---
abstract: |
    In this paper, two conjectures which were proposed in \[Phys. Rev. A **82**, 052122(2010)\] on the correlations in a bipartite state $\rho^{AB}$ are studied. If the mutual information $I{\left(\rho^{AB}\right)}$ between two quantum systems $A$ and $B$ before any measurement is considered as the total amount of correlations in the state $\rho^{AB}$, then it can be separated into two parts: classical correlations and quantum correlations. The so-called classical correlations $C{\left(\rho^{AB}\right)}$ in the state $\rho^{AB}$, defined by the maximizing mutual information between two quantum systems $A$ and $B$ after von Neumann measurements on system $B$, we show that it is upper bounded by the von Neumann entropies of both subsystems $A$ and $B$, this answered the conjecture on the classical correlation. If the quantum correlations $Q{\left(\rho^{AB}\right)}$ in the state $\rho^{AB}$ is defined by $Q{\left(\rho^{AB}\right)}= I{\left(\rho^{AB}\right)} -
    C{\left(\rho^{AB}\right)}$, we show also that it is upper bounded by the von Neumann entropy of subsystem $B$. It is also obtained that $Q{\left(\rho^{AB}\right)}$ is upper bounded by the von Neumann entropy of subsystem $A$ for a class of states.
author:
- Lin Zhang and Junde Wu
title: On Conjectures of Classical and Quantum Correlations in Bipartite States
---

Introduction
============

In quantum information theory, each realizable physical set-up that processes states of quantum system is described by a quantum operation @Kraus which is mathematically represented by a linear, completely positive super-operator from a set of quantum states to another. The information encoded in a given quantum state is quantified by its von Neumann entropy. In general, the decoherence will be induced in the quantum system when the quantum state is acted by a quantum operation. There are few general and quantitative investigation on the decorrelating capabilities of quantum operations although the decoherent effects of quantum operations are popularly realized.

In order to investigate the decorrelating capabilities of quantum operations, Luo @Luo suggested that the decorrelating capabilities of quantum operations should be separated into classical and quantum parts, and the decoherence involved should be related to the quantum part. By the duality of quantum operations and quantum states, each quantum operation can be identified with a bipartite state via the well-known Choi-Jamio<span>ł</span>kowski isomorphism @Choi. Thus the study of the decorrelating capabilities of quantum operations may be transformed into the investigation of correlations of its corresponding Choi-Jamio<span>ł</span>kowski bipartite states. In view of this, the total correlations in a bipartite state play an essential role in the study of the decorrelating capabilities of quantum operations. In order to get some finer quantitative results, after the total correlation was separated into classical and quantum parts, two related conjectures were proposed in @Luo with some supporting examples. In this paper, the two conjectures are investigated.

Classical and quantum correlations in bipartite states {#sec:correlation}
======================================================

Let $\sH^1$ be a finite dimensional complex Hilbert space. A quantum operation $\Phi$ on $\sH^1$ is a completely positive linear super-operator defined on the set of the quantum states on $\sH^1$. It follows from (@Watrous, Prop. 5.2 and Cor. 5.5) that there exists linear operators $\{M_\mu\}_{\mu=1}^K$ on $\sH^1$ such that $\sum_{\mu=1}^K M^\dagger_\mu M_\mu = {\mathbbm{1}}^1$ and for each quantum state $\rho$ on $\sH^1$, we have the Kraus representation $$\begin{aligned}
\Phi(\rho) = \sum_{\mu=1}^K M_\mu \rho M^\dagger_\mu.\end{aligned}$$ Moreover, let $\sH^2=\mathbb{C}^K$ and ${\{{|\mu\rangle}\}}_{\mu=1}^K$ be the standard orthonormal basis of $\sH^2$. If we define $V: \sH^1
\longrightarrow \sH^1 \ot \sH^2$ by $$\begin{aligned}
 V{|\psi\rangle} &
= & \sum_{\mu=1}^K M_\mu {|\psi\rangle} \ot {|\mu\rangle},\quad {|\psi\rangle}
\in \sH^1,\end{aligned}$$ then $V$ is an isometry and for each quantum state $\rho$ on $\sH^1$, we have the Stinespring representation $$\begin{aligned}
\Phi(\rho) = {\operatorname{Tr}_{2}{\left(V\rho V^\dagger\right)}}.\end{aligned}$$ It is easy to see that $$\begin{aligned}
 V\rho V^\dagger & = &
\sum_{\mu,\nu} M_\mu \rho M^\dagger_\nu \ot {| \mu\rangle\langle \nu |}.\end{aligned}$$ On the other hand, note that for each state $\rho$ on $\sH^1$, ${\operatorname{Tr}_{1}{(V \rho V^\dagger)}}$ is a state on $\sH^2$, thus, the map $$\begin{aligned}
\widehat{\Phi} : \rho \mapsto {\operatorname{Tr}_{1}{\left(V \rho V^\dagger\right)}} =
\sum_{\mu,\nu} {{\operatorname{Tr}_{}{\left(M_\mu \rho M^\dagger_\nu\right)}}} {| \mu\rangle\langle \nu |}\end{aligned}$$ is a quantum operation from quantum system $\sH^1$ to quantum system $\sH^2$, we call it *complementary* to $\Phi$.

