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Suppose that you're given a non-perturbative $S$-matrix that corresponds to some Wightmanian QFT. By this I mean that you're given a Hilbert space and a unitary operator $S$ that acts on the Hilbert space, and transforms appropriately under the Poincare symmetries.

Is it possible to reconstruct the QFT from its $S$-matrix? By reconstructing the QFT I mean obtaining the Wightman functions, or equivalently, the operator distributions corresponding to the quantum fields.

I'm not asking for a reconstruction algorithm – I'm aware that it isn't known presently. I'm asking about the possibility of such reconstruction in general. For example, if there is a counterexample of two different Wightmanian QFTs having the same $S$-matrix, that would answer my question in the negative.

It is frequently stated that onshell amplitudes contain less information than offshell correlation functions, which is not obvious to me. It would be nice to have a concrete example demonstrating this.

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  • $\begingroup$ There are non-QFT models which produce S-matrices which satisfy the expected analyticity properties, e.g. the tree-level S-matrix of the bosonic string. $\endgroup$
    – Ultima
    Commented Dec 29, 2019 at 3:43

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Maybe a non-rigorous answer formulated in the language of perturbative QFT helps.

Examples can be produced in this context by performing non-trivial redefinitions of the fields. The off-shell Green functions in the original QFT and in the one with redefined fields will not be equal in general, but the S-matrices will, by the equivalence theorem.

The simplest explicit example I can think of is the two Lagrangians: $$ \mathcal{L}_1 = (\partial_\mu \phi)^2, \qquad \mathcal{L}_2 = (\partial_\mu \phi)^2 - 2 \lambda \phi^2 \partial^2 \phi - \frac{4\lambda^2}{3} \phi^3 \partial^2 \phi. $$ related by $\phi \to \phi + \lambda \phi^2$. Consider for example the 4-point function for $\mathcal{L}_2$. It receives contributions from both the second and third term, and it is not equivalent to the trivial one generated by $\mathcal{L}_1$. However, it can be checked that (at tree level) it takes the form $$ G(p_1, p_2, p_3, p_4) = \frac{p_1^2 + p_2^2 + p_3^2 + p_4^2}{p_1^2 \,p_2^2 \, p_3^2 \, p_4^2}f(p_1, p_2, p_3, p_4), $$ with $f$ an analytic function at $p^2_i \to 0$, so that the corresponding on-shell amplitude vanishes. In this case, it does not matter that the coefficients of the two interaction terms are correlated, but it will for other Green functions. Of course, the equivalence theorem will still hold, but it will become more and more difficult to check it explicitly.

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  • $\begingroup$ This is not an answer that I expected, but now that I think of it, it is a good example. Thanks! $\endgroup$ Commented Dec 29, 2019 at 3:18

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