# Sequentially witnessing entanglement by independent observer pairs

Mao-Sheng Li<EMAIL_ADDRESS>School of Mathematics, South China
University of Technology, Guangzhou 510641, China  Yan-Ling Wang
<EMAIL_ADDRESS>School of Computer Science and Technology, Dongguan
University of Technology, Dongguan, 523808, China

(August 28, 2024)

###### Abstract

This study investigates measurement strategies in a scenario where multiple
pairs of Alices and Bobs independently and sequentially observe entangled
states. The aim is to maximize the number of observer pairs $(A_{k},B_{l})$
that can witness entanglement. Prior research has demonstrated that arbitrary
pairs $(A_{k},B_{k})$ ($k\leq n$) can observe entanglement in all pure
entangled states and a specific class of mixed entangled states [Phys. Rev. A
106 032419 (2022)]. However, it should be noted that other pairs
$(A_{k},B_{l})$ with $(k\neq l\leq n)$ may not observe entanglement using the
same strategy. Moreover, a novel strategy is presented, enabling every pair of
arbitrarily many Alices and Bobs to witness entanglement regardless of the
initial state being a Bell state or a particular class of mixed entangled
states. These findings contribute to understanding measurement strategies for
maximizing entanglement observation in various contexts.

##  I Introduction

Quantum entanglement QE_HHHH09 ; QE_GT09  is of paramount importance in the
field of quantum mechanics and finds numerous applications in various quantum
information processing tasks, including quantum cryptography QC_BB84 ;
QC_Ekert91 ; QC_GRTZ02 , quantum key distributionQK_MY98 ; QK_BHK05 ;
QK_ABGMPS07 , quantum metrology and sensing QM_PS09 ; QM_GLM11 , quantum
simulation QS_GAN14 , and quantum computing QC_Preskill18 . Consequently,
investigating efficient utilization of this entanglement resource is of great
importance. One intriguing area of research in this field focuses on the
recycling of entanglement that is shared among sequential observers, examining
how effectively it can be reused.

Bell nonlocality, as a powerful manifestation of entanglement, has attracted
significant attention. In the scenario introduced by Silva et al. Silva15 ,
Alice and Bob share an entangled state. Bob measures his qubit and then passes
it to a second Bob, who performs another measurement, and so on. The objective
is to maximize the number of Bobs who can exhibit an expected violation of the
Clauser-Horne-Shimony-Holt (CHSH) Bell inequality with a single Alice. For
bipartite quantum systems, it was shown in Silva15 ; Mal16  that, when each
Bob performs sharp measurements with equal probability, at most two Bobs can
achieve expected CHSH violations with a single Alice. However, by employing
unsharp measurements, Brown and Colbeck Brown20  demonstrated that an
arbitrarily large number of independent Bobs can share Bell nonlocality of the
Bell state with a single Alice. Subsequently, these findings have been
extended to higher dimensional bipartite systems Zhang21 ; Zhang22  and
scenarios involving multiple Alices and Bobs Cheng21 ; Cheng22 . It has also
been conjectured that, in general, at most pair of Alices and Bobs can detect
Bell-nonlocal correlations Cheng21 . Furthermore, several studies have
explored the recyclability of nonlocal correlations Mal16 ; Das19 ; Saha19 ;
Maity20 ; Srivastava20 ; Cabello21 ; Ren22 ; Zhu22 ; Das22 ; Hu22 ; Xi23 .
These investigations contribute to our understanding of how nonlocal
correlations can be effectively utilized in various settings, shedding light
on the potential applications and implications of Bell nonlocality.

Apart from studying entanglement through the violation of Bell inequalities,
the use of entanglement witnesses (EWs) has proven to be valuable in the
analysis and characterization of entanglement QE_HHHH09 ; QE_GT09 ; EW_HHH96 ;
EW_T00 . An entanglement witness operator, denoted as $W$, is an Hermitian
operator that satisfies the following two conditions (1) $\langle
W\rangle_{\rho_{s}}\geq 0$ for all separable states $\rho_{s},$ (2) there
exists an entangled state $\rho_{e}$ such that $\langle
W\rangle_{\rho_{e}}<0$. Here we use the notation $\langle
W\rangle_{\rho}=\mathrm{Tr}[W\rho]$.

Several works Bera18 ; Srivastava22 ; Pandit22  have investigated whether
entanglement can be sequentially witnessed by an arbitrary number of
observers. In contrast to the case of Bell nonlocality, Pandit et al. Pandit22
demonstrated that entanglement can be detected and recycled an arbitrary
number of times. Specifically, they considered a scenario where a bipartite
entangled state is initially shared between the first Alice-Bob pair, denoted
as $(A_{1},B_{1})$. To witness entanglement, the first pair, along with
subsequent Alice-Bob pairs, performs local measurements on their respective
qubits and passes them to the next pair $(A_{2},B_{2})$, and so on. The
objective is to maximize the number of pairs that can independently witness
entanglement. They found that an arbitrarily long sequence of Alice-Bob pairs
can successfully detect entanglement sequentially. Recent works have extended
similar findings to multipartite systems Srivastava22b ; Srivastava22c .
However, it is important to note that the successful witnessing of
entanglement by a pair $(A_{k},B_{k})$ does not imply the witness of
entanglement for every pair $(A_{i},B_{j})$ with $i,j\leq k$. There are some
pairs, as independent observers, may not witness entanglement themselves.
Therefore, it is intriguing to explore whether entanglement can be detected by
every pair of arbitrarily many Alices and Bobs. In this study, we will provide
an affirmative answer to this problem.

The paper is structured as follows. In Section II, we introduce the scenario
under investigation and present the measurement strategies and entanglement
witness employed in this study. In Section III, we demonstrate that every pair
of arbitrarily many Alices and Bobs can sequentially witness the entanglement
of any pure entangled states and a specific class of mixed entangled states.
Finally, we provide a conclusion in Section IV.

