Why are CFTs not usually studied in momentum space? Conformal symmetry in QFT has been extremely useful for physics. However, while most of QFT is usually done in momentum space, CFTs are usually studied in position space or in terms of Mellin transformed variables (as opposed to Fourier transformation to go to momentum space). It seems the reasoning must have something to do with existence of massless particles and bad branch cuts in momentum space. But I want to make these statements precise.
So, what exactly is the reason CFTs (especially 2D CFTs) are not studied in momentum space?
There are some relatively new papers in momentum space CFT (e.g., this one), but still by and large momentum space is usually not the first choice for analysing CFTs.
 A: Another answer points out that scattering amplitudes are not good observables in CFT. This is certainly true. But also in QFT not everything is scattering amplitudes.
One important reason why CFT is usually favoring position space (or Mellin space) is the operator product expansion (OPE). This is the tool that makes CFT so much more tractable than QFT, and it works like a charm in position space: the OPE converges very fast whenever two operators are close to each other.
There is also an OPE in momentum space, but:

*

*it only applies to Wightman functions, not time-ordered products or Euclidean correlators,

*its convergence properties are not as nice as in position space (since the question asks specifically about 2D CFT, see arXiv:1912.05550).

A: In a standard QFT you study scattering amplitudes. The Fourier transform of the two point function $\langle \phi(x) \phi(0) \rangle$ contains a pole at stable particles $p^2 = m^2$. This pole is picked up by the LSZ reduction theorem when you study scattering amplitudes.
However, in a CFT, conformal symmetry implies that you don't have these isolated poles in the complex $p^2$ plane. If you try to apply the LSZ reduction theorem to a CFT, your scattering amplitudes will always be $0$ for this reason, as you won't have this pole to cancel out the factor of $(p^2 - m^2)$.
Thing is, in a CFT, scattering is not the right thing to study. Because you have scaling symmetry, there's no meaningful notion of widely separated wave packets. How can you "widely separate" anything when you can always rescale them close together?
Because of this, the meaningful things to study in CFT take on a very different character than standard QFTs, like QED, where you usually study scattering amplitudes of momentum eigenstates.
