Entropy Decay for Davies Semigroups of a One Dimensional Quantum Lattice

Abstract

Given a finite-range, translation-invariant commuting system Hamiltonian on a spin chain, we show that the Davies semigroup describing the reduced dynamics resulting from the joint Hamiltonian evolution of a spin chain weakly coupled to a large heat bath thermalizes rapidly at any temperature. More precisely, we prove that the relative entropy between any evolved state and the equilibrium Gibbs state contracts exponentially fast with an exponent that scales logarithmically with the length of the chain. Our theorem extends a seminal result of Holley and Stroock (Commun Math Phys 123(1):85–93, 1989) to the quantum setting as well as provides an exponential improvement over the non-closure of the gap proved by Brandao and Kastoryano (Commun Math Phys 344(3):915–957, 2016). This has wide-ranging applications to the study of many-body in and out-of-equilibrium quantum systems. Our proof relies upon a recently derived strong decay of correlations for Gibbs states of one dimensional, translation-invariant local Hamiltonians, and tools from the theory of operator spaces.

Appendices

Appendix A: Proof of Lemma 5.4

We now argue the other direction in Lemma 5.4. Namely,

$$\begin{aligned} \left\| x \right\| _{L_1^\infty (\mathcal {N}\subset \mathcal {M})}\ge \inf _{x=yz}\left\| yy^\dagger \right\| _{L_1^\infty (\mathcal {N}\subset \mathcal {M})}^{1/2}\left\| z^\dagger z \right\| _{L_1^\infty (\mathcal {N}\subset \mathcal {M})}^{1/2}. \end{aligned}$$

(106)

Let us denote

$$\begin{aligned} \left\| x \right\| _{h}:=\inf _{x=yz}\left\| yy^\dagger \right\| _{L_1^\infty }^{1/2}\left\| z^\dagger z \right\| _{L_1^\infty }^{1/2} \end{aligned}$$

where the infimum takes over all factorization $x=yz$. We first show $\left\| \cdot \right\| _{h}$ is a norm. To verify the triangle inequality, it suffices to show that for any $x_1=y_1z_1$, $x_2=y_2z_2$, and $\delta >0$, there exist $y_0,z_0$ such that $x_1+x_2=y_0z_0$ and

$$\begin{aligned} \left\| y_0y^\dagger _0 \right\| _{L_1^\infty }^{1/2}\left\| z^\dagger _0z_0 \right\| _{L_1^\infty }^{1/2}\le \left\| y_1y_1^\dagger \right\| _{L_1^\infty }^{1/2}\left\| z^\dagger _1z_1 \right\| _{L_1^\infty }^{1/2}+\left\| y_2y_2^\dagger \right\| _{L_1^\infty }^{1/2}\left\| z^\dagger _2z_2 \right\| _{L_1^\infty }^{1/2}+\delta . \end{aligned}$$

By rescaling $\mathcal {X}_1=t\mathcal {Y}_1\cdot t^{-1}z_1$, we can assume

$$\begin{aligned}&\left\| y_1 y_1^\dagger \right\| _{L_1^\infty }^{1/2}\left\| z^\dagger _1z_1 \right\| _{L_1^\infty }^{1/2}+ \left\| y_2y_2^\dagger \right\| _{L_1^\infty }^{1/2}\left\| z^\dagger _2z_2 \right\| _{L_1^\infty }^{1/2}\nonumber \\&\quad =\Big (\left\| y_1 y_1^\dagger \right\| _{L_1^\infty }+\left\| y_2y_2^\dagger \right\| _{L_1^\infty }\Big )^{1/2} \Big (\left\| z^\dagger _1z_1 \right\| _{L_1^\infty }+\left\| z^\dagger _2z_2 \right\| _{L_1^\infty }\Big )^{1/2} \end{aligned}$$

(107)

Take $\delta >0$ and

$$\begin{aligned}&y=(y_1 y_1^\dagger +y_2y_2^\dagger +\delta \mathbbm {1})^{1/2}, z=(z_1^\dagger z_1+z_2^\dagger z_2+\delta \mathbbm {1})^{1/2},\\&x=y^{-1} (y_1z_1+y_2z_2)z^{-1}=y^{-1}(x_1+x_2)z^{-1}. \end{aligned}$$

