# Entanglement measures for detectability

Masahito Hayashi<EMAIL_ADDRESS>School of Data Science, The Chinese
University of Hong Kong, Shenzhen, Longgang District, Shenzhen, 518172, China
International Quantum Academy (SIQA), Futian District, Shenzhen 518048, China
Graduate School of Mathematics, Nagoya University, Nagoya, 464-8602, Japan
Yuki Ito<EMAIL_ADDRESS>Graduate School of Mathematics,
Nagoya University, Nagoya, 464-8602, Japan

###### Abstract

We propose new entanglement measures as the detection performance based on the
hypothesis testing setting. We clarify how our measures work for detecting an
entangled state by extending the quantum Sanov theorem. Our analysis covers
the finite-length setting. Exploiting this entanglement measure, we present
how to derive entanglement witness to detect the given entangled state by
using the geometrical structure of this measure. We derive their calculation
formulas for maximally correlated states, and propose their algorithms that
work for general entangled states. In addition, we investigate how our
algorithm works for solving the membership problem for separability.

##  I Introduction

Entanglement is a valuable resource for quantum information processing. It is
important to distinguish a given entangled state from separable states, i.e.,
non-entangled states. However, detecting an entanglement state has two
meanings, the physical detection and the mathematical detection. The physical
detection means the experimental detection of a generated entangled state. The
mathematical detection means identifying whether a given density matrix has a
separable form or not. Since the mathematical detection is often called the
membership problem of separability, in this paper, we consider that
entanglement detection means the physical one while this paper address both
problems.

Several practical methods to achieve entanglement detection have been proposed
[1, 2, 3, 4, 5, 6, 7], but they did not discuss the optimal performance for
the detection. On the other hand, to characterize the amount of entanglement,
many entangled measures have been proposed. However, these proposed entangled
measures have less relation with the experimental detectability of
entanglement. In fact, when Vedral et al [8, 9] introduced entanglement of
relative entropy, they pointed out the relation between the hypothesis testing
and the relative entropy. They employed the minimum relative entropy between
the given entangled state and a separable state. However, this measure
expresses the performance of the discrimination between the given entangled
state and the closest separable state in the sense of relative entropy. When
one hypothesis is chosen to be the set of all separable states, this measure
does not express the performance of this discrimination [10]. Until now, no
study discussed the optimal detection efficiency for entanglement
verification.

In this paper, to resolve this problem, we propose a new entanglement measure
that expresses the optimal performance of the discrimination between the given
entangled state and the set of separable states. In this characterization, in
the first step, we employ the quantum Sanov theorem [11, 12] for hypothesis
testing, and consider the same minimization as entanglement of relative
entropy by exchanging the first and second inputs in the relative entropy. The
proposed quantity satisfies the entanglement monotone. To enhance this
characterization, we derive a much simpler upper bound of the error
probability in the finite-length setting of this kind of hypothesis testing,
which can be considered another proof of the quantum Sanov theorem. In
addition, we derive the strong converse exponent of the same problem setting
as the quantum Sanov theorem. However, when we simply apply the quantum Sanov
theorem, we assume that the separable states are limited to the iid states of
separable states. To remove this assumption, in the second step, we derive the
optimal performance of this discrimination when possible separable states are
allowed to take a form of a general separable state.

This entanglement measure is useful for finding an entanglement witness to
detect the target entangled state. This relation is characterized by
information geometry [13]. Also. we calculate the proposed entanglement
measures when a state is a maximally correlated state. This class contains all
pure states. For this calculation, we introduce a notable relation with
coherence measures, and several calculation formulas for coherence measures.

In the next step, we propose an efficient algorithm to compute the proposed
entanglement measure on the joint system
${\mathcal{H}}_{A}\otimes{\mathcal{H}}_{B}$. It is quite difficult to directly
calculate entanglement measures because the calculations of entanglement
measures are directly related to the membership problem of separability, which
is known as NP-hard [14, 15, 16, 17]. Similar to the paper [17, Section 3], we
employ $\delta$-nets on the second system ${\mathcal{H}}_{B}$. In the
following, we assume that the first system ${\mathcal{H}}_{A}$ has a larger
dimension. That is, to optimize the choice of a separable state, we fix as
many states on ${\mathcal{H}}_{B}$ as possible, and optimize the choice of
states on ${\mathcal{H}}_{A}$. Then, we define an approximate value of the
proposed measure. Employing the quantum Arimoto-Blahut algorithm proposed by
the paper [18], we propose an algorithm for the approximate value. This
algorithm is an iterative algorithm, and we show its convergence to the global
minimum and its convergence speed.

As the third step, we apply our algorithm to the membership problem of
separability. Although it is known that this problem is NP-hard [14, 15, 16,
17], several algorithms have been proposed for this problem. As a typical
method, Doherty et al. [19, 20] proposed a method to identify whether a given
state is a separable state or not by considering “symmetric extensions” and
“PPT-symmetric extensions” in the second system ${\mathcal{H}}_{B}$. This kind
of extension is called DPS hierarchy. Later, Navascués et al. [21] evaluated
its calculation complexity. DPS hierarchy was discussed in [22, (14)], [23,
24]. The paper [17] proposed another algorithm for this problem by using
$\delta$-nets on the second system ${\mathcal{H}}_{B}$. The paper [17] also
discussed its calculation complexity. It was shown that the calculation
complexity of these preceding algorithms are exponential only for the
dimension of the second system ${\mathcal{H}}_{B}$. The papers [21, 17]
evaluated the exponential component for the dimension of the second system
${\mathcal{H}}_{B}$ in their algorithms. The paper [21] evaluated the
exponential component for the dimension of the first system
${\mathcal{H}}_{A}$, but the paper [17] evaluated the polynomial component for
the dimension of the first system ${\mathcal{H}}_{A}$. This paper discussed
our calculation complexity, and showed that the calculation complexity of our
algorithm is exponential only for the dimension of the second system
${\mathcal{H}}_{B}$. In addition, we evaluated the exponential component for
the dimension of the second system ${\mathcal{H}}_{B}$ in our algorithm, and
evaluated the polynomial component for the dimension of the first system
${\mathcal{H}}_{A}$. As the comparison with the evaluation by [21], we find
that our exponential component for the dimension of the second system
${\mathcal{H}}_{B}$ is larger than that in the analysis [21], but our
polynomial component for the dimension of the first system ${\mathcal{H}}_{A}$
is smaller than that in the analysis [21]. This fact shows that our algorithm
has an advantage over the algorithm based on “symmetric extensions” and “PPT-
symmetric extensions” when the dimension of the second system
${\mathcal{H}}_{B}$ is fixed and only the dimension of the first system
${\mathcal{H}}_{A}$ increases.

Finally, we apply our algorithm to finding a separation hyperplane. When we
solve a conic linear programming with a general cone, it is a key step to find
a separation hyperplane for an element that does not belong to the cone [25,
26, 27, 28]. Recently, a conic linear programming with the separable cone has
been applied to several problems in quantum information [29, 30]. Therefore,
it is desired to find a separation hyperplane for the separable cone. In
quantum information, a separation hyperplane for the separable cone is called
an entanglement witness [31, 32, 33, 34, 35]. In this paper, based on our
algorithm, we also propose an algorithm to find an entanglement witness, i.e.,
a separation hyperplane for an entangled state.

The remainder of this paper is organized as follows. Section II formulates the
detection of an entangled state as a problem of hypothesis testing under the
iid setting. Section III addresses this problem under the non-iid setting.
Section IV explains information geometrical characterization of our proposed
measure and its relation to entanglement witness. Section V proposes and
studies new coherence measures as the preparation of the calculation of our
entanglement measures. Section VI derives several calculation formulas for our
entanglement measures for maximally entangled states based on the analysis in
Section V. Section VII proposes algorithms for our entanglement measures by
using $\delta$-net. Section VIII applies our results to the membership problem
for separability and separation hyperplane. Section IX numerically calculates
our entanglement measure, and check how our method works for the membership
problem. Section IV explains that our method finds an entanglement witness.
Section X presents our exponential version of quantum Sanov theorem. Section
XI makes conclusions and discussions. Appendix A presents concrete
constructions of $\delta$-nets. Appendix B proves the theorem for the relation
between our entanglement measures and our proposed coherence measure. Appendix
C proves theorems for the convergence for our algorithms. Appendix D proves
our exponential version of quantum Sanov theorem. Appendix E proves our
exponential evaluation of error probabilities in our problem of hypothesis
testing under the non-iid setting.

##  II Detecting entangled state with iid setting

We consider the joint system ${\mathcal{H}}_{A}\otimes{\mathcal{H}}_{B}$, and
discuss how efficiently an entangled state $\rho_{AB}$ on
${\mathcal{H}}_{A}\otimes{\mathcal{H}}_{B}$ is distinguished from the set
${\cal S}_{AB}$ of separable states on
${\mathcal{H}}_{A}\otimes{\mathcal{H}}_{B}$. This problem can be formulated as
a hypothesis testing problem to distinguish $\rho_{AB}$ from ${\cal S}_{AB}$.
For this detection, we assume the $n$-fold independently and identical
distributed (iid) condition. That is, we have the following two hypotheses for
the possible state as follows.

$\displaystyle H_{0}:$ $\displaystyle\hbox{The state is }\rho_{AB}^{\otimes
n}.$ (1) $\displaystyle H_{1}:$ $\displaystyle\hbox{The state is
}\sigma_{AB}^{\otimes n}\hbox{ with }\sigma_{AB}\in{\cal S}_{AB}.$ (2)

To address this problem, we focus on the following quantity.

$\displaystyle\beta_{\epsilon,n}(\rho_{AB})$ $\displaystyle:=$
$\displaystyle\min_{0\leq T\leq
I}\Big{\\{}\operatorname{\mathrm{Tr}}T\rho_{AB}^{\otimes
n}\Big{|}\max_{\sigma_{AB}\in{\cal
S}_{AB}}\operatorname{\mathrm{Tr}}(I-T)\sigma_{AB}^{\otimes
n}\leq\epsilon\Big{\\}}.$ (3)

In the above setting, we do not impose the LOCC restriction to our measurement
unlike [1, 33]. However, these existing studies did not discuss the optimal
detection efficiency for the entanglement verification. Hence, the above
quantity expresses the performance of discrimination at the fundamental level.

Here, it is better to explain the relation with the entanglement verification
[36, 37, 38, 39, 40, 41]. Entanglement verification is the task of identifying
a given entangled state within a certain error. Since entanglement
verification is a more difficult task, the detection efficiency of the
entanglement verification is worse than the detection efficiency of the
entanglement detection. Existing studies for entanglement verification are
limited to the verification for several special entangled states. No general
theory covers the entanglement verification for general entangled states.

The application of quantum Sanov theorem [11, 12] guarantees the following

$\displaystyle\lim_{n\to\infty}-\frac{1}{n}\log\beta_{\epsilon,n}(\rho_{AB})=E^{*}(\rho_{AB})$
$\displaystyle:=$ $\displaystyle\min_{\sigma_{AB}\in{\cal
S}_{AB}}D(\sigma_{AB}\|\rho_{AB}),$ (4)

where
$D(\sigma\|\rho):=\operatorname{\mathrm{Tr}}\sigma(\log\sigma-\log\rho)$. This
value is different from entanglement of relative entropy
$E(\rho_{AB}):=\min_{\sigma_{AB}\in{\cal S}_{AB}}D(\rho_{AB}\|\sigma_{AB})$.
This fact shows that our measure $E^{*}(\rho_{AB})$ expresses the limit of the
performance to distinguish the entangled state $\rho_{AB}$ from ${\cal
S}_{AB}$. In contrast, when the choice of $H_{0}$ and $H_{1}$ is exchanged,
$E(\rho_{AB})$ does not have the same meaning [10]. The proof of (4) is given
as follows. The part $\leq$ in (4) follows from the strong converse part of
quantum Stein’s lemma. The paper [11] showed the part $\geq$ in (4). However,
they did not consider the case when $\epsilon$ is exponentially small. To
enhance the meaning of the quantity $E^{*}(\rho_{AB})$, it is needed to
evaluate $\beta_{\epsilon,n}(\rho_{AB})$ in such a case.

Table 1: List of mathematical symbols for detecting entangled state from Section II onwards. Symbol | Description | Eq. number
---|---|---
${\cal S}_{AB}$ | Set of separable states on |
${\mathcal{H}}_{A}\otimes{\mathcal{H}}_{B}$ |
$\beta_{\epsilon,n}(\rho_{AB})$ | Minimum error probability | (3)
with constraint $\epsilon$ when
$H_{1}$ is given as (2)
$E^{*}(\rho_{AB})$ | Minimum relative entropy | (4)
$E_{\alpha}^{*}(\rho_{AB})$ | Minimum relative Rényi | (5)
entropy of order $\alpha$
${\cal S}_{AB}^{n}$ | Set of separable states on |
${\mathcal{H}}_{A}^{\otimes n}\otimes{\mathcal{H}}_{B}^{\otimes n}$ |
$\overline{\beta}_{\epsilon,n}(\rho_{AB})$ | Minimum error probability | (3)
with constraint $\epsilon$ when
$H_{1}$ is given as (10)
$\overline{E}^{*}(\rho_{AB})$ | Normalized minimum | (12)
relative entropy
$\overline{E}_{\alpha}^{*}(\rho_{AB})$ | Normalized minimum relative | (13)
Rényi entropy of order $\alpha$
$\tilde{E}^{*}(\rho_{AB})$ | The above value with | (18)
$\alpha=1-0$
$\sigma_{AB}^{*}$ | Closest separable state | (19)
${\cal S}_{D}$ | Divergence sphere | (25)
${\cal M}(\sigma_{AB}^{*},\rho_{AB})$ | Separation hyperplane | (20)
$|\Phi\rangle$ | Maximally entangled state |
$\frac{1}{\sqrt{d}}\sum_{j=0}^{d-1}|j\rangle|j\rangle$ |
$E_{{\cal D}}^{*}(\rho_{AB})$ | Minimum relative entropy | (58)
with fixed finite set ${\cal D}$
$\epsilon_{2}({\cal D})$ |  | (59)
$E_{{\alpha,\cal D}}^{*}(\rho_{AB})$ | Minimum relative Rényi | (70)
entropy of order $\alpha$
with fixed finite set ${\cal D}$

For this aim, using the sandwich relative entropy
$D_{\alpha}(\sigma\|\rho):=\frac{1}{\alpha-1}\log\operatorname{\mathrm{Tr}}(\rho^{\frac{1-\alpha}{2\alpha}}\sigma\rho^{\frac{1-\alpha}{2\alpha}})^{\alpha}$,
we generalize $E^{*}(\rho_{AB})$ as

$\displaystyle E_{\alpha}^{*}(\rho_{AB}):=\min_{\sigma_{AB}\in{\cal
S}_{AB}}D_{\alpha}(\sigma_{AB}\|\rho_{AB}).$ (5)

As presented in Section X, we have the following theorem.

###### Theorem 1

We have the following relations

$\displaystyle\lim_{n\to\infty}-\frac{1}{n}\log\beta_{e^{-nr},n}(\rho_{AB})$
$\displaystyle\geq$
$\displaystyle\sup_{0\leq\alpha<1}\frac{(1-\alpha)E_{\alpha}^{*}(\rho_{AB})-\alpha
r}{1-\alpha},$ (6)
$\displaystyle-\frac{1}{n}\log\beta_{1-e^{-nr},n}(\rho_{AB})$
$\displaystyle\leq$ $\displaystyle\inf_{\alpha\geq
1}\frac{(1-\alpha)E_{\alpha}^{*}(\rho_{AB})-\alpha r}{1-\alpha}.$ (7)

More precisely, we have

$\displaystyle-\frac{1}{n}\log\beta_{e^{-nr},n}(\rho_{AB})$
$\displaystyle\geq$
$\displaystyle\sup_{0\leq\alpha<1}\frac{(1-\alpha)E_{\alpha}^{*}(\rho_{AB})-\alpha
r}{1-\alpha}$ $\displaystyle-\frac{(d_{AB}+2)(d_{AB}-1)}{n}\log(n+1),$ (8)

where $d_{AB}$ is the dimension of
${\mathcal{H}}_{A}\otimes{\mathcal{H}}_{B}$.

Under the limit $r\to 0$, the RHS of (6) goes to
$E_{1}^{*}(\rho_{AB}):=E^{*}(\rho_{AB})$. Hence, the relation (6) recovers the
inequality $\geq$ in (4). When a TP-CP map $\Lambda$ preserves the set ${\cal
S}_{AB}$, these quantities satisfy the monotonicity as

$\displaystyle E^{*}(\rho_{AB})\geq E^{*}(\Lambda(\rho_{AB})),\quad
E_{\alpha}^{*}(\rho_{AB})\geq E_{\alpha}^{*}(\Lambda(\rho_{AB})).$ (9)

##  III Detecting entangled state with non-iid setting

The above discussion assumes the iid assumption for the alternative
hypothesis. However, several existing studies for entanglement verification
and entanglement detection assume that the alternative hypothesis includes
general separable states including non-iid separable stets [5, 4, 39, 40, 41],
which is called the non-iid setting. Our analysis can be extended to the non-
iid setting as follows. We denote the set of separable states on
${\mathcal{H}}_{A}^{\otimes n}\otimes{\mathcal{H}}_{B}^{\otimes n}$ by ${\cal
S}_{AB}^{n}$. Then, the alternative hypothesis is changed to

$\displaystyle H_{1}:$ $\displaystyle\hbox{The state is }\sigma\hbox{ with
}\sigma\in{\cal S}_{AB}^{n}$ (10)

while the null hypothesis is kept to (1). To address this problem, we focus on
the following quantity.

$\displaystyle\overline{\beta}_{\epsilon,n}(\rho_{AB})$ $\displaystyle:=$
$\displaystyle\min_{0\leq T\leq
I}\Big{\\{}\operatorname{\mathrm{Tr}}T\rho_{AB}^{\otimes
n}\Big{|}\max_{\sigma_{AB}\in{\cal
S}_{AB}^{n}}\operatorname{\mathrm{Tr}}(I-T)\sigma_{AB}\leq\epsilon\Big{\\}}.$
(11)

The relation (4) yields that

$\displaystyle\lim_{n\to\infty}-\frac{1}{n}\log\overline{\beta}_{\epsilon,n}(\rho_{AB})\leq\overline{E}^{*}(\rho_{AB}):=\inf_{n}\frac{1}{n}{E}^{*}(\rho_{AB}^{\otimes
n})$ $\displaystyle=$
$\displaystyle\lim_{n\to\infty}\frac{1}{n}{E}^{*}(\rho_{AB}^{\otimes
n})=\lim_{n\to\infty}\frac{1}{n}\min_{\sigma\in{\cal
S}_{AB}^{n}}D(\sigma\|\rho_{AB}^{\otimes n}).$ (12)

Extending ${E}_{\alpha}^{*}(\rho_{AB})$, we define

$\displaystyle\overline{E}_{\alpha}^{*}(\rho_{AB}):=$
$\displaystyle\inf_{n}\frac{1}{n}{E}_{\alpha}^{*}(\rho_{AB}^{\otimes
n})=\lim_{n\to\infty}\frac{1}{n}{E}_{\alpha}^{*}(\rho_{AB}^{\otimes n})$
$\displaystyle=$
$\displaystyle\lim_{n\to\infty}\frac{1}{n}\min_{\sigma_{AB}\in{\cal
S}_{AB}^{n}}D_{\alpha}(\sigma_{AB}\|\rho_{AB}^{\otimes n}).$ (13)

As shown in Appendix E, we have the following theorem.

