The $S=\frac{1}{2}$ XY and XYZ models on the two or higher dimensional hypercubic lattice do not possess nontrivial local conserved quantities

TL;DR

Shiraishi and Tasaki prove that the $S= frac{1}{2}$ XY and XYZ spin models on $d\ge2$ hypercubic lattices have no nontrivial local conserved quantities. They extend a shift-based method from one dimension to higher dimensions, reducing the problem to a essentially one-dimensional analysis along a chosen direction and solving a sequence of width-$k$ constraints. For all $3\le k_{\max}\le L/2$, all candidate local conserved quantities vanish; when $k_{\max}=2$ the only possible two-body contribution is proportional to the Hamiltonian, implying no extra local conserved quantities. The results provide strong evidence of non-integrability in higher-dimensional quantum spin systems and offer a framework to study quasi-local conserved quantities and operator growth in chaotic many-body dynamics.

Abstract

We study the $S=\frac{1}{2}$ quantum spin system on the $d$-dimensional hypercubic lattice with $d\ge2$ with uniform nearest-neighbor interaction of the XY or XYZ type and arbitrary uniform magnetic field. By extending the method recently developed for quantum spin chains, we prove that the model possesses no local conserved quantities except for the trivial ones, such as the Hamiltonian. This result strongly suggests that the model is non-integrable. We note that our result applies to the XX model without a magnetic field, which is one of the easiest solvable models in one dimension.

Figures (9)

• Figure 1: The lattice $\Lambda$ with $d=2$ and $L=5$, namely, the $5\times 5$ square lattice. Note that the figure represents the whole lattice. A subset $S$ with four elements is depicted by black disks. Taking into account the periodic boundary conditions, we see $\operatorname{Wid}_1 S=3$ and $\operatorname{Wid}_2S=2$.
• Figure 2: The figure represents a small portion of the square lattice, where the numbers indicate the first and second coordinates. The product $\boldsymbol{\hat{B}}=\hat{X}_{(1,1)}\hat{Y}_{(2,1)}\hat{Z}_{(2,0)}\hat{X}_{(4,0)}$ is generated, for example, from $\boldsymbol{\hat{A}}=\hat{X}_{(1,1)}\hat{X}_{(2,0)}\hat{X}_{(4,0)}$ with $\hat{Y}_{(2,1)}\hat{Y}_{(2,0)}$, from $\boldsymbol{\hat{A}}'=\hat{X}_{(1,1)}\hat{Y}_{(2,1)}\hat{Z}_{(2,0)}\hat{Z}_{(4,0)}$ with $\hat{Y}_{(4,0)}$, and from $\boldsymbol{\hat{A}}"=\hat{X}_{(1,1)}\hat{Y}_{(2,1)}\hat{Y}_{(2,0)}\hat{X}_{(3,0)}\hat{X}_{(4,0)}$ with $\hat{X}_{(2,0)}\hat{X}_{(3,0)}$. Note that $\operatorname{Supp}\boldsymbol{\hat{B}}\supsetneqq\operatorname{Supp}\boldsymbol{\hat{A}}$, $\operatorname{Supp}\boldsymbol{\hat{B}}=\operatorname{Supp}\boldsymbol{\hat{A}}'$, and $\operatorname{Supp}\boldsymbol{\hat{B}}\subsetneqq\operatorname{Supp}\boldsymbol{\hat{A}}"$. For our proof, generation processes in which the support strictly increases, as in the case of $\boldsymbol{\hat{A}}$, are most important. We express the process by the appending operation as $\boldsymbol{\hat{B}}={\cal A}^{\hat{Y}\hat{Y}}_{(2,1)\leadsto(2,0)}(\boldsymbol{\hat{A}})$ or $\boldsymbol{\hat{B}}={\cal A}^{\hat{Y}\hat{Y}\leadsto\hat{X}}_{(2,1)\leadsto(2,0)}(\boldsymbol{\hat{A}})$. See \ref{['e:D1C1E2']} for the latter notation.
• Figure 3: Black disks represent the Pauli matrices. Here, the figure represents a small portion of a bigger lattice. The product $\boldsymbol{\hat{A}}$ with horizontal width $k_\mathrm{max}=3$ has a right-most site $x$ and two left-most sites $y$ and $y'$. The product $\boldsymbol{\hat{B}}$ with horizontal width 4 is obtained by appending $x'=x+\boldsymbol{e}_1$ to $\boldsymbol{\hat{A}}$. One readily sees that $\boldsymbol{\hat{A}}$ is the only product with horizontal width 3 that generates $\boldsymbol{\hat{B}}$.
• Figure 4: The product $\boldsymbol{\hat{A}}$ with horizontal width $k_\mathrm{max}=3$ has a unique right-most site $x$ and a unique left-most site $y$. The product $\boldsymbol{\hat{B}}$ with horizontal width 4 is obtained by appending $x'=x+(1,0)$ to $\boldsymbol{\hat{A}}$. By truncating the left-most site $y$ from $\boldsymbol{\hat{B}}$ (when possible), we get $\boldsymbol{\hat{A}}'={\cal S}(\boldsymbol{\hat{A}})$ with horizontal width 3, which is the shift of $\boldsymbol{\hat{A}}$. Here, the figure represents a small portion of a bigger lattice.
• Figure 5: The products of Pauli operators that play central roles in Section \ref{['S:k=3']}. Here, the figure represents a small portion of a bigger lattice.

Theorems & Definitions (16)

• Theorem 2.1
• Theorem 2.2
• Lemma 3.1
• Lemma 3.2
• Lemma 3.3
• Lemma 3.4
• Lemma 3.5
• Lemma 3.6
• Lemma 3.7
• Lemma 3.8