We have been claiming that the vacuum polarization diagram can only make the Coulomb force stronger - not weaker.
q + k
q ~~~~~ O ~~~~~
-k
We argued that the photon cannot change its phase by colliding with "nothing", that is, with a virtual pair which pops up from empty space. We called such a phase change a Baron Munchausen trick - pulling oneself from a swamp by one's own ponytail.
q ~~~~~~~~~~~
If the phase of the photon q does not change, then the vacuum diagram adds to the probability amplitude of the plain diagram above. That is, there is more scattering, and the Coulomb force must be stronger.
e- ---------------------------------------
| virtual photon
| spatial momentum q
|
/ \
/ \ massive boson
/ k \ q + k 4-momentum
-----------------------------------------
proton
But on September 5, 2021 we wrote that Formula (12.448) in Hagen Kleinert's book looks like the diagram above. In the diagram, the proton sends a massive Klein-Gordon boson whose propagator is something like
1 / (p² + m²),
where p is the 4-momentum of the particle.
The particle can be interpreted as a photon which has the rest mass m. In the diagram, q is spatial momentum and k is arbitrary 4-momentum.
Note that we in this blog use the east coast sign convention (- + + +) in the Minkowski metric. That is why m² has the plus sign.
For the diagram above, we no longer can claim outright that it has to make the Coulomb force stronger. It could be that it makes the force weaker.
What is different in the new diagram? The 4-momentum k no longer pops up from empty space. It comes from the proton. The new diagram makes a lot of sense: it is the proton which is generating polarization around itself. The proton should provide the resources for the process. It should not rely on something coming out from empty space.
Earlier we have suggested that the natural cutoff for |k| is |q|. Could the natural cutoff for |k| in the diagram be the proton mass 1 GeV? For the Lamb shift, that would be high enough to act as an "ultraviolet" cutoff.
If we assume that the new diagram makes the Coulomb force weaker, and put the cutoff at 1 GeV, we will get the exact same numerical results as Hagen Kleinert in Formula (12.543) for the vacuum polarization contribution to the Lamb shift.
The high-momentum case
e- ------------------------------------
| virtual photon
| spatial momemtum q
|
e+ ------------------------------------
Let us then consider electron-positron scattering (Bhabha scattering) which was measured in the LEP at CERN.
In the case of the LEP, the energy of the collision, say, 80 GeV, is the natural cutoff. Is that high enough?
In our new model, a charge gets more polarization around it if the charged particle is heavier, or is a member in a tightly bound system, like the quarks inside the proton.
Question. Why does the hydrogen atom then appear neutral to the outside world? The proton is much heavier. The charge of the proton should be screened by larger polarization. This, of course, depends on the measuring apparatus, too. What is the mass of the system which is used in the measurement?
Could it be that the massive boson cannot reach far? No. Then the charge would appear to be smaller at a short distance.
In a collision, the effective measured charge will depend on both the mass-energy of the charge and the exchanged momentum q.
We need to check the results from the LEP and other tests of the running coupling constant. Do they measure the effective charge as a function of q only?
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