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

Question

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 $i\epsilon$ (with the correct sign) 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?

This post imported from StackExchange Physics at 2014-04-07 12:30 (UCT), posted by SE-user Jia Yiyang\

UPDATE(27-Jun-2015): I recently came across the following paragraph in Faddeev and Slavnov's book (Gauge Fields: An Introduction to Quantum Theory, 2nd Edition), page 94:

I take it as they are claiming that for ghosts (and other non-propagating fields), it doesn't matter whether you use $+i\epsilon$ or $-i\epsilon$. The claim would be a very desirable one if true, but I cannot prove it in any simple-minded way by inspecting the Feynman graphs, for example, in the following graph (where wavy lines represent gluons and dashed lines ghosts)

the sign in front of $i\epsilon$ of the gluon propagator must be kept fixed, and changing the sign for that of the ghosts seems to induce a very nontrivial change of the loop integral.

(I would've uploaded the whole page of the book in case someone wants to see the context, but the upload limit is only 1MB.)

EDIT: Let me add some context here. I met Professor Faddeev on a conference meeting and asked him this question during a coffee break. He promptly agreed that the $i\epsilon$ for ghost doesn't appear naturally in the path integral, since there's no in and out states for them. But due to the limited time window of the coffee break, he only pointed me to his book with Slavnov. So far I've only found the quoted paragraph which vaguely makes the assertion that the boundary condition doesn't matter, which seems to be suggesting either sign for $i\epsilon$ is fine.

Comments

The $i\epsilon$ prescription doesn't seem to depend on which propagator you are talking about. It is naturally introduced when calculating the free Feynman propagator for any field. We don't need to refer to in and out states at all. It arises when writing (scalar field example) $\langle 0 |T\{ \phi_1(x) \phi_2(y) \}| 0\rangle$ as a Fourier transform of the momentum space result. That is, you calculate in position space and rearrange to get it in the form $\int \frac{d^4k}{(2\pi)^4} (propagator)$.

This post imported from StackExchange Physics at 2014-04-07 12:30 (UCT), posted by SE-user Will

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What I am referring to applies for the operator approach to QFT - I'm not sure how you get the $i\epsilon$ in the path integral, but given that they are equivalent methods, you should be able to get the same result, somehow? This seems like a fun little paradox.

This post imported from StackExchange Physics at 2014-04-07 12:30 (UCT), posted by SE-user Will

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@Will - In the Path Integral approach, you do in fact get the $i \epsilon$ prescription as a contribution from the In and Out states. The two methods are equivalent and therefore we should be able to deduce the $i \epsilon$ prescription for the ghosts without having to invoke the operator approach at all, right?

This post imported from StackExchange Physics at 2014-04-07 12:30 (UCT), posted by SE-user Prahar

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@Prahar the OP's problem is that there shouldn't be ghost in and out states. Well at least, that's what I think the problem is?

This post imported from StackExchange Physics at 2014-04-07 12:30 (UCT), posted by SE-user Will

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(Because they aren't physical particles).

This post imported from StackExchange Physics at 2014-04-07 12:30 (UCT), posted by SE-user Will

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Hmmm... even in the operator approach we are assuming that the ghosts are in and out states in the free theory. It seems that they only way to get the $i\epsilon$ term is to do this, but make the restriction that ghosts are never in and out states in the full vacuum (that is, don't use them as external states). Can others comment on this?

This post imported from StackExchange Physics at 2014-04-07 12:30 (UCT), posted by SE-user Will

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Related physics.stackexchange.com/q/44250

This post imported from StackExchange Physics at 2014-04-07 12:30 (UCT), posted by SE-user drake

Answer 1

Bosonic path integrals :

$$Z = \int D\phi ~e^{-i \large \int ~ dx [\frac{1}{2}\phi (\square+m^2)\phi]}$$

or Femionic path integrals (like Fadeev-Popov ghosts) :

$$Z = \int D\eta D \tilde \eta ~e^{-i \large \int ~ dx [\tilde \eta^a \square \eta^a]}$$

are not mathematically well-defined, because of the presence of the imaginary unit in the exponential.

To ensure convergence and meaning of these expressions, the prescription is then : $$\square + m^2 \rightarrow \square + m^2 - i\epsilon$$ When $m=0$, this simply gives the prescription : $$\square \rightarrow \square - i\epsilon$$

Obviously, the form of the propagators comes direcly from this prescription.

This post imported from StackExchange Physics at 2014-04-07 12:30 (UCT), posted by SE-user Trimok

Comments

@JiaYiyang: The point is that since ghosts do not appear as an external line, the sign of the ghost propagator $i\epsilon$ doesn't seem to matter. But for the purpose of analytic continuation to imaginary time, the standard sign is needed. (The opposite sign would lead to an analytic continuation across the branch cut, and one would end up in the nonphysical second sheet.)

Faddeev and Slavnov's ''it is convenient'' possibly refers to just this advantage of the ''correct'' sign.

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@JiaYiyang: I don't know about ''not necessary'' - can't follow the argument in detail. Approximate calculations in QFT are rarely completely transparent unless they are done by yourself or in endless places. 

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That one necessarily ends up in the nonphysical sheet is because of the way the analytic continuation has to be done. Frequencies make physical sense only if their imaginary part is negative, and the continuation to imaginary time must go through the physical domain to be meaningful. If you now do the same analytic continuation on the Green's function with the wrong sign then it is clear that one immediately passes the branch cut and then is automatically (by continuity on the Riemann surface) in the unphysical sheet.

Because of this I also wouldn't trust the statement that the choice of the sign for ghosts is irrelevant.

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@ArnoldNeumaier,

Because of this I also wouldn't trust the statement that the choice of the sign for ghosts is irrelevant.

Well, the statement is vague and that sign doesn't matter is my own interpretation of it, which might not be what Faddeev meant. However I still think it's a reasonable guess since ghost was introduced to represent a functional determinant det$M$ which has no $\epsilon$ dependence at all, unlike physical fields.

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@JiaYiyang: Always remember that functional integrals are themselves ill-defined and need a concrete interpretation to give proper sense to them.

Even though the sense is approximate only (except for quadratic actions), it does not justify arbitrary formal manipulations - it is easy to give ''sensible'' recipes that produce arbitrarily wrong results. One must always select the correct manipulations to extract a physical meaning. Unfortunately, without a proper mathematical foundation that doesn't yet exist, what is correct is difficult to tell before you finished a calculation and can compare with experiment or alternative approximate methods such as lattice simulations.

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@Trimok: Let me put it this way: what's the reason for performing a Wick rotation? Does the same reason apply to ghost fields?

This post imported from StackExchange Physics at 2014-04-07 12:30 (UCT), posted by SE-user Jia Yiyang

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In fact, the prescription $-i\epsilon$ allow a Wick rotation, and a Wick rotation corresponds to euclidean path integrals : $Z = \int D\phi ~e^{- \large \int ~ dx [\frac{1}{2}\phi (P_0^2+\vec P^2+m^2)\phi]}$ and $Z = \int D\eta D \tilde \eta ~e^{- \large \int ~ dx [\tilde \eta^a (P_0^2+\vec P^2)+ \eta^a]}$, where $P_0, \vec P$ are operators.

This post imported from StackExchange Physics at 2014-04-07 12:30 (UCT), posted by SE-user Trimok