##  II Scenario, measurement strategy and entanglement Witness

Throughout this paper, we denote $\\{|0\rangle,|1\rangle\\}$ as the
computational basis of a qubit system. Under the computational basis, the
Pauli operators as $\sigma_{i},i\in\\{1,2,3\\}$ are defined as follows

$\sigma_{1}=\left[\begin{array}[]{cc}0&1\\\ 1&0\\\ \end{array}\right],\ \
\sigma_{2}=\left[\begin{array}[]{cc}0&\mathrm{i}\\\ -\mathrm{i}&0\\\
\end{array}\right],\ \ \sigma_{3}=\left[\begin{array}[]{cc}1&0\\\ 0&-1\\\
\end{array}\right],$

where $\mathrm{i}$ is the square root of unit. And we will use the maximally
entangled state

$|\Psi\rangle=\frac{1}{\sqrt{2}}(|01\rangle+|10\rangle)$

several times. Now we introduce the scenario that will consider and our
measurement strategy.

Figure 1: ($(m,n)$-sequential scenario) Sequentially witnessing entanglement
by multiple independent Alices and Bobs. Each pair $(A_{k},B_{l})$ ($1\leq
k\leq m,1\leq l\leq n$) of Alices and Bobs are intended to detect the
entanglement.

$(m,n)$-sequential scenario: We investigate a generalized version of the
sequential scenario described in Silva15 . Consider a scenario where there are
$m$ sequential Alices and $n$ sequential Bobs, with an initial entangled state
shared between the first Alice and Bob. The first Alice performs local
measurements randomly on her qubit and then passes it to the next Alice, who
performs subsequent measurements, and so on (see Fig. 1). The same sequential
process is followed by the Bobs.

The objective is to determine measurement strategies for both Alices and Bobs,
such that each pair of Alices and Bobs can witness the entanglement. The
ultimate goal is to maximize both $m$ and $n$ in order to successfully
accomplish the aforementioned task. We will prove that by employing the weak
measurement strategy presented below, one can achieve this task for
arbitrarily long sequences of Alices and Bobs, regardless of whether the
initial state is a pure entangled state or a specific class of mixed entangled
states.

Measurement strategy: All the measurements we use are two-outcome positive
operator-valued measurements (POVM) $\\{M^{(0)},M^{(1)}\\}$ where $M^{(i)}\geq
0$ and $M^{(0)}+M^{(1)}=\mathbb{I}_{2}$. For each
$P\in\\{\sigma_{1},\sigma_{2},\sigma_{3}\\}$ and $\lambda\in[0,1]$, we can
define a POVM $\\{P^{(0)}(\lambda),P^{(1)}(\lambda)\\}$ where

$P^{(0)}(\lambda)=\frac{\mathbb{I}_{2}+\lambda P}{2},\ \
P^{(1)}(\lambda)=\frac{\mathbb{I}_{2}-\lambda P}{2}.$

Suppose that each Alices and Bobs performs three types of POVMs
$\\{\sigma_{i}^{(0)}(\lambda),\sigma_{i}^{(1)}(\lambda)\\}_{i=1}^{3}$ and
$\\{\sigma_{j}^{(0)}(\gamma),\sigma_{j}^{(1)}(\gamma)\\}_{j=1}^{3}$
respectively with equal probability. The $k$-th parameter of Alice is denoted
as $\lambda^{(i)}_{k}$, and the $l$-th parameter of Bob is denoted as
$\gamma^{(j)}_{l}$, where $i,j=1,2,3$. Moreover, throughout this paper, we
assume that

$\begin{array}[]{c}\lambda^{(3)}_{k}=1\leavevmode\nobreak\
\leavevmode\nobreak\ {\rm and}\leavevmode\nobreak\ \leavevmode\nobreak\
\lambda^{(1)}_{k}=\lambda^{(2)}_{k}=\lambda_{k},\\\\[5.69054pt]
\gamma^{(3)}_{l}=1\leavevmode\nobreak\ \leavevmode\nobreak\ {\rm
and}\leavevmode\nobreak\ \leavevmode\nobreak\
\gamma^{(1)}_{l}=\gamma^{(2)}_{l}=\gamma_{l}.\end{array}$ (1)

Let $\rho_{A_{k},B_{l}}$ denote the state held by the $(A_{k},B_{l})$ pair.
Now the pair $(A_{k},B_{l})$ try to calculate the total probability $p$ of the
following three cases.

  1. (a)

Both of Alice and Bob are taking measurements
$\\{\sigma_{3}^{(0)}(1),\sigma_{3}^{(1)}(1)\\}$ and with the same outcomes (00
or 11). The corresponding probability $p_{1}$ is

$\frac{1}{9}\mathrm{Tr}\left[\left(\frac{(\mathbb{I}_{2}+\sigma_{3})\otimes(\mathbb{I}_{2}+\sigma_{3})}{4}+\frac{(\mathbb{I}_{2}-\sigma_{3})\otimes(\mathbb{I}_{2}-\sigma_{3})}{4}\right)\rho_{A_{k},B_{l}}\right].$

  2. (b)

Alice and Bob are taking measurements
$\\{\sigma_{1}^{(0)}(\lambda_{k}),\sigma_{1}^{(1)}(\lambda_{k})\\}$ and
$\\{\sigma_{1}^{(0)}(\gamma_{l}),\sigma_{1}^{(1)}(\gamma_{l})\\}$ respectively
but with different outcomes (01 or 10). The corresponding probability $p_{2}$
is

$\frac{1}{9}\mathrm{Tr}\left[\left(\frac{(\mathbb{I}_{2}+\lambda_{k}\sigma_{1})\otimes(\mathbb{I}_{2}-\gamma_{l}\sigma_{1})}{4}+\frac{(\mathbb{I}_{2}-\lambda_{k}\sigma_{1})\otimes(\mathbb{I}_{2}+\gamma_{l}\sigma_{1})}{4}\right)\rho_{A_{k},B_{l}}\right].$

  3. (c)

Alice and Bob are taking measurements
$\\{\sigma_{2}^{(0)}(\lambda_{k}),\sigma_{2}^{(1)}(\lambda_{k})\\}$ and
$\\{\sigma_{2}^{(0)}(\gamma_{l}),\sigma_{2}^{(1)}(\gamma_{l})\\}$ respectively
but with different outcomes (01 or 10). The corresponding probability $p_{3}$
is

