Rotational properties of single-particle states in a $D^T_2$-symmetric potential

In this Appendix, we discuss elementary rotational properties of s.p. eigenstates of a mean-field Hamiltonian, which is symmetric with respect to the $D^T_2$ group. The group $D^T_2$ comprises the time-reversal operation, $\hat{T}$, three signature operations, $\hat{R}_x$, $\hat{R}_y$, $\hat{R}_y$, which are rotations through 180$^\circ$ about the three Cartesian axes, and products of the time-reversal and signature operations, which are called $T$-signatures, cf. Ref. [45] for more information about this group in the context of mean-field calculations. In most cranking solutions corresponding to quadrupole deformation, the group $D^T_2$ is a symmetry group of the s.p. Hamiltonian, $\hat{h}$, of Eq. (3). For a single Kramers pair in a fixed potential, we investigate the response of the s.p. angular momenta to a cranking frequency applied in an arbitrary direction. Our conclusions are based only on symmetry arguments, and are thus independent on the particular implementation of the mean field.

Irrespective of spatial symmetries, whenever the s.p. Hamiltonian is invariant under the time reversal, its spectrum exhibits the two-fold Kramers degeneracy. We consider a single Kramers pair, whose states are denoted as $\vert\mu\rangle$ and $\vert\bar\mu\rangle$, where

$$\hat{T}\vert\mu\rangle=s_\mu\vert\bar\mu\rangle~, \qquad \hat{T}\vert\bar\mu\rangle=s_{\bar\mu}\vert\mu\rangle~;$$ (29)

$s_\mu$ is an arbitrary phase factor and $s_{\bar\mu}=-s_\mu$.

All information about the matrix elements of the angular-momentum operator, $\hat{\vec {J}}$, between the states $\vert\mu\rangle$ and $\vert\bar\mu\rangle$ can be represented in a convenient way in terms of the real alignment vector, $\vec {J}^\mu$, and the complex decoupling vector, $\vec {D}^\mu$, of the state $\vert\mu\rangle$. They are defined as

$$\vec {J}^\mu=\langle\mu\vert\hat{\vec {J}}\vert\mu\rangle~, \qquad \vec {D}^\mu=\langle\mu\vert\hat{\vec {J}}\vert\bar\mu\rangle~.$$ (30)

Although the components of the decoupling vector change their phases when $\vert\mu\rangle$ and $\vert\bar\mu\rangle$ change theirs, the relative phases of those components do not depend on the phase convention. Since the angular-momentum operator is odd under the time reversal, it can be easily verified that

$$\vec {J}^{\bar\mu}=\langle\bar\mu\vert\hat{\vec {J}}\vert\bar\mu\... ...}^{\bar\mu}=\langle\bar\mu\vert\hat{\vec {J}}\vert\mu\rangle=\vec {D}^{\mu~*}~.$$ (31)

If the s.p. Hamiltonian is symmetric with respect to the $D^T_2$ group, it is possible to chose the states forming the Kramers pair as eigenstates of any of the three signature operators, $\hat{R}_i$, where $i=x,y,z$, but only one at a time, because the signature operators do not commute among themselves, i.e., for $i\neq j$

$$\hat{R}_i\hat{R}_j=\sum_{k=x,y,z}\epsilon_{ijk}\hat{R}_k.$$ (32)

This results, respectively, in three formally different pairs, ( $\vert\mu_i\rangle$,  $\vert\bar\mu_i\rangle$), which are just three different bases in the same two-dimensional eigenspace of $\hat{h}$.

We choose states $\vert\mu_i\rangle$ so that they correspond to eigenvalues $-i$ under the action of $\hat{R}_i$, while the eigenvalues of $\vert\bar\mu_i\rangle$ are $+i$. Multiplication rules (32) allow to easily express eigenstates $\vert\mu_x\rangle$, $\vert\bar\mu_x\rangle$, $\vert\mu_y\rangle$, and $\vert\bar\mu_y\rangle$ through linear combinations of eigenstates $\vert\mu_z\rangle$ and $\vert\bar\mu_z\rangle$. By fixing the relative phase between states $\vert\mu_z\rangle$ and $\vert\bar\mu_z\rangle$ we obtain the following expressions:

