# Mixed-permutation channel with its application to estimate quantum coherence

Lin Zhang1111E-mail<EMAIL_ADDRESS>and Ming-Jing Zhao2222E-mail:
<EMAIL_ADDRESS>
1Institute of Mathematics, Hangzhou Dianzi University, Hangzhou 310018, PR
China
2School of Science, Beijing Information Science and Technology University,
Beijing, 100192, PR China

###### Abstract

Quantum channel, as the information transmitter, is an indispensable tool in
quantum information theory. In this paper, we study a class of special quantum
channels named the _mixed-permutation channels_. The properties of these
channels are characterized. The mixed-permutation channels can be applied to
give a lower bound of quantum coherence with respect to any coherence measure.
In particular, the analytical lower bounds for $l_{1}$-norm coherence and the
relative entropy of coherence are shown respectively. The extension to
bipartite systems is presented for the actions of the mixed-permutation
channels.

##  1 Introduction

Quantum channel is a deterministic quantum operation and it transmits the
information from the input to the output. It is indispensable in various
information processing tasks, such as quantum communication [1], quantum
computation [2, 3] and quantum cryptography [4, 5]. So the study about quantum
channel is an interesting and attractive topic. Explicitly, the resource
theory of quantum channels deals with understanding the properties and
capabilities of quantum channels in terms of the resources they consume and
produce [6, 7, 8, 9, 10, 11]. It also has important implications for the
design of quantum protocols, as it provides a unified framework [12] for
characterizing the capabilities and limitations of different types of quantum
channels. For quantum channels, the uncertainty relations, a fundamental topic
in quantum mechanics, are also established recently. They describe the
theoretical restrictions about two or more quantum channels [13, 14, 15], just
like that about two or more observables. In addition, quantum channels are
also intimately related to the dynamical behavior of quantum states and the
investigation of open quantum system [16, 17, 18].

One special kind of the most commonly-used quantum channels is the projective
measurement. Within the framework of the projective measurement, many quantum
features are manifested. For instance, quantum coherence is such a quantum
feature and it is of practical significance in quantum computation and quantum
communication [19, 20, 21]. The quantum coherence is further quantified by
coherence measures such as the $l_{1}$-norm coherence [22], the relative
entropy of coherence [22], the coherence concurrence [23], the geometric
coherence [24] and the robustness of coherence [25] and so on. However, the
evaluation of the coherence measure is not always easy for arbitrary quantum
states, especially for mixed states.

In this paper, we investigate an interesting quantum channel called the
_mixed-permutation channel_ , which maps an input state to an average of their
permutation conjugations. This mixed-permutation channel can be viewed as a
special mixed-unitary channels with equal weights [26]. More than that, it is
also an example of a covariant quantum channel [27], meaning that it commutes
with certain symmetry operations on the input states. In this case, this
mixed-permutation channel is covariant with respect to the conjugated action
of the symmetric group $S_{d}$.

This work is factually a follow-up study of [28], which investigates the
application of the mixed-permutation channel in quantifying quantum coherence.
Here we aim to explore the mixed-permutation channel further and characterize
it systematically. Moreover, we find the application of the mixed-permutation
channel to evaluate the quantum coherence. This evaluation is applicable to
all coherence measures and is tight for some special quantum states.

The paper is organized as follows. In Sect. 2, we give the definition of the
mixed-permutation channels and list their basic properties. We also present
the specific form of the mixed-permutation channels, and characterize the
image states. The applications of the mixed-permutation channel in estimating
quantum coherence is also presented. Subsequently, in Sect. 3, we extend the
action of the mixed-permutation channels to bipartite systems. We conclude
this paper in Sect. 4.

##  2 Mixed-permutation channel and its application in estimating quantum
coherence

In this section, we consider the $d$-dimensional quantum system with its
Hilbert space $\mathbb{C}^{d}$ and fix the reference basis as
$\\{|i\rangle\\}_{i=1}^{d}$. The so-called quantum state is represented by the
density matrices/operators acting on the Hilbert space $\mathbb{C}^{d}$. We
denote the set of all density matrices by

$\mathrm{D}\left(\mathbb{C}^{d}\right)=\left\\{\rho=\sum^{d}_{i,j=1}\rho_{ij}|i\rangle\langle
j|:\rho\geqslant 0,\operatorname{Tr}\left(\rho\right)=1\right\\}.$

Any quantum channel $\Lambda$ acting on the density matrices/operators in
$\mathrm{D}\left(\mathbb{C}^{d}\right)$ can be expressed with Kraus
representation $\Lambda(\cdot)=\sum_{\mu}K_{\mu}(\cdot)K_{\mu}^{\dagger}$
where $\sum_{\mu}K_{\mu}^{\dagger}K_{\mu}=\mathbb{1}_{d}$, the identity
operator on $\mathbb{C}^{d}$. Next we shall introduce the mixed-permutation
channels and characterize it both analytically and geometrically. Then we
explain its application in estimating quantum coherence.

###  2.1 Mixed-permutation channels

######  Definition 1.

For any matrix $\boldsymbol{O}$ acting on $\mathbb{C}^{d}$, we define the
_mixed-permutation channel_ $\Delta$ as

$\displaystyle\Delta(\boldsymbol{O}):=\frac{1}{d!}\sum_{\pi\in
S_{d}}\boldsymbol{P}_{\pi}\boldsymbol{O}\boldsymbol{P}^{\dagger}_{\pi},$ (1)

where $S_{d}$ is the symmetric group of permutations of $d$ elements, and the
permutation matrix $\boldsymbol{P}_{\pi}$ is

$\displaystyle\boldsymbol{P}_{\pi}=\sum^{d}_{i=1}|\pi(i)\rangle\langle i|.$
(2)

induced by $\pi\in S_{d}$.

By this definition we see for any matrix $\boldsymbol{O}$ acting on
$\mathbb{C}^{d}$, the action of the mixed-permutation channel is the average
action over all possible permutation matrices. It is easily seen that
$\boldsymbol{P}_{\pi}\boldsymbol{P}_{\sigma}=\boldsymbol{P}_{\pi\sigma}$ and
$\boldsymbol{P}^{\dagger}_{\pi}=\boldsymbol{P}^{\scriptscriptstyle\mathsf{T}}_{\pi}=\boldsymbol{P}_{\pi^{-1}}$
for any $\pi$ and $\sigma$ in $S_{d}$. Moreover,
$\Delta(\boldsymbol{O})=\frac{1}{d!}\sum_{\pi\in
S_{d}}\boldsymbol{P}^{\dagger}_{\pi}\boldsymbol{O}\boldsymbol{P}_{\pi}$. The
properties of the mixed-permutation channel $\Delta$ can be listed below:

  1. (i)

$\Delta\circ\Delta=\Delta$, that is
$\Delta\circ\Delta(\boldsymbol{O})=\Delta(\boldsymbol{O})$ for any matrix
$\boldsymbol{O}$ acting on $\mathbb{C}^{d}$.

  2. (ii)

$\Delta$ is unital, that is $\Delta(\mathbb{1}_{d})=\mathbb{1}_{d}$, where
$\mathbb{1}_{d}$ is the identity matrix on $\mathbb{C}^{d}$.

  3. (iii)

$\Delta$ is self-adjoint with respect to Hilbert-Schmidt inner product, i.e.,
$\Delta=\Delta^{\dagger}$ in the sense of
$\left\langle\Delta^{\dagger}(\boldsymbol{X}),\boldsymbol{Y}\right\rangle=\left\langle\boldsymbol{X},\Delta(\boldsymbol{Y})\right\rangle$
for all matrices $\boldsymbol{X},\boldsymbol{Y}$ acting on $\mathbb{C}^{d}$,
that is, $\Delta^{\dagger}(\boldsymbol{O})=\frac{1}{d!}\sum_{\pi\in
S_{d}}\boldsymbol{P}^{\dagger}_{\pi}\boldsymbol{O}\boldsymbol{P}_{\pi}=\Delta(\boldsymbol{O})$.

  4. (iv)

$[\Delta(\boldsymbol{O})]^{\scriptscriptstyle\mathsf{T}}=\Delta(\boldsymbol{O}^{{\scriptscriptstyle\mathsf{T}}})$,
that is
$[\Delta(\boldsymbol{O})]^{\scriptscriptstyle\mathsf{T}}=(\frac{1}{d!}\sum_{\pi\in
S_{d}}\boldsymbol{P}_{\pi}\boldsymbol{O}\boldsymbol{P}^{\dagger}_{\pi})^{\scriptscriptstyle\mathsf{T}}=\frac{1}{d!}\sum_{\pi\in
S_{d}}\boldsymbol{P}_{\pi}\boldsymbol{O}^{\scriptscriptstyle\mathsf{T}}\boldsymbol{P}^{\dagger}_{\pi}=\Delta(\boldsymbol{O}^{{\scriptscriptstyle\mathsf{T}}})$.

