††thanks: These two authors contributed equally. Part of this work was done
when they were at Baidu Research.††thanks: These two authors contributed
equally. Part of this work was done when they were at Baidu Research.

# Estimate distillable entanglement and quantum capacity by squeezing useless
entanglement

Chengkai Zhu  Chenghong Zhu  Xin Wang<EMAIL_ADDRESS>Institute
for Quantum Computing, Baidu Research, Beijing, China.  Thrust of Artificial
Intelligence, Information Hub, Hong Kong University of Science and Technology
(Guangzhou), Nansha, China

###### Abstract

Quantum Internet relies on quantum entanglement as a fundamental resource for
secure and efficient quantum communication, reshaping data transmission. In
this context, entanglement distillation emerges as a crucial process that
plays a pivotal role in realizing the full potential of the quantum internet.
Nevertheless, it remains challenging to accurately estimate the distillable
entanglement and its closely related essential quantity, the quantum capacity.
In this work, we consider a general resource measure known as the reverse
divergence of resources which quantifies the minimum divergence between a
target state and the set of free states. Leveraging this measure, we propose
methods for evaluating both quantities by squeezing out useless entanglement
within a state or a quantum channel, whose contributions are expected to be
ignored for the distillable entanglement or the quantum capacity,
respectively. Our method has practical applications for purifying maximally
entangled states under practical noises, such as depolarizing and amplitude
damping noises, leading to improvements in estimating the one-way distillable
entanglement. Furthermore, we provide valuable benchmarks for evaluating the
quantum capacities of qubit quantum channels, including the Pauli channels and
the random mixed unitary channels.

###### Contents

  1. I Introduction
    1. I.1 Background
    2. I.2 Main contributions
  2. II Reverse divergence of resources
    1. II.1 Preliminaries
    2. II.2 Reverse divergence of resources
  3. III Applications on distillable entanglement
    1. III.1 Upper bound on the one-way distillable entanglement
    2. III.2 Continuity bounds of the one-way distillable entanglement
    3. III.3 Examples of less-entangled states
    4. III.4 Extending the method to the two-way distillable entanglement
  4. IV Applications on quantum channel capacity
    1. IV.1 Quantum capacity of qubit channels
  5. V Concluding remarks
  6. A Dual SDP for ${\cal R}_{\max,{\rm ADG}}(\rho_{AB})$ and $\widetilde{{\cal R}}_{\max,{\rm ADG}}({\cal N})$
  7. B Two-way distillable entanglement
  8. C Proof of Proposition 3 and Proposition 4
  9. D Proof of Proposition 9

##  I Introduction

###  I.1 Background

Quantum Internet [1, 2, 3, 4], a transformative paradigm in data transmission,
leverages the extraordinary properties of quantum entanglement to
revolutionize communication. At the heart of this groundbreaking innovation
lies quantum entanglement, which is the most nonclassical manifestation of
quantum mechanics and serves as the cornerstone resource for quantum
information processing [5, 6, 7, 8]. Simultaneously, quantum communication
emerges as a unique task, utilizing the power of entanglement for secure and
efficient information transfer [7]. To fully unlock the potential of quantum
technologies and achieve a secure, efficient, and globally connected quantum
internet, it is desirable to push the boundaries of our understanding and
capabilities in entanglement and quantum communication theories.

Typically, the entanglement resource is assumed to be copies of ideal
maximally entangled states. In practical scenarios, noise inevitably occurs,
resulting in mixed entangled states that require distillation or purification.
A natural question is how to obtain the maximally entangled states from a
source of less entangled states using well-motivated operations, known as the
entanglement distillation. One fundamental measure for characterizing the
entanglement distillation is the one-way distillable entanglement [9], denoted
by $E_{D,\to}$, which is also one of the most important entanglement measures
motivated by operational tasks. It captures the highest rate at which one can
obtain the maximally entangled states from less entangled states by one-way
local operations and classical communication (LOCC):

$E_{D,\to}(\rho_{AB})=\sup\\{r\mathrel{\mathop{\mathchar
58\relax}}\lim_{n\to\infty}[\inf_{\Lambda}\|\Lambda(\rho_{AB}^{\otimes
n})-\Phi(2^{rn})\|_{1}]=0\\},$ (1)

where $\Lambda$ ranges over one-way LOCC operations and
$\Phi(d)=1/d\sum_{i,j=1}^{d}|ii\rangle\\!\langle jj|$ is the standard
$d\otimes d$ maximally entangled state. Likewise, the two-way distillable
entanglement $E_{D,\leftrightarrow}(\rho_{AB})$ is defined by the supremum
over all achievable rates under two-way LOCC. We have for all bipartite states
$\rho_{AB}$ that $E_{D,\to}(\rho_{AB})\leq E_{D,\leftrightarrow}(\rho_{AB})$.
Notably, the distillable entanglement is closely connected to the fundamental
notion of quantum capacity in quantum communication [10], which is central to
quantum Shannon theory. Consider modeling the noise in transmitting quantum
information from Alice to Bob as a quantum channel ${\cal N}_{A\to B}$. The
quantum capacity $Q({\cal N}_{A\to B})$ is the maximal achievable rate at
which Alice can reliably transmit quantum information to Bob by asymptotically
many uses of the channel. By the state–channel duality, if the distillation
protocol of the Choi state [11] $J_{AB}^{\cal N}$ of ${\cal N}_{A\to B}$
yields the maximally entangled states at a positive rate, then Alice may apply
the standard teleportation scheme to send arbitrary quantum states to Bob at
the same rate. Thus, one has $Q({\cal N}_{A\to B})\geq E_{D,\to}(J_{AB}^{\cal
N})$ since classical forward communication in teleportation does not affect
the channel capacity. For the teleportation-simulable channels [12, 5, 13],
the equality here holds.

Despite many efforts that have been made in the past two decades, computing
$E_{D,\to}(\cdot)$ and $Q(\cdot)$ still generally remains a challenging task.
Even for the qubit isotropic states and the depolarizing channels, it remains
unsolved. Therefore, numerous studies try to estimate them by deriving lower
and upper bounds (see, e.g., [9, 14, 15, 16, 17, 18, 19] for the distillable
entanglement, e.g., [20, 21, 22, 23] for the quantum capacity). For the
distillable entanglement, a well-known lower bound dubbed Hashing bound is
established by Devetak and Winter [9], i.e., $E_{D,\to}(\rho_{AB})\geq
I_{\textup{\tiny C}}(A\rangle B)_{\rho}$, where $I_{\textup{\tiny C}}(A\rangle
B)_{\rho}$ is the coherent information of the bipartite quantum state
$\rho_{AB}$. Considering upper bounds, the Rains bound [14] is arguably the
best-known efficiently computable bound for the two-way distillable
entanglement of general states. Recent works [17, 19] utilize the techniques
that involve constructing meaningful extended states to find upper bounds. For
quantum capacity, many useful upper bounds for general quantum channels are
studied for benchmarking arbitrary quantum noise [24, 25, 26, 27, 28, 29, 30,
31]. When considering some specific classes of quantum channels, useful upper
bounds are also developed to help us better understand quantum communication
via these channels [32, 22, 33, 34, 19, 23, 21].

In specific, due to the regularization in the characterizations of
$E_{D,\to}(\cdot)$ and $Q(\cdot)$, one main strategy to establish efficiently
computable upper bounds on them is to develop single-letter formulae. For
example, one common approach is to decompose a state (resp. a quantum channel)
into degradable parts and anti-degradable parts [22], or use approximate
degradability (anti-degradability) [25]. Another recent fruitful technique
called flag extension optimization [23, 19, 21] relies on finding a degradable
extension of the state or the quantum channel. However, the performance of
these methods is limited by the absence of a good decomposition strategy. It
is unknown how to partition a general state or quantum channel to add flags or
how to construct a proper and meaningful convex decomposition on them. Thus,
the flag extension optimization is only effective for the states and channels
with symmetry or known structures.

###  I.2 Main contributions

This work considers a family of resource measures called reverse divergence of
resources. With a specific construction, we define the reverse max-relative
entropy of entanglement for quantum states, which has applications for
estimating the distillable entanglement. In the meantime, we introduce reverse
max-relative entropy of anti-degradability for quantum channels as a
generalization of the concept of that for states, which can be applied to
bound the quantum capacity. All these bounds can be efficiently computed via
semidefinite programming [35]. Furthermore, drawing on the idea of [25], we
thoroughly analyze different continuity bounds on the one-way distillable
entanglement of a state in terms of its anti-degradability. Finally, we
investigate the distillation of the maximally entangled states under practical
noises and focus on the quantum capacity of qubit channels. We show that the
bound obtained by the reverse max-relative entropy of entanglement outperforms
other known bounds in a high-noise region, including the Rains bound and the
above continuity bounds. The upper bound offered by the reverse max-relative
entropy of anti-degradability also provides an alternative interpretation of
the no-cloning bound of the Pauli channel [32], and notably outperforms the
continuity bounds on random unital qubit channels.

This paper is structured as follows. Section II.1 provides preliminaries used
throughout the paper. Section II.2 introduces a family of resource measures
called the reverse divergence of resources. Section III shows the application
of this concept on upper bounding the distillable entanglement, with practical
examples provided. Section IV demonstrates the application of the method in
deriving upper bounds on quantum capacity, including analytical results for
Pauli channels and comparison with continuity bounds in subsection IV.1 for
random mixed unitary channels. The paper concludes with a summary and outlooks
for future research in section V.

##  II Reverse divergence of resources

###  II.1 Preliminaries

Let ${\cal H}$ be a finite-dimensional Hilbert space, and ${\cal L}({\cal H})$
be the set of linear operators acting on it. We consider two parties Alice and
Bob with Hilbert space $\mathcal{H}_{A},\mathcal{H}_{B}$, whose dimensions are
$d_{A},d_{B}$, respectively. A linear operator $\rho\in{\cal L}({\cal H})$ is
called a density operator if it is Hermitian and positive semidefinite with
trace one. We denote the trace norm of $\rho$ as
$\|\rho\|_{1}=\operatorname{Tr}\sqrt{\rho^{\dagger}\rho}$ and let ${\cal
D}({\cal H})$ denote the set of density operators. We call a linear map CPTP
if it is both completely positive and trace-preserving. A CPTP map that
transforms linear operators in ${\cal L}({\cal H}_{A})$ to linear operators in
${\cal L}({\cal H}_{B})$ is also called a quantum channel, denoted as ${\cal
N}_{A\to B}$. For a quantum channel ${\cal N}_{A\to B}$, its Choi-Jamiołkowski
state is given by $J_{AB}^{{\cal
N}}\equiv\sum_{i,j=0}^{d_{A}-1}|i\rangle\\!\langle j|\otimes\mathcal{N}_{A\to
B}(|i\rangle\\!\langle j|)$, where $\\{|i\rangle\\}_{i=0}^{d_{A}-1}$ is an
orthonormal basis in ${\cal H}_{A}$. The von Neumann entropy of a state
$\rho_{A}$ is $S(A)_{\rho}\mathrel{\mathop{\mathchar
58\relax}}=-\operatorname{Tr}(\rho_{A}\log\rho_{A})$ and the coherent
information of a bipartite state $\rho_{AB}$ is defined by $I_{c}(A\rangle
B)_{\rho}\mathrel{\mathop{\mathchar 58\relax}}=S(B)_{\rho}-S(AB)_{\rho}$. The
entanglement of formation of a state $\rho_{AB}$ is given by

$E_{F}(\rho_{AB})=\min_{\\{p_{i},|\phi_{i}\rangle\\}}\sum_{i}p_{i}S(A)_{\phi_{i}},$
(2)

where $\rho_{AB}=\sum_{i}p_{i}|\phi_{i}\rangle\\!\langle\phi_{i}|_{AB}$ and
the minimization ranges over all pure state decomposition of $\rho_{AB}$. We
introduce the generalized divergence $\bm{D}(\rho_{A}\|\sigma_{A})$ as a map
$\bm{D}\mathrel{\mathop{\mathchar 58\relax}}{\cal D}({\cal H}_{A})\times{\cal
D}({\cal H}_{A})\mapsto\mathbb{R}\cup\\{+\infty\\}$ that obeys:

  1. 1.

