The absence of the selfaveraging property of the entanglement entropy of disordered free fermions in one dimension

TL;DR

This paper demonstrates that in one-dimensional disordered free fermion systems, the entanglement entropy does not self-average, meaning its distribution remains broad even for large system sizes, unlike higher-dimensional cases.

Contribution

It proves the absence of the self-averaging property of entanglement entropy in 1D disordered free fermions, contrasting with higher-dimensional systems where self-averaging occurs.

Findings

Variance of entanglement entropy remains bounded away from zero as system size increases.

Entanglement entropy's distribution is non-trivial and not characterized by its mean in 1D.

Contrasts with higher dimensions where variance vanishes, indicating self-averaging.

Abstract

We consider the macroscopic system of free lattice fermions in one dimensions assuming that the one-body Hamiltonian of the system is the one dimensional discrete Schr\"odinger operator with independent identically distributed random potential. We show that the variance of the entanglement entropy of the segment $[-M,M]$ of the system is bounded away from zero as $% M\rightarrow \infty $. This manifests the absence of the selfaveraging property of the entanglement entropy in our model, meaning that in the one-dimensional case the complete description of the entanglement entropy is provided by its whole probability distribution. This also may be contrasted the case of dimension two or more, where the variance of the entanglement entropy per unit surface area vanishes as $M\rightarrow \infty $ \cite{El-Co:17}, thereby guaranteing the representativity of its mean for large $M$ in the multidimensional case.

Equations

$$H=-{\Delta }_d+V,$$

$$({\Delta }_{d}u)(x)=-\sum _{|x-y|=1}u(y)+2du(x),x\in {\mathbb{Z}}^d$$

$$V=\{V(x){\}}_{x\in {\mathbb{Z}}^d}$$

$$P=\{P(x,y){\}}_{x,y\in {\mathbb{Z}}^d}=\{{\mathcal{E}}_H(x,y;(-\infty ,E)){\}}_{x,y\in {\mathbb{Z}}^d}$$

$$\mathbf{E}\{|P(x,y)|\}\le C{e}^{-\gamma |x-y|},x,y\in {\mathbb{Z}}^d$$

$$\Lambda =[-M,M{]}^d\subset {\mathbb{Z}}^d$$

$$S_{\Lambda }=\mathrm{Tr}h(P_{\Lambda }),$$

$$h(t)=-t\log t-(1-t)\log (1-t),t\in [0,1]$$

$$P_{\Lambda }=\{P(x,y){\}}_{x,y\in \Lambda }$$

$$\underset{M\to \infty }{\lim }\mathbf{E}\{L^{(d-1)}{S}_{\Lambda }\}=2d\mathbf{E}\{\mathrm{Tr}h(P_{{\mathbb{Z}}_{+}^d})\},L=2M+1,$$

$${\mathbb{Z}}_{+}^d={\mathbb{Z}}_{+}\times {\mathbb{Z}}^{d-1},{\mathbb{Z}}_{+}=[0,1,..).$$

$$\mathbf{Var}\{L^{(d-1)}{S}_{\Lambda }\}:=$$

$$\le C_d{L}^{-a_d},d\ge 2,$$

$$S_{[-M,M]}=S_{+}(T^M\omega )+S_{-}(T^{-M}\omega )+o(1),M\to \infty ,$$

$$S_{\pm }(\omega )=\mathrm{Tr}h(P_{{\mathbb{Z}}_{\mp }}(\omega )),\mathbf{E}\{S_{\pm }\}>0.$$

$${\int }_0^{\infty }{v}^{\kappa }f(v)dv<\infty ;$$

$$F(t)=J(t)-1,J(t)={\int }_0^{\infty }\frac{f^2(v-t)}{f(v)}dv$$

$$\mathbf{Var}\{S_{[-M,M]}\}:=\mathbf{E}\{S_{[-M,M]}^2\}-{\mathbf{E}}^2\{S_{[-M,M]}\}\ge A>0,$$

