Two-nucleon binding from lattice QCD at physical quark masses
In plain words
The deuteron (a bound proton and neutron) is held together by only 2.2 MeV, while two neutrons do not bind at all. Lattice QCD calculations at heavier-than-real quark masses disagree with each other on whether two nucleons bind.
Precise statement
Compute the $I=0$ (deuteron) and $I=1$ (dineutron) S-wave nucleon-nucleon phase shifts and pole positions in lattice QCD at the physical pion mass, with continuum and infinite-volume extrapolation. Answer: deuteron binding energy (experiment $2.2246\,\mathrm{MeV}$) and the $1S0$ nn scattering length (experiment about $-18.9\,\mathrm{fm}$), plus resolution of the dependence of results on the choice of interpolating operators found at $m_{\pi}$ above $300\,\mathrm{MeV}$.
What would settle it
Variational lattice QCD spectra with complete two-nucleon operator bases at several pion masses down to physical, converted to scattering amplitudes with the Luscher finite-volume method (which maps energies in a finite box to phase shifts).
Status in the literature
Unverified note
A 2025 calculation with a large operator basis at $m_{\pi}$ about $714\,\mathrm{MeV}$ found no bound dinucleons (BaSc collaboration, arXiv 2505.05547), contradicting earlier deep-binding results obtained with compact hexaquark operators.