# Robust Multipartite Entanglement Without Entanglement Breaking

Ming-Xing Luo1, Shao-Ming Fei2,3 1\. School of Information Science and
Technology, Southwest Jiaotong University, Chengdu 610031, China;
2\. School of Mathematical Sciences, Capital Normal University, Beijing
100048, China;
3\. Max-Planck-Institute for Mathematics in the Sciences, 04103 Leipzig,
Germany

###### Abstract

Entangled systems in experiments may be lost or offline in distributed quantum
information processing. This inspires a general problem to characterize
quantum operations which result in breaking of entanglement or not. Our goal
in this work is to solve this problem both in single entanglement and network
scenarios. We firstly propose a local model for characterizing all entangled
states that are breaking for losing particles. This implies a simple criterion
for witnessing single entanglement such as generalized GHZ states and Dicke
states. It further provides an efficient witness for entangled quantum
networks depending on its connectivity such as $k$-independent quantum
networks, completely connected quantum networks, and $k$-connected quantum
networks. These networks are universal resources for measurement-based quantum
computations. The strong nonlocality can be finally verified by using
nonlinear inequalities. These results show distinctive features of both single
entangled systems and entangled quantum networks.

##  I Introduction

As one of the remarkable features in quantum mechanics, quantum entanglement
has attracted great attentions EPR . The quantum correlations generated by
local measurements on entangled two-spin systems can not be reproduced from
classical physics. These nonlocal quantum correlations are verified by
violating Bell inequalities Bell ; CHSH ; Gis . For multipartite scenarios the
quantum correlations may be verified in specific local models under different
assumptions GHZ ; Merm ; Sy ; BCPS . The entangled states have become
important resources in various ongoing studies LJL ; ZZH ; HHH .

Experimentally states based on atomic ensembles provide attractive systems for
both the storage of quantum information and the coherent conversion of quantum
information between atomic and optical degrees of freedom HSS . However, under
the evolution $U_{t}=e^{it\mathbf{H}}$ with Hamiltonian $\mathbf{H}$, the
nucleus of an unstable isotope may lose one of several particles including
neutrons, alpha particles, electrons or positrons PKGR , see Fig. 1(a). The
particle-lose channel ${\cal E}_{S}={\rm Tr}_{S}$, where the partial trace
goes over the lost particles contained in set $S$, may be an entanglement-
breaking channel Holevo ; HSR . The output is given by $\rho_{out}={\cal
E}_{S}(\rho_{in})$ with input $\rho_{in}$. It is marvelous that some states
like Dicke states keep entangled after passing through the particle-lose
channel GZ ; QABZ ; BPB ; SZD . This inspires a natural problem of
characterizing multipartite quantum systems in term of particle-lose channels.

Compared with entangled qubit systems EPR ; GHZ , high-dimensional entangled
systems may inherent local tensor decompositions, see Fig.1 (b), which
intrigue distributed experiments in preparing quantum networks Kim ; LJL . One
example is the cluster states that are universal resources for measurement-
based quantum computations Cluster . In this case, some experimental devices
may be not online in large-scale or remote tasks, regarded as party-lose
noises in which all the local particles shared by one party are taken as one
unavailable high-dimensional particle or quantum sources. Different from the
permutationally symmetric states GZ ; QABZ ; BPB ; SZD , these entangled
systems depend on network configurations, which give rise to an interesting
problem in characterizing distributed quantum resources for quantum
information processing.

Figure 1: (Color online) (a) Atomic ensembles with losing particles. (b)
Quantum networks with offline parties. High-dimensional states may be prepared
by distributed experiments with small entangled systems such as Affleck-
Kennedy-Lieb-Tasaki (AKLT) system AKLT .

In this work, we explore the multipartite entangled systems both in Bell and
network scenarios. The main idea is to characterize the strong correlations
against particle loss. We firstly propose a local model for featuring fragile
entangled states, that is, the entanglement breaks for some particle-lose
channels. This model is stronger than the biseparable model Sy  because GHZ
states GHZ  are not entangled in the present model. It is also different from
the network local model NWR ; Kraft ; Luo2020  that can witness GHZ states GHZ
. Especially, the generalized GHZ states are the unique kind of fragile
entangled states that become fully separable under any one particle-loss
channel. Instead, the generalized Dicke states Toth  keep entangled after
particle loss. This shows distinctive features beyond previous models Sy ; GHZ
; Cluster ; SAS ; GS . Moreover, it is extended to witness distributed
resources such as $k$-independent quantum networks Luo2018 , completely
connected quantum networks, or $k$-connected quantum networks AKLT ; Cluster
according to the connectivity. The strong nonlocality is verified by using
nonlinear Bell inequalities. The results may shed new light on network
entanglement and quantum information processing such as distributed quantum
computation.

##  II Fragile multipartite entanglement

Let $\mathcal{H}_{A_{j}}$ be the Hilbert space associated with the particle
${A_{j}}$, and $\rho_{A_{1}\cdots A_{n}}$ an $n$-partite state in
$\otimes_{j=1}^{n}\mathcal{H}_{A_{j}}$. Denote $S$ as a subset of
$\\{A_{1},\cdots,A_{n}\\}$, and $\overline{S}$ as the complement set of $S$.
Let ${\cal E}_{S}(\cdot{})$ be a complete positive trace preserving (CPTP)
mapping of the particle-lose channel associated with the particle set $S$,
that is, ${\cal E}_{S}(\rho)=\sum_{j\in K_{S}}E_{j}\rho{}E^{\dagger}_{j}$ for
any state $\rho$, where $K_{S}$ denotes the Kraus operator decomposition of
${\cal E}_{S}(\cdot{})={\rm Tr}_{A_{j}\in S}(\cdot{})$ and ${\rm Tr}_{A_{j}\in
S}(\cdot{})$ denotes the partial trace operator of particles in $S$. For
instance, when $S=\\{A_{1}\\}$, we have $E_{j}=\langle
j|_{A_{1}}\otimes\mathbbm{1}_{A_{2}\cdots A_{n}}$, where
$\mathbbm{1}_{A_{2}\cdots A_{n}}$ is the identity operator on particles
$A_{2},\cdots,A_{n}$. Let $\rho_{(S)}$ denote the output state after $\rho$
passing through the particle-lose channel ${\cal E}_{S}(\cdot{})$, that is,
$\rho_{(S)}={\cal E}_{S}(\rho)$. The main motivation here is to explore new
features of multipartite entangled systems both in single source and network
configurations, which cannot be carried out by using the biseparable model Sy
, network Bell inequalities Luo2018  or network local model NWR ; Kraft ;
Luo2020 .

Our main result in what follows is based on the biseparable model Sy , that
is, an $n$-particle state of $\rho_{A_{1}\cdots A_{n}}$ on Hilbert space
$\otimes_{j=1}^{n}\mathcal{H}_{A_{j}}$ is genuinely $n$-partite entanglement
if it cannot be decomposed into as

$\displaystyle\rho_{bs}=\sum_{S}\sum_{j}p_{j;S}\rho_{j}^{(S)}\otimes\rho_{j}^{(\overline{S})}$
(1)

where $S$ and $\overline{S}$ are any bipartition of
$\\{A_{1},\cdots,A_{n}\\}$, $\\{p_{j;S}\\}$ is a probability distribution,
$\rho_{j}^{(S)}$ and $\rho_{j}^{(\overline{S})}$ are respectively states of
particles in $S$ and $\overline{S}$. Here, $\rho_{bs}$ in Eq.(1) is named as
biseparable state Sy .

Definition 1. An $n$-partite state $\rho$ is particle-lose separable if
$\rho_{(S)}$ is biseparable for some set $S$ with no more than $n-2$
particles. Otherwise, $\rho$ is robust entanglement.

Consider the GHZ state
$|\Phi\rangle=\frac{1}{\sqrt{2}}(|000\rangle+|111\rangle)$ GHZ , the reduced
state is given by $\rho=\frac{1}{2}(|00\rangle\langle 00|+|11\rangle\langle
11|)$ which is separable. This implies the GHZ state is particle-lose
separable. Instead, for Dicke state of
$|\Phi\rangle=\frac{1}{\sqrt{3}}(|001\rangle+|010\rangle+|100\rangle)$ Toth ,
we will prove it is robust entanglement in the next section. This means that
Definition 1 gives stronger conditions than the $k$-uniform entanglement GHZ
because all the $k$-uniform $n$-particle entangled states which require all
its reductions to $k$-qudits being maximally mixed are particle-lose separable
for $k\leq n-2$. Moreover, different from the lose channel with fixed
particles for symmetric Dicke states BPB ; SZD  of general symmetric states,
the present model may lose any particles with the only assumption of particle
number.

###  II.1 Qubit states

Consider the generalized GHZ state GHZ ; Ghose09  given by

$\displaystyle|GHZ\rangle_{A_{1}\cdots A_{n}}=\cos\theta|0\rangle^{\otimes
n}+\sin\theta|1\rangle^{\otimes n},$ (2)

where $\theta\in(0,\frac{\pi}{4})$. Note that $|GHZ\rangle$ is genuinely
multipartite entangled Sy  for $\theta\in(0,\frac{\pi}{4})$. However, it
reduces to a fully separable state after losing any one qubit. Namely, it is
weakly entangled with respect to the particle-loss noises. Interestingly, this
kind of states is unique under local unitary operations.

Proposition 1. For a genuinely pure entangled qubit state
$|\Phi\rangle_{A_{1}\cdots A_{n}}$ on Hilbert space
$\otimes_{j=1}^{n}\mathbbm{H}_{A_{j}}$, it is equivalent to the generalized
GHZ state in Eq.(2) if and only if the reduced state of
$\rho_{(A_{j})}={\cal{}E}_{A_{j}}(|\Phi\rangle\langle\Phi|)$ is fully
separable for any $j\in\\{1,\cdots,n\\}$, that is, there are local unitary
operations $U_{j}$ such that

$\displaystyle(\otimes_{j=1}^{n}U_{j})|\Phi\rangle_{A_{1}\cdots
A_{n}}=|GHZ\rangle_{A_{1}\cdots A_{n}}.$ (3)

Proof. For a generalized GHZ state in Eq.(2), it is easy to prove that
$\rho_{(A_{j})}=\cos^{2}\theta(|0\rangle\langle 0|)^{\otimes
n-1}+\sin^{2}\theta(|1\rangle\langle 1|)^{\otimes n-1}$, which is fully
separable. This has proved the sufficient condition.

In what follows, we prove the necessary condition, that is, for a genuinely
pure entangled qubit state $|\Phi\rangle_{A_{1}\cdots A_{n}}$ on Hilbert space
$\otimes_{j=1}^{n}\mathbbm{H}_{A_{j}}$, if the reduced state of
$\rho_{(A_{j})}={\cal{}E}_{A_{j}}(|\Phi\rangle\langle\Phi|)$ is fully
separable for any $j\in\\{1,\cdots,n\\}$ there are local unitary operations
$U_{j}$ satisfying Eq.(55). From this assumption, we get
$|\Phi\rangle_{A_{1}\cdots A_{n}}$ is a genuinely $n$-partite entanglement
while $\rho_{(A_{j})}$ is fully separable.

The followed proof is completed by induction on $j$.

For $j=1$, we consider the bipartition $\\{A_{1}\\}$ and
$\\{A_{2},\cdots,A_{n}\\}$ of $\\{A_{1}$, $\cdots$, $A_{n}\\}$. The Schmidt
decomposition of $|\Phi\rangle$ is given by

$\displaystyle|\Phi\rangle_{A_{1}\cdots
A_{n}}=\sum_{i=0}^{1}\sqrt{p_{i}}|\psi_{i}\rangle_{A_{1}}|\Phi_{i}\rangle_{A_{2}\cdots
A_{n}},$ (4)

where $\\{|\psi_{i}\rangle,\forall i\\}$ are orthogonal states of qubit
$A_{1}$, and $\\{|\Phi_{i}\rangle,\forall i\\}$ are orthogonal states of
qubits $A_{2},\cdots,A_{n}$, and $p_{i}$ are positive Schmidt coefficients.

For $j=2$, we will prove that there two qubits $A_{1}$ and $A_{2}$ are
symmetric under local unitary operations, that is, there are unitary
operations $U_{A_{t}}^{(t)}$ on the qubits $A_{1}$ and $A_{2}$ satisfying

$\displaystyle(U_{A_{1}}^{(1)}\otimes{}U_{A_{2}}^{(2)}\otimes\mathbbm{1}_{3})|\Phi\rangle=\sum_{s=0}^{1}\sqrt{p_{s}}|ss\rangle_{A_{1}A_{2}}|\Psi_{s}\rangle_{A_{3}\cdots
A_{n}},$ (5)

where $\mathbbm{1}_{3}$ denotes the identity operator on qubits
$A_{3},\cdots{},A_{n}$, and $|\Psi_{s}\rangle_{A_{3}\cdots A_{n}}$ are
orthogonal states on qubits $A_{3},\cdots{},A_{n}$. In fact, from the
assumption of necessary condition,
${\cal{}E}_{A_{1}}(|\Phi\rangle\langle\Phi|)$ is fully separable. Combined
with Eq.(4) the reduced state of $\varrho_{(A_{1})}$ is given by

$\displaystyle\varrho_{(A_{1})}$ $\displaystyle=$
$\displaystyle{\cal{}E}_{A_{1}}(|\Phi\rangle\langle\Phi|)$ $\displaystyle=$
$\displaystyle{\cal{}E}_{A_{1}}((U_{A_{1}}^{(1)}\otimes\mathbbm{1}_{2})|\Phi\rangle\langle\Phi|((U_{A_{1}}^{(1)})^{\dagger}\otimes\mathbbm{1}_{2}))$
$\displaystyle=$
$\displaystyle\sum_{s=0}^{1}\langle\psi_{s}|_{A_{1}}|\Phi\rangle\langle\Phi|\psi_{s}\rangle_{A_{1}}$
$\displaystyle=$
$\displaystyle\sum_{s=0}^{1}p_{s}|\Phi_{s}\rangle_{A_{2}\cdots{}A_{n}}\langle\Phi_{s}|,$
(7)

where $\mathbbm{1}_{2}$ is the identity operator on qubits
$A_{2},\cdots,A_{n}$. Eq.(II.1) is followed from the invariance of the
particle-lose channel ${\cal{}E}_{A_{1}}(\cdot)$ under local unitary
operations in the orthogonal states $|\psi_{s}\rangle_{A_{1}}$. Another method
to derive Eq.(7) is using the normal form of multipartite pure entanglement
under local unitary operations Kraus2009 , that is, one firstly performs the
unitary operation $U_{A_{1}}^{(1)}$ on $|\Phi\rangle$ in Eq.(4) and then
implements the particle-lose channel ${\cal{}E}_{A_{1}}(\cdot)$.

From the assumption of necessary condition, $\varrho_{(A_{1})}$ in Eq.(7) is
fully separable. So, there are four cases for $|\Phi_{0}\rangle$ and
$|\Phi_{1}\rangle$:

  * (C1)

Both states of $|\Phi_{0}\rangle$ and $|\Phi_{1}\rangle$ given in Eq.(7) are
fully separable.

  * (C2)

Only one state $|\Phi_{0}\rangle$ (for example) given in Eq.(7) is fully
separable.

  * (C3)

Both states of $|\Phi_{0}\rangle$ and $|\Phi_{1}\rangle$ given in Eq.(7) are
entangled.

  * (C4)

One state $|\Phi_{0}\rangle$ (for example) in Eq.(7) is biseparable Sy .

The case (C1) will be proved in Lemma 1 shown in Appendix A. The case (C2)
will be proved in Lemma 2 shown in Appendix B. The case (C3) will be proved in
Lemma 3 shown in Appendix C. Moreover, the case (C4) will be reduced to the
case (C2) in the following paragraph.

Lemma 1. If both states of $|\Phi_{0}\rangle$ and $|\Phi_{1}\rangle$ given in
Eq.(7) are fully separable, then $|\Phi\rangle$ in Eq.(4) is equivalent to the
generalized GHZ state in Eq.(2) under local unitary operations.

Lemma 2. If one state $|\Phi_{0}\rangle$ (for example) given in Eq.(7) is
fully separable, then $|\Phi\rangle$ in Eq.(4) is equivalent to the
generalized GHZ state in Eq.(2) under local unitary operations.

Lemma 3. If both states of $|\Phi_{0}\rangle$ and $|\Phi_{1}\rangle$ given in
Eq.(7) are genuinely entangled, then the qubits $A_{1}$ and $A_{2}$ in
$|\Phi\rangle$ in Eq.(4) are symmetric under local unitary operations.

Now, continue the proof. For the case (C1), from Lemma 1 $|\Phi\rangle$ in
Eq.(4) is equivalent to the generalized GHZ state in Eq.(2) under local
unitary operations, that is, all the qubits are symmetric under specific local
unitary operations. For the case (C2), from Lemma 2 the state $|\Phi\rangle$
in Eq.(4) is equivalent to the generalized GHZ state in Eq.(2) under local
unitary operations. This has completed the proof. In both cases we have proved
the necessary condition of Proposition 1.

For the case (C3), from Lemma 3 there are two qubits $A_{1}$ and $A_{2}$ of
$|\Phi\rangle$ in Eq.(4) are symmetric under local unitary operations. This
has proved Eq.(5).

In what follows, we prove the case (C4) will be reduced to the case (C3), that
is, assume that $|\Phi_{0}\rangle_{A_{2}\cdots A_{n}}$ given in Eq.(7) is
biseparable for one bipartition $\\{A_{2},\cdots,A_{k}\\}$ and
$\\{A_{k+1},\cdots,A_{n}\\}$ of $\\{A_{2},\cdots,A_{n}\\}$ (for simplicity) Sy
, that is,

$\displaystyle|\Phi_{0}\rangle_{A_{2}\cdots
A_{n}}=|\Phi_{00}\rangle_{A_{2}\cdots
A_{k}}|\Phi_{01}\rangle_{A_{k+1}\cdots{}A_{n}},$ (8)

where $|\Phi_{00}\rangle$ is an genuinely entangled state of qubits
$A_{2},\cdots,A_{k}$, and $|\Phi_{01}\rangle$ is a general state (which may be
not entangled) of qubits $A_{k+1},\cdots,A_{n}$. In this case,
$|\Phi_{1}\rangle$ given in Eq.(7) is biseparable in terms of the same
bipartition, that is,

$\displaystyle|\Phi_{1}\rangle_{A_{2}\cdots
A_{n}}=|\Phi_{10}\rangle_{A_{2}\cdots
A_{k}}|\Phi_{11}\rangle_{A_{k+1}\cdots{}A_{n}},$ (9)

where $|\Phi_{10}\rangle$ is a general state of qubits $A_{2},\cdots,A_{k}$,
and $|\Phi_{11}\rangle$ is a general state of qubits $A_{k+1},\cdots,A_{n}$.
Otherwise, $|\Phi_{1}\rangle_{A_{2}\cdots A_{n}}$ given in Eq.(7) is a
genuinely $n-1$-partite entanglement in the biseparable model Sy . This
implies that $\varrho_{(A_{1})}$ in Eq.(7) is a genuinely $n-1$-partite
entanglement from the definition of the biseparable model Sy , that is,
$\varrho_{(A_{1})}$ is a mixed state of one biseparable state
$|\Phi_{0}\rangle_{A_{2}\cdots A_{n}}$ and one genuinely $n-1$-partite
entanglement $|\Phi_{1}\rangle_{A_{2}\cdots A_{n}}$, which cannot be
decomposed into a mixture of all the biseparable states. This contradicts to
the assumption that $\varrho_{(A_{1})}$ is fully separable. Hence,
$|\Phi_{1}\rangle_{A_{2}\cdots A_{n}}$ has the decomposition in Eq.(9).

