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?

• First of all: in real life (accelerators), you are not interested in observables of the form $< \phi(x_1) \phi(x_2) \phi(x_3) >$ etc. but on scattering amplitudes in terms of incoming momenta $< \phi(p_1) \ldots >.$ Furthermore, it is much more intuitive: you can 'see' the flow of momentum through propagators into vertices and relate the momenta (through the Mandelstam variables) to the masses of the particles you're scattering. Feb 12 '13 at 14:20
• @Vibert, (+1)!, Why don't you write that as an answer? Feb 12 '13 at 23:39
• I cannot post this as a question, it would be [CLOSED] but, why Peskin&Schroeder. has such horrible notations? For example they use the Newton dot for the derivative with respect to $x^0$, that happens to be $t$ only if you are using the signature (+---) but then you run intro trouble if you work with (-+++) and happily derive with respect to $t$ because you remember to have seen the dot in Peskin... I guess it is a beginner issue but, as such, I am finding Srednicki more helpful and clear... Feb 12 '13 at 23:45
• I am more comfortable with the (+---) sign. I have done courses on relativistic quantum mechanics and classical field theory in this signature. I read peskin because I found it pretty elaborate and the problems at the end of the chapter are pretty fun to solve.
– venu
Feb 13 '13 at 9:26
• Of course, srednicki is a brilliant book too. I read a little of it at the start. But there are only so many books one can read. I am also referring to sidney coleman's lecture notes and david tong's videos.
– venu
Feb 13 '13 at 9:27

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.

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.

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.

• Are you sure this is true as well for $|0\rangle$ in an interacting (renormalized) theory? Feb 13 '13 at 8:53
• That's why I specified "in the context of quantized Klein-Gordon field" which refers to a free scalar. Feb 13 '13 at 19:13