# Electric dipole moment of electron: about what point is the moment taken?

There is a lot of experimental research activity into whether the electron has an electric dipole moment. The electron, however, has a net charge, and so its dipole moment $${\bf \mu}= \int ({\bf r}- {\bf r}_0)\rho({\bf r}) \,d^3r$$ depends on the chosen origin ${\bf r}_0$. Indeed, if one takes moments about the center of charge, then - by definition - the electric dipole moment is zero.

Now I know that what is really meant by the experimentalists is that their "electric dipole moment" corresponds to adding to the Dirac Lagrangian a term propertional to $$\frac 12 \bar \psi \sigma_{\mu \nu}\psi\, ^*F^{\mu\nu},$$ where $^*F^{\mu\nu} = \frac 12 \epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}$ is the dual Maxwell field. So I have two questions:

a) What point ${\bf r}_0$ does this correspond to? I'd guess that it it is something like the center of energy of the electron's wavepacket measured in its rest frame. Is there a way to see this?

b) If my guess in (a) is correct what would happen if the electron were massless? There is then no rest frame, and the center of energy is frame-dependent. I imagine therefore that the electric dipole moment would have to be zero. Is this correct? Certainly $\bar \psi \sigma_{\mu \nu}\psi$ is zero for a purely left or right helicity particle obeying a Weyl equation as $\gamma_0 [\gamma_\mu,\gamma_\nu]$ is off-diagonal in the helicity basis

• It seems to me that the interesting and hard part of this question is how to relate all this to the formalism of QFT. But I think the answer to (b) is classical and more straightforward. See physics.stackexchange.com/q/74366 . – user4552 Jul 24 '18 at 20:13
• @Ben Crowell A massless charged chiral particle has a magnetic moment of exactly $\mu= \pm e /(2E) \times {\bf k}/|{\bf k}|$ where the $\pm$ is the helicity and $E$ the energy. It's the electric dipole moment that I find problematic. – mike stone Jul 24 '18 at 21:01
• If so, then my argument in the answer to the other question must be wrong. But I still don't see why you think (b) needs a non-classical answer. Surely all electric dipoles have the same transformation properties. If you make an electric dipole by gluing charges $\pm q$ to the ends of a popsicle stick of length $L$, then under a boost $v$ parallel to the stick, we have $qL\rightarrow 0$ as $v\rightarrow c$. What is wrong with this as a purely classical proof that a massless particle has zero electric dipole moment parallel to its motion? – user4552 Jul 24 '18 at 21:30
• @Ben Crowell Mmmm... a nice argument. – mike stone Jul 24 '18 at 22:11
• – rob Jul 25 '18 at 21:59

A careful analysis of the Dirac equation coupled to an exterior electromagnetic field shows that a massive Dirac particle possesses an oscillating electric dipole moment of magnitude $\frac{q \hbar}{2 m}$ rotating with the zitterbewegung angular frequency $\omega = \frac{2mc^2}{\hbar}$, please see Hestenes' work: Zitterbewegung in Quantum Mechanics. Thus the electric dipole moment averages to zero. However, since its second moment is non-vanishing, it may be possible to measure it experimentally.
The limit at $m \rightarrow 0$ of the average electric dipole is undefined. I don't know how to generalize the result to this limit.