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arXiv 1408.0459nucl-thnucl-ex
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He同位素及其镜像核的镜像对称性破缺

Mirror symmetry breaking in He isotopes and their mirror nuclei

Takayuki Myo, Kiyoshi Kato

摘要:

本文采用$^4$ He + $X$ N($X$ =2,4)团簇模型研究了$^6$ He-$^6$ Be和$^8$ He- $^8$ C的镜像对称破缺.多体共振的正确的边界条件,使用复尺度法处理。我们发现由于$^8$ C中的库仑排斥作用,$^8$ C的基态半径大于$^8$ He的基态半径。另一方面,两个核的$0^+_2$共振呈现相反的关系; $^8$ C半径小于$^8$ He半径。这是由于价质子在$^8$ C中的$^4$ He团簇核心周围的库仑势垒破坏了两个核半径的镜像对称性。在镜像核$^6$ He和$^6$ Be中也得到类似的半径变化。在$^8$ He和$^8$ C的$0^+_2$态中观察到价核子的极大空间伸展。这种性质与四个价核子的$(p_{3/2})^2(p_{1/2})^2$组态的优势有关,这是从涉及价核子的空间扩展分量而使耦合强度降低到其他组态来理解的。

英文摘要:

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.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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