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This is a general question, but what is meant when people refer to the S-Matrix of $\mathcal{N}=4$ Super Yang Mills? The way I understood it is the S-Matrix is only well defined for theories with a mass gap so we can consider the asymptotic states to be non interacting and then apply the LSZ formalism. The idea breaks down for general CFTs and the observables should just be the correlation functions.

Based on what I've seen in the literature, this does not seem to be the case and people talk about the S-Matrix for $\mathcal{N}=4$ Super Yang Mills, a superconformal field theory. Is it that we consider a deformed CFT so there exists a gap in the spectrum and take the limit as the deformation goes to zero? Or is there a way to define an S-matrix in an exactly conformal theory?

Edit: For anyone who finds this question the following reference (the introduction at least) is of use in showing how the normal logic breaks down: http://arxiv.org/abs/hep-th/0610251

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    $\begingroup$ From what I understand, the S-matrix for $N=4$ is IR-divergent, as expected for any CFT, but there are some terms in it that are IR-finite and can be unambiguously computed. I would elaborate but that's where my "knowledge" ends. $\endgroup$ – Matthew Dec 25 '13 at 2:24
  • $\begingroup$ People indeed compute the amplitudes in a deformed theory: you can use a flavour of dim reg that is compatible with SUSY. But I've never done any work in this field, so really some expert should answer (there are many!) and get the proper bounty. $\endgroup$ – Vibert Jan 12 '14 at 20:52
  • $\begingroup$ Have a look at the following paper by Steve Weinberg -- inspirehep.net/record/1190706. The claim in the abstract seems to take on your question. $\endgroup$ – Siva Jan 14 '14 at 3:59
  • $\begingroup$ @Siva that is an interesting result but I don't understand why having free massless particles would be interesting for calculating scattering amplitudes $\endgroup$ – David M Jan 15 '14 at 2:04
  • $\begingroup$ Minor comment to the post (v5): In the future please link to abstract pages rather than pdf files, e.g., arxiv.org/abs/hep-th/0610251 $\endgroup$ – Qmechanic May 9 '14 at 23:03
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As you note the construction of asymptotic states breaks down in a CFT since there is no mass gap. It is therefore necessary to introduce an IR regulator by using, eg, dimensional regularisation. The full scattering amplitude will then depend on this regulator. However, it is possible to construct physical observables that do not depend on the regulator. Furthermore, the amplitude contains subleading terms that also are independent of the regulator, and these will be the same in any regularisation scheme. As an example of this the four particle scattering amplitude takes the form $$ \mathcal{A}_4 = \mathcal{A}_4^{\text{tree}} \exp\big[(\text{IR div.}) + \frac{f(\lambda)}{8} (\log(s/t))^2 + (\text{const}) \big] $$ The coefficient $f(\lambda)$, the cusp anomalous dimension, is independent of the IR regulator and hence universal.

The AdS/CFT dual of a field theory scattering amplitude is the expectation value of a polygonal light-like Wilson loop, see, eg, arxiv:0705.0303, which also contains a bit of discussion about IR divergences and a bunch of useful references.

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    $\begingroup$ Thanks for you answer and I'll definitely take a look at the reference. I have to ask but do you know if or how these results carry over to more general CFTs? That is, with a proper IR regulator when are we guaranteed that the S-Matrix of a CFT contains some useful regulator independent terms? $\endgroup$ – David M Jan 15 '14 at 4:29
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    $\begingroup$ I think it should be similar at least in any Yang-Mills type gauge theory, but I can't remember seeing this discussed in detail anywhere. $\endgroup$ – Olof Jan 15 '14 at 8:26
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    $\begingroup$ Also note that the general $AdS_{d+1}$ dual to a gluon scattering amplitude is an S-matrix for scattering of open strings localised on some $(d-1)$-dimensional surface (in The $AdS_5 \times S^5$ case this would be a $D3$ brane). However, in the maximally supersymmetric case you can relate this to the Wilson loop picture I mention in my answer using T-duality. $\endgroup$ – Olof Jan 15 '14 at 8:30
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Type II B string theory on $ AdS_5\times S^5 $ gives, through the AdS/CFT duality, $\mathcal{N}= 4, D=4 $ super yang-mills theory. Therefore, if one derives the S-matrix elements of type II B string theory on $ AdS_5\times S^5 $, then the S-matrix for $\mathcal {N}=4, D=4$ super yang mills arises. At least, that is my way of thinking of it. I suggest you read the paper by Giddings, "Stephen B. Giddings, The boundary S-matrix and the AdS to CFT dictionary, hep-th/9903048".

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    $\begingroup$ Hi Sanath, Thank you for the reference, I am aware of the AdS/CFT dictionary but probably not enough about the subtleties of the S-Matrix in AdS space. To be honest though I was looking for an answer on the field theory side. I brought up $\mathcal{N}$=4 SYM because I know its a CFT and has an S matrix and was looking for where the usual lore that CFTs don't have S-matrices breaks down (for example CFTs are exceptions to the Coleman-Mandula theorem because they assumed an S-matrix). $\endgroup$ – David M Jan 12 '14 at 17:33
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    $\begingroup$ I believe now it has to do with if the massless particles are noninteracting at asymptotic infinity an S-matrix can be unambiguously defined (based on a footnote in a paper by Maldacena) and hopefully the AdS/CFT dictionary further clarifies this idea. Edit: The paper was arxiv.org/pdf/1112.1016v1.pdf on page 2: " [2]also mentions the conformal group in the case of massless particles. However, [2] assumed that these massless particles are free in the IR, so that an S-matrix exists" [2] references the Haag-Lopuszanski-Sohnius theorem. I'm not sure if this is a complete answer though. $\endgroup$ – David M Jan 12 '14 at 17:48
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    $\begingroup$ I'm also unsure that the s-matrix in AdS gives rise to one on the cft side. Rather its the correlation functions of the cft that should give the scattering amplitudes in the bulk. The question is the LSZ prescription and if you have to deform your cft in the IR to properly do this. $\endgroup$ – David M Jan 12 '14 at 20:50

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