Geometric phase of open paths and a geodesic-selection rule at a level degeneracy

To construct a connection suitable for treating open paths, we choose a base point $\hat{\mathbf{n}}_{0}\in S^{2}$ and seek a one-form $\mathcal{A}=\mathbf{A}\cdot d\hat{\mathbf{n}}$ whose circulation around a closed curve equals the enclosed solid angle. Since $\hat{\mathbf{n}}$ is a unit vector, $\mathrm{d}\hat{\mathbf{n}}$ lies in the tangent plane of $S^{2}$ and satisfies $\hat{\mathbf{n}}\cdot\mathrm{d}\hat{\mathbf{n}}=0$. Consequently, only the tangential component of $\boldsymbol{A}$ contributes to $\mathcal{A}$: adding an arbitrary radial term $h(\hat{\mathbf{n}})\,\hat{\mathbf{n}}$ leaves the one-form unchanged, so we may choose $\boldsymbol{A}\cdot\hat{\mathbf{n}}=0$ without loss of generality. Two tangential directions remain, the polar (meridional) direction $\hat{\mathbf{e}}_{\theta}$ and the azimuthal one $\hat{\mathbf{e}}_{\phi}$, and we now show that only the latter carries physical content.

Consider first the component of $\boldsymbol{A}$ along $\hat{\mathbf{e}}_{\theta}$. Rotational symmetry about the $\hat{\mathbf{n}}_{0}$ axis restricts this component to the form $g(\theta)\,\hat{\mathbf{e}}_{\theta}$. Since $\hat{\mathbf{e}}_{\theta}\cdot\mathrm{d}\hat{\mathbf{n}}=\mathrm{d}\theta$, its contribution to the one-form is $g(\theta)\,\mathrm{d}\theta$, which is pure gauge: it leaves the curvature unchanged, since $\mathrm{d}[g(\theta)\,\mathrm{d}\theta]=0$. Setting $g\equiv 0$ is thus a choice of gauge rather than a restriction. We adopt it here because it makes the great circles through $\pm\hat{\mathbf{n}}_{0}$ parallel-transport paths: on such a circle $\mathrm{d}\hat{\mathbf{n}}$ points along $\hat{\mathbf{e}}_{\theta}$, so with $g\equiv 0$ a geodesic joining any point $\hat{\mathbf{n}}$ to the base point $\hat{\mathbf{n}}_{0}$ contributes nothing to the line integral. This property is established in Lemma 3.1 and underlies the geodesic-closure construction for open paths developed below.

The remaining tangential component lies along the azimuthal direction $\hat{\mathbf{e}}_{\phi}$ , which is parallel to $\hat{\mathbf{n}}_{0}\times\hat{\mathbf{n}}$. Unlike $g$, its coefficient does affect the curvature and cannot be chosen freely; it is fixed uniquely by Stokes’ theorem below. Rotational symmetry about the $\hat{\mathbf{n}}_{0}$ axis further requires this coefficient to depend only on the polar angle $\theta$, defined by $\cos\theta\equiv\hat{\mathbf{n}}_{0}\cdot\hat{\mathbf{n}}$. We therefore write

To determine $f$, take $C_{\theta}$ to be the circle at constant polar angle $\theta$ about the $\hat{\mathbf{n}}_{0}$ axis. The spherical cap it bounds has the solid angle

On $C_{\theta}$ the magnitude $|\hat{\mathbf{n}}_{0}\times\hat{\mathbf{n}}|=\sin\theta$ is constant, $\boldsymbol{A}$ is parallel to $\mathrm{d}\hat{\mathbf{n}}$, and the circumference is $2\pi\sin\theta$, so

Equating (5) and (6) through Stokes’ theorem gives $f(\theta)=(1-\cos\theta)/\sin^{2}\theta=1/(1+\cos\theta)$, and restoring $\cos\theta=\hat{\mathbf{n}}_{0}\cdot\hat{\mathbf{n}}$ yields the coordinate-free form,

