# Quantum Bell nonlocality cannot be shared under a special kind of bilateral
measurements for high-dimensional quantum states

Tinggui Zhang1,2,†, Qiming Luo1 and Xiaofen Huang1,2 ${1}$ School of
Mathematics and Statistics, Hainan Normal University, Haikou, 571158, China
${2}$ Key Laboratory of Data Science and Smart Education, Ministry of
Education, Hainan Normal University, Haikou, 571158, China
† Correspondence to<EMAIL_ADDRESS>

###### Abstract

Quantum Bell nonlocality is an important quantum phenomenon. Recently, the
shareability of Bell nonlocality under unilateral measurements has been widely
studied. In this study, we consider the shareability of quantum Bell
nonlocality under bilateral measurements. Under a specific class of projection
operators, we find that quantum Bell nonlocality cannot be shared for a
limited number of times, as in the case of unilateral measurements. Our proof
is analytical and our measurement strategies can be generalized to higher
dimension cases.

Keywords: Quantum Bell nonlocality; Shareability; Unilateral measurements;
Bilateral measurements

###### pacs:

03.67.-a, 02.20.Hj, 03.65.-w

##  I Introduction

As the source of paradoxes such as the Einstein, Podolsky, Rosen paradox qwer
quantum nonlocal correlation was a controversial phenomenon in quantum
mechanics. Nowadays it has become a key resource in the blooming areas of
quantum information and computing qwet ; qwey ; qweu ; qwei ; qweo . Realizing
quantum violations of the Bell-CHSH inequality chsh  in various quantum
systems has acquired great interest as evidenced by a wide range of
studiesqwep ; qwea ; qwes ; qwed ; qwef ; qweg ; qweh ; qwej ; qwek .
According to quantum physics, measurement outcomes cannot be predicted with
certainty in general hjki . Quantum nonlocality implies that the correlations
between the probabilities of measurement outcomes from two distant systems
cannot be described by classical probability correlation models. Such nonlocal
correlations in multipartite systems have been identified as useful resources
in device-independent quantum information processing aaaa , such as key
distribution cccc ; dddd , randomness expansion eeee ; ffff ; gggg  and
randomness amplification hhhh .

Recently, the shareability of quantum Bell nonlocality has been extensively
studiedmnbc ; mnbx ; ddag ; ctdh ; mnbv ; ztfs ; scll . By constructing an
explicit measurement strategy, the authors in mnbv  show that, contrary to
previous expectations mnbc ; mnbx , there is no limit on the number of
independent Bobs that can have an expected violation of the CHSH inequality
with only one Alice. A class of initial two-qubit states, including all pure
two-qubit entangled states, that are capable of achieving an unlimited number
of CHSH inequality violations has been presented. This fact has recently been
illustrated for higher dimensional bipartite pure states ztfs . Furthermore,
in mnbv , the open question of whether quantum nonlocality can be shared under
bilateral measurements was been raised.

In this study, we focus on quantum Bell nonlocality shareability under
bilateral measurements. We consider the following scenario: a nonlocal
correlated bipartite state $\rho_{AB}$ is initially shared by the first Alice
and first Bob. The first Bob performs a randomly selected measurement, records
the measurement outcome, and passes the post-measurement qubit to the second
Bob. Then, the first Alice performs a randomly selected measurement, records
the measurement outcome, and passes the post-measurement qubit to the second
Alice. The problem of interest is whether the quantum state between the second
Alice and Bob is still nonlocal. In fact, there have been some numerical and
experimental studies on this topic. In Ref.scll , the authors used 17
parameters to verify numerically that two-qubit quantum states do not share
quantum nonlocality. Moreover, in Ref. cslb , they have studied the sequential
generation of Bell nonlocality between independent observers via recycling the
components of entangled systems. They obtained the stronger one-sided monogamy
relations than scll . In Ref.jzmh , using entangled photon pairs, the authors
experimentally verified the case of two Alices and two Bobs where Alice(1) and
Bob(1) performed optimal weak measurements and Alice(2) and Bob(2) performed
projective measurements. To adopt the same measurement strength for Alice(1)
and Bob(1), they observed double EPR steering simultaneously and showed that
double Bell-CHSH inequality violations cannot be obtained. But for high-
dimensional quantum states, the method used in scll  is not efficient as too
many parameters are involved. Here, we find that Bell nonlocality cannot be
shared under bilateral measurements for a specific class of projection
measurement operators.

##  II Bipartite state under bilateral measurement

We considered a measurement scenario where the second Alice (Alice(2))
attempts to share nonlocal correlations of an entangled pure state with the
second Bob (Bob(2))First, Alice(1) and Bob(1) share an arbitrary entangled
bipartite pure state $|\psi\rangle\in H_{A}\otimes H_{B}$, where
$dim(H_{A})=s$ and $dim(H_{B})=t$ ($s\leq t$). The state has the Schmidt
decomposition form,
$|\psi\rangle=\sum_{i=1}^{s}c_{i}|i_{A}\rangle\otimes|i_{B}\rangle$, where
$c_{i}\in[0,1]$ and $\sum_{i}^{s}c_{i}^{2}=1$ and $\\{i_{A}\\}_{1}^{s}$ and
$\\{i_{B}\\}_{1}^{t}$ are the orthonormal bases of $H_{A}$ and $H_{B}$,
respectively. $|\psi\rangle$ is entangled if and only if at least two $c_{i}$s
are nonzero. Without loss of generality, below we assume that the $c_{i}$ are
arranged in descending order. The density matrix corresponding to
$|\psi\rangle$ is denoted as $\rho_{A^{1}B^{1}}=|\psi\rangle\langle\psi|$.

