A silly question about bound states in QED (or QCD...)

Question

My question concerns the formation of bound states in QED, let's say the formation of a hydrogen atom. I know bound states are a difficult (unresolved?) subject in general QFT, so my question is perhaps not so silly, but it has been in my head for some time and I've not figured out the answer for sure yet. 

Even if the actual computation of a bound state may be intractable, the usual answer to the formation of an atom is well known: we are told that a proton p and an electron e might form an atom H by exchanging photons together. Indeed if they can exchange photons from sufficiently close and for sufficiently long time, and if the resulting exchanges in momentum can be shown to yield an effective Coulomb attraction "on average", the resulting configuration would deserve to be called a bound state. 

What puzzles me is that direct exchanges from the electron to the proton and back seem to be unfit to the task of gluing the two particles together. The naive picture of such an exchange implies (by conservation of momentum) that the proton and electron will deflect away from each other. Of course this picture is naive (hence my question silly). But it seems to imply that the photons that are indeed exchanged need to follow rather non-trivial indirect trajectories, starting say of the p in a direction opposite to the direction where e lies, making a kind of loop and coming back to e from the other side (I hope the picture I try to describe is not completely unclear). The 'direct' exchanges would seem to contribute very little to the overall quantum interaction in a bound state. 

Does my concern make any sense? Or can it be solved in a trivial (or non-trivial) way? And more generally, what is the status of the generation of bound states "from scratch" in QED?

Comments

If you want to speak in terms of "photon exchanges", note that these photons are virtual, not real ones. They are never radiated. In Classical Electrodynamics it is known as the "near field" of a charge. Such a field may be time-dpendent for a moving charge ("retarded"), but it never becomes radiated since it does not propagate too far, unlike a free field (compare fields $\propto 1/R^2$ and  $1/R$ and their energy flows at long distances).

Each charge "feels" (has) the fields of the other charges as external fields in the charge equations. That are the guiding ideas of how to build the bound states.

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QED has no bound states.

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@ArnoldNeumier: Yes, it has. Otherwise the positronium highly excited (Rydberg) states would have decayed with a continuous rather than with a Hydrogen-like "discrete" spectra.

As well, the bound states manifest themselves as poles in S-matrix. There are many approaches to get some equations to describe the bound states in QED.

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It depends on the bound state theory and definition :)

To describe electrons bound by an external electrostatic potential generated by nuclei, one can treat the continuous nucleus-electron and electron-electron interactions as infinitely many scattering events (defined above) between each pair of particles. This treatment is valid, as the lifetime of a bound state is infinitely long compared to the duration of a scattering event. This theory of bound states is quite different from the treatment of scattering free particles from before. The key equation of this theory is the Bethe–Salpeter equation (BS equation), which defines a bound state as a pair of particles scattering infinitely many times. Some manipulation and reorganization of this equation yields the exact equal-time Bethe–Salpeter equation (eBS equation), or Salpeter–Sucher equation

Bound-state relativistic quantum electrodynamics: a perspective for precision physics with atoms and molecules ( Introduction )  by Ádám Nonn, Ádám Margócsy & Edit Mátyus

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Positronium is a resonance, not a bound state. 

Moreover, the QED S-matrix has no poles at all. Since it has massless degrees of freedom, one has branch cuts in place of poles - and only at the masses of the photon and the electron.

However, QED in an external classical potential may have bound states; the standard example is the hydrogen atom treated as QED in a Coulomb field.

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@ArnoldNeumier:

Positronium is a resonance, not a bound state. 

Right, but I prefer calling the excited states of positronium "unstabe bound states", just like any excited atomic states.

("The butterfly counts not months but moments, and has time enough.") (C)