# Do we wick rotate momentum axis on correlation function?

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?

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

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. $$(a)$$.
• 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. Commented Jan 30, 2023 at 3:31