# A-unital Operations and Quantum Conditional Entropy

Mahathi Vempati  Centre for Quantum Science and Technology,
International Institute of Information Technology-Hyderabad, Gachibowli,
Telangana-500032, India.  Center for Computational Natural Sciences and
Bioinformatics,
International Institute of Information Technology-Hyderabad, Gachibowli,
Telangana-500032, India<EMAIL_ADDRESS>Saumya Shah
Department of Physics,
Indian Institute of Technology Guwahati, Guwahati-781039, Assam, India.
<EMAIL_ADDRESS>Nirman Ganguly  Department of Mathematics,
Birla Institute of Technology and Science Pilani, Hyderabad Campus,
Telangana-500078, India<EMAIL_ADDRESS>Indranil Chakrabarty  Centre
for Quantum Science and Technology,
International Institute of Information Technology-Hyderabad, Gachibowli,
Telangana-500032, India.  Center for Security, Theory and Algorithmic
Research,
International Institute of Information Technology-Hyderabad, Gachibowli,
Telangana-500032, India<EMAIL_ADDRESS>
###### Abstract

Negative quantum conditional entropy states are key ingredients for
information theoretic tasks such as superdense coding, state merging and one-
way entanglement distillation. In this work, we ask: how does one detect if a
channel is useful in preparing negative conditional entropy states? We answer
this question by introducing the class of A-unital channels, which we show are
the largest class of conditional entropy non-decreasing channels. We also
prove that A-unital channels are precisely the completely free operations for
the class of states with non-negative conditional entropy. Furthermore, we
study the relationship between A-unital channels and other classes of channels
pertinent to the resource theory of entanglement. We then prove similar
results for ACVENN: a previously defined, relevant class of states and also
relate the maximum and minimum conditional entropy of a state with its von
Neumann entropy.

The definition of A-unital channels naturally lends itself to a procedure for
determining membership of channels in this class. Thus, our work is valuable
for the detection of resourceful channels in the context of conditional
entropy.

##  1 Introduction

The myriad information-theoretic applications of entanglement [1] have
elevated it from being merely an interesting physical phenomenon to that of a
resource. In the resource-theoretic study of entanglement, the separable
states (quantum states that can be written as convex combinations of pure
product states) are called the free states, and those outside this set, the
entangled states, are known as the resource states [2]. The resource theory
also defines various classes of free operations, that is, quantum operations
that cannot convert a free state to a resourceful one. Local Operations and
Classical Communication (LOCC) is a physically defined class of free
operations [3, 2, 1], characterized by the practical difficulty of
communicating quantum information. A superset of LOCC is the class of
separable channels [4, 2], which has a more appealing mathematical structure
than LOCC and is therefore used to prove properties of LOCC itself [4, 5].
Entanglement-breaking channels form another important class of channels. These
channels destroy all entanglement between a system and its environment [6].

Once these classes of channels have been defined, it is important to be able
to decide whether or not a given channel belongs to any of these free classes.
The quantum channel detection problem asks: given a channel and a well-defined
class of channels, does the channel belong to the class [7, 8, 9, 10]? The
channel itself maybe provided as a blackbox, as in a laboratory setting, where
it can be characterized by measuring its output on various input states, or as
a full description (for instance, via its Kraus decomposition or Choi matrix).
Unfortunately, for some classes defined with respect to entanglement, this
problem is notoriously difficult. For example, it is currently unknown how to
decide whether a given channel belongs to LOCC [2], and the problem of
membership of a channel in the entanglement-breaking class is NP Hard [11].
There have been witness-based probabilistic approaches to deal with this
problem, which are useful when some apriori information about a channel is
known [7].

The interest in defining free operations for entanglement and studying
mechanisms for detection of these operations is in part due to the
resourcefulness of entanglement. While entanglement is certainly important in
the characterization of quantum states for information-theoretic tasks, there
are several phenomena for which merely possessing entanglement is not a
sufficient condition to determine usefulness of a quantum state. Often, a
state needs to have other properties beyond being entangled in order to be
useful. One such property is the negativity of the quantum conditional entropy
(also known as the conditional von Neumann entropy) of a quantum state [12].
The quantum conditional entropy (henceforth referred to as conditional entropy
where there is no ambiguity) of a quantum state $\rho_{AB}$ is defined as

$S_{A|B}(\rho_{AB})=S(\rho_{AB})-S(\rho_{B}),$ (1)

where $S(\rho)$ is the von Neumann entropy (henceforth referred to as entropy
where there is no ambiguity) of the state $\rho$, given by
$S(\rho)=-\Tr(\rho\log\rho)$. All logarithms are taken to the base 2.
Conditional entropy is used to define the information-theoretic quantities
_coherent information_ [13], which is just the negative of conditional
entropy, and _quantum discord_ [14, 15, 16], which is a measure of non-
classical correlations in a quantum state. While all negative conditional
entropy states are entangled, not all entangled states possess negative
conditional entropy. Negative conditional entropy states are important for the
information-theoretic primitive quantum state merging [17, 18], and provide
quantum advantage in superdense coding [19, 20, 21]. Negativity of conditional
entropy is a necessary feature for the one-way distillation of entanglement
[22]. It is important in maximizing the rates of distributed private
randomness distillation [23], as well as in the reduction of uncertainty in
predicting the outcomes of two incompatible measurements [24]. Previously, the
set of states that are not useful for any of the above tasks, that is, the set
of states possessing non-negative conditional entropy (we refer to this set as
$\mathcal{F}^{c}(A|B)$) was characterized as convex and compact [25,
26].111This is proven for the the case where $\dim A=\dim B=d$ [25], but a
similar argument can be used to extend this proof to the case where $\dim
A\neq\dim B$.The set of states for which conditional entropy remains non-
negative even under the application of global unitary channels (referred to as
$\mathcal{F}^{ac}(A|B)$ in this work) was also similarly characterized [27].

In this work, taking $\mathcal{F}^{c}(A|B)$ to be a set of free states, we
ask: what is the class of free operations for $\mathcal{F}^{c}(A|B)$, and what
are its properties? How does this class of operations relate to operations
defined in the context of entanglement: separable channels and entanglement-
breaking channels? How does one detect if a given channel belongs to the class
of free operations for $\mathcal{F}^{c}(A|B)$? These questions are important
in the context of preparation of quantum states for the aforementioned tasks,
in order to determine if an operation one has access to is capable of
generating resource.

This paper is organized as follows: In Section 2, we introduce the notation,
and recap the relevant preliminaries. In Section 3, we define a class of free
operations $\operatorname{NCVE}(A|B)$, the largest class of conditional
entropy non-decreasing channels. We prove that this class is exactly equal to
$\mathcal{O}^{c}_{cmax}(A|B)$, the largest class of completely free channels
for $\mathcal{F}^{c}(A|B)$, and is elegantly characterized by a property we
call $A$-unitality. We discuss properties of $A$-unital channels in Section 4.
In Section 5, we discuss the relation between $A$-unital channels, separable
channels and entanglement-breaking channels. While the set of separable states
and $\mathcal{F}^{c}(A|B)$ share a subset-superset relationship, we show that
surprisingly, this relationship does not hold for $A$-unital channels and
separable channels—there exist $A$-unital channels that are not separable,
separable channels that are not $A$-unital, as well as channels in the
intersection. In Section 6, we extend the class $\mathcal{F}^{ac}(A|B)$,
previously characterized only for the $2\otimes 2$ case to the $d\otimes d$
case, and prove that unital channels are the largest class of completely free
channels for $\mathcal{F}^{ac}(A|B)$. Since both entropy and conditional
entropy are heavily discussed in this manuscript, in Section 7, we provide
upper and lower bounds for the conditional entropy of a state for a given
entropy. We conclude in Section 8 with some concrete open questions that
emerge from our work.

##  2 Preliminaries

In this section, we review relevant definitions and concepts from literature.
$\operatorname{\mathcal{H}}^{A}$ denotes a finite dimensional Hilbert space
for physical system $A$. The dimension of the space
$\operatorname{\mathcal{H}}^{A}$ is given by $d_{A}$. The set of bounded
linear operators on the Hilbert space $\operatorname{\mathcal{H}}^{A}$ is
denoted by $\operatorname{\mathcal{B}}(A)$. The set of bounded linear
operators over Hilbert space
$\operatorname{\mathcal{H}}^{A}\otimes\operatorname{\mathcal{H}}^{B}$ is
$\operatorname{\mathcal{B}}(AB)$. The set of bounded linear operators that are
Hermitian, positive semi-definite and have unit trace, that is, the density
matrices (or quantum states) over $\operatorname{\mathcal{H}}^{A}$ is denoted
by $\operatorname{\mathcal{D}}(A)$. A channel
$\operatorname{\mathcal{N}}:\operatorname{\mathcal{D}}(A)\rightarrow\operatorname{\mathcal{D}}(B)$
is a linear map from $\operatorname{\mathcal{D}}(A)$ to
$\operatorname{\mathcal{D}}(B)$ that is completely positive and trace
preserving (CPTP).

A quantum state $\rho_{AB}$ is called separable if it can be written as a
convex combination of pure product states. That is, $\rho_{AB}$ is separable
if $\rho_{AB}=\sum_{i}p_{i}|\alpha_{i}\rangle\langle\alpha_{i}|$ where
$p_{i}\in[0,1]$, $\sum_{i}p_{i}=1$ and
$|\alpha_{i}\rangle=|\phi_{i}\rangle_{A}\otimes|\psi_{i}\rangle_{B}$ for some
$|\phi_{i}\rangle\in\operatorname{\mathcal{H}}^{A}$ and
$|\psi_{i}\rangle\in\operatorname{\mathcal{H}}^{B}$. Separable states are the
free states in the resource theory of entanglement. We denote the set of
separable states in $\operatorname{\mathcal{D}}(AB)$ by
$\mathcal{F}^{s}(A|B)$. The set of states
$\rho_{AB}\in\operatorname{\mathcal{D}}(AB)$ for which $S_{A|B}(\rho_{AB})\geq
0$, that is, the set of states possessing non-negative conditional entropy
(Eq. 1) is denoted by $\mathcal{F}^{c}(A|B)$.

