In this answer by Arturo don Juan, and also, in section $7.8$ of Sakurai's Modern Quantum Mechanics, it is argued that resonances in the scattering cross-section at certain energies are due to the existence of quasi-bound states supported by the effective potential. The occurrence of such quasi-bound states is, in turn, caused by a repulsive tail of an otherwise attractive potential. This repulsive tail is either due to an angular momentum barrier $\frac{\hbar^2\ell(\ell+1)}{2mr^2}$ for $\ell\neq 0$ or could be entirely due to the shape of the potential itself. For example, it is caused by the Coulomb barrier seen by charged particles (e.g., protons) accelerating towards a nucleus.

If, on the other hand, a beam of slow neutrons is bombarded at a target nucleus, the neutrons do not experience a Coulomb barrier. Moreover, if we consider slow neutrons i.e. $s$-wave scattering, there is also no angular momentum barrier. Therefore, the potential will have no repulsive tail. Therefore, there exist no quasi-bound states. Thus, one would expect to observe no $s$-wave resonances. However, contrary to expectation, $s-$wave resonances are indeed observed in $s$-wave neutron scattering. What is going on here?

  • $\begingroup$ reference for the neutron statement, are they the same resonances as with protons? neutrons themselves, as protons, are a conglomerate of quarks, and quarks are charged and will see a barrier. look at the quark composition of protons, neutrons will just have overall zero charge, but the quarks will be still there. profmattstrassler.com/articles-and-posts/largehadroncolliderfaq/… $\endgroup$
    – anna v
    Nov 14, 2023 at 8:53
  • $\begingroup$ @annav For resonances in slow neutron capture cross-section, see this: researchgate.net/figure/… . $\endgroup$ Nov 14, 2023 at 9:09
  • 2
    $\begingroup$ I think the crucial observation is that neutron scattering off a nucleus is not simply potential scattering. The fact that the nucleus is composite, and made of A=N+Z neutrons and protons, is essential. As a result, there are "compound nucleus" resonances, long-lived states of the A+1 systems, that couple to the asymptotic scattering states. $\endgroup$
    – Thomas
    Nov 18, 2023 at 16:54
  • $\begingroup$ @Thomas Your comment is useful to me. Can you write it as an answer? $\endgroup$ Dec 1, 2023 at 3:54

1 Answer 1


The occurrence of s-wave resonances in neutron scattering, despite the absence of a repulsive tail in the potential, can be attributed to the complex nature of quantum mechanics.

In quantum mechanics, a resonance state is one that lives long enough to have a discretely identifiable nature, but which eventually decays1. The “s-wave” part of “s-wave resonance” just indicates that the resonance has total angular momentum ℓ = 01. This property is completely unrelated to whether or not a state is a resonance1.

In the low energy case, we obtain maximum scattering when K0R = (n + 1/2)π, when the scattering cross section is σ = 4π/K²2. This is an example of s-wave resonance2.

In observed neutron resonances, long believed to be a form of quantum chaos, regular family structures are found in the s-wave resonances of many even-even nuclei in the tens keV to MeV region3.

So, even though slow neutrons do not experience a Coulomb barrier and there is no angular momentum barrier in s-wave scattering, the complex interactions at the quantum level can still lead to the formation of s-wave resonances.

  • $\begingroup$ This does not explain the physical origin of the neutron s-wave resonances. $\endgroup$ Nov 30, 2023 at 15:26
  • $\begingroup$ It does, read it properly. I think I have already answered your question aptly. $\endgroup$
    – Alex Aboda
    Nov 30, 2023 at 17:03

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