# Aharonov-Bohm phase picked up when a magnetic dipole goes around a charge

When a particle with charge $q$ traverses a loop that encloses a magnetic flux $\Phi$, it picks up a phase $e^{iq\Phi}$ (I have set $c$ and $\hbar=1$). This is the usual Aharonov-Bohm phase. Now, let us look at a different scenario - consider a particle with a magnetic dipole moment that gives rise to a flux $\Phi$ through a plane. Move this particle along a loop enclosing a charge $q$ in the same plane. What is the phase picked up in this case?

According to Section 2.1.3 of Intro. to Topological Quantum Computation by Pachos, this phase is also $e^{iq\Phi}$. I am not able to see why, though.

The interaction energy between the two particles is given by: $$E_{int} = \vec{\mu} \cdot \vec{B}$$ Where $B$ is the magnetic field felt by the neutral particle in its rest frame. This magnetic field originates from a Lorentz transformation of the charge electric field:
$$\vec{B} = − \gamma \frac{\vec{v}}{c} \times \vec{E} \approx -\frac{\vec{v}}{c} \times \vec{E}$$ (In the non-relativistic approximation).
Thus the interaction energy becomes: $$E_{int} = -\vec{\mu} \cdot \frac{\vec{v}}{c} \times \vec{E} = \frac{1}{c} \vec{\mu} \times \vec{E} \cdot \vec{v} = \frac{q}{c} \frac{\vec{\mu} \times \vec{r} }{|\mathbf{r}|^3}\cdot \vec{v}$$ This term has the form: $$E_{int} = \frac{q}{c} \vec{A}(r)\cdot \vec{v}$$ Which is just the standard form of the Lorentz-force Lagrangian, giving rise to the Aharonov-Bohm phase, with: $$\vec{A}(r) = \frac{\vec{\mu} \times \vec{r} }{|\mathbf{r}|^3}$$ Suppose, that the motion is circular on a plane and the magnetic moment is perpendicular to the plane, then the Aharonov-Casher phase is given by: $$\phi_{AC} = \oint \frac{q}{c} \frac{\vec{\mu} \times \vec{r} }{|\mathbf{r}|^3}\cdot d\vec{r} = q \frac{2 \pi \mu}{rc}$$
As can be observed, the phase depends on the radius of the circular trajectory. This is an example of a geometric phase. If we replace the point charge by an infinite wire of charge per unit length $\lambda$, then the electric potential becomes: $$V(\vec{r}) =\lambda \ln(|\mathbf{r}|)$$ And the electric field $$\vec{E} (\vec{r}) = \frac{\lambda \vec{r}}{|\mathbf{r}|^2}$$ In this case, the Aharonov-Casher phase becomes: $$\phi_{AC} = \lambda \frac{2 \pi \mu}{c}$$ As we can see, this phase does not depend on the radius of the trajectory. In fact, we can prove that it does not depend on the shape of the trajectory, but only on the number of times the trajectory encircles the charge. This is an example of a topological phase. In contrast the geometric phase in first case depends on the trajectory.