The binary input and output of Alice(k) (Bob(k)) are denoted by $X^{(k)}$
($Y^{(k)}$) and $A^{(k)}$ ($B^{(k)}$) and $k=1,2$, respectively. Suppose that
Bob(1) performs the measurement according to $Y^{(1)}=y$ with the outcome
$B^{(1)}=b$. Averaged over the inputs and outputs of Bob(1), the state shared
between Alice(1) and Bob(2) is given by the Lüders rule mnbv

$\rho_{A^{1}B^{2}}=\frac{1}{2}\Sigma_{b,y}(I_{s}\otimes\sqrt{B_{b|y}^{(1)}})\rho_{A^{1}B^{1}}(I_{s}\otimes\sqrt{B_{b|y}^{(1)}}),$

where $B^{(1)}_{b|y}$ is the positive operator-valued measure (POVM) effect
corresponding to outcome $b$ of Bob(1)’s measurement for input $y$, and
$I_{s}$ is the $s\times s$ identity matrix. Next Alice(1) similarly performs
the measurement on subsystem A. Then, the state $\rho_{A^{2}B^{2}}$ shared
between Alice(2) and Bob(2) is acquired.

To detect the Bell nonlocality of a state $\rho$ we employ the CHSH inequality
chsh , $I_{CHSH}=|\langle\mathbb{B}\rangle|\leq 2$, where
$\langle\mathbb{B}\rangle=Tr(\mathbb{B}\rho)$, $\mathbb{B}=A_{0}\otimes
B_{0}+A_{0}\otimes B_{1}+A_{1}\otimes B_{0}-A_{1}\otimes B_{1}$, $A_{i}$, and
$B_{i}$ and $i=0,1$ are Hermitian operators with eigenvalues of $\in[-1,1]$.
If for some binary observables $A_{i}^{(k)}$ and $B_{i}^{(k)}$, $i=0,1$,
$I_{CHSH}^{(k)}\equiv Tr(\mathbb{B}\rho_{A^{k}B^{k}})>2$, then the state
$\rho_{A^{k}B^{k}}$ is nonlocally correlated.

###  II.1 Two-qubit pure states

We first assume that the initial bipartite pure quantum state is a two-qubit
state, $|\psi\rangle\in H_{2}\otimes H_{2}$, with Schmidt decomposition
$|\psi\rangle=\sum^{2}_{i=1}c_{i}|i_{A}\rangle|i_{B}\rangle$. Namely,
$\rho_{A^{1}B^{1}}=|\psi\rangle\langle\psi|$. We employ the POVMs with
measurement operators $\\{E,I-E\\}$, where $E$ has the form
$E=\frac{1}{2}(I+\gamma\cdot{\sigma_{r}})$, $r\in R^{3}$ with $\|r\|=1$,
$\sigma_{r}=r_{1}\sigma_{1}+r_{2}\sigma_{2}+r_{3}\sigma_{3}$, $\sigma_{i}$ for
$i=1,2,3$ are the standard Pauli matrices, and $\gamma\in[0,1]$ is the
sharpness of the measurement.

We set the POVM of Alice(1) to

$A_{0|0}=\frac{1}{2}(I+(\cos\theta\sigma_{1}+\sin\theta\sigma_{3})),$ (1)
$A_{0|1}=\frac{1}{2}(I+(\cos\theta\sigma_{1}-\sin\theta\sigma_{3}))$ (2)

for $\theta\in(0,\frac{\pi}{4}]$. We also let the POVM of Bob(1) be given by

$B^{(1)}_{0|0}=\frac{1}{2}(I+\sigma_{1}),$ (3)
$B^{(1)}_{0|1}=\frac{1}{2}(I+\gamma_{1}\sigma_{3}),$ (4)

where $0\leq\gamma_{1}\leq 1$. Further, we defined the expectation operators
$A_{x}=A_{0|x}-A_{1|x}$ and $B_{y}=B_{0|y}-B_{1|y}$ and reached the following
conclusions:

###### Lemma 1

For the quantum state $\rho_{A^{2}B^{2}}$, we have

$\displaystyle Tr[{\rho_{A^{2}B^{2}}}{({\sigma_{1}}{\otimes}{\sigma_{1}})}]$
$\displaystyle=$
$\displaystyle{\frac{1+{\sqrt{1-\gamma_{1}^{2}}}}{2}}{\cos^{2}{(\theta)}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{1}}{\otimes}{\sigma_{1}})}]$

and

$\displaystyle Tr[{\rho_{A^{2}B^{2}}}{({\sigma_{3}}{\otimes}{\sigma_{3}})}]$
$\displaystyle=$
$\displaystyle{\frac{1}{2}}{\sin^{2}{(\theta)}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{3}}{\otimes}{\sigma_{3}})}].$

Proof: First, after the Bob(1)’s measurement we have

$\rho_{A^{1}B^{2}}={\frac{2+{\sqrt{1-\gamma_{1}^{2}}}}{4}}\rho_{A^{1}B^{1}}+{\frac{1}{4}}{(I\otimes\sigma_{1})}\rho_{A^{1}B^{1}}{(I\otimes\sigma_{1})}+{\frac{1-{\sqrt{1-\gamma_{1}^{2}}}}{4}}{(I\otimes\sigma_{3})}\rho_{A^{1}B^{1}}{(I\otimes\sigma_{3})}.$

