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A few closely related questions regarding the physical interpretation of the S-matrix in QFT: I am interested in both heuristic and mathematically precise answers.

Given a quantum field theory when can you define an S-matrix? Given an S-matrix when can you define a quantum field theory from it? (e.g. I have heard conformal field theories do not have S-matrices, is there a simple heuristic way to understand this? Also, I have heard that it is not possible to define a traditional S-matrix in AdS space.)

How does one, in principle, completely specify the S-matrix of a physical theory and what information is encoded in it? (Do I have to give you both a list of all the stable bound states and single-particle states, as well as the probability amplitudes for scattering from initial and final states? or can I get a way with just telling you some single particle states and from poles in the scattering amplitudes deduce the existence of additional bound states?)

Can two inequivalent QFT's give rise to the same S-matrix? E.g., from the S-matrix of the standard model can one, in principle, read off energy levels of the hydrogen atom and other atoms?

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related: physics.stackexchange.com/q/41206 –  Arnold Neumaier Nov 24 '12 at 16:23
    
related: physics.stackexchange.com/q/41439 –  Arnold Neumaier Nov 24 '12 at 16:23
    
related; physics.stackexchange.com/q/27620 –  Arnold Neumaier Nov 24 '12 at 16:24
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2 Answers 2

An S-matrix describes the unitary tranformation between asymptotic states at $t=-\infty$ and $t=+\infty$. It can exist only in a QFT in which space-time is noncompact, in particular in flat Minkowski space (where QED and the standard model are defined).

This rules out most constructed conformal field theories, where space-time is a compact Riemann surface. But as user1504 remarked, there are other, noncompact conformal field theories. The free massless field in $R^n$ is conformal and has a (trivial) S-matrix.

If an S-matrix exists, it does not determine the dynamics at finite times; thus different QFTs may lead to the same S-matrix; one then says these are distinct ''interpolating'' field theories. (See, e.g., Section 5 of http://ls.poly.edu/~jbain/papers/lsz.pdf, and references there.)

Specification of the S-matrix is usually done perturbatively. Traditionally by LSZ theory due to Lehmann, Symanzik and Zimmermann,
http://en.wikipedia.org/wiki/LSZ_reduction_formula,
in a more rigorous way by causal perturbation theory in the spirit of Epstein and Glaser
http://en.wikipedia.org/wiki/Causal_perturbation_theory.

One gets the probability aplitudes as outputs, as these are just the matrix elements of the S-matrix. But as input, one needs the action (or Hamiltonian), and a complete list of corresponding bound states, as these define the asymptotic Hilbert space. (But in the standard textbooks, this is essentially swept under the carpet, resulting in severe infrared problems once bound states exist. Bound state in QFT are quite poorly understood; see Different kinds of S-matrices?.)

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Mild correction: A conformal field theory need not be defined on a compact Riemann surface. There are conformal field theories on $\mathbb{R}^n$ (e.g., massless free fields). –  user1504 Nov 24 '12 at 16:44
    
@user1504: Thanks, corrected. –  Arnold Neumaier Nov 24 '12 at 16:54
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The S-matrix (scattering matrix) is the unitary operator $S$ that determines the evolution of the initial state at $t=-\infty$ to the final state at $t=+\infty$. $$|\psi(t=+\infty)\rangle = S |\psi(t=-\infty)\rangle$$ This matrix/operator is therefore a collection of complex numbers that are ready to calculate the probabilities of various scattering processes, e.g. those at particle accelerators if we talk about a realistic quantum field theory.

So the matrix acts on the Hilbert (sub)space of all collections of stable particles with arbitrary momenta – which may be assumed to be infinitely separated from each other so that their mutual interactions are negligible at $t=\pm \infty$ – and the result of the action belongs to the same space. In particular, the S-matrix of the Standard model may contain arbitrary combinations of electrons, photons, and other stable particles with arbitrary 3-momenta. However, it should also allow arbitrary protons, hydrogen atoms, or other stable composite particles! It's harder to compute the S-matrix elements involving composite states in the final state or the initial state but in principle, they're as good parts of the S-matrix as the S-matrix for electrons and photons.