Free and completely free channels for a given set of free states were defined
in [2]. We define these here for $\mathcal{F}^{c}(A|B)$.

######  Definition 1.

A channel $\operatorname{\mathcal{N}}_{AB\rightarrow
CD}:\operatorname{\mathcal{D}}(AB)\rightarrow\operatorname{\mathcal{D}}(CD)$
is free for $\mathcal{F}^{c}(A|B)$ if for all $\rho\in\mathcal{F}^{c}(A|B)$,
we have $\operatorname{\mathcal{N}}_{AB\rightarrow
CD}(\rho)\in\mathcal{F}^{c}(C|D)$. The class of all such free channels mapping
$\operatorname{\mathcal{D}}(AB)$ to $\operatorname{\mathcal{D}}(CD)$ is
denoted by $\mathcal{O}^{c}_{max}(A|B\rightarrow C|D)$. When the input and
output spaces are the same, the class is denoted by
$\mathcal{O}^{c}_{max}(A|B)\equiv\mathcal{O}^{c}_{max}(A|B\rightarrow A|B)$.

######  Definition 2.

A channel $\operatorname{\mathcal{N}}_{AB\rightarrow
CD}:\operatorname{\mathcal{D}}(AB)\rightarrow\operatorname{\mathcal{D}}(CD)$
is completely free for $\mathcal{F}^{c}(A|B)$ if for all finite dimensional
Hilbert spaces $\operatorname{\mathcal{H}}^{A^{\prime}}$ and
$\operatorname{\mathcal{H}}^{B^{\prime}}$ and
$\rho\in\mathcal{F}^{c}(A^{\prime}A|B^{\prime}B)$, we have
$\left(\operatorname{id}_{A^{\prime}B^{\prime}}\otimes\operatorname{\mathcal{N}}_{AB\rightarrow
CD}\right)(\rho)\in\mathcal{F}^{c}(A^{\prime}C|B^{\prime}D)$. The class of all
such completely free channels mapping $\operatorname{\mathcal{D}}(AB)$ to
$\operatorname{\mathcal{D}}(CD)$ is denoted by
$\mathcal{O}^{c}_{cmax}(A|B\rightarrow C|D)$. When the input and output spaces
are the same, the class is denoted by
$\mathcal{O}^{c}_{cmax}(A|B)\equiv\mathcal{O}^{c}_{cmax}(A|B\rightarrow A|B)$.

The free and completely free channels for $\mathcal{F}^{s}(A|B)$, denoted by
$\mathcal{O}^{s}_{max}(A|B\rightarrow C|D)$ and
$\mathcal{O}^{s}_{cmax}(A|B\rightarrow C|D)$ respectively are defined
analogously, replacing $\mathcal{F}^{c}(A|B)$ in the above definitions by
$\mathcal{F}^{s}(A|B)$.

Separable channels [4] form an important class of channels in the entanglement
resource theory. We define these channels here:

######  Definition 3.

A channel
$N:\operatorname{\mathcal{D}}(AB)\rightarrow\operatorname{\mathcal{D}}(CD)$ is
called separable if and only if its action on a state
$\rho\in\operatorname{\mathcal{D}}(AB)$ can be expressed as

$\operatorname{\mathcal{N}}(\rho)=\sum_{i=1}^{n}(K^{i}_{A}\otimes
L^{i}_{B})\rho(K^{i}_{A}\otimes L^{i}_{B})^{\dagger}$ (2)

for some integer $n\geq 1$ and linear maps
$K^{i}_{A}:\operatorname{\mathcal{H}}^{A}\rightarrow\operatorname{\mathcal{H}}^{C}$
and
$L^{i}_{B}:\operatorname{\mathcal{H}}^{B}\rightarrow\operatorname{\mathcal{H}}^{D}$
such that $\sum_{i=1}^{n}(K^{i}_{A}\otimes
L^{i}_{B})^{\dagger}(K^{i}_{A}\otimes L^{i}_{B})=I_{AB}$. The class of all
separable channels mapping $\operatorname{\mathcal{D}}(AB)$ to
$\operatorname{\mathcal{D}}(CD)$ is denoted by
$\operatorname{SEP}(A|B\rightarrow C|D)$. When the input and output spaces are
the same, the class is denoted by
$\operatorname{SEP}(A|B)\equiv\operatorname{SEP}(A|B\rightarrow A|B)$.

It turns out that $\operatorname{SEP}(A|B)$ is exactly equal to
$\mathcal{O}^{s}_{cmax}(A|B)$ [2]. That is, the class of separable channels
form the largest class of completely free channels for $\mathcal{F}^{s}(A|B)$.

Entanglement-breaking channels constitute another relevant class of channels
[13]:

######  Definition 4.

A channel
$\operatorname{\mathcal{N}}:\operatorname{\mathcal{D}}(A)\rightarrow\operatorname{\mathcal{D}}(B)$
is called entanglement-breaking if the quantum state
$(\operatorname{id}_{R}\otimes\operatorname{\mathcal{N}})(\rho_{RA})$ is
separable for all $\rho_{RA}$ where $R$ is a reference system of arbitrary
dimension. The class of all entanglement-breaking channels mapping
$\operatorname{\mathcal{D}}(A)$ to $\operatorname{\mathcal{D}}(B)$ is denoted
by $\operatorname{EB}(A\rightarrow B)$. When the input and output spaces are
the same, the class is denoted by
$\operatorname{EB}(A)\equiv\operatorname{EB}(A\rightarrow A)$.

Unital channels play an important role in the context of entropy, as the class
of unital channels mapping $\operatorname{\mathcal{D}}(A)$ to itself are
exactly the ones that do not decrease the entropy of any state
$\rho\in\operatorname{\mathcal{D}}(A)$ [13]. We define them as

######  Definition 5.

A channel
$\operatorname{\mathcal{N}}:\operatorname{\mathcal{D}}(A)\rightarrow\operatorname{\mathcal{D}}(A)$
is called unital if
$\operatorname{\mathcal{N}}\left(\frac{I_{A}}{d_{A}}\right)=\frac{I_{A}}{d_{A}}$.
The class of all unital channels mapping $\operatorname{\mathcal{D}}(A)$ to
$\operatorname{\mathcal{D}}(A)$ is denoted by $\operatorname{UNI}(A)$.

Having reviewed the definitions, we now characterize the class of completely
free operations for $\mathcal{F}^{c}(A|B)$.

##  3 Completely Free Operations for Conditional Entropy

Since the negativity of conditional entropy implies usefulness for various
tasks, a natural candidate for a class of operations that are free for
$\mathcal{F}^{c}(A|B)$ is the class of channels that does not decrease the
conditional entropy of any state. Conditional entropy non-decreasing channels
are defined as follows:

######  Definition 6.

A channel $\operatorname{\mathcal{N}}_{AB\rightarrow
CD}:\operatorname{\mathcal{D}}(AB)\rightarrow\operatorname{\mathcal{D}}(CD)$
is said to be conditional von Neumann entropy (CVE) non-decreasing, if for all
$\rho\in\operatorname{\mathcal{D}}(AB)$, we have
$S_{C|D}\left(\operatorname{\mathcal{N}}_{AB\rightarrow CD}(\rho)\right)\geq
S_{A|B}(\rho)$. The class of CVE non-decreasing channels mapping
$\operatorname{\mathcal{D}}(AB)$ to $\operatorname{\mathcal{D}}(CD)$ is
denoted by $\operatorname{NCVE}(A|B\rightarrow C|D)$. When the input and
output spaces are the same, the class is denoted by
$\operatorname{NCVE}(A|B)\equiv\operatorname{NCVE}(A|B\rightarrow A|B)$.

It turns out that this class of channels is exactly equal to
$\mathcal{O}^{c}_{cmax}(A|B)$ which was introduced in Def. 2. We prove this
equivalence by showing that both $\operatorname{NCVE}(A|B)$ and
$\mathcal{O}^{c}_{cmax}(A|B)$ are in fact equal to a third class of
operations, defined as follows:

######  Definition 7.

A channel
$\operatorname{\mathcal{N}}:\operatorname{\mathcal{D}}(AB)\rightarrow\operatorname{\mathcal{D}}(AB)$
is called A-unital if for all $\rho_{B}\in\operatorname{\mathcal{D}}(B)$,
$\operatorname{\mathcal{N}}$ satisfies

$\operatorname{\mathcal{N}}\left(\frac{I_{A}}{d_{A}}\otimes\rho_{B}\right)=\frac{I_{A}}{d_{A}}\otimes\rho^{\prime}_{B}$
(3)

for some $\rho^{\prime}_{B}\in\operatorname{\mathcal{D}}(B)$. We denote the
set of A-unital channels on $\operatorname{\mathcal{D}}(AB)$ as
$\operatorname{UNI}(A|B)$. Likewise, the channel $\operatorname{\mathcal{N}}$
is called $B$-unital if for all $\rho_{A}\in\operatorname{\mathcal{D}}(A)$,
$\operatorname{\mathcal{N}}$ satisfies

$\operatorname{\mathcal{N}}\left(\rho_{A}\otimes\frac{I_{B}}{d_{B}}\right)=\rho^{\prime}_{A}\otimes\frac{I_{B}}{d_{B}}$
(4)

for some $\rho^{\prime}_{A}\in\operatorname{\mathcal{D}}(A)$, and the set of
$B$-unital channels on $\operatorname{\mathcal{D}}(AB)$ is denoted as
$\operatorname{UNI}(B|A)$.

Not only does the class of $A$-unital channels serve to show that both
$\operatorname{NCVE}(A|B)$ and $\mathcal{O}^{c}_{cmax}(A|B)$ are equal, but
its definition also makes this class amenable to detection, as discussed in
Section 4.3. We now prove the equivalence of these three classes.

