Investigation of the determination of nuclear deformation using high-energy heavy-ion scattering

Abstract

Background: Nuclear deformation provides a crucial characteristic of nuclear structure. Conventionally, the quadrupole deformation length of a nucleus, ${\delta }_2$, has often been determined based on a macroscopic model through a deformed nuclear potential with the deformation length ${\delta }_2^{(\mathrm{pot})}$, which is determined to reproduce the nuclear scattering data. This approach assumes ${\delta }_2={\delta }_2^{(\mathrm{pot})}$ although there is no theoretical foundation. Purpose: We clarify the relationship between ${\delta }_2$ and ${\delta }_2^{(\mathrm{pot})}$ for high-energy heavy-ion scattering systematically to evaluate the validity of the conventional approach to determine the nuclear deformation. Method: The deformation lengths for the ${}^{12}$C inelastic scattering by ${}^{12}$C, ${}^{16}$O, ${}^{40}$Ca, and ${}^{208}$Pb targets at $E/A$ = 50--400 MeV are examined. First, we perform microscopic coupled-channel (CC) calculations to relate ${\delta }_2$ of the deformed density into the inelastic scattering cross section. Second, we use the deformed potential model to determine ${\delta }_2^{(\mathrm{pot})}$ so as to reproduce the microscopic CC result. We then compare ${\delta }_2^{(\mathrm{pot})}$ with ${\delta }_2$. Results: We find that ${\delta }_2^{(\mathrm{pot})}$ is about 20--40 \% smaller than presumed ${\delta }_2$, showing strong energy and target dependence. Further analysis, which considers higher-order deformation effects beyond the derivative model, reveals that ${\delta }_2^{(\mathrm{pot})}$ is still about 15--35 \% smaller than ${\delta }_2$. Conclusion: Our results suggest that one needs to be careful when the deformed potential model for the high-energy heavy-ion scattering is used to extract the nuclear deformation. The conventional approach may underestimate the deformation length ${\delta }_2$ systematically.

arXiv comment7 pages, 5 figures

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