Faithfulness: $\bm{D}(\rho_{A}\|\sigma_{A})=0$ iff $\rho_{A}=\sigma_{A}$.

  2. 2.

Data processing inequality:
$\bm{D}(\rho_{A}\|\sigma_{A})\geq\bm{D}[\mathcal{N}_{A\to
A^{\prime}}(\rho_{A})\|\mathcal{N}_{A\to A^{\prime}}(\sigma_{A})]$, where
$\mathcal{N}_{A\to A^{\prime}}$ is an arbitrary quantum channel.

The generalized divergence is intuitively some measure of distinguishability
of the two states, e.g., Bures metric, quantum relative entropy. Another
example of interest is the sandwiched Rényi relative entropy [36, 37] of
$\rho,\sigma$ that is defined by

$D_{\alpha}(\rho\|\sigma)\mathrel{\mathop{\mathchar
58\relax}}=\frac{1}{\alpha-1}\log\operatorname{Tr}\left[\left(\sigma^{\frac{1-\alpha}{2\alpha}}\rho\sigma^{\frac{1-\alpha}{2\alpha}}\right)^{\alpha}\right],$
(3)

if $\operatorname{supp}(\rho)\subset\operatorname{supp}(\sigma)$ and it is
equal to $+\infty$ otherwise, where $\alpha\in(0,1)\cup(1,\infty)$. In the
case that $\alpha\to\infty$, one can find the max-relative entropy [38] of
$\rho$ with respect to $\sigma$ by

$D_{\max}(\rho||\sigma)=\inf\\{\lambda\in\mathbb{R}\mathrel{\mathop{\mathchar
58\relax}}\rho\leq 2^{\lambda}\sigma\\}.$ (4)

###  II.2 Reverse divergence of resources

In the usual framework of quantum resource theories [39], there are two main
notions: i) subset ${\cal F}$ of free states, i.e., the states that do not
possess the given resource; ii) subset ${\cal O}$ of free operations, i.e.,
the quantum channels that are unable to generate the resource. Meanwhile, two
axioms for a quantity being a resource measure ${\cal R}(\cdot)$ are
essential:

  * 1).

Vanishing for free states: $\rho\in{\cal F}\Rightarrow{\cal R}(\rho)=0$.

  * 2).

Monotonicity: ${\cal R}({\cal O}(\rho))\leq{\cal R}(\rho)$ for any free
operation ${\cal O}$. ${\cal R}(\cdot)$ is called a resource monotone.

Let us define a family of resource measures called reverse divergence of
resources:

${\cal R}_{\mathcal{F}}(\rho_{AB})\mathrel{\mathop{\mathchar
58\relax}}=\min_{\tau\in\mathcal{F}}\bm{D}(\tau_{AB}||\rho_{AB}),$ (5)

where $\mathcal{F}$ is some set of free states. By the definition of the
reverse divergence of resources in Eq. (5), one can easily check it satisfies
condition 1). Whenever the free state set ${\cal F}$ is closed, by the data-
processing inequality of $\bm{D}(\cdot\|\cdot)$, condition 2) will be
satisfied. Thus ${\cal R}_{\mathcal{F}}(\cdot)$ is a resource measure.
Specifically, in the resource theory of entanglement, some representative free
state sets are the separable states (SEP) and the states having a positive
partial transpose (PPT). Examples of free sets of operations are LOCC and PPT.
We note that the "reverse" here means minimizing the divergence over a free
state set in the first coordinate, rather than the second one which has helped
one define the relative entropy of entanglement [40, 41] and the max-relative
entropy of entanglement [38]. For some divergence of particular interest,
e.g., the quantum relative entropy $D(\cdot||\cdot)$, relevant discussion of
the coordinate choices can be traced back to work in [41]. In [42], the
authors further studied properties of the quantity $\min_{\tau\in{\cal
F}}D(\tau||\rho)$. Here, we try to further investigate meaningful applications
of some reverse divergence of resources.

In the following, we consider the generalized divergence as the max-relative
entropy and study a measure called reverse max-relative entropy of resources,

${\cal R}_{\max,\mathcal{F}}(\rho_{AB})\mathrel{\mathop{\mathchar
58\relax}}=\min_{\tau\in\mathcal{F}}D_{\max}(\tau_{AB}||\rho_{AB}),$ (6)

where $\mathcal{F}$ is some set of free states. If there is no such state
$\tau_{AB}\in{\cal F}$ that satisfies $\tau_{AB}\leq 2^{\lambda}\rho_{AB}$ for
any $\lambda\in\mathbb{R}$, ${\cal R}_{\max,\mathcal{F}}(\rho_{AB})$ is set to
be 0. This measure bears many nice properties. First, it can be efficiently
computed via semidefinite programming (SDP) in many cases. Second, Eq. (6)
gives the closest free state $\tau_{AB}\in{\cal F}$ to $\rho_{AB}$, w.r.t. the
max-relative entropy. Third, ${\cal R}_{\max,\mathcal{F}}(\cdot)$ is
subadditive w.r.t the tensor product of states. In fact, ${\cal
R}_{\max,\mathcal{F}}(\rho_{AB})$ is closely related to the weight of resource
$W(\rho_{AB})$ [43, 44, 45, 46] and the free component $\Gamma(\rho_{AB})$
[47], both of which have fruitful properties and applications [48, 49, 50], as
follows

$2^{-{\cal
R}_{\max,\mathcal{F}}(\rho_{AB})}=1-W(\rho_{AB})=\Gamma(\rho_{AB}).$ (7)

We note that each part of Eq. (7) quantifies the largest weight where a free
state can take in a convex decomposition of $\rho_{AB}$. When moving on to
operational tasks that the free state can be ignored, what is left in a convex
decomposition becomes our main concern. Optimization of the weight in the
decomposition can be visualized as squeezing out all free parts of the given
state. Thus, we further introduce the ${\cal F}$-squeezed state of $\rho_{AB}$
as follows.

###### Definition 1

For a bipartite quantum state $\rho_{AB}$ and a free state set ${\cal F}$, If
${\cal R}_{\max,\mathcal{F}}(\rho_{AB})$ is non-zero, the ${\cal F}$-squeezed
state of $\rho_{AB}$ is defined by

$\omega_{AB}=\frac{\rho_{AB}-2^{-{\cal
R}_{\max,\mathcal{F}}(\rho_{AB})}\cdot\tau_{AB}}{1-2^{-{\cal
R}_{\max,\mathcal{F}}(\rho_{AB})}},$ (8)

where $\tau_{AB}$ is the closest free state to $\rho_{AB}$ in terms of the
max-relative entropy, i.e., the optimal solution in Eq. (6). If ${\cal
R}_{\max,\mathcal{F}}(\rho_{AB})=0$, the ${\cal F}$-squeezed state of
$\rho_{AB}$ is itself.

In the following sections, we will illustrate the applications of the reverse
max-relative entropy of resources as well as the squeezing idea on example
tasks. One is giving upper bounds on the distillable entanglement of arbitrary
quantum states. The other is to derive upper bounds on the quantum capacity of
channels.

##  III Applications on distillable entanglement

In this section, we investigate the information-theoretic application of the
reverse max-relative entropy of resources in deriving efficiently computable
upper bounds on the distillable entanglement. To showcase the advantage of our
bounds, we compare the results with different continuity bounds and the Rains
bound on the maximally entangled states with practical noises.

###  III.1 Upper bound on the one-way distillable entanglement

Recall that the one-way distillable entanglement has a regularized formula
[9]:

$\displaystyle
E_{D,\to}(\rho_{AB})=\lim_{n\rightarrow\infty}\frac{1}{n}E^{(1)}_{D,\to}(\rho_{AB}^{\otimes
n}),$ (9)

where $E^{(1)}_{D,\to}(\rho_{AB})\mathrel{\mathop{\mathchar
58\relax}}=\max_{T}I_{\textup{\tiny C}}(A^{\prime}\rangle MB)_{T(\rho_{AB})}$,
and the maximization ranges over all quantum instruments
$T\mathrel{\mathop{\mathchar 58\relax}}A\to A^{\prime}M$ on Alice’s system.
The regularization in Eq. (9) for $E_{D,\to}(\rho_{AB})$ is intractable to
compute in most cases. However, there are some categories of states whose
$E_{D,\to}$ can be reduced to single-letter formulae. Two important classes
are called degradable states and anti-degradable states.

Let $\rho_{AB}$ be a bipartite state with purification $|\phi\rangle_{ABE}$.
$\rho_{AB}$ is called degradable if there exists a CPTP map
$\mathcal{M}_{B\rightarrow E}$ such that $\mathcal{M}_{B\rightarrow
E}(\rho_{AB})=\operatorname{Tr}_{B}(\phi_{ABE})$, and is called anti-
degradable if there exists a CPTP map $\mathcal{M}_{E\rightarrow B}$ such that
$\mathcal{M}_{E\rightarrow B}(\rho_{AE})=\operatorname{Tr}_{E}(\phi_{ABE})$.
Equivalently, a state is anti-degradable if and only if it has a symmetric
extension [51], thus is also called a two-extendible state. For the degradable
states, it is shown that [17]

$\displaystyle E^{(1)}_{D,\to}(\rho_{AB}^{\otimes
n})=nE^{(1)}_{D,\to}(\rho_{AB})=nI_{\textup{\tiny C}}(A\rangle
B)_{\rho},\forall n\in\mathbb{N},$ (10)

resulting in $E_{D,\to}(\rho_{AB})=I_{\textup{\tiny C}}(A\rangle B)_{\rho}$.
For the anti-degradable states, consisting of a compact and convex set, it
always holds

$E^{(1)}_{D,\to}(\rho_{AB})=E_{D,\to}(\rho_{AB})=0.$ (11)

Moreover, $E_{D,\to}(\cdot)$ is convex on decomposing a state into degradable
and anti-degradable parts [17]. To better exploit this convexity, we introduce
the reverse max-relative entropy of unextendible entanglement to help identify
the anti-degradable (two-extendible)baobao part of a given bipartite state
$\rho_{AB}$:

${\cal R}_{\max,{\rm ADG}}(\rho_{AB})\mathrel{\mathop{\mathchar
58\relax}}=\min_{\tau\in{\rm ADG}}D_{\max}(\tau_{AB}||\rho_{AB}),$ (12)

where ${\rm ADG}$ is the set of all anti-degradable states. In this resource
theory, the resource we consider is the extendible entanglement, and the free
states are bipartite states that are possibly shareable between $A$ and a
third party $E$, where $E$ is isomorphic to $B$. Notably, the extendibility of
entanglement is a key property in entanglement theory with many existing
applications [18, 52]. Here, combined with the idea of entanglement of
formation, ${\cal R}_{\max,{\rm ADG}}(\rho_{AB})$ can be applied to derive an
upper bound on the one-way distillable entanglement of an arbitrary state
$\rho_{AB}$ as shown in Theorem 1.

Fig 1: Illustration of the ${\cal F}$-squeezed state of $\rho_{AB}$ and
Theorem 1. The brown region corresponds to the sub-state $\tau_{AB}$ belonging
to the free state set ${\cal F}$, which is squeezed out via a convex
decomposition of $\rho_{AB}$. $\omega_{AB}$ is the ${\cal F}$-squeezed state.

###### Theorem 1

For any bipartite state $\rho_{AB}$, it satisfies

$E_{D,\to}(\rho_{AB})\leq E_{\rm rev}^{u}(\rho_{AB})\mathrel{\mathop{\mathchar
58\relax}}=[1-2^{-{\cal R}_{\max,{\rm ADG}}(\rho_{AB})}]\cdot
E_{F}(\omega_{AB}),$ (13)

where $\omega_{AB}$ is the ${\rm ADG}$-squeezed state of $\rho_{AB}$, ${\cal
R}_{\max,{\rm ADG}}(\cdot)$ is the reverse max-relative entropy of
unextendible entanglement, and $E_{F}(\cdot)$ is the entanglement of
formation.

