Abstract
We study the mirror symmetry breaking of $^6$He-$^6$Be and $^8$He-$^8$C using the $^4$He + $X$N ($X$=2, 4) cluster model. The many-body resonances are treated for the correct boundary condition using the complex scaling method. We find that the ground state radius of $^8$C is larger than that of $^8$He due to the Coulomb repulsion in $^8$C. On the other hand, the $0^+_2$ resonances of the two nuclei exhibit the inverse relation; the $^8$C radius is smaller than the $^8$He radius. This is due to the Coulomb barrier of the valence protons around the $^4$He cluster core in $^8$C, which breaks the mirror symmetry of the radius in the two nuclei. A similar variation in the radius is obtained in the mirror nuclei, $^6$He and $^6$Be. A very large spatial extension of valence nucleons is observed in the $0^+_2$ states of $^8$He and $^8$C. This property is related to the dominance of the $(p_{3/2})^2(p_{1/2})^2$ configuration for four valence nucleons, which is understood from the reduction in the strength of the couplings to other configurations by involving the spatially extended components of valence nucleons.
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