######  Theorem 1.

The classes $\operatorname{UNI}(A|B)$, $\operatorname{NCVE}(A|B)$ and
$\mathcal{O}^{c}_{cmax}(A|B)$ are all equal to each other.

Figure 1: Equivalence of $\operatorname{UNI}(A|B),\operatorname{NCVE}(A|B)$
and $\mathcal{O}^{c}_{cmax}(A|B)$.

###### Proof.

The structure of the proof is depicted in Fig. 1. Let
$\operatorname{\mathcal{N}}:\operatorname{\mathcal{D}}(AB)\rightarrow\operatorname{\mathcal{D}}(AB)$
be a quantum channel. In Lemma 1, we show that
$\operatorname{\mathcal{N}}\in\operatorname{UNI}(A|B)\implies\operatorname{\mathcal{N}}\in\operatorname{NCVE}(A|B)$,
labelled as (a). In Lemma 2, we show that
$\operatorname{\mathcal{N}}\in\operatorname{NCVE}(A|B)\implies\operatorname{\mathcal{N}}\in\operatorname{UNI}(A|B)$,
which is labelled as (b). In Lemma 3, we show that
$\operatorname{\mathcal{N}}\in\operatorname{UNI}(A|B)\implies(\operatorname{id}_{A^{\prime}B^{\prime}}\otimes\operatorname{\mathcal{N}})\in\operatorname{UNI}(A^{\prime}A|B^{\prime}B)$.
We then use Lemma 1 and Lemma 3 to prove Lemma 4, which is
$\operatorname{\mathcal{N}}\in\operatorname{UNI}(A|B)\implies\operatorname{\mathcal{N}}\in\mathcal{O}^{c}_{cmax}(A|B)$,
labelled as (c) in the figure. Finally, we prove Lemma 5, which is
$\operatorname{\mathcal{N}}\in\mathcal{O}^{c}_{cmax}(A|B)\implies\operatorname{\mathcal{N}}\in\operatorname{NCVE}(A|B)$,
labelled as (d) in the figure. These lemmas establish the equivalence between
the three classes.

∎

######  Lemma 1.

$\operatorname{\mathcal{N}}\in\operatorname{UNI}(A|B)\implies\operatorname{\mathcal{N}}\in\operatorname{NCVE}(A|B)$.

###### Proof.

Let $\operatorname{\mathcal{N}}\in\operatorname{UNI}(A|B)$. Thus, for all
$\rho_{B}\in\operatorname{\mathcal{D}}(B)$, we have
$\operatorname{\mathcal{N}}\left(\frac{I_{A}}{d_{A}}\otimes\rho_{B}\right)=\frac{I_{A}}{d_{A}}\otimes\rho^{\prime}_{B}$.
To prove the Lemma, we require two results related to quantum relative
entropy. The first result is a characterization of the conditional entropy in
terms of relative entropy [28, 13]:

$\displaystyle\begin{split}S_{A|B}(\rho_{AB})&=\log
d-S\left(\rho_{AB}\,\middle\|\,\frac{I_{A}}{d_{A}}\otimes\rho_{B}\right)\\\
&=\underset{\sigma_{B}\in\operatorname{\mathcal{D}}(B)}{\max}\log
d-S\left(\rho_{AB}\,\middle\|\,\frac{I_{A}}{d_{A}}\otimes\sigma_{B}\right).\end{split}$
(5)

Next, we require the monotonicity of relative entropy, which is

$\displaystyle
S\left(\operatorname{\mathcal{N}}(\rho)\,\middle\|\,N(\sigma)\right)\leq
S\left(\rho\,\middle\|\,\sigma\right).$ (6)

From Eq. 5, for all $\rho\in\operatorname{\mathcal{D}}(AB)$ and
$\sigma_{B}\in\operatorname{\mathcal{D}}(B)$, we have

$S_{A|B}\left(\operatorname{\mathcal{N}}(\rho)\right)\geq\log
d-S\left(\operatorname{\mathcal{N}}(\rho)\,\middle\|\,\frac{I_{A}}{d_{A}}\otimes\sigma_{B}\right).$
(7)

In particular, we must have

$S_{A|B}\left(\operatorname{\mathcal{N}}(\rho)\right)\geq\log
d-S\left(\operatorname{\mathcal{N}}(\rho)\,\middle\|\,\frac{I_{A}}{d_{A}}\otimes\rho^{\prime}_{B}\right),$
(8)

where
$\operatorname{\mathcal{N}}\left(\frac{I_{A}}{d_{A}}\otimes\rho_{B}\right)=\frac{I_{A}}{d_{A}}\otimes\rho^{\prime}_{B}$.
Therefore,

$S_{A|B}\left(\operatorname{\mathcal{N}}(\rho)\right)\geq\log
d-S\left(\operatorname{\mathcal{N}}(\rho)\,\middle\|\,\operatorname{\mathcal{N}}\left(\frac{I_{A}}{d_{A}}\otimes\rho_{B}\right)\right).$
(9)

From Eq. 6, we have

$\log
d-S\left(\operatorname{\mathcal{N}}(\rho)\,\middle\|\,\operatorname{\mathcal{N}}\left(\frac{I_{A}}{d_{A}}\otimes\rho_{B}\right)\right)\geq\log
d-S\left(\rho\,\middle\|\,\frac{I_{A}}{d_{A}}\otimes\rho_{B}\right)=S_{A|B}(\rho).$
(10)

Thus, for all $\rho\in\operatorname{\mathcal{D}}(AB)$, we have
$S_{A|B}\left(\operatorname{\mathcal{N}}(\rho)\right)\geq S_{A|B}(\rho)$,
which implies that $\operatorname{\mathcal{N}}\in\operatorname{NCVE}(A|B)$.

∎

######  Lemma 2.

$\operatorname{\mathcal{N}}\in\operatorname{NCVE}(A|B)\implies\operatorname{\mathcal{N}}\in\operatorname{UNI}(A|B)$.

###### Proof.

For any $\rho_{AB}\in\operatorname{\mathcal{D}}(AB)$, the maximum value of
$S_{A|B}(\rho_{AB})$ is $\log d_{A}$ and is attained if and only if
$\rho_{AB}=\frac{I_{A}}{d_{A}}\otimes\sigma_{B}$ for some
$\sigma_{B}\in\operatorname{\mathcal{D}}(B)$ [13].

We use the above fact to prove the lemma by contradiction. Let
$\operatorname{\mathcal{N}}\in\operatorname{NCVE}(A|B)$. Assume that
$\operatorname{\mathcal{N}}\notin\operatorname{UNI}(A|B)$. Then, there exists
some $\sigma_{B}\in\operatorname{\mathcal{D}}(B)$ such that
$\operatorname{\mathcal{N}}\left(\frac{I_{A}}{d_{A}}\otimes\sigma_{B}\right)\neq\frac{I_{A}}{d_{A}}\otimes\sigma_{B}^{\prime}$
for all $\sigma_{B}^{\prime}\in\operatorname{\mathcal{D}}(B)$. This implies
that $S_{A|B}\left(\frac{I_{A}}{d_{A}}\otimes\sigma_{B}\right)=\log d_{A}$,
but
$S_{A|B}\left(\operatorname{\mathcal{N}}\left(\frac{I_{A}}{d_{A}}\otimes\sigma_{B}\right)\right)<\log
d_{A}$. However, this is a contradiction since
$\operatorname{\mathcal{N}}\in\operatorname{NCVE}(A|B)$. Therefore, the
assumption that $\operatorname{\mathcal{N}}\notin\operatorname{UNI}(A|B)$ must
be false, and we have
$\operatorname{\mathcal{N}}\in\operatorname{NCVE}(A|B)\implies\operatorname{\mathcal{N}}\in\operatorname{UNI}(A|B)$.
∎

######  Lemma 3.

$\operatorname{\mathcal{N}}\in\operatorname{UNI}(A|B)\implies(\operatorname{id}_{A^{\prime}B^{\prime}}\otimes\operatorname{\mathcal{N}})\in\operatorname{UNI}(A^{\prime}A|B^{\prime}B)$.

###### Proof.

Let $\operatorname{\mathcal{N}}_{AB}\in\operatorname{UNI}(A|B)$. Then, for all
$\rho_{B}\in\operatorname{\mathcal{D}}(B)$, we have
$\operatorname{\mathcal{N}}\left(I_{A}\otimes\rho_{B}\right)=I_{A}\otimes\rho^{\prime}_{B}$
for some $\rho^{\prime}_{B}\in\operatorname{\mathcal{D}}(B)$. It is always
possible to find a basis for any $\operatorname{\mathcal{D}}(X)$, in which
each basis element is a density matrix, that is, each basis element also
belongs to $\operatorname{\mathcal{D}}(X)$ [13]. Let
$\\{\beta^{\prime}_{i}\\}_{i}$ be such a basis for
$\operatorname{\mathcal{D}}(B^{\prime})$ and $\\{\beta_{j}\\}_{j}$ be such a
basis for $\operatorname{\mathcal{D}}(B)$. Thus,
$\\{\beta^{\prime}_{i}\otimes\beta_{j}\\}_{ij}$ is a basis of density matrices
for $\operatorname{\mathcal{D}}(B^{\prime}B)$. We have,

