Λ single particle energies

Abstract

The \ensuremath{\Lambda} single-particle energies ${B}_{\ensuremath{\Lambda}}$ of hypernuclei (HN) are calculated microscopically using the Fermi hypernetted chain method to obtain for our $\ensuremath{\Lambda}N$ and $\ensuremath{\Lambda}\mathrm{NN}$ potentials the \ensuremath{\Lambda} binding $D(\ensuremath{\rho})$ to nuclear matter, and the effective mass ${m}_{\ensuremath{\Lambda}}^{*}(\ensuremath{\rho})$ at densities $\ensuremath{\rho}<~{\ensuremath{\rho}}_{0}$ $({\ensuremath{\rho}}_{0}$ is normal nuclear density), and also the corresponding effective $\ensuremath{\Lambda}N$ and $\ensuremath{\Lambda}\mathrm{NN}$ potentials. The \ensuremath{\Lambda} core-nucleus potential ${U}_{\ensuremath{\Lambda}}(r)$ is obtained by suitably folding these into the core density. The Schr\"odinger equation for ${U}_{\ensuremath{\Lambda}}$ and ${m}_{\ensuremath{\Lambda}}^{*}$ is solved for ${B}_{\ensuremath{\Lambda}}.$ The fringing field (FF) due to the finite range of the effective potentials is theoretically required. We use a dispersive $\ensuremath{\Lambda}\mathrm{NN}$ potential but also include a phenomenological \ensuremath{\rho} dependence allowing for less repulsion for $\ensuremath{\rho}<{\ensuremath{\rho}}_{0},$ i.e., in the surface. The best fits to the data with a FF give a large \ensuremath{\rho} dependence, equivalent to an A dependent strength consistent with variational calculations of ${}_{\mathrm{\ensuremath{\Lambda}}}^{5}\mathrm{He},$ indicating an effective $\ensuremath{\Lambda}\mathrm{NN}$ dispersive potential increasingly repulsive with A whose likely interpretation is in terms of dispersive plus two-pion-exchange $\ensuremath{\Lambda}\mathrm{NN}$ potentials. The well depth is $29\ifmmode\pm\else\textpm\fi{}1\mathrm{MeV}.$ The $\ensuremath{\Lambda}N$ space-exchange fraction corresponds to ${m}_{\ensuremath{\Lambda}}^{*}(\ensuremath{\rho})\ensuremath{\approx}0.75--0.80$ and a ratio of p- to s-state potentials of $\ensuremath{\approx}0.5\ifmmode\pm\else\textpm\fi{}0.1.$ Charge symmetry breaking (CSB) is significant for heavy HN with a large neutron excess; with a FF the strength agrees with that obtained from the $A=4\mathrm{HN}.$ The fits without FF are excellent but inconsistent with the requirement for a FF, with ${}_{\mathrm{\ensuremath{\Lambda}}}^{5}\mathrm{He},$ and also with the CSB sign for $A=4.$