From Eqs.(8) and (9), the reduced state of $\rho_{(A_{1}A_{k+1}\cdots A_{n})}$
after $\rho_{(A_{1})}$ in Eq.(7) passing through the particle-lose channel
${\cal E}_{A_{k+1}\cdots A_{n}}(\cdot)$ has the following decomposition as

$\displaystyle\rho_{(A_{1}A_{k+1}\cdots A_{n})}$ $\displaystyle=$
$\displaystyle{\cal E}_{A_{k+1}\cdots A_{n}}(\rho_{(A_{1})})$ (10)
$\displaystyle:=$
$\displaystyle\sum_{s=0}^{1}q_{s}|\Phi_{s0}\rangle_{A_{2}\cdots
A_{k}}\langle\Phi_{s0}|,$

where $\\{q_{s}\\}$ is a probability distribution which may be different from
$\\{p_{s}\\}$ in Eq.(9). From the separability assumption of $\rho_{(A_{1})}$,
$\rho_{(A_{1}A_{k+1}\cdots A_{n})}$ should be fully separable. Moreover, from
assumption in Eq.(8), $|\Phi_{00}\rangle_{A_{2}\cdots A_{k}}$ is an genuinely
entangled state of qubits $A_{2},\cdots,A_{k}$. It follows that
$|\Phi_{10}\rangle_{A_{2}\cdots A_{k}}$ should be entangled. Otherwise, that
is, $|\Phi_{10}\rangle_{A_{2}\cdots A_{k}}$ is biseparable. This follows an
entangled state $\rho_{(A_{1}A_{k+1}\cdots A_{n})}$ in Eq.(10) because it is a
mixed state of the genuinely entangled state $|\Phi_{00}\rangle_{A_{2}\cdots
A_{k}}$ and the biseparable state $|\Phi_{10}\rangle_{A_{2}\cdots A_{k}}$. It
contradicts to the assumption of the necessary condition that states
$\rho_{(A_{1}A_{k+1}\cdots A_{n})}$ should be fully separable. Hence,
$\rho_{(A_{1}A_{k+1}\cdots A_{n})}$ in Eq.(10) is the mixed state of two
genuinely entangled states. From Lemma 3, three are two qubits $A_{1}$ and
$A_{v}$ (with $v=2$ for simplicity) in $\rho_{(A_{1}A_{k+1}\cdots A_{n})}$ in
Eq.(10) are symmetric under local unitary operations, that is, two qubits
$A_{1}$ and $A_{2}$ in $|\Phi\rangle_{A_{1}\cdots A_{n}}$ in Eq.(4) are
symmetric under specific local unitary operations. We have completed the proof
for the case (C4).

For any $j=k-1$, assume that qubits $A_{1},\cdots,A_{j}$ are symmetric under
local operations, that is, there are unitary operations $U_{A_{t}}^{(t)}$ on
qubits $A_{t}$ such that

$\displaystyle(\otimes_{t=1}^{k-1}U_{A_{t}}^{(t)}\otimes\mathbbm{1}_{k})|\Phi\rangle_{A_{1}\cdots{}A_{n}}$
(11) $\displaystyle=$
$\displaystyle\sum_{s=0}^{1}\sqrt{p_{s}}|s\cdots{}s\rangle_{A_{1}\cdots
A_{k-1}}|\hat{\Phi}_{s}\rangle_{A_{k}\cdots A_{n}},$

where $\mathbbm{1}_{k}$ is the identity operator on qubits
$A_{k},\cdots{},A_{n}$, and $|\hat{\Phi}_{s}\rangle_{A_{k}\cdots A_{n}}$ are
orthogonal states. We will prove the result for $j=k$, that is, there is at
least one qubit $A_{\ell_{k}}$ such that $A_{1},\cdots,A_{k-1}$ and
$A_{\ell_{k}}$ are symmetric under specific local unitary operations. In fact,
from Eq.(11) we get the reduced state

$\displaystyle\rho_{(A_{1}\cdots{}A_{k-1})}$ $\displaystyle=$
$\displaystyle{\cal{}E}_{A_{1}\cdots{}A_{k-1}}(|\Phi\rangle\langle\Phi|)$ (12)
$\displaystyle=$
$\displaystyle\sum_{s=0}^{1}p_{s}|\hat{\Phi}_{s}\rangle_{A_{k}\cdots
A_{n}}\langle\hat{\Phi}_{s}|$

The followed proof is similar to the proof after Eq.(7) by considering the
four cases (C1)-(C4) for two states $|\hat{\Phi}_{0}\rangle_{A_{k}\cdots
A_{n}}$ and $|\hat{\Phi}_{1}\rangle_{A_{j}\cdots A_{n}}$. Hence, there is at
least one qubit $A_{\ell_{k}}$ ($\ell_{k}=k$ for simplicity) such that
$A_{1},\cdots,A_{k-1}$ and $A_{\ell_{k}}$ are symmetric under specific local
unitary operations, that is, there are unitary operations $U_{A_{t}}^{(t)}$ on
qubit $A_{t}$ such that

$\displaystyle(\otimes_{t=1}^{k}U_{A_{t}}^{(t)}\otimes\mathbbm{1}_{k+1})|\Phi\rangle$
(13) $\displaystyle=$
$\displaystyle\sum_{s=0}^{1}\sqrt{p_{s}}|s\cdots{}s\rangle_{A_{1}\cdots
A_{k}}|\tilde{\Phi}_{s}\rangle_{A_{k+1}\cdots A_{n}},$

where $\mathbbm{1}_{k+1}$ is the identity operator on qubits
$A_{k+1},\cdots{},A_{n}$, and $|\tilde{\Phi}_{s}\rangle_{A_{k+1}\cdots A_{n}}$
are orthogonal states. This has completed the proof for $j=k$.

This procedure will be ended when $j=n-1$. Finally, all the qubits
$A_{1},\cdots,A_{n}$ are symmetric under specific local unitary operations.
From Eq.(13), $|\Phi\rangle$ in Eq.(4) is equivalent to the generalized
$n$-qubit GHZ state in Eq.(2) under local unitary operations. This completes
the proof. $\Box$

Proposition 1 implies the uniqueness of such weakly entangled qubit systems.
It shows one difference between the present model and the biseparable model Sy
or the network model NWR ; Kraft ; Luo2020  in which the generalized GHZ
states are entangled. Such property may hold for high-dimensional entangled
pure states including the absolute maximal entangled states GW ; HCL ; HGS ,
or the symmetric states which will be proved in the next section.

###  II.2 High-dimensional entangled symmetric pure states

In this subsection, we extend Proposition 1 to high-dimensional
permutationally symmetric states. For an $n$-partite $d$-dimensional state
$|\Phi\rangle_{A_{1}\cdots A_{n}}$ on Hilbert space
$\otimes_{j=1}^{n}\mathbb{H}_{A_{j}}$, it is permutationally symmetric if
$|\Phi\rangle_{A_{1}\cdots A_{n}}$ is invariant under any permutation
operation (swapping the particles). Define a $d$-dimensional generalized GHZ
state as

$\displaystyle|GHZ_{d}\rangle=\sum_{j=0}^{d-1}\alpha_{j}|j\cdots
j\rangle_{A_{1}\cdots A_{n}},$ (14)

where $\alpha_{j}$’s satisfy $\sum_{j=0}^{d-1}\alpha_{j}^{2}=1$ and
$\theta\in(0,\frac{\pi}{4}]$.

Proposition 2. For a permutationally symmetric $n$-partite entanglement
$|\Phi\rangle_{A_{1}\cdots A_{n}}$, it is equivalent to the generalized GHZ
state in Eq.(14) if and only if
$\rho_{(A_{j})}={\cal{}E}_{A_{j}}(|\Phi\rangle\langle\Phi|)$ is fully
separable for any $j\in\\{1,\cdots,n\\}$, that is, there are local unitary
operations $U_{A_{j}}^{(j)}$ on particle $A_{j}$ such that

$\displaystyle(\otimes_{j=1}^{n}U_{A_{j}}^{(j)})|\Phi\rangle_{A_{1}\cdots
A_{n}}=|GHZ_{d}\rangle.$ (15)

Proof. The sufficient condition is easily followed from Eq.(14). In what
follows, we prove that a permutationally symmetric $n$-partite entanglement
$|\Phi\rangle$ is equivalent to the generalized GHZ state in Eq.(14) if
$\rho_{(A_{j})}={\cal{}E}_{A_{j}}(|\Phi\rangle\langle\Phi|)$ is fully
separable for any $j\in\\{1,\cdots,n\\}$. Note that
$\\{|GHZ_{d}\rangle,|D_{k,n}\rangle,\forall k\\}$ consists the orthogonal
basis of permutationally symmetric subspace of
$\otimes_{j=1}^{n}\mathbb{H}_{A_{j}}$, where $|D_{k,n}\rangle$ denotes the
maximally entangled $n$-particle Dicke states Toth . It means that any
permutationally symmetric state $|\Phi_{sy}\rangle_{A_{1}\cdots A_{n}}$ can be
decomposed into

$\displaystyle|\Phi_{sy}\rangle=\sum_{k}\beta_{k}|D_{k,n}\rangle_{A_{1}\cdots
A_{n}}+\beta|GHZ_{d}\rangle_{A_{1}\cdots A_{n}}$ (16)

where $\beta_{k}$’s and $\beta$ satisfy $\sum_{i}|\beta_{k}|^{2}+\beta^{2}=1$.

Note that

$\displaystyle\sum_{k}\beta_{k}|D_{k,n}\rangle_{A_{1}\cdots A_{n}}$ (17)
$\displaystyle=$
$\displaystyle\sum_{j_{1}=0}^{d-1}|j_{1}\rangle_{A_{1}}\sum_{k}\frac{\beta_{k}}{\sqrt{N_{k-j_{1},n}}}|D_{k-j_{1}}\rangle_{A_{2}\cdots
A_{n}},$

where $|D_{k-j_{1},n}\rangle$ denotes $n-1$-particle Dicke state with
excitations $k-j_{1}$ defined by

$\displaystyle|D_{k-j_{1},n}\rangle=\frac{1}{\sqrt{N_{k-j_{1},n}}}\sum_{j_{2}+\cdots+j_{n}=k-j_{1}}|j_{2}\cdots{}j_{n}\rangle$
(18)

and $N_{k-j_{1},n}$ denotes the normalization constant of
$|D_{k-j_{1},n}\rangle$.

The reduced state after $|\Phi_{sy}\rangle$ in Eq.(16) passing through the
particle-lose channel ${\cal{}E}_{A_{1}}(\cdot)$ is given by

$\displaystyle\rho_{(A_{1})}$ $\displaystyle=$
$\displaystyle{{\cal{}E}}_{A_{1}}(|\Phi_{sy}\rangle\langle\Phi_{sy}|)$ (19)
$\displaystyle=$ $\displaystyle\beta^{2}\sum_{j=0}^{d-1}\alpha_{j}^{2}|j\cdots
j\rangle_{A_{2}\cdots{}A_{n}}\langle j\cdots j|$
$\displaystyle+{\cal{}E}_{A_{1}}(\sum_{k,s}\beta_{k}\beta_{s}|D_{k,n}\rangle_{A_{2}\cdots{}A_{n}}\langle
D_{s,n}|)$ $\displaystyle=$
$\displaystyle\sum_{k}\frac{\beta_{k}^{2}}{N_{k-j_{1},n}}|D_{k-j_{1},n-1}\rangle_{A_{2}\cdots{}A_{n}}\langle
D_{k-j_{1},n-1}|$
$\displaystyle+\beta^{2}\sum_{j=0}^{d-1}\alpha_{j}^{2}|j\cdots
j\rangle_{A_{2}\cdots{}A_{n}}\langle j\cdots j|$

From Eq.(18) the states $|D_{k_{1}-j_{1},n-1}\rangle$ and
$|D_{k_{2}-j_{1},n-1}\rangle$ have different excitations for any
$k_{1}\not=k_{2}$. This implies that all the states of
$\\{|D_{k-j_{1},n-1}\rangle,\forall k\\}$ are defined on different orthogonal
subspaces ${\cal H}_{k_{1}}$ and ${\cal H}_{k_{2}}$ of
$\otimes_{j=2}^{n}\mathbb{H}_{A_{j}}$, where ${\cal H}_{k_{s}}$ is spanned by
all the states $|j_{2}\cdots{}j_{n}\rangle$ with
$j_{2}+\cdots+j_{n}=k_{s}-j_{1}$, $s=1,2$. Hence,
$\\{|D_{k-j_{1},n-1}\rangle,\forall k\\}$ are orthogonal states. Moreover, the
state of $|D_{k-j_{1},n-1}\rangle$ is a genuinely entangled Sy . It follows
that

$\displaystyle|\Psi\rangle_{A_{2}\cdots{}A_{n}}=\sum_{k}\frac{\beta_{k}^{2}}{N_{k-j_{1},n}}|D_{k-j_{1},n-1}\rangle\langle
D_{k-j_{1},n-1}|$ (20)

is genuinely entangled DPR ; Dicke4  if $\sum_{k}\beta_{k}^{2}\not=0$.
Moreover, from the assumption, $\rho_{(A_{1})}$ is fully separable. Note that
$\sum_{j=0}^{d-1}\alpha_{j}^{2}|j\cdots j\rangle_{A_{2}\cdots{}A_{n}}\langle
j\cdots j|$ is fully separable. This implies that $\sum_{k}\beta_{k}^{2}=0$,
that is, $\beta_{k}=0$ for all $k$’s. It follows from Eq.(19) that

$\displaystyle\rho_{(A_{1})}=\beta^{2}\sum_{j=0}^{d-1}\alpha_{j}^{2}|j\cdots
j\rangle_{A_{2}\cdots{}A_{n}}\langle j\cdots j|$ (21)

Hence, from Eq.(21) and the symmetry assumption, it follows that
$|\Phi_{sy}\rangle_{A_{1}\cdots A_{n}}$ in Eq.(16) is equivalent to a
generalized GHZ state $|GHZ_{d}\rangle$ defined in Eq.(14). This completes the
proof. $\Box$

Proposition 2 has considered the permutationally symmetric $n$-partite
entangled states. A natural problem is to consider general high-dimensional
entangled states. One example is
$|\Phi\rangle_{A_{1}A_{2}A_{3}}=\frac{1}{2}(|000\rangle+|011\rangle+|120\rangle+|131\rangle)$.
The reduced state after $|\Phi\rangle_{A_{1}A_{2}A_{3}}$ passing through the
particle-lose channel ${\cal E}_{A_{2}}(\cdot)$ is separable but not for other
channels ${\cal E}_{A_{1}}(\cdot)$ and ${\cal E}_{A_{3}}(\cdot)$. This means
that the reduced states after passing particle-losing channel have different
properties. Hence, it may be difficult to get general result for all high-
dimensional entangled states.

##  III Robust multipartite entanglement

Let ${\cal B}_{d}$ consist of all the particle-lose separable states. For any
state $\rho\not\in{\cal B}_{d}$, Definition 1 shows strong robustness in terms
of particle-lose channels. One example is W state DVC ; AR ; Sy , which can be
witnessed by using the PPT criterion PPT . Different from entanglement
theories Bell ; Sy ; Luo2020 , ${\cal B}_{d}$ is only star-convex, see
Appendix D. The star-convex means that there is one center state
$\rho_{0}\in{\cal B}_{d}$ (such as the maximally mixed state or diagonal
states
$\rho=\sum_{j_{1},\cdots,j_{n}}p_{j_{1}\cdots{}j_{n}}|j_{1}\cdots{}j_{n}\rangle\langle
j_{1}\cdots{}j_{n}|$ with the probability distribution
$\\{p_{j_{1}\cdots{}j_{n}}\\}$) satisfying $p\rho_{0}+(1-p)\rho\in{\cal
B}_{d}$ for any $\rho\in{\cal B}_{d}$ and $p\in(0,1)$. Without the convexity
Rudin , it is difficult to verify a general robust entanglement. The problem
may be simplified for special states. One is the generalized $d$-dimensional
Dicke state on Hilbert space $\otimes_{j=1}^{n}\mathbb{H}_{A_{j}}$ Toth  given
by

$\displaystyle|D_{k,n}\rangle_{A_{1}\cdots{}A_{n}}=\sum_{j_{1}+\cdots{}+j_{n}=k}\alpha_{j_{1}\cdots{}j_{n}}|j_{1}\cdots{}j_{n}\rangle,$
(22)

where $k$ denotes the number of excitations, $n$ denotes the number of
particles, $\alpha_{j_{1}\cdots{}j_{n}}$’s satisfy
$\sum_{j_{1},\cdots{},j_{n}}\alpha_{j_{1}\cdots{}j_{n}}^{2}=1$ and
$\alpha_{j_{1}\cdots{}j_{n}}\not=0$ for any $j_{1},\cdots,j_{n}$.

Result 1. Any generalized Dicke state in Eq.(22) is robust entanglement.

Proof. Note that $|D_{k,n}\rangle_{A_{1}\cdots{}A_{n}}$ is a generalized
symmetric state which means that after any permutation $S$ of particles the
final state $S|D_{k,n}\rangle$ has the form similar to Eq.(22) (which may have
different superposed amplitudes). This is weaker than the permutationally
symmetry that requires the state to be invariant under any permutations of
particles. From the generalized symmetry of Dicke states $|D_{k,n}\rangle$, it
only needs to consider the reduced state of
$\rho_{(A_{j+1}\cdots{}A_{n})}={\cal E}_{A_{j+1}\cdots
A_{n}}(|D_{k,n}\rangle\langle D_{k,n}|)$ for $j=2,\cdots,n-1$.

Take $n=3$ as an example. Note that

$\displaystyle|D_{k,3}\rangle_{A_{1}A_{2}A_{3}}$ $\displaystyle=$
$\displaystyle\sum_{j_{1}+j_{2}=k-j_{3}}\alpha_{j_{1}j_{2}}|j_{1}j_{2}\rangle_{A_{1}A_{2}}\sum_{j_{3}}\alpha_{j_{3}}|j_{3}\rangle_{A_{3}}$
(23) $\displaystyle=$
$\displaystyle\sum_{j_{3}}\alpha_{j_{3}}|j_{3}\rangle_{A_{3}}|D_{k-j_{3},2}\rangle_{A_{1}A_{2}},$

where $|D_{k-j_{3},2}\rangle$ is a generalized Dicke state on Hilbert space
$\mathbb{H}_{A_{1}}\otimes\mathbb{H}_{A_{2}}$ and is defined by

$\displaystyle|D_{k-j_{3},2}\rangle_{A_{1}A_{2}}=\sum_{j_{1}+j_{2}=k-j_{3}}\alpha_{j_{1}j_{2}}|j_{1}j_{2}\rangle.$
(24)

The density matrix of $\rho_{(A_{3})}$ is given by

$\displaystyle\rho_{(A_{3})}=\sum_{j_{3}}\alpha_{j_{3}}^{2}|D_{k-j_{3},2}\rangle_{A_{1}A_{2}}\langle
D_{k-j_{3},2}|.$ (25)

From Eq.(25), for any $j_{3}$ and $j_{3}^{\prime}$ with
$j_{3}\not=j_{3}^{\prime}$, both the states of $|D_{k-j_{3},2}\rangle$ and
$|D_{k-j_{3}^{\prime},2}\rangle$ are generalized Dicke states with excitations
$k-j_{3}$ and $k-j_{3}^{\prime}$ satisfying $k-j_{3}\not=k-j_{3}^{\prime}$.
This implies that $|D_{k-j_{3},2}\rangle$ and $|D_{k-j_{3}^{\prime},2}\rangle$
are defined on different subspaces
$\mathcal{H}_{1}\subseteq\mathbb{H}_{A_{1}}\otimes\mathbb{H}_{A_{2}}$ and
$\mathcal{H}_{2}\subseteq\mathbb{H}_{A_{1}}\otimes\mathbb{H}_{A_{2}}$
respectively, where $\mathcal{H}_{1}$ is spanned by the basis states
$\\{|j_{1}j_{2}\rangle,j_{1}+j_{2}=k-j_{3},j_{1},j_{2}\geq 0\\}$,
$\mathcal{H}_{2}$ is spanned by the basis states
$\\{|j_{1}j_{2}\rangle|j_{1}+j_{2}=k-j_{3}^{\prime},j_{1},j_{2}\geq 0\\}$, and
$\mathcal{H}_{1}\cap\mathcal{H}_{2}=\emptyset$. Hence, $\rho_{(A_{3})}$ in
Eq.(25) is a bipartite entanglement if there is at least one integer $j_{3}$
such that $|D_{k-j_{3},2}\rangle$ is a bipartite entanglement from Eq.(1).
This can be proved for any $d\geq 2$ ($d$ is the local dimension of each
system $A_{j}$). In fact, for $d=2$,
$|D_{1,2}\rangle=\alpha_{01}|01\rangle+\alpha_{10}|10\rangle)$ is a bipartite
entanglement for any $\alpha_{01}\alpha_{10}\not=0$. For $d>2$, we get that
$|D_{k-j_{3},2}\rangle$ is an entangled Dicke state for any $k$ with
$j_{3}<k\leq d$ Luo2020 . This completes the proof for $n=3$.

