On October 22, 2025 we were able to analyze the QED vertex correction to some extent. We continued our analysis on August 5, 2026.
Our hypothesis is that the formula
F₁(0) - F₁(q²)
calculates the "long-lived" virtual photons which are "detached" from the electron for a long time during the scattering process.
e- • ----------------
\
● v
proton+
The classical vertex correction is due to the fact that the far electric field of the electron does not "follow" the electron as the electron makes a sharp turn. The far field is "detached". The inertial mass of the electron is reduced, and it will pass the proton closer, gaining more momentum. This increases the scattering amplitude.
The question: why would the Feynman diagram and the integral calculate this classical process?
In the link we have a paper which calculates the QED vertex correction. The paper first handles the ultraviolet divergence in electron self-energy. Then it proceeds to calculate the vertex correction. The infrared divergence is handled by assuming a small photon mass λ.
The self-energy integral diverges for large virtual photons |k|. Our hypothesis in this blog is that those k do not exist at all. They are wiped out by destructive interference.
Similarly, we expect large |k| in the vertex correction to be wiped out by destructive interference.
Classical action
If we study the classical behavior through an action, we accept off-shell electrons, and so on. Any history, or a path, is legal in an action. We have to find a locally extremal value for the action integral, in order to find a legal physical history.
Note that since the electric field of a classical electron can be bent or distorted, a classical electron does not always obey the energy-momentum relation (we set c = 1):
E² = p² + mₑ².
The field may be missing some energy, or it can have too much energy. A classical electron can be off-shell.
From this point of view, the Feynman vertex correction diagram describes also a classical electron.
-k virtual photon
~~~~~~~~
/ \
/ A B \
e- -----------------------------------------
p k + p | k + p + q
|
|
| q virtual photon
proton+ -----------------------------------------
The Feynman integral F₁(0) corresponds to an infinitesimal q. The integral is not zero. In QED, that integral is considered the "null case", which does not increase the scattering amplitude of the electron. This makes sense in the classical context, too. The big question is that, when we let q grow, why does the change in the integral value describe the added scattering amplitude?
Recall that the electron propagator essentially is:
1 / how much off-shell is the electron.
In our previous blog post, we already noted that if k is orthogonal to the spatial component of p and p + q, then the value of the integral does not change as q changes. This is because then the value of k determines uniquely how much off-shell the electron is in the internal electron lines.
Only when k is a spatial momentum which lies in the plane determined by spatial momenta of p and q, then value of the integral changes.
Let the spatial momentum in p be p' and in q let it be q'.
----> p'
|
v p' + q'
If q is infinitesimal, then the most off-shell we get from a value of k which points to the direction of p'. Both internal electron lines are a lot off-shell then. Also a value k antiparallel to p' makes the lines a lot off-shell.
Such a value of k reduces the value of the integral. But it does not reduce the integral value "maximally", because the formula is
reduced value * reduced value.
We could reduce more, if p' and p' + q' would point to different directions. Then we would have
unreduced value * reduced value.
We see that the integral becomes smaller when q increases. That makes sense.
We can interpret the smaller integral value as saying that, of the virtual photons produced by the hammer strike as the electron approaches the proton, fewer are absorbed "easily" by the electron. All those virtual photons must eventually be absorbed, but the process may be more complicated, and slower.
Let us compare q infinitesimal and q such that p' + q' makes a 90 degree turn.
The integral value grows for k pointing to the direction of p'. We interpret that the electron is able to absorb more virtual photons k of such type.
The integral value decreases for k pointing to the direction of p' + q'. It also decreases if k is antiparallel with p' + q', but the decrease is less. We interpret that more such k will "live long" before being absorbed by the electron.
^ -k
p |
e- • ------
\ | p' + q'
\ v
v
●
proton+
If the electron emits a virtual photon -k which is antiparallel to p' + k', then the electron will pass the proton closer. The impulse from the proton, q, increases. The scattering amplitude increases. It makes sense that we add something to the scattering amplitude. But why should it be F₁(0) - F₁(q²) that we add?
When the electron in the diagram turns tightly down, pulled by the proton, then intuitively, a virtual photon -k sent straight upward can "break loose". The virtual photon will live longer. Effectively, the electron will turn down more than if we ignore its electric field.
Now we are approaching the classical vertex correction. A part of the field of the electron breaks loose.
*** WORK IN PROGRESS ***
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