Competition between $T$=1 and $T$=0 pairing in $pf$-shell nuclei with $N=Z$

TL;DR

This study investigates the competition between T=1 and T=0 pairing interactions in pf-shell nuclei with N=Z, revealing conditions under which T=0 pairing can dominate and its implications for ground state configurations.

Contribution

It provides a detailed analysis of T=1 and T=0 pairing correlations in pf-shell nuclei, highlighting the impact of spin-orbit splitting and pairing strength differences on nuclear ground states.

Findings

T=0 pairing is weakened in 1f orbits but dominates in 2p configurations with smaller spin-orbit splitting.

Spin-triplet T=0 pairing can overcome T=1 pairing if its strength exceeds T=1 by 50% or more.

Ground states of odd-odd pf-shell nuclei are likely J^π=0^+ except near the shell middle, consistent with experimental data.

Abstract

The pairing correlation energy for two-nucleon configurations with the spin-parity and isospin of $J^\pi=0^+$, $T$=1 and $J^\pi=1^+$, $T$=0 are calculated with $T$=1 and $T$=0 pairing interactions, respectively. To this end, we consider the $(1f2p)$ shell model space, including single-particle angular momenta of $l=3$ and $l=1$. It is pointed out that a two-body matrix element of the spin-triplet $T$=0 pairing is weakened substantially for the $1f$ orbits, even though the pairing strength is much larger than that for the spin-singlet $T$=1 pairing interaction. In contrast, the spin-triplet pairing correlations overcome the spin-singlet pairing correlations for the $2p$ configuration, for which the spin-orbit splitting is smaller than that for the $1f$ configurations, if the strength for the T=0 pairing is larger than that for the T=1 pairing by 50% or more. Using the Hartree-Fock wave functions, it is also pointed out that the mismatch of proton and neutron radial wave functions is at most a few % level, even if the Fermi energies are largely different in the proton and neutron mean-field potentials. These results imply that the configuration with $J^\pi=0^+$, $T$=1 is likely in the ground state of odd-odd $pf$ shell nuclei even under the influence of the strong spin-triplet $T$=0 pairing, except at the middle of the $pf$ shell, in which the odd proton and neutron may occupy the $2p$ orbits. These results are consistent with the observed spin-parity $J^{\pi}=0^+$ for all odd-odd $pf$ shell nuclei except for $^{58}_{29}$Cu, which has $J^{\pi}=1^+$.