## Introduction

Quantum entanglement is a non-classical nature of quantum mechanics, which is a useful resource in many quantum information processing tasks such as quantum teleportation, dense coding and quantum cryptography1,2,3. Because of its important roles in the field of quantum information and computation theory, there has been a significant amount of research focused on quantification of entanglement in bipartite quantum systems. Entanglement of formation(EoF) is the most well-known bipartite entanglement measure with an operational meaning that asymptotically quantifies how many bell states are needed to prepare the given state using local quantum operations and classical communications4. Although EoF is defined in any bipartite quantum systems of arbitrary dimension, its definition for mixed states is based on ‘convex-roof extension’, which takes the minimum average over all pure-state decompositions of the given state. As such an optimization is hard to deal with, analytic evaluation of EoF is known only in two-qubit systems5 and some restricted cases of higher-dimensional systems so far.

In multi-party quantum systems, entanglement shows a distinct behavior that does not have any classical counterpart; if a pair of parties in a multi-party quantum system is maximally entangled, then they cannot be entanglement, not even classically correlated, with the rest parties. This restriction of sharing entanglement in multi-party quantum systems is known as the monogamy of entanglement(MoE)6,7. MoE plays an important role such as the security proof of quantum key distribution in quantum cryptography2,8 and the N-representability problem for fermions in condensed-matter physics9.

Mathematically, MoE can be characterized using monogamy inequality; for a three-party quantum state $$\rho _{ABC}$$ and its two-party reduced density matrices $$\rho _{AB}$$ and $$\rho _{AC}$$,

\begin{aligned} E\left( \rho _{A(BC)}\right) \ge E\left( \rho _{AB}\right) +E\left( \rho _{AC}\right) \end{aligned}
(1)

where $$E\left( \rho _{XY}\right)$$ is an entanglement measure quantifying the amount of entanglement between subsystems X and Y of the bipartite quantum state $$\rho _{XY}$$. Inequality (1) shows the mutually exclusive nature of bipartite entanglement $$E\left( \rho _{AB}\right)$$ and $$E\left( \rho _{AC}\right)$$ shared in three-party quantum systems so that their summation cannot exceeds the total entanglement $$E\left( \rho _{A(BC)}\right)$$.

Using tangle10 as the bipartite entanglement measure, Inequality (1) was first shown to be true for all three-qubit states, and generalized for multi-qubit systems as well as some cases of higher-dimensional quantum systems11,12. However, not all bipartite entanglement measures can characterize MoE in forms of Inequality (1), but only few measures are known so far satisfying such monogamy inequality13,14,15. Although EoF is the most natural bipartite entanglement measure with the operational meaning in quantum state preparation, EoF is known to fail in characterizing MoE as the monogamy inequality in (1) even in three-qubit systems; there exists quantum states in three-qubit systems violating Inequality (1) if EoF is used as the bipartite entanglement measure. Thus, a natural question we can ask is ‘On what condition does the monogamy inequality hold in terms of the given bipartite entanglement measure?’.

Here, we provide a sufficient condition that monogamy inequality of quantum entanglement holds in terms of EoF in multi-party, arbitrary dimensional quantum systems. For a three-party quantum state, we first consider the classical–classical–quantum (ccq) states whose quantum parts are obtained from the two-party reduced density matrices of the three-party state. By evaluating quantum mutual information of the ccq states as well as their reduced density matrices, we show that the additivity of the mutual information of the ccq states guarantees the monogamy inequality of the three-party quantum state in terms of EoF. We provide some examples of three-party pure state to illustrate our result, and we generalize our result of three-party systems into any multi-party systems of arbitrary dimensions.

This paper is organized as follows. First we briefly review the definitions of classical and quantum correlations in bipartite quantum systems and recall their trade-off relation in three-party quantum systems. After providing the definition of ccq states as well as their mutual information between classical and quantum parts, we establish the monogamy inequality of three-party quantum entanglement in arbitrary dimensional quantum systems in terms of EoF conditioned on the additivity of the mutual information for the ccq states. We also illustrate our result of monogamy inequality in three-party quantum systems with some examples, and we generalize our result of entanglement monogamy inequality into multi-party quantum systems of arbitrary dimensions. Finally, we summarize our results.

## Results

### Correlations in bipartite quantum systems

For a bipartite pure state $${\left| \psi \right\rangle }_{AB}$$, its entanglement of formation (EoF) is defined by the entropy of a subsystem, $${E}_\mathbf{f}\left( {\left| \psi \right\rangle }_{AB} \right) =S(\rho _A)$$ where $$\rho _A={\text {tr}}_{B} {\left| \psi \right\rangle }_{AB}{\left\langle \psi \right| }$$ is the reduced density matrix of $${\left| \psi \right\rangle }_{AB}$$ on subsystem A, and $$S\left( \rho \right) =-{\text {tr}}\rho \ln \rho$$ is the von Neumann entropy of the quantum state $$\rho$$. For a bipartite mixed state $$\rho _{AB}$$, its EoF is defined by the minimum average entanglement

\begin{aligned} E_{{\mathbf{f}}}\left( \rho _{AB} \right) =\min \sum _i p_i E_{{\mathbf{f}}}({\left| \psi _i \right\rangle }_{AB}), \end{aligned}
(2)

over all possible pure state decompositions of $$\rho _{AB}=\sum _{i} p_i |\psi _i\rangle _{AB}\langle \psi _i|$$.

