# Nonlocal correlations in the asymmetric quantum network

Souradeep Sasmal<EMAIL_ADDRESS>National Institute of Technology
Patna, Ashok Rajpath, Patna, Bihar 800005, India  Department of Physics,
Indian Institute of Technology Hyderabad, Telengana-502284, India  Shyam
Sundar Mahato<EMAIL_ADDRESS>National Institute of Technology
Patna, Ashok Rajpath, Patna, Bihar 800005, India  A. K. Pan
<EMAIL_ADDRESS>National Institute of Technology Patna, Ashok Rajpath,
Patna, Bihar 800005, India  Department of Physics, Indian Institute of
Technology Hyderabad, Telengana-502284, India

###### Abstract

The nonlocality revealed in a multiparty multi-source network Bell experiment
is conceptually different than the standard multiparty Bell nonlocality
involving a single common source. Here, by introducing variants of asymmetric
bilocal as well as trilocal network scenarios, we go beyond the typical
bilocal network scenario where both the edge parties have an equal number of
measurement settings. We first introduce an asymmetric bilocal network where
one of the edge parties (say, Alice) receives $2^{n-1}$ inputs and the other
edge party (say, Charlie) receives $n$ inputs. We derive two variants of
asymmetric bilocality inequalities and demonstrate their optimal quantum
violations. Further, we explore two types of asymmetric trilocal scenarios-
(i) when two edge parties receive $2^{n-1}$ inputs each and the other edge
party receives $n$ inputs, and (ii) when one edge party receives $2^{n-1}$
inputs and the other two edge parties have $n$ inputs each. We use an elegant
sum-of-squares technique that enables us to evaluate the quantum optimal
values of the proposed network inequalities without assuming the dimension of
the systems for both the asymmetric bilocal as well as the trilocal scenarios.
Further, we demonstrate the robustness of the quantum violations of the
proposed inequalities in the presence of white noise.

##  I Introduction

The study of quantum nonlocality in the network scenario [1] has recently been
receiving considerable attention. Such a form of nonlocality is conceptually
different from the standard Bell nonlocality [2]. While a multiparty Bell
experiment involves a single common source, the multiparty network Bell
experiment involves several independent sources. Each source distributes a
physical system to subsequent parties, and each party performs a measurement
on their subsystem prepared from different sources.

The simplest non-trivial network scenario [3, 4, 5] features three parties and
two independent sources, commonly referred to as the bilocality scenario. The
quantum nonlocality in a network is demonstrated through the quantum violation
of suitably formulated nonlinear bilocality inequality. A straightforward
generalization of the bilocality scenario is the $n$-locality scenario [6, 7,
8] involving an arbitrary $n$ number of sources. For example, a star network
may have $n$ number of sources and edge parties. Each edge party shares the
physical system with a central party. In recent times, the network nonlocality
has been extensively studied in various topologies [9, 10, 8, 11, 12, 13, 14,
15, 16, 17, 18, 19, 20, 21, 22, 23].

The reported interesting results such as possibility of observing quantum
nonlocality without inputs [24, 25] or showing the nonlocality of certain
entangled states which do not exhibit nonlocality in the usual Bell scenario
[26, 27] establishes the fundamental importance of viewing nonlocality in
terms of the symmetric network scenario in contrast to the standard Bell
scenario. Characterization of network nonlocality and its correspondence with
the bipartite Bell nonlocality has been studied [7]. Several theoretical
proposals have also been experimentally verified [28, 29, 7, 30, 31, 32, 33,
34]. Recently, genuine network nonlocality has also been introduced that
cannot be traced back to Bell nonlocality [35, 36, 23]. Self-testing protocols
using the quantum network have recently been proposed [37, 38, 39, 40].
Further, by using a quantum network, it has been established [38, 41, 42] that
the real quantum theory can be experimentally falsified, i.e., quantum theory
inevitably needs complex numbers. To this end, different forms of network
scenarios like the star-network [6], chain-shaped network [19] and cycle
network [24] have been explored.

We note here that while most of the studies concerning the star-network
scenario have been investigated for the symmetric input scenario, i.e., each
edge party performs the same number of measurements, a generalised study of
network nonlocality in asymmetric input scenarios remains unexplored. In this
regard, by introducing the asymmetric bilocal network scenario that comprises
two edge parties performing three and six measurements respectively, and the
central party performing four measurements, a couple of recent works [38, 42]
have shown that complex numbers are necessary for quantum predictions. It is
crucial to remark here that such a study is based on the two-qubit system.
Here, the purpose of the present work is to probe hitherto unexplored
generalised asymmetric network nonlocality in a device-independent way. In
particular, by considering the bilocality scenario, we first derive asymmetric
non-linear inequality for the scenario in which one of the two edge parties
(say, Alice) has four measurement settings and the other edge party (say,
Charlie) has three measurement settings. In addition to that, we extend the
study from the asymmetric bilocality scenario to the asymmetric trilocality
scenario involving three independent sources. In the trilocal network
scenario, we explore two particular variants of asymmetric trilocality. First,
we consider one edge party (Alice) has four measurement settings and the other
two edge parties (Charlie and Diana) have three measurement settings each.
Then we consider one edge party has three measurement settings (Charlie) and
the other two edge parties have four measurement settings each (Alice and
Diana).

Then, by using an elegant sum-of-squares (SOS) approach developed in [43, 44,
45], we analytically obtain the optimal quantum violations of the asymmetric
bilocality as well as the asymmetric trilocality inequalities. It is important
to note that we evaluate such optimal quantum bounds without assuming the
dimension of the systems. In this process, we evaluate the required
constraints on the observables of each party for achieving the optimal quantum
violation. For the asymmetric bilocality scenario, we demonstrate that the
quantum optimal value will be achieved if each of the both Alice-Bob and Bob-
Charlie shares at least a single copy of maximally entangled two-qubit state.
Moreover, for the considered asymmetric trilocality scenario, as similar to
the bilocality scenario, we find that the optimal quantum value will be
obtained if each of all the edge parties shares at least a single copy of
maximally entangled two-qubit state with the central party Bob.

Furthermore, we extend our study for any arbitrary number of measurement
settings. In particular, we consider an asymmetric bilocal network where one
of the edge parties receives $2^{n-1}$ inputs and the other edge party
receives $n$ inputs. We explore two types of asymmetric trilocal scenarios for
arbitrary inputs. (i) when two edge parties receive $2^{n-1}$ inputs each and
the other edge party receives $n$ inputs, and (ii) when one edge party
receives $2^{n-1}$ inputs and the other two edge parties have $n$ inputs each.
Finally, we illustrate the robustness of the quantum violations of the
proposed inequalities in the presence of the white noise for both the cases of
bilocality and trilocality scenarios. We find that the proposed asymmetric
inequality is most robust to white noise in the simplest bilocality network
scenario.

This paper is organized as follows. To begin with, in Sec. II.1, by invoking
the SOS approach [43, 44], we derive the optimal quantum violation of the
standard bilocal network inequality without assuming the dimension of the
system. Next, in Secs. III and IV, we introduce two variants of asymmetric
bilocal scenario and propose two different bilocal network inequalities. Then,
we obtain the optimal quantum bounds along with the states and observables
corresponding to the optimal quantum values (Secs. III.1 and III.2). In Secs.
V and VI, by going beyond the bilocality network scenario, we introduce
asymmetricity in the trilocal network and propose two different types of
asymmetric trilocal inequalities. We also evaluate the corresponding optimal
quantum bounds as well as the states and observables required for attaining
such optimal quantum values (Secs. V.1 and V.2). In particular, we illustrate
that in order to achieve the optimal quantum bound for both the asymmetric
bilocal and trilocal cases, each of all the edge parties must share at least a
single copy of maximally entangled state with the central party Bob. Then, in
Sec. VII, we have generalised the asymmetric bilocal and trilocal network
scenario for arbitrary number of measurement settings. Further, in Sec.
(VIII), we provide an analysis regarding the robustness of quantum violations
of the proposed inequalities to white noise. Finally, in Sec. IX, we discuss
the salient features of our work and propose some interesting open questions.

##  II Preliminaries: The standard bilocal network scenario

The standard bilocal network scenario comprises of three spatially separated
parties, Alice, Bob, and Charlie. Two independent sources $S_{1}$ and $S_{2}$
prepare a bipartite physical system for Alice-Bob and Bob-Charlie,
respectively. Upon receiving the system from the respective source, Alice
performs one of $m_{A}$ local measurements, denoted by
$A_{n,x}\in\\{A_{n,1},A_{n,2},...,A_{n,m_{A}}\\}$. Similarly for Bob and
Charlie the respective measurements are denoted by
$B_{n,j}\in\\{B_{n,1},B_{n,2},...,B_{n,m_{B}}\\}$ and
$C_{n,z}\in\\{C_{n,1},C_{n,2},...,C_{n,m_{C}}\\}$. The outcomes for Alice, Bob
and Charlie are denoted by $a,b,c\in\\{0,1\\}$. The standard bilocal scenario
is a symmetric scenario that implies an equal number of measurement settings
for the edge parties, i.e., $m_{A}=m_{C}$. On the other hand, in our present
work we consider two types of asymmetric bilocal network scenario - (i) Alice
performs one of $m_{A}=2^{n-1}$ measurements; Charlie and the central party
Bob perform one of $m_{C}=m_{B}=n$ measurements. (ii) Alice and Bob perform
one of $m_{A}=m_{B}=2^{n-1}$ measurements; Charlie performs one of $m_{C}=n$
measurements. The index ‘$n$’ appearing in the subscript denotes the scenario
involving the number of measurement settings considered. For example, $n=2$
corresponds the standard bilocal network scenario comprising two measurement
settings for each party.

Now, in the ontological model of the tripartite standard Bell scenario, it is
assumed that the source prepares a common hidden variable $\lambda$. Then the
reproducibility condition is given by

$P(a,b,c|x,j,z)=\int d\lambda\
\mu(\lambda)P(a|x,\lambda)P(b|j,\lambda)P(c|z,\lambda)$ (1)

Figure 1: The standard bilocal scenario featuring two edge parties (Alice and
Charlie) and the central party Bob. The source $S_{1}(S_{2})$ emits physical
system for Alice (Charlie) and Bob. The sources are assumed to be independent
to each other.

In contrast to the ontological model of tripartite standard Bell scenario, the
ontological model of bilocality scenario is that each source $S_{1}$ and
$S_{2}$ prepares the physical system in the state $\lambda_{1}\in\Lambda_{1}$
and $\lambda_{2}\in\Lambda_{2}$ with a probability distribution
$\mu(\lambda_{1})$ and $\mu(\lambda_{2})$ respectively, with
$\int_{\Lambda_{k}}\mu(\lambda_{k})d\lambda_{k}=1$. The crucial assumption
here is that the sources $S_{1}$ and $S_{2}$ are independent to each other.
This means that the joint probability distribution
$\mu(\lambda_{1},\lambda_{2})$ can be written in the factorized form as
$\mu(\lambda_{1},\lambda_{2})=\mu(\lambda_{1})\mu(\lambda_{2})$. Now, in order
to reproduce the quantum theoretical prediction from the bilocal ontic model,
the following reproducibility condition need to be satisfied.

$\displaystyle P(a,b,c|x,j,z)$ $\displaystyle=$ $\displaystyle\iint
d\lambda_{1}d\lambda_{2}\ \mu(\lambda_{1})\mu(\lambda_{2})P(a|x,\lambda_{1})$
$\displaystyle\hskip 28.45274pt\times\
P(b|j,\lambda_{1},\lambda_{2})P(c|z,\lambda_{2})$

Note that in the 2 measurements per party scenario $(m_{A}=m_{B}=m_{C}=2)$, it
has been shown [5] that any operational theory satisfying the above Eq.(II)
satisfies the following nonlinear inequality

$\mathcal{B}_{2}\equiv\sqrt{|\mathcal{I}_{2,1}|}+\sqrt{|\mathcal{I}_{2,2}|}\leq(\mathcal{B}_{2})_{bl}=2$
(3)

where $\mathcal{I}_{2,2}$ are suitably defined linear combinations of the
tripartite correlations, given by

$\displaystyle\mathcal{I}_{2,1}$ $\displaystyle=$ $\displaystyle\Big{\langle}\
\quantity(A_{2,1}+A_{2,2})\otimes B_{2,1}\otimes\quantity(C_{2,1}+C_{2,2})\
\Big{\rangle}$ (4) $\displaystyle\mathcal{I}_{2,2}$ $\displaystyle=$
$\displaystyle\Big{\langle}\ \quantity(A_{2,1}-A_{2,2})\otimes
B_{2,2}\otimes\quantity(C_{2,2}-C_{2,1})\ \Big{\rangle}$ (5)

with111From now on we will denote $\langle A_{n,x}\otimes B_{n,j}\otimes
C_{n,z}\rangle$ as $\langle A_{n,x}B_{n,j}C_{n,z}\rangle$ $\langle
A_{n,x}B_{n,j}C_{n,z}\rangle=\sum_{a,b,c}(-1)^{a+b+c}P(a,b,c|x,j,z)$ and
$P(a,b,c|x,j,z)$ is the joint probability for obtaining the outcomes $(a,b,c)$
corresponding to the dichotomic measurements performed by Alice, Bob, and
Charlie. In quantum theory, the joint probability $P(a,b,c|x,j,z)$ is given as

$P(a,b,c|x,j,z)=\Tr[\quantity(\rho_{AB}\otimes\rho_{BC})\
\Pi_{A_{n,x}}^{a}\otimes\Pi_{B_{n,j}}^{b}\otimes\Pi_{C_{n,z}}^{c}]$ (6)

where $\rho_{AB}$ and $\rho_{BC}$ are bipartite states produced from two
independent sources $S_{1}$ and $S_{2}$ respectively.