After Alice(1)’s measurement we get

$\displaystyle\rho_{A^{2}B^{2}}$ $\displaystyle=$
$\displaystyle{\frac{1}{2}}{\sum\limits_{a,x}}{(\sqrt{A_{a|x}}{\otimes}I)}\rho_{A^{1}B^{2}}{(\sqrt{A_{a|x}}{\otimes}I)}$
$\displaystyle=$
$\displaystyle{\frac{1}{2}}\rho_{A^{1}B^{2}}+{\frac{1}{2}}{({\cos{(\theta)}\sigma_{1}}{\otimes}I)}\rho_{A^{1}B^{2}}{({\cos{(\theta)}\sigma_{1}}{\otimes}I)}+{\frac{1}{2}}{({\sin{(\theta)}\sigma_{3}}{\otimes}I)}\rho_{A^{1}B^{2}}{({\sin{(\theta)}\sigma_{3}}{\otimes}I)}$
$\displaystyle=$
$\displaystyle{\frac{2+{\sqrt{1-\gamma_{1}^{2}}}}{8}}\rho_{A^{1}B^{1}}+{\frac{1}{8}}{(I\otimes\sigma_{1})}\rho_{A^{1}B^{1}}{(I\otimes\sigma_{1})}+{\frac{1-{\sqrt{1-\gamma_{1}^{2}}}}{8}}{(I\otimes\sigma_{3})}\rho_{A^{1}B^{1}}{(I\otimes\sigma_{3})}$
$\displaystyle+{\frac{2+{\sqrt{1-\gamma_{1}^{2}}}}{8}}{({\cos{(\theta)}\sigma_{1}}{\otimes}I)}\rho_{A^{1}B^{1}}{({\cos{(\theta)}\sigma_{1}}{\otimes}I)}$
$\displaystyle+{\frac{1}{8}}{({\cos{(\theta)}\sigma_{1}}{\otimes}I)}{(I\otimes\sigma_{1})}\rho_{A^{1}B^{1}}{(I\otimes\sigma_{1})}{({\cos{(\theta)}\sigma_{1}}{\otimes}I)}$
$\displaystyle+{\frac{1-{\sqrt{1-\gamma_{1}^{2}}}}{8}}{({\cos{(\theta)}\sigma_{1}}{\otimes}I)}{(I\otimes\sigma_{3})}\rho_{A^{1}B^{1}}{(I\otimes\sigma_{3})}{({\cos{(\theta)}\sigma_{1}}{\otimes}I)}$
$\displaystyle+{\frac{2+{\sqrt{1-\gamma_{1}^{2}}}}{8}}{({\sin{(\theta)}\sigma_{3}}{\otimes}I)}\rho_{A^{1}B^{1}}{({\sin{(\theta)}\sigma_{3}}{\otimes}I)}$
$\displaystyle+{\frac{1}{8}}{({\sin{(\theta)}\sigma_{3}}{\otimes}I)}{(I\otimes\sigma_{1})}\rho_{A^{1}B^{1}}{(I\otimes\sigma_{1})}{({\sin{(\theta)}\sigma_{3}}{\otimes}I)}$
$\displaystyle+{\frac{1-{\sqrt{1-\gamma_{1}^{2}}}}{8}}{({\sin{(\theta)}\sigma_{3}}{\otimes}I)}{(I\otimes\sigma_{3})}\rho_{A^{1}B^{1}}{(I\otimes\sigma_{3})}{({\sin{(\theta)}\sigma_{3}}{\otimes}I)}.$

Then

$\displaystyle Tr[{\rho_{A^{2}B^{2}}}{({\sigma_{1}}{\otimes}{\sigma_{1}})}]$
$\displaystyle=$
$\displaystyle{\frac{2+{\sqrt{1-\gamma_{1}^{2}}}}{8}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{1}}{\otimes}{\sigma_{1}})}]+{\frac{1}{8}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{1}}{\otimes}{\sigma_{1}})}]-{\frac{1-{\sqrt{1-\gamma_{1}^{2}}}}{8}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{1}}{\otimes}{\sigma_{1}})}]$
$\displaystyle+{\frac{2+{\sqrt{1-\gamma_{1}^{2}}}}{8}}{\cos^{2}{(\theta)}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{1}}{\otimes}{\sigma_{1}})}]+{\frac{1}{8}}{\cos^{2}{(\theta)}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{1}}{\otimes}{\sigma_{1}})}]$
$\displaystyle-{\frac{1-{\sqrt{1-\gamma_{1}^{2}}}}{8}}{\cos^{2}{(\theta)}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{1}}{\otimes}{\sigma_{1}})}]-{\frac{2+{\sqrt{1-\gamma_{1}^{2}}}}{8}}{\sin^{2}{(\theta)}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{1}}{\otimes}{\sigma_{1}})}]$
$\displaystyle-{\frac{1}{8}}{\sin^{2}{(\theta)}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{1}}{\otimes}{\sigma_{1}})}]+{\frac{1-{\sqrt{1-\gamma_{1}^{2}}}}{8}}{\sin^{2}{(\theta)}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{1}}{\otimes}{\sigma_{1}})}]$
$\displaystyle=$
$\displaystyle{\frac{1+{\sqrt{1-\gamma_{1}^{2}}}}{4}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{1}}{\otimes}{\sigma_{1}})}]+{\frac{1+{\sqrt{1-\gamma_{1}^{2}}}}{4}}{\cos^{2}{(\theta)}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{1}}{\otimes}{\sigma_{1}})}]$
$\displaystyle-{\frac{1+{\sqrt{1-\gamma_{1}^{2}}}}{4}}{\sin^{2}{(\theta)}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{1}}{\otimes}{\sigma_{1}})}]$
$\displaystyle=$
$\displaystyle{\frac{1+{\sqrt{1-\gamma_{1}^{2}}}}{2}}{\cos^{2}{(\theta)}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{1}}{\otimes}{\sigma_{1}})}].$