$$\vert\mu_x\rangle = \sqrt{\frac{-i}{2}}(\vert\mu_z\rangle+\vert\bar\mu_z\rangle)~,$$ (33)

$$\vert\bar\mu_x\rangle = -\sqrt{\frac{i}{2}}(\vert\mu_z\rangle-\vert\bar\mu_z\rangle)~,$$ (34)

$$\vert\mu_y\rangle = \sqrt{\frac{i}{2}}(\vert\mu_z\rangle+i\vert\bar\mu_z\rangle)~,$$ (35)

$$\vert\bar\mu_y\rangle = \sqrt{\frac{-i}{2}}(i\vert\mu_z\rangle+\vert\bar\mu_z\rangle)~,$$ (36)

where $\sqrt{i}=\exp(i\pi/4)$ and $\sqrt{-i}=\exp(-i\pi/4)$. These formulae allow to write $\vec {J}^{\mu_x}$, $\vec {D}^{\mu_x}$, $\vec {J}^{\mu_y}$, and $\vec {D}^{\mu_y}$ in terms of $\vec {J}^{\mu_z}$ and $\vec {D}^{\mu_z}$, i.e.,

$$\vec {J}^{\mu_x} = \mathrm{Re~}\vec {D}^{\mu_z}~,$$ (37)

$$\vec {D}^{\mu_x} = -i\vec {J}^{\mu_z}-\mathrm{Im~}\vec {D}^{\mu_z}~,$$ (38)

$$\vec {J}^{\mu_y} = -\mathrm{Im~}\vec {D}^{\mu_z}~,$$ (39)

$$\vec {D}^{\mu_y} = \vec {J}^{\mu_z}-i\mathrm{Re~}\vec {D}^{\mu_z}~.$$ (40)

The fact that $\vert\mu_i\rangle$ and $\vert\bar\mu_i\rangle$ are eigenstates of $\hat{R}_i$, together with the transformation rules of the components, $\hat{J}_j$, of the angular momentum operator under the three signatures,

$$\hat{R}_i^+\hat{J}_j\hat{R}_i=\left\{ \begin{array}{rcl} +\hat{J}_j & \mbox{for} & j=i \\ -\hat{J}_j & \mbox{for} & j\neq i \\ \end{array}\right.~,$$ (41)

induces limitations on the components, $J^{\mu_i}_j$ and $D^{\mu_i}_j$, of the alignment and decoupling vectors. Namely,

$$J^{\mu_i}_j = \left\{ \begin{array}{rcl} \mbox{non-zero} & \mbox{for} & j=i \\ 0 & \mbox{for} & j\neq i \\ \end{array}\right.~,$$ (42)

$$D^{\mu_i}_j = \left\{ \begin{array}{rcl} 0 & \mbox{for} & j=i \\ \mbox{non-zero} & \mbox{for} & j\neq i \\ \end{array}\right.~.$$ (43)

In other words, $\vec {J}^{\mu_i}$ is confined to the axis $i$ and $\vec {D}^{\mu_i}$ to the plane perpendicular to that axis.

From these relations and for the eigenstates defined as in Eqs. (33)-(36), it follows that all the quantities $J^{\mu_i}_j$ and $D^{\mu_i}_j$ can be expressed through the three ``diagonal'' components, $J^{\mu_i}_i$, of the alignment vector,

$$\vec {J}^{\mu_x}=(J^{\mu_x}_x,0,0)~, \vec {D}^{\mu_x}=(0,J^{\mu_y}_y,-iJ^{\mu_z}_z)~,$$ (44)

$$\vec {J}^{\mu_y}=(0,J^{\mu_y}_y,0)~, \vec {D}^{\mu_y}=(-iJ^{\mu_x}_x,0,J^{\mu_z}_z)~,$$ (45)

$$\vec {J}^{\mu_z}=(0,0,J^{\mu_z}_z)~, \vec {D}^{\mu_z}=(J^{\mu_x}_x,-iJ^{\mu_y}_y,0)~.$$ (46)

Thus, all the information about the angular-momentum matrix elements within a Kramers pair in the spectrum of a $D^T_2$-symmetric s.p. Hamiltonian is contained in three real numbers, $J^{\mu_x}_x$, $J^{\mu_y}_y$, and $J^{\mu_z}_z$.

