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Qmechanic
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Jia Yiyang
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How to get the $i\epsilon$ prescription for a Faddeev-Popov ghost propagator?

In path integral formalism, for a physical field there will be an $i\epsilon$ term in the action, which comes from identifying the in and out vacuum, and in turn this $i\epsilon$ will naturally appear in the denominator of the corresponding propagator. However for FP ghost, it is only introduced to rewrite the functional determinant in an exponential form, and the issue of identifying an in and out ghost vacuum never enters the picture, thus no $i\epsilon$ term in the ghost part of the action. Yet all ghost propagators I've seen do have an $i\epsilon$ in the denominator, so where does it come from?