Introduce spherical coordinates $(\theta,\phi)$ with $\hat{\mathbf{n}}_{0}$ at the north pole, i.e., $\hat{\mathbf{n}}_{0}=\hat{\mathbf{z}}$, so that $\hat{\mathbf{n}}=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$. With the right-handed orthonormal frame $\{\hat{\mathbf{n}},\hat{\mathbf{e}}_{\theta},\hat{\mathbf{e}}_{\phi}\}$ one has $\hat{\mathbf{n}}\times\hat{\mathbf{e}}_{\theta}=\hat{\mathbf{e}}_{\phi}$, $\hat{\mathbf{n}}\times\hat{\mathbf{e}}_{\phi}=-\hat{\mathbf{e}}_{\theta}$, and $\mathrm{d}\hat{\mathbf{n}}=\hat{\mathbf{e}}_{\theta}\,\mathrm{d}\theta+\sin\theta\,\hat{\mathbf{e}}_{\phi}\,\mathrm{d}\phi$. A short computation using $\hat{\mathbf{z}}\cdot\hat{\mathbf{e}}_{\phi}=0$ and $\hat{\mathbf{z}}\cdot\hat{\mathbf{e}}_{\theta}=-\sin\theta$ gives the one-form connection

The exterior derivative of this is the coordinate-free two-form curvature, $\mathrm{d}\big[(1-\cos\theta)\,\mathrm{d}\phi\big]=\sin\theta\,\mathrm{d}\theta\wedge\mathrm{d}\phi\equiv\mathrm{d}\Omega$, so for a closed curve $C=\partial\Sigma$,

Before proceeding we fix the relation to the familiar Berry phase. For the two-level Hamiltonian in equation (1), we follow the eigenstate $|\psi_{-}(\hat{\mathbf{n}})\rangle$ of energy $-|\boldsymbol{d}|$ aligned with $\hat{\mathbf{n}}$. A standard computation of its Berry connection in the gauge with base point $\hat{\mathbf{n}}_{0}$ gives the familiar monopole form $a_{B}=-\tfrac{1}{2}(1-\cos\theta)\,\mathrm{d}\phi$, which by equation (8) is nothing but

The Berry connection and the one-form studied here therefore differ only by the constant factor $-\tfrac{1}{2}$, and we work with $\mathcal{A}$ in this paper.

The curvature derived above is that of a monopole located at $\mathbf{d}=0$. A gauge potential for a monopole cannot be chosen smoothly over the entire sphere. In the gauge defined by $\hat{\mathbf{n}}_{0}$, the connection (7) is regular everywhere except at $\hat{\mathbf{n}}=-\hat{\mathbf{n}}_{0}$, where its denominator vanishes. This point is the intersection of the unit sphere with the Dirac string [4], an unphysical line singularity of the monopole gauge potential in the full parameter space. Changing $\hat{\mathbf{n}}_{0}$ moves this singularity and therefore corresponds to a change of gauge. For a closed curve, $\Omega$ is independent of $\hat{\mathbf{n}}_{0}$ by equation (9) (exactly so if the curve does not cross the string, and up to an integer multiple of $4\pi$ otherwise). For an open curve, however, a change of base point acts as a gauge transformation $a_{B}\to a_{B}+\mathrm{d}\chi$ that leaves a boundary term $\int_{C}\mathrm{d}\chi=\chi(\text{end})-\chi(\text{start})\neq 0$. The choice of base point is thus an additional gauge—a path-dependent term—for open curves, and the value of $\Omega[C]$ depends on it. It is therefore natural to tie the base point to the curve itself. In the situation of interest this choice is essentially forced: taking $\hat{\mathbf{n}}_{0}=+\hat{\mathbf{v}}$ (the starting direction) places the string at the opposite endpoint $-\hat{\mathbf{v}}$, so that $C$ terminates exactly on the string. As we show in section 3, this is precisely what renders the non-uniqueness of the closing geodesic harmless, and we adopt $\hat{\mathbf{n}}_{0}=\pm\hat{\mathbf{v}}$ henceforth (Definition 3.2).

