Why do we know that an elementary particle is a point particle?

[fxdung]

Elementary particle can be consider as a "wave packet" of the field,but a "packet" of field must have a size.Why do we know elementary particle is point particle?

 

[hilbert2]

If you make finite sized particles collide with high enough speed for the de Broglie wavelength to be smaller than the particle diameter, you will begin seeing evidence of internal structure in the scattering results. That way you can at least find an upper limit for the particle diameter.

 

[DarMM]

Elementary particle can be consider as a "wave packet" of the field,but a "packet" of field must have a size.Why do we know elementary particle is point particle?

They're not really point particles like in idealized classical physics, i.e. a single point carrying a mass, charge, etc

The proper details of particles in Quantum Field Theory is quite above a B-level thread, but roughly we model them as a sort of collection of properties that have certain chances of being detected at certain values with specific pieces of equipment.

 

[fxdung]

Please explain at A level

 

[DarMM]

Please explain at A level

Well in Quantum Field Theory a "particle" is simply a space of states that transform into each other under the Poincaré group and other internal symmetry groups. This space must be irreducible, i.e. no subspaces within it transform separately without mixing.

That is a particle is a fundamental (in the sense of not decomposable) group transformation "unit" in the Hilbert space of the field theory's states. Nothing really related to being a "point" or anything like it.

In a free theory you can combine these irreducible representations via the tensor product to get states with 2, 3, 4, etc particles

In an interacting field theory this is generally not possible. The ability to combine particle states only emerges at asymptotic time, so multiparticle particles only enter as an idealization in the far future and past.

 

[fxdung]

Is there any "suggestion" it must be Poincare group?

 

[DarMM]

Is there any "suggestion" it must be Poincare group?

Well if you are doing it in a Minkowksi background it would be. However the same results essentially apply to QFT in curved space time since most spacetimes are asymptotically flat.

 

[PeterDonis]

Please explain at A level

Moderator's note: Thread level changed to "A".

 

[PeterDonis]

the same results essentially apply to QFT in curved space time since most spacetimes are asymptotically flat.

I think you mean all spacetimes are locally flat, i.e., the group of local transformations whose irreducible representations define "particles" is still the Poincare group. No curved spacetime has the Poincare group as a global group of transformations, not even if it is asymptotically flat (and I'm not sure I would agree that "most" spacetimes are asymptotically flat anyway, since I don't know how you would define a measure on the set of all possible spacetimes).

 

[DarMM]

I think you mean all spacetimes are locally flat, i.e., the group of local transformations whose irreducible representations define "particles" is still the Poincare group. No curved spacetime has the Poincare group as a global group of transformations, not even if it is asymptotically flat

It's to do with formulating scattering theory in curved backgrounds, the theory becomes more complex if the spacetime is not asymptotically flat and in most cases the S-matrix is not known to exist. It's a complicated subject I shouldn't have tried to summarize it in one line. The "most" was loose, i.e. "most spacetimes one sees in practice" rather than a formal measure theoretic statement. As you said the set of spacetimes is a poorly understood problem.

 

[PeterDonis]

It's to do with formulating scattering theory in curved backgrounds

Ah, ok, so for this particular case the relevant group is given by the asymptotic spacetime.

 

[DarMM]

Ah, ok, so for this particular case the relevant group is given by the asymptotic spacetime.

Let me gather the details. I don't have all the theorems and references to mind off the top of my head. I'll do a bit of reading of my old notes and post something detailed.

 

[A. Neumaier]

It's to do with formulating scattering theory in curved backgrounds, the theory becomes more complex if the spacetime is not asymptotically flat and in most cases the S-matrix is not known to exist.

I believe that in spacetimes that are bounded at a fixed time, a conventional S-matrix cannot exist because of the Poincare recurrence theorem.

 

[fxdung]

Now let us consider with another way of seeing:When field is excited then it create a particle.A mode of excited field can consider as a particle(a quantum of field).But in the book QFT of Zee, he say that a particle is a "packet wave" of field.So I do not understand because a "packet wave" is a set of many modes.Is it correct that only when"particle" enters the measure machine it become "packet"(point particle) so when we observe the experiment we see "point particle"?Is that Zee want to say?

 

[vanhees71]

In relativistic QFT the wave-function picture is misleading. There a particle is defined as a single-particle Fock state of an asymptotic free field (see the postings by @DarMM above). I don't think that there is a good particle description for "transient states" of interacting fields at all.

 

[Heikki Tuuri]

When field is excited then it create a particle. A mode of excited field can consider as a particle (a quantum of field). But in the book QFT of Zee, he say that a particle is a "packet wave" of field. So I do not understand because a "packet wave" is a set of many modes. Is it correct that only when "particle" enters the measure machine it become "packet" (point particle) so when we observe the experiment we see "point particle"? Is that Zee want to say?

Let us use a radio transmitter to send a wave packet containing just one photon of a wavelength 1 meter to space. Suppose that an observer in a rocket approaches us almost at the speed of light. He might measure the photon wavelength as 1 micrometer. He is holding a photographic plate where a very small dot appears.

Did he prove that the photon is a "point particle"? The wave function of the photon "collapsed" into a small dot on the photographic plate.

The photon is a "point particle" just in this sense, that in a measurement, the energy of the photon is found in a small patch of space.

If we point a red laser towards a slit of width 100 nanometers, some red light will pass the slit. We were able to measure the horizontal position of a 700 nanometer photon at the accuracy of 100 nanometers. If the photon arrived in a wave packet, after this measurement, the new wave packet has a new form. It will diffract strongly when it comes out of the slit. In this case, a spatially large wave packet "collapsed" into a narrow wave packet which will spread very rapidly after the slit.