  5. (v)

$\Delta$ is invariant under any permutation matrices, that is,
$\boldsymbol{P}_{\tau}\Delta(\boldsymbol{O})\boldsymbol{P}^{\dagger}_{\tau}=\Delta(\boldsymbol{O})$
for any permutation matrix $\boldsymbol{P}_{\tau}$, induced by the permutation
$\tau\in S_{d}$.

Furthermore, the explicit form of the mixed-permutation channel on the
$\mathbb{C}^{d}$ can be obtained analytically.

######  Theorem 1.

For any $d\times d$ matrix $\boldsymbol{O}=(o_{ij})$, the mixed-permutation
channel can be characterized as

$\displaystyle\Delta(\boldsymbol{O})=\operatorname{Tr}\left(\boldsymbol{O}\right)\frac{\mathbb{1}_{d}}{d}+\operatorname{Tr}\left(\boldsymbol{O}(\boldsymbol{E}-\mathbb{1}_{d})\right)\frac{(\boldsymbol{E}-\mathbb{1}_{d})}{d(d-1)},$
(3)

where $\boldsymbol{E}:=|\boldsymbol{e}\rangle\\!\langle\boldsymbol{e}|$ for
$d$-dimensional vector
$\boldsymbol{e}=(1,1,\ldots,1)^{\scriptscriptstyle\mathsf{T}}$ with all
entries being one.

###### Proof.

In fact, this result will be established by linearity once we show it holds
for any Hermitian matrix $\boldsymbol{O}$ because every complex square matrix
$\boldsymbol{M}$ can be represented as a complex linear combination of two
Hermitian matrices, i.e.,
$\boldsymbol{M}=\boldsymbol{H}+\mathrm{i}\boldsymbol{K}$, where
$\boldsymbol{H}:=\frac{\boldsymbol{M}+\boldsymbol{M}^{\dagger}}{2}$ and
$\boldsymbol{K}=\frac{\boldsymbol{M}-\boldsymbol{M}^{\dagger}}{2\mathrm{i}}$.
In what follows, it suffices to consider Hermitian case.

Let us assume that $\boldsymbol{O}$ is Hermitian. If $i\neq j$, then

$\displaystyle\left\langle i\left|\Delta(\boldsymbol{O})\right|j\right\rangle$
$\displaystyle=$ $\displaystyle\frac{1}{d!}\sum_{\pi\in S_{d}}\left\langle
i\left|\boldsymbol{P}^{\dagger}_{\pi}\boldsymbol{O}\boldsymbol{P}_{\pi}\right|j\right\rangle=\frac{1}{d!}\sum_{\pi\in
S_{d}}\left\langle\pi(i)\left|\boldsymbol{O}\right|\pi(j)\right\rangle=\frac{1}{d!}\sum_{\pi\in
S_{d}}o_{\pi(i)\pi(j)}.$

Now there exists a permutation $\tau_{0}\in S_{d}$ such that $i=\tau_{0}(1)$
and $j=\tau_{0}(2)$ due to the assumption $i\neq j$. Then

$\displaystyle\sum_{\pi\in S_{d}}o_{\pi(i)\pi(j)}=\sum_{\pi\in
S_{d}}o_{\pi\tau_{0}(1)\pi\tau_{0}(2)}=\sum_{\pi\in S_{d}}o_{\pi(1)\pi(2)},$

implying that, whenever $i\neq j$, then

$\displaystyle\left\langle
i\left|\Delta(\boldsymbol{O})\right|j\right\rangle=\left\langle
1\left|\Delta(\boldsymbol{O})\right|2\right\rangle.$

Similarly, if $i=j$, then

$\displaystyle\left\langle
i\left|\Delta(\boldsymbol{O})\right|i\right\rangle=\frac{1}{d!}\sum_{\pi\in
S_{d}}o_{\pi(i)\pi(i)}=\frac{1}{d!}\sum_{\pi\in
S_{d}}o_{\pi(1)\pi(1)}=\left\langle
1\left|\Delta(\boldsymbol{O})\right|1\right\rangle.$

Note that
$\operatorname{Tr}\left(\boldsymbol{O}\right)=\operatorname{Tr}\left(\Delta(\boldsymbol{O})\right)=\sum^{d}_{i=1}\left\langle
i\left|\Delta(\boldsymbol{O})\right|i\right\rangle=d\left\langle
1\left|\Delta(\boldsymbol{O})\right|1\right\rangle$, which means that

$\displaystyle\left\langle
i\left|\Delta(\boldsymbol{O})\right|i\right\rangle=\frac{\operatorname{Tr}\left(\boldsymbol{O}\right)}{d}.$

From the above discussion, let
$\boldsymbol{E}:=|\boldsymbol{e}\rangle\\!\langle\boldsymbol{e}|$, where
$\boldsymbol{e}=(1,\ldots,1)^{\scriptscriptstyle\mathsf{T}}$. We see that

$\displaystyle\Delta(\boldsymbol{O})=\frac{\operatorname{Tr}\left(\boldsymbol{O}\right)}{d}\mathbb{1}_{d}+\zeta(\boldsymbol{E}-\mathbb{1}_{d}).$

Then, via $\boldsymbol{E}^{2}=d\cdot\boldsymbol{E}$,

$\displaystyle\boldsymbol{E}\Delta(\boldsymbol{O})\boldsymbol{E}=\frac{\operatorname{Tr}\left(\boldsymbol{O}\right)}{d}\boldsymbol{E}^{2}+\zeta(\boldsymbol{E}^{3}-\boldsymbol{E}^{2})=[\operatorname{Tr}\left(\boldsymbol{O}\right)+\zeta
d(d-1)]\boldsymbol{E}.$

Because $\boldsymbol{P}_{\pi}\boldsymbol{E}=\boldsymbol{E}$, we get that
$\boldsymbol{E}\Delta(\boldsymbol{O})\boldsymbol{E}=\boldsymbol{E}\boldsymbol{O}\boldsymbol{E}=\left\langle\boldsymbol{e}\left|\boldsymbol{O}\right|\boldsymbol{e}\right\rangle\boldsymbol{E}$.
From these, we see that

$\displaystyle\left\langle\boldsymbol{e}\left|\boldsymbol{O}\right|\boldsymbol{e}\right\rangle=\operatorname{Tr}\left(\boldsymbol{O}\right)+\zeta
d(d-1),$

from which we get

$\displaystyle\zeta=\frac{\left\langle\boldsymbol{e}\left|\boldsymbol{O}\right|\boldsymbol{e}\right\rangle-\operatorname{Tr}\left(\boldsymbol{O}\right)}{d(d-1)}=\frac{2\mathrm{Re}(o_{ij})}{d(d-1)}=\binom{d}{2}^{-1}\mathrm{Re}(o_{ij}).$

Therefore

$\displaystyle\Delta(\boldsymbol{O})$ $\displaystyle=$
$\displaystyle\frac{\operatorname{Tr}\left(\boldsymbol{O}\right)}{d}\mathbb{1}_{d}+\zeta(\boldsymbol{E}-\mathbb{1}_{d})$
$\displaystyle=$
$\displaystyle\frac{\operatorname{Tr}\left(\boldsymbol{O}\right)}{d}\mathbb{1}_{d}+\frac{\left\langle\boldsymbol{e}\left|\boldsymbol{O}\right|\boldsymbol{e}\right\rangle-\operatorname{Tr}\left(\boldsymbol{O}\right)}{d(d-1)}(\boldsymbol{E}-\mathbb{1}_{d})$
$\displaystyle=$
$\displaystyle\frac{\operatorname{Tr}\left(\boldsymbol{O}\right)}{d}\mathbb{1}_{d}+\frac{\operatorname{Tr}\left(\boldsymbol{O}\boldsymbol{E}\right)-\operatorname{Tr}\left(\boldsymbol{O}\right)}{d(d-1)}(\boldsymbol{E}-\mathbb{1}_{d}).$

After simplifying it then we get the desired expression. This completes the
proof. ∎

In Theorem 1, when $\boldsymbol{O}$ is taken as a quantum state
$\rho\in\mathrm{D}\left(\mathbb{C}^{d}\right)$, i.e., if a quantum state
$\rho$ goes through the mixed-permutation channel $\Delta$, then the output
state $\Delta(\rho)$ can be characterized analytically.

