Timeline for How to get the $i\epsilon$ prescription for a Faddeev-Popov ghost propagator?
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| Jul 8, 2013 at 15:26 | comment | added | Trimok | 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. | |
| Jul 8, 2013 at 15:16 | comment | added | Jia Yiyang | @Trimok: Let me put it this way: what's the reason for performing a Wick rotation? Does the same reason apply to ghost fields? | |
| Jul 8, 2013 at 14:46 | comment | added | Trimok | @JiaYiyang : It is not the same thing, see Feyman propagators, because the idea is to be allowed to perform a Wick rotation, so, if you define the Wick rotation to be a rotation with a positive angle $90°$, it is possible with the prescription I gave. With the other prescription, you are stuck. | |
| Jul 8, 2013 at 13:46 | comment | added | Jia Yiyang | @Trimok: I thought you were talking about the convergence of the Gaussian integral, but if you were talking about the convergence of the propagator, then both ${(\square+i\epsilon)}^{-1}$ and ${(\square-i\epsilon)}^{-1}$ look equally valid to me. | |
| Jul 8, 2013 at 7:20 | comment | added | Trimok | @Prahar : The path integral formalism is the more fundamental one, while it is true, that the operator formalism is more practical in a lot of cases. The presentation by Zee (Quantum Field Theory in a nutshell) is very clear and very impressive about that. | |
| Jul 8, 2013 at 7:16 | comment | added | Trimok | @JiaYiyang : The propagator for ghosts is $\square^{-1}$, while it is $(\square + m^2)^{-1}$ for a scalar field, so it is the same kind of problem. | |
| Jul 8, 2013 at 2:21 | comment | added | Jia Yiyang | @Trimok: Prahar understands me correctly. Besides, I'm a bit skeptical about convergence argument, for bosonic fields of course no problem, but for grassman fields I'm not sure how one defines convergence. | |
| Jul 7, 2013 at 20:07 | comment | added | Prahar | @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. | |
| Jul 7, 2013 at 20:01 | comment | added | Prahar | @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. | |
| Jul 7, 2013 at 18:21 | comment | added | Trimok | @Prahar : "While its true that the iϵ prescription ensures convergence, it is not introduced ad hoc just to ensure convergence" . Not at all, this is precisely to ensure convergence that the $i\epsilon$ prescription is introduced. Without that, the coherence of QFT would be just wrong. This is the same trick that the Wick rotation which brings you to an Euclidean Action $S_E$ which has to be positive. | |
| Jul 7, 2013 at 18:11 | comment | added | Will | 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. | |
| Jul 7, 2013 at 18:03 | comment | added | Prahar | 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? | |
| Jul 7, 2013 at 17:57 | comment | added | Will | Ah!!! I had just worked that out and was about to write my solution. :) +1 | |
| Jul 7, 2013 at 17:53 | history | answered | Trimok | CC BY-SA 3.0 |