We have $\left\| \left[ \begin{array}{cc}y^{-1}y_1&{} y^{-1}y_2\\ 0&{}0 \end{array}\right] \right\| _{}\le 1$ and $\left\| \left[ \begin{array}{cc}z_1z^{-1}&{} 0 \\ z_2z^{-1} &{} 0\end{array}\right] \right\| {}\le 1$. Thus

$$\begin{aligned} \left\| x \right\| =\left\| \left[ \begin{array}{cc}x&{}0\\ 0&{}0 \end{array}\right] \right\| {}\ \le \ \left\| \left[ \begin{array}{cc}y^{-1}y_1&{} y^{-1}y_2\\ 0&{}0 \end{array}\right] \right\| {}\left\| \left[ \begin{array}{cc}z_1z^{-1}&{} 0\\ z_2z^{-1}&{} 0 \end{array}\right] \right\| {}\ \le \ 1. \end{aligned}$$

Then we have $x=x_1+x_2=yXz$ and

$$\begin{aligned}&\left\| yy^\dagger \right\| _{L_1^\infty }\left\| z^\dagger x^\dagger xz \right\| _{L_1^\infty }\\&\quad \le \left\| yy^\dagger \right\| _{L_1^\infty }\left\| z^\dagger z \right\| _{L_1^\infty }\\&\quad \le \left\| y_1 y_1^\dagger +y_2y_2^\dagger +\delta \right\| _{L_1^\infty }\left\| z_1^\dagger z_1+z_2^\dagger z_2+\delta \right\| _{L_1^\infty }\\&\quad \le \Big (\left\| y_1 y_1^\dagger \right\| _{L_1^\infty }+\left\| y_2y_2^\dagger \right\| _{L_1^\infty }+\delta \Big )\Big (\left\| z_1^\dagger z_1 \right\| _{L_1^\infty }+\left\| z_2^\dagger z_2 \right\| _{L_1^\infty }+\delta \Big )\ \end{aligned}$$

Since $\delta >0$ is arbitrary, we have by (107)

$$\begin{aligned}&\left\| yy^\dagger \right\| _{L_1^\infty }^{1/2}\left\| z^\dagger x^\dagger x z \right\| _{L_1^\infty }^{1/2}\\&\quad \le \left\| y_1 y_1^\dagger \right\| _{L_1^\infty }^{1/2}\left\| y_2y_2^\dagger \right\| _{L_1^\infty }^{1/2}+\left\| z_1^\dagger z_1 \right\| _{L_1^\infty }^{1/2}\left\| z_2^\dagger z_2 \right\| _{L_1^\infty }^{1/2} \end{aligned}$$

This proves the triangular inequality and also

$$\begin{aligned} \left\| x \right\| _{h}=\inf _{x=\sum _{i}y_iz_i}\left\| \sum _{i}y_iy_i^\dagger \right\| _{L_1^\infty }^{1/2}\left\| \sum _{i}z^\dagger _iz_i \right\| _{L_1^\infty }^{1/2}, \end{aligned}$$

where the supremum is over all finite families $\{y_i\}$ and $\{z_i\}$ such that $\sum _{i=1}^ky_iz_i=x$. We now use a standard Grothendieck–Pietsch factorization to show $\left\| \cdot \right\| _{h}=\left\| \cdot \right\| _{L_1^\infty }$. Suppose $\left\| x \right\| _{h}=1$. By Hahn–Banach Theorem, there exists a linear functional $\phi :\mathcal {M}\rightarrow \mathbb {C}$ such that $\phi (x)=\left\| x \right\| _{h}=1$ and for any finite families $\{y_i\}$ and $\{z_i\}$,

$$\begin{aligned} |\sum _{i=1}^k\phi (y_iz_i)|\le \sup _{ \left\| a \right\| _{L_\infty ^1}=1}\langle a,\sum _{i}y_iy_i^\dagger \rangle _{\sigma _{{\text {tr}}}}^{1/2}\sup _{\left\| b \right\| _{L_\infty ^1}=1}\langle b, \sum _{i}z_i^\dagger z_i\rangle _{\sigma _{{\text {tr}}}}^{1/2}. \end{aligned}$$