Similar proof holds for general integers of $n$ and $k$. In fact, from Eq.(23)
we have the following decomposition

$\displaystyle|D_{k,n}\rangle$ $\displaystyle=$
$\displaystyle\sum_{\sum_{i=1}^{s}j_{i}=k-\sum_{t=s+1}^{n}j_{t}}\\!\\!\\!\\!\\!\\!\\!\\!\alpha_{j_{1}\cdots{}j_{s}}|j_{1}\cdots{}j_{s}\rangle_{A_{1}\cdots{}A_{s}}$
(26)
$\displaystyle\otimes\sum_{j_{s+1},\cdots,j_{n}}\alpha_{j_{s+1}\cdots{}j_{n}}|j_{s+1}\cdots{}j_{n}\rangle_{A_{s+1}\cdots{}A_{n}}$
$\displaystyle=$
$\displaystyle\sum_{j_{s+1},\cdots,j_{n}}\alpha_{j_{s+1}\cdots{}j_{n}}|j_{s+1}\cdots{}j_{n}\rangle_{A_{s+1}\cdots{}A_{n}}$
$\displaystyle\otimes|D_{k-\sum_{t=s+1}^{n}j_{t},s}\rangle_{A_{1}\cdots{}A_{s}}$

where $|D_{k-\sum_{t=s+1}^{n}j_{t},s}\rangle_{A_{1}\cdots{}A_{s}}$ are
generalized Dicke states on Hilbert space
$\otimes_{j=1}^{s}\mathbb{H}_{A_{j}}$ defined by

$\displaystyle|D_{k-\sum_{t=s+1}^{n}j_{t},s}\rangle_{A_{1}\cdots{}A_{s}}=\sum_{k-\sum_{t=s+1}^{n}j_{t}=\ell_{s}}\\!\\!\\!\\!\\!\\!\\!\\!\alpha_{j_{1}\cdots{}j_{s}}|j_{1}\cdots{}j_{s}\rangle$
(27)

with $\sum_{i=1}^{s}j_{i}=\ell_{s}$.

After $|D_{k,n}\rangle$ passing through the particle-lose channel ${\cal
E}_{A_{s+1}\cdots{}A_{n}}(\cdot)$, the reduced state of
$\rho_{(A_{s+1}\cdots{}A_{n})}$ is given by

$\displaystyle\rho_{(A_{s+1}\cdots{}A_{n})}$ $\displaystyle=$
$\displaystyle{\cal E}_{A_{s+1}\cdots{}A_{n}}(|D_{k,n}\rangle\langle
D_{k,n}|)$ $\displaystyle=$
$\displaystyle\\!\\!\\!\\!\\!\\!\sum_{j_{s+1},\cdots,j_{n}}\\!\\!\\!\\!\\!\\!\alpha_{j_{s+1}\cdots{}j_{n}}^{2}|D_{k-\sum_{t=s+1}^{n}j_{t},s}\rangle_{A_{1}\cdots{}A_{s}}\langle
D_{k-\sum_{t=s+1}^{n}j_{t},s}|$

Similar to the case of $n=3$, a simple fact is as follows. For each pair of
$\ell$ and $\ell^{\prime}$ with $\ell\not=\ell^{\prime}$, the nonnegative
solutions of $i_{s+1}+\cdots+i_{n}=\ell$ and
$j_{s+1}+\cdots+j_{n}=\ell^{\prime}$ are different, that is,

$\displaystyle{\cal S}_{\ell}\cap{\cal S}_{\ell^{\prime}}=\emptyset,$ (29)

where ${\cal S}_{\ell}$ and ${\cal S}_{\ell^{\prime}}$ are given respectively
by

$\displaystyle{\cal
S}_{\ell}=\\{(i_{s+1},\cdots,i_{n})|\sum_{t=s+1}^{n}i_{t}=\ell\\},$ (30)
$\displaystyle{\cal
S}_{\ell^{\prime}}=\\{(j_{s+1},\cdots,j_{n})|\sum_{t=s+1}^{n}j_{t}=\ell^{\prime}\\}.$
(31)

This yields to $k-\ell\not=k-\ell^{\prime}$. So, all the nonnegative solutions
of $i_{1}+\cdots+i_{s}=k-\ell$ and $j_{1}+\cdots+j_{s}=k-\ell^{\prime}$ are
different, that is,

$\displaystyle{\cal S}_{k-\ell}\cap{\cal S}_{k-\ell^{\prime}}=\emptyset,$ (32)

where ${\cal S}_{\ell}$ and ${\cal S}_{\ell^{\prime}}$ are given respectively
by

$\displaystyle{\cal
S}_{k-\ell}=\\{(i,\cdots,i_{s})|\sum_{t=1}^{s}i_{t}=k-\ell\\},$ (33)
$\displaystyle{\cal
S}_{k-\ell^{\prime}}=\\{(j_{1},\cdots,j_{s})|\sum_{t=1}^{s}j_{t}=k-\ell^{\prime}\\}.$
(34)

It follows $\langle i\cdots{}i_{s}|j\cdots{}j_{s}\rangle=0$ for
$(i,\cdots,i_{s})\in{\cal S}_{k-\ell}$ and $(j,\cdots,j_{s})\in{\cal
S}_{k-\ell^{\prime}}$. Denote $\mathcal{H}_{k}$ as the subspace spanned by the
basis states
$\\{|j_{1}\cdots{}j_{s}\rangle|\sum_{t=s+1}^{n}j_{t}=\ell,j_{1},\cdots,j_{s}\geq
0\\}$ and $\mathcal{H}_{k}^{\prime}$ as the subspace spanned by the basis
states
$\\{|j_{1}\cdots{}j_{s}\rangle|\sum_{t=1}^{s}j_{t}=\ell^{\prime},j_{1},\cdots,j_{s}\geq
0\\}$. We have proved $\mathcal{H}_{k}\cap\mathcal{H}_{k}^{\prime}=\emptyset$
for each pair of $\ell\not=\ell^{\prime}$. This implies that for
$\ell\not=\ell^{\prime}$ the states of $|D_{k-\ell,s}\rangle$ and
$|D_{k-\ell^{\prime},s}\rangle$ in Eq.(LABEL:BB4) are generalized Dicke states
defined on different subspaces $\mathcal{H}_{k}$ and
$\mathcal{H}_{k}^{\prime}$, respectively. Hence,
$\rho_{(A_{s+1}\cdots{}A_{n})}$ is a genuinely $s$-partite entanglement if and
only if there is at least one integer $\ell$ with $\ell\leq k$ such that
$|D_{k-\ell,s}\rangle$ is a genuinely $s$-partite entanglement Sy . This
proves the result for $d\geq 2$ and $k\geq 1$ by using the recent result
Luo2020  that states any generalized Dicke state of $|D_{k-\ell,s}\rangle$ is
genuinely multipartite entangled in the biseparable model Sy . Hence, it has
proved the result.$\Box$

In what follows, consider the superposition of generalized Dicke states as

$\displaystyle|\Phi\rangle_{A_{1}\cdots{}A_{n}}=\sum_{k}\beta_{k}|D_{k,n}\rangle,$
(35)

where $\beta_{k}$’s satisfies $\sum_{k}\beta_{k}^{2}=1$ and $|D_{k,n}\rangle$
are defined in Eq.(22). Note that a fully separable state of
$|\phi\rangle^{\otimes n}$ has the decomposition in Eq.(35) for any single-
particle state $|\phi\rangle$, where $\beta_{k}$’s depend on the specific form
of $|\phi\rangle$. This kind of states is not entangled and then not robust
entanglement. Hence, in what follows, we assume that the state in Eq.(35) is
an $n$-partite genuinely entanglement Sy  (not any fully separable state).
With this assumption, from Result 1 we can prove the state in Eq.(35) is
robust entanglement in the present model for any $\beta_{k}$’s.

Corollary 1. For a given state in Eq.(35), it is robust entanglement if it is
a genuinely $n$-partite entanglement in the biseparable model.

Proof. From Eq.(22), $|D_{k,n}\rangle$’s in Eq.(35) are generalized Dicke
states with different excitations. This means that $|D_{k,n}\rangle$’s are
defined on different subspaces of $\otimes_{j=1}^{n}\mathbb{H}_{A_{j}}$. From
Eqs.(26) and (35), it follows that

$\displaystyle\rho_{(A_{s+1}\cdots{}A_{n})}$ $\displaystyle=$
$\displaystyle\sum_{k}\beta_{k}^{2}\sum_{j_{s+1},\cdots,j_{n}}\alpha_{j_{s+1}\cdots{}j_{n}}^{2}|D_{k-\sum_{t=s+1}^{n}j_{t},s}\rangle$
(36) $\displaystyle\times\langle D_{k-\sum_{t=s+1}^{n}j_{t},s}|$

where $\rho_{(A_{s+1}\cdots{}A_{n})}$ is the density matrix of the particles
$A_{1},\cdots,A_{s}$ after the state in Eq.(35) passing through the particle-
lose channel ${\cal E}_{A_{s+1}\cdots{}A_{n}}(\cdot)$, and
$|D_{k-\sum_{t=s+1}^{n}j_{t},s}\rangle$ are defined in Eq.(27). Similar to the
analysis after Eq.(LABEL:BB4), for any pair of $k$ and $k^{\prime}$ with
$k\not=k^{\prime}$, the states of $|D_{k-\ell,s}\rangle$ and
$|D_{k^{\prime}-\ell,s}\rangle$ have different excitations
$k-\ell\not=k^{\prime}-\ell$. This means $|D_{k-\ell,s}\rangle$ and
$|D_{k^{\prime}-\ell,s}\rangle$ are generalized Dicke states defined on
subspaces
$\mathcal{H}_{k},\mathcal{H}_{k}^{\prime}\subseteq\otimes_{j=1}^{n}\mathbb{H}_{A_{j}}$
with $\mathcal{H}_{k}\cap\mathcal{H}_{k}^{\prime}=\emptyset$, where
$\mathcal{H}_{k}$ denotes the subspace spanned by the basis states
$\\{|j_{1}\cdots{}j_{s}\rangle|j_{1}+\cdots+j_{s}=k-\ell,j_{1},\cdots,j_{s}\geq
0\\}$ and $\mathcal{H}_{k}^{\prime}$ denotes the subspace spanned by the basis
states
$\\{|j_{1}\cdots{}j_{s}\rangle|j_{1}+\cdots{}+j_{s}=k-\ell^{\prime},j_{1},\cdots,j_{s}\geq
0\\}$. This implies that $\rho_{(A_{s+1}\cdots{}A_{n})}$ in Eq.(36) is a
genuinely $s$-partite entanglement if and only if one of Dicke states
$|D_{k-\ell,s}\rangle$ is a genuinely $n$-partite entanglement for some
integers $\ell$ and $k$ from the definition in Eq.(1). This is proved by using
two facts as follows. One is from Result 1 which states all the generalized
Dicke states of $|D_{k-\ell,s}\rangle$ in Eq.(27) are genuinely $s$-partite
entangled states in the biseparable model Sy . The other is from the
assumptions in Eqs.(22) and (35), that is, all the parameters of
$\alpha_{j_{s+1}\cdots{}j_{n}}$ in Eq.(22) satisfy
$\alpha_{j_{s+1}\cdots{}j_{n}}\not=0$; and there is at least one parameter
$\beta_{k}$ in Eq.(35) with $\beta_{k}\not=0$. Hence, from Eq.(36), we have
completed the proof. $\Box$

When $\alpha_{j_{1}\cdots{}j_{n}}$’s are all equal, $|D_{k,n}\rangle$ in
Eq.(35) become Dicke states Toth  which are genuinely multipartite entangled
DPR ; Dicke4 ; TDS ; Sy . Generally, $|D_{k,n}\rangle$ cannot be generated by
using entangled states which have no more than $n-1$ particles assisted by
CPTP mappings. This implies that they are genuinely network entangled Luo2020
and genuinely $n$-partite entangled Sy . Result 1 and Corollary 1 provide two
kinds of non-symmetric states that are robust against particle-loss.

##  IV Robust entangled quantum networks

An $n$-partite quantum network ${\cal N}_{q}$ consists of independent
entangled states such as Affleck-Kennedy-Lieb-Tasaki (AKLT) system AKLT  with
small number of particles, as shown in Fig.2. These entangled systems show
great convenience for large-scale quantum tasks Kim  with short-range
experimental settings. The independence assumption of ${\cal N}_{q}$ is the
key to activate new non-localities depending on network configurations BGP ;
SBP ; RBB ; Chav ; Luo2018 ; Luo2019 ; CPV ; Fri ; RBBB . Nevertheless, it may
rule out specific scenarios such as cyclic networks RBBB ; Luo2018 . Our goal
is to characterize general quantum networks under local unitary operations or
generalized CPTP mappings. This allows for regarding all the particles shared
by one party as one combined particle in large Hilbert space. In this case,
the particle-lose in Definition 1 means network nodes or parties in
applications may be lost or offline.

###  IV.1 $k$-independent quantum networks

Figure 2: (Color online) Schematic $k$-independent quantum networks consisting
of bipartite entangled states. (a) Chain-type network. (b) Star-type network.
(c) Cyclic network. (d) Planer network used in measurement-based quantum
computation Cluster . These networks are particle-lose separable.

Consider a general network ${\cal N}_{q}$ consisting of parties
$\textsf{A}_{1},\cdots,\textsf{A}_{n}$, where each party may share some
entangled states with other. ${\cal N}_{q}$ is a $k$-independent quantum
network if there are $k$ number of parties
$\textsf{A}_{1},\cdots,\textsf{A}_{k}$ (for example) any pair of them have not
pre-shared entangled states. Although this kind of quantum networks show
nonlocality according to the specific Bell-type inequalities Luo2018 ; Luo2019
, we can prove they are not robust against node-loss. These include the chain-
type network ZZH , the star-type network and general networks Luo2018 , as
shown in Fig.2.

Result 2. Any $k$-independent quantum network ${\cal N}_{q}$ with $k\geq 2$ is
particle-lose separable.

Proof. Consider an $n$-partite $k$-independent quantum network ${\cal N}_{q}$
consisting of $n$ parties $\textsf{A}_{1},\cdots,\textsf{A}_{n}$, where
$\textsf{A}_{1},\cdots,\textsf{A}_{k}$ (for example) are independent, that is,
they have no prior-shared entanglement Luo2018 . Suppose that a
$k$-independent quantum network ${\cal N}_{q}$ consists of entangled pure
states $|\Phi_{1}\rangle,\cdots,|\Phi_{m}\rangle$, where each party
$\textsf{A}_{j}$ may share some particles in $|\Phi_{l}\rangle$’s with other
parties. The total state of ${\cal N}_{q}$ is given by

$\displaystyle|\Omega\rangle_{\textsf{A}_{1}\cdots\textsf{A}_{n}}=\otimes_{j=1}^{m}|\Phi_{j}\rangle,$
(37)

where $\textsf{A}_{1},\cdots,\textsf{A}_{k}$ may share some entangled states
with other parties $\textsf{A}_{k+1},\cdots,\textsf{A}_{n}$, who can share
some entangled states with each other.

After losing all the particles shared by the parties
$\textsf{A}_{k+1},\cdots,\textsf{A}_{n}$, the joint state shared by
$\textsf{A}_{1},\cdots,\textsf{A}_{k}$ is given by

$\displaystyle\rho_{(\textsf{A}_{k+1}\cdots\textsf{A}_{n})}={\cal
E}_{\textsf{A}_{k+1}\cdots{}\textsf{A}_{n}}(|\Omega\rangle\langle\Omega|),$
(38)

where ${\cal E}_{\textsf{A}_{k+1}\cdots{}\textsf{A}_{n}}(\cdot)$ denotes the
particle-lose channel associated with all the particles owned by
$\textsf{A}_{k+1},\cdots{},\textsf{A}_{n}$.

A simple fact is that the local operations of all the parties
$\textsf{A}_{k+1},\cdots,\textsf{A}_{n}$ do not change the joint state of
$\textsf{A}_{1},\cdots,\textsf{A}_{k}$, that is,

$\displaystyle\rho_{(\textsf{A}_{k+1}\cdots\textsf{A}_{n})}$ $\displaystyle=$
$\displaystyle{\cal
E}_{\textsf{A}_{k+1}\cdots{}\textsf{A}_{n}}(\mathbbm{1}_{\textsf{A}_{1}\cdots\textsf{A}_{k}}\otimes(\otimes_{j=k+1}^{n}U_{j})|\Omega\rangle\langle\Omega|$
(39)
$\displaystyle\times\mathbbm{1}_{\textsf{A}_{1}\cdots\textsf{A}_{k}}\otimes(\otimes_{j=k+1}^{n}U^{\dagger}_{j}))$

where $\mathbbm{1}_{\textsf{A}_{1}\cdots\textsf{A}_{k}}$ denotes the identity
operator on the joint system of $\textsf{A}_{1},\cdots,\textsf{A}_{k}$, and
$U_{j}$ denotes the local unitary operation performed by the party
$\textsf{A}_{j}$, $j\in\\{k+1,\cdots,n\\}$. From Eq.(39), it follows that
$\rho_{(\textsf{A}_{k+1}\cdots\textsf{A}_{n})}$ is fully separable because
$\textsf{A}_{k+1},\cdots,\textsf{A}_{n}$ do not share any entanglement from
the assumption of $k$-independence. This implies that $|\Omega\rangle$ in
Eq.(37) is robust entangled in the present model.

Similar proof holds for any $k$-independent quantum network consisting of
general entangled states including mixed entangled states. This completes the
proof. $\Box$

Example 1. Consider an $n$-partite chain-type network ${\cal N}_{q}$ shown in
Fig.2(a), where each pair of two adjacent parties $\textsf{A}_{j}$ and
$\textsf{A}_{j+1}$ shares one EPR state
$|\Phi_{j}\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$. The total state
of ${\cal N}_{q}$ is given by
$|\Omega\rangle_{\textsf{A}_{1}\cdots\textsf{A}_{n}}=\otimes_{j=1}^{n-1}|\Phi_{j}\rangle$
which can be regarded as a pure state on Hilbert space
$\otimes_{j=1}^{n}\mathbbm{H}_{\textsf{A}_{j}}$, where
$\mathbbm{H}_{\textsf{A}_{1}}$ and $\mathbbm{H}_{\textsf{A}_{n}}$ are
2-dimensional spaces while $\mathbbm{H}_{\textsf{A}_{j}}$ with
$j\in\\{2,\cdots,n-1\\}$ are 4-dimensional spaces. Take $n=3$ as example, the
total state is given by

$\displaystyle|\Omega\rangle_{\textsf{A}_{1}\textsf{A}_{2}\textsf{A}_{3}}$
$\displaystyle=$
$\displaystyle\frac{1}{2\sqrt{2}}(|0\rangle_{\textsf{A}_{1}}|00\rangle_{\textsf{A}_{2}}|0\rangle_{\textsf{A}_{3}}+|0\rangle_{\textsf{A}_{1}}|01\rangle_{\textsf{A}_{2}}|1\rangle_{\textsf{A}_{3}}$
(40)
$\displaystyle+|1\rangle_{\textsf{A}_{1}}|10\rangle_{\textsf{A}_{2}}|0\rangle_{\textsf{A}_{3}}+|1\rangle_{\textsf{A}_{1}}|11\rangle_{\textsf{A}_{2}}|1\rangle_{\textsf{A}_{3}})$

Here, $\textsf{A}_{1}$ and $\textsf{A}_{3}$ are independent parties. After
passing through the particle-lose channel ${\cal
E}_{\textsf{A}_{2}}(\cdot{})$, the reduced state of $\textsf{A}_{1}$ and
$\textsf{A}_{3}$ is given by

$\displaystyle\rho_{(\textsf{A}_{2})}=\frac{1}{4}\sum_{i,j=0}^{1}|ij\rangle\langle{}ij|)$
(41)

which is the maximally mixed state. Similarly, for general case of $n\geq 3$,
all the parties $\textsf{A}_{j}$’s with even $j$ (or odd $j$) are independent.
The reduced state after ${\cal N}_{q}$ passing through all the particle-lose
channels ${\cal E}_{\textsf{A}_{j}}(\cdot{})$ with even $j$’s is fully
separable.

