All the explanations of the Aharonov-Casher effect seem to imply it only "works" for neutral particles with a magnetic moment. This seems to stem from the duality of the A-C effect with the more known Aharonov-Bohm effect. The particle being neutral is also mentioned in the original paper.

Is it strictly necessary to consider the A-C particle to be neutral? Can't the A-B and A-C phases just add up in the case of an electron in an electromagnetic field?

I know for sure that in e.g. the Pauli-Breit Hamiltonian (a non-relativistic limit of the Dirac Hamiltonian) contains a $\boldsymbol \mu\times\boldsymbol E$ A-C term, together with the $\boldsymbol A\cdot \boldsymbol p$ A-B term.


1 Answer 1


The Aharonov-Casher effect occurs due to a non-minimal interaction term of a spinning particle to an electromagnetic field. This interaction term can arise when the spinning particle is composite having an anomalous magnetic moment. In this case, even when the particle is neutral, it can interact with the electromagnetic field. The interaction is governed by the Pauli-Dirac equation:

$(\gamma^{\nu}\partial_{\nu}-\frac{\mu}{2}\sigma^{\alpha\beta}F_{\alpha\beta})\Psi = 0$

($\mu$ is the anomalous magnetic moment). But, if in addition the particle is charged, then it can couple minimally to the electromagnetic field in addition to possessing an anomalous magnetic moment and its equation of motion takes the form:

$(\gamma^{\nu}(\partial_{\nu}-eA_{\nu})-\frac{\mu}{2}\sigma^{\alpha\beta}F_{\alpha\beta})\Psi = 0$

Thus in principle, there can be two contributions to the topological phase, one proportional to the charge $e$ and one proportional to the anomalous magnetic moment $\mu$. However, the first contribution (proportional to $e$) is an ordinary Aharonov-Bohm phase.

Thus a composite charged particle can, in principle, have an Aharonov-Casher phase in adition to an Aharonov-Bohm phase.

It should be noticed that the conditions for the existence of the Aharonov-Casher phase are much more strict than that of the Aharonov-Bohm phase. In the case of the Aharonov-Bohm case, the particle needs to move in a force free region, i.e., in which the vector potential is locally a pure gauge $A_{\nu} = \partial_{\nu}\theta$, but its line integral over the close trajectory is nonzero due to non-simple connectedness of the configuration space. In the case of the Aharonov-Casher effect. The particle must be spinning; otherwise no anomalous magnetic moment can exist in the first place. Secondly, the trajectory must be two dimensional making the problem 2+1 dimensional. In this case only there exists a "dual potential"

$\tilde{A}^{\nu} =\epsilon^{\alpha\beta\nu}F_{\alpha\beta}$,

with respect to which the "Pauli-Dirac" equation has the form of a minimally coupled equation.


In relativistic mechanics, the anomalous magnetic moment couples differently from the standard magnetic moment. It's coupling involves the spin generators (cf. the anomalous magnetic moment term of the Dirac equation $\frac{\mu}{2}\sigma^{\alpha\beta}F_{\alpha\beta}\Psi$).

As a consequence, a scalar particle would not possess an anomalous magnetic moment. The standard magnetic moment stems from the usual minimal coupling term.

The origin of the anomalous magnetic moment is either the interaction of a fundamental particle with the quantized electromagnetic field, as in the case of the electron , or due to the internal motion of the constituents as in the case of the Proton and Neutron.

The anomalous magnetic moment adds up to the standard magnetic moment in the effect of spin precession in a magnetic field, but introduces a further interaction with the electromagnetic field proportional to the velocity of the particle (and the anomalous magnetic moment alone). This last contribution vanishes in the non-relativistic limit.

This is the reason that in the Aharonov-Casher article, in the first heuristic derivation of the interaction term they worked with a non-relativistic model, thus, the two magnetic moments appeared additively. But in their second derivation, they used the anomalous magnetic moment term. I think that they did not emphasize the fact that this interaction term is solely due to the anomalous magnetic moment because they referred to a neutral particle not possessing a standard magnetic moment.

  • $\begingroup$ If I understand correctly, you say the A-C effect "works" only on particles with anomalous magnetic moment, but reading the original A-C paper it seems any magnetic moment will do. Which is it? $\endgroup$
    – rubenvb
    Mar 28, 2013 at 14:36
  • $\begingroup$ @rubenvb:I have added an update answering your question. $\endgroup$ Mar 31, 2013 at 10:40
  • $\begingroup$ @DavidBarMoshe: thank you for this excellent answer. Would you mind having a look at this post here to clarify the relation between the AC-effect and spin-orbit interaction? $\endgroup$
    – flaudemus
    Feb 19, 2019 at 21:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.