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

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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?

asked Jul 9 in Theoretical Physics by jmonvel (5 points) [ no revision ]

1 Answer

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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.

answered Jul 9 by Vladimir Kalitvianski (112 points) [ revision history ]
edited Jul 10 by Vladimir Kalitvianski

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