The symmetry group $D^T_2$ itself does not impose any conditions on the``diagonal'' components, $J^{\mu_i}_i$. However, these values can be further restricted if some other symmetry is present. For example, if $\hat{h}$ is axially symmetric, say with respect to the $z$ axis, then the states $\vert\mu_z\rangle$, $\vert\bar\mu_z\rangle$ are eigenstates of $\hat{J}_z$, which leads to quantization of $J^{\mu_z}_z$. In fact, $J^{\mu_z}_z=+1/2,-3/2,...$, because $\hat{R}_z=\exp(-i\pi\hat{J}_z)$, while in the adopted convention $\hat{R}_z\vert\mu_z\rangle=-i\vert\mu_z\rangle$. For states $\vert\mu_x\rangle$ and $\vert\mu_y\rangle$, defined by (33) and (35), one easily finds

$$J^{\mu_x}_x = \frac{1}{2}\mathrm{Re}\langle\mu_z\vert\hat{J}_++\hat{J}_-\vert\bar\mu_z\rangle~,$$ (47)

$$J^{\mu_y}_y = \frac{1}{2}\mathrm{Re}\langle\mu_z\vert\hat{J}_+-\hat{J}_-\vert\bar\mu_z\rangle~,$$ (48)

where $\hat{J}_+=\hat{J}_x+i\hat{J}_y$ and $\hat{J}_-=\hat{J}_x-i\hat{J}_y$ are the ladder operators, that increment and decrement the magnetic quantum number, $J_z$, of an eigenstate, $\vert J_z\rangle$, of $\hat{J}_z$,

$$\hat{J}_+\vert J_z\rangle\sim\vert J_z+1\rangle~, \qquad \hat{J}_-\vert J_z\rangle\sim\vert J_z-1\rangle~.$$ (49)

One can see, therefore, that the matrix elements in (47) and (48) can be non-zero only if $\vert\mu_z\rangle$ and $\vert\bar\mu_z\rangle$ differ in $J_z$ by one, that is if $J^{\mu_z}_z=1/2$. In such a case, $\langle\mu_z\vert\hat{J}_-\vert\bar\mu_z\rangle=0$, and $J^{\mu_x}_x=J^{\mu_y}_y$. These results can be summarized as

$$(J^{\mu_x}_x,J^{\mu_y}_y,J^{\mu_z}_z)=\left\{ \begin{array}{lll} ... ...allel) & \mbox{for} & J^{\mu_\parallel}_\parallel=3/2,... \\ \end{array}\right.$$ (50)

The parameter

$$J^{\mu_\perp}_\perp=\frac{1}{2}\mathrm{Re}\langle\mu_z\vert\hat{J}_+\vert\bar\mu_z\rangle$$ (51)

is not restricted by the above kinematic conditions. It is related to the standard decoupling parameter, $a=-2J^{\mu_\perp}_\perp$, considered by Bohr and Mottelson [14]. They take such phases for the states $\vert\mu_z\rangle$ and $\vert\bar\mu_z\rangle$ that the $T$-signature-$y$, $\hat{R}_y^T=\hat{T}\hat{R}_y$, is the complex conjugation in the basis formed by these states. Since $\hat{J}_+$ is even under $\hat{R}_y^T$, in that convention the matrix element in Eq. (51) is real.

We now consider the TAC for a single Kramers pair in a $D^T_2$-symmetric potential. We make two simplifying assumptions. First, that the Hamiltonian $\hat{h}$ in the Routhian $\hat{h}^\prime$ of Eq. (3) does not change with rotational frequency (non-selfconsistent cranking). Second, that the considered Kramers pair has no coupling to other eigenstates of $\hat{h}$ through the angular-momentum operator (isolated pair). The TAC under such conditions becomes a two-dimensional diagonalization problem, that can be solved analytically.

