Why use Fourier expansion in Quantum Field Theory? I have just begun studying quantum field theory and am following the book by Peskin and Schroeder for that.
So while quantising the Klein Gordon field, we Fourier expand the field and then work only in the momentum space. What is the need for this expansion?
 A: Free equations are linear, so exponentials are their solutions. Thus we construct a linear superposition of exponentials to embrace a general case.
Interactions are supposed to change amplitudes of particular waves in these superpositions.
A: First of all, this is just a change of basis, which is up to us to make. Furthermore we should always choose a basis that makes our calculations easier, and hopefully makes things more intuitive. For a simpler example - just try finding the volume of a sphere in cartesian coordinates, its just a bad choice. 
Second of all, you don't have to use a Fourier basis, to my knowledge everything -loops renormalization etc can be done in a position basis. 
Now as to why the Fourier basis is a convenient choice:
(1) It simplifies derivative terms in the Lagrangian - as usual the Fourier basis turns derivative expressions into algebraic ones, which are much easier to manipulate.
(2) It it more intuitive - written in terms of a Fourier basis the Feynman rules are in terms of momentum. So for example at the vertices momentum is conserved - its just a nice tidy way to think about whats happening at the vertex.
(3) Even if you start in position space, one method for doing the integrals you will encounter when writing for your loop expressions will be going to momentum space - so you sort of cut this step out from the outset.
(4) (following up on Vibert's comment) Plane waves are the basis we do the experiment in. That is, we send in wave packets highly localized in p space, i.e. this is the exact solution we perturb around. 
A: I think it's also important to emphasize the physical significance of the Fourier modes in the context of QFT.  The Fourier modes $a^\dagger(\mathbf k)$ and $a(\mathbf k)$ in the context of the quantized Klein-Gordon field, for example, create and destroy particles with momentum $\mathbf k$ respectively.  Namely, if $|\emptyset\rangle$ is the vacuum of the theory, then
$$
  a^\dagger(\mathbf k)|\emptyset\rangle
$$
gives a state with a single particle of momentum $\mathbf k$, and more generally
$$
  a^\dagger(\mathbf k_1)a^\dagger(\mathbf k_2)\cdots a^\dagger(\mathbf k_N)|\emptyset\rangle
$$
represents a state with $N$ particles with momentum $\mathbf k_1, \mathbf k_2, \dots, \mathbf k_N$ respectively.
