In quantum field theory, calculations generally are made by using a perturbation approximation with the aid of Feynman diagrams. The theory is not well suited for bound states as Feynman diagrams refer to situations where initially and finally free fields are present, which obviously is not the case for bound states (like the hydrogen atom).
If not perturbatively, how calculations on bound states can be made? Does it even make sense to say that there are perturbations in the case of bound states? Does QFT exist non-perturbatively?
I saw this article (MURRAY GELL-MANN AND FRANCIS Low Institute for Advanced Study, Princeton, Eem Jersey (Received June 13, 1951)), which treats the bound state of a proton and a neutron interacting through a meson field (as the old strong force is handled) but I'm not sure if this is a good example where QFT can be perturbatively used, as this refers to an old fashioned strong force (the meson field acting between nucleons is replaced by the gluon field acting between quarks in the modern theory of the strong force).
In the comments, reference is made to lattice quantum chromodynamics. It is said that this is a non-perturbative procedure. That begs the question though. Why stepping over to discrete spacetime points? And this doesn't answer the question of why non-perturbative calculations can't be made in real spacetime. In fact, the example is a hint that non-perturbative calculations are not possible in continuous spacetime.
To put it differently: is perturbation inherent to the QFT?
Can't we use (perturbative) QFT in solving bound state problems by connecting (summing?) an infinite (or very high) amount of disconnected (in time) free interactions? Each of these interactions can be treated perturbatively and the total of these interactions represents the bound state). Or do these disconnected free interactions somehow overlap?