For a probability ensemble $${\mathscr {E}} = \{p_i, \rho _i\}$$ realizing a quantum state $$\rho$$ such that $$\rho =\sum _{i}p_i\rho _i$$, its Holevo quantity is defined as

\begin{aligned} \chi \left( {\mathscr {E}}\right) =S\left( \rho \right) -\sum _{i}p_i S\left( \rho _i\right) . \end{aligned}
(3)

Given a bipartite quantum state $$\rho _{AB}$$, each measurement $$\{M^x_B\}$$ applied on subsystem B induces a probability ensemble $${\mathscr {E}} = \{p_x, \rho _A^x\}$$ of the reduced density matrix $$\rho _A={\text {tr}}_A\rho _{AB}$$ in the way that $$p_x={\text {tr}}[(I_A\otimes M_B^x)\rho _{AB}]$$ is the probability of the outcome x and $$\rho ^x_A={\text {tr}}_B[(I_A\otimes {M_B^x})\rho _{AB}]/p_x$$ is the state of system A when the outcome was x. The one-way classical correlation (CC)16 of a bipartite state $$\rho _{AB}$$ is defined by the maximum Holevo quantity

\begin{aligned} {{\mathscr {J}}}^{\leftarrow }(\rho _{AB})&= \max _{{\mathscr {E}}} \chi \left( {\mathscr {E}}\right) \end{aligned}
(4)

over all possible ensemble representations $${\mathscr {E}}$$ of $$\rho _A$$ induced by measurements on subsystem B.

The following proposition shows a trade-off relation between classical correlation and quantum entanglement (measured by CC and EoF, respectively) distributed in three-party quantum systems.

### Proposition 1.

17 For a three-party pure state $${\left| \psi \right\rangle }_{ABC}$$ with reduced density matrices $$\rho _{AB}={\text {tr}}_C{\left| \psi \right\rangle }_{ABC}{\left\langle \psi \right| }$$, $$\rho _{AC}={\text {tr}}_B{\left| \psi \right\rangle }_{ABC}{\left\langle \psi \right| }$$ and $$\rho _{A}={\text {tr}}_{BC}{\left| \psi \right\rangle }_{ABC}{\left\langle \psi \right| }$$, we have

\begin{aligned} S(\rho _A)={{\mathscr {J}}}^{\leftarrow }(\rho _{AB})+E_\mathbf{f}\left( \rho _{AC}\right) . \end{aligned}
(5)

### Classical–classical–quantum (CCQ) states

In this section, we consider a four-party ccq states obtained from a bipartite state $$\rho _{AB}$$, and provide detail evaluations of their mutual information. Without loss of generality, we assume that any bipartite state as a two-qudit state by taking d as the dimension of larger dimensional subsystem.

For a two-qudit state $$\rho _{AB}$$, let us consider the reduced density matrix $$\rho _B={\text {tr}}_{A}\rho _{AB}$$ and its spectral decomposition

\begin{aligned} \rho _B=\sum _{i=0}^{d-1}\lambda _{i}{\left| e_i \right\rangle }_B{\left\langle e_i \right| }. \end{aligned}
(6)

Let

\begin{aligned} {\mathscr {E}}_0=\{\lambda _i,\sigma _A^i\}_i \end{aligned}
(7)

be the probability ensemble of $$\rho _A={\text {tr}}_{B}\rho _{AB}$$ from the measurement $$\{{\left| e_i \right\rangle }_B{\left\langle e_i \right| }\}_{i=1}^{d-1}$$ on subsystem B of $$\rho _{AB}$$, in a way that

\begin{aligned} \lambda _i={\text {tr}}\left[ (I_A \otimes {\left| e_i \right\rangle }_B{\left\langle e_i \right| })\rho _{AB}\right] \end{aligned}
(8)

and

\begin{aligned} \sigma _A^i=\frac{1}{\lambda _i}{\text {tr}}_B \left[ (I_A \otimes {\left| e_i \right\rangle }_B{\left\langle e_i \right| })\rho _{AB}\right] . \end{aligned}
(9)

Based on the eigenvectors $$\{ {\left| e_j \right\rangle }_{B}\}$$ of $$\rho _B$$, we also consider the d-dimensional Fourier basis elements $$|{\tilde{e}}_j \rangle _B = \frac{1}{\sqrt{d}}\sum _{k=0}^{d-1}\omega _d^{jk}{\left| e_k \right\rangle }_B$$ for each $$j=0,\ldots ,d-1$$, where $$\omega _d = e^{\frac{2\pi i}{d}}$$ is the dth-root of unity. Let

\begin{aligned} {\mathscr {E}}_1=\left\{\frac{1}{d},\tau _A^j\right \}_j, \end{aligned}
(10)

be the probability ensemble of $$\rho _A={\text {tr}}_{B}\rho _{AB}$$ obtained by measuring subsystem B in terms of the Fourier basis $$\{{\left| \tilde{e}_j \right\rangle }_B{\left\langle \tilde{e}_j \right| }\}_{j=1}^{d-1}$$ where

\begin{aligned} \frac{1}{d}={\text {tr}}\left[ (I_A \otimes {\left| \tilde{e}_j \right\rangle }_B{\left\langle \tilde{e}_j \right| })\rho _{AB}\right] \end{aligned}
(11)

and

\begin{aligned} \tau _A^j=d{\text {tr}}_B\left[ (I_A\otimes |{\tilde{e}}_j\rangle _B \langle {\tilde{e}}_j|)\rho _{AB}\right] . \end{aligned}
(12)

Now we define the generalized d-dimensional Pauli operators based on the eigenvectors of $$\rho _B$$ as

\begin{aligned} Z=\sum _{j=0}^{d-1}\omega _d^j{\left| e_j \right\rangle }_B{\left\langle e_j \right| },~X=\sum _{j=0}^{d-1} {\left| e_{j+1} \right\rangle }_B{\left\langle e_j \right| }=\sum _{j=0}^{d-1} \omega _d^{-j}|\tilde{e}_j \rangle _B \langle {\tilde{e}}_j |, \end{aligned}
(13)

and consider a four-qudit ccq state $$\Gamma _{XYAB}$$

\begin{aligned} \Gamma _{XYAB}=\frac{1}{d^2}\sum _{x,y=0}^{d-1}{\left| x \right\rangle }_X {\left\langle x \right| }\otimes {\left| y \right\rangle }_Y{\left\langle y \right| } \otimes \left[ (I_A\otimes X^x_BZ^y_B)\rho _{AB}(I_A\otimes Z^{-y}_BX^{-x}_B)\right] , \end{aligned}
(14)

for some d-dimensional orthonormal bases $$\{{\left| x \right\rangle }_X\}$$ and $$\{{\left| y \right\rangle }_Y\}$$ of the subsystems X and Y, respectively. From Eqs. (8), (9), (11), (12) and (13), the reduced density matrices of $$\Gamma _{XYAB}$$ are obtained as

