Lower Bound to the Entanglement Entropy of the XXZ Spin Ring

Abstract

We study the free XXZ quantum spin model defined on a ring of size L and show that the bipartite entanglement entropy of certain eigenstates belonging to the first energy band above the vacuum ground state satisfies a logarithmically corrected area law. This applies in particular to eigenstates corresponding to the lowest eigenenergy above the ground state. To this end, we develop a new perturbational approach, which allows us to control the eigenvalues of reduced states in the XXZ model in terms of the corresponding reduced states in the Ising model. Along the way, we show a Combes–Thomas estimate for fiber operators which can also be applied to discrete many-particle Schrödinger operators on more general translation-invariant graphs.

Appendix A

1.1 A.1. Uniqueness of Fiber Operator Ground States

Lemma A.1

Let $\Delta >2$, let $L,N\in \mathbb {N}$ with $1<N<L-1$. Then, for any $\gamma \in \mathcal V_L$, the operator $\hat{H}_{L,\gamma }^N$ has exactly one eigenvalue in $[1-\frac{1}{\Delta },1]$ and no eigenvalue in $(1,2-\frac{2}{\Delta })$.

Proof

By Lemma 3.5, we get

$$\begin{aligned} \hat{H}_{L,\gamma }^N=-\frac{1}{2\Delta }\hat{A}^N_{L,\gamma }+\hat{W}_{L,\gamma }^N\ge \left( 1-\frac{1}{\Delta }\right) \hat{W}_{L,\gamma }^N\,. \end{aligned}$$

(A.1)

There exists exactly one element $\hat{x}_0\in \widehat{\mathcal {V}}_L^N\cap \mathcal V_{L,1}^N$. It satisfies $W(\hat{x}_0)=1$. For any other $\hat{x}\in \widehat{\mathcal {V}}_L^N{\setminus }\{\hat{x}_0\}$, we have $W(\hat{x})\ge 2$. Let $\phi _{L,\gamma ,0}^N\in \mathbb S_{L,\gamma }^N$ be defined by $\phi _{L,\gamma ,0}^N(\sigma ,{\hat{x}}):=\delta _{\gamma ,\sigma }\delta _{{\hat{x}}_0,{\hat{x}}}$.

Hence, the operator

$$\begin{aligned}{} & {} \hat{H}_{L,\gamma }^N+\left( 1-\frac{1}{\Delta }\right) |\phi _{L,\gamma ,0}^N\rangle \!\langle \phi _{L,\gamma ,0}^N| \nonumber \\{} & {} \quad \ge \left( 1-\frac{1}{\Delta }\right) \left( \hat{W}_{L,\gamma }^N+ |\phi _{L,\gamma ,0}^N\rangle \!\langle \phi _{L,\gamma ,0}^N|\right) \ge \left( 2-\frac{2}{\Delta }\right) \end{aligned}$$

(A.2)

is a rank-one perturbation of $\hat{H}_{L,\gamma }^N$. Therefore, the unperturbed operator ${\hat{H}}_{L,\gamma }^N$ has at most one eigenvalue below $(2-\frac{2}{\Delta })$. On the other hand, since $\langle \phi _{L,\gamma ,0}^N,\hat{H}_{L,\gamma }^N\phi _{L,\gamma ,0}^N\rangle =1$, there exists at least one eigenvalue which is less than or equal to 1. Since for $\Delta >2$, we get $1<2-\frac{2}{\Delta }$, this concludes the proof. $\square$

For $\gamma =0$, it follows from the explicit structure of the fiber operator $\hat{H}_{L,0}^N$ that it has a unique ground state $\hat{\varphi }_{L,0}^N$ which can be chosen to be strictly positive. The same is true for the original operator $H_L^N$. This will allow us to conclude that $\varphi _{L,0}^N:=(\mathfrak {F}_L^N)^*\hat{\varphi }_{L,0}^N$ is the ground state of $H_L^N$. The main tool for our result will be an idea presented in [31], where the existence of a strictly positive ground state for the XXZ model on the ring was established; however, let us also point out that this piece of the proof also follows from the Allegretto–Piepenbrink theorem shown in [11].

Lemma A.2

Let $N,L\in \mathbb {N}$, $0<N<L$. Moreover, let $E_0\equiv E_0(L,N,\Delta )=\inf \sigma (\hat{H}_{L,0})$. Then, $E_0$ is non-degenerate and the corresponding eigenvector $\hat{\varphi }_{L,0}^N\in \mathbb {S}_{L,0}^N$ can be chosen such that $\Vert \hat{\varphi }_{L,0}^N\Vert =1$ and $\hat{\varphi }_{L,0}^N(0,\hat{x})>0$ for all $\hat{x}\in \widehat{\mathcal {V}}_L^N$. In addition, $\varphi _{L,0}^N:=(\mathfrak {F}_L^N)^*\hat{\varphi }_{L,0}^N$ is the unique ground state of $H_L^N$.