###### Proof.

Suppose the ADG-squeezed state of $\rho_{AB}$ is $\omega_{AB}$ and the optimal
solution in Eq. (6) for $\rho_{AB}$ is $\tau_{AB}$. It follows
$\rho_{AB}=[1-2^{-{\cal R}_{\max,{\rm ADG}}(\rho_{AB})}]\omega_{AB}+2^{-{\cal
R}_{\max,{\rm ADG}}(\rho_{AB})}\tau_{AB}$, where $\tau_{AB}$ is anti-
degradable. Since any pure state is degradable, we can decompose $\rho_{AB}$
into degradable and anti-degradable parts as $\rho_{AB}=[1-2^{-{\cal
R}_{\max,{\rm
ADG}}(\rho_{AB})}]\sum_{i}p_{i}|\omega_{i}\rangle\\!\langle\omega_{i}|_{AB}+2^{-{\cal
R}_{\max,{\rm ADG}}(\rho_{AB})}\tau_{AB}$, where
$\omega_{AB}=\sum_{i}p_{i}|\omega_{i}\rangle\\!\langle\omega_{i}|_{AB}$ is any
pure state decomposition of $\omega_{AB}$. Based on the essential convexity of
$E_{D,\to}(\cdot)$ on decomposing a state into degradable and anti-degradable
parts [17], we have

$\displaystyle E_{D,\to}(\rho_{AB})$ $\displaystyle\leq[1-2^{-{\cal
R}_{\max,{\rm
ADG}}(\rho_{AB})}]\cdot\sum_{i}p_{i}E_{D,\to}(\omega_{i})+2^{-{\cal
R}_{\max,{\rm ADG}}(\rho_{AB})}E_{D,\to}(\tau_{AB})$ (14)
$\displaystyle\leq[1-2^{-{\cal R}_{\max,{\rm
ADG}}(\rho_{AB})}]\cdot\sum_{i}p_{i}I_{c}(A\rangle B)_{\omega_{i}},$

where the second inequality is due to the properties in Eq. (10) and Eq. (11)
for degradable states and anti-degradable states, respectively. After taking
the minimization over all possible decomposition of $\omega_{AB}$, we arrived
at $E_{D,\to}(\rho_{AB})\leq[1-2^{-{\cal R}_{\max,{\rm ADG}}(\rho_{AB})}]\cdot
E_{F}(\omega_{AB})$.  $\sqcap$$\sqcup$

Remark 1 The bound $E_{\rm rev}^{u}(\rho_{AB})$ has a cartoon illustration as
shown in Fig. 1. The main insight of it is to squeeze out as much of the free
or useless part, the anti-degradable state here, as possible. We point out
that squeezing all useless parts out does not necessarily give the tightest
upper bound in terms of the one-way distillable entanglement, e.g., isotropic
state [17]. Instead of squeezing out all the useless parts, there may be room
for exploring more appropriate partitions when we try to decompose a specific
quantum state. However, the approach we present in Theorem 1 is an invaluable
method for general states as shown in subsection III.3 and can be seen as a
variant of the continuity bound in terms of the anti-degradability of the
state.

###### Corollary 2

For any bipartite state $\rho_{AB}$, it satisfies

$E_{D,\to}(\rho_{AB})\leq\hat{E}_{\rm
rev}^{u}(\rho_{AB})\mathrel{\mathop{\mathchar 58\relax}}=[1-2^{-{\cal
R}_{\max,{\rm ADG}}(\rho_{AB})}]\cdot\sum_{i}\lambda_{i}S(B)_{\psi_{i}},$ (15)

where $\omega_{AB}=\sum_{i}\lambda_{i}|\psi_{i}\rangle\\!\langle\psi_{i}|$ is
the spectral decomposition of the ADG-squeezed state $\omega_{AB}$ of
$\rho_{AB}$.

Corollary 2 is followed by the fact that $E_{F}(\omega_{AB})$ has a trivial
upper bound as $E_{F}(\omega_{AB})\leq\sum_{i}\lambda_{i}S(B)_{\psi_{i}}$. We
note that any other good upper bounds on the entanglement of formation can
also be applied to Theorem 1. In particular, the bound $\hat{E}_{\rm
rev}^{u}(\rho_{AB})$ is efficiently computable since ${\cal R}_{\max,{\rm
ADG}}(\rho_{AB})$ and $2^{-{\cal R}_{\max,{\rm ADG}}(\rho_{AB})}$ can be
efficiently computed via an SDP. By Slater’s condition [53], the following two
optimization programs satisfy the strong duality, and both evaluate to
$1-2^{-{\cal R}_{\max,{\rm ADG}}(\rho_{AB})}$. We denote $P_{BE}$ as a
permutation operator on the system $B$ and $E$ and remain the derivation of
the dual program in Appendix A.

$\begin{aligned} &\underline{\textbf{Primal Program}}\\\
\min_{\omega_{AB},\tau_{AB},\tau_{ABE}}&\;\operatorname{Tr}[\omega_{AB}],\\\
{\rm s.t.}&\;\rho_{AB}=\omega_{AB}+\tau_{AB},\\\ &\;\omega_{AB}\geq
0,\tau_{AB}\geq 0,\tau_{ABE}\geq 0,\\\
&\;\operatorname{Tr}_{E}[\tau_{ABE}]=\operatorname{Tr}_{B}[\tau_{ABE}]=\tau_{AB},\end{aligned}\begin{aligned}
&\underline{\textbf{Dual Program}}\\\
\max_{M_{AB},N_{AB},K_{AE}}&\;\operatorname{Tr}[M_{AB}\rho_{AB}],\\\ {\rm
s.t.}&\;M_{AB}\leq I_{AB},\\\ &\;M_{AB}+N_{AB}+K_{AB}\leq 0,\\\
&\;N_{AB}\otimes I_{E}+P_{BE}(K_{AB}\otimes I_{E})P_{BE}^{\dagger}\geq
0.\end{aligned}$

It is worth mentioning that this new bound is related to the bound $E_{\rm
DA}(\cdot)$ proposed in [17] utilizing the convexity of $E_{D,\to}(\cdot)$ on
decomposing a state into degradable and anti-degradable parts. Remarkably,
constructing such a decomposition is challenging due to the non-convex nature
of the degradable state set. Thus it is difficult to compute $E_{\rm
DA}(\cdot)$ in practice due to the hardness of tracking all possible
decompositions. For example, when considering maximally entangled states
affected by general noises, we currently lack strategies to compute the exact
value of $E_{\rm DA}(\cdot)$. In contrast, $\hat{E}_{\rm rev}^{u}(\cdot)$
overcomes this difficulty and can be efficiently computed for general states,
serving as a valuable approximation of $E_{\rm DA}(\cdot)$. It outperforms the
known computable upper bounds for many maximally entangled states under
practical noises in a high-noise region, as shown in subsection III.3.
Moreover, our method offers flexibility in selecting other decomposition
strategies, i.e., the object functions in the SDP, beyond simply calculating
the ratio at the sub-state decomposition.

###  III.2 Continuity bounds of the one-way distillable entanglement

Note that the insight of the bound above is considering the distance of a
given state between the anti-degradable states set. With different distance
measures, the authors in [25] derived continuity bounds on quantum capacity in
terms of the (anti)degradability of the channel. Thus for self-contained, we
introduce some distance measures between a state and the set ${\rm ADG}$ and
prove the continuity bounds for the state version as a comparison with
$\hat{E}_{\rm rev}^{u}(\rho_{AB})$.

###### Definition 2

Let $\rho_{AB}$ be a bipartite quantum state, the anti-degradable set distance
is defined by

$d_{\rm set}(\rho_{AB})\mathrel{\mathop{\mathchar
58\relax}}=\min_{\sigma_{AB}\in{\rm
ADG}}\frac{1}{2}\|\rho_{AB}-\sigma_{AB}\|_{1},$ (16)

where the minimization ranges over all anti-degradable states on system AB.

Similarly, one also has the anti-degradable map distance as follows.

######  Definition 3 ( [17])

Let $\rho_{AB}$ be a bipartite quantum state with purification $\phi_{ABE}$,
the anti-degradable map distance is defined by

$d_{\rm map}(\rho_{AB})\mathrel{\mathop{\mathchar
58\relax}}=\min_{\mathcal{D}\mathrel{\mathop{\mathchar 58\relax}}E\rightarrow
B}\frac{1}{2}\|\rho_{AB}-\mathcal{D}(\rho_{AE})\|_{1}\\\ ,$ (17)

where $\rho_{AE}=\operatorname{Tr}_{B}(\phi_{ABE})$ and the minimization
ranges over all CPTP maps $\mathcal{D}$.

Both parameters can be computed via SDP, ranging from $[0,1]$, and are equal
to $0$ iff $\rho_{AB}$ is anti-degradable. Similar to the idea in [25] for
channels and the proof techniques in [17], we utilize the continuity bound of
the conditional entropy in Lemma S2 proved by Winter [54] to derive two
continuity upper bounds on the one-way distillable entanglement, concerning
the distance measures above. The proofs can be found in Appendix C. We denote
$h(p)\mathrel{\mathop{\mathchar 58\relax}}=-p\log p-(1-p)\log(1-p)$ as the
binary entropy and $g(p)\mathrel{\mathop{\mathchar
58\relax}}=(1+p)h(\frac{p}{1+p})$ as the bosonic entropy.

###### Proposition 3

For any bipartite state $\rho_{AB}$ with an anti-degradable set distance
$\varepsilon_{\rm set}$, it satisfies

$E_{D,\to}(\rho_{AB})\leq E_{\rm SCB}(\rho_{AB})\mathrel{\mathop{\mathchar
58\relax}}=2\varepsilon_{\rm set}\log(|A|)+g(\varepsilon_{\rm set}).$ (18)

###### Proposition 4

For any bipartite state $\rho_{AB}$ with an anti-degradable map distance
$\varepsilon_{\rm map}$, it satisfies

$E_{D,\to}(\rho_{AB})\leq E_{\rm MCB}(\rho_{AB})\mathrel{\mathop{\mathchar
58\relax}}=4\varepsilon_{\rm map}\log(|B|)+2g(\varepsilon_{\rm map}).$ (19)

###  III.3 Examples of less-entangled states

We now compare the performance of different continuity bounds and the Rains
bound with $\hat{E}_{\rm rev}^{u}(\cdot)$ by some concrete examples. Due to
noise and other device imperfections, one usually obtains some less entangled
states in practice rather than the maximally entangled ones. Such a
disturbance can be characterized by some CPTP maps appearing in each party.
Thus for the task of the distillation of the maximally entangled states under
practical noises, we consider Alice and Bob are sharing pairs of maximally
entangled states affected by bi-local noisy channels, i.e.,

$\rho_{A^{\prime}B^{\prime}}={\cal N}_{A\to A^{\prime}}\otimes{\cal N}_{B\to
B^{\prime}}(\Phi_{AB}).$ (20)

#### Qubit system

Suppose Alice’s qubit is affected by the qubit amplitude damping channel with
Kraus operators $K_{0}=|0\rangle\\!\langle
0|+\sqrt{1-\gamma}|1\rangle\\!\langle 1|$,
$K_{1}=\sqrt{\gamma}|0\rangle\\!\langle 1|$, and Bob’s qubit is affected by
the depolarizing channel ${\cal D}(\rho)=(1-p)\rho+p{I_{2}}/{2}$. Set
$\gamma=0.1$ and the noise parameter of depolarizing noise varies in the range
$p\in[0.15,0.3]$. $E_{D,\to}(\rho_{A^{\prime}B^{\prime}})$ has upper bounds as
functions of $p$ shown in Fig. 2 (a).