\begin{aligned} \Gamma _{XAB}&=\frac{1}{d}\sum _{x=0}^{d-1}{\left| x \right\rangle }_X{\left\langle x \right| }\otimes \left[ \left( I_A \otimes X^x_B\right) \left( \sum _{i=0}^{d-1} \sigma _A^i \otimes \lambda _i{\left| e_i \right\rangle }_B{\left\langle e_i \right| }\right) \left( I_A \otimes X_B^{-x}\right) \right] , \end{aligned}
(15)
\begin{aligned} \Gamma _{YAB}&=\frac{1}{d}\sum _{y=0}^{d-1}{\left| y \right\rangle }_Y{\left\langle y \right| }\otimes \left[ \left( I_A \otimes Z_B^y \right) \left( \sum _{j=0}^{d-1} \tau _A^j \otimes \frac{1}{d}|{\tilde{e}}_j \rangle _B \langle {\tilde{e}}_j|\right) \left( I_A \otimes Z_B^{-y}\right) \right] , \end{aligned}
(16)

and

\begin{aligned} \Gamma _{AB}=\rho _A\otimes \frac{I_B}{d}, \end{aligned}
(17)

where $$I_A$$ and $$I_B$$ are d-dimensional identity operators of subsystems A and B, respectively.

Here we note that the mutual information between the classical and quantum parts of the ccq state in Eq. (14) as well as its reduced density matrices in Eqs. (15) and (16) are

\begin{aligned}{}&{I}\left( \Gamma _{XY:AB}\right) =\ln d+S(\rho _A)-S(\rho _{AB}), \end{aligned}
(18)
\begin{aligned}{}&{I}(\Gamma _{X:AB})=\ln d-S(\rho _B)+\chi ({\mathscr {E}}_0), \end{aligned}
(19)

and

\begin{aligned} {I}(\Gamma _{Y:AB})=\chi ({\mathscr {E}}_1), \end{aligned}
(20)

where the detail calculation can be found in “Methods” section.

### Monogamy inequality of multi-party entanglement in terms of EoF

It is known that quantum mutual information is superadditive for any ccq state of the form

\begin{aligned} \Xi _{XYAB}=\frac{1}{d^2}\sum _{x,y=0}^{d-1}{\left| x \right\rangle }_X {\left\langle x \right| }\otimes {\left| y \right\rangle }_Y{\left\langle y \right| }\otimes \sigma ^{xy}_{AB}, \end{aligned}
(21)

that is, $${I}\left( \Xi _{XY:AB}\right) \ge {I}\left( \Xi _{X:AB}\right) +{I}\left( \Xi _{Y:AB}\right)$$18. The following theorem shows that the additivity of quantum mutual information for ccq states guarantees the monogamy inequality of three-party quantum entanglement in therms of EoF.

### Theorem 1.

For any three-party pure state $${\left| \psi \right\rangle }_{ABC}$$ with its two-qudit reduced density matrices $${\text {tr}}_C {\left| \psi \right\rangle }_{ABC}{\left\langle \psi \right| }=\rho _{AB}$$ and $${\text {tr}}_B {\left| \psi \right\rangle }_{ABC}{\left\langle \psi \right| }=\rho _{AC}$$, we have

\begin{aligned} E_\mathbf{f}\left( {\left| \psi \right\rangle }_{A(BC)}\right) \ge&E_\mathbf{f}\left( \rho _{AB}\right) +E_\mathbf{f}\left( \rho _{AC}\right) , \end{aligned}
(22)

conditioned on the additivity of quantum mutual information

\begin{aligned} {I}\left( \Gamma _{XY:AB}\right) = {I}\left( \Gamma _{X:AB}\right) +{I}\left( \Gamma _{Y:AB}\right) \end{aligned}
(23)

and

\begin{aligned} {I}\left( \Gamma _{XY:AC}\right) = {I}\left( \Gamma _{X:AC}\right) +{I}\left( \Gamma _{Y:AC}\right) \end{aligned}
(24)

where $$\Gamma _{XYAB}$$ and $$\Gamma _{XYAC}$$ are the ccq states of the form in Eq. (14) obtained by $$\rho _{AB}$$ and $$\rho _{AC}$$, respectively.

Conditioned on the additivity of quantum mutual information for ccq states, Theorem 1 shows that EoF can characterize the monogamous nature of bipartite entanglement shared in three-party quantum systems, which is illustrateed in Fig. 1.

### Example 1.

Let us consider three-qubit GHZ state19,

\begin{aligned} {\left| GHZ \right\rangle }_{ABC}=\frac{1}{\sqrt{2}}\left( {\left| 000 \right\rangle }_{ABC}+{\left| 111 \right\rangle }_{ABC}\right) , \end{aligned}
(25)

with its reduced density matrices

\begin{aligned} \rho _{AB}=\frac{1}{2}\left( {\left| 00 \right\rangle }_{AB}{\left\langle 00 \right| } +{\left| 11 \right\rangle }_{AB}{\left\langle 11 \right| }\right) ,~\rho _{A}=\frac{1}{2}\left( {\left| 0 \right\rangle }_{A}{\left\langle 0 \right| } +{\left| 1 \right\rangle }_{A}{\left\langle 1 \right| }\right) ,~\rho _{B}=\frac{1}{2}\left( {\left| 0 \right\rangle }_{B}{\left\langle 0 \right| } +{\left| 1 \right\rangle }_{B}{\left\langle 1 \right| }\right) . \end{aligned}
(26)

The eigenvalues of $$\rho _B$$ are $$\lambda _0=\lambda _1=\frac{1}{2}$$ with corresponding eigenvectors $${\left| e_0 \right\rangle }_B={\left| 0 \right\rangle }_B$$ and $${\left| e_1 \right\rangle }_B={\left| 1 \right\rangle }_B$$ respectively. Thus the ensemble of $$\rho _A$$ induced by measuring subsystem B of $$\rho _{AB}$$ in terms of the eigenvectors of $$\rho _B$$, that is, $$\{ M_B^0={\left| 0 \right\rangle }_B{\left\langle 0 \right| }, M_B^1={\left| 1 \right\rangle }_B{\left\langle 1 \right| }\}$$ is

\begin{aligned} {\mathscr {E}}_0= \{ \lambda _0=\frac{1}{2}, \sigma _A^0={\left| 0 \right\rangle }_A {\left\langle 0 \right| }, \lambda _1=\frac{1}{2}, \sigma _A^1={\left| 1 \right\rangle }_A{\left\langle 1 \right| }\}. \end{aligned}
(27)

