Why destructive interference does not cancel high 4-momentum in the vacuum polarization loop?

In our blog we have previously claimed that a physical phenomenon with an associated 4-momentum p cannot produce any phenomena of a higher absolute absolute value of the 4-momentum.

In the wave interpretation, p is associated with a wavelength λ = 1 / |p| in a spacetime diagram.

time
^      wavelength λ
|    \     \
|      \     \   -----> 4-momentum p
|        \     \
 ------------------------------> space

We can develop the wave forward in time and space in the diagram through the Huygens principle: let each point act as a point source of new waves, and calculate the interference of the new waves at a new point.

Let

       D(ψ) = 0

be a wave equation. If the right side is strictly zero at all spacetime points, we say that it is a homogeneous equation.

If the right side is not zero everywhere, then we call the non-zero part a source.

The Green's function method calculates the response of a wave equation to a Dirac delta source in the equation. That is, we assume that the wave equation

         D(ψ) = 0

has on the right side, instead of 0, a Dirac delta term at a certain spacetime point x. The Green's function is the solution of the new equation. We say that it is the response to an impulse source of the wave equation.

If we have a tense string, then pressing the string briefly at a certain point with a finger with a force F, is an impulse F × Δt, and the resulting wave is the response to an impulse source.

If we have a source which is not concentrated to one spacetime point, but is continuous, we can build an approximate solution by summing the response to an impulse source at each each spacetime point x.

Suppose that the source is cyclic and has a certain wavelength λ in the spacetime diagram.

The response to a Dirac delta impulse contains waves for all kinds of 4-momenta p.

It is obvious that there tends to be a destructive interference for all waves where p does not match the cycle (wavelength λ) of the source.

In particular, all high 4-momenta p will have a total destructive interference.

This is the reason why tree-like Feynman diagrams have strictly restricted 4-momenta p at every part of the diagram.

However, if we allow an imagined wave, as in the previous blog post, to have any 4-momentum p, then the imagined wave introduces an arbitrarily high p, or an arbitrarily short wavelength λ, to the diagram. Destructive interference does not cancel it.

Diverging of the vacuum polarization loop integral

          q + p -->
     ~~~O~~~~~~~~~~~~
q -->   <-- p            q -->

The vacuum polarization loop carries the photon 4-momentum q, as well as an arbitrary 4-momentum p which circles around the loop.

The impulse response to a Dirac delta impulse at a spacetime point x contains a certain spectrum (= propagator) of various 4-momenta. The intensity depends on the sum q + p. That is, the probability amplitude of the diagram above depends on both q and p.

If we allow any p, then there exists no sensible probability distribution for p. The integral of probability amplitudes over all p diverges, or alternatively, we may say that the integral is not defined.

The causality of a Feynman diagram

The imagined wave with an arbitrarily high 4-momentum p does not follow "causally" from the input waves to the Feynman diagram.

The diverging of the integral seems to be the result of this acausality.