$\displaystyle\begin{split}(\operatorname{id}_{A^{\prime}B^{\prime}}&\otimes\operatorname{\mathcal{N}}_{AB})\left(\frac{I_{A^{\prime}A}}{d_{A^{\prime}A}}\otimes\left(\beta^{\prime}_{i}\otimes\beta_{j}\right)_{B^{\prime}B}\right)\\\
&=(\operatorname{id}_{A^{\prime}B^{\prime}}\otimes\operatorname{\mathcal{N}}_{AB})\left(\frac{I_{A^{\prime}}}{d_{A^{\prime}}}\otimes\frac{I_{A}}{d_{A}}\otimes\beta^{\prime}_{iB^{\prime}}\otimes\beta_{jB}\right)\\\
&=\frac{I_{A^{\prime}}}{d_{A^{\prime}}}\otimes\beta^{\prime}_{iB^{\prime}}\otimes\operatorname{\mathcal{N}}_{AB}\left(\frac{I_{A}}{d_{A}}\otimes\beta_{jB}\right)\\\
&=\frac{I_{A^{\prime}}}{d_{A^{\prime}}}\otimes\beta^{\prime}_{iB^{\prime}}\otimes\frac{I_{A}}{d_{A}}\otimes\gamma_{jB}\\\
&=\frac{I_{A^{\prime}A}}{d_{A^{\prime}A}}\otimes\left(\beta^{\prime}_{i}\otimes\gamma_{j}\right)_{B^{\prime}B}.\end{split}$
(11)

Now, for all $\sigma\in\operatorname{\mathcal{D}}(B^{\prime}B)$, we have

$\displaystyle\begin{split}(\operatorname{id}_{A^{\prime}B^{\prime}}&\otimes\operatorname{\mathcal{N}}_{AB})\left(\frac{I_{A^{\prime}A}}{d_{A^{\prime}A}}\otimes\sigma_{B^{\prime}B}\right)\\\
&=(\operatorname{id}_{A^{\prime}B^{\prime}}\otimes\operatorname{\mathcal{N}}_{AB})\left(\frac{I_{A^{\prime}A}}{d_{A^{\prime}A}}\otimes\left(\sum_{ij}b_{ij}\beta^{\prime}_{i}\otimes\beta_{j}\right)_{B^{\prime}B}\right)\\\
&=\sum_{ij}b_{ij}\,(\operatorname{id}_{A^{\prime}B^{\prime}}\otimes\operatorname{\mathcal{N}}_{AB})\left(\frac{I_{A^{\prime}A}}{d_{A^{\prime}A}}\otimes\left(\beta^{\prime}_{i}\otimes\beta_{j}\right)_{B^{\prime}B}\right)\\\
&=\sum_{ij}b_{ij}\,\frac{I_{A^{\prime}A}}{d_{A^{\prime}A}}\otimes\left(\beta^{\prime}_{i}\otimes\gamma_{j}\right)_{B^{\prime}B}\\\
&=\frac{I_{A^{\prime}A}}{d_{A^{\prime}A}}\otimes\sum_{ij}b_{ij}\left(\beta^{\prime}_{i}\otimes\gamma_{j}\right)_{B^{\prime}B}.\\\
\end{split}$ (12)

Thus,
$\operatorname{id}_{A^{\prime}B^{\prime}}\otimes\operatorname{\mathcal{N}}_{AB}\in\operatorname{UNI}(A^{\prime}A|B^{\prime}B)$.
∎

######  Lemma 4.

$\operatorname{\mathcal{N}}\in\operatorname{UNI}(A|B)\implies\operatorname{\mathcal{N}}\in\mathcal{O}^{c}_{cmax}(A|B)$.

###### Proof.

Let $\operatorname{\mathcal{N}}_{AB}\in\operatorname{UNI}(A|B)$. From Lemma 3,
we have
$\operatorname{id}_{A^{\prime}B^{\prime}}\otimes\operatorname{\mathcal{N}}_{AB}\in\operatorname{UNI}(A^{\prime}A|B^{\prime}B)$.
Hence, using Lemma 1, we have
$\operatorname{id}_{A^{\prime}B^{\prime}}\otimes\operatorname{\mathcal{N}}_{AB}\in\operatorname{NCVE}(A^{\prime}A|B^{\prime}B)$.
Thus, for all $\rho\in\mathcal{F}^{c}(A^{\prime}A|B^{\prime}B)$, we have
$S_{A|B}\left(\left(\operatorname{id}_{A^{\prime}B^{\prime}}\otimes
N_{AB}\right)(\rho)\right)\geq S_{A|B}(\rho)\geq 0$, which implies
$\left(\operatorname{id}_{A^{\prime}B^{\prime}}\otimes\operatorname{\mathcal{N}}_{AB}\right)(\rho)\in\mathcal{F}^{c}(A^{\prime}A|B^{\prime}B)$.
Now, using Definition 2, we have
$\operatorname{\mathcal{N}}_{AB}\in\mathcal{O}^{c}_{cmax}(A|B)$. ∎

######  Lemma 5.

$\operatorname{\mathcal{N}}\in\mathcal{O}^{c}_{cmax}(A|B)\implies\operatorname{\mathcal{N}}\in\operatorname{NCVE}(A|B)$.

###### Proof.

In order to prove the lemma, we prove the contrapositive statement, that is
$\operatorname{\mathcal{N}}_{AB}\notin\operatorname{NCVE}(A|B)\implies\operatorname{\mathcal{N}}_{AB}\notin\mathcal{O}^{c}_{cmax}(A|B)$.
Let
$\operatorname{\mathcal{N}}_{AB}:\operatorname{\mathcal{D}}(AB)\rightarrow\operatorname{\mathcal{D}}(AB)$
be a quantum channel that does not belong to $\operatorname{NCVE}(A|B)$. Thus
there exists $\rho\in\operatorname{\mathcal{D}}(AB)$ such that
$S_{A|B}\left(\operatorname{\mathcal{N}}(\rho)\right)<S_{A|B}(\rho)$. Consider
Hilbert spaces $\operatorname{\mathcal{H}}^{A^{\prime}}$ and
$\operatorname{\mathcal{H}}^{B^{\prime}}$, with $d_{A^{\prime}}=d_{A}$ and
$d_{B^{\prime}}=d_{B}$. Now, we select a state
$\sigma\in\operatorname{\mathcal{D}}(A^{\prime}B^{\prime})$ such that

$-S_{A|B}(\rho)\leq
S_{A^{\prime}|B^{\prime}}(\sigma)<-S_{A|B}\left(\operatorname{\mathcal{N}}(\rho)\right).$
(13)

Note that because
$-S_{A|B}(\rho)<-S_{A|B}\left(\operatorname{\mathcal{N}}(\rho)\right)$ and
both these values lie in $[-\log d_{A},\log d_{A}]=[-\log d_{A^{\prime}},\log
d_{A^{\prime}}]$, the state $\sigma$ always exists.

Now, consider the state
$\sigma_{A^{\prime}B^{\prime}}\otimes\rho_{AB}\in\operatorname{\mathcal{D}}(A^{\prime}B^{\prime}AB)$.
Using Eq. 13, we have
$S_{A^{\prime}A|B^{\prime}B}(\sigma_{A^{\prime}B^{\prime}}\otimes\rho_{AB})=S_{A^{\prime}|B^{\prime}}(\sigma)+S_{A|B}(\rho)\geq-
S_{A|B}(\rho)+S_{A|B}(\rho)=0$. Thus,
$\sigma_{A^{\prime}B^{\prime}}\otimes\rho_{AB}\in\mathcal{F}^{c}(A^{\prime}A|B^{\prime}B)$.

We now apply the operation
$\operatorname{id}_{A^{\prime}B^{\prime}}\otimes\operatorname{\mathcal{N}}_{AB}$
on the state $\sigma_{A^{\prime}B^{\prime}}\otimes\rho_{AB}$. Once again,
using Eq. 13, we have
$S_{A^{\prime}A|B^{\prime}B}(\operatorname{id}_{A^{\prime}B^{\prime}}\otimes\operatorname{\mathcal{N}}_{AB}(\sigma_{A^{\prime}B^{\prime}}\otimes\rho_{AB}))=S_{A^{\prime}A|B^{\prime}B}(\sigma_{A^{\prime}B^{\prime}}\otimes\operatorname{\mathcal{N}}_{AB}(\rho_{AB}))$
$=S_{A^{\prime}|B^{\prime}}(\sigma_{AB})+S_{A|B}(\operatorname{\mathcal{N}}_{AB}(\rho_{AB}))<-S_{A|B}(\operatorname{\mathcal{N}}_{AB}(\rho_{AB}))+S_{A|B}(\operatorname{\mathcal{N}}_{AB}(\rho_{AB}))=0$.
This implies that
$\operatorname{id}_{A^{\prime}B^{\prime}}\otimes\operatorname{\mathcal{N}}_{AB}(\sigma_{A^{\prime}B^{\prime}}\otimes\rho_{AB})\notin\mathcal{F}^{c}(A^{\prime}A|B^{\prime}B)$.
Therefore, the channel
$\operatorname{id}_{A^{\prime}B^{\prime}}\otimes\operatorname{\mathcal{N}}_{AB}$
takes a state from inside $\mathcal{F}^{c}(A^{\prime}A|B^{\prime}B)$ to
outside, and using Def. 2,
$\operatorname{\mathcal{N}}_{AB}\notin\mathcal{O}^{c}_{cmax}(A|B)$. ∎

Thus, Lemmas 1-5 prove all the implications depicted in Fig. 1, and Theorem 1
follows.

