Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

It seems to me that the notion of an "S-matrix" refers to several different objects

One construction you can find in the literature is allowing the coupling constant to adiabatically approach 0 in the asymptotic past and future. This means the incoming and outgoing states belong to the free theory and the S-matrix is an operator on the free theory state space (Fock space). It is the ratio between the time evolution operators of the interacting (but adiabatically "turned off") and the free theory over an infinite time span

This construction already yields potentially several objects since there can be several free field limits. For example if the model admits something like S-duality then both the g -> 0 and the 1/g -> 0 limits are free. It is then possible to consider 2 kinds of S-matrices: in one we adiabatically set g -> 0, in the other 1/g -> 0

But it seems that there also should be something like a "fully interacting" S-matrix. This S-matrix should be an operator on the space $Fock(H_{discrete})$ where $H_{discrete}$ is the subspace of the full state space corresponding to the discrete part of mass spectrum

Is it indeed the case there are several different S-matrices? If so, why is it "carefully hidden" in the literature? Is it possible to compute the "fully interacting" S-matrix in perturbation theory using the Bethe-Salpeter equation to extract the bound state spectrum?

share|improve this question

4 Answers 4

I don't think the S-Matrix depends on the existence of a free limit of the theory. Rather, I think it depends on cluster decomposition, which identifies the (non-perturbative) asymptotic states as approximately non-intercating multi-particle states, whose relationship to a specific Lagrangian may be complicated, e.g. they may be solitons or bound states etc. There are two dimensional models where the S-matrices can be computed exactly using algebraic methods, both for perturbative and solitonic states, which demonstrates the distinction nicely.

On the other hand, if you want to calculate the S-matrix in perturbation theory using the LSZ reduction, you need a concrete identification of that Fock space and those asymptotic states. In perturbation theory, for perturbative states, this is the Hilbert space of the free theory. Note that this may no be the complete Hilbert space of the theory, and there are known examples where the perturbative S-matrix is not unitary since there is some non-zero probability to creating non-perturbative states. This is also demonstrated very concretely in two-dimensional examples.

I vaguely recall Weinberg having a nice discussion of the general definition in his QFT course, I am assuming he covers this in the first volume of his QFT series.

share|improve this answer
So you mean that what I call the "fully interacting" S-matrix is the only one which is defined nonperturbatively? Do I identify the space on which it acts correctly? –  Squark Feb 5 '12 at 19:57
That seems to capture the gist of it, but precise formulation already exists in the literature (I suspect Weinberg's text is a good entry point). –  user566 Feb 5 '12 at 21:43

There is only one S-matrix, but it is frequently introduced in sloppy ways.

The S-matrix is a unitary matrix between two isomorphic Fock spaces whose 1-particle sector contains precisely one particle for each bound state of the system. It can be constructed by the usual adiabatic textbook procedure if and only if there are no bound states (which is a standard requirement for ordinary perturbation theory already for nonrelativistic QM without fields).

See a somewhat cryptic remark in Volume 1 of Weinberg's QFT book p.110 (at the anchor of the ** footnote), who is the only textbook author I know of who mentions this and hints at how to do the more general case perturbatively, though not with enough detail to be really telling. (He says a little bit more on p.461/2, but again it is quite cryptic.)

In particular, in case of QCD, only the S-matrix where the asymptotic states are hadrons, makes physical (and hence nonperturbative) sense. But this is not tractable perturbatively anyway, as it is part of the unsolved infrared problem for QCD.

The problem is carefully hidden from textbooks because nobody really knows how to treat bound states in QFT, and talk about ignorance isn't very suitable for textbooks.

Weinberg treats bound states in Chapter 14, but only for an electron in an external field, which begs the real question. He barely mentions the oldest (and quite unreliable) method for bound states, the Bethe-Salpeter equation, which figures on p.560, where one can find the remark ''It must be said that the theory of relativistic effects and radiative corrections in bound states is not yet in entirely satisfactory state.'' - an euphemism for the fact that it is a real mess, and nobody knows how to treat it well.

On a more positive note, people work nowadays with

share|improve this answer
Please upvote this comment to encourage Arnold Neumaier to link to abstract pages rather than pdf files in his (in all other respect) nice answer. –  Qmechanic Oct 22 '12 at 18:20
@Qmechanic: I will not do that, no matter how many upvotes such comments get. My strategy is to provide the relevant information in a way that is quick for me and informative for the readers. I even find linking to abstracts worse than linking to the pdf as this requires me to search for the link on the abstract page, slowing down the fetching process. Whereas I can read the abstract also in the pdf itself. –  Arnold Neumaier Oct 22 '12 at 18:50
Why not? It is straightforward to change http:/ /arxiv.org/pdf/XXXX.XXXXvX.pdf into http:/ /arxiv.org/abs/XXXX.XXXX –  Qmechanic Oct 22 '12 at 18:57
@Qmechanic: This requires extra work that from my perspective doesn't add any value. –  Arnold Neumaier Oct 22 '12 at 19:28

In the literature which is old enough, the existence of free limits in $g\to\infty$ or elsewhere – if the moduli space is more complicated – is neglected because S-duality and the multiplicity of the free limits wasn't known.

However, it's not too big an omission because all the matrices you mention differ at most by phases that depend on the external particle masses; the interacting part of the information in the S-matrix is identical. The interactions really occur in a region of the spacetime where the coupling constant has a particular finite value. The adiabatic turning-on is just a way to define the S-matrix rigorously and its details don't really matter.

