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

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?

## 1 Answer

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.

• 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! – P. C. Spaniel Jan 30 at 22:42
• I updated the answer. – Qmechanic Jan 31 at 21:52