Example 2. Consider an $n+1$-partite star-type network ${\cal N}_{q}$ shown in
Fig.2(b), where each outer party $\textsf{A}_{j}$ shares one EPR state
$|\Phi\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$ with the center party
B. The total state of ${\cal N}_{q}$ is given by

$\displaystyle|\Omega\rangle_{\textsf{A}_{1}\cdots\textsf{A}_{n}\textsf{B}}=\frac{1}{\sqrt{2^{n}}}\\!\\!\\!\\!\sum_{j_{1},\cdots,j_{n}=0}^{1}\\!\\!\\!\\!|j_{1}\cdots{}j_{n}\rangle_{\textsf{A}_{1}\cdots\textsf{A}_{n}}|j_{1}\cdots{}j_{n}\rangle_{\textsf{B}}$
(42)

which can be regarded as a pure state on Hilbert space
$\mathbbm{H}_{B}\otimes(\otimes_{j=1}^{n}\mathbbm{H}_{\textsf{A}_{j}})$, where
$\mathbbm{H}_{\textsf{A}_{j}}$ are 2-dimensional spaces while
$\mathbbm{H}_{\textsf{B}}$ is $2^{n}$-dimensional space. Here, all the parties
of $\textsf{A}_{j}$’s are independent parties. After passing through the
particle-lose channel ${\cal E}_{\textsf{B}}(\cdot{})$, the reduced state of
$\textsf{A}_{1},\cdots,\textsf{A}_{n}$ is given by

$\displaystyle\rho_{(\textsf{B})}=\frac{1}{2^{n}}\sum_{j_{1},\cdots,j_{n}=0}^{1}|j_{1}\cdots{}j_{n}\rangle_{\textsf{A}_{1}\cdots\textsf{A}_{n}}\langle
j_{1}\cdots{}j_{n}|$ (43)

which is the maximally mixed state.

Another example is cyclic network in Fig.2(c) or planer networks Fig.2(d).
Result 2 shows new feature of all the $k$-independent quantum networks, which
depends only on the independence assumption of the parties, but not the shared
entangled states. The lack of robustness against particle-loss may rule out
some specific applications such as measurement-based quantum computation
Cluster  using $k$-independent quantum networks.

Figure 3: (Color online) Schematic completely connected quantum networks
consisting of bipartite entangled states. (a) Triangle network. (b) Completed
network with six parties.

###  IV.2 Completely connected quantum networks

Different from these scenarios in Result 2 there are other networks which are
robust against particle-lose, as shown in Fig.3.

Definition 2. An $n$-partite quantum network ${\cal N}_{q}$ is a completely
connected if each pair of two parties shares at least one entanglement.

Suppose that ${\cal N}_{q}$ consists of bipartite entangled pure states EPR ;
Gis  given by

$\displaystyle|\phi\rangle_{AB}=\sum_{j}\beta_{j}|jj\rangle$ (44)

on a finite-dimensional Hilbert space $\mathbbm{H}_{A}\otimes\mathbbm{H}_{B}$,
and generalized Dicke states in Eq.(22), where $\beta_{j}$’s satisfy
$\sum_{j}\beta_{j}^{2}=1$ and $\beta_{j}\not=0$ for at least two integers $j$.
We have the following result.

Result 3. Any completely connected quantum network is robust entanglement.

Proof. We firstly prove that the total state of any completely connected
quantum network ${\cal N}_{q}$ is an $n$-partite genuinely entanglement in the
biseparable model Sy  if ${\cal N}_{q}$ consists of bipartite entangled pure
states and generalized Dicke states. This can be proved by using the following
fact.

Fact 1. An $m$-partite quantum network is genuinely $m$-partite entanglement
if it consists of bipartite entangled pure states.

Fact 1 can be proved by using a recent method assisted by local operations and
classical communication ZDBS , that is, an $n$-partite state is genuinely
entangled if each pair of two parties can share one bipartite pure
entanglement with the help of others’ local measurements and classical
communication.

For any $m$-particle Dicke state $|D_{k,m}\rangle$ defined similar to Eq.(22),
it is genuinely multipartite entanglement in the biseparable model Sy  from
Result 1. Moreover, the final joint state of any two particles after
$|D_{k,m}\rangle$ being measured using proper projections on other particles
is a bipartite entanglement ZDBS . Hence, from the generalized entanglement
swapping ZZH , each pair of two parties $\textsf{A}_{i}$ and $\textsf{A}_{j}$
in ${\cal N}_{q}$ can share one bipartite entanglement with the help of
others’ local measurement and classical communication. This means the total
state of ${\cal N}_{q}$ is a genuinely $n$-partite entanglement. The following
proof is completed by two cases.

Case 1. An $n$-partite completely connected quantum network consists of
bipartite entangled pure states.

In this case, ${\cal N}_{q}\cup\\{|\phi\rangle\\}$ is robust entanglement if
${\cal N}_{q}$ is robust entanglement, where ${\cal
N}_{q}\cup\\{|\phi\rangle\\}$ denotes a new network by adding one bipartite
entanglement $|\phi\rangle$ shared by one party or two parties. With this
fact, it only needs to consider the simplest case of the completely connected
network ${\cal N}_{q}$ on which each pair of $\textsf{A}_{i}$ and
$\textsf{A}_{j}$ shares one bipartite entanglement $|\phi_{i;j}\rangle$
defined by

$\displaystyle|\phi_{i;j}\rangle=\sum_{k}\alpha_{k;ij}|kk\rangle,$ (45)

where the local dimension of each system satisfies $d\geq 2$,
$\alpha_{k;ij}$’s satisfy $\sum_{k}\alpha_{k;ij}^{2}=1$ and
$\alpha_{k;ij}\not=0$ for at least two different integers $k$. The total
system of ${\cal N}_{q}$ is given by

$\displaystyle|\Phi\rangle_{\textsf{A}_{1}\cdots\textsf{A}_{n}}=\otimes_{i,j,i\not=j}|\phi_{i;j}\rangle$
(46)

Consider a subset
$S=\\{\textsf{A}_{i_{1}},\cdots,\textsf{A}_{i_{k}}\\}\subset\\{\textsf{A}_{1},\cdots,\textsf{A}_{n}\\}$
and $\overline{S}$ being the complement set of $S$. The joint state after
${\cal N}_{q}$ passing through the particle-lose channel ${\cal
E}_{\overline{S}}(\cdot):={\cal E}_{\textsf{A}_{i}\in\overline{S}}(\cdot)$ is
given by

$\displaystyle\rho_{(\overline{S})}=(\otimes_{s\not=t,s,t\in\\{i_{1},\cdots,i_{k}\\}}|\phi_{s;t}\rangle\langle\phi_{s;t}|)\otimes_{j=1}^{k}\rho_{i_{j}},$
(47)

where $\rho_{i_{j}}$ denotes the local state shared by $\textsf{A}_{i_{j}}$
after losing all the particles entangled with $\textsf{A}_{i_{j}}$, which is
given by $\rho_{i_{j}}=\otimes_{\tau\not\in\\{i_{1},\cdots,i_{k}\\}}{\rm
Tr}_{\textsf{A}_{i_{\tau}}}(|\phi_{i_{j};\tau}\rangle\langle\phi_{i_{j};\tau}|)$,
and ${\rm Tr}_{\textsf{A}_{i}}(\cdot)$ denotes the trace operator by tracing
out all the particles shared by $\textsf{A}_{i}$. From Eq.(47) it only needs
to consider
$\otimes_{s\not=t,s,t\in\\{i_{1},\cdots,i_{k}\\}}|\phi_{s;t}\rangle\langle\phi_{s;t}|$
while $\otimes_{j=1}^{k}\rho_{i_{j}}$ is fully separable.

In fact, for any $S$ and $S^{\prime}$ with $S\not=S^{\prime}$ and
$|S|=|S^{\prime}|$, the density matrices of $\rho_{(\overline{S})}$ and
$\rho_{(\overline{S^{\prime}})}$ have decompositions similar to Eq.(47). This
implies the generalized symmetry of $|\Phi\rangle$ in Eq.(46). Hence, it only
needs to consider $S=\\{\textsf{A}_{1},\cdots,\textsf{A}_{k}\\}$ for
$k=2,\cdots,n-1$. Note that the subnetwork consisting of
$\textsf{A}_{1},\cdots,\textsf{A}_{k}$ is connected from Definition 2 after
losing all the particles owned by $\textsf{A}_{k+1},\cdots,\textsf{A}_{n}$.
This implies a chain-type subnetwork consisting of
$\textsf{A}_{1},\cdots,\textsf{A}_{k}$. From Fact 1, $\rho_{(\overline{S})}$
in Eq.(47) is a genuinely $k$-partite entanglement ZDBS . It means that the
total state of ${\cal N}_{q}$ is robust entanglement.

Case 2. An $n$-partite completely connected quantum network consists of
bipartite entangled pure states and generalized Dicke states.

Similar to Case 1, it only needs to consider the simplest case that each pair
of $\textsf{A}_{i}$ and $\textsf{A}_{j}$ shares one bipartite entanglement
$|\phi_{ij}\rangle$ defined in Eq.(45) or generalized Dicke state
$|D_{k_{ij},m_{ij}}\rangle$ in Eq.(22). The total system of ${\cal N}_{q}$ is
given by

$\displaystyle|\Psi\rangle_{\textsf{A}_{1}\cdots\textsf{A}_{n}}=\otimes_{i,j}|\phi_{i;j}\rangle\otimes_{s,t}|D_{k_{st},m_{st}}\rangle$
(48)

For a given set
$S=\\{\textsf{A}_{j_{1}},\cdots,\textsf{A}_{j_{k}}\\}\subset\\{\textsf{A}_{1},\cdots,\textsf{A}_{n}\\}$
and its complement set $\overline{S}$, the joint state after ${\cal N}_{q}$
passing through the particle-lose channel ${\cal
E}_{\overline{S}}(\cdot)={\cal E}_{\textsf{A}_{i}\in\overline{S}}(\cdot)$ is
given by

$\displaystyle\rho_{(\overline{S})}$ $\displaystyle=$ $\displaystyle{\cal
E}_{\overline{S}}(|\Psi\rangle\langle\Psi|)$ (49) $\displaystyle=$
$\displaystyle\otimes_{s\not=t,s,t\in\\{j_{1},\cdots,j_{k}\\}}|\phi_{s;t}\rangle\langle\phi_{s;t}|$
$\displaystyle\otimes(\otimes_{u\not=v,u,v\in\\{j_{1},\cdots,j_{k}\\}}\rho_{u,v})\otimes(\otimes_{\ell=1}^{k}\varrho_{(j_{\ell})}),$

where $\rho_{u,v}$ as bipartite states shared by the parties $\textsf{A}_{s}$
and $\textsf{A}_{t}$ are the remained states of generalized Dicke states after
losing all the particles being not shared by $\textsf{A}_{s}$ and
$\textsf{A}_{t}$ and $\varrho_{j_{\ell}}$ denotes the local state shared by
$\textsf{A}_{j_{\ell}}$ after losing all the particles entangled with
$\textsf{A}_{j_{\ell}}$. From Eq.(26) it is sufficient to consider
$(\otimes_{s\not=t,s,t\in\\{j_{1},\cdots,j_{k}\\}}|\phi_{s;t}\rangle\langle\phi_{s;t}|)\otimes(\otimes_{u\not=v,u,v\in\\{j_{1},\cdots,j_{k}\\}}\rho_{u,v})$
while $\otimes_{\ell=1}^{k}\varrho_{j_{\ell}}$ is fully separable.

From Result 1, any generalized Dicke state is robust entanglement. After
passing through a complete positive preserve trace channel ${\cal
E}_{\overline{S}}(\cdot)$, ${\cal N}_{q}$ is connected, where each pair of
$\textsf{A}_{s},\textsf{A}_{t}\in S$ shares one bipartite entangled pure state
$|\phi_{s;t}\rangle$ or the remained entanglement $\rho_{s,t}$ from
generalized Dicke states. From Fact 1, $\rho_{\overline{S}}$ is a genuinely
$k$-partite entanglement ZDBS . It means that the total state of ${\cal
N}_{q}$ is robust entanglement in the present model. This has completed the
proof. $\Box$

Similar result may be proved for any completely connected quantum network
consisting of other pure entangled states or mixed entangled states ZDBS .

Example 3. Consider an $3$-partite completely connected quantum network ${\cal
N}_{q}$ EPR ; RBBB  shown in Fig.3(a), where each pair shares one EPR state
$|\phi\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$. The total state of
${\cal N}_{q}$ is given by

$\displaystyle|\Omega\rangle_{\textsf{A}_{1}\textsf{A}_{2}\textsf{A}_{3}}$
$\displaystyle=$
$\displaystyle\frac{1}{2\sqrt{2}}(|00\rangle_{\textsf{A}_{1}}|00\rangle_{\textsf{A}_{2}}|00\rangle_{\textsf{A}_{3}}$
(50)
$\displaystyle+|10\rangle_{\textsf{A}_{1}}|00\rangle_{\textsf{A}_{2}}|01\rangle_{\textsf{A}_{3}}$
$\displaystyle+|00\rangle_{\textsf{A}_{1}}|01\rangle_{\textsf{A}_{2}}|10\rangle_{\textsf{A}_{3}}$
$\displaystyle+|10\rangle_{\textsf{A}_{1}}|01\rangle_{\textsf{A}_{2}}|11\rangle_{\textsf{A}_{3}}$
$\displaystyle+|01\rangle_{\textsf{A}_{1}}|10\rangle_{\textsf{A}_{2}}|00\rangle_{\textsf{A}_{3}}$
$\displaystyle+|11\rangle_{\textsf{A}_{1}}|10\rangle_{\textsf{A}_{2}}|01\rangle_{\textsf{A}_{3}}$
$\displaystyle+|01\rangle_{\textsf{A}_{1}}|11\rangle_{\textsf{A}_{2}}|10\rangle_{\textsf{A}_{3}}$
$\displaystyle+|11\rangle_{\textsf{A}_{1}}|11\rangle_{\textsf{A}_{2}}|11\rangle_{\textsf{A}_{3}})$

which can be regarded as a pure state on Hilbert space
$\otimes_{j=1}^{3}\mathbbm{H}_{A_{j}}$, where $\mathbbm{H}_{A_{j}}$ are
4-dimensional spaces. Here, all the parties of $\textsf{A}_{j}$’s are
independent parties. The reduced state $\rho_{(\textsf{A}_{1})}$ is given by

$\displaystyle\rho_{(\textsf{A}_{1})}=\frac{1}{4}\mathbbm{1}_{\textsf{A}_{2}}\otimes|\phi\rangle_{\textsf{A}_{2}\textsf{A}_{3}}\langle\phi|\otimes\mathbbm{1}_{\textsf{A}_{3}}$
(51)

which is an entanglement for two parties $\textsf{A}_{2}$ and
$\textsf{A}_{3}$, where $\mathbbm{1}_{\textsf{A}_{i}}$ denotes an identity
operator. Similar proofs hold for other reduced states
$\rho_{(\textsf{A}_{2})}$ and $\rho_{(\textsf{A}_{3})}$. Hence, the total
state of ${\cal N}_{q}$ is robust entanglement in the present model. Moreover,
this can be extended to other quantum network shown in Fig.3(b).

Result 3 shows new insight on quantum networks going beyond the
$k$-independent networks Luo2018 . Note that ${\cal N}_{q}$ is separable in
the recent model NWR ; Kraft ; Luo2020  if each Dicke state is at most
$n-1$-partite entangled. This presents another feature of completely connected
quantum networks Luo2020 .

##  V Strong nonlocality of robust multipartite entanglement

The present model is useful for witnessing robust entanglement such as
generalized Dicke states Toth  or completed quantum networks. A natural
problem is how to verify its nonlocality from Bell experiments, similar to
single entanglement of EPR, GHZ or Dicke states CHSH ; GHZ ; Sy  (without
local tensor decompositions or a rigid definition of network local states NWR
; Kraft ; Luo2020 ) or entangled networks RBBB ; Luo2018 ; Luo2019  as shown
in Figs.2 and 3. Compared with the biseparable model Sy , one may expect
stronger nonlocality in the present model. It is difficult to verify the
strong nonlocality of robust entanglement due to the non-convexity of ${\cal
B}_{d}$ and that the output state depends on the particle-loss channel ${\cal
E}_{S}(\cdot)$ associated with specific set $S$. Our goal here is to address
this problem by using new Bell-like inequalities. The main idea is inspired by
the following set of Bell inequalities Sy ,

$\displaystyle\left\\{\begin{array}[]{ll}{\cal L}_{n}(\rho_{A_{1},\cdots
A_{n}})\leq c_{n},\\\ {\cal L}_{k}({\cal E}_{S}(\rho))\leq
c_{k},\forall|S|\leq k,\end{array}\right.$ (54)

where ${\cal L}_{m}(\cdot)$ is an $m$-partite Bell operator for verifying the
genuinely $m$-partite nonlocality Sy , $c_{n}$ and $c_{k}$ are constants. It
is necessary to violate all the inequalities in order to verify the strong
nonlocality of a robust entanglement.

Example 4. Consider a triangle network consisting three EPR states
$|\phi\rangle=\cos\theta|00\rangle+\sin\theta|11\rangle$ with
$\theta\in(0,\frac{\pi}{4}]$, as shown in Fig.3(a). The nonlocality can be
verified by using recent method LuoDV  with multiple measurement settings. The
final state after passing through particle-lose channel ${\cal
E}_{\textsf{A}_{i}}(\cdot)$ can be verified by violating the CHSH inequality
CHSH ; Loub  for any $\theta\in(0,\frac{\pi}{4}]$.

Figure 4: (Color online) The violations of the inequality (56) by using Pauil
observable. Here, $\alpha=\sin\theta\cos\phi,\beta=\sin\theta\sin\phi$ and
$\gamma=\cos\theta$.

Example 5. Consider a generalized W state DVC  as

$\displaystyle|W\rangle_{A_{1}A_{2}A_{3}}=\alpha|001\rangle+\beta|010\rangle+\gamma|100\rangle$
(55)

with $\alpha^{2}+\beta^{2}+\gamma^{2}=1$. Its genuinely tripartite nonlocality
can be verified by using the Svetlichny inequality Sy ; AR . Surprisingly, all
the linear inequalities such as the CHSH inequality CHSH  with dichotomic
settings are useless for verifying all the output states of $\rho_{(A_{i})}$’s
Fei ; Cheng ; Fine , see Appendix E. Here, we construct a new nonlinear
inequality as follows (see Appendix F),

$\displaystyle(\langle X_{1}X_{2}\rangle-\langle
Y_{1}Y_{2}\rangle)^{2}+(\langle X_{1}Y_{2}\rangle+\langle
Y_{1}X_{2}\rangle)^{2}$ $\displaystyle-(1-\langle{}Z_{1}Z_{2}\rangle)^{2}\leq
0,$ (56)

which holds for any separable qubit states, where $X_{j}$, $Y_{j}$ and $Z_{j}$
are Pauli observables on the $j$-th qubit, $j=1,2$. The maximal quantum bound
is $4$. The inequality (56) provides a nonlinear entanglement witness for
verifying the non-locality of $\rho_{(A_{j})}$, $j=1,2,3$, as shown in Fig. 4.
It can be further regarded as a Hardy-type inequality Hardy . The new
inequality is applicable for verifying the strong nonlocality of the
generalized W states.

Figure 5: (Color online) The visibilities of $v$. Svetlichny inequality Sy ;
AR  is used for verifying noisy state $\rho$ while the inequality (56) is used
for the reduced states $\rho_{(A_{i})}$. Here,
$\alpha=\sin\theta\cos\phi,\beta=\sin\theta\sin\phi$ and $\gamma=\cos\theta$.

For the noisy state Werner  of

$\displaystyle\rho=v|W\rangle\langle W|+\frac{1-v}{8}\mathbbm{1}_{8},$ (57)

the Svetlichny inequality Sy ; AR  detects the non-locality of $\rho$ for
$v\geq\min\\{\frac{4}{\max_{\theta}\\{2\Delta(\sin 3\theta+\sin\theta)-\sin
3\theta+3\sin\theta\\}},1\\}$ with
$\Delta=\alpha\beta+\alpha\gamma+\beta\gamma$, where $\mathbbm{1}_{8}$ denotes
the identity operator on three qubits and $v\in(0,1)$. Fig. 5 shows the
visibilities of $v$ for verifying the strong nonlocalities using the
Svetlichny inequality Sy ; AR  and the inequality (56). The inequality (56)
can be further extended for multipartite scenarios as

$\displaystyle 4\langle(iX+Y)^{\otimes n}\rangle\langle(iX-Y)^{\otimes
n}\rangle-(2^{n}-\langle(\mathbbm{1}+Z)^{\otimes n}\rangle$
$\displaystyle-\langle(\mathbbm{1}-Z)^{\otimes n}\rangle)^{2}\leq 0,$ (58)

which holds for any biseparable qubit state Sy . The present inequality (58)
is useful for verifying some entangled states Ghose09 ; AR  which cannot be
verified by using the Svetlichny inequality Sy  under the assumption of each
particle is qubit state in local hidden state model. Hence, the assumption of
the present method for the nonlocality is stronger than Bell nonlocality. The
proof is shown in Appendix G.