###### Theorem 2

We have

$\displaystyle\lim_{n\to\infty}-\frac{1}{n}\log\overline{\beta}_{e^{-nr},n}(\rho_{AB})$
$\displaystyle\geq$ $\displaystyle\sup_{0\leq\alpha\leq
1}\frac{(1-\alpha)\overline{E}_{\alpha}^{*}(\rho_{AB})-\alpha r}{1-\alpha},$
(14)
$\displaystyle\lim_{n\to\infty}-\frac{1}{n}\log\overline{\beta}_{1-e^{-nr},n}(\rho_{AB})$
$\displaystyle=$ $\displaystyle\inf_{\alpha\geq
1}\frac{(1-\alpha)\overline{E}_{\alpha}^{*}(\rho_{AB})-\alpha r}{1-\alpha}.$
(15)

More precisely, we have

$\displaystyle-\frac{1}{n}\log\beta_{e^{-nr},n}(\rho_{AB})$
$\displaystyle\geq$ $\displaystyle\sup_{0\leq\alpha\leq
1}\frac{\frac{1}{n}(1-\alpha)E_{\alpha}^{*}(\rho_{AB}^{\otimes n})-\alpha
r}{1-\alpha}$ $\displaystyle-\frac{(d_{AB}+2)(d_{AB}-1)}{n}\log(n+1),$ (16)
$\displaystyle-\frac{1}{n}\log\beta_{1-e^{-nr},n}(\rho_{AB})$
$\displaystyle\leq$ $\displaystyle\inf_{\alpha\geq
1}\frac{\frac{1}{n}(1-\alpha){E}_{\alpha}^{*}(\rho_{AB}^{\otimes n})-\alpha
r}{1-\alpha}.$ (17)

As an inequality opposite to (4), the inequality (14) implies the following
inequality.

$\displaystyle\lim_{n\to\infty}-\frac{1}{n}\log\overline{\beta}_{\epsilon,n}(\rho_{AB})\geq\tilde{E}^{*}(\rho_{AB}):=\lim_{\delta\to+0}\overline{E}_{1-\delta}^{*}(\rho_{AB}).$
(18)

While $\overline{E}^{*}(\rho_{AB})\geq\tilde{E}^{*}(\rho_{AB})$, it is not
clear whether equality holds in general. When the quantity
$E_{\alpha}^{*}(\rho_{AB})$ satisfies the additivity
$E_{\alpha}^{*}(\rho_{AB,1}\otimes\rho_{AB,2})=E_{\alpha}^{*}(\rho_{AB,1})+E_{\alpha}^{*}(\rho_{AB,2})$,
the equations
${E}^{*}(\rho_{AB})=\overline{E}^{*}(\rho_{AB})=\tilde{E}^{*}(\rho_{AB})$ and
${E}_{\alpha}^{*}(\rho_{AB})=\overline{E}_{\alpha}^{*}(\rho_{AB})$ hold.
Therefore, it is crucial to clarify when the above additivity holds.

##  IV Information-geometrical characterization

Given a non-separable state $\rho_{AB}\notin{\cal S}_{AB}$, we discuss how to
find an entanglement witness to distinguish $\rho_{AB}$ and ${\cal S}_{AB}$.
For this aim, we employ the minimizer of $E^{*}(\rho_{AB})$, i.e., the state

$\displaystyle\sigma_{AB}^{*}:=\mathop{\rm argmin}\limits_{\sigma_{AB}\in{\cal
S}_{AB}}D(\sigma_{AB}\|\rho_{AB}).$ (19)

Then, using the operator $(\log\sigma_{AB}^{*}-\log\rho_{AB})$, we define the
hyperplane ${\cal M}(\sigma_{AB}^{*},\rho_{AB})$ as (See Fig. 1.)

$\displaystyle{\cal M}(\sigma_{AB}^{*},\rho_{AB})$ $\displaystyle:=$
$\displaystyle\\{\rho^{\prime}_{AB}|\operatorname{\mathrm{Tr}}\rho_{AB}^{\prime}(\log\sigma_{AB}^{*}-\log\rho_{AB})=D(\sigma_{AB}^{*}\|\rho_{AB})\\}.$
(20)

Figure 1: Entanglement witness: This figure shows the relation among the
entangled state $\rho_{AB}$, the set of separable states ${\cal S}_{AB}$, and
the separation hyperplane ${\cal M}(\sigma_{AB}^{*},\rho_{AB})$.

The following theorem guarantees that ${\cal M}(\sigma_{AB}^{*},\rho_{AB})$
separate $\rho_{AB}$ from ${\cal S}_{AB}$. Hence, the operator
$(\log\sigma_{AB}^{*}-\log\rho_{AB})-D(\sigma_{AB}^{*}\|\rho_{AB})$ works as
entanglement witness for $\rho_{AB}$.

###### Theorem 3

(i) For any separable state $\sigma_{AB}\in{\cal S}$, we have

$\displaystyle\operatorname{\mathrm{Tr}}\sigma_{AB}(\log\sigma_{AB}^{*}-\log\rho_{AB})\geq
D(\sigma_{AB}^{*}\|\rho_{AB}).$ (21)

(ii) Also, we have

$\displaystyle\operatorname{\mathrm{Tr}}\rho_{AB}(\log\sigma_{AB}^{*}-\log\rho_{AB})=-D(\rho_{AB}\|\sigma_{AB}^{*})<0.$
(22)

To grasp the geometrical meaning of the above fact, we employ information
geometry. Given a Hermitian matrix $X$, we often focus on the set ${\cal
E}:=\\{\rho_{t}\\}$, where $\rho_{t}:=\exp(tX+\log\rho_{0})/c_{t}$ with
$c_{t}:=\operatorname{\mathrm{Tr}}\exp(tX+\log\rho_{0})$. The set
$\\{\rho_{t}\\}$ is called an exponential family generated by $X$. Also, the
set ${\cal M}:=\\{\rho|\operatorname{\mathrm{Tr}}\rho
X=\operatorname{\mathrm{Tr}}\rho_{0}X\\}$ is called a mixture family generated
by $X$. Any elements $\sigma\in{\cal M}$ and $\rho\in{\cal E}$ satisfy

$\displaystyle D(\sigma\|\rho)=D(\sigma\|\rho_{0})+D(\rho_{0}\|\rho),$ (23)

which is called Pythagorean theorem [13], [42, Theorem 2.3]. We apply (23) to
the case with $X=-(\log\sigma_{AB}^{*}-\log\rho_{AB})$ and
$\rho_{0}=\sigma_{AB}^{*}$. Any element $\sigma_{AB}^{\prime}\in{\cal
M}(\sigma_{AB}^{*},\rho_{AB})$ satisfies

$\displaystyle
D(\sigma_{AB}^{\prime}\|\rho_{AB})=D(\sigma_{AB}^{\prime}\|\sigma_{AB}^{*})+D(\sigma_{AB}^{*}\|\rho_{AB}).$
(24)

Proof:  Since (ii) is trivial, we show only (i) by contradiction. We assume
that (i) does not hold. Hence, there exists a separable state
$\sigma_{AB}\in{\cal S}$ that does not satisfy (21).

We define the divergence sphere

$\displaystyle{\cal
S}_{D}:=\\{\rho_{AB}^{\prime}|D(\rho_{AB}^{\prime}\|\rho_{AB})=D(\sigma_{AB}^{*}\|\rho_{AB})\\}.$
(25)

Then, ${\cal M}(\sigma_{AB}^{*},\rho_{AB})$ is the tangent space of the sphere
${\cal S}_{D}$ at $\sigma_{AB}^{*}$. If a separable state $\sigma_{AB}$
satisfies

$\displaystyle\operatorname{\mathrm{Tr}}\sigma_{AB}(\log\sigma_{AB}^{*}-\log\rho_{AB})<D(\sigma_{AB}^{*}\|\rho_{AB}),$
(26)

there exists $\lambda\in(0,1)$ such that
$\sigma_{AB}^{\prime}:=\lambda\sigma_{AB}+(1-\lambda)\sigma_{AB}^{*}$ is
inside of the sphere ${\cal S}_{D}$ as Fig. 2. Hence,

$\displaystyle
D(\sigma_{AB}^{\prime}\|\rho_{AB})<D(\sigma_{AB}^{*}\|\rho_{AB}).$ (27)

Since $\lambda\sigma_{AB}+(1-\lambda)\sigma_{AB}^{*}$ is also a separable
state, the relation (27) contradicts with the choice of $\sigma_{AB}^{*}$.

Figure 2: Divergence sphere: This figure shows the relation among the
entangled state $\rho_{AB}$, the divergence sphere ${\cal S}_{D}$, and the
separation hyperplane ${\cal M}(\sigma_{AB}^{*},\rho_{AB})$.

##  V Coherence measure

For further analysis on our entanglement measures, we study the coherence
measures defined below, because, as explained in the next section, our
entanglement measures are related to these coherence measures. We consider a
state $\sum_{i,j}\theta_{i,j}|j\rangle\langle i|$ on a $d$-dimensional Hilbert
space${\mathcal{H}}$ spanned by $\\{|j\rangle\\}_{j=0}^{d-1}$, and denote the
set of incoherent states by ${\cal I}_{c}$, which is written as
$\\{\sigma(p)\\}$ with $\sigma(p):=\sum_{i}p_{i}|i\rangle\langle i|$, where
$p=(p_{i})$ is a distribution. Then, we define the following coherence
measures.

$\displaystyle C^{*}(\rho):=\min_{\sigma\in{\cal I}_{c}}D(\sigma\|\rho),\quad
C_{\alpha}^{*}(\rho):=\min_{\sigma\in{\cal I}_{c}}D_{\alpha}(\sigma\|\rho).$
(28)

Although the papers [43, 44] considered the coherence measures based on Rényi
relative entropy, our measures are different from their coherence measures
because their measures are based on the quantities with exchanging the two
input states from ours. In fact, $C^{*}(\rho)$ has a geometrical
characterization similar to Section IV by replacing ${\cal S}_{AB}$ by ${\cal
I}_{c}$.

For $\alpha\in(0,1)$ and a pure state $|\phi\rangle\langle\phi|$ with
$|\phi\rangle=\sum_{j}e^{\theta_{j}\sqrt{-1}}p_{j}|j\rangle$, we have

$\displaystyle
e^{(\alpha-1)C_{\alpha}^{*}(|\phi\rangle\langle\phi|)}=\max_{\sigma\in{\cal
I}_{c}}\operatorname{\mathrm{Tr}}(|\phi\rangle\langle\phi|^{\frac{1-\alpha}{2\alpha}}\sigma|\phi\rangle\langle\phi|^{\frac{1-\alpha}{2\alpha}})^{\alpha}$
$\displaystyle=$ $\displaystyle\max_{\sigma\in{\cal
I}_{c}}\operatorname{\mathrm{Tr}}(|\phi\rangle\langle\phi|\sigma|\phi\rangle\langle\phi|)^{\alpha}$
$\displaystyle=$ $\displaystyle\max_{\sigma(q)\in{\cal
I}_{c}}(\langle\phi|\sigma(q)|\phi\rangle)^{\alpha}=\max_{q}(\sum_{j}q_{j}p_{j})^{\alpha}=(\max_{i}p_{i})^{\alpha}.$
(29)

Hence, we have

$\displaystyle
C_{\alpha}^{*}(|\phi\rangle\langle\phi|)=\frac{\alpha}{\alpha-1}\log(\max_{i}p_{i}).$
(30)

Also, $C^{*}(|\phi\rangle\langle\phi|)$ and
$C_{\alpha}^{*}(|\phi\rangle\langle\phi|)$ take infinity value for $\alpha>1$
unless it is an incoherent state.

In the following discussion, we employ the operators
$Z:=\sum_{j=0}^{d-1}\omega^{j}|j\rangle\langle j|$ with
$\omega:=e^{2\pi\sqrt{-1}/d}$ and $X:=\sum_{j=0}^{d-1}|j+1\rangle\langle j|$.
We define $\rho(p):=\sum_{j=0}^{d-1}p_{j}Z^{j}|+\rangle\langle+|Z^{-j}$ with
$|+\rangle:=\frac{1}{\sqrt{d}}\sum_{j=0}^{d-1}|j\rangle$. Then, we have

$\displaystyle C^{*}(\rho(p))$ $\displaystyle=D(p_{\mathop{\rm
mix}}\|p)=-\frac{1}{d}\sum_{j=0}^{d-1}\log p_{j}-\log d$ (31) $\displaystyle
C_{\alpha}^{*}(\rho(p))$ $\displaystyle=D_{\alpha}(p_{\mathop{\rm mix}}\|p)$
$\displaystyle=\frac{1}{\alpha-1}\log\sum_{j=0}^{d-1}p_{j}^{1-\alpha}-\frac{\alpha}{\alpha-1}\log
d,$ (32)

where $p_{mix}$ is the uniform distribution. Since the above second equations
follow from the direct calculation, the following analysis shows the first
equations.

We have $X\rho(p)X^{-1}=\rho(p).$ Thus,
$D(\sigma(q)\|\rho(p))=D(X^{j}\sigma(q)X^{-j}\|\rho(p))$. Hence, the joint
convexity of the relative entropy implies

$\displaystyle
D(\sigma(q)\|\rho(p))=\sum_{j=0}^{d-1}\frac{1}{d}D(X^{j}\sigma(q)X^{-j}\|\rho(p))$
$\displaystyle\geq$ $\displaystyle
D(\sum_{j=0}^{d-1}\frac{1}{d}X^{j}\sigma(q)X^{-j}\|\rho(p))$ $\displaystyle=$
$\displaystyle
D(\rho_{mix}\|\rho(p))=D(\rho(p_{mix})\|\rho(p))=D(p_{mix}\|p),$ (33)

where $\rho_{mix}$ is the completely mixed state. The above relations imply
the first equation (31). Similarly, since $x\mapsto x^{\alpha}$ is matrix
convex for $\alpha>1$,

$\displaystyle
e^{(\alpha-1)D_{\alpha}(\sigma(q)\|\rho(p))}=\sum_{j=0}^{d-1}\frac{1}{d}e^{(\alpha-1)D_{\alpha}(X^{j}\sigma(q)X^{j}\|\rho(p))}$
$\displaystyle\geq$ $\displaystyle
e^{(\alpha-1)D_{\alpha}(\sum_{j=0}^{d-1}\frac{1}{d}X^{j}\sigma(q)X^{j}\|\rho(p))}$
$\displaystyle=$ $\displaystyle
e^{(\alpha-1)D_{\alpha}(\rho_{mix}\|\rho(p))}=e^{(\alpha-1)D_{\alpha}(p_{mix}\|p)},$
(34)

which implies the first equation (32) for $\alpha>1$. Similarly, since
$x\mapsto x^{\alpha}$ is matrix concave for $\alpha\in[0,1)$, an analysis
similar to (34) yields the first equation (32) for $\alpha\in[0,1)$.

To calculate $C^{*}(\rho)$ for a general state, we define the state
$I_{c}(\rho):=\frac{1}{t(\rho)}\exp((\log\rho)^{\rm diag})$ with
$t(\rho):=\operatorname{\mathrm{Tr}}\exp((\log\rho)^{\rm diag})$ and
$(\log\rho)^{\rm diag}$ is the diagonal matrix whose diagonal elements are the
same as $\log\rho$. Then, $C^{*}(\rho)$ is calculated as follows.

###### Lemma 4

The relation

$\displaystyle C^{*}(\rho)=D(I_{c}(\rho)\|\rho)$ (35)

holds.

Proof:  For $\sigma\in{\cal I}_{c}$, we have

$\displaystyle
D(\sigma\|\rho)=\operatorname{\mathrm{Tr}}\sigma(\log\sigma-\log\rho)$
$\displaystyle=$
$\displaystyle\operatorname{\mathrm{Tr}}\sigma(\log\sigma-(\log\rho)^{\rm
diag}))$ $\displaystyle=$
$\displaystyle\operatorname{\mathrm{Tr}}\sigma(\log\sigma-\log
I_{c}(\rho)-\log t(\rho)$ $\displaystyle=$ $\displaystyle
D(\sigma\|I_{c}(\rho))-\log t(\rho).$ (36)

Hence, the minimization is realized when $\sigma=I_{c}(\rho)$.

###### Lemma 5

The relation

$\displaystyle C_{\alpha}^{*}(\rho)$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{l}\max_{\tau}\frac{\alpha}{\alpha-1}\log\|(\rho^{\frac{1-\alpha}{2\alpha}}\tau^{-\frac{1-\alpha}{\alpha}}\rho^{\frac{1-\alpha}{2\alpha}})^{\rm
diag}\|\\\ \hskip 150.69397pt\hbox{ for }\alpha\in(0,1)\\\
\max_{\tau}\frac{-\alpha}{\alpha-1}\log\|((\rho^{\frac{1-\alpha}{2\alpha}}\tau^{-\frac{1-\alpha}{\alpha}}\rho^{\frac{1-\alpha}{2\alpha}})^{\rm
diag})^{-1}\|\\\ \hskip 150.69397pt\hbox{ for }\alpha>1\end{array}\right.$
(41)

holds.

Proof:  The idea of this proof comes from the proof of [48, Lemma 6 and (B4)].
We choose $\beta$ such that $\frac{1}{\beta}+\frac{1}{\alpha}=1$ so that
$\beta=\frac{\alpha-1}{\alpha}$. We fix $\alpha\in(0,1)$ so that $\beta<0$.
For any state $\tau$ and any incoherent state $\sigma\in{\cal I}_{c}$, the
reverse matrix Hölder inequality [51, Theorem 2.7] implies

$\displaystyle\operatorname{\mathrm{Tr}}\tau^{\beta}\rho^{\frac{1-\alpha}{2\alpha}}\sigma\rho^{\frac{1-\alpha}{2\alpha}}\geq(\operatorname{\mathrm{Tr}}\tau)^{\beta}(\operatorname{\mathrm{Tr}}(\rho^{\frac{1-\alpha}{2\alpha}}\sigma\rho^{\frac{1-\alpha}{2\alpha}})^{\alpha})^{\frac{1}{\alpha}}$
$\displaystyle=$
$\displaystyle(\operatorname{\mathrm{Tr}}(\rho^{\frac{1-\alpha}{2\alpha}}\sigma\rho^{\frac{1-\alpha}{2\alpha}})^{\alpha})^{\frac{1}{\alpha}}.$
(42)

Considering the equality condition, we have

$\displaystyle\min_{\tau}\operatorname{\mathrm{Tr}}\tau^{\beta}\rho^{\frac{1-\alpha}{2\alpha}}\sigma\rho^{\frac{1-\alpha}{2\alpha}}=(\operatorname{\mathrm{Tr}}(\rho^{\frac{1-\alpha}{2\alpha}}\sigma\rho^{\frac{1-\alpha}{2\alpha}})^{\alpha})^{\frac{1}{\alpha}}.$
(43)

Hence,

$\displaystyle e^{(\alpha-1)C_{\alpha}^{*}(\rho)}=\max_{\sigma\in{\cal
I}_{c}}\operatorname{\mathrm{Tr}}(\rho^{\frac{1-\alpha}{2\alpha}}\sigma\rho^{\frac{1-\alpha}{2\alpha}})^{\alpha}$
$\displaystyle=$ $\displaystyle(\max_{\sigma\in{\cal
I}_{c}}\min_{\tau}\operatorname{\mathrm{Tr}}\rho^{\frac{1-\alpha}{2\alpha}}\tau^{\beta}\rho^{\frac{1-\alpha}{2\alpha}}\sigma)^{\alpha}$
$\displaystyle\stackrel{{\scriptstyle(a)}}{{=}}$
$\displaystyle(\min_{\tau}\max_{\sigma\in{\cal
I}_{c}}\operatorname{\mathrm{Tr}}\rho^{\frac{1-\alpha}{2\alpha}}\tau^{\beta}\rho^{\frac{1-\alpha}{2\alpha}}\sigma)^{\alpha}$
$\displaystyle=$
$\displaystyle\min_{\tau}\|(\rho^{\frac{1-\alpha}{2\alpha}}\tau^{\beta}\rho^{\frac{1-\alpha}{2\alpha}})^{\rm
diag}\|^{\alpha}.$ (44)

Here, $(a)$ follows from the minimax theorem because
$\operatorname{\mathrm{Tr}}\rho^{\frac{1-\alpha}{2\alpha}}\tau^{\beta}\rho^{\frac{1-\alpha}{2\alpha}}\sigma$
is linear for $\sigma$ and is convex for $\tau$. Hence, we obtain the desired
equation with $\alpha\in(0,1)$.