It has been shown [5, 44] that the maximum quantum value,
$(\mathcal{B}_{2})_{Q}^{opt}=2\sqrt{2}$ is obtained when Alice’s and Charlie’s
observables are mutually anticommuting and Bob’s observables are mutually
commuting. An example of such choices of observables in two-dimensional
Hilbert space $(\mathcal{H}^{2})$ is given as follows

$\displaystyle A_{2,1}=C_{2,1}=\left(\sigma_{z}+\sigma_{x}\right)/\sqrt{2}\ ;\
\ B_{2,1}=\sigma_{z}\otimes\sigma_{z}$ $\displaystyle
A_{2,2}=C_{2,2}=\left(\sigma_{z}-\sigma_{x}\right)/\sqrt{2}\ ;\ \
B_{2,2}=\sigma_{x}\otimes\sigma_{x}$ (7)

Note that while the optimal quantum value of $(\mathcal{B}_{2})_{Q}^{opt}$ was
earlier derived [5] by taking a pair of the two-qubit entangled state, the
dimension-independent derivation of $(\mathcal{B}_{2})_{Q}^{opt}$ has recently
been proposed [44]. Throughout this work, we adopt the SOS approach introduced
in [43] and derive the optimal quantum bound without assuming the dimension of
the system. Thus, our optimal value possess the the potential to be used as
device-independent certification of quantum correlations.

###  II.1 Optimal quantum bound for $\mathcal{B}_{2}$

Here, by invoking the elegant SOS approach [44], we evaluate the optimal
quantum value of $(\mathcal{B}_{2})_{Q}$. Without loss of generality, one can
always construct a suitable operator $\gamma_{2}$ satisfying
$\langle\gamma_{2}\rangle=\beta_{2}-(\mathcal{B}_{2})_{Q}$ such that
$\langle\gamma_{2}\rangle\geq 0$. The existence of such operator $\gamma_{2}$
can be shown by suitably considering a set of operators $M_{2,j},\ \forall
j\in\\{1,2\\}$ which are polynomial functions of $A_{2,x}$, $B_{2,j}$ and
$C_{2,z}$.

$\langle\gamma_{2}\rangle=\sum\limits_{j=1}^{2}\frac{\sqrt{\omega_{2,j}}}{2}\left|M_{2,j}\ket{\psi}\right|^{2}$
(8)

where $\omega_{2,j}\geq 0$ is suitable positive numbers that will be specified
soon. We choose the operator $M_{2,j}$ as the following way222From now on, we
will write $A_{n,x}\otimes\mathbb{I}\otimes C_{n,z}$ as $A_{n,x}\otimes
C_{n,z}$ and $\mathbb{I}_{d}\otimes B_{n,j}\otimes\mathbb{I}_{d}$ as $B_{n,j}$

$\displaystyle\Big{|}M_{2,1}\ket{\psi}\Big{|}=\sqrt{\bigg{|}\left(\frac{{A}_{2,1}+{A}_{2,2}}{\omega_{2,1}^{A}}\otimes\mathbb{I}_{d}\otimes\frac{C_{2,1}+C_{2,2}}{\omega_{2,1}^{C}}\right)\ket{\psi}\bigg{|}}$
$\displaystyle\hskip 56.9055pt-\sqrt{\Big{|}\mathbb{I}_{d}\otimes
B_{2,1}\otimes\mathbb{I}_{d}\ \ket{\psi}\Big{|}}$
$\displaystyle\Big{|}M_{2,2}\ket{\psi}\Big{|}=\sqrt{\bigg{|}\left(\frac{{A}_{2,1}-{A}_{2,2}}{\omega_{2,2}^{A}}\otimes\mathbb{I}_{d}\otimes\frac{C_{2,1}-C_{2,2}}{\omega_{2,2}^{C}}\right)\ket{\psi}\bigg{|}}$
$\displaystyle\hskip 56.9055pt-\sqrt{\Big{|}\mathbb{I}_{d}\otimes
B_{2,2}\otimes\mathbb{I}_{d}\ \ket{\psi}\Big{|}}$ (9)
$\displaystyle\omega_{2,1}^{A}=||({A}_{2,1}+{A}_{2,2})\ket{\psi}||_{2}=\sqrt{2+\langle\\{A_{2,1},A_{2,2}\\}\rangle}$
$\displaystyle\omega_{2,2}^{A}=||({A}_{2,1}-{A}_{2,2})\ket{\psi}||_{2}=\sqrt{2-\langle\\{A_{2,1},A_{2,2}\\}\rangle}$
$\displaystyle\omega_{2,1}^{C}=||({C}_{2,1}+{C}_{2,2})\ket{\psi}||_{2}=\sqrt{2+\langle\\{C_{2,1},C_{2,2}\\}\rangle}$
$\displaystyle\omega_{2,2}^{C}=||({C}_{2,1}-{C}_{2,2})\ket{\psi}||_{2}=\sqrt{2-\langle\\{C_{2,1},C_{2,2}\\}\rangle}$
(10)

Where $||\cdot||_{2}$ denotes the Frobenious norm given by $||\ \mathcal{O}\ \
||_{2}=\sqrt{\bra{\psi}\mathcal{O}^{\dagger}\mathcal{O}\ket{\psi}}$.

Now, since Alice, Bob and Charlie are space-like separated their observables
are mutually commuting. Thus the operators
($A_{2,j}\otimes\mathbb{I}_{d}\otimes\mathbb{I}_{d}$), ($\mathbb{I}_{d}\otimes
B_{2,j}\otimes\mathbb{I}_{d}$) and
($\mathbb{I}_{d}\otimes\mathbb{I}_{d}\otimes C_{2,j}$) are also mutually
commuting. Hence, these three observables must have at least one common
eigenstate. Without loss of generality, $\ket{\psi}$ is taken to be one of the
common eigenstate. Therefore, by evaluating the quantity
$|M_{2,j}\ket{\psi}|^{2}$ from Eq. (9), a simple algebraic manipulation gives
us from Eq. (8) the following

$\displaystyle\langle\gamma_{2}\rangle$ $\displaystyle=$
$\displaystyle\quantity(\sqrt{\omega_{2,1}}+\sqrt{\omega_{2,2}})-(\mathcal{B}_{2})_{Q}$
(11)

where $\omega_{2,j}=\omega_{2,j}^{A}\ \omega_{2,j}^{C}$.

Now, since by construction $\langle\gamma_{2}\rangle\geq 0$, it is evident
that from the above Eq. (11) that the optimal quantum value of
$\mathcal{B}_{2}$ corresponds to $\langle\gamma_{2}\rangle=0$. Therefore, the
quantum optimal value is given as follows

$(\mathcal{B}_{2})_{Q}^{opt}=\sqrt{\omega_{2,1}^{A}\
\omega_{2,1}^{C}}+\sqrt{\omega_{2,2}^{A}\ \omega_{2,2}^{C}}$ (12)

Next, using the inequality,
$\sqrt{r_{1}s_{1}}+\sqrt{r_{2}s_{2}}\leq\sqrt{r_{1}+r_{2}}\sqrt{s_{1}+s_{2}}\
\ \forall r_{1},s_{1},r_{2},s_{2}\geq 0$, we can write Eq.(12) as

$\displaystyle(\mathcal{B}_{2})_{Q}^{opt}$ $\displaystyle=$
$\displaystyle\sqrt{\left(\omega_{2,1}^{A}+\omega_{2,2}^{A}\right)\left(\omega_{2,1}^{C}+\omega_{2,2}^{C}\right)}$
$\displaystyle=$
$\displaystyle\sqrt{\quantity(\sqrt{2+\langle\\{A_{2,1},A_{2,2}\\}\rangle}+\sqrt{2-\langle\\{A_{2,1},A_{2,2}\\}\rangle})}$
$\displaystyle\times$
$\displaystyle\sqrt{\quantity(\sqrt{2+\langle\\{C_{2,1},C_{2,2}\\}\rangle}+\sqrt{2-\langle\\{C_{2,1},C_{2,2}\\}\rangle})}$

The above Eq. (II.1) is optimised when
$\\{A_{2,1},A_{2,2}\\}=\\{C_{2,1},C_{2,2}\\}=0$ and the optimal quantum value
is then given by

$(\mathcal{B}_{2})_{Q}^{opt}=2\sqrt{2}$ (14)

Note that the evaluated optimal quantum value
$(\mathcal{B}_{2})_{Q}^{opt}=2\sqrt{2}$ is same as that obtained when the
state between three parties are assumed to be pair of maximally entangled two-
qubit state. This means that for the 2-settings bilocality scenario, the
optimal quantum violation of the bilocal inequality remain same even if one
consider higher dimensional maximally entangled states.

##  III Asymmetric bilocal network: scenario-I

Figure 2: Asymmetric bilocal network scenario featuring two edge parties
(Alice and Charlie) and a central party Bob. The independent sources $S_{1}$
and $S_{2}$ emit physical systems for Alice-Bob and Charlie-Bob respectively.

Here we introduce a variant of asymmetric bilocal network scenario for $n=3$,
in which Alice performs one of the four dichotomic measurements and the other
edge party (Charlie) perform one of three dichotomic measurements. For our
purpose, we consider that the central party performs one of three dichotomic
measurements. Note that for the 2-settings bilocality scenario, is same as the
symmetric bilocal scenario, where each party has equal number (two) of
measurement settings. In the following, we derive a family of non-linear
asymmetric bilocality inequalities and also evaluate their optimal quantum
violations.

In this scenario, let us consider the nonlinear bilocal inequality of the
following form

$\mathcal{B}_{3}=\sum_{j=1}^{3}\sqrt{\quantity|\mathcal{I}_{3,j}|}\ \leq\
\left(\mathcal{B}_{3}\right)_{bl}$ (15)

where $\left(\mathcal{B}_{3}\right)_{bl}$ is the bilocal bound of
$\mathcal{B}_{3}$ and the quantity
$\mathcal{I}_{3,j}=\Big{\langle}\mathcal{\tilde{A}}_{3,j}B_{3,j}\mathcal{\tilde{C}}_{3,j}\Big{\rangle}$.
We define the quantities $\mathcal{\tilde{A}}_{3,j}$ and
$\mathcal{\tilde{C}}_{3,j}$ as

$\displaystyle\mathcal{\tilde{A}}_{3,1}$ $\displaystyle=$ $\displaystyle
A_{3,1}+A_{3,2}+A_{3,3}-A_{3,4}$ $\displaystyle\mathcal{\tilde{A}}_{3,2}$
$\displaystyle=$ $\displaystyle A_{3,1}+A_{3,2}-A_{3,3}+A_{3,4}$
$\displaystyle\mathcal{\tilde{A}}_{3,3}$ $\displaystyle=$ $\displaystyle
A_{3,1}-A_{3,2}+A_{3,3}+A_{3,4}$ $\displaystyle\mathcal{\tilde{C}}_{3,j}$
$\displaystyle=$ $\displaystyle C_{3,j}+C_{3,j+1}\ \ \text{with}\
C_{3,4}=-C_{3,1}$ (16)

In an ontological model $\lambda_{1}\in\Lambda_{1}$ and
$\lambda_{2}\in\Lambda_{2}$ completely determines the statistics of all the
measurements. This means that in the ontological model, we can write the
following

$\displaystyle\langle A_{3,x}\rangle_{\lambda_{1}}$ $\displaystyle=$
$\displaystyle\sum_{a}(-1)^{a}P(a|x,\lambda_{1})\ \ \ \ \forall
x\in\\{1,2,3,4\\}$ $\displaystyle\langle C_{3,z}\rangle_{\lambda_{2}}$
$\displaystyle=$ $\displaystyle\sum_{c}(-1)^{c}P(c|z,\lambda_{2})\ \ \ \
\forall z\in\\{1,2,3\\}$ $\displaystyle\langle
B_{3,j}\rangle_{\lambda_{1},\lambda_{2}}$ $\displaystyle=$
$\displaystyle\sum_{b}(-1)^{b}P(b|j,\lambda_{1},\lambda_{2})\ \ \forall
j\in\\{1,2,3\\}$ (17)