##  VI Robustness-depth of multipartite entanglement

Let ${\cal B}_{b}$ consist of all biseparable states Sy . It is easy to show
that ${\cal B}_{b}\subset{\cal B}_{d}$ on the same Hilbert space
$\otimes_{j=1}^{n}\mathbbm{H}_{A_{j}}$, where ${\cal B}_{d}$ is defined at the
beginning of Sec.III. Denote ${\cal B}_{n}$ as the set consisting of all the
$n$-partite network local states NWR ; Kraft ; Luo2020  that can be generated
by using $m$-partite entangled states with $m<n$, shared randomness and local
operations. For each $n$-partite state $\rho\in{\cal B}_{b}$, it is
biseparable Sy . It implies that $\rho\in{\cal B}_{d}$ from Definition 1 with
$|S|=n$. We get that ${\cal B}_{b}\subseteq{\cal B}_{d}\cap{\cal B}_{n}$. Note
there are states (Examples 1 and 2) that are genuinely multipartite entangled
states but not network entanglement or robust entanglement. This means ${\cal
B}_{b}\subset{\cal B}_{d}\cap{\cal B}_{n}$. This inspires one natural problem
to verify entangled states in terms of the robustness-depth.

Definition 3. An $n$-partite state $\rho$ has robustness-depth $k$ if ${\cal
E}_{S}(\rho)$ is entangled in the biseparable model for any $S$ with $|S|\leq
k-1$.

Let ${\cal B}_{d}^{(k)}$ be the set consisting of all the states with the
robustness-depth less than $k$. ${\cal B}_{d}^{(k)}$ is star-convex with one
center subset consisting of fully separable states, see Appendix H. For any
$n$-partite state $\rho\not\in{\cal B}_{d}^{(k)}$, $\rho$ has the robustness-
depth no less than $k$, that is, $\rho$ is robust entanglement against the
loss of no more than $k-1$ particles. This model is useful for witnessing the
robustness depth in terms of the particle loss.

Definition 4. An $n$-partite quantum network is $k$-connected if each party
shares at most $k$ bipartite entangled states with other parties.

Proposition 3. The robustness-depth of any $n$-partite $k$-connected quantum
network is at most $k-1$.

Proof. It is sufficient to show that there is one subset
$S=\\{\textsf{A}_{j_{1}},\cdots,\textsf{A}_{j_{k}}\\}$ such that the reduced
state of $\rho_{(S)}={\cal{}E}_{S}(\rho)$ is biseparable, where $\rho$ denotes
the total state of ${\cal N}_{q}$.

The main idea is from the $k$-connectedness of ${\cal N}_{q}$ in Definition 4.
In detail, consider any party $\textsf{A}_{j}$ who shares at most $k$
bipartite entangled states with the parties
$\textsf{A}_{j_{1}},\cdots,\textsf{A}_{j_{k}}$. Now, define the subset
$S=\\{\textsf{A}_{j_{1}},\cdots,\textsf{A}_{j_{k}}\\}$. Consider the final
state $\rho_{(S)}={\cal{}E}_{S}(\rho)$ after passing through the particle-loss
channel ${\cal{}E}_{S}(\cdot{})$, which is associated with the particles owned
by $\textsf{A}_{j_{1}},\cdots,\textsf{A}_{j_{k}}$. From Definition 4, there
are at most $k$ bipartite entangled states shared by $\textsf{A}_{j}$. After
${\cal N}_{q}$ passing through the particle-loss channel
${\cal{}E}_{S}(\cdot{})$, $\textsf{A}_{j}$ does not share entangled states
with others, where each party $\textsf{A}_{j_{s}}$ in $S$ shares at most one
entanglement with $\textsf{A}_{j}$. Here, the particle-loss channel
${\cal{}E}_{S}(\cdot{})$ is to cut all the entangled states shared with the
party $\textsf{A}_{j}$. So, we get that

$\displaystyle\rho_{(S)}=\rho_{\textsf{A}_{j}}\otimes\varrho_{(S)|\textsf{A}_{j}},$
(59)

where
$\rho_{\textsf{A}_{j}}={\cal{}E}_{B_{t_{1}}\cdots{}B_{t_{k}}}(\otimes_{s=1}^{k}|\Phi_{s}\rangle_{A_{t_{s}}B_{t_{s}}}\langle\Phi_{s}|)$
denotes the state owned by $\textsf{A}_{j}$, and
$\varrho_{(S)|\textsf{A}_{j}}$ denotes the total state (after the particle-
lose channel ${\cal E}_{\textsf{A}_{j_{s}}\not\in S}(\cdot{})$ being
performed) shared by all the parties except for $\textsf{A}_{j}$ and all
parties not in $S$. Since $\rho_{(S)}$ is separable, from Definition 4, the
robustness-depth of $\rho$ is at most $k-1$. This completes the proof. $\Box$

For an $n$-partite $k$-connected quantum network ${\cal N}_{q}$, it is
genuinely $n$-partite entangled Sy ; ZDBS , where each pair can recover one
bipartite entanglement assisted by other parties’ local operations and
classical communication. These networks are inherent nonlocal because of the
connectedness CPV . Nevertheless, ${\cal N}_{q}$ is not robust against the
loss of at most $k$ particles. This provides interesting examples included in
${\cal B}_{d}^{(k)}$ and yields to a general hierarchy of multipartite states
in terms of the robustness-depth as

$\displaystyle{\cal B}_{s}\subset{\cal B}_{b}\subset{\cal
B}_{d}^{(1)}\subset{\cal B}_{d}^{(2)}\subset\cdots\subset{\cal
B}_{d}^{(n-1)},$ (60)

where ${\cal B}_{s}$ consists of all fully separable states. Due to the non-
convexity of ${\cal B}_{d}^{(k)}$, it is difficult to verify the robustness-
depth for general state $\rho\not\in{\cal B}_{d}^{(k)}$. Interestingly,
quantum networks provide an easy example. Suppose that ${\cal N}_{q}$ is an
$n$-partite network consisting of bipartite entangled states. The robustness-
depth of ${\cal N}_{q}$ is then determined by its minimum degree.

Definition 5. For a given quantum network ${\cal N}_{q}$ consisting of all
bipartite entangled states, the degree $\deg(\textsf{A}_{j})$ of one party
$\textsf{A}_{j}$ is the number of parties $\textsf{A}_{k}$ with $k\not=j$ such
that $\textsf{A}_{j}$ and $\textsf{A}_{k}$ share at least one entanglement.
The minimum degree of ${\cal N}_{q}$ is the minimal degrees of all parties,
that is, $\deg({\cal N}_{q})=\min\\{\deg(\textsf{A}_{j}),\forall j\\}$.

Result 4. The robustness-depth of ${\cal N}_{q}$ is given by $d=\deg({\cal
N}_{q})-1$, where $\deg({\cal N}_{q})$ denotes its minimum degree.

Proof. Consider an $n$-partite quantum network ${\cal N}_{q}$ with $\deg({\cal
N}_{q})=k$, that is, each party shares a bipartite entangled state with at
least $k$ parties in ${\cal N}_{q}$. We need the following lemma, see Appendix
H.

Lemma 4. For each pair of parties in ${\cal N}_{q}$ there are at least $k$
different chain-type subnetworks for connecting them.

From Lemma 4 after ${\cal N}_{q}$ losing all the particles owned by at most
$k-1$ parties, there are at least one chain-type subnetwork connected by any
two parties $\textsf{A}_{i}$ and $\textsf{A}_{j}$. Hence, after ${\cal N}_{q}$
passing through the channel ${\cal E}_{S}(\cdot{})$ with
$S=\\{\textsf{A}_{j_{1}},\cdots,\textsf{A}_{j_{k-1}}\\}$, the final subnetwork
is 1-connected from Definition 4. By using the recent method ZDBS , we can
prove that the final state after ${\cal N}_{q}$ passing through the particle-
lose channel ${\cal E}_{S}(\cdot{})$ is genuinely entangled. From Definitions
3 and 5 the robustness-depth of ${\cal N}_{q}$ is given by $d=\deg({\cal
N}_{q})-1$. This completes the proof. $\Box$

Example 6. Cluster states have inherent network decompositions Cluster . As
any linear cluster state (chain-type network as shown Fig.2(a)) is associated
with a specific linear graph with degree no more than 2 Cluster , the limited
connectedness rules out the robustness against particle loss. The second
example is the 2D cluster states Jozsa2006  as shown Fig.2(d), or 3D cluster
states which are associated with planar graph or cubic graph, respectively.
The robustness-depth of these universal resources is no more than its minimum
degree. This implies strong restrictions on the specific computation tasks
without error correction. Other examples including the honeycomb networks
consisting of GHZ states Wei2011 ; GHZ ; Luo2018  may be considered similarly.

##  VII Losing channel associated with single particles

So far, each node in entangled networks is regarded as a combined particle.
One may consider the lose of partial particles owned by each party, that is,
$S$ consists of a single particle in Definition 1. The final state after
particle-loss channels may be an $m$-partite state with $m\geq n-|S|$.

Definition 6. Consider an $n$-partite entangled network ${\cal N}_{q}$ in the
state $\rho$ on Hilbert space $\otimes_{i=1}^{n}\mathbbm{H}_{A_{i}}$. It is
robust entanglement if the output state given by

$\displaystyle\varrho$ $\displaystyle=$ $\displaystyle{\cal E}_{S}(\rho)$ (61)
$\displaystyle=$ $\displaystyle\sum_{i\in K_{S}}E_{i}\rho{}E^{\dagger}_{i}$

is $m$-partite entangled for any subset $S$ (consisting of particles)
satisfying $|S|\leq N-2$, where $\varrho$ is an $m$-partite state, $K_{S}$
denotes the Kraus operator decomposition of ${\cal E}_{S}(\cdot{})$, $N$
denotes the number of all the particles and $m\leq N-|S|$.

Different from Definition 1, $S$ in Definition 6 may contain particles owned
by more than $n-2$ parties, but cannot contain all the particles owned by
$n-1$ parties. Here, the remained particles after ${\cal N}_{q}$ passing
through ${\cal E}_{S}(\cdot{})$ should be owned by different parties. Take
triangle network ${\cal N}_{q}$ shown in Fig.3(a) as an example. Suppose that
${\cal N}_{q}$ consists of three EPR states
$|\phi\rangle_{B_{1}B_{2}},|\phi\rangle_{B_{3}B_{4}}$ and
$|\phi\rangle_{B_{5}B_{6}}$. One can consider the set $S$ with $|S|=3$ such as
$S=\\{B_{1},B_{2},B_{3}\\}$ or $\\{B_{2},B_{3},B_{4}\\}$, or $|S|=4$ such as
$S=\\{B_{1},B_{2},B_{3},B_{4}\\}$ or $\\{B_{2},B_{3},B_{4},B_{6}\\}$. With
this definition, we can extend Results 2-4 as follows.

Result 5. The total state of any $k$-independent ($k\geq 2$) quantum network
${\cal N}_{q}$ is not robust in the present model.

For a given $k$-independent quantum network ${\cal N}_{q}$ with $k\geq 2$,
there are at least two parties who do not share entanglement after ${\cal
N}_{q}$ passing through the particle-lose channel associated with all the
particles owned by other parties. So, the total state of ${\cal N}_{q}$ is not
robust entanglement from Definition 6.

Result 6. Suppose ${\cal N}_{q}$ is an $n$-partite completely connected
network consisting of bipartite entangled pure states and generalized Dicke
states. The total state of ${\cal N}_{q}$ is robust entanglement if one of the
following conditions satisfies

  * (i)

$|S|\leq n-2$;

  * (ii)

$n-2<|S|\leq N-2$ and the remained subnetwork after ${\cal N}_{q}$ passing
through the channel ${\cal E}_{S}(\cdot{})$ is connected.

The proof is easily followed from the connectedness of final networks after
passing through particle-lose channels assisted by the recent method ZDBS . In
fact, from the assumption of Result 6 and Definition 2, each pair of two
parties shares at least one bipartite pure entanglement or generalized Dicke
state. The connectedness implies that all the final states in Eq.(61) are at
least $k$-partite entanglement with $k=n-|S|$ for $|S|\leq n-2$. For
$n-2<|S|\leq N-2$, from the assumption we have a connected network after
${\cal N}_{q}$ passing through the channel ${\cal E}_{S}(\cdot{})$. The proof
is completed by using recent method ZDBS .

For the case of $n-2<|S|\leq N-2$, Result 6 is different from Result 3. In
fact, the remained $m$-partite state $\varrho$ has the entanglement depth less
than $m$. Take the triangle network stated above as an example. The remained
state is fully separable if $S=\\{B_{1},B_{3},B_{5}\\}$. Hence, the remained
subnetwork after passing through the channel ${\cal E}_{S}(\cdot{})$ may be
disconnected. In this case, the final state is not entangled. Hence, with the
assumption of connectedness, the remained network is entangled. This implies
special restrictions on $S$.

Definition 7. An $n$-partite entangled network $\rho$ is robust entanglement
with depth $k$ if ${\cal E}_{S}(\rho)$ is entangled for any $S$ with $|S|\leq
k-1$, where ${\cal E}_{S}(\cdot)$ is defined in Definition 6.

From Result 4 we get the following result.

Result 7. Suppose that ${\cal N}_{q}$ is an $n$-partite network consisting of
bipartite entangled states. The robustness-depth of ${\cal N}_{q}$ is given by
$d=\deg({\cal N}_{q})-1$ for $|S|\leq d-1$, where $\deg({\cal N}_{q})$ denotes
the minimum degree of ${\cal N}_{q}$.

The proof is similar to its for Result 4. Different from Result 4, ${\cal
N}_{q}$ may be robust against losing more than $\deg({\cal N}_{q})-1$ number
of particles. One example is the triangle network stated above, where the
final state is entangled for $S=\\{B_{1},B_{2},B_{3}\\}$ while $\deg({\cal
N}_{q})=2$. However, this does not hold for all the subsets $S$ with $|S|=3$.
One example is $S=\\{B_{1},B_{3},B_{5}\\}$. Similarly, there exists one subset
$S$ with $|S|=\deg({\cal N}_{q})$ such that the final state $\varrho={\cal
E}_{S}(\rho)$ is biseparable for ${\cal N}_{q}$. For each party
$\textsf{A}_{j}$, one can choose $S$ consisting of all the particles entangled
with the party $\textsf{A}_{j}$.

##  VIII Conclusions

It is generally difficult to verify all the outputs of particle-loss channels
by using entanglement witnesses HHH  or Bell inequalities Sy ; BCPS . The
problems can be solved for special states such as generalized symmetric states
DVC ; Toth  which have inherent encoding meanings in experiment, and the
entangled networks for lang-distant quantum communications or distributed
quantum computation. Additionally, the present model is also related to the
relaxed absolute maximal entanglement by assuming all the final states after
passing particle-losing channels being maximally mixed states.

In conclusion, we have proposed a local model to verify strongly correlated
multipartite entanglement robust against particle-loss. This provides an
interesting way to characterize single entangled systems or network scenarios
going beyond the biseparable model or network local models. It has been used
to explore new generic features of multipartite entangled qubit states. This
model is useful for characterizing different entangled quantum networks. It is
applicable for witnessing the robustness-depth under particle-loss. The
present results may highlight further studied on entanglement theory, quantum
information processing and measurement-based quantum computation.

## Acknowledgements

This work was supported by the National Natural Science Foundation of China
(Grant Nos.62172341,61772437,12075159), Sichuan Youth Science and Technique
Foundation (Grant No.2017JQ0048), Fundamental Research Funds for the Central
Universities (Grant No. 2018GF07), Beijing Natural Science Foundation (Grant
No.Z190005), Academy for Multidisciplinary Studies, Capital Normal University,
Shenzhen Institute for Quantum Science and Engineering, Southern University of
Science and Technology (Grant No.SIQSE202001), and the Academician Innovation
Platform of Hainan Province.

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##  Appendix A Proof of Lemma 1

The proof is completed by two subcases.

Subcase 1. $\\{|\phi_{j}^{(i)}\rangle_{A_{j}},\forall i\\}$ are orthogonal
states for all $j$’s.

In this case, the state of $|\Phi\rangle_{A_{1}\cdots{}A_{n}}$ in Eq.(4) is
equivalent to a generalized GHZ state
$\sum_{j=1}^{2}\sqrt{\lambda_{j}}|j\cdots{}j\rangle_{A_{1}\cdots{}A_{n}}$
under local operations, that is,

$\displaystyle(\otimes_{j=1}^{n}W_{A_{j}})|\Phi\rangle_{A_{1}\cdots{}A_{n}}=\sqrt{p_{1}}|0\cdots
0\rangle+\sqrt{1-p_{1}}|1\cdots 1\rangle$ (62)

where $W_{A_{j}}$ are unitary operations on the qubit $A_{j}$ defined by

$\displaystyle
W_{A_{j}}:|\phi_{k}^{(j)}\rangle_{A_{j}}\mapsto|k\rangle_{A_{j}}$ (63)

for $j=1,\cdots,n$.

Subcase 2. $\\{|\phi_{j}^{(k)}\rangle_{A_{j}},\forall k\\}$ are not orthogonal
states for some $k$’s.

Define the local unitary operation $W_{A_{k}}$ as

$\displaystyle W_{A_{k}}:|\phi_{j}^{(k)}\rangle\mapsto|j\rangle,j=0,1$ (64)

for $k=1,\cdots,n$. Since local unitary operations do not change the
entanglement, from Eq.(4) it is sufficient to consider the following state

$\displaystyle|\Psi\rangle_{A_{1}\cdots{}A_{n}}$ $\displaystyle:=$
$\displaystyle(\otimes_{i\not=2}W_{A_{i}})|\Phi\rangle$ (65) $\displaystyle=$
$\displaystyle\sqrt{1-p_{0}}|1\rangle_{A_{1}}|1\rangle_{A_{2}}(\otimes_{j=3}^{n}|\psi_{j}\rangle_{A_{j}})$
$\displaystyle+\sqrt{p_{0}}|0\cdots 0\rangle_{A_{1}\cdots{}A_{n}}$

Without changing of notations, assume that
$\\{|\phi_{2}^{(i)}\rangle_{A_{2}},\forall i\\}$ are orthogonal states from
the orthogonality of $|\Phi_{0}\rangle$ and $|\Phi_{1}\rangle$ in Eq.(4), and
$|\psi_{j}\rangle:=W_{A_{j}}|\phi_{j}^{(2)}\rangle_{A_{j}}$ for
$j=3,\cdots,n$. For the simplicity, suppose that

$\displaystyle|\psi_{j}\rangle=a_{0j}|0\rangle+a_{1j}|1\rangle$ (66)

for $j=3,\cdots,n$.

