Numerical Results and Discussion

It is well known that the isospin symmetry is only weakly broken in atomic nuclei and the concepts of the isospin conservation and isospin quantum number prevail even in the presence of the Coulomb interaction. In order to control the approximate isospin conservation we have employed, as already mentioned, the isocranking method, which corresponds to the lowest-order isospin projection. We parametrize the isocranking frequency $\vec{\lambda}$ as

$$\vec \lambda=(\lambda\sin \theta,0,\lambda\cos\theta) =(\l... ...ime,0,\lambda^{\prime}\cos\theta^\prime+\lambda_{\rm off}).$$ (1)

Even in the calculation with the Coulomb term, our Hamiltonian is invariant under rotation about the $T_z$ axis. Therefore, we set $\lambda_y=0$ and consider the isocranking only in the $T_x - T_z$ plane. The procedure of the isocranking calculations is as follows [4]. First, we perform the standard Hartree-Fock (HF) calculation for the isoaligned ($\vert T_z\vert=T$) states and thus we find the corresponding neutron and proton Fermi energies $\lambda_n$ and $\lambda_p$. Next, we determine values of $\lambda_{\rm off}$ and $\lambda^{\prime}$ as

$$(\lambda_{\rm off}, \lambda^\prime)= \frac{1}{2}(\lambda_... ...}^{T_z=-T}, \lambda_{np}^{T_z=T} -\lambda_{np}^{T_z=-T} ),$$ (2)

where $\lambda_{np}^{T_z=\pm T}\equiv \lambda_n-\lambda_p$ is the difference of the neutron and proton Fermi energies in the $T_z=\pm T$ isobars. Finally, we vary the tilting angle $\theta^\prime$ with $\lambda^{\prime}$ and $\lambda_{\rm off}$ fixed. In the calculations without the Coulomb interaction, we set $\lambda_{\rm off}=0$ and $\theta=\theta^\prime$. Due to the Coulomb interaction, the s.p. levels vary in function of $\langle \hat T_z \rangle$. Therefore, in varying the tilting angle $\theta$ from 0$^\circ$ to 180$^\circ$, level crossings may take place. The choice of $\vec \lambda$ in Eq. (1) helps avoiding the level crossings and smooths a way to obtain the isobaric analogue states (IASs) from $\theta^\prime=0^\circ$ to 180$^\circ$.

In Fig. 1(a), we show the total energies of the $T\simeq 4$ IASs in $A=40$ isobars calculated with and without the Coulomb interaction. We have used $(\lambda_{\rm off}, \lambda^\prime)=(0, 12.5)$ MeV and $(\lambda_{\rm off}, \lambda^\prime)=(-6.8, 13.6)$ MeV for the calculations without and with the Coulomb interaction, respectively. They are determined from the difference of the proton and neutron Fermi energies $\lambda_{n}-\lambda_{p}$ in the standard HF solution for $^{40}$S and $^{40}$Cr. When the Coulomb interaction is switched off, our EDF is invariant under the isospin rotation. The total energy calculated without the Coulomb interaction is independent of the direction of the isospin, which constitutes a test of the code.

When the Coulomb interaction is switched on, the total energy depends on the expectation value $\langle \hat T_z \rangle$. One can see that the total energy depends on $\langle \hat T_z \rangle$ almost linearly. The effect comes predominantly from the Coulomb energy which exhibits almost the same dependence on $\langle \hat T_z \rangle$ as that of the total energy as shown in Fig. 1(a). Its linearity results from proportionality of the Coulomb energy to $Z^2 = T_z^2-AT_z+A^2/4$, that clearly enhances the linear term by a factor of $A$ as compared to the quadratic term for small $T_z$.

Figure: (a) Total energies of $T \approx 4$ IASs in $A=40$ isobars calculated with and without the Coulomb interaction in function of $\langle \hat T_z \rangle$. (b) Coulomb energies of $T \approx 4$ IASs in $A=40$ in function of $\langle \hat T_z \rangle$. The results of isocranking calculations for every $10^\circ$ of $\theta^\prime$ between $\theta^\prime=0^\circ$ and $180^\circ$ are plotted.

In Fig. 2, we show the same results as in Fig. 1, but calculated for the $T\simeq 8$ states in the $A=40$ isobars. For these calculations, we have used $(\lambda_{\rm off}, \lambda^\prime)=(0, 27.4)$ MeV and $(\lambda_{\rm off}, \lambda^\prime)=(-6.0, 28.6)$ MeV for the calculations without and with the Coulomb interaction, respectively. The values are determined from the standard HF ground state solutions in $^{40}$Mg and $^{40}$Ni. Again, the total energy calculated without the Coulomb interaction is independent on $\langle \hat T_z \rangle$. However, in Fig. 2(b) one can see traces of the quadratic dependence of the Coulomb energy on $\langle \hat T_z \rangle$, although the contribution from the linear term is dominant.

