$\displaystyle\langle\psi_{m}|\mathcal{L}_{O_{A}}\rho\mathcal{L}_{O_{B}}|\psi_{m}\rangle=\langle\psi_{m}|\tilde{\alpha}_{1,m}\rho\tilde{\alpha}_{2,m}+\delta\mathcal{L}_{O_{A}}\rho\mathcal{L}_{O_{B}}|\psi_{m}\rangle,$
$\displaystyle\langle\psi_{m}|\mathcal{L}_{O_{A}}\mathcal{L}_{O_{B}}\rho|\psi_{m}\rangle=\langle\psi_{m}|\tilde{\alpha}_{1,m}\tilde{\alpha}_{2,m}\rho+\mathcal{L}_{O_{A}}\delta\mathcal{L}_{O_{B}}\rho|\psi_{m}\rangle,$
$\displaystyle\langle\psi_{m}|\mathcal{L}_{O_{B}}\rho\mathcal{L}_{O_{A}}|\psi_{m}\rangle=\langle\psi_{m}|\tilde{\alpha}_{1,m}\rho\tilde{\alpha}_{2,m}+\delta\mathcal{L}_{O_{B}}\rho\mathcal{L}_{O_{A}}|\psi_{m}\rangle.$
(180)

Using the above equations, we reduce Eq. (D.3) to

$\displaystyle\langle\phi_{m}|O_{A}O_{B}|\phi_{m}\rangle$
$\displaystyle=\tilde{\alpha}_{1,m}\tilde{\alpha}_{2,m}+\frac{1}{4p_{m}}\langle\psi_{m}|(\rho\delta\mathcal{L}_{O_{A}}\mathcal{L}_{O_{B}}+\delta\mathcal{L}_{O_{A}}\rho\mathcal{L}_{O_{B}}+\mathcal{L}_{O_{A}}\delta\mathcal{L}_{O_{B}}\rho+\delta\mathcal{L}_{O_{B}}\rho\mathcal{L}_{O_{A}})|\psi_{m}\rangle$
$\displaystyle\quad+\frac{1}{4}\langle\phi_{m}|\left[\left(\rho^{-1/2}\mathcal{L}_{O_{A}}\rho^{1/2}\right),\left(\rho^{1/2}\mathcal{L}_{O_{B}}\rho^{-1/2}\right)\right]|\phi_{m}\rangle,$
(181)

where we use $\langle\psi_{m}|\rho|\psi_{m}\rangle=p_{m}$. Thus, we obtain

$\displaystyle\sum_{m}p_{m}|\langle\phi_{m}|O_{A}O_{B}|\phi_{m}\rangle-\tilde{\alpha}_{1,m}\tilde{\alpha}_{2,m}|$
$\displaystyle\leq\frac{1}{2}\left(\left\|\delta\mathcal{L}_{O_{A}}\right\|\cdot\left\|\mathcal{L}_{O_{B}}\right\|+\left\|\mathcal{L}_{O_{A}}\right\|\cdot\left\|\delta\mathcal{L}_{O_{B}}\right\|\right)+\frac{1}{4}\left\|\left[\left(\rho^{-1/2}\mathcal{L}_{O_{A}}\rho^{1/2}\right),\left(\rho^{1/2}\mathcal{L}_{O_{B}}\rho^{-1/2}\right)\right]\right\|,$
(182)

where we use similar analyses to those in inequality (D.3). Using the
condition (141) and $\left\|\mathcal{L}_{O_{A}}\right\|\leq\|O_{A}\|=1$, which
is proved as the inequality (145) in Lemma 19, we reach the inequality of

$\displaystyle\sum_{m}p_{m}|\langle\phi_{m}|O_{A}O_{B}|\phi_{m}\rangle-\tilde{\alpha}_{1,m}\tilde{\alpha}_{2,m}|\leq\frac{\delta_{1}+\delta_{2}}{2}+\frac{1}{4}\left\|\left[\left(\rho^{-1/2}\mathcal{L}_{O_{A}}\rho^{1/2}\right),\left(\rho^{1/2}\mathcal{L}_{O_{B}}\rho^{-1/2}\right)\right]\right\|.$
(183)

By combining the inequalities (176) and (183), we finally obtain

$\displaystyle\sum_{m}p_{m}|\langle\phi_{m}|O_{A}O_{B}|\phi_{m}\rangle-\langle\phi_{m}|O_{A}|\phi_{m}\rangle\langle\phi_{m}|O_{B}|\phi_{m}\rangle|$
$\displaystyle=$
$\displaystyle\sum_{m}p_{m}\left|\langle\phi_{m}|O_{A}O_{B}|\phi_{m}\rangle-\tilde{\alpha}_{1,m}\tilde{\alpha}_{2,m}-\langle\phi_{m}|O_{A}|\phi_{m}\rangle\langle\phi_{m}|O_{B}|\phi_{m}\rangle+\tilde{\alpha}_{1,m}\tilde{\alpha}_{2,m}\right|$
$\displaystyle\leq$
$\displaystyle\sum_{m}p_{m}\left(\left|\langle\phi_{m}|O_{A}O_{B}|\phi_{m}\rangle-\tilde{\alpha}_{1,m}\tilde{\alpha}_{2,m}\right|+p_{m}\left|\langle\phi_{m}|O_{A}|\phi_{m}\rangle\langle\phi_{m}|O_{B}|\phi_{m}\rangle-\tilde{\alpha}_{1,m}\tilde{\alpha}_{2,m}\right|\right)$
$\displaystyle\leq$
$\displaystyle\frac{3\delta_{1}+3\delta_{2}}{2}+\frac{1}{4}\left\|\left[\left(\rho^{-1/2}\mathcal{L}_{O_{A}}\rho^{1/2}\right),\left(\rho^{1/2}\mathcal{L}_{O_{B}}\rho^{-1/2}\right)\right]\right\|.$
(184)

By applying the above inequality to (171), we prove the main inequality (18).
This completes the proof. $\square$

###  D.4 Proof of Lemma 19

We first rewrite the eigenvalues $\\{\lambda_{s}\\}$ and the eigenstates
$\\{|\lambda_{s}\rangle\\}$ as

$\displaystyle\lambda_{s}=e^{-\beta
E_{s}},\quad|\lambda_{s}\rangle=|E_{s}\rangle,$ (185)

where $H|E_{s}\rangle=E_{s}|E_{s}\rangle$. Then, for an arbitrary operator
$O$, the definition (137) gives

$\displaystyle\mathcal{L}_{O}$
$\displaystyle=\sum_{s,s^{\prime}}\frac{2\sqrt{e^{-\beta(E_{s}-E_{s}^{\prime})}}}{1+e^{-\beta(E_{s}-E_{s}^{\prime})}}\langle
E_{s}|O|E_{s^{\prime}}\rangle|E_{s}\rangle\langle E_{s^{\prime}}|$
$\displaystyle=\int_{-\infty}^{\infty}\frac{2\sqrt{e^{-\beta\omega}}}{1+e^{-\beta\omega}}O_{\omega}d\omega,$
(186)

where we use the notation of Eq. (92).

Using Eq. (93), we reduce the above form to

$\displaystyle\mathcal{L}_{O}$
$\displaystyle=\int_{-\infty}^{\infty}\frac{2\sqrt{e^{-\beta\omega}}}{1+e^{-\beta\omega}}\frac{1}{2\pi}\int_{-\infty}^{\infty}O(t)e^{-i\omega
t}dtd\omega$ $\displaystyle=\int_{-\infty}^{\infty}f_{\beta}(t)O(t)dt,$ (187)

with

$\displaystyle
f_{\beta}(t):=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2\sqrt{e^{-\beta\omega}}}{1+e^{-\beta\omega}}e^{-i\omega
t}d\omega$ (188)

By following the same analysis in Sec. C.3.1, we can prove that $f_{\beta}(t)$
is given by

$\displaystyle f_{\beta}(t)$ $\displaystyle=\begin{cases}&\displaystyle
i\sum_{m=1}^{\infty}{\rm Res}_{\omega=(2\pi
im-i\pi)/\beta}\left(\frac{2\sqrt{e^{-\beta\omega}}}{1+e^{-\beta\omega}}e^{-i\omega
t}\right)\\\ &{\rm for}\quad t<0,\\\ &\displaystyle-i\sum_{m=1}^{\infty}{\rm
Res}_{\omega=(-2\pi
im+i\pi)/\beta}\left(\frac{2\sqrt{e^{-\beta\omega}}}{1+e^{-\beta\omega}}e^{-i\omega
t}\right)\\\ &{\rm for}\quad t\geq 0,\end{cases}$
$\displaystyle=\begin{cases}\displaystyle-\sum_{m=1}^{\infty}\frac{2(-1)^{m}}{\beta}e^{\pi(2m-1)t/\beta}&\quad\textrm{for}\quad
t<0,\\\
\displaystyle-\sum_{m=1}^{\infty}\frac{2(-1)^{m}}{\beta}e^{-\pi(2m-1)t/\beta}&\quad\textrm{for}\quad
t\geq 0,\end{cases}$
$\displaystyle=\sum_{m=1}^{\infty}\frac{2(-1)^{m}}{\beta}e^{-\pi(2m-1)|t|/\beta}=\frac{1}{\beta\cosh(\pi|t|/\beta)}.$

This completes the proof of Eq. (143).

The proof of the inequality (145) is simply given as follows. Because of
$f_{\beta}(t)\geq 0$, we have

$\displaystyle\|\mathcal{L}_{O}\|$
$\displaystyle\leq\int_{-\infty}^{\infty}f_{\beta}(t)\|O(t)\|dt\leq\|O\|\int_{-\infty}^{\infty}f_{\beta}(t)dt.$
(189)

Using the inverse Fourier transform

$\displaystyle\int_{-\infty}^{\infty}f_{\beta}(t)e^{i\omega
t}dt=\frac{2\sqrt{e^{-\beta\omega}}}{1+e^{-\beta\omega}}.$ (190)

with $\omega=0$, we reduce the inequality (189) to (145). This completes the
proof. $\square$

###  D.5 Proof of Lemma 20

Figure 4: Approximations of $\mathcal{L}_{O_{A}}$ and $\mathcal{L}_{O_{B}}$.
In order to obtain the approximations $\tilde{\mathcal{L}}_{O_{A}}$ and
$\tilde{\mathcal{L}}_{O_{B}}$ which commute with each other, we approximate
$\mathcal{L}_{O_{A}}$ and $\mathcal{L}_{O_{B}}$ onto the extended regions
$A[r_{1}]$ and $B[r_{2}]$ ($r_{1}+r_{2}<R$), respectively. In Eqs. (191) and
(192), we show the explicit forms of $\mathcal{L}_{O_{A}}$ and
$\mathcal{L}_{O_{B}}$.

We first show the explicit construction of $\tilde{\mathcal{L}}_{O_{A}}$ and
$\tilde{\mathcal{L}}_{O_{B}}$ such that
$[\tilde{\mathcal{L}}_{O_{A}},\tilde{\mathcal{L}}_{O_{B}}]=0$. For this
purpose, we use Eq. (143) in Lemma 19 and approximate the time-evolved
operator $O_{A}(t)$ on $A[r_{1}]$, which yields

$\displaystyle\tilde{\mathcal{L}}_{O_{A}}=\int_{-\infty}^{\infty}f_{\beta}(t)O_{A}(t,A[r_{1}])dt,$
(191)

where the notation of $O_{A}(t,A[r_{1}])$ has been given in Eq. (16), and we
choose $r_{1}$ appropriately. Note that $\tilde{\mathcal{L}}_{O_{A}}$ is now
supported on the subset $A[r_{1}]$. In the same way, we define
$\tilde{\mathcal{L}}_{O_{A}}$ as

$\displaystyle\tilde{\mathcal{L}}_{O_{B}}=\int_{-\infty}^{\infty}f_{\beta}(t)O_{B}(t,B[r_{2}])dt.$
(192)

Then, if we set $r_{1}+r_{2}<d_{A,B}=R$, we have
$[\tilde{\mathcal{L}}_{O_{A}},\tilde{\mathcal{L}}_{O_{B}}]=0$. Therefore, in
the following sections, we choose $r_{1}=r_{2}=\lceil R/2-1\rceil$.

