Classical model

The classical model of chiral rotation was briefly introduced in Ref. [32]; here we give its more detailed description and discussion. In order to define the model, we begin with elementary considerations related to the dynamics of rigid bodies. By a classical gyroscope we understand an axial-shape rigid body with moments of inertia, $\mathcal{J}_{\parallel}$ and $\mathcal{J}_{\perp}$, with respect to the symmetry axis and an axis perpendicular to it, respectively. Such a body spins with fixed angular frequency $\Omega$ around its symmetry axis; one can imagine that this motion is ensured by a motor and frequency regulator such that $\Omega$ is strictly constant in time.

Furthermore, let us imagine that the spinning body is rigidly mounted on another rigid body that has triaxial inertia tensor and three moments of inertia $\mathcal{J}'_s$, $\mathcal{J}'_m$, and $\mathcal{J}'_l$, with respect to its short, medium, and long axis, respectively. Then, the angular frequency $\Omega$ is maintained fixed with respect to the triaxial body, irrespective of how the whole device moves, and the angular frequency vector $\vec {\Omega}$ has by definition three time-independent components $\Omega_s$, $\Omega_m$, and $\Omega_l$. To simplify our considerations let us assume that the axis of the gyroscope coincides with the short axis of the triaxial body, i.e., $\Omega_s$=$\Omega$, $\Omega_m$=0, and $\Omega_l$=0. In this case the principal axes of the tensor of inertia of the whole device coincide with those of the triaxial body, and the three moments of inertia of the device read,

$$\begin{array}{rcl} \mathcal{J}_s &=& \mathcal{J}'_s + \mathca... ...{J}_l &=& \mathcal{J}'_l + \mathcal{J}_{\perp} . \\ \end{array}$$ (7)

We assume that the device rotates with the total angular frequency vector $\vec {\omega}$, which in the moving frame of the triaxial body has components $\omega_s$, $\omega _m$, and $\omega_l$. In general, these components may vary with time, although later we study only such a motion of the device when they are time-independent. The kinetic energy of the device is the sum of that of the triaxial body and gyroscope,

$$T=T_{\text{triax}}+T_{\text{gyro}},$$ (8)

where

$$T_{\text{triax}} = \tfrac{1}{2}\left(\mathcal{J}'_s\omega^2_s +\mathcal{J}'_m\omega^2_m +\mathcal{J}'_l\omega^2_l\right),$$ (9)

$$T_{\text{gyro}} = \tfrac{1}{2}\left(\mathcal{J}_{\parallel}(\omega_s+\Omega)^2 +\mathcal{J}_{\perp}\omega^2_m +\mathcal{J}_{\perp}\omega^2_l\right),$$ (10)

hence

$$T = \tfrac{1}{2}\left(\mathcal{J}_s\omega^2_s +\mathcal{J}_m\omeg... ...cal{J}_{\parallel}\Omega\omega_s + \tfrac{1}{2}\mathcal{J}_{\parallel}\Omega^2.$$ (11)

We see that the total kinetic energy is a sum of three terms. The first one represents the rotation of the entire device irrespective of the fact that it contains a spinning gyroscope; it depends only on the total moments of inertia. The second one represents the additional energy coming from the spinning gyroscope and depends on its time-independent spin $s_s=\mathcal{J}_{\parallel}\Omega$, while the third one is a constant which can be dropped from further considerations.

It is obvious that if we add two other gyroscopes aligned with the medium and long axes and spinning with spins $s_m$ and $s_l$, respectively, the second term can be simply written as a scalar product $\vec {\omega}\cdot\vec {s}$, where vector $\vec {s}$ has components $s_s$, $s_m$, and $s_l$. In this case, the total moment of inertia $\mathcal{J}$ will be a sum of contributions from the triaxial body and three gyroscopes. We note in passing that exactly the same result is obtained for a single spherical gyroscope, $\mathcal{J}_{\parallel}=\mathcal{J}_{\perp}$, tilted with respect to the principal axes of the triaxial body in such a way that it has spin components equal to $s_s$, $s_m$, and $s_l$.