If we consider $\sH^2$ to be the environment, then $\widehat{\Phi}(\rho)$ is the state of the environment after the interaction and is called a *correlation matrix*. If the initial state $\rho$ is pure, then the von Neumann entropy $$\S{\left(\widehat{\Phi}(\rho)\right)} = -
{{\operatorname{Tr}_{}{\left(\widehat{\Phi}(\rho)\log_2\widehat{\Phi}(\rho)\right)}}}$$ of $\widehat{\Phi}(\rho)$ describes the entropy exchanged between the system and the environment. Therefore, $\S{\left(\widehat{\Phi}(\rho)\right)}$ is called the *exchange entropy*. The relationship among the $\S(\Phi(\rho)), \S(\rho)$, and $\S{\left(\widehat{\Phi}(\rho)\right)}$ is connected by the well-known Lindblad's entropy inequality @Lindblad: $$\left|\S{\left(\widehat{\Phi}(\rho)\right)} - \S(\rho)\right| \leqslant \S(\Phi(\rho))
\leqslant \S{\left(\widehat{\Phi}(\rho)\right)} + \S(\rho). \eqno (1)$$ It follows from $\sum_{\mu=1}^K M^\dagger_\mu M_\mu = {\mathbbm{1}}^1$ that ${\{ M_\mu\}}_{\mu=1}^K$ describes a measurement which transforms the initial state $\rho$ into one of the output states $$\rho'_\mu = \frac{1}{q_\mu} M_\mu \rho M^\dagger_\mu$$ with probability $q_\mu = {{\operatorname{Tr}_{}{\left(M_\mu \rho M^\dagger_\mu\right)}}}$. Thus, ${\left\{q_\mu,\rho'_\mu\right\}}$ is a quantum ensemble and its *Holevo quantity* is defined by $$\begin{aligned}
\chi{\left({\left\{q_\mu,\rho'_\mu\right\}}\right)} = \S{\left(\sum_\mu q_\mu \rho'_\mu\right)} - \sum_\mu q_\mu \S{\left(\rho'_\mu\right)}.\end{aligned}$$ Let $\H{\left({\{q_\mu\}}\right)} = - \sum_{\mu=1}^Kq_\mu\log_2q_\mu$ be the Shannon entropy of the probability distribution ${\{q_\mu\}}$. Then we have the following inequality @Roga1: $$\chi{\left({\left\{q_\mu,\rho'_\mu\right\}}\right)} \leqslant \S{\left(\widehat{\Phi}(\rho)\right)}
\leqslant \H{\left({\{q_\mu\}}\right)}. \eqno (2)$$

