As in Ref. [27], we diagonalize the s.p. Routhian,
For systems obeying the time-reversal symmetry, vanishes [17] and the rotation in isospace is described by a two-dimensional isocranking, that is,
In the absence of the Coulomb interaction, choosing MeV, MeV, and varying from to , generates all the and IASs. The angles , , and correspond to the HF solutions for Ni (), Y ( in the odd-odd system), and Sn (), respectively. Our example involves very exotic nuclei, including those beyond the proton dripline. We find this case interesting because the nuclei at both ends of the isobaric chain are heavy and doubly magic, thus spherical.
As discussed in Ref. [27], with the Coulomb term off, the value of is roughly equal to the absolute value of difference between the proton and neutron Fermi energies in Ni or Sn, MeV. Then, the isocranking term makes the Fermi energies of neutrons and protons almost equal. In the presence of the Coulomb interaction, however, a large asymmetry between develops between Ni (12.31MeV) and Sn (33.62MeV). Therefore, to offset the difference of Fermi energies at and with Coulomb interaction present, we set the values:
Using these expressions, MeV and MeV.
Figure 1 shows that in the absence of the Coulomb interaction, the total energy is independent of . This should be the case, as the pnEDF is isospin-invariant and thus the energy must be independent of the direction of the isospin vector. This also turned out to be an important test on the derived expressions and numerical code, as different terms of pnEDF become active for different values of . For and , solutions are unmixed and the densities are block-diagonal in neutron and proton subspaces. At intermediate values of , the solutions are p-n mixed. For the special case of , proton and neutron densities are equally mixed. When the Coulomb interaction is turned on, the total energy increases with (Fig. 1), because more and more protons replace neutrons and the Coulomb repulsion grows.
The degree of p-n mixing can be directly inferred from the expectation values of plotted in Fig. 2(a). As expected, the p-n mixing increases with and reaches its maximum value for , and then drops again. In Fig. 2(b), we show and it is seen that the values of , , and do correspond to Ni, Y, and Sn, respectively. The behavior of and weakly depends on whether the Coulomb term is included or not. This is entirely due to our choice of the shifted semicircle (18), whereupon the linear constraint absorbs the major part of the isovector component of the Coulomb interaction.
The Coulomb interaction breaks isospin and thus induces the isospin mixing in the HF wave function. To illustrate this, Fig. 3 shows the average value of for the converged HF solutions. For the considered case of the systems, should be exactly equal to in the absence of isospin mixing. However, as shown in Fig. 3, even in the absence of the Coulomb interaction, slightly deviates from this value. At the origin of this effect is the spurious isospin mixing [35,36,37]. Indeed, within the mean-field approximation, the isospin symmetry is broken spontaneously as the HF wave function is not an isospin eigenstate. However, since the Skyrme EDF is isospin covariant [38,15], the HF solutions corresponding to different orientations in the isospin space are degenerate in energy. While the neutron-proton mixing changes with the angle , must remain the same in the absence of the Coulomb interaction. In the presence of the Coulomb term, the isospin mixing is very small in the isospin-stretched configurations (for and ) and reaches its maximum around for [37,39].
The s.p. Routhians as functions of are shown in Figs. 4 (without Coulomb) and 5 (with Coulomb). Eleven spherical neutron levels and seven proton levels are occupied at , and the neutron and proton Fermi energies are shifted in such a way that the gaps in the s.p. spectra appear at around MeV (Fig. 4) and MeV (Fig. 5). Our choice of guarantees that, in the presence of the Coulomb interaction, the s.p. Routhians near the Fermi surface do not cross as functions of ; this would have caused a drastic structural changes of the mean-field and made the adiabatic tracing of the IAS as a function of extremely difficult. At and , the s.p. states have pure values of . At , most of the s.p. Routhians have close to zero, that is, they are fully p-n mixed.
Figure 6 displays s.p. HF energies, that is, s.p. Routhians with the isocranking term removed. Note that these are not eigenvalues but the average values of the HF Hamiltonian, calculated for states that are eigenstates of the Routhian (16). With increasing , owing to the increasing Coulomb field, s.p. states that increase proton (neutron) component gradually increase (decrease) in energy.
To better visualize the relative shifts of s.p. levels with , in Fig. 7 we show s.p. energies relative to the energy of the 1 shell. The figure nicely illustrates the effect of the Coulomb interaction on the proton components of the s.p. orbits: the relative level shifts correlate with their binding energies and values [40,41]. Indeed, the deeply (loosely) bound levels, which have smaller (larger) rms radii and thus experience stronger (weaker) Coulomb repulsion, are shifted up (down) in energy relative to the high- 1 shell.
Some of the calculated , IASs are predicted to appear beyond the proton drip line. As seen in Fig. 6, energy of the 1g level (which is neutron at and proton at ) becomes positive at around , where . At , energies of the 1, 1, 2, and 2 shells are positive. However, all these states are well localized by the Coulomb barrier, and thus correspond to narrow resonances, whose energies can be reasonably well described within the HO basis expansion [41].
To investigate properties of the unbound proton orbits, for the ground states of , nuclei, we performed the HFBRAD [43] calculations (without p-n mixing). In Fig. 10, we show results obtained for the 1 and 2 proton states. For each isobar, a dot is placed at the values of s.p. energies and radii, and lines show standard total effective HF proton potentials. The total effective HF proton potential consists of the standard central, spin-orbit, centrifugal, and Coulomb terms. The proton 1 orbit in Zr is bound, and in Mo it becomes slightly unbound. This result is consistent with the experimental observation, whereby the last bound nucleus of the isobaric chain, which is experimentally known, is Zr . The rms radii of the proton 1 orbits are about 5fm, and the s.p. wave functions are still localized, even if the orbits become unbound. This is because the 1 and 2 protons occupy states well below the potential barrier, which pushes the proton continuum up in energy, thus effectively extending the range of nuclear landscape into the proton-unstable region [44,21,22].
It is worth noting that the 2 orbit, which has a small centrifugal barrier, is bound up to around ( ) in Fig. 6. This is consistent with Fig. 10 that shows that the s.p. energy of 2 is unbound in Ru . In the presence of p-n mixing, the proton components of the s.p. states are smaller than those in the pure proton states, and this effectively reduces the repulsive Coulomb energies of the 1 orbits.
Jacek Dobaczewski 2014-12-07