Functional derivative for the same function expressed before and after Wick rotation

Question

This question arises when I'm reading section "3.3.1 Minkowski Space" of page 16-17 of the following document: http://www-thphys.physics.ox.ac.uk/people/JohnCardy/qft/qftcomplete.pdf

On page 17, they took a functional derivative of $Z[J]$ with respect to $iJ$ to obtain an expression for $G_{(0)}(x_1,x_2)$. We're supposed to take derivatives with respect to $J(x)$, but on page 17 the document took derivatives with respect to $J(x')$, where $x_0=ix_0'$ (the subscript 0 indicates the first element of $x$; the other elements remain equivalent).

Is the results the same or did the document made a mistake?

Note: The definition of functional derivative the document is using is a delta function as the test function, as explained in section 4 of the following Wikipedia article: https://en.wikipedia.org/wiki/Functional_derivative#Using_the_delta_function_as_a_test_function

Answer 1

Ultimately, it is a matter of conventions, but here is one line of reasoning:

1. There exists two types of real integration measures $d^nx$: an un-signed (signed), which transforms under change of coordinates with (without) an absolute value $|\cdot|$ of the Jacobian factor, respectively. We will consider the latter, since this can naturally be continued to complex coordinates, as is needed for a Wick rotation.

2. The product $ d^nx~\frac{\delta}{\delta \phi(x)}$ remains invariant under coordinate transformations $x\to x^{\prime}$. So since the integral measure factor $d^nx$ transforms with a Jacobian factor (without absolute value), the functional derivative $\frac{\delta}{\delta \phi(x)}$ transforms with an inverse Jacobian factor.

3. The above reasoning suggests that one should assign$^1$ $$ x^0_E~=~ix^0_M, \qquad d^nx_E~=~i d^nx_M, \qquad \frac{\delta^M}{\delta \phi(x)}~=~i\frac{\delta^E}{\delta \phi(x)}.$$

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$^1$ Be aware that additional $i$-factors can arise for objects that transform non-trivially under Wick-rotation. E.g. the Lagrangian density transforms as a double time derivative: ${\cal L}_M~=~i^2{\cal L}_E$.

Answer 2

The answer actually stays unchanged. You're still taking functional derivative with respect to the same point. If you refer back to the definition of functional derivative, such as the one in Wikipedia, you'll see that it's the coefficient that matters.