Because the Fourier basis elements of subsystem B with respect to the eigenvectors of $$\rho _B$$ are

\begin{aligned} |{\tilde{e}}_0 \rangle _B=\frac{1}{2}\left( {\left| 0 \right\rangle }_B+{\left| 1 \right\rangle }_B\right) , ~|{\tilde{e}}_1 \rangle _B= \frac{1}{2}\left( {\left| 0 \right\rangle }_B-{\left| 1 \right\rangle }_B\right) , \end{aligned}
(28)

the ensemble of $$\rho _A$$ induced by measuring subsystem B of $$\rho _{AB}$$ in terms of the Fourier basis in Eq. (28) is

\begin{aligned} {\mathscr {E}}_1= \{ \frac{1}{2}, \tau _A^0=\frac{1}{2}I_A, \frac{1}{2}, \tau _A^1=\frac{1}{2}I_A\}. \end{aligned}
(29)

Now we consider the additivity of mutual information of the ccq state $$\Gamma _{XYAB}$$ obtained from $$\rho _{AB}$$ in Eq. (26). Due to Eq. (18), the mutual information of $$\Gamma _{XYAB}$$ between XY and AB is

\begin{aligned} I\left( \Gamma _{XY:AB}\right) =&\ln 2+S(\rho _A)-S(\rho _{AB})=\ln 2 \end{aligned}
(30)

because $$S(\rho _A)=S(\rho _{AB})=\ln 2$$ from Eqs. (26). For the mutual information of $$\Gamma _{XAB}$$ between X and AB, Eq. (19) leads us to

\begin{aligned} I\left( \Gamma _{X:AB}\right) =&\ln 2-S(\rho _B)+\chi ({\mathscr {E}}_0)=\ln 2 \end{aligned}
(31)

where the second equality is from $$S(\rho _B)= \ln 2$$ and

\begin{aligned} \chi ({\mathscr {E}}_0)=S(\rho _A)-\frac{1}{2}S({\left| 0 \right\rangle }_A {\left\langle 0 \right| })-\frac{1}{2}S({\left| 1 \right\rangle }_A {\left\langle 1 \right| })=\ln 2, \end{aligned}
(32)

for the ensemble $${\mathscr {E}}_0$$ in Eq. (27). For the mutual information of $$\Gamma _{YAB}$$ between Y and AB, Eq. (20) leads us to

\begin{aligned} I\left( \Gamma _{Y:AB}\right) =&\chi ({\mathscr {E}}_1) =S(\rho _A)-\frac{1}{2}S\left( \frac{1}{2}I_A \right) -\frac{1}{2}S\left( \frac{1}{2}I_A \right) =0 \end{aligned}
(33)

where the second equality is due to the ensemble $${\mathscr {E}}_1$$ in Eq. (29).

From Eqs. (30), (31) and (33), we note that the mutual information of the ccq state $$\Gamma _{XYAB}$$ obtained from $$\rho _{AB}$$ is additive as in Eq. (23). Moreover, the symmetry of GHZ state assures that the same is also true for the reduced density matrix $$\rho _{AC}={\text {tr}}_B {\left| GHZ \right\rangle }_{ABC}{\left\langle GHZ \right| }$$. Thus Theorem 1 guarantees the monogamy inequality of the three-qubit GHZ state in Eq. (25) in terms of EoF. In fact, we have $$E_\mathbf{f}\left( {\left| GHZ \right\rangle }_{A(BC)}\right) =S(\rho _A)=\ln 2$$, whereas the two-qubit reduced density matrices $$\rho _{AB}$$ and $$\rho _{AC}$$ are separeble. Thus $$E_\mathbf{f}\left( \rho _{AB}\right) =E_\mathbf{f}\left( \rho _{AC}\right) =0$$ and this implies the monogamy inequality in (22).

Let us consider another example of three-qubit state.

### Example 2.

Three-qubit W-state is defined as20

\begin{aligned} {\left| W \right\rangle }_{ABC}=\frac{1}{\sqrt{3}}\left( {\left| 100 \right\rangle }_{ABC}+{\left| 010 \right\rangle }_{ABC}+{\left| 001 \right\rangle }_{ABC}\right) . \end{aligned}
(34)

The two-qubit reduced density matrix of $${\left| W \right\rangle }_{ABC}$$ on subsystem AB is obtained as

\begin{aligned} \rho _{AB}=\frac{2}{3}{\left| \psi ^+ \right\rangle }_{AB}{\left\langle \psi ^+ \right| }+\frac{1}{3}{\left| 00 \right\rangle }_{AB}{\left\langle 00 \right| } \end{aligned}
(35)

where $${\left| \psi ^+ \right\rangle }_{AB}=\frac{1}{\sqrt{2}}\left( {\left| 01 \right\rangle }_{AB}+{\left| 10 \right\rangle }_{AB}\right)$$ is the two-qubit Bell state. The one-qubit reduced density matrices of $$\rho _{AB}$$ are

\begin{aligned} \rho _{A}=\frac{2}{3}{\left| 0 \right\rangle }_{A}{\left\langle 0 \right| }+\frac{1}{3}{\left| 1 \right\rangle }_{A}{\left\langle 1 \right| },~ \rho _{B}=\frac{2}{3}{\left| 0 \right\rangle }_{B}{\left\langle 0 \right| }+\frac{1}{3}{\left| 1 \right\rangle }_{B}{\left\langle 1 \right| }. \end{aligned}
(36)

