# Wick rotation vs. Feynman $i\varepsilon$-prescription

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

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

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

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

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

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

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?

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.