Let $\sH^R$ and $\sH^Q$ be two finite dimensional complex Hilbert spaces. If $\Phi^Q$ is a quantum operation on $\sH^Q$, then ${\mathbbm{1}}^R \ot \Phi^Q$ is a quantum operation on $\sH^R\ot \sH^Q$, moreover, if $\rho^{RQ}$ is a state on $\sH^R\ot \sH^Q$ and $\rho^Q
= {\operatorname{Tr}_{R}{\left(\rho^{RQ}\right)}}$, then we have @Roga2: $$\S{\left(\widehat{{\mathbbm{1}}^R \ot \Phi^Q}{\left(\rho^{RQ}\right)}\right)} =
\S{\left(\widehat{\Phi}^Q{\left(\rho^Q\right)}\right)}. \eqno (3)$$ Let $\sH^A$ and $\sH^B$ be two finite dimensional complex Hilbert spaces, $\rho^{AB}$ is a state on $\sH^A\ot \sH^B$, $\rho^A =
{\operatorname{Tr}_{B}{\left(\rho^{AB}\right)}}$, $\rho^B = {\operatorname{Tr}_{A}{\left(\rho^{AB}\right)}}$. Then the total correlations in $\rho^{AB}$ are usually quantified by the quantum mutual information $$\begin{aligned}
I{\left(\rho^{AB}\right)} = \S{\left(\rho^A\right)} + \S{\left(\rho^B\right)} - \S{\left(\rho^{AB}\right)}.\end{aligned}$$

In @Luo, the author separated the total correlations $I{\left(\rho^{AB}\right)}$ into classical correlations $C{\left(\rho^{AB}\right)}$ and quantum correlations @Ollivier $Q{\left(\rho^{AB}\right)} = I{\left(\rho^{AB}\right)} - C{\left(\rho^{AB}\right)}$, where $C{\left(\rho^{AB}\right)}$ was defined by $$C{\left(\rho^{AB}\right)} = \sup_{\Pi^B}I{\left(\Pi^B{\left(\rho^{AB}\right)}\right)},$$ the sup is taken over all von Neumann measurements $\Pi^B =
{\left\{\Pi^B_j\right\}}$ on $\sH^B$, and $$\begin{aligned}
\Pi^B{\left(\rho^{AB}\right)} = \sum_j {\left({\mathbbm{1}}^A \ot \Pi^B_j\right)}\rho^{AB}{\left({\mathbbm{1}}^A \ot \Pi^B_j\right)}\end{aligned}$$ is the output state after executing the nonselective measurement $\Pi^B = {\left\{\Pi^B_j\right\}}$; ${\mathbbm{1}}^A$ is the identity operator on $\sH^A$.

In @Luo, the following conjectures are proposed : $$\begin{aligned}
C{\left(\rho^{AB}\right)} & \leqslant & \min {\left\{\S{\left(\rho^A\right)}, \S{\left(\rho^B\right)}\right\}},\,\,\,\,\,\,\,\,\,\,\,\, (I)\\
Q{\left(\rho^{AB}\right)} & \leqslant & \min {\left\{\S{\left(\rho^A\right)}, \S{\left(\rho^B\right)}\right\}}.\,\,\,\,\,\,\,\,\,\,\,\, (II)\end{aligned}$$

In this paper, we prove the conjecture (I). Moreover, we show that $Q{\left(\rho^{AB}\right)} \leqslant \S{\left(\rho^B\right)}$ is always valid, and the conjecture (II) is true if $\S{\left(\rho^B\right)} \leqslant \S{\left(\rho^A\right)}$ or $\rho^{AB}$ is separable. It is obtained that $Q{\left(\rho^{AB}\right)}$ is upper bounded by the von Neumann entropy of subsystem $A$ for a class of states. So the conjecture (II) is also true for these states.

The proof of the conjecture {#sec:mainresult}
===========================

Our main results are the following:

\[th:main\] Let $\rho^{AB}$ be a quantum state on $\sH^A \ot \sH^B$. Then we have

(i) $C{\left(\rho^{AB}\right)} \leqslant \min {\left\{\S{\left(\rho^A\right)}, \S{\left(\rho^B\right)}\right\}}$,

(ii) $Q{\left(\rho^{AB}\right)} \leqslant
    \S{\left(\rho^B\right)}$, and $Q(\rho^{AB}) \leqslant \min {\{\S{\left(\rho^A\right)},
    \S{\left(\rho^B\right)}\}}$ whenever $\S{\left(\rho^B\right)} \leqslant \S{\left(\rho^A\right)}$ or $\rho^{AB}$ is separable.