I like to think that a photon is a wave, but when we do an experiment, we may force the wave to print, or output, a precise position in space.

Bell's inequality bans simple hidden variable theories. The photon did not have any precise position before the measurement. It was not a point particle.

What about the electron? We can measure the location of a high-energy electron very precisely. But since the electron has a spin and a magnetic moment, we cannot imagine it as a classical point particle. A point does not spin.

In my physics blog, I have described my efforts during this spring to find a Feynman path integral type description for the electron as a point particle or a billiard ball. I was not able to find any intuitive model. I had to settle with the fact that the electron is a wave which is described by the Dirac equation. The wave can under a measurement output a precise location in space, but it is not possible to think of it as a classical point particle. No one has found an intuitive physical model for the Dirac equation. The equation is a result of a mathematical trick which seems to lack a physical interpretation.

 

[PeterDonis]

In my physics blog

Your blog is not a valid source. Please do not reference it here.

 

[fxdung]

Why do proton, neutron... have sizes, but electron,neutrino... are point particles?

 

[PeterDonis]

Why do proton, neutron... have sizes, but electron,neutrino... are point particles?

Protons and neutrons are composite objects--they are made of three quarks. Electrons and neutrinos, as far as we can tell, are not composite objects.

 

[akhmeteli]

Elementary particle can be consider as a "wave packet" of the field,but a "packet" of field must have a size.Why do we know elementary particle is point particle?

My understanding is: when people say that elementary particles, such as electrons, are point particles, they mean that they are described by the Dirac equation / QED with great accuracy. The Nobel prize winner Dehmelt, who established strong experimental limitations on the size of the electron, wrote (Physica Scripta. Vol. T22, 102-110, 1988): "an elementary Dirac particle, such as the electron, is the closest laboratory approximation of a point particle."

Why do proton, neutron... have sizes, but electron,neutrino... are point particles?​

Because proton and neutron are not described by the Dirac equation well enough, they manifest some additional structure at small distances. Also note that the anomalous magnetic moment of proton/neutron is very high.

 

[DarMM]

Why do proton, neutron... have sizes, but electron,neutrino... are point particles?

As I said above the electron and neutrino aren't really point particles in any sense in modern quantum field theory.

Particles are states forming group transformation units, that is irreducible representations. Collections of them only appear as idealizations at asymptotic times.

Beyond even this, the electron doesn't possesses a sharp mass. Nonperturbatively the electron's two point function has no poles. So an electron is in fact sort of an integral over irreps.

If you look at rigorous treatments like that of Haag or Steinmann you ultimately define an electron as the class of states which excite certain types of probes. I'll have more details in a while.

 

[fxdung]

Please give the details, I wish I will be able to understand.

 

[Dale]

Several forum feedback posts have been moved to the feedback section. They are off topic here. Please keep to the technical discussion here. Further off topic posts here will be deleted, not moved.

 

[Heikki Tuuri]

In physics, we do not have a clear picture of what being a "point particle" means.

If the electron would be a classical point particle, then its electric field would have an infinite energy: integrating r^2 * 1/r^4 down to r = 0 gives an infinite value.

What we know is that you can measure the position of a photon or an electron with a great precision. But that does not imply that they are "point particles" in any sense of classical mechanics.

 

[DennisN]

Hi, @DarMM

Beyond even this, the electron doesn't possesses a sharp mass. Nonperturbatively the electron's two point function has no poles. So an electron is in fact sort of an integral over irreps.

This is way beyond my knowledge, but it sounds very interesting, and if you could explain this further I would be very interested. Particularly what "sharp mass" means.

 

[DarMM]

Hi, @DarMM

This is way beyond my knowledge, but it sounds very interesting, and if you could explain this further I would be very interested. Particularly what "sharp mass" means.

I am currently gathering all the information required to write this up. It will be a very long set of posts. It was the last thing I promised before going inactive, so I'm working on it.

 

[vanhees71]

Since when does an electron decay? Within the Standard Model the electron as the lightest charged lepton cannot decay and thus has a sharp mass. It's a stable particle and its Green's function thus has a pole on the real axis of $s=p_{\mu} p^{\mu}$, which defines its mass.

 

[A. Neumaier]

Since when does an electron decay? Within the Standard Model the electron as the lightest charged lepton cannot decay and thus has a sharp mass. It's a stable particle and its Green's function thus has a pole on the real axis of $s=p_{\mu} p^{\mu}$, which defines its mass.

In QED, the electron mass is a branch point, not a pole, becaiuse of Imfrared effects coming from the zero mass of photons. Thus the electron mass spectrum is continuous.

 

[vanhees71]

True, but does this imply that the free electron is in fact unstable? If so what's the (theoretical) decay mode, and why isn't this observed?

 

[A. Neumaier]

True, but does this imply that the free electron is in fact unstable? If so what's the (theoretical) decay mode, and why isn't this observed?

It is a stable infraparticle, which means that it has an additional mass degree of freedom, which formally behaves like an additional momentum dof - the latter generates the continuous spectrum of the energy. In the QM treatment of multielectronic systems, this dof is generally suppressed. Indeed, infrared problems are not much addressed in the literature.

 

[A. Neumaier]

It is a stable infraparticle, which means that it has an additional mass degree of freedom, which behaves like an additional momentum dof.

Here is more on infraparticles. For more on the branch point of the electron propagator, see, e.g., section II of

• T. Appelquist and J. Carazzone, Infrared singularities and massive fields, Phys. Rev. D 11, 2856–2861 (1975).