Feynman $i\varepsilon$-prescription in path integral by adding an imaginary part to time

It is known that the well-definiteness of the path integral leads to the Feynman's $$i\varepsilon$$-prescription for the field propagator. I've found many ways of showing this in the literature, but it is precisely the way that I have learned in my QFT course (and which I have not found in literature) that I do not understand.

Context of the problem

Considering the case of a real scalar field for simplicity, one has that the following path integral (evaluated at asymptotic times)

$$\begin{equation} \lim_{T \rightarrow \infty}\int_{\phi(-T, \vec{x})}^{\phi(T,\vec{x})} \mathcal{D}\phi \ \text{exp} \left( i \int^T_T dt \int d³ x \ ( \mathcal{L}+J \phi) \right)\tag{1} \end{equation}$$

can be expressed as

$$\begin{equation} \lim_{T \rightarrow \infty} \sum_{m, n} e^{-i\left(E_n+E_m \right)T} <\phi, T|n, T>_J _J \tag{2} \end{equation}$$

where $$|n>_J$$ are eigenstates of the hamiltonian $$H$$ in the pressence of the source $$J$$. In order to make this oscillatory exponential converge (and properly define the path integral) one adds to $$T$$ a small imaginary part $$T \rightarrow T(1-i\varepsilon)$$. With this, one writes the vacuum persistence amplitude as $$\begin{equation} <0|0>_J = \frac{1}{N} \lim_{\varepsilon \rightarrow 0} \ \lim_{T \rightarrow \infty(1-i\varepsilon)} \int \mathcal{D} \phi \ \text{exp} \left( i \int_{-T}^T dt \int d³x (\mathcal{L}+J\phi)\ \right) \equiv \frac{1}{N} Z[J]\tag{3} \end{equation}$$ where the constant $$N$$ is typically taken to be $$N=Z$$.

My problem

In order to relate $$Z[J]$$ with the Feynman propagator $$D_{F}(x-y)$$, one typically writes the argument of the exponential in the Fourier space, then makes the change of variable $$\hat{\phi}(p)'=\hat{\phi}(p)+(p²-m²)^{-1} \hat{J}(p)\tag{4}$$ (which leaves $$\mathcal{D}\phi'=\mathcal{D}\phi$$) to get

$$\begin{equation} Z[J]=Z \text{exp} \left(-\frac{i}{2} \int \frac{d⁴ p}{(2\pi)⁴} \hat{J}(p) \frac{1}{p²-m²} \hat{J}(-p)\right).\tag{5} \end{equation}$$ Here I'm using the $$(+,-,-,-)$$ Minkowski sign convention.

Now here I've been told that the fact of replacing $$T\rightarrow (1-i\varepsilon)T$$ to define the path integral is equivalent in Fourier space as $$p⁰ \rightarrow (1+i\varepsilon)p⁰$$. With this, one gets the correct $$i \varepsilon$$ Feynman prescription for the propagator $$D_F(x-y)$$

$$\begin{equation} Z[J]=Z \text{exp} \left(-\frac{i}{2} \int \frac{d⁴ p}{(2\pi)⁴} \hat{J}(p) \frac{1}{p²-m²+i\varepsilon} \hat{J}(-p) \right) = Z \text{exp} \left(-\frac{1}{2} \int d⁴y \int d⁴x \ J(x) D_F(x-y) J(y) \right). \tag{6} \end{equation}$$

And this is the part that I don't get at all, I've tried but I don't see how the fact that $$T\rightarrow (1-i\varepsilon)T$$ leads in the previous approach to $$p⁰ \rightarrow (1+i\varepsilon)p⁰$$ and therefore to the Feynman prescription. I'm having nightmares with this, any help would be really appreciated.

NOTE: I use the following convention for the Fourier transform $$\begin{equation} \hat{\phi}(p)=\int d⁴ x \ \phi(x) e^{-ip \cdot x}\tag{7} \end{equation}$$

so that $$\begin{equation} \phi(x)= \int \frac{d⁴p}{(2 \pi)⁴} \ \hat{\phi}(p) e^{+ip \cdot x}.\tag{8} \end{equation}$$

1. In this answer we would like to understand the Wick rotation as an analytic continuation, i.e. as an (almost) $$90^{\circ}$$ continuous rotation in the complex plane. The before and after scenario is not enough: We would like to trace every step along the way of the Wick rotation.

