In algebraic Bethe ansatz procedure, one of the central objects is the R-matrix satisfying the Yang-Baxter equation, but all the papers/books give directly its expression without deriving it, so my question is How can i derive the R-matrix for XYZ/XXZ Heisenberg model?


1 Answer 1


This is essentially an answer to your questions R-matrix for spin chains, Elliptic R-matrix and Yang Baxter solution for XYZ model, $R$ matrix for XYZ spin chain, Algebraic Bethe Ansatz and $R$-matrices, which all basically ask the same question anyway.

In short: to the best of my knowlegde, coming up with an R-matrix is an art, not a derivation. (Cf. the notion of a Lax pair in classical integrability.)

The quantum inverse-scattering method (QISM) was developed as a synthesis of the classical ISM, spin chains and lattice models. The best way to understand it is from this multi-topic point of view, rather than focussing just on spin chains. Faddeev's How Algebraic Bethe Ansatz works for integrable model [arXiv:hep-th/9605187] focusses on spin chains mostly, which makes several constructions --- such as the introduction of an auxiliary space --- appear somewhat ad hoc; at least it certainly felt so to me when I first read them. The vertex-model point of view makes these constructions much more natural; this is also why I organized my lecture notes A pedagogical introduction to quantum integrability, with a view towards theoretical high-energy physics, [arXiv:1501.06805] in the way I did.

Some more comments:

  • Once you know the Lax matrix (containing the vertex weights) of the six- or eight-vertex model you can solve for the R-matrix (solving the "RTT-relation" with $T=L$ for the case of one site), see e.g. Sections 9.6 and 10.4 in Baxter, Exactly solved models in statistical mechanics (or Appendix C in my lecture notes mentioned above).

  • Alternatively, you can look for solutions of the Yang--Baxter equation, and then interpret each R-matrix you get as a vertex model or see which spin chain it yields by computing the logarithmic derivative of the associated transfer matrix.

  • It might be instructive to read up on another example: Shastry's R-matrix for the Hubbard model. See e.g. Section 12.2 in Essler, Frahm, Göhmann, Klümper, Korepin, The one-dimensional Hubbard model [e-print].

  • $\begingroup$ Thank you for your answer ! i understand what you're telling me, and i've seen your notes which seems very interesting, But : suppose that you don't know that XYZ model is integrable, so it's contradictory to find its R-matrix by using the "RTT" relation because the "RTT" is valid only if your model is integrable. In another hand if you look for solutions of Yang-Baxter equation directly, you can not know for which model you're solving. What i was thinking was : In order to find R-matrix for XYZ you have to find $Z_2 $ solutions of YBE, but i don't know if it's possible to do? $\endgroup$
    – Giuseppe
    Commented Apr 6, 2018 at 11:11
  • $\begingroup$ @Giuseppe You're welcome. I don't know of examples of models that were found to have an R-matrix without already expecting (or constructing) them to be quantum integrable. Here hints for the integrability (loosely: the existence of many conserved quantities) may for example come from factorized scattering in a Bethe-ansatz type of solution. $\endgroup$ Commented Apr 6, 2018 at 11:17
  • $\begingroup$ @JulesLamers Yang-Baxter eq (YBE) seems to be a sufficient condition, i.e. if you have an R-matrix satisfying YBE, then the model is integrable. But how about the reverse? Do all 1+1d integrable spin-chain models have an R-matrix satisfying YBE? For example, spin-chain models having discrete (say $\mathbb{Z}_2$) on-site symmetry, do these models have an R-matrix satisfying YBE? $\endgroup$
    – QGravity
    Commented Sep 14, 2023 at 10:49
  • $\begingroup$ @QGravity: That is a good question, but really beyond the OP. If you ask it as a separate question I'd be happy to write an answer. This will make it easier to find, which I think will be more useful for others $\endgroup$ Commented Sep 14, 2023 at 11:36
  • $\begingroup$ @JulesLamers thank you. I just did. $\endgroup$
    – QGravity
    Commented Sep 14, 2023 at 15:13

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