######  Corollary 1.

For any quantum state $\rho=\sum^{d}_{i,j=1}\rho_{ij}|i\rangle\langle
j|\in\mathrm{D}\left(\mathbb{C}^{d}\right)$, the corresponding output state
$\rho_{\star}:=\Delta(\rho)$ from the mixed-permutation channel is given by

$\displaystyle\rho_{\star}=\left(\begin{array}[]{cccccccc}d^{-1}&\mathfrak{a}&\cdots&\mathfrak{a}\\\
\mathfrak{a}&d^{-1}&\cdots&\mathfrak{a}\\\ \vdots&\vdots&\ddots&\vdots\\\
\mathfrak{a}&\mathfrak{a}&\cdots&d^{-1}\end{array}\right),$ (8)

where

$\displaystyle\mathfrak{a}:=\frac{1}{d(d-1)}\sum_{i\neq j}\rho_{ij}$ (9)

for the given reference basis $\\{|i\rangle:1,\ldots,d\\}$.

######  Remark 1.

We remark here that
$\mathfrak{a}\in\left[-\frac{1}{d(d-1)},\frac{1}{d}\right]$. Indeed,

$\displaystyle\sum_{i\neq
j}\rho_{ij}=\sum_{i,j=1}^{d}\rho_{ij}-1=\operatorname{Tr}\left(\rho\boldsymbol{E}\right)-1=\left\langle\boldsymbol{e}\left|\rho\right|\boldsymbol{e}\right\rangle-1.$

Therefore, combining
$d\lambda_{\min}(\rho)\leqslant\left\langle\boldsymbol{e}\left|\rho\right|\boldsymbol{e}\right\rangle\leqslant
d\lambda_{\max}(\rho)$, where $\lambda_{\min}(\rho)$ and
$\lambda_{\max}(\rho)$ are the minimum and maximum eigenvalues of the quantum
state $\rho$ respectively, we have,

$\displaystyle-\frac{1}{d(d-1)}\leqslant\frac{d\lambda_{\min}(\rho)-1}{d(d-1)}\leqslant\mathfrak{a}=\frac{1}{d(d-1)}\sum_{i\neq
j}\rho_{ij}\leqslant\frac{d\lambda_{\max}(\rho)-1}{d(d-1)}\leqslant\frac{1}{d}.$
(10)

In order to see clearly what is the image states $\Delta(\rho)$ of the channel
$\Delta$ when input state is $\rho$, let us focus on qubit systems. In fact,
any qubit state can be represented as

$\displaystyle\rho=\frac{1}{2}(\mathbb{1}_{2}+\boldsymbol{r}\cdot\boldsymbol{\sigma})=\frac{1}{2}\left(\begin{array}[]{cc}1+r_{3}&r_{1}-\mathrm{i}r_{2}\\\
r_{1}+\mathrm{i}r_{2}&1-r_{3}\end{array}\right),$ (13)

where $r_{i}\in\mathbb{R}$,
$r:=\left\lvert\mspace{1.0mu}\boldsymbol{r}\mspace{1.0mu}\right\rvert=\sqrt{r_{1}^{2}+r_{2}^{2}+r_{3}^{2}}\leqslant
1$, and
$\boldsymbol{r}\cdot\boldsymbol{\sigma}:=\sum^{3}_{k=1}r_{k}\sigma_{k}$,
$\boldsymbol{\sigma}:=(\sigma_{1},\sigma_{2},\sigma_{3})$ is the vector of the
Pauli matrices, given by

$\displaystyle\sigma_{1}=\left(\begin{array}[]{cc}0&1\\\
1&0\end{array}\right),\quad\sigma_{2}=\left(\begin{array}[]{cc}0&-\mathrm{i}\\\
\mathrm{i}&0\end{array}\right),\quad\sigma_{3}=\left(\begin{array}[]{cc}1&0\\\
0&-1\end{array}\right).$ (20)

It is easily seen that the action of $\Delta$ is

$\displaystyle\Delta(\rho)=\frac{1}{2}(\mathbb{1}_{2}+\boldsymbol{r}^{\prime}\cdot\boldsymbol{\sigma})=\frac{1}{2}\left(\begin{array}[]{cc}1&r_{1}\\\
r_{1}&1\end{array}\right).$ (23)

So the action of the mixed-permutation channel on the set of quantum states
can be simulated by the transformation on the Bloch ball, i.e.,

$\displaystyle\boldsymbol{r}=(r_{1},r_{2},r_{3})\rightarrow\boldsymbol{r}^{\prime}=(r_{1},0,0).$
(24)

From Figure 1, we see that the Bloch ball is mapped into a red line segment by
the action of the mixed-permutation channel.

Figure 1: (Color Online) The whole yellow Bloch ball stands for the family of
all qubit states in Eq. (13); the red line segment stands for the image states
$\Delta(\rho)$ in Eq. (23) when qubit states $\rho$ through the mixed-
permutation channel.

###  2.2 Applying the mixed-permutation channel to estimate quantum coherence

In order to estimate quantum coherence using the mixed-permutation channel, we
need recall some notions about quantum coherence. Given the reference basis
$\\{|i\rangle\\}_{i=1}^{d}$, a diagonal quantum state
$\varrho=\sum_{i=1}^{d}\lambda_{i}|i\rangle\\!\langle i|$ is called the
incoherent state. The set of incoherent states is denoted as $\mathcal{I}$.
The so-called incoherent operation is a completely positive linear mapping
$\Phi$ such that its Kraus decomposition $\Phi(\rho)=\sum_{\nu}K_{\nu}\rho
K_{\nu}^{\dagger}$ with $\sum_{\nu}K_{\nu}^{\dagger}K_{\nu}=\mathbb{1}$
fulfills $K_{\nu}\delta K_{\nu}^{\dagger}/\operatorname{Tr}\left(K_{\nu}\delta
K_{\nu}^{\dagger}\right)\in\mathcal{I}$ for all $\delta\in\mathcal{I}$ and for
all $\nu$.

The quantum coherence is quantified by a nonnegative function named the
coherence measure. A coherence measure is required to satisfy the following
conditions [22]:

  1. (A1)

Nonnegativity: $C(\rho)\geqslant 0$ for
$\rho\in\mathrm{D}\left(\mathbb{C}^{d}\right)$, and moreover $C(\rho)=0$ for
all $\rho\in\mathcal{I}$;

  2. (A2)

Monotonicity: $C(\Phi(\rho))\leqslant C(\rho)$ for any incoherent operation
$\Phi$;

  3. (A3)

Strong monotonicity: $\sum_{\nu}p_{\nu}C(K_{\nu}\rho
K_{\nu}^{\dagger}/p_{\nu})\leqslant C(\rho)$ for any incoherent operation
$\Phi(\rho)=\sum_{\nu}K_{\nu}\rho K_{\nu}^{\dagger}$ with
$p_{\nu}=\operatorname{Tr}\left(K_{\nu}\rho K_{\nu}^{\dagger}\right)$ and
$\rho_{\nu}=K_{\nu}\rho K_{\nu}^{\dagger}/p_{\nu}$;

  4. (A4)

Convexity: $C(\rho)\leqslant\sum_{k}p_{k}C(\rho_{k})$ for any quantum state
$\rho=\sum_{k}p_{k}\rho_{k}$ with $p_{k}\geqslant 0$ and $\sum_{k}p_{k}=1$.

Two commonly used coherence measures are the $l_{1}$-norm coherence and the
relative entropy of coherence [22]. The $l_{1}$-norm coherence of the quantum
state $\rho=\sum_{i,j}\rho_{ij}|i\rangle\langle j|$ is the sum of the
magnitudes of all the off-diagonal entries

$\displaystyle C_{l_{1}}(\rho):=\sum_{i\neq
j}\left\lvert\mspace{1.0mu}\rho_{ij}\mspace{1.0mu}\right\rvert.$ (25)

The relative entropy of coherence is the difference of von Neumann entropy
between the density matrix and the diagonal matrix given by its diagonal
entries,

$\displaystyle C_{r}(\rho):=S(\Pi(\rho))-S(\rho),$ (26)

where $\Pi(\rho)=\mathrm{diag}(\rho_{11},\rho_{22},\cdots,\rho_{dd})$ is the
diagonal matrix obtained by the diagonal entries of $\rho$ and
$S(\rho):=-\operatorname{Tr}\left(\rho\ln\rho\right)$ is the von Neumann
entropy of the state $\rho$.

Now we come back to the mixed-permutation channel. It is obvious that
$\boldsymbol{P}_{\pi}\rho\boldsymbol{P}^{\dagger}_{\pi}$ is incoherent for any
incoherent state $\rho$. So the mixed-permutation channel is an incoherent
operation. By the monotonicity of the coherence measures, we know the
coherence of input state $\rho$ is nonincreasing under this channel. Since the
output states of the mixed-permutation channel have an analytical form
$\rho_{\star}$ in Eq. (8), we can utilize it to estimate the coherence of the
input state $\rho$.