Using Eq. (191), we can estimate $\delta_{1}$ as

$\displaystyle\delta_{1}\leq\int_{-\infty}^{\infty}f_{\beta}(t)\|O_{A}(t)-O_{A}(t,A[r_{1}])\|dt.$
(193)

For the estimation of the integral, we use a similar approach in Sec. C.3.2.
We start from

$\displaystyle\int_{-\infty}^{\infty}f_{\beta}(t)\|O_{A}(t)-O_{A}(t,A[r_{1}])\|dt$
$\displaystyle=\int_{|t|>t_{0}}f_{\beta}(t)\|O_{A}(t)-O_{A}(t,A[r_{1}])\|dt$
$\displaystyle+\int_{|t|\leq
t_{0}}f_{\beta}(t)\|O_{A}(t)-O_{A}(t,A[r_{1}])\|dt$ (194)

where we choose $t_{0}:=\mu r_{1}/(2v)$. Because of

$\displaystyle
f_{\beta}(t)=\frac{1}{\beta\cosh(\pi|t|/\beta)}\leq\frac{2}{\beta}e^{-\pi|t|/\beta},$
$\displaystyle\|O_{A}(t)-O_{A}(t,A[r_{1}])\|\leq 2\|O_{A}\|=2,$ (195)

the first term is upper-bounded as

$\displaystyle\int_{|t|>t_{0}}f_{\beta}(t)\|O_{A}(t)-O_{A}(t,A[r_{1}])\|dt$
$\displaystyle\leq\frac{4}{\beta}\int_{|t|>t_{0}}e^{-\pi|t|/\beta}dt\leq\frac{8}{\pi}e^{-\pi\mu
r_{1}/(2v\beta)},$ (196)

The quantity $\|O_{A}(t)-O_{A}(t,A[r_{1}])\|$ is upper-bounded using the
inequality (18), and hence the second term is upper-bounded as

$\displaystyle\int_{|t|\leq
t_{0}}f_{\beta}(t)\|O_{A}(t)-O_{A}(t,A[r_{1}])\|dt$
$\displaystyle\leq\frac{2}{\beta}\int_{|t|\leq
t_{0}}e^{-\pi|t|/\beta}C|\partial A|\left(e^{v|t|}-1\right)e^{-\mu r_{1}}dt$
$\displaystyle\leq\frac{4C}{\beta}|\partial A|\int_{0}^{t_{0}}e^{v|t|}e^{-\mu
r_{1}}dt$ $\displaystyle\leq\frac{4C}{v\beta}|\partial A|e^{-\mu
r_{1}+vt_{0}}=\frac{4C}{v\beta}|\partial A|e^{-\mu r_{1}/2}$ (197)

where we use $(e^{xy}-1)/x\leq y(1+e^{xy})$ ($y\geq 0$) in the third
inequality. By applying the inequalities (D.5) and (D.5), we reduce Eq. (D.5)
to

$\displaystyle\delta_{1}$
$\displaystyle\leq\int_{-\infty}^{\infty}f_{\beta}(t)\|O_{A}(t)-O_{A}(t,A[r_{1}])\|dt$
$\displaystyle\leq\left(\frac{8}{\pi}+\frac{4C}{v\beta}\right)|\partial
A|e^{-\min[\mu r_{1}/2,\pi\mu r_{1}/(2v\beta)]}$
$\displaystyle\leq\left(\frac{8}{\pi}+\frac{4C}{v\beta}\right)|\partial
A|e^{-\mu r_{1}/(2+2v\beta/\pi)},$ (198)

where we use $r_{1}\geq 1$. In the same way, we obtain

$\displaystyle\delta_{2}$
$\displaystyle\leq\int_{-\infty}^{\infty}f_{\beta}(t)\|O_{B}(t)-O_{B}(t,B[r_{2}])\|dt$
$\displaystyle\leq\left(\frac{8}{\pi}+\frac{4C}{v\beta}\right)|\partial
B|e^{-\mu r_{2}/(2+2v\beta/\pi)}.$ (199)

By applying $r_{1}=r_{2}=\lceil R/2-1\rceil$, we prove the inequality (20).
This completes the proof. $\square$

###  D.6 Proof of Lemma 21

We first show the integral expression of $\rho^{\pm
1/2}\mathcal{L}_{O}\rho^{\mp 1/2}$ for an arbitrary operator $O$. Using

$\displaystyle e^{\pm\beta H/2}O_{\omega}e^{\mp\beta
H/2}=e^{\pm\beta\omega/2}O_{\omega},$ (200)

we have from Eq. (D.4)

$\displaystyle\rho^{\pm 1/2}\mathcal{L}_{O}\rho^{\mp
1/2}=\int_{-\infty}^{\infty}\frac{2\sqrt{e^{-\beta\omega}}}{1+e^{-\beta\omega}}e^{\pm\beta\omega/2}O_{\omega}d\omega.$
(201)

Using Eq. (93), we reduce the above equation to

$\displaystyle\rho^{\pm 1/2}\mathcal{L}_{O}\rho^{\mp 1/2}$
$\displaystyle=\int_{-\infty}^{\infty}\frac{2\sqrt{e^{-\beta\omega}}}{1+e^{-\beta\omega}}e^{\pm\beta\omega/2}\frac{1}{2\pi}\int_{-\infty}^{\infty}O(t)e^{-i\omega
t}dtd\omega$ $\displaystyle=\int_{-\infty}^{\infty}g_{\beta,\pm}(t)O(t)dt,$
(202)

where we define $g_{\beta}(t)$ as

$\displaystyle
g_{\beta,\pm}(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2\sqrt{e^{-\beta\omega}}}{1+e^{-\beta\omega}}e^{\pm\beta\omega/2}e^{-i\omega
t}d\omega.$ (203)

We have

$\displaystyle g_{\beta,\pm}(t)$
$\displaystyle=\frac{1}{2\pi}\int_{-\infty}^{\infty}\left[\pm\tanh(\beta\omega/2)+1\right]e^{-i\omega
t}d\omega$ $\displaystyle=\delta(t)\pm g_{\beta}(t),$ (204)

where $\delta(t)$ is the delta function and $g_{\beta}(t)$ is the Fourier
transform of $\tanh(\beta\omega/2)$.

As in Sec. C.3.1, we have

$\displaystyle g_{\beta}(t)$ $\displaystyle=\begin{cases}\displaystyle
i\sum_{m=1}^{\infty}{\rm Res}_{\omega=(2\pi
im-i\pi)/\beta}\left(\tanh(\beta\omega/2)e^{-i\omega
t}\right)&\quad\textrm{for}\quad t<0,\\\
\displaystyle-i\sum_{m=1}^{\infty}{\rm Res}_{\omega=(-2\pi
im+i\pi)/\beta}\left(\tanh(\beta\omega/2)e^{-i\omega
t}\right)&\quad\textrm{for}\quad t\geq 0,\end{cases}$
$\displaystyle=\begin{cases}\displaystyle
i\sum_{m=1}^{\infty}\frac{2}{\beta}e^{\pi(2m-1)t/\beta}&\quad\textrm{for}\quad
t<0,\\\
\displaystyle-i\sum_{m=1}^{\infty}\frac{2}{\beta}e^{-\pi(2m-1)t/\beta}&\quad\textrm{for}\quad
t\geq 0,\end{cases}$ $\displaystyle=\frac{-2i}{\beta}{\rm
sign}(t)\sum_{m=1}^{\infty}e^{-\pi(2m-1)|t|/\beta}=-i\frac{{\rm
sign}(t)}{\beta\sinh(\pi|t|/\beta)}=\frac{-i}{\beta\sinh(\pi t/\beta)}.$ (205)

We thus obtain

$\displaystyle\rho^{\pm 1/2}\mathcal{L}_{O}\rho^{\mp 1/2}$
$\displaystyle=O\pm\int_{-\infty}^{\infty}g_{\beta}(t)O(t)dt.$ (206)

For the proof of the lemma, we must prove the following two claims:

######  Claim 22.

Let $O$ be an arbitrary operator supported on a subset $X\subset\Lambda$.
Then, the norm of $\rho^{\pm 1/2}\mathcal{L}_{O}\rho^{\mp 1/2}$ is upper-
bounded as

$\displaystyle\left\|\rho^{\pm 1/2}\mathcal{L}_{O}\rho^{\mp 1/2}\right\|$
$\displaystyle\leq\|O\|\log\left(1+\frac{\beta\|{\rm
ad}_{H}(O)\|}{\|O\|}\right)+2\|O\|.$ (207)

######  Claim 23.

Let $O$ be the operator defined in Claim 22. Then, for $\|O\|=1$, the operator
$\rho^{\pm 1/2}\mathcal{L}_{O}\rho^{\mp 1/2}$ is approximated on $X[r]$ with
an error of

$\displaystyle\left\|\rho^{\pm 1/2}\mathcal{L}_{O}\rho^{\mp
1/2}-\left(\rho^{\pm 1/2}\mathcal{L}_{O}\rho^{\mp
1/2}\right)_{X[r]}\right\|\leq|\partial
X|\left[\frac{8}{\pi}\left(1+\frac{\xi_{\beta}}{2r}\right)+4C\left(\frac{1}{\pi}+\frac{1}{v\beta}\right)\right]e^{-2r/\xi_{\beta}},$
(208)

where $\left(\rho^{\pm 1/2}\mathcal{L}_{O}\rho^{\mp 1/2}\right)_{X[r]}$ is
supported on $X[r]$ and is chosen appropriately.

Using these claims, we can immediately upper-bound the norm of (149). Let us
approximate

$\displaystyle\mathfrak{O}_{1}:=\rho^{-1/2}\mathcal{L}_{O_{A}}\rho^{1/2}\approx\mathfrak{O}_{1,A[r_{1}]},$
$\displaystyle\mathfrak{O}_{2}:=\rho^{1/2}\mathcal{L}_{O_{B}}\rho^{-1/2}\approx\mathfrak{O}_{2,B[r_{2}]},$
(209)

where $r_{1}+r_{2}<R$. Then, from
$[\mathfrak{O}_{1,A[r_{1}]},\mathfrak{O}_{2,B[r_{2}]}]=0$, we obtain

$\displaystyle\left\|[\mathfrak{O}_{1},\mathfrak{O}_{2}]\right\|$
$\displaystyle=\left\|[\mathfrak{O}_{1}-\mathfrak{O}_{1,A[r_{1}]},\mathfrak{O}_{2}]+[\mathfrak{O}_{1,A[r_{1}]},\mathfrak{O}_{2}-\mathfrak{O}_{2,B[r_{2}]}]\right\|$
$\displaystyle\leq
2\left\|\delta\mathfrak{O}_{1}\right\|\cdot\left\|\mathfrak{O}_{2}\right\|+2\left\|\delta\mathfrak{O}_{2}\right\|\cdot\left\|\mathfrak{O}_{1}\right\|+2\left\|\delta\mathfrak{O}_{1}\right\|\cdot\left\|\delta\mathfrak{O}_{2}\right\|,$
(210)

where we define
$\delta\mathfrak{O}_{1}:=\mathfrak{O}_{1}-\mathfrak{O}_{1,A[r_{1}]}$ and
$\delta\mathfrak{O}_{2}:=\mathfrak{O}_{2}-\mathfrak{O}_{2,B[r_{2}]}$, and use
$\left\|\mathfrak{O}_{1,A[r_{1}]}\right\|\leq\left\|\mathfrak{O}_{1}\right\|+\left\|\delta\mathfrak{O}_{1}\right\|$
in the inequality. For
$\left\|\delta\mathfrak{O}_{s}\right\|>\left\|\mathfrak{O}_{s}\right\|$
($s=1,2$), the above inequality is worse than the trivial inequality, i.e.,
$\left\|[\mathfrak{O}_{1},\mathfrak{O}_{2}]\right\|\leq
2\left\|\mathfrak{O}_{1}\right\|\cdot\left\|\mathfrak{O}_{2}\right\|$. Hence,
we only must consider
$\left\|\delta\mathfrak{O}_{s}\right\|\leq\left\|\mathfrak{O}_{s}\right\|$,
which yields

$\displaystyle\left\|[\mathfrak{O}_{1},\mathfrak{O}_{2}]\right\|\leq
3\left(\left\|\delta\mathfrak{O}_{1}\right\|\cdot\left\|\mathfrak{O}_{2}\right\|+\left\|\delta\mathfrak{O}_{2}\right\|\cdot\left\|\mathfrak{O}_{1}\right\|\right).$
(211)

By choosing $r_{1}=r_{2}=\lceil R/2-1\rceil$ and applying Claims 22 and 23, we
obtain the main inequality (21) as follows:

$\displaystyle\left\|[\mathfrak{O}_{1},\mathfrak{O}_{2}]\right\|\leq$
$\displaystyle
3e^{2/\xi_{\beta}}\left[\frac{8}{\pi}\left(1+\frac{\xi_{\beta}}{R-2}\right)+4C\left(\frac{1}{\pi}+\frac{1}{v\beta}\right)\right]e^{-R/\xi_{\beta}}$
$\displaystyle\times\bigl{[}|\partial A|(2+\log(1+\beta\|{\rm
ad}_{H}(O_{B})\|))+|\partial B|(2+\log(1+\beta\|{\rm
ad}_{H}(O_{A})\|))\bigr{]},$ (212)

where we use $\|O_{A}\|=\|O_{B}\|=1$. This completes the proof of Lemma 21.

####  D.6.1 Proof of Claim 22

From the integral expression (206), we first obtain

$\displaystyle\left\|\rho^{\pm 1/2}\mathcal{L}_{O}\rho^{\mp 1/2}\right\|$
$\displaystyle\leq\|O\|+\left\|\int_{-\infty}^{\infty}g_{\beta}(t)O(t)dt\right\|.$
(213)

In a standard approach, we use

$\displaystyle\left\|\int_{-\infty}^{\infty}g_{\beta}(t)O(t)dt\right\|\leq\|O\|\int_{-\infty}^{\infty}|g_{\beta}(t)|dt.$
(214)

However, the integral of $|g_{\beta}(t)|$ does not converge because
$|g_{\beta}(t)|\propto 1/t$ for $t\ll 1$.

