Unitarity of S-matrix in QFT

Question

I am a beginner in QFT, and my question is probably very basic.

As far as I understand, usually in QFT, in particular in QED, one postulates existence of IN and OUT states. Unitarity of the S-matrix is also essentially postulated. On the other hand, in more classical and better understood non-relativistic scattering theory unitarity of S-matrix is a non-trivial theorem which is proved under some assumptions on the scattering potential, which are not satisfied automatically in general. For example, unitarity of the S-matrix may be violated if the potential is too strongly attractive at small distances: in that case a particle (or two interacting with each other particles) may approach each other from infinity and form a bound state. (However the Coulomb potential is not enough attractive for this phenomenon.)

The first question is why this cannot happen in the relativistic situation, say in QED. Why electron and positron (or better anti-muon) cannot approach each other from infinity and form a bound state?

As far as I understand, this would contradict the unitarity of S-matrix. On the other hand, in principle S-matrix can be computed, using Feynmann rules, to any order of approximation in the coupling constants. Thus in principle unitarity of S-matrix could be probably checked in this sense to any order.

The second question is whether such a proof, for QED or any other theory, was done anywhere? Is it written somewhere?

Answer 1

In principle, bound states are possible in a QFT. In this case, their states must be part of the S-matrix in- and out- state space in order that the S-matrix is unitary. (Weinberg, QFT I, p.110)

However, for QED proper (i.e., without any other species of particles apart from photon, electron, and positron) it happens that there are no bound states; electron and positron only form positronium, which is unstable, and decays quickly into two photons. http://en.wikipedia.org/wiki/Positronium

[Edit: Positronium is unstable: http://arxiv.org/abs/hep-ph/0310099 - muonium is stable electromagnetically (i.e., in QED + muon without weak force), but decays via the weak interaction, hence is unstable, too: http://arxiv.org/abs/nucl-ex/0404013. About how to make muonium, see page 3 of this article, or the paper discovering muonium, Phys. Rev. Lett. 5, 63–65 (1960). There is no obstacle in forming the bound state; due to the attraction of unlike charges, an electron is easily captured by an antimuon.]

Note that the current techniques for relativistic QFT do not handle bound states well. Bound states of two particles are (in the simplest approximation) described by Bethe-Salpeter equations. The situation is technically difficult because such bound states always have multiparticle contributions.

Answer 2

Unitarity of the S-matrix can be checked perturbatively. Bound states tend to be non-perturbative effects, so may not show up naive perturbative calculations. Unfortunately, the datailed proof is not discussed in many places. One book that has it is Scharf's book on QED. When looking through other books you should look for keywords like optical theorem and Cutkosky rules. Bound states are usefully discussed in the last chapter of vol.1 of Weinberg's tretease on QFT.

Answer 3

In and Out states are not obligatory free states, but can be bound states too, so transitions from free to bound states are also possible. In case of QED with electron-antimuon bound state, its formation is accompanied with photon emission present in the final system state. It does not contradict the unitarity.

Problems with proofs in QED and other QFTs are due to wrong coupling term like $jA$ which is not correct alone and is corrected with counterterms. In addition, these counterterms cannot be treated exactly but only perturbatively so the true interaction of true constituents is not seen.