Effective field theory for deformed odd-mass nuclei

Many odd-mass deformed nuclei can be viewed as an even-even deformed nucleus to which the extra nucleon is coupled. We take ²³⁹Pu as an example. The corresponding even-even nucleus is ²³⁸Pu, and Fig. 1 shows all its levels below 800 keV. At sufficiently low energies the spectrum of ²³⁸Pu is essentially that of an axially symmetric rigid rotor: The excitation energies $E(I)$ versus angular momentum $I$ obey $E(I)=AI(I+1)$. Here $A$ is a rotational constant of about 7 keV, and $\xi\approx 40$ keV (the energy of the $I=2$ state) sets the low-energy scale. Only even spins enter because the ground state is invariant under rotations by $\pi$ around any axis that is perpendicular to the symmetry axis. This symmetry is usually denoted as $R$ symmetry Bohr and Mottelson (1975). At energy $\Lambda\approx 600$ keV a second rotational band with a $K^{\pi}=1^{-}$ band head occurs, followed by more rotational bands at higher energies. In this work we will, however, consider only the lowest energies and restrict ourselves to the description of the ground-state rotational band. Then, the energy of the $K^{\pi}=1^{-}$ band head sets the breakdown scale $\Lambda$ of our EFT, because a new degree of freedom enters at this energy. We have a separation of scale $\xi\ll\Lambda$. The analysis of Ref. Coello Pérez and Papenbrock (2015b) shows that the ground-state band will exhibit noticeable deviations from the leading-order $E(I)=AI(I+1)$ rule for spins $I\gtrsim\Omega/\xi$. In the EFT this is due to subleading interactions that couple the ground-state band to other bands. While the interaction between the positive-parity ground-state band and the shown negative-parity band is suppressed, a positive-parity band enters at about 940 keV. These arguments suggest that the breakdown scale is properly chosen. An EFT for the lowest energies in deformed nuclei was presented in Refs. Papenbrock (2011); Coello Pérez and Papenbrock (2015b), and we briefly review its essential features.

To gain insight into how to construct the EFT, we consider the odd-mass nucleus ²³⁹Pu. Figure 2 shows all levels below 800 keV that can be grouped into rotational bands (omitting the few exceptions). The ground-state rotational band is built on a $K^{\pi}={1\over 2}^{+}$ state, i.e., a $K^{\pi}={1\over 2}^{+}$ neutron coupled to the ²³⁸Pu ground state. Rotations of this nucleon-nucleus state then produce the rotational band on top of the $1/2^{+}$ ground state. The first excited neutron state yields the $K^{\pi}={5\over 2}^{+}$ state at $\Omega\approx 300$ keV, and its rotations produce the corresponding rotational bands. Thus, the fermion single-particle excitation energy is about half the breakdown scale in this nucleus, and the condition $\Omega\ll\Lambda$ is fulfilled only marginally. The $K^{\pi}={1\over 2}^{-}$ band head at about 470 keV could be due either to a single-neutron excitation or to the coupling of the $K^{\pi}={1\over 2}^{+}$ neutron with the excited $1^{-}$ state (at the breakdown energy $\Lambda$) in ²³⁸Pu. Therefore, that rotational band is beyond the breakdown scale of the EFT we present in this paper.

The rotational bands depicted in Fig. 2 all follow the pattern

$$E(I,K)=E_{0}+A\left[I(I+1)+a\delta_{K,1/2}(-1)^{I+{1\over 2}}\left(I+{1\over 2}\right)\right]\ .$$

(22)

Here, $E_{0}$ is an energy offset, $A$ the rotational constant, and $a$ the decoupling parameter (that occurs only for $K={1\over 2}$ bands). These constants depend on the band under consideration. Typically, we have $E_{0}\sim\Omega$, $A\sim\xi/6$, and $a\sim{\cal O}(1)$. Equation (22) is well known from a variety of models Rowe (2010); Eisenberg and Greiner (1970b); Bohr and Mottelson (1975); Iachello and Arima (1987); Rowe and Wood (2010). As shown below, it is also the leading-order result of the EFT we develop in this paper.

We use the insight gained in the previous Subsection and request that the Lagrangian of the nucleon be invariant under SO(2) rotations in the body-fixed system. That guarantees invariance under SO(3) rotations in the space-fixed system.

The field operator $\hat{\psi}_{s}(\mathbf{x}^{\prime})$ creates a fermion at position $\mathbf{x}^{\prime}$ with spin projection $s=\pm{1\over 2}$ onto the $z^{\prime}$-axis in the body-fixed frame. Denoting the vacuum as $|0\rangle$ we thus have

$$\hat{\psi}_{s}^{\dagger}(\mathbf{x}^{\prime})|0\rangle=\chi_{{1\over 2}s}|\mathbf{x}^{\prime}\rangle\ .$$

(23)

Here $\chi_{{1\over 2}s}$ denotes a spin state of spin-${1\over 2}$ fermion with $z^{\prime}$ projection $s$ Varshalovich et al. (1988), and $|\mathbf{x}^{\prime}\rangle$ is an eigenstate of the position operator. The corresponding annihilation operator is $\hat{\psi}_{s}(\mathbf{x}^{\prime})$ and we have the usual anti-commutation relation for fermions

$\displaystyle\left\{\hat{\psi}_{s}(\mathbf{x}^{\prime}),\hat{\psi}^{\dagger}_{\sigma}(\mathbf{y}^{\prime})\right\}=\delta_{\sigma}^{s}\delta(\mathbf{x}^{\prime}-\mathbf{y}^{\prime})\ ,$

