# 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:

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

2. 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.

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Well, any complex number $z=a+ib$ consist of two independent real numbers $a$ and $b$. So one should definitely treat them as independent. I guess that the discretized scalar field $\phi(x_i)$ is real right? That implies that the fourier components are not all independent since $\phi(k_n) = \phi(-k_n)$. So even though the fields are doubled (a real field is turned into a complex one), you only have a product over $k_n>0$ in the momentum space integration measure and thereby kill half of the degrees of freedom. – Heidar Nov 2 '12 at 22:46
I don't have the book here right now and only spend two second thinking about this. So I'm sorry if my comments are trivial and your question is something slightly different. – Heidar Nov 2 '12 at 22:47
Yep, that makes sense. And yea, the fields $\phi(x_i)$ are real. – k-selectride Nov 3 '12 at 0:59
Just to add to this discussion: I think the importance of the transformation being unitary might be the fact that unitary transformations don't alter the integration measure. So before the Fourier transformation we had $\mathcal{D} \phi(x)$ and after the transformation we still have $\mathcal{D} \phi(x)$, it just so happens that this can be expressed in terms of real and imaginary $\phi(k)$. Non-unitary transformations may have altered the integration measure so that $\mathcal{D} \phi(x)$ can't be used. It would be helpful if somebody could either confirm or reject this reasoning however. – Siraj R Khan Jan 10 '15 at 13:49

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.

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Thanks for the response already. I think I follow your explanation, but I can't quite apply it in this case. By transforming to complex $\phi(k_n)$ are we doubling the number of integrals being performed? If it's not too much trouble could you fill in the steps? I'd be happy to mark your post as the answer. – k-selectride Nov 3 '12 at 1:34
@k-selectride: You aren't duplicating anything. Just think of the integral over a complex variable and its conjugate as equal to the integral over the real and imaginary parts by definition. In the context of quantum field theory, it's just a definition. You aren't deforming any contours. – Ron Maimon Nov 3 '12 at 3:27
@Ron (I might be wrong, but) I think the confusion of k-selectride is that he considers a path integral measure for a REAL scalar on a lattice $\phi(x_i)$, where there is one integral for each lattice point. But after fourier transformation the field is complex and you have two integrals per k-space point (real and imaginary part), thus it seems like the number of integrals are doubled. But due to the symmetry property $\phi(k_n) = \phi(-k_n)$, we only keep half of the k-space points and the number of integrals are the same. – Heidar Nov 3 '12 at 3:59
That's pretty much it, I understand intuitively that by transforming to complex field variables there needs to be two variables per lattice point to be integrated over, and that requiring $k_n^0$ gets rid of the redundancies. I was wondering if there was a more precise, or rigorous way to describe that. – k-selectride Nov 3 '12 at 4:34
@k-selectride: I got it now (thanks Heidar), the answer is what Heidar says, there is a nonlocal linear constraint in k-space that makes the number of integrals the same. You should do the Fourier transform explicitly on a 5 site lattice with a real function to see how the counting works. – Ron Maimon Nov 3 '12 at 13:52