Here we use the duality $L_1^\infty (\mathcal {N}\subset \mathcal {M})^*=L_\infty ^1(\mathcal {N}\subset \mathcal {M})$ and for positive $Y\ge 0$,

$$\begin{aligned} \left\| Y \right\| _{L_1^\infty }=\sup _{\left\| a \right\| _{L_\infty ^1}=1}|\langle Y, a\rangle _{\sigma _{{\text {tr}}}}|=\sup _{a\ge 0, \left\| a \right\| _{L_\infty ^1}=1}\langle Y, a\rangle _{\sigma _{{\text {tr}}}}. \end{aligned}$$

By modifying the phase factor and by the arithmetic–geometric mean inequality, we have

$$\begin{aligned} \sup _{a\ge 0, \left\| a \right\| _{L_\infty ^1}=1}\langle a,\sum _{i}y_iy_i^\dagger \rangle _{\sigma _{{\text {tr}}}}+\sup _{b\ge 0, \left\| b \right\| _{L_\infty ^1}=1}\langle b, \sum _{i}z_i^\dagger z_i\rangle _{\sigma ,{{\text {tr}}}}-2\sum _{i=1}^k \text {Re }\phi (y_iz_i)\ge 0. \end{aligned}$$

(108)

Denote $B_{+}$ as the positive unit ball of $L_\infty ^1(\mathcal {N}\subset \mathcal {M})$ and $C(B_{+}\times B_{+})$ as the real continuous function space. For each pair of finite families $\textbf{y}=\{y_i\}$ and $\textbf{z}=\{z_i\}$, we define the function

$$\begin{aligned} f_{\textbf{y}, \textbf{z}}:B_+\times B_+\rightarrow \mathbb {R}, f_{\textbf{y}, \textbf{z}}(a,b)=\langle a,\sum _{i}y_iy_i^\dagger \rangle _{\sigma _{{\text {tr}}}}+\langle b, \sum _{i}z_i^\dagger z_i\rangle _{\sigma ,{{\text {tr}}}}-\sum _{i=1}^k \text {Re} \phi (y_iz_i). \end{aligned}$$

We define the following cones in $C(B_{+}\times B_{+})$

$$\begin{aligned}&C=\{f_{\textbf{y}, \textbf{z}} \ | \ \{y_i\}, \{z_i\}\subset \mathcal {M}\}\\&C_-=\{f\in C(B_{+}\times B_{+},\mathbb {R}) \ | \sup f<0 \}. \end{aligned}$$

Note that both C and $C_-$ are convex and $C_-$ is furthermore open. Moreover, $C\cap C_-=\emptyset$ because of (108). By Hahn–Banach separation theorem, there exists a linear functional $\psi : C(B_{+}\times B_{+})\rightarrow \mathbb {R}$ such that

$$\begin{aligned} \psi (f_-) \le \lambda \le \psi (f_{\textbf{y}, \textbf{z}}) \end{aligned}$$

for any $f_-\in C_-$ and $f_{\textbf{y}, \textbf{z}}\in C$. Since $C_-$ is a cone, $\lambda \ge 0$ and hence $\psi$ is a positive linear function. Up to normalization, there exists a probability measure $\mu$ on $B_+\times B_+$ such that $\psi (f)=\int _{B_+\times B_+}f(a,b)\ d\mu (a,b)$. Take

$$\begin{aligned} a_0=\int _{B_+\times B_+} a\ d\mu (a,b),\ b_0=\int _{B_+\times B_+} b\ d\mu (a,b). \end{aligned}$$