(24)

and all other anti commutators vanish. It will be useful to combine the two spin components of the field operator into the spinor

$\displaystyle\hat{\Psi}(\mathbf{x}^{\prime})\equiv\left(\begin{array}[]{c}\hat{\psi}_{+{1\over 2}}(\mathbf{x}^{\prime})\\ \hat{\psi}_{-{1\over 2}}(\mathbf{x}^{\prime})\end{array}\right)\ .$

(27)

The nucleon Lagrangian is

$\displaystyle L_{\Psi}=\int{\rm d}^{3}\mathbf{x}^{\prime}\hat{\Psi}^{\dagger}(\mathbf{x}^{\prime})\left(i\partial_{t}+{\hbar^{2}\Delta^{\prime}\over 2m}-V\right)\hat{\Psi}(\mathbf{x}^{\prime})\ .$

(28)

Here, $V$ is an axially symmetric potential which may also depend on spin, i.e., be a $2\times 2$ matrix. The potential of the Nilsson model Nilsson (1955) is an example. The Lagrangian (28) exhibits axial symmetry. The construction (28) is not only mandated by the nonlinear realization of rotational symmetry Papenbrock (2011). It is also consistent with an adiabatic approach where the light nucleon is much faster than the heavy and slowly rotating core and able to follow the core’s motion quasi instantaneously. The canonical momentum is

$$\hat{\Pi}(\mathbf{x}^{\prime})=\frac{\delta L}{\delta\partial_{t}\hat{\Psi}(\mathbf{x}^{\prime})}=i\hat{\Psi}^{\dagger}(\mathbf{x}^{\prime})\ .$$

(29)

The Legendre transform of the Lagrangian (28) yields the Hamiltonian

$\displaystyle H_{\Psi}$

$\displaystyle=$

$\displaystyle\int{\rm d}^{3}\mathbf{x}^{\prime}\hat{\Psi}^{\dagger}(\mathbf{x}^{\prime})\left(-{\hbar^{2}\Delta^{\prime}\over 2m}+V\right)\hat{\Psi}(\mathbf{x}^{\prime})\ .$

(30)

Here, we introduced the Laplacian $\Delta^{\prime}=\nabla^{\prime}\cdot\nabla^{\prime}$. The total angular momentum of the fermion,

$$\mathbf{K}=\int{\rm d}^{3}\mathbf{x}^{\prime}\hat{\Psi}^{\dagger}(\mathbf{x}^{\prime})\left(-i\mathbf{x}^{\prime}\times\mathbf{\nabla}^{\prime}+\hat{\mathbf{S}}\right)\hat{\Psi}(\mathbf{x}^{\prime})\ ,$$

(31)

is the sum of orbital angular momentum and spin

$\displaystyle\hat{\mathbf{S}}={1\over 2}\left(\begin{array}[]{c}\sigma_{x^{\prime}}\\ \sigma_{y^{\prime}}\\ \sigma_{z^{\prime}}\end{array}\right)\ .$

(35)

Here, $\sigma_{x^{\prime},y^{\prime},z^{\prime}}$ denote the usual Pauli matrices. These act with respect to the axes of the body-fixed system. The action of the general operators ($J_{x^{\prime}}$, $J_{y^{\prime}}$, $J_{z^{\prime}})$ on the space- and spin-degrees of freedom of the nucleon coincides with that of the corresponding angular momentum plus spin operators in Eq. (31). Thus, for $k^{\prime}=x^{\prime},y^{\prime},z^{\prime}$,

$\displaystyle K_{k^{\prime}}\equiv\mathbf{e}^{\prime}_{k}\cdot\mathbf{K}=\int{\rm d}^{3}\mathbf{x}^{\prime}\hat{\Psi}^{\dagger}(\mathbf{x}^{\prime})J_{k^{\prime}}\hat{\Psi}(\mathbf{x}^{\prime})$

(36)

are the projections of the fermion’s angular momentum onto the body-fixed axes.

Fermion states of axially-symmetric Hamiltonians $H_{\Psi}$ are written as $|K,\alpha\rangle$. Here $K$ denotes the projection of the angular momentum onto the nuclear symmetry axis, while $\alpha$ comprises the remaining quantum numbers energy, parity, and third component of isospin. Kramers’ degeneracy (i.e., time-reversal invariance) implies that the single-Fermion states come in degenerate pairs $|{\pm K},\alpha\rangle$, carrying identical quantum numbers $\alpha$ and, in particular, sharing the same energy $E_{|K|,\alpha}$. Thus, we have

	$\displaystyle H_{\Psi}|K,\alpha\rangle$	$\displaystyle=E_{|K|,\alpha}|K,\alpha\rangle\ ,$
	$\displaystyle\hat{K}_{z^{\prime}}|K,\alpha\rangle$	$\displaystyle=K|K,\alpha\rangle\ .$		(37)

We understand the band heads in Fig. 2 simply as energy eigenvalues of the fermion Hamiltonian $H_{\Psi}$ with a suitably chosen potential $V$. The $R$ symmetry of the nuclear ground state ensures that only a suitable linear combination of the states $|{\pm K},\alpha\rangle$ enters. The energies $E_{|K|,\alpha}$ of the band heads are of the scale $\Omega$. Thus, if we had spontaneous rather than emergent symmetry breaking, $\xi\to 0$ would hold, and the rotational bands on top of each band head would collapse.