Now, invoking the reproducibility condition given by Eq. (II) it follows that

$\displaystyle\mathcal{I}_{3,1}$ $\displaystyle=$ $\displaystyle\iint
d\lambda_{1}d\lambda_{2}\ \mu(\lambda_{1})\mu(\lambda_{2})\ \quantity[\langle
C_{3,1}\rangle_{\lambda_{2}}+\langle C_{3,2}\rangle_{\lambda_{2}}]\ \langle
B_{3,1}\rangle_{\lambda_{1},\lambda_{2}}$ (18) $\displaystyle\ \times\ \
\quantity[\langle A_{3,1}\rangle_{\lambda_{1}}+\langle
A_{3,2}\rangle_{\lambda_{1}}+\langle A_{3,3}\rangle_{\lambda_{1}}-\langle
A_{3,4}\rangle_{\lambda_{1}}]$

Since $|\langle B_{3,1}\rangle_{\lambda_{1},\lambda_{2}}|\leq 1$, we obtain

$\displaystyle\quantity|\mathcal{I}_{3,1}|$ $\displaystyle\leq$
$\displaystyle\iint d\lambda_{1}d\lambda_{2}\
\mu(\lambda_{1})\mu(\lambda_{2})\ \quantity|\langle
C_{3,1}\rangle_{\lambda_{2}}+\langle C_{3,2}\rangle_{\lambda_{2}}|$ (19)
$\displaystyle\ \times\ \ \quantity|\langle
A_{3,1}\rangle_{\lambda_{1}}+\langle A_{3,2}\rangle_{\lambda_{1}}+\langle
A_{3,3}\rangle_{\lambda_{1}}-\langle A_{3,4}\rangle_{\lambda_{1}}|$

The terms $|\mathcal{I}_{3,2}|$ and $|\mathcal{I}_{3,3}|$ can also be written
in a similar manner as Eq. (19). Now for our purpose, by utilizing the
inequality333 The inequality proved in Appendix A of [6]
$\sum\limits_{i=1}^{t}\bigg{(}\prod\limits_{k=1}^{r}z_{k}^{i}\bigg{)}^{\frac{1}{r}}\leq\prod\limits_{k=1}^{r}\bigg{(}\sum\limits_{i=1}^{t}z_{k}^{i}\bigg{)}^{\frac{1}{r}}\
\ \ \forall z_{k}^{i}\geq 0$ (20) Here r is the number of edge party in a
star-network. Note that for our bilocal case $r=2$. proved in [6], we obtain
the following

$(\mathcal{B}_{3})_{bl}\leq\quantity(\iint d\lambda_{1}d\lambda_{2}\
\mu(\lambda_{1})\mu(\lambda_{2})\ \delta_{1}\ \delta_{2})^{\frac{1}{2}}$ (21)

where
$\delta_{1}=\big{[}|\langle{A_{3,1}}\rangle_{\lambda_{1}}+\langle{A_{3,2}}\rangle_{\lambda_{1}}+\langle{A_{3,3}}\rangle_{\lambda_{1}}-\langle{A_{3,4}}\rangle_{\lambda_{1}}|+|\langle{A_{3,1}}\rangle_{\lambda_{1}}+\langle{A_{3,2}}\rangle_{\lambda_{1}}-\langle{A_{3,3}}\rangle_{\lambda_{1}}+\langle{A_{3,4}}\rangle_{\lambda_{1}}|+|\langle{A_{3,1}}\rangle_{\lambda_{1}}-\langle{A_{3,2}}\rangle_{\lambda_{1}}+\langle{A_{3,3}}\rangle_{\lambda_{1}}+\langle{A_{3,4}}\rangle_{\lambda_{1}}|\big{]}$
and
$\delta_{2}=\big{[}|\langle{C_{3,1}}\rangle_{\lambda_{2}}+\langle{C_{3,2}}\rangle_{\lambda_{2}}|+|\langle{C_{3,2}}\rangle_{\lambda_{2}}+\langle{C_{3,3}}\rangle_{\lambda_{2}}|+|\langle{C_{3,3}}\rangle_{\lambda_{2}}-\langle{C_{3,1}}\rangle_{\lambda_{2}}|\big{]}$.

Since all the observables are dichotomic with eigenvalues $\pm 1$, it is
straightforward to derive that $\delta_{1}\leq 6$ and $\delta_{2}\leq 4$.
Therefore, from Eq. (21), integrating over $\lambda_{1}$ and $\lambda_{2}$ we
obtain

$(\mathcal{B}_{3})_{bl}\leq 2\sqrt{6}\approx 4.89$ (22)

We show that there are suitable states and observables for which the bilocal
bound can be violated in quantum theory. Now, in the following, we evaluate
the quantum optimal bound of $\mathcal{B}_{3}$.

###  III.1 Optimal quantum bound of the asymmetric bilocality inequality in
scenario-I

To derive the optimal quantum bound of $\mathcal{B}_{3}$ without assuming the
dimension of the system, we again invoke the SOS approach discussed in the
preceding Sec. II.1. Following the similar argument presented earlier in the
Sec. II.1, we first show that there exists a suitable operator $\gamma_{3}$
satisfying $\langle\gamma_{3}\rangle=\beta_{3}-(\mathcal{B}_{3})_{Q}$ such
that $\langle\gamma_{3}\rangle\geq 0$. Note that the power of SOS approach in
evaluating the optimal quantum bound of a particular Bell functional lies in
the suitable construction of the operator $\gamma_{3}$ in such a way so that
it is a semi-definite operator and can be reduced to the form of the concerned
Bell functional. Now, in order to evaluate the quantum optimal bound of
$\mathcal{B}_{3}$, we construct $\gamma_{3}$ in terms of a set of operators
$M_{3,j}\ \forall j\in\\{1,2,3\\}$ in the following way

$\langle\gamma_{3}\rangle=\sum\limits_{j=1}^{3}\frac{\sqrt{\omega_{3,j}}}{2}\quantity|M_{3,j}\ket{\psi}|^{2}$
(23)

where $\omega_{3,j}\geq 0$ is suitable positive numbers and
$\omega_{3,j}=\omega^{A}_{3,j}\cdot\omega^{C}_{3,j}$ that will be specified
soon. In the following $\forall j\in\\{1,2,3\\}$, we choose the operator
$M_{3,j}$ and the quantity $\omega_{3,j}$.

$\displaystyle\quantity|M_{3,j}\ket{\psi}|=\sqrt{\quantity|\quantity(\frac{\mathcal{\tilde{A}}_{3,j}}{\omega_{3,j}^{A}}\otimes\frac{\mathcal{\tilde{C}}_{3,j}}{\omega_{3,j}^{C}})\ket{\psi}|}-\sqrt{\quantity|B_{3,j}\ket{\psi}|}$
(24)
$\displaystyle\omega_{3,j}^{A}=||\mathcal{\tilde{A}}_{3,j}\ket{\psi}||_{2}\ ;\
\ \omega_{3,j}^{C}=||\mathcal{\tilde{C}}_{3,j}\ket{\psi}||_{2}$ (25)

Putting $M_{3,j}$ from Eq. (24) into the Eq. (23), after a simple algebraic
evaluation, we obtain
$\langle\gamma_{3}\rangle=\sum\limits_{j=1}^{3}\sqrt{\omega_{3,j}}-(\mathcal{B}_{3})_{Q}$.
Then, it follows that the quantum optimal value corresponds
$\langle\gamma_{3}\rangle=0$. Therefore,

$\displaystyle\left(\mathcal{B}_{3}\right)_{Q}^{opt}$ $\displaystyle=$
$\displaystyle\max\sum\limits_{j=1}^{3}\sqrt{\omega_{3,j}^{A}\cdot\omega_{3,j}^{C}}$
(26)

which in turn gives the optimisation condition as follows

$\quantity|M_{3,j}\ket{\psi}|=0\ \implies M_{3,j}\ket{\psi}=0\ \ \forall
j\in\\{1,2,3\\}$ (27)

Now, in order to evaluate the optimal quantum value
$\left(\mathcal{B}_{3}\right)_{Q}^{opt}$ and thus the quantity
$\sum\limits_{j=1}^{3}\sqrt{\omega_{3,j}^{A}\cdot\omega_{3,j}^{C}}$ , we
invoke the inequality given in the footnote 3. Then the right-hand-side of Eq.
(26) reduces to the following

$\sum\limits_{j=1}^{3}\left(\prod\limits_{k=A,C}\omega^{k}_{3,j}\right)^{\frac{1}{2}}\leq\prod\limits_{k=A,C}\left(\sum\limits_{j=1}^{3}\omega^{k}_{3,j}\right)^{\frac{1}{2}}$
(28)

Further, by applying the convex inequality444 From the Jensen’s inequality
given by
$f(\sum\limits_{k=1}^{t}r_{k}x_{k})\leq\sum\limits_{k=1}^{t}r_{k}f(x_{k})$
where $\sum\limits_{k=1}^{t}r_{k}=1$, the following inequality can be derived
$\sum\limits_{k=1}^{t}\omega_{k}\leq\sqrt{\
t\sum\limits_{k=1}^{t}\omega^{2}_{k}}$ (29) , the quantity
$\sum\limits_{j=1}^{3}\omega^{k}_{3,j}$ can be written as

$\sum\limits_{j=1}^{3}(\omega^{k}_{3,j})\leq\sqrt{3\sum\limits_{j=1}^{3}\bigg{(}\omega^{k}_{3,j}\bigg{)}^{2}}$
(30)

Then, by combining Eq. (28) and (30), from Eq. (26 ) we obtain

$\displaystyle\left(\mathcal{B}_{3}\right)_{Q}^{opt}=\max\left[\prod_{k=A,C}\left(3\sum_{j=1}^{3}\left(\omega^{k}_{3,j}\right)^{2}\right)\right]^{\frac{1}{4}}$
(31)

where each $\left(\omega^{k}_{3,j}\right)^{2}$ is evaluated from Eq. (25) as
follwos

$\displaystyle\left(\omega^{A}_{3,1}\right)^{2}$ $\displaystyle=$
$\displaystyle\bra{\psi}(4+\\{A_{3,1},(A_{3,2}+A_{3,3}-A_{3,4})\\}$
$\displaystyle+$
$\displaystyle\\{A_{3,2},(A_{3,3}-A_{3,4})\\}-\\{A_{3,3},A_{3,4}\\})\ket{\psi}$
$\displaystyle\left(\omega^{A}_{3,2}\right)^{2}$ $\displaystyle=$
$\displaystyle\bra{\psi}(4+\\{A_{3,1},(A_{3,2}-A_{3,3}+A_{3,4})\\}$
$\displaystyle+$
$\displaystyle\\{A_{3,2},(-A_{3,3}+A_{3,4})\\}-\\{A_{3,3},A_{3,4}\\})\ket{\psi}$
$\displaystyle\left(\omega^{A}_{3,3}\right)^{2}$ $\displaystyle=$
$\displaystyle\bra{\psi}(4+\\{A_{3,1},(-A_{3,2}+A_{3,3}+A_{3,4})\\}$ (32)
$\displaystyle-$
$\displaystyle\\{A_{3,2},(A_{3,3}+A_{3,4})\\}+\\{A_{3,3},A_{3,4}\\})\ket{\psi}$
$\displaystyle\left(\omega^{C}_{3,1}\right)^{2}$ $\displaystyle=$
$\displaystyle\bra{\psi}(2+\\{C_{3,1},C_{3,2}\\})\ket{\psi}$
$\displaystyle\left(\omega^{C}_{3,2}\right)^{2}$ $\displaystyle=$
$\displaystyle\bra{\psi}(2+\\{C_{3,2},C_{3,3}\\})\ket{\psi}$
$\displaystyle\left(\omega^{C}_{3,3}\right)^{2}$ $\displaystyle=$
$\displaystyle\bra{\psi}(2-\\{C_{3,1},C_{3,3}\\})\ket{\psi}$ (33)

Now, in the following we calculate
$\sum_{j=1}^{3}\left(\omega^{A}_{3,j}\right)^{2}$ from Eq. (32) and
$\sum_{j=1}^{3}\left(\omega^{C}_{3,j}\right)^{2}$ from Eq. (33) separately.

