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Size of an elementary particle

+ 1 like - 0 dislike
3513 views

Do we have a well defined mathematical expression denoting the size of a fundamental particle with no internal structure (electron for example) ? If we do, how does it fit in with the uncertainty principle ? And if we don't then what exactly do the experiments reveal claiming radius of electron is of the order $10^{-18} m$ or something like that ? Also in this particular case, does this mean that the probability of the electron being within that radius is 1 and zero outside ? A proper clarification would be very helpful.

This post imported from StackExchange Physics at 2014-08-07 15:36 (UCT), posted by SE-user smiley06
asked Aug 6, 2014 in Theoretical Physics by smiley06 (40 points) [ no revision ]
possible duplicate of What is the mass density distribution of an electron?

This post imported from StackExchange Physics at 2014-08-07 15:36 (UCT), posted by SE-user John Rennie
The difference I see is the question about the uncertainty principle. And the answer to that probably is: there is no mass operator. I am not familiar with the Higgs-mechanism, but as I understand it, mass is a parameter (or emerges from one) of the theory. So there is nothing prohibiting you from measuring the mass (indirectly via position or momentum) to arbitrary precision.

This post imported from StackExchange Physics at 2014-08-07 15:36 (UCT), posted by SE-user M.Herzkamp
And about the mass measurement we have: physics.stackexchange.com/q/19424

This post imported from StackExchange Physics at 2014-08-07 15:36 (UCT), posted by SE-user jinawee
@M.Herzkamp But I am not concerned about mass, I just wanted to ask how exactly the size or radius of these point like particles are defined.

This post imported from StackExchange Physics at 2014-08-07 15:36 (UCT), posted by SE-user smiley06

3 Answers

+ 3 like - 0 dislike

Elementary particles are considered to be point-like, but not point particles.

''QED, or relativistic quantum field theory in general, is not based on the notion of ''point particles'', as one sees stated so often and yet so erroneously.'' (emphasis as in the original)

This quote from p.2 of the book
    O. Steinmann, 
    Perturbative quantum electrodynamics and axiomatic field theory, 
    Springer, Berlin 2000.
tells everything. 

A point particle is the idealization of a real particle seen from so far away that scattering of other particles is as if the given particle were a point. Specifically, a relativistic charged particle is considered to be a point particle at the energies of interest if its interaction with an external electromagnetic field can be accurately described by the Dirac equation.

Both electrons and neutrinos are considered to be pointlike because of the way they appear in the standard model. Pointlike means that the associated bare particles are points. But these bare particles are very 
strange objects. According to renormalization theory, the basis of modern QED and other relativistic field theories, bare electrons have no associated electromagnetic field although they have an infinite charge (and an infinite mass) -- something inconsistent with real physics. They do not exist.

The bare particles are points = structureless formal building blocks of the theories with which (after renormalization = dressing) the physical = real = dressed = observable particles are described. The latter have a nontrivial electromagnetic structure encoded in their form factors. (The term ''dresses'' comes from an intuitive picture form the early days of quantum field theory, where a dressed particle was viewed as 
the corresponding bare particle dressed in a shirt made of infinitely many soft bare photons and bareparticle-antiparticle pairs.)

Physical, measurable particles are not points but have extension. By definition, an electron without extension would be described exactly by the 1-particle Dirac equation, which has a degenerate spectrum. 
But the real electron is described by a modified Dirac equation, in which the so-called form factors figure. These are computable from QED, resulting in an anomalous magnetic moment and a nonzero Lamb shift removing the degeneracy of the spectrum. Both are measurable to high accuracy, and are not present for point particles, which by definition satisfy the Dirac equation exactly.

The size of a particle is determined by how the particle responds to scattering experiments, and therefore is (like the size of a balloon) somewhat context-dependent. (The context is given by a wave function and determines the detailed state of the particle.)

On the other hand, the deviations from being a point are usually described by means of  context-independent
form factors that would be constant for a point particle but become momentum-dependent for particles in general. They characterize the particle as a state-independent entity. Together with a particle's state, the form factors contain all information that can be observed about single particles in an electromagnetic field.
An electron has two form factors, a magnetic and an electric one.

''The electric form factor is the Fourier transform of electric charge distribution in space.''
 (from Wikipedia)

The electric form factor determines the charge radius of a particle, defined as the number $r$ such that 
the electric form factor has an expansion of the form $F_1(q^2) = 1-(r^2/6) q^2$  if  $r^2q^2\ll 1$.
(Here units are such that $c=1$ and $\hbar=1$.) This definition is motivated by the fact that the average over $e^{i q dot x}$ over a spherical shell of radius r has this asymptotic behavior. To probe the electric form factors, one usually uses scattering experiments and fits their results to phenomenological expressions for the form factor.

But the form factor contains nothing at all about interaction- or state-dependent information. The interaction-dependent information is instead coded in an external potential or a multiparticle formulation, and the state-dependent information is coded in the wave function or density matrix, which (at any given time) is independent of the Hamiltonian.

Also, the information contained in the form factor is only about the free particle in the rest system, defined by a state in which momentum and orbital angular momentum vanish identically. In an external potential, or in a state where momentum (or orbital angular momentum) doesn't vanish, the charge density (and the 
resulting charge radius) can differ arbitrarily much from the charge density (and charge radius) at rest. 