The line integral (11) admits a purely geometric reading. Discretising $C$ as $\{\hat{\mathbf{n}}^{(0)}=\hat{\mathbf{n}}_{0},\allowbreak\hat{\mathbf{n}}^{(1)},\allowbreak\dots,\allowbreak\hat{\mathbf{n}}^{(M)}\}$ and letting $\Delta_{i}$ be the signed solid angle, seen from the centre of the sphere, of the spherical triangle $(\hat{\mathbf{n}}_{0},\hat{\mathbf{n}}^{(i)},\hat{\mathbf{n}}^{(i+1)})$, one has $\Omega[C]=\lim_{\max|\mathrm{d}\hat{\mathbf{n}}|\to 0}\sum_{i}\Delta_{i}$. Indeed, the signed solid angle of a spherical triangle is given by the van Oosterom–Strackee formula [13],

and inserting $\boldsymbol{\alpha}=\hat{\mathbf{n}}_{0}$, $\boldsymbol{\beta}=\hat{\mathbf{n}}$, $\boldsymbol{\gamma}=\hat{\mathbf{n}}+\mathrm{d}\hat{\mathbf{n}}$ and expanding to first order reproduces the integrand of equation (11). The “solid angle seen from the centre”, the “base-point triangulation”, and the “one-form line integral” therefore coincide.

The point $\boldsymbol{d}(t_{0})=\mathbf{0}$ (the degeneracy) is, as noted in section 1, where $\hat{\mathbf{n}}$ jumps between the antipodes $\mp\hat{\mathbf{v}}$; the “degeneracy” and the “antipodal endpoints of the $\hat{\mathbf{n}}$-trajectory” are the same event. Set $\boldsymbol{d}(t_{0})=\mathbf{0}$, $\boldsymbol{v}\equiv\dot{\boldsymbol{d}}(t_{0})\neq\mathbf{0}$, $\boldsymbol{a}\equiv\ddot{\boldsymbol{d}}(t_{0})$ and $s\equiv t-t_{0}$. Then $\boldsymbol{d}(t_{0}+s)=\boldsymbol{v}s+\tfrac{1}{2}\boldsymbol{a}s^{2}+O(s^{3})$,

where

so that $\hat{\mathbf{n}}\to\pm\hat{\mathbf{v}}$ as $s\to 0^{\pm}$: the direction field jumps between antipodes.

Assume $\boldsymbol{a}_{\perp}\neq\mathbf{0}$ (equivalently $\boldsymbol{a}\nparallel\boldsymbol{v}$), i.e., that the curve $\mathbf{d}(t)$ has nonzero curvature at $t_{0}$. Introduce the Frenet frame at $t_{0}$, with orthonormal basis consisting of the tangent $\hat{\mathbf{v}}$, principal normal $\hat{\mathbf{a}}_{\perp}=\boldsymbol{a}_{\perp}/|\boldsymbol{a}_{\perp}|$, and binormal $\hat{\mathbf{b}}=\hat{\mathbf{v}}\times\hat{\mathbf{a}}_{\perp}$, and let $\Pi_{\rm osc}\equiv\operatorname{span}\{\hat{\mathbf{v}},\hat{\mathbf{a}}_{\perp}\}=\operatorname{span}\{\boldsymbol{v},\boldsymbol{a}\}$ be the osculating plane. The geometry is illustrated in figure 1. Since the $\hat{\mathbf{b}}$-component of $\boldsymbol{d}$ is $O(s^{3})$, the curve lies in $\Pi_{\rm osc}$ to second order. Decomposing $\boldsymbol{d}$ in the Frenet frame, $\boldsymbol{d}(s)=\xi(s)\,\hat{\mathbf{v}}+\eta(s)\,\hat{\mathbf{a}}_{\perp}+\zeta(s)\,\hat{\mathbf{b}}$, with components

To leading order the curve is the parabola

with vertex at the origin, opening towards $+\hat{\mathbf{a}}_{\perp}$.

