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I've seen across several posts, that Wick rotation is not an ordinary substitution. Instead we're rotating the contour of integral and analytically continuing time $t$ to include imaginary time $-i\tau$.

However if one looks at this post to a related question in Physics SE, one can see that Wick rotation just seems to be a trivial variable change in this example.

Is there something obvious that I'm missing here? Moreover, using complex integration to show wick rotations seems to work out fine when time goes from $0$ to $\infty$. However, how can we justify Wick rotations for finite time.

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Well, $t$ is a real variable so you are not allowed to simply set $t = - i \tau$. The proper way to do this is via a rotation of the contour (as you have read). There are three problems that could potentially occur as we are changing the contour:

  1. The integrand is not holomorphic in $t$. In this case, changing the contour is strictly prohibited. PS - This never actually happens, but I'm mentioning this for completeness.

  2. The change of contour crosses a pole of the integrand. If this happens, then as we deform the contour, we should also pick up the residue of that pole.

  3. The value of the integrand at infinity may not be zero. In this case, the integral receives extra contributions from the integral at infinity.

If none of the above problems occur, then Wick rotation is simply a replacement $t \to - i \tau$. If not, a more careful analysis of the integral must be done.

For example, in GR, the conformal mode has the opposite sign so when we Wick rotation in the traditional way, this mode does not vanish at infinity and we encounter problem number 3. If you try to do the Wick rotation the other way, then the conformal mode is fine, but all other modes now have the wrong sign. This is known as the "conformal mode problem".

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  • $\begingroup$ Okay, so if none of the three problems appear, then Wick rotation is identical to a substitution. However, what actually happens is, we first analytically continue $t$ to say $z$. Then we deform this contour to bring it on the $Im(z)$ axis and integrate from $-i\infty$ to $i\infty$. Then we make the change $z=-i \tau$ and so the limits become real again. Now we integrate over this $d\tau$. $\endgroup$ Commented Mar 20, 2023 at 17:38
  • $\begingroup$ However suppose I was integrating over a finite time $0$ to $T$, as in the link provided in the question. Imagine I consider $t$ to be $z$ as before and integrate $z$ along the imaginary axis from $iT$ to $0$. Then we set $z=-i\tau$, and integrate. My limits become real numbers $0$ and say $\beta$. My final answer is now in terms of the $\beta$, right ? Then I have to analytically continue this back to the entire complex place by setting $Z=i\beta$. Then to recover my original integral, I consider the real axis, where $Z=T$. Is this what happens ? $\endgroup$ Commented Mar 20, 2023 at 17:43
  • $\begingroup$ If it is not too much to ask, can you please do the exact problem that I've linked above, in this contour integral way as opposed to the direct substitution? I guess that would be the best or only way this would clear things for me, as I need to see explicitly what is happening at each step. $\endgroup$ Commented Mar 20, 2023 at 17:50

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