To obtain a refined bound, we parametrize $O(t)$ as $O(\lambda t)$ using a
parameter $\lambda$. We then obtain

$\displaystyle O(t)$ $\displaystyle=O+\int_{0}^{1}\frac{d}{d\lambda}O(\lambda
t)d\lambda=O+it\int_{0}^{1}{\rm ad}_{H}(O)(\lambda t)d\lambda,$ (215)

which yields

$\displaystyle\left\|\int_{-\infty}^{\infty}g_{\beta}(t)O(t)dt\right\|$
$\displaystyle\leq\left\|\int_{|t|>\delta
t}g_{\beta}(t)O(t)dt\right\|+\left\|\int_{|t|\leq\delta
t}g_{\beta}(t)Odt+\int_{|t|\leq\delta t}it\int_{0}^{1}g_{\beta}(t){\rm
ad}_{H}(O)(\lambda t)d\lambda dt\right\|$ $\displaystyle\leq
2\|O\|\int_{t>\delta t}\frac{1}{\beta\sinh(\pi t/\beta)}dt+2\|{\rm
ad}_{H}(O)\|\int_{0}^{\delta t}\frac{t}{\beta\sinh(\pi t/\beta)}dt$
$\displaystyle\leq\frac{-2\|O\|}{\pi}\log\left[\tanh\left(\frac{\pi\delta
t}{2\beta}\right)\right]+\frac{2\|{\rm ad}_{H}(O)\|}{\pi}\delta
t\leq\frac{2\|O\|}{\pi}\log\left(1+\frac{2\beta}{\pi\delta
t}\right)+\frac{2\|{\rm ad}_{H}(O)\|}{\pi}\delta t,$ (216)

where we use $\int_{|t|\leq\delta t}g_{\beta}(t)dt=0$, $1/\sinh(x)\leq 1/x$
and $-\log[\tanh(x)]\leq\log(1+1/x)$ in the second, third, and fourth
inequalities, respectively. Note that $g_{\beta}(-t)=-g_{\beta}(t)$. Thus, by
choosing $\delta t=\|O\|/\|{\rm ad}_{H}(O)\|$, we have

$\displaystyle\left\|\int_{-\infty}^{\infty}g_{\beta}(t)O(t)dt\right\|$
$\displaystyle\leq\frac{2\|O\|}{\pi}\log\left(1+\frac{2\beta\|{\rm
ad}_{H}(O)\|}{\pi\|O\|}\right)+\frac{2\|O\|}{\pi},$ (217)

By combining the inequalities (213) and (217) with $2/\pi\leq 1$, we prove the
inequality (207). $\square$

####  D.6.2 Proof of Claim 23

As in the proof of Lemma 20, we consider the approximation as Using the
integral expression (206), we have

$\displaystyle\left(\rho^{\pm 1/2}\mathcal{L}_{O}\rho^{\mp
1/2}\right)_{X[r]}:=O\pm\int_{-\infty}^{\infty}g_{\beta}(t)O(t,X[r])dt,$ (218)

which yields

$\displaystyle\left\|\rho^{\pm 1/2}\mathcal{L}_{O}\rho^{\mp
1/2}-\left(\rho^{\pm 1/2}\mathcal{L}_{O}\rho^{\mp
1/2}\right)_{X[r]}\right\|\leq\int_{-\infty}^{\infty}|g_{\beta}(t)|\cdot\left\|O(t,X[r])-O(t)\right\|dt.$
(219)

Using $1/\sinh(x)\leq 2e^{-x}(1+1/x)$ ($x\geq 0$), we have

$\displaystyle|g_{\beta}(t)|=\frac{1}{\beta\sinh(\pi|t|/\beta)}\leq\frac{2e^{-\pi|t|/\beta}}{\beta}\left(1+\frac{1}{\pi|t|/\beta}\right).$
(220)

In addition, the Lieb-Robinson bound (18), we have

$\displaystyle\left\|O(t)-O_{X}(t,X[r])\right\|\leq\min\left(C|\partial
X|\left(e^{v|t|}-1\right)e^{-\mu r},2\right)$ (221)

Then, we can apply the same analyses as (C.3.2), (C.3.2) and (C.3.2). For
$t_{0}=\mu r/(2v)$, we can obtain

$\displaystyle\int_{-\infty}^{\infty}|g_{\beta}(t)|\cdot\left\|O(t,X[r])-O(t)\right\|dt$
$\displaystyle\leq\int_{|t|>t_{0}}\frac{2e^{-\pi|t|/\beta}}{\beta}\left(1+\frac{1}{\pi|t|/\beta}\right)\cdot
2dt+\int_{|t|\leq
t_{0}}\frac{2e^{-\pi|t|/\beta}}{\beta}\left(1+\frac{1}{\pi|t|/\beta}\right)\cdot
C|\partial X|\left(e^{v|t|}-1\right)e^{-\mu r}dt$
$\displaystyle\leq\frac{8e^{-\pi t_{0}/\beta}}{\pi}\left(1+\frac{1}{\pi
t_{0}/\beta}\right)+\frac{4C}{\beta}|\partial
X|\left(\frac{1}{v}+\frac{1}{\pi/\beta}\right)e^{-\mu R+vt_{0}}$
$\displaystyle\leq|\partial X|\left[\frac{8}{\pi}\left(1+\frac{2v\beta}{\pi\mu
r}\right)e^{-\pi\mu
r/(2v\beta)}+4C\left(\frac{1}{\pi}+\frac{1}{v\beta}\right)e^{-\mu
r/2}\right]\leq|\partial
X|\left[\frac{8}{\pi}\left(1+\frac{\xi_{\beta}}{2r}\right)+4C\left(\frac{1}{\pi}+\frac{1}{v\beta}\right)\right]e^{-2r/\xi_{\beta}},$

where we use the definition of $\xi_{\beta}:=4/\mu(1+v\beta/\pi)$ in the last
inequality. This completes the proof of Claim 23. $\square$

##  Appendix E Proof of Proposition 9

We here show the proof of Proposition 9 which connects the PPT relative
entanglement and the quantum correlation. When the quantum correlation
satisfies the following:

$\displaystyle{\rm
QC}_{\rho_{AB}}(O_{A},O_{B})\leq\epsilon\|O_{A}\|\cdot\|O_{B}\|$ (222)

for arbitrary two operators $O_{A}$ and $O_{B}$, Proposition 9 gives

$\displaystyle E_{R}^{{\rm PPT}}(\rho_{AB})\leq
4\mathcal{D}_{AB}\bar{\delta}\log(1/\bar{\delta})\leq
4\mathcal{D}_{AB}\bar{\delta}^{1/2},$ (223)

where $\bar{\delta}:=4\epsilon\min(\mathcal{D}_{A},\mathcal{D}_{B})$.

###  E.1 Proof

Let us define the eigenstates of $\rho^{T_{A}}_{AB}$ with negative eigenvalues
as $\\{|\eta_{i}\rangle\\}_{i=1}^{M_{0}}$. Then, the proof of Proposition 9 is
immediately obtained by the following lemma:

######  Lemma 24.

For quantum state $\rho_{AB}$ given in Prop. 9, the minimum negative
eigenvalue of $\rho_{AB}^{T_{A}}$ satisfies

$\displaystyle\delta:-\min_{i\in[M_{0}]}\langle\eta_{i}|\rho_{AB}^{T_{A}}|\eta_{i}\rangle\leq
4\epsilon\min(\mathcal{D}_{A},\mathcal{D}_{B}).$ (224)

where the parameter $\epsilon$ has been defined in (222).

To prove the inequality (9), we define a quantum state $\tilde{\sigma}_{AB}$
as follows:

$\displaystyle\tilde{\sigma}_{AB}=(1-\mathcal{D}_{AB}\delta)\rho_{AB}+\delta\cdot\hat{1}_{AB},$
(225)

where ${\rm tr}(\tilde{\sigma}_{AB})=1$ since ${\rm
tr}(\delta\hat{1}_{AB})=\mathcal{D}_{AB}\delta$. Because of the definition of
$\delta$ in (224), we have $\tilde{\sigma}_{AB}^{T_{A}}\succeq 0$ (i.e.,
$\tilde{\sigma}_{AB}\in{\rm PPT}$). We then obtain

$\displaystyle E_{R}^{{\rm PPT}}(\rho_{AB})\leq
S(\rho_{AB}||\tilde{\sigma}_{AB}).$ (226)

Using the continuity bound on the relative entropy (Audenaert and Eisert,
2005, Theorem 3) (or Ref. Donald and Horodecki (1999)), we have

$\displaystyle S(\rho_{AB}||\tilde{\sigma}_{AB})\leq$
$\displaystyle\delta_{AB}\log(\mathcal{D}_{AB})-\delta_{AB}\log(\delta_{AB})$
$\displaystyle-\delta_{AB}\log[\lambda_{\min}(\tilde{\sigma}_{AB})],$ (227)

under the assumption of $\delta_{AB}\leq 1/e$, where we define
$\delta_{AB}:=\|\rho_{AB}-\tilde{\sigma}_{AB}\|_{1}$ and
$\lambda_{\min}(\tilde{\sigma}_{AB})$ is the minimum eigenvalue of
$\tilde{\sigma}_{AB}$. From the definition (225), we have
$\lambda_{\min}(\tilde{\sigma}_{AB})\geq\delta$ and

$\displaystyle\delta_{AB}\leq 2\mathcal{D}_{AB}\delta,$ (228)

and hence the inequality (E.1) reduces to

$\displaystyle S(\rho_{AB}||\tilde{\sigma}_{AB})$
$\displaystyle\leq-2\mathcal{D}_{AB}\delta\log(2\delta^{2})$
$\displaystyle\leq-4\mathcal{D}_{AB}\delta\log(\delta),$ (229)

which gives a trivial upper bound for $\delta>1/e$. By combining the above
inequality with (226) and the upper bound (224), we prove the main inequality
(9). This completes the proof. $\square$

###  E.2 Proof of Lemma 24

Our task now is to estimate

$\displaystyle\min_{i}\langle\eta_{i}|\rho_{AB}^{T_{A}}|\eta_{i}\rangle=\inf_{|\eta\rangle}{\rm
tr}\left(\rho_{AB}^{T_{A}}P_{\eta}\right).$ (230)

under the assumption of (222), where $P_{\eta}:=|\eta\rangle\langle\eta|$.
Therefore, we first rewrite

$\displaystyle{\rm tr}\left(\rho_{AB}^{T_{A}}P_{\eta}\right)$
$\displaystyle={\rm tr}\left(\rho_{AB}P_{\eta}^{T_{A}}\right)$
$\displaystyle={\rm tr}\left(\rho_{AB}P_{\eta}\right)+{\rm
tr}\left[\rho_{AB}(P_{\eta}^{T_{A}}-P_{\eta})\right],$

and prove the second term is approximately equal to zero for an arbitrary
quantum state $|\eta\rangle$. Because the eigenvalues of $\rho_{AB}^{T_{A}}$
do not depend on the choice of basis Peres (1996), we choose the basis which
gives the Schmidt decomposition of $|\eta\rangle$ as follows:

$\displaystyle|\eta\rangle=\sum_{s=1}^{\mathcal{D}_{A}}\nu_{s}|s_{A},s_{B}\rangle,\quad\sum_{s}|\nu_{s}|^{2}=1,$
(231)

where we assume $\mathcal{D}_{A}<\mathcal{D}_{B}$ without loss of generality.

To verify the point, we first consider the qubit case, i.e.,
$\mathcal{D}_{A}=\mathcal{D}_{B}=2$. We can prove the following lemma:

######  Lemma 25.

When $\mathcal{D}_{A}=\mathcal{D}_{B}=2$, we have

$\displaystyle\left|{\rm
tr}\left[\rho_{AB}(P_{\eta}^{T_{A}}-P_{\eta})\right]\right|\leq 2\epsilon,$
(232)

where the parameter $\epsilon$ is given in (222).