#### Qutrit system

For the system with a dimension $d>2$, we consider the multilevel versions of
the amplitude damping channel (MAD) [55] as a local noise for Alice. The Kraus
operators of a MAD channel in a $d$-dimensional system are defined by

$\displaystyle\hat{K}_{ij}$
$\displaystyle\equiv\sqrt{\gamma_{ji}}|i\rangle\\!\langle j|,\quad\forall
i,j\text{ s.t. }0\leq i\leq j\leq d-1,$ (21) $\displaystyle\hat{K}_{0}$
$\displaystyle\equiv|0\rangle\\!\langle 0|+\sum_{1\leq j\leq
d-1}\sqrt{1-\xi_{j}}|j\rangle\\!\langle j|,$

with $\gamma_{ji}$ real quantities describing the decay rate from $j$-th to
$i$-th level that fulfill the conditions

$\begin{cases}0\leq\gamma_{ji}\leq 1,&\forall i,j\text{ s.t. }0\leq i<j\leq
d-1,\\\ \xi_{j}\equiv\sum_{0\leq i<j}\gamma_{ji}\leq 1,&\forall
j=1,\cdots,d-1.\end{cases}$ (22)

Suppose Alice qutrit is affected by a MAD channel with
$\gamma_{10}=\gamma_{20}=0.1,\gamma_{21}=0$. Bob’s qutrit is affected by a
qutrit depolarizing channel with the noise parameter $p$. Then
$E_{D,\to}(\rho_{A^{\prime}B^{\prime}})$ has upper bounds as functions of $p$
shown in Fig. 2 (b).

#### Qudit system

For the qudit system, we consider Alice’s qudit is affected by a MAD channel
with $d=4$ defined in Eq. (21) and Eq. (22), where
$\gamma_{10}=\gamma_{20}=\gamma_{30}=\gamma_{21}=0.1,\gamma_{31}=\gamma_{32}=0$.
Let Bob’s qudit be affected by a qudit depolarizing channel with noise
parameter $p$, then $E_{D,\to}(\rho_{A^{\prime}B^{\prime}})$ has upper bounds
as functions of $p$ shown in Fig. 2 (c).

Fig 2: Upper bounds on the one-way distillable entanglement of less entangled
states in a qubit, qutrit, and qudit system. Panel (a-c) depict the cases
where the maximally entangled states are affected by bi-local noises in
different systems as stated in Eq. (20). The $x$-axis represents the change of
the depolarizing noise $p$. The state’s coherent information $I_{c}$ provides
a lower bound. $R$ is the Rains bound. $\hat{E}_{\rm rev}^{u}$ is the upper
bound derived in Corollary 2. $E_{\rm SCB}$ and $E_{\rm MCB}$ are continuity
bounds derived in Proposition 3 and 4, respectively. It shows that
$\hat{E}_{\rm rev}^{u}$ outperforms all other upper bounds on these less
entangled states.

###  III.4 Extending the method to the two-way distillable entanglement

Similar to the reverse max-relative entropy of unextendible entanglement, for
a given bipartite state $\rho_{AB}$, we introduce the reverse max-relative
entropy of NPT entanglement as

${\cal R}_{\max,{\text{\rm PPT}}}(\rho_{AB})\mathrel{\mathop{\mathchar
58\relax}}=\min_{\tau\in{\text{\rm PPT}}}D_{\max}(\tau_{AB}||\rho_{AB}),$ (23)

where the minimization ranges over all PPT states. The term NPT refers to
states whose partial transpose has negative eigenvalues. Based on the
convexity of $E_{D,\leftrightarrow}(\cdot)$ on decomposing a state into
maximally correlated (MC) states and PPT states [17] (see Appendix B for more
details), we can utilize the reverse max-relative entropy of NPT entanglement
to establish an upper bound on the two-way distillable entanglement, as
outlined in Theorem 5.

###### Theorem 5

For any bipartite state $\rho_{AB}$, it satisfies

$E_{D,\leftrightarrow}(\rho_{AB})\leq E_{\rm
rev}^{npt}(\rho_{AB})\mathrel{\mathop{\mathchar 58\relax}}=[1-2^{-{\cal
R}_{\max,{\text{\rm PPT}}}(\rho_{AB})}]\cdot E_{F}(\omega_{AB}),$ (24)

where $\omega_{AB}$ is the PPT-squeezed state of $\rho_{AB}$ and ${\cal
R}_{\max,{\text{\rm PPT}}}(\cdot)$ is the reverse max-relative entropy of NPT
entanglement.

It also follows an efficiently computable relaxation as $\hat{E}_{\rm
rev}^{npt}(\rho_{AB})=[1-2^{-{\cal R}_{\max,{\text{\rm
PPT}}}(\rho_{AB})}]\sum_{i}\lambda_{i}S(B)_{\psi_{i}}$, where
$\omega_{AB}=\sum_{i}\lambda_{i}|\psi_{i}\rangle\\!\langle\psi_{i}|$ is the
spectral decomposition of the PPT-squeezed state $\omega_{AB}$ of $\rho_{AB}$.

In fact, $\hat{E}_{\rm rev}^{npt}(\cdot)$ can be interpreted as an easily
computable version of the bound $E_{\rm MP}(\cdot)$ in [17], utilizing the
convexity of $E_{D,\leftrightarrow}(\rho_{AB})$ on the convex decomposition of
$\rho_{AB}$ into MC states and PPT states. Since the set of all MC states is
not convex, tracking all possible decompositions to compute $E_{\rm
MP}(\cdot)$ is generally hard. However, $\hat{E}_{\rm rev}^{npt}(\cdot)$ is
efficiently computable by SDP and provides a remarkably tighter approximation
of $E_{\rm MP}(\cdot)$ for the example states presented in [17]. The
comparison in detail can be found in Appendix B. We note that
$R(\rho_{AB})\leq E_{\rm MP}(\rho_{AB})\leq\hat{E}_{\rm rev}^{npt}(\rho_{AB})$
where $R(\cdot)$ is the Rains bound for the two-way distillable entanglement.
Nevertheless, $\hat{E}_{\rm rev}^{npt}(\cdot)$ connects the reverse max-
relative entropy of NPT entanglement with the entanglement of formation, and
we believe such connection would shed light on the study of other quantum
resource theories as well.

##  IV Applications on quantum channel capacity

For a general quantum channel ${\cal N}_{A\rightarrow B}$, its quantum
capacity has a regularized formula proved by Lloyd, Shor, and Devetak [56, 57,
58]:

$Q({\cal N})=\lim_{n\rightarrow\infty}\frac{1}{n}Q^{(1)}({\cal N}^{\otimes
n}),$ (25)

where $Q^{(1)}({\cal N})\mathrel{\mathop{\mathchar
58\relax}}=\max_{|\phi\rangle_{A^{\prime}A}}I_{c}(A^{\prime}\rangle B)_{({\cal
I}\otimes{\cal N})(\phi)}$ is the channel coherent information. Similar to the
one-way distillable entanglement of a state, the regularization in Eq. (25)
makes the quantum capacity of a channel intractable to compute generally.
Substantial efforts have been made to establish upper bounds. One routine is
inspired by the Rains bound in entanglement theory [14]. Tomamichel et al.
introduced Rains information [59] for a quantum channel as an upper bound on
the quantum capacity. Then some efficiently computable relaxations or
estimations are given in [28, 31]. Another routine is to consider the
(anti)degradability of the quantum channels and to construct flag extensions,
which gives the currently tightest upper bound for quantum channels with
symmetry or known structures [25, 19, 23, 21].

A channel ${\cal N}_{A\rightarrow B}$ is called degradable if there exits a
CPTP map ${\cal D}_{B\rightarrow E}$ such that ${\cal N}^{c}={\cal
D}\circ{\cal N}$, and is called anti-degradable if there exits a CPTP map
${\cal A}_{E\rightarrow B}$ such that ${\cal N}={\cal A}\circ{\cal N}^{c}$. It
is known that ${\cal N}$ is (anti)degradable if and only if its Choi state
$J_{{\cal N}}$ is (anti)degradable. The quantum capacity of an anti-degradable
channel is zero and the coherent information of a degradable channel is
additive which leads to $Q({\cal N})=Q^{(1)}({\cal N})$. Concerning the
(anti)degradability of a channel, the authors in [25] called a channel
$\varepsilon$-degradable channel if there is a CPTP map ${\cal
D}_{B\rightarrow E}$ such that $||{\cal N}^{c}-{\cal D}\circ{\cal
N}||_{\diamond}\leq 2\varepsilon$. A channel is called $\varepsilon$-anti-
degradable channel if there is a CPTP map ${\cal A}_{E\rightarrow B}$ such
that $||{\cal N}-{\cal A}\circ{\cal N}^{c}||_{\diamond}\leq 2\varepsilon$.
Based on these, one has continuity bounds of the quantum capacity as follows.

######  Theorem 6 ([25])

Given a quantum channel ${\cal N}_{A\to B}$, if it is
$\varepsilon$-degradable, then it satisfies $Q({\cal N})\leq Q^{(1)}({\cal
N})+\varepsilon\log(d_{E}-1)+h(\varepsilon)+2\varepsilon\log
d_{E}+g(\varepsilon)$. If ${\cal N}_{A\to B}$ is $\varepsilon$-anti-
degradable, it satisfies $Q({\cal
N})\leq\varepsilon\log(|B|-1)+2\varepsilon\log|B|+h(\varepsilon)+g(\varepsilon)$.

With a similar spirit of the reverse max-relative entropy of unextendible
entanglement in Eq. (12), we define the reverse max-relative entropy of anti-
degradability of the channel as

$\widetilde{{\cal R}}_{\max,{\rm ADG}}({\cal N}_{A\to
B})\mathrel{\mathop{\mathchar 58\relax}}=\min_{{\cal N}^{\prime}_{A\to
B}\in{\cal C}_{{\rm ADG}}}D_{\max}({\cal N}^{\prime}_{A\to B}||{\cal N}_{A\to
B}),$ (26)

where ${\cal C}_{{\rm ADG}}$ is the set of all anti-degradable channels and
the max-relative entropy of ${\cal N}^{\prime}_{A\to B}$ with respect to
${\cal N}_{A\to B}$ is defined by

$D_{\max}({\cal N}^{\prime}_{A\to B}||{\cal N}_{A\to
B})\mathrel{\mathop{\mathchar
58\relax}}=\inf\\{\lambda\in\mathbb{R}\mathrel{\mathop{\mathchar
58\relax}}J_{AB}^{{\cal N}^{\prime}}\leq 2^{\lambda}J_{AB}^{{\cal N}}\\}.$
(27)

If there is no such a channel ${\cal N}^{\prime}_{A\to B}\in{\cal C}_{{\rm
ADG}}$ that satisfies $J_{AB}^{{\cal N}^{\prime}}\leq 2^{\lambda}J_{AB}^{{\cal
N}}$, $\widetilde{{\cal R}}_{\max,{\rm ADG}}({\cal N}_{A\to B})$ is set to be
0. Similar to the state case, $\widetilde{{\cal R}}_{\max,{\rm ADG}}({\cal
N}_{A\to B})$ has a geometric implication analogous to the distance between
${\cal N}_{A\to B}$ to the set of all anti-degradable channels. We can
introduce the ${\rm ADG}$-squeezed channel of ${\cal N}_{A\to B}$ as follows.

###### Definition 4

For a quantum channel ${\cal N}_{A\rightarrow B}$ and the anti-degradable
channel set ${\cal C}_{{\rm ADG}}$, if $\widetilde{{\cal R}}_{\max,{\rm
ADG}}({\cal N})$ is non-zero, the ${\rm ADG}$-squeezed channel of ${\cal
N}_{A\rightarrow B}$ is defined by

${\cal S}_{A\to B}=\frac{{\cal N}_{A\to B}-2^{-\widetilde{{\cal R}}_{\max,{\rm
ADG}}({\cal N})}\cdot{\cal N}^{\prime}_{A\to B}}{1-2^{-\widetilde{{\cal
R}}_{\max,{\rm ADG}}({\cal N})}},$ (28)

where ${\cal N}^{\prime}_{A\to B}$ is the closest anti-degradable channel to
${\cal N}_{A\to B}$ in terms of the max-relative entropy, i.e., the optimal
solution in Eq. (26). If $\widetilde{{\cal R}}_{\max,{\rm ADG}}({\cal N})$ is
zero, the ${\rm ADG}$-squeezed channel of ${\cal N}_{A\rightarrow B}$ is
itself.

Notably, $\widetilde{{\cal R}}_{\max,{\rm ADG}}({\cal N})$ can be efficiently
computed via SDP shown in Appendix A. The conceptual idea we used here is
similar to that for the state case in Eq. (6) and Definition 1, which is to
squeeze or isolate out as much part of anti-degradable channel as possible via
a convex decomposition of the original channel. The insight here is that one
can ignore the contribution from the anti-degradable part for the quantum
capacity, and the quantum capacity admits convexity on the decomposition into
degradable and anti-degradable parts. In this way, the following Theorem 7
gives an upper bound $Q_{\rm sqz}({\cal N})$ for the quantum capacity of
${\cal N}$.