By convexity of $B_+$, we have $a_0,b_0\in B_+$ and moreover for every $y,z\in \mathcal {M}$,

$$\begin{aligned} \psi (f_{\{y\}, \{z\}})=&\int _{B_+\times B_+}f(a,b)d\mu (a,b)\\ =&\int _{B_+\times B_+}\langle a,yy^\dagger \rangle _{\sigma _{{\text {tr}}}}d\mu (a,b)+\int _{B_+\times B_+}\langle b z^\dagger z\rangle _{\sigma ,{{\text {tr}}}}d\mu (a,b)-2 \text {Re} \phi (yz)\\ =&\langle a_0,yy^\dagger \rangle _{\sigma _{{\text {tr}}}}+\langle b_0,z^\dagger z\rangle _{\sigma ,{{\text {tr}}}}- \text {Re} \phi (yz)\ge 0. \end{aligned}$$

Rescaling $y_i$ and $z_i$ again, we have

$$\begin{aligned} |\phi (yz)|\le {{\text {tr}}}(a_0\sigma _{{\text {tr}}}^{1/2}yy^\dagger \sigma _{{\text {tr}}}^{1/2})^{1/2} {{\text {tr}}}(b_0\sigma _{{\text {tr}}}^{1/2}z^\dagger z\sigma _{{\text {tr}}}^{1/2})^{1/2}=\left\| a_0^{1/2}\sigma _{{\text {tr}}}^{1/2}y \right\| _{2,{{\text {tr}}}}\left\| z\sigma _{{\text {tr}}}^{1/2}b_0^{1/2} \right\| _{2,{{\text {tr}}}}, \end{aligned}$$

where $\Vert .\Vert _{2,{{\text {tr}}}}$ denotes the Hilbert Schmidt norm. One can further find invertible $a_1,b_1\in (1+\epsilon )B_+$ such that

$$\begin{aligned} |\phi (yz)|\le \left\| a_1^{1/2}\sigma _{{\text {tr}}}^{1/2}y \right\| _{2,{{\text {tr}}}}\left\| z\sigma _{{\text {tr}}}^{1/2}b_1^{1/2} \right\| _{2,{{\text {tr}}}} \end{aligned}$$

Because of the invertibility of $a_1,b_1$, there exists a contraction u such that

$$\begin{aligned} \phi (yz)={{\text {tr}}}( ua_1^{1/2}\sigma _{{\text {tr}}}^{1/2}yz\sigma _{{\text {tr}}}^{1/2}b_1^{1/2})=\langle a_1^{1/2}u^\dag b_1^{1/2}, yz \rangle _{\sigma _{{\text {tr}}}} \end{aligned}$$

Note that by Hölder inequality,

$$\begin{aligned} \left\| a_1^{1/2}u^\dag b_1^{1/2} \right\| _{L_\infty ^1}=&\sup _{\left\| X \right\| _{2}=\left\| Y \right\| _{2}=1}\left\| Xa_1^{1/2}u^\dag b_1^{1/2}Y \right\| _{1}\\ =&\sup _{X,Y}\left\| \sigma _{{\text {tr}}}^{1/2}Xa_1^{1/2} \right\| _{2,{{\text {tr}}}} \left\| u^\dag b_1^{1/2}Y \sigma _{{\text {tr}}}^{1/2} \right\| _{2,{{\text {tr}}}}\\ =&\sup _{X,Y}{{\text {tr}}}(\sigma _{{\text {tr}}}Xa_1X^\dag )^{1/2} {{\text {tr}}}(Y^\dag b_1^{1/2}uu^\dag b_1^{1/2}Y\sigma _{{\text {tr}}})\\ \le&\sup _{X,Y}\left\| Xa_1X^\dag \right\| _{1}^{1/2} \left\| Y^\dag b_1 Y \right\| _{1}^{1/2}\le \left\| a_1 \right\| _{L_\infty ^1}^{1/2}\left\| b_1 \right\| _{L_\infty ^1}^{1/2}=1+\epsilon . \end{aligned}$$

Therefore, for any factorization $x=yz$,

$$\begin{aligned} \left\| x \right\| _{h}=\phi (x)= \langle a_1^{1/2}u^\dag b_1^{1/2}, x \rangle _{\sigma _{{\text {tr}}}}\le \left\| a_1^{1/2}u^\dag b_1^{1/2} \right\| _{L_\infty ^1}\left\| x \right\| _{L_1^\infty }\le (1+\epsilon ) \left\| x \right\| _{L_1^\infty }. \end{aligned}$$

Since $\epsilon$ is arbitrary, that concludes the proof.