######  Theorem 2.

For any coherence measure $C$ and any quantum state
$\rho\in\mathrm{D}\left(\mathbb{C}^{d}\right)$, the quantum coherence of
$\rho$ is bounded from below by the quantum coherence of $\rho_{\star}$,
namely,

$\displaystyle C(\rho)\geqslant C(\rho_{\star}).$ (27)

###### Proof.

First, since any permutation matrices $\boldsymbol{P}_{\pi}$ and
$\boldsymbol{P}_{\pi^{-1}}$ are incoherent operations, so

$C(\rho)=C[\boldsymbol{P}^{\dagger}_{\pi}(\boldsymbol{P}_{\pi}\rho\boldsymbol{P}^{\dagger}_{\pi})\boldsymbol{P}_{\pi}]\leqslant
C(\boldsymbol{P}_{\pi}\rho\boldsymbol{P}^{\dagger}_{\pi})\leqslant C(\rho).$

Therefore $C(\boldsymbol{P}_{\pi}\rho\boldsymbol{P}^{\dagger}_{\pi})=C(\rho)$
for any permutation matrcix $\boldsymbol{P}_{\pi}$. Then by the convexity of
coherence measure, we have further that

$\displaystyle C(\rho_{\star})=C(\Delta(\rho))$ $\displaystyle=$
$\displaystyle C\left(\frac{1}{d!}\sum_{\pi\in
S_{d}}\boldsymbol{P}_{\pi}\rho\boldsymbol{P}^{\dagger}_{\pi}\right)$
$\displaystyle\leqslant$ $\displaystyle\frac{1}{d!}\sum_{\pi\in
S_{d}}C\left(\boldsymbol{P}_{\pi}\rho\boldsymbol{P}^{\dagger}_{\pi}\right)$
$\displaystyle=$ $\displaystyle\frac{1}{d!}\sum_{\pi\in S_{d}}C(\rho)$
$\displaystyle=$ $\displaystyle C(\rho),$

which completes the proof. ∎

From the proof above, we see that Theorem 2 is universal in the sense that Eq.
(30) is independent of the coherence measures. So Theorem 2 can be applied to
all coherence measures. Additionally, the output states $\rho_{\star}$ is of
one parameter, so its coherence is much easy to calculate compared with the
input state. In fact, for any quantum state
$\rho=\sum^{d}_{i,j=1}\rho_{ij}|i\rangle\langle j|$, the output state
$\rho_{\star}$ in Corollary 1 can be decomposed as

$\displaystyle\rho_{\star}=(1-\mathfrak{p})\mathbb{1}_{d}/d+\mathfrak{p}|\Phi_{d}\rangle\\!\langle\Phi_{d}|,$
(28)

where the weight is relevant to the quantum state as

$\displaystyle\mathfrak{p}:=\frac{1}{d-1}\sum_{i\neq
j}\rho_{ij}\in\left[-\frac{1}{d-1},1\right]$ (29)

and $|\Phi_{d}\rangle:=\frac{1}{\sqrt{d}}\sum^{d}_{i=1}|i\rangle$ is the
maximally coherent state [29]. This state $\rho_{\star}$ is also called the
_maximally coherent mixed state_ in [30]. By this decomposition in Eq. (28),
we get the eigenvalues of $\rho_{\star}$ is
$\mathfrak{p}+\frac{1-\mathfrak{p}}{d}$ and $(1-\mathfrak{p})/d$ with
multiplicity $d-1$. In view of this, we get the following result:

######  Corollary 2.

For any quantum state $\rho\in\mathrm{D}\left(\mathbb{C}^{d}\right)$, the
$l_{1}$-norm coherence and the relative entropy of coherence can be valuated
from below by

$\displaystyle C_{l_{1}}(\rho)$ $\displaystyle\geqslant$ $\displaystyle
d(d-1)\left\lvert\mspace{1.0mu}\mathfrak{a}\mspace{1.0mu}\right\rvert,$ (30)
$\displaystyle C_{r}(\rho)$ $\displaystyle\geqslant$
$\displaystyle\left(1-\frac{1}{d}\right)(1-\mathfrak{p})\ln(1-\mathfrak{p})+\frac{1}{d}[(d-1)\mathfrak{p}+1]\ln[(d-1)\mathfrak{p}+1],$
(31)

respectively, where $\mathfrak{a}$ and $\mathfrak{p}$ are depended by $\rho$
as Eqs. (9) and (29).

For $l_{1}$ norm coherence, the lower bound in Eq. (30) is tight for _real_
quantum states, that is,

$\displaystyle C_{l_{1}}(\rho)=C_{l_{1}}(\rho_{\star})$ (32)

for any _real_ quantum state $\rho$. Now we consider the $l_{1}$-norm
coherence and the relative entropy of coherence in qubit system.

######  Example 1.

In qubit systems, any qubit state $\rho$ can be represented as in Eq. (13)
with the spectral decomposition as

$\rho=\lambda_{1}|\phi_{1}\rangle\\!\langle\phi_{1}|+\lambda_{2}|\phi_{2}\rangle\\!\langle\phi_{2}|$

with the eigenvalues $\lambda_{1}=\frac{1}{2}(1+r)$,
$\lambda_{2}=\frac{1}{2}(1-r)$ and the corresponding eigenvectors

$|\phi_{1}\rangle=\frac{(r_{1}-\mathrm{i}r_{2})|0\rangle-(r_{3}-r)|1\rangle}{\sqrt{2r(r-r_{3})}},\quad|\phi_{2}\rangle=\frac{(r_{1}-\mathrm{i}r_{2})|0\rangle-(r_{3}+r)|1\rangle}{\sqrt{2r(r+r_{3})}}.$

Let $r_{2}=r_{3}=\frac{1}{10}$, we calculate the $l_{1}$-norm coherence
$C_{l_{1}}(\rho)$ and the relative entropy coherence $C_{r}(\rho)$ as well as
the lower bound the $l_{1}$-norm coherence $C_{l_{1}}(\rho_{\star})$ and the
relative entropy coherence $C_{r}(\rho_{\star})$ and make a comparison in
Figures 2 and 2.

Figure 2: (Color Online) The estimation of the coherence of $\rho$ by that of
$\rho_{\star}$ using the $l_{1}$-norm coherence measure in subfigure (a) and
the relative entropy of coherence in subfigure (b).

From another point of view, it is interesting to estimate quantum coherence
from above. Let us turn to another quantity called the _coherence of
assistance_ , induced by any coherence measure $C$, which is defined in the
form of concave bottom extension,

$\displaystyle C_{a}(\rho)=\max\sum_{k}p_{k}C(|\psi_{k}\rangle),$ (33)

where the maximization is taken over all pure state decompositions of
$\rho=\sum_{k}p_{k}|\psi_{k}\rangle\\!\langle\psi_{k}|$. Coherence of
assistance quantifies the coherence that can be extracted assisted by another
party under local measurements and classical communication [31]. Suppose Alice
holds a state $\rho^{A}=\rho=\sum_{k}p_{k}|\psi_{k}\rangle\\!\langle\psi_{k}|$
with coherence $C(\rho)$. Bob holds another part of the purified state of
$\rho$. The joint state between Alice and Bob is
$\sum_{k}p_{k}|\psi_{k}\rangle_{A}\otimes|k\rangle_{B}$. Bob performs local
projective measurements $\left\\{|k\rangle\\!\langle k|\right\\}$ along the
given basis and informs Alice the measurement outcomes by classical
communication. Alice’s system will be in a pure state ensemble
$\left\\{p_{k},|\psi_{k}\rangle\\!\langle\psi_{k}|\right\\}$ with average
coherence $\sum_{k}p_{k}C(|\psi_{k}\rangle\\!\langle\psi_{k}|)$. The process
is called assisted coherence distillation. The maximum average coherence is
called the coherence of assistance which quantifies the one-way coherence
distillation rate [31].

The coherence of assistance of any quantum state $\rho$ can be estimated by
the output state $\rho_{\star}=\Delta(\rho)$.

