Abstract

The first $0^+$ resonant state of the $^{12}$C nucleus ${}^{12}$C$(0_2^+)$, so called the Hoyle state, is investigated in a three-$\alpha$-particle (3-$\alpha$) model. A wave function for the photodisintegration reaction of a $^{12}$C bound state to 3-$\alpha$ final states is defined and calculated by the Faddeev three-body formalism, in which three-body bound- and continuum states are treated consistently. From the wave function at the Hoyle state energy, I calculated distributions of outgoing $\alpha$-particles and density distributions at interior region of the Hoyle state. Results show that a process through a two-$\alpha$ resonant state is dominant in the decay and contributions of the rest process are very small, less than 1 \%. There appear some peaks in the interior density distribution corresponding to configurations of an equilateral- and an isosceles triangles. It turns out that these results are obtained independently of the choice of $\alpha$-particle interaction models, when they are made to reproduce the Hoyle state energy.

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