##  4 Properties of A-Unital channels

###  4.1 Serial and Parallel Concatenation

Since we have $\operatorname{UNI}(A|B)=\mathcal{O}^{c}_{cmax}(A|B)$, it
follows that serial concatenation of two $A$-unital channels is $A$-unital,
and parallel concatenation of an $A^{\prime}$-unital channel and an $A$-unital
channel is $A^{\prime}A$-unital [2]. However, this is also readily seen from
the definition of $\operatorname{UNI}(A|B)$. Let
$\operatorname{\mathcal{M}}_{AB},\operatorname{\mathcal{N}}_{AB}\in\operatorname{UNI}(A|B)$.
Then, we have

$\operatorname{\mathcal{M}}_{AB}\circ\operatorname{\mathcal{N}}_{AB}\left(\frac{I_{A}}{d_{A}}\otimes\rho_{B}\right)=\operatorname{\mathcal{M}}_{AB}\left(\frac{I_{A}}{d_{A}}\otimes\sigma_{B}\right)=\frac{I_{A}}{d_{A}}\otimes\gamma_{B}.$
(14)

Thus,
$\operatorname{\mathcal{M}}_{AB}\circ\operatorname{\mathcal{N}}_{AB}\in\operatorname{UNI}(A|B)$.
Similarly, if
$\operatorname{\mathcal{M}}_{A^{\prime}B^{\prime}}\in\operatorname{UNI}(A^{\prime}|B^{\prime})$,
and $\operatorname{\mathcal{N}}_{AB}\in\operatorname{UNI}(A|B)$, from Lemma 3,
we have
$(\operatorname{id}_{A^{\prime}B^{\prime}}\otimes\operatorname{\mathcal{N}}_{AB}),(\operatorname{\mathcal{M}}_{A^{\prime}B^{\prime}}\otimes\operatorname{id}_{AB})\in\operatorname{UNI}(A^{\prime}A|B^{\prime}B)$.
Then, from the closedness under serial concatenation above, it follows that

$(\operatorname{id}_{A^{\prime}B^{\prime}}\otimes\operatorname{\mathcal{N}}_{AB})\circ(\operatorname{\mathcal{M}}_{A^{\prime}B^{\prime}}\otimes\operatorname{id}_{AB})=\operatorname{\mathcal{M}}_{A^{\prime}B^{\prime}}\otimes\operatorname{\mathcal{N}}_{AB}\in\operatorname{UNI}(A^{\prime}A|B^{\prime}B).$
(15)

###  4.2 Convexity

######  Theorem 2.

A convex combination of $A$-unital channels is an $A$-unital channel.

###### Proof.

Let
$\operatorname{\mathcal{N}}_{1},\operatorname{\mathcal{N}}_{2}\in\operatorname{UNI}(A|B)$.
Then we have,

$\displaystyle\begin{split}\left(p\operatorname{\mathcal{N}}_{1}+(1-p)\operatorname{\mathcal{N}}_{2}\right)&\left(\frac{I_{A}}{d_{A}}\otimes\beta_{B}\right)\\\
&=p\operatorname{\mathcal{N}}_{1}\left(\frac{I_{A}}{d_{A}}\otimes\beta_{B}\right)+(1-p)\operatorname{\mathcal{N}}_{2}\left(\frac{I_{A}}{d_{A}}\otimes\beta_{B}\right)\\\
&=p\left(\frac{I_{A}}{d_{A}}\otimes\gamma_{1B}\right)+(1-p)\left(\frac{I_{A}}{d_{A}}\otimes\gamma_{2B}\right)\\\
&=\frac{I_{A}}{d_{A}}\otimes\left(p\gamma_{1}+(1-p)\gamma_{2}\right)_{B}\end{split}$
(16)

Thus,
$p\operatorname{\mathcal{N}}_{1}+(1-p)\operatorname{\mathcal{N}}_{2}\in\operatorname{UNI}(A|B)$.
∎

###  4.3 Detection

The characterization of both conditional entropy non-decreasing channels and
completely free channels for $\mathcal{F}^{c}(A|B)$ as $A$-unital channels
makes apparent how these channels can be detected. For any
$\operatorname{\mathcal{D}}(B)$, it is always possible to find $d_{B}^{2}$
density matrices $\\{\beta_{i}\\}_{i}$ from $\operatorname{\mathcal{D}}(B)$
that form a basis for for $\operatorname{\mathcal{D}}(B)$ [13]. For some
unknown channel $\operatorname{\mathcal{N}}$ given as a blackbox, a necessary
and sufficient condition for $A$-unitality is
$\operatorname{\mathcal{N}}\left(\frac{I_{A}}{d_{A}}\otimes\beta_{iB}\right)=\frac{I_{A}}{d_{B}}\otimes\beta^{\prime}_{iB}$
for all the $d_{B}^{2}$ states $\beta_{i}$ from the basis. This follows using
the same proof technique as in Lemma 3.

##  5 Relationship between A-Unital Channels and Other Classes

We now turn our attention to the relationship between $A$-unital channels and
other classes. If we detect that a channel is $A$-unital (or not $A$-unital),
can anything be said about the membership of the channel in
$\operatorname{SEP}(A|B)$? It turns out that this is not the case, as there
exist $A$-unital channels that are not separable, separable channels that are
not $A$-unital, as well as channels in the intersection. We discuss this in
the following subsections. We also discuss the relationship between
entanglement breaking channels and $A$-unital channels. These relationships
are depicted in Fig. 2. Throughout this section, we assume $d_{A}=d_{B}=d.$

Figure 2: The relationship between $\operatorname{UNI}(A|B)$,
$\operatorname{SEP}(A|B)$ and
$\operatorname{id}_{A}\otimes\operatorname{EB}(B)$ when $d_{A}=d_{B}$.

###  5.1 A-Unital Non-Separable Channels

We prove the existence of channels that belong to $\operatorname{UNI}(A|B)$
but not $\operatorname{SEP}(A|B)$ by constructing such a channel. Let
$\\{\ket{i}\\}_{i}$ be an orthonormal basis for
$\operatorname{\mathcal{H}}^{A}$, and $\\{\ket{j}\\}_{j}$ be an orthonormal
basis for $\operatorname{\mathcal{H}}^{B}$.

######  Definition 8.

The Swap and Prepare channel, denoted by
$\operatorname{\mathcal{P}}_{AB}:\operatorname{\mathcal{D}}(AB)\rightarrow\operatorname{\mathcal{D}}(AB)$
is defined as the composition of the following channels:

  * •

The Swap Channel
$[Sw_{AB}:\operatorname{\mathcal{D}}(AB)\rightarrow\operatorname{\mathcal{D}}(AB)]$.
The unitary channel that swaps systems $A$ and $B$. Its Kraus decomposition
contains just one Kraus operator which is the unitary matrix
$U_{s}\equiv\sum_{ij}\ket{ji}\bra{ij}$. We have

$Sw_{AB}(\rho)=U_{s}\rho U_{s}^{\dagger}.$ (17)

  * •

The Partial Trace Channel
$[\Tr_{A}:\operatorname{\mathcal{D}}(AB)\rightarrow\operatorname{\mathcal{D}}(B)]$.
The Kraus operators for this channel are $\left\\{\bra{i}_{A}\otimes
I_{B}\right\\}_{i}$. We have

$\Tr_{A}(\rho)=\sum_{i}\left(\bra{i}_{A}\otimes
I_{B}\right)(\rho_{AB})\left(\ket{i}_{A}\otimes I_{B}\right).$ (18)

  * •

The Preparation of $\frac{I_{A}}{d}$ Channel
$[Pr_{A}:\operatorname{\mathcal{D}}(B)\rightarrow\operatorname{\mathcal{D}}(AB)]$.
The Kraus operators for this channel are
$\left\\{\frac{1}{\sqrt{d}}\ket{i}_{A}\otimes I_{B}\right\\}_{i}$. We have

$\displaystyle Pr_{A}(\rho_{B})$
$\displaystyle=\sum_{i}\frac{1}{d}\left(\ket{i}_{A}\otimes
I_{B}\right)(\rho_{B})\left(\bra{i}_{A}\otimes I_{B}\right)$ (19)
$\displaystyle=\frac{I_{A}}{d}\otimes\rho_{B}$ (20)

Finally, we define the Swap and Prepare Channel:

$\operatorname{\mathcal{P}}_{AB}(\rho_{AB})=\left(Pr_{A}\circ\Tr_{A}\circ
Sw_{AB}\right)(\rho_{AB})$ (21)

######  Theorem 3.

There exist A-unital non-separable channels.

###### Proof.

In Lemma 6, we show that the channel
$\operatorname{\mathcal{P}}_{AB}\in\operatorname{UNI}(A|B)$, and in Lemma 7,
we show that $\operatorname{\mathcal{P}}_{AB}\notin\operatorname{SEP}(A|B)$. ∎

######  Lemma 6.

$\operatorname{\mathcal{P}}_{AB}\in\operatorname{UNI}(A|B)$.

###### Proof.

Let $\sigma_{B}\in\operatorname{\mathcal{B}}(B)$. We have

$\displaystyle\begin{split}\operatorname{\mathcal{P}}_{AB}\left(\frac{I_{A}}{d}\otimes\sigma_{B}\right)&=Pr_{A}\circ\Tr_{A}\circ
Sw_{AB}\left(\frac{I_{A}}{d}\otimes\sigma_{B}\right)=Pr_{A}\circ\Tr_{A}\left(\sigma_{A}\otimes\frac{I_{B}}{d}\right)\\\
&=Pr_{A}\left(\frac{I_{B}}{d}\right)=\frac{I_{A}}{d}\otimes\frac{I_{B}}{d}.\end{split}$
(22)

Thus, using Def. 7,
$\operatorname{\mathcal{P}}_{AB}\in\operatorname{UNI}(A|B)$. ∎

######  Lemma 7.

$\operatorname{\mathcal{P}}_{AB}\notin\operatorname{SEP}(A|B)$.

###### Proof.

From the discussion following Def. 3, we have that
$\operatorname{SEP}(A|B)=\mathcal{O}^{s}_{cmax}(A|B)$. Thus, it is sufficient
to prove that
$\operatorname{\mathcal{P}}_{AB}\notin\mathcal{O}^{s}_{cmax}(A|B)$. Consider
Hilbert spaces $\operatorname{\mathcal{H}}^{A^{\prime}}$ and
$\operatorname{\mathcal{H}}^{B^{\prime}}$ where $d_{A^{\prime}}$ and
$d_{B^{\prime}}$ are also equal to $d$. Take the state
$\chi\in\operatorname{\mathcal{D}}(A^{\prime}AB^{\prime}B)$ defined as
$\chi\equiv\outerproduct{\phi^{+}}{\phi^{+}}_{A^{\prime}A}\otimes\outerproduct{00}{00}_{B^{\prime}B}$,
where $\outerproduct{\phi^{+}}{\phi^{+}}_{A^{\prime}A}$ is the entangled state
$\frac{1}{d}\sum_{ij}\outerproduct{i}{j}_{A^{\prime}}\otimes\outerproduct{i}{j}_{A}$.
The state $\chi$ is a product state across the $A^{\prime}A|B^{\prime}B$ cut,
and hence $\chi\in\mathcal{F}^{s}(A^{\prime}A|B^{\prime}B).$ We now apply
$\operatorname{id}_{A^{\prime}B^{\prime}}\otimes\operatorname{\mathcal{P}}_{AB}$
to the state $\chi$.