To generalize the results of two qubit to two qudit systems, we consider

$\displaystyle P_{\eta}^{T_{A}}-P_{\eta}$
$\displaystyle=\sum_{s,s^{\prime}:s\neq
s^{\prime}}^{\mathcal{D}_{A}}\nu_{s}\nu_{s^{\prime}}(-|s_{A},s_{B}\rangle\langle
s^{\prime}_{A},s^{\prime}_{B}|+|s^{\prime}_{A},s_{B}\rangle\langle
s_{A},s^{\prime}_{B}|)$

and

$\displaystyle\nu_{s}\nu_{s^{\prime}}(-|s_{A},s_{B}\rangle\langle
s^{\prime}_{A},s^{\prime}_{B}|+|s^{\prime}_{A},s_{B}\rangle\langle
s_{A},s^{\prime}_{B}|)+{\rm h.c.}$
$\displaystyle=(\nu_{s}^{2}+\nu_{s^{\prime}}^{2})\left(|\eta_{s,s^{\prime}}\rangle\langle\eta_{s,s^{\prime}}|^{T_{A}}-|\eta_{s,s^{\prime}}\rangle\langle\eta_{s,s^{\prime}}|\right),$
(233)

where
$|\eta_{s,s^{\prime}}\rangle:=(\nu_{s}^{2}+\nu_{s^{\prime}}^{2})^{-1/2}(\nu_{s}|s_{A},s_{B}\rangle+\nu_{s^{\prime}}|s^{\prime}_{A},s^{\prime}_{B}\rangle)$.
Now, the quantum state $|\eta_{s,s^{\prime}}\rangle$ is reduced to a quantum
state with two qubits. We thus obtain from Lemma 25

$\displaystyle\left|{\rm
tr}\left[\rho_{AB}\left(|\eta_{s,s^{\prime}}\rangle\langle\eta_{s,s^{\prime}}|^{T_{A}}-|\eta_{s,s^{\prime}}\rangle\langle\eta_{s,s^{\prime}}|\right)\right]\right|\leq
2\epsilon,$ (234)

which yields

$\displaystyle\left|{\rm
tr}\left[\rho_{AB}(P_{\eta}^{T_{A}}-P_{\eta})\right]\right|$
$\displaystyle=\sum_{1\leq
s<s^{\prime}\leq\mathcal{D}_{A}}(\nu_{s}^{2}+\nu_{s^{\prime}}^{2})$
$\displaystyle\quad\quad\times\left|{\rm
tr}\left[\rho_{AB}\left(|\eta_{s,s^{\prime}}\rangle\langle\eta_{s,s^{\prime}}|^{T_{A}}-|\eta_{s,s^{\prime}}\rangle\langle\eta_{s,s^{\prime}}|\right)\right]\right|$
$\displaystyle\leq 2\epsilon\sum_{1\leq
s<s^{\prime}\leq\mathcal{D}_{A}}(\nu_{s}^{2}+\nu_{s^{\prime}}^{2})\leq
4\epsilon\mathcal{D}_{A}.$ (235)

We thus obtain

$\displaystyle{\rm tr}\left(\rho_{AB}P_{\eta}^{T_{A}}\right)\geq{\rm
tr}\left(\rho_{AB}P_{\eta}\right)-4\epsilon\mathcal{D}_{A}\geq-4\epsilon\mathcal{D}_{A},$
(236)

where we use ${\rm tr}\left(\rho_{AB}P_{\eta}\right)\geq 0$ in the second
inequality. By using the above inequality, we have

$\displaystyle\inf_{|\eta\rangle}{\rm
tr}\left(\rho_{AB}^{T_{A}}P_{\eta}\right)\geq-4\epsilon\mathcal{D}_{A}.$ (237)

When $\mathcal{D}_{B}\leq\mathcal{D}_{A}$, the above upper bound is replaced
by $\inf_{|\eta\rangle}{\rm
tr}\left(\rho_{AB}^{T_{A}}P_{\eta}\right)\geq-4\epsilon\mathcal{D}_{B}$.
Therefore, the parameter $\delta$
($=-\min_{i}\langle\eta_{i}|\rho_{AB}^{T_{A}}|\eta_{i}\rangle$) is upper-
bounded by

$\displaystyle\delta\leq 4\epsilon\min(\mathcal{D}_{A},\mathcal{D}_{B}).$
(238)

By using it, we reduce the inequality (G) to the main inequality (9). This
completes the proof. $\square$

####  E.2.1 Proof of Lemma 25

When $\mathcal{D}_{A}=\mathcal{D}_{B}=2$, an arbitrary operator $O_{AB}$ is
described in the form of

$\displaystyle
O_{AB}=\sum_{P=x,y,z}(J_{P}\hat{\sigma}_{1,P}\hat{\sigma}_{2,P}+h_{1,P}\hat{\sigma}_{1,P}+h_{2,P}\hat{\sigma}_{2,P})$
(239)

by appropriately choosing the bases (see Ref. (Kuwahara and Hatano, 2011,
Lemma 1) for example), where $A=\\{1\\}$ and $B=\\{2\\}$ and
$\\{\hat{\sigma}_{x},\hat{\sigma}_{y},\hat{\sigma}_{z}\\}$ are the Pauli
matrices. Then, the partial transpose $T_{A}$ only changes
$\hat{\sigma}_{1,y}\to-\hat{\sigma}_{1,y}$, and hence,

$\displaystyle O_{AB}-O_{AB}^{T_{A}}$
$\displaystyle=2(J_{y}\hat{\sigma}_{1,y}\hat{\sigma}_{2,y}+h_{1,y}\hat{\sigma}_{1,y})$
$\displaystyle=2\hat{\sigma}_{1,y}\otimes(J_{y}\hat{\sigma}_{2,y}+h_{1,y}).$
(240)

In this way, we can always write

$\displaystyle P_{\eta}^{T_{A}}-P_{\eta}=\Phi_{A}\otimes\Phi_{B},$ (241)

where we can make $\|\Phi_{A}\|\leq 2$ and $\|\Phi_{B}\|=1$ because of
$\|P_{\eta}^{T_{A}}-P_{\eta}\|\leq 2$. Then, from the condition (222) and the
inequality (37) in Lemma 5, we get

$\displaystyle{\rm QC}_{\rho_{AB}}(\Phi_{A},\Phi_{B})\leq{\rm
QC}_{\rho}(\Phi_{A},\Phi_{B})\leq\epsilon\|\Phi_{A}\|\cdot\|\Phi_{B}\|,$

which yields

$\displaystyle\left|{\rm
tr}\left[\rho_{AB}(P_{\eta}^{T_{A}}-P_{\eta})\right]\right|=\left|{\rm
tr}\left(\rho_{AB}\Phi_{A}\otimes\Phi_{B}\right)\right|$
$\displaystyle\leq\left|\sum_{s}p_{s}{\rm tr}(\rho_{s,A}\Phi_{A}){\rm
tr}(\rho_{s,B}\Phi_{B})\right|$ $\displaystyle+\left|\sum_{s}p_{s}\left({\rm
tr}(\rho_{s,AB}\Phi_{A}\Phi_{B})\right)-{\rm tr}(\rho_{s,A}\Phi_{A}){\rm
tr}(\rho_{s,B}\Phi_{B})\right|$ $\displaystyle\leq\sum_{s}p_{s}\left|{\rm
tr}(\rho_{s,A}\otimes\rho_{s,B}\Phi_{A}\otimes\Phi_{B})\right|+{\rm
QC}_{\rho_{AB}}(\Phi_{A},\Phi_{B})$ $\displaystyle\leq\sum_{s}p_{s}\left|{\rm
tr}\left[\rho_{s,A}\otimes\rho_{s,B}(P_{\eta}^{T_{A}}-P_{\eta})\right]\right|+2\epsilon,$
(242)

where $\\{\rho_{s,A}\\}_{s}$ and $\\{\rho_{s,B}\\}_{s}$ are the reduced
density matrices of $\\{\rho_{s,AB}\\}_{s}$ which are appropriately chosen.

We here aim to prove

$\displaystyle{\rm
tr}\left[\rho_{A}\otimes\rho_{B}(P_{\eta}^{T_{A}}-P_{\eta})\right]=0$ (243)

for arbitrary $\rho_{A}$ and $\rho_{B}$. Let $u_{A}$ and $u_{B}$ be unitary
matrices which diagonalize $\rho_{A}$ and $\rho_{B}$, respectively. We then
obtain

$\displaystyle{\rm
tr}\left[\rho_{A}\otimes\rho_{B}(P_{\eta}^{T_{A}}-P_{\eta})\right]$
$\displaystyle={\rm
tr}\left[\tilde{\rho}_{A}\otimes\tilde{\rho}_{B}(u_{A}\otimes
u_{B})(P_{\eta}^{T_{A}}-P_{\eta})(u_{A}\otimes u_{B})^{\dagger}\right]$
$\displaystyle={\rm
tr}\left[\tilde{\rho}_{A}\otimes\tilde{\rho}_{B}(\tilde{P}_{\eta}^{\dagger_{A}}-\tilde{P}_{\eta})\right],$
(244)

where $\tilde{\rho}_{A}:=u_{A}\rho_{A}u_{A}^{\dagger}$,
$\tilde{\rho}_{B}:=u_{B}\rho_{B}u_{B}^{\dagger}$,
$\tilde{P}_{\eta}:=(u_{A}\otimes u_{B})P_{\eta}(u_{A}\otimes
u_{B})^{\dagger}$. Note that by using the form (231), we have
$P_{\eta}^{T_{A}}=P_{\eta}^{\dagger_{A}}$ with $\dagger_{A}$ being the partial
conjugate transpose, which gives

$\displaystyle(u_{A}\otimes u_{B})P_{\eta}^{\dagger_{A}}(u_{A}\otimes
u_{B})^{\dagger}=\tilde{P}_{\eta}^{\dagger_{A}}.$ (245)

In Eq. (E.2.1), only the diagonal terms of
$(\tilde{P}_{\eta}^{\dagger_{A}}-\tilde{P}_{\eta})$ contribute to the value
since $\tilde{\rho}_{A}\otimes\tilde{\rho}_{B}$ is a diagonal matrix. It is
easy to see that all the diagonal terms in
$(\tilde{P}_{\eta}^{\dagger_{A}}-\tilde{P}_{\eta})$ is equal to zero, and
hence, we conclude that Eq. (E.2.1) reduces to Eq. (243). By applying Eq.
(243) to the inequality (E.2.1), we obtain the main inequality (232). This
completes the proof. $\square$

##  Appendix F Proof of Theorem 12

We here give the full proof of Theorem 12, where the following inequality has
been obtained for one-dimensional quantum Gibbs states:

$\displaystyle E_{R}^{{\rm PPT}}(\rho_{\beta,AB})$
$\displaystyle\leq\bar{C}_{\beta}\log(\mathcal{D}_{AB})e^{-R/[16\log(d_{0})\xi^{2}_{\beta}]+7gk\beta},$
(246)

where
$\bar{C}_{\beta}:=20\left(\tilde{C}_{\beta}+16d_{0}^{4}C_{\beta}\right)^{1/2}$
with $C_{\beta}$ and $\tilde{C}_{\beta}$ defined in Eqs. (10) and (57),
respectively. Here, the assumption of the finite interaction length has been
imposed for Hamiltonian $H$.

###  F.1 Proof

Figure 5:  For the proof, we decompose subsets $A$ and $B$ into three pieces.
The decomposed subsets $A_{1}$, $A_{2}$, $B_{2}$, and $B_{1}$ are taken such
that they have the same cardinality, i.e., $A_{1}=A_{2}=B_{0}=B_{1}=\ell$. We
denote the interactions between the subsystems $A_{1}$ and $A_{2}$ ($B_{1}$
and $B_{2}$) by $h_{\partial A_{1}}$ ($h_{\partial B_{1}}$). Then, in the
Hamiltonian $H-h_{\partial A_{1}}-h_{\partial B_{1}}$, the regions
$A_{0}A_{1}$, $A_{2}CB_{2}$, and $B_{1}B_{0}$ are decoupled. Using the quantum
belief propagation, we prove that the effect of the regions $A_{0}$ and
$B_{0}$ does not influence the entanglement value. Then, roughly speaking, the
entanglement between $A$ and $B$ is characterized by the entanglement between
$A_{1}A_{2}$ and $B_{1}B_{2}$. Because the size of these regions is $2\ell$,
the dependence on the Hilbert space dimension in (55) is significantly
improved.