<!-- -->

(i) Let ${\left\{{\left|\psi^B_j\right\rangle}\right\}}_{j=1}^k$ be a orthonormal basis of $\sH^B$ and $\Pi^B_j = {\left|\psi^B_j\right\rangle}{\left\langle\psi^B_j\right|}$. Then ${{\operatorname{Tr}_{}{\left({\mathbbm{1}}^A \ot \Pi^B_j \rho^{AB} {\mathbbm{1}}^A \ot \Pi^B_j\right)}}}
    = {\left\langle\psi^B_j\right|} \rho^B {\left|\psi^B_j\right\rangle}$. If we denote ${\left\langle\psi^B_j\right|} \rho^B {\left|\psi^B_j\right\rangle}$ by $p_j$, then $p_j\geqslant
    0$ and $\sum_jp_j=1$. Without loss of generality, we assume that all $p_j>0$. Now, we define $$\rho^A_j = \frac{
    {\left\langle\psi^B_j\right|} \rho^{AB} {\left|\psi^B_j\right\rangle}}{p_j},$$ then $\rho^A_j$ is a state on $\sH^A$ and $$\Pi^B{\left(\rho^{AB}\right)}  = \sum_j p_j \rho^A_j \ot \Pi^B_j, \quad \Pi^B{\left(\rho^B\right)}  =  \sum_j \Pi^B_j\rho^B\Pi^B_j  =  \sum_j p_j
    \Pi^B_j,\quad
    \rho^A  = \sum_j p_j \rho^A_j.$$ Thus, $$\S{\left(\Pi^B{\left(\rho^{AB}\right)}\right)}  =  \H{\left({\{p_j\}}\right)} + \sum_j p_j \S{\left(\rho^A_j\right)},\quad \S{\left(\Pi^B{\left(\rho^B\right)}\right)}  =  \H{\left({\{p_j\}}\right)},$$ and $$\begin{aligned}
    I{\left(\Pi^B{\left(\rho^{AB}\right)}\right)} & = & \S{\left(\rho^A\right)} + \S{\left(\Pi^B{\left(\rho^B\right)}\right)} - \S{\left(\Pi^B{\left(\rho^{AB}\right)}\right)} \\
     & = & \S{\left(\rho^A\right)} - \sum_j p_j \S{\left(\rho^A_j\right)} = \chi{\left({\left\{p_j,\rho^A_j\right\}}\right)}.\end{aligned}$$ Note that $\sum_j p_j \S{\left(\rho^A_j\right)} \geqslant 0$. Hence $I{\left(\Pi^B{\left(\rho^{AB}\right)}\right)}  \leqslant \S{\left(\rho^A\right)}$. Thus $C{\left(\rho^{AB}\right)} =
    \sup_{\Pi^B}I{\left(\Pi^B{\left(\rho^{AB}\right)}\right)}  \leqslant \S{\left(\rho^A\right)}$. On the other hand, it follows from $C{\left(\rho^{AB}\right)} =
    \sup_{\Pi^B}I{\left(\Pi^B{\left(\rho^{AB}\right)}\right)}$ and $I{\left(\Pi^B{\left(\rho^{AB}\right)}\right)}  = \chi{\left({\left\{p_j,\rho^A_j\right\}}\right)}$ that in order to prove $C{\left(\rho^{AB}\right)} \leqslant \S{\left(\rho^B\right)}$, we only need to prove $\chi{\left({\left\{p_j,\rho^A_j\right\}}\right)} \leqslant \S{\left(\rho^B\right)}$. Note that the quantum ensemble ${\left\{p_j,\rho^A_j\right\}}$ is obtained from the quantum operation of taking partial trace over $\cH^B$ from the quantum state $\rho^{AB}$, this inspired us to define the following quantum operation $\Psi$ on the quantum system $\sH^A \ot \sH^B$: Let ${\left|\omega^B\right\rangle} \in \sH^B$ be a fixed unit vector, for each quantum state $\sigma^{AB}$ on $\sH^A \ot \sH^B$, $$\begin{aligned}
    \Psi{\left(\sigma^{AB}\right)} & = & \sum_{j} {\left({\mathbbm{1}}^A \ot {\left|\omega^B\right\rangle}{\left\langle\psi^B_j\right|}\right)} \sigma^{AB} {\left({\mathbbm{1}}^A \ot {\left|\omega^B\right\rangle}{\left\langle\psi^B_j\right|}\right)} \\
    & = & {\operatorname{Tr}_{B}{\left(\sigma^{AB}\right)}} \ot {\left|\omega^B\right\rangle}{\left\langle\omega^B\right|}.\end{aligned}$$

