virtual e-
________
/ \
/ \
~~~~~~ A B ~~~~~
virtual \ /
photon \_________/
virtual e+
time ------------>
The above vacuum polarization loop is a part of a larger Feynman diagram.
The energy E of the experiment gives us some bounds on how precisely we can know the position of A in spacetime. Let the length scale be L, for example, L = 10^-12 m for a 1 MeV collision.
Let |p| be the 4-momentum of the virtual electron e-.
If we can hypothetically measure the momentum of the virtual positron e+, then we can know p almost precisely.
How much does hypothetical destructive interference of various e- paths suppress the magnitude of the wave function of e- at B?
Since e- is virtual, it can move to any direction from A. The direction is not dictated by p.
In one dimension, if we have each point in a line segment of length L sending waves in unison, that is, the phase of the created wave is the same at each point at any time t_0, then there is a total destructive interference for each line segment of the length λ, where λ is the wavelength. The suppression by the destructive interference goes as ~ 1 / |p|.
What about n dimensions and a smooth "brightness" distribution of each sender in an area of size L^n? There is certainly a lot of destructive interference, but does the suppression go as
~ 1 / |p|
or even better?
The Feynman vacuum polarization integral is very roughly
1 / |p|^2
over the 4-dimensional p space. It diverges badly. If we integrate over 3D thin spherical shells of a radius r in the p space, we get an integral of
|r|^3 / |r|^2
over r that runs from 0 to infinity. To make the integral to converge, we would need a suppression factor better than 1 / |p|^2 from destructive interference.
We need to study the impact of various dimensions and brightness distributions on the amount of destructive interference.
But is there destructive interference? The standard way to calculate Feynman diagrams ignores any destructive interference at the internal vertices of the diagram. On the other hand, the probability amplitudes for a final outcome are summed in the calculation, and there we do have destructive interference.
In our vacuum polarization loop, the phases of e- and e+ run to opposite directions. There is no destructive interference at all caused by a large momentum p in e- and the opposite momentum -p in e+, in the virtual photon which is produced at B, because the effects of e- and e+ cancel each other out. A question: does it make sense that the probability amplitude of the particles coming to B is much less than the amplitude of the particle flying off?
We may hypothesize about measuring the position of the virtual positron e+, and reducing the uncertainty in the position of the spacetime point A. But if we accept the principle that a virtual particle can move any way regardless of its momentum p, then our measurement would not help to pinpoint the location of A.
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