$\displaystyle\begin{split}\chi^{\prime}&=\left(\operatorname{id}_{A^{\prime}B^{\prime}}\otimes\operatorname{\mathcal{P}}_{AB}\right)(\chi)\\\
&=\left(\operatorname{id}_{A^{\prime}B^{\prime}}\otimes\operatorname{\mathcal{P}}_{AB}\right)\left(\frac{1}{d}\sum_{ij}\outerproduct{i}{j}_{A^{\prime}}\otimes\outerproduct{i}{j}_{A}\otimes\outerproduct{0}{0}_{B^{\prime}}\otimes\outerproduct{0}{0}_{B}\right)\\\
&=\left(\operatorname{id}_{A^{\prime}B^{\prime}}\otimes\>Pr_{A}\circ\Tr_{A}\circ
Sw_{AB}\right)\left(\frac{1}{d}\sum_{ij}\outerproduct{i}{j}_{A^{\prime}}\otimes\outerproduct{i}{j}_{A}\otimes\outerproduct{0}{0}_{B^{\prime}}\otimes\outerproduct{0}{0}_{B}\right)\\\
&=\left(\operatorname{id}_{A^{\prime}B^{\prime}}\left(\frac{1}{d}\sum_{ij}\outerproduct{i}{j}_{A^{\prime}}\otimes\outerproduct{0}{0}_{B^{\prime}}\right)\right)\otimes\left(\left(Pr_{A}\circ\Tr_{A}\circ
Sw_{AB}\right)\left(\outerproduct{i}{j}_{A}\otimes\outerproduct{0}{0}_{B}\right)\right)\\\
&=\left(\frac{1}{d}\sum_{ij}\outerproduct{i}{j}_{A^{\prime}}\otimes\outerproduct{0}{0}_{B^{\prime}}\right)\otimes\left((Pr_{A}\circ\Tr_{A})\left(\outerproduct{0}{0}_{A}\otimes\outerproduct{i}{j}_{B}\right)\right)\\\
&=\left(\frac{1}{d}\sum_{ij}\outerproduct{i}{j}_{A^{\prime}}\otimes\outerproduct{0}{0}_{B^{\prime}}\right)\otimes\left(Pr_{A}\left(\outerproduct{i}{j}_{B}\right)\right)\\\
&=\frac{1}{d}\sum_{ij}\outerproduct{i}{j}_{A^{\prime}}\otimes\outerproduct{0}{0}_{B^{\prime}}\otimes\frac{I_{A}}{d}\otimes\outerproduct{i}{j}_{B}\\\
&=\outerproduct{\phi^{+}}{\phi^{+}}_{A^{\prime}B}\otimes\frac{I_{A}}{d}\otimes\outerproduct{0}{0}_{B^{\prime}}.\end{split}$
(23)

Observe that $\chi^{\prime}$ is entangled across the $A^{\prime}A|B^{\prime}B$
cut. Thus, $\chi^{\prime}\notin\mathcal{F}^{s}(A^{\prime}A|B^{\prime}B)$. From
the definition of $\mathcal{O}^{s}_{cmax}(A|B)$, this implies that
$\operatorname{\mathcal{P}}_{AB}\notin\mathcal{O}^{s}_{cmax}(A|B)$, hence
$\operatorname{\mathcal{P}}_{AB}\notin\operatorname{SEP}(A|B)$. The action of
$\operatorname{\mathcal{P}}_{AB}$ is depicted in Fig. 3. ∎

From Lemmas 6 and 7, we have that
$\operatorname{\mathcal{P}}_{AB}\in\operatorname{UNI}(A|B)$ but
$\operatorname{\mathcal{P}}_{AB}\notin\operatorname{SEP}(A|B)$, and thus,
Theorem 3 follows.

Figure 3: Demonstration of $\operatorname{\mathcal{P}}_{AB}$ acting on $\chi$.
Subfigure $(1)$ depicts the state $\chi$, which is separable across the
$A^{\prime}A|B^{\prime}B$ cut. Upon applying
$\operatorname{id}_{A^{\prime}B^{\prime}}\otimes Sw_{AB}$, the states of
subsystems $A$ and $B$ are swapped, as shown in subfigure $(2)$. Application
of $\Tr_{A}$ is depicted in subfigure $(3)$, and finally, $Pr_{A}$ prepares
the state $\frac{I_{A}}{d}$ and this is depicted in subfigure $(4)$. Thus,
$\operatorname{\mathcal{P}}_{AB}$ is $A$-unital, but not separable, as the
channel generates entanglement across the $A^{\prime}A|B^{\prime}B$ cut.

###  5.2 Non-A-Unital Separable Channels

Examples of channels that are non-A-unital but separable are replacement
channels [2] $R$ which trace out a given state and prepare a product state
$\rho_{A}\otimes\rho_{B}$ where $\rho_{A}\neq\frac{I_{A}}{d}$. Since for all
$\sigma_{B}\in\operatorname{\mathcal{D}}(B)$, we have
$R\left(\frac{I_{A}}{d}\otimes\sigma_{B}\right)\neq\frac{I_{A}}{d}\otimes\gamma_{B}$
for some $\gamma\in\operatorname{\mathcal{D}}(B)$, $R$ is not an $A$-unital
channel. However, since both tracing out as well as preparation of a product
state are separable channels, $R$ is a separable channel. A specific example
of a channel which prepares the state $\outerproduct{00}{00}_{AB}$ is

$R_{0}(\rho_{AB})=\sum_{ij}\left(\outerproduct{0}{i}_{A}\otimes\outerproduct{0}{j}_{B}\right)(\rho_{AB})\left(\outerproduct{i}{0}_{A}\otimes\outerproduct{j}{0}_{B}\right),$
(24)

where $\\{\ket{i}\\}_{i}$ and $\\{\ket{j}\\}_{j}$ are orthonormal bases for
$\operatorname{\mathcal{H}}^{A}$ and $\operatorname{\mathcal{H}}^{B}$
respectively.

###  5.3 A-Unital Separable Channels

A trivial example of a channel that is both $A$-unital and separable is the
identity channel. A more non-trivial example is any channel of the form

$C_{AB}(\rho)=\sum_{i}\left(K^{i}_{A}\otimes
L^{i}_{B}\right)\rho\left(K^{i}_{A}\otimes L^{i}_{B}\right)^{\dagger}$ (25)

where $\sum_{i}\left(K^{i}_{A}\otimes
L^{i}_{B}\right)^{\dagger}\left(K^{i}_{A}\otimes L^{i}_{B}\right)=I_{AB}$ and
each $K^{i}_{A}$ is a unitary channel. $C_{AB}$ is separable by definition. It
is also $A$-unital as shown below. Let
$\sigma\in\operatorname{\mathcal{D}}(B)$. We have

$\displaystyle\begin{split}C_{AB}\left(\frac{I_{A}}{d}\otimes\sigma_{B}\right)&=\sum_{i}\left(K^{i}_{A}\otimes
L^{i}_{B}\right)\left(\frac{I_{A}}{d}\otimes\sigma_{B}\right)\left(K^{i}_{A}\otimes
L^{i}_{B}\right)^{\dagger}\\\
&=\sum_{i}\frac{1}{d}K^{i}_{A}I_{A}K^{i\dagger}_{A}\otimes
L^{i}_{B}\sigma_{B}L^{i\dagger}_{B}=\frac{I_{A}}{d}\otimes\sum_{i}L^{i}_{B}\sigma_{B}L^{i\dagger}_{B}.\end{split}$
(26)

###  5.4 A-Unital and Entanglement-Breaking Channels

Both $\operatorname{UNI}(A|B)$ and $\operatorname{SEP}(A|B)$ are concerned
with correlations across the $A|B$ split. Thus, we relate both these classes
with channels of the form $\operatorname{id}_{A}\otimes\Phi_{B}$ where
$\Phi_{B}\in\operatorname{EB}(B)$, that is $\Phi_{B}$ is an entanglement-
breaking channel (Def. 4), as they destroy all entanglement across the $A|B$
split. These channels are $A$-unital, as
$(\operatorname{id}_{A}\otimes\Phi_{B})\left(\frac{I_{A}}{d}\otimes\sigma_{B}\right)=\frac{I_{A}}{d}\otimes\Phi(\sigma)_{B}$
for all $\sigma_{B}\in\operatorname{\mathcal{D}}(B)$, as well as separable, as
the Kraus operators for the channel are $\\{I_{A}\otimes K^{i}_{B}\\}_{i}$
where $\\{K^{i}_{B}\\}_{i}$ are the Kraus operators for $\Phi_{B}$. This is
depicted in Figure 2. On the contrary, channels of the form
$\Phi_{A}\otimes\operatorname{id}_{B}$ where $\Phi_{A}\in\operatorname{EB}(A)$
are not necessarily $A$-unital, although they are separable. As an example,
consider the following classical-quantum channel
$\operatorname{\mathcal{M}}_{A}:\operatorname{\mathcal{D}}(A)\rightarrow\operatorname{\mathcal{D}}(A)$,
that acts as

$\operatorname{\mathcal{M}}_{A}(\rho)=\sum_{i}\bra{i}\rho\ket{i}\outerproduct{0}{0}_{A}.$
(27)

All classical-quantum channels are entanglement-breaking [13], however, the
channel $\operatorname{\mathcal{M}}_{A}\otimes\operatorname{id}_{B}$ is not
$A$-unital, as we have
$\left(\operatorname{\mathcal{M}}_{A}\otimes\operatorname{id}_{B}\right)\left(\frac{I_{A}}{d}\otimes\sigma_{B}\right)=\outerproduct{0}{0}\otimes\sigma_{B}$
which is not of the form $\frac{I_{A}}{d}\otimes\gamma_{B}$ for some
$\gamma_{B}\in\operatorname{\mathcal{D}}(B)$.