The degrees of freedom of the rotor do not appear explicitly in Eq. (30). Conversely, the potential $V$ has no impact on the degrees of freedom of the rotor. The potential $V$ constitutes an implicit interaction between the rotor and the particle which is solely based on the fact that the potential is axially symmetric and defined in the body-fixed frame. That is consistent with emergent symmetry breaking which allows only a coupling to derivatives of Nambu-Goldstone bosons, in our case: the angular velocity. Such interactions – not yet contained in the Hamiltonian (30) – will appear as gauge couplings of the nucleon to the rotor. These are partly constrained by the nonlinear realization of rotational symmetry. They appear in universal form as a covariant derivative or as Coriolis terms. They can also be understood within an adiabatic approach that involves Berry phases.

The power-counting procedure for the rotor was worked out in Refs. Papenbrock (2011); Papenbrock and Weidenmüller (2014). We briefly present the arguments. We associate the low-energy scale $\xi$ with the rotor. Thus, the angular velocity scales as $\xi$,

	$\displaystyle|\mathbf{v}|$	$\displaystyle\sim$	$\displaystyle\xi\ ,$
	$\displaystyle\dot{\theta}$	$\displaystyle\sim$	$\displaystyle\xi\ ,$
	$\displaystyle\dot{\phi}$	$\displaystyle\sim$	$\displaystyle\xi\ ,$		(38)

and so does the Lagrangian (11) of the free rotor. That implies that its low-energy constant scales as

$$C_{0}\sim\xi^{-1}.$$

(39)

The spectrum of the free axially symmetric rotor forms a rotational band, see Eq. (16), and $C_{0}$ is the moment of inertia. Let us give examples. $C_{0}^{-1}\approx 1$ MeV for a light rotor such as ⁸Be, 0.5 MeV in ²⁴Mg, 0.2 MeV in the neutron-rich nucleus ³⁴Mg, 30 keV for a rare earth nucleus, and only 15 keV for actinides. These are the smallest energy scales in the nuclei we consider. The breakdown energy $\Lambda$ for the rotor is set by excitations that are not part of its ground-state rotational band. This energy is about 17 MeV in ⁸Be, 4 MeV in ²⁴Mg, 1 MeV in rare earth nuclei, and about 0.5 MeV in actinides. Thus, $\Lambda\gg\xi$ in all cases.

The subleading correction to the rotor Lagrangian (11) is

$$C_{2}(\mathbf{v}^{2})^{2}\ .$$

(40)

At the breakdown scale, i.e., when $|\mathbf{v}|\sim\Lambda$, the term (40) is by definition equal in importance to the leading-order Lagrangian (11). That yields

$$C_{2}\sim\xi^{-1}\Lambda^{-2}\ .$$

(41)

At low energy where $|\mathbf{v}|\sim\xi$, the term (40) yields a contribution $\sim\xi^{3}/\Omega^{2}$ to the total Lagrangian, and this is suppressed by $\xi^{2}/\Omega^{2}\ll 1$ compared to the leading term (11). That argument establishes the power-counting procedure for the rotor: the energy scale $\xi$ is associated with rotational bands. Corrections to the leading-order term come in powers of $\xi/\Lambda$.

We turn to the energy scales of the Hamiltonian (30) of the nucleon. The energy scale $\Omega$ is set by the mean level spacing of the single-nucleon states, i.e., by the spacing of band-head energies $E_{|K|\alpha}$ in odd-mass nuclei, see Eq. (II.2). That scale is about 1.7 MeV in ⁹Be, (given by the energy difference of the $3/2^{-}$ ground state and the excited $1/2^{+}$ band head), 0.6 MeV in ²⁵Mg, (given by the energy difference between the $5/2^{+}$ ground state and the excited $1/2^{+}$ band head), and amounts to some hundreds of keV in rare earth nuclei, and to tens to hundreds of keV for actinides. In most odd-mass deformed nuclei we have $\Omega\gtrsim\xi$, and in many cases one even finds $\Omega\gg\xi$, see ²³⁹Pu in Fig. 2 as an example. In such cases $\Omega$ is only about a factor two or three away from the breakdown scale $\Lambda$, and the separation between $\Omega$ and $\Lambda$ is marginal so that $\Omega\lesssim\Lambda$ but not $\Omega\ll\Lambda$. In such cases, the power counting uses both small expansion parameters $\xi/\Lambda$ and $\xi/\Omega$, and it is difficult to decide which is the more important one. If we had the ambition to construct an EFT for the nucleon potential in the Hamiltonian (30) we would have to deal, in addition, with an expansion in powers of $\Omega/\Lambda$, but we do not pursue this task in the present paper. In some nuclei such as ¹⁸⁷Os discussed below, we have $\Omega\sim\xi$ so that $\xi,\Omega\ll\Lambda$. Then, the power counting is in $\xi/\Lambda$. [We avoid the equivalent parameter $\Omega/\Lambda$ as that might incorrectly suggest a systematic expansion of the nucleon potential in the Hamiltonian (30).]

In any case, the separation of scales allows us to construct an EFT that systematicaly improves the energies and states in rotational bands. We note that there are many states in odd-mass nuclei that do not result from coupling a nucleon to the ground-state band of the even-even nucleus (but rather from coupling to excited band heads of the even-even nucleus). Such states fall outside the purview of the EFT we aim to construct. Including such effects would require us to introduce fields that describe the non-rotational excitations of the rotor.

