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Almost everything from the wikipedia page you link is just false, or at best very misleading. IMHO, that page was written by someone that doesn't know anything about quantum mechanics beyond what one could find in TV documentaries. "Not even wrong" came into my mind many times as I was reading the article. In quantum physics, a quantum fluctuation (or ...

3

You have drawn a Feynman diagram. Feynman diagrams are iconic shorthand for integrals over the variables of the problem. The calculation gives the probability for the reaction to happen, in this case the decay of a neutron . The observables are the four vectors of the initial (neutron) and final particles. The integral is over the variables . Here is a ...

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It can be stated this way: In this particular diagram, the W boson is in a state named off-shell i.e. we say that this boson is virtual. Virtual particles are allowed to have any mass value. They can't although violate charge conservation at the vertex. This 80 GeV mass of the W boson, is for a real W boson, which is on the on-shell state. The real ...

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So the entire electromagnetic force can be described as having these objects which interact by exchanging virtual photons. Photons -- light -- in some sense are the electromagnetic force. It is therefore unsurprising that light can be absorbed by a particle -- an electron, say -- as a sudden "push" which launches the electron in some new direction with ...

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I'm not a physicist but based on studies I did : all types of elementary particles and forcrs also have fields in the whole universe. and fields always are there so even if we don't see any particle in a place it does't mean that there is "nothing" in that place because fields are always there. based on the uncertainty principle Since we can't accurately ...

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A can be photon, Z or gluon. B can be u or d quark.

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As AccidentalFourierTransform has pointed out, there is no change of flavour in A, because the $u$ and $\bar{u}$ are annihilating, so it is a photon. For the $\pi^0$ decay, B is a virtual $u$ or $d$.

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According to the rules of qft there are virtual photons in the vacuüm. No, according to QFT the vacuum is static, in the sense that $P^\mu|\Omega\rangle=0$. Or put it another way, The vacuum at a time $t$ is exactly the same vacuum at a time $t+\Delta t$ for any $\Delta t$. This means that the picture of particles constantly appearing and disappearing ...

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