#### Evaluation of $\sum_{j=1}^{3}\left(\omega^{A}_{3,j}\right)^{2}$:

$\displaystyle\sum_{j=1}^{3}\left(\omega^{A}_{3,j}\right)^{2}$
$\displaystyle=$
$\displaystyle\bra{\psi}(12+\\{A_{3,1},(A_{3,2}-A_{3,3}+A_{3,4})\\}$ (34)
$\displaystyle-$
$\displaystyle\\{A_{3,2},(A_{3,3}+A_{3,4})\\}-\\{A_{3,3},A_{3,4}\\})\ket{\psi}$
$\displaystyle=$ $\displaystyle\bra{\psi}(12+\Delta_{3})\ket{\psi}$

where
$\Delta_{3}=\\{A_{3,1},(A_{3,2}-A_{3,3}+A_{3,4})\\}-\\{A_{3,2},(A_{3,3}+A_{3,4})\\}-\\{A_{3,3},A_{3,4}\\}$.
Without loss of generality we can always write
$\ket{\psi^{\prime}}=(A_{3,1}+A_{3,2}+A_{3,3}-A_{3,4})\ket{\psi}$ such that
$\ket{\psi}\neq 0$. Therefore,
$\braket{\psi^{\prime}}{\psi^{\prime}}=\bra{\psi}(4-\Delta_{3})\ket{\psi}$
implies $\braket{\Delta_{3}}=4-\braket{\psi^{\prime}}{\psi^{\prime}}$. Then it
immediately follows that $\braket{\Delta_{3}}_{max}$ is obtained iff
$\braket{\psi^{\prime}}{\psi^{\prime}}=0$. Since $\ket{\psi}\neq 0$, then the
following relation must satisfy

$A_{3,1}-A_{3,2}-A_{3,3}-A_{3,4}=0$ (35)

Hence, in order to obtain the quantum optimal value of
$(\mathcal{B}_{3})^{opt}_{Q}$, observables of Alice must satisfy the linear
condition given by Eq. (35). Therefore, $\langle\Delta_{3}\rangle_{max}=4$
leads to

$\sum_{j=1}^{3}\left(\omega^{A}_{3,j}\right)^{2}=\bra{\psi}(12+\Delta_{3})\ket{\psi}\leq
16$ (36)

#### Evaluation of $\sum_{j=1}^{3}\left(\omega^{C}_{3,j}\right)^{2}$:

$\displaystyle\sum_{j=1}^{3}\left(\omega^{C}_{3,j}\right)^{2}$
$\displaystyle=$
$\displaystyle\bra{\psi}(6+\\{C_{3,2},(C_{3,1}+C_{3,3})\\}-\\{C_{3,1},C_{3,3}\\})\ket{\psi}$
(37) $\displaystyle=$
$\displaystyle\bra{\psi}(6+3\mathbb{I}-(C_{3,1}-C_{3,2}+C_{3,3})^{2})\ket{\psi}\leq
9$

The above Eq. (37) is maximized when

$C_{3,1}-C_{3,2}+C_{3,3}=0$ (38)

Finally, we obtain the optimal quantum value from Eqs. (26), (36) and (37) as

$\displaystyle\left(\mathcal{B}_{3}\right)_{Q}^{opt}=6$ (39)

It is important to remark here that the optimal quantum value
$\left(\mathcal{B}_{3}\right)_{Q}^{opt}=6$ is evaluated without specifying the
dimension of both the system and observables. The optimal value fixes the
states, and the observables are the following.

###  III.2 The state and observables for the optimal quantum violation of
$\mathcal{B}_{3}$

We further obtain relationships between the observables of all the parties for
achieving the optimal quantum violation. Such relationships are given in Eqs.
(35) and Eq. (38), which in turn provide the relationship between the
observables in terms of the anti-commuting relations.

$\displaystyle\\{A_{3,1},A_{3,2}\\}=\\{A_{3,1},A_{3,3}\\}=\\{A_{3,1},A_{3,4}\\}=\frac{2}{3}\mathbb{I}_{d}$
$\displaystyle\\{A_{3,2},A_{3,3}\\}=\\{A_{3,2},A_{3,4}\\}=\\{A_{3,3},A_{3,4}\\}=-\frac{2}{3}\mathbb{I}_{d}$
$\displaystyle\\{C_{3,1},C_{3,2}\\}=\\{C_{3,2},C_{3,3}\\}=-\\{C_{3,1},C_{3,3}\\}=\mathbb{I}_{d}$
(40)

By using the above relations between the observables given by Eq. (40) on the
observables, one can always construct a set of observables for Alice and
Charlie in the Hilbert space dimension $\mathcal{H}^{d},\ \ \forall d\geq 2$.

Next, we recall the optimization condition obtained in the SOS method from the
Eq. (27) to find the constraints on Bob’s observable. The specific condition
$M_{3,j}\ket{\psi}=0\ ;\ \forall j\in\\{1,2,3\\}$ implies the following

$\displaystyle B_{3,j}$ $\displaystyle=$
$\displaystyle\frac{\mathcal{\tilde{A}}_{3,j}}{\omega_{3,j}^{A}}\otimes\frac{\mathcal{\tilde{C}}_{3,j}}{\omega_{3,j}^{C}}$
(41)

In the following, we then explicitly construct a set of observables for the
Hilbert space dimension $\mathcal{H}^{2}$

$\displaystyle C_{3,1}$ $\displaystyle=$ $\displaystyle\sigma_{z}\ ;\
C_{3,2}=\left(\frac{\sqrt{3}}{2}\sigma_{x}+\frac{\sigma_{z}}{2}\right);\
C_{3,3}=\left(\frac{\sqrt{3}}{2}\sigma_{x}-\frac{\sigma_{z}}{2}\right)$
$\displaystyle A_{3,1}$ $\displaystyle=$
$\displaystyle\frac{\sigma_{x}+\sigma_{y}+\sigma_{z}}{\sqrt{3}}\ \ ;\ \ \ \
A_{3,2}=\frac{\sigma_{x}+\sigma_{y}-\sigma_{z}}{\sqrt{3}}$ $\displaystyle
A_{3,3}$ $\displaystyle=$
$\displaystyle\frac{\sigma_{x}-\sigma_{y}+\sigma_{z}}{\sqrt{3}}\ \ ;\ \ \ \
A_{3,4}=\frac{-\sigma_{x}+\sigma_{y}+\sigma_{z}}{\sqrt{3}}$ (42)

Note that employing the above-mentioned observables, we find that the quantum
optimal value $\left(\mathcal{B}_{3}\right)_{Q}^{opt}=6$ is achieved when two
maximally entangled two-qubit states are shared between Alice-Bob and Bob-
Charlie.

##  IV Asymmetric bilocal network: scenario-II

Figure 3: Asymmetric bilocal network scenario featuring two edge parties
(Alice and Charlie) and a central party Bob. The independent sources $S_{1}$
and $S_{2}$ emit physical systems for Alice-Bob and Charlie-Bob respectively.

Here we consider the central party Bob performs equal number of measurements
(four) as Alice in contrast to the preceding scenario discussed in Sec. III
where number of measurements for Bob and Charlie were considered to be equal.
In this scenario, let us consider the nonlinear bilocal inequality of the
following form

$\mathcal{B}_{3}^{\prime}=\sum_{j=1}^{4}\sqrt{\ \quantity|\
\Big{\langle}\mathcal{\tilde{A}}_{3,j}^{\prime}B_{3,j}\mathcal{\tilde{C}}_{3,j}^{\prime}\Big{\rangle}\
|\ }\leq(\mathcal{B}_{3}^{\prime})_{bl}$ (43)

where $(\mathcal{B}_{3}^{\prime})_{bl}$ is the bilocal bound of
$\mathcal{B}_{3}^{\prime}$ and
$\mathcal{\tilde{A}}_{3,j}^{\prime}=A_{3,j}+A_{3,j+1}$ satisfying
$A_{3,4}=-A_{3,1}$;
$\mathcal{\tilde{C}}_{3,1}^{\prime}=C_{3,1}+C_{3,2}+C_{3,3}$;
$\mathcal{\tilde{C}}_{3,2}^{\prime}=C_{3,1}+C_{3,2}-C_{3,3}$;
$\mathcal{\tilde{C}}_{3,3}^{\prime}=C_{3,1}-C_{3,2}+C_{3,3}$;$\mathcal{\tilde{C}}_{3,4}^{\prime}=-C_{3,1}+C_{3,2}+C_{3,3}$.

We find (see appendix A.1) that the bilocal bound in this scenario is
$(\mathcal{B}_{3}^{\prime})_{bl}=6$. It can be shown that there are suitable
states and observables for which quantum correlations violate the bilocal
bound. In the following, we evaluate the quantum optimal bound of
$\mathcal{B}_{3}^{\prime}$.

The optimal quantum bound of $\mathcal{B}_{3}^{\prime}$ without assuming the
dimension of the system is derived by suitably invoking the SOS approach. The
explicit construction of the SOS method is provided in Appendix (A.2). The
optimal quantum bound is found to be

$\displaystyle\quantity(\mathcal{\mathcal{B}}_{3}^{\prime})_{Q}^{opt}=4\left[3\left(2+\sqrt{2}\right)\right]^{\frac{1}{4}}\approx
7.16$ (44)

It is to be noted here that the obtained optimal quantum bound
$\quantity((\mathcal{\mathcal{B}}_{3}^{\prime})_{Q}^{opt}\approx 7.16)$ in
this scenario is greater than that obtained
$\quantity((\mathcal{\mathcal{B}}_{3})_{Q}^{opt}=6)$ in the earlier
scenario-I. Thus, the bilocal as well as the quantum optimal bound depends on
how the asymmetry invoked in the bilocal scenario. The corresponding state and
observables for which the quantum optimal bound is achieved is given in
Appendix (A.3).

##  V Asymmetric trilocal network: scenario-I

Figure 4: Asymmetric trilocal network scenario featuring three edge parties
(Alice, Charlie, and Diana) and a central party Bob. The independent sources
$S_{1}$, $S_{2}$ and $S_{3}$ emit physical systems for Alice-Bob, Charlie-Bob,
and Diana-Bob respectively.

Here we introduce a variant of asymmetric trilocal network scenario in which
one of the edge party Charlie performs one of three dichotomic measurements
and rest of the edge parties Alice and Diana as well as the central party Bob
perform four dichotomic measurements. In this scenario, let us introduce the
nonlinear trilocal inequality of the following form

$\mathcal{T}_{3}=\sum_{j=1}^{4}|\mathcal{J}_{3,j}|^{\frac{1}{3}}\leq(\mathcal{T}_{3})_{tl}$
(45)

where $\left(\mathcal{T}_{3}\right)_{tl}$ is the local bound of
$\mathcal{T}_{3}$ and the quantity
$\mathcal{J}_{3,j}=\Big{\langle}\mathscr{\tilde{A}}_{3,j}B_{3,j}\mathscr{\tilde{C}}_{3,j}\mathscr{\tilde{D}}_{3,j}\Big{\rangle}$.
We define the quantities $\mathscr{\tilde{A}}_{3,j}$,
$\mathscr{\tilde{C}}_{3,j}$ and $\mathscr{\tilde{D}}_{3,j}$ as follows

$\displaystyle\mathscr{\tilde{A}}_{3,j}=A_{3,j}+A_{3,j+1}\ \ \text{with}\
A_{3,5}=-A_{3,1}$
$\displaystyle\mathscr{\tilde{C}}_{3,1}=C_{3,1}+C_{3,2}+C_{3,3}\ ;\ \
\mathscr{\tilde{C}}_{3,2}=C_{3,1}+C_{3,2}-C_{3,3}$
$\displaystyle\mathscr{\tilde{C}}_{3,3}=C_{3,1}-C_{3,2}+C_{3,3}\ ;\ \
\mathscr{\tilde{C}}_{3,4}=-C_{3,1}+C_{3,2}+C_{3,3}$
$\displaystyle\mathscr{\tilde{D}}_{3,1}=D_{3,1}+D_{3,2}+D_{3,3}+D_{3,4}$
$\displaystyle\mathscr{\tilde{D}}_{3,2}=D_{3,1}+D_{3,2}+D_{3,3}-D_{3,4}$
$\displaystyle\mathscr{\tilde{D}}_{3,3}=D_{3,1}+D_{3,2}-D_{3,3}-D_{3,4}$
$\displaystyle\mathscr{\tilde{D}}_{3,4}=D_{3,1}-D_{3,2}-D_{3,3}-D_{3,4}$ (46)

Now, invoking the reproducibility condition given by Eq. (II) it follows that

$\displaystyle\mathcal{J}_{3,1}$ $\displaystyle=$ $\displaystyle\iiint
d\lambda_{1}d\lambda_{2}d\lambda_{3}\
\mu(\lambda_{1})\mu(\lambda_{2})\mu(\lambda_{3})\ \quantity[\langle
A_{3,1}\rangle_{\lambda_{1}}+\langle A_{3,2}\rangle_{\lambda_{1}}]$ (47)
$\displaystyle\ \ \ \ \ \ \times\ \ \quantity[\langle
B_{3,1}\rangle_{\lambda_{1},\lambda_{2},\lambda_{3}}]\ \ \quantity[\langle
C_{3,1}\rangle_{\lambda_{2}}+\langle C_{3,2}\rangle_{\lambda_{2}}+\langle
C_{3,3}\rangle_{\lambda_{2}}]$ $\displaystyle\ \ \ \ \ \ \times\ \
\quantity[\langle D_{3,1}\rangle_{\lambda_{3}}+\langle
D_{3,2}\rangle_{\lambda_{3}}+\langle D_{3,3}\rangle_{\lambda_{3}}+\langle
D_{3,4}\rangle_{\lambda_{3}}]$

