Rescaled Fourier Transforms

The ordinary Fourier transform of a field $f$ is defined as

$$F_k \equiv \int d^3x f(x) e^{-i k x}.$$ (5.5)

The problem with using this definition for a classical scalar field is that if the Fourier components $F_k$ are given fixed values the resultant field values $f(x)$ become dependent on the overall size of the box within which the theory is defined. For example

$$\int d^3x f(x)^2 = {1 \over (2 \pi)^3} \int d^3k \vert F_k\vert^2$$ (5.6)

which implies

$$\left< f(x)^2\right> = {1 \over (2 \pi)^3 \mathcal{V}} \int d^3k \vert F_k\vert^2.$$ (5.7)

The size of the box does not affect the integral, except by turning it into a discrete sum. So to keep $f(x)$ (and by extension all intensive quantities) independent of the box size we define a modified Fourier transform

$$\tilde{F}_k \equiv {1 \over \sqrt{\mathcal{V}}} F_k = {1 \over L^{3/2}} F_k$$ (5.8)

where $L$ is the size of the box. This modified transform takes on the same value regardless of the box size, while the actual Fourier transform must be rescaled. Note that the units of $F_k$ are $[M]^{-2}$ while those of $\tilde{F}_k$ are $[M]^{-1/2}$.

The Fourier transform used by the program is neither of these, however, but rather the discrete Fourier transform $f_k$, related to the usual, continuous, one by

$$f_k = {1 \over dx^3} F_k = {L^{3/2} \over dx^3} \tilde{F}_k.$$ (5.9)

All physical quantities should be defined in terms of $\tilde{F}_k$ and the Fourier transform $f_k$ used by the program should be adjusted accordingly. For example, the initial vacuum state $\tilde{F}_k^2 = {1 \over 2 \omega_k}$ becomes $f_k^2 = {L^3 \over 2 dx^6 \omega_k}$. See section 6.3.2 for more details.