We compute the momentum-space probability density of ${}^6$He at leading order in Halo EFT. In this framework, the ${}^6$He nucleus is treated as a three-body problem with a ${}^4$He core ($c$) and two valence neutrons ($n$). This requires the $nn$ and $nc$ t-matrices as well as a $cnn$ force as input in the Faddeev equations. Since the $nc$ t-matrix corresponds to an energy-dependent potential, we consider the consequent modifications to the standard normalization and orthogonality conditions. We find that these are small for momenta within the domain of validity of Halo EFT. In this regime, the ${}^6$He probability density is regulator independent, provided the cutoff is significantly above the EFT breakdown scale.

中文翻译：

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光环有效场论中$${}^6$$ 6 He的动量空间概率密度

我们在 Halo EFT 中以领先顺序计算 ${}^6$He 的动量空间概率密度。在此框架中，${}^6$He 核被视为具有 ${}^4$He 核 ($c$) 和两个价中子 ($n$) 的三体问题。这需要 $nn$ 和 $nc$ t 矩阵以及 $cnn$ 力作为 Faddeev 方程的输入。由于 $nc$ t 矩阵对应于一个依赖于能量的势，我们考虑对标准归一化和正交条件的后续修改。我们发现这些对于 Halo EFT 有效性域内的动量来说很小。在这种情况下，${}^6$He 概率密度与调节器无关，前提是截止值显着高于 EFT 击穿量表。