Consider the one-dimensional quantum XXZ model defined by the Hamiltonian: $$ H = J \sum_{i} \left (X_i X_{i+1} + Y_i Y_{i+1} + \Delta Z_i Z_{i+1} \right). $$

First, let us focus at zero temperatures, $T = 0$. For $\Delta >1$, the ground state is a Néel antiferromagnet; this phase spontaneously breaks a $\mathbb{Z}_2$ symmetry associated with the transformation $Z_i \rightarrow -Z_i$ as discussed in this answer. For $-1< \Delta <1$, the system is in a gapless XY phase where quantum fluctuations destroy the long-range order. This phase breaks the U(1) symmetry associated with spin rotations about the $\hat{z}$ axis as discussed here. The quantum phase transition (at $T=0$) between these two phases occurs at $\Delta = 1$. On one side, we are breaking a $U(1)$ symmetry, so I would expect this transition to be of the BKT type. On the other side, we are breaking a $\mathbb{Z}_2$ symmetry, so I would expect an Ising transition. So, which is the correct universality class for this quantum phase transition at $\Delta =1, T=0$ and why?

Next, consider finite temperatures, $T>0$. Since we are in (1+1) dimensions, the Mermin-Wagner theorem prevents true long-range order for the $\lvert \Delta\rvert < 1$ side. However, we can still have quasi-long-range order, so I would expect a finite-temperature BKT transition as we increase $T$ at a fixed $\lvert \Delta\rvert < 1$. On the $\Delta > 1$ side, do we have a finite temperature phase transition or does the ordered Néel phase exist only at $T=0$? I understand that in a classical spin system, there can be no finite-temperature phase transition in one dimension, but since we have a (1+1)-dimensional quantum system---which corresponds to a 2D classical system---should we expect a finite-$T$ transition?

Any thoughts are appreciated!

  • $\begingroup$ You probably meant $+J$ in the Hamiltonian, or you're considering negative $J$. Am I right? $\endgroup$ Commented Mar 14 at 2:46
  • $\begingroup$ Yes, I fixed that to avoid confusion. Thanks! $\endgroup$ Commented Mar 14 at 2:47
  • 3
    $\begingroup$ The Tomonaga-Luttinger Liquid <-> Neel transition is BKT. At finite temperature in 1d, neither the XY QLRO nor the Ising order are stable, and correlations decay exponentially. There is no finite temperature transition. For reference see "Quantum Phase Transitions," Sachdev; "Quantum Physics in One Dimension," Giamarchi. $\endgroup$
    – mbintz
    Commented Mar 16 at 18:25

1 Answer 1


For the second question, what should be the case is the moment you get to $T>0$, the "quantum-classical correspondence" with $d+1$ dimension breaks down, and it becomes simply like a $d$ dimensional (almost classical) system in terms of the stat-mech. In this case, because the system is 1d, if you go to finite temperature the system simply loses both order Ising and XY. So, you wouldn't see a transition but rather a cross-over between Ising-like and XY-like.

For the first question, because the Hamiltonian becomes the isotropic Heisenberg model exactly at $\Delta=1$, the critical exponent should be neither the Ising criticality nor the XY criticality. As a matter of fact, I'm pretty sure I heard that the theory of Luttinger-Liquid is supposed to describe the entire XY phase here but also up to the $\Delta=1$ point.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.