Why can you re-write the functional measure of a real-valued field $\phi(x)$ as $\mathcal{D}\phi=\prod_{k_n^0>0}dRe \phi(k_n) d Im \phi(k_n)$? This happens in Peskin and Schroeder, An Introduction to QFT, on page 285. They set out to calculate correlation functions for the free real-valued Klein-Gordon field $\phi(x)\in \mathbb{R}$. They define a 4D space-time lattice with lattice spacing $\epsilon$, and define
$$\mathcal{D}\phi~=~\prod_id\phi(x_i).$$
Next they write the Fourier series of $\phi(x_i)$ as
$$\phi(x_i)~=~\frac{1}{V}\sum_n e^{-i k_n\cdot x_i}\phi(k_n).$$
My questions are:


*

*Why do they treat the real and imaginary parts of $\phi(k_n)$ as independent variables?

*Why does the fact that the change of variables is unitary let them write the measure as
$$\mathcal{D}\phi(x)=\prod_{k_n^0>0}dRe \phi(k_n) dIm \phi(k_n)?$$
I've never seen an integration measure split into its real and imaginary part, so maybe I'm missing something obvious.
 A: This is an old notational trick, with absolutely no significance. When you have a two-component object, you can write it as a pair of real numbers x,y , or as one complex number
$$ z = x + i y $$
and it's conjugate
$$ \bar{z} = x - iy $$
If you imagine for a moment that x and y are complex numbers, then the transformation from x,y to $z,\bar{z}$ makes two independent variables. If you have any function of x,y, you can pretend that you have transformed to z and \bar{z}, but the condition that x and y are real becomes the condition that z is equal to the conjugate of $\bar{z}$.
The differentiation with respect to z and $\bar{z}$ is found by changing variables from complex x,y to complex independent $z,\bar{z}$, and using the appropriate differential operators for this change of variables
$$ \partial_z = {1\over 2} (\partial_x - i\partial_y)$$
$$ \partial_\bar{z} = {1\over 2} (\partial_x + i \partial_y)$$
Note that the derivative with respect to z of $\bar{z}$ is zero, etc,etc, all the obvious properties are ok.
The integral over several complex variables is, when you are looking at holomorphic stuff, the integral over half the dimensions. For one complex variable, you integrate over a contour, and the contour doesn't matter. For two complex variables, you integrate over a 2d surface which is locally compatible with the complex structure (locally, it is a product of contours in some complex coordinate pair). One such surface is the surface where x is real and y is real, so this is the integral over the surface z-bar equals the complex conjugate of z. 
The notation is not hard to unravel, it can always be translated to real variable language, and then it is obvious--- you integrate over all the real fields.
