Absence of nontrivial local conserved quantities in the Hubbard model on the two or higher dimensional hypercubic lattice

TL;DR

This work proves that the standard Hubbard model on a $d$-dimensional hypercubic lattice with $d\ge 2$ has no nontrivial local conserved quantities, strengthening the view that the model is non-integrable in higher dimensions. The authors extend Shiraishi's fermionic framework to a two- or higher-dimensional Hubbard system, revealing a three-step width-based proof that crucially relies on partial knowledge of the one-dimensional Hubbard conserved quantities. The argument combines a careful analysis of commutators and a generalized Shiraishi shift across widths $k+1$, $k$, and $k-1$ to show all candidate conserved-quantity coefficients vanish for $3 \le k \le L/2$. This result supports the expectation that integrability is unique to one dimension and suggests the method can be generalized to other fermionic lattice models, including non-Hermitian hopping cases. The work thereby provides a rigorous fermionic analog of absence-of-conserved-quantities results in higher-dimensional quantum spin systems and clarifies the landscape of local conservation laws in strongly correlated electron models.

Abstract

By extending the strategy developed by Shiraishi in 2019, we prove that the standard Hubbard model on the $d$-dimensional hypercubic lattice with $d\ge2$ does not admit any nontrivial local conserved quantities. The theorem strongly suggests that the model is non-integrable. To our knowledge, this is the first extension of Shiraishi's proof of the absence of conserved quantities to a fermionic model. Although our proof follows the original strategy of Shiraishi, it is essentially more subtle compared with the proof by Shiraishi and Tasaki of the corresponding theorem for $S=\tfrac12$ quantum spin systems in two or higher dimensions; our proof requires three steps, while that of Shiraishi and Tasaki requires only two steps. It is also necessary to partially determine the conserved quantities of the one-dimensional Hubbard model to accomplish our proof.

Figures (8)

• Figure 1: An example of the Shiraishi shift for $k = 4$. Here, $\hat{\mathsf{A}} = \hat{c}^{-} _ { ( 1 , 2 ) , \downarrow } \hat{c}^{+} _ { ( 2 , 2 ) , \downarrow } \hat{n} _ { ( 3 , 1 ) , \downarrow } \hat{c}^{+} _ { ( 3 , 2 ) , \downarrow } \hat{c}^{-} _ { ( 4 , 1 ) , \uparrow }$ is a product of width $\operatorname{Wid} \hat{\mathsf{A}} = 4$ that satisfies conditions (i) and (ii) of Lemma 3.2. The product $\hat{\mathsf{B}} = \hat{c}^{-} _ { ( 1 , 2 ) , \downarrow } \hat{c}^{+} _ { ( 2 , 2 ) , \downarrow } \hat{n} _ { ( 3 , 1 ) , \downarrow } \hat{c}^{+} _ { ( 3 , 2 ) , \downarrow } \hat{c}^{-} _ { ( 5 , 1 ) , \uparrow }$ is defined from the commutator \ref{['definition of sfB from cimmutator']}. Since $\hat{c}^{-}$ is added to the right-most site, the width becomes $\operatorname{Wid} \hat{\mathsf{B}} = 5$. Next, by removing $\hat{c}^{-}$ at the left-most site, we obtain $\hat{\mathsf{A}} ^ { \prime }$ such that the commutation relation \ref{['Shift fulfillment relationship']} holds. The resulting product $\hat{\mathsf{A}} ^ { \prime }$ with $\operatorname{Wid} \hat{\mathsf{A}} ^ { \prime } = 4$ is the Shiraishi shift $\mathcal{S} ( \hat{\mathsf{A}} )$. Here, $\operatorname{Supp} \mathcal{S} ( \hat{\mathsf{A}} )$ has a unique left-most site, but condition (i) is not satisfied because $\hat{A} ^ { \prime } _ { \mathrel{\mathop{x}\limits^{{\raisebox{-0.5ex}{$\leftharpoonup$}}}} } \neq \hat{c} _ { \mathrel{\mathop{x}\limits^{{\raisebox{-0.5ex}{$\leftharpoonup$}}}} } ^ { \pm }$. Therefore, $\mathcal{S} ^ { 2 } ( \hat{\mathsf{A}} )$ does not exist, and $q _ { \mathcal{S} ( \hat{\mathsf{A}} ) } = 0$. Hence, from \ref{['E:proportional relation']}, it follows that $q_{\hat{\mathsf{A}}} = 0$.
• Figure 2: An example for $k = 3$ with $m = 2$, $\alpha = +$, $\beta = -$, $\sigma = \uparrow , \tau = \downarrow$. Here, $\hat{\mathsf{D}} _ { 1 } ^ { \prime }$ is generated only from $\hat{\mathsf{A}}$ and $\hat{\mathsf{E}} _ { 2 } ^ { \prime }$. $\hat{\mathsf{D}} _ { 2 } ^ { \prime }$ is generated only from $\hat{\mathsf{E}} _ { 2 } ^ { \prime }$ and $\hat{\mathsf{E}} _ { 3 } ^ { \prime }$. However, $\hat{\mathsf{D}} _ { 3 } ^ { \prime }$ is generated only from $\hat{\mathsf{E}} _ { 3 }$.
• Figure 3: An example for $k = 3$ with $\alpha = +$, $\beta = -$, and $\sigma = \uparrow$. Here, $\hat{\mathsf{D}} _ { 1 } ^ { \prime \prime }$ is generated only from $\hat{\mathsf{A}}$ and $\hat{\mathsf{E}} _ { 2 } ^ { \prime \prime }$, and $\hat{\mathsf{D}} _ { 2 } ^ { \prime \prime }$ is generated only from $\hat{\mathsf{E}} _ { 2 } ^ { \prime \prime }$.
• Figure 4: The products appearing in the proof of Lemma \ref{['creation-annihilation pair']} for $k = 3$.
• Figure 5: The products appearing in the proof of Lemma \ref{['creation-annihilation pair']} for $k = 4$.