The essential motivation to transform to momentum space is that the translation invariance of the dynamics and of the vacuum state requires the 2-point Vacuum Expectation Value to be diagonal in the momentum variable. Thus, for the free field we have
$$\left<0\right|\hat\phi(x)\hat\phi(y)\left|0\right>=G(x-y),$$ for some function $G(x-y)$ of the separation $x-y$, where the Fourier transform of this is $$\tilde G(k)=2\pi\delta(k^2-m^2)\theta(k_0).$$
$G$ has to be a function of $x-y$, not of $x$ and $y$ separately, for it to be translation invariant.
In contrast, in real space,
$$G(x-y)=\int 2\pi\delta(k^2-m^2)\theta(k_0)\mathrm{e}^{-\mathrm{i}k\cdot(x-y)}\frac{\mathrm{d}^4k}{(2\pi)^4},$$
which is manifestly not diagonal in the $x$ and $y$ "indices".
If one considers two general "vectors" in the space of operators, $\hat\phi_f=\int\hat\phi(x)f(x)\mathrm{d}^4x$ and a similarly defined $\hat\phi_g=\int\hat\phi(y)g(y)\mathrm{d}^4y$, in the $f$ and $g$ "directions", one obtains
$$\left<0\right|\hat\phi_f^\dagger\hat\phi_g\left|0\right>=\int f^*(x) G(x-y)g(y)\mathrm{d}^4x\mathrm{d}^4y=\int \tilde f^*(k)\tilde G(k)\tilde g(k)\frac{\mathrm{d}^4k}{(2\pi)^4},$$
which hopefully makes clear the Hermitian structure of the inner product this gives us.
Another fact that is vitally important to the Hilbert space structure is manifest in the Fourier space presentation, that the function $G(x-y)$ is positive semi-definite in the Fourier co-ordinates (and hence it is positive semi-definite, but not diagonal, in all coordinates).

Working in the appropriate co-ordinates as always makes computations more compact, and keeps the computations relatively closer to the underlying conceptual structure, but one can use whatever coordinates one likes, or, better, that one can justify using.

This post imported from StackExchange Physics at 2015-03-30 13:53 (UTC), posted by SE-user Peter Morgan