Similarly,

$\displaystyle Tr[{\rho_{A^{2}B^{2}}}{({\sigma_{3}}{\otimes}{\sigma_{3}})}]$
$\displaystyle=$
$\displaystyle{\frac{2+{\sqrt{1-\gamma_{1}^{2}}}}{8}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{3}}{\otimes}{\sigma_{3}})}]-{\frac{1}{8}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{3}}{\otimes}{\sigma_{3}})}]+{\frac{1-{\sqrt{1-\gamma_{1}^{2}}}}{8}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{3}}{\otimes}{\sigma_{3}})}]$
$\displaystyle-{\frac{2+{\sqrt{1-\gamma_{1}^{2}}}}{8}}{\cos^{2}{(\theta)}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{3}}{\otimes}{\sigma_{3}})}]+{\frac{1}{8}}{\cos^{2}{(\theta)}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{3}}{\otimes}{\sigma_{3}})}]$
$\displaystyle-{\frac{1-{\sqrt{1-\gamma_{1}^{2}}}}{8}}{\cos^{2}{(\theta)}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{3}}{\otimes}{\sigma_{3}})}]+{\frac{2+{\sqrt{1-\gamma_{1}^{2}}}}{8}}{\sin^{2}{(\theta)}}Tr[\rho_{AB^{1}}{({\sigma_{3}}{\otimes}{\sigma_{3}})}]$
$\displaystyle-{\frac{1}{8}}{\sin^{2}{(\theta)}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{3}}{\otimes}{\sigma_{3}})}]+{\frac{1-{\sqrt{1-\gamma_{1}^{2}}}}{8}}{\cos^{2}{(\theta)}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{3}}{\otimes}{\sigma_{3}})}]$
$\displaystyle=$
$\displaystyle{\frac{1}{4}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{3}}{\otimes}{\sigma_{3}})}]-{\frac{1}{4}}{\cos^{2}{(\theta)}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{3}}{\otimes}{\sigma_{3}})}]$
$\displaystyle+{\frac{1}{4}}{\sin^{2}{(\theta)}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{3}}{\otimes}{\sigma_{3}})}]$
$\displaystyle=$
$\displaystyle{\frac{1}{2}}{\sin^{2}{(\theta)}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{3}}{\otimes}{\sigma_{3}})}].$

$\Box$

Using the Lemma above, we have the following Theorem:

###### Theorem 1

For any initial entangled bipartite pure quantum state $|\psi\rangle\in
H_{2}\otimes H_{2}$,
$|\psi\rangle=\sum^{2}_{i=1}c_{i}|i_{A}\rangle|i_{B}\rangle$. After the
bilateral measurements, the expected CHSH value of $\rho_{A^{2}B^{2}}$ is less
than or equal to 2, that is,

$\displaystyle I_{CHSH}$ $\displaystyle=$ $\displaystyle
Tr[{\rho_{A^{2}B^{2}}}((A_{0}+A_{1}){\otimes}B_{0})]$
$\displaystyle+Tr[{\rho_{A^{2}B^{2}}}((A_{0}-A_{1}){\otimes}B_{1})\leq 2.$

Proof:

$\displaystyle I_{CHSH}$ $\displaystyle=$ $\displaystyle
Tr[{\rho_{A^{2}B^{2}}}((A_{0}+A_{1}){\otimes}B_{0})]$
$\displaystyle+Tr[{\rho_{A^{2}B^{2}}}((A_{0}-A_{1}){\otimes}B_{1})]$
$\displaystyle=$ $\displaystyle
2{\cos{(\theta)}}Tr[{\rho_{A^{2}B^{2}}}{({\sigma_{1}}{\otimes}{\sigma_{1}})}]$
$\displaystyle+2{\gamma_{1}}{\sin{(\theta)}}Tr[{\rho_{A^{2}B^{2}}}{({\sigma_{3}}{\otimes}{\sigma_{3}})}]$
$\displaystyle=$
$\displaystyle{\cos^{3}{(\theta)}}{(1+{\sqrt{1-\gamma_{1}^{2}}})}Tr[\rho_{A^{1}B^{1}}{({\sigma_{1}}{\otimes}{\sigma_{1}})}]$
$\displaystyle+{\gamma_{1}}{\sin^{3}{(\theta)}}Tr[\rho_{A^{1}B^{1}}{({\sigma_{3}}{\otimes}{\sigma_{3}})}].$

Since $Tr[\rho_{A^{1}B^{1}}{({\sigma_{1}}{\otimes}{\sigma_{1}})}]\leq 1$ and
$Tr[\rho_{A^{1}B^{1}}{({\sigma_{3}}{\otimes}{\sigma_{3}})}]\leq 1$ we have

$\displaystyle I_{CHSH}$ $\displaystyle=$
$\displaystyle{\cos^{3}{(\theta)}}{(1+{\sqrt{1-\gamma_{1}^{2}}})}+{\gamma_{1}}{\sin^{3}{(\theta)}}$
$\displaystyle\leq$ $\displaystyle
2{\cos^{3}{(\theta)}}+{\sin^{3}{(\theta)}}.$

Using

$f(\theta)=2{\cos^{3}{(\theta)}}+{\sin^{3}{(\theta)}},~{}~{}~{}0<\theta\leq{\frac{\pi}{4}},$

we have
$f^{\prime}(\theta)=3{\sin{(\theta)}}{\cos{(\theta)}}{[\sin{(\theta)}-2\cos{(\theta)}]}<0$,
as $\sin{(\theta)}<2\cos{(\theta)}$ for $0<\theta\leq{\frac{\pi}{4}}$. Hence,
$f(\theta)$ is a decreasing function of $\theta$ with $f(\theta)\leq f(0)=2$.
Therefore, $I_{CHSH}\leq 2$. $\Box$

The above Theorem shows that the second Bob shares no quantum nonlocality with
the second Alice.