######  Theorem 3.

For any coherence measure $C$ and any quantum state
$\rho\in\mathrm{D}\left(\mathbb{C}^{d}\right)$, the coherence of assistance of
$\rho$ is bounded from above by the coherence of assistance of
$\rho_{\star}=\Delta(\rho)$,

$\displaystyle C_{a}(\rho)\leqslant C_{a}(\rho_{\star}).$ (34)

###### Proof.

Note that

$\displaystyle C_{a}(\rho_{\star})$ $\displaystyle=$ $\displaystyle
C_{a}(\Delta(\rho))$ $\displaystyle=$ $\displaystyle
C_{a}\left(\frac{1}{d!}\sum_{\pi\in
S_{d}}\boldsymbol{P}_{\pi}\rho\boldsymbol{P}^{\dagger}_{\pi}\right)$
$\displaystyle\geqslant$ $\displaystyle\frac{1}{d!}\sum_{\pi\in
S_{d}}C_{a}(\boldsymbol{P}_{\pi}\rho\boldsymbol{P}^{\dagger}_{\pi})$
$\displaystyle=$ $\displaystyle C_{a}(\rho).$

Here the inequality is the concavity of the coherence of assistance. The last
equality is due to the equality
$C_{a}(\boldsymbol{P}_{\pi}\rho\boldsymbol{P}^{\dagger}_{\pi})=C_{a}(\rho)$
for any permutation $\pi\in S_{d}$. ∎

##  3 Mixed-permutation channel on bipartite systems

In this section we study the mixed-permutation channel on bipartite systems
$\mathbb{C}^{d_{A}}\otimes\mathbb{C}^{d_{B}}$. Without loss of generality, we
suppose $d_{A}\leqslant d_{B}$. Suppose that $\\{|i\rangle\\}^{d_{A}}_{i=1}$
and $\\{|j\rangle\\}^{d_{B}}_{j=1}$ are rothonormal bases of the subsystems
$\mathbb{C}^{d_{A}}$ and $\mathbb{C}^{d_{B}}$, respectively, we denote by

$\mathrm{D}\left(\mathbb{C}^{d_{A}}\otimes\mathbb{C}^{d_{B}}\right)=\left\\{\rho_{AB}=\sum^{d_{A}}_{i,i^{\prime}=1}\sum^{d_{B}}_{j,j^{\prime}=1}\rho_{i^{\prime}j^{\prime},ij}|i^{\prime}j^{\prime}\rangle\langle
ij|:\rho_{AB}\geqslant 0,\operatorname{Tr}\left(\rho_{AB}\right)=1\right\\}$

the set of density operators acting on
$\mathbb{C}^{d_{A}}\otimes\mathbb{C}^{d_{B}}$. The identical channel on
operator spaces are denoted by $\mathrm{id}$. By Theorem 1, we derive the
explicit form of the mixed-permutation channel on bipartite systems
$\mathbb{C}^{d_{A}}\otimes\mathbb{C}^{d_{B}}$.

######  Theorem 4.

For any bipartite operator $\boldsymbol{X}_{AB}$ acting on
$\mathbb{C}^{d_{A}}\otimes\mathbb{C}^{d_{B}}$, denote
$\boldsymbol{E}_{A(B)}=|\boldsymbol{e}_{A(B)}\rangle\\!\langle\boldsymbol{e}_{A(B)}|$
where $\boldsymbol{e}_{A(B)}$ is a $d_{A(B)}$-dimensional column vector with
all entries being one, we have

$\displaystyle(\Delta_{A}\otimes\mathrm{id}_{B})(\boldsymbol{X}_{AB})=\frac{\mathbb{1}_{A}}{d_{A}}\otimes\operatorname{Tr}_{A}\left(\boldsymbol{X}_{AB}\right)+\frac{\boldsymbol{E}_{A}-\mathbb{1}_{A}}{d_{A}(d_{A}-1)}\otimes\operatorname{Tr}_{A}\left(\boldsymbol{X}_{AB}(\boldsymbol{E}_{A}-\mathbb{1}_{A})\otimes\mathbb{1}_{B}\right)$
(35)

and thus

$\displaystyle(\Delta_{A}\otimes\Delta_{B})(\boldsymbol{X}_{AB})$
$\displaystyle=$ $\displaystyle
c_{0}(\boldsymbol{X}_{AB})\mathbb{1}_{AB}+c_{1}(\boldsymbol{X}_{AB})\mathbb{1}_{A}\otimes(\boldsymbol{E}_{B}-\mathbb{1}_{B})+c_{2}(\boldsymbol{X}_{AB})(\boldsymbol{E}_{A}-\mathbb{1}_{A})\otimes\mathbb{1}_{B}$
(36)
$\displaystyle+c_{3}(\boldsymbol{X}_{AB})(\boldsymbol{E}_{A}-\mathbb{1}_{A})\otimes(\boldsymbol{E}_{B}-\mathbb{1}_{B}).$

where

$\displaystyle\begin{cases}c_{0}(\boldsymbol{X}_{AB})=\frac{1}{d_{A}d_{B}}\operatorname{Tr}\left(\boldsymbol{X}_{AB}\right),\\\
c_{1}(\boldsymbol{X}_{AB})=\frac{1}{d_{A}d_{B}(d_{B}-1)}\operatorname{Tr}\left(\boldsymbol{X}_{AB}(\boldsymbol{E}_{A}-\mathbb{1}_{A})\otimes\mathbb{1}_{B}\right),\\\
c_{2}(\boldsymbol{X}_{AB})=\frac{1}{d_{A}d_{B}(d_{A}-1)}\operatorname{Tr}\left(\boldsymbol{X}_{AB}\mathbb{1}_{A}\otimes(\boldsymbol{E}_{B}-\mathbb{1}_{B})\right),\\\
c_{3}(\boldsymbol{X}_{AB})=\frac{1}{d_{A}d_{B}(d_{A}-1)(d_{B}-1)}\operatorname{Tr}\left(\boldsymbol{X}_{AB}(\boldsymbol{E}_{A}-\mathbb{1}_{A})\otimes(\boldsymbol{E}_{B}-\mathbb{1}_{B})\right).\end{cases}$
(37)

###### Proof.

The result can be also established by linearity once we show it holds for any
Hermitian matrix $\boldsymbol{X}_{AB}$. So it suffices to consider the
Hermitian case.

  1. (i)

In fact, for $\boldsymbol{X}_{AB}=|\Psi_{AB}\rangle\\!\langle\Psi_{AB}|$ where
$|\Psi_{AB}\rangle\in\mathbb{C}^{d_{A}}\otimes\mathbb{C}^{d_{B}}$, we suppose
its Schmidt decomposition is

$|\Psi_{AB}\rangle=\sum_{i=1}^{d_{A}}\sqrt{\lambda_{i}}|\boldsymbol{a}_{i}\rangle_{A}\otimes|\boldsymbol{b}_{i}\rangle_{B},$

with $\\{|\boldsymbol{a}_{i}\rangle_{A}\\}_{i=1}^{d_{A}}$ and
$\\{|\boldsymbol{b}_{j}\rangle_{B}\\}_{j=1}^{d_{B}}$ the orthnormal bases of
the subsystems $\mathbb{C}^{d_{A}}$ and $\mathbb{C}^{d_{B}}$ respectively,
then the density operator of $|\Psi_{AB}\rangle\\!\langle\Psi_{AB}|$ can be
expressed as

$|\Psi_{AB}\rangle\\!\langle\Psi_{AB}|=\sum_{i,j=1}^{d_{A}}\sqrt{\lambda_{i}\lambda_{j}}|\boldsymbol{a}_{i}\rangle\langle\boldsymbol{a}_{j}|_{A}\otimes|\boldsymbol{b}_{i}\rangle\langle\boldsymbol{b}_{j}|_{B}.$

In the following reasoning, we omit the subindexes $A$ and $B$ when no
confusing arises. The action of the mixed-permutation channel on the first
subsystem is then

$\displaystyle(\Delta_{A}\otimes\mathrm{id}_{B})(|\Psi_{AB}\rangle\\!\langle\Psi_{AB}|)$
$\displaystyle=$
$\displaystyle\sum_{i,j=1}^{d_{A}}\sqrt{\lambda_{i}\lambda_{j}}\Delta_{A}(|\boldsymbol{a}_{i}\rangle\langle\boldsymbol{a}_{j}|)\otimes|\boldsymbol{b}_{i}\rangle\langle\boldsymbol{b}_{j}|$
$\displaystyle=$
$\displaystyle\sum_{i,j=1}^{d_{A}}\sqrt{\lambda_{i}\lambda_{j}}\left[\operatorname{Tr}\left(|\boldsymbol{a}_{i}\rangle\langle\boldsymbol{a}_{j}|\right)\frac{\mathbb{1}_{d_{A}}}{d_{A}}+\operatorname{Tr}\left(|\boldsymbol{a}_{i}\rangle\langle\boldsymbol{a}_{j}|(\boldsymbol{E}_{A}-\mathbb{1}_{d_{A}})\right)\frac{(\boldsymbol{E}_{A}-\mathbb{1}_{d_{A}})}{d_{A}(d_{A}-1)}\right]\otimes|\boldsymbol{b}_{i}\rangle\langle\boldsymbol{b}_{j}|$
$\displaystyle=$
$\displaystyle\frac{\mathbb{1}_{d_{A}}}{d_{A}}\otimes\left(\sum_{i,j=1}^{d_{A}}\sqrt{\lambda_{i}\lambda_{j}}\operatorname{Tr}\left(|\boldsymbol{a}_{i}\rangle\langle\boldsymbol{a}_{j}|\right)|\boldsymbol{b}_{i}\rangle\langle\boldsymbol{b}_{j}|\right)$
$\displaystyle+\frac{(\boldsymbol{E}_{A}-\mathbb{1}_{d_{A}})}{d_{A}(d_{A}-1)}\otimes\left(\sum_{i,j=1}^{d_{A}}\sqrt{\lambda_{i}\lambda_{j}}\operatorname{Tr}\left(|\boldsymbol{a}_{i}\rangle\langle\boldsymbol{a}_{j}|(\boldsymbol{E}_{A}-\mathbb{1}_{d_{A}})\right)|\boldsymbol{b}_{i}\rangle\langle\boldsymbol{b}_{j}|\right)$
$\displaystyle=$
$\displaystyle\frac{\mathbb{1}_{d_{A}}}{d_{A}}\otimes\operatorname{Tr}_{A}(|\Psi_{AB}\rangle\\!\langle\Psi_{AB}|)+\frac{(\boldsymbol{E}_{A}-\mathbb{1}_{d_{A}})}{d_{A}(d_{A}-1)}\otimes\operatorname{Tr}_{A}(|\Psi_{AB}\rangle\\!\langle\Psi_{AB}|(\boldsymbol{E}_{A}-\mathbb{1}_{d_{A}})\otimes\mathbb{1}_{d_{B}}).$