Appendix B: Temperature Dependence of the Mixing Condition

Here we discuss the dependence of $\mathcal {K}_\beta$ from Eq. (42), and thus that of $\mathcal {C}_\beta$ from Lemma 3.3 on the inverse temperature $\beta$. This can be studied by carefully reviewing the estimates presented in [16], jointly with the fact that we are focusing on commuting Hamiltonians. A complete proof of the exponential dependence of $\mathcal {C}_\beta$ with $\beta$ can be found in [47], but we include here a short sketch of the proof for completeness.

We consider a local interaction on 1D, i.e., a family $\Phi = (\Phi _{X})_{X \in \mathcal {P}_0(\mathbb {Z})}$, where $\Phi _X \in \mathcal {B}(\mathcal {H}_X)$ and $\Phi _X = \Phi _X^*$ for every $X \in \mathcal {P}_0(\mathbb {Z})$. We further assume that this interaction is translation invariant, commuting, and define for every finite subset $\Lambda \subset \subset \mathbb {Z}$ the corresponding Hamiltonian by

$$\begin{aligned} H_{\Lambda }:= \sum _{X \subset \Lambda }{\Phi _{X}}. \end{aligned}$$

Given two segments $A,B \subset \Lambda$, we use the following notation for the expansional:

$$\begin{aligned} E_{A,B}(\beta ):=e^{-\beta H_{AB}}e^{\beta (H_{A}+H_B)}. \end{aligned}$$

With this notation at hand, and dropping $\beta$ from the expressions below for better readability, we can derive the dependence of the mixing condition from Eq. (44) with $\beta$, namely can show that given an interval $\Lambda \subset \mathbb {Z}$ split as $\Lambda =XYZ$, the following holds:

$$\begin{aligned} \left\| \sigma _{XZ}\sigma _X^{-1} \otimes \sigma _Z^{-1} - \mathbbm {1}_{XZ} \right\| \le \mathcal {O}(e^{\beta }). \end{aligned}$$

(109)

Indeed, by [16, Proof of Proposition 8.1], and denoting by $\sigma ^\xi := e^{-H_\xi }/Z_\xi$ with $Z_\xi :={{\text {tr}}}[e^{-H_\xi }]$ for any $\xi \subset \Lambda$, we can write:

$$\begin{aligned}&\left\| \sigma _{XZ}\sigma _X^{-1} \otimes \sigma _Z^{-1} - \mathbbm {1}_{XZ} \right\| \\&\quad \le \left\| {{\text {tr}}}_{YZ}\big (\sigma ^{YZ} E_{X,YZ}^\dagger \big )^{-1} \right\| \left\| {{\text {tr}}}_{XY}\big (\sigma ^{XY} E_{XY,Z}^\dagger \big )^{-1} \right\| \\&\qquad \Bigg ( \left\| {{\text {tr}}}_{X}\big (\sigma ^{X} E_{X,Y}^\dagger E_{XY,Z}^\dagger \big ) \right\| \underbrace{\left| \lambda _{XYZ} - 1 \right| }_{\chi _1} \\&\qquad + \underbrace{\left\| {{\text {tr}}}_{YZ}\big (\sigma ^{YZ} E_{X,YZ}^\dagger \big ) {{\text {tr}}}_{XY}\big (\sigma ^{XY} E_{XY,Z}^\dagger \big ) - {{\text {tr}}}_{X}\big (\sigma ^{X} E_{X,Y}^\dagger E_{XY,Z}^\dagger \big ) \right\| }_{\chi _2} \Bigg ), \end{aligned}$$

where

$$\begin{aligned} \lambda _{XYZ}:= \frac{Z_{XYZ}Z_Y}{Z_{XY} Z_{YZ}}, \end{aligned}$$

and

$$\begin{aligned}{} & {} {{\text {tr}}}_{YZ}\big (\sigma ^{YZ} E_{X,YZ}^\dagger \big )^{-1} \nonumber \\{} & {} = {{\text {tr}}}_{YZ} \big ( e^{-H_{YZ}}e^{H_X+H_{YZ}} e^{-H_{XYZ}} \big )^{-1} Z_{YZ} =e^{-H_X} {{\text {tr}}}_{YZ} \big ( e^{-H_{XYZ}} \big )^{-1} Z_{YZ}, \end{aligned}$$

