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I have a system where the two-particle scattering matrix $S_{12}(p_1,p_2)$ depends on the momentum difference $p_1-p_2$, and also on the total momentum $P=p_1+p_2$ in some non-trivial way. One can use parametrization of energies and momenta $E=m\cosh(\alpha)$ and $p=m\sinh(\alpha)$ to express everything in terms of the rapidities $\alpha$. Then, I obtain $S_{12}(\alpha,\beta)$ as a function of $\alpha-\beta$ and $\alpha+\beta$. It is known that if $S_{12}$ satisfies the Yang-Baxter equation (TBE), then the $N$-body $S$ matrix factorizes as a product of 2-body $S$ matrices.

But if $S$ depends on the combination of $\alpha+\beta$, not only on $\alpha-\beta$, can we already rule out the possibility for such $S$ matrix to satisfy the YBE? Or, is it possible that even depending on $\alpha+\beta$ (and $\alpha-\beta$), the YBE can still be satisfied?

The quantity $p_1+p_2$ can always be expressed in terms of the invariant:

$$s^{2}=(E_1+E_2)^{2}-(p_1+p_2)^{2}$$

But I don't know if that is of any help in such cases.

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    $\begingroup$ Lets assume for simplicity there is only one particle. How can I prove that this certain $S(\alpha,\beta)$ does (or does not) satisfy the YBE? Is the general recipe just to substitute the form of $S$ in the YBE? For instance, assuming $S=e^{i\Phi_{12}}$, i.e. it is just a simple phase, but $\Phi(p_1,p_2)\in\mathbb{C}$. $\endgroup$
    – Zarathustra
    Commented Jun 30, 2022 at 13:38
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    $\begingroup$ Thanks for the question, I promoted the comment to an answer, including some formulas etc. $\endgroup$
    – Cream
    Commented Jun 30, 2022 at 13:51
  • $\begingroup$ As a more generic question, if $S_{12}$ satisfies the YBE, even in the simplest scalar case, does that necessarily mean the model is integrable in the Bethe Ansatz sense? Or what are the true conditions that need to be satisfied then? I always thought YBE is a necessary and sufficient condition for a model to be considered integrable, but I might be completely wrong. $\endgroup$
    – Zarathustra
    Commented Jun 30, 2022 at 15:55
  • $\begingroup$ What do you mean by "in the Bethe Ansatz sense"? Integrability holds when there are infinitely many integrals of motion, connected to conserved quantities. In ch. 3.1 of these (pdf) lecture notes, it is shown that this implies a scattering without particle production or momentum exchange. I would think that this works the other way around, with the conserved quantities being functions of the momenta which are unchanged by scattering. $\endgroup$
    – Cream
    Commented Jul 1, 2022 at 6:49
  • $\begingroup$ Yes, I understand the integrability from the algebraic terms, namely that the transfer matrix derivatives give all these conserved quantities. But I am referring to the original coordinate Bethe Ansatz, where the wavefunction coefficients can be expressed in terms of the S-matrix of the problem. In the coordinate Bethe Ansatz, no explicit mention to the infinite set of conserved laws is done; one just solves the two-body problem and from there, one makes the full Ansatz for the N-body wavefunction. The question is, given the solution of a 2-body problem, how do we know if this is possible. $\endgroup$
    – Zarathustra
    Commented Jul 1, 2022 at 7:06

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It can certainly not be ruled out, that such scattering matrices exist.

An example:
The YBE (Yang-Baxter Equation) is a matrix equation. So, if you consider a scalar or diagonal scattering matrix (i.e. no exchange of charge/change of particle type), then it is satisfied for any $S(p,q)$ due to commutativity: $$ S(p_1,p_2) S(p_1,p_3) S(p_2,p_3) = S(p_2,p_3) S(p_1,p_3) S(p_1,p_2). $$ If $S$ is non-diagonal (particle type can change in scattering), this is a non-trivial requirement. Even if $S(p,q) = S(\alpha-\beta)$ depends only on differences in rapidities.

Non-diagonal scattering matrices:
I see no reason why it should be incompatible with general $S(p,q)$, even though the constraint is stronger and it might be more difficult to find an example. WP states:

$$S(p,q) = S(\alpha-\beta)$$

as a "common Ansatz", but not as a requirement.

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