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Are point particles the reason for 'infinities' in QFT?

+ 5 like - 0 dislike
214 views

One of my professors told us this semester, that the 'infinities' that arise in QFT are partly due to the use of the $\delta$-distribution in the commutator relations which read (for fermions)

$\left\{\Psi(r'), \Psi^\dagger(r)\right\} = \delta(r-r')$

In reality we would not have such a $\delta$-distribution but an extended version of it.

Is this view correct? And if definitely yes, is my following view wrong?

As far as I understand it, the $\delta$-distribution is due to the fact that we deal with point particles. If e.g. the electron was an extended particle, then the $\delta$-distribution would be 'finite'.

Since experiments pin down the extension of a particle to $R < 10^{-18}m$ it is also likely that the $\delta$-distribution should really be there.

This post imported from StackExchange Physics at 2014-08-07 15:40 (UCT), posted by SE-user physicsGuy
asked Mar 7, 2014 in Theoretical Physics by PhysicsGuy (25 points) [ no revision ]
The fact that $\delta (0)=\infty$ is really enough to see that this will yield some infinities when (anti-)commutation relations have to be employed. Since we don't measure $\infty$ in experiments, I think it's reasonable to say that this mathematical framework is an idealization.

This post imported from StackExchange Physics at 2014-08-07 15:40 (UCT), posted by SE-user Danu
I guess what my professor meant was infinities that we get when calculating cross sections. The $\delta$-distribution is of course infinit at $r = r'$, but it is well defined if we assume that we integrate over it to get the cross section. What I am more interested in is, what is the physics behind the $\delta$-distribution? Is it's occurrence really due to the concept of point particles in QFT?

This post imported from StackExchange Physics at 2014-08-07 15:40 (UCT), posted by SE-user physicsGuy
I'm afraid you will get things like $\delta^2(x)$ or even higher power to integrate, in such case if you calculate cross section it is not just that at certain point you get a infinite density, it is that the total cross section also become infinite, which is unphysical.

This post imported from StackExchange Physics at 2014-08-07 15:40 (UCT), posted by SE-user Jia Yiyang

As you know, point particles in QM create bound states without problem or scatter from each other without problem because we use their interaction in the wave equations. As soon as you include their "self-actions", you encounter those problems. In Classical Electrodynamics the electron self-action includes unnecessary self-induction whose contribution is removed (subtracted) within the renormalization program.

I'll just comment on a single aspect of your question: it's important to keep hold of the "partly due to the use of the δ-distribution" that you mention in your Question. Discussing ways to approach understanding the relationship between the point-like, distributional structure of free fields and what is often called the infraparticle structure of interacting fields is too much for here.

I'll specialize to a scalar field in 1+3 dimensions so that I can be explicit. As well as the delta function at light-like separation, the 2-point Vacuum Expectation Values also diverge inverse quadratically at small space-like and time-like separation, which also causes problems. Correspondingly, the Fourier transform is asymptotically inverse quadratic in the 4-momentum. If the Fourier transform were asymptotically inverse quartic in the 4-momentum (which is what Pauli-Villars regularization achieves by subtracting the effect of a second field, which is unphysical because only addition is physically valid here(da-da-da-unitarity-da-da-da-so it's too big an if)), then there would be no delta function at light-like separation and the 2-point Vacuum Expectation Values would only diverge logarithmically (and, if there are no derivative couplings, all loop integrals would converge --- again, this for the special case of a scalar field, in 1+3 dimensions).

1 Answer

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In reality, there are no point particles. Particles in relativistic QFT are only point-like; see this answer. Point particles only appear in the unrenormalized version of QFT, which is divergent due to nonexistent products of distributions.

The physical, renormalized version ''dresses'' the particles with nontrivial form factors which describe their extension; then the infinities are gone. During the renormalization process, one has to work with distributions, but using microlocal theory (about the wave fronts of distributions) one can rearrange the computations such that one multiplies only distributions regular enough that their product is defined. This is the content of causal perturbation theory.

answered Nov 6, 2014 by Arnold Neumaier (12,355 points) [ revision history ]
edited Nov 6, 2014 by Arnold Neumaier

The physical, renormalized version ''dresses'' the particles with nontrivial form factors which describe their extension; then the infinities are gone.

What an "extension" has the dressed electron and why?

A charge radius, for example. We discussed this before; it is in the answer I had linked to.

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