Vacuum polarization in QED: the virtual pair is a dipole particle

We found a new way to interpret the QED vacuum polarization diagram.

   e-  ------------------------------------

                            |  q  dipole interaction

                            |

                  ~~~~~~~~~    virtual pair (dipole)

                /          |            \       

              / k         | q           \ k

        ------------------------------------

   proton

The virtual pair (wavy line) in the diagram is an electric dipole particle, and the interactions k and q in the diagram are dipole interactions with an electric charge. The dipole particle is a boson.

The first 4-momentum transfer k creates the dipole particle. The second transfer k destroys it.

The Fourier decomposition of a dipole potential is of the form

       p • P / p²,

where p is a spatial momentum vector and P is the dipole momentum of the dipole.

Let us assume that the polarization vector always aligns parallel to the momentum vector. Then the propagator of the dipole potential is of the form

       1 / |p|.

In the diagram, the dipole absorbs and emits the spatial momentum q at the same time. Why at the same time?

Otherwise, the dipole would gain angular momentum which the proton cannot absorb. The proton interacts with the dipole through its symmetric 1 / r Coulomb potential. The proton cannot absorb angular momentum through that potential.

In classical polarization, the dipole in the diagram would exert forces on the electron and the proton simultaneously.

In the diagram we only count the propagator of the dipole particle once because the absorption and the emission of the momentum q is simultaneous. The motion of the dipole itself does not change at all.

We assume that the propagator of the proton is 1. It is almost on-shell in the diagram. In Feynman diagrams, outgoing on-shell particles are conceptually assigned a "propagator" value 1, that is, no propagator is written for them.

The propagator of the dipole particle is the massive Klein-Gordon propagator for a tunneling particle:

       1 / ( |k|² + m² ),

where |k|² is the euclidean square of the 4-momentum k. See our previous blog post for an explanation why a tunneling particle must live in the euclidean metric.

The high-momentum q case

For high momenta q, the value of the Feynman integral over all 4-momenta|k| < q for our diagram is something like

       ~ 1 / q * 1 / q * log(q² / m²).

But what is the power of the electron charge e in the integral?

There are 8 vertices. The electron and the proton couple to the dipole through their charge e. The dipole moment depends on e and the length of the axis of the dipole. We have not specified any length in the diagram.

Let us assume that polarization is only slightly superlinear. Then doubling e approximately doubles the force. The vertices which connect to the dipole do not carry the factor e on them. The power of e in the integral is only 4.

The low-momentum q case

The integral is over

        1 / q * 1 / q * 1 / |k| * 1 / |k| * 1 / m².

We integrate over all 4-momenta |k| < q.

The result is

       ~ 1 / m²,

like it should be.

Conclusions

We assumed that the virtual pair behaves like a massive boson electric dipole particle which is tunneling, and that its dipole moment is slightly superlinear on the electric field strength.

We assumed that the dipole absorbs and emits the momentum q simultaneously. We claimed that the propagator of the dipole particle must therefore be counted only once.

Our model produces formulae which have the right form in terms of q and m, save some constant coefficients. We do not even try to calculate the coefficients now. We have to be able to derive the properties of the dipole from basic Feynman diagrams, or from classical arguments, first.