What is the right insight or intuition regarding difference (if any) between a resonance and a bound but unstable state?

Question

When we introduce quantum mechanics we might ask students to calculate the bound states for a potential well of some finite depth. In an introductory example the well might be $V(r)=0$ for $r < a$ and $V(r)=V_0 > 0$ for $r>a$, for example, and then an energy eigenstate with $E < V_0$ is strictly bound.

Later on we might consider something like a Coulomb barrier with a strong-force well in the middle. Suppose the potential $V(r)$ goes to zero as $r \rightarrow \infty$ and has a maximum $V_{\rm max}$ at $r=a$. You can have energy eigenstates with energy $0< E < V_{\rm max}$ and they have a character intermediate between bound and unbound. They are like bound states at small $r$, with exponential decay in the classically forbidden region, and then at large $r$ they extend to infinity like unbound states.

I think there may be an inconsistency in standard terminology here. In nuclear physics an unstable nucleus is said to be in a 'bound state', albeit one that is unstable, so this suggests the kind of state I just described would be called 'bound'. But I think people studying quantum mechanics in the abstract would want to say such states are 'not bound'. Similar considerations apply to a molecule which can spontaneously dissociate or an atomic ion with an extra electron. A natural sense of the physics invites us to say 'bound but unstable', but a strict study of the eigenstates says 'not bound'. (If I am wrong about this confusion of terminology I would be glad to be corrected, but this is not my question here).

In the "Coulomb + well" example there will be a continuum of states with $E > V_{\rm max}$ and within this continuum the density of states may have one or more pronounced peaks for $E$ larger than but near to $V_{\rm max}$. This suggests that when we see a resonance in a scattering cross-section it is not possible to say, from that evidence alone, whether the state in question was bound or unbound. Is that right?

My main question is, is there any difference (beyond the mere assertion $E < V_{\rm max}$ or $E > V_{\rm max}$) between a bound state with short lifetime and an unbound state whose energy is near such a resonance?

Answer 1

For simplicity, I'll consider nonrelativistic-style QM with position and momentum operators and a classical potential energy function $V(x)$.

I think that the best definition for a bound state $|\psi(t)\rangle$ is one where the quantity $\langle \psi(t) | \hat{X}^2 | \psi(t) \rangle$ remains bounded as $t \to \infty$. (Or we could consider $\langle |\hat{X}|\rangle$ - it doesn't matter too much conceptually.) But time-dependent states are messy to work with, and by the usual argument where we take the momentum-eigenstate limit of the incoming scattering wave packet, we can gain qualitative insight into the time-dependent behavior by only considering energy eigenstates. Eigenstates that don't have $\psi(x) \to 0$ as $x \to \pm \infty$ will generically lead to unbounded $\langle \hat{X} \rangle$ at long times, because (heuristically) there's some probability of the particle tunneling out of its well with every bounce off of the finite potential barrier.

We can understand the behavior of a slowly evolving near-energy-eigenstate with energy localized around some central value $E$ by considering the exact eigenstates with energy $E$.

Assume that $V(x) \to 0$ as $x \to \pm \infty$. Then

1. A near-eigenstate with energy(s) around $E < 0$ is truly bound, because $\langle \hat{X}^2 \rangle$ will stay finite for all time.
2. If $0 < E \ll V_\text{max}$, then $|\psi_\text{outside}(x)| \ll |\psi_\text{inside}(x)|$ for the eigenstate, so the near-eigenstate will take a very very long time to escape the well. (Heuristically, every time it bounces off the wall, only a tiny fraction of its probability will leak out of the well, so it will take many bounces before there's an appreciable probability of measuring the particle outside the well.) So $\langle \hat{X}^2 \rangle$ eventually diverges, but only after a very long time. I'd say that this state is technically unbound, but can naturally be thought of as a metastable "bound state" because it takes so long for the particle to escape.
3. The higher $E$ goes toward $V_\text{max}$, the more of the total probability will be outside of the well in the eigenstate picture, and the faster the particle will escape the well (the fewer bounces before it leaks out) in the near-eigenstate picture.
4. I don't think there's any qualitative change as $E$ passes $V_\text{max}$, because if $E$ is just a tiny bit under $V_\text{max}$ then the particle's very unlikely to bounce off the barrier at all anyway. (I think the more precise statement is that $\sqrt{2 m (V_\text{max} - E)} \times \text{(barrier width)} \ll \hbar$ or something like that.) There might be a peak in the density of states, as you said, but no qualitative change.

