According to the Wikipedia page on the electron:

The electron has no known substructure. Hence, it is defined or assumed to be a point particle with a point charge and no spatial extent.

Does point particle mean the particle should not have a shape, surface area or volume?

But when I searched Google for the "electron shape" I got many results (like this and this) that says electrons are round in shape.

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    $\begingroup$ Please read this link. It says physicsist found shape of electron, any comment or update by you people? medium.com/predict/… $\endgroup$ – Anubhav Goel May 25 '19 at 15:08
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    $\begingroup$ @AnubhavGoel I understand that this has to do with the distribution of the electron's wavefunction in space. It's not "shape" strictly speaking, i.e. not the shape of an object, but the distribution of where it can be found. (I'll read it more, though.) $\endgroup$ – Helen May 25 '19 at 15:17
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    $\begingroup$ @AnubhavGoel That is some awful coverage -- it's basically regurgitating the press release from the University of Basel PR office. As Helen just mentioned, the work it (pretends to) report on has nothing to do with the intrinsic shape of the electron (which is the subject of this Q&A thread), but with its distribution in space when placed in some specific arrangements of semiconductors. If you want some actual coverage, the piece in Physics does get it right. It lacks the shiny headlines... $\endgroup$ – Emilio Pisanty May 25 '19 at 15:25
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    $\begingroup$ ... but that's because there's no shiny-headline-worthy material in that paper. The bounty banner's claim that the current answers are out of date and require revision given recent changes does not hold -- there is nothing in the paper that has any bearing on the current Q&A, to the extent that any answers that try to address the Camenzind et al. paper would be off-topic here. If you want to see a discussion of what that paper does and does not do, then you should ask it as a separate question. And, frankly, your bounty should be refunded so you can spend it on that separate thread. $\endgroup$ – Emilio Pisanty May 25 '19 at 15:26
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    $\begingroup$ @AnubhavGoel I'm refunding your bounty because the "shape" that press release is about is an entirely different notion of "shape" than the one being asked about here, therefore any answer that addresses your reason for offering the bounty would simultaneously have to be deleted as not answering this question. If you have a particular question you want to ask about that press release, please ask a new question. If you think the bounty should have been allowed to stay, please post to Physics Meta. $\endgroup$ – ACuriousMind May 25 '19 at 15:39

As far as we know the electron is a point particle - this is addressed in the question Qmechanic suggested: What is the mass density distribution of an electron?

However an electron is surrounded by a cloud of virtual particles, and the experiments in the links you provided have been studying the distribution of those virtual particles. In particular they have been attempting to measure the electron electric dipole moment, which is determined by the distribution of the virtual particles. In this context the word shape means the shape of the virtual particle cloud not the shape of the electron itself.

The Standard Model predicts that the cloud of virtual particles is spherically symmetric to well below current experimental error. However supersymmetry predicts there are deviations from spherical symmetry that could be measurable. The recent experimentals have found the electric dipole moment to be zero, i.e. the virtual particle cloud spherically symmetric, to an accuracy that is challenging the supersymmetric calculations.

However there are many different theories based upon supersymmetry, so the result doesn't prove supersymmetry doesn't exist - it just constrains it.

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    $\begingroup$ "However an electron is surrounded by a cloud of virtual particles". Can you please explain to me what those virtual particles would be? Quarks? $\endgroup$ – Árpád Szendrei Nov 22 '16 at 21:57
  • $\begingroup$ @ÁrpádSzendrei mainly virtual photons building up the static 1/r coulomb potentuial $\endgroup$ – anna v May 25 '19 at 17:29

The shape of a distribution of charges is described in terms of multipole expansion, which you can think of as similar to Fourier expansion but in two dimensions. The total charge gives you the "monopole term", whose interaction is spherically symmetric. If there's an offset between the center of the mass distribution and the center of the charge distribution, you have a dipole moment. A coin-shaped or cigar-shaped distribution has nonzero quadrupole moment, a pear-shaped distribution has an octupole moment, and so on. As in Fourier analysis, it's possible to represent any charge distribution in terms of multipole moments, though a shape with sharp edges (like, say, a cube) would require an infinite number of terms.

The electron cannot be cube-shaped, or even coin- or cigar-shaped, due to a theorem relating multipolarity and spin. A spinless particle may have a monopole moment, but not a dipole moment; a spin-half particle may have monopole and dipole moments, but not a quadrupole moment; a spin-one particle may have monopole, dipole, and quadrupole moments, but not octupole moments. A handwavy, cartoony way to think of this is that any such moments must be quantized along the direction of the particle's spin — otherwise, as the particle spins, they'd average to zero. If you wanted the electron's charge distribution to be cigar-shaped, like a uranium nucleus is, you'd need to specify that a polarized electron has more charge near its poles than it does near its middle. But a spin-half particle doesn't have any spin projection near its middle — there are only "up" and "down." An electron may have monopole and dipole moments, but doesn't have enough degrees of freedom to have any more complicated shape.