###### Theorem 7

Given a quantum channel ${\cal N}_{A\to B}$, if it has an ADG-squeezed channel
${\cal S}_{A\to B}$, we denote $\widehat{{\cal S}}_{A\to BB^{\prime}}$ as an
extended channel of ${\cal S}_{A\to B}$ such that
$\operatorname{Tr}_{B^{\prime}}[\widehat{{\cal S}}_{A\to
BB^{\prime}}(\rho_{A})]={\cal S}_{A\to B}(\rho_{A}),\forall\rho_{A}\in{\cal
D}({\cal H}_{A})$. Then it satisfies

$Q({\cal N})\leq Q_{\rm sqz}({\cal N})\mathrel{\mathop{\mathchar
58\relax}}=[1-2^{-\widetilde{{\cal R}}_{\max,{\rm ADG}}({\cal
N})}]\cdot\min\left\\{Q^{(1)}(\widehat{{\cal S}})|\;\widehat{{\cal S}}_{A\to
BB^{\prime}}\mbox{\rm\; is degradable}\right\\},$ (29)

where the minimization is over all possible extended channels of $S_{A\to B}$.
If there is no such a degradable $\widehat{{\cal S}}_{A\to BB^{\prime}}$
exists, the value of this bound is set to be infinity.

###### Proof.

By the definition of the ${\rm ADG}$-squeezed channel of ${\cal N}_{A\to B}$,
we have

${\cal N}_{A\to B}=[1-2^{-\widetilde{{\cal R}}_{\max,{\rm ADG}}({\cal
N})}]\cdot{\cal S}_{A\to B}+2^{-\widetilde{{\cal R}}_{\max,{\rm ADG}}({\cal
N})}\cdot{\cal N}^{\prime}_{A\to B},$ (30)

where ${\cal N}^{\prime}_{A\to B}$ is anti-degradable. We write an extended
channel of ${\cal N}^{\prime}_{A\to B}$ as $\widehat{{\cal N}^{\prime}}_{A\to
BB^{\prime}}(\rho_{A})={\cal N}_{A\to B}(\rho_{A})\otimes|0\rangle\\!\langle
0|_{B^{\prime}}$, which is obviously anti-degradable. Then we can construct a
quantum channel $\widehat{{\cal N}}_{A\to BB^{\prime}}$ as $\widehat{{\cal
N}}_{A\to BB^{\prime}}=[1-2^{-\widetilde{{\cal R}}_{\max,{\rm ADG}}({\cal
N})}]\cdot\widehat{{\cal S}}_{A\to BB^{\prime}}+2^{-\widetilde{{\cal
R}}_{\max,{\rm ADG}}({\cal N})}\cdot\widehat{{\cal N}^{\prime}}_{A\to
BB^{\prime}},$ such that $\operatorname{Tr}_{B^{\prime}}[\widehat{{\cal
S}}_{A\to BB^{\prime}}(\rho_{A})]={\cal S}_{A\to B}(\rho_{A})$ for any state
$\rho_{A}$ and $\widehat{{\cal S}}_{A\to BB^{\prime}}$ is degradable. This
means after discarding the partial environment $B^{\prime}$, the receiver can
obtain the original quantum information sent through ${\cal N}_{A\to B}$. In
this case, $\widehat{{\cal N}}$ can certainly convey more quantum information
than the ${\cal N}$, i.e., $Q({\cal N})\leq Q(\widehat{{\cal N}})$. Note that
the quantum capacity admits convexity on the decomposition into degradable
parts and anti-degradable parts [22]. We conclude that

$\displaystyle Q({\cal N})\leq Q(\widehat{{\cal N}})$
$\displaystyle\leq[1-2^{-\widetilde{{\cal R}}_{\max,{\rm ADG}}({\cal
N})}]\cdot Q(\widehat{{\cal S}})+2^{-\widetilde{{\cal R}}_{\max,{\rm
ADG}}({\cal N})}\cdot Q(\widehat{{\cal N}^{\prime}})$ (31)
$\displaystyle=[1-2^{-\widetilde{{\cal R}}_{\max,{\rm ADG}}({\cal N})}]\cdot
Q^{(1)}(\widehat{{\cal S}}),$

where the equality is followed by the quantum capacity is additive on
degradable channels and is zero for anti-degradable channels. Considering the
freedom of the choice of $\widehat{{\cal S}}_{A\to BB^{\prime}}$, we obtain

$Q_{\rm sqz}({\cal N})\mathrel{\mathop{\mathchar
58\relax}}=[1-2^{-\widetilde{{\cal R}}_{\max,{\rm ADG}}({\cal
N})}]\cdot\min\left\\{Q^{(1)}(\widehat{{\cal S}})|\;\widehat{{\cal S}}_{A\to
BB^{\prime}}\mbox{\rm\; is degradable}\right\\}$ (32)

as an upper bound on $Q({\cal N})$.  $\sqcap$$\sqcup$

Remark 2 Theorem 7 can be seen as a channel version of Theorem 1. However, in
order to utilize the convexity of the quantum capacity after the squeezing
process, it is challenging to decompose the ADG-squeezed channel into the sum
of degradable ones. An alternative approach here is to use the idea of the
extension channel. For the qubit channels specifically, this bound is
efficiently computable and effective, as shown in subsection IV.1.

###  IV.1 Quantum capacity of qubit channels

For a quantum channel with dimension two in both input and output systems, we
prove that the ADG-squeezed channel is always degradable. Thus, we give an
efficiently computable upper bound on the quantum capacity using the idea of
the reverse max-relative entropy of anti-degradability.

###### Proposition 8

For any qubit channel ${\cal N}_{A\to B}$, it is either anti-degradable or
satisfies

$Q({\cal N})\leq[1-2^{-\widetilde{{\cal R}}_{\max,{\rm ADG}}({\cal
N})}]\cdot\max_{p\in[0,1]}I_{\mathrm{c}}\Big{(}p|0\rangle\\!\langle
0|+(1-p)|1\rangle\\!\langle 1|,{\cal S}_{A\to B}\Big{)},$ (33)

where $I_{\mathrm{c}}(\rho,\mathcal{N})\equiv
H(\mathcal{N}(\rho))-H\left(\mathcal{N}^{c}(\rho)\right)$ and ${\cal S}_{A\to
B}$ is the ADG-squeezed channel of ${\cal N}_{A\to B}$.

###### Proof.

By the definition of the ${\rm ADG}$-squeezed channel of ${\cal N}_{A\to B}$,
we have

${\cal N}_{A\to B}=[1-2^{-\widetilde{{\cal R}}_{\max,{\rm ADG}}({\cal
N})}]\cdot{\cal S}_{A\to B}+2^{-\widetilde{{\cal R}}_{\max,{\rm ADG}}({\cal
N})}\cdot{\cal N}^{\prime}_{A\to B},$ (34)

where ${\cal N}^{\prime}_{A\to B}$ is anti-degradable and ${\cal S}_{A\to B}$
is not anti-degradable. If ${\cal S}_{A\to B}$ is also not degradable, we can
further decompose ${\cal S}_{A\to B}$ into ${\cal S}=c{\cal S}_{0}+(1-c)S_{1}$
such that ${\cal S}_{0}$ is degradable and ${\cal S}_{1}$ is anti-degradable
since the extreme points of the set of all qubit channels have been shown to
be degradable or anti-degradable channels [60, 25]. This conflicts with the
definition of $2^{-\widetilde{{\cal R}}_{\max,{\rm ADG}}({\cal N})}$, which
implies ${\cal S}_{A\to B}$ is degradable. Thus,

$\displaystyle Q({\cal N})$ $\displaystyle\leq[1-2^{-\widetilde{{\cal
R}}_{\max,{\rm ADG}}({\cal N})}]\cdot Q({\cal S})+2^{-\widetilde{{\cal
R}}_{\max,{\rm ADG}}({\cal N})}\cdot Q({\cal N}^{\prime})$ (35)
$\displaystyle=[1-2^{-\widetilde{{\cal R}}_{\max,{\rm ADG}}({\cal N})}]\cdot
Q^{(1)}({\cal S})$ $\displaystyle=[1-2^{-\widetilde{{\cal R}}_{\max,{\rm
ADG}}({\cal
N})}]\cdot\max_{p\in[0,1]}I_{\mathrm{c}}\Big{(}p|0\rangle\\!\langle
0|+(1-p)|1\rangle\\!\langle 1|,{\cal S}_{A\to B}\Big{)}.$

Note that the last equality is because ${\cal S}_{A\to B}$ is degradable, and
diagonal input states outperform non-diagonal states during the optimization
of the channel coherent information [22].  $\sqcap$$\sqcup$

Fig 3: Upper bounds on the quantum capacity of the random mixed unitary
channels. In Panel (a) and (b), we compare our bound $Q_{\rm sqz}$ in
Proposition 8 with other upper bounds on the quantum capacity of 1000 randomly
generated mixed unitary channels with fixed parameters $p_{0}=0.58$,
$p_{1}=0.22$, $p_{2}=0.15$, $p_{3}=0.05$. The $x$-axis represents the distance
between two bounds, and the $y$-axis represents the distribution of these
channels. $Q_{\epsilon-\rm adg}$ is the continuity bound in Theorem 6
regarding the anti-degradability. $\hat{R}_{\alpha(10)}$ is the bound in [31].
Panel (c) and (d) depict the same comparison for another 1000 randomly
generated mixed unitary channels with fixed parameters $p_{0}=0.6$,
$p_{1}=0.2$, $p_{2}=0.1$, $p_{3}=0.1$. In (a) and (c), we see that our bound
always outperforms $Q_{\epsilon-\rm adg}$. In (b) and (d), we can see for many
cases, our bound is tighter than $\hat{R}_{\alpha(10)}$.

#### Mixed unitary channels

To compare the performance of our method with some best-known computable
bounds, e.g., the continuity bound in Theorem 6 and the bound
$\hat{R}_{\alpha}$ [31] generalized from the max-Rain information [28], we
consider the mixed unitary channel ${\cal U}_{A\rightarrow
B}(\rho)=\sum_{i=0}^{k}p_{i}U_{i}\rho U_{i}^{\dagger}$, where
$\sum_{i=0}^{k}p_{i}=1$ and $U_{i}$ are unitary operators on a qubit system.
In specific, we choose some fixed set of parameters and sample 1000 channels
with randomly generated unitaries according to the Haar measure. We compute
the distance between $Q_{\rm sqz}$ and other bounds, then obtain statistics on
the distribution of these channels according to the distance value. The
distribution results are depicted in Fig. 3 where the purple region
corresponds to the cases $Q_{\rm sqz}$ is tighter, and the pink region
corresponds to the cases $Q_{\rm sqz}$ is looser. We can see that in Fig. 3(a)
and Fig. 3(c), $Q_{\rm sqz}$ always outperforms the continuity bound of anti-
degradability and in Fig. 3(b) and Fig. 3(d), our bound is tighter than
$\hat{R}_{\alpha(10)}$ for many cases.

Fig 4: Upper bounds on quantum capacity of the covariant Pauli channels. The
parameters are $p_{3}=0.05,p_{0}\in[0.75,0.85]$. The single-shot coherent
information $Q^{(1)}$ [61, 62] provides a lower bound on the quantum capacity.
$\hat{R}_{\alpha(10)}$ is the upper bound in [31]. $Q_{\epsilon-\rm adg}$ is
the continuity bound of anti-degradability. $Q_{\rm flag}$ is the upper bound
by flag extension method in [63]. $Q_{\rm sqz}$ is our bound in Corollary 11
and $Q_{\rm nc}$ is the no-cloning bound [32]. It shows that $Q_{\rm sqz}$ can
recover the no-cloning bound and outperform other computable bounds.