Next, we fix $\alpha>1$ so that $1>\beta>0$. The matrix Hölder inequality
[52],[51, Theorem 2.6] implies

$\displaystyle\operatorname{\mathrm{Tr}}\tau^{\beta}\rho^{\frac{1-\alpha}{2\alpha}}\sigma\rho^{\frac{1-\alpha}{2\alpha}}\leq(\operatorname{\mathrm{Tr}}\tau)^{\beta}(\operatorname{\mathrm{Tr}}(\rho^{\frac{1-\alpha}{2\alpha}}\sigma\rho^{\frac{1-\alpha}{2\alpha}})^{\alpha})^{\frac{1}{\alpha}}$
$\displaystyle=$
$\displaystyle(\operatorname{\mathrm{Tr}}(\rho^{\frac{1-\alpha}{2\alpha}}\sigma\rho^{\frac{1-\alpha}{2\alpha}})^{\alpha})^{\frac{1}{\alpha}}.$
(45)

Considering the equality condition, we have

$\displaystyle\max_{\tau}\operatorname{\mathrm{Tr}}\tau^{\beta}\rho^{\frac{1-\alpha}{2\alpha}}\sigma\rho^{\frac{1-\alpha}{2\alpha}}=(\operatorname{\mathrm{Tr}}(\rho^{\frac{1-\alpha}{2\alpha}}\sigma\rho^{\frac{1-\alpha}{2\alpha}})^{\alpha})^{\frac{1}{\alpha}}.$
(46)

Hence,

$\displaystyle e^{(\alpha-1)C_{\alpha}^{*}(\rho)}=\min_{\sigma\in{\cal
I}_{c}}\operatorname{\mathrm{Tr}}(\rho^{\frac{1-\alpha}{2\alpha}}\sigma\rho^{\frac{1-\alpha}{2\alpha}})^{\alpha}$
$\displaystyle=$ $\displaystyle(\min_{\sigma\in{\cal
I}_{c}}\max_{\tau}\operatorname{\mathrm{Tr}}\rho^{\frac{1-\alpha}{2\alpha}}\tau^{\beta}\rho^{\frac{1-\alpha}{2\alpha}}\sigma)^{\alpha}$
$\displaystyle\stackrel{{\scriptstyle(a)}}{{=}}$
$\displaystyle(\max_{\tau}\min_{\sigma\in{\cal
I}_{c}}\operatorname{\mathrm{Tr}}\rho^{\frac{1-\alpha}{2\alpha}}\tau^{\beta}\rho^{\frac{1-\alpha}{2\alpha}}\sigma)^{\alpha}$
$\displaystyle=$
$\displaystyle(\max_{\tau}\|((\rho^{\frac{1-\alpha}{2\alpha}}\tau^{\beta}\rho^{\frac{1-\alpha}{2\alpha}})^{\rm
diag})^{-1}\|^{-1})^{\alpha}$ $\displaystyle=$
$\displaystyle\max_{\tau}\|((\rho^{\frac{1-\alpha}{2\alpha}}\tau^{\beta}\rho^{\frac{1-\alpha}{2\alpha}})^{\rm
diag})^{-1}\|^{-\alpha}.$ (47)

Here, $(a)$ follows from the minimax theorem because
$\operatorname{\mathrm{Tr}}\rho^{\frac{1-\alpha}{2\alpha}}\tau^{\beta}\rho^{\frac{1-\alpha}{2\alpha}}\sigma$
is linear for $\sigma$ and is concave for $\tau$. Hence, we obtain the desired
equation with $\alpha>1$.

Coherence measures satisfy the additivity.

###### Lemma 6

The relations

$\displaystyle\begin{split}C^{*}(\rho_{1}\otimes\rho_{2})&=C^{*}(\rho_{1})+C^{*}(\rho_{2}),\\\
C_{\alpha}^{*}(\rho_{1}\otimes\rho_{2})&=C_{\alpha}^{*}(\rho_{1})+C_{\alpha}^{*}(\rho_{2})\end{split}$
(48)

hold.

Proof:  Since the first equation in (48) follows from the second equation in
(48) with taking the limit $\alpha\to 1$, we show only the second equation in
(48). The idea of this proof is similar to the proof of [48, Lemma 7]. For
$\alpha\in(0,1)$, using (41), we have

$\displaystyle C_{\alpha}^{*}(\rho_{1})+C_{\alpha}^{*}(\rho_{2})\geq
C_{\alpha}^{*}(\rho_{1}\otimes\rho_{2})$ $\displaystyle=$
$\displaystyle\max_{\tau}\frac{\alpha}{\alpha-1}\log\|((\rho_{1}\otimes\rho_{2})^{\frac{1-\alpha}{2\alpha}}\tau^{-\frac{1-\alpha}{\alpha}}(\rho_{1}\otimes\rho_{2})^{\frac{1-\alpha}{2\alpha}})^{\rm
diag}\|$ $\displaystyle\geq$
$\displaystyle\max_{\tau_{1},\tau_{2}}\frac{\alpha}{\alpha-1}\log\|((\rho_{1}\otimes\rho_{2})^{\frac{1-\alpha}{2\alpha}}(\tau_{1}\otimes\tau_{2})^{-\frac{1-\alpha}{\alpha}}$
$\displaystyle\hskip
86.11084pt\cdot(\rho_{1}\otimes\rho_{2})^{\frac{1-\alpha}{2\alpha}})^{\rm
diag}\|$ $\displaystyle=$
$\displaystyle\max_{\tau_{1},\tau_{2}}\frac{\alpha}{\alpha-1}(\log\|(\rho_{1}^{\frac{1-\alpha}{2\alpha}}\tau_{1}^{-\frac{1-\alpha}{\alpha}}\rho_{1}^{\frac{1-\alpha}{2\alpha}})^{\rm
diag}\|$
$\displaystyle+\log\|(\rho_{2}^{\frac{1-\alpha}{2\alpha}}\tau_{2}^{-\frac{1-\alpha}{\alpha}}\rho_{2}^{\frac{1-\alpha}{2\alpha}})^{\rm
diag}\|)$ $\displaystyle=$ $\displaystyle
C_{\alpha}^{*}(\rho_{1})+C_{\alpha}^{*}(\rho_{2}),$ (49)

which implies the second equation in (48). The case with $\alpha>1$ can be
shown in the same way.

Table 2: List of mathematical symbols for coherence. Symbol | Description | Eq. number
---|---|---
${\cal I}_{c}$ | Set of incoherent states |
$C^{*}(\rho)$ | Minimum relative entropy | (28)
$C_{\alpha}^{*}(\rho)$ | Minimum relative Rényi | (28)
entropy of order $\alpha$
$I_{c}(\rho)$ | Closet incoherent state | (35)
$\sigma(p)$ | $\sum_{i}p_{i}|i\rangle\langle i|$ |
$\rho(p)$ | $\sum_{j=0}^{d-1}p_{j}Z^{j}|+\rangle\langle+|Z^{-j}$ |

###### Remark 1

It is possible to make the same discussion for the coherence as Sections II –
IV. Since the coherence measures satisfy the additivity as Lemma 6, the
detectability even with the non-iid setting like Section III can be
characterized by the quantities $C^{*}(\rho)$ and $C_{\alpha}^{*}(\rho)$.

##  VI Maximally correlated states

The discussion in Section II shows the operational meaning of
$E_{\alpha}^{*}(\rho_{AB})$ including $E^{*}(\rho_{AB})$. We calculate
$E^{*}(\rho_{AB})$ and $E^{*}_{\alpha}(\rho_{AB})$ for several special cases.
When $\rho_{AB}$ is a pure state $|\phi\rangle\langle\phi|$ $E^{*}(\rho_{AB})$
and $E^{*}_{\alpha}(\rho_{AB})$ with $\alpha>1$ are infinity. However,
$E^{*}_{\alpha}(\rho_{AB})$ with $\alpha\in[0,1)$ takes a finite value as
follows. When the Schmidt coefficients of $|\phi\rangle\langle\phi|$ are
$\sqrt{p_{i}}$ with $i=1,\ldots,d$, $\max_{\sigma_{AB}\in{\cal
S}_{AB}}\langle\phi|\sigma|\phi\rangle=\max_{i}p_{i}$ [44, (40) and (97)].
Hence,

$\displaystyle
e^{(\alpha-1)E_{\alpha}^{*}(\rho_{AB})}=\max_{\sigma_{AB}\in{\cal
S}_{AB}}\operatorname{\mathrm{Tr}}(\rho^{\frac{1-\alpha}{2\alpha}}\sigma_{AB}\rho^{\frac{1-\alpha}{2\alpha}})^{\alpha}$
$\displaystyle=$ $\displaystyle\max_{\sigma_{AB}\in{\cal
S}_{AB}}\operatorname{\mathrm{Tr}}(|\phi\rangle\langle\phi|\sigma_{AB}|\phi\rangle\langle\phi|)^{\alpha}=\max_{\sigma_{AB}\in{\cal
S}_{AB}}\langle\phi|\sigma_{AB}|\phi\rangle^{\alpha}$ $\displaystyle=$
$\displaystyle(\max_{\sigma_{AB}\in{\cal
S}_{AB}}\langle\phi|\sigma_{AB}|\phi\rangle)^{\alpha}=(\max_{i}p_{i})^{\alpha}.$
(50)

Hence, we have

$\displaystyle
E_{\alpha}^{*}(\rho_{AB})=\frac{\alpha}{\alpha-1}\log(\max_{i}p_{i}).$ (51)

More generally, the calculation of these quantities can be converted to the
calculation of the coherence measure for special states $\rho_{AB}$. Similar
to the paper [44], we consider maximally correlated states. A state
$\rho_{AB}$ on ${\mathcal{H}}_{A}\otimes{\mathcal{H}}_{B}$ is called a
maximally correlated state when there exist bases $\\{|u_{j,A}\rangle\\}_{j}$
and $\\{|u_{j,B}\rangle\\}_{j}$ of ${\mathcal{H}}_{A}$ and ${\mathcal{H}}_{B}$
such that the support $\rho_{AB}$ is included in the subspace spanned by
$\\{|u_{j,A}\rangle|u_{j,B}\rangle\\}_{j}$ on
${\mathcal{H}}_{A}\otimes{\mathcal{H}}_{B}$ [49, 50], [42, Section 8.8]. In
this case, the state $\rho_{AB}$ is written as
$\sum_{i,j}\theta_{i,j}|u_{j,A}\rangle|u_{j,B}\rangle\langle u_{i,A}|\langle
u_{i,B}|$. In this case, the calculation of $E^{*}(\rho_{AB})$ and
$E^{*}_{\alpha}(\rho_{AB})$ can be simplified as follows. Similar to Theorem 1
in [44], the calculation of $E^{*}(\rho_{AB})$ and $E^{*}_{\alpha}(\rho_{AB})$
are reduced to coherence measures.

###### Lemma 7

The relations

$\displaystyle
E^{*}\Big{(}\sum_{i,j}\theta_{i,j}|u_{j,A}\rangle|u_{j,B}\rangle\langle
u_{i,A}|\langle u_{i,B}|\Big{)}=$ $\displaystyle
C^{*}\Big{(}\sum_{i,j}\theta_{i,j}|j\rangle\langle i|\Big{)}$ (52)
$\displaystyle
E_{\alpha}^{*}\Big{(}\sum_{i,j}\theta_{i,j}|u_{j,A}\rangle|u_{j,B}\rangle\langle
u_{i,A}|\langle u_{i,B}|\Big{)}=$ $\displaystyle
C_{\alpha}^{*}\Big{(}\sum_{i,j}\theta_{i,j}|j\rangle\langle i|\Big{)}$ (53)

hold.

Lemma 7 is shown in Appendix B. Since a pure state is a maximally correlated
state, the calculation (51) can be recovered by Lemma 7 and (30). Also,
$E^{*}\Big{(}\sum_{i,j}\theta_{i,j}|u_{j,A}\rangle|u_{j,B}\rangle\langle
u_{i,A}|\langle u_{i,B}|\Big{)}$ can be calculated by Lemma 4 as

$\displaystyle
E^{*}\Big{(}\sum_{i,j}\theta_{i,j}|u_{j,A}\rangle|u_{j,B}\rangle\langle
u_{i,A}|\langle u_{i,B}|\Big{)}=D(I_{c}(\rho_{\theta})\|\rho_{\theta}),$ (54)

where $\rho_{\theta}:=\sum_{i,j}\theta_{i,j}|j\rangle\langle i|$. The
combination of Lemmas 6 and 7 guarantees that any maximally entangled state
$\rho_{AB}$ satisfies

$\displaystyle
E^{*}(\rho_{AB})=\overline{E}^{*}(\rho_{AB})=\tilde{E}^{*}(\rho_{AB}),~{}E_{\alpha}^{*}(\rho_{AB})=\overline{E}_{\alpha}^{*}(\rho_{AB}).$
(55)

The iid setting and the non-iid setting have the same bounds.

To find another example, we set ${\mathcal{H}}_{A}$ and ${\mathcal{H}}_{B}$ to
be $d$-dimensional spaces, and define the maximally entangled state
$|\Phi\rangle:=\frac{1}{\sqrt{d}}\sum_{j=0}^{d-1}|j\rangle|j\rangle$. We
define the state $\rho_{AB}(p):=\sum_{j=0}^{d-1}p_{j}Z^{j}\otimes
I|\Phi\rangle\langle\Phi|Z^{-j}\otimes I$. The combination of (31), (32),
Lemma 7 implies

$\displaystyle E^{*}(\rho_{AB}(p))$ $\displaystyle=D(p_{\mathop{\rm
mix}}\|p)=-\frac{1}{d}\sum_{j=0}^{d}\log p_{j}-\log d$ (56) $\displaystyle
E_{\alpha}^{*}(\rho_{AB}(p))$ $\displaystyle=D_{\alpha}(p_{\mathop{\rm
mix}}\|p)$
$\displaystyle=\frac{1}{\alpha-1}\log\sum_{j=0}^{d-1}p_{j}^{1-\alpha}-\frac{\alpha}{\alpha-1}\log
d.$ (57)

However, it is not easy to calculate these quantities for general states.
Therefore, it is needed to develop an efficient method for their calculation
in the general case.

##  VII Algorithms

###  VII.1 Algorithm for $E^{*}(\rho_{AB})$

To calculate the proposed quantities for general states, we introduce
algorithms for their calculation that can be applied to general states. Since
the calculation of $E^{*}(\rho_{AB})$ is hard, we modify $E^{*}(\rho_{AB})$ as
follows. We choose a finite set ${\cal D}$ of pure states on ${\cal H}_{B}$ as
a discretization. That is, we denote ${\cal D}$ as
$\\{\rho_{B|c}^{*}\\}_{c\in{\cal X}}$. Then, we define

$\displaystyle E_{{\cal
D}}^{*}(\rho_{AB}):=\min_{P(c),\rho_{A|c}}D\Big{(}\sum_{c}P(c)\rho_{A|c}\otimes\rho_{B|c}^{*}\Big{\|}\rho_{AB}\Big{)}.$
(58)

We define $\epsilon_{2}({\cal D})$ as

$\displaystyle\epsilon_{2}({\cal
D}):=2\sqrt{1-\min_{\psi_{1}}\max_{\psi_{2}\in{\cal
D}}|\langle\psi_{1}|\psi_{2}\rangle|^{2}}.$ (59)

We have

$\displaystyle\max_{P^{\prime}(c),\rho_{A|c}^{\prime},\rho_{B|c}^{\prime}}\min_{P(c),\rho_{A|c}}\Big{\|}\sum_{c}P(c)\rho_{A|c}\otimes\rho_{B|c}^{*}$
$\displaystyle-\sum_{c}P^{\prime}(c)\rho_{A|c}^{\prime}\otimes\rho_{B|c}^{\prime}\Big{\|}_{1}$
$\displaystyle\leq$ $\displaystyle\max_{\psi_{1}}\min_{\psi_{2}\in{\cal
D}}\||\psi_{1}\rangle\langle\psi_{1}|-|\psi_{2}\rangle\langle\psi_{2}|\|_{1}$
$\displaystyle=$ $\displaystyle 2\sqrt{1-\min_{\psi_{1}}\max_{\psi_{2}\in{\cal
D}}|\langle\psi_{1}|\psi_{2}\rangle|^{2}}=\epsilon_{2}({\cal D}).$ (60)

Therefore, Fannes inequality implies

$\displaystyle\max_{P^{\prime}(c),\rho_{A|c}^{\prime},\rho_{B|c}^{\prime}}\min_{P(c),\rho_{A|c}}\Big{|}D\Big{(}\sum_{x}P(x)\rho_{A|x}\otimes\rho_{B|x}^{*}\Big{\|}\rho_{AB}\Big{)}$
$\displaystyle\hskip
86.11084pt-D\Big{(}\sum_{x}P^{\prime}(x)\rho_{A|x}^{\prime}\otimes\rho_{B|x}^{\prime}\Big{\|}\rho_{AB}\Big{)}\Big{|}$
$\displaystyle\leq$ $\displaystyle\epsilon_{2}({\cal
D})(\|\log\rho_{AB}\|+\log d_{A}d_{B})+\eta_{0}(\epsilon_{2}({\cal D})),$ (61)

where $\eta_{0}$ is defined as

$\displaystyle\eta_{0}(x):=\left\\{\begin{array}[]{ll}-x\log x&\hbox{when
}0\leq x\leq 1/e\\\ 1/e&\hbox{otherwise.}\end{array}\right.$ (64)

Therefore, we have

$\displaystyle E_{{\cal D}}^{*}(\rho_{AB})$
$\displaystyle-\Big{(}\epsilon_{2}({\cal D})(\|\log\rho_{AB}\|+\log
d_{A}d_{B})+\eta_{0}(\epsilon_{2}({\cal D}))\Big{)}$ $\displaystyle\leq$
$\displaystyle E^{*}(\rho_{AB})\leq E_{{\cal D}}^{*}(\rho_{AB}).$ (65)

We rewrite $P(c)\rho_{A|c}$ to $X_{c}$. Using the idea of the algorithm by
[18], we propose an iterative algorithm to calculate $E_{{\cal
D}}^{*}(\rho_{AB})$, in which matrices $X_{A|c}$ is updated as follows. Given
the set $\\{X_{c}^{(t)}\\}_{c}$, we define
$V_{c}[\\{X_{A|c^{\prime}}^{(t)}\\}_{c^{\prime}}]$ and
$\kappa[\\{X_{A|c}^{(t)}\\}_{c}]$ as

$\displaystyle V_{c}[\\{X_{A|c^{\prime}}^{(t)}\\}_{c^{\prime}}]$
$\displaystyle:=$
$\displaystyle\operatorname{\mathrm{Tr}}_{B}\Big{(}\log\Big{(}\sum_{c^{\prime}\in{\cal
C}}X_{A|c^{\prime}}^{(t)}\otimes\rho_{B|c^{\prime}}^{*}\Big{)}$
$\displaystyle\hskip
86.11084pt-\log\rho_{AB}\Big{)}(I_{A}\otimes\rho_{B|c}^{*})\Big{)},$ (66)
$\displaystyle\kappa[\\{X_{A|c}^{(t)}\\}_{c}]:=\sum_{c}\operatorname{\mathrm{Tr}}\exp(X_{A|c}^{(t)}-V_{c}[\\{X_{A|c^{\prime}}^{(t)}\\}_{c^{\prime}}]).$
(67)

Then, we choose the updated matrix ${X}_{A|c}^{(t+1)}$

$\displaystyle{X}_{A|c}^{(t+1)}:=$
$\displaystyle\frac{1}{\kappa[\\{X_{A|c}^{(t)}\\}_{c}]}\exp(X_{A|c}^{(t)}-V_{c}[\\{X_{A|c^{\prime}}^{(t)}\\}_{c^{\prime}}]).$
(68)

This algorithm is summarized as Algorithm 1.

Algorithm 1 Minimization of
$D\Big{(}\sum_{c}P(c)\rho_{A|c}\otimes\rho_{B|c}^{*}\Big{\|}\rho_{AB}\Big{)}$

1:Choose the initial value $\\{X_{A|c}^{(t)}\\}_{c}$;

2:repeat

3: Calculate
${X}_{A|c}^{(t+1)}:=\frac{1}{\kappa[\\{X_{A|c}^{(t)}\\}_{c}]}\exp(X_{A|c}^{(t)}-V_{c}[\\{X_{A|c^{\prime}}^{(t)}\\}_{c^{\prime}}])$;

4:until convergence.