Since $|\langle B_{3,1}\rangle_{\lambda_{1},\lambda_{2},\lambda_{3}}|\leq 1$,
we obtain

$\displaystyle\quantity|\mathcal{J}_{3,1}|$ $\displaystyle=$
$\displaystyle\iiint d\lambda_{1}d\lambda_{2}d\lambda_{3}\
\mu(\lambda_{1})\mu(\lambda_{2})\mu(\lambda_{3})\ \quantity|\langle
A_{3,1}\rangle_{\lambda_{1}}+\langle A_{3,2}\rangle_{\lambda_{1}}|$ (48)
$\displaystyle\ \ \ \ \ \ \times\ \ \quantity|\langle
C_{3,1}\rangle_{\lambda_{2}}+\langle C_{3,2}\rangle_{\lambda_{2}}+\langle
C_{3,3}\rangle_{\lambda_{2}}|$ $\displaystyle\ \ \ \ \ \ \times\ \
\quantity|\langle D_{3,1}\rangle_{\lambda_{3}}+\langle
D_{3,2}\rangle_{\lambda_{3}}+\langle D_{3,3}+\langle
D_{3,4}\rangle_{\lambda_{3}}|$

The terms $|\mathcal{J}_{3,2}|$, $|\mathcal{J}_{3,3}|$, and
$|\mathcal{J}_{3,4}|$ can also be written in a similar manner as Eq. (48).
Then, we obtain the following

$(\mathcal{T}_{3})_{tl}\leq\quantity(\iiint
d\lambda_{1}d\lambda_{2}d\lambda_{3}\
\mu(\lambda_{1})\mu(\lambda_{2})\mu(\lambda_{3})\
\eta_{1}\eta_{2}\eta_{3})^{\frac{1}{3}}$ (49)

where
$\eta_{1}=\big{[}|\langle{A_{3,1}}\rangle_{\lambda_{1}}+\langle{A_{3,2}}\rangle_{\lambda_{1}}|+|\langle{A_{3,2}}\rangle_{\lambda_{1}}+\langle{A_{3,3}}\rangle_{\lambda_{1}}|+|\langle{A_{3,3}}\rangle_{\lambda_{1}}+\langle{A_{3,4}}\rangle_{\lambda_{1}}|+|\langle{A_{3,4}}\rangle_{\lambda_{1}}-\langle{A_{3,1}}\rangle_{\lambda_{1}}|\big{]}$,
$\eta_{2}=\big{[}|\langle{C_{3,1}}\rangle_{\lambda_{2}}+\langle{C_{3,2}}\rangle_{\lambda_{2}}+\langle{C_{3,3}}\rangle_{\lambda_{2}}|+|\langle{C_{3,1}}\rangle_{\lambda_{2}}+\langle{C_{3,2}}\rangle_{\lambda_{2}}-\langle{C_{3,3}}\rangle_{\lambda_{2}}|+|\langle{C_{3,1}}\rangle_{\lambda_{2}}-\langle{C_{3,2}}\rangle_{\lambda_{2}}+\langle{C_{3,3}}\rangle_{\lambda_{2}}|+|-\langle{C_{3,1}}\rangle_{\lambda_{2}}+\langle{C_{3,2}}\rangle_{\lambda_{2}}+\langle{C_{3,3}}\rangle_{\lambda_{2}}|\big{]}$,
and $\eta_{3}=\big{[}|\langle D_{3,1}\rangle_{\lambda_{3}}+\langle
D_{3,2}\rangle_{\lambda_{3}}+\langle D_{3,3}\rangle_{\lambda_{3}}+\langle
D_{3,4}\rangle_{\lambda_{3}}|+|\langle D_{3,1}\rangle_{\lambda_{3}}+\langle
D_{3,2}\rangle_{\lambda_{3}}+\langle D_{3,3}\rangle_{\lambda_{3}}-\langle
D_{3,4}\rangle_{\lambda_{3}}|+|\langle D_{3,1}\rangle_{\lambda_{3}}+\langle
D_{3,2}\rangle_{\lambda_{3}}-\langle D_{3,3}\rangle_{\lambda_{3}}-\langle
D_{3,4}\rangle_{\lambda_{3}}|+|\langle D_{3,1}\rangle_{\lambda_{3}}-\langle
D_{3,2}\rangle_{\lambda_{3}}-\langle D_{3,3}\rangle_{\lambda_{3}}-\langle
D_{3,4}\rangle_{\lambda_{3}}|\big{]}$.

Since all the observables are dichotomic with eigenvalues $\pm 1$, it is
straightforward to derive that $\eta_{1}\leq 6$, $\eta_{2}\leq 6$, and
$\eta_{3}\leq 8$. Therefore, from Eq. (49), integrating over $\lambda_{1}$,
$\lambda_{2}$ and $\lambda_{3}$ we get

$(\mathcal{T}_{3})_{tl}\leq 2\left(6\right)^{\frac{2}{3}}\approx 6.60$ (50)

i.e., the trilocal bound
$(\mathcal{T}_{3})_{tl}=2\left(6\right)^{\frac{2}{3}}\approx 6.60$. There are
suitable states and observables for which the trilocal bound can be violated
in quantum theory. Now, in the following, we evaluate the quantum optimal
bound of $\mathcal{T}_{3}$.

###  V.1 Optimal quantum bound of the asymmetric trilocality inequality for
scenario-I

Here, by invoking the SOS approach, we evaluate the optimal quantum bound of
$\mathcal{T}_{3}$ without assuming the dimension of the system. We first show
that there exists a positive semi-definite operator $\Gamma_{3}$ satisfying
$\langle\Gamma_{3}\rangle=\zeta_{3}-\left(\mathcal{T}_{3}\right)_{Q}$. The
existence of such operator can be proved by considering a set of operators
$L_{3,j}\ \forall j\in\\{1,2,3,4\\}$ such that

$\langle\Gamma_{3}\rangle=\sum\limits_{j=1}^{4}\dfrac{\left(\omega_{3,j}\right)^{\frac{1}{3}}}{2}\quantity|L_{3,j}\ket{\psi}|^{2}$
(51)

where $\omega_{3,j}\geq 0$ and
$\omega_{3,j}=\omega^{A}_{3,j}\cdot\omega^{C}_{3,j}\cdot\omega^{D}_{3,j}$. We
choose $L_{3,j}$ and $\omega_{3,j}$ as the following way

$\displaystyle\quantity|L_{3,j}\ket{\psi}|=\quantity|\left(\frac{\mathscr{\tilde{A}}_{3,j}}{\omega_{3,j}^{A}}\otimes\frac{\mathscr{\tilde{C}}_{3,j}}{\omega_{3,j}^{C}}\otimes\frac{\mathscr{\tilde{D}}_{3,j}}{\omega_{3,j}^{D}}\right)\ket{\psi}|^{\frac{1}{3}}-\quantity|B_{3,j}\ket{\psi}|^{\frac{1}{3}}$
(52)
$\displaystyle\omega_{3,j}^{A}=||\mathscr{\tilde{A}}_{3,j}\ket{\psi}||_{2}\ ;\
\ \omega_{3,j}^{C}=||\mathscr{\tilde{C}}_{3,j}\ket{\psi}||_{2}\ ;\ \
\omega_{3,j}^{D}=||\mathscr{\tilde{D}}_{3,j}\ket{\psi}||_{2}$

Now, putting $\quantity|L_{3,j}\ket{\psi}|$ from Eq. (52) into the Eq. (51),
after a simple algebraic evaluation, we obtain
$\langle\Gamma_{3}\rangle=\sum\limits_{j=1}^{3}\quantity(\omega_{3,j})^{\frac{1}{3}}-(\mathcal{T}_{3})_{Q}$.
Then, it follows that the quantum optimal value corresponds
$\langle\Gamma_{3}\rangle=0$. Therefore,

$\displaystyle\left(\mathcal{T}_{3}\right)_{Q}^{opt}$ $\displaystyle=$
$\displaystyle\sum\limits_{j=1}^{4}\quantity(\omega_{3,j}^{A}\cdot\omega_{3,j}^{C}\cdot\omega_{3,j}^{D})^{\frac{1}{3}}$
(54)

Such optimal quantum value will occur under the following optimisation
condition

$L_{3,j}\ \ket{\psi}=0\ \ \forall j\in\\{1,2,3,4\\}$ (55)

Hence, from Eq. (54) and by using the inequalities given in footnotes (3) and
(4), we obtain the quantum optimal value as follows

$\displaystyle\left(\mathcal{T}_{3}\right)_{Q}^{opt}=2\
\quantity[\prod_{k=A,C,D}\quantity(\max\sum_{j=1}^{4}\quantity(\omega^{k}_{3,j})^{2})]^{\frac{1}{6}}$
(56)

Note that the quantities $\sum_{j=1}^{3}\left(\omega^{A}_{3,j}\right)^{2}\leq
4(2+\sqrt{2})$ and $\sum_{j=1}^{3}\left(\omega^{C}_{3,j}\right)^{2}\leq 12$
have already derived in Eqs. (A.2) and (A.2). Now, we evaluate the quantity
$\sum_{j=1}^{4}\left(\omega^{D}_{3,j}\right)^{2}$ from Eq. (V.1) as follows

$\displaystyle\sum\limits_{j=1}^{4}\quantity(\omega^{D}_{3,j})^{2}$
$\displaystyle=$
$\displaystyle\bra{\psi}(16+2\big{(}\\{D_{3,1},(D_{3,2}-D_{3,4})\\}$ (57)
$\displaystyle\ \ \ \ +\\{D_{3,3},(D_{3,2}+D_{3,4})\\}\big{)}\ket{\psi}$

Since the quantities $D_{3,1}$ and $D_{3,3}$ appeared independently with
$\\{D_{3,1},(D_{3,2}-D_{3,4})\\}$ and $\\{D_{3,3},(D_{3,2}+D_{3,4})\\}$
respectively, without loss of generality, we chose
$D_{3,1}=(D_{3,2}-D_{3,4})/\nu_{1}$ and $D_{3,3}=(D_{3,2}+D_{3,4})/\nu_{1}$,
where $\nu_{1}=||(D_{3,2}-D_{3,4})\ket{\psi}||_{2}$ and
$\nu_{2}=||(D_{3,2}+D_{3,4})\ket{\psi}||_{2}$. Therefore, the above Eq. (57)
reduces to

$\sum\limits_{j=1}^{4}\bigg{(}\omega^{D}_{4,j}\bigg{)}^{2}=16+4\left[\sqrt{2-\langle\\{D_{3,2},D_{3,4}\\}\rangle}+\sqrt{2-\langle\\{D_{3,2},D_{3,4}\\}\rangle}\right]$
(58)

The maximum value of
$\sum\limits_{j=1}^{4}\quantity(\omega^{D}_{4,j})^{2}=8(2+\sqrt{2})$ is then
achieved when $\\{D_{3,2},D_{3,4}\\}=0$ which automatically implies
$\nu_{1}=\nu_{2}=\sqrt{2}$. Therefore, for the optimal quantum violation, the
linear constraints on Diana’s observables are given as

$\displaystyle D_{3,4}-\sqrt{2}\ D_{3,1}-D_{3,2}=D_{3,2}+D_{3,4}-\sqrt{2}\
D_{3,3}=0$ (59)

Therefore, we obtain the quantum optimal value as given by

$\left(\mathcal{T}_{3}\right)^{opt}_{Q}=4\
\quantity[2\sqrt{3}\quantity(1+\sqrt{2})]^{\frac{1}{3}}\approx 8.12$ (60)

It is important to remark here that the optimal quantum value
$\left(\mathcal{T}_{3}\right)_{Q}^{opt}=4\
\quantity[2\sqrt{3}\quantity(1+\sqrt{2})]^{\frac{1}{3}}\approx 8.12$ is
evaluated without specifying the dimension of both the system and observables.
The states, and the observables for which the optimal value will be achieved
are given in the following.

###  V.2 The state and observables for the optimal quantum violation of
$(\mathcal{T})_{3}$

We further obtain relationships between the observables of all the parties in
terms of the anti-commuting relations for achieving the optimal quantum
violation. The anti-commutations relations for Alice’s and Charlie’s
observables are evaluated from Eqs. (103) and (105). The anti-commutations
relations for Diana’s observables are evaluated from Eq. (59). All the
relations are given as follows

$\displaystyle\\{A_{3,1},A_{3,2}\\}=\\{A_{3,2},A_{3,3}\\}=\\{A_{3,3},A_{3,4}\\}=-\\{A_{3,1},A_{3,4}\\}=\sqrt{2}\
\mathbb{I}_{d}$ $\displaystyle\\{A_{3,1},A_{3,3}\\}=\\{A_{3,2},A_{3,4}\\}=0;\
\\{D_{3,1},D_{3,3}\\}=\\{D_{3,2},D_{3,4}\\}=0$
$\displaystyle\\{D_{3,1},D_{3,2}\\}=\\{D_{3,2},D_{3,3}\\}=\\{D_{3,3},D_{3,4}\\}=-\\{D_{3,1},D_{3,4}\\}=\sqrt{2}\
\mathbb{I}_{d}$
$\displaystyle\\{C_{3,1},C_{3,2}\\}=\\{C_{3,2},C_{3,3}\\}=-\\{C_{3,1},C_{3,3}\\}=\mathbb{I}_{d}$
(61)

By using the above relations between the observables given by Eq. (V.2) on the
observables, one can always construct a set of observables for Alice and
Charlie in the Hilbert space dimension $\mathcal{H}^{d},\ \ \forall d\geq 2$.