$$A={\mathbf{E}}^2\{S_{-}\}/F(t_0)$$

$$J(t)$$

$$\mathbf{E}\{S_{[-M,M]}\}$$

$$\mathbf{E}\{S_{[-M,M]}^2\}$$

$$\mathbf{E}\{S_{+}(T^{2M}\omega )S_{-}(\omega )\}$$

$$\mathbf{Var}\{S_{[-M,M]}\}=2\mathbf{Var}\{S_{-}\}+o(1),M\to \infty .$$

$$\mathbf{Var}\{\phi (\xi ,\eta )\}\ge \mathbf{Var}\{\mathbf{E}\{\phi (\xi ,\eta )|\xi \}\}$$

$$\mathbf{Var}\{S_{-}\}\ge \mathbf{Var}\{\mathbf{E}\{S_{-}|V(0)\}\}.$$

$$\mathbf{Var}\{S_{-}\}\ge (\mathbf{E}\{S_{-}\}-\mathbf{E}\{S_{-}^t\}{)}^2/F(t),$$

$$H^t=H{|}_{V(0)\to V(0)+t}$$

$$\underset{t\to \infty }{\lim }\mathbf{E}\{S_{-}^t\}=0.$$

$$\mathbf{E}\{S_{-}\}-\mathbf{E}\{S_{-}^t\})=(1-\epsilon (t))\mathbf{E}\{S_{-}\}>0,t\to \infty$$

$$P={\mathcal{E}}_H((0,E)),P^t={\mathcal{E}}_{H^t}((0,E)),E>0.$$

Full text

PACS numbers 03.67.Mn, 03.67, 05.30.Fk, 72.15.Rn

1 Problem and Results

This note is an addition to the paper [6] by A. Elgart, M. Shcherbina and the present author in which it is proved the following. Consider the macroscopic system of free disordered fermions living on the $d$-dimensional lattice $\mathbb{Z}^{d}$ and having as the one-body Hamiltonian the discrete Schrödinger operator

[TABLE]

where

[TABLE]

is the $d$-dimensional discrete Laplacian and

[TABLE]

is the random ergodic potential. Assume that the Fermi energy $E$ of the system lies in the exponentially localized part of spectrum of $H$. This means that the Fermi projection

[TABLE]

of $H$, i.e., its spectral projection measure $\mathcal{E}_{H}((-\infty,E))$ corresponding to the spectral interval  $(-\infty,E)$, admits the bound

[TABLE]

for some $C<\infty$ and $\gamma>0$. Here and below the symbol $\mathbf{E}\{...\}$ denotes the expectation with respect to the random potential.

We refer the reader to [6] for the discussion of the cases where the bound (1.5) holds and guaranties the pure point spectrum of (1.1) with exponentially decaying eigenfunctions (exponential localization). It is important for us in this paper that in the one-dimensional case $d=1$ the bound holds on the whole spectrum of $H$ if the potential (1.3) is the collection of independent identically distributed (i.i.d.) random variables.

Given the lattice cube (the block)

[TABLE]

of the system, define the entanglement entropy of free fermions as

[TABLE]

where

[TABLE]

and

[TABLE]

is the restriction of the Fermi projection 1.4) to the block $\Lambda$.

It is proved in [6] that for any ergodic potential satisfying condition (1.5) of the exponential localization the entanglement entropy satisfies the area law in the mean, i.e., there exists the limit

[TABLE]

where $P_{\mathbb{Z}_{+}^{d}}=\{P(x,y)\}_{x,y\in\mathbb{Z}_{+}^{d}}$ is the restriction of the Fermi projection (1.4) to the $d$-dimensional lattice half-space

[TABLE]

See [2, 5, 8, 9] for various results on the validity of the area law and its violation in translation invariant (non-random) systems.