2. Advice: Use the $$(-,+,+,+)$$ Minkowski signature convention$$^1$$, cf. my Phys.SE answer here. [The opposite sign convention is also possible but there are more surprises in store along the way.]

3. The Feynman $$i\epsilon$$-prescription can be viewed as stopping the Wick rotation just before a full $$\frac{\pi}{2}$$-rotation in the complex plane, cf. my Phys.SE answer here.

4. The Wick rotation treats in principle all contravariant 4-vectors the same way. (However, see section 8 below!) In particular, time $$x^0$$ and energy $$p^0$$ rotate in the same direction, cf. my Phys.SE answer here. $$x^0_E~=~ie^{-i\epsilon}x^0_M, \qquad p^0_E~=~ie^{-i\epsilon}p^0_M. \tag{A}$$ This means that covariant components, such as e.g. $$p_0$$, rotate in the opposite direction. $$p_0^M~=~ie^{-i\epsilon}p_0^E .\tag{B}$$ In particular, the metric component $$g_{00}$$ rotates twice as fast in the opposite direction. $$g_{00}^M~=~-e^{-2i\epsilon}g_{00}^E, \qquad g_{00}^E~=~1.\tag{C}$$

5. Let us for simplicity consider a real scalar field. The Fourier transform (and inverse Fourier transform) read $$\begin{array}{rcl} \widetilde{\phi}_{\! M}(k_M)&=&\int \! d^4x^{\bullet}_M ~e^{-ik_M \cdot x_M}\phi(x_M),\cr &&\phi(x_M)~=~\int \! \frac{d^4k_{\bullet M}}{(2\pi)^4}~e^{ik_M \cdot x_M}\widetilde{\phi}_{\! M}(k_M) \cr \widetilde{\phi}_{\! E}(k_E)&=&\int_{\mathbb{R}^4} \! d^4x_E ~e^{-ik_E \cdot x_E}\phi(x_E)~=~ ie^{-i\epsilon}\widetilde{\phi}_{\! M}(k_M), \cr &&\phi(x_E)~=~\int_{\mathbb{R}^4} \! \frac{d^4k_E}{(2\pi)^4}~e^{ik_E \cdot x_E}\widetilde{\phi}_{\! E}(k_E)~=~\phi(x_M).\end{array}\tag{D}$$ The bullet $$\bullet$$ in the integration measure indicates the position the spacetime index.

6. We want to perform the Gaussian integration in the Euclidean formulation: \begin{align}Z_0[J] =~~~~~&\int\! {\cal D}\phi \exp\left\{ \frac{i}{\hbar}\int \! d^4x^{\bullet}_M \left( \rule[1.5ex]{0ex}{1ex} J(x_M)\phi(x_M)\right.\right.\cr & \qquad\qquad \left.\left.+\frac{1}{2}\phi(x_M)(\Box_M-m^2/\hbar^2 +i\epsilon)\phi(x_M) \right)\right\}\cr \stackrel{x^0_E=ie^{-i\epsilon}x^0_M}{=}&\int\! {\cal D}\phi \exp\left\{ \frac{1}{\hbar}\int_{\mathbb{R}^4} \! d^4x_E \left( \rule[1.5ex]{0ex}{1ex}J(x_E)\phi(x_E)\right.\right.\cr & \qquad\qquad \left.\left.- \frac{1}{2}\phi(x_E)(-\Box_E+m^2/\hbar^2)\phi(x_E) \right)\right\}\cr \stackrel{\text{Gauss. int.}}{\sim}&\exp\left\{ \frac{1}{2\hbar}\int_{\mathbb{R}^4} \! d^4x_E ~ J(x_E)\frac{1}{-\Box_E+m^2/\hbar^2}J(x_E) \right\}\cr \stackrel{\text{Fourier}}{=}~~&\exp\left\{ \frac{1}{2\hbar}\int_{\mathbb{R}^4} \! \frac{d^4k_E}{(2\pi)^4} ~ \frac{\widetilde{J}_{\! E}(k_E)\widetilde{J}_{\! E}(-k_E)}{k^2_E+m^2/\hbar^2}\right\}. \end{align}\tag{E} Here is an alternative derivation of the same: \begin{align}Z_0[J] =~~~~~&\int\! {\cal D}\phi \exp\left\{ \frac{i}{\hbar}\int \! d^4x^{\bullet}_M \left( J(x_M)\phi(x_M)\right.\right.\cr & \qquad\qquad \left.\left.+\frac{1}{2}\phi(x_M)(\Box_M-m^2/\hbar^2 +i\epsilon)\phi(x_M) \right)\right\}\cr \stackrel{\text{Fourier}}{=}~~&\int\! {\cal D}\phi ~\exp\left\{ \frac{i}{2\hbar}\int \! \frac{d^4k_{\bullet M}}{(2\pi)^4} \left(\widetilde{J}_{\! M}(-k_M)\widetilde{\phi}_{\! M}(k_M)\right.\right.\cr & \qquad\qquad+\widetilde{J}_{\! M}(k_M)\widetilde{\phi}_{\! M}(-k_M) \cr & \qquad\qquad \left.\left. -\widetilde{\phi}_{\! M}(k_M)(k^2_M+m^2/\hbar^2 -i\epsilon)\widetilde{\phi}_{\! M}(-k_M)\right)\right\} \cr \stackrel{\text{Gauss. int.}}{\sim}&\exp\left\{ \frac{i}{2\hbar}\int \! \frac{d^4k_{\bullet M}}{(2\pi)^4} ~ \frac{\widetilde{J}_{\! M}(k_M)\widetilde{J}_{\! M}(-k_M)}{k^2_M+m^2/\hbar^2 -i\epsilon}\right\}\cr \stackrel{k^M_0=ie^{-i\epsilon}k^E_0}{=}&\exp\left\{ \frac{1}{2\hbar}\int_{\mathbb{R}^4} \! \frac{d^4k_E}{(2\pi)^4} ~ \frac{\widetilde{J}_{\! E}(k_E)\widetilde{J}_{\! E}(-k_E)}{k^2_E+m^2/\hbar^2}\right\}.\end{align}\tag{F}

