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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!

PhysicsNerd
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1 Answers1

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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.

Jun_Gitef17
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