From to the spectral decomposition of $$\rho _B$$ in Eq. (36) with the eigenvalues $$\lambda _0=\frac{2}{3}, \lambda _1=\frac{1}{3}$$ and corresponding eigenvectors $${\left| e_0 \right\rangle }_B={\left| 0 \right\rangle }_B$$ and $${\left| e_1 \right\rangle }_B={\left| 1 \right\rangle }_B$$, respectively, it is straightforward to check that the ensemble of $$\rho _A$$ induced from measuring subsystem B of $$\rho _{AB}$$ by the eigenvectors of $$\rho _B$$ is

\begin{aligned} {\mathscr {E}}_0= \{ \lambda _0=\frac{2}{3}, \sigma _A^0=\frac{1}{2}I_A, \lambda _1=\frac{1}{3}, \sigma _A^1={\left| 0 \right\rangle }_A{\left\langle 0 \right| }\}. \end{aligned}
(37)

Because the Fourier basis of subsystem A is the same as Eq. (28), it is also straightforward to obtain the ensemble of $$\rho _A$$ induced by measuring subsystem B of $$\rho _{AB}$$ in terms of the Fourier basis,

\begin{aligned} {\mathscr {E}}_1= \{ \frac{1}{2}, \tau _A^j\}_{j=1,2}, \end{aligned}
(38)

where

\begin{aligned} \tau _A^0=&\frac{1}{3}\left( 2{\left| 0 \right\rangle }_{A}{\left\langle 0 \right| }+{\left| 0 \right\rangle }_{A}{\left\langle 1 \right| } +{\left| 1 \right\rangle }_{A}{\left\langle 0 \right| }+{\left| 1 \right\rangle }_{A}{\left\langle 1 \right| }\right) ,~ \tau _A^1=\frac{1}{3}\left( 2{\left| 0 \right\rangle }_{A}{\left\langle 0 \right| }-{\left| 0 \right\rangle }_{A}{\left\langle 1 \right| } -{\left| 1 \right\rangle }_{A}{\left\langle 0 \right| }+{\left| 1 \right\rangle }_{A}{\left\langle 1 \right| }\right) . \end{aligned}
(39)

For the mutual information of the ccq state $$\Gamma _{XYAB}$$ obtained from $$\rho _{AB}$$ in Eq. (35), Eq. (18) together with Eqs. (35) and (36) lead us to

\begin{aligned} I\left( \Gamma _{XY:AB}\right) =&\ln 2+S(\rho _A)-S(\rho _{AB})=\ln 2. \end{aligned}
(40)

For the mutual information of $$\Gamma _{XAB}$$ between X and AB, Eq. (19) leads us to

\begin{aligned} I\left( \Gamma _{X:AB}\right) =\ln 2-S(\rho _B)+\chi ({\mathscr {E}}_0), \end{aligned}
(41)

where Eq. (37) impies

\begin{aligned} \chi ({\mathscr {E}}_0)=S(\rho _A)-\frac{2}{3}S\left( \frac{1}{2}I_A \right) -\frac{1}{3}S\left( {\left| 0 \right\rangle }_A{\left\langle 0 \right| }\right) =S(\rho _A)-\frac{2}{3}\ln 2. \end{aligned}
(42)

Due to Eq. (36), we have $$S(\rho _A)=S(\rho _B)$$, therefore Eqs. (41) and (42) lead us to

\begin{aligned} I\left( \Gamma _{X:AB}\right) =\frac{1}{3}\ln 2. \end{aligned}
(43)

For the mutual information of $$\Gamma _{YAB}$$ between Y and AB, Eq. (20) leads us to

\begin{aligned} I\left( \Gamma _{Y:AB}\right) =&\chi ({\mathscr {E}}_1)=S(\rho _A) -\frac{1}{2}S\left( \tau _A^0 \right) -\frac{1}{2}S\left( \tau _A^1 \right) , \end{aligned}
(44)

where the second equality is from the ensemble $${\mathscr {E}}_1$$ in Eq. (38). Here we note that $$\tau _A^0$$ and $$\tau _A^1$$ in Eq. (39) have the same eigenvalues, that is $$\mu _0=\frac{3+\sqrt{5}}{6}$$ and $$\mu _0=\frac{3-\sqrt{5}}{6}$$, therefore we have $$S\left( \tau _A^0 \right) =S\left( \tau _A^1 \right)$$. From the spectral decomposition of $$\rho _A$$ in Eq. (36), we have $$S\left( \rho _A\right) =\ln 3 -\frac{2}{3}\ln 2$$, and this turns Eq. (44) into

\begin{aligned} I\left( \Gamma _{Y:AB}\right) =\ln 3 -\frac{2}{3}\ln 2-S\left( \tau _A^0 \right) . \end{aligned}
(45)

From Eqs. (40), (43) and (45), we have

\begin{aligned} I\left( \Gamma _{XY:AB}\right) -I\left( \Gamma _{X:AB}\right) -I\left( \Gamma _{Y:AB}\right) =\frac{4}{3}\ln 2 -\ln 3 +S\left( \tau _A^0 \right) , \end{aligned}
(46)

where $$\ln 2 \approx 0.693147$$, $$\ln 3 \approx 1.098612$$ and $$S\left( \tau _A^0 \right) \approx 0.381264$$. Thus we have

\begin{aligned} I\left( \Gamma _{XY:AB}\right) -I\left( \Gamma _{X:AB}\right) -I\left( \Gamma _{Y:AB}\right) \approx 0.206848 >0, \end{aligned}
(47)

which implies the nonadditivity of mutual information fo r the ccq state $$\Gamma _{XYAB}$$ obtained from $$\rho _{AB}$$ in Eq. (35). We also note that the symmetry of W state in Eq. (34) would imply the nonadditivity of mutual information for the ccq state $$\Gamma _{XYAC}$$ obtained from the two-qubit reduced density matrix $$\rho _{AC}$$ of W state in Eq. (34).