7. To raise the bullet $$k^M_0=-k^0_M$$ in the Fourier transform (D) or in the second last expression of eq. (F) cost a minus sign, which we remove again by implicitly interchanging the corresponding integration limits: $$~=~\exp\left\{ \frac{i}{2\hbar}\int \! \frac{d^4k^{\bullet}_M}{(2\pi)^4} ~ \frac{\widetilde{J}_{\! M}(k_M)\widetilde{J}_{\! M}(-k_M)}{k^2_M+m^2/\hbar^2 -i\epsilon}\right\}.\tag{G}$$ Warning: Interchanging integration limits in Minkowskian formulation implies interchanging integration limits in the Euclidian formulation.

8. In the rest of this answer we consider a formulation in momentum space where the metric $$g^M_{00}=-1$$ is kept fixed without a rotation a la eq. (C). The Minkowski propagator has $$\text{denominator}~=~p^2_M+m^2 -i\epsilon ~=~-(p^M_0)^2+\omega_{\bf p}^2-i\epsilon,\tag{H}$$ and therefore poles at $$-p_M^0~=~p^M_0~=~\pm(\omega_{\bf p}-i\epsilon), \qquad \omega_{\bf p}~:=~\sqrt{{\bf p}^2+m^2}~\geq~0,\tag{I}$$ in the complex $$p^M_0$$ plane.

$$\downarrow$$ Figure from Ref. 1. The covariant $$p^M_0$$ poles are not crossed during the Wick rotation (B), but the contravariant $$p^0_M$$ poles are crossed during the Wick rotation (A). Therefore, to get a consistent formulation for contravariant $$p^0$$, we should do an opposite Wick rotation $$p^0_M~=~ie^{-i\epsilon}p^0_E .\tag{J}$$ as compared to eq. (A), cf. Refs. 2 & 3.

References:

1. J. Cardy, Intro to QFT, 2010; p. 17.

2. M. Srednicki, QFT, 2007; p. 55 + p. 99. A prepublication draft PDF file is available here.

3. S. Weinberg, Quantum Theory of Fields, Vol. 1, 1995; Section 11.2 p. 476.

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$$^1$$ Conventions & notations: In this answer we use the $$(-,+,+,+)$$ Minkowski signature convention and the speed of light is $$c=1$$. The subscripts $$E$$ and $$M$$ means Euclidean and Minkowskian, respectively.

• Wow, thank you very much for writting such an elaborated answer, I had lost the chance to find the solution, now I will sleep well :) So if I am right, taking the integral limit $T \rightarrow \infty (1-i\varepsilon)$ is equivalent to considering a term like $+i\varepsilon \phi (x_M)²$ inside the lagrangian? In any case, I will have a look at Srednicki and your other answers ^^ – Guillermo Franco Abellán Feb 11 at 23:02