Then, we have the following theorem.

###### Theorem 8

We choose ${X}_{A|c}^{(1)}$ such that $\sum_{c\in{\cal C}}|c\rangle\langle
c|\otimes{X}_{A|c}^{(1)}$ is the completely mixed state $\rho_{AC,\mathop{\rm
mix}}$. Then, we have

$\displaystyle
D\Big{(}\sum_{c}X_{A|c}^{(t+1)}\otimes\rho_{B|c}^{*}\Big{\|}\rho_{AB}\Big{)}-E_{{\cal
D}}^{*}(\rho_{AB})\leq\frac{\log d_{A}|{\cal D}|}{t}.$ (69)

The above theorem is shown in Appendix C by using [18, Theorem 3.3].

###  VII.2 Algorithm for $E_{\alpha}^{*}(\rho_{AB})$

Next, we consider the algorithm for $E_{\alpha}^{*}(\rho_{AB})$. Similarly,
instead of $E_{\alpha}^{*}(\rho_{AB})$, we calculate

$\displaystyle E_{\alpha,{\cal
D}}^{*}(\rho_{AB}):=\min_{X_{A|c}}D_{\alpha}\Big{(}\sum_{c}X_{A|c}\otimes\rho_{B|c}^{*}\Big{\|}\rho_{AB}\Big{)}.$
(70)

The following algorithm minimizes $\mathop{\rm
sgn}(\alpha-1)\operatorname{\mathrm{Tr}}\Big{(}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{(}\sum_{c^{\prime}\in{\cal
C}}X_{A|c^{\prime}}\otimes\rho_{B|c^{\prime}}^{*}\Big{)}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{)}^{\alpha}$.
For this aim, we iteratively choose matrices $X_{A|c}$ as follows. Given the
set $\\{X_{c}^{(t)}\\}_{c}$, we define
$V_{\alpha,c}[\\{X_{A|c^{\prime}}^{(t)}\\}_{c^{\prime}}]$ and
$\kappa_{\alpha}[\\{X_{A|c}^{(t)}\\}_{c}]$ as

$\displaystyle V_{\alpha,c}[\\{X_{A|c^{\prime}}^{(t)}\\}_{c^{\prime}}]$
$\displaystyle:=$
$\displaystyle\operatorname{\mathrm{Tr}}_{B}\Big{(}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{(}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{(}\sum_{c^{\prime}\in{\cal
C}}X_{A|c^{\prime}}^{(t)}\otimes\rho_{B|c^{\prime}}^{*}\Big{)}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{)}^{\alpha-1}$
$\displaystyle\cdot\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{)}(I_{A}\otimes\rho_{B|c}^{*}),$
(71) $\displaystyle\kappa_{\alpha}[\\{X_{A|c}^{(t)}\\}_{c}]$ $\displaystyle:=$
$\displaystyle\sum_{c}\operatorname{\mathrm{Tr}}\exp(X_{A|c}^{(t)}-\frac{\mathop{\rm
sgn}(\alpha-1)}{\gamma}V_{\alpha,c}[\\{X_{A|c^{\prime}}^{(t)}\\}_{c^{\prime}}]).$
(72)

Then, we choose the updated matrix ${X}_{A|c}^{(t+1)}$

$\displaystyle{X}_{A|c}^{(t+1)}$ $\displaystyle:=$
$\displaystyle\frac{1}{\kappa[\\{X_{A|c}^{(t)}\\}_{c}]}\exp(X_{A|c}^{(t)}-\frac{\mathop{\rm
sgn}(\alpha-1)}{\gamma}V_{\alpha,c}[\\{X_{A|c^{\prime}}^{(t)}\\}_{c^{\prime}}]).$
(73)

Algorithm 2 Minimization of $\mathop{\rm
sgn}(\alpha-1)e^{(\alpha-1)D_{\alpha}\Big{(}\sum_{c}X_{A|c}\otimes\rho_{B|c}^{*}\Big{\|}\rho_{AB}\Big{)}}$

1:Choose the initial value $\\{X_{A|c}^{(t)}\\}_{c}$;

2:repeat

3: Calculate
${X}_{A|c}^{(t+1)}:=\frac{1}{\kappa[\\{X_{A|c}^{(t)}\\}_{c}]}\exp(X_{A|c}^{(t)}-\frac{\mathop{\rm
sgn}(\alpha-1)}{\gamma}V_{\alpha,c}[\\{X_{A|c^{\prime}}^{(t)}\\}_{c^{\prime}}])$;

4:until convergence.

Then, we have the following theorem.

###### Theorem 9

We choose ${X}_{A|c}^{(1)}$ as $\sum_{c\in{\cal C}}|c\rangle\langle
c|\otimes{X}_{A|c}^{(1)}$ as the completely mixed state $\rho_{AC,\mathop{\rm
mix}}$. For $\alpha\in(0,1)$, and a sufficiently large real number $\gamma>0$,
we have

$\displaystyle\mathop{\rm
sgn}(\alpha-1)\operatorname{\mathrm{Tr}}\Big{(}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{(}\sum_{c^{\prime}\in{\cal
C}}X_{A|c^{\prime}}^{(t+1)}\otimes\rho_{B|c^{\prime}}^{*}\Big{)}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{)}^{\alpha}$
$\displaystyle-\mathop{\rm sgn}(\alpha-1)e^{(\alpha-1)E_{\alpha,{\cal
D}}(\rho_{AB})}\leq\frac{\gamma\log d_{A}|{\cal D}|}{t}.$ (74)

The above theorem is shown in Appendix C by using [18, Theorem 3.3].

##  VIII Application of our algorithm

###  VIII.1 Membership problem

Next, we apply our algorithm to the membership problem of the set ${\cal
S}_{AB}$. That is, we employ the quantity $E^{*}(\rho_{AB})$ for the
membership problem. This measure will be used for constructing the separation
hyperplane. In the following, we consider how to distinguish the following two
cases.

$\displaystyle H_{0}:$ $\displaystyle E^{*}(\rho_{AB})=0$ (75) $\displaystyle
H_{1}:$ $\displaystyle E^{*}(\rho_{AB})\geq\epsilon_{1}.$ (76)

Due to (65), when $E^{*}(\rho_{AB})=0$, we have

$\displaystyle E_{{\cal D}}^{*}(\rho_{AB})\leq\epsilon_{2}({\cal
D})(\|\log\rho_{AB}\|+\log d_{A}d_{B})+\eta_{0}(\epsilon_{2}({\cal D})).$ (77)

Also, when $E^{*}(\rho_{AB})>\epsilon$, we have

$\displaystyle E_{{\cal D}}^{*}(\rho_{AB})>\epsilon.$ (78)

Therefore, if the following two cases are distinguished, $H_{0}$ and $H_{1}$
are distinguished.

$\displaystyle H_{0}^{\prime}:E_{{\cal D}}^{*}(\rho_{AB})\leq$
$\displaystyle\epsilon_{2}({\cal D})(\|\log\rho_{AB}\|+\log d_{A}d_{B})$
$\displaystyle+\eta_{0}(\epsilon_{2}({\cal D})),$ (79) $\displaystyle
H_{1}^{\prime}:E_{{\cal D}}^{*}(\rho_{AB})>$ $\displaystyle\epsilon_{1}$ (80)

under the assumption that $\epsilon_{2}({\cal D})(\|\log\rho_{AB}\|+\log
d_{A}d_{B})+\eta_{0}(\epsilon_{2}({\cal D}))\leq\epsilon_{1}$.

We choose the initial state according to the assumption of Theorem 8. Then,
when $H_{0}^{\prime}$ holds, i.e., $E_{{\cal
D}}^{*}(\rho_{AB})\leq\epsilon_{2}({\cal D})(\|\log\rho_{AB}\|+\log
d_{A}d_{B})+\eta_{0}(\epsilon_{2}({\cal D}))$, we have

$\displaystyle
D\Big{(}\sum_{c}X_{A|c}^{(t+1)}\otimes\rho_{B|c}^{*}\Big{\|}\rho_{AB}\Big{)}$
$\displaystyle\leq$ $\displaystyle\epsilon_{2}({\cal
D})(\|\log\rho_{AB}\|+\log d_{A}d_{B})+\eta_{0}(\epsilon_{2}({\cal
D}))+\frac{\log d_{A}|{\cal D}|}{t}.$ (81)

Also, when $E_{{\cal D}}^{*}(\rho_{AB})>\epsilon$, we have

$\displaystyle
D\Big{(}\sum_{c}X_{A|c}^{(t+1)}\otimes\rho_{B|c}^{*}\Big{\|}\rho_{AB}\Big{)}>\epsilon.$
(82)

Therefore, if the following two cases are distinguished, $H_{0}^{\prime}$ and
$H_{1}^{\prime}$, i.e. $H_{0}$ and $H_{1}$ are distinguished.

$\displaystyle H_{0}^{\prime\prime}:$ $\displaystyle
D\Big{(}\sum_{c}X_{A|c}^{(t+1)}\otimes\rho_{B|c}^{*}\Big{\|}\rho_{AB}\Big{)}$
$\displaystyle\leq\epsilon_{2}({\cal D})(\|\log\rho_{AB}\|+\log
d_{A}d_{B})+\eta_{0}(\epsilon_{2}({\cal D}))$ $\displaystyle\quad+\frac{\log
d_{A}|{\cal D}|}{t},$ (83) $\displaystyle H_{1}^{\prime\prime}:$
$\displaystyle\operatorname{\mathrm{Tr}}\rho_{AC}^{(t+1)}\Omega_{F}[\rho_{AC}^{(t+1)}]>\epsilon_{1}$
(84)

under the assumption that $\epsilon_{2}({\cal D})(\|\log\rho_{AB}\|+\log
d_{A}d_{B})+\eta_{0}(\epsilon_{2}({\cal D}))+\frac{\log d_{A}|{\cal
D}|}{t}\leq\epsilon_{1}$. That is, the number $t$ of iterations needs to
satisfy $t\geq\log d_{A}|{\cal D}|/\Big{(}\epsilon_{1}-(\epsilon_{2}({\cal
D})(\|\log\rho_{AB}\|+\log d_{A}d_{B})+\eta_{0}(\epsilon_{2}({\cal
D})))\Big{)}$.

Next, we evaluate the computation complexity to distinguish $H_{0}$ and
$H_{1}$ under the above method when we choose ${\cal D}_{n,d_{B}}$ as ${\cal
D}$ in the way as Appendix A and we fix $\epsilon_{1}$ and move $d_{A},d_{B}$
with $d_{A}\geq d_{B}$.

In this case, we have $|{\cal
D}_{n,d_{B}}|\leq(\frac{2d_{B}^{1/2}\pi}{\epsilon_{2}})^{d_{B}-1}$ with
$\epsilon_{2}({\cal D}_{n,d_{B}})=\epsilon_{2}$. The above condition is
rewritten as

$\displaystyle t\geq\frac{\log
d_{A}+(d_{B}-1)\log\frac{2d_{B}^{1/2}\pi}{\epsilon_{2}}}{\epsilon_{1}-(\epsilon_{2}(\|\log\rho_{AB}\|+\log
d_{A}d_{B})+\eta_{0}(\epsilon_{2}))}$ (85)

Since $\epsilon_{1}$ is fixed, we need to choose $\epsilon_{2}$ to be
$O(\frac{\epsilon_{1}}{\log d_{A}d_{B}})$, which realizes
$\epsilon_{1}-(\epsilon_{2}(\|\log\rho_{AB}\|+\log
d_{A}d_{B})+\eta_{0}(\epsilon_{2}))=O(\epsilon_{1})$. The above condition is
rewritten as

$\displaystyle t\geq$ $\displaystyle\frac{\log
d_{A}+(d_{B}-1)\log\frac{2d_{B}^{1/2}\pi}{O(\frac{\epsilon_{1}}{\log
d_{A}d_{B}})}}{O(\epsilon_{1})}$ $\displaystyle=$ $\displaystyle
O\Big{(}\frac{\log d_{A}+(d_{B}-1)\log\frac{d_{B}^{1/2}\log
d_{A}d_{B}}{\epsilon_{1}}}{\epsilon_{1}}\Big{)}$ $\displaystyle=$
$\displaystyle O\Big{(}\frac{\log d_{A}+d_{B}\log d_{B}+d_{B}(\log\log
d_{A}-\log\epsilon_{1})}{\epsilon_{1}}\Big{)}.$ (86)

However, each step has various calculations. In each step, we need to
calculate ${\cal F}[\rho_{AC}]$ from $\rho_{AC}$. For this calculation, we
need to calculate $\log\rho_{AC}$ and $\Omega_{R,D}[\rho_{AC}]$. $\rho_{AC}$
is a block diagonal matrix. That is, it is sufficient to calculate every
component. Every component has a matrix function on a $d_{A}d_{B}$-dimensional
space, and a factor whose complexity is linear to $|{\cal D}_{n,d_{B}}|$. But,
the latter part is common for all $c$-components. The farmer can be done by
diagonalization. The complexity of the diagonalization on $d$-dimensional
matrix is $O(d^{3})$ with error $\epsilon$ [45, 46, 47]. Therefore, the
calculation complexity of ${\cal F}[\rho_{AC}]$ from $\rho_{AC}$ is $O(|{\cal
D}_{n,d_{B}}|(d_{A}^{3}d_{B}^{3}))\leq
O((\frac{2d_{B}^{1/2}\pi}{\epsilon_{2}})^{d_{B}-1}(d_{A}^{3}d_{B}^{3}))=O((\frac{d_{B}^{1/2}\log
d_{A}d_{B}}{\epsilon_{1}})^{d_{B}-1}(d_{A}^{3}d_{B}^{3}))$. The total
complexity is

$\displaystyle O\Big{(}$ $\displaystyle\frac{\log d_{A}+d_{B}\log
d_{B}+d_{B}(\log\log d_{A}-\log\epsilon_{1})}{\epsilon_{1}}$
$\displaystyle\cdot(\frac{d_{B}^{1/2}\log
d_{A}d_{B}}{\epsilon_{1}})^{d_{B}-1}(d_{A}^{3}d_{B}^{3})\Big{)}.$ (87)

On the other hand, the method by [21] has the following calculation complexity

$\displaystyle O(d_{A}^{6}\epsilon^{-4d_{B}})$ (88)

In the above evaluation, $\epsilon$ takes the same role as $\epsilon_{1}$. For
$d_{B}$, our bound has the factor $d_{B}^{(d_{B}-1)/2}$, which is larger than
$\frac{1}{\epsilon^{4d_{B}}}$. However, when $d_{B}$ is also fixed and only
$d_{A}$ increases, our bound is simplified to $O(\max(\log
d_{A},-\log\epsilon_{1})(\log
d_{A})^{d_{B}-1}d_{A}^{3}\epsilon_{1}^{-d_{B}})$. When $\epsilon_{1}$ and
$\epsilon$ are fixed, our bound is simplified to $O((\log
d_{A})^{d_{B}}d_{A}^{3})$, and their bound is simplified to $O(d_{A}^{6})$. In
this case, our bound is smaller than their bound.

###  VIII.2 Separation hyperplane

Next, we discuss how to obtain separation hyperplane by using the above
approximation and the geometrical characterization by Theorem 3. This is
because it is a key step for conic linear programming with a general cone to
find a separation hyperplane for an element that does not belong to the cone
[25, 26, 27, 28]. For this aim, we employ the minimizer of $E_{{\cal
D}}^{*}(\rho_{AB})$, i.e., the state

$\displaystyle\sigma_{AB}^{*}({\cal D}):=\mathop{\rm
argmin}\limits_{\sigma_{AB}\in{\cal S}({\cal D})}D(\sigma_{AB}\|\rho_{AB}).$
(89)

Then, the following theorem means that the operator
$(\log\sigma_{AB}^{*}({\cal D})-\log\rho_{AB})$ gives a separation hyperplane
to distinguish $\rho_{AB}$ and ${\cal S}$.

###### Theorem 10

(i) For any separable state $\sigma_{AB}\in{\cal S}$, we have

$\displaystyle\operatorname{\mathrm{Tr}}\sigma_{AB}(\log\sigma_{AB}^{*}({\cal
D})-\log\rho_{AB})$ $\displaystyle\geq$ $\displaystyle D(\sigma_{AB}^{*}({\cal
D})\|\rho_{AB})-\|\log\sigma_{AB}^{*}({\cal
D})-\log\rho_{AB}\|\epsilon_{2}({\cal D}).$ (90)

(ii) Also, we have

$\displaystyle\operatorname{\mathrm{Tr}}\rho_{AB}(\log\sigma_{AB}^{*}({\cal
D})-\log\rho_{AB})=-D(\rho_{AB}\|\sigma_{AB}^{*}({\cal D}))<0.$ (91)

Proof:  Since (ii) is trivial, we show only (i) by contradiction. Since the
set ${\cal S}({\cal D})$ is the convex set, we have the following in the same
way as Theorem 3. (i) For any separable state $\sigma_{AB}\in{\cal S}({\cal
D})$, we have

$\displaystyle\operatorname{\mathrm{Tr}}\sigma_{AB}(\log\sigma_{AB}^{*}({\cal
D})-\log\rho_{AB})\geq D(\sigma_{AB}^{*}({\cal D})\|\rho_{AB}).$ (92)

Due to the combination of (92) and (60), any separable state
$\sigma_{AB}\in{\cal S}$ satisfies (90).

##  IX Numerical analysis

For a typical example, we focus on the case when ${\mathcal{H}}_{A}$ is two
qubits and ${\mathcal{H}}_{B}$ is one qubit. We parametrize a qubit pure state
as
$|\phi_{\theta}\rangle:=\frac{1}{\sqrt{2}}(|0\rangle+e^{\sqrt{-1}\theta}|1\rangle)$.
Consider the entangled state
$|\Psi_{p}\rangle:=\sqrt{p}|\phi_{0}\rangle|\phi_{0}\rangle|0\rangle+\sqrt{1-p}|\phi_{\pi/2}\rangle|\phi_{\pi/2}\rangle|1\rangle)$.
Given a real number $\lambda$ with $0<\lambda<1$, we focus on the phase-
damping channel ${\cal E}_{\lambda}$ defined as

$\displaystyle{\cal
E}_{\lambda}\left(\left(\begin{array}[]{cc}a_{0,0}&a_{0,1}\\\
a_{1,0}&a_{1,1}\end{array}\right)\right):=\left(\begin{array}[]{cc}a_{0,0}&\lambda
a_{0,1}\\\ {\lambda}a_{1,0}&a_{1,1}\end{array}\right).$ (97)

This channel acts on the state $|\phi_{\theta}\rangle$ as

$\displaystyle{\cal
E}_{\lambda}(\rho)=\frac{1+\lambda}{2}\rho+\frac{1-\lambda}{2}\left(\begin{array}[]{cc}1&0\\\
0&-1\end{array}\right)\rho\left(\begin{array}[]{cc}1&0\\\
0&-1\end{array}\right).$ (102)

That is, the phase-damping channel is replaced by the above over the state
$|\Psi_{p}\rangle$. Then, we define the state with two parameters $\lambda$
and $\delta$ as

$\displaystyle\rho_{p,\lambda,\delta}:=(1-\delta)({\cal
E}_{\lambda}\otimes{\cal E}_{\lambda}\otimes
Id)(|\Psi_{p}\rangle\langle\Psi_{p}|)+\delta\rho_{mix},$ (103)

where $\delta\rho_{mix}$ is the completely mixed state on
${\mathcal{H}}_{A}\otimes{\mathcal{H}}_{B}$.

