No. Given an S-matrix to find a corresponding Hamiltonian is called the inverse scattering problem. It leads to a unique solution only under appropriate additional constraints - otherwise there are infinitely many solutions.gonadas91 said:Hello, I was wondering if two models show the same S matrix by a direct relation between their parameters, does that necessarily mean that both models are exactly equivalent? My idea is that this is true, but would like to know about a solid argument about it if possible, thank you!
Sometimes, as in the case of sine-Gordon and Thirring, this is the case. But one needs to find a separate argument to give a conclusive proof, as the general answer is negative. Physicists generally assume equivalence if integrable systems have identical S-matrices, but a proof of that in any particular case may be quite difficult as there is no general theorem.gonadas91 said:Say that the parameter "a" for the first model has a relation with "b" in the second one, this is a relation between their coupling constants just like the sine-Gordon and Thirring models. Then, when finding the rapidities distribution under Bethe ansatz, the kernel of the integral equation is the same, the rapidities distribution is the same for both models, and then the eigenvalues are the same. Doesn't this reflect an equivalence class between the two models then?