    Let $\sH^C=\mathbb{C}^k$ and ${\{{|i\rangle}\}}_{i=1}^k$ be the standard orthonormal basis of $\sH^C$. Then the correlation matrix $\widehat{\Psi}{\left(\rho^{AB}\right)}$ is given by $$\begin{aligned}
    \widehat{\Psi}{\left(\rho^{AB}\right)} & = & \sum_{i,j} {{\operatorname{Tr}_{}{\left({\mathbbm{1}}^A \ot {\left|\omega^B\right\rangle}{\left\langle\psi^B_i\right|} \rho^{AB} {\mathbbm{1}}^A \ot {\left|\psi^B_j\right\rangle}{\left\langle\omega^B\right|}\right)}}}{| i\rangle\langle j |}\\
    & = & \sum_{i,j}{\left\langle\psi^B_i\right|}\rho^B{\left|\psi^B_j\right\rangle}{| i\rangle\langle j |},\end{aligned}$$

    If we define $W = \sum_j{\left|j\right\rangle}{\left\langle\psi^B_j\right|}$, then $W^\dagger W =
    {\mathbbm{1}}^B, WW^\dagger = {\mathbbm{1}}^C$, that is, $W$ is an unitary operator from $\cH^B$ to $\cH^C$. It follows from $\widehat{\Psi}{\left(\rho^{AB}\right)} = W \rho^B W^\dagger$ that $\S{\left(\widehat{\Psi}{\left(\rho^{AB}\right)}\right)} = \S{\left(\rho^B\right)}$. Note that the quantum ensemble ${\left\{p_j,\rho^A_j \ot {\left|\omega^B\right\rangle}{\left\langle\omega^B\right|}\right\}}$ can be obtained by the quantum operation $\Psi$ and $\chi{\left({\left\{p_j,\rho^A_j\right\}}\right)} = \chi{\left({\left\{p_j,\rho^A_j \ot
    {\left|\omega^B\right\rangle}{\left\langle\omega^B\right|}\right\}}\right)}$. By using the inequality (2) we have $$\begin{aligned}
    \chi{\left({\left\{p_j,\rho^A_j\right\}}\right)} &=& \chi{\left({\left\{p_j,\rho^A_j \ot
    {\left|\omega^B\right\rangle}{\left\langle\omega^B\right|}\right\}}\right)} \leqslant \S{\left(\widehat{\Psi}{\left(\rho^{AB}\right)}\right)} = \S{\left(\rho^B\right)}.\end{aligned}$$ Thus, we have proved $C{\left(\rho^{AB}\right)} \leqslant \min {\left\{\S{\left(\rho^A\right)}, \S{\left(\rho^B\right)}\right\}}$.

