I was learning about Feynman diagrams some years ago and I was addressing a proton turning into a neutron. In my notes I wrote that the energy of the boson released by the proton was greater than that of the proton's energy, and the reason this was possible was because of the uncertainty principle of energy and time. I then noted $\sigma_E \sigma_t \geq \hbar/2$. Now when I look back I was wondering how does the uncertainty principle of energy and time relate to the proton being able to attain such energy to dispel a W boson?
Answer
This usage of the "time-energy uncertainty principle" is nonsense, although sadly fairly common as a heuristic. For the correct usage of the time-energy uncertainty principle, see this answer from joshphysics.
The Feynman diagrams are human-readable encodings of terms in a perturbation series (in this case the one that computes the probability of a proton turning into a neutron). They do not encode an actual "process" in the sense that there was ever an actual W boson produced during the process - internal lines in such diagrams, often termed "virtual particles", do not correspond to actual particle states in the theory.
If you want to see them as particles for heuristic purposes, then you must allow them to be off-shell, that is, they can be "produced" with momenta that do not match the mass/energy that they should have if they existed as actual free particles. It is here that the time-energy uncertainty principle is often cited to "allow" such off-shell conditions for a brief amount of time, but there is no rigorous underpinning of this point of view, since there is no particle state associated with the internal line to begin with.
In particular, to compute the amplitude, we integrate over all possible momenta of the virtual particles, and there is no way to decide which of these was "actually" "produced", just as there is no way to ask of a quantum particle that you only detected at A and B which path it "actually" "took" between A and B.
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