# How does the Feynman's $i\epsilon$-prescription make the Feynman propagator causal?

The Feynman propagator is non-vanishing outside the light cone, but still manages to be in accord with causality. How is this achieved? What does the $i\epsilon$-prescription have to do with this?

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## 1 Answer

The $i\epsilon$ prescription is so that you get the time-ordered product of operators, which is the correct interpretation of such an expectation value from the perspective of the path integral.

If you think about the propagator as a correlation function, one expects some spacelike correlation simply due to uncertainty. In some sense, if you just look locally, this looks like there are virtual particles travelling faster than the speed of light. You cannot actually send messages with these, however.

This is explained nicely here http://en.wikipedia.org/wiki/Propagator#Faster_than_light.3F . See also the EPR paradox mentioned there.

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