It was also shown in [6] that if the random potential is a collection of i.i.d. random variables and (1.5) holds, then there exist some $C_{d}<\infty$ and $a_{d}>0$ such that

[TABLE]

i.e., that the fluctuations of the entanglement entropy per unit surface area vanish as $L\rightarrow\infty$.

The relations (1.9) and (1.10) imply that in dimension two and higher the entanglement entropy per unit surface area possesses the selfaveraging property (see, e.g., [3, 4, 10, 12] for discussion and use of the property in the condensed matter theory, spectral theory and the quantum information theory where it is known as the typicality).

On the other hand, it follows from the numerical results of [13] that for $d=1$ the fluctuations of the entanglement entropy of the lattice segment $\Lambda=[-M,M]$ do not vanish as $M\rightarrow\dot{\infty}$ and according to [6] we have for every typical realization (with probability 1)

[TABLE]

where

[TABLE]

Here and below $\omega=\{V(x)\}_{x\in\mathbb{Z}}$ denotes a realization of random ergodic potential and $T$ is the shift operator acting in the space of realizations of potential as $T\omega=\{V(x+1)\}_{x\in\mathbb{Z}}$. This suggests that for $d=1$ and for i.i.d. potential the entanglement entropy of  disordered free fermions is not a selfaveraging quantity.

In this note we confirm the suggestion by establishing an $M$-independent and strictly positive lower bound on the variance of the entanglement entropy for $d=1$. Unfortunately, the class of random i.i.d. potentials, for which this results is established, is somewhat limited (see, e.g. Remark 1.2). However, since the absence of selfaveraging property is not completely common and sufficiently studied in the theory of disordered systems, we believe that our result is of certain interest.

Result 1.1

Consider the macroscopic system of free lattice fermions in one dimension whose one-body Hamiltonian is the discrete Schrödinger operator (1.1) with i.i.d. potential (1.3). Assume that the common probability distribution of $V(x),\;x\in\mathbb{Z}$ has a bounded density $f$ such that

(i) $\mathrm{supp\ }f=[0,\infty)$ and for some $\kappa>0$

[TABLE]

(ii) the quantity

[TABLE]

is finite for all sufficiently large $t>0.$

Then there exist a sufficiently large $t_{0}>0$ and $M_{0}>0$ such that we have for the entanglement entropy (1.6) uniformly in $M>M_{0}$

[TABLE]

[TABLE]

and $S_{-}$ is defined in (1.12).

Remark 1.2

(i) It is easy to show (see (2) below) that $F\geq 0$. Moreover, $F$ is unbounded as $t\rightarrow\infty$. Indeed, we have from (1.14) and the Jensen inequality

[TABLE]

Thus, $A$ in (1.15) – (1.16) can be rather small. Note that above lower bound for $I$ is exact for the density $f(v)=a^{-1}e^{-av}$

(ii) Condition (i) of the result can be replaced by that for the support of $f$ to be bounded from below. However, a compact support is not allowed, since in this case $J(t)$ in (1.14) is not well defined for large $t$, since the supports of the numerator and denominator in $J(t)$ do not intersect. Moreover, even if the support of $f$ is the positive semiaxis, $f$ should not have zeros of the order 1 and higher.

Proof of result. It follows from (1.9) for $d=1$ (or from (1.11)) that

[TABLE]

and in obtaining the second equality we used the shift and the reflection invariance of the probability distribution of the infinite sequence $V=\{V(x)\}_{x\in\mathbb{Z}}$ of i.i.d. random variables, see [6].

Likewise, repeating almost literally the proof of (1.9) for $d=1$ in [6], we obtain

[TABLE]

Since the infinite sequence $V=\{V(x)\}_{x\in\mathbb{Z}}$ of i.i.d. random variables is a mixing stationary process (see e.g. [14], Section V.2 ), we have

[TABLE]

Combining (1) – (1), we obtain

[TABLE]

Thus, it suffices to show that $\mathbf{Var}\{S_{-}\}$ is strictly positive.