As the additivity of mutual information in Theorem 1 is only a sufficient condition for monogamy inequality in terms of EoF, nonadditivity does not directly imply violation of Inequality (22) for the W state in Eq. (34). However, we note that $$\rho _{AB}$$ in Eq. (35) is a two-qubit state, therefore its EoF can be analytically evaluated as5

\begin{aligned} E_\mathbf{f}\left( \rho _{AB}\right) \approx 0.3812. \end{aligned}
(48)

Moreover, the symmetry of the W state assures that the EoF of $$\rho _{AC}={\text {tr}}_B {\left| W \right\rangle }_{ABC}{\left\langle W \right| }$$ is the same,

\begin{aligned} E_\mathbf{f}\left( \rho _{AC}\right) \approx 0.3812, \end{aligned}
(49)

whereas

\begin{aligned} E_\mathbf{f}\left( {\left| W \right\rangle }_{A(BC)}\right) =S(\rho _A)\approx 0.6365. \end{aligned}
(50)

As Eqs. (48), (49) and (50) imply the violation of Inequality (22), W state in Eq. (34) can be considered as an example for the contraposition of Theorem 1; violation of monogamy inequality in (22) implies nonadditivity of quantum mutual information for the ccq state.

Now, we generalize Theorem 1 for multi-party quantum states of arbitrary dimension.

### Theorem 2.

For any multi-party quantum state $$\rho _{A_1A_2\cdots A_n}$$ with two-party reduced density matrices $$\rho _{A_1A_i}$$ for $$i=2, \cdots , n$$, we have

\begin{aligned} E_\mathbf{f}\left( \rho _{A_1(A_2\cdots A_n)}\right) \ge&\sum _{i=2}^{n}E_\mathbf{f}\left( \rho _{A_1A_i}\right) , \end{aligned}
(51)

conditioned on the additivity of quantum mutual information

\begin{aligned} {I}\left( \Gamma _{XY:A_1A_i}\right) = {I}\left( \Gamma _{X:A_1A_i}\right) +{I}\left( \Gamma _{Y:A_1A_i}\right) \end{aligned}
(52)

where $$\Gamma _{XYA_1A_i}$$ is the ccq state of the form in Eq. (14) obtained by $$\rho _{A_1A_i}$$ for $$i=2, \ldots , n$$.

## Discussion

We have considered possible conditions for monogamy inequality of multi-party quantum entanglement in terms of EoF, and shown that the additivity of mutual information of the ccq states implies the monogamy inequality of three-party quantum entanglement in terms of EoF. We have also provided examples of three-qubit GHZ and W states to illustrate our result in three-party case, and generalized our result into any multi-party systems of arbitrary dimensions.

Most monogamy inequalities of quantum entanglement deal with bipartite entanglement measures based on the minimization over all possible pure state ensembles. As analytic evaluation of such entanglement measure is generally hard especially in higher dimensional quantum systems more than qubits, the situation becomes far more difficult in investigating and establishing entanglement monogamy of multi-party quantum systems of arbitrary dimensions. The sufficient condition provided here deals with the quantum mutual information of the ccq states to guarantee the monogamy inequality of entanglement in terms of EoF in arbitrary dimensions. As the sufficient condition is not involved with any minimization process, our result can provide a useful methodology to understand the monogamy nature of multi-party quantum entanglement in arbitrary dimensions. We finally remark that it would be an interesting future task to investigate if the condition provided here is also necessary.

## Methods

### Evaluation for the quantum mutual information of the ccq states

Here we evaluate the mutual information of the ccq state in Eq. (14) as well as the reduced density matrices in Eqs. (15) and (16); the classical parts of the four-qudit ccq state $$\Gamma _{XYAB}$$ in Eq. (14) is $$\Gamma _{XY}=\frac{1}{d^2}\sum _{x,y=0}^{d-1}{\left| x \right\rangle }_X {\left\langle x \right| }\otimes {\left| y \right\rangle }_Y{\left\langle y \right| }$$, which is the maximally mixed state in $$d^2$$-dimensional quantum system, therefore its von Neumann entropy is

\begin{aligned} S\left( \Gamma _{XY}\right) =-\sum _{x,y=0}^{d-1}\frac{1}{d^2}\ln \left( \frac{1}{d^2}\right) =2\ln d. \end{aligned}
(53)

We also note that Eq. (17) leads us to

\begin{aligned} S\left( \Gamma _{AB} \right) =S(\rho _A)+\ln d. \end{aligned}
(54)

From the joint entropy theorem21,22, we have

\begin{aligned} S(\Gamma _{XYAB})=&2\ln d+\frac{1}{d^2}\sum _{x,y=0}^{d-1}S\left( (I_A\otimes X^x_BZ^y_B)\rho _{AB}(I_A\otimes Z^{-y}_BX^{-x}_B)\right) =2\ln d+S\left( \rho _{AB}\right) , \end{aligned}
(55)

where the second equality is due to the unitary invariance of von Neumann entropy. Thus Eqs. (53), (54) and (55) give us the mutual information of the four-qudit ccq state $$\Gamma _{XYAB}$$ with respect to the bipartition between XY and AB in Eq. (18).

For the von Neumann entropy of $$\Gamma _{XAB}$$ in Eq. (15), we have

\begin{aligned} S\left( \Gamma _{XAB}\right) =&\ln d +\frac{1}{d}\sum _{x=0}^{d-1} S\left( \left( I_A \otimes X^x_B\right) \left( \sum _{i=0}^{d-1} \sigma _A^i \otimes \lambda _i{\left| e_i \right\rangle }_B{\left\langle e_i \right| }\right) \left( I_A \otimes X_B^{-x}\right) \right) \nonumber \\ =&\ln d+S\left( \sum _{i=0}^{d-1} \sigma _A^i \otimes \lambda _i{\left| e_i \right\rangle }_B {\left\langle e_i \right| }\right) \nonumber \\ =&\ln d+H(\Lambda )+\sum _{i=1}^{d-1}\lambda _i S\left( \sigma _A^i \right) , \end{aligned}
(56)

where the first equality is from the joint entropy theorem, the second equality is due to the unitary invariance of von Neumann entropy and the last equality is due to the joint entropy theorem together with $$H(\Lambda )=-\sum _{i}\lambda _i \ln \lambda _i$$ that is the shannon entropy of the spectrum $$\Lambda =\{\lambda _i\}$$ of $$\rho _B$$ in Eq. (6). Thus we can rewrite the von Neumann entropy of $$\Gamma _{XAB}$$ as

\begin{aligned} S\left( \Gamma _{XAB}\right) =\ln d+S\left( \rho _B\right) +\sum _{i=1}^{d-1}\lambda _i S\left( \sigma _A^i \right) . \end{aligned}
(57)