Applying the unitary matrix
$V=\frac{1}{2}\left(\begin{array}[]{cc}1+\sqrt{-1}&1-\sqrt{-1}\\\
1-\sqrt{-1}&1+\sqrt{-1}\end{array}\right)$ on both qubits in
${\mathcal{H}}_{A}$, the above model is converted to the following model. We
define
$|\tilde{\Psi}_{p}\rangle:=\sqrt{p}|\phi_{\pi}\rangle|\phi_{\pi}\rangle|0\rangle+\sqrt{1-p}|0\rangle|0\rangle|1\rangle)$.
We define the channel $\tilde{\cal E}_{\lambda}$ as

$\displaystyle\tilde{\cal
E}_{\lambda}(\rho):=\frac{1+\lambda}{2}\rho+\frac{1-\lambda}{2}\left(\begin{array}[]{cc}0&-1\\\
1&0\end{array}\right)\rho\left(\begin{array}[]{cc}0&-1\\\
1&0\end{array}\right).$ (108)

Then, we define the state with two parameters $\lambda$ and $\delta$ as

$\displaystyle\tilde{\rho}_{p,\lambda,\delta}:=(1-\delta)(\tilde{\cal
E}_{\lambda}\otimes\tilde{\cal E}_{\lambda}\otimes
Id)(|\tilde{\Psi}_{p}\rangle\langle\tilde{\Psi}_{p}|)+\delta\rho_{mix}.$ (109)

That is, we have $(V^{\otimes 2}\otimes I){\rho}_{p,\lambda,\delta}(V^{\otimes
2}\otimes I)^{\dagger}=\tilde{\rho}_{p,\lambda,\delta}$. Hence, this model is
essentially written with real matrix elements.

We make numerical experiments when $n$ is set to be $16$. That is, $|{\cal
D}_{n,d_{B}}|$ is $16^{2}$. By setting $p=1/2$, Fig. 3 plots the minimum
relative entropy $E_{{\cal D}}^{*}({\rho}_{1/2,\lambda,\delta})$ with
$\delta=10^{-1},10^{-2},10^{-4}$. When $\lambda$ is close to zero, the state
${\rho}_{p,\lambda,\delta}$ approaches to the set ${\cal S}_{AB}$ of separable
states.

Figure 3: Minimum relative entropy $E_{{\cal
D}}^{*}({\rho}_{1/2,\lambda,\delta})$: The vertical axis shows the minimum
relative entropy. The horizontal axis shows the parameter $\lambda$. Blue
(orange, green) curve presents the case with $\delta=10^{-1}$,
($10^{-2},10^{-4}$).

Next, we apply our algorithm to the membership problem of separability by
choosing $p=0.1$. Figure 4 plots the boundary to satisfy the condition
$E_{{\cal D}}^{*}({\rho}_{0.1,\lambda,\delta})=10^{-8}$. To compare our method
with the partial transpose negative test, this figure plots the boundary to
satisfy the condition that the minimum eigenvalue of the partial transposed
state is $-10^{-8}$. These comparisons show that our method can detect
entangled states that cannot be detected by the partial transpose negative
test even when all the matrix components of the density matrix of a $4\times
2$ system are written in real numbers.

Figure 4: Comparison between our method and the partial transpose negative
test: The vertical axis shows the parameter $\lambda$. The horizontal axis
shows the parameter $\delta$. Blue (Orange) curve shows our method (the
partial transpose negative test). The upper regions of both curves are
composed of states that are guaranteed to be entangled by the respective
methods.

##  X Exponential quantum Sanov theorem

We consider a $d$-dimensional Hilbert space ${\cal H}$ and a state $\rho$ on
${\mathcal{H}}$. We assume the $n$-fold independently and identical
distributed (iid) condition and consider the following two hypotheses.

$\displaystyle H_{0}:$ $\displaystyle\hbox{The state is }\rho^{\otimes n},$
(110) $\displaystyle H_{1}:$ $\displaystyle\hbox{The state is }\sigma^{\otimes
n}\hbox{ with }\sigma\in S.$ (111)

To address this problem, we focus on the following quantity.

$\displaystyle\beta_{\epsilon,n}(\rho):=\min_{0\leq T\leq
I}\Big{\\{}\operatorname{\mathrm{Tr}}T\rho^{\otimes n}\Big{|}\max_{\sigma\in
S}\operatorname{\mathrm{Tr}}(I-T)\sigma^{\otimes n}\leq\epsilon\Big{\\}}.$
(112)

The relation

$\displaystyle\lim_{n\to\infty}\frac{1}{n}-\log\beta_{\epsilon,n}(\rho)=E^{*}(\rho):=\min_{\sigma\in
S}D(\sigma\|\rho).$ (113)

is known as quantum Sanov theorem [11, 12]. To get an exponential evaluation,
as the extension of $E^{*}(\rho)$, we define the minimum sandwich relative
entropy as

$\displaystyle E_{\alpha}^{*}(\rho):=\min_{\sigma\in
S}D_{\alpha}(\sigma\|\rho).$ (114)

As an exponential version of quantum Sanov theorem, the following theorem
holds.

###### Theorem 11

We have

$\displaystyle\lim_{n\to\infty}-\frac{1}{n}\log\beta_{e^{-nr},n}(\rho)$
$\displaystyle\geq\sup_{0\leq\alpha\leq
1}\frac{(1-\alpha)E_{\alpha}^{*}(\rho)-\alpha r}{1-\alpha}$ (115)
$\displaystyle-\frac{1}{n}\log\beta_{1-e^{-nr},n}(\rho)$
$\displaystyle\leq\inf_{\alpha\geq
1}\frac{(1-\alpha)E_{\alpha}^{*}(\rho)-\alpha r}{1-\alpha}.$ (116)

More precisely, we have

$\displaystyle-\frac{1}{n}\log\beta_{e^{-nr},n}(\rho)$ $\displaystyle\geq$
$\displaystyle-\frac{(d+2)(d-1)}{n}\log(n+1)$
$\displaystyle+\max_{0\leq\alpha\leq
1}\frac{(1-\alpha)E_{\alpha}^{*}(\rho)-\alpha r}{1-\alpha}.$ (117)

Our bound is the optimal when the number of elements of $S$ is one and the
unique element of $S$ is commutative with $\rho$. Although the paper [12]
derived exponential evaluation for both error probabilities, its exponent does
not equal to the optimal bound even when the number of elements of $S$ is one
and the unique element of $S$ is commutative with $\rho$. In this sense, our
evaluation is better than the upper bound by [12].

The proof is given in Appendix D. Theorem 11 proves Theorem 1. Under the limit
$r\to 0$, the RHS of (6) goes to $E_{1}^{*}(\rho)=E^{*}(\rho)$. Hence, the
relation (6) recovers the inequality $\geq$ in (4) while the opposite
inequality in (4) follows from the converse part of quantum Stein’s lemma.

##  XI Discussion

We have proposed new entanglement measures based on the distinguishability of
the given entangled state from the set of separable states with two kinds of
formulations by using quantum relative entropy and sandwich Rényi relative
entropy. That is, we have proposed two kinds of measures based on quantum
relative entropy. This measure is different from the entanglement of relative
entropy [8, 9] because the choice of two input states are opposite to each
other.

In the first formulation, the possible separable states are limited to tensor
product states in this discrimination problem. This case can be considered as
a special case of quantum Sanov theorem [11, 12]. Our analysis realizes an
exponential evaluation for both kinds of error probabilities, which is better
than the existing results [12]. In the second formulation, the possible
separable states are not limited to tensor product states. This problem
setting cannot be considered as a special case of quantum Sanov theorem. Even
in this setting, we have derived an exponential evaluation for both kinds of
error probabilities. Based on these two formulations, we have proposed two
kinds of entanglement measures. The former measure is not smaller than the
latter measure.

Also, the proposed quantity is useful for finding entanglement witness based
on its information-geometrical structure. In the next step, we have derived
several useful calculation formulas for these measures. In particular, we have
shown that these two kinds of measures coincide with each other when the
entangled state is maximally correlated. In particular, the entanglement
measure based on the quantum relative entropy has been concretely derived.
However, it is not easy to calculate these measures for general entangled
states. To resolve this problem, we have proposed algorithms to calculate
these measures. In the first step of this calculation, we have approximated
these values by introducing the $\delta$-net on the second quantum system,
which was introduced to derive an entanglement witness [17, Section 3]. Then,
we apply the quantum Arimoto-Blahut algorithm [18] to these approximating
values. Then, we have obtained an efficient algorithm for these approximating
values. These algorithms are shown to converge to the global minimum, and
their errors are evaluated in a simple form.

Finally, we have applied our algorithm to the membership problem of
separability. Although it is known to be NP-hard when the dimensions of both
Hilbert spaces increase [14, 15, 16, 17], the calculation complexity is not so
large when the dimension of the second Hilbert space is fixed [21]. In this
case, our algorithm has a smaller calculation complexity than the algorithm by
[21]. Also, we have applied our algorithm to find an entanglement witness.

As a future problem, we can consider the following algorithm. We choose a
$\delta$ net on the second system ${\mathcal{H}}_{B}$, and apply our
algorithm. Then, by fixing the obtained states on the system
${\mathcal{H}}_{A}$, our algorithm exchanges the roles of ${\mathcal{H}}_{A}$
and ${\mathcal{H}}_{B}$ so that we derive states on ${\mathcal{H}}_{B}$. In
this way, we iteratively improve the states on ${\mathcal{H}}_{A}$ and
${\mathcal{H}}_{B}$. We might expect that this algorithm works well even when
the $\delta$ is not so small. However, this algorithm solves the membership
problem of separability, which is known to be NP-hard when the dimensions of
both Hilbert spaces increase. Hence, it is impossible to choose so small
$\delta$ that the calculation time complexity is polynomial for the local
dimension. It is an interesting problem whether such an iterative algorithm
can improve our proposed algorithm.

## Acknowledgement

MH was supported in part by the National Natural Science Foundation of China
under Grant 62171212. MH is thankful to Professor Hiroshi Hirai for helpful
discussions on the membership problem for separability and the importance of
separation hyperplane, and informing the references [15, 20, 25, 21, 23, 24].
MH is thankful to Professor Marco Tomamichel for informing the reference [22].
MH and YI are grateful for Professor Masaki Owari to a helpful discussion on
membership problem of separability and informing the reference [17].

##  Appendix A Construction of discrete subset

###  A.1 Real discrete subset

First, similar to [29, Section 5.4], we construct the real discrete subset
${\cal D}_{n,d,R}\subset{\cal S}_{d}$, where ${\cal S}_{d}$ is the
$d$-dimensional sphare in $\mathbb{R}^{d+1}$. We define ${\cal D}_{n,1,R}$ as

$\displaystyle{\cal D}_{n,1,R}:=\Big{\\{}\Big{(}\cos\frac{2\pi
j}{n},\sin\frac{2\pi j}{n}\Big{)}\Big{\\}}_{j=0}^{n-1}.$ (118)

Then, we inductively define ${\cal D}_{n,d,R}$ as

$\displaystyle{\cal D}_{n,d,R}:=\Big{\\{}\Big{(}\cos\frac{2\pi
j}{n}v,\sin\frac{2\pi j}{n}\Big{)}\Big{|}v\in{\cal
D}_{n,d-1,R},j=0,\ldots,n-1\Big{\\}}.$ (119)

We have

$\displaystyle F({\cal D}_{n,1,R}):=$ $\displaystyle\min_{\psi_{1}\in{\cal
S}_{2}}\max_{\psi_{2}\in{\cal
D}_{n,2,R}}|\langle\psi_{1}|\psi_{2}\rangle|=\cos\frac{\pi}{n}.$ (120)

In general, we have

$\displaystyle F({\cal D}_{n,d,R})=$ $\displaystyle\min_{\psi_{1}\in{\cal
S}_{d}}\max_{\psi_{2}\in{\cal
D}_{n,d,R}}|\langle\psi_{1}|\psi_{2}\rangle|=\cos^{d}\frac{\pi}{n}.$ (121)

This can be inductively shown as follows.

$\displaystyle F({\cal D}_{n,d})=\min_{\psi_{1}\in{\cal
S}_{d}}\max_{\psi_{2}\in{\cal D}_{n,d}}|\langle\psi_{1}|\psi_{2}\rangle|$
$\displaystyle=$ $\displaystyle\min_{\theta}\min_{\psi_{1}\in{\cal
S}_{d-1}}\max_{j=0,\ldots,n-1}\max_{\psi_{2}\in{\cal
D}_{n,d-1}}\Big{|}\langle\psi_{1}|\psi_{2}\rangle\cos\theta\cos\frac{2\pi
j}{n}+\sin\theta\sin\frac{2\pi j}{n}\Big{|}$ $\displaystyle=$
$\displaystyle\min_{\theta}\min_{\psi_{1}\in{\cal
S}_{d-1}}\max_{j=0,\ldots,n-1}\max_{\psi_{2}\in{\cal
D}_{n,d-1}}\Big{|}\langle\psi_{1}|\psi_{2}\rangle\Big{(}\cos\theta\cos\frac{2\pi
j}{n}+\frac{1}{\langle\psi_{1}|\psi_{2}\rangle}\sin\theta\sin\frac{2\pi
j}{n}\Big{)}\Big{|}$ $\displaystyle=$
$\displaystyle\min_{\theta}\min_{\psi_{1}\in{\cal
S}_{d-1}}\max_{\psi_{2}\in{\cal
D}_{n,d-1}}|\langle\psi_{1}|\psi_{2}\rangle|\Big{(}\max_{j=0,\ldots,n-1}\Big{|}\cos\theta\cos\frac{2\pi
j}{n}+\frac{1}{\langle\psi_{1}|\psi_{2}\rangle}\sin\theta\sin\frac{2\pi
j}{n}\Big{|}\Big{)}$ $\displaystyle\geq$
$\displaystyle\min_{\theta}\min_{\psi_{1}\in{\cal
S}_{d-1}}\max_{\psi_{2}\in{\cal D}_{n,d-1}}|\langle\psi_{1}|\psi_{2}\rangle|$
$\displaystyle\cdot\Big{(}\max_{j=0,\ldots,n-1}\max\Big{(}\Big{|}\cos\theta\cos\frac{2\pi
j}{n}+\sin\theta\sin\frac{2\pi j}{n}\Big{|},\Big{|}\cos\theta\cos\frac{2\pi
j}{n}-\sin\theta\sin\frac{2\pi j}{n}\Big{|}\Big{)}\Big{)}$ $\displaystyle=$
$\displaystyle\min_{\theta}\min_{\psi_{1}\in{\cal
S}_{d-1}}\max_{\psi_{2}\in{\cal
D}_{n,d-1}}|\langle\psi_{1}|\psi_{2}\rangle|\Big{(}\max_{j=0,\ldots,n-1}\max\Big{(}\Big{|}\sin(\frac{2\pi
j}{n}+\theta)\Big{|},\Big{|}\sin(\frac{2\pi
j}{n}-\theta)\Big{|}\Big{)}\Big{)}$ $\displaystyle=$
$\displaystyle\Big{(}\min_{\theta}\max_{j=0,\ldots,n-1}\max\Big{(}\Big{|}\sin(\frac{2\pi
j}{n}+\theta)\Big{|},\Big{|}\sin(\frac{2\pi
j}{n}-\theta)\Big{|}\Big{)}\Big{(}\min_{\psi_{1}\in{\cal
S}_{d-1}}\max_{\psi_{2}\in{\cal
D}_{n,d-1}}|\langle\psi_{1}|\psi_{2}\rangle|\Big{)}$ $\displaystyle\geq$
$\displaystyle\cos\frac{\pi}{n}F({\cal D}_{n,d-1})=\cos^{d}\frac{\pi}{n}.$
(122)

###  A.2 Complex discrete subset

Next, we construct the complex dscrete subset ${\cal
D}_{n,d}\subset\mathbb{C}^{d}$, where all elements of ${\cal D}_{n,d}$ are
normalized vectors. Using ${\cal D}_{n,2d-2,R}$, we define ${\cal D}_{n,d}$ as

$\displaystyle{\cal D}_{n,d}:=\\{$
$\displaystyle(x_{1},x_{2}+x_{3}i,x_{4}+x_{5}i,\ldots,x_{2d-2}+x_{2d-1}i)|$
$\displaystyle x=(x_{1},\ldots,x_{2d-1})\in{\cal D}_{n,2d-2,R}\\}.$ (123)

The number of elements $|{\cal D}_{n,d}|$ is $n^{2d-2}$.

Then, we have

$\displaystyle\frac{1}{4}\epsilon_{2}({\cal
D}_{n,d})^{2}=1-\min_{\psi_{1}\in\mathbb{C}_{n}^{d}}\max_{\psi_{2}\in{\cal
D}_{n,d}}|\langle\psi_{1}|\psi_{2}\rangle|^{2}$ $\displaystyle\leq$
$\displaystyle 1-(\min_{\psi_{1}\in\mathbb{C}_{n}^{d}}\max_{\psi_{2}\in{\cal
D}_{n,d}}\Re|\langle\psi_{1}|\psi_{2}\rangle|)^{2}$ $\displaystyle=$
$\displaystyle 1-(\min_{\psi_{1}\in S_{2d-2}}\max_{\psi_{2}\in{\cal
D}_{n,2d-2,R}}\langle\psi_{1}|\psi_{2}\rangle)^{2}$ $\displaystyle=$
$\displaystyle 1-(\min_{\psi_{1}\in S_{2d-2}}\max_{\psi_{2}\in{\cal
D}_{n,2d-2,R}}\langle\psi_{1}|\psi_{2}\rangle)^{2}$ $\displaystyle\leq$
$\displaystyle 1-\cos^{2d}\frac{\pi}{n}=1-(1-\sin^{2}\frac{\pi}{n})^{d}\leq
d\sin^{2}\frac{\pi}{n}\leq d\frac{\pi^{2}}{n^{2}}.$ (124)

When we choose $n$ as $4d\frac{\pi^{2}}{n^{2}}=\epsilon_{2}^{2}$, i.e.,
$n=\sqrt{\frac{4d\pi^{2}}{\epsilon_{2}^{2}}}$, the number of elements $|{\cal
D}_{n,d}|$ is $(\frac{2d^{1/2}\pi}{\epsilon_{2}})^{d-1}$.

##  Appendix B Proof of Lemma 7

Since the relation (52) follows from (53) with taking the limit $\alpha\to 1$,
we show only (53). Define the projection
$P:=\sum_{j}|u_{j,A}\rangle|u_{j,B}\rangle\langle u_{j,A}|\langle u_{j,B}|$
and the TP-CP map

$\displaystyle\Lambda^{*}(\rho^{\prime}):=\sum_{j}P\rho^{\prime}P+(I-P)\rho^{\prime}(I-P).$
(125)

We choose a product pure state $|v_{A}\rangle|v_{B}\rangle$. The relation
$\|(I-P)|v_{A}\rangle|v_{B}\rangle\|=0$ holds if and only if the relations

$\displaystyle\begin{split}&|\\{j:|\langle u_{j,A}|v_{A}\rangle|\neq
0\\}|=1,\\\ &\\{j:|\langle u_{j,A}|v_{A}\rangle|\neq 0\\}=\\{j:|\langle
u_{j,B}|v_{B}\rangle|\neq 0\\}\end{split}$ (126)

hold. When the above condition holds, the relation
$P|v_{A}\rangle|v_{B}\rangle=|u_{j,A}\rangle|u_{j,B}\rangle$ holds for
$j\in\\{j:|\langle u_{j,A}|v_{A}\rangle|\neq 0\\}\cap\\{j:|\langle
u_{j,B}|v_{B}\rangle|\neq 0\\}$.

