The idea is that the electron sends a random virtual photon to a random direction, which "smears" the location of its probability distribution close to the proton.
The classical vertex correction is not random. That is a big difference from Welton. Why would the classical correction produce the same result? Does the classical correction displace the electron in such a way that δr² is roughly the same as for Welton?
The mass reduction fraction is:
M(r) = 1 / (4 π ε₀)¹⋅⁵ * e³ / (mₑ¹⋅⁵ c³)
* sqrt(2) / r¹⋅⁵.
Let the electron move some distance Δr toward the proton. If it moves 1% faster, it will in the same time move 1.01 Δr. If the electron comes from a large distance close to the proton, we obtain the correction simply by integrating:
∞
∫ -1 / r¹⋅⁵ dr * constant
r
= 2 / sqrt(r) * constant.
A reasonable assumption is that we should calculate the displacement when
r = λₑ = h / (mₑ c).
But the exponents for e and ε₀ will differ from the Welton formula. This is not going to work. The classical correction integrates over a wide volume around the proton, while the Welton approach only considers the vicinity of the proton. The displacement is fundamentally different in these two approaches.
The Feynman approach and the classical vertex correction
Above is the Feynman integral from Peskin and Schroeder (1995). The reduction in the scattering amplitude is calculated this way:
1. cut off the infrared divergence by introducing a "mass" λ for the photon p - k;
2. remove the ultraviolet divergence with a cutoff, or with Pauli-Villars, or with dimensional regularization;
3. do a "renormalization" and declare that the difference in the integral value for q infinitesimal and q (the studied value) is the correction to the scattering amplitude. The difference reduces the amplitude. That is, the correction is the missing part of the integral when we let q grow.
Let us assume that the electron scatters to a very large angle, close to 180 degrees, from a proton. In the classical vertex correction, what is "missing", is the far field of the electron, which does not have time to bounce back from the field of the proton.
The Feynman integral calculates that part of the electric field which does bounce back with the electron? In the Feynman diagram, the virtual photon looks like that it misses the bounce, but that is misleading. It probably describes the part of the electric field which does bounce back from the proton. For very large |k|, the Feynman integral value depends little on q. These very large |k| definitely stick tightly attached to the electron, and bounce back together with the electron. Large |k| describe the inner electric field of the electron.
Bremsstrahlung does not bounce back. Bremsstrahlung can be understood as the electric field which does not stick to the electron. In the previous blog post we calculated that it is a fixed fraction 1/7.5 of the effective mass reduction of the electron.
The Feynman diagram calculates the "flux" of electrons which moves to the same direction as the simplest elastic scattering diagram says, that is, the direction of the movement has a momentum change q. It makes very much sense that the integral value describes a flux which must be added to the flux of the simplest elastic scattering diagram.
These observations do not yet explain why the missing Feynman integral would calculate the effect of the mass reduction.
Classical vertex correction calculates how much q is reduced if we assume a certain fixed distance R at which the electron passes the proton? Yes. If the mass is reduced, then the bounce from the proton transfers less q, this is clear. But is this argument bound to some R? Yes, since if the electron in the Schrödinger solution for an s orbital is at some distance r from the proton, then it typically does not come very close to the proton. This is the "fuzziness" in the Schrödinger solution.
The Feynman diagram calculates the probability amplitude for a fixed q. This is clear, and is different from the classical calculation. The amplitude is reduced for each fixed q, because of the vertex correction.
If the Feynman diagram calculates anything sensible at all, then it has to calculate the classical vertex correction which is corrected by the Schrödinger solution of the s orbital. The classical limit of the vertex correction has to obey the classical vertex correction. The Feynman diagram must, to some extent, obey the classical limit. The classical vertex correction for a very narrow ellipse around the proton would be very large. To obtain a reasonable result, we must modify the classical vertex correction according to the Schrödinger wave function for an s orbital. That is what we did in the previous blog post.
The semiclassical vertex correction which includes the Schrödinger s orbital solution is not "perturbative", like the Feynman integral is. This may be one of the reasons why there are no divergences in the semiclassical correction. An extremely narrow ellipse would in the classical vertex correction lead to divergence-like behavior: the far field which does not follow the electron could be everything whose distance from the electron is larger than infinitesimal, and have an infinite mass-energy. Schrödinger saves us from that behavior. The Schrödinger solution is not perturbative, but a "full" solution. The effective mass reduction of the electron from the far field is not perturbative, either. Thus, the semiclassical vertex correction is not perturbative.
Divergences in the classical vertex correction
As we mentioned in the previous section, the classical electric field of a point charge contains an infinite mass-energy. When calculating the effective mass reduction of the electron, it might happen that the reduction is larger than the electron mass, or even infinite. Then we have a classical divergence.
If a point-like electron passes a point-like proton at a distance much less than rₑ, then the classical mass reduction is likely to exceed mₑ, which is a nonsensical result. Classically, the electron could easily come this close to the proton.
The Schrödinger s orbital solution removes this problem, if we accept that we can cut off the miniature probability that the electron is closer than rₑ from the proton. But is that cut-off acceptable?