$\frac{1}{9}\mathrm{Tr}\left[\left(\frac{(\mathbb{I}_{2}+\lambda_{k}\sigma_{2})\otimes(\mathbb{I}_{2}-\gamma_{l}\sigma_{2})}{4}+\frac{(\mathbb{I}_{2}-\lambda_{k}\sigma_{2})\otimes(\mathbb{I}_{2}+\gamma_{l}\sigma_{2})}{4}\right)\rho_{A_{k},B_{l}}\right].$

So $p$ is equal to $p_{1}+p_{2}+p_{3}$ which can be simplified as

$\begin{split}\frac{1}{9}\mathrm{Tr}\Big{[}\Big{(}\frac{3}{2}\mathbb{I}_{4}+\frac{1}{2}(\sigma_{3}\otimes\sigma_{3}-\lambda_{k}\gamma_{l}\sigma_{1}\otimes\sigma_{1}-\lambda_{k}\gamma_{l}\sigma_{2}\otimes\sigma_{2})\rho_{A_{k},B_{l}}\Big{)}\Big{]}.\end{split}$
(2)

Note that $\mathrm{Tr}[\mathbb{I}_{4}\rho_{A_{k},B_{l}}]=1$. Therefore,
$18(p_{1}+p_{2}+p_{3}-\frac{1}{9})$ is equal to

$\mathrm{Tr}\left[(\mathbb{I}_{4}+\sigma_{3}\otimes\sigma_{3}-\lambda_{k}\gamma_{l}\sigma_{1}\otimes\sigma_{1}-\lambda_{k}\gamma_{l}\sigma_{2}\otimes\sigma_{2})\rho_{A_{k},B_{l}}\right].$

Hence, we could define an observable
$W_{k,l}:=\mathbb{I}_{4}+\sigma_{3}\otimes\sigma_{3}-\lambda_{k}\gamma_{l}\sigma_{1}\otimes\sigma_{1}-\lambda_{k}\gamma_{l}\sigma_{2}\otimes\sigma_{2}.$
Surprisingly, one could show that

$\langle W_{k,l}\rangle_{\rho}:=\mathrm{Tr}[W_{k,l}\rho]\geq 0$

for all separable states when $0\leq\lambda_{k},\gamma_{l}\leq 1$ (the proof
is very similar to that in Srivastava22 . So we omit it here). So $W_{k,l}$
can be looked as an entanglement witness when $0\leq\lambda_{k},\gamma_{l}\leq
1$. We wish to use this witness $W_{k,l}$ to observe the entanglement of
$\rho_{A_{k},B_{l}}.$ Therefore, we would like to find the condition of
$\lambda_{k},\gamma_{l}$ such that

$\begin{split}\displaystyle\langle
W_{k,l}\rangle_{\rho_{A_{k},B_{l}}}=\mathrm{Tr}[(\mathbb{I}_{4}+\sigma_{3}\otimes\sigma_{3}-\lambda_{k}\gamma_{l}\sigma_{1}\otimes\sigma_{1}-\lambda_{k}\gamma_{l}\sigma_{2}\otimes\sigma_{2})\rho_{A_{k},B_{l}}]<0.\end{split}$
(3)

Now we want to express $\langle W_{k,l}\rangle_{\rho_{A_{k},B_{l}}}$ in term
of the initial state $\rho_{A_{1},B_{1}}.$ Note that $\rho_{A_{k},B_{l}}$ can
be seem as a postmeasurement states of $\rho_{A_{k-1},B_{l}}$ or
$\rho_{A_{k},B_{l-1}}$. By von Neumann-Lüder’s rule Busch96 , we have the
following recursive relations

$\displaystyle\rho_{A_{k}B_{l}}$ $\displaystyle=$
$\displaystyle\frac{1}{3}\sum_{i=1}^{3}\sum_{a=0}^{1}\left(\sqrt{\sigma_{i}^{(a)}({\lambda^{(i)}_{k-1}})}\otimes\mathbb{I}_{2}\right)\rho_{A_{k-1}B_{l}}\left(\sqrt{\sigma_{i}^{(a)}({\lambda^{(i)}_{k-1}})}\otimes\mathbb{I}_{2}\right),$
$\displaystyle\rho_{A_{k}B_{l}}$ $\displaystyle=$
$\displaystyle\frac{1}{3}\sum_{j=1}^{3}\sum_{b=0}^{1}\left(\mathbb{I}_{2}\otimes\sqrt{\sigma_{j}^{(b)}{(\gamma^{(j)}_{l-1}})}\right)\rho_{A_{k}B_{l-1}}\left(\mathbb{I}_{2}\otimes\sqrt{\sigma_{j}^{(b)}{(\gamma^{(j)}_{l-1}})}\right).$
(4)

By some calculation, Eqs. (II) could be tranferred to

$\displaystyle\rho_{A_{k}B_{l}}$ $\displaystyle=$
$\displaystyle\frac{1}{6}\sum_{i=1}^{3}\Big{[}\left(1+\Lambda_{k-1}^{(i)}\right)\rho_{A_{k-1}B_{l}}+\left(1-\Lambda_{k-1}^{(i)}\right)\sigma_{i}\otimes\mathbb{I}_{2}\rho_{A_{k-1}B_{l}}\sigma_{i}\otimes\mathbb{I}_{2}\Big{]},$
$\displaystyle\rho_{A_{k}B_{l}}$ $\displaystyle=$
$\displaystyle\frac{1}{6}\sum_{j=1}^{3}\Big{[}\left(1+\Gamma_{l-1}^{(j)}\right)\rho_{A_{k}B_{l-1}}+\left(1-\Gamma_{l-1}^{(j)}\right)\mathbb{I}_{2}\otimes\sigma_{j}\rho_{A_{k}B_{l-1}}\mathbb{I}_{2}\otimes\sigma_{j}\Big{]}$
(5)

where $\Lambda_{k}^{(i)}=\sqrt{1-\lambda^{(i)\leavevmode\nobreak\ 2}_{k}}$ and
$\Gamma_{l}^{(j)}=\sqrt{1-\gamma^{(j)\leavevmode\nobreak\ 2}_{l}}$. Therefore,
$\Lambda_{k}^{(3)}=\Gamma_{l}^{(3)}=0$ and
$\Lambda_{k}^{(i)}=\sqrt{1-\lambda^{2}_{k}}$ and
$\Gamma_{l}^{(j)}=\sqrt{1-\gamma^{2}_{l}}$ for $i,j\in\\{1,2\\}$ which will be
denoted as $\Lambda_{k}$ and $\Gamma_{l}$ respectively for simplicity. So the
key three terms of $\langle W_{k,l}\rangle_{\rho_{A_{k},B_{l}}}$ could be
expressed by

