Berry's phase for an electron in a two-level system

Question

I am trying to reconcile two seemingly contradictory statements about Berry's phase for a two-level system in a magnetic field. Consider the Hamiltonian $H(t) = -\mathbf{B}(t)\cdot\pmb{\sigma}$ where $\mathbf{B}= B(\sin(\theta)\cos(\phi),\sin(\theta)\sin(\phi),\cos(\theta))$, and the time dependence is embedded in the strength $B=B(t)$ and the two angular variables $\phi=\phi(t)$ and $\theta=\theta(t)$.

Consider the (instantaneous) "spin-up" eigenstate $$ |\uparrow_{\hat{\mathbf{n}}}\rangle = \begin{bmatrix} \cos(\theta/2) \\ \sin(\theta/2) e^{i\phi} \end{bmatrix},\quad \hat{\mathbf{n}} = (\sin(\theta)\cos(\phi),\sin(\theta)\sin(\phi),\cos(\theta))$$

For simplicity, let's keep the strength $B$ fixed and consider varying $\phi(t)$ and $\theta(t)$ slowly enough so that the adiabatic theorem holds. The first statement, which is found in various resources (e.g., Griffiths Eq. 10.62), says that the Berry phase accumulated during a closed trajectory on the sphere $|\mathbf{B}(t)| = B$ is proportional to half the solid angle subtended by said trajectory. This is consistent with the idea that the degeneracy at $B=0$ acts like a monopole of strength $-1/2$ at the origin (eliciting a 'flux' proportional to half the solid angle).

Let's put that result to the test. Consider a trajectory along the vertical great circle in the $xz$-plane. This circle subtends a solid angle of $4\pi/2 = 2\pi$ and so the Berry's phase should be proportional to $2\pi/2 = \pi$. One way we can compute the Berry's phase is by parameterizing $\theta(t) = 2\pi t/T$ and $\phi(t)=0$ and take $t=0$ to $t=T$. Then, we have the second statement: $$ \gamma_\uparrow = i\int_0^T \langle\uparrow_{\hat{\mathbf{n}}}(s)|\frac{d}{ds}|\uparrow_{\hat{\mathbf{n}}}(s)\rangle\,ds = i\int_0^T 0\,ds = 0$$ This can also be seen by integrating the $\theta$-component of the Berry connection $A_\theta = i\langle\uparrow_{\hat{\mathbf{n}}}|\frac{\partial}{\partial \theta}|\uparrow_{\hat{\mathbf{n}}}\rangle$ from $\theta=0$ to $2\pi$. This result is clearly not equal to $-\pi$ as one would expect from the subtended-angle point of view. On the other hand, doing the same computation for any other great circle that divides the sphere into two hemispheres and does not intersect the north or south poles (e.g., around the equator) does give $\gamma_\uparrow = -\pi$.

My attempt at resolving this is the following: I suspect that the result $\gamma_\uparrow = (-1/2)\cdot(\Omega/2) = -\Omega/4$ with $\Omega$ the solid angle subtended by the trajectory is only valid for trajectories where $|\uparrow_{\hat{\mathbf{n}}}\rangle$ is single-valued. The vertical great circle passes through the south pole where $|\uparrow_{\hat{\mathbf{n}}}\rangle = [0,e^{i\phi}]^T$, and $\phi$ is ill-defined ($x=y=0$). However, I don't see how this requirement is reflected in the derivation of said result. Moreover, I cannot reconcile this with the simple interpretation of the Berry's phase as the 'flux' due to a monopole (with strength -1/2, in this case) situated at the origin. In particular, this seems at odds with spherical symmetry of the problem. Any help would be greatly appreciated!

Answer 1

The problem is that when you use this parametrization of the "spin-up" state, the wavefunction is not single-valued in $\theta$. Namely, $|\theta+2\pi,\phi\rangle=-|\theta,\phi\rangle$. The usual formula for the Berry phase assumes that the state vector returns exactly to itself following the path in the parameter space, not up to a phase. If the state vector is not single-valued along the path, then one has to add this "jump" in phase as well to get a gauge-invariant Berry phase. More precisely, if we have a path $C$ from $a$ to $b$ in the parameter space, such that $|\psi_b\rangle=e^{i\alpha}|\psi_a\rangle$, then we can define a gauge-invariant phase

$$ \gamma=\int_{C}\vec{A}\cdot d\vec{s}+\alpha $$ The first term is the usual Berry phase, $\alpha$ is often called the "monodromy".

In your example, with the parametrization, $\int \vec{A}\cdot d\vec{s}=0, \alpha=\pi$, so $\gamma=\pi$ now in agreement with the textbook result.