By the linearity of both sides, this equality holds for any positive semi-
definite bipartite operator, and thus for a general bipartite operator.

  2. (ii)

The proof goes similarly.

We have done the proof. ∎

If we focus on the bipartite quantum states, then we get an output state of
the one-sided and two-sided mixed-permutation channels as follows.

######  Corollary 3.

For any quantum state
$\rho_{AB}\in\mathrm{D}\left(\mathbb{C}^{d_{A}}\otimes\mathbb{C}^{d_{B}}\right)$,
if the first subsystem goes through the mixed-permutation channel, then the
output state is

$\displaystyle(\Delta_{A}\otimes\mathrm{id}_{B})(\rho_{AB})=\frac{\mathbb{1}_{d_{A}}}{d_{A}}\otimes\rho_{B}+\frac{d_{A}|\Phi_{d_{A}}\rangle\\!\langle\Phi_{d_{A}}|-\mathbb{1}_{d_{A}}}{d_{A}(d_{A}-1)}\otimes\operatorname{Tr}_{A}\left(\rho_{AB}(d_{A}|\Phi_{d_{A}}\rangle\\!\langle\Phi_{d_{A}}|-\mathbb{1}_{d_{A}})\otimes\mathbb{1}_{d_{B}}\right),$
(38)

where $\rho_{B}=\operatorname{Tr}_{A}(\rho_{AB})$. If both subsystems go
through the separate mixed-permutation channels, the output state is

$\displaystyle(\Delta_{A}\otimes\Delta_{B})(\rho_{AB})$ $\displaystyle=$
$\displaystyle\frac{\mathbb{1}_{d_{A}}}{d_{A}}\otimes\frac{\mathbb{1}_{d_{B}}}{d_{B}}+\gamma_{1}(\rho_{AB})\frac{\mathbb{1}_{d_{A}}}{d_{A}}\otimes\frac{d_{B}|\Phi_{d_{B}}\rangle\\!\langle\Phi_{d_{B}}|-\mathbb{1}_{d_{B}}}{d_{B}(d_{B}-1)}$
$\displaystyle+\gamma_{2}(\rho_{AB})\frac{d_{A}|\Phi_{d_{A}}\rangle\\!\langle\Phi_{d_{A}}|-\mathbb{1}_{d_{A}}}{d_{A}(d_{A}-1)}\otimes\frac{\mathbb{1}_{B}}{d_{B}}$
$\displaystyle+\gamma_{3}(\rho_{AB})\frac{d_{A}|\Phi_{d_{A}}\rangle\\!\langle\Phi_{d_{A}}|-\mathbb{1}_{A}}{d_{A}(d_{A}-1)}\otimes\frac{d_{B}|\Phi_{d_{B}}\rangle\\!\langle\Phi_{d_{B}}|-\mathbb{1}_{d_{B}}}{d_{B}(d_{B}-1)}.$

Here
$|\Phi_{d_{A}}\rangle=\frac{1}{\sqrt{d_{A}}}\sum_{i=1}^{d_{A}}|\boldsymbol{a}_{i}\rangle$
and
$|\Phi_{d_{B}}\rangle=\frac{1}{\sqrt{d_{B}}}\sum_{j=1}^{d_{B}}|\boldsymbol{b}_{j}\rangle$.
Moreover,

$\displaystyle\gamma_{1}(\rho_{AB})$ $\displaystyle:=$
$\displaystyle\operatorname{Tr}\left(\rho_{AB}(d_{A}|\Phi_{d_{A}}\rangle\\!\langle\Phi_{d_{A}}|-\mathbb{1}_{d_{A}})\otimes\mathbb{1}_{d_{B}}\right),$
$\displaystyle\gamma_{2}(\rho_{AB})$ $\displaystyle:=$
$\displaystyle\operatorname{Tr}\left(\rho_{AB}\mathbb{1}_{d_{A}}\otimes(d_{B}|\Phi_{d_{B}}\rangle\\!\langle\Phi_{d_{B}}|-\mathbb{1}_{d_{B}})\right),$
$\displaystyle\gamma_{3}(\rho_{AB})$ $\displaystyle:=$
$\displaystyle\operatorname{Tr}\left(\rho_{AB}(d_{A}|\Phi_{d_{A}}\rangle\\!\langle\Phi_{d_{A}}|-\mathbb{1}_{d_{A}})\otimes(d_{B}|\Phi_{d_{B}}\rangle\\!\langle\Phi_{d_{B}}|-\mathbb{1}_{d_{B}})\right).$

By the form of the output states
$(\Delta_{A}\otimes\mathrm{id}_{B})(\rho_{AB})$ and
$(\Delta_{A}\otimes\Delta_{B})(\rho_{AB})$ in the above Corollary 3, we get
both $(\Delta_{A}\otimes\mathrm{id}_{B})(\rho_{AB})$ and
$(\Delta_{A}\otimes\Delta_{B})(\rho_{AB})$ are positive under partial
transpositions (PPT) for all states $\rho_{AB}$. Since PPT states in qubit-
qubit systems and qubit-qutrit systems are all separable, so the action of the
local mixed-permutation channel(s) on two-qubit systems will completely erase
the entanglement between two subsystems.

######  Example 2.

For any quantum state $\rho_{AB}$ in
$\mathrm{D}\left(\mathbb{C}^{2}\otimes\mathbb{C}^{2}\right)$,

$\rho_{AB}=\left(\begin{array}[]{cccc}\rho_{11}&\rho_{12}&\rho_{13}&\rho_{14}\\\
\rho_{21}&\rho_{22}&\rho_{23}&\rho_{24}\\\
\rho_{31}&\rho_{32}&\rho_{33}&\rho_{34}\\\
\rho_{41}&\rho_{42}&\rho_{43}&\rho_{44}\end{array}\right),$

if the first qubit goes through the mixed-permutation channel, then the output
state is

$\displaystyle(\rho_{AB})_{\star,I}=(\Delta_{A}\otimes\mathrm{id}_{B})(\rho_{AB})=\frac{\mathbb{1}_{d_{A}}}{d_{A}}\otimes\rho_{B}+\frac{\boldsymbol{E}_{A}-\mathbb{1}_{d_{A}}}{d_{A}(d_{A}-1)}\otimes\operatorname{Tr}_{A}\left(\rho_{AB}(\boldsymbol{E}_{A}-\mathbb{1}_{d_{A}})\otimes\mathbb{1}_{d_{B}}\right),$
(39)

with

$\rho_{B}=\operatorname{Tr}_{A}(\rho_{AB})=\left(\begin{array}[]{cc}\rho_{11}+\rho_{22}&\rho_{13}+\rho_{24}\\\
\rho_{31}+\rho_{42}&\rho_{33}+\rho_{44}\end{array}\right).$

More explicitly,

$\displaystyle(\rho_{AB})_{\star,I}=\left(\begin{array}[]{cccc}\rho_{11}+\rho_{22}&\rho_{12}+\rho_{21}&\rho_{13}+\rho_{24}&\rho_{14}+\rho_{23}\\\
\rho_{12}+\rho_{21}&\rho_{11}+\rho_{22}&\rho_{14}+\rho_{23}&\rho_{13}+\rho_{24}\\\
\rho_{31}+\rho_{42}&\rho_{32}+\rho_{41}&\rho_{33}+\rho_{44}&\rho_{34}+\rho_{43}\\\
\rho_{32}+\rho_{41}&\rho_{31}+\rho_{42}&\rho_{34}+\rho_{43}&\rho_{33}+\rho_{44}\\\
\end{array}\right)$ (44)

with
$(\rho_{AB})^{{\scriptscriptstyle\mathsf{T}}_{A}}_{\star,I}=(\rho_{AB})_{\star,I}$.
Furthermore, if both two qubits go through the mixed-permutation channel, then
the output state is