Next, we recall the optimization condition obtained in the SOS method from the
Eq. (55) to find the constraints on Bob’s observable. The specific condition
$L_{3,j}\ket{\psi}=0\ \ \forall j\in\\{1,2,3,4\\}$ implies the following

$\displaystyle B_{3,j}$ $\displaystyle=$
$\displaystyle\frac{\mathscr{\tilde{A}}_{3,j}}{\omega_{3,j}^{A}}\otimes\frac{\mathscr{\tilde{C}}_{3,j}}{\omega_{3,j}^{C}}\otimes\frac{\mathscr{\tilde{D}}_{3,j}}{\omega_{3,j}^{C}}$
(62)

We explicitly construct a set of observables of Alice, Charlie, and Diana for
the Hilbert space dimension $\mathcal{H}^{2}$ are the following.

$\displaystyle A_{3,1}$ $\displaystyle=$ $\displaystyle r\
\sigma_{x}+\sqrt{1-r^{2}}\sigma_{z}\ \ ;\ \ A_{3,2}=t\
\sigma_{x}+\sqrt{1-t^{2}}\sigma_{z}$ $\displaystyle\ A_{3,3}$ $\displaystyle=$
$\displaystyle t\ \sigma_{x}-\sqrt{1-t^{2}}\ \sigma_{z}\ \ ;\ \ A_{3,4}=r\
\sigma_{x}-\sqrt{1-r^{2}}\ \sigma_{z}\ \ $ $\displaystyle D_{3,1}$
$\displaystyle=$ $\displaystyle-t\ \sigma_{x}+\sqrt{1-t^{2}}\sigma_{z}\ \ ;\ \
D_{3,2}=-t\ \sigma_{x}-\sqrt{1-t^{2}}\sigma_{z}$ $\displaystyle\ D_{3,3}$
$\displaystyle=$ $\displaystyle-r\ \sigma_{x}-\sqrt{1-r^{2}}\ \sigma_{z}\ \ ;\
\ D_{3,4}=r\ \sigma_{x}-\sqrt{1-r^{2}}\ \sigma_{z}\ \ $ $\displaystyle
C_{3,1}$ $\displaystyle=$ $\displaystyle\sigma_{x}\ \ ;\ \ C_{3,2}=\sigma_{y}\
\ ;\ \ C_{3,3}=\sigma_{z}$ (63)

where $r=\frac{1}{2}\sqrt{2-\sqrt{2}}$ and $t=\frac{1}{2}\sqrt{2+\sqrt{2}}$.
Note that employing the above-mentioned observables, we find that the quantum
optimal value $\left(\mathcal{T}_{3}\right)_{Q}^{opt}\approx 7.23$ is achieved
when three maximally entangled two-qubit states are shared between Alice-Bob,
Charlie-Bob and Diana-Bob.

##  VI Asymmetric trilocal network: scenario-II

Figure 5: Asymmetric trilocal network scenario featuring three edge parties
(Alice, Charlie, and Diana) and a central party Bob. The independent sources
$S_{1}$, $S_{2}$ and $S_{3}$ emit physical systems for Alice-Bob, Charlie-Bob,
and Diana-Bob respectively.

Here we present another asymmetric trilocal network scenario in which one of
the edge party Alice performs one of four dichotomic measurements and rest of
the edge parties Charlie and Diana as well as the central party Bob perform
three dichotomic measurements. In this scenario, let us consider the nonlinear
trilocal inequality of the following form

$\mathcal{T}_{3}^{\prime}=\sum_{j=1}^{3}\quantity|\
\Big{\langle}\mathscr{\tilde{A}}_{3,j}^{\prime}B_{3,j}\mathscr{\tilde{C}}_{3,j}^{\prime}\mathscr{\tilde{D}}_{3,j}^{\prime}\Big{\rangle}\
|^{\frac{1}{3}}\ \leq\ (\mathcal{T}_{3}^{\prime})_{tl}$ (64)

where $(\mathcal{T}_{3}^{\prime})_{tl}$ is the local bound of
$\mathcal{T}_{3}^{\prime}$ and the quantities
$\mathscr{\tilde{A}}_{3,1}^{\prime}=A_{3,1}+A_{3,2}+A_{3,3}-A_{3,4}$;
$\mathscr{\tilde{D}}_{3,1}^{\prime}=D_{3,1}+D_{3,2}+D_{3,3}$;
$\mathscr{\tilde{A}}_{3,2}^{\prime}=A_{3,1}+A_{3,2}-A_{3,3}+A_{3,4}$;
$\mathscr{\tilde{D}}_{3,2}^{\prime}=D_{3,1}-D_{3,2}+D_{3,3}$;
$\mathscr{\tilde{A}}_{3,3}^{\prime}=A_{3,1}-A_{3,2}+A_{3,3}+A_{3,4}$;
$\mathscr{\tilde{D}}_{3,3}^{\prime}=D_{3,1}-D_{3,2}-D_{3,3}$;
$\mathscr{\tilde{C}}_{3,1}^{\prime}=C_{3,j}+C_{3,j+1}$ satisfying
$C_{3,4}=-C_{3,1}$. We find (see appendix B.1) that the trilocal bound in this
scenario is given by

$(\mathcal{T}_{3}^{\prime})_{tl}=2(15)^{\frac{1}{3}}\approx 4.93$ (65)

It can be shown that there are suitable states and observables for which
quantum correlations violate the trilocal bound. In the following, we evaluate
the quantum optimal bound of $\mathcal{T}_{3}^{\prime}$.

The optimal quantum bound of $\mathcal{T}_{3}^{\prime}$ without assuming the
dimension of the system is derived by suitably invoking the SOS approach as
discussed earlier. The explicit construction of the SOS method is provided in
Appendix (B.2). The optimal quantum bound is found to be

$\displaystyle\quantity(\mathcal{\mathcal{T}}_{3}^{\prime})_{Q}^{opt}=6$ (66)

It is to be noted here that the obtained optimal quantum bound
$\quantity((\mathcal{\mathcal{T}}_{3}^{\prime})_{Q}^{opt}=6)$ in this scenario
is lesser than that obtained
$\quantity((\mathcal{\mathcal{T}}_{3})_{Q}^{opt}\approx 7.23)$ in the earlier
scenario-I. Thus, the trilocal as well as the quantum optimal bound depends on
how the asymmetry invoked in the trilocal scenario.

The corresponding state and observables for which the quantum optimal bound is
achieved is given in Appendix (B.3).

##  VII Asymmetric Network for arbitrary $n$

Let us now provide a sketch of the result in an asymmetric bilocal network
scenario for arbitrary $n$. Alice performs one of $m_{A}=2^{n-1}$ dichotomic
measurements and both the central party (Bob) and Charlie perform one of
$m_{B}=n$ and $m_{c}=n$ dichotomic measurements respectively. In this
scenario, we consider the non-linear asymmetric bilocality inequality for
arbitrary input $n$ of the following form

$\mathcal{B}_{n}=\sum_{j=1}^{n}\sqrt{|\mathcal{I}_{n,j}|}\ \leq\
\left(\mathcal{B}_{n}\right)_{bl}$ (67)

where, $\left(\mathcal{B}_{n}\right)_{bl}$ is the bilocal bound of
$\mathcal{B}_{n}$ and
$\mathcal{I}_{n,j}=\Big{\langle}\tilde{\mathcal{A}}_{n,j}\otimes
B_{n,j}\otimes\tilde{\mathcal{C}}_{n,j}\Big{\rangle}$ with
$\tilde{\mathcal{A}}_{n,j}$ and $\tilde{\mathcal{C}}_{n,j}$ are unnormalised
observables given as follows

$\displaystyle\tilde{\mathcal{A}}_{n,j}=\sum\limits_{x=1}^{2^{n-1}}(-1)^{y^{x}_{j}}A_{n,x}\
;\ \ \tilde{\mathcal{C}}_{n,j}=\sum_{z=1}^{n}\left(C_{n,z}+C_{n,z+1}\right)$
(68)

with $C_{n,n+1}=-C_{n,1}$ and $y^{x}_{j}\in\\{0,1\\}\ \forall j\in[n]$. For
our purpose, by using the encoding scheme which was earlier introduced in the
context of Random Access Codes (RACs) [46, 47, 43, 44, 45] protocol, we fix
the values of $y^{x}_{j}$ in the following way. Let us consider a random
variable $y^{\alpha}\in\\{0,1\\}^{n}$ with $\alpha\in\\{1,2...2^{n}\\}$. Each
element of the bit string can be written as
$y^{\alpha}=y^{\alpha}_{j=1}y^{\alpha}_{j=2}y^{\alpha}_{j=3}....y^{\alpha}_{j=n}$.
For example, if $y^{\alpha}=011...00$ then $y^{{\alpha}}_{j=1}=0$,
$y^{{\alpha}}_{j=2}=1$, $y^{{\alpha}}_{j=3}=1$ and so on. We denote the
$n$-bit binary strings as $y^{x}$. Here we consider the bit strings such that
for any two $x$ and $x^{\prime}$, $y^{x}\oplus_{2}y^{x^{\prime}}=11\cdots 1$.
Clearly, we have $x\in\\{1,2...2^{n-1}\\}$ constituting the inputs for Alice.
If $x=1$, we get all the first bit of each bit string $y_{j}$ for every
$j\in\\{1,2\cdots n\\}$.

Now it follows from the reproducibility condition given by Eq. (II) that

$\mathcal{I}_{n,1}=\iint\mu(\lambda_{1})\mu(\lambda_{2})d\lambda_{1}d\lambda_{2}\langle\tilde{\mathcal{A}}_{n,1}\rangle_{\lambda_{1}}\langle
B_{n,1}\rangle_{\lambda_{1},\lambda_{2}}\langle\tilde{\mathcal{C}}_{n,1}\rangle_{\lambda_{1}}$
(69)

Since $|\langle B_{n,1}\rangle_{\lambda_{1},\lambda_{2}}|\leq 1$, we obtain
from the above Eq. (69)

$\displaystyle|\mathcal{I}_{n,j}|$ $\displaystyle\leq$
$\displaystyle\iint\mu(\lambda_{1})\mu(\lambda_{2})d\lambda_{1}d\lambda_{2}\ \
\Big{|}\big{\langle}\tilde{\mathcal{A}}_{n,j}\big{\rangle}_{\lambda_{1}}\Big{|}\
\ \Big{|}\langle\tilde{\mathcal{C}}_{n,j}\rangle_{\lambda_{2}}\Big{|}$ (70)
$\displaystyle\ \ \ \ ;\forall j\in\\{1,2,...,n\\}$

By putting the values of Eq. (70) in Eq. (67) and, then by applying the
property of the inequality given in footnote 3, we obtain

$\left(\mathcal{B}_{n}\right)_{bl}\leq\sqrt{\int d\lambda_{1}\mu(\lambda_{1})\
\big{|}\big{\langle}\tilde{\mathcal{A}}_{n,j}\big{\rangle}_{\lambda_{1}}\big{|}}\times\sqrt{\int
d\lambda_{2}\mu(\lambda_{2})\
\big{|}\big{\langle}\tilde{\mathcal{C}}_{n,j}\big{\rangle}_{\lambda_{2}}\big{|}}$
(71)

Note that all the observables $A_{n,x}$ and $C_{n,z}$ are dichotomic with
eigenvalues $\pm 1$ . In [44, 45] it was derived that the value of
$\big{|}\big{\langle}\tilde{\mathcal{A}}_{n}\big{\rangle}_{\lambda_{1}}\big{|}=n\
\binom{n-1}{\lfloor\frac{n-1}{2}\rfloor}$ and
$\big{|}\big{\langle}\tilde{\mathcal{C}}_{n}\big{\rangle}_{\lambda_{2}}\big{|}=(2n-2)$.
Hence, the bilocal bound is given by

$\left(\mathcal{B}_{n}\right)_{bl}=\sqrt{2n(n-1)\
\binom{n-1}{\lfloor\frac{n-1}{2}\rfloor}}$ (72)

where $\lfloor x\rfloor$ denotes the largest integer smaller or equal to $x$.