Because the classical parts of $$\Gamma _{XAB}$$ is the d-dimensional maximally mixed state $$\Gamma _X=\frac{1}{d}\sum _{x=0}^{d-1}{\left| x \right\rangle }_X {\left\langle x \right| }$$, we have the mutual information of $$\Gamma _{XAB}$$ with respect to the bipartition between X and AB as

\begin{aligned} {I}(\Gamma _{X:AB})=&S(\Gamma _X)+S(\Gamma _{AB})-S(\Gamma _{XAB}) =\ln d-S(\rho _B)+\chi ({\mathscr {E}}_0). \end{aligned}
(58)

For the von Neumann entropy of $$\Gamma _{YAB}$$ in Eq. (16), we have

\begin{aligned} S\left( \Gamma _{XAB}\right) =&\ln d +\frac{1}{d}\sum _{y=0}^{d-1} S \left( \left( I_A \otimes Z_B^y \right) \left( \sum _{j=0}^{d-1} \tau _A^j \otimes \frac{1}{d}|{\tilde{e}}_j \rangle _B \langle {\tilde{e}}_j|\right) \left( I_A \otimes Z_B^{-y}\right) \right) \nonumber \\ =&\ln d+S\left( \sum _{j=0}^{d-1} \tau _A^j \otimes \frac{1}{d}|{\tilde{e}}_j \rangle _B \langle {\tilde{e}}_j|\right) \nonumber \\ =&2\ln d+\frac{1}{d}\sum _{j=1}^{d-1} S\left( \tau _A^j \right) , \end{aligned}
(59)

where the first and third equalities are due to the joint entropy theorem and the second equality is from the unitary invariance of von Neumann entropy. Thus the mutual information of $$\Gamma _{YAB}$$ with respect to the bipartition between Y and AB is

\begin{aligned} {I}(\Gamma _{Y:AB})=&S(\Gamma _Y)+S(\Gamma _{AB})-S(\Gamma _{YAB})\nonumber \\ =&\ln d + S(\rho _A)+\ln d-2\ln d-\frac{1}{d}\sum _{j=1}^{d-1} S \left( \tau _A^j \right) \nonumber \\ =&\chi ({\mathscr {E}}_1). \end{aligned}
(60)

### Proof of Theorem 1

Let us first consider the four-qudit ccq state $$\Gamma _{XYAB}$$ of the form in Eq. (14) obtained by the two-qudit reduced density matrix $$\rho _{AB}$$ of $${\left| \psi \right\rangle }_{ABC}$$. From Eqs. (18), (19) and (20), the additivity condition of quantum mutual information for $$\Gamma _{XYAB}$$ in Eq. (23) can be rewritten as

\begin{aligned} \chi ({\mathscr {E}}_0)+\chi ({\mathscr {E}}_1)=S\left( \rho _A\right) +S\left( \rho _B\right) -S\left( \rho _{AB}\right) ={I}\left( \rho _{AB}\right) , \end{aligned}
(61)

where $${\mathscr {E}}_0$$ and $${\mathscr {E}}_1$$ are the probability ensembles of $$\rho _{A}$$ in Eqs. (7) and (10), respectively.

Because $${\mathscr {E}}_0$$ and $${\mathscr {E}}_1$$ can be obtained from measuring subsystem B of $$\rho _{AB}$$ by the rank-1 measurement $$\{{\left| e_i \right\rangle }_B{\left\langle e_i \right| }\}_{i=1}^{d-1}$$ and $$\{{\left| {\tilde{e}}_j \right\rangle }_B{\left\langle {\tilde{e}}_j \right| }\}_{j=1}^{d-1}$$, respectively, the definition of CC in Eq. (4) leads us to $${{\mathscr {J}}}^{\leftarrow }(\rho _{AB})\ge \chi ({\mathscr {E}}_0)$$ and $${{\mathscr {J}}}^{\leftarrow }(\rho _{AB})\ge \chi ({\mathscr {E}}_1)$$, therefore

\begin{aligned} {{\mathscr {J}}}^{\leftarrow }(\rho _{AB})\ge \frac{1}{2}\left( \chi ({\mathscr {E}}_0)+\chi ({\mathscr {E}}_1)\right) =\frac{1}{2} {I}\left( \rho _{AB}\right) . \end{aligned}
(62)

By considering the ccq state $$\Gamma _{XYAC}$$ obtained by $$\rho _{AC}$$ as well as Eq. (24), we can analogously have

\begin{aligned} {{\mathscr {J}}}^{\leftarrow }(\rho _{AC})\ge&\frac{1}{2} {I}\left( \rho _{AC}\right) . \end{aligned}
(63)

As the trade-off relation of Eq. (5) in Proposition 1 is universal with respect to the subsystems, we also have $$S(\rho _A)={{\mathscr {J}}}^{\leftarrow }(\rho _{AC})+E_\mathbf{f}\left( \rho _{AB}\right)$$ for the given two-qudit state $${\left| \psi \right\rangle }_{ABC}$$, therefore

\begin{aligned} E_\mathbf{f}\left( \rho _{AB}\right) +E_\mathbf{f}\left( \rho _{AC}\right) =&2S(\rho _A)-\left( {{\mathscr {J}}}^{\leftarrow }(\rho _{AB}) +{{\mathscr {J}}}^{\leftarrow }(\rho _{AC})\right) . \end{aligned}
(64)

Now Inequalities (62), (63) as well as Eq. (64) lead us to

\begin{aligned} E_\mathbf{f}\left( \rho _{AB}\right) +E_\mathbf{f}\left( \rho _{AC}\right) \le&2S(\rho _A)-\frac{1}{2}\left( {I}(\rho _{AB})+{I}(\rho _{AC})\right) \nonumber \\ =&2S(\rho _A)-\frac{1}{2}\left( S(\rho _A)+S(\rho _B)-S(\rho _{AB})+S(\rho _A)+S(\rho _C)-S(\rho _{AC})\right) \nonumber \\ =&S(\rho _A)\nonumber \\ =&E_\mathbf{f}\left( {\left| \psi \right\rangle }_{A(BC)}\right) , \end{aligned}
(65)

where the second equality is due to $$\rho _{AC}=\rho _B$$ and $$\rho _{AB}=\rho _C$$ for three-party pure state $${\left| \psi \right\rangle }_{ABC}$$.