We now fix the angular variables used below. The pole of the spherical coordinates has already been tied to the curve, $\hat{\mathbf{n}}_{0}=\hat{\mathbf{v}}$; we further choose the azimuthal origin in the transverse plane spanned by $(\hat{\mathbf{a}}_{\perp},\hat{\mathbf{b}})$ so that $\phi=0$ along $+\hat{\mathbf{a}}_{\perp}$ and $\phi=\pi/2$ along $+\hat{\mathbf{b}}$. The transverse part of $\hat{\mathbf{n}}$—its projection onto this plane, $\hat{\mathbf{n}}-(\hat{\mathbf{n}}\cdot\hat{\mathbf{v}})\,\hat{\mathbf{v}}=(\eta\,\hat{\mathbf{a}}_{\perp}+\zeta\,\hat{\mathbf{b}})/|\boldsymbol{d}|$ with $\eta,\zeta$ as in equation (15)—then has azimuth $\phi=\atantwo(\zeta,\eta)$, which is what we mean by the transverse azimuth of $\hat{\mathbf{n}}$. We call the trajectory of the unregularized direction field $\hat{\mathbf{n}}(s)$ of equation (13) the bare curve, in contradistinction to the regularized field introduced later, and write $\phi_{\rm bare}$ for its azimuth.

The coefficient $b_{3}$ is proportional to the Frenet torsion of $\boldsymbol{d}$ at the crossing, so the azimuth is frozen at $0$ precisely for the planar (reflection-symmetric) case of Corollary 4.4. The parabola in equation (16) and Lemma 3.5 together carry the two facts on which the rest of the paper turns. First, both branches ($s\gtrless 0$) of the curve lie on the $+\hat{\mathbf{a}}_{\perp}$ side, since the coefficient $\tfrac{1}{2}s^{2}$ in $\eta$ is positive irrespective of the sign of $s$; the transverse direction of $\hat{\mathbf{n}}$ therefore agrees with $+\hat{\mathbf{a}}_{\perp}$ to leading order on both sides of the degeneracy, deviating only at $O(s)$ through the torsion, and by Lemma 3.5 the jump of $\hat{\mathbf{n}}$ from $-\hat{\mathbf{v}}$ to $+\hat{\mathbf{v}}$ leaves no concentrated contribution to the solid angle. Second, the tangent directions of the parabola at $\pm\hat{\mathbf{v}}$ single out, among the infinitely many geodesics joining the antipodes, the one great circle actually tangent to the curve—the osculating-plane great circle; this tangency is a second-order statement, unaffected by the torsion. The identity of the “geodesic that realizes the enclosed solid angle”, anticipated in Remark 3.4, is thus supplied by the curvature at the degeneracy.

Let us introduce the regularization $\boldsymbol{d}_{\epsilon}\equiv\boldsymbol{d}+\epsilon\hat{\mathbf{u}}$ with $\epsilon>0$ and $\hat{\mathbf{u}}\in S^{2}$ having a nonzero transverse component, i.e., $\hat{\mathbf{u}}\nparallel\hat{\mathbf{v}}$. Then $\boldsymbol{d}_{\epsilon}(t_{0})=\epsilon\hat{\mathbf{u}}\neq\mathbf{0}$ avoids the origin, and $\hat{\mathbf{n}}_{\epsilon}$ is a smooth closed curve. Decomposing as in equation (15),

where $\hat{\mathbf{u}}_{\perp}\equiv\hat{\mathbf{u}}-(\hat{\mathbf{u}}\cdot\hat{\mathbf{v}})\hat{\mathbf{v}}$. The transverse plane is spanned by $(\hat{\mathbf{a}}_{\perp},\hat{\mathbf{b}})$ with coordinates $(\eta_{\epsilon},\zeta_{\epsilon})$. Writing

so that $\hat{\mathbf{u}}_{\perp}=u(\cos\alpha\,\hat{\mathbf{a}}_{\perp}+\sin\alpha\,\hat{\mathbf{b}})$, the transverse components become

Geometrically, $\alpha$ is the angle specifying the direction of the transverse regularization $\hat{\mathbf{u}}_{\perp}$ within the $(\hat{\mathbf{a}}_{\perp},\hat{\mathbf{b}})$ plane, measured from the principal normal $\hat{\mathbf{a}}_{\perp}$; as we now show, it also parametrizes the closing geodesic: $\alpha=0$ corresponds to the osculating-plane great circle (the canonical geodesic), and $\alpha\neq 0$ to a great circle rotated by $\alpha$ about the $\hat{\mathbf{v}}$ axis (see figure 1).

