Abstract
An \({\alpha}\)–n–n three-cluster model of the \({^6}\)He nucleus is studied by solving the Faddeev equations, where the cluster potential between \({\alpha}\) and n takes into account the Pauli exclusion correction, using the Fish-Bone Optical Model (Schmid in Z Phys A 297:105, 1980). The resulting binding energy of the ground state (\({0^+}\)) is 0.831 MeV and the resonance energy of the first excited state (\({2^+}\)), 0.60–i0.012 MeV, is extracted from the three-cluster break-up threshold. These theoretical values are in reasonable agreement with the experimental data: 0.973 MeV and 0.824–i0.056 MeV, respectively. In order to investigate the structure of these states, we calculate the angle density matrix for the \({\angle n_1 \alpha n_2}\) angle in the triangle formed by the three clusters. The angle density matrix of the ground state has two peaks and the configuration of \({0^+}\) wave function corresponding to the peaks constitutes a mixture of an acute-angled triangle structure and an obtuse-angled one. This finding is consistent with the former result from a variational approach (Hagino and Sagawa in Phys Rev C 72:044321, 2005). On the other hand, in the case of \({2^+}\) state only a single peak is obtained.
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