# Is there a published upper limit on the electron's electric quadrupole moment?

I understand an electric quadrupole moment is forbidden in the standard electron theory. In this paper considering general relativistic corrections (Kerr-Newman metric around the electron), however, there is a claim that it could be on the order of $Q=-124 \, \mathrm{eb}$. That seems crazy large to me, but I can't find any published upper limits to refute it. Surely someone has tested this? Maybe it's hidden in some dipole moment data? If not, is anyone planning to measure it soon?

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Excellent question :-) (Unfortunately I wouldn't know where to look for the answer) –  David Z Oct 6 '12 at 8:05
Dear Grant, the first sentence "I understand an electric quadrupole moment is forbidden in the standard electron theory." that you inserted is both dishonest and it makes your question internally inconsistent. Could you please remove it again? If you understand that the paper is wrong, why are you asking about it? The only correct answer to any question related to that paper is one that explains that the paper is wrong and $Q_{ij}=0$ for spin-1/2 systems. It looks like you are trying to push people to write some different, "more pleasing" answer. There isn't any non-standard electron theory. –  Luboš Motl Oct 6 '12 at 15:36
If "it" has a nonzero electric quadrupole moment, it can't be an electron. An electron is defined as a particle with spin 1/2 (and other properties), so if it doesn't have spin 1/2, it's not electron, but something else, and it has completely different properties that depend on what "it" actually is. If it is electron, $j=1/2$ and $Q_{ij}=0$ immediately follows. You can't even write a candidate effect: there just isn't any possible "electron quadrupole correction to the S-matrix" just like there isn't any even prime greater than 6. –  Luboš Motl Oct 6 '12 at 16:28
To imagine that the electron with $j=1/2$ has a quadrupole effect, you would have to violate postulates of quantum mechanics such as the superposition principle. If this violation of quantum mechanics existed anywhere in Nature, it would affect all experiments that have ever been done. There isn't any way to violate quantum mechanics' postulates "mildly" so the scales don't matter. Your question is meaningless. You can't ask about experimental verifications of a mathematically inconsistent theory. –  Luboš Motl Oct 6 '12 at 16:31
@LubošMotl There does not seem to be an edit history, how could something be inserted? It is only in the first five minutes that edit history is not kept, treating the text as a preview. –  anna v Oct 7 '12 at 5:21

I think that the paper is completely wrong and the conclusions are preposterous. The paper argues that when one models the vicinity of the electron as a rotating black hole, he will get new effects.

However, the black hole corresponding to the electron mass – which is much lighter than the Planck mass – would have a much smaller radius than the Planck length. It really means that the Einstein-Hilbert action can't be trusted and all the quantum corrections are important. It also implies that the typical distance scale in any hypothetical electric quadrupole moment of the electron would be much shorter than the Planck scale – surely not a femtometer. Also, the black holes with masses, charges, and spins similar to those of electrons would heavily violate the extremality bound – something that isn't a problem because the classical general theory of relativity can't be trusted for such small systems.

The facts in the previous paragraphs are just different perspectives on the universal facts that gravity may be neglected in any observable particle physics, a fact that the author of the paper tries to deny.

Proof of the vanishing of the quadrupole moment

More seriously, one may prove from quantum mechanics that the quadrupole moment for an electron, a spin-1/2 particle, has to vanish because of the rotational symmetry. The quadrupole moment is a traceless symmetric tensor and because the electron's spin is the only quantum number of the particle that breaks the rotational symmetry, one would have to express the quadrupole moment as a function of the spin, i.e. as $$Q_{ij} = \gamma\cdot (3S_i S_j+3S_j S_i - 2S^2 \delta_{ij})$$ However, in the rest frame, $S_i$ simply act as multiples of Pauli matrices (with respect to the up/down basis vectors of the electron's spin) and the anticommutator $\{S_i,S_j\}$ above – needed for the symmetry of the tensor – is nothing else than the multiple of the Kronecker delta symbol, so it cancels against the last term. $Q_{ij}=0$ for all spin-1/2 objects (and similarly, of course, for all spin-0 objects). Only particles (nuclei) with the spin at least equal to $j=1$ (the case of deuteron) may have a nonzero electric quadrupole moment. This simple group-theoretical selection rule is the reason why you won't find any experiments trying to measure the electron's (or proton's or neutron's or other spin-1/2 particles') electric quadrupole moment. Such experiments would be as nonsensical as the paper quoted by the OP.

Note that unlike the case of the electron's dipole moment, one doesn't have to rely on any C, P, or CP-symmetry (which are broken) to show that the quadrupole vanishes. To deny the vanishing, one would have to reject the rotational symmetry.

Let me wrap by saying that the quadrupole moment may always be interpreted as some "elliptical shape" of the object or particle. This ellipsoid would be stretched along some axes and shrunk along other axes. However, the electron's spin-up and spin-down state really pick the same preferred axis in space – the sign doesn't matter for the quadrupole – so they can't have different values of the quadrupole moment. In other words, the quadrupole moment doesn't depend on the spin, and because the spin is the only rotational-symmetry-breaking quantum number that the electron has, the quadrupole moment has to be zero. (A Pauli-matrix-free proof.)

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Dear Grant, these comments are completely nonsensical. My proof doesn't make any assumption whatsoever about space's being flat. It is completely universal - it only assumes that one may define generators of rotations. To do so, from a GR viewpoint, it's enough if the space is flat asymptotically at infinity, infinitely far from the electron, and it surely is. That's also the reason why we can talk about spin states of the electron at all: they transform as representations of the generators. –  Luboš Motl Oct 6 '12 at 16:17
You still misunderstand how completely wrong those claims are. You can't even design a hypothetical effect that could be seen in an experiment. This is what the proof shows. There isn't any algebraic structure that would be interpreted as electron's quadrupole moment and that would influence the observations in a measurable way. It's mathematically nonsensical to try to measure such a quadrupole moment - just like it's nonsensical to measure the deviation of 2+3 from 5; how would you measure that? One can prove it's zero. –  Luboš Motl Oct 6 '12 at 16:19
The maths is just a little bit more abstract than for the $j=0$ case - I needed the anticommutator of Pauli matrices - but the strength of the result is exactly the same as it is for $j=0$. Take a spinless particle, like the alpha particle, and measure its quadrupole moment. What does it even mean? The alpha particle is spherically symmetric, otherwise it would have several spin states remembering the orientation, but it clearly doesn't because $j=0$, so no direction is preferred. The $j=1/2$ particles are doublets which is still too little and the situation for them is completely analogous. –  Luboš Motl Oct 6 '12 at 16:23
Attempts to model particles as classical solutions in classical fields - in contradiction with QM - are not "new physics" by any stretch of imagination. They're old physics, ancient pseudophysics superstitions from past centuries. I am constantly amazed how someone is ready to deny everything that's been learned to be wrong about the previous ideas about Nature and call it "new physics". Is geocentrism and creationism also new physics? If it's not, what's the difference? Why do crackpots fighting QM have so much more affirmative action than creationists or flat earthers? –  Luboš Motl Oct 7 '12 at 5:56
Good grief. I know full well that this is in contradiction with QM. As anna v noted above, I stated this at the very beginning. All I was asking was what numerical value I could place on this particular contradiction. That seemed to me a rather vanilla question. –  Grant Teply Oct 7 '12 at 7:43

This question is logically related to the question "what is the radius of the electron?"