In QFT written by Peskin and Schroeder, it is discussed how correlation function is evaluated in Euclidean space, on page 292 to 293, In (9.48) $$<\phi (x_{E1})\phi(x_{E2})>=\int \frac{d^4k_E}{(2\pi)^4}\frac{e^{ik_E\Delta x_E}}{k_E^2+m^2}.\tag{9.48}$$In (9.27) $$I=\int \frac{d^4k}{(2\pi)^4}\frac{ie^{-ik\Delta x}}{k^2-m^2+i\epsilon}.\tag{9.27}$$ At first, PS told us time axis is wick rotated in clockwise direction, $x^o \rightarrow -ix^o_E$ which is fine and $$I=\int \frac{d^4k}{(2\pi)^4}\frac{ie^{-ik^o(-i\Delta x^o_E)+i\vec{k}\vec{\Delta x}}}{k^2-m^2+i\epsilon}$$ To proceed, I do not wick rotate the $k^o$ but just define $k^o_E=ik^o$, hence $k^o_E$ runs from $-i\infty$ to $+i\infty$ and define $k^j_E=k^j$ $$I=-i\int_{-i\infty}^{+i\infty} \frac{dk^o_E}{2\pi} \int \frac{d^3k_E}{(2\pi)^3}\frac{ie^{ik^o_E\Delta x^o_E+i\vec{k}\vec{\Delta x}}}{-(k^o_E)^2-(\vec{k_E})^2+m^2-i\epsilon}$$ $$I=\int_{-i\infty}^{+i\infty} \frac{dk^o_E}{2\pi} \int \frac{d^3k_E}{(2\pi)^3}\frac{e^{ik^o_E\Delta x^o_E+i\vec{k}\vec{\Delta x}}}{-(k^o_E)^2-(\vec{k_E})^2+m^2-i\epsilon}.\tag{a}$$ It seems correct but the $k^o_E$ is along imaginary axis, if I tried to wick rotate $k^o_E$, since poles of $k^o$ are $\pm E_k \mp i\epsilon$, poles of $k^o_E$ are $\pm iE_k \pm\epsilon$, thus $k^o_E$ should be rotated in anticlockwise direction, and it gives $$I=\int_{\infty}^{-\infty} \frac{dk^o_E}{2\pi} \int \frac{d^3k_E}{(2\pi)^3}\frac{e^{ik^o_E\Delta x^o_E+i\vec{k}\vec{\Delta x}}}{-(k_E^2-m^2+i\epsilon)}$$ After I flipped the upper and lower limit of $k^o_E$, it gives $$I=\int \frac{d^4k_E}{(2\pi)^3}\frac{e^{-ik^o_E\Delta x^o_E+i\vec{k}\vec{\Delta x}}}{k_E^2-m^2+i\epsilon}$$ I modified after answered. This integral is after wick rotation of momentum axis. $$I=\int \frac{d^4k_E}{(2\pi)^3}\frac{e^{-ik^o_E\Delta x^o_E+i\vec{k}\vec{\Delta x}}}{k_E^2-m^2}.\tag{b}$$ I personally doubt 2 things:

  1. do we need to wick rotate both $x^o$-axis and $k^o$-axis? If we do not wick rotate $k^o$, how to interpret $k_E$?
  2. In PS working, $x^o$ is rotated in clockwise direction, while $k^o$ is rotated in anti-clockwise direction, is it permitted?

1 Answer 1


The answer to OP's title question is: Yes, we do Wick rotate the 4-momentum.

Further justification and comments:

  1. P&S uses the $(+,-,-,-)$ Minkowski sign convention, which makes Wick rotation somewhat awkward, cf. my Phys.SE answer here. Consider to redo the setup in the $(-,+,+,+)$ Minkowski sign convention.

  2. As a sanity check, note that the (massive) Euclidean propagator should be free of poles on (and near) the real $k^0_E$-axis. Compare with OP's eq. $(b)$.

    In particular the Feynman $i\epsilon$ prescription is not needed in the Euclidean formulation. Compare with OP's eq. $(a)$.

  3. OP raises an interesting question about the coexistence of Wick rotation in spacetime and its Fourier transformed momemtum space. This is e.g. discussed in my Phys.SE answer here.

  • $\begingroup$ Thanks for your reply, I have two points still not clear. If we need to wick rotate $p^o$ axis, which I now rotate $k^o_E$ in equation (a), Let $k^o_E=\rho(cos\theta+isin\theta)$, the exp in (a) involve $e^{i\rho cos\theta-\rho sin\theta}$, and rotation is clockwise from imaginary axis to real axis, the integral in quad-II is fine, but the integral in quad-IV diverge, how this is possible? Secondly, I do not understand why the euclidean metric of the exponent in eqn (b) is (-+++) after rotation, if I do not make any mistake. $\endgroup$
    – Li Chiyan
    Commented Jan 30, 2023 at 3:31

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