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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 <n|m>_J <m,-T|\phi, -T>\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 $N=Z[0]$.

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[0] \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[0] \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[0] \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}

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TL;DR: No need to lose sleep & having nightmares: The Wick rotation & the Feynman $i\epsilon$-prescription work.

  1. 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.]

  2. The Wick rotation treats all contravariant 4-vectors the same way. In particular, time $x^0$ and energy $p^0$ rotate in the same direction, cf. my Phys.SE answer here.

  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. E.g., the energy transforms as follows $$ p^0_E~=~ie^{-i\epsilon}p^0_M \qquad \Rightarrow \qquad -(p^0_M)^2~=~ e^{2i\epsilon}\underbrace{(p^0_E)^2}_{\geq 0} ~=~(p^0_E)^2+i\epsilon $$ $$\qquad \Leftrightarrow \qquad (p^0_E)^2~=~-(p^0_M)^2-i\epsilon \qquad \Leftrightarrow \qquad p^2_E~=~p^2_M-i\epsilon.\tag{A} $$

  4. Perhaps it is most convincing to perform the Gaussian integration in the Euclidean formulation: $$\begin{align}Z_0[J]&~~~~~~=~\int\! {\cal D}\phi \exp\left\{ \frac{i}{\hbar}\int \! d^4x_M \left( J(x_M)\phi(x_M)+\frac{1}{2}\phi(x_M)(\Box_M-m^2/\hbar^2 +i\epsilon)\phi(x_M) \right)\right\} \cr ~&\stackrel{x^0_E=ie^{-i\epsilon}x^0_M}{=}~\int\! {\cal D}\phi \exp\left\{ \frac{1}{\hbar}\int_{\mathbb{R}^4} \! d^4x_E \left( J(x_E)\phi(x_E)- \frac{1}{2}\phi(x_E)(-\Box_E+m^2/\hbar^2)\phi(x_E) \right)\right\}\cr &~\stackrel{\text{Gauss. int.}}{\sim} ~\exp\left\{ \frac{1}{2\hbar}\int_{\mathbb{R}^4} \! d^4x_E ~ J(x_E)\frac{1}{-\Box_E+m^2/\hbar^2}J(x_E) \right\}\cr &~~~\stackrel{\text{Fourier}}{=}~\exp\left\{ \frac{1}{2\hbar}\int_{\mathbb{R}^4} \! \frac{d^4k_E}{(2\pi)^4} ~ \frac{\widetilde{J}(k_E)\widetilde{J}(-k_E)}{k^2_E+m^2/\hbar^2}\right\} \cr &\stackrel{k^0_E=ie^{-i\epsilon}k^0_M}{=}~\exp\left\{ \frac{i}{2\hbar}\int \! \frac{d^4k_M}{(2\pi)^4} ~ \frac{\widetilde{J}(k_M)\widetilde{J}(-k_M)}{k^2_M+m^2/\hbar^2 -i\epsilon}\right\}.\end{align}\tag{B}$$

References:

  1. M. Srednicki, QFT, 2007; Chapter 8, p. 55. A prepublication draft PDF file is available here.

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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.

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  • $\begingroup$ 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 ^^ $\endgroup$ – Guillermo Franco Abellán Feb 11 at 23:02
  • $\begingroup$ @Qmechanic, can you please shed some light on the 'surprises', if $(+,-,-,-)$ sign convention is adopted. $\endgroup$ – MadMax Apr 18 at 14:11
  • $\begingroup$ See e.g. my Phys.SE answer here. $\endgroup$ – Qmechanic Apr 18 at 20:14

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