$A$-unital channels are closely related to unital channels. Indeed, as given
in Def. 5, while $A$-unital channels map states of the form
$\frac{I_{A}}{d_{A}}\otimes\sigma_{B}$ to
$\frac{I_{A}}{d_{A}}\otimes\sigma^{\prime}_{B}$, unital channels map
$\frac{I_{AB}}{d_{AB}}$ to $\frac{I_{AB}}{d_{AB}}$. In the next section, we
explore this connection further. We show that while $A$-unital channels are
the largest class of completely free channels for $\mathcal{F}^{c}(A|B)$,
unital channels form the largest class of completely free channels for a
previously defined, related set $\mathcal{F}^{ac}(A|B)$, the set of all states
whose conditional entropy remains non-negative under global unitary
operations. The relationship between $A$-unital, $B$-unital and unital
channels is depicted in Fig. 4.

##  6 Unital Operations and Absolute Non-Negative Conditional Entropy

###  6.1 Absolutely Non-Negative Conditional Entropy States

Previously, the absolute version of $\mathcal{F}^{c}(A|B)$ (analogous to
absolutely separable states [29, 30]) was defined and characterized for the
case where $d_{A}=d_{B}=2$ [27]. Here, we extend the definition by
characterizing it for the case where $d_{A}=d_{B}=d$.

######  Definition 9.

A quantum state $\rho\in\operatorname{\mathcal{D}}(AB)$, where
$d_{A}=d_{B}=d$, which satisfies $S_{A|B}(U\rho U^{\dagger})\geq 0$ for all
unitary channels
$U:\operatorname{\mathcal{D}}(AB)\rightarrow\operatorname{\mathcal{D}}(AB)$ is
called an absolutely non-negative conditional entropy state. The set of
absolutely non-negative conditional entropy states in
$\operatorname{\mathcal{D}}(AB)$ is denoted by $\mathcal{F}^{ac}(A|B)$.

The set $\mathcal{F}^{ac}(A|B)$ is well-characterized in terms of the von
Neumann entropy, as shown in the following theorem.

######  Theorem 4.

$\rho_{AB}\in\mathcal{F}^{ac}(A|B)$ if and only if $S(\rho_{AB})\geq\log d$.

###### Proof.

Let $\rho_{AB}\in\operatorname{\mathcal{D}}(AB)$ and $S(\rho_{AB})\geq\log d$.
As the maximum von Neumann entropy of the subsystem $B$ is upperbounded by
$\log d$, we have $\forall\,U,\,S(U\rho_{AB}U^{\dagger})=S(\rho_{AB})\geq\log
d\geq S\left((U\rho_{AB}U^{\dagger})_{B}\right)$. Therefore,
$\forall\,U,\,S_{A|B}(U\rho U^{\dagger})=S(U\rho
U^{\dagger})-S\left((U\rho_{AB}\rho^{\dagger})_{B}\right)\geq 0$ and
$\rho_{AB}\in\mathcal{F}^{ac}(A|B)$.

To prove the “only if" part of the statement, consider the following set of
vectors $\\{\ket{\psi_{nm}}\\}_{nm}$ which is a basis for the vector space
$\operatorname{\mathcal{H}}^{A}\otimes\operatorname{\mathcal{H}}^{B}$ [31]

$\ket{\psi_{nm}}=\sum_{j}e^{2\pi ijn/d}\ket{j}\otimes\ket{(j+m)\mod
d}/\sqrt{d}.$ (28)

Notice that for each $\ket{\psi_{nm}}$, we have
$\left(\ket{\psi_{nm}}\bra{\psi_{nm}}\right)_{B}=\frac{I}{d}$. Now, let
$\sigma_{AB}\in\mathcal{F}^{ac}(A|B)$. Let the set of vectors
$\\{\ket{\gamma_{nm}}\\}_{nm}$ be the eigenbasis of $\sigma_{AB}$. Consider
the unitary channel $U_{e}\equiv\sum_{nm}\ket{\psi_{nm}}\bra{\gamma_{nm}}$.
The state $U_{e}\sigma_{AB}U_{e}^{\dagger}$ is diagonal in the
$\ket{\psi_{nm}}$ basis, hence,
$\left(U_{e}\sigma_{AB}U_{e}^{\dagger}\right)_{B}=\frac{I}{d}$, and
$S\left((U_{e}\sigma_{AB}U_{e}^{\dagger})_{B}\right)=\log d$.

By definition, we have $S_{A|B}(U\sigma_{AB}U^{\dagger})\geq 0$ for all $U$,
in particular for $U_{e}$. Therefore,
$S_{A|B}(U_{e}\sigma_{AB}U_{e}^{\dagger})=S(U_{e}\sigma_{AB}U_{e}^{\dagger})-S\left((U_{e}\sigma_{AB}U_{e}^{\dagger})_{B}\right)\geq
0$, which implies that $S(\sigma_{AB})-\log d\geq 0$. Thus,
$\,\forall\,\sigma_{AB}\in\mathcal{F}^{ac}(A|B)$, we have
$S(\sigma_{AB})\geq\log d$. ∎

Figure 4: The relationship between $A$-unital, $B$-unital and unital channels
when $d_{A}=d_{B}$.

###  6.2 Completely Free Operations for Absolute Non-Negative Conditional
Entropy States

Here, we define the class of completely free operations for
$\mathcal{F}^{ac}(A|B)$, respecting the $d_{A}=d_{B}=d$ condition in the
definition of $\mathcal{F}^{ac}(A|B)$.

######  Definition 10.

A channel
$\operatorname{\mathcal{N}}_{AB}:\operatorname{\mathcal{D}}(AB)\rightarrow\operatorname{\mathcal{D}}(AB)$
is completely free for $\mathcal{F}^{ac}(A|B)$ if for all finite dimensional
Hilbert spaces $\operatorname{\mathcal{H}}^{A^{\prime}}$ and
$\operatorname{\mathcal{H}}^{B^{\prime}}$, where
$d_{A^{\prime}}=d_{B^{\prime}}=d^{\prime}$ and
$\rho\in\mathcal{F}^{ac}(A^{\prime}A|B^{\prime}B)$, we have
$\left(\operatorname{id}_{A^{\prime}B^{\prime}}\otimes\operatorname{\mathcal{N}}_{AB}\right)(\rho)\in\mathcal{F}^{ac}(A^{\prime}A|B^{\prime}B)$.
The class of all such completely free operations mapping
$\operatorname{\mathcal{D}}(AB)$ to itself is denoted by
$\mathcal{O}^{ac}_{cmax}(A|B)$.

We now prove that for the above definition of $\mathcal{O}^{ac}_{cmax}(A|B)$,
unitality is a necessary and sufficient condition for complete freeness.

######  Theorem 5.

A channel $\operatorname{\mathcal{N}}\in\mathcal{O}^{ac}_{cmax}(A|B)$ if and
only if it is unital.

###### Proof.

Let
$\operatorname{\mathcal{H}}^{A},\operatorname{\mathcal{H}}^{B},\operatorname{\mathcal{H}}^{A^{\prime}}$
and $\operatorname{\mathcal{H}}^{B^{\prime}}$ be finite dimensional Hilbert
spaces such that $d_{A}=d_{B}=d$ and
$d_{A^{\prime}}=d_{B^{\prime}}=d^{\prime}$. Let
$\operatorname{\mathcal{N}}_{AB}$ be a unital channel mapping
$\operatorname{\mathcal{D}}(AB)$ to $\operatorname{\mathcal{D}}(AB)$. Then,
notice that the channel
$\operatorname{\mathcal{N}}^{\prime}=\operatorname{id}_{A^{\prime}B^{\prime}}\otimes\operatorname{\mathcal{N}}_{AB}$
is also a unital channel mapping
$\operatorname{\mathcal{D}}(A^{\prime}AB^{\prime}B)$ to itself. This unitality
implies we have $S\left(\operatorname{\mathcal{N}}^{\prime}(\rho)\right)\geq
S\left(\rho\right)\,\forall\,\rho\in\operatorname{\mathcal{D}}(A^{\prime}AB^{\prime}B)$
[13]. Hence, $\forall\,\sigma\in\mathcal{F}^{ac}(A^{\prime}A|B^{\prime}B)$, we
have $S\left(\operatorname{\mathcal{N}}^{\prime}(\sigma)\right)\geq
S\left(\sigma\right)\geq\log dd^{\prime}$, using Theorem 4. Thus,
$\operatorname{\mathcal{N}}^{\prime}(\sigma)\in\mathcal{F}^{ac}(A^{\prime}A|B^{\prime}B)$.
This implies that $\operatorname{\mathcal{N}}^{\prime}$ is a free operation,
which in turn implies that
$\operatorname{\mathcal{N}}_{AB}\in\mathcal{O}^{ac}_{cmax}(A|B)$.