$\displaystyle(\rho_{AB})_{\star,\star}$ $\displaystyle=$
$\displaystyle(\Delta_{A}\otimes\Delta_{B})(\rho_{AB})$ $\displaystyle=$
$\displaystyle
c_{0}(\rho)\mathbb{1}_{d_{A}d_{B}}+c_{1}(\rho)\mathbb{1}_{d_{A}}\otimes(\boldsymbol{E}_{B}-\mathbb{1}_{d_{B}})$
$\displaystyle+c_{2}(\rho)(\boldsymbol{E}_{A}-\mathbb{1}_{d_{A}})\otimes\mathbb{1}_{d_{B}}+c_{3}(\rho)(\boldsymbol{E}_{A}-\mathbb{1}_{d_{A}})\otimes(\boldsymbol{E}_{B}-\mathbb{1}_{d_{B}})$
$\displaystyle=$
$\displaystyle\left(\begin{array}[]{cccc}c_{0}(\rho_{AB})&c_{1}(\rho_{AB})&c_{2}(\rho_{AB})&c_{3}(\rho_{AB})\\\
c_{1}(\rho_{AB})&c_{0}(\rho_{AB})&c_{3}(\rho_{AB})&c_{2}(\rho_{AB})\\\
c_{2}(\rho_{AB})&c_{3}(\rho_{AB})&c_{0}(\rho_{AB})&c_{1}(\rho_{AB})\\\
c_{3}(\rho_{AB})&c_{2}(\rho_{AB})&c_{1}(\rho_{AB})&c_{0}(\rho_{AB})\\\
\end{array}\right)=(\rho_{AB})^{{\scriptscriptstyle\mathsf{T}}_{A}}_{\star,\star}\geqslant
0,$

where

$\displaystyle c_{0}(\rho_{AB})$ $\displaystyle=$ $\displaystyle\frac{1}{4},$
$\displaystyle c_{1}(\rho_{AB})$ $\displaystyle=$
$\displaystyle\frac{1}{4}\operatorname{Tr}\left(\rho_{A}(\boldsymbol{E}_{B}-\mathbb{1}_{d_{B}})\right),$
$\displaystyle c_{2}(\rho_{AB})$ $\displaystyle=$
$\displaystyle\frac{1}{4}\operatorname{Tr}\left(\rho_{B}(\boldsymbol{E}_{A}-\mathbb{1}_{d_{A}})\right),$
$\displaystyle c_{3}(\rho_{AB})$ $\displaystyle=$
$\displaystyle\frac{1}{4}\operatorname{Tr}\left(\rho_{AB}(\boldsymbol{E}_{A}-\mathbb{1}_{d_{A}})\otimes(\boldsymbol{E}_{B}-\mathbb{1}_{d_{B}})\right),$

implying that $(\rho_{AB})_{\star,\star}$ must be separable. Moreover, the
eigenvalues are given by

$\displaystyle
c_{0}(\rho_{AB})+c_{1}(\rho_{AB})\pm(c_{2}(\rho_{AB})+c_{3}(\rho_{AB})),\quad
c_{0}(\rho_{AB})-c_{1}(\rho_{AB})\pm(c_{2}(\rho_{AB})-c_{3}(\rho_{AB})).$

From the above, we see that $(\rho_{AB})_{\star,I}$ and
$(\rho_{AB})_{\star,\star}$ are PPT states. Thus by Peres-Horodecki criterion
[32, 33], the output states $(\rho_{AB})_{\star,I}$ and
$(\rho_{AB})_{\star,\star}$ are separable ones for all two-qubit states
$\rho_{AB}$.

In what follows, let us recall the notion of the entanglement-breaking channel
[34]. The so-called entanglement-breaking channel
$\mathcal{T}\equiv\mathcal{T}_{A}$ means that
$(\mathcal{T}_{A}\otimes\mathrm{id}_{B})(\rho_{AB})$ is separable for every
bipartite state
$\rho_{AB}\in\mathrm{D}\left(\mathbb{C}^{d_{A}}\otimes\mathbb{C}^{d_{B}}\right)$.
As already known in [35], in qubit systems, any quantum channel is
entanglement breaking if and only if its Choi representation is separable.
Because $(\Delta_{A}\otimes\mathrm{id}_{B})(\rho_{AB})$ is separable for any
two-qubit state $\rho_{AB}$, a fortiori for maximally entangled two-qubit
state. Thus $\Delta$ is entanglement breaking in qubit systems. In fact, the
mixed-permutation channel $\Delta$ is entanglement-breaking channel in qudit
systems, which can be summarized into the following result:

######  Corollary 4.

The mixed-permutation channel $\Delta$ characterized in Theorem 1 is
entanglement-breaking channel in qudit systems.

###### Proof.

According to [34, Theorem 4, (B)$\Longleftrightarrow$(C)], it suffices to show
that the Choi-representation of the mixed-permutation channel $\Delta$,
$J(\Delta):=(\Delta\otimes\mathrm{id})(|\Omega\rangle\\!\langle\Omega|)$, is
separable, where $|\Omega\rangle:=\frac{1}{\sqrt{d}}\sum^{d}_{i=1}|ii\rangle$.
Indeed, using Eq. (38),

$\displaystyle J(\Delta)$ $\displaystyle=$
$\displaystyle\frac{1}{d^{2}}\mathbb{1}_{d}\otimes\mathbb{1}_{d}+\frac{1}{d^{2}(d-1)}(d|\Phi_{d}\rangle\\!\langle\Phi_{d}|-\mathbb{1}_{d})^{\otimes
2}$ $\displaystyle=$
$\displaystyle\frac{1}{d}|\Phi_{d}\rangle\\!\langle\Phi_{d}|\otimes|\Phi_{d}\rangle\\!\langle\Phi_{d}|+\left(1-\frac{1}{d}\right)\left(\frac{\mathbb{1}_{d}-|\Phi_{d}\rangle\\!\langle\Phi_{d}|}{d-1}\right)\otimes\left(\frac{\mathbb{1}_{d}-|\Phi_{d}\rangle\\!\langle\Phi_{d}|}{d-1}\right),$

a separable state. Here
$|\Phi_{d}\rangle=\frac{1}{\sqrt{d}}\sum^{d}_{i=1}|i\rangle$. This indicates
that $\Delta$ is entanglement-breaking. ∎

######  Example 3 (The family of Bell-diagonal states).

The Bell-diagonal states [36] in two-qubit system can be written as

$\displaystyle\rho_{\text{Bell}}=\frac{1}{4}(\mathbb{1}_{4}+\sum_{i=1}^{3}t_{i}\sigma_{i}\otimes\sigma_{i})$
(46)

with $\sigma_{i}$ three Pauli operators in Eq. (20). So a Bell-diagonal state
is specified by three real variables $t_{1},t_{2}$, and $t_{3}$ such that

$\displaystyle\begin{cases}1-t_{1}-t_{2}-t_{3}\geqslant 0,\\\
1-t_{1}+t_{2}+t_{3}\geqslant 0,\\\ 1+t_{1}-t_{2}+t_{3}\geqslant 0,\\\
1+t_{1}+t_{2}-t_{3}\geqslant 0.\end{cases}$

Denote the set of all above such tuples $(t_{1},t_{2},t_{3})$ by
$D_{\text{Bell}}$. Because the four eigenvalues of $\rho$ are in $[0,1]$, we
see that $t_{i}\in[-1,1](i=1,2,3)$. That is,
$D_{\text{Bell}}\subset[-1,1]^{3}$. The Bell-diagonal states can be
geometrically described by a tetrahedron. One can show that a Bell-diagonal
state is separable if and only if
$\left\lvert\mspace{1.0mu}t_{1}\mspace{1.0mu}\right\rvert+\left\lvert\mspace{1.0mu}t_{2}\mspace{1.0mu}\right\rvert+\left\lvert\mspace{1.0mu}t_{3}\mspace{1.0mu}\right\rvert\leqslant
1$ holds. Geometrically, the set of Bell-diagonal states is a tetrahedron and
the set of separable Bell-diagonal states is an octahedron [36], which is
denoted by $D_{\text{Bellsep}}$. That is,
$D_{\text{Bellsep}}=\\{(t_{1},t_{2},t_{3})\in
D_{\text{Bell}}:\sum^{3}_{i=1}\left\lvert\mspace{1.0mu}t_{i}\mspace{1.0mu}\right\rvert\leqslant
1\\}$. We find that

$\displaystyle\Delta\otimes\mathrm{id}(\rho_{\mathrm{Bell}})=\frac{1}{4}(\mathbb{1}_{4}+t_{1}\sigma_{1}\otimes\sigma_{1}).$
(47)

We make a plot in arguments $(t_{1},t_{2},t_{3})$ about the image states in
(47) of the one-sided action of the mixed-permutation channel on the family
(46) of Bell diagonal states in the following Figure 3.

