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?

  • $\begingroup$ Why do you say that two particles can't form a bound state in QFT? I'm pretty sure there are two-dimensional integrable field theories with scattering $A+B \to C$ and where $A$, $B$ and $C$ are perfectly stable particle states. $\endgroup$
    – Sidious Lord
    Feb 26, 2012 at 15:36
  • $\begingroup$ @Sidious Lord: Can I read somewhere about such examples? Can it happen in QED? (As far as I heard, the 2d case is somewhat exceptional in QED: in the Schwinger model polarization of vacuum has an effect of creation of a bound state of electron-positron pair which is a free boson. But I might be wrong about this, I do not really know this.) $\endgroup$
    – MKO
    Feb 26, 2012 at 18:56
  • $\begingroup$ Hi @Dilaton: Concerning the tag edit(v3) I would suggest the unitarity tag and the s-matrix-theory tag instead of the qed tag (because OP is really asking about qft) and the research-level tag (because the question is textbook material). $\endgroup$
    – Qmechanic
    Jan 1, 2013 at 16:27
  • $\begingroup$ Thanks @Qmechanic, it never hurts when you hava a look at it too when I retag, since you are much much much more knowledgable. I change the tags as you suggest. And happy new year to you :-) $\endgroup$
    – Dilaton
    Jan 1, 2013 at 17:33

3 Answers 3


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.


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.


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.

  • $\begingroup$ Thanks for the comment. I realize that In and Out states may be bound states in principle. However in QED bound states are not taken into account. That means that free electrons, muons etc. cannot become a bound state (am I wrong?). Also I realize that when one uses Feynmann rules to compute S-matrix, one should include all counterterms. So I think it does not really answer the question. $\endgroup$
    – MKO
    Feb 26, 2012 at 12:40
  • $\begingroup$ There is a cross section of producing bound states when two opposite-charge particles collide. All what is necessary is to emit the excess of energy-momentum as photons that is quite possible. Also it is possible to create a pair in the final state that is in a bound state, not free electron and positron. In QED there is no problem with unitarity in this respect. Renormalized and infra-red fixed QED is adequate theory. Feynman rules can include bound states in In and Out states, as a matter of fact. $\endgroup$ Feb 26, 2012 at 16:47
  • $\begingroup$ If I understand correctly, in QED in 4d space-time there is no cross-section of producing bound states when two opposite-charge particles collide. Definitely in the non-relativistic setting two particle coming from infinity and interacting according to Coulomb law (at short distances) cannot collide. $\endgroup$
    – MKO
    Feb 26, 2012 at 18:50
  • $\begingroup$ A pair of non interacting electron and positron is described with a product of two plane waves. A bound state is described with a product of a plane wave (center of mass) and a wave function of a bound state, easy. $\endgroup$ Feb 26, 2012 at 19:53

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