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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!

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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.

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  • $\begingroup$ Thanks, I'm starting to see, but am still missing some gaps. I agree, my spin-up state is not single-valued. Define $|\uparrow_{\hat{\mathbf{n}}}'\rangle = e^{i\theta/2}|\uparrow_{\hat{\mathbf{n}}}\rangle$. This new state is single-valued on the trajectory. Computing Berry's phase in this case does indeed give $-\pi$, as I would expect from the textbook result. However, it is my understanding that Berry's phase should be gauge-invariant. These states are related via a gauge-transformation, but yield different Berry phases. How do I see the need for single-valuedness in the standard derivation? $\endgroup$
    – vpflynn19
    May 27, 2022 at 18:18
  • $\begingroup$ @vpflynn19 In the standard derivation, if $C$ is an open path from $a$ to $b$, doing gauge transformation $|{\mathbf{n}}\rangle \rightarrow e^{i\beta(\mathbf{n})}|\mathbf{n}\rangle$ changes the Berry phase by $\beta(\mathbf{n}_b)-\beta(\mathbf{n}_a)$. Now usually we would say when the path is closed, so $a$ and $b$ are the same point, we must have $\beta(\mathbf{n}_b)-\beta(\mathbf{n}_a)=0$. However, this is only true if we demand single-valuedness, because if we allow state vector to be multi-valued, then there is no reason to restrict $\beta(\mathbf{n}_b)=\beta(\mathbf{n}_a)$. $\endgroup$
    – Meng Cheng
    May 27, 2022 at 18:35
  • $\begingroup$ Thank you! I think I understand now: a multi-valued choice of the instantaneous eigenstates will lead to a Berry's phase that is not, in general, gauge invariant (as in my original post). So single-valuedness is a key ingredient for gauge-invariance. $\endgroup$
    – vpflynn19
    May 28, 2022 at 1:50
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    $\begingroup$ Right. I would say that an even better thing to do is the definition in my answer, where the sum of Berry phase and monodromy is the gauge-invariant phase. Then you can use single-valued or multi-valued state. $\endgroup$
    – Meng Cheng
    May 28, 2022 at 1:59

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