#### Pauli channels

A representative qubit channel is the Pauli channel describing bit-flip errors
and phase-flip errors with certain probabilities in qubits. A qubit Pauli
channel $\Lambda(\cdot)$ is defined as:

$\Lambda(\rho)=p_{0}\rho+p_{1}X\rho X+p_{2}Y\rho Y+p_{3}Z\rho Z,$ (36)

where $X,Y,Z$ are the Pauli operators and $\sum_{i=0}^{3}p_{i}=1$ are
probability parameters. Note that for the quantum capacity, we only need to
consider the cases where $p_{0}$ dominates. Since if, for example, bit flip
error $X$ happens with probability larger than $p_{0}$, one can first apply a
$X$ flip, mapping that channel back into the case where $p_{0}>p_{1}$. After
utilizing our method on Pauli channels, the ADG-squeezed parameter is
characterized in Proposition 9, whose proof can be found in Appendix D. Thus,
combined with Proposition 8, we can recover the no-cloning bound [32] on the
quantum capacity of qubit Pauli channels.

###### Proposition 9

For a qubit Pauli channel $\Lambda(\cdot)$ with $p_{0}\geq p_{i}>0$
$(i=1,2,3)$, it is either anti-degradable or satisfies

$2^{-\widetilde{{\cal R}}_{\max,{\rm
ADG}}(\Lambda)}=(\sqrt{p_{1}}+\sqrt{p_{2}})^{2}+(\sqrt{p_{2}}+\sqrt{p_{3}})^{2}+(\sqrt{p_{1}}+\sqrt{p_{3}})^{2}$
(37)

with an ADG-squeezed channel as the identity channel.

######  Theorem 10 ( [32])

For a qubit Pauli channel $\Lambda(\cdot)$, its quantum capacity is either
vanishing or satisfies

$Q(\Lambda)\leq
1-(\sqrt{p_{1}}+\sqrt{p_{2}})^{2}-(\sqrt{p_{2}}+\sqrt{p_{3}})^{2}-(\sqrt{p_{1}}+\sqrt{p_{3}})^{2}.$
(38)

One recent work [63] studies the capacities of a subclass of Pauli channels
called the covariant Pauli channel, where the parameters are set $p_{1}=p_{2}$
with $p_{0}+2p_{1}+p_{3}=1$, i.e., $\Lambda_{\rm
cov}(\rho)=p_{0}\rho+p_{1}(X\rho X+Y\rho Y)+p_{3}Z\rho Z$. Applying Theorem 10
on the covariant Pauli channels, we can bound their quantum capacity as
follows.

###### Corollary 11

For a covariant Pauli channel $\Lambda_{\rm cov}(\cdot)$, it is either anti-
degradable with a zero quantum capacity or satisfies $Q(\Lambda_{\rm cov})\leq
Q_{\rm sqz}(\Lambda_{\rm cov})$, where

$Q_{\rm sqz}(\Lambda_{\rm
cov})=3p_{0}+p_{3}-\sqrt{8(p_{3}-p_{0}p_{3}-p_{3}^{2})}-2.$ (39)

In Fig. 4, we compare our bound with the upper bounds given in [63] and the
continuity bound of anti-degradability in Theorem 6. It can be seen that our
bound in the orange line, coinciding with the no-cloning bound, outperforms
previous bounds, and thus can better characterize the quantum capacity of
$\Lambda_{\rm cov}(\cdot)$ when it is close to being anti-degradable.

##  V Concluding remarks

The design and implementation of a quantum internet [1, 2, 3, 4] involve a
unique set of challenges and require advancements in the realms of
entanglement generation, manipulation, and distribution, as well as the
development of robust quantum communication protocols. By addressing the
challenges associated with entanglement distillation and reliable quantum
communication, our work contributes to the advancement of the field of quantum
internet.

We have introduced the reverse max-relative entropy of entanglement which is
related to the "weight of resource" in general resource theory. From a
conceptual and technical side, the concept helps us to quantify how much
useless entanglement we can squeeze out from a state or a channel, which is
meaningful in the distillable entanglement and the quantum capacity,
respectively. Importantly, these quantities can be efficiently computed via
SDP, providing upper bounds on the distillable entanglement and quantum
capacity that can be easily computed.

To further investigate entanglement distillation, we have derived various
continuity bounds on the one-way distillable entanglement using the anti-
degradability of the state. In particular, our bound derived from the reverse
max-relative entropy of unextendible entanglement outperforms both continuity
bounds and the Rains bound in estimating the one-way distillable entanglement
for maximally entangled states under certain relevant noises. These bounds are
also the only known computable ones for general states. We further introduced
the reverse max-relative entropy of NPT entanglement and established
connections to prior results on the two-way distillable entanglement. For the
quantum capacity of noisy channels, our bounds based on the reverse max-
relative entropy of anti-degradability deliver improved results for random
mixed unitary qubit channels. Also, the analytical bound obtained from our
method recovers the no-cloning bound on Pauli channels [32].

These results open a novel way to connect valuable quantum resource measures
with quantum communication tasks. Except for the existing applications of the
reverse max-relative entropy of resources [47, 48, 45, 46, 64, 49, 50], we
expect the reverse divergence of resources will find more applications in
quantum resource theories in both asymptotic and non-asymptotic regimes [65,
66, 67, 68, 69, 70]. Our method may also be generalized to quantum
entanglement and quantum communication theories in continuous-variable quantum
information [71]. These advancements hold promise to facilitate the
development of practical, scalable, and efficient techniques for generating,
preserving, and utilizing entangled states over long distances, thereby
advancing the field of quantum internet.

## Acknowledgements

Part of this work was done when C. Z., C. Z. and X. W. were at Baidu Research.
We would like to thank Bartosz Regula and Ludovico Lami for helpful comments.

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##  Appendix A Dual SDP for ${\cal R}_{\max,{\rm ADG}}(\rho_{AB})$ and
$\widetilde{{\cal R}}_{\max,{\rm ADG}}({\cal N})$

The primal SDP for calculating $2^{-{\cal R}_{\max,{\rm ADG}}(\rho_{AB})}$ of
the state $\rho_{AB}$ can be written as:

$\displaystyle 1-2^{-{\cal R}_{\max,{\rm
ADG}}(\rho_{AB})}=\min_{\omega_{AB},\tau_{AB},\tau_{ABE}}$
$\displaystyle\operatorname{Tr}[\omega_{AB}],$ (A.1a) $\displaystyle{\rm
s.t.}$ $\displaystyle\;\rho_{AB}=\omega_{AB}+\tau_{AB},$ (A.1b)
$\displaystyle\;\omega_{AB}\geq 0,\tau_{AB}\geq 0,\tau_{ABE}\geq 0,$ (A.1c)
$\displaystyle\;\operatorname{Tr}_{E}[\tau_{ABE}]=\operatorname{Tr}_{B}[\tau_{ABE}]=\tau_{AB},$
(A.1d)

where Eq. (A.1d) corresponds to the anti-degradable condition of $\tau_{AB}$.
The Lagrange function of the primal problem is

$\displaystyle L(\omega_{AB},\tau_{AB},\tau_{ABE},M,N,K)$ (A.2)
$\displaystyle=\operatorname{Tr}[\omega_{AB}]+\langle
M,\rho_{AB}-\omega_{AB}-\tau_{AB}\rangle+\langle
N,\operatorname{Tr}_{E}[\tau_{ABE}]-\tau_{AB}\rangle+\langle
K,\operatorname{Tr}_{B}[\tau_{ABE}]-\tau_{AB}\rangle$ $\displaystyle=\langle
M,\rho_{AB}\rangle+\langle I-M,\omega_{AB}\rangle+\langle-M-
N-K,\tau_{AB}\rangle+\langle N\otimes I_{E}+P_{BE}(K_{AB}\otimes
I_{E})P_{BE}^{\dagger},\tau_{ABE}\rangle,$

where $M_{AB},N_{AB},K_{AB}$ are Lagrange multipliers and $P_{BE}$ is the
permutation operator between $B$ and $E$. The corresponding Lagrange dual
function is

$g(M,N,K)=\inf_{\omega_{AB},\tau_{AB},\tau_{ABE}\geq
0}L(\omega_{AB},\tau_{AB},\tau_{ABE},M,N,K).$ (A.3)

Since $\omega_{AB}\geq 0,\tau_{AB}\geq 0,\tau_{ABE}\geq 0$, it must hold that
$I-M_{AB}\geq 0,-M-N-K\geq 0,N\otimes I+P_{BE}(K\otimes I)P_{BE}^{\dagger}\geq
0$. Thus the dual SDP is

$\displaystyle\max_{M_{AB},N_{AB},K_{AE}}$
$\displaystyle\operatorname{Tr}[M_{AB}\rho_{AB}],$ (A.4) $\displaystyle{\rm
s.t.}$ $\displaystyle\;M_{AB}\leq I_{AB},$
$\displaystyle\;M_{AB}+N_{AB}+K_{AB}\leq 0,$ $\displaystyle\;N_{AB}\otimes
I_{E}+P_{BE}(K_{AB}\otimes I_{E})P_{BE}^{\dagger}\geq 0.$

The primal SDP for calculating $2^{-\widetilde{{\cal R}}_{\max,{\rm
ADG}}({\cal N})}$ of the channel ${\cal N}_{A\to B}$ is:

$\displaystyle 1-2^{-\widetilde{{\cal R}}_{\max,{\rm ADG}}({\cal
N})}=\min_{\Gamma_{AB}^{\cal S},\Gamma_{AB}^{{\cal N}^{\prime}},\gamma_{ABE}}$
$\displaystyle\operatorname{Tr}[\Gamma_{AB}^{\cal S}],$ (A.5a)
$\displaystyle{\rm s.t.}$ $\displaystyle\;J_{AB}^{\cal N}=\Gamma_{AB}^{\cal
S}+\Gamma_{AB}^{{\cal N}^{\prime}},$ (A.5b) $\displaystyle\;\Gamma_{AB}^{\cal
S}\geq 0,\Gamma_{AB}^{{\cal N}^{\prime}}\geq 0,\gamma_{ABE}\geq 0,$ (A.5c)
$\displaystyle\;\operatorname{Tr}_{B}[\Gamma_{AB}^{{\cal
N}^{\prime}}]=\operatorname{Tr}[\Gamma_{AB}^{{\cal N}^{\prime}}]/d_{A}\cdot
I_{A},$ (A.5d)
$\displaystyle\;\operatorname{Tr}_{E}[\gamma_{ABE}]=\operatorname{Tr}_{B}[\gamma_{ABE}]=\Gamma_{AB}^{{\cal
N}^{\prime}},$ (A.5e)

where Eq. (A.5e) corresponds to the anti-degradable condition of the
unnormalized Choi state $\Gamma_{AB}^{{\cal N}^{\prime}}$. The Lagrange
function of the primal problem is

$\displaystyle L(\Gamma_{AB}^{\cal S},\Gamma_{AB}^{{\cal
N}^{\prime}},\gamma_{ABE},M,N,K,R)$ (A.6)
$\displaystyle=\operatorname{Tr}[\Gamma_{AB}^{\cal S}]+\langle M,J_{AB}^{\cal
N}-\Gamma_{AB}^{\cal S}-\Gamma_{AB}^{{\cal N}^{\prime}}\rangle+\langle
N,\operatorname{Tr}_{E}[\gamma_{ABE}]-\Gamma_{AB}^{{\cal
N}^{\prime}}\rangle+\langle
K,\operatorname{Tr}_{B}[\gamma_{ABE}]-\Gamma_{AB}^{{\cal N}^{\prime}}\rangle$
$\displaystyle\quad+\langle R,\operatorname{Tr}_{B}[\Gamma_{AB}^{\cal
S}]-\operatorname{Tr}[\Gamma_{AB}^{\cal S}]/d_{A}\cdot I_{A}\rangle$
$\displaystyle=\langle M,J_{AB}^{\cal
N}\rangle+\langle(1-\operatorname{Tr}R/d_{A})I-M+R\otimes
I_{B},\Gamma_{AB}^{\cal S}\rangle+\langle-M-N-K,\Gamma_{AB}^{{\cal
N}^{\prime}}\rangle$ $\displaystyle\quad+\langle N\otimes
I_{E}+P_{BE}(K_{AE}\otimes I_{B})P_{BE}^{\dagger},\gamma_{ABE}\rangle,$

where $M_{AB},N_{AB},K_{AB}$ are Lagrange multipliers and $P_{BE}$ is the swap
operator between the system $B$ and $E$. The corresponding Lagrange dual
function is

