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}


1 Answer 1

  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. $$\begin{align} x^0_E~=~&ie^{-i\epsilon}x^0_M, \cr p^0_E~=~&ie^{-i\epsilon}p^0_M. \end{align}\tag{A}$$ 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. $$\begin{align} g_{00}^M~=~& -e^{-2i\epsilon}g_{00}^E, \cr g_{00}^E~=~&1.\end{align}\tag{C}$$

  5. Let us for simplicity consider a real scalar field. The Fourier transform (and inverse Fourier transform) read $$\begin{array}{rcl} \widetilde{\phi}_{\! M}(k_M)&=&\int \! d^4x^{\bullet}_M ~e^{-ik_M \cdot x_M}\phi(x_M),\cr \phi(x_M)&=&\int \! \frac{d^4k_{\bullet M}}{(2\pi)^4}~e^{ik_M \cdot x_M}\widetilde{\phi}_{\! M}(k_M) \cr \widetilde{\phi}_{\! E}(k_E)&=&\int_{\mathbb{R}^4} \! d^4x_E ~e^{-ik_E \cdot x_E}\phi(x_E)\cr &=& ie^{-i\epsilon}\widetilde{\phi}_{\! M}(k_M), \cr \phi(x_E)&=&\int_{\mathbb{R}^4} \! \frac{d^4k_E}{(2\pi)^4}~e^{ik_E \cdot x_E}\widetilde{\phi}_{\! E}(k_E)\cr &=&\phi(x_M).\end{array}\tag{D}$$ 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 $$\begin{align} \text{denominator} ~~=~&p^2_M+m^2 -i\epsilon \cr ~=~& -(p^M_0)^2+\omega_{\bf p}^2-i\epsilon,\end{align}\tag{H} $$ and therefore poles at $$\begin{align} -p_M^0~=~& p^M_0~=~\pm(\omega_{\bf p}-i\epsilon), \cr \omega_{\bf p}~:=~&\sqrt{{\bf p}^2+m^2}~\geq~0, \end{align}\tag{I}$$ 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.


  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.


$^1$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.