integer multiple of the wavelength.
For a finite square well, the $E<0$ bound states still fit approximately a half-integer number of wavelengths. The approximation is better for the most deeply bound states, where the probability of finding the bound particle outside of the well is the smallest.
For the unbound states of a finite square well, you could reasonably expect some funny business to happen with reflection/transmission coefficients if the wavelength within the well happens to be near a half-integer multiple of the well’s size.  For a one-dimensional well there might be a magic unbound energy where the reflection coefficient goes to zero or to one, or where the phase shift of the transmitted wave vanishes. It would be reasonable to refer to such magic energies as “states” of the well, even though they are unbound.
For a three-dimensional finite square well, such “unbound states” would correspond to energies where the scattering length becomes very large; see this hand-waving derivation of the scattering length for an approach which might let you get the idea by sketching some wavefunctions rather than getting lost in a bunch of mathematics.  Such an energy would correspond to a “scattering resonance.”
A: I'm quite confident the article is simply referring to the usual quasi-bound states which @Newbie linked to in the comments, especially since this is in the context of nuclear physics. The picture you gave is a little misleading. A quasi-bound state is generally one that has energy above the nuclear binding energy, but below the Coulomb barrier, and therefore tunnels through the Coulomb barrier with some partial width. The lifetime of this quasi-bound state of course depends on how low the energy is, i.e. how much of the Coulomb mountain it needs to tunnel through. A better picture would be the following (the black line represents the binding potential).

You can see that if the potential were instead only a Coulomb well, i.e. a deep well given by $V\sim -1/r$, no such quasi-bound state could exist.
