Fourier transform of integration measure (Peskin and Schroeder)

[center o bass]

At page 285 in Peskin and Schroeder's Introduction to quantum field theory the author defines the integration measure $D\phi = \Pi_i d\phi(x_i)$ where space-time is being discretised into a square lattice of volume L^4. He proceeds by Fourier-transforming

$$\phi(k_n) = \frac{1}{V} \sum_n e^{-i k_n \cdot x_i} \phi(k_n)$$

and considering only $k^0_n >0$ as independent variables he concludes that since the Fourier transformation is unitary, the measure $D\phi = \Pi_i d\phi(x_i)$ can equivalently be expressed as (equation 9.22)

$$D\phi = \Pi_{k_n^0 > 0} d Re \phi(k_n) d I am \phi(k_n).$$

Why is he only considering $k^0_n >0$ and how does he arrive at this conclusion? What is the relevance of the Fourier transform being unitary?

 

[azntoon]

The author is considering only k^0_n > 0 because they are the independent variables. In order to understand why, it is helpful to look at the Fourier transformation as a change of basis from the x_i space to the k_n space. In this change of basis, some of the k_n may become dependent on each other, meaning that not all of them are independent variables. By considering only k^0_n > 0, the author is ensuring that he is only working with the independent variables.The relevance of the Fourier transform being unitary is that it ensures that the measure D\phi = \Pi_i d\phi(x_i) is equivalent to the measure D\phi = \Pi_{k_n^0 > 0} d Re \phi(k_n) d I am \phi(k_n). This is because the unitarity of the Fourier transformation means that the inverse transformation is the same as the forward transformation, which means that the two measures are equivalent.