(In fact, there was a whole S-matrix or Bootstrap program, initiated by Heisenberg and followed in the late 1960s by Chew and others – this program was a branch of research that inspired the birth of string theory. A basic philosophy of this program was that our calculations shouldn't be constructive; they shouldn't make a qualitative distinction between elementary and composite particles; all the allowed states and mass spectrum should be determined by some consistency conditions, "by itself", and that's where the "bootstrap" terminology comes from.)

One of the answers to your questions is therefore Yes, the full S-matrix knows about all the energy levels of the hydrogen atom and other things, too. In fact, the complete spectrum of stable particles in the QFT should be given before we try to determine the S-matrix elements in between them. Even unstable particles and unstable levels are "known" to the S-matrix because they manifest themselves as poles in the scattering of stable particles or objects.

Some quantum theories only have the S-matrix but they don't admit local "off-shell" Green's functions. Quantum gravity is a typical example. Off-shell Green's functions $G(x^\mu,y^\mu,z^\mu)$ etc. are not gauge-invariant because gauge invariance in gravity includes the reparameterization of the coordinates. So only the S-matrix for graviton (and other) states with particles obeying the on-shell condition $p_\mu p^\mu = m^2$ ($m^2=0$ for gravitons) are among the quantities that are manifestly Lorentz-covariant and well-defined in quantum gravity theories (such as string theory). So in the case of some theories, the S-matrix is the only calculable quantity among the conventional ones that makes sense. (Quantum gravity, i.e. string theory, isn't quite a [local] quantum field theory in the same spacetime; it's a generalization, and the fact that Green's functions aren't well-defined is an example of this fact.)

On the contrary, some quantum field theories (obeying all the locality etc. conditions of quantum field theories) don't have an S-matrix because they don't admit multiparticle states with widely separated and virtually non-interacting particles. All conformal field theories belong to this class. Conformal field theories don't have any preferred distance scale $L$, so if we ask what's the minimum distance between particles at which the interactions become negligible, the only answer may be $L=\infty$. It follows that the interactions in conformal field theories are always "long-range forces" – they act at arbitrarily long distances. In the case of weakly coupled theories with short-range (or Coulomb) forces, the possible scattering states didn't differ from the Fock space of all the excitations much but in CFTs, they do.

In CFTs, one can't define multiparticle states with isolated, nearly non-interacting, particles, and because the S-matrix is a matrix between such states and there are none in CFTs, there is no S-matrix in CFTs. (Or, less usefully but correctly when it comes to the literal rules, we could say that $|0\rangle$ is the only allowed asymptotic state for the S-matrix of a CFT and $1$ is the only scattering amplitude between two such vacua.) Instead, we usually want to calculate various correlation functions of local operators in CFTs. CFTs are always local; they can't suffer from the problem linked to quantum gravity in the previous paragraphs because a preferred scale similar to the Planck scale is needed for the locality to break down.

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Nice question/answer. With respect to "Can two inequivalent QFT's give rise to the same S-matrix?", is it true to say then that a "conventional" QFT is fully characterized by its S matrix? "Conventional" isn't a good word, but I mean to just include QFTs for which asymptotically free particle states make sense. Presumably yes. –  twistor59 Nov 24 '12 at 9:05
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Thanks and oops, I forgot about this part of the question. Well, I would be interested about examples of inequivalent QFTs with the same S-matrix. –  Luboš Motl Nov 24 '12 at 12:05
    
@LubošMotl: For examples of inequivalent QFTs with the same S-matrix, see the reference in my answer. –  Arnold Neumaier Nov 24 '12 at 16:36
    
Well, @arnold, this is a trivial class of examples that was discussed in my answer, too. I am only interested in nontrivial examples in which the S-matrix acts between initial- and final-state Hilbert spaces that know about all the degrees of freedom in the QFT. –  Luboš Motl Nov 26 '12 at 13:06
    
thanks for the answer, it was pretty complete. –  newbie Nov 26 '12 at 17:16
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