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(3) |
In a nucleus such as Hg, with no intrinsic octupole deformation, many
intermediate states contribute to the sum in Eq. (2). By
contrast, the asymmetric shape of
Ra implies the existence of a very
low-energy
state, in this case 55 keV above the ground state
, that dominates the sum because of the
corresponding small denominator. To very good approximation,
then,
The amount of the enhancement is not easy to calculate accurately, however.
The
reason is that the matrix element of the two-body spin-dependent operator
in Eq. (5) depends sensitively on the behavior
of a few valence particles,
which carry most of the spin. In the approximation that particles (or
quasiparticles) move in independent orbits generated by a mean field, the
potential can be written as an effective density-dependent one-body operator
that we will denote
, defined implicitly by
The authors of Refs. [Spevak et al.(1997)Spevak, Auerbach,
and
Flambaum,Auerbach et al.(1996)Auerbach,
Flambaum,
and Spevak] used a version of the
particle-rotor model [Leander and Sheline(1984)] to represent the odd- nucleus. In
this
model, all but one of the nucleons are treated as a rigid core, and the last
valence nucleon occupies a deformed single-particle orbit, obtained by
solving
a Schrödinger equation for a Nilsson or deformed Wood-Saxon potential. The
model implies that the core carries no intrinsic spin whatever, that the
neutron and proton densities are proportional, and that the exchange terms on
the right-hand side of Eq. (7) are negligible. Under these
assumptions,
, which now acts only on the single valence
nucleon, reduces to [Sushkov et al.(1984)Sushkov, Flambaum,
and Khriplovich]
Ref. [Engel et al.(1999)Engel, Friar, and
Hayes] confirmed the collectivity of the intrinsic Schiff
moments
obtained in Refs. [Spevak et al.(1997)Spevak, Auerbach,
and
Flambaum,Auerbach et al.(1996)Auerbach,
Flambaum,
and Spevak], but questioned the accuracy of
some of the
assumptions used to evaluate the matrix element of , suggesting
that either core-spin polarization or self-consistency in the
nuclear wave function might reduce laboratory Schiff moments. The zero-range
approximation and the neglect of exchange in
are also open
to question. As a result, it is not clear whether the Schiff moment of
Ra is 1000 times that of
Hg or 100 times, or even less. In
what
follows, we provide a (tentative) answer by moving
beyond the particle-rotor model. Our calculation is not the final word on
Schiff moments in octupole-deformed nuclei -- we only do mean-field
theory, neglecting in particular to project onto states with good parity, and
do not fully account for the pion's nonzero range -- but is a major step
forward.