For the proof, we first decompose the subsystems $A$ and $B$ as follows (Fig.
5):

$\displaystyle A=A_{0}\sqcup A_{1}\sqcup A_{2},\quad B=B_{0}\sqcup B_{1}\sqcup
B_{2},$ (247)

where we set $A_{1}=A_{2}=B_{0}=B_{1}=\ell$. Let $h_{\partial A_{1}}$
($h_{\partial B_{1}}$) be the interactions between $A_{1}$ and $A_{2}$
($B_{1}$ and $B_{2}$):

$\displaystyle h_{\partial A_{1}}=\sum_{Z:Z\cap A_{1}\neq\emptyset,Z\cap
A_{2}\neq\emptyset}h_{Z},$ $\displaystyle h_{\partial B_{1}}=\sum_{Z:Z\cap
B_{1}\neq\emptyset,Z\cap B_{2}\neq\emptyset}h_{Z}.$ (248)

Then, we formally describe the quantum Gibbs state $\rho_{\beta}$ as

$\displaystyle\rho_{\beta}=\Phi e^{-\beta(H-h_{\partial A_{1}}-h_{\partial
B_{1}})}\Phi^{\dagger},$ (249)

where $\Phi$ is an appropriate operator. We can prove that $\Phi$ is given by
a quasi-local operator and is approximated by
$\Phi_{A_{1},A_{2}}\otimes\Phi_{B_{1},B_{2}}$, which is formulated by the
following lemma:

######  Lemma 26.

The operator $\Phi$ in Eq. (249) is approximated as follows:

$\displaystyle\tilde{\Phi}=\Phi_{A_{1},A_{2}}\otimes\Phi_{B_{1},B_{2}}\quad{\rm
s.t.}$
$\displaystyle\left\|\rho_{\beta}-\left(\tilde{\Phi}e^{-\beta(H-h_{\partial
A_{1}}-h_{\partial B_{1}})}\tilde{\Phi}^{\dagger}\right)\right\|_{1}$
$\displaystyle\leq\tilde{C}_{\beta}e^{-2\ell/\xi_{\beta}+14gk\beta}=:\delta_{1,\ell},$
(250)

where we define
$\delta_{1,\ell}:=\tilde{C}_{\beta}e^{-2\ell/\xi_{\beta}+14gk\beta}$, and
$\xi_{\beta}$ is defined in Eq. (10) and

$\displaystyle\tilde{C}_{\beta}:=1280\left(\frac{5+2Ce^{\mu
k}}{\pi^{2}}+\frac{2Ce^{\mu k}}{\pi v\beta}\right)^{2}$ (251)

We also have

$\displaystyle\left\|\tilde{\Phi}\right\|\leq e^{2gk\beta}.$ (252)

In the following, we prove the main inequality (246) based on the above two
lemmas. For the purpose, we first define $\tilde{\rho}_{\beta}$ and
$\tilde{Z}$ as follows:

$\displaystyle\tilde{\rho}_{\beta}=\frac{e^{-\beta(H-h_{\partial
A_{1}}-h_{\partial B_{1}})}}{\tilde{Z}},$ $\displaystyle\tilde{Z}:={\rm
tr}\left(e^{-\beta(H-h_{\partial A_{1}}-h_{\partial B_{1}})}\right),$ (253)

Because we have

$\displaystyle e^{-\beta(H-h_{\partial A_{1}}-h_{\partial
B_{1}})}=e^{-\beta(H_{A_{0}A_{1}}+H_{A_{2}CB_{2}}+H_{B_{1}B_{0}})},$ (254)

we obtain $\tilde{\rho}_{\beta,AB}$ in the form of

$\displaystyle\tilde{\rho}_{\beta,AB}=\tilde{\rho}_{A_{0}A_{1}}\otimes\tilde{\rho}_{A_{2}B_{2}}\otimes\tilde{\rho}_{B_{0}B_{1}}.$
(255)

We here define $\tilde{\delta}$ for $\tilde{\rho}_{A_{2}B_{2}}$ in the same
way as (224), and define $\tilde{\sigma}_{A_{2}B_{2}}$ as

$\displaystyle\tilde{\sigma}_{A_{2}B_{2}}=\tilde{\rho}_{A_{2}B_{2}}+\tilde{\delta}\cdot\hat{1}_{A_{2}B_{2}}.$
(256)

Using $\tilde{\sigma}_{A_{2}B_{2}}$ above, we define $\tilde{\sigma}_{AB}$ as

$\displaystyle\tilde{\sigma}_{AB}=\frac{\tilde{Z}}{Z_{\tilde{\sigma}}}\tilde{\Phi}\tilde{\rho}_{A_{0}A_{1}}\otimes\tilde{\sigma}_{A_{2}B_{2}}\otimes\tilde{\rho}_{B_{0}B_{1}}\tilde{\Phi}^{\dagger}$
$\displaystyle=\frac{\tilde{Z}}{Z_{\tilde{\sigma}}}\left(\tilde{\Phi}\tilde{\rho}_{\beta,AB}\tilde{\Phi}^{\dagger}+\tilde{\delta}\cdot\tilde{\Phi}\tilde{\rho}_{A_{0}A_{1}}\otimes\hat{1}_{A_{2}B_{2}}\otimes\tilde{\rho}_{B_{0}B_{1}}\tilde{\Phi}^{\dagger}\right),$
(257)

where $Z_{\tilde{\sigma}}$ is the normalization factor to make ${\rm
tr}(\tilde{\sigma}_{AB})=1$.

We can prove $\tilde{\sigma}_{AB}^{T_{A}}\succeq 0$ as follows. Because of
$\tilde{\sigma}_{A_{2}B_{2}}^{T_{A}}\succeq 0$, we get

$\displaystyle\left(\tilde{\rho}_{A_{0}A_{1}}\otimes\tilde{\sigma}_{A_{2}B_{2}}\otimes\tilde{\rho}_{B_{0}B_{1}}\right)^{T_{A}}\succeq
0.$ (258)

Hence, by representing the spectral decomposition of the above operator as

$\displaystyle\tilde{\rho}_{A_{0}A_{1}}\otimes\tilde{\sigma}_{A_{2}B_{2}}\otimes\tilde{\rho}_{B_{0}B_{1}}=\sum_{i}\tilde{\lambda}_{i}|\tilde{\lambda}_{i}\rangle\langle\tilde{\lambda}_{i}|$
(259)

with $\tilde{\lambda}_{i}\geq 0$, we have

$\displaystyle\left(\tilde{\Phi}\tilde{\rho}_{A_{0}A_{1}}\otimes\tilde{\sigma}_{A_{2}B_{2}}\otimes\tilde{\rho}_{B_{0}B_{1}}\tilde{\Phi}^{\dagger}\right)^{T_{A}}$
$\displaystyle=\sum_{i}\tilde{\lambda}_{i}\left(\Phi_{A_{1},A_{2}}^{\ast}\otimes\Phi_{B_{1},B_{2}}\right)|\tilde{\lambda}_{i}\rangle\langle\tilde{\lambda}_{i}|\left(\Phi_{A_{1},A_{2}}^{T_{A}}\otimes\Phi_{B_{1},B_{2}}^{\dagger}\right)$
$\displaystyle\succeq 0.$ (260)

In the following section, we aim to estimate the upper bound of
$\left\|\tilde{\sigma}_{AB}-\rho_{\beta,AB}\right\|_{1}$. We have

$\displaystyle\left\|\tilde{\sigma}_{AB}-\rho_{\beta,AB}\right\|_{1}$
$\displaystyle\leq\left\|\tilde{Z}\tilde{\Phi}\tilde{\rho}_{\beta,AB}\tilde{\Phi}^{\dagger}-\rho_{\beta,AB}\right\|_{1}+\left\|\tilde{Z}\tilde{\Phi}\tilde{\rho}_{\beta,AB}\tilde{\Phi}^{\dagger}-\tilde{\sigma}_{AB}\right\|_{1}.$
(261)

For the first term, because $\tilde{\Phi}$ supported on $A_{1}A_{2}\cup
B_{1}B_{2}$, we have

$\displaystyle\left\|\tilde{Z}\tilde{\Phi}\tilde{\rho}_{\beta,AB}\tilde{\Phi}^{\dagger}-\rho_{\beta,AB}\right\|_{1}=\left\|\tilde{Z}{\rm
tr}_{C}\left(\tilde{\Phi}\tilde{\rho}_{\beta}\tilde{\Phi}^{\dagger}-\rho_{\beta}\right)\right\|$
$\displaystyle\leq\left\|\rho_{\beta}-\left(\tilde{\Phi}e^{-\beta(H-h_{\partial
A_{1}}-h_{\partial
B_{1}})}\tilde{\Phi}^{\dagger}\right)\right\|_{1}\leq\delta_{1,\ell},$ (262)

where $\delta_{1,\ell}$ has been defined in Lemma 26. For the second term,
from the definition (F.1), we have

$\displaystyle\left\|\tilde{\Phi}\tilde{\rho}_{\beta,AB}\tilde{\Phi}^{\dagger}-\tilde{\sigma}_{AB}\right\|_{1}$
$\displaystyle\leq\left\|\left(1-\frac{1}{Z_{\tilde{\sigma}}}\right)\tilde{Z}\tilde{\Phi}\tilde{\rho}_{\beta,AB}\tilde{\Phi}^{\dagger}\right\|_{1}$
$\displaystyle\quad+\frac{\tilde{\delta}\tilde{Z}}{Z_{\tilde{\sigma}}}\left\|\tilde{\Phi}\tilde{\rho}_{A_{0}A_{1}}\otimes\hat{1}_{A_{2}B_{2}}\otimes\tilde{\rho}_{B_{0}B_{1}}\tilde{\Phi}^{\dagger}\right\|_{1}$
$\displaystyle\leq\left(1-\frac{1}{Z_{\tilde{\sigma}}}\right)\left(1+\delta_{1,\ell}\right)+\frac{\tilde{\delta}\tilde{Z}}{Z_{\tilde{\sigma}}}\left\|\tilde{\Phi}\right\|^{2},$
(263)

where we use the inequality (F.1) with $\|\rho_{\beta,AB}\|_{1}=1$ in deriving
the first term of the RHS.

The remaining task is to estimate the parameters $\tilde{Z}$, $\tilde{\delta}$
and $Z_{\tilde{\sigma}}$. We can prove the following inequalities:

$\displaystyle\tilde{Z}\leq e^{4gk\beta},\quad\tilde{\delta}\leq\delta_{2,R},$
$\displaystyle\frac{1}{Z_{\tilde{\sigma}}}\leq
1+\delta_{1,\ell}+\delta_{2,R}d_{0}^{2\ell}e^{8gk\beta},$ (264)

where $\delta_{2,R}:=16C_{\beta}e^{-R/\xi_{\beta}+2\ell\log(d_{0})}$.

Proof of inequalities in (F.1). First, the first inequality in (F.1) for the
partition function $\tilde{Z}$ can be immediately derived using the
Golden–Thompson inequality:

$\displaystyle\tilde{Z}$ $\displaystyle={\rm tr}\left(e^{-\beta(H-h_{\partial
A_{1}}-h_{\partial B_{1}})}\right)$ $\displaystyle\leq{\rm tr}\left(e^{-\beta
H}e^{\beta(h_{\partial A_{1}}+h_{\partial B_{1}})}\right)$
$\displaystyle\leq{\rm tr}\left(e^{-\beta H}\right)e^{\beta(\|h_{\partial
A_{1}}\|+\|h_{\partial B_{1}}\|)}\leq e^{4gk\beta},$ (265)

where we use ${\rm tr}(e^{-\beta H})=1$, and the norm of $\|h_{\partial
A_{1}}\|+\|h_{\partial B_{1}}\|$ is upper-bounded in (F.1.1).

In addition, for $\tilde{\delta}$, we apply Lemma 24 with Theorem 10 to
$\tilde{\rho}_{A_{2}B_{2}}$, which yields the second inequality in (F.1):

$\displaystyle\tilde{\delta}$ $\displaystyle\leq
4\min(\mathcal{D}_{A_{2}},\mathcal{D}_{B_{2}})\times C_{\beta}(|\partial
A_{2}|+|\partial B_{2}|)$
$\displaystyle\quad\times\left(1+\log|A_{2}B_{2}|\right)e^{-R/\xi_{\beta}}$
$\displaystyle\leq
16C_{\beta}e^{-R/\xi_{\beta}+2\ell\log(d_{0})}=\delta_{2,R},$ (266)

where we use $|A_{2}|=|B_{2}|=\ell$, $|\partial A_{2}|=|\partial B_{2}|=2$ and
$1+\log|A_{2}B_{2}|=1+\log(2\ell)\leq d_{0}^{\ell}$ for $d_{0}\geq 2$.

