Why do we only sum probability amplitudes of Feynman diagrams, never subtract?

Let us consider an electron-electron collision, Møller scattering.

https://en.wikipedia.org/wiki/Møller_scattering

   ^              ^
     \           /
       \       /
        |~~|
       /       \
     /           \
   e-            e-

If the energy of the collision is less than 1.022 MeV, then no new electron-positron pair can be produced.

If the energy is larger, then new pairs will be created.

Intuitively, the production of a pair should reduce the probability amplitude of an elastic collision where the electrons exchange a large amount p of 4-momentum.

But the possibility of pair production is not explicitly apparent in the Feynman formula which sums the t- and u-channels of elastic scattering.

Could it be that the Feynman formula for elastic scattering in some implicit way "knows" about the possibility of pair production? That is probably not true. We may imagine new physics where a lighter variant of electron exists and can produce a large number of new pairs. How could the Feynman formula for ordinary electrons be aware of such new physics?

Furthermore, the electron-electron collision may produce a photon. That is, the collision is not elastic. How could the elastic collision formula be aware of the (complex) process of photon radiation in the collision?

The simplest pair production diagram in Møller scattering is the following:


 e-      e+             e-     e+
  ^     ^               ^     ^
    \     \              /     /
      \     \          /     /
       |~~\____/~~|
      /                      \
    /                          \
   e-                          e-

The diagram contains four photon-electron vertices:

        |~~

while the elastic scattering only has two such vertices. If the integral formula for the pair production has a much smaller probability amplitude (or cross section) than the formula for elastic scattering, then in the first approximation we can ignore the effect which pair production has on the elastic probability amplitude.

We have not yet found the pair production cross section from literature.

A semiclassical treatment of pair production

Let us consider electrons and positrons as classical objects which obey special relativity.

For each classical trajectory of particles we associate an integral over a lagrangian density. The integral gives the phase, or the probability amplitude, of that history.

If an electron and a positron come to the distance 3 * 10^-15 m from each other, then their combined energy is zero, assuming that they have no kinetic energy.

We assume that we can add such a zero-energy pair to any history where the existing other particles bump into the particles in the pair, giving the pair a 4-momentum which makes them real, a 511 keV electron and positron.

That is, classical collisions can create new pairs by tearing apart an electron and a positron which exist as a zero-energy pair.

Our assumption is somewhat similar to the hole theory of Dirac. In Dirac's hole theory, an electron with an energy -511 keV gets excited to a state of an energy +511 keV, leaving behind a hole, which is the positron.

The zero-energy pair can be considered a zero-energy state of a positronium "atom". The atom gets excited by other particles, and goes into a state where the pair will annihilate again (= virtual pair), or goes into a state where the electron and the positron escape as real particles.

Is our model deterministic? Suppose that the electrons exchange more than 1.022 MeV of energy in a collision. What determines if a pair is produced, or if the collision is elastic?

Or should we make the model probabilistic?

The two electrons which enter the experiment can be considered as uncorrelated. Pair production can be seen as a positron moving backward in time, colliding with both electrons, and scattering forward in time.

But is the positron moving backward in time uncorrelated with the two electrons?

Suppose that the initial state of electrons is such that they would collide and exchange more than 1.022 keV of kinetic energy. Is there always some positron trajectory which will rob some of the energy and produce a real pair?

We believe that empirical tests show that elastic collisions are possible at large energies. Pairs are not always produced.

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.