We only considered the fact that a reduced electron mass helps it to come closer to the proton: more Coulomb focusing, resulting in more momentum exchange.
But we forgot the fact that the reduced mass also makes the electron to move faster, so that it will be a shorter time close to the proton: less momentum exchange.
Our blog post on March 13, 2021 contained a very rough calculation of the effect of a shorter time close to the proton: the effect raises the energy level of 2s by 5 - 7 μeV, while the vertex correction contribution to the Lamb shift is 4 μeV.
Which process wins in which case?
Opposing classical vertex corrections in a fast fly-by: scattering increases
e- • --> v large << c
|
v F
1 impact factor
● proton+
^ y
|
-----> x
We assume that the electron moves very fast and that the electron arrives along a straight line. The impact factor is 1. We assume that its mass becomes reduced during the length 2 of its path closest to the proton. We assume that the electron moves the closest to the proton in the time 1 / v.
Let us reduce the mass of the electron by a fraction
0 < f << 1
during that path of length 2. Let the electron move to its closest position to the proton:
1 Δx
e- • ------------- ---
| Δy
1 × corrected position
● proton+
The reduced mass makes the acceleration in the x direction by the fraction f larger, and also the acceleration in the y direction. We have marked the corrected position of the electron after these changes.
The component of F in the x direction is, on the average,
Fx ≈ 0.35 F.
The force downward is, on the average,
Fy ≈ 0.85 F.
Thus,
Δy / Δx ≈ 0.85 / 0.35 ≈ 2.4.
The time to the closest position is reduced by, say 0.35%, but the force downward increases by 2 * 0.85% = 1.7%. The momentum gained downward increases by 1.7% - 0.35% = 1.35%.
In this case, the classical vertex correction increases the scattering. The effect of the electron coming closer to the proton dominates. We can almost ignore the fact that the time for the fly-by decreases.
Opposing classical vertex corrections in the hydrogen atom in the 2s orbital: scattering decreases
The electron moves very slowly when it is far from the proton. Then it dives toward the proton, and its velocity greatly increases as it gains kinetic energy from the electric potential.
e-
• ^
| |
| |
| ● | proton+
-----
The reduced mass electron is "reflected" from the proton. It first goes down, and then comes up almost the same way.
If we reduce the mass of the electron as it descends, then less momentum is "reflected" back upward. The momentum exchange q is smaller. In this case, mass reduction reduces the momentum exchange: it reduces scattering.
Our March 13, 2021 post calculated this effect very crudely. The order of magnitude agreed with the Lamb shift. We got a rise of the energy level by 5 - 7 μeV.
A summary of the problems in the QED vertex correction
Let us recapitulate what we have learned about the QED vertex correction in the past years.
1. The correction is ambiguous. By setting the infrared cutoff (the photon mass λ) to a suitable value, we can get almost any value for the QED correction.
2. The vertex correction seems to be canceled somewhat by the correction caused by bremsstrahlung, but it is not clear how much. Is it canceled entirely?
3. The QED vertex seems to produce an infinite number of infrared photons during every fly-by, but QED uses a clearly flawed infrared cutoff λ to sweep that under the rug.
4. The correction uses a suspicious ad hoc regularization scheme for the ultraviolet divergence.
5. The QED vertex correction claims that there is less scattering because of it, while the classical vertex correction says that in many cases, there is more scattering.
6. The classical limit of the QED vertex correction is wrong.
7. If we let h → 0, the QED vertex correction becomes infinite, while in quantum mechanics, the system should approach the classical limit, i.e., the classical behavior.
8. The QED vertex correction only depends on the momentum exchange q. If we let the electron move very slowly very far away from the proton, it will gather the momentum exchange q. Intuitively, the vertex correction under such peaceful circumstances should be zero, not the value predicted by QED.
A lot of problems. As if the QED vertex correction would be an entirely useless construct.
The following can be said to be an advantage of the QED vertex correction:
- It can reproduce the Bethe term in the Lamb shift, at least if the infrared cutoff λ is tuned in a suitable way.
Reanalysis of the far field reabsorption in the QED vertex correction
On August 21, 2026 we outlined why the QED vertex correction might calculate the same thing as the classical vertex correction, in a special setting. But we had the wrong impression that the vertex correction increases the scattering probability, while it decreases that. Let us do a new analysis. We study the hydrogen 2s orbital mentioned above.
*** WORK IN PROGRESS ***