We derive the optimal quantum value of $\left(\mathcal{B}_{n}\right)_{Q}$ of
the bilocality inequality proposed in Eq. (67) by again invoking the SOS
approach. We consider a positive semi-definite operator
$\langle\gamma_{n}\rangle\geq 0$, that can be expressed as
$\langle\gamma_{n}\rangle=\beta_{n}\mathbb{I}-\left(\mathcal{B}_{n}\right)_{Q}$.
Where $\beta_{n}$ is the optimal value that can be obtained when
$\langle\gamma_{n}\rangle$ is equal to zero. This can be proved by considering
a set of positive operators $M_{n,j}$ which is polynomial functions of
$A_{n,x}$, $C_{n,z}$, $B_{n,j}$ such that

$\langle\gamma_{n}\rangle=\sum_{j=1}^{n}\dfrac{(\omega_{n,j})^{\frac{1}{2}}}{2}\langle\psi|(M_{n,j})^{\dagger}(M_{n,j})|\psi\rangle$
(73)

where $\omega_{n,j}$ is suitable positive numbers and
$\omega_{n,j}=\quantity(\omega^{A}_{n,j})\ \quantity(\omega^{C}_{n,j})$. The
optimal quantum value of $\left(\mathcal{B}_{n}\right)_{Q}$ is obtained if
$\langle\gamma_{n}\rangle=0$, implying that $M_{n,j}|\psi\rangle=0$. We choose
a suitable set of positive operators $M_{n,j}$ (with $j\in[n]$), such that

$\displaystyle M_{n,j}|\psi\rangle$ $\displaystyle=$
$\displaystyle\sqrt{\bigg{|}\left(\mathcal{A}_{n,j}\otimes\mathcal{C}_{n,j}\right)|\psi\rangle\bigg{|}}-\sqrt{\left|B_{n,j}|\psi\rangle\right|}$
(74)

where $\mathcal{A}_{n,j}=\frac{\tilde{\mathcal{A}}_{n,j}}{\omega^{A}_{n,j}}$
and $\mathcal{C}_{n,j}=\frac{\tilde{\mathcal{C}}_{n,j}}{\omega^{C}_{n,j}}$
with $\omega_{n,j}^{A}=||\tilde{\mathcal{A}}_{n,j}|\psi\rangle||_{2}$ and
$\omega_{n,j}^{C}=||\tilde{\mathcal{C}}_{n,j}|\psi\rangle||_{2}$. By inserting
Eq. (74) in Eq. (73), we obtain
$\langle\gamma_{n}\rangle=-\left(\mathcal{B}_{n}\right)_{Q}+\sum\limits_{j=1}^{n}\left(\omega_{n,j}\right)^{\frac{1}{2}}$.
The optimal value of $\left(\mathcal{B}_{n}\right)_{Q}$ is obtained if
$\langle\gamma_{n}\rangle=0$. Therefore the optimal value is given as follows

$\displaystyle\left(\mathcal{B}_{n}\right)_{Q}^{opt}=\max\left(\sum\limits_{j=1}^{n}(\omega_{n,j})^{\frac{1}{2}}\right)$
(75)

Now, by using the inequality given in footnote 3 along with the convex
inequality given in the footnote 4, we obtain

$\displaystyle\left(\mathcal{B}_{n}\right)_{Q}^{opt}=\max\left[\prod_{k=A,C}\left(n\sum_{j=1}^{n}\left(\omega^{k}_{n,j}\right)^{2}\right)^{\frac{1}{2}}\right]^{\frac{1}{2}}$
(76)

We can always evaluate each term $(\omega_{n,j}^{A})^{2}$ and
$(\omega_{n,j}^{C})^{2}$ by following the procedure discussed for $n=3$ case.
However, such evaluation will take rigorous algebraic treatment which we are
skipping here. Thus, instead of following the same path, in the following, we
present an argument that leads to the optimal quantum value of
$(\mathcal{B}_{n})_{Q}$.

To begin with, let us revisit the expression $\mathcal{B}_{n}$ given by Eq.
(67). Since all $\mathcal{I}_{n,j}$ are real numbers and positive by
construction, we can always invoke the convex inequality (as mentioned in the
footnote 4) to obtain the following

$(\mathcal{B}_{n})_{Q}\leq\left(n\sum\limits_{j=1}^{n}\mathcal{I}_{n,j}\right)^{\frac{1}{2}}$
(77)

Then the quantity $\sum\limits_{j=1}^{n}\mathcal{I}_{n,j}$ is evaluated from
the Eq. (III).

$\sum\limits_{j=1}^{n}\mathcal{I}_{n,j}=\left\langle\sum\limits_{j=1}^{n}\left(\tilde{\mathcal{A}}_{n,j}\otimes
B_{n,j}\otimes\tilde{\mathcal{C}}_{n,j}\right)\right\rangle$ (78)

Now, by noting the optimization condition $M_{n,j}|\psi\rangle=0$ form the SOS
method, we write

$B_{n,j}=\mathcal{A}_{n,j}\otimes\mathcal{C}_{n,j}\ \ \ \forall
j\in\\{1,2,...,n\\}$ (79)

Hence, from Eq. (79), we conclude that for achieving the quantum optimal
value, it is sufficient to assume that Bob measures his system on a product
basis. Thus, Bob’s observable are given by $B_{n,j}=B_{n,j}^{A}\otimes
B_{n,j}^{C}$, where $B_{n,j}^{A}$ and $B_{n,j}^{C}$ are normalised observable.
The ability to express Bob’s observable in such a way buoyed up the fact that
for the optimal value, the quantity $\sum\limits_{j=1}^{n}\mathcal{I}_{n,j}$
can be expressed as follows

$\max\sum\limits_{j=1}^{n}\mathcal{I}_{n,j}=\quantity(\mathcal{S}_{n})^{opt}_{Q}\
\quantity(\mathcal{L}_{n})^{opt}_{Q}$ (80)

where $\mathcal{S}_{n}$ is the Bell functional proposed in [47] and
$\mathcal{L}_{n}$ is the $n$-settings Chain-Bell functional [48]. Note that
the optimal quantum values of $\mathcal{S}_{n}$ and $\mathcal{L}_{n}$ have
already been derived as $\left(\mathcal{S}\right)_{Q}^{opt}=2^{n-1}\sqrt{n}$
and $\left(\mathcal{L}_{n}\right)_{Q}^{opt}=2n\cos\frac{\pi}{2n}$ respectively
[47, 49]. By combining these results, it is straightforward to obtain the
quantum optimal bound of $(\mathcal{B}_{n})_{Q}$.

$\displaystyle\quantity(\mathcal{B}_{n})_{Q}^{opt}\ =\ \quantity(2^{n}\
n^{\frac{3}{2}}\ \cos\frac{\pi}{2n})^{\frac{1}{2}}$ (81)

which violates the local bound $\left(\mathcal{B}_{n}\right)$ for any
arbitrary $n$.

###  VII.1 The state and observables for the optimal quantum violation of
$(\mathcal{B})_{n}$

Note that the optimal quantum value $\left(\mathcal{B}_{n}\right)_{Q}^{opt}$
is evaluated without specifying the dimension of both the system and
observables. Importantly, from the argument presented in the preceding
section, the optimal value is achieved when both the quantity
$\mathcal{S}_{n}$ and $\mathcal{L}_{n}$ are optimized simultaneously. It has
earlier been shown [47] that optimal of $\mathcal{B}_{n}$ implies the
observables $B_{n,j}^{A}$ are mutually anti-commuting, i.e.,
$\\{B_{n,j}^{A},B_{n,i}^{A}\\}=0\ \forall i,j$.

Thus, it is crucial to remark that to achieve the optimal quantum violation of
the asymmetric bilocality inequality, there should be $n$ mutually anti-
commuting operators. Further it follows that
$\mathcal{A}_{n,j}\in\mathcal{H}^{d}_{A}$ with necessarily
$(d_{A})_{min}=2^{\lfloor n/2\rfloor}$. Thus, the Bob’s observables
$B_{n,j}=\mathcal{A}_{n,j}\otimes\mathcal{C}_{n,j}$ must belong to
$\mathcal{H}^{d}_{B}$ with necessarily $(d_{B})_{min}>2^{\lfloor n/2\rfloor}$.
We can also conclude that the optimal value cannot be achieved if Alice-Bob
and Bob-Charlie share a single copy of the maximally entangled two-qubit
state.

Hence, taking a cue from the preceding discussions, we find that if there
should be at least $N=\lfloor n/2\rfloor$ copies of maximally entangled two-
qubit states between Alice-Bob and at least a single copy of maximally
entangled two-qubit state between Bob-Charlie, then the quantum optimal value
$\left(\mathcal{B}_{n}\right)_{Q}^{opt}$ will achieve for the observables in
total dimension $d_{min}=2^{N}\times 2^{N+1}\times 2=4^{N+1}$.

Now, following a similar argument, it is straightforward to obtain the bilocal
as well as the quantum optimal bound of asymmetric bilocality inequality for
Scenario-2 where Alice and Bob perform $2^{n-1}$ number of measurements and
Charlie performs $n$ number of measurements. In this scenario, the bilocal
bound of the $n$-settings asymmetric bilocal inequality is given by

$\mathcal{B}_{n}^{\prime}=\sum\limits_{j=1}^{2^{n-1}}\sqrt{\quantity|\Big{\langle}\tilde{\mathcal{A}}_{n,j}^{\prime}\otimes
B_{n,j}\otimes\tilde{\mathcal{C}}_{n,j}^{\prime}\Big{\rangle}|}\leq\sqrt{2n(2^{n-1}-1)\
\binom{n-1}{\lfloor\frac{n-1}{2}\rfloor}}$ (82)

where
$\tilde{\mathcal{A}}_{n,j}=\sum\limits_{x=1}^{2^{n-1}}\left(C_{n,z}+C_{n,z+1}\right)$
with $A_{n,n+1}=-A_{n,1}$ and
$\tilde{\mathcal{C}}_{n,j}=\sum_{z=1}^{n}(-1)^{y^{z}_{j}}C_{n,z}$. The optimal
quantum bound in this case is given by

$(\mathcal{B}_{n})_{Q}^{opt}\ =\ \quantity(2^{2n-1}\ \sqrt{n}\
\cos\frac{\pi}{2^{n}})^{\frac{1}{2}}$ (83)

Next, for the asymmetric trilocal network scenario-I, Alice, Bob and Diana
perform $2^{n-1}$ measurements and Charlie performs $n$ measurements. In this
scenario, the trilocal bound is given by

$(\mathcal{T}_{n})_{tl}=\quantity[2n\ (2^{n-1}-1)\
\binom{n-1}{\lfloor\frac{n-1}{2}\rfloor}\ \left\lfloor
2^{2n-3}+\frac{1}{2}\right\rfloor]^{\frac{1}{3}}$ (84)

The optimal quantum value can be written in the form
$(\mathcal{T}_{n})_{tl}^{opt}=(\mathcal{L}_{2^{n-1}})^{opt}_{Q}\
(\mathcal{G}_{2^{n-1}})^{opt}_{Q}\ (\mathcal{S}_{n})^{opt}_{Q}$. The Bell
functional $\mathcal{G}_{n}$ is the family of n-settings Bell inequalities
proposed in [50] with the optimal quantum value is given by
$(\mathcal{G}_{n})_{Q}^{opt}=2n\cos\frac{\pi}{2n}/\sin\frac{\pi}{n}$. Thus,
the corresponding optimal quantum bound of $(\mathcal{T}_{n})_{tl}$ is given
by

$\displaystyle(\mathcal{T}_{n})_{Q}^{opt}\ $ $\displaystyle=$ $\displaystyle
2^{n-1}\ \quantity(2\ \sqrt{n}\ \cot\frac{\pi}{2^{n}})^{\frac{1}{3}}$ (85)

Now, for the asymmetric trilocal network scenario-II, the respective trilocal
and optimal quantum bound is given as follows

$(\mathcal{T}_{n}^{\prime})_{tl}=\quantity(2n(n-1)\binom{n-1}{\lfloor\frac{n-1}{2}\rfloor}\left\lfloor\frac{n^{2}+1}{2}\right\rfloor)^{\frac{1}{3}}$
(86)
$(\mathcal{T}_{n}^{\prime})_{Q}^{opt}=\quantity(2^{n}n^{\frac{5}{2}}\cot\frac{\pi}{2n})^{\frac{1}{3}}$
(87)

The optimal quantum values
$(\mathcal{T}_{n})_{Q}^{opt}>(\mathcal{T}_{n})_{tl}$ and
$(\mathcal{T}_{n}^{\prime})_{Q}^{opt}>(\mathcal{T}_{n}^{\prime})_{tl}$ for any
value of $n$, thereby demonstrating the nonlocality in trilocal network
featuring arbitrary inputs.