The key structural fact is that the regularization contributes an $s$-independent transverse shift $\epsilon\hat{\mathbf{u}}_{\perp}$. To leading order the transverse trajectory is then the horizontal line $\zeta\simeq\epsilon u\sin\alpha$ traced as $s$ runs through the crossing: the point $(\eta_{\epsilon},\zeta_{\epsilon})$ comes in from large $\eta$, reaches its closest approach to the origin at $s=0$, where $\eta_{\epsilon}=\epsilon u\cos\alpha$, and recedes back to large $\eta$, staying at fixed height $\zeta\simeq\epsilon u\sin\alpha$ throughout. The azimuth is accordingly

According to equation (8), an azimuthal change $\mathrm{d}\phi$ contributes $(1-\cos\theta)\mathrm{d}\phi$ to the solid angle. Its contribution therefore depends strongly on the polar angle: the weight vanishes at the north pole $(\theta=0)$ and reaches its maximum value of $2$ at the south pole $(\theta=\pi)$. To determine the net contribution of the azimuthal turn in equation (21), we compare the characteristic $s$-scales over which $\theta_{\epsilon}$ and $\phi_{\epsilon}$ vary. The transverse vector entering equation (21) can be written as

which shows that the quadratic contribution becomes comparable to the magnitude of the regularization-induced transverse displacement at

The polar angle is governed by a different scale, which is best seen through its deviation from the nearer pole, $\delta\equiv\min(\theta_{\epsilon},\pi-\theta_{\epsilon})$. From $\tan\theta_{\epsilon}=\sqrt{\eta_{\epsilon}^{2}+\zeta_{\epsilon}^{2}}\,/\,|\xi_{\epsilon}|$ and $\tan\delta\simeq\delta$ for small $\delta$, $|\xi_{\epsilon}|\simeq|\boldsymbol{v}||s|$ to leading order, and the transverse part dominated by the regularization for $|s|\ll s_{\phi}$, $\sqrt{\eta_{\epsilon}^{2}+\zeta_{\epsilon}^{2}}\simeq\epsilon u$, the deviation obeys

Equation (24) yields the polar scale: the deviation is small precisely while $|s|\gg s_{\theta}\equiv\epsilon u/|\boldsymbol{v}|$, so that the polar angle stays pinned at its pole, $\theta_{\epsilon}\approx 0$ for $s>0$ and $\theta_{\epsilon}\approx\pi$ for $s<0$, whereas for $|s|\lesssim s_{\theta}$ the deviation is $O(1)$ and $\theta_{\epsilon}$ sweeps continuously from $\pi$ to $0$ as $s$ crosses the origin. The two scales are therefore well separated, $s_{\theta}\sim\epsilon\ll\sqrt{\epsilon}\sim s_{\phi}$, and this separation splits the crossing into three regions.

The three regions are distinguished by which of $\theta$ and $\phi$ is in motion. In region $\mathrm{I}$ ($s_{\phi}\ll|s|\ll 1$) the polar angle has already reached its pole while the azimuth still sits at $0$; in region $\mathrm{II}$ ($s_{\theta}\ll|s|\ll s_{\phi}$) the azimuth turns from $0$ to $\alpha$ while $\theta_{\epsilon}$ stays pinned at its pole; and in region $\mathrm{III}$ ($|s|\lesssim s_{\theta}$) $\theta_{\epsilon}$ sweeps from $\pi$ to $0$, the azimuth being already fixed at $\alpha$.

The essential point is that the azimuthal turn of region $\mathrm{II}$ takes place where the regularized curve passes the two poles $\pm\hat{\mathbf{v}}$, with $\theta$ settled at $0$ or $\pi$. Evaluating equation (24) at the azimuthal scale $|s|\sim s_{\phi}$ gives $\delta\sim\epsilon u/(|\boldsymbol{v}|\sqrt{\epsilon})=O(\sqrt{\epsilon})\to 0$, so expanding the monopole weight to second order in $\delta$, with $\theta_{\epsilon}=\delta$ on the $s>0$ branch and $\theta_{\epsilon}=\pi-\delta$ on the $s<0$ branch,

The same turn $\alpha$ is thus counted with weight $O(\epsilon)$ near one pole and weight $2-O(\epsilon)$ near the other; this asymmetry is what produces the net $2\alpha$.