Untitled

Abstract

A new symmetry of the Core-Particle-Hole Coupling model of odd-odd nuclei is discussed. This symmetry is used to explain a staggering pattern of the M 1 and E2 transitions within and between partner bands.

Electromagnetic Properties of Chiral Bands in $^{124}$Cs

“…Presented results agree with a spontaneous chiral symmetry breaking scenario, analogous to that observed in 126,128 Cs [2,3]. Moreover, the B(M1) staggering in the yrast band suggests that the s-symmetry [5] is useful in description of the 124 Cs nucleus. A more detailed study at the level energies in the yrast and side bands shows that some properties of these bands change below spin around 15 .…”

“…If the core exhibits some additional property (e.g. γ = 30 • in the rigid case), one can expect another symmetry -the so-called s-symmetry [4,5], which leads to characteristic staggering of M1 transitions inside and between bands. Some properties of 124 Cs bands suggest that this nucleus is a good case for investigating existence of the critical frequency ω c (or critical spin) below which the effects of the chiral symmetry breaking disappear, see [6] for the theoretical background.…”

Spontaneous time-reversal symmetry breaking in [sup 124]Cs

“…Characteristic B(M1) staggering in both bands have been observed. The B(M1) staggering has been predicted as a distinctive feature of strong chiral symmetry breaking [9]. Similar staggering has been found also in 126 Cs [6] and 124 Cs (see Ref.…”

Odd-odd nuclei as the core-particle-hole systems and chirality

“…In consequence, the M 1 transition operator is approximately odd with respect to C πν , and thus to S, which means that the M 1 reduced matrix elements within the states of the opposite s should be much bigger than those within the states of the same s (see the figure in [28]). …”

“…There exists a large class of the Bohr Hamiltonians invariant under P α alone. General conditions for the Bohr Hamiltonian to be even under inversion P α are mentioned in [28]. Hamiltonian (4) is the P α invariant provided b 1 = h 1 = 0.…”