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Why doesn't neutron scattering have selection rules?

According to Wikipedia, it seems that the value of the transition moment integral will be non-zero, but I guess another way of phrasing my question is, What does the transition moment integral for neutron scattering represent?

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  • $\begingroup$ What selection rules for neutron scattering are you expecting? Polarized or unpolarized neutrons? $\endgroup$
    – Jon Custer
    Jan 14 at 20:51
  • $\begingroup$ @JonCuster I didn't know what to expect. I just wondered why, unlike with light where there are selection rules (e.g., for Raman vs. IR) why isn't that the case with neutrons? $\endgroup$ Jan 14 at 21:01
  • $\begingroup$ Well, the neutron doesn't have to be absorbed and re-emitted for one. $\endgroup$
    – Jon Custer
    Jan 14 at 21:02
  • $\begingroup$ Right, but Raman is a scattering phenomenon for example. I am curious about what the fundamental reason is for the lack of a selection rule in inelastic neutron scattering. $\endgroup$ Jan 14 at 22:58
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Selection rules in atomic transitions (including the virtual transitions which characterize Raman scattering) occur because the incident photon is changing the internal state of the atom, and must do so in ways which conserve energy, angular momentum, parity, and perhaps some other quantum numbers.

Neutron scattering is generally done with milli-eV neutrons, but the lowest excitations in nuclei typically have mega-eV energies above the ground state. So a scattered neutron typically can’t drive any excitations in the target nucleus, and therefore doesn’t care about any of the low-lying nuclear quantum numbers. To the extent that there are quantized states, they are in the virtual compound nucleus $(\text{target}+\text n)$ whose decay towards its ground state is the mechanism for neutron capture. However, most nuclei have an extremely dense set of very broad, short-lived states that overlap with their neutron separation energy, so you don’t end up with selection rules there, either. If you need some quantum numbers in a compound nucleus, there’s almost certainly an available state which has them.

If your neutrons are energetic enough to scatter from nuclear resonances, their wavelengths are probably too short to be interesting for materials science. And there are selection rules for some capture reactions, such as the $\rm^3He(n,p)^3H$ reaction whose spin dependence is important for neutron polarimetry. But that’s nucleon scattering, which is much more exciting than neutron scattering.

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