Exact solution of the pairing problem for spherical and deformed systems

There has been increasing interest in studying the Richardson model from which one can derive the exact solution for certain pairing Hamiltonians. However, it is still a numerical challenge to solve the nonlinear equations involved. In this paper we tackle this problem by employing a simple hybrid polynomial approach. The method is found to be robust and is valid for both deformed and nearly spherical nuclei. It also provides important and convenient initial guesses for spherical systems with large degeneracy. As an example, we apply the method to study the shape coexistence in neutron-rich Ni isotopes.

Figure 1

Correlation energies for each pair, $x_i^{\prime }=x_i-2{\varepsilon }_i+G$, as a function of the pairing strength $G$ within the Richardson model for a fully occupied system (left) and a half-occupied (right) system with $N=6$ pairs and, respectively, 6 and 12 equally spaced, doubly degenerate orbitals separated by one unit. The kinks correspond to where the transition from real to complex numbers of the Richardson variables occurs. In the left case, the summation over all $x_i^{\prime }$ equals zero.

Figure 2

Correlation energies for each pair (solid line) for two half-occupied systems with $N=6$ (left) and 25 (right) pairs within two $\mathrm{\Omega }=6$ (left) and 25 (right) orbitals separated by one unit. The red dashed lines correspond to those of the fully occupied systems in the single-$j$ orbital.

Figure 3

Comparison between the exact (circles) and approximate (black solid line) solutions for total correlation energy of a half-filled system with $N=6$ pairs in two $\mathrm{\Omega }=6$ orbitals separated by one unit. The approximate solution is derived by solving Eq. (9) by slightly breaking the degeneracy of the first orbit by 1%.

Figure 4

Correlation energy for each pair for a half-filled doubly degenerate system with $N=6$ pairs as studied in Fig. (1) by gradually decreasing the splitting between the filled orbitals. The $G$ values are taken as $G=0.3$, 0.65, and 0.9 for the left, middle, and right panels, respectively.

Figure 5

Relative energies of ${}^{70}\mathrm{Ni}$ as a function of quadrupole deformation ${\beta }_2$ for calculations without pairing (HF) and with pairing with three different strengths. The strengths of the proton and neutron pairings are supposed to be the same for simplicity.

Figure 6

Correlation energies for the four proton pairs in the $f_{7/2}$ orbital as a function of ${\beta }_2$ for $G=0.2$ (left), 0.25 (middle), and 0.3 (right).