Finally, From Eq. (F.1), we have

$\displaystyle Z_{\tilde{\sigma}}$ $\displaystyle={\rm
tr}\left(\tilde{Z}\tilde{\Phi}\tilde{\rho}_{\beta,AB}\tilde{\Phi}^{\dagger}+\tilde{\delta}\cdot\tilde{Z}\cdot\tilde{\Phi}\tilde{\rho}_{A_{0}A_{1}}\otimes\hat{1}_{A_{2}B_{2}}\otimes\tilde{\rho}_{B_{0}B_{1}}\tilde{\Phi}^{\dagger}\right)$
$\displaystyle\geq\left\|\rho_{\beta,AB}\right\|_{1}-\left\|\tilde{Z}\tilde{\Phi}\tilde{\rho}_{AB}\tilde{\Phi}^{\dagger}-\rho_{\beta,AB}\right\|_{1}$
$\displaystyle\quad-\tilde{\delta}\cdot\tilde{Z}\cdot\left\|\tilde{\Phi}\right\|^{2}\mathcal{D}_{A_{2}B_{2}}$
$\displaystyle\geq 1-\delta_{1,\ell}-\delta_{2,R}d_{0}^{2\ell}e^{8gk\beta},$
(267)

where in the last inequality, we use $\mathcal{D}_{A_{2}B_{2}}=d_{0}^{2\ell}$,
$\tilde{Z}\leq e^{4gk\beta}$ and $\left\|\tilde{\Phi}\right\|\leq
e^{2gk\beta}$ in (252). Using $1/(1-x)\leq 1+x$ for $x\geq 0$, we can prove
the third inequality in (F.1) from the above inequality. This completes the
proof of the inequalities (F.1). $\square$

By combining the inequalities (F.1) and (F.1), we obtain

$\displaystyle\left\|\tilde{\Phi}\tilde{\rho}_{\beta,AB}\tilde{\Phi}^{\dagger}-\tilde{\sigma}_{AB}\right\|_{1}\leq\left(\delta_{1,\ell}+\delta_{2,R}d_{0}^{2\ell}e^{8gk\beta}\right)\left(1+\delta_{1,\ell}\right)+\delta_{2,R}e^{8gk\beta}\left(1+\delta_{1,\ell}+\delta_{2,R}d_{0}^{2\ell}e^{8gk\beta}\right).$
(268)

We then apply the inequalities (F.1) and (268) to (F.1) and finally obtain

$\displaystyle\left\|\tilde{\sigma}_{AB}-\rho_{\beta,AB}\right\|_{1}$
$\displaystyle\leq\min\left[2,\left(\delta_{1,\ell}+\delta_{2,R}d_{0}^{2\ell}e^{8gk\beta}\right)\left(2+\delta_{1,\ell}+\delta_{2,R}d_{0}^{2\ell}e^{8gk\beta}\right)\right]$
$\displaystyle\leq
3\left(\delta_{1,\ell}+\delta_{2,R}d_{0}^{2\ell}e^{8gk\beta}\right),$ (269)

where we use the trivial inequality of
$\left\|\tilde{\sigma}_{AB}-\rho_{\beta,AB}\right\|_{1}\leq 2$. By choosing
$\ell=\lceil R/(16\log(d_{0})\xi_{\beta})\rceil$, we obtain

$\displaystyle\delta_{1,\ell}+\delta_{2,R}d_{0}^{2\ell}e^{8gk\beta}$
$\displaystyle=\tilde{C}_{\beta}e^{-2\ell/\xi_{\beta}+14gk\beta}+16C_{\beta}e^{-R/\xi_{\beta}+4\ell\log(d_{0})+8gk\beta}$
$\displaystyle\leq\left(\tilde{C}_{\beta}+16d_{0}^{4}C_{\beta}\right)e^{-R/[8\log(d_{0})\xi^{2}_{\beta}]+14gk\beta}=:\tilde{\delta}_{AB}$
(270)

Finally, we consider

$\displaystyle\tilde{\sigma}_{AB}^{\prime}=(1-\tilde{\delta}_{AB})\tilde{\sigma}_{AB}+\tilde{\delta}_{AB}\mathcal{D}_{AB}^{-1}\hat{1}_{AB},$
(271)

which yields
$\lambda_{\min}(\tilde{\sigma}_{AB}^{\prime})=\tilde{\delta}_{AB}\mathcal{D}_{AB}^{-1}$.
Note that $\tilde{\sigma}_{AB}^{\prime}\in{\rm PPT}$. We then obtain

$\displaystyle\left\|\tilde{\sigma}_{AB}^{\prime}-\rho_{\beta,AB}\right\|_{1}$
$\displaystyle\leq
3\tilde{\delta}_{AB}+\left\|\tilde{\sigma}_{AB}^{\prime}-\tilde{\sigma}_{AB}\right\|_{1}$
$\displaystyle\leq 5\tilde{\delta}_{AB}.$ (272)

The inequality (E.1) on the relative entropy gives

$\displaystyle S(\rho_{\beta,AB}||\tilde{\sigma}^{\prime}_{AB})$
$\displaystyle\leq
5\tilde{\delta}_{AB}\log(\mathcal{D}_{AB})-5\tilde{\delta}_{AB}\log(5\tilde{\delta}_{AB})$
$\displaystyle\quad-5\tilde{\delta}_{AB}\log[\lambda_{\min}(\tilde{\sigma}^{\prime}_{AB})]$
$\displaystyle\leq
10\tilde{\delta}_{AB}\log\left(\mathcal{D}_{AB}\tilde{\delta}_{AB}^{-1}\right)$
$\displaystyle\leq 20\sqrt{\tilde{\delta}_{AB}}\log(\mathcal{D}_{AB}),$ (273)

where we use $x\log(z/x)\leq 2\sqrt{x}\log(z)$ for $0\leq x\leq 2$ and $z\geq
2$. Because of $E_{R}^{{\rm PPT}}(\rho_{AB})\leq
S(\rho_{\beta,AB}||\tilde{\sigma}^{\prime}_{AB})$, we prove the main
inequality (246) by applying the definition of $\tilde{\delta}_{AB}$ in (F.1)
to (F.1). This completes the proof. $\square$

####  F.1.1 Proof of Lemma 26

Using the quantum belief propagation Kim (2012), we can describe $\Phi$ as
follows:

$\displaystyle\Phi:=\mathcal{T}e^{\int_{0}^{1}\phi(\tau)d\tau},$
$\displaystyle\phi(\tau):=-\frac{\beta}{2}\int_{-\infty}^{\infty}F_{\beta}(t)[h_{\partial
A_{1}}(H_{\tau},t)+h_{\partial B_{1}}(H_{\tau},t)]dt,$ $\displaystyle
H_{\tau}:=H-(1-\tau)h_{\partial A_{1}}-(1-\tau)h_{\partial B_{1}}$ (274)

where $\mathcal{T}$ is the time ordering operator, $h_{\partial
A_{1}}(H_{\tau},t)=e^{iH_{\tau}t}h_{\partial A_{1}}e^{-iH_{\tau}t}$, and we
define $F_{\beta}(t)$ as

$\displaystyle
F_{\beta}(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\tilde{F}_{\beta}(\omega)e^{-i\omega
t}d\omega,\quad\tilde{F}(\omega):=\frac{\tanh(\beta\omega/2)}{\beta\omega/2}.$

The explicit form of $f_{\beta}(t)$ can be calculated as follows (Anshu  _et
al._ , 2021, Eq. (103) in Supplementary Information):

$\displaystyle F_{\beta}(t)$
$\displaystyle=\frac{2}{\beta\pi}\log\left(\frac{e^{\pi|t|/\beta}+1}{e^{\pi|t|/\beta}-1}\right)\leq\frac{4/(\beta\pi)}{e^{\pi|t|/\beta}-1}$
$\displaystyle\leq\frac{4}{\beta\pi}e^{-\pi|t|/\beta}\left(1+\frac{1}{\pi|t|/\beta}\right),$
(275)

where we use $\log\left(\frac{e^{x}+1}{e^{x}-1}\right)\leq 2/(e^{x}-1)$ and
$1/(e^{x}-1)\leq e^{-x}(1+x^{-1})$ for $x\geq 0$.

In the following sections, we adopt the following approximation:

$\displaystyle\tilde{\Phi}:=\mathcal{T}e^{\int_{0}^{1}\tilde{\phi}(\tau)d\tau},$
$\displaystyle\tilde{\phi}(\tau):=-\frac{\beta}{2}\int_{-\infty}^{\infty}F_{\beta}(t)\bigl{[}h_{\partial
A_{1}}(H_{\tau},t,A_{1}A_{2})$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad+h_{\partial
B_{1}}(H_{\tau},t,B_{1}B_{2})\bigr{]}dt,$ (276)

where we use the notation of Eq. (16). Here, $h_{\partial
A_{1}}(H_{\tau},t,A_{1}A_{2})$ and $h_{\partial
B_{1}}(H_{\tau^{\prime}},t,B_{1}B_{2})$ are supported on $A_{1}A_{2}$ and
$B_{1}B_{2}$, respectively. Because of $[h_{\partial
A_{1}}(H_{\tau},t,A_{1}A_{2}),h_{\partial
B_{1}}(H_{\tau^{\prime}},t,B_{1}B_{2})]=0$, $\tilde{\Phi}$ is given by the
form of

$\displaystyle\tilde{\Phi}=\Phi_{A_{1},A_{2}}\otimes\Phi_{B_{1},B_{2}}.$ (277)

We here consider the norm of $\Phi-\tilde{\Phi}$, which is upper-bounded as

$\displaystyle\left\|\Phi-\tilde{\Phi}\right\|\leq
e^{\int_{0}^{1}\left(\left\|\phi(\tau)\right\|+\left\|\tilde{\phi}(\tau)\right\|\right)d\tau}\int_{0}^{1}\left\|\phi(\tau)-\tilde{\phi}(\tau)\right\|d\tau,$
(278)

where we use the analysis in Ref. (Kuwahara  _et al._ , 2021, Claim 25). To
estimate the RHS of (278), we first consider

$\displaystyle\int_{0}^{1}\left(\left\|\phi(\tau)\right\|+\left\|\tilde{\phi}(\tau)\right\|\right)d\tau$
$\displaystyle\leq\beta(\|h_{\partial A_{1}}\|+\|h_{\partial
B_{1}}\|)\int_{-\infty}^{\infty}F_{\beta}(t)dt$
$\displaystyle=\beta(\|h_{\partial A_{1}}\|+\|h_{\partial B_{1}}\|),$ (279)

where we use $\|h_{\partial A_{1}}(H_{\tau},t,A_{1}A_{2})\|\leq\|h_{\partial
A_{1}}(H_{\tau},t)\|=\|h_{\partial A_{1}}\|$ and
$\int_{-\infty}^{\infty}F_{\beta}(t)dt=\tilde{F}(0)=1$. Second, using the
Lieb-Robinson bound (18), we have

$\displaystyle\left\|h_{\partial A_{1}}(H_{\tau},t)-h_{\partial
A_{1}}(H_{\tau},t,A_{1}A_{2})\right\|$ $\displaystyle\leq\|h_{\partial
A_{1}}\|\min\left(2,2C\left(e^{v|t|}-1\right)e^{-\mu(\ell-k)}\right),$ (280)

where we use $|\partial{\rm Supp}(h_{\partial A_{1}})|=2$ on a 1D chain, and
we use the assumption that $h_{\partial A_{1}}$ has the interaction length $k$
(i.e., $|{\rm Supp}(h_{\partial A_{1}})|\leq 2k$). Note that
$\left\|h_{\partial A_{1}}(H_{\tau},t)-h_{\partial
A_{1}}(H_{\tau},t,A_{1}A_{2})\right\|$ is trivially smaller than
$2\|h_{\partial A_{1}}\|$.

By combining the above inequality with Eqs. (F.1.1) and (F.1.1), we obtain

$\displaystyle\left\|\phi(\tau)-\tilde{\phi}(\tau)\right\|\leq\frac{\beta(\|h_{\partial
A_{1}}\|+\|h_{\partial
B_{1}}\|)}{2}\int_{-\infty}^{\infty}F_{\beta}(t)\min\left(2,2C\left(e^{v|t|}-1\right)e^{-\mu(\ell-k)}\right)dt.$
(281)

Because of the form of $F_{\beta}(t)$ in Eq. (F.1.1), we can apply the same
calculations as in Sec. C.3.2. For $t_{0}=\mu\ell/(2v)$, we have

$\displaystyle\frac{\left\|\phi(\tau)-\tilde{\phi}(\tau)\right\|}{\beta(\|h_{\partial
A_{1}}\|+\|h_{\partial B_{1}}\|)}$
$\displaystyle\leq\int_{t_{0}}^{\infty}\frac{4}{\beta\pi}e^{-\pi
t/\beta}\left(1+\frac{1}{\pi t/\beta}\right)\cdot
2dt+\int_{0}^{t_{0}}\frac{4}{\beta\pi}e^{-\pi t/\beta}\left(1+\frac{1}{\pi
t/\beta}\right)\cdot 2C\left(e^{vt}-1\right)e^{-\mu(\ell-k)}dt$
$\displaystyle\leq\frac{8}{\pi^{2}}\left(1+\frac{1}{\pi
t_{0}/\beta}\right)e^{-\pi t_{0}/\beta}+\frac{8Ce^{\mu
k}}{\beta\pi}e^{-\mu\ell}\int_{0}^{t_{0}}\left(1+\frac{1}{\pi
t/\beta}\right)\left(e^{vt}-1\right)dt$
$\displaystyle\leq\frac{8}{\pi^{2}}\left(1+\frac{1}{\pi
t_{0}/\beta}\right)e^{-\pi t_{0}/\beta}+\frac{8Ce^{\mu
k}}{\beta\pi}e^{-\mu\ell}\left(\frac{e^{vt_{0}}}{v}+\frac{e^{vt_{0}}}{\pi/\beta}\right)$
$\displaystyle=\frac{8}{\pi^{2}}\left(1+\frac{2\beta
v}{\pi\mu\ell}\right)e^{-\pi\mu\ell/(2\beta v)}+\frac{8Ce^{\mu
k}}{\beta\pi}\left(\frac{1}{v}+\frac{\beta}{\pi}\right)e^{-\mu\ell/2}$
$\displaystyle\leq\left[\frac{8}{\pi^{2}}\left(1+\frac{\xi_{\beta}}{2\ell}\right)+\frac{8Ce^{\mu
k}}{\pi}\left(\frac{1}{\pi}+\frac{1}{v\beta}\right)\right]e^{-2\ell/\xi_{\beta}},$
(282)

where we use the definition of
$\xi_{\beta}:=\frac{4}{\mu}\left(1+\frac{v\beta}{\pi}\right)$. Because of
inequality (F.1.1), the LHS of (F.1.1) is trivially smaller than $1$. In
contrast, for $\ell\leq\xi_{\beta}/3$, the RHS of (F.1.1) is larger than
$20e^{-2/3}/\pi^{2}$, which is worse than the trivial upper bound. Hence, we
need consider only the case of $\ell\geq\xi_{\beta}/3$, which reduces (F.1.1)
to