To this end we start with the inequality

[TABLE]

involving the conditional expectation and valid for any random (multi-component in general) variables $\xi$ and $\eta$ and a function $\varphi$. Choosing here $\xi=V(0)$, $\eta=\{V(x)\}_{x\neq 0}$ and $\varphi=S_{-}$, we obtain

[TABLE]

Next, we will use Lemma 2.1 with $\xi=V(0)$ and $\varphi(\xi)=$ $\mathbf{E}\{S_{-}|V(0)=\xi\}$ yielding

[TABLE]

where $S_{-}^{t}$ is the entanglement entropy (1.6) – (1.8) corresponding to the Schrödinger operator $H^{t}$ (see (1.1) – (1.3)) in which the potential $V(0)$ at the origin is replaced by $V(0)+t$

[TABLE]

We will prove below that

[TABLE]

Thus, there exists $\varepsilon(t)\to 0,\;t\to\infty$ (see, e.g (1.26)) such that we have in view of (1.12)

[TABLE]

and then (1.20) – (1.21) yield (1.15) – (1.16) upon choosing sufficiently large $t_{0}$ and $M_{0}$ and assuming that $M\geq M_{0}$.

Let us prove (1.23). Since the potential is a collection of i.i.d. random variables satisfying condition (1.13), the spectrum of $H$ is the positive semiaxis (see Corollary 4.23 in [12]). The same is true for the spectrum of $H^{t}$ since $t>0$. Hence, we have in view of (1.4)

[TABLE]

We have from the proof of Lemma 4.5 of [6] for some $t$-independent $C_{0}<\infty$ and any $\alpha\in(0,1)$:

[TABLE]

where we took into account the inequality $(a+b)^{\alpha}\leq a^{\alpha}+b^{\alpha},\;\alpha\in(0,1)$ and that $P^{t}$ is an orthogonal projection, hence

[TABLE]

Now, (1.25) and Lemma 2.3 below yield

[TABLE]

where $C_{\alpha,s}$ does not depend on $t$. This implies (1.23) $\blacksquare$.

2 Auxiliary results

Lemma 2.1

Let $\xi$ be a non-negative random variable, $\varphi:\mathbb{R}_{+}\rightarrow\mathbb{R}$ be a function and $\mathbf{E}\{|\varphi(\xi)|\}<\infty$. Assume that the probability law of $\xi$ has a bounded density $f$ such that

(i) $\mathrm{supp}\,f=[0,\infty);$

(ii) the quantity $F(t)$ in (1.14) is well defined for some $t>0$. Then we have

[TABLE]

Proof. Consider the random variables $\varphi(\xi)$ and $\psi(\xi)=(f(\xi-t)-f(\xi))/f(\xi)$. It follows from the normalization condition

[TABLE]

that

[TABLE]

Thus,

[TABLE]

On the other hand, we have from the Schwarz inequality for the expectations:

[TABLE]

Combining these two relations and using the definition of $\psi$, according to which

[TABLE]

we get (2.1).

Remark 2.2

The inequality is, in fact, the Hammersley-Chapman-Robbins inequality (see [7], Section 2.5.1) and is a version of the Cramér-Rao inequality of statistics.

Lemma 2.3

Let $H$ be the one-dimensional Schrödinger operator (see (1.1) – (1.3) for $d=1$) with an i.i.d. non-negative random potential $V=\{V(x)\}_{x\in\mathbb{Z}}$ whose common probability law has a density $f$ satisfying (1.13). Denote $P^{t}=\mathcal{E}_{H^{t}}((0,E))=\{P^{t}(x,y)\}_{x,y\in\mathbb{Z}}$ the Fermi projection of $H^{t}$ of (1.22) corresponding to the spectral interval $(0,E)$. Then there exist $s\in(0,1),\;C<\infty$ and $\gamma>0$ that do not depend on $t$ and are such that we have for $E<t$:

(i)  $\mathbf{E}\{|P^{t}(0,0)|\}\leq C(t-E)^{-s}$;

(ii) $\mathbf{E}\{|P^{t}(x,y)|\}\leq C(t-E)^{-s}e^{-\gamma(x-y)},$ for all $x\geq 1$ and $y\leq-1$.

