LAUSR

Background: The band structure of the negative-parity states of $^{24}\mathrm{Mg}$ has not yet been clarified. The ${K}^{\ensuremath{\pi}}={0}^{\ensuremath{-}}$, ${K}^{\ensuremath{\pi}}={1}^{\ensuremath{-}}$, and ${K}^{\ensuremath{\pi}}={3}^{\ensuremath{-}}$ bands have been suggested, but the assignments have been inconsistent between experiments and theories.Purpose: Negative-parity states of $^{24}\mathrm{Mg}$ are investigated by microscopic structure and reaction calculations via proton and $\ensuremath{\alpha}$ inelastic scattering to clarify the band assignment for the observed negative-parity spectra.Method: The structure of $^{24}\mathrm{Mg}$ was calculated using the antisymmetrized molecular dynamics (AMD). Proton and $\ensuremath{\alpha}$ inelastic reactions were calculated using microscopic coupled-channel (MCC) calculations by folding the Melbourne $g$-matrix $NN$ interaction with the AMD densities of $^{24}\mathrm{Mg}$.Results: The member states of the ${K}^{\ensuremath{\pi}}={0}^{+}$, ${K}^{\ensuremath{\pi}}={2}^{+}$, ${K}^{\ensuremath{\pi}}={0}^{\ensuremath{-}}$, ${K}^{\ensuremath{\pi}}={1}^{\ensuremath{-}}$, and ${K}^{\ensuremath{\pi}}={3}^{\ensuremath{-}}$ bands of $^{24}\mathrm{Mg}$ were obtained through the AMD result. In the $\mathrm{MCC}+\mathrm{AMD}$ results for proton and $\ensuremath{\alpha}$ elastic and inelastic cross sections, reasonable agreements were obtained with existing data, except in the case of the ${4}_{1}^{+}$ state.Conclusions: The ${3}^{\ensuremath{-}}$ state of the ${K}^{\ensuremath{\pi}}={3}^{\ensuremath{-}}$ band and the ${1}^{\ensuremath{-}}$ and ${3}^{\ensuremath{-}}$ states of the ${K}^{\ensuremath{\pi}}={0}^{\ensuremath{-}}$ bands were assigned to the ${3}_{1}^{\ensuremath{-}}$(7.62 MeV), ${1}_{1}^{\ensuremath{-}}$(7.56 MeV), and ${3}_{2}^{\ensuremath{-}}$(8.36 MeV) states, respectively. The present AMD calculation is the first microscopic structure calculation to reproduce the energy ordering of the ${K}^{\ensuremath{\pi}}={0}^{\ensuremath{-}}$, ${K}^{\ensuremath{\pi}}={1}^{\ensuremath{-}}$, and ${K}^{\ensuremath{\pi}}={3}^{\ensuremath{-}}$ bands of $^{24}\mathrm{Mg}$.