###  II.2 Generation for Higher dimensional pure states

For general $d\otimes d$ ($d\geq 3$) entangled pure state $\rho$ is given by
$\rho_{A^{1}B^{1}}=|\varphi\rangle\langle\varphi|$, where
$|\varphi\rangle=\sum_{i=1}^{d}c_{i}|ii\rangle$ with
$\sum_{i=1}^{d}c_{i}^{2}=1$. Let

$\displaystyle A_{0|0}$ $\displaystyle=$
$\displaystyle{\frac{1}{2}}[I_{d}+\left(\begin{array}[]{cc}{\cos(\theta)}\sigma_{3}+{\sin(\theta)}\sigma_{1}&0\\\
0&I_{d-2}\\\ \end{array}\right)],$ $\displaystyle A_{0|1}$ $\displaystyle=$
$\displaystyle{\frac{1}{2}}[I_{d}+\left(\begin{array}[]{cc}{\cos(\theta)}\sigma_{3}-{\sin(\theta)}\sigma_{1}&0\\\
0&I_{d-2}\\\ \end{array}\right)],$ $\displaystyle B_{0|0}$ $\displaystyle=$
$\displaystyle{\frac{1}{2}}[I_{d}+\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\sigma_{3}\\\ \end{array}\right)],$ $\displaystyle B_{0|1}$ $\displaystyle=$
$\displaystyle{\frac{1}{2}}[I_{4}+\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\gamma_{1}\sigma_{1}\\\ \end{array}\right)],$

i.e.

$\displaystyle A_{0}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}{\cos(\theta)}\sigma_{3}+{\sin(\theta)}\sigma_{1}&0\\\
0&I_{d-2}\\\ \end{array}\right)$ $\displaystyle A_{1}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}{\cos(\theta)}\sigma_{3}-{\sin(\theta)}\sigma_{1}&0\\\
0&I_{d-2}\\\ \end{array}\right)$ $\displaystyle B_{0}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}I_{d-2}&0\\\ 0&\sigma_{3}\\\
\end{array}\right)$ $\displaystyle B_{1}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}I_{d-2}&0\\\ 0&\gamma_{1}\sigma_{1}\\\
\end{array}\right).$

Suppose Bob and Alice each perform the measurement above and we write it as
$\rho_{A^{2}B^{2}}$. We can easily obtain the following Lemma with its proof
given in the Appendix:.

###### Lemma 2

For the quantum state $\rho_{A^{2}B^{2}}$, we have

$\displaystyle Tr[\rho_{A^{2}B^{2}}((A_{0}+A_{1}){\otimes}B_{0})]$
$\displaystyle=$ $\displaystyle
2\cos^{3}(\theta)c_{1}^{2}-2\cos^{3}(\theta)c_{2}^{2}-(1+{\sqrt{1-\gamma_{1}^{2}}})c_{d}^{2}$
(13) $\displaystyle+$
$\displaystyle(1+{\sqrt{1-\gamma_{1}^{2}}})[c_{3}^{2}+c_{4}^{2}+\cdots+c_{d-1}^{2}]\leq
2.$ $Tr[\rho_{A^{2}B^{2}}((A_{0}-A_{1}){\otimes}B_{1})]=0.$

We can naturally draw the following conclusion.

###### Theorem 2

For any initial entangled bipartite pure quantum state
$\rho_{A^{1}B^{1}}=|\varphi\rangle\langle\varphi|$. After the bilateral
measurements, the expected CHSH value of $\rho_{A^{2}B^{2}}$ satisfies

$\displaystyle
I_{CHSH}^{(2)}=Tr[\rho_{A^{2}B^{2}}((A_{0}+A_{1}){\otimes}B_{0})]$
$\displaystyle+Tr[\rho_{A^{2}B^{2}}((A_{0}-A_{1}){\otimes}B_{1})]\leq 2.$

Remark: In Lemma 3, there are only the first three terms for $d=3$, the last
term will appear only when $d\geq 4$.

##  III conclusion and discussion

In this article, we explored the ability to share the quantum nonlocality of
bipartite quantum states under specific measurements. It has been shown that
in these cases, quantum nonlocality cannot be shared under bilateral
measurements. We have made an attempt in verifying the shareability of quantum
nonlocality for high-dimensional quantum states. But now our analysis is only
true under the kind of quantum measurements we give. We don’t know whether
they are optimal or not. Next, we can discuss the selection of optimal
measurements for bipartite quantum pure states and some mixed states. For
multipartite quantum states, the sharing ability of nonlocality in unilateral
POVM measurement is already very weakztfs ; sdsd , so it should be weaker than
bipartite quantum state in bilateral measurement, and we can continue to study
it. In the latest literature mhsc , by characterising two-valued qubit
observables in terms of strength, bias, and directional parameters, the
authors investigated generalising the Horodecki criterion to nonprojective
qubit observables. Therefore, we may continue to think about a series of
problems such as the sharing of network nonlocality wxcr  or other quantum
resources under nonprojective measurement. Ref.scll  discussed that for the
qubit case CHSH nonlocality can be shared by bilateral measurements when there
is a bias on the measurements made by Alice and Bob. It is an interesting
question whether there are similar results for higher-dimensional quantum
states.

Data Availability Statement: Our manuscript has no associated data.

Acknowledgments: We thank Shao-Ming Fei, Naihuan Jing and Yuan-Hong Tao for
their helpful discussions. This work was supported by Hainan Provincial
Natural Science Foundation of China under Grant No.121RC539 and the National
Natural Science Foundation of China under Grant
Nos.12126314,12126351,11861031. This project is also supported by the specific
research fund of the Innovation Platform for Academicians of Hainan Province
under Grant No.YSPTZX202215.