$\displaystyle\text{Tr}[\sigma_{1}\otimes\sigma_{1}\rho_{A_{k}B_{l}}]=\text{Tr}[\sigma_{1}\otimes\sigma_{1}\rho_{A_{1}B_{1}}]\displaystyle\prod_{i=1}^{k-1}\frac{\left(1+\Lambda_{i}\right)}{3}\prod_{j=1}^{l-1}\frac{\left(1+\Gamma_{j}\right)}{3},$
$\displaystyle\text{Tr}[\sigma_{2}\otimes\sigma_{2}\rho_{A_{k}B_{l}}]=\text{Tr}[\sigma_{2}\otimes\sigma_{2}\rho_{A_{1}B_{1}}]\displaystyle\prod_{i=1}^{k-1}\frac{\left(1+\Lambda_{i}\right)}{3}\prod_{j=1}^{l-1}\frac{\left(1+\Gamma_{j}\right)}{3},$
$\displaystyle\text{Tr}[\sigma_{3}\otimes\sigma_{3}\rho_{A_{k}B_{l}}]=\text{Tr}[\sigma_{3}\otimes\sigma_{3}\rho_{A_{1}B_{1}}]\displaystyle\prod_{i=1}^{k-1}\frac{\left(1+2\Lambda_{i}\right)}{3}\displaystyle\prod_{j=1}^{l-1}\frac{\left(1+2\Gamma_{j}\right)}{3}.$
(6)

In Ref. Pandit22 , the authors presented that arbitrary many pairs of
$(A_{k},B_{k})$ could detect the entanglement of
$\rho_{A_{1}B_{1}}=|\Psi\rangle\langle\Psi|$ sequentially . In fact, the
measurement strategies are similar as above except that
$\gamma_{k}=\lambda_{k}$. They found that the pair $(A_{k},B_{k})$ will able
to witness the entanglement if and only if ${\rm
Tr}[\rho_{A_{k}B_{k}}W_{k,k}]<0$, i.e.,

$\lambda_{k}^{2}>\displaystyle\frac{1-\prod_{l=1}^{k-1}\displaystyle\frac{\left(1+2\Lambda_{l}\right)^{2}}{9}}{2\prod_{l=1}^{k-1}\displaystyle\frac{\left(1+\Lambda_{l}\right)^{2}}{9}}.$
(7)

Therefore, they defined the sequence, $\lambda^{2}_{k}$, for $k\in\mathbb{N}$
as follows,

$\lambda^{2}_{k}=\left\\{\begin{array}[]{cc}(1+\epsilon)\displaystyle\frac{1-\displaystyle\prod_{l=1}^{k-1}\frac{\left(1+2\Lambda_{l}\right)^{2}}{9}}{2\displaystyle\prod_{l=1}^{k-1}\displaystyle\frac{\left(1+\Lambda_{l}\right)^{2}}{9}}&\mbox{if
}\lambda^{2}_{k-1}\in(0,1)\\\ \\\ {\rm\infty},&{\rm
otherwise},\end{array}\right.$ (8)

with $0<\lambda^{2}_{1}<1$ and where $\epsilon>0$.

Choosing $\lambda_{1}=0.005$ and $\epsilon=4$, the parameters
$(\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},\lambda_{5})$ defined by Eq.
(8) are $(0.0050,0.0097,0.0317,0.1459,0.9844).$ Define the difference matrix
$D=(d_{kl})$ whose $(k,l)$ entry is

$d_{kl}:=\lambda_{k}\lambda_{l}-\displaystyle\frac{1-\displaystyle\prod_{i=1}^{k-1}\frac{\left(1+2\Lambda_{i}\right)}{3}\displaystyle\prod_{j=1}^{l-1}\frac{\left(1+2\Lambda_{j}\right)}{3}}{2\displaystyle\prod_{i=1}^{k-1}\frac{\left(1+\Lambda_{i}\right)}{3}\displaystyle\prod_{j=1}^{l-1}\frac{\left(1+\Lambda_{j}\right)}{3}}.$

One finds that the pair $(A_{k},B_{l})$ will able to witness the entanglement
if and only if ${\rm Tr}[\rho_{A_{k}B_{l}}W_{k,l}]<0$, i.e., $d_{kl}>0.$ A
numerical calculation shows that

$100D\approx\left[\begin{array}[]{rrrrr}0.0025&0.0042&0.0114&0.0099&-1.4184\\\\[5.69054pt]
0.0042&0.0075&0.0226&0.0446&-1.9159\\\\[5.69054pt]
0.0114&0.0226&0.0802&0.3049&-1.2054\\\\[5.69054pt]
0.0099&0.0446&0.3049&1.7031&7.5946\\\\[5.69054pt]
-1.4184&-1.9159&-1.2054&7.5946&77.5252\end{array}\right]$

which indicates that although $A_{5}B_{5}$ can detect the entanglement through
$W_{55}$ but $A_{k}B_{l}$ could not through $W_{kl}$ for
$(k,l)\in\\{(1,5),(2,5),(3,5),(5,1),(5,2),(5,3)\\}$ (see Fig. 2 for an
intuition). Therefore, it is interesting to change the measurement strategies
such that all the corresponding pairs could detect the entanglement by
themselves without communication with other members.

Figure 2: Sequentially witnessing entanglement by multiple independent Alices
and Bobs in Example. The pair $(A_{k},B_{l})$ with blue line indicates the
corresponding pair could witness the entanglement while the pair
$(A_{k},B_{l})$ with red line indicates the corresponding pair could not
witness the entanglement.

##  III Sequential entanglement detection between arbitrarily many Alices and
Bobs

In this section, we aim at constructing some measurement strategies such that
all the pairs $(A_{k},B_{l})$ ($1\leq k\leq m,1\leq l\leq n$) could detect the
entanglement regardless of the initial state being a Bell state or a
particular class of mixed entangled states.

