Abstract
In the reaction of $\overline{K}$-meson nuclear absorption, three fundamental processes are considered: a direct (nonresonant) $\overline{K}+N\ensuremath{\rightarrow}{\ensuremath{\pi}}^{\ensuremath{-}}+{\ensuremath{\Lambda}}^{0}$ reaction, the same reaction with the ${{Y}_{1}}^{*}$ resonance formation in the intermediate state, and the reaction with a $\ensuremath{\Sigma}$ production in the first stage and its subsequent conversion into a ${\ensuremath{\Lambda}}^{0}$ in a successive collision with another nucleon. The "zero-range" impulseapproximation is assumed. The initial $\overline{K}$ state is an $\mathrm{nS}$ or an $\mathrm{mP}$ Bohr mesoatomic orbit. Several forms of the ${\ensuremath{\Lambda}}^{0}$-nucleus final-state interaction are considered. For the case of the $\overline{K}+^{4}\mathrm{He}\ensuremath{\rightarrow}{\ensuremath{\pi}}^{\ensuremath{-}}+{\ensuremath{\Lambda}}^{0}+^{3}\mathrm{He}$ reaction the recoiling-$^{3}\mathrm{He}$ momentum distribution, the pion momentum distribution, and the ${\ensuremath{\pi}}^{\ensuremath{-}}+{\ensuremath{\Lambda}}^{0}$$^{3}\mathrm{He}$ angular distribution are analyzed. It turns out that in order to explain the $^{3}\mathrm{He}$ momentum distribution no elastic scattering distortion of the ${\ensuremath{\Lambda}}^{0}$-$^{3}\mathrm{He}$ wave is sufficient, and one has to introduce the $\ensuremath{\Sigma}\ensuremath{-}{\ensuremath{\Lambda}}^{0}$ conversion amplitude, which is most important at high $^{3}\mathrm{He}$-momenta, and which also improves our pion momentum distribution.
Published Version
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