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It's announced that researchers at Imperial College London has found that the electron is almost a perfect sphere. The popular articles all have a nice photo of a billiard ball, etc. It is reported that they found this by measuring the "wobble" as the electron spins (and finding none).

What does it exactly mean that the electron is a sphere? Is it the wave function that is spherical? Measuring the spin wobble brings to mind a solid object, which I think the electron surely isn't. So, what would have been wobbling if it didn't have perfect spherical symmetry?

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It means people still think of the electron in terms of classical charge distribution. It means they learnt nothing from QM. In QM the charge distribution is state-dependent. – Vladimir Kalitvianski May 26 '11 at 16:45
Chad Orzel has a delightful writeup of the experiment here:… – Anonymous Coward May 27 '11 at 18:15

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There's a Nature article that describes the experiment and the results,, but that's behind a paywall. The experiment is described in some detail in "Prospects for the measurement of the electron electric dipole moment using YbF", (I've only scanned the latter, but it looks to be quite informative).

From the Nature article, "In an atom or molecule with an unpaired valence electron, the interaction of the electron EDM [Electric Dipole Moment] with an applied electric field results in an energy difference between two states that differ only in their spin orientation. This energy difference is proportional to $d_e$ and changes sign when the direction of the field is reversed. A sensitive method of measuring this energy difference is to align the spin perpendicular to the field and measure its precession rate, which is proportional to the energy difference. An alternative description of the method is in terms of an interferometer. There is quantum interference between the two spin states, and the EDM appears as an interferometer phase shift that changes sign when the electric field is reversed."

An electron is not either a sphere or not-a-sphere, but we can introduce more or fewer internal degrees of freedom into the quantum fields that are used to describe experiments results that we attribute to the electron field. Introducing different degrees of freedom has consequences for the geometrical configurations of recorded experimental results. The Nature article is explicit in saying that this is intended to distinguish between different speculative quantum field theories, "many extensions to the standard model naturally predict much larger values of $d_e$ that should be detectable". This is an experimentalists' article, however, so they link to a theory paper on the subject (which I cannot access directly). If these fields give better descriptions than the standard model of particle physics, we expect to see different, less geometrically symmetrical statistics of events.

Many of the problems of reference here can be avoided if we talk about electron fields instead of about electrons. An "electron field" is less likely to be misrepresented as spherical or not spherical, but it can be associated with (representation spaces of) space-time symmetry groups (which describe in a systematic way how something deviates from being symmetrical). Care is needed because a quantum field is a more elaborate mathematical object than a classical field, but we can loosely think of a quantum field as a way to generate probabilities that the configuration of a classical field is one thing or another at any single time, while the details of quantum measurement are such that we can't talk about such probabilities at multiple times.

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I think the distinction that you point out between electrons and (quantum) electron fields is crucial for a clear understanding of the result. Yet, this raises an interesting question: if these two are different things, than what does the word "electron" refer to? – recipriversexclusion May 26 '11 at 18:45
We could say that "electron" is an adjective. An electron field as opposed to a "green" field. We see an "electron event" in a detector, not an electron, caused by the preparation of a specific quantum field state, and the detector being put where it is, not by an electron. Would you like a yellow? A yellow what? There are other ways to go with the interpretation of QM/QFT, of course. – Peter Morgan May 26 '11 at 18:56

Well, one strong piece of evidence for this is the fact that the electric dipole moment

$${\vec D} \equiv \int d^{3}x' \rho({\vec r}-{\vec r_{0}})$$

of the electron is nearly zero. The electric dipole moment would measure a reflection asymmetry in a potential charge distribution of the electron. The fact that it takes on a zero means that the electron charge distribution is symmetric about all planes. Now, it is still possible that the electron may have nonzero values for it's higher multipole moments, while having a zero electric dipole moment which would indicate non-sphericity as well, but as far as I know, there is no evidence for this either, and finding these moments would be an even harder experiment.

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Jerry Schirmer is correct, but just to rephrase his answer so it answers the question as asked: "What does it mean that the electron is [almost a perfect] sphere?", it means that the electric dipole moment has been measured to be very small. – Anonymous Coward May 26 '11 at 17:34
Yes, they should have state this clearly, without mentioning the electron shape. – Vladimir Kalitvianski May 26 '11 at 17:51
But they have to dumb it down so people that don't understand Quantum Mech or spherical harmonics can follow it. – Omega Centauri May 26 '11 at 17:58
Maybe you should say what $\vec r_0$ is, @Jerry, because it does affect the result - because the total charge is nonzero. Is that the center of mass? Also, it's not clear why you integrated over $x'$ even though no $x'$ appears later. One more thing, the dipole moment has a direction that must be proportional to the spin because it's the only preferred direction of the electron. Finally, a dipole moment is prohibited by CP-symmetry, so a (large) dipole moment requires a (new) source of CP-violation. – Luboš Motl May 26 '11 at 18:28
Vladimir: if the dipole moment wasn't parallel to the spin, the electron would have an additional degree of freedom. This would change the entire periodic table, since we could fit 4 electrons in the 1s orbital, etc. etc. – Anonymous Coward May 26 '11 at 23:45

As the comment to your post says: This spherical charge distribution picture hasn't really much to do with modern QM physics.

In fact we are talking about different quantities. The charge distribution is given mathematically by $\bar{\psi}\gamma^o\psi$ while the relevant quantities are the components of the polarization/magnetization tensor $\bar{\psi}\sigma^{\mu\nu}\psi$

The electric polarization components are zero in the current (Dirac) theory (in the local restframe) while the magnetization components are non zero. The latter represent the magnetic moment of the electron. It is the Dirac equation which mathematically describes the precession of the electron searched for in the experiment.

A non-zero electric dipole moment would give the electron the very special property of either going one of two different ways in a non-homogeneous electric field. Just like the famous Stern Gerlach experiment shows in the case of the magnetic moment.

Such a behavior can of course never be explained by a classical non spherical charge distribution...

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It is probably better to talk about deviation from spherical symmetry. If you want to think of it as a sphere then the classical radius computed from equating the mass-energy of the electron to its field $$ mc^2 = \frac{1}{4\pi\epsilon_0}\frac{e^2}{r} $$ and with variations on this way of computing this. If there were some deviation from spherical symmetry it turns out to be on part $10^{16}$ of this classical electron radius.

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