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Based on this wiki page: https://en.wikipedia.org/wiki/Heisenberg_model_(quantum) The XXZ model is exactly Bethe ansatz solvable, but based on this paper (pape 5): https://arxiv.org/abs/1011.0380, the XZ model is non-integrable. The only difference between the two models is that the XZ model does not contain the $\sum \sigma^y_i\sigma^y_{i+1}$ term. This is very much confusing to me, because intuitively the XZ model should be much simpler, even though I understand that simplicity does not necessarily mean solvability or integrability. So given that I currently have no background knowledge of Bethe ansatz solvability, is there any intuitive way to see the non-integrability of the XZ model?

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At a quick glance it seems that at the value $g=1$ of the "non-integrability parameter", this XZ spin chain is quantum integrable. Indeed, then (and only then) it's just the special case $\Delta=0$, known as the free-fermion point, of the XXZ spin chain -- with external magnetic field, but that is compatible with the integrability.

(In case you'd like to know a bit more about the quantum integrability of the XXZ spin chain you could take a look at my notes on arXiv or the references therein.)

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  • $\begingroup$ but the $\Delta=0$ case of the XXZ model is in fact the XY model, which is very different from the XZ model. Indeed the XY model is exactly solvable, but the XZ model seems not. $\endgroup$ – M. Zeng Jul 2 '17 at 4:38
  • $\begingroup$ Exactly, XY = "XX0", XZ = "XY0". $\endgroup$ – Jules Lamers Jul 4 '17 at 14:50
  • $\begingroup$ Write the XXZ Hamiltonian via spin ladder operators $S^\pm = S^x \pm i S^y$ to see that the XY model describes the hopping of spin-z excitations (kinetic term), while the $\Delta$-dependent part of the XXZ chain is a sort of potential term. The XZ model seems to have a strange kinetic term, so from that point of view it's perhaps not too surprising that the inclusion of $g$ destroys the integrability. $\endgroup$ – Jules Lamers Jul 4 '17 at 14:58
  • $\begingroup$ what do you mean by XZ=XY0? $\endgroup$ – M. Zeng Jul 5 '17 at 16:31
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    $\begingroup$ Just to resolve the above issue: indeed, the model in equation (12) is non-integrable even for $g=1$ (if $h\neq 0$; of course if $h=0$ then the model can be mapped to free fermions for any value of $g$). Note that this is not a special case of the integrable XXZ chain, cause the latter requires the $U(1)$ symmetry around the $z$-axis (or more generally, whatever direction the eternal field is pointing in), which is absent in equation (12). BTW, @JulesLamers, those are some beautiful notes, thanks! $\endgroup$ – Ruben Verresen Feb 6 '18 at 10:19

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