$g(M,N,K)=\inf_{\Gamma_{AB}^{\cal S},\Gamma_{AB}^{{\cal
N}^{\prime}},\gamma_{ABE}\geq 0}L(\Gamma_{AB}^{\cal S},\Gamma_{AB}^{{\cal
N}^{\prime}},\gamma_{ABE},M,N,K).$ (A.7)

Since $\Gamma_{AB}^{\cal S}\geq 0,\Gamma_{AB}^{{\cal N}^{\prime}}\geq
0,\gamma_{ABE}\geq 0$, it must hold that

$\displaystyle(1-\operatorname{Tr}R_{A}/d_{A})I_{AB}-M_{AB}+R_{A}\otimes
I_{B}$ $\displaystyle\geq 0$ (A.8) $\displaystyle-M_{AB}-N_{AB}-K_{AB}$
$\displaystyle\geq 0$ $\displaystyle N_{AB}\otimes I_{E}+P_{BE}(K_{AB}\otimes
I_{E})P_{BE}^{\dagger}$ $\displaystyle\geq 0.$

Thus the dual SDP is

$\displaystyle\max_{M_{AB},N_{AB},K_{AB},R_{A}}$
$\displaystyle\operatorname{Tr}[M_{AB}J_{AB}^{\cal N}],$ (A.9)
$\displaystyle{\rm s.t.}$
$\displaystyle\;M_{AB}\leq(1-\operatorname{Tr}R_{A}/d_{A})I_{AB}+R_{A}\otimes
I_{B},$ $\displaystyle\;M_{AB}+N_{AB}+K_{AB}\leq 0,$
$\displaystyle\;N_{AB}\otimes I_{E}+P_{BE}(K_{AB}\otimes
I_{E})P_{BE}^{\dagger}\geq 0.$

##  Appendix B Two-way distillable entanglement

We first start with the definition of the maximally correlated (MC) state.

###### Definition S1

A bipartite state $\rho_{AB}$ on $\mathbb{C}^{d}\times\mathbb{C}^{d}$ is said
to be maximally correlated (MC), if there exist bases
$\left\\{|i\rangle_{A}\right\\}_{i=0}^{d-1}$ and
$\left\\{|i\rangle_{B}\right\\}_{i=0}^{d-1}$ such that

$\rho_{AB}=\sum_{i,j=0}^{d-1}\alpha_{ij}|i\rangle\\!\langle
j|_{A}\otimes|i\rangle\\!\langle j|_{B},$ (B.1)

where $\left(\alpha_{ij}\right)$ is a positive semidefinite matrix with trace
1 .

We note that every pure state is an MC state. Then, by the following lemma,
one can easily arrive at the upper bound in Theorem 5.

###### Lemma S1

The two-way distillable entanglement is convex on convex combinations of MC
and PPT states.

We recall the example state given in [17]. Consider a $d$ dimensional Hilbert
space, the generalized Pauli operators $X$ and $Z$ are defined by their action
on a computational basis $\\{|k\rangle\\}_{k=0}^{d-1}$ as:

$X|k\rangle\mathrel{\mathop{\mathchar 58\relax}}=|k+1(\bmod d)\rangle\quad
Z|k\rangle\mathrel{\mathop{\mathchar 58\relax}}=\omega^{k}|k\rangle,$ (B.2)

where $\omega\mathrel{\mathop{\mathchar 58\relax}}=\exp(2\pi i/d)$ is a $d$-th
root of unity. The generalized Pauli operators satisfy $XZ=\omega ZX$. Then
the generalized Bell basis
$\left\\{|\Phi_{n,m}\rangle\right\\}_{n,m=0,\ldots,d-1}$ are defined as

$\left|\Phi_{n,m}\right\rangle\mathrel{\mathop{\mathchar
58\relax}}=\left(I_{d}\otimes X^{m}Z^{n}\right)\left|\Phi\right\rangle,$ (B.3)

where $|\Phi\rangle\mathrel{\mathop{\mathchar
58\relax}}=\frac{1}{\sqrt{d}}\sum_{i=0}^{d-1}|ii\rangle$. Now set $d=3$ and
denote $B=\\{1,6,8\\}$. After numbering
$|\Phi_{n,m}\rangle=|\Phi_{3n+m+1}\rangle$, we define the state

$\omega_{AB}=\sum_{i,j=1}^{d}\alpha_{i,j}|\Phi_{B[i]}\rangle\\!\langle\Phi_{B[j]}|,$
(B.4)

where
$\alpha=|\psi\rangle\\!\langle\psi|,|\psi\rangle=\frac{1}{3}(|0\rangle+|1\rangle+|2\rangle)$.
Then consider the states of the form

$\rho_{AB}=(1-p)\omega_{AB}+p\tau_{AB},$ (B.5)

where $p\in[0,1]$, the state $\tau_{AB}$ is the following PPT entangled state
with $a=\frac{1}{2}$:

$\tau_{AB}=\frac{1}{8a+1}\left(\begin{array}[]{ccccccccc}a&0&0&0&a&0&0&0&a\\\
0&a&0&0&0&0&0&0&0\\\ 0&0&a&0&0&0&0&0&0\\\ 0&0&0&a&0&0&0&0&0\\\
a&0&0&0&a&0&0&0&a\\\ 0&0&0&0&0&a&0&0&0\\\
0&0&0&0&0&0&\frac{1+a}{2}&0&\frac{\sqrt{1-a^{2}}}{2}\\\ 0&0&0&0&0&0&0&a&0\\\
a&0&0&0&a&0&\frac{\sqrt{1-a^{2}}}{2}&0&\frac{1+a}{2}\end{array}\right).$ (B.6)

We plot different upper bounds on $E_{D,\leftrightarrow}(\rho_{AB})$ in Fig.
S1. Note that the result presented in [17] is actually an approximation of
$E_{\rm MP}(\cdot)$, as it does not consider the minimization over all
possible decompositions. From the plot, we observe that our bound
$\hat{E}_{\rm rev}^{npt}(\rho_{AB})$ is tighter than both $E_{\rm W}(\cdot)$
and the approximation of $E_{\rm MP}(\cdot)$ as a function of $p$.

Fig S1: Upper bounds on the two-way distillable entanglement.
$\hat{E}_{\mathrm{rev}}^{npt}$ is our bound derived from the reverse max-
relative entropy of NPT entanglement. $E_{\rm W}$ is the SDP bound in [15].
$E_{\rm MP}$ is the upper bound in [17] based on the direct division of
$\rho_{AB}$ into the MC states and the PPT states.

##  Appendix C Proof of Proposition 3 and Proposition 4

######  Lemma S2 ([54])

For any states $\rho_{AB}$ and $\sigma_{AB}$ such that
$\frac{1}{2}\|\rho_{AB}-\sigma_{AB}\|_{1}\leq\varepsilon\leq 1$, it satisfies

$|S(A|B)_{\rho}-S(A|B)_{\sigma}|\leq
2\varepsilon\log(|A|)+(1+\varepsilon)h(\frac{\varepsilon}{1+\varepsilon}).$
(C.1)

###### Proposition 3

For any bipartite state $\rho_{AB}$ with an anti-degradable set distance
$\varepsilon_{\rm set}$, it satisfies

$E_{D,\to}(\rho_{AB})\leq 2\varepsilon_{\rm set}\log(|A|)+(1+\varepsilon_{\rm
set})h(\frac{\varepsilon_{\rm set}}{1+\varepsilon_{\rm set}}).$ (C.2)

###### Proof.

Since $\rho_{AB}$ has a anti-degradable set distance $\varepsilon_{\rm set}$,
we denote $\sigma_{AB}$ the anti-degradable state with
$\frac{1}{2}\|\rho_{AB}-\sigma_{AB}\|_{1}\leq\varepsilon_{\rm set}$. Let
$T\mathrel{\mathop{\mathchar 58\relax}}A^{n}\rightarrow A^{\prime}M$ be an
instrument with isometry $U_{n}\mathrel{\mathop{\mathchar
58\relax}}A^{n}\rightarrow A^{\prime}MN$ and denote $\Delta=2\varepsilon_{\rm
set}\log(|A|)+(1+\varepsilon_{\rm set})h(\frac{\varepsilon_{\rm
set}}{1+\varepsilon_{\rm set}})$. Then we have

$\displaystyle E^{(1)}_{D,\to}(\rho_{AB}^{\otimes n})$
$\displaystyle=\max_{U_{n}}I_{c}(A^{\prime}\rangle B^{n}M)_{U_{n}\rho^{\otimes
n}U_{n}^{\dagger}}$ (C.3a)
$\displaystyle=\max_{U_{n}}-S(A^{\prime}|B^{n}M)_{U_{n}\rho^{\otimes
n}U_{n}^{\dagger}}$ (C.3b)
$\displaystyle\leq\max_{U_{n}}-S(A^{\prime}|B^{n}M)_{U_{n}(\sigma\otimes\rho^{\otimes
n-1})U_{n}^{\dagger}}+\Delta$ (C.3c)
$\displaystyle\leq\max_{U_{n}}-S(A^{\prime}|B^{n}M)_{U_{n}\sigma^{\otimes
n}U_{n}^{\dagger}}+n\Delta$ (C.3d) $\displaystyle=n\Delta,$ (C.3e)

where Eq. (C.3c) follows by the fact that
$\|\rho_{0}\otimes\rho_{1}-\sigma_{0}\otimes\sigma_{1}\|_{1}\leq\|\rho_{0}-\sigma_{0}\|_{1}+\|\rho_{1}-\sigma_{1}\|_{1}$
and Lemma S2. The inequality in Eq. (C.3d) follows by applying the same
argument $n$ times considering $\sigma^{\otimes i}\rho^{\otimes n-i}$ for
$i=1,...,n$. After dividing Eq. (C.3) by $n$ and taking the limit
$n\rightarrow\infty$, we arrive at
$E_{D,\to}(\rho_{AB})=\lim_{n\rightarrow\infty}\frac{1}{n}E^{(1)}_{D,\to}(\rho_{AB}^{\otimes
n})\leq\Delta$.  $\sqcap$$\sqcup$

###### Proposition 4

For any bipartite state $\rho_{AB}$ with an anti-degradable map distance
$\varepsilon_{\rm map}$, it satisfies

$E_{D,\to}(\rho_{AB})\leq 4\varepsilon_{\rm map}\log(|B|)+2(1+\varepsilon_{\rm
map})h(\frac{\varepsilon_{\rm map}}{1+\varepsilon_{\rm map}}).$ (C.4)

###### Proof.

Let $\phi_{ABE}$ be a purification of $\rho_{AB}$, ${\cal
D}\mathrel{\mathop{\mathchar 58\relax}}E\rightarrow B$ be the CPTP map such
that $\frac{1}{2}\|\rho_{AB}-{\cal D}(\rho_{AE})\|_{1}\leq\varepsilon_{\rm
map}$ with an isometry $W\mathrel{\mathop{\mathchar 58\relax}}E\rightarrow
B^{\prime}G$. Let $T\mathrel{\mathop{\mathchar 58\relax}}A^{n}\rightarrow
A^{\prime}M$ be an instrument with isometry $U_{n}\mathrel{\mathop{\mathchar
58\relax}}A^{n}\rightarrow A^{\prime}MN$ and denote $\Delta=2\varepsilon_{\rm
map}\log(|B|)+(1+\varepsilon_{\rm map})h(\frac{\varepsilon_{\rm
map}}{1+\varepsilon_{\rm map}})$. For $t=1,2,...,n$, we can define pure states

$\displaystyle\psi_{A^{n}B^{n}B^{\prime}_{1}G_{1}...B^{\prime}_{t}G_{t}E_{t+1}...E_{n}}^{t}$
$\displaystyle=(W_{1}\otimes\cdots\otimes W_{t})\phi_{ABE}^{n}$ (C.5)
$\displaystyle\theta_{A^{\prime}MNB^{n}B^{\prime}_{1}G_{1}...B^{\prime}_{t}G_{t}E_{t+1}...E_{n}}^{t}$
$\displaystyle=U_{n}\psi^{t},\quad\omega_{A^{\prime}MNB^{n}E^{n}}=U_{n}\phi_{ABE}^{n}$