Also, we have $P|v_{A}\rangle|u_{j,B}\rangle=(\langle
u_{j,A}|v_{A}\rangle)|u_{j,A}\rangle|u_{j,B}\rangle$. Hence, we have

$\displaystyle P(\sigma_{A}\otimes
I_{B})P=\sum_{j}P(\sigma_{A}\otimes|u_{j,B}\rangle\langle u_{j,B}|)P$
$\displaystyle=$ $\displaystyle\sum_{j}\langle
u_{j,A}|\sigma_{A}|u_{j,A}\rangle)|u_{j,A}\rangle|u_{j,B}\rangle\langle
u_{j,A}|\langle u_{j,B}|.$ (127)

Therefore,

$\displaystyle C_{\alpha}^{*}\Big{(}\sum_{i,j}\theta_{i,j}|j\rangle\langle
i|\Big{)}$ $\displaystyle=$
$\displaystyle\min_{p}D_{\alpha}\Big{(}\sum_{j}p_{j}|j\rangle\langle
j|\Big{\|}\sum_{i,j}\theta_{i,j}|u_{j,A}\rangle|u_{j,B}\rangle\langle
u_{i,A}|\langle u_{j,B}|\Big{)}$ $\displaystyle\geq$ $\displaystyle
E_{\alpha}^{*}\Big{(}\sum_{i,j}\theta_{i,j}|u_{j,A}\rangle|u_{j,B}\rangle\langle
u_{i,A}|\langle u_{j,B}|\Big{)}.$ (128)

Now, we fix $\alpha\in(0,1]$. For any separable state $\sigma_{AB}\in{\cal
S}_{AB}$, we have

$\displaystyle\sigma_{AB}=\sum_{j}p_{j}\rho_{A,j}\otimes\rho_{B,j}\leq\sum_{j}p_{j}\rho_{A,j}\otimes
I_{B}=\sigma_{A}\otimes I_{B},$ (129)

where $\sigma_{A}:=\operatorname{\mathrm{Tr}}_{B}\sigma_{AB}$. Hence, we have

$\displaystyle\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\sigma_{AB}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\leq\rho_{AB}^{\frac{1-\alpha}{2\alpha}}(\sigma_{A}\otimes
I_{B})\rho_{AB}^{\frac{1-\alpha}{2\alpha}}.$ (130)

Since $x\mapsto x^{\alpha}$ is matrix monotone, we have

$\displaystyle\operatorname{\mathrm{Tr}}(\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\sigma_{AB}\rho_{AB}^{\frac{1-\alpha}{2\alpha}})^{\alpha}\leq\operatorname{\mathrm{Tr}}(\rho_{AB}^{\frac{1-\alpha}{2\alpha}}(\sigma_{A}\otimes
I_{B})\rho_{AB}^{\frac{1-\alpha}{2\alpha}})^{\alpha}.$ (131)

Thus,

$\displaystyle
E_{\alpha}^{*}(\rho_{AB})\geq\min_{\sigma_{A}}D_{\alpha}(\sigma_{A}\otimes
I_{B}\|\rho_{AB}).$ (132)

We have

$\displaystyle D_{\alpha}\Big{(}\sigma_{A}\otimes
I_{B}\Big{\|}\sum_{i,j}\theta_{i,j}|u_{j,A}\rangle|u_{j,B}\rangle\langle
u_{i,A}|\langle u_{i,B}|\Big{)}$ $\displaystyle\geq$ $\displaystyle
D_{\alpha}\Big{(}\Lambda^{*}(\sigma_{A}\otimes
I_{B})\Big{\|}\sum_{i,j}\theta_{i,j}|u_{j,A}\rangle|u_{j,B}\rangle\langle
u_{i,A}|\langle u_{i,B}|\Big{)}$ $\displaystyle=$ $\displaystyle
D_{\alpha}\Big{(}P(\sigma_{A}\otimes
I_{B})P\Big{\|}\sum_{i,j}\theta_{i,j}|u_{j,A}\rangle|u_{j,B}\rangle\langle
u_{i,A}|\langle u_{i,B}|\Big{)}$ $\displaystyle=$ $\displaystyle
D_{\alpha}\Big{(}\sum_{j}\langle
u_{j,A}|\sigma_{A}|u_{j,A}\rangle)|u_{j,A}\rangle|u_{j,B}\rangle\langle
u_{j,A}|\langle u_{j,B}|$
$\displaystyle\Big{\|}\sum_{i,j}\theta_{i,j}|u_{j,A}\rangle|u_{j,B}\rangle\langle
u_{i,A}|\langle u_{i,B}|\Big{)}$ $\displaystyle=$ $\displaystyle
D_{\alpha}\Big{(}\sum_{j}\langle
u_{j,A}|\sigma_{A}|u_{j,A}\rangle)|j\rangle\langle
j|\Big{\|}\sum_{i,j}\theta_{i,j}|j\rangle\langle i|\Big{)}.$ (133)

Taking the minimum for $\sigma_{A}$, we have

$\displaystyle\min_{\sigma_{A}}D_{\alpha}\Big{(}\sigma_{A}\otimes
I_{B}\Big{\|}\sum_{i,j}\theta_{i,j}|u_{j,A}\rangle|u_{j,B}\rangle\langle
u_{i,A}|\langle u_{i,B}|\Big{)}$ $\displaystyle\geq$ $\displaystyle
C_{\alpha}^{*}\Big{(}\sum_{i,j}\theta_{i,j}|j\rangle\langle i|\Big{)}.$ (134)

The combination of (128), (132), and (134) yields (53) with $\alpha\in(0,1]$.

Next, we fix $\alpha>1$. When a separable state $\sigma_{AB}\in{\cal S}_{AB}$
satisfies $(I-P)\sigma_{AB}(I-P)\neq 0$,
$D_{\alpha}\Big{(}\sigma_{AB}\Big{\|}\sum_{i,j}\theta_{i,j}|u_{j,A}\rangle|u_{j,B}\rangle\langle
u_{i,A}|\langle u_{i,B}|\Big{)}=\infty$. Hence, we have

$\displaystyle
E_{\alpha}^{*}\Big{(}\sum_{i,j}\theta_{i,j}|u_{j,A}\rangle|u_{j,B}\rangle\langle
u_{i,A}|\langle u_{i,B}|\Big{)}$ $\displaystyle=$
$\displaystyle\min\Big{\\{}D_{\alpha}\Big{(}\sigma_{AB}\Big{\|}\sum_{i,j}\theta_{i,j}|u_{j,A}\rangle|u_{j,B}\rangle\langle
u_{i,A}|\langle u_{i,B}|\Big{)}$ $\displaystyle\hskip
43.05542pt\Big{|}\sigma_{AB}\in{\cal S}_{AB},(I-P)\sigma_{AB}(I-P)=0\Big{\\}}$
$\displaystyle\stackrel{{\scriptstyle(a)}}{{=}}$
$\displaystyle\min_{p}D_{\alpha}\Big{(}\sum_{j}p_{j}|u_{j,A}\rangle|u_{j,B}\rangle\langle
u_{j,A}|\langle u_{j,B}|$ $\displaystyle\hskip
43.05542pt\Big{\|}\sum_{i,j}\theta_{i,j}|u_{j,A}\rangle|u_{j,B}\rangle\langle
u_{i,A}|\langle u_{i,B}|\Big{)}.$ (135)

The relation $(a)$ follows from the condition (126). The combination of (128)
and (135) yields (53) with $\alpha>1$.

##  Appendix C Proof of Theorems 8 and 9

###  C.1 Generalized Arimoto-Blahut algorithm

Extending Arimoto-Blahut algorithm [53, 54], the paper [18] proposed a general
algorithm. Recently, this general algorithm was generalized to another form
[55]. We consider a finite-dimensional Hilbert space ${\cal H}$ and focus on
the set ${\cal S}({\cal H})$ of density matrices on ${\cal H}$. In this
section, as a preparation, we consider a minimization problem over the set
${\cal S}({\cal H})$. Given a continuous function $\Omega$ from the set ${\cal
S}({\cal H})$ to the set of Hermitian matrices on ${\cal H}$, $B({\cal H})$,
we consider the minimization $\min_{\rho\in{\cal S}({\cal H})}{\cal G}(\rho)$;

$\displaystyle{\cal G}(\rho):=\operatorname{\mathrm{Tr}}\rho\Omega[\rho].$
(136)

In the following, we discuss the calculation of the following two problems;

$\displaystyle\overline{{\cal G}}(a):=\min_{\rho\in{\cal S}({\cal H})}{\cal
G}(\rho),\quad\rho_{*}:=\mathop{\rm argmin}\limits_{\rho\in{\cal S}({\cal
H})}{\cal G}(\rho).$ (137)

For this aim, with $\gamma>0$, we define the density matrix ${\cal F}[\sigma]$
as ${\cal
F}[\sigma]:=\frac{1}{\kappa[\sigma]}\exp(\log\sigma-\frac{1}{\gamma}\Omega[\sigma])$,
where $\kappa[\sigma]$ is the normalization factor
$\operatorname{\mathrm{Tr}}\exp(\log\sigma-\frac{1}{\gamma}\Omega[\sigma])$.
The paper [18] proposed Algorithm 3.

Algorithm 3 Minimization of ${\cal G}(\rho)$

1:Choose the initial value $\rho^{(1)}\in\mathcal{M}$;

2:repeat

3: Calculate $\rho^{(t+1)}:={\cal F}[\rho^{(t)}]$;

4:until convergence.

For any two densities $\rho$ and $\sigma$, we define

$\displaystyle
D_{\Omega}(\rho\|\sigma):=\operatorname{\mathrm{Tr}}\rho(\Omega[\rho]-\Omega[\sigma]).$
(138)

The paper [18, Theorem 3.3] showed the following theorem, which discusses the
convergence to the global minimum and the convergence speed.

###### Theorem 12

When any two densities $\rho^{(t)}$ and $\rho^{(t+1)}$ in satisfy the
condition

$\displaystyle D_{\Omega}(\rho^{(t+1)}\|\rho^{(t)})\leq\gamma
D(\rho^{(t+1)}\|\rho^{(t)}),$ (139)

and a state $\rho_{*}$ satisfies

$\displaystyle 0\leq D_{\Omega}(\rho_{*}\|\rho)$ (140)

with any state $\rho$, Algorithm 3 satisfies the condition

$\displaystyle{\cal G}(\rho^{(t_{0}+1)})-{\cal G}(\rho_{*})\leq\frac{\gamma
D(\rho_{*}\|\rho^{(1)})}{t_{0}}$ (141)

with any initial state $\rho^{(1)}$.

###  C.2 Proof of Theorem 8

We choose a classical-quantum state $\rho_{AC}$ as

$\displaystyle\rho_{AC}:=\sum_{c\in{\cal C}}|c\rangle\langle c|\otimes
X_{A|c}.$ (142)

We choose the function $\Omega_{R,{\cal D}}$ as

$\displaystyle\Omega_{{\cal D}}[\rho_{AC}]:=\sum_{c\in{\cal
C}}|c\rangle\langle
c|\otimes\Big{(}\operatorname{\mathrm{Tr}}_{B}\Big{(}\log\Big{(}\sum_{c^{\prime}\in{\cal
C}}X_{A|c^{\prime}}\otimes\rho_{B|c^{\prime}}^{*}\Big{)}-\log\rho_{AB}\Big{)}(I_{A}\otimes\rho_{B|c}^{*})\Big{)}.$
(143)

Then, we have

$\displaystyle\operatorname{\mathrm{Tr}}\rho_{AC}\Omega_{\cal
D}[\rho_{AC}]=\operatorname{\mathrm{Tr}}_{AC}\rho_{AC}\sum_{c\in{\cal
C}}|c\rangle\langle
c|\otimes\Big{(}\operatorname{\mathrm{Tr}}_{B}\Big{(}\log\Big{(}\sum_{c^{\prime}\in{\cal
C}}X_{A|c^{\prime}}\otimes\rho_{B|c^{\prime}}^{*}\Big{)}-\log\rho_{AB}\Big{)}(I_{A}\otimes\rho_{B|c}^{*})\Big{)}$
$\displaystyle=$
$\displaystyle\operatorname{\mathrm{Tr}}_{AC}(\sum_{c^{\prime\prime}\in{\cal
C}}|c^{\prime\prime}\rangle\langle c^{\prime\prime}|\otimes
X_{A|c^{\prime\prime}})\sum_{c\in{\cal C}}|c\rangle\langle
c|\otimes\Big{(}\operatorname{\mathrm{Tr}}_{B}\Big{(}\log\Big{(}\sum_{c^{\prime}\in{\cal
C}}X_{A|c^{\prime}}\otimes\rho_{B|c^{\prime}}^{*}\Big{)}-\log\rho_{AB}\Big{)}(I_{A}\otimes\rho_{B|c}^{*})\Big{)}$
$\displaystyle=$ $\displaystyle\operatorname{\mathrm{Tr}}_{A}\sum_{c\in{\cal
C}}X_{A|c}\Big{(}\operatorname{\mathrm{Tr}}_{B}\Big{(}\log\Big{(}\sum_{c^{\prime}\in{\cal
C}}X_{A|c^{\prime}}\otimes\rho_{B|c^{\prime}}^{*}\Big{)}-\log\rho_{AB}\Big{)}(I_{A}\otimes\rho_{B|c}^{*})\Big{)}$
$\displaystyle=$ $\displaystyle\operatorname{\mathrm{Tr}}_{AB}\sum_{c\in{\cal
C}}X_{A|c}\otimes\rho_{B|c}^{*}\Big{(}\log\Big{(}\sum_{c^{\prime}\in{\cal
C}}X_{A|c^{\prime}}\otimes\rho_{B|c^{\prime}}^{*}\Big{)}-\log\rho_{AB}\Big{)}=D\Big{(}\sum_{c\in{\cal
C}}X_{A|c}\otimes\rho_{B|c}^{*}\Big{\|}\rho_{AB}\Big{)}.$ (144)

That is, we can apply Algorithm 3. In this case, when we set
$\sigma_{AC}[\\{X_{A|c}\\}_{c}]$ as

$\displaystyle\sigma_{AC}[\\{X_{A|c}\\}_{c}]:=\sum_{c\in{\cal
C}}|c\rangle\langle c|\otimes X_{A|c},$ (145)

${\cal F}[\sigma_{AC}[\\{X_{A|c}\\}_{c}]]$ is written as

$\displaystyle{\cal F}[\sigma_{AC}[\\{X_{A|c}\\}_{c}]]=$
$\displaystyle\frac{1}{\kappa[\\{X_{A|c}\\}_{c}]}\sum_{c\in{\cal
C}}|c\rangle\langle
c|\otimes\exp(X_{A|c}-V_{c}[\\{X_{A|c^{\prime}}\\}_{c^{\prime}}]).$ (146)

Algorithm 1 is the same as the special case of Algorithm 3 with the above
choice and $\gamma=1$.

Also, we have

$\displaystyle D_{\Omega_{{\cal D}}}(\rho_{AC}\|\rho_{AC}^{\prime})$
$\displaystyle=$
$\displaystyle\operatorname{\mathrm{Tr}}_{AC}\rho_{AC}\sum_{c\in{\cal
C}}|c\rangle\langle
c|\otimes\Big{(}\operatorname{\mathrm{Tr}}_{B}\Big{(}\log\Big{(}\sum_{c^{\prime}\in{\cal
C}}X_{A|c^{\prime}}\otimes\rho_{B|c^{\prime}}^{*}\Big{)}-\Big{(}\log\Big{(}\sum_{c^{\prime\prime}\in{\cal
C}}X_{A|c^{\prime\prime}}^{\prime}\otimes\rho_{B|c^{\prime\prime}}^{*}\Big{)}\Big{)}(I_{A}\otimes\rho_{B|c}^{*})\Big{)}$
$\displaystyle=$ $\displaystyle\operatorname{\mathrm{Tr}}_{AB}(\sum_{c\in{\cal
C}}P(c)\rho_{A|c}\otimes\rho_{B|c}^{*})\Big{(}\log\Big{(}\sum_{c^{\prime}\in{\cal
C}}X_{A|c^{\prime}}\otimes\rho_{B|c^{\prime}}^{*}\Big{)}-\Big{(}\log\Big{(}\sum_{c^{\prime\prime}\in{\cal
C}}X_{A|c^{\prime\prime}}^{\prime}\otimes\rho_{B|c^{\prime\prime}}^{*}\Big{)}\Big{)}$
$\displaystyle=$ $\displaystyle D\Big{(}\sum_{c^{\prime}\in{\cal
C}}X_{A|c^{\prime}}\otimes\rho_{B|c^{\prime}}^{*}\Big{\|}\sum_{c^{\prime\prime}\in{\cal
C}}X_{A|c^{\prime\prime}}^{\prime}\otimes\rho_{B|c^{\prime\prime}}^{*}\Big{)}$
(147) $\displaystyle\leq$ $\displaystyle D\Big{(}\sum_{c^{\prime}\in{\cal
C}}|c^{\prime}\rangle\langle c^{\prime}|\otimes
X_{A|c^{\prime}}\otimes\rho_{B|c^{\prime}}^{*}\Big{\|}\sum_{c^{\prime\prime}\in{\cal
C}}|c^{\prime\prime}\rangle\langle c^{\prime\prime}|\otimes
X_{A|c^{\prime\prime}}^{\prime}\otimes\rho_{B|c^{\prime\prime}}^{*}\Big{)}$
$\displaystyle=$ $\displaystyle D\Big{(}\sum_{c^{\prime}\in{\cal
C}}|c^{\prime}\rangle\langle c^{\prime}|\otimes
X_{A|c^{\prime}}\Big{\|}\sum_{c^{\prime\prime}\in{\cal
C}}|c^{\prime\prime}\rangle\langle c^{\prime\prime}|\otimes
X_{A|c^{\prime\prime}}^{\prime}\Big{)}=D(\rho_{AC}\|\rho_{AC}^{\prime}).$
(148)

Also, (147) guarantees that the relation $D_{\Omega_{R,{\cal
D}}}(\rho_{AC}\|\rho_{AC}^{\prime})\geq 0$ holds. Hence, the condition of
Theorem 12 holds. That is, Theorem 8 holds.

###  C.3 Proof of Theorem 9

We choose the function $\Omega_{\alpha,{\cal D}}$ as

$\displaystyle\Omega_{\alpha,{\cal D}}[\rho_{AC}]:=\mathop{\rm
sgn}(\alpha-1)\sum_{c\in{\cal C}}|c\rangle\langle
c|\otimes\operatorname{\mathrm{Tr}}_{B}\Big{(}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{(}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{(}\sum_{c^{\prime}\in{\cal
C}}X_{A|c^{\prime}}\otimes\rho_{B|c^{\prime}}^{*}\Big{)}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{)}^{\alpha-1}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{)}(I_{A}\otimes\rho_{B|c}^{*})\Big{)}.$
(149)

Then, we have

$\displaystyle\operatorname{\mathrm{Tr}}\rho_{AC}\Omega_{\alpha,\cal
D}[\rho_{AC}]$ $\displaystyle=$ $\displaystyle\mathop{\rm
sgn}(\alpha-1)\operatorname{\mathrm{Tr}}_{AC}\rho_{AC}\sum_{c\in{\cal
C}}|c\rangle\langle
c|\otimes\operatorname{\mathrm{Tr}}_{B}\Big{(}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{(}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{(}\sum_{c^{\prime}\in{\cal
C}}X_{A|c^{\prime}}\otimes\rho_{B|c^{\prime}}^{*}\Big{)}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{)}^{\alpha-1}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{)}(I_{A}\otimes\rho_{B|c}^{*})\Big{)}$
$\displaystyle=$ $\displaystyle\mathop{\rm
sgn}(\alpha-1)\operatorname{\mathrm{Tr}}_{AC}(\sum_{c^{\prime\prime}\in{\cal
C}}|c^{\prime\prime}\rangle\langle c^{\prime\prime}|\otimes
X_{A|c^{\prime\prime}})\sum_{c\in{\cal C}}|c\rangle\langle c|$
$\displaystyle\otimes\operatorname{\mathrm{Tr}}_{B}\Big{(}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{(}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{(}\sum_{c^{\prime}\in{\cal
C}}X_{A|c^{\prime}}\otimes\rho_{B|c^{\prime}}^{*}\Big{)}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{)}^{\alpha-1}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{)}(I_{A}\otimes\rho_{B|c}^{*})\Big{)}$
$\displaystyle=$ $\displaystyle\mathop{\rm
sgn}(\alpha-1)\operatorname{\mathrm{Tr}}(\sum_{c\in{\cal
C}}X_{A|c}\otimes\rho_{B|c}^{*})\Big{(}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{(}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{(}\sum_{c^{\prime}\in{\cal
C}}X_{A|c^{\prime}}\otimes\rho_{B|c^{\prime}}^{*}\Big{)}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{)}^{\alpha-1}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{)}\Big{)}$
$\displaystyle=$ $\displaystyle\mathop{\rm
sgn}(\alpha-1)\operatorname{\mathrm{Tr}}\Big{(}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{(}\sum_{c^{\prime}\in{\cal
C}}X_{A|c^{\prime}}\otimes\rho_{B|c^{\prime}}^{*}\Big{)}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{)}^{\alpha}\Big{)}.$
(150)

That is, we can apply Algorithm 3. In this case, when we set
$\sigma_{AC}[\\{X_{A|c}\\}_{c}]$ as (145), ${\cal
F}[\sigma_{AC}[\\{X_{A|c}\\}_{c}]]$ is written as
$\frac{1}{\kappa[\\{X_{A|c}\\}_{c}]}\exp(X_{A|c}-\frac{\mathop{\rm
sgn}(\alpha-1)}{\gamma}V_{\alpha,c}[\\{X_{A|c^{\prime}}^{(t)}\\}_{c^{\prime}}])$.
Algorithm 2 is the same as the special case of Algorithm 3 with the above
choice.