Proof. Let

[TABLE]

be the resolvent of $H^{t}$. It is shown below that the bounds

[TABLE]

and

[TABLE]

are valid for some $s\in(0,1)$, all $\lambda\in(0,E),\;E<t,\;\eta\neq 0$ and $x\geq 1$, $y\leq-1$ with $C<\infty$ and $\gamma>0$ which are independent of $t,\;\lambda$ and $\eta\neq 0$.

It follows from a slightly modified version of proof of Theorem 13.6 of [1], based on the contour integral representation of $P^{t}$ via $G^{t}$ and the Combes-Thomas theorem, that the assertion of the lemma can be deduced from (2.4) – (2.5).

Hence, it suffices to prove (2.4) and (2.5). To this end we introduce the restrictions $H_{-}$ and $H_{+}$ of $H^{t}$ (or $H$) to the integer-valued intervals $(-\infty,-1]$ and $[1,\infty)$ and the rank one operator $\widehat{V}_{0}^{t}$ of multiplication by $V(0)+2+t$. Let

[TABLE]

be the double infinite block matrix consisting of the $(-\infty,-1]\times(-\infty,-1]$ semi-infinite block $H_{-}$, $1\times 1$ ”central” block $\widehat{V}_{0}^{t}$ and the $[1,\infty)\times[1,\infty)$ semi-infinite block $H_{+}$. In other words, $\widehat{H}^{t}$ is obtained from $H^{t}$ by replacing the four entries (equal $-1)$ with indices $(0,\pm 1)$ and $(\pm 1,0)$ by zero. Denote

[TABLE]

the corresponding resolvents, where we omit the complex spectral parameters $z,\;\Im z\neq 0$ in the r.h.s. We have in view of (2.6):

[TABLE]

By using the resolvent identity $G^{t}=\widehat{G}^{t}-\widehat{G}^{t}(H^{t}-\widehat{H}^{t})G^{t}$, we obtain for all $x,y\in\mathbb{Z}$

[TABLE]

This and (2.8) imply

[TABLE]

and

[TABLE]

Likewise, we have from the resolvent identity $G^{t}=\widehat{G}^{t}-G^{t}(H^{t}-\widehat{H}^{t})\widehat{G}^{t}$:

[TABLE]

and

[TABLE]

We have then from (2.9) and the Schwarz inequality for any $s>0$ and all $y\leq 0$

[TABLE]

where $C_{s}$ depends only on $s>0$ and

[TABLE]

Choosing here $\zeta=z-2-t,\;z=\lambda+i\eta$ and using (2.2), we have for $\lambda\leq E<t$

[TABLE]

and then (2) implies for any $s>0$ and all $y\leq 0$

[TABLE]

We will use now Theorem 8.7 of [1], according to which if $A_{0}$ is a selfadjointe operator in $l^{2}(\mathbb{Z}^{d})$, $U=\{U(x)\}_{x\in\mathbb{Z}^{d}}$ is a collection of independent random variables whose probability densities $f_{x},\,x\in\mathbb{Z}$ are bounded uniformly in $x$, i.e., $\sup_{x\in\mathbb{Z}}\sup_{v\in\mathbb{R}}<\infty$ and if $\mathcal{G}(z)=(A-z)^{-1}=\{\mathcal{G}(x,y;z)\}_{x,y\in\mathbb{Z}^{d}}$ is the resolvent of $A=A_{0}+U$, then for any $s\in(0,1)$ there exists $C_{s}^{\prime}<\infty$ such that the bound

[TABLE]

holds uniformly in $z\in\mathbb{C\setminus R}$ for all $x,y\in\mathbb{Z}^{d}$.