Furthermore, we have the observation that the electron's interactions are very nearly invariant under the symmetries of parity conjugation, $P$, and charge conjugation, $C$. This further restricts the moments that are available, because the electron's spin, to which the dipole moments must be coupled, is an axial vector and does not change sign under $P$. To a very good approximation, then, the electron's mass and charge distributions may carry only a monopole moment, while its magnetic field (another axial vector quantity) may carry only a dipole moment. This gives us the usual toy-model picture of an electron as a spherical, spinning bar magnet.

However, the electron's interactions are not quite invariant under conjugation of parity and charge at the same time. This transformation, $CP$, is the operator that transforms an electron into a positron. Our strongest evidence that the universe treats electrons and positrons differently is that the universe is quite full of electrons, but contains only incidental positrons. To reach this state requires, among other things, $CP$ violation. But essentially every model that contains enough $CP$ violation to predict our observed matter/antimatter asymmetry also predicts permanent electric dipole moments for the proton, electron, and neutron which are much larger than the current limits. This is what the Hudson and DeMille groups have measured in the news stories you found. I thought that DeMille's explanation in your first link was quite nice.

  • $\begingroup$ Can you provide me some references related to your article. It is too compact for the guys like me. Hope you will help. $\endgroup$ – L.K. Jul 1 '14 at 14:49
  • $\begingroup$ @lavkush For recent electron EDM searches, try here, and references therein. For multipole expansion see any advanced textbook on electricity and magnetism. For this particular consequence of the Wigner-Eckart theorem, the original EDM search proposal refers to Hans Bethe's nuclear physics textbook, which I have not read; I learned the argument from folks doing EDM searches. $\endgroup$ – rob Jul 1 '14 at 15:21
  • $\begingroup$ @rob Could you comment on my question: physics.stackexchange.com/questions/168741/… ? It has been linked to this question as already providing an answer, but my question is quite specific and your reply here seems the most relevant. I show how even a highly non-spherical charge distribution can have zero dipole moment - which suggests measurements of the dipole moment do not tell you about the lack of sphericity of a charge distribution... $\endgroup$ – kotozna Oct 4 '15 at 10:17
  • $\begingroup$ @kotozna, the distribution in your question has nonzero quadrupole moment; see my second paragraph here. $\endgroup$ – rob Oct 4 '15 at 17:02
  • $\begingroup$ @rob yes I see that. My question is specifically asking why it is said (in the papers I link) that measuring the dipole moment tells you something about the sphericity. Would you agree that a measurement of a very small dipole moment does not tell you the charge distribution is very close to spherical (as the papers are suggesting)? Thx. $\endgroup$ – kotozna Oct 4 '15 at 20:54

Electrons, and such small things :-) are handled by quantum mechanics. Quantum mechanics differs very, very much from the classical, Newtonian mechanics and from our intuition based on our experience.

In QM, although the electron is handled as if it were a point-like body, it doesn't have an exact location. Instead of it, its location is described by a wave function named $\psi(r)$. This is a complex scalar field interpreted in the space, thus we can describe this as $\mathbb{R}^3\rightarrow\mathbb{C}$. What makes the picture really interesting, this wave function has complex values. The quadratic absolute value ($\psi\psi^*$) is the same is the probability distribution of the location of the electron on a specific place.

The integration of $\psi\psi^*$ on a volume gives the probability to the electron exists in that volume.

As a classical intuition we could imagine that as if the electron were some like a "cloud", with different densities in the space. As a possible interpretation of the "shape of the electron", we can imagine the wave function or the probability distribution, or we could even imagine this "cloud".

Well, we could even calculate this, although they aren't the simplest calculations. And from the calculated density images, we can generate visible pictures. So:

enter image description here

These are electron shapes around atomic nuclei. But there are also very different distributions as well, for example a free-electron in a double slit experiment has a very different wave function.

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    $\begingroup$ This is the perspective (electron as probability distribution) that I imagined when I read the question. It's good that you made the point that this distribution can be distorted by context. $\endgroup$ – Ryan Reich Jun 18 '14 at 15:12
  • $\begingroup$ actually this has been measured for the hydrogen atom io9.com/… $\endgroup$ – anna v Jun 18 '14 at 17:05
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    $\begingroup$ Isn't that a bit like answering the question, "What shape is a cow?" by providing the probability distribution of finding a cow in any particular part of the pasture? $\endgroup$ – David Richerby Jun 19 '14 at 8:52
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    $\begingroup$ @DavidRicherby I stated explicitly, that it is only a single possible interpretation of the question. I also have to remark, in the case of the cow there is a very complex shape information and practically total precise location info. The case of the electron is exactly opposite (point-like thing in not exact location). Think about the Heisenberg-relation. $\endgroup$ – peterh - Reinstate Monica Jun 19 '14 at 11:58
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    $\begingroup$ @DavidRicherby: The two questions are connected. The shape of a cow is given by the probability of a point P+dP being inside the cow, given that a point P lies inside the cow. And the shape of that cow influences the chance of finding a cow at any point in the pasture (You won't find the spherical cow in the corner of the pasture ;) ) $\endgroup$ – MSalters Jun 19 '14 at 15:28

