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Unitarity of S-matrix in QFT

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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?

This post has been migrated from (A51.SE)
asked Feb 26, 2012 in Theoretical Physics by MKO_2 (50 points) [ no revision ]
retagged Mar 18, 2014 by dimension10
@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.)

This post has been migrated from (A51.SE)
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.

This post has been migrated from (A51.SE)

2 Answers

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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.

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answered Mar 3, 2012 by Igor Khavkine (420 points) [ no revision ]
Thanks a lot. I will have a look at these references.

This post has been migrated from (A51.SE)
+ 3 like - 0 dislike

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.

This post has been migrated from (A51.SE)
answered Mar 12, 2012 by Arnold Neumaier (11,395 points) [ no revision ]
Thanks for the answer. Can I read somewhere a proof that positronium is unstable? Another related question, let us consider QED with photon, electron, muon and their anti-particles. I have heard that electron and anti-muon can form a bound state (muonium). Is there a good place to read a proof that muonium is stable? Also in fact the question I asked is more specialized: even if in QED bound states exist, electron and anti-muon coming from infinity may not collide to form this bound state, like in non-relativistic situation. That was the question.

This post has been migrated from (A51.SE)
Thanks a lot. The reference seems to be very relevant. I will have a look.

This post has been migrated from (A51.SE)
I edited my answer to reply to this.

This post has been migrated from (A51.SE)

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