(The above is taken from the entry Are electrons pointlike/structureless? of my theoretical physics FAQ, where more information and references can be found.)

answered Aug 11, 2014 by Arnold Neumaier (12,365 points) [ revision history ]
Most voted comments show all comments

The OP talked of experiments, not of theory. Who is ''they''? How do you know what ''they'' mean? What would be the difference between observing a radius of electron of the order $10^{−18}$m and one of the order $10^{−15}$m?

One doesn't need QED to find out that the proton has a finite radius - comparison with the classical scattering is enough. The electron also has a very accurately known finite radius of 2.8179403267(27) fm; see CODATA Recommended Values of the Fundamental Physical Constants: 2010.

The difference is that the electron radius is explained by QED since the physical electron is different from the bare Dirac electron (which has charge but no e/m field), while the proton radius is far too big for a point-like particle of the proton's mass and charge.

@ArnoldNeumaier: Sorry, Arnold, I did not know that the electron had a finite radius of 2.82 fm. I thought these data were not used in QED calculations.

QED doesn't assume an electron radius. Rather it is calculated; see Weinberg's QFT book, vol. 1,  (11.3.33). He doesn't complete the calculation, though, due to an infrared divergence since soft photons remain unaccounted for. A more thorough discussion is here, where the authors say: ''According to QED an electron continuously emits and absorbs virtual photons (see the leading order diagram in Fig. 8) and as a result its electric charge is spread over a finite volume instead of being pointlike.''

Strictly speaking, quantum objects cannot have sizes since their boundary is always fuzzy. Not even big quantum objects such as the sun. But the sun has already the same problem as a classical object; its radius is not well-defined. In both cases one describes the radius by some number obtained by comparison with the exact radius a comparison object would have.

The form factors of the electron are computable from QED. Reporting a charge radius is just a simple way of summarizing some information on the extension of the electron contained in the form factor. It gives the radius of a classical particle that would generate a (to low order) equivalent form factor, but it is a measure of the quantum size. 

Most recent comments show all comments

@ArnoldNeumaier: A finite radius of electron $r_e\approx 2.82 fm$ you refer above is just a classical electron radius. It has nothing to do with QED effects and is not measurable.

The value $\langle r_e^2\rangle_n$ due to "emission and absorption of virtual photons" depends on $|n\rangle$, i.e., it is state-dependent. When $n\to\infty$, it diverges (which is physical). In other words, a free real electron is rather "extended" in the elastic channel.

Yes, I agree to a great extent. I just insist on importance of distinguishing elastic, inelastic, and inclusive pictures. Factually, we speak (as we think) of an objective (elastic) size, whereas experiments give inclusive picture and we mistake them.

EDIT: I do not understand why people vote down. It is often said that one cannot "localize" the electron in a region smaller than its Compton length because of pair creation. In scattering problems it corresponds to the "distances", so the Compton length is a specific distance for creation pairs. But below this threshold, photons are created. It is as inelastic process as the pair creation. All scattering processes from an electron are inelastic. If the probability of elastic process is zero, it means a huge quantum mechanical charge smearing of the target size. And point-like picture is an inclusive one. What is wrong with these QED results if you vote down?

+ 2 like - 0 dislike

“[P]robability of the [particle] being within that radius” is about probabilistic interpretation of the wave function of the said particle. Internal structure is another question (it is related somehow to wave functions, but of constituent parts, not the particle as a whole). See what happens to particles that we discovered to be composite, such as proton. What means that its radius has an order of one femtometre? Nothing about uncertainty of position of its center of mass, that may be wider or narrower for a specific wave function. And for the wave functions of, say, a hydrogen atom in a highly excited state, we can easily have a wave packet with much narrower position uncertainty (for the center of mass) than the atom’s size. I repeat: uncertainty of the position and the (intrinsic) size are two different questions.

The question in this form, as far as I understand, is an oxymoron. It says that the particle must have “no internal structure”, whereas implicitly refers to concepts of Copenhagen quantum mechanics for its (non-existent?) constituents. Wouldn’t quantum field theory be a better paradigm to address the problem of the size of fundamental objects? Doesn’t the notion of size itself become progressively less meaningful when we use deeper and deeper explanations of the structure of matter?

answered Aug 11, 2014 by Incnis Mrsi (-15 points) [ revision history ]
edited Aug 12, 2014 by Incnis Mrsi
+ 1 like - 0 dislike

Just another (much shorter) answer, which might be all wanted by the OP.

The smallest radius where the electron in a hydrogen atom is with high probability is of the order of the Compton wave length and roughly defines the radius of the hydrogen atom, not that of its electron.

The size of the electron is instead defined to be its charge radius, which (by QED) is nonzero but extremely tiny - far below the resolution of experiments which only give upper bounds. The charge radius is determined by means of scattering experiments and matching the results with those of scattering at a charged sphere of the right radius, the latter computed with classical electrodynamics.

 An experimental result that the radius of electron is $<r_0$ means that, with high confidence, the experimental set-up would have produced observable effects for a hypothetical charged sphere of radius $r_0$ following classical electrodynamics. (Compare the discussion for composite particles in http://en.wikipedia.org/wiki/Charge_radius.)

answered Aug 12, 2014 by Arnold Neumaier (12,365 points) [ revision history ]
edited Aug 12, 2014 by Arnold Neumaier

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