### Proof of Theorem 2

We first prove the theorem for any three-party mixed state $$\rho _{ABC}$$, and inductively show the validity of the theorem for any n-party quantum state $$\rho _{A_1A_2\cdots A_n}$$. For a three-party mixed state $$\rho _{ABC}$$, let us consider an optimal decomposition of $$\rho _{ABC}$$ realizing EoF with respect to the bipartition between A and BC, that is,

\begin{aligned} \rho _{ABC}=\sum _i p_i{\left| \psi _i \right\rangle }_{ABC}{\left\langle \psi _i \right| }, \end{aligned}
(66)

with $$E_\mathbf{f}\left( \rho _{A(BC)}\right) =\sum _i p_i E_\mathbf{f}\left( {\left| \psi _i \right\rangle }_{A(BC)}\right)$$. From Theorem 1, each pure state $${\left| \psi _i \right\rangle }_{ABC}$$ of the decomposition (66) satisfies $$E_\mathbf{f}\left( {\left| \psi _i \right\rangle }_{A(BC)}\right) \ge E_\mathbf{f}\left( \rho ^i_{AB}\right) +E_\mathbf{f}\left( \rho ^i_{AC}\right)$$ with $$\rho ^i_{AB}={\text {tr}}_C {\left| \psi _i \right\rangle }_{ABC}{\left\langle \psi _i \right| }$$ and $$\rho ^i_{AC}={\text {tr}}_B {\left| \psi _i \right\rangle }_{ABC}{\left\langle \psi _i \right| }$$, therefore,

\begin{aligned} E_\mathbf{f}\left( \rho _{A(BC)}\right) =&\sum _i p_i E_\mathbf{f}\left( {\left| \psi _i \right\rangle }_{A(BC)}\right) \ge \sum _i p_i E_\mathbf{f}\left( \rho ^i_{AB}\right) + \sum _i p_i E_\mathbf{f}\left( \rho ^i_{AC}\right) . \end{aligned}
(67)

For each i and the two-party reduced density matrices $$\rho _{AB}^i$$, let us consider its optimal decomposition $$\rho _{AB}^i=\sum _{j}r_{ij}{\left| \mu _j^i \right\rangle }_{AB}{\left\langle \mu _j^i \right| }$$ realizing EoF, that is, $$E_\mathbf{f}\left( \rho _{AB}^i\right) =\sum _{j}r_{ij} E_\mathbf{f}\left( {\left| \mu _j^i \right\rangle }_{AB}\right)$$. Now we have

\begin{aligned} \sum _i p_i E_\mathbf{f}\left( \rho ^i_{AB}\right) =\sum _{i,j} p_i r_{ij} E_\mathbf{f}\left( {\left| \mu _j^i \right\rangle }_{AB}\right) \ge E_\mathbf{f}\left( \rho _{AB}\right) , \end{aligned}
(68)

where the inequality is due to $$\rho _{AB}=\sum _i p_i \rho _{AB}^i=\sum _{i,j}p_i r_{ij}{\left| \mu _j^i \right\rangle }_{AB}{\left\langle \mu _j^i \right| }$$ and the definition of EoF.

For each i, we also consider an optimal decomposition $$\rho _{AC}^i=\sum _{l}s_{il}{\left| \nu _l^i \right\rangle }_{AC}{\left\langle \nu _l^i \right| }$$ such that $$E_\mathbf{f}\left( \rho _{AC}^i \right) =\sum _{l}s_{il} E_\mathbf{f}\left( {\left| \nu _l^i \right\rangle }_{AC}\right)$$. We can analogously have

\begin{aligned} \sum _i p_i E_\mathbf{f}\left( \rho ^i_{AC}\right) \ge&E_\mathbf{f}\left( \rho _{AC}\right) , \end{aligned}
(69)

due to $$\rho _{AC}=\sum _i p_i \rho _{AC}^i=\sum _{i,l}p_i s_{il}{\left| \nu _l^i \right\rangle }_{AC}{\left\langle \nu _l^i \right| }$$ and the definition of EoF in Eq. (2). From Inequalities (67), (68) and (69), we have

\begin{aligned} E_\mathbf{f}\left( \rho _{A(BC)}\right) \ge E_\mathbf{f}\left( \rho _{AB}\right) +E_\mathbf{f}\left( \rho _{AC}\right) , \end{aligned}
(70)

which proves the theorem for three-party mixed states.

For general multi-party quantum system, we use the mathematical induction on the number of parties n; let us assume Inequality (51) is true for any k-party quantum state, and consider an $$k+1$$-party quantum state $$\rho _{A_1A_2\cdots A_{k+1}}$$ for $$k \ge 3$$. By considering $$\rho _{A_1A_2\cdots A_{k+1}}$$ as a three-party state with respect to the tripartition $$A_1$$, $$A_2\cdots A_k$$ and $$A_{k+1}$$, Inequality (70) leads us to

\begin{aligned} E_\mathbf{f}\left( \rho _{A_1(A_2\cdots A_{k+1})}\right) \ge&E_\mathbf{f}\left( \rho _{A_1(A_2\cdots A_k)}\right) +E_\mathbf{f}\left( \rho _{A_1A_{k+1}}\right) . \end{aligned}
(71)

As $$\rho _{A_1A_2\cdots A_k}$$ in Inequality (71) is a k-party quantum state, the induction hypothesis assures that

\begin{aligned} E_\mathbf{f}\left( \rho _{A_1(A_2\cdots A_k)}\right) \ge E_\mathbf{f}\left( \rho _{A_1A_2}\right) +\cdots +E_\mathbf{f}\left( \rho _{A_1A_k}\right) . \end{aligned}
(72)

Now Inequalities (71) and (72) lead us to the monogamy inequality in (51), which completes the proof.