# Can real particles be annihilated with virtual antiparticles and the other virtual particle pair replaces the real? Can this explain uncertainty?

Time goes from left to right.

An electron is staying still, so from left to right, it's a straight line. A virtual photon got emitted from it, becomes a virtual pair of positron, electron, the virtual position annihilates with the original electron, then the resulting photon got absorbed by the virtual electron, which became real, position of original electron changed.

Is the Feynmann diagram above valid? If so, it should happen all the time. What's not forbidden happens. Might this explain the uncertainty principle of position and momentum?

• The diagram looks the same as an electron propagator with two photon bubbles (four vertices). If you stretch it horizontally. Isn't the e an e-? Feb 10, 2022 at 10:22
• As @Felicia points out, if you pull the ends of your diagram you get a straight electron line, with two virtual photons interlacing in their attachments, a fine diagram. The bogus spacetime interpretation is what is confusing you in attempting to come up with a pseudo-"physical story" about it. It is a valid and useful piece of mathematics already incorporating the uncertainty principle. How you are attempting to "explain" that UP this way is totally unclear, but why should it matter? Feb 10, 2022 at 14:40
• @Felicia Sorry, forgot to mention that time goes from left to right. The e is a positron then. Say if this happens all the time, there's always uncertainty of position of the electron, and also uncertainty of the direction of travel of the electron (momentum), I dunno if there's any derivation that can be done to come out with the uncertainty principle relationship. Anyway, thanks. As long as the Feynman diagram is valid, I think I got my answer. Feb 11, 2022 at 1:01
• I am not sure how you are relating this to the uncertainty principle (which is a property of the basic principles of quantum mechanics that go into QFT and it's not QFT's place to explain the uncertainty principle) but you should note that Feynman diagrams are not processes, they are terms in a perturbative expansion. All such diagrams of $e\to e$ scattering add up and give the correction to the free-electron propagator. There is no rigorous sense in which they literally represent physical processes that are happening.
– user87745
Feb 11, 2022 at 1:50

The amplitude of the diagram from your question depends on the electron-photon coupling (aka electron charge, $$e$$, or the fine structure constant $$\alpha \sim e^2$$). This is a free parameter in QED that describes the strength of the electromagnetic interaction. The dependence of the amplitude of your diagram on this parameter is $$e^4 \sim \alpha^2$$ (this is determined by the number of interaction vertices in the diagram which in your case is 4).
The observable effects of the uncertainty principle do not in any way depend on $$e$$ (or on $$\alpha$$): they only depend on $$\hbar$$.
A particularly relevant example is a theory of neutral particles (mesons, neutrinos, etc.) that don't interact with electromagnetism at all (modulo the higher order moments for compound particles coming from internal structure): $$e = \alpha = 0$$. The diagram's contribution to the amplitude is exactly zero, but the uncertainty principle still applies.