Shell-model test of the rotational-model relation between static quadrupole momentsQ(21+),B(E2)'s, and orbitalM1transitions

Abstract

In this work, we examine critically the relation between orbital magnetic dipole (scissors mode) strength and quadrupole deformation properties. Assuming a simple $K=0$ ground-state band in an even-even nucleus, the quantities $Q({2}_{1}^{+})$ (i.e., the static quadrupole moment) and $B(E2){}_{{0}_{1}\ensuremath{\rightarrow}{2}_{1}}$ both are described by a single parameter---the intrinsic quadrupole moment ${Q}_{0}$. In the shell model, we can operationally define ${Q}_{0}(\mathrm{static})$ and ${Q}_{0}(\mathit{BE}2)$ and see if they are the same. Following a brief excursion to the sd shell, we perform calculations in the fp shell. The nuclei we consider ($^{44,46,48}\mathrm{Ti}$ and $^{48,50}\mathrm{Cr}$) are far from being perfect rotors, but we find that the calculated ratios ${Q}_{0}(\mathrm{static})/{Q}_{0}(\mathit{BE}2)$ with an FPD6 interaction are often very large (very close to unity) and far from the simple vibrational limit of zero. The experimental ratios for $^{46}\mathrm{Ti}$ and $^{48}\mathrm{Ti}$ are somewhat smaller ($~0.75$), but the $^{50}\mathrm{Cr}$ value is larger, exceeding unity. We also discuss the quadrupole collectivity of orbital magnetic dipole transitions. We find that the large orbital $B(M1)$ strength in $^{44}\mathrm{Ti}$ relative to $^{46}\mathrm{Ti}$ and $^{48}\mathrm{Ti}$ cannot be explained by simple deformation arguments.