In Griffiths, a state with energy $E$ is said to be "bound" if $$E < \min\left(\lim_{x\to\infty} V(x), \lim_{x\to-\infty} V(x)\right)$$ (i.e. $E$ is less than both of those quantities). My confusion arises from how Griffiths explains these bound states.

He uses the usual "trolley on a hill" visualisation, whereby if the trolley's energy is smaller than the potential at two distinct points, the trolley will stay between those two points and not be able to leave (let's disregard tunnelling for now).

But with this explanation, do "locally bound" states exist? Consider the potential $-x^4 + x^2$. This potential is concave up in some neighbourhood of zero; so if my "trolley" is in that neighbourhood and doesn't have enough energy to leave this concave up area, then isn't it "bound"? Clearly, no "fully bound" states exist for this potential, since it goes to $-\infty$ as $x\to\pm \infty$, but do such "locally bound" states exist near $0$? Or does "locally bound" make no sense?

An explanation would be appreciated.

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    $\begingroup$ quantum theory allows for quantum tunnelling, and indeed, the potential you considered does not have any bound state. It is a standard problem in QFT. There is a lot of literature on this subject. $\endgroup$ Commented Apr 17 at 2:38
  • $\begingroup$ @naturallyInconsistent Ah, that's good to know, thank you. But does this apply to all potentials? As in, do "locally bound" states never exist? Or is this a peculiarity of just the example I gave? $\endgroup$
    – Trisztan
    Commented Apr 17 at 2:43
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    $\begingroup$ Quantum tunnelling is simply a universal phenomenon of quantum theory, so if it can tunnel out, it will. There are definitions of states of the type that you are considering, all of them temporary, but they are all in advanced texts. $\endgroup$ Commented Apr 17 at 2:49
  • $\begingroup$ @naturallyInconsistent Alright, I think that answers my question then. Thank you. $\endgroup$
    – Trisztan
    Commented Apr 17 at 2:50

1 Answer 1


I think that the keyword you are looking for is resonance or metastable state (locally bound state is not standard). The idea is that those trapped regions do leave their marks on the quantum system even if they do not correspond to any energy eigenstate. In some sense, they are generalised eigenstates with complex eigenvalues, the imaginary part describing their decay time. Rigorously, they are not true eigenstates so will not be square integrable, but are detectable as poles in the $S$ matrix. They leave the signature Lorentzian resonance in non relativistic quantum mechanics. In fact, if you modify the potential by raising the barriers, the imaginary part can vanish, and you can recover a genuine eigenstate with real energy because the particle cannot tunnel out anymore. In this limit, the decay time follows the quantum analogue of Arrhenius law in statistical mechanics.

Your example is a bit pathological as the Hamiltonian is not selfadjoint, due to the finite time escape to infinity. You can take a look at Tong's lecture notes Topics in Quantum Mechanics (section 6.1.5 "Resonances"). He uses instead a double Dirac delta, which is conceptually the same as your approach, but has the advantage of being better posed and analytically solvable. Indeed, the two deltas act as the barriers classically trapping the particle between them. Quantum mechanically, the particle can tunnel out, but the trapping leaves signature resonances. In the limit of infinitely strong peaks, you recover the particle in a box with its eigenstates.

Hope this helps.


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