Let $\operatorname{\mathcal{M}}_{AB}$ be a non-unital channel in
$\operatorname{\mathcal{D}}(AB)$. Hence,
$\operatorname{\mathcal{M}}_{AB}\left(\frac{I_{AB}}{d^{2}}\right)=\sigma_{AB}$
where $\sigma_{AB}\neq\frac{I_{AB}}{d^{2}}$. Thus, $S(\sigma_{AB})<\log
d^{2}$, since the maximum entropy for
$\rho\in\operatorname{\mathcal{D}}(AB)=\log d^{2}$ and is achieved only at
$\frac{I_{AB}}{d^{2}}$. Consider the state
$\chi=\gamma_{A^{\prime}B^{\prime}}\otimes\frac{I_{AB}}{d^{2}}$ where $\gamma$
is a pure state and $d^{\prime}=d$. Hence $S(\chi)=\log d^{2}$ which implies
that $\chi\in\mathcal{F}^{ac}(A^{\prime}A|B^{\prime}B)$. We have
$\chi^{\prime}=\left(\operatorname{id}_{A^{\prime}B^{\prime}}\otimes\operatorname{\mathcal{M}}_{AB}\right)\left(\chi\right)=\gamma_{A^{\prime}B^{\prime}}\otimes\sigma_{AB}$.
Therefore, $S(\chi^{\prime})<\log d^{2}$ and
$\chi^{\prime}\notin\mathcal{F}^{ac}(A^{\prime}A|B^{\prime}B)$. This implies
that
$\operatorname{id}_{A^{\prime}B^{\prime}}\otimes\operatorname{\mathcal{M}}_{AB}$
is not a free operation, and therefore,
$\operatorname{\mathcal{M}}_{AB}\notin\mathcal{O}^{ac}_{cmax}(A|B).$

∎

Figure 5: Upper and lower bounds for $S_{A|B}(\rho)$ given $S(\rho)$.

##  7 Upper and Lower Bounds for Conditional Entropy given Entropy

The characterization of $\mathcal{F}^{ac}(A|B)$ via entropy suggests that the
entropy of a state can determine the values of conditional entropy that the
state can possess. In the following section, we discuss upper and lower bounds
for conditional entropy, given the entropy of a state $\rho_{AB}$, where
$d_{A}=d_{B}=d$.

######  Theorem 6.

Let
$E_{x}\equiv\\{S_{A|B}(\rho)\;|\;S(\rho)=x\;\forall\;\rho\in\operatorname{\mathcal{D}}(AB)\\}$.
Then, the tight lower bound of $E_{x}$ is $x-\log d$ and the tight upper bound
of $E_{x}$ is $\max(x,\log d).$

###### Proof.

Fig. 5 depicts the bounds of the above theorem. The lower bound is proven in
Lemma 8 and the upper bound is proven in Lemmas 9 and 10. ∎

######  Lemma 8.

$\forall x\in[0,2\log d]$, the tight lower bound of
$E_{x}\equiv\\{S_{A|B}(\rho)\mid
S(\rho)=x\land\rho\in\operatorname{\mathcal{D}}(AB)\\}$ is $x-\log d$.

###### Proof.

Consider $\rho\in\operatorname{\mathcal{D}}(AB)\mid S(\rho)=x$. Since
$S(\rho_{B})\leq\log d$ (Theorem 11.8 in [28]), we have $S_{A|B}(\rho)\geq
x-\log d$. Thus,

$\text{a lower bound of }E_{x}\text{ is }x-\log d.$ (29)

It now remains to show that $\forall x\in[0,2\log d]$, $x-\log d\in E_{x}$.
Consider the function
$\operatorname{\mathcal{R}}:[0,1]\rightarrow\operatorname{\mathcal{D}}(AB):p\rightarrow
p|\phi^{+}\rangle\langle\phi^{+}|+(1-p)\frac{I}{d^{2}}$ where
$|\phi^{+}\rangle=\sum_{i=0}^{d-1}\frac{|ii\rangle}{\sqrt{d}}$. Note that
$\forall p\in[0,1]$, we have $\operatorname{\mathcal{R}}(p)_{B}=\frac{I}{d}$.
Thus,

$S(\operatorname{\mathcal{R}}(p)_{B})=\log d.$ (30)

As $S:\operatorname{\mathcal{D}}(AB)\rightarrow[0,2\log d]$ is a continuous
function (Box 11.2 in [28]) and $\operatorname{\mathcal{R}}$ can be shown to
be a continuous function 222This can be shown using the $\epsilon-\delta$
definition of continuity., we have
$S\circ\operatorname{\mathcal{R}}:[0,1]\rightarrow[0,2\log d]$ is a continuous
function. We have $S(\operatorname{\mathcal{R}}(0))=2\log d$ and
$S(\operatorname{\mathcal{R}}(1))=0$. Thus, by the intermediate value theorem,

$\forall x\in[0,2\log d],\;\exists\;p_{x}\in[0,1]\mid
S(\operatorname{\mathcal{R}}(p_{x}))=x.$ (31)

Using Eq. 30 and 31, we have

$S_{A|B}(\operatorname{\mathcal{R}}(p_{x}))=S(\operatorname{\mathcal{R}}(p_{x}))-S(\operatorname{\mathcal{R}}(p_{x})_{B})=x-\log
d.$ (32)

$S(\operatorname{\mathcal{R}}(p_{x}))=x$ (Eq. 31) implies that
$S_{A|B}(\operatorname{\mathcal{R}}(p_{x}))=x-\log d\in E_{x}$ (Eq. 32). Thus,
as $x-\log d$ is a lower bound of $E_{x}$, and $x-\log d\in E_{x}$, it is the
tight lower bound. ∎

######  Lemma 9.

$\forall x\in[0,\log d]$, the tight upper bound of
$E_{x}\equiv\\{S_{A|B}(\rho)\mid
S(\rho)=x\land\rho\in\operatorname{\mathcal{D}}(AB)\\}$ is $x$.

###### Proof.

Consider $\rho\in\operatorname{\mathcal{D}}(AB)\mid S(\rho)=x$. Since
$S(\rho_{B})\geq 0$, we have $S_{A|B}(\rho)\leq x$. Thus, an upper bound of
$E_{x}$ is $x$. It now remains to show that $\forall x\in[0,\log d]$, $x\in
E_{x}$. For a given $x\in[0,\log d]$, consider states of the form
$\sigma=\sigma_{A}\otimes\sigma_{B}$, where $S(\sigma_{A})=x$ and $\sigma_{B}$
is any pure state. We have $S(\sigma)=x+0=x$, which implies
$S_{A|B}(\sigma)=x-0=x\in E_{x}$.

Therefore, since $x$ is an upperbound of $E_{x}$ and $x\in E_{x}$, it is a
tight upper bound. ∎

######  Lemma 10.

$\forall x\in[\log d,2\log d]$, the tight upper bound of
$E_{x}\equiv\\{S_{A|B}(\rho)\mid
S(\rho)=x\land\rho\in\operatorname{\mathcal{D}}(AB)\\}$ is $\log d$.

###### Proof.

From Theorem 11.4.1 in [13] and Theorem 11.8 in [28], we have
$\forall\rho\in\operatorname{\mathcal{D}}(AB),\;S_{A|B}(\rho)\leq
S(\rho_{A})\leq\log d$. Therefore, $\log d$ is also an upper bound for
$E_{x}$. It now remains to show that $\forall x\in[\log d,2\log d],\;\log d\in
E_{x}$. Consider states of the form $\sigma=\frac{I}{d}\otimes\sigma_{B}$
where $S(\sigma_{B})=x-\log d$. We have $S(\sigma)=x$ which implies that
$S_{A|B}(\sigma)=\log d\in E_{x}$. Therefore, since $\log d$ is an upper bound
for $E_{x}$ and $\log d\in E_{x}$, it is a tight upper bound. ∎

Theorem 6 follows from the lower bound proved in Lemma 8 and the upper bounds
proved in Lemmas 9 and 10.

##  8 Conclusion

In this paper, we introduce the class of $A$-unital channels, and show that
this is equal to the largest class of completely free channels for the set of
states with non-negative conditional entropy
$\left(\mathcal{F}^{c}(A|B)\right)$. We also show that this is the largest
class of channels that do not decrease the conditional entropy of a state. We
discuss the relation between $A$-unital channels, separable channels and
entanglement-breaking channels. Furthermore, we extended the characterization
of the set of states that possess non-negative conditional entropy even under
global unitary transformations ($\mathcal{F}^{ac}(A|B)$) from the $2\otimes 2$
case to the $d\otimes d$ case. We demonstrate that unital channels are the
largest class of completely free operations for this set. Finally, we provide
upper and lower bounds for the conditional entropy of a state as a function of
the entropy of a state. We conclude with some questions for further research
that emerge from our work.

  1. 1.

Characterization of $\mathcal{F}^{ac}(A|B)$ for the $d_{A}\otimes d_{B}$ case.
In this work, we show that when $d_{A}=d_{B}=d$, $\mathcal{F}^{ac}(A|B)$ is
the set of states in $\operatorname{\mathcal{D}}(AB)$ for which
$S(\rho_{AB})\geq\log d$. An interesting problem is the extension of this
characterization to the case $d_{A}\neq d_{B}$. This is important for the full
characterization of completely free operations for $\mathcal{F}^{ac}(A|B)$.

  2. 2.

Decomposition of A-unital channels and a Choi matrix representation. A
necessary and sufficient condition for the unitality of a channel is for it to
be expressed as an affine combination of unitary quantum channels [32].
Providing a similar decomposition for A-unital channels is an open problem.
Moreover, separable channels, completely PPT-preserving channels and
entanglement-breaking channels all have elegant characterizations in terms of
Choi matrices [4, 33, 34, 35, 13, 36]. A similar characterization remains to
be given for $A$-unital channels.

  3. 3.

State conversion under A-unital channels. According to Uhlmann’s theorem, a
state $\rho$ can be converted to state $\sigma$ under unital channels if and
only if $\sigma\prec\rho$, that is $\rho$ majorizes $\sigma$ [37, 38]. What
are the conditions for state conversion under $A$-unital channels?

#### Note added:

Also see related independent work by Brandsen et al. [39]

## Acknowledgements

We gratefully acknowledge Prof. Arun K. Pati for his enriching discussions,
and Mohammad Alhejji for his fruitful comments.

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