$\displaystyle\frac{\left\|\phi(\tau)-\tilde{\phi}(\tau)\right\|}{\beta(\|h_{\partial
A_{1}}\|+\|h_{\partial B_{1}}\|)}$ $\displaystyle\leq\left(\frac{20+8Ce^{\mu
k}}{\pi^{2}}+\frac{8Ce^{\mu k}}{\pi v\beta}\right)e^{-2\ell/\xi_{\beta}},$
(283)

Using the upper bound of

$\displaystyle\|h_{\partial A_{1}}\|\leq\sum_{i\in{\rm Supp}(h_{\partial
A_{1}})}\sum_{Z:Z\ni i}\|h_{Z}\|$ $\displaystyle\leq|{\rm Supp}(h_{\partial
A_{1}})|g$ $\displaystyle\leq 2gk,$ (284)

we can reduce the inequalities (F.1.1) and (F.1.1) to

$\displaystyle\int_{0}^{1}\left(\left\|\phi(\tau)\right\|+\left\|\tilde{\phi}(\tau)\right\|\right)d\tau\leq
4gk\beta,$ $\displaystyle\left\|\phi(\tau)-\tilde{\phi}(\tau)\right\|$
$\displaystyle\leq 4gk\beta\left(\frac{20+8Ce^{\mu k}}{\pi^{2}}+\frac{8Ce^{\mu
k}}{\pi v\beta}\right)e^{-2\ell/\xi_{\beta}},$ (285)

respectively. By applying the above inequalities to (278), we obtain

$\displaystyle\left\|\Phi-\tilde{\Phi}\right\|\leq 16gk\beta
e^{4gk\beta}\left(\frac{5+2Ce^{\mu k}}{\pi^{2}}+\frac{2Ce^{\mu k}}{\pi
v\beta}\right)e^{-2\ell/\xi_{\beta}}.$ (286)

Therefore, by letting $\tilde{\rho}_{\beta}:=e^{-\beta(H-h_{\partial
A_{1}}-h_{\partial B_{1}})}/\tilde{Z}$ with $\tilde{Z}={\rm
tr}(e^{-\beta(H-h_{\partial A_{1}}-h_{\partial B_{1}})})$, we obtain

$\displaystyle\rho_{\beta}-\tilde{\Phi}e^{-\beta(H-h_{\partial
A_{1}}-h_{\partial B_{1}})}\tilde{\Phi}^{\dagger}$
$\displaystyle=\rho_{\beta}-\tilde{\Phi}\Phi^{-1}\rho_{\beta}(\tilde{\Phi}\Phi^{-1})^{\dagger}$
$\displaystyle=(1-\tilde{\Phi}\Phi^{-1})\rho_{\beta}\left[1-(\tilde{\Phi}\Phi^{-1})^{\dagger}\right]$
$\displaystyle+\tilde{\Phi}\Phi^{-1}\rho_{\beta}\left[1-(\tilde{\Phi}\Phi^{-1})^{\dagger}\right]+(1-\tilde{\Phi}\Phi^{-1})\rho_{\beta}(\tilde{\Phi}\Phi^{-1})^{\dagger},$

where we use Eq. (249), i.e., $e^{-\beta(H-h_{\partial A_{1}}-h_{\partial
B_{1}})}=\Phi^{-1}\rho_{\beta}(\Phi^{\dagger})^{-1}$. Using the above
equation, we have

$\displaystyle\left\|\rho_{\beta}-\left(\tilde{\Phi}e^{-\beta(H-h_{\partial
A_{1}}-h_{\partial B_{1}})}\tilde{\Phi}^{\dagger}\right)\right\|_{1}$
$\displaystyle\leq\left\|\rho_{\beta}\right\|_{1}\left\|1-\tilde{\Phi}\Phi^{-1}\right\|\left(\left\|1-\tilde{\Phi}\Phi^{-1}\right\|+2\left\|\tilde{\Phi}\Phi^{-1}\right\|\right)$
$\displaystyle\leq
3\left\|1-\tilde{\Phi}\Phi^{-1}\right\|^{2}+2\left\|1-\tilde{\Phi}\Phi^{-1}\right\|,$
(287)

where we use the triangle inequality to obtain
$\left\|\tilde{\Phi}\Phi^{-1}\right\|\leq\left\|1-\tilde{\Phi}\Phi^{-1}\right\|+1$.
From the inequality of $\|\Phi^{-1}\|\leq e^{2gk\beta}$, which is derived in
the same way as (F.1.1), we have

$\displaystyle\left\|1-\tilde{\Phi}\Phi^{-1}\right\|\leq\left\|\Phi^{-1}\right\|\cdot\left\|\Phi-\tilde{\Phi}\right\|$
$\displaystyle\leq 16gk\beta e^{6gk\beta}\left(\frac{5+2Ce^{\mu
k}}{\pi^{2}}+\frac{2Ce^{\mu k}}{\pi v\beta}\right)e^{-2\ell/\xi_{\beta}}$
$\displaystyle\leq 16e^{7gk\beta}\left(\frac{5+2Ce^{\mu
k}}{\pi^{2}}+\frac{2Ce^{\mu k}}{\pi v\beta}\right)e^{-2\ell/\xi_{\beta}}$
(288)

from the inequality (286), where we use $xe^{6x}\leq e^{7x}$ for $x\geq 0$.

Therefore, by combining the inequalities (F.1.1) and (F.1.1), we obtain the
inequality (26) as follows:

$\displaystyle\left\|\rho_{\beta}-\left(\tilde{\Phi}e^{-\beta(H-h_{\partial
A_{1}}-h_{\partial B_{1}})}\tilde{\Phi}^{\dagger}\right)\right\|_{1}$
$\displaystyle\leq 1280\left(\frac{5+2Ce^{\mu k}}{\pi^{2}}+\frac{2Ce^{\mu
k}}{\pi v\beta}\right)^{2}e^{-2\ell/\xi_{\beta}+14gk\beta}.$

Finally, on the norm $\left\|\tilde{\Phi}\right\|$, we can immediately obtain
from (F.1.1)

$\displaystyle\left\|\tilde{\Phi}\right\|\leq
e^{\int_{0}^{1}\left\|\tilde{\phi}(\tau)\right\|d\tau}\leq
e^{\frac{\beta}{2}(\|h_{\partial A_{1}}\|+\|h_{\partial B_{1}}\|)}\leq
e^{2gk\beta},$

which gives the inequality (252). This completes the proof. $\square$

##  Appendix G Remark on entanglement negativity

The PPT relative entanglement in Eq. (50) is relevant to another definition of
quantum entanglement. We here consider the entanglement negativity which is
given by Vidal and Werner (2002)

$\displaystyle E_{N}(\rho_{AB}):=\log\|\rho_{AB}^{T_{A}}\|_{1}.$ (289)

Using the Proposition 9, we can immediately obtain the following corollary:

######  Corollary 27.

Let $\rho$ be an arbitrary quantum state such that

$\displaystyle{\rm QC}_{\rho}(O_{A},O_{B})\leq\epsilon\|O_{A}\|\cdot\|O_{B}\|$
(290)

for two arbitrary operators $O_{A}$ and $O_{B}$; we then obtain

$\displaystyle E_{N}(\rho_{AB})\leq\|\rho_{AB}^{T_{A}}\|_{1}-1\leq
8\epsilon\min(\mathcal{D}_{A},\mathcal{D}_{B})\mathcal{D}_{AB},$ (291)

where the first inequality is trivially derived from $\log(1+x)\leq x$ for
$x\geq 0$. Recall that $\mathcal{D}_{AB}$ is the Hilbert space dimension in
the region $AB$. By applying Theorem 10 to the inequality (291), we can derive
a similar inequality to (55).

Proof of Corollary 27. First, because of ${\rm tr}(\rho_{AB}^{T_{A}})=1$, we
get

$\displaystyle\|\rho_{AB}^{T_{A}}\|_{1}=1+\sum_{i=1}^{M_{0}}2|\langle\eta_{i}|\rho_{AB}^{T_{A}}|\eta_{i}\rangle|$
$\displaystyle\leq 1+2\delta M_{0}$ $\displaystyle\leq
1+2\delta\mathcal{D}_{AB}$ (292)

with $\delta:=-\min_{i}\langle\eta_{i}|\rho_{AB}^{T_{A}}|\eta_{i}\rangle$,
where we use $M_{0}\leq\mathcal{D}_{AB}$. We note that the value $M_{0}$ can
be as large as $(\mathcal{D}_{A}-1)(\mathcal{D}_{B}-1)$ in general (see Ref.
Rana (2013)). Thus, using the upper bound on $\delta$ in Lemma 24, we prove
the inequality (291). This completes the proof. $\square$

In contrast, we cannot derive a similar inequality to (246) for 1D quantum
Gibbs states if we consider the entanglement negativity. This point is
explained as follows. As shown in Lemma 26, we have derived

$\displaystyle\left\|\rho_{\beta}-\tilde{\Phi}e^{-\beta(H-h_{\partial
A_{1}}-h_{\partial B_{1}})}\tilde{\Phi}\right\|_{1}\leq
e^{-\ell/\mathcal{O}(\beta)+\mathcal{O}(\beta)},$ (293)

where $\tilde{\Phi}$ has been supported on $A_{1}A_{2}\cup B_{1}B_{2}$. We
thus conclude that for $\ell\gtrsim\beta^{2}$, we have

$\displaystyle\rho_{\beta}\approx\tilde{\Phi}e^{-\beta(H-h_{\partial
A_{1}}-h_{\partial B_{1}})}\tilde{\Phi}.$ (294)

The primary difficulty here is that the entanglement negativity cannot satisfy
a convenient continuity inequality. In Ref. (Lu and Grover, 2020b, Ineq.
(16)), it has been proved that for arbitrary quantum states $\rho_{AB}$ and
$\rho^{\prime}_{AB}$

$\displaystyle|E_{N}(\rho_{AB})-E_{N}(\rho^{\prime}_{AB})|$
$\displaystyle\leq\log\left(1+\sqrt{\mathcal{D}_{AB}}\|\rho_{AB}-\rho_{AB}^{\prime}\|_{2}\right)$
$\displaystyle\leq\log\left(1+\sqrt{\mathcal{D}_{AB}}\|\rho_{AB}-\rho_{AB}^{\prime}\|_{1}\right).$
(295)

Hence, even for $\|\rho-\rho^{\prime}\|_{1}=e^{-\mathcal{O}(n^{z})}$
($0<z<1$), the difference of the entanglement negativity can be significantly
large *8*8*8For example, let us consider the case in which
$\rho_{AB}=|0^{\otimes n}\rangle\langle 0^{\otimes n}|$ with $AB=\Lambda$ and
$\rho_{AB}^{\prime}=(1-\epsilon)\rho_{AB}+\epsilon\rho_{\rm r}$. Here, the
state $\rho_{\rm r}=|\rm r\rangle\langle\rm r|$ is a random pure state on
$\Lambda$ which is orthogonal to $|0^{\otimes n}\rangle$. We then obtain
$E_{N}(\rho_{AB})=0$ and
$E_{N}(\rho_{AB})=\log\left(1-\epsilon+\epsilon\bigl{\|}\rho_{\rm
r}^{T_{A}}\bigr{\|}_{1}\right)$. Here, $\bigl{\|}\rho_{\rm
r}^{T_{A}}\bigr{\|}_{1}=e^{\Theta(n)}$ Datta (2010), and hence, as long as
$\epsilon=e^{-\mathcal{O}(n^{z})}$ ($0<z<1$), we have
$|E_{N}(\rho_{AB})-E_{N}(\rho^{\prime}_{AB})|\propto n$. . Therefore, we
cannot utilize the error estimation (293) for our purpose.

To take the same steps as in Sec. F, we need to calculate

$\left\|\left(\rho_{\beta}-\tilde{\Phi}e^{-\beta\left(H-h_{\partial
A_{1}}-h_{\partial
B_{1}}\right)}\tilde{\Phi}^{\dagger}\right)^{T_{A}}\right\|_{1}$

instead of

$\left\|\rho_{\beta}-\tilde{\Phi}e^{-\beta\left(H-h_{\partial
A_{1}}-h_{\partial B_{1}}\right)}\tilde{\Phi}\right\|_{1}$

to obtain a meaningful upper bound for the entanglement negativity. However,
in general, the partial-transpose operation can significantly increase the
operator norm, i.e.,
$\|O^{T_{A}}\|_{1}\leq\min(\mathcal{D}_{A},\mathcal{D}_{A^{\rm c}})\|O\|_{1}$
as shown in Ref. Tomiyama (1983); Ando and Sano (2008). Because of this
difficulty, it is an open question whether we can derive a similar statement
to Theorem 12 for the entanglement negativity (289). We expect that we would
be able to prove it for the entanglement negativity by employing a similar
analysis to Ref. Sherman  _et al._ (2016).

