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I'm following a class on QFT. I'm having a hard time understanding the rotation to Euclidean of the generating functional $W[J]$ of some scalar theory $L(\phi, x)$. $$ W[J] := \mathcal{N} \int [D\phi] \exp\left\{ i \int_{\mathbb{R}^4} L(\phi, x) + J(x)\phi(x) \; \text{d}^4 x \right\} $$ My understanding is that in order to go to Euclidean we substitute the integral over $\mathbb{R}^4$ with an integral over $i\mathbb{R} \times \mathbb{R}^3$, which is the Wick rotated version of the original domain*. $$ W_E[J] := \mathcal{N} \int [D\phi] \exp\left\{i \int_{i\mathbb{R} \times \mathbb{R}^3} L(\phi, x) + J(x)\phi(x) \; \text{d}^4 x \right\} $$ Performing a change of integration variable and renaming it, I get $$ W_E[J] = \mathcal{N} \int [D\phi] \exp\left\{- \int_{\mathbb{R}^4} L(\phi, x_E) + J(x_E)\phi(x_E) \; \text{d}^4 x \right\} $$ where $x_E(x) = (i x^0, \vec{x})$.

I can't decide if this is the correct definition/procedure because books seem to gloss over a lot of things and abuse of notation. Take Cheng-Li for example, in equation (1.75) the integration domain is not specified making the equation useless since I don't known whether the Wick rotation was "inverted" or not with a change of variables**.

The point: I'm not looking for fancy answers, just the bare minimum to perform the jump to Euclidean correctly.

I'm very new to this things so any help is appreciated.

Notes

*When we do this rotation to the causal Green's function of the free real scalar field $\Delta_F(x)$ the integral stays the same, thanks to the Feynman $i\varepsilon$ prescription which moves the poles away from the deformation region. This time we don't know, a priori, if the integral will stay the same, therefore we give a new name to the result: $W_E[J]$. See Is the Euclidean generating functional $Z_{E}[J]$ identified with original Minkowskian generating functional $Z[J]$?

**Also Cheng-Li says that $W_E[J]$ is an analytical continuation of $W[J]$ for complex times, which at first seems complete nonsense as $W$ is not a function of spacetime coordinate ($x$ is just a mute variable). I guess that the time-dependence is hidden inside the boundary conditions of the path-integral but I can't find it written anywhere. If this is the case, I guess the complete procedure should include a change in boundary conditions. Anyway you don't need to answer to this secondary question if you don't have time.

Addendum

I often read/hear that the jump to euclidean is done by simply making a change of variables in the $\text{d}^4 x$ integral. In my opinion, this must be incorrect since a trivial change of variables cannot turn a not-well-defined quantity into a well-defined quantity. For something like this to happen, there must be some step which does not involve a mathematical equality. Analytical continuations are an example of such a step.

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  • $\begingroup$ You should probably look at the equivalent treatment in statistical thermodynamics, where it is Euclidean from the start. Quite a lot of the transition between Minkowski and Euclidean versions are done by comparing the equations from both sides and then seeing what minimal steps can be done to do this conversion. The proper thing to do is analytical continuation and shifting of contours, but we actually are not doing that; we are doing a shortcut that is justified by the fact that it must be possible to do that analytical continuation. $\endgroup$ Commented Jun 4 at 16:06
  • $\begingroup$ Yes, disregarding mathematical details, is the procedure correct then? $\endgroup$ Commented Jun 4 at 16:09
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    $\begingroup$ Related: physics.stackexchange.com/q/106292/2451 and links therein. $\endgroup$ Commented Jun 4 at 16:50
  • $\begingroup$ This question is similar to: How to Perform Wick Rotation in the Lagrangian of a Gauge Theory (like QCD)?. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. $\endgroup$ Commented Jun 4 at 16:54
  • $\begingroup$ Oh, Qmechanic's answer is way more comprehensive. I was also already rather wary of your Wick rotation, because your middle equation expression still had $i$ in there. Your last expression should have a $\mathrm d^4x_E$ too. The details, just check Qmechanic's answer. Sooo good. $\endgroup$ Commented Jun 4 at 16:56

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The point: I'm not looking for fancy answers, just the bare minimum to perform the jump to Euclidean correctly.

