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.
Best Answer
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.