Let $|\Psi\rangle_{A_{1}\cdots{}A_{n}}$ in Eq.(65) pass through the particle-
lose channels ${\cal{}E}_{A_{3}\cdots{}A_{n}}(\cdot)$. The remained state is
given by

$\displaystyle\rho_{(A_{3}\cdots{}A_{n})}$ $\displaystyle=$
$\displaystyle{\cal{}E}_{A_{3}\cdots{}A_{n}}(|\Psi\rangle_{A_{1}\cdots{}A_{n}}\langle\Psi|)$
(67) $\displaystyle=$ $\displaystyle
c|\hat{\phi}\rangle_{A_{1}A_{2}}\langle\hat{\phi}|+(1-c)|11\rangle_{A_{1}A_{2}}\langle
11|,$

where $|\hat{\phi}\rangle_{A_{1}A_{2}}$ is given by

$\displaystyle|\hat{\phi}\rangle_{A_{1}A_{2}}=\frac{1}{\sqrt{c}}(\sqrt{p_{0}}|00\rangle+\sqrt{1-p_{0}}\prod_{j=3}^{n}a_{0j}|11\rangle)$
(68)

and $c=p_{0}+(1-p_{0})\prod_{j=3}^{n}a_{0j}$. $\rho_{(A_{3}\cdots{}A_{n})}$ in
Eq.(67) is a bipartite entanglement from the PPT criterion PPT  for any
$\prod_{j=3}^{n}a_{0j}\not=0$. This contradicts to the assumption that
$\rho_{(A_{i})}$ is fully separable for any $i$. Hence, there is $j$ such that
$a_{0j}=0$. Take $j=3$ for an example. From Eq.(65) it follows that

$\displaystyle|\Psi\rangle$ $\displaystyle=$
$\displaystyle\sqrt{1-p_{0}}|111\rangle_{A_{1}A_{2}A_{3}}\otimes_{j=4}^{n}|\psi_{j}\rangle_{A_{j}}$
(69) $\displaystyle+\sqrt{p_{0}}|0\cdots{}0\rangle_{A_{1}\cdots{}A_{n}}$

Let $|\Psi\rangle$ in Eq.(69) pass through the particle-lose channel
${\cal{}E}_{A_{4}\cdots{}A_{n}}(\cdot)$. The remained state is given by

$\displaystyle\rho_{(A_{4}\cdots{}A_{n})}$ $\displaystyle=$ $\displaystyle{\rm
Tr}_{A_{4}\cdots{}A_{n}}(|\Psi\rangle\langle\Psi|)$ $\displaystyle=$
$\displaystyle
c^{\prime}|\tilde{\phi}\rangle_{A_{1}A_{2}A_{3}}\langle\tilde{\phi}|+(1-c^{\prime})|111\rangle_{A_{1}A_{2}A_{3}}\langle
111|,$

where $|\tilde{\phi}\rangle_{A_{1}A_{2}A_{3}}$ is given by

$\displaystyle|\tilde{\phi}\rangle_{A_{1}A_{2}A_{3}}=\frac{1}{\sqrt{c}}(\sqrt{p_{0}}|000\rangle+\sqrt{1-p_{0}}\prod_{j=4}^{n}a_{0j}|111\rangle)$

and $c^{\prime}=p_{0}+(1-p_{0})\prod_{j=4}^{n}a_{0j}$.

Note that $\rho_{(A_{4}\cdots{}A_{n})}$ is a genuinely tripartite entanglement
Sy  for any $\prod_{j=4}^{n}a_{0j}\not=0$, where
$|\tilde{\phi}\rangle_{A_{1}A_{2}A_{3}}$ is genuinely tripartite entangled GHZ
. This contradicts to the assumption that $\rho_{(A_{i})}$ is fully separable
for any $i$. Hence, there is an integer $j$ with $a_{0j}=0$. Take $j=4$ for an
example. From Eq.(69) we get

$\displaystyle|\Psi\rangle_{A_{1}\cdots{}A_{n}}$ $\displaystyle=$
$\displaystyle\sqrt{1-p_{0}}\otimes_{s=1}^{4}|1\rangle_{A_{s}}\otimes_{j=5}^{n}|\psi_{j}\rangle_{A_{j}}$
(72) $\displaystyle+\sqrt{p_{0}}|0\cdots 0\rangle.$

The procedure stated above can be iteratively performed for $j=5,\cdots,n$.
This implies that

$\displaystyle|\Psi\rangle_{A_{1}\cdots{}A_{n}}=\sqrt{p_{0}}|0\cdots
0\rangle+\sqrt{1-p_{0}}|1\cdots 1\rangle$ (73)

under local operations. Hence, from Eq.(65), the state of
$|\Phi\rangle_{A_{1}\cdots{}A_{n}}$ is equivalent to a generalized $n$-partite
GHZ state under local unitary operations. This completes the proof for the
subcase 2.

##  Appendix B Proof of Lemma 2

By using the normal form Kraus2009  of $|\Phi\rangle_{A_{1}\cdots{}A_{n}}$,
from Eq.(4) assume that

$\displaystyle|\Phi\rangle=\sqrt{p_{0}}|0\cdots
0\rangle_{A_{1}\cdots{}A_{n}}+\sqrt{1-p_{0}}|1\rangle_{A_{1}}|\hat{\Phi}\rangle_{A_{2}\cdots{}A_{n}}$
(74)

where $|\hat{\Phi}\rangle$ is an $n-1$-partite state of qubits
$A_{2},\cdots,A_{n}$, which is orthogonal to the state $|0\cdots
0\rangle_{A_{2}\cdots{}A_{n}}$.

In what follows, the main goal is to prove $|\hat{\Phi}\rangle=|1\cdots
1\rangle_{A_{2}\cdots{}A_{n}}$. Consider the subsystem after $|\Phi\rangle$
passing through the particle-lose channel ${\cal E}_{A_{1}}(\cdot)$ as follows

$\displaystyle\rho_{(A_{1})}=p_{0}\otimes_{j=2}^{n}|0\rangle\langle
0|_{A_{j}}+(1-p_{0})|\hat{\Phi}\rangle_{A_{2}\cdots{}A_{n}}\langle\hat{\Phi}|$
(75)

Note that $\rho_{(A_{1})}$ is fully separable. Assume that

$\displaystyle|\hat{\Phi}\rangle_{A_{2}\cdots{}A_{n}}=\otimes_{j=2}^{n}|\psi_{j}\rangle_{A_{j}}.$
(76)

Moreover, $|\hat{\Phi}\rangle_{A_{2}\cdots{}A_{n}}$ and $|0\cdots
0\rangle_{A_{2}\cdots{}A_{n}}$ are orthogonal states from Eq.(74). From
Eqs.(65)-(73), we get that $|\hat{\Phi}\rangle=|1\cdots
1\rangle_{A_{2}\cdots{}A_{n}}$. So, the state of $|\Phi\rangle$ in Eq.(B1) is
equivalent to a generalized GHZ state in Eq.(2) under local unitary
operations. This completes the proof.

##  Appendix C Proof of Lemma 3

We prove that $p=\frac{1}{2}$ and $|\Phi_{0}\rangle_{A_{2}\cdots A_{n}}$ and
$|\Phi_{1}\rangle_{A_{2}\cdots A_{n}}$ have special decompositions, that is,
the qubits $A_{1}$ and $A_{2}$ are symmetric. From the Schmidt decomposition,
by using the normal form Kraus2009  suppose that

$\displaystyle|\Phi_{0}\rangle=\alpha_{0}|0\rangle_{A_{2}}|\phi_{0}\rangle_{A_{3}\cdots
A_{n}}+\beta_{0}|1\rangle_{A_{2}}|\phi_{1}\rangle_{A_{3}\cdots A_{n}}$ (77)

where $|\phi_{0}\rangle$ and $|\phi_{1}\rangle$ are orthogonal states.

Define $W$ as an $n-2$-partite unitary operation given by

$\displaystyle W:|\phi_{i}\rangle_{A_{3}\cdots{}A_{n}}\mapsto|i\cdots
i\rangle_{A_{3}\cdots{}A_{n}},i=0,1.$ (78)

This can be extended to general unitary operation on Hilbert space
$\otimes_{j=3}^{n}\mathbb{H}_{A_{j}}$. From Eqs.(77) and (78), we get that

$\displaystyle(\mathbbm{1}_{A_{2}}\otimes{}W)|\Phi_{0}\rangle_{A_{2}\cdots{}A_{n}}$
$\displaystyle=$ $\displaystyle\alpha_{0}|0\cdots 0\rangle+\beta_{0}|1\cdots
1\rangle,$
$\displaystyle(\mathbbm{1}_{A_{2}}\otimes{}W)|\Phi_{1}\rangle_{A_{2}\cdots{}A_{n}}$
$\displaystyle=$ $\displaystyle\alpha_{1}|0\cdots 0\rangle+\beta_{1}|1\cdots
1\rangle$ (79)
$\displaystyle+\delta|\hat{\Phi}_{1}\rangle_{A_{2}\cdots{}A_{n}},$

where $|\hat{\Phi}_{1}\rangle_{A_{2}\cdots{}A_{n}}$ and
$\mathbbm{1}_{A_{2}}\otimes{}W|\Phi_{0}\rangle_{A_{2}\cdots{}A_{n}}$ are
orthogonal states. Assume that $|\hat{\Phi}_{1}\rangle_{A_{2}\cdots{}A_{n}}$
is on the subspace spanned by all the $n-1$ qubit states $|i_{2}\cdots
i_{n}\rangle_{A_{2}\cdots{}A_{n}}$ except for $|0\cdots
0\rangle_{A_{2}\cdots{}A_{n}}$ and $|1\cdots 1\rangle_{A_{2}\cdots{}A_{n}}$.

In what follows, we prove $\delta=0$ and $p=1/2$. In fact, consider the
bipartition of $\\{A_{2}\\}$ and $\\{A_{3},\cdots,A_{n}\\}$.
$(\mathbbm{1}_{A_{2}}\otimes{}W_{A_{3}\cdots
A_{n}})\rho_{(A_{1})}(\mathbbm{1}_{A_{2}}\otimes W^{\dagger}_{A_{3}\cdots
A_{n}})$ is biseparable because $\rho_{(A_{1})}$ is fully separable, where
$\mathbbm{1}_{A_{2}}$ is the identity operator on qubit $A_{2}$. Hence, by
using the PPT criterion PPT  we get

$\displaystyle\hat{\rho}^{T}_{A_{2}}\geq 0,$ (80)

where $\hat{\rho}^{T}_{A_{2}}$ is defined by
$\hat{\rho}^{T}_{A_{2}}=(x_{ij,k\ell})$ with $x_{ij,k\ell}=y_{kj,i\ell}$ and
$\hat{\rho}_{(A_{1})}=(y_{ij,k\ell})$ in terms of the bipartition
$\\{A_{2}\\}$ and $\\{A_{3},\cdots,A_{n}\\}$.

Consider the principal minor $D_{0\overline{1}}$ of $\hat{\rho}^{T}_{A_{2}}$
defined by the matrix basis
$\\{|00\rangle\langle\overline{1}\,\overline{1}|,|0\overline{1}\rangle\langle\overline{1}0|,|\overline{1}0\rangle\langle
0\overline{1}|,|\overline{1}\,\overline{1}\rangle\langle 00|\\}$ with
$\overline{1}=1\cdots 1$ ($n-2$ number of 1). It follows that

$\displaystyle p\alpha_{0}\beta_{0}+(1-p)\alpha_{1}\beta_{1}=0.$ (81)

Combined with $\alpha_{0}\alpha_{1}+\beta_{0}\beta_{1}=0$, that is, the
orthogonality of $|\Phi_{i}\rangle$’s, we get that

$\displaystyle\alpha_{1}=\pm\alpha_{0}\gamma,\beta_{1}=\mp\beta_{0}\gamma,p=\frac{1}{1+\gamma^{2}}.$
(82)

Now, we will prove that $\gamma=1$ or $\delta=0$, that is, there does not
exist $|\hat{\Phi}_{1}\rangle$ in Eq.(79). The proof is completed by
contradiction. In fact, consider the principal minor of
$\hat{\rho}^{T}_{A_{2}}$ defined on the subspace associated with
$|\hat{\Phi}_{1}\rangle$. Since $|\hat{\Phi}_{1}\rangle$ and
$\alpha_{1}|0\cdots 0\rangle_{A_{2}\cdots{}A_{n}}+\beta_{1}|1\cdots
1\rangle_{A_{2}\cdots{}A_{n}}$ are states on different subspaces, the minor
depends only on the state of $|\hat{\Phi}_{1}\rangle_{A_{2}\cdots{}A_{n}}$.
From Eq.(80), it follows that $\gamma=0$, or
$|\hat{\Phi}_{1}\rangle_{A_{2}\cdots{}A_{n}}$ is biseparable Sy  in terms of
the bipartition $\\{A_{2}\\}$ and $\\{A_{3},\cdots,A_{n}\\}$, that is,
$|\hat{\Phi}_{1}\rangle=|\phi\rangle_{A_{2}}|\hat{\Phi}_{2}\rangle_{A_{3}\cdots
A_{n}}$. In what follows, it only needs to prove the second case.

Generally, suppose that

$\displaystyle|\hat{\Phi}_{1}\rangle_{A_{2}\cdots{}A_{n}}=|\phi\rangle_{A_{2}}|\hat{\Phi}_{2}\rangle_{A_{3}\cdots
A_{n}}.$ (83)

Note that $|\hat{\Phi}_{1}\rangle_{A_{2}\cdots{}A_{n}}$ does not contain the
terms of $|0\rangle^{\otimes n-1}$ and $|1\rangle^{\otimes n-1}$. There are
three subcases.

  * (i)

$|\phi\rangle_{A_{2}}=|0\rangle$. We get $|\hat{\Phi}_{2}\rangle_{A_{3}\cdots
A_{n}}$ does not contain the terms of $|0\rangle^{\otimes n-2}$. Suppose that

$\displaystyle|\hat{\Phi}_{2}\rangle_{A_{3}\cdots
A_{n}}=\alpha_{2}|\hat{\Phi}_{3}\rangle_{A_{3}\cdots A_{n}}+\beta_{2}|1\cdots
1\rangle_{A_{3}\cdots A_{n}}.$ (84)

Define $H$ as an $n-2$-partite unitary operation $H$ given by

$\displaystyle H:$ $\displaystyle|i\cdots{}i\rangle_{A_{3}\cdots
A_{n}}\mapsto|i\cdots{}i\rangle_{A_{3}\cdots A_{n}},$ (85)
$\displaystyle|\hat{\Phi}_{3}\rangle_{A_{3}\cdots
A_{n}}\mapsto|0\rangle_{A_{3}}|1\cdots 1\rangle_{A_{4}\cdots A_{n}}.$

From Eqs.(79), (82) and (85), it follows that

$\displaystyle(\mathbbm{1}_{A_{2}}+HW)|\Phi_{0}\rangle$ $\displaystyle=$
$\displaystyle(\mathbbm{1}_{A_{2}}\otimes W)|\Phi_{0}\rangle,$
$\displaystyle(\mathbbm{1}_{A_{2}}+HW)|\Phi_{1}\rangle$ $\displaystyle=$
$\displaystyle\alpha_{1}|0\cdots 0\rangle_{A_{2}\cdots A_{n}}$
$\displaystyle+\beta_{1}|1\cdots 1\rangle_{A_{2}\cdots A_{n}}$
$\displaystyle+\delta\alpha_{2}|0\rangle_{A_{2}}|1\cdots 1\rangle_{A_{3}\cdots
A_{n}}$ $\displaystyle+\delta\beta_{2}|00\rangle_{A_{2}A_{3}}|1\cdots
1\rangle_{A_{3}\cdots A_{n}}.$

Now, consider the remained state $\rho_{(A_{1}A_{4}\cdots{}A_{n})}$ as

$\displaystyle\rho_{(A_{1}A_{4}\cdots{}A_{n})}$ (87) $\displaystyle=$
$\displaystyle{\cal{}E}_{A_{4}\cdots
A_{n}}((\mathbbm{1}_{A_{2}}+HW)\rho_{(A_{1})}(\mathbbm{1}_{A_{2}}+HW)^{\dagger})$
$\displaystyle=$
$\displaystyle(p\alpha_{0}^{2}+(1-p)\alpha_{1}^{2})|00\rangle_{A_{2}A_{3}}\langle
00|$ $\displaystyle+p\beta_{0}^{2}|11\rangle_{A_{2}A_{3}}\langle
11|+(1-p)\tau|\psi\rangle_{A_{2}A_{3}}\langle\psi|,$

where $|\psi\rangle_{A_{2}A_{3}}$ is given by

$\displaystyle|\psi\rangle_{A_{2}A_{3}}=\frac{1}{\tau}(\delta\beta_{2}|00\rangle+\delta\alpha_{2}|01\rangle+\beta_{1}|11\rangle)$
(88)

and $\tau=\sqrt{\delta^{2}+\beta_{1}^{2}}$. Note that
$\rho_{(A_{1}A_{4}\cdots{}A_{n})}$ should be separable because
$\rho_{(A_{1})}$ is fully separable, where $H$ and $W$ are not performed on
qubit $A_{2}$. Hence, it follows that $\delta=0$ or $\beta_{2}=0$ by using the
PPT criterion PPT , that is, $|\psi\rangle_{A_{2}A_{3}}$ should be separable.
From Eq.(84) it yields to

$\displaystyle|\hat{\Phi}_{2}\rangle_{A_{3}\cdots
A_{n}}=\alpha_{2}|\hat{\Phi}_{3}\rangle.$ (89)

Similarly, suppose that $H$ in Eq.(85) is replaced by the following operation

$\displaystyle H:$ $\displaystyle|\hat{\Phi}_{3}\rangle_{A_{3}\cdots
A_{n}}\mapsto|1\rangle_{A_{3}}|0\cdots 0\rangle_{A_{4}\cdots A_{n}},$ (90)
$\displaystyle|i\cdots i\rangle_{A_{3}\cdots A_{n}}\mapsto|i\cdots
i\rangle_{A_{3}\cdots A_{n}}.$

We can prove that $\alpha_{2}=0$ or $\delta=0$ by using the separability of
the reduced density matrix $\rho_{(A_{1}A_{4}\cdots{}A_{n})}={\cal
E}_{A_{4}\cdots
A_{n}}((\mathbbm{1}_{A_{2}}+HW)\rho_{(A_{1})}(\mathbbm{1}_{A_{2}}+HW)^{\dagger})$.
So, we get $\delta=0$.

  * (ii)

$|\phi\rangle_{A_{2}}=|1\rangle$. We get that
$|\hat{\Phi}_{2}\rangle_{A_{3}\cdots A_{n}}$ does not contain the terms of
$|1\cdots 1\rangle_{A_{3}\cdots A_{n}}$. Suppose that

$\displaystyle|\hat{\Phi}_{2}\rangle=\alpha_{2}|\hat{\Phi}_{3}\rangle_{A_{3}\cdots{}A_{n}}+\beta_{2}|0\cdots
0\rangle_{A_{3}\cdots A_{n}}.$ (91)

Similar to Eqs.(85)-(90), we get $\delta=0$.

  * (iii)

$|\phi\rangle_{A_{2}}=\alpha_{2}|0\rangle+\beta_{2}|1\rangle$. In this case,
$|\hat{\Phi}_{2}\rangle_{A_{3}\cdots A_{n}}$ does not contain the terms of
$|0\cdots 0\rangle_{A_{3}\cdots A_{n}}$ and $|1\cdots 1\rangle_{A_{3}\cdots
A_{n}}$. Define $H$ as a unitary operation given by

$\displaystyle H:$ $\displaystyle|\hat{\Phi}_{2}\rangle_{A_{3}\cdots
A_{n}}\mapsto|1\rangle_{A_{3}}|0\cdots 0\rangle_{A_{4}\cdots A_{n}},$ (92)
$\displaystyle|i\cdots i\rangle_{A_{3}\cdots A_{n}}\mapsto|i\cdots
i\rangle_{A_{3}\cdots A_{n}}.$

From Eqs.(79), (82) and (92) we get that

$\displaystyle(\mathbbm{1}_{A_{2}}+HW)|\Phi_{0}\rangle_{A_{2}\cdots{}A_{n}}$
$\displaystyle=$ $\displaystyle(\mathbbm{1}_{A_{2}}+W)|\Phi_{0}\rangle,$ (93)

and

$\displaystyle(\mathbbm{1}_{A_{2}}+HW)|\Phi_{1}\rangle_{A_{2}\cdots{}A_{n}}$
(94) $\displaystyle=$ $\displaystyle\alpha_{1}|0\cdots
0\rangle_{A_{2}\cdots{}A_{n}}+\beta_{1}|1\cdots 1\rangle_{A_{2}\cdots{}A_{n}}$
$\displaystyle+\delta(\alpha_{2}|0\rangle+\beta_{2}|1\rangle)_{A_{2}}|1\rangle_{A_{3}}|0\cdots
0\rangle_{A_{4}\cdots{}A_{n}}.$

Consider the remained state $\rho_{(A_{1}\cdots{}A_{4}\cdots{}A_{n})}$ after
passing through the particle-lose channel ${\cal E}_{A_{4}\cdots A_{n}}$ as

$\displaystyle\rho_{(A_{1}\cdots{}A_{4}\cdots{}A_{n})}$ (95) $\displaystyle=$
$\displaystyle{\cal E}_{A_{4}\cdots
A_{n}}(\mathbbm{1}_{A_{2}}+HW)\rho_{(A_{1})}(\mathbbm{1}_{A_{2}}+HW)^{\dagger})$
$\displaystyle=$
$\displaystyle(p\beta_{0}^{2}+(1-p)\beta_{1}^{2})|11\rangle_{A_{2}A_{3}}\langle
11|$ $\displaystyle p\alpha_{0}^{2}|00\rangle_{A_{2}A_{3}}\langle
00|+(1-p)\tau|\psi^{\prime}\rangle_{A_{2}A_{3}}\langle\psi^{\prime}|.$

where $|\psi^{\prime}\rangle_{A_{2}A_{3}}$ is given by

$\displaystyle|\psi^{\prime}\rangle=\frac{1}{\sqrt{\tau^{\prime}}}(\alpha_{1}|00\rangle+\delta(\alpha_{2}|0\rangle+\beta_{2}|1\rangle)_{A_{2}}|1\rangle_{A_{3}})$
(96)

and $\tau^{\prime}=\sqrt{\alpha_{1}^{2}+\delta^{2}}$.
$\rho_{(A_{1}A_{4}\cdots{}A_{n})}$ should be separable since $\rho_{(A_{1})}$
is fully separable, where $H$ and $W$ are not performed on qubit $A_{2}$.
Hence, it follows that $\delta=0$ or $\alpha_{1}=0$ by using the PPT criterion
PPT . However, from the assumption in Case (iii), we have $\alpha_{1}\not=0$.
It follows that $\delta=0$.