Figure 3: (Color Online) The yellow tetrahedron stands for the family of Bell-
diagonal states in Eq. (46); the orange octahedron stands for separable Bell-
diagonal states; the red line segment stands for image states
$\Delta\otimes\mathrm{id}(\rho_{\mathrm{Bell}})$ in Eq. (47) of all diagonal
Bell-diagonal states through the one-sided mixed-permutation channel.

##  4 Conclusions and discussions

In summary, we studied one kind of special channels, i.e., the mixed-
permutation channels. The properties of this channel was characterized. The
mixed-permutation channel can be used in bounding quantum coherence with
respect to all coherence measures. The action of the mixed-permutation channel
on bipartite systems is also discussed and can be generalized to multipartite
systems. Beyond that, there are some interesting problems which can be further
considered in the future:

  * •

The first question is what is the average collective action of the
permutations on the bipartite quantum states,

$\displaystyle\frac{1}{d!}\sum_{\pi\in
S_{d}}(\boldsymbol{P}_{\pi}\otimes\boldsymbol{P}_{\pi})\boldsymbol{X}(\boldsymbol{P}_{\pi}\otimes\boldsymbol{P}_{\pi})^{\dagger}=?.$

We are expecting a closed-form formula of the above expression. With this
potential formula, we can extend symmetrized variance for pure states in [28]
to mixed states.

  * •

The second question is what is the correlation quantifiers [37] or coherence
measure [38] induced by the mixed-permutation channel.

In light of the special properties and the closed form of the output state of
the mixed-permutation channel, we expect the corresponding correlation measure
and coherence measure could characterize the quantum features of quantum
states from a permutation-invariant way.

#### Acknowledgments

This research is supported by Zhejiang Provincial Natural Science Foundation
of China under Grant No. LZ23A010005 and by NSFC under Grant Nos.11971140 and
12171044.

#### Data Availability Statement

No Data associated in the manuscript.

## References

  * [1] R. Takagi, K. Wang, and M. Hayashi, Application of the resource theory of channels to communication scenarios, Phys. Rev. Lett. 124, 120502 (2020).
  * [2] D.P. Divincenzo, Quantum computation, Science 270(5234), 255-261(1995).
  * [3] A. Ekert and R. Jozsa, Quantum computation and Shor’s factoring algorithm, Rev. Math. Phys. 68, 733 (1996).
  * [4] N. Gisin, G. Ribordy, W. Tittel, and H. Zbinden, Quantum cryptography, Rev. Math. Phys. 74, 145 (2002).
  * [5] S. Pirandola et al., Advances in quantum cryptography, Advances in Optics and Photonics 12(4), 1012-1236 (2020).
  * [6] E. Chitambar and G. Gour, Quantum resource theories, Rev. Math. Phys. 91, 025001 (2019).
  * [7] X. Wang and M.M. Wilde, Resource theory of asymmetric distinguishability for quantum channels, Phys. Rev. Research 1, 033169 (2019).
  * [8] X. Wang, M.M. Wilde, and Y. Su, Quantifying the magic of quantum channels, New. J. Phys. 21, 103002 (2019).
  * [9] F.H. Kamin, F.T. Tabesh, S. Salimi, and F. Kheirandish, The resource theory of coherence for quantum channels, Quant Inf Process 19, 210 (2020).
  * [10] Y. Liu and X. Yuan, Operational resource theory of quantum channels, Phys. Rev. Research 2, 012035(R) (2020).
  * [11] H. Zhou, T. Gao, F. Yan, Entanglement resource theory of quantum channel, Phys. Rev. Research 4, 013200 (2022).
  * [12] R. Takagi, Operational Quantum Resource Theories: Unified Framework and Applications, PhD Thesis (2020).
  * [13] S. Fu, Y. Sun, S. Luo, Skew information-based uncertainty relations for quantum channels, Quantum Inf. Process. 18, 258 (2019).
  * [14] N. Zhou, M. J. Zhao, Z. Wang, T. Li, The uncertainty relation for quantum channels based on skew information, Quantum Inf. Process. 22, 6 (2023).
  * [15] Q-H. Zhang, J.F. Wu, S-M. Fei, A note on uncertainty relations of arbitrary N quantum channels, Laser Phys. Lett. 18, 095204(2021).
  * [16] T.R. Bromley, M. Cianciaruso, and G. Adesso, Frozen quantum coherence, Phys. Rev. Lett. 114, 210401 (2015).
  * [17] X.D. Yu, D.J. Zhang, C.L. Liu, and D.M. Tong, Measure-independent freezing of quantum coherence, Phys. Rev. A 93, 060303 (2016).
  * [18] M.L. Hu, and H. Fan, Evolution equation for quantum coherence, Sci. Rep. 6, 29260 (2016).
  * [19] A. Streltsov, G. Adesso, and M. B. Plenio, Colloquium: Quantum coherence as a resource, Rev. Math. Phys. 89, 041003 (2017).
  * [20] E. Chitambar and G. Gour, Quantum resource theories, Rev. Math. Phys. 91, 025001 (2019).
  * [21] M.L. Hu, X. Hu, J. Wang, Y. Peng, Y.R. Zhang, H. Fan, Quantum coherence and geometric quantum discord, Phys. Rep. 762-764, 1-100 (2018).
  * [22] T. Baumgratz, M. Cramer, and M. B. Plenio, Quantifying coherence, Phys. Rev. Lett. 113, 140401 (2014).
  * [23] X. Qi, T. Gao, and F.L. Yan, Measuring coherence with entanglement concurrence, J. Phys. A 50, 285301 (2017).
  * [24] A. Streltsov, U. Singh, H. S. Dhar, M. N. Bera, and G. Adesso, Measuring Quantum Coherence with Entanglement, Phys. Rev. Lett. 115, 020403 (2015).
  * [25] C. Napoli, T. R. Bromley, M. Cianciaruso, M. Piani, N. Johnston, and G. Adesso, Robustness of coherence: an operational and observable measure of quantum coherence, Phys. Rev. Lett. 116, 150502 (2016).
  * [26] J. I de Vicente and A. Streltsov, Genuine quantum coherence, J. Phys. A: Math. Theor. 50, 045301 (2017).
  * [27] A.S. Holevo, Additive conjecture and covariant channels, Int. J. Quantum Information, 3(1),41-47 (2005).
  * [28] M. J. Zhao, L. Zhang, and S. M. Fei, Standard symmetrized variance with applications to coherence, uncertainty, and entanglement, Phys. Rev. A 106, 012417 (2022).
  * [29] Y. Peng, Y. Jiang, and H. Fan, Maximally coherent states and coherence-preserving operations, Phys. Rev. A 93, 032326 (2016).
  * [30] U. Singh, M. N. Bera, H. S. Dhar, and A. K. Pati, Maximally coherent mixed states: Complementarity between maximal coherence and mixedness, Phys. Rev. A 91, 052115 (2015).
  * [31] E. Chitambar, A. Streltsov, S. Rana, M.N. Bera, G. Adesso, and M. Lewenstein, Assisted Distillation of Quantum Coherence, Phys. Rev. Lett. 116, 070402 (2016).
  * [32] A. Peres, Separability Criterion for Density Matrices, Phys. Rev. Lett. 77, 1413 (1996).
  * [33] M. Horodecki and P. Horodecki, Reduction criterion of separability and limits for a class of distillation protocols, Phys. Rev. A 59, 4206 (1999).
  * [34] M. Horodecki, P.W. Shor, and M.B. Ruskai, Entanglement breaking channels, Rev. Math. Phys. 15(6), 629-641 (2003).
  * [35] M.B. Ruskai, Qubit entanglement breaking channels, Rev. Math. Phys. 15(6), 643(2003).
  * [36] R. Horodecki, M. Horodecki, Information-theoretic aspects of inseparability of mixed states, Phys. Rev. A 54(3), 1838 (1996).
  * [37] N. Li, S. Luo, Y. Sun, Quantifying correlations via local channels, Phys. Rev. A 105, 032436 (2022).
  * [38] Y. Sun, N. Li, S. Luo, Quantifying coherence relative to channels via metric-adjusted skew information, Phys. Rev. A  106, 012436 (2022).