Choosing here $A=H^{t}$ with $H^{t}$ of (1.22) and noting that for the potential $V^{t}$ the conditions of the theorem are satisfied (all $V^{t}(x)=V(x),\,x\neq 0$ are i.i.d random variables with a bounded common probability density and $V^{t}(0)=V(0)+t$ has the density $f(v-t)$ also bounded), we obtain for any $s\in(0,1),\;\lambda<E,\;\eta\neq 0$ and all $x,y\in\mathbb{Z}^{d}$

[TABLE]

where $C_{s}^{\prime\prime}$ does not depend on $t,E$ and $\eta$.

Plugging this bound with $x=\pm 1$ into (2.15), we get for any $s_{1}\in(0,1/2),\;\lambda<E<t$ and all $y\leq 0$

[TABLE]

where $B_{s_{1}}$ does not depend on $t,E$ and $\eta$.

Analogous argument yields for $s_{1}\in(0,1/2),\;\lambda<E<t$ and all $x\geq 0$

[TABLE]

We obtain (2.4), hence assertion (i) of the lemma, from (2.17) with $y=0$ (or from (2.18) with $x=0$).

To prove (2.5), hence assertion (ii) of the lemma, we combine (2.10) and (2.12) to write for any $s>0$ and all $x\geq 1$ and $y\leq 1$:

[TABLE]

and then, by Hölder inequality for expectations,

[TABLE]

Using here (2.17) and (2.18), we get for $s_{1}\in(0,1/2)$, and all $x\geq 1$ and$\;y\leq-1$

[TABLE]

To bound the two last factors on the right, we will use a result from [11] according to which if $H_{+}$ is the discrete one dimensional Schrödinger operator in $l^{2}([1,\infty))$ with i.i.d. potential whose common probability law is such that $\mathbf{E}\{|V(0)|^{\kappa}\}<\infty$ for some $\kappa>0$, then for any spectral interval $I$ there exist $C(I)<\infty,\,\ \gamma(I)>0$ and $s_{2}\leq\kappa/2$ such that

[TABLE]

The same is valid for the Hamiltonian $H_{-}$ acting in $l^{2}((-\infty,-1])$, see (2.6):

[TABLE]

The bounds are the basic ingredient of the proof of (1.5) for the one dimensional case [11].

Using these bounds in the r.h.s. of (2.20), we obtain assertion (ii) of the lemma with $s=\min\{s_{1},s_{2}\}$, where $s_{1}$ and $s_{2}$ are defined in (2.17) – (2.18) and (2.21) – (2.22).

Remark 2.4

By using the standard facts of spectral theory, it is easy to prove the weaker version of Lemma 2.3

[TABLE]

which, however, does not allow us to justifies the limiting transition $t\rightarrow\infty$ in the second term in the r.h.s. of (1.25).

Indeed, according to the spectral theorem

[TABLE]

The formula, the continuity properties of the Stieltjes transform of a bounded signed measure and (1.4) imply that (2.23) follows from the analogous limiting relation for the resolvent with $\Im z\neq 0$:

[TABLE]

Viewing the part $(t-1)V(0)$ of the $(0,0)$th entry of $H^{t}$ as a rank one perturbation of $H=H^{t}|_{t=1}$, we obtain

[TABLE]

hence

[TABLE]

Recall now the Weyl formula for the resolvent of the discrete Schrödinger operator (see, e.g. [15], Section 1.2):

[TABLE]

where $\psi^{\pm}$ are the solutions of the corresponding discrete Schrödinger equation which belong to $l^{2}(\mathbb{Z}_{\pm})$ for $\Im z\neq 0$ and satisfy the condition $\psi^{\pm}(0)=1$. Combining the formula with (2.25), we obtain (2.24), hence (2.23).

Acknowledgment The work is supported in part by the grant 4/16-M of the National Academy of Sciences of Ukraine.