Hence, we have proved $\delta=0$. Combining with Eqs.(77) and (79),
$|\Phi_{i}\rangle$ in Eq.(4) is written into

$\displaystyle|\Phi_{1}\rangle=\beta_{0}|0\rangle_{A_{2}}|\phi_{0}\rangle_{A_{3}\cdots
A_{n}}-\alpha_{0}|1\rangle_{A_{2}}|\phi_{1}\rangle_{A_{3}\cdots A_{n}}$ (97)

From Eqs.(4), (77) and (97), it follows that

$\displaystyle|\Phi\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{2}(\alpha_{0}|0\rangle\pm\beta_{0}|1\rangle)_{A_{1}}|0\rangle_{A_{2}}|\phi_{0}\rangle_{A_{3}\cdots
A_{n}}$ (98)
$\displaystyle+\frac{1}{2}(\beta_{0}|0\rangle\mp\alpha_{0}|1\rangle)_{A_{1}}|1\rangle_{A_{2}}|\phi_{1}\rangle_{A_{3}\cdots
A_{n}}$

Define $U_{A_{1}}^{\pm}$ as local unitary operations given by

$\displaystyle U_{A_{1}}^{\pm}:$
$\displaystyle\alpha_{0}|0\rangle_{A_{1}}\pm\beta_{0}|1\rangle_{A_{1}}\mapsto|0\rangle,$
(99)
$\displaystyle\beta_{0}|0\rangle_{A_{1}}\mp\alpha_{0}|1\rangle_{A_{1}}\mapsto|1\rangle.$

It follows that

$\displaystyle(U_{A_{1}}^{\pm}+\mathbbm{1}_{A_{2}\cdots{}A_{n}})|\Phi\rangle$
$\displaystyle=$
$\displaystyle\frac{1}{2}(|00\rangle_{A_{1}A_{2}}|\phi_{0}\rangle_{A_{3}\cdots{}A_{n}}$
(100)
$\displaystyle+|11\rangle_{A_{1}A_{2}}|\phi_{1}\rangle_{A_{3}\cdots{}A_{n}})$

where $\mathbbm{1}_{A_{2}\cdots{}A_{n}}$ denotes the identity operator on
qubits $A_{2},\cdots,A_{n}$. So, the qubits $A_{1}$ and $A_{2}$ in
$|\Phi\rangle$ are symmetric under local unitary operations.

##  Appendix D The star-convexity of ${\cal B}_{d}$

Consider Hilbert space
$\mathbb{H}:=\mathbb{H}_{A_{1}}\otimes\cdots\otimes\mathbb{H}_{A_{n}}$. The
goal is to construct a new state $\rho=p\rho_{1}+(1-p)\rho_{2}$ such that
$\rho\not\in{\cal B}_{d}$ for some $p$, and $\rho_{1},\rho_{2}\in{\cal B}_{d}$
on $\mathbb{H}$. Define

$\displaystyle\rho_{1}$ $\displaystyle=$
$\displaystyle|\Phi_{1}\rangle_{A_{1}A_{2}A_{3}}\langle\Phi_{1}|,$
$\displaystyle\rho_{2}$ $\displaystyle=$
$\displaystyle|\Phi_{1}\rangle_{A_{1}A_{2}A_{3}}\langle\Phi_{2}|$ (101)

with

$\displaystyle|\Phi_{1}\rangle=\frac{1}{2}(|000\rangle+|011\rangle+|120\rangle+|131\rangle),$
$\displaystyle|\Phi_{2}\rangle=\frac{1}{2}(|000\rangle+|101\rangle+|210\rangle+|311\rangle).$
(102)

It is easy to prove that $|\Phi_{1}\rangle$ and $|\Phi_{2}\rangle$ are
genuinely tripartite entanglement Sy . Moreover, let $\rho_{(A_{i})}^{(j)}$ be
the output state of $\rho_{j}$ after passing through the particle-lose channel
${\cal{}E}_{A_{i}}(\cdot)$, $i=1,\cdots,3;j=1,2$, where the density matrices
of $\rho_{(A_{i})}^{(j)}$ are defined by

$\displaystyle\rho_{(A_{1})}^{(1)}$ $\displaystyle=$
$\displaystyle\rho_{(A_{2})}^{(2)}=\frac{1}{2}|\phi_{00}\rangle\langle\phi_{00}|+\frac{1}{2}|\psi_{20}\rangle\langle\psi_{20}|,$
$\displaystyle\rho_{(A_{2})}^{(1)}$ $\displaystyle=$
$\displaystyle\rho_{(A_{1})}^{(2)}=\frac{1}{4}(|00\rangle\langle
00|+|01\rangle\langle 01|$ $\displaystyle+|10\rangle\langle
10|+|11\rangle\langle 11|),$ $\displaystyle\rho_{(A_{3})}^{(1)}$
$\displaystyle=$
$\displaystyle\rho_{(A_{3})}^{(2)}=\frac{1}{2}|\psi_{12}\rangle\langle\psi_{12}|+\frac{1}{2}|\psi_{01}\rangle\langle\psi_{01}|,$
(103)

and

$\displaystyle|\phi_{00}\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle),|\psi_{20}\rangle=\frac{1}{\sqrt{2}}(|20\rangle+|31\rangle),$
$\displaystyle|\psi_{12}\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{2}}(|00\rangle+|12\rangle),|\psi_{01}\rangle=\frac{1}{\sqrt{2}}(|01\rangle+|13\rangle).$

Note that $|\psi_{ij}\rangle$’s are equivalent to the EPR state EPR  under
local unitary operations. Moreover,
$\rho_{(A_{1})}^{(1)},\rho_{(A_{2})}^{(2)},\rho_{(A_{3})}^{(1)}$ and
$\rho_{(A_{3})}^{(2)}$ are bipartite entangled states, where
$|\psi_{ij}\rangle$’s are defined on different subspaces which rule out
separable decompositions. Instead, $\rho_{(A_{2})}^{(1)}$ and
$\rho_{(A_{1})}^{(2)}$ are separable. So, we get $\rho_{1},\rho_{2}\in{\cal
B}_{d}$.

Now, define

$\displaystyle\varrho=\frac{1}{2}\rho_{1}+\frac{1}{2}\rho_{2}.$ (104)

It follows from Eq.(103) that

$\displaystyle\varrho_{(A_{1})}$ $\displaystyle=$
$\displaystyle\varrho_{(A_{2})}$ $\displaystyle=$
$\displaystyle\frac{1}{4}|\phi_{00}\rangle\langle\phi_{00}|+\frac{1}{4}|\psi_{20}\rangle\langle\psi_{20}|+\frac{1}{8}(|00\rangle\langle
00|$ $\displaystyle+|01\rangle\langle 01|+|10\rangle\langle
10|+|11\rangle\langle 11|)$ $\displaystyle\varrho_{(A_{3})}$ $\displaystyle=$
$\displaystyle\frac{1}{2}|\psi_{12}\rangle\langle\psi_{12}|+\frac{1}{2}|\psi_{01}\rangle\langle\psi_{01}|.$
(105)

Since $|\psi_{20}\rangle$ is defined on specific subspace spanned by
$\\{|20\rangle,|31\rangle\\}$, which is different from the associated subspace
of
$\frac{1}{4}|\phi_{00}\rangle\langle\phi_{00}|+\frac{1}{8}(|00\rangle\langle
00|+|01\rangle\langle 01|+|10\rangle\langle 10|+|11\rangle\langle 11|)$.
$\varrho_{(A_{1})}$ and $\varrho_{(A_{2})}$ are bipartite entangled states.
So, $\varrho_{(A_{i})}$’s are bipartite entangled. Moreover, $\varrho$ is a
genuinely tripartite entanglement PPT , where $\rho_{1}$ can be decomposed
into the chain-shape network as shown in Fig.2(a) Luo2018 . It means that
$\rho\not\in{\cal B}_{d}$. Hence, ${\cal B}_{d}$ is not convex.

The star convexity of ${\cal B}_{d}$ can be followed by choosing the maximally
mixed state $\rho_{0}=\frac{1}{2^{n}}\mathbbm{1}$ on Hilbert space
$\otimes_{i}\mathbb{H}_{A_{i}}$ as the center point. For any state
$\rho\in{\cal B}_{d}$, it is easy to prove that $p\rho+(1-p)\rho_{0}\in{\cal
B}_{d}$ for any $p\in(0,1)$. This completes the proof.

##  Appendix E The useless of CHSH inequality

We prove that the CHSH inequality CHSH  cannot be applied for verifying the
strong nonlocality of the generalized W state DVC ; Fei ; Cheng  as

$\displaystyle|W\rangle_{A_{1}A_{2}A_{3}}=\alpha|001\rangle+\beta|010\rangle+\gamma|100\rangle$
(106)

with $\alpha^{2}+\beta^{2}+\gamma^{2}=1$. The proof is completed for three
states $\rho_{(A_{3})}$, $\rho_{(A_{2})}$ and $\rho_{(A_{1})}$. In fact, from
Eq.(106) it follows that

$\displaystyle\rho_{(A_{3})}=\alpha^{2}|00\rangle_{A_{1}A_{2}}\langle
00|+(1-\alpha^{2})|\psi_{1}\rangle_{A_{1}A_{2}}\langle\psi_{1}|,$
$\displaystyle\rho_{(A_{2})}=\beta^{2}|00\rangle_{A_{1}A_{3}}\langle
00|+(1-\beta^{2})|\psi_{2}\rangle_{A_{1}A_{3}}\langle\psi_{2}|,$
$\displaystyle\rho_{(A_{1})}=\gamma^{2}|00\rangle_{A_{2}A_{3}}\langle
00|+(1-\gamma^{2})|\psi_{3}\rangle_{A_{2}A_{3}}\langle\psi_{3}|,$ (107)

where $|\psi_{i}\rangle$ are given by
$|\psi_{1}\rangle=\frac{1}{\sqrt{1-\alpha^{2}}}(\beta|01\rangle+\gamma|10\rangle)$,
$|\psi_{2}\rangle=\frac{1}{\sqrt{1-\beta^{2}}}(\alpha|01\rangle+\gamma|10\rangle)$
and
$|\psi_{3}\rangle=\frac{1}{\sqrt{1-\gamma^{2}}}(\alpha|01\rangle+\beta|10\rangle)$.

Figure 6: (Color online) The violation of CHSH inequality for the state of
$\rho_{(A_{3})}$. Here, $\Delta(\rho_{(A_{3})})=4\sqrt{2}\beta\gamma-2$,
$\alpha=\sin\theta\cos\phi,\beta=\sin\theta\sin\phi$, and $\gamma=\cos\theta$.
The maximal violation of $2.8284$ is unavailable because of $\alpha=0$.

There are two cases to achieve the maximal violation of the CHSH inequality
CHSH . One is from local measurements on the $x$-$y$ plane of Pauli sphere.
The other is from local measurements on $x$-$z$ plane of Pauli sphere.
Specially, take local observables $A_{1}=X$ and $A_{2}=Y$ for the first party,
$B_{1}=\cos\theta{}X+i\sin\theta{}Y$ and $B_{2}=\cos\theta{}X-i\sin\theta{}Y$
for the second party. For the state of $\rho_{(A_{3})}$, it follows that

$\displaystyle\langle A_{1}B_{1}\rangle+\langle A_{1}B_{2}\rangle+\langle
A_{2}B_{1}\rangle-\langle A_{2}B_{2}\rangle$ (108) $\displaystyle=$
$\displaystyle 4\cos\theta\beta\gamma+4\sin\theta\beta\gamma$ $\displaystyle=$
$\displaystyle 4\sqrt{2}\beta\gamma$

when $\theta=\frac{\pi}{4}$. This means that the statistics generated from
local measurements on $\rho_{(A_{3})}$ violates the CHSH inequality CHSH  if
$|\beta\gamma|>\frac{1}{2\sqrt{2}}$, as shown in Fig.6.

Figure 7: (Color online) The violations of the CHSH inequality with
observablei from the $x$-$y$ plane of Pauli sphere. Here, the red region is
for $\rho_{(A_{3})}$. The blue region is for $\rho_{(A_{2})}$. The green
region is for $\rho_{(A_{1})}$. It shows that there is no choice of
$\alpha,\beta$ and $\gamma$ such that its statistics of $\rho_{(A_{j})}$’s
violate the CHSH inequality.

Figure 8: (Color online) The violations of the CHSH inequality with
observables from the $x$-$z$ plane of Pauli sphere. Here, the red region is
for $\rho_{(A_{3})}$. The blue region is for $\rho_{(A_{2})}$. The green
region is for $\rho_{(A_{1})}$. It shows that there is no choice of
$\alpha,\beta$ and $\gamma$ such that the statistics of $\rho_{(A_{j})}$’s
violate the CHSH inequality.

Similarly, take local observable $A_{1}$ and $A_{2}$ for the first party,
$C_{1}=\cos\theta_{1}{}X+i\sin\theta_{1}{}Y$ and
$C_{2}=\cos\theta_{1}{}X-i\sin\theta_{1}{}Y$ for the third party. The
statistics generated by local measurements on the state of $\rho_{(A_{2})}$
violates the CHSH inequality CHSH  if $|\alpha\gamma|>\frac{1}{2\sqrt{2}}$,
where $\theta_{1}=\frac{\pi}{4}$. Finally, by taking local observable
$B_{1}^{\prime}=Y$ and $B_{2}^{\prime}=X$ for the second party,
$C_{1}=\cos\theta_{2}{}X+i\sin\theta_{2}{}Y$ and
$C_{2}=\cos\theta_{2}{}X-i\sin\theta_{2}{}Y$ for the third party, the
statistics generated by local measurements on the state of $\rho_{(A_{1})}$
violates the CHSH inequality CHSH  if $|\alpha\beta|>\frac{1}{2\sqrt{2}}$,
where $\theta_{2}=\frac{\pi}{2}$. Now, the values of $\alpha,\beta$ and
$\gamma$ which satisfy all the violation conditions are shown in Fig.7. Hence,
there is no choice of $\alpha,\beta$ and $\gamma$ such that the correlations
derived from local measurements on $\rho_{(A_{3})}$, $\rho_{(A_{2})}$ and
$\rho_{(A_{1})}$ can violate the CHSH inequality CHSH  simultaneously. Here,
each party can use different local observables dependent on the verified
states Fei .

Moreover, take observables from the $x$-$z$ plane of Pauli sphere, that is,
Pauli matrices $X$ and $Z$. All the violations are shown in Fig.8 vias two
phases. It shows that there is no choice of $\alpha,\beta$ and $\gamma$ such
that the quantum correlations derived from local measurements on
$\rho_{(A_{i})}$’s violate the CHSH inequality CHSH . This means that the CHSH
inequality cannot be used for verifying the strong nonlocality of any W state
in Eq.(E1). While the strong nonlocality of W states can be verified by
violating the inequality (54), as shown in Fig. 9.

##  Appendix F Proof of the inequality (56)

Consider a separable state on Hilbert space
$\mathbbm{H}_{A}\otimes\mathbbm{H}_{B}$ as

$\displaystyle\rho_{AB}=\sum_{j}p_{j}|\phi_{j}\rangle_{A}\langle\phi_{j}|\otimes|\varphi_{j}\rangle_{B}\langle\varphi_{j}|,$
(109)

where $|\phi_{j}\rangle=\alpha_{j0}|0\rangle+\alpha_{j1}|1\rangle$,
$|\varphi_{j}\rangle=\beta_{j0}|0\rangle+\beta_{j1}|1\rangle$, and
$\\{p_{j}\\}$ is a probability distribution. Let
$\rho=(\rho_{s_{1}s_{2};j_{1}j_{2}})$ be density matrix of $\rho$. We firstly
prove that

$\displaystyle 4\rho_{00;11}\rho_{11;00}-(\rho_{01;01}+\rho_{10;10})^{2}\leq
0.$ (110)

If $\rho_{AB}$ is given by a product state
$|\phi_{j}\rangle_{A}|\varphi_{j}\rangle_{B}$, we have
$4\rho_{00;11}\rho_{11;00}=4|\alpha_{j0}\beta_{j0}\alpha_{j1}\beta_{j1}|^{2}$,
and
$(\rho_{01;01}+\rho_{10;10})^{2}=(|\alpha_{j0}\beta_{j1}|^{2}+|\alpha_{j1}\beta_{j0}|^{2})^{2}$.
By using the Cauchy-Schwartz inequality of $2xy\leq x^{2}+y^{2}$, it follows
the inequality (110). From the Hermitian symmetry of density matrix $\rho$
that $4\rho_{00;11}\rho_{11;00}=4|\rho_{00;11}|^{2}$. It follows that

$\displaystyle 4\rho_{00;11}\rho_{11;00}$ $\displaystyle=$ $\displaystyle
4|\rho_{00;11}|^{2}$ (111) $\displaystyle=$ $\displaystyle
4|\sum_{j}p_{j}\rho^{(j)}_{00;11}|^{2}$ $\displaystyle\leq$
$\displaystyle(\sum_{j}2p_{j}|\rho^{(j)}_{00;11}|)^{2}$ $\displaystyle=$
$\displaystyle(\sum_{j}2p_{j}|\alpha_{j0}\beta_{j0}\alpha_{j1}\beta_{j1}|)^{2}$
$\displaystyle\leq$
$\displaystyle(\sum_{j}p_{j}|\alpha_{j0}\beta_{j1}|^{2}+\sum_{j}p_{j}|\alpha_{j1}\beta_{j0}|^{2})^{2}$
$\displaystyle=$
$\displaystyle(\sum_{j}p_{j}(\rho_{01;01}^{(j)}+\rho_{10;10}^{(j)}))^{2}$
(114) $\displaystyle=$ $\displaystyle(\rho_{01;01}+\rho_{10;10})^{2}.$ (115)

Here, $\rho^{(j)}_{00;11}$ in Eq.(111) is given by
$\rho^{(j)}_{00;11}=\alpha_{j0}\beta_{j0}\alpha_{j1}\beta_{j1}$. The
inequality (F) follows from the triangle inequality of $|x+y|\leq|x|+|y|$. The
inequality (LABEL:L9) follows from the Cauchy-Schwartz inequality of
$x^{2}+y^{2}\geq 2xy$. $\rho_{01;01}^{(j)}$ and $\rho_{10;10}^{(j)}$ in
Eq.(114) are given respectively by
$\rho_{01;01}^{(j)}=|\alpha_{j0}\beta_{j1}|^{2}$ and
$\rho_{10;10}^{(j)}=|\alpha_{j1}\beta_{j0}|^{2}$. This proves the inequality
(110).

Figure 9: (Color online) The violations of the inequality (108) by measuring
the states $\rho_{(A_{1})},\rho_{(A_{2})},\rho_{(A_{3})}$ with observable from
Pauli sphere.

Now, by using the definitions of Pauli matrices $X,Y,Z$, we get
$\rho_{01;01}=\frac{1}{4}\langle(\mathbbm{1}+Z)(\mathbbm{1}-Z)\rangle$,
$\rho_{10;10}=\frac{1}{4}\langle(\mathbbm{1}-Z)(\mathbbm{1}+Z)\rangle$,
$\rho_{11;00}=-\frac{1}{4}\langle(iX-Y)(iX-Y)\rangle$ and
$\rho_{00;11}=-\frac{1}{4}\langle(iX+Y)(iX+Y)\rangle$ with $i=\sqrt{-1}$. It
follows the inequality (56).