###  III.1 Maximally entangled state as initial state

We assume that the initial state is $|\Psi\rangle$, i.e., it’s density matrix
is $\rho_{A_{1}B_{1}}=|\Psi\rangle\langle\Psi|$. As
$\\{\mathbb{I}_{2}\otimes\mathbb{I}_{2},\sigma_{1}\otimes\sigma_{1},\sigma_{2}\otimes\sigma_{2},\sigma_{3}\otimes\sigma_{3}\\}$
form a basis of $4\times 4$ matrices, it can be easily checked that

$\rho_{A_{1}B_{1}}=\frac{1}{4}\left(\mathbb{I}_{2}\otimes\mathbb{I}_{2}+\sigma_{1}\otimes\sigma_{1}+\sigma_{2}\otimes\sigma_{2}-\sigma_{3}\otimes\sigma_{3}\right).$
(9)

As the detector is

$W_{k,l}=\mathbb{I}_{4}+\sigma_{3}\otimes\sigma_{3}-\lambda_{k}\sigma_{1}\otimes\gamma_{l}\sigma_{1}-\lambda_{k}\sigma_{2}\otimes\gamma_{l}\sigma_{2},$
(10)

then by Eq. (II), we have $\langle W_{k,l}\rangle_{\rho_{A_{k}B_{l}}}={\rm
Tr}[W_{k,l}\rho_{A_{k}B_{l}}]$ which is equal to

$\begin{array}[]{l}\displaystyle
1-\displaystyle\prod_{i=1}^{k-1}\frac{\left(1+2\Lambda_{i}\right)}{3}\displaystyle\prod_{j=1}^{l-1}\frac{\left(1+2\Gamma_{j}\right)}{3}-2\lambda_{k}\gamma_{l}\displaystyle\prod_{i=1}^{k-1}\frac{\left(1+\Lambda_{i}\right)}{3}\displaystyle\prod_{j=1}^{l-1}\frac{\left(1+\Gamma_{j}\right)}{3}.\end{array}$

Therefore, the first pair will observe entanglement if
$-\frac{1}{2}\lambda_{1}\gamma_{1}={\rm Tr}[\rho_{A_{1}B_{1}}W_{1,1}]<0$,
which implies

$\lambda_{1}\gamma_{1}>0.$ (11)

And the pair $A_{k}B_{l}$ will able to witness the entanglement if ${\rm
Tr}[\rho_{A_{k}B_{l}}W_{k,l}]<0$, which implies

$\lambda_{k}\gamma_{l}>\displaystyle\frac{1-\displaystyle\prod_{i=1}^{k-1}\frac{\left(1+2\Lambda_{i}\right)}{3}\displaystyle\prod_{j=1}^{l-1}\frac{\left(1+2\Gamma_{j}\right)}{3}}{2\displaystyle\prod_{i=1}^{k-1}\frac{\left(1+\Lambda_{i}\right)}{3}\displaystyle\prod_{j=1}^{l-1}\frac{\left(1+\Gamma_{j}\right)}{3}}.$
(12)

Let us define $\lambda_{k}(\theta)$, and $\gamma_{l}(\theta)$ for
$k,l\in\mathbb{N}$ as functions of a parameter $\theta$ on the interval
$(0,1)$. Let $\lambda_{1}(\theta)=\gamma_{1}(\theta)=\theta$ and for $k\geq 2$

$\lambda_{k}(\theta)=\left\\{\begin{array}[]{cc}\epsilon\frac{1-\displaystyle\prod_{i=1}^{k-1}\frac{\left(1+2\Lambda_{i}(\theta)\right)}{3}}{L\gamma_{1}(\theta)\displaystyle\prod_{i=1}^{k-1}\frac{\left(1+\Lambda_{i}(\theta)\right)}{3}}&\mbox{if
}\lambda_{k-1}(\theta)\in(0,1)\\\ \\\ {\rm\infty},&{\rm
otherwise},\end{array}\right.$ (13)

and

$\gamma_{l}(\theta)=\left\\{\begin{array}[]{cc}\epsilon\frac{1-\displaystyle\prod_{j=1}^{l-1}\frac{\left(1+2\Gamma_{j}(\theta)\right)}{3}}{L\lambda_{1}(\theta)\displaystyle\prod_{j=1}^{l-1}\frac{\left(1+\Gamma_{j}(\theta)\right)}{3}}&\mbox{if
}\gamma_{l-1}(\theta)\in(0,1)\\\ \\\ {\rm\infty},&{\rm
otherwise},\end{array}\right.$ (14)

where $\epsilon>2$ and $L=2$ in this setting.

######  Lemma 1.

Let $L\leq 2<\epsilon$ and let
$\big{(}\lambda_{k}(\theta)\big{)}_{k\in\mathbb{N}}$
and$\big{(}\gamma_{l}(\theta)\big{)}_{l\in\mathbb{N}}$ be the sequences
defined by Eqs. (13) and (14) respectively with
$\lambda_{1}(\theta)=\gamma_{1}(\theta)=\theta.$ Then the sequences
$\big{(}\lambda_{k}(\theta)\big{)}_{k\in\mathbb{N}}$
and$\big{(}\gamma_{l}(\theta)\big{)}_{l\in\mathbb{N}}$ are the same sequence
and they are increasing. Moreover, the limit $\lim_{\theta\rightarrow
0+}(\lambda_{k}(\theta)/\theta)$ exists for any $k\in\mathbb{N}.$ Suppose that
$a_{k}:=\lim_{\theta\rightarrow 0+}(\lambda_{k}(\theta)/\theta)$, then we have
the following recursive relation

$a_{k}=\frac{\epsilon}{3L}(\frac{3}{2})^{k-1}(\sum_{i=1}^{k-1}a_{i}^{2}).$

######  Theorem 1.

Let $L=2$ and $\epsilon\geq 4.$ For any $m,n\in\mathbb{N},$ there exists a
$\theta_{m,n}\in(0,1)$ such that the sequences
$\big{(}\lambda_{k}:=\lambda_{k}(\theta_{m,n})\big{)}_{k=1}^{m}$ and
$\big{(}\gamma_{l}:=\gamma_{l}(\theta_{m,n})\big{)}_{l=1}^{n}$ satisfies
inequalities (12) for each pair $(k,l)$ ($1\leq k\leq m,$ $1\leq l\leq n$),
that is, the detector $W_{k,l}$ defined by Eq. (10) can witness the
entanglement of $\rho_{A_{k}B_{l}}$ where $\rho_{A_{k}B_{l}}$ is defined by
the recursive relation (II) with $\rho_{A_{1}B_{1}}$ given by Eq. (9).