We further define $\hat{\rho}_{AB^{\prime}}={\cal D}(\rho_{AE})$ which shares
the same purification with $\rho_{AB}$ listed above, thus an anti-degradable
state. Then for $t=n$ we have $\theta^{n}=U_{n}(W_{1}\otimes\cdots\otimes
W_{n})U_{n}^{\dagger}\omega$, it yields

$\displaystyle I_{c}(A^{\prime}\rangle B^{n}M)_{\omega}$
$\displaystyle=I_{c}(A^{\prime}\rangle
B^{n}M)_{\theta^{n}_{A^{\prime}MNB^{n}B^{\prime n}G^{n}}}$ (C.6)
$\displaystyle=S(B^{n}M)_{\theta}-S(A^{\prime}B^{n}M)_{\theta}=S(A^{\prime}NB^{\prime
n}G^{n})_{\theta}-S(A^{\prime}B^{n}M)_{\theta}$
$\displaystyle=S(A^{\prime}NB^{\prime n}G^{n})_{\theta}-S(A^{\prime}B^{\prime
n}M)_{\theta}+S(A^{\prime}B^{\prime
n}M)_{\theta}-S(A^{\prime}B^{n}M)_{\theta}$
$\displaystyle=S(G^{n}|A^{\prime}B^{\prime n}M)_{\theta}+S(A^{\prime}B^{\prime
n}M)_{\theta}-S(A^{\prime}B^{n}M)_{\theta}$

where we abbreviate $\theta=\theta^{n}_{A^{\prime}MNB^{n}B^{\prime n}G^{n}}$.
Applying the same technique in the proof of Theorem 2.12 in [17], we can bound
$S(A^{\prime}B^{\prime n}M)_{\theta}-S(A^{\prime}B^{n}M)_{\theta}\leq
n\Delta$. Consequently, it follows that

$\displaystyle I_{c}(A^{\prime}\rangle B^{n}M)_{\omega}$ $\displaystyle\leq
S(G^{n}|A^{\prime}B^{\prime n}M)_{\theta}+n\Delta$ (C.7) $\displaystyle\leq
S(G^{n}|B^{\prime n})_{\theta}+n\Delta=S(G^{n}B^{\prime
n})_{\theta}-S(B^{\prime n})_{\theta}+n\Delta$ $\displaystyle\leq
S(G^{n}B^{\prime n})_{\theta}-S(A^{\prime}MNG^{n}B^{\prime
n})_{\theta}+2n\Delta$ $\displaystyle\leq I(A^{\prime}MN\rangle G^{n}B^{\prime
n})_{\theta}+2n\Delta$ $\displaystyle=I(A^{n}\rangle B^{\prime
n})_{\hat{\rho}_{AB^{\prime}}^{\otimes n}}+2n\Delta=n[I(A\rangle
B)_{\hat{\rho}_{AB^{\prime}}}+2\Delta]=2n\Delta,$

where the last equality is due to the anti-degradability of
$\hat{\rho}_{AB^{\prime}}$. After dividing Eq. (C.7) by $n$ and taking the
limit $n\rightarrow\infty$, we arrive at
$E_{D,\to}(\rho_{AB})=\lim_{n\rightarrow\infty}\frac{1}{n}E^{(1)}_{D,\to}(\rho_{AB}^{\otimes
n})\leq 2\Delta$.  $\sqcap$$\sqcup$

##  Appendix D Proof of Proposition 9

######  Lemma S3 ([72])

A two qubit state $\rho_{AB}$ is anti-degradable if and only if,

$\operatorname{Tr}\left(\rho_{B}^{2}\right)\geq\operatorname{Tr}\left(\rho_{AB}^{2}\right)-4\sqrt{\operatorname{det}\left(\rho_{AB}\right)}.$
(D.1)

###### Proposition 9

For a qubit Pauli channel $\Lambda(\cdot)$ with $p_{0}\geq p_{i}>0(i=1,2,3)$,
it is either anti-degradable or satisfies

$2^{-\widetilde{{\cal R}}_{\max,{\rm
ADG}}(\Lambda)}=(\sqrt{p_{1}}+\sqrt{p_{2}})^{2}+(\sqrt{p_{2}}+\sqrt{p_{3}})^{2}+(\sqrt{p_{1}}+\sqrt{p_{3}})^{2}$
(D.2)

with an ADG-squeezed channel as the identity channel.

###### Proof.

We first will prove

$2^{-\widetilde{{\cal R}}_{\max,{\rm
ADG}}(\Lambda)}\geq(\sqrt{p_{1}}+\sqrt{p_{2}})^{2}+(\sqrt{p_{2}}+\sqrt{p_{3}})^{2}+(\sqrt{p_{1}}+\sqrt{p_{3}})^{2},$
(D.3)

by using the SDP in Eq. (A.5). We show that $\Gamma_{AB}^{{\cal
S}}=\frac{\alpha}{2}\left(|0\rangle\\!\langle 0|+|0\rangle\\!\langle
1|+|1\rangle\\!\langle 0|+|1\rangle\\!\langle 1|\right)$, is a feasible
solution where
$\alpha=1-[(\sqrt{p_{1}}+\sqrt{p_{2}})^{2}+(\sqrt{p_{2}}+\sqrt{p_{3}})^{2}+(\sqrt{p_{1}}+\sqrt{p_{3}})^{2}]$.
Note that the Choi state of the Pauli channel is

$J_{AB}^{\Lambda}=\frac{1}{2}\left(\begin{array}[]{cccc}p_{0}+p_{3}&0&0&p_{0}-p_{3}\\\
0&p_{1}+p_{2}&p_{1}-p_{2}&0\\\ 0&p_{1}-p_{2}&p_{1}+p_{2}&0\\\
p_{0}-p_{3}&0&0&p_{0}+p_{3}\\\ \end{array}\right),$ (D.4)

and the unnormalized state $\Gamma_{AB}^{{\cal
N}^{\prime}}=J_{AB}^{\Lambda}-\Gamma_{AB}^{\cal S}$ is

$\Gamma_{AB}^{{\cal
N}^{\prime}}=\frac{1}{2}\left(\begin{array}[]{cccc}p_{0}+p_{3}-\alpha&0&0&p_{0}-p_{3}-\alpha\\\
0&p_{1}+p_{2}&p_{1}-p_{2}&0\\\ 0&p_{1}-p_{2}&p_{1}+p_{2}&0\\\
p_{0}-p_{3}-\alpha&0&0&p_{0}+p_{3}-\alpha\\\ \end{array}\right).$ (D.5)

Recalling that $p_{0}+p_{1}+p_{2}+p_{3}=1$, it is then straightforward to
check that $\operatorname{Tr}_{B}[\Gamma_{AB}^{{\cal
N}^{\prime}}]=\operatorname{Tr}[\Gamma_{AB}^{{\cal N}^{\prime}}]/d_{A}\cdot
I_{A}$. The constraint in Eq. (A.5e) corresponds to the anti-degradable
condition of $\Gamma_{AB}^{{\cal N}^{\prime}}$. By direct calculation, we have

$\displaystyle\operatorname{Tr}[(\Gamma_{B}^{{\cal N}^{\prime}})^{2}]$
$\displaystyle=\frac{1}{2}(1-\alpha)^{2},\quad\det(\Gamma_{AB}^{{\cal
N}^{\prime}})=p_{1}p_{2}p_{3}(p_{0}-\alpha),$ (D.6)
$\displaystyle\operatorname{Tr}[(\Gamma_{AB}^{{\cal N}^{\prime}})^{2}]$
$\displaystyle=\alpha^{2}-2\alpha
p_{0}+p_{0}^{2}+p_{1}^{2}+p_{2}^{2}+p_{3}^{2}.$

Then Eq. (D.1) holds and $\Gamma_{AB}^{{\cal N}^{\prime}}$ is anti-degradable
by Lemma S3, which satisfy the constraint in Eq. (A.5e). Thus, we have proven
that $\Gamma_{AB}^{\cal S}$ is a feasible solution to the primal SDP, which
yields

$2^{-\widetilde{{\cal R}}_{\max,{\rm
ADG}}(\Lambda)}\geq(\sqrt{p_{1}}+\sqrt{p_{2}})^{2}+(\sqrt{p_{2}}+\sqrt{p_{3}})^{2}+(\sqrt{p_{1}}+\sqrt{p_{3}})^{2}.$
(D.7)

Second, we will use the dual SDP in Eq. (A.9) to prove

$2^{-\widetilde{{\cal R}}_{\max,{\rm
ADG}}(\Lambda)}\leq(\sqrt{p_{1}}+\sqrt{p_{2}})^{2}+(\sqrt{p_{2}}+\sqrt{p_{3}})^{2}+(\sqrt{p_{1}}+\sqrt{p_{3}})^{2}.$
(D.8)

We show that $\\{M_{AB},N_{AB},K_{AB},R_{A}\\}$ is a feasible solution to the
dual problem, where

$M_{AB}=\left(\begin{array}[]{cccc}\eta&0&0&-\eta+1\\\ 0&\xi&\zeta&0\\\
0&\zeta&\xi&0\\\ -\eta+1&0&0&\eta\\\
\end{array}\right),N_{AB}=K_{AB}=-\frac{1}{2}M_{AB},R_{A}=0,$ (D.9)
$\eta=-\frac{\sqrt{p_{1}}+\sqrt{p_{2}}}{2\sqrt{p_{3}}},\xi=-\frac{\sqrt{p_{1}}+\sqrt{p_{3}}}{2\sqrt{p_{2}}}-\frac{\sqrt{p_{2}}+\sqrt{p_{3}}}{2\sqrt{p_{1}}}-1,\zeta=\frac{\sqrt{p_{1}}+\sqrt{p_{3}}}{2\sqrt{p_{2}}}-\frac{\sqrt{p_{2}}+\sqrt{p_{3}}}{2\sqrt{p_{1}}}.$
(D.10)

It is easy to check that when $p_{0}\geq p_{i}>0(i=1,2,3)$, we have
$M_{AB}+N_{AB}+K_{AB}=0$,

$\displaystyle M_{AB}\leq(1-\operatorname{Tr}R_{A}/d_{A})I_{AB}+R_{A}\otimes
I_{B}\Leftrightarrow I_{AB}-M_{AB}$ $\displaystyle\geq 0,\;\text{and}$ (D.11)
$\displaystyle N_{AB}\otimes I_{E}+P_{BE}(K_{AB}\otimes
I_{E})P_{BE}^{\dagger}=\frac{1}{2}M_{AB}\otimes
I_{E}+P_{BE}(\frac{1}{2}M_{AB}\otimes I_{E})P_{BE}^{\dagger}$
$\displaystyle\geq 0.$

It also satisfies
$\operatorname{Tr}[M_{AB}J_{AB}^{\Lambda}]=1-[(\sqrt{p_{1}}+\sqrt{p_{2}})^{2}+(\sqrt{p_{2}}+\sqrt{p_{3}})^{2}+(\sqrt{p_{1}}+\sqrt{p_{3}})^{2}]$.
Thus we have proven that $\\{M_{AB},N_{AB},K_{AB},R_{A}\\}$ is a feasible
solution to the dual SDP in Eq. (A.9), which yields

$2^{-\widetilde{{\cal R}}_{\max,{\rm
ADG}}(\Lambda)}\leq(\sqrt{p_{1}}+\sqrt{p_{2}})^{2}+(\sqrt{p_{2}}+\sqrt{p_{3}})^{2}+(\sqrt{p_{1}}+\sqrt{p_{3}})^{2}.$
(D.12)

Thus, we arrive at $2^{-\widetilde{{\cal R}}_{\max,{\rm
ADG}}(\Lambda)}=(\sqrt{p_{1}}+\sqrt{p_{2}})^{2}+(\sqrt{p_{2}}+\sqrt{p_{3}})^{2}+(\sqrt{p_{1}}+\sqrt{p_{3}})^{2}$.
Since $\Gamma_{AB}^{{\cal S}}$ is the Bell state after normalization, we know
the ADG-squeezed channel is the identity channel.

$\sqcap$$\sqcup$