Microscopic study of4α-particle condensation with inclusion of resonances

Abstract

The $4\ensuremath{\alpha}$ condensate state for $^{16}\mathrm{O}$ is discussed with the Tohsaki-Horiuchi-Schuck-R\"opke (THSR) wave function which has $\ensuremath{\alpha}$-particle condensate character. With a treatment of resonances, it is found that the $4\ensuremath{\alpha}$ THSR wave function yields a fourth ${0}^{+}$ state in the continuum above the $4\ensuremath{\alpha}$-breakup threshold, in addition to the three ${0}^{+}$ states obtained in a previous analysis. It is shown that this fourth ${0}^{+}$ $[({0}_{4}^{+}){}_{\mathrm{THSR}}]$ state has a structure analogous to that of the Hoyle state because it has a very dilute density and a large component of $\ensuremath{\alpha}+^{12}\mathrm{C}({0}_{2}^{+})$ configuration. Furthermore, single-$\ensuremath{\alpha}$ motions are extracted from the microscopic 16-nucleon wave function, and the condensate fraction and momentum distribution of $\ensuremath{\alpha}$ particles are quantitatively discussed. It is found that for the $({0}_{4}^{+}){}_{\mathrm{THSR}}$ state a large $\ensuremath{\alpha}$-particle occupation probability concentrates on a single-$\ensuremath{\alpha}$ $0S$ orbit and the $\ensuremath{\alpha}$-particle momentum distribution has a $\ensuremath{\delta}$-function-like peak at zero momentum, both indicating that the state has a strong $4\ensuremath{\alpha}$ condensate character. It is argued that the $({0}_{4}^{+}){}_{\mathrm{THSR}}$ state is the counterpart of the ${0}_{6}^{+}$ state which was obtained as the $4\ensuremath{\alpha}$ condensate state in the previous $4\ensuremath{\alpha}$ orthogonality condition model calculation and therefore is likely to correspond to the ${0}_{6}^{+}$ state observed at $15.1$ MeV. The necessity of including $\ensuremath{\alpha}+^{12}\mathrm{C}$ configurations in the THSR wave function is pointed out.