Proof. By Lemma 1, for each $k\in\mathbb{N},$
$\lambda_{k}(\theta)=\gamma_{k}(\theta)$ and the limit
$\lim_{\theta\rightarrow 0+}(\lambda_{k}(\theta)/\theta)$ exists. Suppose
$a_{k}:=\lim_{\theta\rightarrow 0+}(\lambda_{k}(\theta)/\theta)$ we also have

$a_{k}=\frac{\epsilon}{3L}(\frac{3}{2})^{k-1}(\sum_{i=1}^{k-1}a_{i}^{2})$

with $a_{1}=1.$ As $\epsilon\geq 4$, we can deduce that

$a_{k}\geq\sum_{i=1}^{k-1}a_{i}^{2}\geq a_{k-1}^{2}\geq a_{k-1}\geq 1.$

Set

$f_{kl}(\theta):=\displaystyle\frac{1-\displaystyle\prod_{i=1}^{k-1}\frac{\left(1+2\Lambda_{i}(\theta)\right)}{3}\displaystyle\prod_{j=1}^{l-1}\frac{\left(1+2\Gamma_{j}(\theta)\right)}{3}}{L\displaystyle\prod_{i=1}^{k-1}\frac{\left(1+\Lambda_{i}(\theta)\right)}{3}\displaystyle\prod_{j=1}^{l-1}\frac{\left(1+\Gamma_{j}(\theta)\right)}{3}}.$

With similar argument as the proof of Eq. (A2) in Appendix A, one finds

$\lim_{\theta\rightarrow
0+}\frac{f_{kl}(\theta)}{\theta^{2}}=\frac{1}{3L}(\frac{3}{2})^{k+l-2}(\sum_{i=1}^{k-1}a_{i}^{2}+\sum_{j=1}^{l-1}a_{j}^{2}).$

Therefore, for $2\leq k\leq m,2\leq l\leq n,$ the limit
$L_{kl}:=\lim_{\theta\rightarrow
0+}\frac{\lambda_{k}(\theta)\gamma_{l}(\theta)-f_{kl}(\theta)}{\theta^{2}}$ is
equal to

$\frac{1}{3L}(\frac{3}{2})^{k+l-2}\left[\frac{\epsilon^{2}}{3L}(\sum_{i=1}^{k-1}a_{i}^{2})(\sum_{j=1}^{l-1}a_{j}^{2})-(\sum_{i=1}^{k-1}a_{i}^{2}+\sum_{j=1}^{l-1}a_{j}^{2})\right]$

which is positive provided the condition $\frac{\epsilon^{2}}{3L}>2.$ Let
$\mathcal{M}=\max\\{a_{m},a_{n}\\}$ and
$\mathcal{L}=\min\\{L_{k,l}\\}_{k=2,l=2}^{m,n}.$ By the properties of limit,
there exists a small enough $\theta_{0}>0$ (which is assumed to be less than
$\frac{1}{\mathcal{M}+1}$ here) such that for all
$\theta_{m,n}\in(0,\theta_{0})$ are satisfies

$\begin{array}[]{l}\displaystyle\lambda_{i}(\theta_{m,n})=\gamma_{i}(\theta_{m,n})<(\mathcal{M}+1)\theta_{m,n}<1,\\\\[5.69054pt]
\displaystyle\lambda_{k}(\theta_{m,n})\gamma_{l}(\theta_{m,n})-f_{kl}(\theta_{m,n})>\frac{\mathcal{L}}{2}{\theta_{m,n}^{2}}>0\end{array}$

for all $1\leq i\leq\max\\{m,n\\},2\leq k\leq m,2\leq l\leq n$ which implies
the validity of inequalities of (12) for these $k,l.$ On the other hand, for
the pairs $(A_{1},B_{l})$ and $(A_{k},B_{1})$, the inequality (12) holds by
the definition of $\lambda_{k}(\theta_{m,n})$ and $\gamma_{l}(\theta_{m,n})$
as $\epsilon\geq 4.$ ∎

######  Corollary 1.

For any positive integers $m,n$, if the first pair of Alices and Bobs in an
$(m,n)$-sequential scenario share a Bell state, then there exist some
measurement strategies such that all pairs $(A_{k},B_{l})$ ($1\leq k\leq
m,1\leq l\leq n$) could witness the entanglement sequentially.

###  III.2 Pure entangled states and a class of mixed entangled states as
initial state

Now we explore whether the aforementioned property holds true for other
initial states. It is well known that any two qubits entangled state could be
locally unitary equivalent to the following form

$|\Psi_{\alpha}\rangle=\sqrt{\alpha}|01\rangle+\sqrt{1-\alpha}|10\rangle,$

for some $\alpha\in(0,1]$. Now we assume that the initial state
$\rho_{A_{1},B_{1}}$ is of the following form

$\rho_{A_{1}B_{1}}=p_{1}|\Psi_{\alpha}\rangle\langle\Psi_{\alpha}|+p_{2}|01\rangle\langle
01|+p_{3}|10\rangle\langle 10|$ (15)

where $p_{1}>0$, $p_{2},p_{3}\geq 0$ and $p_{1}+p_{2}+p_{3}=1$. Note that

$\begin{array}[]{l}\mathrm{Tr}[\sigma_{1}\otimes\sigma_{1}|01\rangle\langle
01|]=\mathrm{Tr}[\sigma_{1}\otimes\sigma_{1}|10\rangle\langle
10|]=0,\\\\[5.69054pt]
\mathrm{Tr}[\sigma_{2}\otimes\sigma_{2}|01\rangle\langle
01|]=\mathrm{Tr}[\sigma_{2}\otimes\sigma_{2}|10\rangle\langle
10|]=0,\\\\[5.69054pt]
\mathrm{Tr}[\sigma_{3}\otimes\sigma_{3}|01\rangle\langle
01|]=\mathrm{Tr}[\sigma_{3}\otimes\sigma_{3}|10\rangle\langle
10|]=-1.\end{array}$

