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?
$\begingroup$
$\endgroup$
7
-
$\begingroup$ 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 $\int \frac{d^4k}{(2\pi)^4} (propagator)$. $\endgroup$– WillCommented Jul 7, 2013 at 15:30
-
$\begingroup$ 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. $\endgroup$– WillCommented Jul 7, 2013 at 15:40
-
$\begingroup$ @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? $\endgroup$– PraharCommented Jul 7, 2013 at 15:45
-
$\begingroup$ @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? $\endgroup$– WillCommented Jul 7, 2013 at 15:47
-
1$\begingroup$ Related physics.stackexchange.com/q/44250 $\endgroup$– Diego MazónCommented Jul 7, 2013 at 19:31
|
Show 2 more comments
1 Answer
$\begingroup$
$\endgroup$
13
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.
-
1$\begingroup$ Ah!!! I had just worked that out and was about to write my solution. :) +1 $\endgroup$– WillCommented Jul 7, 2013 at 17:57
-
1$\begingroup$ I don't think that's right. While its true that the $i \epsilon$ prescription ensures convergence, it is not introduced ad hoc just to ensure convergence. In fact, the In and Out states precisely provide the extra contribution of $+i \epsilon$ which in the end makes it all work. Now, when doing the ghost path integral it is not clear where a similar contribution of $+i \epsilon$ should come from since one does not have In and Out ghost states. My argument for this was that we do indeed have In and Out ghost states but that they do not contribute to any physical amplitudes. Any comment? $\endgroup$– PraharCommented Jul 7, 2013 at 18:03
-
1$\begingroup$ Oh no! This seems right on the surface, but I agree with Prahar in that you are effectively using in and out states to get this $i\epsilon$ prescription as defined by the path integral. I think a precise answer will require a careful derivation from the ground up, beginning with the method FP gauge fixing. $\endgroup$– WillCommented Jul 7, 2013 at 18:11
-
1$\begingroup$ @Trimok - I agree that the $i\epsilon$ prescription is required for the path integral to converge. I am not contesting that. Further, Wick rotation to a Euclidean action is also possible only due to the presence of the $i\epsilon$. However, I don't think it is introduced "by hand". It follows from the derivation of the path integral from the operator formalism. It's the time ordering in the operator side that tells us exactly which prescription of $i\epsilon$ to use and a derivation of this prescription can be done. So, no ad hoc introduction of $i\epsilon$ is required. $\endgroup$– PraharCommented Jul 7, 2013 at 20:01
-
1$\begingroup$ @Trimok - In fact, I think that is precisely the OPs question. While the $i\epsilon$ prescription can be derived for usual fields, it does seem to come out naturally using the FP procedure. Either we are not being careful or it must be introduced by hand this time. The second option does not sound to appealing to me. But maybe that's what's required to be done. Note that one often DEFINEs the theory using the gauged fixed path integral (with the correct $i\epsilon$ prescription) without any reference to the original action. In this case, this question does not arise. $\endgroup$– PraharCommented Jul 7, 2013 at 20:07