It would be desirable to construct the potential $V$ in the fermion Hamiltonian (30) in a similarly systematic fashion. We briefly illuminate the difficulties in doing so for halo rotors, i.e., odd-mass nuclei where the nucleon is weakly bound to the even-even core. Examples are ⁹Be (with a neutron separation energy of about 1.7 MeV) and neutron-rich magnesium isotopes with separation energies below 1 MeV. In these cases, the fermion’s de-Broglie wave length exceeds the rotor’s size, and a derivative expansion of the potential seems appropriate. The potential $V$ must be axially symmetric. Total spin $\hat{\mathbf{S}}^{2}=3/4$ and its projection $\hat{S}_{z^{\prime}}^{2}=1/4$ are trivial constants, while $\hat{K}_{z^{\prime}}$ is the nontrivial conserved quantity and can be used to classify the fermion’s wave functions. Thus, we can parameterize the potential as

	$\displaystyle V$	$\displaystyle=$	$\displaystyle v_{01}\delta(\mathbf{r}^{\prime})$		(42)
		$\displaystyle+$	$\displaystyle v_{11}\nabla_{\perp}\delta(\mathbf{r}^{\prime})\cdot\nabla_{\perp}+v_{12}\partial_{z^{\prime}}\delta(\mathbf{r}^{\prime})\partial_{z^{\prime}}$
		$\displaystyle+$	$\displaystyle v_{13}\left[\nabla^{2}_{\perp}\delta(\mathbf{r}^{\prime})+\delta(\mathbf{r}^{\prime})\nabla^{2}_{\perp}\right]$
		$\displaystyle+$	$\displaystyle v_{14}\left[\partial_{z^{\prime}}^{2}\delta(\mathbf{r}^{\prime})+\delta(\mathbf{r}^{\prime})\partial_{z^{\prime}}^{2}\right]$
		$\displaystyle+$	$\displaystyle v_{15}\left(\hat{\sigma}_{x^{\prime}}\partial_{x^{\prime}}+\hat{\sigma}_{y^{\prime}}\partial_{y^{\prime}}\right)\delta(\mathbf{r}^{\prime})\left(\hat{\sigma}_{x^{\prime}}\partial_{x^{\prime}}+\hat{\sigma}_{y^{\prime}}\partial_{y^{\prime}}\right)$
		$\displaystyle+$	$\displaystyle v_{16}\big{[}\left(\hat{\sigma}_{x^{\prime}}\partial_{x^{\prime}}+\hat{\sigma}_{y^{\prime}}\partial_{y^{\prime}}\right)\hat{\sigma}_{z^{\prime}}\partial_{z^{\prime}}\delta(\mathbf{r}^{\prime})$
			$\displaystyle\,\,\,+\,\,\,\,\delta(\mathbf{r}^{\prime})\hat{\sigma}_{z^{\prime}}\partial_{z^{\prime}}\left(\hat{\sigma}_{x^{\prime}}\partial_{x^{\prime}}+\hat{\sigma}_{y^{\prime}}\partial_{y^{\prime}}\right)\big{]}$
		$\displaystyle+$	$\displaystyle\cdots\ .$

Here, $\nabla_{\perp}\equiv r^{-1}\nabla_{\Omega}$. In Eq. (42) we did not present all second-order derivatives, and higher-order derivatives are missing as well. If the fermion has quantum numbers $J^{\pi}={1\over 2}^{+}$, the leading-order contribution consists solely of the $v_{01}$ contact coupling. For $J^{\pi}={3\over 2}^{-}$ or ${1\over 2}^{-}$ states (e.g. for the ground state and excited band head at about 2.8 MeV, respectively, in ⁹Be), second-order derivatives in the potential $V$ must enter. In the latter case, one also needs to employ a potential that breaks spherical symmetry down to axial symmetry, thus lifting the four-fold degeneracy of a $p_{3/2}$ orbital in the body-fixed frame. The considerable number of low-energy coefficients then requires that a significant amount of data is available. In practice, one would like to adjust to scattering data (and make predictions for spectra and transitions), but those are rare. Odd-mass neutron-rich isotopes of magnesium, for instance, are expected to have ${5\over 2}^{+}$ ground states. This would require us to carry the expansion (42) to even higher order, and the scarcity of data in rare isotopes would prohibit us to follow such an EFT approach. It is, therefore, more practical to assume that the Hamiltonian (30) is already in diagonal form, with low-energy eigenstates as given in Eq. (II.2) fitted to the data. In other words, we take the single-particle energies of the fermion from data. This is somewhat similar in spirit to halo EFT Hammer et al. (2017) where each state of the core is represented as a separate field and simply adjusted to data.

The resulting EFT involves – in leading order – terms of order $\xi$ and $\Omega$. Subleading corrections are suppressed by factors of $\xi/\Lambda$ (or $\xi/\Omega$ provided that $\xi\ll\Omega$ holds). This EFT does not provide us with an expansion in powers of $\Omega/\Lambda$, because we do not construct such an expansion of the potential $V$.

We deal with emergent symmetry breaking. Thus, the nucleon can couple to the rotor only derivatively, i.e., via the angular velocity $\mathbf{v}$. All terms allowed by the symmetries must be considered. At face value, the resulting velocity-dependent couplings are well known. They involve – in the body-fixed frame – Coriolis forces. However, the essential physical argument for the couplings is more subtle and profound. The coupling terms are gauge couplings that involve Berry phases (or geometrical phases). Such phases occur in many quantum systems Simon (1983); Berry (1984). While originally conceived for systems that undergo a time-dependent adiabatic motion, they may also occur where “fast” degrees of freedom have been removed or integrated out, and where one is only interested in the remaining “slow” degrees of freedom Kuratsuji and Iida (1985). A well-known example from molecular physics is the Born-Oppenheimer approximation. Here, Berry phases and the corresponding gauge potentials enter the dynamics of the nuclei of the molecule once the electronic degrees of freedom have been removed. That leads to the molecular Aharonov-Bohm effect Mead and Truhlar (1979); Wilczek and Zee (1984). For the diatomic molecule, some details are presented in Refs. Jackiw (1986); Bohm et al. (1992); Rho (1994). In the present case, the fact that the nucleon is much faster than the slowly rotating core shows that gauge potentials play a role. Likewise, gauge potentials are a general feature of systems where a separation between rotational and intrinsic degrees of freedom is being made, and different conventions for this separation differ by gauge transformations Littlejohn and Reinsch (1997).