As
$W_{k,l}=\mathbb{I}_{4}+\sigma_{3}\otimes\sigma_{3}-\lambda_{k}\gamma_{l}\sigma_{1}\otimes\sigma_{1}-\lambda_{k}\gamma_{l}\sigma_{2}\otimes\sigma_{2},$
we always have

$\mathrm{Tr}[W_{k,l}|01\rangle\langle
01|]=\mathrm{Tr}[W_{k,l}|01\rangle\langle 01|]=0.$

Therefore,
$\mathrm{Tr}[W_{k,l}\rho_{A_{1},B_{1}}]=p_{1}\mathrm{Tr}[W_{k,l}|\Psi_{\alpha}\rangle\langle\Psi_{\alpha}|].$
Moreover, $|\Psi_{\alpha}\rangle\langle\Psi_{\alpha}|$ has similar
decomposition as of $|\Psi\rangle\langle\Psi|$ (see Eq. (9))

$\begin{split}|\Psi_{\alpha}\rangle\langle\Psi_{\alpha}|=\frac{1}{4}\Big{[}\mathbb{I}_{4}+2\sqrt{\alpha(1-\alpha)}\Big{(}\sigma_{1}\otimes\sigma_{1}+\sigma_{2}\otimes\sigma_{2}-\sigma_{3}\otimes\sigma_{3}\Big{)}\Big{]}.\end{split}$
(16)

Therefore, the first pair will observe entanglement if
$-p_{1}4\sqrt{\alpha(1-\alpha)}\lambda_{1}\gamma_{1}={\rm
Tr}[\rho_{A_{1}B_{1}}W_{1,1}]<0$, which implies $\lambda_{1}\gamma_{1}>0.$ And
the pair $A_{k}B_{l}$ will able to witness the entanglement if ${\rm
Tr}[\rho_{A_{k}B_{l}}W_{k,l}]<0$, which implies

$\lambda_{k}\gamma_{l}>\frac{1-\displaystyle\prod_{i=1}^{k-1}\frac{\left(1+2\Lambda_{i}\right)}{3}\displaystyle\prod_{j=1}^{l-1}\frac{\left(1+2\Gamma_{j}\right)}{3}}{4p_{1}\sqrt{\alpha(1-\alpha)}\displaystyle\prod_{i=1}^{k-1}\frac{\left(1+\Lambda_{i}\right)}{3}\displaystyle\prod_{j=1}^{l-1}\frac{\left(1+\Gamma_{j}\right)}{3}}.$
(17)

Let us define $\lambda_{k}(\theta)$, and $\gamma_{l}(\theta)$ for
$k,l\in\mathbb{N}$ as functions of a parameter $\theta$ on the interval
$(0,1)$. Let $\lambda_{1}(\theta)=\gamma_{1}(\theta)=\theta$ and for $k\geq 2$
we define $\lambda_{k}(\theta)$ and $\gamma_{l}(\theta)$ by using the same
recursive relations as Eqs. (13) and (14) but we assume
$L=4p_{1}\sqrt{\alpha(1-\alpha)}$ here. We also have every possible pairs
$(A_{k},B_{l})$ of arbitrarily many Alices and Bobs can observe the
entanglement by the sequentially witness defined in the above setting.

######  Theorem 2.

Let $L=4p_{1}\sqrt{\alpha(1-\alpha)}$ and $\epsilon\geq 4.$ For any
$m,n\in\mathbb{N},$ there exists a $\theta_{m,n}\in(0,1)$ such that the
sequences $\big{(}\lambda_{k}:=\lambda_{k}(\theta_{m,n})\big{)}_{k=1}^{m}$ and
$\big{(}\gamma_{l}:=\gamma_{l}(\theta_{m,n})\big{)}_{l=1}^{n}$ satisfies
inequalities (12) for each pair $(k,l)$ ($1\leq k\leq m,$ $1\leq l\leq n$),
that is, the detector $W_{k,l}$ defined by Eq. (10) can witness the
entanglement of $\rho_{A_{k}B_{l}}$ where $\rho_{A_{k}B_{l}}$ is defined by
the recursive relation (II) with $\rho_{A_{1}B_{1}}$ given by Eq. (15).

The proof is very similar to that of Theorem 1 once one notice that $L\leq 2$
and $\epsilon^{2}/3L>2.$ So we omit the proof here.

######  Corollary 2.

For any positive integers $m,n$, if the first pair of Alices and Bobs in an
$(m,n)$-sequential scenario share a mix entangled state given by Eq. (15),
then there exist some measurement strategies such that all pairs
$(A_{k},B_{l})$ ($1\leq k\leq m,1\leq l\leq n$) could witness the entanglement
sequentially.

##  IV Conclusion

In this study, we investigated the capability of sequential and independent
observers to witness entanglement. It was shown that an arbitrary number of
pairs of Alices and Bobs $(A_{k},B_{k})$ could witness entanglement
sequentially. We first shown that there are some cases that $(A_{k},B_{l})$
could not witness the entanglement even that $(A_{K},B_{K})$
($K\geq\max\\{k,l\\}$) could witness. Furthermore, we demonstrated the
feasibility of a measurement strategy that enables all possible pairs
$(A_{k},B_{l})$ of arbitrarily many Alices and Bobs to observe entanglement in
both pure entangled states and a specific class of mixed entangled states.
Although it has been conjectured that at most a single pair of Alices and Bobs
can detect Bell-nonlocal correlations in the context of CHSH Bell nonlocality
Cheng21 ; Zhu22 , our work shows that quantum entanglement is comparatively
easier to observe than Bell nonlocality.

It has been discovered that an arbitrary number of triples consisting of
Alices, Bobs, and Charlies, denoted as $(A_{k},B_{k},C_{k})$, can sequentially
witness the entanglement of Greenberger-Horne-Zeilinger (GHZ) states
Srivastava22b . This finding has also been extended to multipartite systems
Srivastava22c . Consequently, an interesting question arises: Is it possible
for all triples $(A_{k},B_{l},C_{m})$, comprising arbitrarily many Alices,
Bobs, and Charlies, to sequentially witness entanglement? Additionally,
exploring similar problems concerning multipartite systems adds further
intrigue to the investigation.