Figure: Same as Fig. 1 but calculated for the $T\simeq 8$ IASs in $A=40$ isobars. The results of isocranking calculations for every $5^\circ$ of $\theta^\prime$ between $\theta^\prime=0^\circ$ and $180^\circ$ are plotted.

In Fig. 3, we plot the expectation values of the s.p. Routhian $\hat h^\prime=\hat h - \vec \lambda \cdot \hat \vec t$, calculated for the $T\simeq 8$ IASs with $A=40$. At $\theta^\prime=0$, the Fermi surface appears around $-12.5$MeV, below which 14 neutron and 6 proton orbitals are occupied. The s.p. Routhians vary as functions of $\theta^\prime$, and, unlike in the case of $A=48$ [4], there is no large shell gap above the Fermi surface. Nevertheless, with our choice of $\vec \lambda$, the level crossings are avoided. While the s.p. states are pure proton or neutron states at $\theta=0^{\circ}$ and 180$^{\circ}$, which means that the $\vert T_z\vert=T$ states are nothing but the standard HF states without the p-n mixing, at all other tilting angles, the s.p. states are p-n mixed. In particular, the proton and neutron components are almost equally mixed at $\theta^\prime=90^\circ$, which corresponds to $T_z \approx 0$.

Figure: Single-particle Routhians of the $T\simeq 8$ states in $A=40$ isobars calculated with the Coulomb interaction included. The arrows at the upper left and upper right indicate the positions of the Fermi energies at $\theta^\prime=0^\circ$ and 180$^\circ$, respectively.

Figure: $\langle \hat T^2 \rangle$ calculated for $T\simeq 8$ IASs with and without the Coulomb interaction.

Figure: Root-mean-square radii of the $T =8$ IASs with $A=54$ as functions of $\langle \hat T_z \rangle$.

Fig. 4 shows the expectation values of $\langle \hat T^2 \rangle$ calculated for the $T\simeq 8$ states in $A=40$ isobars. In case of rigorous isospin conservation one should obtain $\langle \hat T^2 \rangle$=72. The Coulomb interaction breaks the isospin symmetry and gives a deviation from this value. However, even in the case without the Coulomb interaction, the calculated $\langle \hat T^2 \rangle$ deviates from the exact value 72 due to the spurious isospin mixing within the mean-field approximation [7,8,9]. Note that around $T_z=8$ the spurious deviation is even larger than in the case with the Coulomb interaction.

Figure: Quadrupole deformation $\beta_{2}$ calculated for the $T\simeq 4$ and $T\simeq 8$ IASs in $A=40$ isobars with the Coulomb interaction included.

Figure: Energies of $T \simeq 1$ isobaric analog states in $A=54$ isobars in comparison with the experimental data [10]. The results of isocranking calculations for every $30^\circ$ of $\theta^\prime$ between $\theta^\prime=0^\circ$ and $180^\circ$ are plotted.

Fig. 5 shows the proton, neutron and total root-mean-square (rms) radii calculated with the Coulomb interaction for the $T\simeq 8$ states in $A=40$ isobars, together with the total rms radius calculated without the Coulomb interaction. The neutron (proton) rms radius increases with increasing (decreasing) $\langle \hat T_z \rangle$, that is, increasing the neutron (proton) components. With the Coulomb interaction, the total rms radius increases with increasing the proton components due to the Coulomb repulsion among protons. Without the Coulomb interaction, it stays constant as a function of $\langle \hat T_z \rangle$. In Fig. 6, we depict the quadrupole deformation parameter $\beta_2$ calculated for the $T\simeq 4$ and $T\simeq 8$ IASs in $A=40$ isobars. In both of the IAS chains, the quadrupole deformation $\beta_2$ is nearly constant, which illustrates the fact that the s.p. configuration for all IASs stays the same.

In the $A=4n$ nuclei, such as the $A$=40 systems discussed above, even-$T$ states are the ground states of even-even nuclei and their IASs. We also performed calculations for $A=4n+2$ nuclei, in which odd-$T$ states are the ground states of even-even nuclei. As an example of those calculations, in Fig. 7, we depict the calculated energies of the $T=1$ triplet in $A=54$ isobars in comparison with the experimental data. Here, the $I=0^+, T \simeq \vert T_z\vert=1$ states are the ground states of $^{54}$Fe and $^{54}$Ni and are described by the standard HF solutions without the p-n mixing. On the other hand, the $T_z=0$ IAS, the lowest $I=0^+$ state in $^{54}$Co, is obtained by the isocranking calculation, and it consists of the p-n mixed s.p. states. It is gratifying to see that both the energy of the $T_z=0$ state as well as those of the $\vert T_z\vert=1$ states are well reproduced by the theory. It is worth stressing that the $T_z=0$ IAS in an odd-odd nucleus is described here by means of a single time-even Slater determinant. This is at variance with single-reference p-n unmixed EDF models, wherein such states do not exist at all [11].