Performing a Wick rotation over an integral is not equivalent to just a change of variable $t \to \mathrm{i}t = \tau$, after that we rotate the complex plane so that

$$\mathrm{i} \int_{-\infty}^{\infty} f(t) \mathrm{d}t = \int_{-\mathrm{i} \infty}^{\mathrm{i} \infty} f(-\mathrm{i} \tau ) \mathrm{d}\tau \; \stackrel{\mathrm{Wick}}{\longrightarrow} \; \int_{-\infty}^{\infty} f(-\mathrm{i} \tau ) \mathrm{d}\tau.$$

Is that right?

If it is, my concern is with the argument in the function of the integrand. How can we guarantee convergence if in the argument there are still possible imaginary functions?

My guess is that in the path integrals we are concerned with there are functions that are quadratic, however their dependence in time is not explicit, for example, we have

$$\exp\left[ \frac{i}{\hbar} \int_{-\infty}^{\infty} \frac{m}{2} \dot{x}^2(t) - \frac{1}{2}m\omega^2 x^2(t) + F(t)x(t) \; \mathrm{d} t \right],$$

but we will have to sum over all paths after, including for example $x(t) = t^{3/2}$, after Wick-rotating we would get

$$\exp\left[ -\frac{1}{\hbar} \int_{-\infty}^{\infty} \frac{9m}{8} (\sqrt{-\mathrm{i}\tau})^2 + \frac{1}{2}m\omega^2 (\sqrt{-\mathrm{i}\tau})^6 - F(-\mathrm{i} \tau)(\sqrt{-\mathrm{i}\tau})^3 \; \mathrm{d} \tau \right],$$

which is non convergent.

This is why I think that I have some misconception about how Wick rotation works, and that the first expression might be wrong. So as a recap:

Is the first expression right? And if it is, isn't it problematic the change in the argument in the integrand?

  • $\begingroup$ You might also ask about the portion of the contour at infinity. The integrand is well behaved in QED with well known propagators, but what about QCD with confined color? $\endgroup$ Jul 3, 2019 at 11:06
  • $\begingroup$ @BertBarrois sorry I don't understand what you mean with "portion of the contour at infinity". And regarding your point on QED and QCD, this question is at a much more elementary level. $\endgroup$ Jul 3, 2019 at 13:06
  • $\begingroup$ I think that the error is after the Wick rotation, where that $f(-i\tau)$ should be $f(\tau)$. There everything works well. But I still don't know how it works $\endgroup$ Jul 3, 2019 at 14:47
  • $\begingroup$ My point is that contours must be closed for Cauchy's theorem to apply, and you cannot rotate the axis without changing the answer unless the contour at infinity vanishes. Propagators like ${{({{k}^{2}}-{{m}^{2}})}^{-1}}$ die off fast enough at infinity, but the presumably pole-free propagators of confined particles are a wild card. $\endgroup$ Jul 4, 2019 at 11:34

1 Answer 1


My view on Wick rotation is the following:

It most definitely isn't a change of variable. A change of variable like $ t = it_E $ can be done for any function (since it is invertible), but is often useless because it works only if $t_E$ is imaginary (if you started from t real).

Wick rotation, instead, comes from analytic continuation of a function $F$ of a real variable $t$ to the whole complex plane. If your function $F$ does not have poles in the I and III quadrants of the complex plane and if it goes to zero sufficiently fast at complex infinity, then you can show with Cauchy theorem that: $ \int_{-\infty}^{+\infty} F(t) dt \ = \ i \int_{-\infty}^{+\infty} F(i t_E) d t_E $ where now $t_E$ is a real variable. If $F$ has no poles in II and IV quadrants instead, you get a minus sign (if it has no poles at all and it goes to zero at infinity as requested, then the integral on the real axis is zero).

This essentially means that your first expression should be wrong by one or two minus signs (depending on the assumptions on the poles of the analytic continuation of $f$).

As for the second part of your question, I'm puzzled too and I found your question exactly because of this. Clearly the time derivatives become $\partial_t \to \partial_{i t_E} = -i \partial_{t_E}$ after Wick rotation. But the dependence of the path $x[t]$ on $t$ is arbitrary, so you can't even be sure that $x(i t_E)$ is still real for $t_E$ real.


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