We denote the minimizer by $\sigma_{AC}[\\{X_{A|c}^{*}\\}_{c}]$. For
$\alpha\in[0,1]$, We choose
$Z_{*}=\Big{(}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{(}\sum_{c\in{\cal
C}}X_{A|c}^{*}\otimes\rho_{B|c}^{*}\Big{)}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{)}$
and
$Z=\Big{(}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{(}\sum_{c^{\prime}\in{\cal
C}}X_{A|c^{\prime}}\otimes\rho_{B|c^{\prime}}^{*}\Big{)}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{)}$.
Since $\sigma_{AC}[\\{X_{A|c}^{*}\\}_{c}]$ is the minimizer, we have

$\displaystyle\operatorname{\mathrm{Tr}}Z_{*}^{\alpha}\geq\operatorname{\mathrm{Tr}}Z^{\alpha}.$
(151)

Since $Z_{*}^{\alpha}/\operatorname{\mathrm{Tr}}Z_{*}^{\alpha})$ and
$(Z^{\alpha}/\operatorname{\mathrm{Tr}}Z^{\alpha})$ are density matrices, we
have

$\displaystyle\operatorname{\mathrm{Tr}}(Z_{*}^{\alpha}/\operatorname{\mathrm{Tr}}Z_{*}^{\alpha})^{\frac{1}{\alpha}}(Z^{\alpha}/\operatorname{\mathrm{Tr}}Z^{\alpha})^{\frac{\alpha-1}{\alpha}}\geq
1.$ (152)

Thus, we have

$\displaystyle D_{\Omega_{\alpha,{\cal
D}}}(\sigma_{AC}[\\{X_{A|c}^{*}\\}_{c}]\|\sigma_{AC}[\\{X_{A|c}\\}_{c}])$
$\displaystyle=$
$\displaystyle-\operatorname{\mathrm{Tr}}_{AC}\sigma_{AC}[\\{X_{A|c}^{*}\\}_{c}]\sum_{c\in{\cal
C}}|c\rangle\langle
c|\otimes\Big{(}\operatorname{\mathrm{Tr}}_{B}\Big{(}\Big{(}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{(}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{(}\sum_{c\in{\cal
C}}X_{A|c}^{*}\otimes\rho_{B|c}^{*}\Big{)}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{)}^{\alpha-1}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{)}$
$\displaystyle-\Big{(}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{(}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{(}\sum_{c^{\prime}\in{\cal
C}}X_{A|c^{\prime}}\otimes\rho_{B|c^{\prime}}^{*}\Big{)}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{)}^{\alpha-1}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{)}\Big{)}(I_{A}\otimes\rho_{B|c}^{*})\Big{)}$
$\displaystyle=$
$\displaystyle-\operatorname{\mathrm{Tr}}\Big{(}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{(}\sum_{c\in{\cal
C}}X_{A|c}^{*}\otimes\rho_{B|c}^{*}\Big{)}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{)}\Bigg{(}\Big{(}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{(}\sum_{c\in{\cal
C}}X_{A|c}^{*}\otimes\rho_{B|c}^{*}\Big{)}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{)}^{\alpha-1}$
$\displaystyle-\Big{(}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{(}\sum_{c^{\prime}\in{\cal
C}}X_{A|c^{\prime}}\otimes\rho_{B|c^{\prime}}^{*}\Big{)}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{)}^{\alpha-1}\Bigg{)}$
$\displaystyle=$
$\displaystyle\operatorname{\mathrm{Tr}}Z_{*}Z^{\alpha-1}-\operatorname{\mathrm{Tr}}Z_{*}^{\alpha}$
$\displaystyle=$
$\displaystyle(\operatorname{\mathrm{Tr}}Z_{*}Z^{\alpha-1})(\operatorname{\mathrm{Tr}}Z_{*}^{\alpha})^{\frac{\alpha-1}{\alpha}}((\operatorname{\mathrm{Tr}}Z_{*}^{\alpha})^{\frac{1-\alpha}{\alpha}}-(\operatorname{\mathrm{Tr}}Z^{\alpha})^{\frac{1-\alpha}{\alpha}})+\operatorname{\mathrm{Tr}}Z_{*}^{\alpha}(\operatorname{\mathrm{Tr}}(Z_{*}^{\alpha}/\operatorname{\mathrm{Tr}}Z_{*}^{\alpha})^{\frac{1}{\alpha}}(Z^{\alpha}/\operatorname{\mathrm{Tr}}Z^{\alpha})^{\frac{\alpha-1}{\alpha}}-1)$
$\displaystyle\geq$ $\displaystyle 0.$ (153)

Hence, the condition (140) holds.

We choose
$Z^{\prime}=\Big{(}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{(}\sum_{c\in{\cal
C}}X_{A|c}^{\prime}\otimes\rho_{B|c}^{*}\Big{)}\rho_{AB}^{\frac{1-\alpha}{2\alpha}}\Big{)}$.
Then, we have

$\displaystyle D_{\Omega_{\alpha,{\cal
D}}}(\sigma_{AC}[\\{X_{A|c}^{\prime}\\}_{c}]\|\sigma_{AC}[\\{X_{A|c}\\}_{c}])$
(154) $\displaystyle=$
$\displaystyle(\operatorname{\mathrm{Tr}}Z^{\prime}Z^{\alpha-1})(\operatorname{\mathrm{Tr}}{Z^{\prime}}^{\alpha})^{\frac{\alpha-1}{\alpha}}((\operatorname{\mathrm{Tr}}{Z^{\prime}}^{\alpha})^{\frac{1-\alpha}{\alpha}}-(\operatorname{\mathrm{Tr}}Z^{\alpha})^{\frac{1-\alpha}{\alpha}})+\operatorname{\mathrm{Tr}}{Z^{\prime}}^{\alpha}(\operatorname{\mathrm{Tr}}({Z^{\prime}}^{\alpha}/\operatorname{\mathrm{Tr}}{Z^{\prime}}^{\alpha})^{\frac{1}{\alpha}}(Z^{\alpha}/\operatorname{\mathrm{Tr}}Z^{\alpha})^{\frac{\alpha-1}{\alpha}}-1).$
(155)

When $Z^{\prime}$ is close to $Z$,
$\operatorname{\mathrm{Tr}}({Z^{\prime}}^{\alpha}/\operatorname{\mathrm{Tr}}{Z^{\prime}}^{\alpha})^{\frac{1}{\alpha}}(Z^{\alpha}/\operatorname{\mathrm{Tr}}Z^{\alpha})^{\frac{\alpha-1}{\alpha}}-1=O(\|Z^{\prime}-Z\|^{2})$
and
$(\operatorname{\mathrm{Tr}}{Z^{\prime}}^{\alpha})^{\frac{1-\alpha}{\alpha}}-(\operatorname{\mathrm{Tr}}Z^{\alpha})^{\frac{1-\alpha}{\alpha}}=O(\|Z^{\prime}-Z\|)$.
However, when ${Z^{\prime}}$ and $Z$ are close to $X_{*}$, the first order
derivative of $\operatorname{\mathrm{Tr}}{Z}^{\alpha}$ is zero. Hence,
$(\operatorname{\mathrm{Tr}}{Z^{\prime}}^{\alpha})^{\frac{1-\alpha}{\alpha}}-(\operatorname{\mathrm{Tr}}Z^{\alpha})^{\frac{1-\alpha}{\alpha}}=O(\|Z^{\prime}-Z\|^{2})$.
Since
$D(\sigma_{AC}[\\{X_{A|c}^{\prime}\\}_{c}]\|\sigma_{AC}[\\{X_{A|c}\\}_{c}])=O(\|Z^{\prime}-Z\|^{2})$,
the value $\frac{D_{\Omega_{\alpha,{\cal
D}}}(\sigma_{AC}[\\{X_{A|c}^{\prime}\\}_{c}]\|\sigma_{AC}[\\{X_{A|c}\\}_{c}])}{D(\sigma_{AC}[\\{X_{A|c}^{\prime}\\}_{c}]\|\sigma_{AC}[\\{X_{A|c}\\}_{c}])}$
takes a finite value when $\sigma_{AC}[\\{X_{A|c}^{\prime}\\}_{c}]$ and
$\sigma_{AC}[\\{X_{A|c}\\}_{c}]$ are close to
$\sigma_{AC}[\\{X_{A|c}^{*}\\}_{c}]$. Since
$\\{\sigma_{AC}[\\{X_{A|c}^{(t)}\\}_{c}]\\}$ converges to an element,
$\frac{D_{\Omega_{\alpha,{\cal
D}}}(\sigma_{AC}[\\{X_{A|c}^{(t+1)}\\}_{c}]\|\sigma_{AC}[\\{X_{A|c}^{(t)}\\}_{c}])}{D(\sigma_{AC}[\\{X_{A|c}^{(t+1)}\\}_{c}]\|\sigma_{AC}[\\{X_{A|c}^{(t)}\\}_{c}])}$
can be bounded by a finite value. That is, when $\gamma>0$ is taken to be a
sufficiently large value, the condition (139) holds.

##  Appendix D Proof of Theorem 11

###  D.1 Proof of (115)

Similar to the papers [56, 57, 12], we employ Schur duality of
${\mathcal{H}}^{\otimes n}$ as

$\displaystyle{\mathcal{H}}^{\otimes n}=\bigoplus_{\lambda\in Y_{d}^{n}}{\cal
U}_{\lambda}\otimes{\cal V}_{\lambda},$ (156)

${\cal U}_{\lambda}$ expresses the irreducible space of $\mathop{\rm SU}(d)$
and ${\cal V}_{\lambda}$ expresses the irreducible space of the representation
$\pi$ of the permutation group $\mathfrak{S}_{n}$. Then, for any state $\rho$,
we have

$\displaystyle\rho^{\otimes n}=\bigoplus_{\lambda\in
Y_{d}^{n}}\rho_{\lambda}\otimes\rho_{\lambda,mix},$ (157)

where $\rho_{\lambda,mix}$ is the completely mixed state on ${\cal
V}_{\lambda}$. Since the system ${\cal V}_{\lambda}$ has no information, it is
sufficient to handle the states $\oplus_{\lambda\in
Y_{d}^{n}}\rho_{\lambda},\oplus_{\lambda\in Y_{d}^{n}}\sigma_{\lambda}$ on
$\oplus_{\lambda\in Y_{d}^{n}}{\cal U}_{\lambda}$. Here, we define the map
$\kappa_{n}$ as

$\displaystyle\kappa_{n}(\rho^{\prime}):=\sum_{\lambda\in
Y_{d}^{n}}\operatorname{\mathrm{Tr}}_{{\cal V}_{\lambda}}P({\cal
U}_{\lambda}\otimes{\cal V}_{\lambda})\rho^{\prime}P({\cal
U}_{\lambda}\otimes{\cal V}_{\lambda})$ (158)

for a general state on ${\mathcal{H}}^{\otimes n}$. We choose a basis
$\\{|u_{\lambda,j}\rangle\\}$ of ${\cal U}_{\lambda}$ such that
$|u_{\lambda,j}\rangle$ is a eigenvector of $\rho_{\lambda}$. We define the
pinching map $\Gamma_{\rho,n}$ as

$\displaystyle\Gamma_{\rho,n}(\rho^{\prime})$ $\displaystyle:=$
$\displaystyle\sum_{\lambda\in Y_{d}^{n},j}(|u_{\lambda,j}\rangle\langle
u_{\lambda,j}|\otimes P({\cal
V}_{\lambda}))\rho^{\prime}(|u_{\lambda,j}\rangle\langle u_{\lambda,j}|\otimes
P({\cal V}_{\lambda}))$ (159)

for a general state $\rho^{\prime}$ on ${\mathcal{H}}^{\otimes n}$. Due to the
above mentioned structure, states $\Gamma_{\rho,n}(\sigma^{\otimes n})$ and
$\rho^{\otimes n}$ are commutative with each other. We denote the dimension of
$\oplus_{\lambda\in Y_{d}^{n}}{\cal U}_{\lambda}$ by $d_{n}$. Since $\dim{\cal
U}_{\lambda}\leq(n+1)^{d(d-1)/2}$ and $|Y_{d}^{n}|\leq(n+1)^{d-1}$ [51, (6.16)
and (6.18)], we have

$\displaystyle d_{n}\leq(n+1)^{(d+2)(d-1)/2}.$ (160)

We have the pinching inequality:

$\displaystyle\rho^{\prime}\leq d_{n}\Gamma_{\rho,n}(\rho^{\prime}).$ (161)

We choose

$\displaystyle T_{n}(\sigma,R):=\\{\Gamma_{\rho,n}(\sigma^{\otimes n})\geq
e^{nR}\rho^{\otimes n}\\}.$ (162)

Given $\alpha\in[0,1]$, we choose $R_{n,\alpha,r}:=\frac{-\max_{\sigma\in
S}\log\operatorname{\mathrm{Tr}}\Gamma_{\rho,n}(\sigma^{\otimes
n})^{\alpha}(\rho^{\otimes n})^{1-\alpha}-nr}{n(1-\alpha)}$ so that
$-nr=\max_{\sigma\in
S}\log\operatorname{\mathrm{Tr}}\Gamma_{\rho,n}(\sigma^{\otimes
n})^{\alpha}(\rho^{\otimes n})^{1-\alpha}+(1-\alpha)nR_{n,\alpha,r}$. Here, we
address several projections that are commutative with each other. For two
commutative projections $P_{1}$ and $P_{2}$, we define

$\displaystyle P_{1}\cup P_{2}:=I-(I-P_{1})(I-P_{2}).$ (163)

Since it is natural to define $(I-P_{1})\cap(I-P_{2}):=(I-P_{1})(I-P_{2})$,
the above definition for $P_{1}\cup P_{2}$ is also natural. When
$\\{|u_{j}\rangle\\}_{j}$ is a common eigenvector system of $P_{1}$ and
$P_{2}$, they can be written as $P_{k}=\sum_{j\in J_{k}}|u_{j}\rangle\langle
u_{j}|$. Then, we have $P_{1}\cup P_{2}=\sum_{j\in J_{1}\cup
J_{2}}|u_{j}\rangle\langle u_{j}|$. Based in this definition, we define the
operator $T_{n,\alpha,r}$ as

$\displaystyle T_{n,\alpha,r}:=\bigcup_{\sigma\in
S}T_{n}(\sigma,R_{n,\alpha,r}).$ (164)

Although $T_{n}(\sigma,R_{n,\alpha,r})$ is an operator on
${\mathcal{H}}^{\otimes n}$, it can be written by using a projection operator
$\oplus_{\lambda\in Y_{d}^{n}}P_{\sigma,n,\lambda}$ on $\oplus_{\lambda\in
Y_{d}^{n}}{\cal U}_{\lambda}$ as

$\displaystyle T_{n}(\sigma,R_{n,\alpha,r})=\bigoplus_{\lambda\in
Y_{d}^{n}}P_{\sigma,n,\lambda}\otimes P({\cal V}_{\lambda}).$ (165)

Hence, we have

$\displaystyle T_{n,\alpha,r}=\bigoplus_{\lambda\in
Y_{d}^{n}}(\bigcup_{\sigma\in S}P_{\sigma,n,\lambda})\otimes P({\cal
V}_{\lambda}).$ (166)

Since all $P_{\sigma,n,\lambda}$ are commutative with each other, we can
choose $\dim{\cal U}_{\lambda}$ elements $\sigma_{\lambda,j}$ in $S$ such that

$\displaystyle\bigcup_{\sigma\in
S}P_{\sigma,n,\lambda}=\bigcup_{j}P_{\sigma_{\lambda,j},n,\lambda}.$ (167)

Therefore, we have

$\displaystyle T_{n,\alpha,r}=\bigoplus_{\lambda\in
Y_{d}^{n}}(\bigcup_{j}P_{\sigma_{\lambda,j},n,\lambda})\otimes P({\cal
V}_{\lambda})$ $\displaystyle=$ $\displaystyle\bigcup_{\lambda^{\prime}\in
Y_{d}^{n},j}\bigoplus_{\lambda\in
Y_{d}^{n}}(P_{\sigma_{\lambda^{\prime},j},n,\lambda})\otimes P({\cal
V}_{\lambda})$ $\displaystyle=$ $\displaystyle\bigcup_{\lambda^{\prime}\in
Y_{d}^{n},j}T_{n}(\sigma_{\lambda^{\prime},j},R_{n,\alpha,r}).$ (168)

We have

$\displaystyle\operatorname{\mathrm{Tr}}\rho^{\otimes
n}T_{n,\alpha,r}\leq\operatorname{\mathrm{Tr}}\rho^{\otimes
n}\sum_{\lambda^{\prime}\in
Y_{d}^{n},j}T_{n}(\sigma_{\lambda^{\prime},j},R_{n,\alpha,r})$
$\displaystyle\leq$ $\displaystyle\sum_{\lambda^{\prime}\in
Y_{d}^{n},j}\operatorname{\mathrm{Tr}}\rho^{\otimes
n}T_{n}(\sigma_{\lambda^{\prime},j},R_{n,\alpha,r})$ $\displaystyle\leq$
$\displaystyle\sum_{\lambda^{\prime}\in
Y_{d}^{n},j}\operatorname{\mathrm{Tr}}\Gamma_{\rho,n}(\sigma_{\lambda^{\prime},j}^{\otimes
n})^{\alpha}(\rho^{\otimes n})^{1-\alpha}e^{-n\alpha R_{n,\alpha,r}}$
$\displaystyle\leq$ $\displaystyle d_{n}\max_{\sigma\in
S}\operatorname{\mathrm{Tr}}\Gamma_{\rho,n}(\sigma^{\otimes
n})^{\alpha}(\rho^{\otimes n})^{1-\alpha}e^{-n\alpha R_{n,\alpha,r}}$
$\displaystyle=$ $\displaystyle d_{n}\big{(}\max_{\sigma\in
S}\operatorname{\mathrm{Tr}}\Gamma_{\rho,n}(\sigma^{\otimes
n})^{\alpha}(\rho^{\otimes n})^{1-\alpha}\big{)}$
$\displaystyle\cdot\big{(}\max_{\sigma\in
S}\operatorname{\mathrm{Tr}}\Gamma_{\rho,n}(\sigma^{\otimes
n})^{\alpha}(\rho^{\otimes
n})^{1-\alpha}e^{nr}\big{)}^{\frac{\alpha}{1-\alpha}}$ $\displaystyle=$
$\displaystyle d_{n}(\max_{\sigma\in
S}\operatorname{\mathrm{Tr}}\Gamma_{\rho,n}(\sigma^{\otimes
n})^{\alpha}(\rho^{\otimes
n})^{1-\alpha})^{\frac{1}{1-\alpha}}e^{n\frac{\alpha r}{1-\alpha}}.$ (169)

For $\sigma\in S$, we have

$\displaystyle\operatorname{\mathrm{Tr}}\sigma^{\otimes
n}(I-T_{n,\alpha,r})=\operatorname{\mathrm{Tr}}\sigma^{\otimes
n}\Gamma_{\rho,n}(I-T_{n,\alpha,r})$ $\displaystyle=$
$\displaystyle\operatorname{\mathrm{Tr}}\Gamma_{\rho,n}(\sigma^{\otimes
n})(I-T_{n,\alpha,r})$ $\displaystyle\leq$
$\displaystyle\operatorname{\mathrm{Tr}}\Gamma_{\rho,n}(\sigma^{\otimes
n})(I-T_{n}(\sigma,R_{n,\alpha,r}))$ $\displaystyle\leq$
$\displaystyle\operatorname{\mathrm{Tr}}\Gamma_{\rho,n}(\sigma^{\otimes
n})^{\alpha}(\rho^{\otimes n})^{1-\alpha}e^{n(1-\alpha)R_{n,\alpha,r}}$
$\displaystyle\leq$ $\displaystyle\max_{\sigma\in
S}\operatorname{\mathrm{Tr}}\Gamma_{\rho,n}(\sigma^{\otimes
n})^{\alpha}(\rho^{\otimes
n})^{1-\alpha}e^{n(1-\alpha)R_{n,\alpha,r}}=e^{-nr}.$ (170)