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## Appendix

Proof of Lemma 2

By straightforward calculation we have

$\rho_{A^{1}B^{2}}={\frac{2+{\sqrt{1-\gamma_{1}^{2}}}}{4}}\rho_{A^{1}B^{1}}+{\frac{1}{4}}{(I\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\sigma_{3}\\\
\end{array}\right))}\rho_{A^{1}B^{1}}{(I\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\sigma_{3}\\\ \end{array}\right))}$
$+{\frac{1-{\sqrt{1-\gamma_{1}^{2}}}}{4}}{(I\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\sigma_{1}\\\
\end{array}\right))}\rho_{A^{1}B^{1}}{(I\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\sigma_{1}\\\ \end{array}\right))}$

and

$\displaystyle\rho_{A^{2}B^{2}}$ $\displaystyle=$
$\displaystyle{\frac{1}{2}}{\sum\limits_{a,x}}{(\sqrt{A_{a|x}}{\otimes}I)}\rho_{A^{1}B^{2}}{(\sqrt{A_{a|x}}{\otimes}I)}$
$\displaystyle=$
$\displaystyle{\frac{1}{8}}([I_{4}+\left(\begin{array}[]{cc}{\cos(\theta)}\sigma_{3}+{\sin(\theta)}\sigma_{1}&0\\\
0&I_{d-2}\\\
\end{array}\right)]\otimes{I})\rho_{A^{1}B^{2}}([I_{4}+\left(\begin{array}[]{cc}{\cos(\theta)}\sigma_{3}+{\sin(\theta)}\sigma_{1}&0\\\
0&I_{d-2}\\\ \end{array}\right)]\otimes{I})$
$\displaystyle+{\frac{1}{8}}([I_{4}-\left(\begin{array}[]{cc}{\cos(\theta)}\sigma_{3}+{\sin(\theta)}\sigma_{1}&0\\\
0&I_{d-2}\\\
\end{array}\right)]\otimes{I})\rho_{A^{1}B^{2}}([I_{4}-\left(\begin{array}[]{cc}{\cos(\theta)}\sigma_{3}+{\sin(\theta)}\sigma_{1}&0\\\
0&I_{d-2}\\\ \end{array}\right)]\otimes{I})$
$\displaystyle+{\frac{1}{8}}([I_{4}+\left(\begin{array}[]{cc}{\cos(\theta)}\sigma_{3}-{\sin(\theta)}\sigma_{1}&0\\\
0&I_{d-2}\\\
\end{array}\right)]\otimes{I})\rho_{A^{1}B^{2}}([I_{4}+\left(\begin{array}[]{cc}{\cos(\theta)}\sigma_{3}-{\sin(\theta)}\sigma_{1}&0\\\
0&I_{d-2}\\\ \end{array}\right)]\otimes{I})$
$\displaystyle+{\frac{1}{8}}([I_{4}-\left(\begin{array}[]{cc}{\cos(\theta)}\sigma_{3}-{\sin(\theta)}\sigma_{1}&0\\\
0&I_{d-2}\\\
\end{array}\right)]\otimes{I})\rho_{A^{1}B^{2}}([I_{4}-\left(\begin{array}[]{cc}{\cos(\theta)}\sigma_{3}-{\sin(\theta)}\sigma_{1}&0\\\
0&I_{d-2}\\\ \end{array}\right)]\otimes{I})$ $\displaystyle=$
$\displaystyle{\frac{1}{2}}\rho_{A^{1}B^{2}}+{\frac{1}{4}}(\left(\begin{array}[]{cc}{\cos(\theta)}\sigma_{3}+{\sin(\theta)}\sigma_{1}&0\\\
0&I_{d-2}\\\
\end{array}\right)\otimes{I})\rho_{A^{1}B^{2}}(\left(\begin{array}[]{cc}{\cos(\theta)}\sigma_{3}+{\sin(\theta)}\sigma_{1}&0\\\
0&I_{d-2}\\\ \end{array}\right)\otimes{I})$
$\displaystyle+{\frac{1}{4}}(\left(\begin{array}[]{cc}{\cos(\theta)}\sigma_{3}-{\sin(\theta)}\sigma_{1}&0\\\
0&I_{d-2}\\\
\end{array}\right)\otimes{I})\rho_{A^{1}B^{2}}(\left(\begin{array}[]{cc}{\cos(\theta)}\sigma_{3}-{\sin(\theta)}\sigma_{1}&0\\\
0&I_{d-2}\\\ \end{array}\right)\otimes{I}).$

Set $P={\cos(\theta)}\sigma_{3}+{\sin(\theta)}\sigma_{1}$ and
$Q={\cos(\theta)}\sigma_{3}-{\sin(\theta)}\sigma_{1}$. $\rho_{A^{2}B^{2}}$ can
be expressed as

