# Why do we use Fourier transforms in QFT? [duplicate]

I ask this question, as someone has recently asked me this and I'm not sure I gave them a satisfactory/correct answer.

I explained that in QFT we describe particles (and there interactions) in terms of fields, which fundamentally have a wave-like nature (indeed, solving the appropriate equations of motion gives rise to exponential solutions, suggesting such a decomposition into Fourier modes). In general, we wish to describe discrete excitations in these fields and this requires us to know hoe the field is "constructed". This can be achieved through expressing the field in terms of its Fourier transform, i.e. an infinite sum of the frequencies that constitute it and the excitation modes (amplitudes) associated with each of those frequencies.

Would this description be correct at all? Any corrections, improvements and/or explanations will be much appreciated.

• I think the important point is that position and momentum are a fourier pair. You can decompose any function into it's component frequencies as a mathematical exercise, but in QFT this transform has added physical meaning. I'm still a beginner in QFT so perhaps someone more experienced can add to this. – MonkeysUncle Mar 27 '15 at 15:47
• This is so close to an exact duplicate, you will find your answer at physics.stackexchange.com/q/53731. – pyramids Mar 27 '15 at 15:51
• Um...Fourier transforms are a very convenient method to solve the differential equations obeyed by the fields, and the Fourier coefficients have a natural interpretation as creation and annihilation operators. Why shouldn't we use them? – ACuriousMind Mar 27 '15 at 15:51