No one has ever directly seen an electron, and it's quite possible that no one ever will. To think of it as a shiny little pinball is as much a mistake as to think of it as an abstract infinitesimally small "point" with certain properties. To add to the confusion, depending on how you are "looking" at the electron, it can appear to be a particle (implying some finite size and a definite shape) or it can appear to be a wave of some sort. As a wave, you can talk about the "clouds" of electron orbitals around an atom, which are not physical things but representations of probabilities. Looking at an electron as a pinball or as a wave/cloud can be useful in certain situations, but is not an absolute truth.

Short answer: no, electrons do not have a "shape", at least in the sense of "it looks like a pinball or ...".


An electron is not a point particle. Point particles don't exist. The world is governed by quantum mechanics, which describes physical systems in terms of quantum mechanical observables, which are represented by Hermitian operators. Different observables represent different ways in which you can interact with a given system and copy information from it. For example, a photomultiplier tube may be useful for telling whether there are more than N photons' worth of energy in some region for some value of N.

If you consider some finite region you can measure an observable that will give you information about whether there is an electron in that region. But that region can't be arbitrarily small. One limitation is that measuring in a smaller region requires putting more energy into that region and at some point the energy required to do that is so large it creates a black hole. There may be other physical limitations that would kick in before you get to that level.

How should we interpret claims about the electron's shape? Such a claim means that when we measure whether an electron where an electron is with some high accuracy, we get a spherical distribution of results to some accuracy sufficient to rule out some supersymmetric theory.

You might think something like "Couldn't we say that the electron really is at some particular point but not at others and we just can't tell exactly where it is?" That idea doesn't match reality because if you want to predict the electron's subsequent evolution you have to to account of observables that don't represent the electron as being at one particular point, like momentum, since those observables appear in the Hamiltonian.

The claim that the electron has no known substructure is correct but what it means is not that the electron is at a particular point but just that it doesn't have subsystems that can be changed independently. A composite system, like a biro, doesn't have that property. You can take the ink tube out of a biro and move it around independently of the plastic shell. But you can't do anything analogous with an electron as far as anybody knows.

For some relevant material see







What is your definition of shape? And on what scale does this information bare relevance to you? If I were to tell you electrons are actually shaped like pyramids, in what way would this change the way you interact with the universe. Albeit analogies being the cause for questions like these (subatomic particles often being depicted as distinctly colored spheres), I will try to point out exactly that by using an analogy.

Buildings can be classified by type. Like a skyscraper, church, castle, bungalow whatever. What would be the smallest structure that assigning the property building type would still make sense to you. What kind of building would you call a single brick? Sure I could imagine a biologist studying ant colonies assigning this property to a brick, because in that context it is relevant information.

All those depictions addressed in previous answers only bare relevance to their respective fields of study and have nothing to do with your concept of shape. So instead of arguing about the definitions of shape you should ask yourself is this a relevant question for you?

On the other hand though, suppose you were to be competing in a regular bowling match and you were to be told to only use cube-shaped balls instead of spherical ones. Even without having tried a cube-shaped bowling ball, it's safe to say this would have (at least some) significant influcnce on how you interact with the ball.

  • $\begingroup$ you are right, this question is not at all relevant for me as I am a software engineer and the shape of electron will not affect my work. But again I asked it for the sake of curiosity. Why newton asked himself a question about the apple ? was it relevant ? $\endgroup$ – Anil Bharadia Jun 19 '14 at 5:19
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    $\begingroup$ -1. There are good reasons to assume shape matters. For instance, the trajectory of an electron tunneling through a crystal may very well depend on its orientation re. the attice. A shapeless electron can't have such an orientation. $\endgroup$ – MSalters Jun 19 '14 at 15:32
  • $\begingroup$ @MSalters No, it has spin, and could have different intrinsic multipole properties (althoigh electron doesn't have the second, other elemental particles have). Your -1 is unjust, I've compensated it with an upvote. $\endgroup$ – peterh - Reinstate Monica Jun 19 '14 at 15:39
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    $\begingroup$ @AnilBharadia By not relevant I don't mean not interesting, au contraire, I understand and admire your curiosity. My point is that the property shape doesn't apply to object electron. Meant here shape in the traditional (macroscopic) sense of way, as I'd assume you mean, since you wouldn't otherwise be asking (by all means no offence). As with properties like pressure or viscosity, shape describes a collective arrangement of particles and by that definition does not apply to single particles. As for Newton, the laws he discovered describe the fall of the apple. So very relevant I'd say. $\endgroup$ – user50718 Sep 30 '14 at 2:25
  • $\begingroup$ I don't understand the downvotes to this question either: it states a very valid argument nevertheless. $\endgroup$ – gented Jun 20 '17 at 15:56