So up to some transformations that are largely trivial, there is just one S-matrix in each physical system.

share|improve this answer
But these matrices live in different spaces, since the free particle spectrum is different from the interacting particle spectrum and the spectra in different free limits are also different –  Squark Feb 5 '12 at 18:35
Dear @Squark, it depends on whether there are lines of marginal stability in the theory (values of couplings at which some stable external states cease to exist) etc. If the "adiabatic" prescription for the S-matrix works, then the $g\to 0$ and $g\to \infty$ Hilbert spaces of free particles are isomorphic and there is a simple isomorphism. Of course, if there exist asymptotic states but this existence depends on the finiteness of $g$, then the isomorphism of the two "free" Hilbert spaces breaks down: but your original adiabatic method won't work, anyway. –  Luboš Motl Feb 5 '12 at 20:46
Take QCD for example. What S-matrix are we computing in perturbation theory? I assumed it is the "adiabatic" S-matrix since the asymptotic states are free quarks and gluons. However in the "fully interacting" S-matrix the asymptotic states are hadrons. So, is it the case that both kinds of S-matrix exist or is that the latter matrix is the only one which makes sense nonperturbatively? –  Squark Feb 5 '12 at 21:23

Is it indeed the case there are several different S-matrices?

The S-matrix is approached in a quite a few different ways. Here are seven (or is it six?) examples. I personally bless only the final three, and only the last of the blessed three is useful in practice.

A. In reslib.com/book/Relativistic_Quantum_Mechanics#88

Bjorken and Drell follow an old approach of Feynman's ("the propagator approach") which they acknowledge is "less compelling" than using QFT. (Understatement!)

B. In reslib.com/book/Quantum_field_theory__Lewis_H_Ryder#175

Ryder may be echoing "the propagator approach". Whatever it is, it differs from the approaches below, and he attributes it to unpublished lectures by Veltman.

C. In reslib.com/book/Introduction_to_the_Theory_of_Quantized_Fields#207

Bogolubov and Shirkov follow an old approach (attributed to Stueckelberg) in which the operator S is a functional of a spacetime-cutoff function g. (Ick.)

In sections II.3 and II.4 of


Haag describes (or touches on) four approaches:

D. S = lim exp(iH₀t₂) exp(-iH(t₂ - t₁)) exp(-iH₀t₁)

Or some such. If you split the middle exponential into two factors, you end up with

S = Ω₊ Ω₋ 

where Ω₊ and Ω₋ are "Moller operators". Haag mentions this only for comparison. It is not suitable for QFT. But I was taught it in my first QFT class.

D. Haag-Ruelle collision theory. Now we are getting somewhere. This is what you would (or should) call a fully-interacting S-matrix. We start with a Wightman field theory with a field that creates single-particle states. [We also assume a mass gap. But let's put infrared issues aside.] By appropriately applying fields to |0> and taking limits, we obtain "in" and "out" states |α out> and |β in>. α and β are descriptive of the Fock states of a free field theory. Indeed, there are free "in" and "out" fields defined on the Hilbert space along with the interacting field. The S-matrix is then defined as:

S[α,β] = <α out | β in>

Good, but computationally intractable.

E. Araki-Haag collision theory. I won't go into the details. But we can now drop the requirement that we have a field that creates single-particle states. This is important to algebraic QFT-ers, because fields that create single fermions are unobservable. Again, computationally intractable.

F. LSZ collision theory.

Bingo! The starting assumption is that the interacting field weakly approaches the "in" and "out" fields (mentioned above) in the two asymptotic limits. Lehmann, Symanzik, and Zimmermann then show how to obtain

S[α,β] = <α out | β in>

from Green functions, ie, from vacuum expectation values of time-ordered products of fields. And Green functions can be obtained in perturbation theory using either the Gellman-Low expansion or Feynman path integrals. (I vote for the latter. In Euclidean space, followed by Wick rotation.)

To summarize: In practice, we (should) obtain Green functions from Feynman diagrams (by one of two methods) and then obtain S-matrix elements from the Green functions à la LSZ. I feel safe in saying that this is the approach to Feynman diagrams and the S-matrix that is found in "better" textbooks.

If so, why is it "carefully hidden" in the literature?

Physics went through a difficult phase between 1930 and 1970. Please read the following quote by Jost:


The recovery may be incomplete.

Is it possible to compute the "fully interacting" S-matrix in perturbation theory using the Bethe-Salpeter equation to extract the bound state spectrum?

What you're asking for, if I understand correctly, is some way to get the infinite sums (that are probed by the B-S equation) into the Green functions for composite fields that create bound states. Then you could, ideally, compute proton-proton cross sections.

I'm not aware of such a method. (But I probably wouldn't be aware even if it existed.)

share|improve this answer
As Arnold Neumaier states above, the Schwinger-Dyson equations are yet a third route to computing Green functions. –  Greg Weeks Apr 13 '12 at 1:46
Thx for your answer, Greg. There's quite a bit here to digest and I'll get back to it sometime –  Squark Apr 28 '12 at 12:14
Btw, if for any reason you want to contact me in the future it is top.squark at google mail dot com (did some obvious anti-spambotting here) –  Squark Apr 28 '12 at 12:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.