The non-linear realization of rotational invariance requires that, in the body-fixed system, we have to replace the time derivative $i\partial_{t}$ in the Lagrangian (28) by the covariant derivative Papenbrock (2011) (see Appendices C and D)

$\displaystyle iD_{t}$

$\displaystyle\equiv$

$\displaystyle i\partial_{t}+\dot{\phi}\cos\theta\left[J_{z^{\prime}},\cdot\right]\ .$

(43)

Here, the commutator’s second argument is left open. The last term of the covariant derivative accounts for Coriolis effects in the body-fixed system. It is present even if the Lagrangian in the body-fixed system does not depend on time explicitly, i.e., even if the partial time derivative vanishes. In the Lagrangian (28) that yields

		$\displaystyle\int{\rm d}^{3}\mathbf{x}^{\prime}\hat{\Psi}^{\dagger}(\mathbf{x}^{\prime})iD_{t}\hat{\Psi}(\mathbf{x}^{\prime})$
		$\displaystyle=\int{\rm d}^{3}\mathbf{x}^{\prime}\hat{\Psi}^{\dagger}(\mathbf{x}^{\prime})i\partial_{t}\hat{\Psi}(\mathbf{x}^{\prime})+\mathbf{v}\cdot\left(\mathbf{e}_{\phi}\cot\theta\hat{K}_{z^{\prime}}\right)\ .$		(44)

We have factored out the angular velocity $\mathbf{v}$, see Eq. (9), and we have used Eq. (36). This naturally introduces the gauge potential

$\displaystyle\mathbf{A}_{\rm a}(\theta,\phi)\equiv\mathbf{e}_{\phi}\cot\theta\hat{K}_{z^{\prime}}$

(45)

which couples the rotor to the nucleon via $\mathbf{v}\cdot\mathbf{A}_{\rm a}$. Here we borrow the expression gauge potential from electrodynamics. In the parlance of differential geometry, the field $\mathbf{A}_{\rm a}$ is a connection. The term $\mathbf{v}\cdot\mathbf{A}_{\rm a}$ scales as $\xi$, i.e. the gauge potential is dimensionless and of order one. Thus, the gauge term is as important as the Lagrangian (11) of the free rotor and enters in leading order.

The gauge potential (45) is singular at the north and south poles of the unit sphere. Single-valuedness of the wave function for the rotor requires that the eigenvalues $K$ of $\hat{K}_{z^{\prime}}$ be integer or half integer. That is obviously guaranteed for the fermion for which $K=\pm{1\over 2}$, $\pm{3\over 2}$, $\cdots$. We compute the corresponding magnetic field (or the Berry curvature in differential geometry) and find

$\displaystyle\mathbf{B}_{\rm a}(\theta,\phi)$

$\displaystyle\equiv$

$\displaystyle\nabla_{\Omega}\times\mathbf{A_{\rm a}}=-\mathbf{e}_{r}\hat{K}_{z^{\prime}}\ .$

(46)

This is the field of a Dirac monopole on the unit sphere and clearly exhibits spherical symmetry Fierz (1944); Wu and Yang (1976), in contrast to the gauge potential (45) whose rotational invariance is not obvious. As shown in App. E.2, the effect of a rotation on the gauge potential (45) can be reversed by a gauge transformation.

To see that the field (46) is indeed a monopole field we take a detour and consider a sphere of radius $R$ of the size of the nucleus, with angular and radial coordinates $(\theta,\phi)$ and $\rho$, respectively. We neglect excitations of the sphere in the radial direction as these relate to vibration with energies at the breakdown scale and put $\rho=R$. Then angular velocities and the vector potential (45) are multiplied with $R$ and $R^{-1}$, respectively. The usual differential operator $\mathbf{e}_{r}\partial_{R}+R^{-1}\nabla_{\Omega}$ then shows that we deal indeed with a monopole field.

The gauge potential (45) is intimately linked to the geometry of the sphere, i.e. the coset space SO(3)/SO(2). To see this, we consider a sequence of three rotations (around space-fixed axes) that take the rotor from a point $A$ on the unit sphere to a point $B$, then from $B$ to a point $C$, and finally from point $C$ back to the point $A$. We assume that the three rotations are around three distinct axes. This ensures that the triangle $ABC$ on the sphere has a finite solid angle (or area). It is clear that at least two of the three rotations will also induce rotations of the fermion field around the body-fixed $z^{\prime}$-axis. While the rotor has returned to its original configuration after the sequence of the three rotations, the fermion’s configuration has been changed by a finite rotation around the body-fixed $z^{\prime}$-axis. Inspection shows that the rotation angle, i.e. the phase acquired by the fermion with spin projection $K$, is equal to the solid angle of the enclosed loop times the spin projection $K$. Dynamically this phase is acquired because of the monopole magnetic field.