So the only really qualitative transition is when $E$ passes the limiting potential energy $\lim_{x \to \pm \infty} V(x)$ - but even there, you need to wait a very long time to see that difference, and to tell whether you have a truly stable bound state or only a metastable "bound state".

Answer 2

First of all: I think your analysis of the terminology is spot on and often confused in certain branches of the literatur. In the following, I will try to provide a minimum math answer to provide intuition.

To answer the central question: To my understanding, a bound state with short lifetime gives you a resonance. For a general resonance, however, one cannot necessarily associate a bound state with it.

Let's discuss briefly what a resonance is, which is one of the central difficulties. In formal resonance theory, the definition usually uses the notion of a pole in the complex energy plane. For simplicity, I will skip the discussion of which quantity exactly has to feature a pole and go with the scattering matrix as an intuitive observable. One can then check if the system has a resonance by analytically continuing $S(E)$ to complex energies $E$. If there is a pole, that's a resonance (although one may distinguish between poles in the lower and upper half of the complex plane).

To understand this, let us look at a typical Lorentzian resonance

$$S(E) \propto \frac{1}{E- E_\mathrm{res} + i\gamma}\,,$$

where $\gamma$ is the linewidth proportional to the inverse lifetime of the resonance. One can see that this features a pole at $\tilde{E}_\mathrm{res}=E_\mathrm{res} - i\gamma$.

For many systems such as the examples mentioned in the question, one can straightforwardly associate a bound state of a slightly different system with such a resonance. For the finite potential well ($V=V_\mathrm{barrier}$ for $r_1<r<r_2$, V=0 elsewhere), for example, one can identify the bound states of the corresponding infinite barrier. Indeed, one will find that for a high barrier, the resonances lie extremely close to the bound state energies ($\tilde{E}_\mathrm{res} \approx E_\mathrm{bound}$). Vice versa, one can think of a bound state as the limit of the pole moving to the real energy axis, where it becomes uncoupled from the far field and does not appear in the scattering matrix any more.

This association works well for systems where you can identify a closed-off region that is well isolated from its environment, which is only weakly coupled to. More generally, the idea can be formalised using the Feshbach projection formalism or related approaches such as R-matrix theory.

However, one then finds that the correspondence between the bound states of the perfectly isolated system and the resonances of the coupled system is somewhat arbitrary, since one can choose the subsystem that one considers perfectly isolated. In the finite potential well example, one could choose the states in the no-potential region or one could include the barrier to get similar results. Which one is the "correct" bound state associated with the resonance is unclear and not a physical question. Practically, one chooses a region which provides a useful basis for further calculations, such as in the presence of additional interactions.

Turning this around, in many systems resonances can also arise from complicated interactions, where identifying an isolated system is highly non-trivial. The best example I can think of are random lasers.

Note that there are some interesting phenomena in resonant systems which somewhat elude the intuition provided above. For example, one can have strict bound states even in the energy region of continuum states ($E > V_\mathrm{max}$ in the question example). This phenomenon is known as a bound state in the continuum. Originally pointed out in the early days of quantum mechanics as a mathematical curiosity, such states have recently been studied extensively in photonics [1, 2]. A related phenomenon are so called exceptional points, where two resonances merge into a non-trivial degeneracy.

Answer 3

Resonances arise from the possibility of transitions between two bound states. This leads for instance to the spectral lines in atomic spectra (be it in emission or absorption). Classically, this can be treated as a damped harmonic oscillator, with the transition probability playing the role of the damping constant (if the transition probability is zero there is thus no resonance).

Transitions between the bound and free energy spectrum (like photo-ionization and recombination for atoms) are sometimes treated as resonances as well though by using the concept of a pseudo-oscillator. This can be done because the overlap integral for the wave functions of the two states involved is still a well defined finite quantity as long as one of the states is bound.