The total angular momentum, $\vec {I}$, of the system reads

$$\vec {I}=\mathcal{J}\vec {\omega}+\vec {s}~,$$ (12)

where $\vec {s}$ is the above vector sum of spins of all the gyroscopes and $\mathcal{J}\vec {\omega}$ stands for the tensor product of the moment of inertia tensor $\mathcal{J}$ with the angular frequency vector $\vec {\omega}$. In absence of potential interactions, the Lagrangian of the system is equal to the total kinetic energy (11), generalized to the case of three gyroscopes, and thus is given by the formula

$$L=\tfrac{1}{2}\vec {\omega}\mathcal{J}\vec {\omega}+\vec {\omega}\cdot\vec {s}~,$$ (13)

where we dropped all constant terms. Taking the laboratory components of $\vec {\omega}$ as generalized velocities, it is easy to check that the generalized momenta are equal to the laboratory components of $\vec {I}$. This fact allows us to write the Legendre transformation [43] and to obtain the Hamiltonian of the system,

$$H=\vec {\omega}\vec {I}-L=\tfrac{1}{2}\vec {\omega}\mathcal{J}\vec {\omega}~.$$ (14)

Since the Lagrangian (13) does not depend explicitly on time, the Hamilton function (14) is a constant of motion, and is identified with the total energy, $E$, of the system. Consider now a particular type of the Routhian [43], $H'$, namely such that no variables undergo the Legendre transformation. In such a case, $H'=-L$, and by rewriting the Routhian in terms of the Hamiltonian one obtains

$$H'=H-\vec {\omega}\cdot\vec {I}~.$$ (15)

Equations of motion for the model can be derived in the following way. As for any vector, the time derivatives of the angular momentum vector, $\partial_{t}\vec {I}$ - taken in the laboratory frame and $\partial^{\omega}_{t}\vec {I}$ - taken in a frame rotating with angular frequency $\vec {\omega}$, are related by the formula $\partial_{t}\vec {I}=\partial^{\omega}_{t}\vec {I}+\vec {\omega}\times\vec {I}$ [43]. Since the angular momentum is conserved in the laboratory frame, $\partial_{t}\vec {I}=0$, one obtains the Euler equations [43] for the time evolution of the angular-momentum vector in the body-fixed frame,

$$\partial^{\omega}_{t}\vec {I}=-\vec {\omega}\times\vec {I}~.$$ (16)

The mean-field cranking approximation can only account for the so-called uniform rotations, in which the mean angular-momentum vector is constant in the intrinsic frame of the nucleus, $\partial^{\omega}_{t}\vec {I}=0$. Because of that, we restrict the classical model studied here to such uniform rotations. Due to Eq. (12), for uniform rotations also the angular frequency vector is constant in the intrinsic frame. The Euler equations (16) now take the form $\vec {\omega}\times\vec {I}=0$, and require that $\vec {\omega}$ and $\vec {I}$ be parallel. The same condition holds for the HF solutions and is known as the Kerman-Onishi theorem; see Section 2 and [37]. The Euler equations can now be easily solved for the considered classical model. However, to show further analogies with the HF method, in what follows we find the uniform solutions by employing a variational principle.

According to the Hamilton's principle, motion of a mechanical system can be found by making the action integral, $\int L~dt$, stationary. The real uniform rotations obviously belong to a wider class of trial motions with $\vec {I}$ and $\vec {\omega}$ being constant in the intrinsic frame, but not necessarily parallel to one another. Within this class, Lagrangian (13) does not depend on time. Therefore, extremizing the action for a given value of $\omega$= $\vert\vec {\omega}\vert$ reduces to finding extrema of the Lagrangian as a function of the intrinsic-frame components of $\vec {\omega}$. Since $H'=-L$, the Routhian (15) can be equally well used for this purpose. This provides us with a bridge between the classical model and the quantum cranking theory, where an analogous Routhian (1) is minimized in the space of the trial wave-functions.

Extrema of $H'$ with respect to the intrinsic-frame components of $\vec {\omega}$ at a given length of $\omega$ can be found by using a Lagrange multiplier, $-\tfrac{1}{2}\mu$, for $\omega^{2}$ (the factor $-\tfrac{1}{2}$ being added for later convenience). We continue further derivations for the case of two gyroscopes aligned along the $s$ and $l$ axes, as dictated by the microscopic results presented in Section 3.4, i.e., we employ the classical model for $s_m$=0. Setting to zero the derivatives of the quantity

$$H'+\tfrac{1}{2}\mu\omega^2 = \tfrac{1}{2} \left[(\mu-\mathcal{J}_s)\omega_s^2+(\mu-\mathcal{J}_m)\omega_m^2\right.$$

$$+ \left.(\mu-\mathcal{J}_l)\omega_l^2\right]-(\omega_s s_s+\omega_l s_l)$$ (17)

with respect to $\omega_{s}$, $\omega_{m}$, $\omega_{l}$, one obtains

$$\omega_s = s_s/(\mu-\mathcal{J}_s)~,$$ (18)

$$\omega_m(\mu-\mathcal{J}_m) = 0~,$$ (19)

$$\omega_l = s_l/(\mu-\mathcal{J}_l)~.$$ (20)

Equation (19) gives either $\omega_{m}=0$ or $\mu={\mathcal{J}}_{m}$, leading to two distinct classes of solutions.