Another gauge coupling, permitted in the framework of our EFT, is

$\displaystyle g\mathbf{v}\cdot\left(\mathbf{e}_{r}\times\mathbf{K}\right)\ .$

(47)

Here, $g$ is a dimensionless coupling constant. Its natural size is of order unity, and the contribution (47) scales as $\xi$ and enters at leading order. The corresponding gauge potential is

$\displaystyle\mathbf{A}_{\rm n}(\theta,\phi)=g\mathbf{e}_{r}\times\mathbf{K}=g\left(\mathbf{e}_{\phi}\hat{K}_{x^{\prime}}-\mathbf{e}_{\theta}\hat{K}_{y^{\prime}}\right)\ .$

(48)

This vector potential contains non-commuting operators and, therefore, constitutes a non-Abelian gauge potential. The corresponding magnetic field is

	$\displaystyle\mathbf{B}_{\rm n}(\theta,\phi)$	$\displaystyle\equiv$	$\displaystyle\nabla_{\Omega}\times\mathbf{A}_{\rm n}-i\mathbf{A}_{\rm n}\times\mathbf{A}_{\rm n}$		(49)
		$\displaystyle=$	$\displaystyle g\mathbf{e}_{r}\cot\theta\hat{K}_{x^{\prime}}+g^{2}\mathbf{e}_{r}\hat{K}_{z^{\prime}}\ .$

Taken by itself, this magnetic field is not invariant under rotations. However, the total gauge potential is

$$\mathbf{A}_{\rm tot}=\mathbf{A}_{\rm a}+\mathbf{A}_{\rm n}\ ,$$

(50)

and the corresponding magnetic field

	$\displaystyle\mathbf{B}_{\rm tot}$	$\displaystyle\equiv$	$\displaystyle\nabla_{\Omega}\times\mathbf{A}_{\rm tot}-i\mathbf{A}_{\rm tot}\times\mathbf{A}_{\rm tot}$		(51)
		$\displaystyle=$	$\displaystyle(g^{2}-1)\mathbf{e}_{r}\hat{K}_{z^{\prime}}$

is spherically symmetric. Again, we deal with a magnetic monopole. However, in contrast to the field (46) its overall strength is not quantized because the non-Abelian vector potential (48) exhibits no singularities on the unit sphere and the coupling $g$ can therefore assume any real value.

To show how this gauge term relates to the Berry phase we observe that the states $|{\pm K},\alpha\rangle$ have the same energy, see Eq. (II.2). When the rotor moves along a closed loop on the unit sphere, a general interaction would mix the two degenerate states while transversing the loop. Thus, the initial and final fermion states could differ by a unitary transformation. The non-Abelian gauge potential (48) generates such a mixing for non-zero values of the coupling $g$.

The choice of the gauge potentials is not unique. At any orientation $(\theta,\phi)$ of the rotor’s symmetry axis, the body-fixed coordinate system is defined up to an arbitrary rotation around the $z^{\prime}$ axis. Thus, the intrinsic degrees of freedom (of the fermion in our case) depend on a convention which is arbitrary. Different choices of the body-fixed system lead to different expressions for the covariant derivative and to different gauge potentials Littlejohn and Reinsch (1997). These are related to each other by gauge transformations. Details are given in Appendix E.

Collecting the leading-order results from the previous Sections, we find that the Lagrangian is given by

	$\displaystyle L$	$\displaystyle=$	$\displaystyle{C_{0}\over 2}\mathbf{v}^{2}+\mathbf{v}\cdot\left(\mathbf{A}_{\rm a}+\mathbf{A}_{\rm n}\right)+L_{\Psi}$		(52)
		$\displaystyle=$	$\displaystyle{C_{0}\over 2}\left(\dot{\theta}^{2}+\dot{\phi}^{2}\sin^{2}\theta\right)+g\left(\dot{\phi}\sin\theta\hat{K}_{x^{\prime}}-\dot{\theta}\hat{K}_{y^{\prime}}\right)$
			$\displaystyle+\dot{\phi}\cos\theta\hat{K}_{z^{\prime}}+L_{\Psi}\ .$

Here, $L_{\Psi}$ is defined in Eq. (28). The Legendre transform of the Lagrangian (52) yields the Hamiltonian. For $L_{\psi}$ that was done in Section II.2. For the remaining variables the transformation is less tedious than might appear at first sight. The Lagrangian (52) is a quadratic form in the velocities $(\dot{\theta},\dot{\phi})$ and can be written as

$$L={1\over 2}\dot{\mathbf{q}}^{T}\hat{M}\dot{\mathbf{q}}+\mathbf{A}\cdot\dot{\mathbf{q}}\ ,$$

(53)

where ^T denotes the transpose. That Lagrangian has the Legendre transform

$$H={1\over 2}\left(\mathbf{q}-\mathbf{A}\right)^{T}\hat{M}^{-1}\left(\mathbf{q}-\mathbf{A}\right)\ .$$

(54)

Here, $\hat{M}$ is a “mass” matrix and $\hat{M}^{-1}$ denotes its inverse. In the present case the canonical momenta are

	$\displaystyle p_{\phi}$	$\displaystyle\equiv$	$\displaystyle{\partial L\over\partial\dot{\phi}}=C_{0}v_{\phi}\sin\theta+\cos\theta\hat{K}_{z^{\prime}}+g\sin\theta\hat{K}_{x^{\prime}}\ ,$
	$\displaystyle p_{\theta}$	$\displaystyle\equiv$	$\displaystyle{\partial L\over\partial\dot{\theta}}=C_{0}v_{\theta}-g\hat{K}_{y^{\prime}}\ .$		(55)

Here, we employed Eq. (II.1.1). The Hamiltonian becomes

	$\displaystyle H$	$\displaystyle=$	$\displaystyle H_{\psi}+\frac{1}{2C_{0}}\left(p_{\theta}+g\hat{K}_{y^{\prime}}\right)^{2}$		(56)
			$\displaystyle+\frac{1}{2C_{0}}\left({p_{\phi}\over\sin\theta}-\cot\theta\hat{K}_{z^{\prime}}-g\hat{K}_{x^{\prime}}\right)^{2}\ .$

