how can the virtual fermionic lines be vertical in kaon mixing Feynman box diagram (or in general)? Wuouldn't this mean that they travel a distance $\neq 0$ in 0 time?
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$\begingroup$ see also physics.stackexchange.com/q/297004/50583 and its linked questions $\endgroup$– ACuriousMind ♦Commented Aug 18, 2022 at 21:40
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That is just a diagrammatical representation for convenient purposes. Note that when we calculate the amplitude of this diagram we end up using the full form of the fermionic propagator, which integrates over all the possible 4-momenta that the virtual particle can take.
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1$\begingroup$ okay I see now, it's a fermionic propagator, so it's like the t-channel of $e+e- \leftarrow e+e-$ having the "vertical" photon propagator. This is possible just with propagators though right? $\endgroup$ Commented Aug 18, 2022 at 21:38
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$\begingroup$ Exactly, it's the same for all diagrams. We use the propagator for the virtual particles since it has all the information about the different ways it will exchange the momenta along the diagram. $\endgroup$ Commented Aug 18, 2022 at 21:41
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