Why do Feynman rules have minus signs?

latentcorpse
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Consider equation 5.46 in these notes:
damtp.cam.ac.uk/user/tong/qft/qft.pdf

(i) why does only one term have the i \epsilon in it? I thought this was because the (p-p')^2-\mu^2 term can never vanish but this doesn't make sense since the \phi particle will have momentum p-p' and since momentum squares to mass squared \mu^2, surely we will actually have (p-p')^2-\mu^2=0 always?

(ii)why is one term plus and the other minus? he talks earlier about how we pick up extra minus signs for fermionic diagrams from statistics but doesn't explicitly tell us how to recognise when they are needed (well he does in the calculation on p120 but quite a common exam question is to write down the amplitude from feynman rules - you aren't going to have time to reproduce p120 every time you need to check a minus sign!)
looking at diagrams 25 and 26 it seems to be that whenever two fermionic legs cross we get a minus sign (this is why one of the terms in figure 25 gets a minus sign but none in figure 26 do since the external legs are bosonic there) but then in (5.46) which corresponds to figure 27, why does that s channel diagram get a minus sign? No external legs get crossed there? There must be an easy way of recognising when we need the minus sign?

Thanks.
 
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