Thus, we have

$\displaystyle-\frac{1}{n}\log\beta_{e^{-nr},n}(\rho)\geq-\frac{1}{n}\log\operatorname{\mathrm{Tr}}\rho^{\otimes
n}T_{n,\alpha,r}.$ (171)

In fact, applying the matrix monotone function to (161), we have

$\displaystyle d_{n}^{\alpha-1}((\rho^{\otimes
n})^{\frac{1-\alpha}{2\alpha}}\Gamma_{\rho,n}(\sigma^{\otimes
n})(\rho^{\otimes n})^{\frac{1-\alpha}{2\alpha}})^{\alpha-1}$ $\displaystyle=$
$\displaystyle\Big{(}d_{n}\Gamma_{\rho,n}\Big{(}(\rho^{\otimes
n})^{\frac{1-\alpha}{2\alpha}}\sigma^{\otimes n}(\rho^{\otimes
n})^{\frac{1-\alpha}{2\alpha}}\Big{)}\Big{)}^{\alpha-1}$ $\displaystyle\leq$
$\displaystyle\Big{(}(\rho^{\otimes
n})^{\frac{1-\alpha}{2\alpha}}\sigma^{\otimes n}(\rho^{\otimes
n})^{\frac{1-\alpha}{2\alpha}}\Big{)}^{\alpha-1}$ (172)

Thus,

$\displaystyle\operatorname{\mathrm{Tr}}\Gamma_{\rho,n}(\sigma^{\otimes
n})^{\alpha}(\rho^{\otimes n})^{1-\alpha}$ $\displaystyle=$
$\displaystyle\operatorname{\mathrm{Tr}}((\rho^{\otimes
n})^{\frac{1-\alpha}{2\alpha}}\Gamma_{\rho,n}(\sigma^{\otimes
n})(\rho^{\otimes n})^{\frac{1-\alpha}{2\alpha}})^{\alpha}$ $\displaystyle=$
$\displaystyle\operatorname{\mathrm{Tr}}\Gamma_{\rho,n}((\rho^{\otimes
n})^{\frac{1-\alpha}{2\alpha}}\sigma^{\otimes n}(\rho^{\otimes
n})^{\frac{1-\alpha}{2\alpha}}))^{\alpha}$ $\displaystyle=$
$\displaystyle\operatorname{\mathrm{Tr}}\big{(}\Gamma_{\rho,n}((\rho^{\otimes
n})^{\frac{1-\alpha}{2\alpha}}\sigma^{\otimes n}(\rho^{\otimes
n})^{\frac{1-\alpha}{2\alpha}})\big{)}$
$\displaystyle\cdot\big{(}\Gamma_{\rho,n}((\rho^{\otimes
n})^{\frac{1-\alpha}{2\alpha}}\sigma^{\otimes n}(\rho^{\otimes
n})^{\frac{1-\alpha}{2\alpha}}))^{\alpha-1}\big{)}$ $\displaystyle=$
$\displaystyle\operatorname{\mathrm{Tr}}((\rho^{\otimes
n})^{\frac{1-\alpha}{2\alpha}}\sigma^{\otimes n}(\rho^{\otimes
n})^{\frac{1-\alpha}{2\alpha}})$
$\displaystyle\cdot\Gamma_{\rho,n}((\rho^{\otimes
n})^{\frac{1-\alpha}{2\alpha}}\sigma^{\otimes n}(\rho^{\otimes
n})^{\frac{1-\alpha}{2\alpha}}))^{\alpha-1}$ $\displaystyle\leq$
$\displaystyle d_{n}^{1-\alpha}\operatorname{\mathrm{Tr}}((\rho^{\otimes
n})^{\frac{1-\alpha}{2\alpha}}\sigma^{\otimes n}(\rho^{\otimes
n})^{\frac{1-\alpha}{2\alpha}})$ $\displaystyle\cdot((\rho^{\otimes
n})^{\frac{1-\alpha}{2\alpha}}\sigma^{\otimes n}(\rho^{\otimes
n})^{\frac{1-\alpha}{2\alpha}})^{\alpha-1}$ $\displaystyle=$ $\displaystyle
d_{n}^{1-\alpha}\operatorname{\mathrm{Tr}}((\rho^{\otimes
n})^{\frac{1-\alpha}{2\alpha}}\sigma^{\otimes n}(\rho^{\otimes
n})^{\frac{1-\alpha}{2\alpha}})^{\alpha}$ $\displaystyle=$ $\displaystyle
d_{n}^{1-\alpha}(\operatorname{\mathrm{Tr}}(\rho^{\frac{1-\alpha}{2\alpha}}\sigma\rho^{\frac{1-\alpha}{2\alpha}})^{\alpha})^{n}.$
(173)

Combining (169) and (173), we have

$\displaystyle\operatorname{\mathrm{Tr}}\rho^{\otimes n}T_{n,\alpha,r}\leq
d_{n}^{2}(\max_{\sigma\in
S}(\operatorname{\mathrm{Tr}}(\rho^{\frac{1-\alpha}{2\alpha}}\sigma\rho^{\frac{1-\alpha}{2\alpha}})^{\alpha})^{\frac{n}{1-\alpha}}e^{n\frac{\alpha
r}{1-\alpha}}$ $\displaystyle=$ $\displaystyle
d_{n}^{2}\exp\Big{(}n\frac{(1-\alpha)E_{\alpha}^{*}(\rho)-\alpha
r}{1-\alpha}\Big{)}.$ (174)

Since arbitrary $\alpha\in[0,1]$ satisfies (171), (174) yields

$\displaystyle-\frac{1}{n}\log\beta_{e^{-nr},n}(\rho)$ $\displaystyle\geq$
$\displaystyle-\frac{1}{n}\log d_{n}^{2}+\max_{0\leq\alpha\leq
1}\frac{(1-\alpha)E_{\alpha}^{*}(\rho)-\alpha r}{1-\alpha}.$ (175)

Hence, using (160), we obtain (117) and (115).

###  D.2 Proof of (116)

Here, we employ the idea invented in [58], whose detail is explained in [42,
Section 3.8]. We choose a test $T_{n}$ such that

$\displaystyle\operatorname{\mathrm{Tr}}\sigma^{\otimes n}T_{n}\geq e^{-nr}$
(176)

for $\sigma\in S$. For $\alpha\geq 1$ and $\sigma\in S$, the information
processing inequality implies

$\displaystyle D_{\alpha}(\sigma^{\otimes n}\|\rho^{\otimes n})$
$\displaystyle\geq$
$\displaystyle\frac{1}{\alpha-1}\log\Big{(}(\operatorname{\mathrm{Tr}}\sigma^{\otimes
n}T_{n})^{\alpha}(\operatorname{\mathrm{Tr}}\rho^{\otimes n}T_{n})^{1-\alpha}$
$\displaystyle+(\operatorname{\mathrm{Tr}}\sigma^{\otimes
n}(I-T_{n}))^{\alpha}(\operatorname{\mathrm{Tr}}\rho^{\otimes
n}(I-T_{n}))^{1-\alpha}\Big{)}$ $\displaystyle\geq$
$\displaystyle\frac{1}{\alpha-1}\log\Big{(}(\operatorname{\mathrm{Tr}}\sigma^{\otimes
n}T_{n})^{\alpha}(\operatorname{\mathrm{Tr}}\rho^{\otimes
n}T_{n})^{1-\alpha}\Big{)}$ $\displaystyle=$
$\displaystyle\frac{\alpha}{\alpha-1}\log\operatorname{\mathrm{Tr}}\sigma^{\otimes
n}T_{n}-\log\operatorname{\mathrm{Tr}}\rho^{\otimes n}T_{n}$
$\displaystyle\geq$
$\displaystyle-\frac{\alpha}{\alpha-1}nr-\log\operatorname{\mathrm{Tr}}\rho^{\otimes
n}T_{n}.$ (177)

Thus, we have

$\displaystyle-\frac{1}{n}\log\operatorname{\mathrm{Tr}}\rho^{\otimes
n}T_{n}\leq\frac{1}{n}D_{\alpha}(\sigma^{\otimes n}\|\rho^{\otimes
n})+\frac{\alpha}{\alpha-1}r.$ (178)

Since any $T_{n}$ satisfying (176) satisfies (178), we have

$\displaystyle-\frac{1}{n}\log\beta_{1-e^{-nr},n}(\rho)\leq\frac{1}{n}D_{\alpha}(\sigma^{\otimes
n}\|\rho^{\otimes n})+\frac{\alpha r}{\alpha-1}$ $\displaystyle=$
$\displaystyle D_{\alpha}(\sigma\|\rho)+\frac{\alpha r}{\alpha-1}.$ (179)

Since any $\sigma\in S$ satisfies (179), we have

$\displaystyle-\frac{1}{n}\log\beta_{1-e^{-nr},n}(\rho)\leq\min_{\sigma\in
S}D_{\alpha}(\sigma\|\rho)+\frac{\alpha r}{\alpha-1}$ $\displaystyle=$
$\displaystyle E_{\alpha}^{*}(\rho)+\frac{\alpha r}{\alpha-1},$ (180)

which implies (116).

##  Appendix E Proof of Theorem 2

###  E.1 Proof of (14)

We employ notations and results in Appendix D when
${\mathcal{H}}={\mathcal{H}}_{A}\otimes{\mathcal{H}}_{B}$ and
$\rho=\rho_{AB}$. We define $\Gamma_{\rho_{AB},n}$ and $d_{n}$ in the same
way.

We define the map ${\cal T}_{n}$ as ${\cal
T}_{n}(\sigma):=\sum_{g\in\mathfrak{S}_{n}}\frac{1}{|\mathfrak{S}_{n}|}\pi(g)\sigma\pi(g)^{\dagger}$
for a state on ${\mathcal{H}}^{\otimes n}$. We define the set of permutation-
invariant separable states as

$\displaystyle{\cal S}_{AB}^{n,sym}:=\\{\sigma\in{\cal
S}_{AB}^{n}|\pi(g)\sigma\pi(g)^{\dagger}=\sigma,\quad
g\in\mathfrak{S}_{n}\\}.$ (181)

Then, we have the following lemma.

###### Lemma 13

$\displaystyle\overline{\beta}_{\epsilon,n}(\rho_{AB})$ $\displaystyle=$
$\displaystyle\min_{0\leq T\leq
I}\Big{\\{}\operatorname{\mathrm{Tr}}T\rho_{AB}^{\otimes
n}\Big{|}\max_{\sigma_{AB}\in{\cal
S}_{AB}^{n,sym}}\operatorname{\mathrm{Tr}}(I-T)\sigma_{AB}\leq\epsilon\Big{\\}}.$
(182)

Proof:  Since ${\cal S}_{AB}^{n,sym}\subset{\cal S}_{AB}^{n}$, we have

$\displaystyle\min_{0\leq T\leq
I}\Big{\\{}\operatorname{\mathrm{Tr}}T\rho_{AB}^{\otimes
n}\Big{|}\max_{\sigma_{AB}\in{\cal
S}_{AB}^{n}}\operatorname{\mathrm{Tr}}(I-T)\sigma_{AB}\leq\epsilon\Big{\\}}$
$\displaystyle\geq$ $\displaystyle\min_{0\leq T\leq
I}\Big{\\{}\operatorname{\mathrm{Tr}}T\rho_{AB}^{\otimes
n}\Big{|}\max_{\sigma_{AB}\in{\cal
S}_{AB}^{n,sym}}\operatorname{\mathrm{Tr}}(I-T)\sigma_{AB}\leq\epsilon\Big{\\}}.$
(183)

Hence, it is sufficient to the opposite inequality.

We choose the test $T_{\epsilon}$ such that any element
$\sigma_{AB}^{\prime}\in{\cal S}_{AB}^{n,sym}$ satisfies

$\displaystyle\operatorname{\mathrm{Tr}}(I-T_{\epsilon})\sigma_{AB}^{\prime}\leq\epsilon,$
(184) $\displaystyle\operatorname{\mathrm{Tr}}T_{\epsilon}\rho_{AB}^{\otimes
n}$ $\displaystyle=$ $\displaystyle\min_{0\leq T\leq
I}\Big{\\{}\operatorname{\mathrm{Tr}}T\rho_{AB}^{\otimes
n}\Big{|}\max_{\sigma_{AB}\in{\cal
S}_{AB}^{n,sym}}\operatorname{\mathrm{Tr}}(I-T)\sigma_{AB}\leq\epsilon\Big{\\}}.$
(185)

Due to the symmetric property, any element $\sigma_{AB}^{\prime}\in{\cal
S}_{AB}^{n,sym}$ satisfies

$\displaystyle\operatorname{\mathrm{Tr}}(I-{\cal
T}_{n}(T_{\epsilon}))\sigma_{AB}^{\prime}\leq\epsilon,$ (186)
$\displaystyle\operatorname{\mathrm{Tr}}{\cal
T}_{n}(T_{\epsilon})\rho_{AB}^{\otimes n}$ $\displaystyle=$
$\displaystyle\min_{0\leq T\leq
I}\Big{\\{}\operatorname{\mathrm{Tr}}T\rho_{AB}^{\otimes
n}\Big{|}\max_{\sigma_{AB}\in{\cal
S}_{AB}^{n,sym}}\operatorname{\mathrm{Tr}}(I-T)\sigma_{AB}\leq\epsilon\Big{\\}}.$
(187)

For any element $\sigma_{AB}\in{\cal S}_{AB}^{n}$, we have
$\operatorname{\mathrm{Tr}}(I-{\cal
T}_{n}(T_{\epsilon}))\sigma_{AB}=\operatorname{\mathrm{Tr}}(I-{\cal
T}_{n}(T_{\epsilon}))\pi(g)\sigma_{AB}\pi(g)^{\dagger}$. Since ${\cal
T}_{n}(\sigma_{AB})\in{\cal S}_{AB}^{n,sym}$, we have
$\operatorname{\mathrm{Tr}}(I-{\cal
T}_{n}(T_{\epsilon}))\sigma_{AB}=\operatorname{\mathrm{Tr}}(I-{\cal
T}_{n}(T_{\epsilon})){\cal T}_{n}(\sigma_{AB})\leq\epsilon$. Thus,

$\displaystyle\min_{0\leq T\leq
I}\Big{\\{}\operatorname{\mathrm{Tr}}T\rho_{AB}^{\otimes
n}\Big{|}\max_{\sigma_{AB}\in{\cal
S}_{AB}^{n}}\operatorname{\mathrm{Tr}}(I-T)\sigma_{AB}\leq\epsilon\Big{\\}}$
$\displaystyle\leq$ $\displaystyle\operatorname{\mathrm{Tr}}{\cal
T}_{n}(T_{\epsilon})\rho_{AB}^{\otimes
n}=\operatorname{\mathrm{Tr}}T_{\epsilon}\rho_{AB}^{\otimes n}$
$\displaystyle=$ $\displaystyle\min_{0\leq T\leq
I}\Big{\\{}\operatorname{\mathrm{Tr}}T\rho_{AB}^{\otimes
n}\Big{|}\max_{\sigma_{AB}\in{\cal
S}_{AB}^{n,sym}}\operatorname{\mathrm{Tr}}(I-T)\sigma_{AB}\leq\epsilon\Big{\\}}.$
(188)

Since any element $\sigma\in{\cal S}_{AB}^{n,sym}$ is written as the form

$\displaystyle\sigma=\bigoplus_{\lambda\in
Y_{d}^{n}}\rho_{\lambda}\otimes\rho_{\lambda,mix}$ (189)

in the same way as (157), replacing the set $\\{\sigma^{\otimes n}|\sigma\in
S\\}$ of states by ${\cal S}_{AB}^{n,sym}$, given $r>0$, we define $T_{n,R}$
and $R_{n,\alpha,r}$. In the same way as (169), we have

$\displaystyle\operatorname{\mathrm{Tr}}\rho_{AB}^{\otimes n}T_{n,\alpha,r}$
$\displaystyle\leq$ $\displaystyle d_{n}(\max_{\sigma\in{\cal
S}_{AB}^{n,sym}}\operatorname{\mathrm{Tr}}\Gamma_{\rho_{AB},n}(\sigma)^{\alpha}(\rho_{AB}^{\otimes
n})^{1-\alpha})^{\frac{1}{1-\alpha}}e^{n\frac{\alpha r}{1-\alpha}}.$ (190)

for $\alpha\in[0,1]$. For $\sigma\in{\cal S}_{AB}^{n,sym}$, in the same way as
(170) we have

$\displaystyle\operatorname{\mathrm{Tr}}\sigma(I-T_{n,\alpha,r})\leq e^{-nr}.$
(191)

Thus, we have

$\displaystyle-\frac{1}{n}\log\overline{\beta}_{e^{-nr},n}(\rho_{AB})\geq-\frac{1}{n}\log\operatorname{\mathrm{Tr}}\rho_{AB}^{\otimes
n}T_{n,\alpha,r}.$ (192)

In the same way as (173), we have

$\displaystyle\operatorname{\mathrm{Tr}}\Gamma_{\rho_{AB},n}(\sigma)^{\alpha}(\rho_{AB}^{\otimes
n})^{1-\alpha}$ $\displaystyle\leq$ $\displaystyle
d_{n}^{1-\alpha}\operatorname{\mathrm{Tr}}((\rho_{AB}^{\otimes
n})^{\frac{1-\alpha}{2\alpha}}\sigma(\rho_{AB}^{\otimes
n})^{\frac{1-\alpha}{2\alpha}})^{\alpha}.$ (193)

Combining (190) and (193), we have

$\displaystyle\operatorname{\mathrm{Tr}}\rho_{AB}^{\otimes
n}T_{n,\alpha,r}\leq
d_{n}^{2}\exp\Big{(}\frac{(1-\alpha){E}_{\alpha}^{*}(\rho_{AB}^{\otimes
n})-n\alpha r}{1-\alpha}\Big{)}.$ (194)

Since an arbitrary real $\alpha\in[0,1]$ satisfies (171), (194) yields

$\displaystyle-\frac{1}{n}\log\beta_{e^{-nr},n}(\rho_{AB})$
$\displaystyle\geq$ $\displaystyle-\frac{1}{n}\log
d_{n}^{2}+\max_{0\leq\alpha\leq
1}\frac{\frac{1}{n}(1-\alpha)E_{\alpha}^{*}(\rho_{AB}^{\otimes n})-\alpha
r}{1-\alpha},$ (195)

using (160), we obtain (14) and (16) by taking the limit.

###  E.2 Proof of (15)

For $\alpha\geq 1$ and $\sigma\in{\cal S}_{AB}^{n}$, in the same way as (179),
we have

$\displaystyle-\frac{1}{n}\log\overline{\beta}_{1-e^{-nr},n}(\rho)\leq\frac{1}{n}D_{\alpha}(\sigma\|\rho^{\otimes
n})+\frac{\alpha r}{\alpha-1}$ (196)

for $\sigma\in{\cal S}_{AB}^{n}$. Since any $\sigma\in{\cal S}_{AB}^{n}$
satisfies (196), we have

$\displaystyle-\frac{1}{n}\log\overline{\beta}_{1-e^{-nr},n}(\rho)\leq$
$\displaystyle\min_{\sigma\in{\cal
S}_{AB}^{n}}\frac{1}{n}D_{\alpha}(\sigma\|\rho^{\otimes n})+\frac{\alpha
r}{\alpha-1}$ $\displaystyle=$
$\displaystyle\frac{1}{n}E_{\alpha}^{*}(\rho^{\otimes n})+\frac{\alpha
r}{\alpha-1},$ (197)

which implies (17) and (15) by taking the limit.

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