Scattering into collection of bound states: relativistic vs non-relativistic case

As far as I understand, in non-relativistic quantum scattering problem there is a possibility (channel) of the following process. Several (in fact, at least 3) particles which are far apart approach each other, interact, and form several non-interacting systems of particles, while within each system the particles form a bound state. A simplest case of such situation is when initially one has 3 non-interacting particles $$a,b,c$$, and after the interaction one gets the particle $$a$$ and a bound state $$(bc)$$ of $$b$$ and $$c$$ which do not interact with each other. (Notice that without $$a$$ the direct process $$b+c\to (bc)$$ is impossible.)

On the other hand, in relativistic QFT the situation seems to be quite different: from a bunch of non-interacting (free) particles one cannot get any bound states of some of these or other particles. The computation of $$S$$-matrix elements, say, in QED and $$\phi^4$$ theories in all standard textbooks I looked shows that one always computes scattering amplitudes of a bunch of free particles to another bunch of free particles.

I am wondering why there is such a strong distinction between relativistic and non-relativistic scattering theories (if my understanding is correct)? Are there a priori reasons to expect that in QFT scattering of a bunch of free particles into a collection of bound states is impossible?

The reason why bound states are not discussed in your typical QFT course is that 1) they do not occur in the models discussed in introductory courses ($$\phi^4$$ theory and QED), 2) there are very few "nice" pedagogical computations of bound states.
However, be it an "approximate bound state" (a resonance) or a proper bound state of 2 particles (the case of more particle is easy to generalize), one can find its properties by considering the Bethe-Salpeter equation (the wiki is not very well written but I link it nonetheless). The Bethe-Salpeter equation was derived by Salpeter and Bethe in 1951 (with some precursors by Gell-Mann and Low) from the requirement that the normalization of the total momentum of the two-particle system is below the sum of the masses the individual particles. One then obtains the following formal integral equation of the two-point wave function $$\psi(x_1,x_2)$$ whose stationary solutions represent bound states and resonances: $$\psi(x_1,x_2) = -i \int G^{\rm free}_1(x_1,x'_1)G^{\rm free}_2(x_1,x'_2)V(x'_1,x'_2,y_1,y_2) \psi(y_1,y_2) d^4x'_1 d^4x'_2 d^4y_1 d^4y_2$$ where I have suppressed space-time indices $$x^\mu$$, and where $$G^{\rm free}$$ are free-field propagators and $$V$$ an interaction kernel consisting of a sum of all 4-point irreducible Feynman diagrams. The energies of the stationary solutions represent the energies of the bound states/resonances. By solving this equation iteratively and including only tree-level diagrams, you will get the "non-relativistic"/weak-field limit equivalent to a Schrödinger type problem in the first iteration. From higher iterations you can get QFT corrections to these types of problems.
If you are interested in the formation of these states, the wave-function actually also represents the transition amplitude of two free particles into a bound state, $$\psi(x_1,x_2) = \langle \Psi_{\rm B}|T[\phi(x_1),\phi(x_2)]|\Omega_{\rm vac} \rangle$$, which you can use to add an effective interaction vertex with a possible free outgoing bound state in your S-matrix computations. (Note, however, that the bound state necessarily has a smaller total momentum than the two particles, so a third particle such as a photon must carry energy away in the processes.)
• Thank you. In that case may I ask why bound states do not occur in QED and $\phi^4$ theory, as you mentioned. Are there any reasons for that? – MKO Jan 10 at 18:39