(ii) Note that equality (3) shows that $\S{\left(\widehat{{\mathbbm{1}}^A \ot \Pi^B}{\left(\rho^{AB}\right)}\right)} =
    \S{\left(\widehat{\Pi}^B{\left(\rho^B\right)}\right)}$. Hence it follows from inequality (1) that $$\S{\left(\Pi^B{\left(\rho^{AB}\right)}\right)} - \S{\left(\rho^{AB}\right)} \leqslant
    \S{\left(\widehat{{\mathbbm{1}}^A \ot \Pi^B}{\left(\rho^{AB}\right)}\right)} =
    \S{\left(\widehat{\Pi}^B{\left(\rho^B\right)}\right)} =
    \H({\{p_j\}}) = \S{\left(\Pi^B{\left(\rho^B\right)}\right)}.\eqno (4)$$ On the other hand, note that $I{\left(\Pi^B{\left(\rho^{AB}\right)}\right)} = \S{\left(\rho^{A}\right)} + \S{\left(\Pi^B{\left(\rho^{B}\right)}\right)} - \S{\left(\Pi^B{\left(\rho^{AB}\right)}\right)}$, by the definition of $Q(\rho^{AB})$ and inequality (4) we have $$Q{\left(\rho^{AB}\right)} = I{\left(\rho^{AB}\right)} - C{\left(\rho^{AB}\right)} \leqslant \S{\left(\Pi^B
    \rho^{AB}\right)} - \S{\left(\rho^{AB}\right)} - \S{\left(\Pi^B \rho^{B}\right)} + \S{\left(\rho^B\right)}
    \leqslant \S{\left(\rho^B\right)}.$$ This showed that $Q{\left(\rho^{AB}\right)} \leqslant \S{\left(\rho^B\right)}$. Clearly, when $\S{\left(\rho^B\right)} \leqslant \S{\left(\rho^A\right)}$, it follows from $Q{\left(\rho^{AB}\right)} \leqslant \S{\left(\rho^B\right)}$ that $Q{\left(\rho^{AB}\right)} \leqslant
    \min {\left\{\S{\left(\rho^A\right)}, \S{\left(\rho^B\right)}\right\}}$. If $\rho^{AB}$ is a separable state, then $\S{\left(\rho^{AB}\right)} \geqslant
    \max {\left\{\S{\left(\rho^A\right)}, \S{\left(\rho^B\right)}\right\}}$ @Fan. Note that $I{\left(\Pi^B{\left(\rho^{AB}\right)}\right)} \geqslant 0$, so $\S{\left(\rho^B\right)} - \S{\left(\rho^{AB}\right)}
    \leqslant I{\left(\Pi^B{\left(\rho^{AB}\right)}\right)}$. Thus, we can prove easily that $Q{\left(\rho^{AB}\right)} \leqslant \min {\left\{\S{\left(\rho^A\right)}, \S{\left(\rho^B\right)}\right\}}$. The theorem is proved.

In what follows, in order to provide a class of states $\rho^{AB}$ such that $Q{\left(\rho^{AB}\right)}
\leqslant \S{\left(\rho^A\right)}$, we need the following:

\[th:main'\] Let $\sH^B$ and $\sH^C$ be two finite dimensional complex Hilbert spaces, $\rho^{BC}$ be a state on $\sH^B\ot \sH^C$, $\rho^B =
{\operatorname{Tr}_{C}{\left(\rho^{BC}\right)}}$, $\rho^C = {\operatorname{Tr}_{B}{\left(\rho^{BC}\right)}}$. Then $\S{\left(\rho^{BC}\right)} = \S{\left(\rho^B\right)} - \S{\left(\rho^C\right)}$ if and only if

(i) $H^B$ can be factorized into the form $\sH^B = \sH^L \ot \sH^R$, and

(ii) $\rho^{BC} = \rho^L \ot {\left|\Psi^{RC}\right\rangle}{\left\langle\Psi^{RC}\right|}$, where ${\left|\Psi^{RC}\right\rangle} \in \sH^R \ot \sH^C$.

$(\Longleftarrow)$ It is trivially.\
$(\Longrightarrow)$ The quantum state $\rho^{BC}$ can be purified into a tripartite state ${\left|\Omega^{ABC}\right\rangle} \in \sH^A \ot \sH^B \ot \sH^C$, where $\sH^A$ is a reference system. If we denote $\rho^{ABC} =
{\left|\Omega^{ABC}\right\rangle}{\left\langle\Omega^{ABC}\right|}$, then $$\begin{aligned}
{\operatorname{Tr}_{AB}{\left(\rho^{ABC}\right)}} = \rho^C, {\operatorname{Tr}_{AC}{\left(\rho^{ABC}\right)}} = \rho^B,\\
{\operatorname{Tr}_{C}{\left(\rho^{ABC}\right)}} = \rho^{AB}, {\operatorname{Tr}_{A}{\left(\rho^{ABC}\right)}} = \rho^{BC}.\end{aligned}$$ Note that $\S{\left(\rho^{ABC}\right)} = 0$, so $\S{\left(\rho^C\right)} = \S{\left(\rho^{AB}\right)}$, thus, we have $$\S{\left(\rho^{AB}\right)} + \S{\left(\rho^{BC}\right)} = \S{\left(\rho^B\right)} = \S{\left(\rho^B\right)} + \S{\left(\rho^{ABC}\right)},$$ it follows from @Hayden that

(i) $\sH^B$ can be factorized into the form $\sH^B = \bigoplus_{k=1}^K \sH_k^L \ot \sH_k^R$,

(ii) $\rho^{ABC} = \bigoplus_{k=1}^K \lambda_k \rho_k^{AL} \ot \rho_k^{RC}$, where $\rho_k^{AL}$ is a state on $\sH^A \ot \sH_k^L$, $\rho_k^{RC}$ is a state on $\sH_k^R \ot \sH^C$, ${\{\lambda_k\}}$ is a probability distribution.