The problem exists in the Schrödinger s orbital, too. What is the interaction like if the electron and the proton come very close? Does the electron see the quarks inside the proton?
Since the energy of the neutron is larger than the hydrogen atom, the proton cannot capture the electron. However, the collision might result in an unexpected scattering of the electron. This unexpected scattering may have a tiny effect on the s orbital energy.
We conclude that for an approximation, we are allowed to cut off the immediate vicinity of the proton. But we simply do not know the exact form of the potential there.
No infrared divergence. The electric field of the electron very far from the hydrogen atom does not have time to take part in the movement of the atom. Theodore Welton cuts off frequencies for which
ν = 2 π c / λ > π c / a₀
<=>
λ > 2 a₀,
where a₀ is the Bohr radius. The cut-off is not logical. Even longer wavelengths than that can displace the electron. Why should they be removed from the smearing? Close to the proton, long wavelenths do not have enough time to displace the electron, so a cut-off is warranted. But that is not related to the Bohr radius a₀.
Classically, a cut-off makes a lot of sense. The far field at a distance greater than
c t
does not have time to take part in the circling of the electron around the proton. The effective mass of the electron is permanently reduced.
The cut-off is ad hoc for Welton, but for the classical vertex correction we have very solid grounds for the cut-off. Also, even if there were no cut-off, the mass-energy of the far field would be finite, and there would be no divergence.
How does the Feynman diagram calculate the effective mass reduction of the electron?
Let us increase |q| from an infinitesimal value. The value of the associated Feynman integral decreases. Here we assume that we have cut off (regularized) the infrared and the ultraviolet divergences. We also assume that the electron "bounces" from the proton to a very large angle, as if it would be "reflected" back.
• e- ^
\ / ^
\ / | q
\ ● /
proton+
The momentum exchange q very roughly is
q ~ 1 / b,
where b is the impact factor. If the effective mass of the electron is reduced by a fraction M << 1, then the momentum exchange q in the "bounce" is also cut by the fraction M. Let M be, say, 1%. The momentum exchange q can be kept the same if we reduce b by 1% – but then the cross section shrinks by 2%.
We see that an effective mass reduction fraction M cuts the cross section by a fraction
2 M.
The Feynman integral method calculates the loss of cross section for a fixed q. The mass reduction method calculates the loss of q in a bounce. These figures are related in the simple linear way shown above: 2 M.
The momentum exchange q does not "disturb" the Feynman integral value much if in the diagram above,
|k| >> |q|.
This, obviously, corresponds to the inner electric field of the electron "sticking" tightly to the electron. The effective mass of the electron is not reduced by the inner field.
Let us assume that q is a result of a
-1 / r
potential field. The momentum exchange q does not happen abruptly.
The following might explain the Feynman integral value. Let us assume that |k| >> |q|. We assume that the emission of the virtual photon k corresponds to a "sharp hammer" blow to the electromagnetic field. If the integral value does not change, that means that destructive interference for earlier blows essentially cancels the k that would be emitted. Nothing changes in the inner field of the electron. The momentum exchange q happening gradually does not affect consecutive hammer blows much.
The following diagram describes the Feynman vertex diagram intuitively better:
^
\ p + q
\
/ \
\ \
\ \ q
k | |------------------
/ /
/ /
\ /
/
/ p
e-
The virtual photon k "sticks" tightly with the electron in the curve caused by the momentum exchange q.
Those k which do not stick with the electron may be absorbed later by the electron, or may escape as bremsstrahlung. An intuitive diagram for those k:
p + q
^
| |
| |
| |
| |
| \
| \ e'
| \
| \
k | |----------------
| / q
| /
| /
| / e'
| /
| /
/
/ p
e-
The mass-energy in k does not "bounce back" with the electron which has a reduced effective mass. We have marked the reduced-mass electron by e'.
When q grows from an infinitesimal value, the second diagram eats the market share of the first diagram. The scattering amplitude is reduced relative to what would happen if the entire electric field of the electron were rigidly bound to the electron.
Mass reduction and the Schrödinger equation
A non-perturbative way to calculate the scattering of an electron from a proton is to use the Schrödinger equation. The de Broglie wavelength
λ = h / p
of the electron shrinks as it approaches the proton, because the momentum
p
of the electron grows.
The shrinking of the wavelength makes the electron wave to turn toward the proton. This is scattering.
If there is some mechanism, through which p grows less as the electron approaches the proton, then the wavelength λ changes less, and the electron wave turns less. The scattering cross section is reduced. This is what happens when the far field of the electron cannot follow the electron, and its effective mass is reduced. The kinetic energy of the electron is
p² / ( 2 mₑ'(r) ) = -V(r) + C,
where mₑ'(r) is the effective mass of the electron, V(r) is the potential in the field of the proton, and C is a constant.
If the effective mass mₑ'(r) becomes smaller as r decreases, then p grows less => the de Broglie wavelength of the electron is longer. Then there is less scattering.
There is a paradox though: if the electron has a high kinetic energy and passes quite far from the proton, then a classical calculation shows that there is more scattering because of the mass reduction. The electron will pass the proton closer, because the electron is lighter. How can we reconcile this with what we claimed above?
*** WORK IN PROGRESS ***
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