Figure 5: (color online). Planar bands A, B, C, D, and the chiral band obtained from the classical model. a) Rotational frequency, $\omega (\mu )$. b) Intrinsic-frame trajectory of the rotational frequency vector, $\vec {\omega}$. The chiral solution corresponds to a straight line perpendicular to the figure plane, and intersecting it at the marked point. That perpendicular direction represents $\omega _m$. c) Angular momentum, $I(\omega )$. d) Energy, $E(I)$.

If $\omega_{m}=0$ then both $\vec {\omega}$ and $\vec {I}$ lie in the $s$-$l$ plane. This gives planar solutions, for which the chiral symmetry is not broken. All values of $\mu$ are allowed, and the Lagrange multiplier must be determined from the length of $\vec {\omega}$ calculated in the obvious way from (18) and (20). Figure 5a shows $\omega$ versus $\mu$ for sample model parameters, extracted from the $^{132}$La HF PAC solutions with the SLy4 force with no time-odd fields, and listed in Table 1. The solutions marked as A and D exist for all values of $\omega$, while above some threshold frequency, $\omega_{\text{thr}}$, two more solutions appear, B and C. This threshold frequency can be determined by finding the minimum of $\omega$ in function of $\mu$, and reads

$$\omega_{\text{thr}}=\frac{\left(s_s^{2/3}+s_l^{2/3}\right)^{3/2}}{\vert\mathcal{J}_l-\mathcal{J}_s\vert}~.$$ (21)

The value of $\omega_{\text{thr}}$ coming from the present HF calculations is rather high, higher than $1\,\mathrm{MeV}/\hbar$. Since bands B and C are situated far above the yrast line (see Fig. 5d) they will not be subject of further analysis.

For $\mu$= $\mathcal{J}_m$, all values of $\omega _m$ are allowed, while components in the $s$-$l$ plane are fixed at $\omega_s=s_s/(\mathcal{J}_m-\mathcal{J}_s)$ and $\omega_l=s_l/(\mathcal{J}_m-\mathcal{J}_l)$. Consequently, the angular momentum has non-zero components along all three axes, and the chiral symmetry is broken. For each value of $\omega$, there are two cases differing by the sign of $\omega_{m}$, and thus giving the chiral doublet. The fact that $\omega_{s}$ and $\omega_{l}$ are constant leads to the principal conclusion that chiral solutions cannot exist for $\omega$ smaller than the critical frequency

$$\omega_{\text{crit}}^{\text{clas}}=\left[\left(\frac{s_s}{\mathca... ...right)^2 +\left(\frac{s_l}{\mathcal{J}_m-\mathcal{J}_l}\right)^2\right]^{1/2}~.$$ (22)

At that frequency, and with $\omega _m$=0, the chiral solution coincides with the planar band A.

In the $\vec {\omega}$ space, the four planar solutions form a hyperbola in the $s$-$l$ plane, while the chiral doublet corresponds to a straight line perpendicular to that plane. These curves are shown in Fig. 5b. Figure 5c gives the angular momentum in function of rotational frequency for all the presented bands. With increasing $\omega$, the so-called dynamical moment, $\mathcal{J}^{(2)}\equiv\mathrm{d}I/\mathrm{d}\omega$, asymptotically approaches $\mathcal{J}_l$ for bands A and B, and $\mathcal{J}_s$ for bands C and D. For the chiral band, $I$ is exactly proportional to $\omega$ with the coefficient $\mathcal{J}_m$. Thus, the critical spin, $I_{\text{crit}}^{\text{clas}}$, corresponding to the critical frequency (22) reads

$$I_{\text{crit}}^{\text{clas}}=\mathcal{J}_m\omega_{\text{crit}}^{\text{clas}}~.$$ (23)

Figure 5d summarizes the energies in function of spin. The spin quantum number, $I$, is related to the length, $I$, of the angular momentum vector, $\vec {I}$, by the condition $I(I+1)=\vert\vec {I}\vert^2$. At low angular momenta, the yrast line coincides with the planar band D. Then it continues along the planar solution A. Since the moment of inertia $\mathcal{J}_m$ is the largest, beyond the critical frequency the chiral solution becomes yrast, thereby yielding good prospects for experimental observation.

Altogether, the classical model described here is defined by five parameters, $\mathcal{J}_s$, $\mathcal{J}_m$, $\mathcal{J}_l$, $s_s$, and $s_l$, which are extracted form the microscopic HF PAC calculations and listed in Table 1. The model can then be applied to predict properties of the planar and chiral TAC bands, and these predictions can be compared with the HF TAC results. Such a comparison is presented and discussed in the following Sections.