##  Appendix H Quantum Fisher information matrix

We here compare our definition (33) for the quantum correlation ${\rm
QC}_{\rho}(O_{A},O_{B})$ to the quantum Fisher information matrix. First, we
note that the quantum Fisher information can be defined in the form of the
convex roof of the variance. If $\rho$ is a pure state, the quantum Fisher
information $\mathcal{F}_{\rho}(K)$ simply reduces to the variance of $K$:

$\displaystyle\mathcal{F}_{\rho}(K)=4\left(\langle\psi|K^{2}|\psi\rangle-\langle\psi|K|\psi\rangle^{2}\right),$
(296)

where $\rho=|\psi\rangle\langle\psi|$. For general state $\rho$, the quantum
Fisher information is known to be equal to the convex roof of the variance
Tóth and Petz (2013); Yu (2013):

$\displaystyle\mathcal{F}_{\rho}(K)=4\inf_{\\{p_{s},|\psi_{s}\rangle\\}}\sum_{s}p_{s}\left(\langle\psi_{s}|K^{2}|\psi_{s}\rangle-\langle\psi_{s}|K|\psi_{s}\rangle^{2}\right),$
(297)

where minimization is taken for all possible decompositions of $\rho$ such
that $\rho=\sum_{s}p_{s}|\psi_{s}\rangle\langle\psi_{s}|$ with $p_{s}>0$. In
this way, the quantum Fisher information has a kind of similarity to the
quantum correlation ${\rm QC}_{\rho}(O_{A},O_{B})$.

To view the similarity in more detail, we consider the following quantum
Fisher information matrix Liu  _et al._ (2019):

$\displaystyle\mathcal{F}_{\rho}(O_{i},O_{j})=\sum_{s,s^{\prime}}\frac{2(\lambda_{s}-\lambda_{s^{\prime}})^{2}}{\lambda_{s}+\lambda_{s^{\prime}}}\langle\lambda_{s}|O_{i}|\lambda_{s^{\prime}}\rangle\langle\lambda_{s^{\prime}}|O_{j}|\lambda_{s}\rangle.$
(298)

We notice that

$\displaystyle\mathcal{F}_{\rho}(K)=\sum_{i,j}\mathcal{F}_{\rho}(O_{i},O_{j}).$
(299)

The quantum Fisher information matrix has been used in multiparameter quantum
estimation theory Helstrom (1969); Liu  _et al._ (2019); Gessner  _et al._
(2018); Fiderer  _et al._ (2021). Then, can we associate the quantum Fisher
information matrix with the convex roof of some observables in the analogy of
Eq. (297)?

The answer is partially yes. The quantum Fisher information matrix is relevant
to the following quantity ${\rm QC}^{\ast}_{\rho}(O_{A},O_{B})$ which is
weaker than (33):

$\displaystyle{\rm
QC}^{\ast}_{\rho}(O_{A},O_{B}):=\inf_{\\{p_{s},\rho_{s}\\}}\left|\sum_{s}p_{s}{\rm
C}_{\rho_{s}}(O_{A},O_{B})\right|,$ (300)

which is the minimization of the absolute value of the average correlation. On
the above quantity, we can prove the following statement, which is similar to
Lemma 17:

######  Lemma 28.

For two arbitrary operators $O_{A}$ and $O_{B}$, if we have

$\displaystyle[\mathcal{L}_{O_{A}},\mathcal{L}_{O_{B}}]=0$ (301)

we upper-bound the quantity ${\rm QC}^{\ast}_{\rho}(O_{A},O_{B})$ in Eq. (300)
as follows:

$\displaystyle{\rm
QC}^{\ast}_{\rho}(O_{A},O_{B})\leq\frac{1}{4}|\mathcal{F}_{\rho}(O_{A},O_{B})|.$
(302)

Here, the operator $\mathcal{L}_{O}$ has been defined in Eq. (137). If the
condition (301) holds only approximately (i.e.,
$[\mathcal{L}_{O_{A}},\mathcal{L}_{O_{B}}]\approx 0$), we need a similar
modification to Lemma 18.

Remark. On the quantity ${\rm QC}^{\ast}_{\rho}(O_{A},O_{B})$ in (300), at
first glance, we find no meaningful constraints on the entanglement structure
since ${\rm C}_{\rho_{s}}(O_{A},O_{B})$ can have a negative value. That is,
even if ${\rm QC}^{\ast}_{\rho}(O_{A},O_{B})$ is equal to zero, ${\rm
QC}_{\rho}(O_{A},O_{B})$ may be still large. Surprisingly though, we can prove
the same statement as Lemma 8 for ${\rm QC}^{\ast}_{\rho}(O_{A},O_{B})$ on the
Peres-Horodecki separability criterion (i.e., the PPT condition):

######  Lemma 29.

We prove the following satatement:

${\rm QC}^{\ast}_{\rho_{AB}}(O_{A},O_{B})=0$ for arbitrary pairs of
$O_{A},O_{B}$ $\displaystyle\longrightarrow\textrm{$\rho_{AB}$ satisfies the
PPT condition}.$ (303)

From the statement (29) and the inequality (302), one can see that the quantum
Fisher information matrix also plays some roles in quantum correlation
measures.

###  H.1 Proof of Lemma 28

We here prove Lemma 28, which implies

$\displaystyle{\rm
QC}^{\ast}_{\rho}(O_{A},O_{B})\leq\frac{1}{4}|\mathcal{F}_{\rho}(O_{A},O_{B})|.$
(304)

with

$\displaystyle{\rm
QC}^{\ast}_{\rho}(O_{A},O_{B}):=\inf_{\\{p_{s},\rho_{s}\\}}\left|\sum_{s}p_{s}{\rm
C}_{\rho_{s}}(O_{A},O_{B})\right|,$ (305)

where $[\mathcal{L}_{O_{A}},\mathcal{L}_{O_{B}}]=0$ is assumed. Note that the
operator $\mathcal{L}_{O}$ has been defined in Eq. (137).

####  H.1.1 Proof

We follow the proof of Lemma 17. We consider the decomposition of $\rho$ as
follows:

$\displaystyle\rho=\sum_{m}p_{m}|\phi_{m}\rangle\langle\phi_{m}|.$
$\displaystyle|\phi_{m}\rangle=\frac{1}{\sqrt{p_{m}}}\sqrt{\rho}|\psi_{m}\rangle,\quad
p_{m}=\langle\psi_{m}|\rho|\psi_{m}\rangle,$ (306)

and choose $|\psi\rangle_{m}$ as the simultaneous eigenstates of
$\mathcal{L}_{O_{A}}$ and $\mathcal{L}_{O_{B}}$ with the corresponding
eigenvalues $\alpha_{1,m}$ and $\alpha_{2,m}$, respectively. We then reach the
same equation as (161):

$\displaystyle\langle\phi_{m}|O_{A}|\phi_{m}\rangle\langle\phi_{m}|O_{B}|\phi_{m}\rangle=\alpha_{1,m}\alpha_{2,m}.$
(307)

We next prove that

$\displaystyle\sum_{m}p_{m}\langle\phi_{m}|O_{A}|\phi_{m}\rangle\langle\phi_{m}|O_{B}|\phi_{m}\rangle=\frac{1}{2}{\rm
tr}(\\{\rho,\mathcal{L}_{O_{A}}\mathcal{L}_{O_{B}}\\}),$ (308)

where $\\{\cdot,\cdot\\}$ is the anticommutator. By expanding the RHS in Eq.
(308), we get

$\displaystyle\frac{1}{2}{\rm
tr}(\\{\rho,\mathcal{L}_{O_{A}}\mathcal{L}_{O_{B}}\\})$
$\displaystyle=\frac{1}{2}\sum_{m}\langle\psi_{m}|\\{\rho,\mathcal{L}_{O_{A}}\mathcal{L}_{O_{B}}\\}|\psi_{m}\rangle$
$\displaystyle=\sum_{m}\langle\psi_{m}|\rho|\psi_{m}\rangle\alpha_{1,m}\alpha_{2,m},$
(309)

which reduces to the LHS in Eq. (308) from
$p_{m}=\langle\psi_{m}|\rho|\psi_{m}\rangle$ and Eq. (307)

In contrast, using the spectral decomposition of
$\rho=\sum_{s}\lambda_{s}|\lambda_{s}\rangle\langle\lambda_{s}|$, we have

$\displaystyle\frac{1}{2}{\rm
tr}(\\{\rho,\mathcal{L}_{O_{A}}\mathcal{L}_{O_{B}}\\})$
$\displaystyle=\sum_{s,s^{\prime}}\frac{2\lambda_{s}\lambda_{s^{\prime}}}{\lambda_{s}+\lambda_{s^{\prime}}}\langle\lambda_{s}|O_{A}|\lambda_{s^{\prime}}\rangle\langle\lambda_{s^{\prime}}|O_{B}|\lambda_{s}\rangle,$
(310)

where we use the form of $\mathcal{L}_{O}$ in Eq. (137). By combining Eqs
(308) and (H.1.1), we have

$\displaystyle\sum_{m}p_{m}\langle\phi_{m}|O_{A}|\phi_{m}\rangle\langle\phi_{m}|O_{B}|\phi_{m}\rangle$
$\displaystyle=\sum_{s,s^{\prime}}\frac{2\lambda_{s}\lambda_{s^{\prime}}}{\lambda_{s}+\lambda_{s^{\prime}}}\langle\lambda_{s}|O_{A}|\lambda_{s^{\prime}}\rangle\langle\lambda_{s^{\prime}}|O_{B}|\lambda_{s}\rangle.$
(311)

Finally, we have

$\displaystyle\sum_{m}p_{m}\langle\phi_{m}|O_{A}O_{B}|\phi_{m}\rangle={\rm
tr}(\rho O_{A}O_{B})$
$\displaystyle=\sum_{s,s^{\prime}}\frac{\lambda_{s}+\lambda_{s^{\prime}}}{2}\langle\lambda_{s}|O_{A}|\lambda_{s^{\prime}}\rangle\langle\lambda_{s^{\prime}}|O_{B}|\lambda_{s}\rangle,$
(312)

where we use $[O_{A},O_{B}]=0$. By subtracting Eq. (H.1.1) from Eq. (H.1.1),
we obtain

$\displaystyle\sum_{m}p_{m}\left(\langle\phi_{m}|O_{A}O_{B}|\phi_{m}\rangle-\langle\phi_{m}|O_{A}|\phi_{m}\rangle\langle\phi_{m}|O_{B}|\phi_{m}\rangle\right)$
$\displaystyle=\sum_{s,s^{\prime}}\frac{(\lambda_{s}-\lambda_{s^{\prime}})^{2}}{2(\lambda_{s}+\lambda_{s^{\prime}})}\langle\lambda_{s}|O_{A}|\lambda_{s^{\prime}}\rangle\langle\lambda_{s^{\prime}}|O_{B}|\lambda_{s}\rangle$
$\displaystyle=\frac{1}{4}\mathcal{F}_{\rho}(O_{i},O_{j}).$ (313)

By applying the above equation to (300), we prove the inequality (302). This
completes the proof. $\square$

###  H.2 Proof of Lemma 29

We here prove the statement of

${\rm QC}^{\ast}_{\rho_{AB}}(O_{A},O_{B})=0$ for arbitrary pairs of
$O_{A},O_{B}$ $\displaystyle\longrightarrow\textrm{$\rho_{AB}$ satisfies the
PPT condition}.$ (314)

This statement can be easily checked by the following discussion.

First, if we can prove the inequality (9) in Proposition 9 by assuming the
inequality (51) for ${\rm QC}^{\ast}_{\rho}(O_{A},O_{B})$ instead of ${\rm
QC}_{\rho}(O_{A},O_{B})$, we obtain the statement (H.2). Second, in the proof
of Proposition 9, the inequality (51) is used only in deriving the upper bound
(E.2.1) for the proof of Lemma 25. From the second line to the third line in
(E.2.1), we use ${\rm QC}_{\rho}(O_{A},O_{B})$ as an upper bound for

$\left|\sum_{s}p_{s}\left({\rm tr}(\rho_{s,A}\Phi_{A}\Phi_{B})\right)-{\rm
tr}(\rho_{s,A}\Phi_{A}){\rm tr}(\rho_{s,B}\Phi_{B})\right|;$

however, in fact, ${\rm QC}^{\ast}_{\rho}(O_{A},O_{B})$ also serves as the
upper bound for the above quantity. This allows us to prove the inequality (9)
only from the constraint on ${\rm QC}^{\ast}_{\rho}(O_{A},O_{B})$. This
completes the proof. $\square$

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