Honoring conservation of the speed of the center of mass removes divergence in vacuum polarization?

In our previous blog post we argued that the energy E of the the positron and the electron in a vacuum polarization loop has to be greater or equal to zero. This is to conserve the speed of the center of mass.

e-  --------------------------

                   |  photon, no energy

                   | 

                   O vacuum polarization

                   |  loop

                   |

Z+ --------------------------

In the simplest Coulomb scattering diagram where the exchanged momentum has E = 0, both particles in the loop must then have E = 0.

This effectively removes one dimension, the time dimension, from the integral for the loop.

Recall that there is no time dimension in the calculation of the diagram without the loop: the "virtual photon" which pushes the colliding particles is a wave which undulates in spatial dimensions but not in time. The wave is a Fourier component of the static Coulomb potential which is time-independent.

Classically, Coulomb scattering does involve exchange of energy: when the electron approaches the nucleus, potential energy is temporarily converted to kinetic energy of the electron. The simple Feynman diagram ignores all this detail and just calculates the end result: spatial momentum was exchanged.

If there exists hypothetical vacuum polarization, it might be that its effect has to be calculated in the same fashion: just look at the end result, where no energy was exchanged. We may drop out the time dimension from the calculation, just as we did for the simplest diagram.

       + /\/\/\/\/\ -

          spring

Classically, an electric dipole in polarization may be modeled with a harmonic spring device. When a charge approaches, some energy is stored into the spring. When the charge recedes, the spring returns the energy back. The spring does not retain any energy.

Also, maybe we should drop one spatial dimension, too? Classically, the electron and the nucleus move in a plane. Why should we need to calculate in 3 spatial dimensions?

Additionally, the momentum exchange lives in a plane. Why would a single virtual pair whose momentum is not in that plane take part in the process?

We have argued that it makes sense to drop a dimension or two in this case from the corresponding Feynman integrals. Does the vacuum polarization loop diverge in that case? Probably not. Removing one dimension is like doing dimensional regularization. If there are less than 4 dimensions, then the integral converges.

https://canvas.harvard.edu/files/936398/download%3Fdownload_frd%3D1%26verifier%3DPAHa18X0trEAYjwcsmQXrhusjECF3ct6JNxDQ41E

The note by Matthew Schwartz (2012) in the link contains calculations about dimensional regularization of the vacuum polarization loop. We need to find a more detailed calculation and check how it is modified if time is dropped.

https://homepages.dias.ie/ydri/qft_chap_five.pdf

Badis Ydri (2011) has written a more detailed calculation.

If we drop the time dimension, the integral Π^μν_2 in formula (31) in Ydri's paper seems to diverge worse than with the normal 4-dimensional space. That is because in the square of the 4-momentum,

        k^2 = -E^2 + q^2,

the negative term -E^2 cancels positive contributions of spatial momentum q^2. If we drop time, we lose -E^2.

The divergence of the integral in Ydri's formula (31) is only logarithmic, because of cancellation effects, if we integrate one thin spherical shell at |k| = r at a time. The divergence would be worse if we would integrate in some other order. Absolute convergence is what we would like to have.

We need to analyze if we can drop also one spatial dimension. Also, we need to check destructive interference again.