# 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)

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

can be expressed as

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

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 $$$$<0|0>_J = \frac{1}{N} \lim_{\varepsilon \rightarrow 0} \ \lim_{T \rightarrow \infty(1-i\varepsilon)} \int \mathcal{D} \phi \ \text{exp} \left( i \int_{-T}^T dt \int d³x (\mathcal{L}+J\phi)\ \right) \equiv \frac{1}{N} Z[J]\tag{3}$$$$ where the constant $$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

$$$$Z[J]=Z[0] \text{exp} \left(-\frac{i}{2} \int \frac{d⁴ p}{(2\pi)⁴} \hat{J}(p) \frac{1}{p²-m²} \hat{J}(-p)\right).\tag{5}$$$$ Here I'm using the $$(+,-,-,-)$$ 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 $$D_F(x-y)$$

$$$$Z[J]=Z[0] \text{exp} \left(-\frac{i}{2} \int \frac{d⁴ p}{(2\pi)⁴} \hat{J}(p) \frac{1}{p²-m²+i\varepsilon} \hat{J}(-p) \right) = Z[0] \text{exp} \left(-\frac{1}{2} \int d⁴y \int d⁴x \ J(x) D_F(x-y) J(y) \right). \tag{6}$$$$

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 $$$$\hat{\phi}(p)=\int d⁴ x \ \phi(x) e^{-ip \cdot x}\tag{7}$$$$

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

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.

--

$$^1$$ 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
• @Qmechanic, can you please shed some light on the 'surprises', if $(+,-,-,-)$ sign convention is adopted. – MadMax Apr 18 at 14:11
• See e.g. my Phys.SE answer here. – Qmechanic Apr 18 at 20:14