In what follows, we construct a generalized Hardy-type inequality with three
inputs Hardy . Consider six distributions $\\{P(ab|xy),a,b\in\\{0,1\\}\\}$ for
any $x,y\in\\{0,1,2\\}$. From the local model, it can be decomposed into

$\displaystyle P(ab|xy)$ $\displaystyle=$ $\displaystyle\int
P(ab|xy,\lambda)d\mu(\lambda)$ (116) $\displaystyle=$ $\displaystyle\int
p(a|x,\lambda)p(b|y,\lambda)d\mu(\lambda),$

where $\lambda$ denotes any shared randomness with the measure $\mu(\lambda)$.
Suppose that these distributions satisfy

$\displaystyle\sum_{a,b}(-1)^{a+b}P(ab|11)$ $\displaystyle=$
$\displaystyle\sum_{a,b}(-1)^{a+b}P(ab|12)$ (117) $\displaystyle=$
$\displaystyle\sum_{a,b}(-1)^{a+b}P(ab|21)$ $\displaystyle=$
$\displaystyle\sum_{a,b}(-1)^{a+b}P(ab|22),$

and

$\displaystyle P(01|00)P(10|00)=0,$ (118) $\displaystyle P(00|11)=P(00|22),$
(119) $\displaystyle P(10|11)\geq P(10|12),$ (120)
$\displaystyle(P(01|11)-P(10|11))^{2}$ $\displaystyle-(P(01|12)-P(10|12))^{2}$
$\displaystyle-(P(00|00)+P(11|00))^{2}\leq 0.$ (121)

Here, the constrictions (117)-(120) are inspired by the joint conditional
distributions associated from Pauli matrices on special separable states. In
fact, from Eq.(117), it follows that

$\displaystyle P(01|11)+P(10|11)=P(01|12)+P(10|12)$ (122)

From the inequality (119), it follows that

$\displaystyle(P(01|11)-P(10|11))^{2}\leq(P(01|12)-P(10|12))^{2}$ (123)

Combined with Eq.(118), it follows the inequality (121).

##  Appendix G Proof of the inequality (58)

Consider a biseparable state on Hilbert space
$\otimes_{j=1}^{n}\mathbbm{H}_{A_{j}}$ as

$\displaystyle\rho_{A_{1}\cdots{}A_{n}}=\sum_{j}p_{j}|\Phi_{j}\rangle_{S}\langle\Phi_{j}|\otimes|\Psi_{j}\rangle_{\overline{S}}\langle\Psi_{j}|,$
(124)

where
$|\Phi_{t}\rangle=\sum_{j_{1},\cdots,j_{k}=0}^{1}\alpha_{t,j_{1}\cdots{}j_{k}}|j_{1}\cdots{}j_{k}\rangle$
denotes the joint system shared by all parties in the subset
$S=\\{A_{j_{1}},\cdots,A_{j_{k}}\\}\subset\\{A_{1},\cdots,A_{n}\\}$,
$|\Psi_{t}\rangle=\sum_{s_{1},\cdots,s_{n-k}=0}^{1}\beta_{t,s_{1}\cdots{}s_{n-k}}|j\rangle$
denotes the joint system shared by all the parties in the complement set
$\overline{S}=\\{A_{s_{1}},\cdots,A_{s_{n-k}}\\}$ of $S$, $s_{t}\not=j_{\ell}$
for any $t,\ell$, $\alpha_{t,j_{1}\cdots{}j_{k}}$ satisfies
$\sum_{j_{1},\cdots,j_{k}=0}^{1}|\alpha_{t,j_{1}\cdots{}j_{k}}|^{2}=1$,
$\beta_{t,j_{1}\cdots{}j_{k}}$ satisfies
$\sum_{s_{1},\cdots,s_{n-k}=0}^{1}|\beta_{t,s_{1}\cdots{}s_{n-k}}|^{2}=1$, and
$\\{p_{i}\\}$ is a probability distribution. Let
$\rho=(\rho_{\vec{j}_{1}\vec{s}_{2};\vec{j}^{\prime}_{1}\vec{s}^{\prime}_{2}})$
be the density matrix of $\rho$ with $\vec{j}_{1}=j_{1}\cdots{}j_{k}$ and
$\vec{s}_{2}=s_{1}\cdots{}s_{n-k}$.

We firstly prove the following inequality

$\displaystyle
4\rho_{\vec{0}_{1}\vec{0}_{2};\vec{1}_{1}\vec{1}_{2}}\rho_{\vec{1}_{1}\vec{1}_{2};\vec{0}_{1}\vec{0}_{2}}-(1-\rho_{\vec{0}_{1}\vec{0}_{2};\vec{0}_{1}\vec{0}_{2}}-\rho_{\vec{1}_{1}\vec{1}_{2};\vec{1}_{1}\vec{1}_{2}})^{2}\leq
0.$ (125)

If $\rho_{A_{1}\cdots{}A_{n}}$ is a product state given by
$|\Phi_{i}\rangle_{S}|\Psi_{i}\rangle_{\overline{S}}$, we have

$\displaystyle
4\rho_{\vec{0}_{1}\vec{0}_{2};\vec{1}_{1}\vec{1}_{2}}\rho_{\vec{1}_{1}\vec{1}_{2};\vec{0}_{1}\vec{0}_{2}}$
$\displaystyle=$ $\displaystyle
4|\alpha_{i,\vec{0}_{1}}\beta_{i,\vec{0}_{2}}\alpha_{i,\vec{1}_{1}}\beta_{i,\vec{1}_{2}}|^{2}$
(126) $\displaystyle\leq$
$\displaystyle(|\alpha_{i,\vec{0}_{1}}\beta_{i,\vec{0}_{2}}|^{2}+|\alpha_{i,\vec{1}_{1}}\beta_{i,\vec{1}_{2}}|^{2})^{2}$
$\displaystyle=$
$\displaystyle(\rho_{\vec{0}_{1}\vec{0}_{2};\vec{0}_{1}\vec{0}_{2}}+\rho_{\vec{1}_{1}\vec{1}_{2};\vec{1}_{1}\vec{1}_{2}})^{2}$

from the Cauchy-Schwartz inequality.

Now, consider the mixed state in Eq.(124). It follows from the inequality
(126) that

$\displaystyle
4\rho_{\vec{0}_{1}\vec{0}_{2};\vec{1}_{1}\vec{1}_{2}}\rho_{\vec{1}_{1}\vec{1}_{2};\vec{0}_{1}\vec{0}_{2}}$
(127) $\displaystyle=$ $\displaystyle
4|\rho_{\vec{0}_{1}\vec{0}_{2};\vec{1}_{1}\vec{1}_{2}}|^{2}$ $\displaystyle=$
$\displaystyle
4|\sum_{j}p_{j}\rho^{(j)}_{\vec{0}_{1}\vec{0}_{2};\vec{1}_{1}\vec{1}_{2}}|^{2}$
$\displaystyle\leq$
$\displaystyle(\sum_{j}2p_{j}|\rho^{(j)}_{\vec{0}_{1}\vec{0}_{2};\vec{1}_{1}\vec{1}_{2}}|)^{2}$
$\displaystyle=$
$\displaystyle(\sum_{j}2p_{j}|\alpha_{j,\vec{0}_{1}}\beta_{j,\vec{0}_{2}}\alpha_{j,\vec{1}_{1}}\beta_{j,\vec{1}_{2}}|)^{2}$
$\displaystyle\leq$
$\displaystyle(\sum_{j}p_{j}|\alpha_{j,\vec{0}_{1}}\beta_{j,\vec{1}_{2}}|^{2}+\sum_{j}p_{j}|\alpha_{j,\vec{1}_{1}}\beta_{j,\vec{0}_{2}}|^{2})^{2}$
$\displaystyle=$
$\displaystyle(\sum_{j}p_{j}(\rho_{\vec{0}_{1}\vec{1}_{2};\vec{1}_{0}\vec{0}_{2}}^{(j)}+\rho_{\vec{1}_{1}\vec{0}_{2};\vec{1}_{1}\vec{0}_{2}}^{(j)}))^{2}$
$\displaystyle\leq$
$\displaystyle(\sum_{\vec{j}_{1}\vec{s}_{2}\not\in\\{\vec{0}_{1}\vec{0}_{2};\vec{1}_{1}\vec{1}_{2}\\}}\rho_{\vec{j}_{1}\vec{s}_{2};\vec{j}_{1}\vec{s}_{2}})^{2}$
(130) $\displaystyle=$
$\displaystyle(1-\rho_{\vec{0}_{1}\vec{0}_{2};\vec{0}_{1}\vec{0}_{2}}-\rho_{\vec{1}_{1}\vec{1}_{2};\vec{1}_{1}\vec{1}_{2}})^{2}.$
(131)

Here, the inequality (127) follows from the triangle inequality of
$|x+y|\leq|x|+|y|$.
$\rho^{(j)}_{\vec{0}_{1}\vec{0}_{2};\vec{1}_{1}\vec{1}_{2}}$ in Eq.(G) is
given by
$\rho^{(j)}_{\vec{0}_{1}\vec{0}_{2};\vec{1}_{1}\vec{1}_{2}}=\alpha_{j,\vec{0}_{1}}\beta_{j,\vec{0}_{2}}\alpha_{j,\vec{1}_{1}}\beta_{j,\vec{1}_{2}}$.
The inequality (G) is followed from the Cauchy-Schwartz inequality of
$x^{2}+y^{2}\geq 2xy$. The inequality (130) is obtained by using the
inequality (126) for all the bipartitions of $S$ and $\overline{S}$. The
equality (131) is followed from the trace equality of ${\rm tr}\rho=1$. This
completes the proof of the inequality (125).

Now, using the definitions of Pauli matrices $X,Y,Z$, we get

$\displaystyle\rho_{\vec{0}_{1}\vec{0}_{2};\vec{0}_{1}\vec{0}_{2}}=\frac{1}{2^{n}}\langle(\mathbbm{1}+Z)^{\otimes
n}\rangle,$
$\displaystyle\rho_{\vec{1}_{1}\vec{1}_{2};\vec{1}_{1}\vec{1}_{2}}=\frac{1}{2^{n}}\langle(\mathbbm{1}-Z)^{\otimes
n}\rangle,$
$\displaystyle\rho_{\vec{0}_{1}\vec{0}_{2};\vec{1}_{1}\vec{1}_{2}}=-\frac{1}{2^{n}}\langle(iX+Y)^{\otimes
n}\rangle,$
$\displaystyle\rho_{\vec{1}_{1}\vec{1}_{2};\vec{0}_{1}\vec{0}_{2}}=-\frac{1}{2^{n}}\langle(iX-Y)^{\otimes
n}\rangle$

with $i=\sqrt{-1}$. It follows the inequality (58).

Moreover, the inequality (58) may be used to construct Hardy-type inequality
similar to the inequalities (117)-(121).

Now, we consider a biseparable state on $d$-dimensional Hilbert space
$\otimes_{j=1}^{n}\mathbbm{H}_{A_{j}}$ with $d\geq 2$ as

$\displaystyle\rho_{A_{1}\cdots{}A_{n}}=\sum_{j}p_{j}|\Phi_{j}\rangle_{S}\langle\Phi_{j}|\otimes|\Psi_{j}\rangle_{\overline{S}}\langle\Psi_{j}|,$
(132)

where

$\displaystyle|\Phi_{t}\rangle=\sum_{j_{1},\cdots,j_{k}=0}^{d-1}\alpha_{t,j_{1}\cdots{}j_{k}}|j_{1}\cdots{}j_{k}\rangle$
(133)

denotes the joint system shared by all parties in the subset
$S=\\{A_{j_{1}},\cdots,A_{j_{k}}\\}\subset\\{A_{1},\cdots,A_{n}\\}$, the state
of

$\displaystyle|\Psi_{t}\rangle=\sum_{s_{1},\cdots,s_{n-k}=0}^{d-1}\beta_{t,s_{1}\cdots{}s_{n-k}}|j\rangle$
(134)

denotes the joint system shared by all the parties in the complement set
$\overline{S}=\\{A_{s_{1}},\cdots,A_{s_{n-k}}\\}$ of $S$, $s_{t}\not=j_{\ell}$
for any $t,\ell$, $\alpha_{t,j_{1}\cdots{}j_{k}}$ satisfies
$\sum_{j_{1},\cdots,j_{k}=0}^{d-1}|\alpha_{t,j_{1}\cdots{}j_{k}}|^{2}=1$,
$\beta_{t,j_{1}\cdots{}j_{k}}$ satisfies
$\sum_{s_{1},\cdots,s_{n-k}=0}^{d-1}|\beta_{t,s_{1}\cdots{}s_{n-k}}|^{2}=1$,
and $\\{p_{i}\\}$ is a probability distribution. Let

$\displaystyle\rho=(\rho_{\vec{j}_{1}\vec{s}_{2};\vec{j}^{\prime}_{1}\vec{s}^{\prime}_{2}})$
(135)

be the density matrix of $\rho$ with $\vec{j}_{1}=j_{1}\cdots{}j_{k}$ and
$\vec{s}_{2}=s_{1}\cdots{}s_{n-k}$.

We firstly prove the following inequality

$\displaystyle\\!\\!\\!\\!4\rho_{\vec{0}_{1}\vec{0}_{2};\overrightarrow{d-1}_{1}\overrightarrow{d-1}_{2}}\rho_{\overrightarrow{d-1}_{1}\overrightarrow{d-1}_{2};\vec{0}_{1}\vec{0}_{2}}$
$\displaystyle\\!\\!\\!\\!-(1-\rho_{\vec{0}_{1}\vec{0}_{2};\vec{0}_{1}\vec{0}_{2}}-\rho_{\overrightarrow{d-1}_{1}\overrightarrow{d-1}_{2};\overrightarrow{d-1}_{1}\overrightarrow{d-1}_{2}})^{2}\leq
0$ (136)

If $\rho_{A_{1}\cdots{}A_{n}}$ is a product state given by
$|\Phi_{i}\rangle_{S}|\Psi_{i}\rangle_{\overline{S}}$, we have

$\displaystyle
4\rho_{\vec{0}_{1}\vec{0}_{2};\overrightarrow{d-1}_{1}\overrightarrow{d-1}_{2}}\rho_{\overrightarrow{d-1}_{1}\overrightarrow{d-1}_{2};\vec{0}_{1}\vec{0}_{2}}$
(137) $\displaystyle=$ $\displaystyle
4|\alpha_{i,\vec{0}_{1}}\beta_{i,\vec{0}_{2}}\alpha_{i,\overrightarrow{d-1}_{1}}\beta_{i,\overrightarrow{d-1}_{2}}|^{2}$
$\displaystyle\leq$
$\displaystyle(|\alpha_{i,\vec{0}_{1}}\beta_{i,\vec{0}_{2}}|^{2}+|\alpha_{i,\overrightarrow{d-1}_{1}}\beta_{i,\overrightarrow{d-1}_{2}}|^{2})^{2}$
$\displaystyle=$
$\displaystyle(\rho_{\vec{0}_{1}\vec{0}_{2};\vec{0}_{1}\vec{0}_{2}}+\rho_{\overrightarrow{d-1}_{1}\overrightarrow{d-1}_{2};\overrightarrow{d-1}_{1}\overrightarrow{d-1}_{2}})^{2}$

from the Cauchy-Schwartz inequality.

Now, consider the mixed state in Eq.(132). It follows from the inequality
(138) that

$\displaystyle
4\rho_{\vec{0}_{1}\vec{0}_{2};\overrightarrow{d-1}_{1}\overrightarrow{d-1}_{2}}\rho_{\overrightarrow{d-1}_{1}\overrightarrow{d-1}_{2};\vec{0}_{1}\vec{0}_{2}}$
(138) $\displaystyle=$ $\displaystyle
4|\rho_{\vec{0}_{1}\vec{0}_{2};\overrightarrow{d-1}_{1}\overrightarrow{d-1}_{2}}|^{2}$
$\displaystyle=$ $\displaystyle
4|\sum_{j}p_{j}\rho^{(j)}_{\vec{0}_{1}\vec{0}_{2};\overrightarrow{d-1}_{1}\overrightarrow{d-1}_{2}}|^{2}$
$\displaystyle\leq$
$\displaystyle(\sum_{j}2p_{j}|\rho^{(j)}_{\vec{0}_{1}\vec{0}_{2};\overrightarrow{d-1}_{1}\overrightarrow{d-1}_{2}}|)^{2}$
$\displaystyle=$
$\displaystyle(\sum_{j}2p_{j}|\alpha_{j,\vec{0}_{1}}\beta_{j,\vec{0}_{2}}\alpha_{j,\overrightarrow{d-1}_{1}}\beta_{j,\overrightarrow{d-1}_{2}}|)^{2}$
$\displaystyle\leq$
$\displaystyle(\sum_{j}p_{j}|\alpha_{j,\vec{0}_{1}}\beta_{j,\overrightarrow{d-1}_{2}}|^{2}+\sum_{j}p_{j}|\alpha_{j,\overrightarrow{d-1}_{1}}\beta_{j,\vec{0}_{2}}|^{2})^{2}$
$\displaystyle=$
$\displaystyle(\sum_{j}p_{j}(\rho_{\vec{0}_{1}\overrightarrow{d-1}_{2};\overrightarrow{d-1}_{0}\vec{0}_{2}}^{(j)}+\rho_{\overrightarrow{d-1}_{1}\vec{0}_{2};\overrightarrow{d-1}_{1}\vec{0}_{2}}^{(j)}))^{2}$
$\displaystyle\leq$
$\displaystyle(\sum_{\vec{j}_{1}\vec{s}_{2}\not\in\\{\vec{0}_{1}\vec{0}_{2};\overrightarrow{d-1}_{1}\overrightarrow{d-1}_{2}\\}\atop{j_{t},s_{\ell}\in\\{0,d-1\\}}}\rho_{\vec{j}_{1}\vec{s}_{2};\vec{j}_{1}\vec{s}_{2}})^{2}.$
(141)

Here, the inequality (138) follows from the triangle inequality of
$|x+y|\leq|x|+|y|$.
$\rho^{(j)}_{\vec{0}_{1}\vec{0}_{2};\overrightarrow{d-1}_{1}\overrightarrow{d-1}_{2}}$
in Eq.(G) is given by
$\rho^{(j)}_{\vec{0}_{1}\vec{0}_{2};\overrightarrow{d-1}_{1}\overrightarrow{d-1}_{2}}=\alpha_{j,\vec{0}_{1}}\beta_{j,\vec{0}_{2}}\alpha_{j,\overrightarrow{d-1}_{1}}\beta_{j,\vec{1}_{2}}$.
The inequality (LABEL:Lh19) is followed from the Cauchy-Schwartz inequality of
$x^{2}+y^{2}\geq 2xy$. The inequality (141) is obtained by using the
inequality (138) for all the bipartitions of $S$ and $\overline{S}$.

Similarly, we get

$\displaystyle
4\rho_{\vec{u}_{1}\vec{u}_{2};\overrightarrow{d-u-1}_{1}\overrightarrow{d-u-1}_{2}}\rho_{\overrightarrow{d-u-1}_{1}\overrightarrow{d-u-1}_{2};\vec{u}_{1}\vec{u}_{2}}$
(142) $\displaystyle\leq$
$\displaystyle(\sum_{\vec{j}_{1}\vec{s}_{2}\not\in\\{\vec{u}_{1}\vec{u}_{2};\overrightarrow{d-u-1}_{1}\overrightarrow{d-u}_{2}\\}\atop{j_{t},s_{\ell}\in\\{u,d-u-1\\}}}\rho_{\vec{j}_{1}\vec{s}_{2};\vec{j}_{1}\vec{s}_{2}})^{2}$

for any $u<\frac{n}{2}$. From the inequalities (141) and (142), it follows
that

$\displaystyle
4\sum_{u=0}^{\lfloor\frac{d-1}{2}\rfloor}(\rho_{\vec{u}_{1}\vec{u}_{2};\overrightarrow{d-u-1}_{1}\overrightarrow{d-u-1}_{2}}\rho_{\overrightarrow{d-u-1}_{1}\overrightarrow{d-u-1}_{2};\vec{u}_{1}\vec{u}_{2}})^{1/2}$
(143) $\displaystyle\leq$