and analogously for the rest. Considering the explicit form of the expansionals, and taking into account that the Hamiltonian is commuting, it is easy to notice that

$$\begin{aligned} \left\| E_{A,B} (\beta ) \right\| \le K = \mathcal {O}(e^\beta ), \end{aligned}$$

for any $A,B \subset \Lambda$, and the same K can be taken for $E_{A,B}^\dagger$ and $E_{A,B}^{-1}$. Moreover, following the steps of [16, Corollary 4.4] and adapting them for the commuting case, we notice that

$$\begin{aligned} \left\| {{\text {tr}}}_{BC}\big (\sigma ^{BC} E_{A,BC}^\dagger \big )^{-1} \right\| \le K, \end{aligned}$$

for $\Lambda = ABC$, and all similar terms behave in an analogous way. Thus, we can upper bound the expression above as

$$\begin{aligned} \left\| \sigma _{XZ}\sigma _X^{-1} \otimes \sigma _Z^{-1} - \mathbbm {1}_{XZ} \right\| \le K^2 ( K \chi _1 + \chi _2 ). \end{aligned}$$

For $\chi _1$, note that by [16, Step 2 of Proposition 8.1], denoting by ${\text {Tr}}_X$ the scalar trace in X (not to be confused with the operator trace denoting by ${\text {tr}}_X$ in X), we have

$$\begin{aligned} \left| \lambda _{XYZ} - 1 \right| \le K \left| {\text {Tr}}_{XY} \big (\sigma ^{XY} E_{X,Y}^{\dagger \ -1} \big ) - {\text {Tr}}_{XYZ} \big (\sigma ^{XYZ} E_{X,YZ}^{\dagger \ -1} \big ) \right| , \end{aligned}$$

and considering now a splitting of Y into $Y_1 Y_2$ with $Y_1:=\partial X$, as well as noticing $E_{X,Y}=E_{X,YZ}=E_{X,Y_1}$ because of commutativity, we obtain:

$$\begin{aligned} \left| \lambda _{XYZ} - 1 \right|&\le K \left| {\text {Tr}}_{XY_1} \Big [ \Big ( {\text {tr}}_{Y_2}\big (\sigma ^{XY} \big ) - {\text {tr}}_{Y_2Z}\big (\sigma ^{XYZ} \big ) \Big ) E_{X,Y_1}^{\dagger \ -1} \Big ] \right| \\&\le K \left\| {\text {tr}}_{Y_2}\big (\sigma ^{XY} \big ) - {\text {tr}}_{Y_2Z}\big (\sigma ^{XYZ} \big ) \right\| \left\| E_{X,Y_1}^{\dagger \ -1} \right\| . \end{aligned}$$

The first norm in the right-hand side can be bounded by the so-called local indistinguishability [17], which is subsequently bounded by the covariance. Since 1D translation-invariant, local, (non-necessarily) commuting systems are known to satisfy exponential decay of correlations [6, 16, 61], $\left\| {\text {tr}}_{Y_2}\big (\sigma ^{XY} \big ) - {\text {tr}}_{Y_2Z}\big (\sigma ^{XYZ} \big ) \right\|$ is known to decay exponentially fast with $|Y_2|$, scaling at most exponentially with $\beta$. Hence, $\left| \lambda _{XYZ} - 1 \right| \le \mathcal {O}(e^\beta )$.