$\displaystyle\rho_{A^{2}B^{2}}$ $\displaystyle=$
$\displaystyle{\frac{2+{\sqrt{1-\gamma_{1}^{2}}}}{8}}\rho_{A^{1}B^{1}}$
$\displaystyle+{\frac{1}{8}}{(I\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\sigma_{3}\\\
\end{array}\right))}\rho_{A^{1}B^{1}}{(I\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\sigma_{3}\\\ \end{array}\right))}$
$\displaystyle+{\frac{1-{\sqrt{1-\gamma_{1}^{2}}}}{8}}{(I\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\sigma_{1}\\\
\end{array}\right))}\rho_{A^{1}B^{1}}{(I\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\sigma_{1}\\\ \end{array}\right))}$
$\displaystyle+{\frac{1}{16}}(\left(\begin{array}[]{cc}P&0\\\ 0&I_{d-2}\\\
\end{array}\right)\otimes{\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\sigma_{3}\\\
\end{array}\right)})\rho_{A^{1}B^{1}}(\left(\begin{array}[]{cc}P&0\\\
0&I_{d-2}\\\ \end{array}\right)\otimes{\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\sigma_{3}\\\ \end{array}\right)})$
$\displaystyle+{\frac{2+{\sqrt{1-\gamma_{1}^{2}}}}{16}}(\left(\begin{array}[]{cc}P&0\\\
0&I_{d-2}\\\
\end{array}\right)\otimes{I})\rho_{A^{1}B^{1}}(\left(\begin{array}[]{cc}P&0\\\
0&I_{d-2}\\\ \end{array}\right)\otimes{I})$
$\displaystyle+{\frac{1-{\sqrt{1-\gamma_{1}^{2}}}}{16}}(\left(\begin{array}[]{cc}P&0\\\
0&I_{d-2}\\\ \end{array}\right)\otimes{{\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\sigma_{1}\\\
\end{array}\right)}})\rho_{A^{1}B^{1}}(\left(\begin{array}[]{cc}P&0\\\
0&I_{d-2}\\\ \end{array}\right)\otimes{{\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\sigma_{1}\\\ \end{array}\right)}})$
$\displaystyle+{\frac{1}{16}}(\left(\begin{array}[]{cc}Q&0\\\ 0&I_{d-2}\\\
\end{array}\right)\otimes{\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\sigma_{3}\\\
\end{array}\right)})\rho_{A^{1}B^{1}}(\left(\begin{array}[]{cc}Q&0\\\
0&I_{d-2}\\\ \end{array}\right)\otimes{\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\sigma_{3}\\\ \end{array}\right)})$
$\displaystyle+{\frac{2+{\sqrt{1-\gamma_{1}^{2}}}}{16}}(\left(\begin{array}[]{cc}Q&0\\\
0&I_{d-2}\\\
\end{array}\right)\otimes{I})\rho_{A^{1}B^{1}}(\left(\begin{array}[]{cc}Q&0\\\
0&I_{d-2}\\\ \end{array}\right)\otimes{I})$
$\displaystyle+{\frac{1-{\sqrt{1-\gamma_{1}^{2}}}}{16}}(\left(\begin{array}[]{cc}Q&0\\\
0&I_{d-2}\\\ \end{array}\right)\otimes{{\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\sigma_{1}\\\
\end{array}\right)}})\rho_{A^{1}B^{1}}(\left(\begin{array}[]{cc}Q&0\\\
0&I_{d-2}\\\ \end{array}\right)\otimes{{\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\sigma_{1}\\\ \end{array}\right)}}),$

where $\rho_{A^{1}B^{1}}=|\varphi\rangle\langle\varphi|$.

Therefore,

$\displaystyle Tr[\rho_{A^{2}B^{2}}((A_{0}+A_{1}){\otimes}B_{0}^{2})]$
$\displaystyle=$ $\displaystyle
2Tr[\rho_{A^{2}B^{2}}(\left(\begin{array}[]{cc}\cos(\theta)\sigma_{3}&0\\\
0&I_{d-2}\\\ \end{array}\right)\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\sigma_{3}\\\ \end{array}\right))]$ $\displaystyle=$ $\displaystyle
2Tr[{\frac{2+{\sqrt{1-\gamma_{1}^{2}}}}{8}}\rho_{A^{1}B^{1}}(\left(\begin{array}[]{cc}\cos(\theta)\sigma_{3}&0\\\
0&I_{d-2}\\\ \end{array}\right)\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\sigma_{3}\\\ \end{array}\right))$ $\displaystyle+$
$\displaystyle{\frac{1}{8}}\rho_{A^{1}B^{1}}(\left(\begin{array}[]{cc}\cos(\theta)\sigma_{3}&0\\\
0&I_{d-2}\\\ \end{array}\right)\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\sigma_{3}\\\ \end{array}\right))$
$\displaystyle+{\frac{1-{\sqrt{1-\gamma_{1}^{2}}}}{8}}\rho_{AB^{1}}(\left(\begin{array}[]{cc}\cos(\theta)\sigma_{3}&0\\\
0&I_{d-2}\\\ \end{array}\right)\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&-\sigma_{3}\\\ \end{array}\right))$
$\displaystyle+{\frac{1}{16}}\rho_{A^{1}B^{1}}(\left(\begin{array}[]{cc}P\cos(\theta)\sigma_{3}P&0\\\
0&I_{d-2}\\\ \end{array}\right)\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\sigma_{3}\\\ \end{array}\right))$
$\displaystyle+{\frac{2+{\sqrt{1-\gamma_{1}^{2}}}}{16}}\rho_{A^{1}B^{1}}(\left(\begin{array}[]{cc}P\cos(\theta)\sigma_{3}P&0\\\
0&I_{d-2}\\\ \end{array}\right)\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\sigma_{3}\\\ \end{array}\right))$
$\displaystyle+{\frac{1-{\sqrt{1-\gamma_{1}^{2}}}}{16}}\rho_{A^{1}B^{1}}(\left(\begin{array}[]{cc}P\cos(\theta)\sigma_{3}P&0\\\
0&I_{d-2}\\\ \end{array}\right)\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&-\sigma_{3}\\\ \end{array}\right))$
$\displaystyle+{\frac{1}{16}}\rho_{A^{1}B^{1}}(\left(\begin{array}[]{cc}Q\cos(\theta)\sigma_{3}Q&0\\\
0&I_{d-2}\\\ \end{array}\right)\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\sigma_{3}\\\ \end{array}\right))$
$\displaystyle+{\frac{2+{\sqrt{1-\gamma_{1}^{2}}}}{16}}\rho_{A^{1}B^{1}}(\left(\begin{array}[]{cc}Q\cos(\theta)\sigma_{3}Q&0\\\
0&I_{d-2}\\\ \end{array}\right)\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\sigma_{3}\\\ \end{array}\right))$
$\displaystyle+{\frac{1-{\sqrt{1-\gamma_{1}^{2}}}}{16}}\rho_{A^{1}B^{1}}(\left(\begin{array}[]{cc}Q\cos(\theta)\sigma_{3}Q&0\\\
0&I_{d-2}\\\ \end{array}\right)\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&-\sigma_{3}\\\ \end{array}\right))]$ $\displaystyle=$ $\displaystyle
Tr[{\frac{3+{\sqrt{1-\gamma_{1}^{2}}}}{2}}\rho_{A^{1}B^{1}}(\left(\begin{array}[]{cc}\cos^{3}(\theta)\sigma_{3}&0\\\
0&I_{d-2}\\\ \end{array}\right)\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\sigma_{3}\\\ \end{array}\right))$
$\displaystyle+{\frac{1-{\sqrt{1-\gamma_{1}^{2}}}}{2}}\rho_{A^{1}B^{1}}(\left(\begin{array}[]{cc}\cos^{3}(\theta)\sigma_{3}&0\\\
0&I_{d-2}\\\ \end{array}\right)\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&-\sigma_{3}\\\ \end{array}\right))]$
$\displaystyle=2\cos^{3}(\theta)c_{1}^{2}-2\cos^{3}(\theta)c_{2}^{2}-(1+{\sqrt{1-\gamma_{1}^{2}}})c_{d}^{2}$
$\displaystyle+(1+{\sqrt{1-\gamma_{1}^{2}}})[c_{3}^{2}+c_{4}^{2}+\cdots+c_{d-1}^{2}].$