We use the basis of states $\vert\mu_z\rangle$ and $\vert\bar\mu_z\rangle$. For a degenerate Kramers pair, $\hat{h}$ reduces to its eigenvalue, $e$. Matrix elements of the angular-momentum operator are defined by Eq. (46). Altogether, matrix of the s.p. Routhian (3) takes the form

$$\hat{h}' = \hat{h}-\vec {\omega}\hat{\vec {J}}$$ (52)

$$= \left[ \begin{array}{cc} e & 0 \\ 0 & e \end{array}\right]-\... ...\omega_yJ^{\mu_y}_y & -\omega_zJ^{\mu_z}_z \end{array}\right]~.$$

It is easy to verify that the two eigenstates of this Routhian have opposite-sign mean angular-momentum vectors $\vec {J}$, whose components read

$$J_i=\pm\frac{\omega_i(J^{\mu_i}_i)^2}{\left(\omega_x^2(J^{\mu_x}_x)^2+\omega_y^2(J^{\mu_y}_y)^2+\omega_z^2(J^{\mu_z}_z)^2\right)^{1/2}}~.$$ (53)

Equation (53) constitutes the central point of discussion, because it defines the sought response of the s.p. angular momenta to rotation under the assumed conditions. Note that the dependence of $J_i$ on $\omega _i$ is non-linear in the general case.

Values of alignments (53) are undefined if, and only if, all the products $\omega_iJ^{\mu_i}_i$ vanish - in particular when $\omega =0$. In such a case, the Routhian (52) is proportional to unity, and the mean angular momenta of its eigenstates depend on their (arbitrary) unitary mixing.

Two extreme cases of the dependence (53) deserve particular attention.

• If $J^{\mu_x}_x=J^{\mu_y}_y=J^{\mu_z}_z=g$, then $J_i=g\omega_i/\omega$, and $\vec {J}$ always orients itself along $\vec {\omega}$, already for infinitesimal $\omega$. This can be dubbed soft alignment.

• When only one of the parameters $J^{\mu_x}_x$, $J^{\mu_y}_y$, $J^{\mu_z}_z$ is non-zero, say $J^{\mu_j}_j$, then $J_i=J^{\mu_j}_j\delta_{ij}$, and $\vec {J}$ is independent of $\vec {\omega}$. We call this stiff alignment on the $j$-th axis.

In axial nuclei, precisely one two-fold degenerate substate of each deformation-split $j$-shell has $J^{\mu_\parallel}_\parallel=1/2$ and $J^{\mu_\perp}_\perp\neq0$, which represents the soft alignment. According to Eq. (50), all other necessarily have a vanishing decoupling parameter, and are thus stiffly aligned with the symmetry axis. For prolate shapes, the lowest-energy substate has $J^{\mu_\parallel}_\parallel=1/2$, and is soft, while for oblate shapes it is the highest substate. In triaxial nuclei, values of the parameters $J^{\mu_i}_i$, where $i=s,m,l$ corresponds to the short, medium, and long principal axes, are equal to the s.p. alignments obtained from the one-dimensional cranking about the three axes. Indeed, for cranking about the axis $i$, the s.p. states are eigenstates of $\hat{R}_i$. For example, from the results of Section 3.4, one can see that for the lowest $h_{11/2}$ substates of a triaxial nucleus only $J^{\mu_s}_s$ is non-zero, while for the highest substates only $J^{\mu_l}_l$ does not vanish. These alignments are thus stiff. Note that there are no states with stiff alignment on the medium axis. The response to rotation of the middle $h_{11/2}$ substates is soft, because all the three parameters, $J^{\mu_s}_s$, $J^{\mu_m}_m$, $J^{\mu_l}_l$, are non-zero.

In realistic cranking calculations the symmetry arguments discussed here interplay with the fact that there is angular-momentum coupling between different Kramers pairs and that the mean field does change with rotational frequency. In the results of the present paper, however, the change of deformation induced by rotation is negligible. The angular-momentum coupling of the lowest and highest $h_{11/2}$ substates to other s.p. states is rather weak, what can be seen from the small curvature of their one-dimensional Routhians in Fig. 2. The stiff character of their alignments is fully confirmed in our self-consistent calculations, as discussed in Section 3.4 and further in the paper. Investigation on how suitable the notion of soft and stiff alignment is in other physical cases remains a subject for further research.