The bound for $\chi _2$ follows similar ideas but is slightly more involved. For $\Lambda =XYZ$, we split now Y into $Y_1 Y_2 Y_3$ such that $Y_1:= \partial X$ and $Y_3:= \partial Z$. Then, we can write:

$$\begin{aligned}&\left\| {{\text {tr}}}_{YZ}\big (\sigma ^{YZ} E_{X,YZ}^\dagger \big ) {{\text {tr}}}_{XY}\big (\sigma ^{XY} E_{XY,Z}^\dagger \big ) - {{\text {tr}}}_{Y}\big (\sigma ^{Y} E_{X,Y}^\dagger E_{XY,Z}^\dagger \big ) \right\| \\&\quad \le \left\| {{\text {tr}}}_{Y}\big (\sigma ^{Y} E_{X,Y_1}^\dagger E_{Y_3,Z}^\dagger \big ) - {{\text {tr}}}_{Y}\big (\sigma ^{Y} E_{X,Y_1}^\dagger \big ) {{\text {tr}}}_{Y}\big (\sigma ^{Y} E_{Y_3,Z}^\dagger \big ) \right\| \\&\qquad + \left\| {{\text {tr}}}_{Y}\big (\sigma ^{Y} E_{X,Y_1}^\dagger \big ) {{\text {tr}}}_{Y}\big (\sigma ^{Y} E_{Y_3,Z}^\dagger \big ) - {{\text {tr}}}_{YZ}\big (\sigma ^{YZ} E_{X,Y_1}^\dagger \big ) {{\text {tr}}}_{XY}\big (\sigma ^{XY} E_{Y_3,Z}^\dagger \big ) \right\| . \end{aligned}$$

The first term in the right-hand side above can be upper bounded following a similar trick as for $\chi _1$. The idea is to split the trace in Y by tracing out first $Y_1$ and $Y_3$, and subsequently $Y_2$. This allows us to decouple the difference inside the partial trace and estimate it after applying Hölder’s inequality by

$$\begin{aligned} \left\| \sigma _{Y_1 Y_3} - \sigma _{Y_1}\otimes \sigma _{Y_3} \right\| _1. \end{aligned}$$

This term is obtained as a supremum over the covariance, and thus it is also known to decay exponentially fast with $|Y_2|$, scaling at most exponentially with $\beta$. For the second term above, we notice that it can be split as

$$\begin{aligned}&\left\| {{\text {tr}}}_{Y}\big (\sigma ^{Y} E_{X,Y_1}^\dagger \big ) {{\text {tr}}}_{Y}\big (\sigma ^{Y} E_{Y_3,Z}^\dagger \big ) - {{\text {tr}}}_{YZ}\big (\sigma ^{YZ} E_{X,Y_1}^\dagger \big ) {{\text {tr}}}_{XY}\big (\sigma ^{XY} E_{Y_3,Z}^\dagger \big ) \right\| \\&\quad \le \left\| {{\text {tr}}}_{Y}\big (\sigma ^{Y} E_{X,Y_1}^\dagger \big ) {{\text {tr}}}_{Y}\big (\sigma ^{Y} E_{Y_3,Z}^\dagger \big ) - {{\text {tr}}}_{YZ}\big (\sigma ^{YZ} E_{X,Y_1}^\dagger \big ) {{\text {tr}}}_{Y}\big (\sigma ^{Y} E_{Y_3,Z}^\dagger \big ) \right\| \\&\qquad + \left\| {{\text {tr}}}_{YZ}\big (\sigma ^{YZ} E_{X,Y_1}^\dagger \big ) {{\text {tr}}}_{Y}\big (\sigma ^{Y} E_{Y_3,Z}^\dagger \big ) - {{\text {tr}}}_{YZ}\big (\sigma ^{YZ} E_{X,Y_1}^\dagger \big ) {{\text {tr}}}_{XY}\big (\sigma ^{XY} E_{Y_3,Z}^\dagger \big ) \right\| \\&\quad \le K \left( \left\| {{\text {tr}}}_{Y}\big (\sigma ^{Y} E_{X,Y_1}^\dagger \big ) - {{\text {tr}}}_{YZ}\big (\sigma ^{YZ} E_{X,Y_1}^\dagger \big ) \right\| + \left\| {{\text {tr}}}_{Y}\big (\sigma ^{Y} E_{Y_3,Z}^\dagger \big ) - {{\text {tr}}}_{XY}\big (\sigma ^{XY} E_{Y_3,Z}^\dagger \big ) \right\| \right) , \end{aligned}$$

and these last two terms can be bounded again by local indistinguishability, yielding thus an scaling with $\beta$ which is at most exponential.