Here, the fermion Hamiltonian $H_{\psi}$ is given in Eq. (30). The momentum $p_{\phi}$ is conserved because $\phi$ does not appear in the Hamiltonian (56). Combining two of the canonical momenta (IV.1) into

$$\mathbf{p}=p_{\theta}\mathbf{e}_{\theta}+{p_{\phi}\over\sin\theta}\mathbf{e}_{\phi}\ ,$$

(57)

we find

$$\mathbf{p}=C_{0}\mathbf{v}+\mathbf{A}_{\rm a}+\mathbf{A}_{\rm n}\ .$$

(58)

Using that, we write the Hamiltonian (56) in compact form as

$\displaystyle H=H_{\psi}+{1\over 2C_{0}}\left(\mathbf{p}-\mathbf{A}_{\rm a}-\mathbf{A}_{\rm n}\right)^{2}=H_{\psi}+\frac{1}{2}C_{0}\mathbf{v}^{2}\ .$

Replacing the canonical momenta by the total angular momentum simplifies the Hamiltonian and establishes the connection to Eq. (2). In the present Subsection we introduce the total angular momentum on an intuitive basis. A derivation based on Noether’s theorem is given in App. B.3.

The total angular momentum

$$\mathbf{I}=\mathbf{I}_{\perp}+\mathbf{I}_{z^{\prime}}$$

(60)

is the sum of the angular momentum of the fermion,

$\displaystyle\mathbf{I}_{z^{\prime}}=\mathbf{e}^{\prime}_{z}\hat{K}_{z^{\prime}}\ ,$

(61)

which points in the direction of the symmetry axis, and that of the rotor,

	$\displaystyle\mathbf{I}_{\perp}$	$\displaystyle=$	$\displaystyle I_{x^{\prime}}\mathbf{e}^{\prime}_{x}+I_{y^{\prime}}\mathbf{e}^{\prime}_{y}$		(62)
		$\displaystyle=$	$\displaystyle p_{\theta}\mathbf{e}^{\prime}_{y}-\left({p_{\phi}\over\sin\theta}-\cot\theta\hat{K}_{z^{\prime}}\right)\mathbf{e}^{\prime}_{x}$
		$\displaystyle=$	$\displaystyle\mathbf{r}\times\left(\mathbf{p}-\mathbf{A}_{\rm a}\right)\ .$

which is perpendicular to it. In the last line of Eq. (62) we have used Eq. (58), see also Refs. Fierz (1944); Wu and Yang (1976). The term $\mathbf{r}\times\mathbf{p}$ is the angular momentum $C_{0}\mathbf{r}\times\mathbf{v}$ of the rotor. The gauge potential $\mathbf{A}_{a}$ is not manifestly invariant under rotations but can be made so via a gauge transformation Fierz (1944). That causes the correction (61) in the direction of $\mathbf{e}^{\prime}_{z}$. The equality

$$I_{z}\equiv\mathbf{e}_{z}\cdot\mathbf{I}=p_{\phi}\ .$$

(63)

shows that the conserved momentum $p_{\phi}$ is the usual angular momentum with respect to the space-fixed $z$ axis. We use Eqs. (62) and (60) to express the angular velocity in terms of the angular momentum. That yields

$\displaystyle\mathbf{v}=-{1\over C_{0}}\left(\mathbf{e}^{\prime}_{z}\times\mathbf{I}+\mathbf{A}_{\rm n}\right)\ .$

(64)

Using that in the rotational energy $(C_{0}/2)\mathbf{v}^{2}$, we arrive at the Hamiltonian

	$\displaystyle H$	$\displaystyle=$	$\displaystyle H_{\Psi}+{g^{2}\over 2C_{0}}\left(\hat{K}_{x^{\prime}}^{2}+\hat{K}_{y^{\prime}}^{2}\right)$		(65)
			$\displaystyle+\frac{\mathbf{I}^{2}-\hat{K}_{z^{\prime}}^{2}}{2C_{0}}+{g\over C_{0}}\left(I_{x^{\prime}}\hat{K}_{x^{\prime}}+I_{y^{\prime}}\hat{K}_{y^{\prime}}\right)\ .$

The term proportional to $g^{2}$ in Eq. (65) might be absorbed into $H_{\Psi}$ (and then be dropped). The rotational part displayed in the second line of Eq. (65) corresponds to the rotor model (2) for the special case of axial symmetry, i.e. for $C_{x^{\prime}}=C_{y^{\prime}}$.

The square of the total angular momentum is given by

	$\displaystyle\mathbf{I}^{2}$	$\displaystyle=$	$\displaystyle p_{\theta}^{2}+{1\over\sin^{2}\theta}\left(p_{\phi}-\cos\theta\hat{K}_{z^{\prime}}\right)^{2}+\hat{K}_{z^{\prime}}^{2}$		(66)
		$\displaystyle=$	$\displaystyle p_{\theta}^{2}+{1\over\sin^{2}\theta}\left(p_{\phi}^{2}-2p_{\phi}\cos\theta\hat{K}_{z^{\prime}}+\hat{K}_{z^{\prime}}^{2}\right)\ .$

Upon quantization this operator, its projection $I_{z^{\prime}}=\hat{K}_{z^{\prime}}$ onto the $z^{\prime}$-axis [see Eq. (61)], and its projection $I_{z}=p_{\phi}$ onto the $z$-axis [see Eq. (63)] form a commuting set of operators. Details are presented in Appendix B.3.

