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.

enter image description here

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?

  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/688782/325035 $\endgroup$
    – Newbie
    Jan 15, 2022 at 21:01
  • $\begingroup$ The quote makes no mention of "discrete", does the image belong to the same source? $\endgroup$
    – ohneVal
    Jan 20, 2022 at 9:38
  • $\begingroup$ @ohneVal Go to the time-stamp 34.36 of this video... The lecturer says "they're discrete states just above the neutron or proton separation energy"... also the diagram draws them as discrete. Also, you can go to the timestamps 34.22 and 35.34. In fact, later the lecturer used the same figure from the source I have mentioned. $\endgroup$ Jan 20, 2022 at 13:46
  • $\begingroup$ which video...? $\endgroup$
    – ohneVal
    Jan 20, 2022 at 14:04
  • $\begingroup$ Sorry... youtu.be/Pa6kl4aasOU See this. $\endgroup$ Jan 20, 2022 at 14:08

2 Answers 2


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).

enter image description here

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.

  • 1
    $\begingroup$ I think you mean V~1/r. Also, a similar effect can be generated with an ‘angular momentum barrier’, since the orbital kinetic energy goes like 1/r^2. $\endgroup$
    – ragnar
    Mar 5, 2022 at 18:27
  • $\begingroup$ @ragnar Thanks, you're right. I corrected my answer. $\endgroup$ Mar 7, 2022 at 18:52
  • $\begingroup$ @ArturodonJuan Sorry but I have a follow-up question. If the quasi-bound states are states above the nuclear potential well but below the Coulomb barrier, is there no quasibound state for neutrons (which are uncharged)? $\endgroup$ Dec 1, 2022 at 17:16
  • $\begingroup$ @Solidification Your logic is correct but the nuclear force is more complicated than that. The lowest-energy state of two neutrons is an isospin triplet $|nn\rangle$, i.e. with aligned isospin. By Pauli's exclusion principle this will imply the normal spin will be anti-aligned, i.e. S=0. This anti-alignment of spin produces an additional repulsion which combats the attraction via color, and surprisingly prohibits a bound state. This is different with a neutron and proton, which can combine to give isospin singlet $\sim |np\rangle - |pn\rangle $ having attractive spin-spin interaction. $\endgroup$ Dec 1, 2022 at 20:34
  • $\begingroup$ @Solidification Upon googling a bit it seems a hot topic of debate whether (quasi-)bound states of neutrons exist, i.e. "dineutron". Naive calculations would suggest no such bound states exist, but quasi-bound states (with very very small lifetimes) potentially could exist. See this paper: sciencedirect.com/science/article/pii/S0370269314005012 $\endgroup$ Dec 1, 2022 at 20:36

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.”


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