Aharonov-Bohm effect of doubly localized wavepacket I want to imagine an exotic situation regarding Aharonov-Bohm effect. The wavefunction $\psi$ of the electron is even ($\psi(\mathbf r) = \psi(-\mathbf r)$) and localized in two spatially separated regions. This situation is described as in the below picture:

The particle is constrained in the black wire, and the red regions are where the wavefunction is localized.
Now, we adiabatically rotate the wavefunction. Since the wavefunction is even, only $\pi$-rotation, rather than $2\pi$-rotation, is required to come back to the original place. We know that for $2\pi$-rotation in the usual situation, the Aharonov-Bohm phase is $e^{2\pi i \Phi/\Phi_0}$ where $\Phi$ is the magnetic flux inside the path and $\Phi_0$ is the flux quanta. How about this exotic case, where we did half rotation?
 A: Similar situations have been extensively studied (theoretically and experimentally) in semiconductor Aharonov-Bohm interferometers with quantum dots in their arms. One important point is phase rigidity, which arises in a closed interferometer, i.e., an interferometer with only two leads (the source and the drain) - the Aharonov-Bohm oscillations become strictly symmetric due to the multiple scattering events - phenomenon more generally known as Onsager-Büttiker relations). This necessitates using so called open interferometers, i.e., the interferometers with particle losses, such as the ones in the groundbreaking papers from the Heiblum group, here and here.
In other words, the Aharonov-Bohm effect in a closed ring is surprisingly different from a naive picture stemming from the two-slit thought experiment, and used in many basic discussions of the Aharonov-Bohm effect (including the original AB paper).
Now, on the top of that you artificially enforce $\psi(r)=\psi(-r)$ symmetry, which probably kills the AB altogether... but this could be checked by directly solving a 1D scattering problem on the ring.
