# 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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it doesn't seem to be, it is, please fix. THe $i\epsilon$ doesn't make it causal, it makes it Wick rotatable. – Ron Maimon Sep 1 '12 at 4:01
Yes, Ron, I fixed it. – Newman Sep 1 '12 at 16:07

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.