Simplifying the notation used in Eq. (II.2) we denote the ground state of the fermionic part of the Hamiltonian (65) as $|K\rangle$. We calculate the eigenfunctions of the rotor part of the Hamiltionian (65) by determining the eigenfunctions of $\mathbf{I}^{2}$, $I_{z}$, and $I_{z^{\prime}}$. For states $|{\pm K}\rangle$ we have

$\displaystyle I_{z^{\prime}}|{\pm K}\rangle=\hat{K}_{z^{\prime}}|{\pm K}\rangle={\pm K}|{\pm K}\rangle\ .$

(67)

The quantization proceeds as in Section II. The eigenfunctions of $I_{z}=p_{\phi}=-i\partial_{\phi}$ are

$\displaystyle I_{z}e^{-iM\phi}=-Me^{-iM\phi}\ .$

(68)

The negative eigenvalue is chosen here to be consistent with chapter 4.2 of Ref. Varshalovich et al. (1988). The eigenfunctions of the square of the total angular momentum operator can be written either in terms of Wigner $d$ functions or in terms of Wigner $D$ functions (see chapter 4 of Ref. Varshalovich et al. (1988)). These are related by

$$D_{M,{M^{\prime}}}^{I}(\phi,\theta,0)=e^{-iM\phi}d^{I}_{M,M^{\prime}}(\theta)\ .$$

(69)

For $I\geq|M|,|K|$ we have

	$\displaystyle\mathbf{I}^{2}D_{M,{\mp K}}^{I}(\phi,\theta,0)|{\pm K}\rangle$		(70)
	$\displaystyle=I(I+1)D_{M,{\mp K}}^{I}(\phi,\theta,0)|{\pm K}\rangle\ .$		(71)

For the Hamiltonian (65) that implies

		$\displaystyle\frac{\mathbf{I}^{2}-\hat{K}_{z^{\prime}}^{2}}{2C_{0}}D_{M,{\mp K}}^{I}(\phi,\theta,0)|{\pm K}\rangle$
		$\displaystyle=\frac{I(I+1)-K^{2}}{2C_{0}}D_{M,{\mp K}}^{I}(\phi,\theta,0)|{\pm K}\rangle\ .$		(72)

Discrete symmetries of the rotor-plus-fermion system may select a definite linear combination of the states $|{\pm K}\rangle$. These share the absolute value $|K|$ and have the same energy $E_{|K|}$ [see Eq. (II.2)]. Combining $E_{|K|}$ with Eq. (IV.3) yields

$$E(I)=E_{|K|}+\frac{I(I+1)-K^{2}}{2C_{0}}\ .$$

(73)

The term linear in $g$ of the Hamiltonian (65) couples states $D^{I}_{M,{-K}}|K\rangle$ and $D^{I}_{M,{-K\mp 1}}|K\pm 1\rangle$. For most heavy nuclei where $\xi\ll\Omega$, the coupling of states with different values of $|K|$ is of subleading order and can be computed perturbatively. For $K=\pm 1/2$, however, the interaction couples the degenerate states $D^{I}_{M,{\mp{1\over 2}}}|{\pm{1\over 2}}\rangle$ and is, thus, of leading order $\xi$. For this case, eigenfunctions and eigenvalues are worked out in Appendix F. The result for the eigenvalues,

	$\displaystyle E(I,K)$	$\displaystyle=$	$\displaystyle E_{|K|}+\frac{I(I+1)-K^{2}}{2C_{0}}$		(74)
		$\displaystyle-$	$\displaystyle{g\over C_{0}}\delta_{|K|,{1\over 2}}(-1)^{I+{1\over 2}}\left(I+{1\over 2}\right)\ ,$

agrees with Eq. (22) when we express the constants $E_{|K|}$, $C_{0}$, and $g$ in terms of $E_{0}$, $A$, and $a$. The last term in Eq. (74) is known as the signature splitting. Equation (74) shows that for $g\neq 0$ the spectrum changes by about $g\xi$. For $|g|\gg 1$, the spectrum would resemble a rotational band only for $I\gtrsim|g|$. This confirms that the natural size of $g$ is of order unity.

States that differ by one unit in $K$ can also be coupled strongly by the term linear in $g$ in the Hamiltonian (65) provided $\Omega\approx\xi$. In that case, the spectrum can be calculated analytically using as a basis the eigenstates obtained for $g=0$ and taking account only of the two bandheads. The diagonalization of the $4\times 4$ matrix spanned by the states $|{\pm K}\rangle$ and $|{\pm(K+1)}\rangle$ yields the eigenvalues Kerman (1956)

	$\displaystyle E(I,K,K+1)$	$\displaystyle=$	$\displaystyle{1\over 2}\left[E(I,K)+E(I,K+1)\right]$
		$\displaystyle\pm$	$\displaystyle{1\over 2}\bigg{\{}\left[E(I,K)-E(I,K+1)\right]^{2}$
			$\displaystyle+4{\tilde{g}^{2}\over C_{0}^{2}}\left[I(I+1)-K(K+1)\right]\bigg{\}}^{1\over 2}\ .$

The energies $E(I,K)$ are given by Eq. (74), and $\tilde{g}\equiv g\langle K|\hat{K}_{-1}|K{+1}\rangle$ is a low-energy constant. The sign on the right-hand side of Eq. (IV.3) has to be chosen such that the energies $E(I,K)$ and $E(I,K+1)$ for the bands with quantum numbers $K$ and $K+1$, respectively, are obtained as $g\to 0$. In nuclei such as ^105,107Mo, groups of more than two band heads are closely spaced and strongly coupled. In such cases, a Hamiltonian matrix of larger dimension needs to be diagonalized.