Proof

Firstly, note that if we choose the constant $C>2N\ge \Vert W_L^N\Vert =\Vert \hat{W}^N_{L,0}\Vert$, we then get that the matrix representations of both operators and have only non–negative entries. Moreover, note that since $A_1$ and $A_2$ are irreducible, we can choose $D\ge \dim (\mathbb H_L^N)$ large enough, such that the matrix entries of $A_1^D$ and $A_2^D$ will all be strictly positive. Hence, by the Perron–Frobenius theorem, the largest eigenvalue of each of these operators $A_1^D$ and $A_2^D$ is positive, non-degenerate and the corresponding eigenfunctions can be chosen to be strictly positive. Let $\varphi _{L,0}^N$ and $\hat{\varphi }_{L,0}^N$ denote the eigenfunctions for $A_1^D$ and $A_2^D$, respectively, that satisfy these properties. Clearly, $\varphi _{L,0}^N$ and $\hat{\varphi }_{L,0}^N$ will then be the eigenfunctions of $H_L^N$ and $\hat{H}_{L,0}^N$ corresponding to the respective minima $E_0$ and $\hat{E}_0$ of the spectra. Now, since $H_L^N$ and $\hat{H}_L^N$ are unitarily equivalent via the Fourier transform $\mathfrak {F}_L^N$, the function $(\mathfrak {F}_L^N)^*\hat{\varphi }_{L,0}^N$ is also an eigenfunction of $H_L^N$ and thus $\hat{E}_0\in \sigma (H_L^N)$. However, from the explicit form of $(\mathfrak {F}_L^N)^*$ as given in (3.13), one sees that since $\hat{\varphi }_{L,0}^N\in \mathbb S_{L,0}^N$ we get that $(\mathfrak {F}_L^N)^*\hat{\varphi }_{L,0}^N$ is a strictly positive eigenfunction of $H_L^N$. Thus, we conclude that $(\mathfrak {F}_L^N)^*\hat{\varphi }_{L,0}^N=\varphi _{L,0}^N$ and consequently $E_0=\hat{E}_0$. $\square$

1.2 A.2 An Auxiliary Result

The following lemma is an adaptation of a similar result in [1, Thm. 6.1].

Lemma A.3

Let $N\in \mathbb {N}$ and

$$\begin{aligned} \mathcal X^N:=\{\chi =(\chi _1,\hdots ,\chi _N)\subseteq \mathbb {N}_0^N:\;\chi _1\le \hdots \le \chi _N\}. \end{aligned}$$

(A.3)

Then, for all $\mu \ge \ln 2$ we have

$$\begin{aligned} \sum _{\chi \in \mathcal X^N\setminus \{0\}}\text {e}^{-\mu |\chi |_1}\le 30\text {e}^{-\mu }, \end{aligned}$$

(A.4)

where $|\cdot |_1$-denotes the $\ell ^1$-norm of $\mathbb {Z}^N$.

Proof

Let $\Psi :\;\mathbb {N}^{N}_0\rightarrow \mathcal X^N$ with

$$\begin{aligned} x\mapsto \Psi (x):=(\Psi _1(x),\hdots ,\Psi _N(x))\,, \end{aligned}$$

(A.5)

where for each $j\in \{1,2,\dots ,N\}$, we have defined $\psi _j(x):=\sum _{i=1}^jx_i$. Note that $\Psi$ is a bijection.

For any $x\in \mathbb {N}^{N}_0$, we therefore get

$$\begin{aligned} \sum _{j=1}^N\Psi _j(x)=\sum _{k=1}^N(N-k+1)x_k \end{aligned}$$

(A.6)

and $\Psi$ is a bijection, we have

$$\begin{aligned} \sum _{\chi \in \mathcal X^N}\text {e}^{-\mu |\chi |_1}= \sum _{x\in \mathbb {N}^N_0}\text {e}^{-\mu |\Psi (x)|_1}=\sum _{x\in \mathbb {N}^N_0}\prod _{k=1}^N\text {e}^{-\mu (N-k+1)x_k}\,, \end{aligned}$$

(A.7)

which yields

$$\begin{aligned} \sum _{\chi \in \mathcal X^N}\text {e}^{-\mu |\chi |_1}=\prod _{k=1}^N\sum _{y=0}^\infty \text {e}^{-\mu y(N-k+1) }=\prod _{k=1}^N\frac{1}{1-\text {e}^{-\mu (N-k+1)}}. \end{aligned}$$

(A.8)

This gives us the following estimate, which is uniform in N:

$$\begin{aligned} \sum _{\chi \in \mathcal X^N}\text {e}^{-\mu |\chi |_1}\le \prod _{m=1}^\infty \frac{1}{1-\text {e}^{-\mu m}}\le \exp \Big (\frac{2\text {e}^{-\mu }}{1-\text {e}^{-\mu }}\Big )\,, \end{aligned}$$

(A.9)

where we have used that $\ln (1-\lambda )^{-1}\le 2\lambda$, whenever $\lambda \in (0,1/2)$. Hence,

$$\begin{aligned} \sum _{\chi \in \mathcal X^N\setminus \{0\}}\text {e}^{-\mu |\chi |_1}\le \exp \Big (\frac{2\text {e}^{-\mu }}{1-\text {e}^{-\mu }}\Big )-1\le 4\text {e}^2\text {e}^{-\mu }, \end{aligned}$$

(A.10)

since $\text {e}^x-1\le x\text {e}^x$ for all $x\ge 0$ and $\text {e}^{-\mu }\le 2^{-1}$ for $\mu \ge \ln 2$. $\square$