###### Acknowledgements.

This work is supported by National Natural Science Foundation of China with
Grant No. 12371458, the Guangdong Basic and Applied Basic Research Foundation
under Grants No. 2023A1515012074, and the Science and Technology Planning
Project of Guangzhou under Grants No. 2023A04J1296.

##  Appendix A Proof of Lemma 1

If $\lambda_{1}(\theta)=\gamma_{1}(\theta)=\theta,$ then the sequences
$\big{(}\lambda_{k}(\theta)\big{)}_{k\in\mathbb{N}}$
and$\big{(}\gamma_{l}(\theta)\big{)}_{l\in\mathbb{N}}$ satisfy the same
recursive relation with the same initial condition. Therefore the two
sequences must be the same. One finds

$\lambda_{2}(\theta)/\lambda_{1}(\theta)\geq\epsilon\frac{1-\frac{1+2(1-\frac{\theta^{2}}{2})}{3}}{2\theta^{2}\frac{1+\sqrt{1-\theta^{2}}}{3}}=\frac{\epsilon}{2(1+\sqrt{1-\theta^{2}})}>\frac{\epsilon}{4}\geq
1.$

Now for $k\geq 2$ and $\lambda_{k}(\theta)\in(0,1)$,

$\frac{\lambda_{k+1}(\theta)}{\lambda_{k}(\theta)}=\frac{3}{(1+\Lambda_{k}(\theta))}\frac{1-\displaystyle\prod_{i=1}^{k}\frac{\left(1+2\Lambda_{i}(\theta)\right)}{3}}{1-\displaystyle\prod_{i=1}^{k-1}\frac{\left(1+2\Lambda_{i}(\theta)\right)}{3}}>\frac{3}{(1+\Lambda_{k}(\theta))}>1,$
(A1)

since $\Lambda_{k}(\theta)=\sqrt{1-\lambda^{2}_{k}(\theta)}\in(0,1)$. So the
sequence $\big{(}\lambda_{k}(\theta)\big{)}_{k\in\mathbb{N}}$ is increasing.
The next important observation about this sequence is that
$\lim_{\theta\rightarrow 0+}\lambda_{k}(\theta)/\theta$ exists for all
$k\in\mathbb{N}$. Now we prove the statement by induction on $k$. For $k=1$,
this statement holds as $\lambda_{1}(\theta)/\theta\equiv 1$ by assumption.
Now we suppose that $\lim_{\theta\rightarrow
0+}\frac{\lambda_{k}(\theta)}{\theta}=a_{k}\geq 1$ for $1\leq k\leq n.$ In the
following, we will show

$\lim_{\theta\rightarrow
0+}\frac{1-\displaystyle\prod_{k=1}^{n}\frac{\left(1+2\Lambda_{k}(\theta)\right)}{3}}{\theta^{2}}=\frac{1}{3}\sum_{k=1}^{n}a_{k}^{2}.$

With this, we could show

$\lim_{\theta\rightarrow
0+}\frac{\lambda_{n+1}(\theta)}{\theta}=\lim_{\theta\rightarrow
0+}\epsilon\frac{1-\displaystyle\prod_{k=1}^{n}\frac{\left(1+2\Lambda_{k}(\theta)\right)}{3}}{L\theta^{2}\displaystyle\prod_{k=1}^{n}\frac{\left(1+\Lambda_{k}(\theta)\right)}{3}}=\frac{\epsilon}{L}(\frac{3}{2})^{n}\lim_{\theta\rightarrow
0+}\frac{1-\displaystyle\prod_{k=1}^{n}\frac{\left(1+2\Lambda_{k}(\theta)\right)}{3}}{\theta^{2}}=\frac{\epsilon}{3L}(\frac{3}{2})^{n}\sum_{k=1}^{n}a_{k}^{2}.$
(A2)

Note that $\lim_{\theta\rightarrow
0+}\Lambda_{k}(\theta)=\lim_{\theta\rightarrow
0+}\sqrt{1-\lambda_{k}(\theta)^{2}}=1.$ Moreover, by the Talor expansion
$\sqrt{1-x}=1-\frac{x}{2}+o(x)$, we have

$\lim_{\theta\rightarrow
0+}\frac{\Lambda_{k}(\theta)-1}{\theta^{2}}=\lim_{\theta\rightarrow
0+}\frac{1-\frac{\lambda_{k}(\theta)^{2}}{2}+o(\lambda_{k}(\theta)^{2})-1}{\theta^{2}}=-\frac{a^{2}_{k}}{2}.$

Define $F_{k}(\theta)=\frac{2}{3}(\Lambda_{k}(\theta)-1).$ Then
$\frac{1+2\Lambda_{k}(\theta)}{3}=1+F_{k}(\theta).$ So

$\displaystyle\prod_{k=1}^{n}\frac{\left(1+2\Lambda_{k}(\theta)\right)}{3}=\displaystyle\prod_{k=1}^{n}(1+F_{k}(\theta))=1+\sum_{k=1}^{n}F_{k}(\theta)+\sum_{1\leq
k<l\leq n}F_{k}(\theta)F_{l}(\theta)+\cdots+F_{1}(\theta)\cdots
F_{n}(\theta).$

If we denote $H_{n}(\theta)=\sum_{1\leq k<l\leq
n}F_{k}(\theta)F_{l}(\theta)+\cdots+F_{1}(\theta)\cdots F_{n}(\theta)$,
clearly, $\lim_{\theta\rightarrow 0+}\frac{H_{n}(\theta)}{\theta^{2}}=0.$
Therefore, we have

$\lim_{\theta\rightarrow
0+}\frac{1-\displaystyle\prod_{k=1}^{n}\frac{\left(1+2\Lambda_{k}(\theta)\right)}{3}}{\theta^{2}}=\lim_{\theta\rightarrow
0+}\frac{1-\displaystyle\prod_{k=1}^{n}(1+F_{k}(\theta))}{\theta^{2}}=-\lim_{\theta\rightarrow
0+}\left(\sum_{k=1}^{n}\frac{F_{k}(\theta)}{\theta^{2}}+\frac{H_{n}(\theta)}{\theta^{2}}\right)=\frac{1}{3}\sum_{k=1}^{n}a_{k}^{2}.$

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