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The generating functional $Z[J]$ of some scalar field theory is

\begin{equation} Z[J(t,\vec{x})]=\int \mathcal{D}\phi e^{i\int (\mathcal{L}+J\phi)d^4x} \end{equation}

This integral is not well defined because it's argument is an oscilating function. In order to be solvable one needs at least a tiny real part so that it eventually decays at large values of $\phi$. At this point there are two ways to solve this (at least I've seen this two ways in the literature).

  1. Wick Rotation

First we change $t \rightarrow i \tau$ which has the following effect

\begin{equation} Z[J(i\tau,\vec{x})]=\int \mathcal{D}\phi e^{-\int (\mathcal{L}_E+J\phi)d^4x} \end{equation}

That is: the $dt=id\tau$ gives a minus sign in the exponential and the Lagrangian becomes euclidean. Since this was only a change of variables the integral is still ill-defined (acording to the change of variables $t=i\tau$ the new variable $\tau$ should be complex so we have the same problem as before. However, we now analitically continuate to the complex plane and make $\tau$ a real number so that the integral actually converges. Then we can do all the calculations and go back to real time evaluating $J$ in the right variable at the end.

  1. $i\epsilon$ Prescription

In this option we make a different change of variables $t\rightarrow t(1-i\epsilon)$ keeping things at first order in $\epsilon$. Without going into much detail, this has the following effect

\begin{equation} \mathcal{L}=\frac{1}{2}\phi\big(\square -m^2+i\epsilon\big)\phi \end{equation}

which gives as that little real part that will make the integral converge. [Here we are using the $(-,+,+,+)$ Minkowski sign convention.] This has the benefit of giving the right Feynman propagator.

Questions

Are these two methods equivalent? Which one is the standard way of calculating the generating functional $[J]$? The second method has the advantage of explicitly giving the Feynman propagator with the correct prescription. How is this achieved with the first method?

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Starting in the Minkowski formulation, the Feynman $i\varepsilon$-prescription is just the first infinitesimal angle $\theta=\varepsilon$ of a Wick rotation

$$ t(\theta) = e^{i\theta} t_M, \qquad \theta~\in~[0,\frac{\pi}{2}], \qquad t(\theta\!=\!0)~=~t_M, \qquad t(\theta\!=\!\frac{\pi}{2})~=~it_M=t_E, $$

in the complex $t$-plane to the Euclidean formulation. Heuristically, on physical grounds, no poles & branch cuts are expected after the first infinitesimal rotation $\theta=\varepsilon$, so this in turn is equivalent to the full $\theta=\frac{\pi}{2}$ Wick rotation.

NB: Wick rotation of spinors is subtle, cf. e.g. arXiv:hep-th/9611043 & this Phys.SE post.

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  • $\begingroup$ Hello! Thanks for the answer, it's clarifying. How does the wick approach give you the i \epsilon prescription? or is it more like this: the epsilon is a trick to solve the path integral and at the end you take the limit to zero just as the wick rotation which is a trick to solve the path integral but at the end you analitically continuate back to real time. is that it? thanks again! $\endgroup$ – P. C. Spaniel Jan 30 at 22:42
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Jan 31 at 21:52

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