That $\S{\left(\rho^{BC}\right)} = \S{\left(\rho^B\right)} - \S{\left(\rho^C\right)}$ implies $\S{\left(\rho^A\right)}
+ \S{\left(\rho^C\right)} = \S{\left(\rho^{AC}\right)}$ is clear, and $\S{\left(\rho^A\right)}
+ \S{\left(\rho^C\right)} = \S{\left(\rho^{AC}\right)}$ if and only if $\rho^{AC} = \rho^A \ot \rho^C$ holds. By the expression form of $\rho^{ABC} = \bigoplus_{k=1}^K
\lambda_k \rho_k^{AL} \ot \rho_k^{RC}$, we have $\rho^{AC} =
\sum_{k=1}^K \lambda_k \rho_k^A \ot \rho_k^C$. Combining these facts we have $K=1$, i.e., the statement (i) of the theorem holds. Hence $\rho^{ABC} = \rho^{AL} \ot \rho^{RC}$, where $\rho^{AL}$ is a state on $\sH^A \ot \sH^L$ and $\rho^{RC}$ is a state on $\sH^R \ot
\sH^C$, it follows from $\rho^{ABC}$ is pure state that both $\rho^{AL}$ and $\rho^{RC}$ are also pure states. Therefore $$\rho^{BC} = {\operatorname{Tr}_{A}{\left(\rho^{AL}\right)}} \ot \rho^{RC} = \rho^L \ot {\left|\Psi^{RC}\right\rangle}{\left\langle\Psi^{RC}\right|}.$$ The statement (ii) holds and the theorem is proved.

*Note added.* After the present work is completed and submitted to the arXiv, Luo *et al.* @Li inform us of the inequality (II) being not valid in general and provide a counter-example while they give another approach to the above inequality (I). We found also that Giorgi @Giorgi gives a proof to the inequalities (I) and (II) in the case of two qubits by monogamy of discord for pure states.

Let $\rho^{AB}$ be a bipartite state on $\sH^A \ot \sH^B$ such that $\S{\left(\rho^{AB}\right)} = \S{\left(\rho^B\right)} - \S{\left(\rho^{A}\right)}$. By Theorem \[th:main'\], we have $\rho^{AB} = {\left|\Phi^{AL}\right\rangle}{\left\langle\Phi^{AL}\right|} \ot
\rho^R$ for ${\left|\Phi^{AL}\right\rangle} \in \sH^A \ot \sH^L$, where $\rho^R$ is a state on $\sH^R$ and $\sH^B = \sH^L \ot \sH^R$. It is easy to show that although $\S{\left(\rho^A\right)} \leqslant \S{\left(\rho^B\right)}$, but $Q{\left(\rho^{AB}\right)} =
\S{\left(\rho^A\right)}$, so the conjecture (II) is true for these states. If $\dim \sH^A = \dim\sH^B = 2$, then $\sH^B$ cannot be factorized. This indicates that $\S{\left(\rho^{AB}\right)} = \S{\left(\rho^B\right)} - \S{\left(\rho^{A}\right)}$ implies that $\rho^{AB}$ is pure state. In other words, if two qubit state $\rho^{AB}$ is not pure, then $\S{\left(\rho^{AB}\right)} > \left|\S{\left(\rho^B\right)} - \S{\left(\rho^{A}\right)}\right|$.

Concluding Remarks
==================

The conjectures and our work are based on the assumption that classical correlations are maximized by von Neumann measurement. In general, it is not true, see @Giorda [@Galve; @Yu].

It follows from Theorem \[th:main\] that von Neumann measurement performed on subsystem $B$ induced the following inequality: $$\chi{\left({\left\{p_j,\rho^A_j\right\}}\right)} \leqslant \S{\left(\rho^B\right)}.$$

By using the above inequality, we studied a conjecture in @Zhang, proposed by W. Roga in @Roga1 [@Roga2].

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{'source': 'arXiv_out/1105'}