Since $2\cos^{3}(\theta)\leq 2$, $-2\cos^{3}(\theta)\leq 2$,
$(1+{\sqrt{1-\gamma_{1}^{2}}})\leq 2$ and $(1+{\sqrt{1-\gamma_{1}^{2}}})\leq
2$, where $0<\gamma_{1}<1,0<\theta\leq{\frac{\pi}{4}}$, and
$\sum_{i=1}^{d}c_{i}^{2}=1$ we have
$Tr[\rho_{A^{2}B^{2}}((A_{0}+A_{1}){\otimes}B_{0})]\leq 2.$

Similarly, we have

$\displaystyle Tr[\rho_{A^{2}B^{2}}((A_{0}-A_{1}){\otimes}B_{1})]$
$\displaystyle=$ $\displaystyle
2Tr[\rho_{A^{2}B^{2}}(\left(\begin{array}[]{cc}\sin(\theta)\sigma_{1}&0\\\
0&0\\\ \end{array}\right)\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\gamma_{1}\sigma_{1}\\\ \end{array}\right))]$ $\displaystyle=$
$\displaystyle
2Tr[{\frac{2+{\sqrt{1-\gamma_{1}^{2}}}}{8}}\rho_{A^{1}B^{1}}(\left(\begin{array}[]{cc}\sin(\theta)\sigma_{1}&0\\\
0&0\\\ \end{array}\right)\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\gamma_{1}\sigma_{1}\\\ \end{array}\right))$
$\displaystyle+{\frac{1}{8}}\rho_{A^{1}B^{1}}(\left(\begin{array}[]{cc}\sin(\theta)\sigma_{1}&0\\\
0&0\\\ \end{array}\right)\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&-\gamma_{1}\sigma_{1}\\\ \end{array}\right))$
$\displaystyle+{\frac{1-{\sqrt{1-\gamma_{1}^{2}}}}{8}}\rho_{A^{1}B^{1}}(\left(\begin{array}[]{cc}\sin(\theta)\sigma_{1}&0\\\
0&0\\\ \end{array}\right)\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\gamma_{1}\sigma_{1}\\\ \end{array}\right))$
$\displaystyle+{\frac{1}{16}}\rho_{A^{1}B^{1}}(\left(\begin{array}[]{cc}P\sin(\theta)\sigma_{1}P&0\\\
0&0\\\ \end{array}\right)\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&-\gamma_{1}\sigma_{1}\\\ \end{array}\right))$
$\displaystyle+{\frac{2+{\sqrt{1-\gamma_{1}^{2}}}}{16}}\rho_{A^{1}B^{1}}(\left(\begin{array}[]{cc}P\sin(\theta)\sigma_{1}P&0\\\
0&0\\\ \end{array}\right)\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\gamma_{1}\sigma_{1}\\\ \end{array}\right))$
$\displaystyle+{\frac{1-{\sqrt{1-\gamma_{1}^{2}}}}{16}}\rho_{A^{1}B^{1}}(\left(\begin{array}[]{cc}P\sin(\theta)\sigma_{1}P&0\\\
0&0\\\ \end{array}\right)\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\gamma_{1}\sigma_{1}\\\ \end{array}\right))$
$\displaystyle+{\frac{1}{16}}\rho_{A^{1}B^{1}}(\left(\begin{array}[]{cc}Q\sin(\theta)\sigma_{1}Q&0\\\
0&0\\\ \end{array}\right)\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&-\gamma_{1}\sigma_{1}\\\ \end{array}\right))$
$\displaystyle+{\frac{2+{\sqrt{1-\gamma_{1}^{2}}}}{16}}\rho_{A^{1}B^{1}}(\left(\begin{array}[]{cc}Q\sin(\theta)\sigma_{1}Q&0\\\
0&0\\\ \end{array}\right)\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\gamma_{1}\sigma_{1}\\\ \end{array}\right))$
$\displaystyle+{\frac{1-{\sqrt{1-\gamma_{1}^{2}}}}{16}}\rho_{A^{1}B^{1}}(\left(\begin{array}[]{cc}Q\sin(\theta)\sigma_{1}Q&0\\\
0&0\\\ \end{array}\right)\otimes\left(\begin{array}[]{cc}I_{d-2}&0\\\
0&\gamma_{1}\sigma_{1}\\\ \end{array}\right))]$ $\displaystyle=$
$\displaystyle 0.$