It may be helpful to think of a specific functional form for $L$ such as $$ L(a, b) = \frac{1}{2}b^2 - \frac{1}{2}a^2\;, $$ where the (infinite time) action is (with the example $L$ being used on the far RHS) $$ S[x] = \int_{-\infty}^{\infty} dt L(x(t), \dot x(t)) = \int dt\left( \frac{1}{2}\dot x(t)^2 - \frac{1}{2}x(t)^2 \right)\;. $$

We define a new complex function of a complex variable $z$ as $$ x_c(z) \equiv x(z)\;. $$ Here too it may be helpful to think of a specific functional form such as $$ x(t) = A\cos(\omega t) $$ and $$ \dot x(t) = -\omega A\sin(\omega t)\;. $$

So for this example $$ x_c(z) = A\cos(\omega z)\;, $$ which is a well known analytic function of $z$. And, we also have $$ \frac{d x_c}{dz} = \dot x(z) =-\omega A \sin(\omega z)\;. $$

In any case, regardless of the form of $x(t)$, we just define $$ x_c(z) \equiv x(t=z)\;. $$

With this introduction, we see that $$ S = \int_{C_1} dz L(x_c(z), \frac{dx_c}{dz}(z))\;, $$ where $C_1$ is a contour that runs directly along the real axis.

Next, if the integrand goes to zero at infinity, and is either zero all along the semicircle at infinite of the first quadrant and the semicircle at infinity of the third quadrant, or those contributions cancel, then we can also write $$ S = \int_{C_2} dz L(x_c(z), \frac{dx_c}{dz}(z))\;, $$ where $C_2$ starts from $-i\infty$, runs along the semicircle of the third quadrant to $-\infty$, runs along $C_1$ to $+\infty$, and runs along the semicircle of the first quadrant up to $+i\infty$.

Next, if the integrand doesn't have poles in the first and third quadrants, the contour can be shifted and we can write $$ S = \int_{C_3} dz L(x_c(z), \frac{dx_c}{dz}(z))\;, $$ where $C_3$ right directly up the imaginary axis from $-i\infty$ to $+i\infty$.

Next, we parametrize the path $C_3$ using $z=i s$, where $s$ is real and runs from $-\infty$ to $\infty$, and find $$ S = \int_{-\infty}^{\infty} ids L(x_c(is), \frac{dx_c}{dz}(is)) $$ $$ =\int_{-\infty}^{\infty} ids L(x(is), \dot x(is))\tag{1} $$ $$ \equiv-i S_E\;, $$ where $$ -S_E[x] \equiv \int_{-\infty}^{\infty} dt L(x(it), x(it))\;, $$ since we can just rename the dummy variable $s\to t$ in Eq. (1). And where I introduced a minus sign in my definition of $S_E$ as a convention.

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  • $\begingroup$ Correct me if I'm wrong. You did exactly what I did but you also required that the deformation preserves the integral. Which suggests that my definition and calculation of W_E is indeed correct. I'd like to be reassured that's the case $\endgroup$ Commented Jun 4 at 17:36
  • $\begingroup$ Seems fine. If you plan to argue that the thing you called $W$ in the first equation of your post is the same as the thing your called $W_E$ in the second equation of your post then you need to make some requirement on the contour integrations and integrand. If you just want to consider $W_E$ as a potentially useful thing that is not necessarily related to $W$ in a specific way, you can do that too (e.g., it might be useful in classical statistical mechanics outside of any relationship with zero temperature quantum field theory...) $\endgroup$ Commented Jun 4 at 18:44

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