##  VIII Resistance to white noise

###  VIII.1 Resistance to white noise for bilocality scenario-I and II

Let us assume that each of both the independent sources $S_{1}$ and $S_{2}$
does not produce maximally entangled two-qubit state, but a mixture of
maximally entangled state with a white noise, known as Werner state [51]. Let
the two sources produce such Werner states with different noise parameters
$v_{1}$ and $v_{2}$. The Werner states between Alice and Bob is
$\rho^{w}_{AB}(v_{1})$ and for Bob and Charlie is $\rho^{w}_{BC}(v_{2})$,
given by
$\rho^{w}_{AB}(v_{k})=v_{k}\ket{\psi}\bra{\psi}+(1-v_{k})\frac{\mathbb{I}}{4}$
with $k\in\\{1,2\\}$ and $\ket{\psi}\bra{\psi}$ is a maximally entangled two-
qubit state and the joint tripartite physical system is given by
$\rho^{w}_{ABC}(v_{1},v_{2})=\rho^{w}_{AB}(v_{1})\otimes\rho^{w}_{BC}(v_{2})$.

For convenience, we first evaluate the robustness of the asymmetric bilocality
scenario-I for arbitrary $n$. Since the optimal quantum violation of
$\mathcal{B}_{n}$ is achieved when $\lfloor n/2\rfloor$ copies of maximally
entangled two-qubit states shared between Alice-Bob and a single copy of
maximally entangled two-qubit state is shared between Bob-Charlie. Thus, we
take the Werner states of the form $\mathcal{W}=\rho_{AB}(v_{1})^{\otimes
N}\otimes\rho_{BC}(v_{2})$ where we take $N=\lfloor n/2\rfloor$. Then, by
invoking the conditions on the observables given in Sec. (VII.1), we obtain

$\quantity(\mathcal{B}_{n})^{\mathcal{W}}_{Q}=\quantity(v_{1}^{N}v_{2})^{\frac{1}{2}}\quantity(2^{n}\
n^{\frac{3}{2}}\ \cos\frac{\pi}{2n})^{\frac{1}{2}}$ (88)

Therefore, in this noisy case if we take all the noise parameter is the same
($v$), the quantum violation will be achieved when

$v^{N+1}>2^{1-n}\ \sqrt{n}\ \quantity(1-\frac{1}{n})\
\binom{n-1}{\lfloor\frac{n-1}{2}\rfloor}\ \sec\frac{\pi}{2n}$ (89)

Note that $n=2,3$ correspond $N=1$. For $n=2$ the critical noise parameter for
each of the Werner state is $v_{c}=1/\sqrt{2}$. This is exactly as to be
expected because this is the critical noise parameter for the Warner state for
violation of the CHSH inequality. On the other hand, for $n=3$ the critical
noise parameter $v_{c}=\sqrt{2/3}\approx 0.82$ which is greater that
$1/\sqrt{2}$. So, there is no advantage has been found over the standard Bell
nonlocality or bilocality scenario for demonstrating the quantum nature of the
noisy maximally entangled bipartite states.

In the asymmetric bilocality scenario-II, for demonstrating nonlocality, the
critical noise parameter is found to be

$v^{N+1}>\sqrt{n}\ 2^{1-n}\ \quantity(1-2^{1-n})\
\binom{n-1}{\lfloor\frac{n-1}{2}\rfloor}\ \sec(\frac{\pi}{2^{n}})$ (90)

For $n=3$, the critical parameter in scenario-II is given by
$v_{c}=\frac{3}{4}\sqrt{3-\frac{3}{\sqrt{2}}}\approx 0.84$. Thus, in the
asymmetric bilocality scenario, scenario-I is more robust against the white
noise than the scenario-II.

###  VIII.2 Resistance to white noise for trilocality scenario-I and II

In the trilocality scenario, all the independent sources produces Werner state
with noise parameters $v_{1}$, $v_{2}$, and $v_{3}$. The joint four-partite
physical system is given by
$\rho^{w}_{ABCD}(v_{1},v_{2},v_{3})=\rho^{w}_{AB}(v_{1})\otimes\rho^{w}_{BC}(v_{2})\otimes\rho^{w}_{DB}(v_{3})$.

In the asymmetric trilocality scenario-I, the critical noise parameter for
demonstrating nonlocality is given by

$v^{N+2}>\sqrt{n}\ 2^{2(1-n)}\ \quantity(1-2^{1-n})\
\binom{n-1}{\lfloor\frac{n-1}{2}\ \rfloor}\left\lfloor
2^{2n-3}+\frac{1}{2}\right\rfloor\ \tan(\frac{\pi}{2^{n}})$ (91)

For $n=3$ critical noise parameter per Werner state is
$v_{c}=\frac{\sqrt{3}}{\sqrt[3]{2\left(\sqrt{2}+2\right)}}\approx 0.92$.

On the other hand, in the asymmetric trilocality scenario-II, the critical
noise parameter is given by

$v^{N+2}>2^{1-n}\ \frac{1}{\sqrt{n}}\ \quantity(1-\frac{1}{n})\
\binom{n-1}{\lfloor\frac{n-1}{2}\
\rfloor}\left\lfloor\frac{n^{2}+1}{2}\right\rfloor\ \tan(\frac{\pi}{2n})$ (92)

For $n=3$, the critical noise parameter per Werner state is
$v_{c}=(\frac{5}{9})^{\frac{1}{3}}\approx 0.82$. Thus, in the asymmetric
trilocality cases, the scenario-II is more robust to white noise than the
Scenario-I.

##  IX Summary and discussion

In this work, we have explored the quantum nonlocality in arbitrary ‘$n$’
input asymmetric bilocal as well as trilocal network scenario. The asymmetric
bilocal scenario proposed here features two edge parties Alice and Charlie who
perform $2^{n-1}$ and $n$ number of measurements, respectively. We derive two
families of bilocality inequalities specifically designed for the asymmetric
scenario when the central party Bob measures $n$ (Fig. 2) and $2^{n-1}$ number
of measurements.

Furthermore, we have extended the asymmetric network scenario into the
trilocal network. In particular, we have introduced two variants of asymmetric
trilocal network - (i) when one edge party, Alice, performs $2^{n-1}$
measurements, the other two edge parties, Charlie and Diana perform $n$
measurements each and the central party Bob performs $2^{n-1}$ measurements.
(ii) When one edge party, Charlie, performs $n$ measurements, the other two
edge parties, Alice and Diana perform $2^{n-1}$ measurements each and the
central party Bob performs $n$ measurements.

In Secs. III and IV, the detailed analytical treatments of the bilocal bounds
for the proposed asymmetric bilocal scenarios have been provided. It has been
found that the bilocal bounds depend on how the asymmetricity is invoked
within the bilocal scenario. In particular, the bilocal bounds have been found
to be $4.89$ and $6$ for scenarios I and II respectively. Subsequently, by
invoking the SOS technique, the DI optimal quantum bound for the asymmetric
bilocality scenario-I $\Big{(}(\mathcal{B}_{3})_{Q}^{opt}=6\Big{)}$ is found
to be less than that obtained
$\Big{(}(\mathcal{B}_{3}^{\prime})_{Q}^{opt}=4\left[3(2+\sqrt{2})\right]^{\frac{1}{4}}\approx
7.16\Big{)}$ in the scenario-II (Sec. III.1). In both the optimization
processes corresponding to scenarios I and II, we obtain the relational
constraints on the observables along with shared states required for Alice-Bob
and Bob-Charlie (Secs III.2 and A.3). The constraints on the observables of
all the parties in terms of the anti-commuting relations have also been
provided in Secs. (III) and (V).

For the asymmetric trilocality scenarios, the trilocal bounds evaluated to be
approximately $6.60$ and $4.93$ for the scenarios I and II, respectively
(Secs. V and VI). Consequently, using the SOS method, while for the scenario
I, the optimal quantum bound is found to be
$(\mathcal{T}_{3})_{Q}^{opt}=4(2\sqrt{3}+\sqrt{6})^{\frac{1}{3}}\approx 7.23$
for scenario-II, it is found to be $(\mathcal{T}_{3}^{\prime})_{Q}^{opt}=6$.
In this process, we have also obtained the constraints relations of all the
parties observables in terms of the anti-commutation relations (Secs. V.2 and
B.3). We have demonstrated that the optimal quantum bound will be achieved if
all the edge parties share three maximally entangled two-qubit states with the
central party Bob.

Moreover, we have generalised our results for arbitrary $n$. For this purpose,
using the SOS method, we first established that it is sufficient for Bob to
measure in the product basis in order to obtain the optimal quantum violations
in both the bilocality and trilocality scenarios. Then an interesting
algebraic manipulation reduces the nonlinear bilocal inequalities into a
product of two standard Bell inequalities (Sec. VII). Importantly, such
reduction in terms of the product of two standard Bell inequalities will only
be possible at the optimal condition. It is crucial to note here that the
rigorous algebraic manipulation for the case of $n=3$ buoyed up such deep-
seated understanding of the quantum optimal bound, which then provides the
necessary intuition for such a simple proof for the case of arbitrary $n$.
Then, following the similar argument of the bilocality scenarios, the optimal
quantum bound of asymmetric trilocality inequalities have also been evaluated.

Finally, for both the asymmetric bilocal and trilocal scenario, we have
demonstrated the robustness of the quantum violations of the proposed
inequalities in the presence of white noise. We found that the proposed
asymmetric inequalities are most robust to white noise in the simplest bilocal
scenario with two measurement settings for each parties. In this case, two
Werner state exhibit nonlocality if $v_{1}v_{2}>\frac{1}{2}$. Although
unfortunately, our proposed inequality becomes less robust with increasing
number of measurement settings or with increasing party (or, source), it
indeed possesses some independent interests.

We conclude by raising some open questions which can be studied in future.

i) Since the optimal quantum violations of the asymmetric bilocality
inequalities cannot always be achieved with the single copy of maximally
entangled two-qubit state. In fact, for the asymmetric bilocality scenario-I,
while Alice-Bob needs to share at least of $\lfloor n/2\rfloor$ copies of
maximally entangled two-qubit state Bob-Charlie need to share at least one
copy of it. Similar results can be proved for other scenarios also. Thus, in
the network scenario, our proposed inequality has the potential to be used as
a dimension witness. Construction of such proof may lead to a wide variety of
interesting results in the field of self-testing of many copies of maximally
entangled two-qubit states, cryptographic applications, or randomness
generation protocols.

ii) A straightforward extension of our proposed bilocality inequality would be
to invoke the 4-outcome measurement scenario for Bob. In this regard, for the
symmetric bilocal scenario, different bilocal inequalities have been tailored
to the $4$-outcome scenario in the context of both the Bell state measurement
and the elegant joint measurement scheme [35] for Bob. Although the inequality
with Bell state measurement does not provide any advantage over the usual
bilocal scenario, the inequality involving elegant joint measurement is more
advantageous in presence of the noise than the earlier bilocal or standard
Bell inequalities. Thus, an extensive study of such a scenario in the
asymmetric case may lead to interesting findings.

iii) Of course, one can always generalize our asymmetric bilocal scenario by
going beyond the four-party three-independent sources into a multiparty multi-
source scenario. From our evaluation of the robustness to white noise, the
asymmetric bilocal scenario does not provide any advantage over the noise
tolerance over the standard Bell scenario. However, with increasing the number
of parties and sources, one can introduce the asymmetry in many ways, which
may lead to a multiparty $n$-locality inequality that may provide such
advantages.

In sum, the essence of our present work lies in constructing a family of
asymmetric bilocality as well as trilocality inequalities and evaluating their
quantum optimal bounds, importantly, by not specifying the dimension of the
system or the dichotomic observables. Our work has the potential to open up
interesting avenues for future research such as self-testing of many copies of
entangled states, sequential sharing of quantum correlations, unbounded
generation of randomness, and secret key sharing in one-to-many scenarios that
calls for further study.

## Acknowledgments

S.S.M. acknowledges the UGC fellowship [Fellowship No. 16-9(June
2018)/2019(NET/CSIR)]. SS and AKP acknowledge the support from the project
DST/ICPS/QuST/Theme 1/2019/4.

##  Appendix A Detailed calculation for the asymmetric bilocality network
scenario-II

###  A.1 Bilocal bound for the asymmetric bilocality network scenario-II

$\mathcal{B}_{3}^{\prime}=\sum_{j=1}^{4}\sqrt{\ \quantity|\
\Big{\langle}\mathcal{\tilde{A}}_{3,j}^{\prime}B_{3,j}\mathcal{\tilde{C}}_{3,j}^{\prime}\Big{\rangle}\
|\ }\ \leq\ (\mathcal{B}_{3}^{\prime})_{bl}$ (93)

where the quantities $\mathcal{\tilde{A}}_{3,j}^{\prime}$ and
$\mathcal{\tilde{C}}_{3,j}^{\prime}$ are defined in Sec. (IV). Now, invoking
the reproducibility condition given by Eq. (II) and since $|\langle
B_{3,1}\rangle_{\lambda_{1},\lambda_{2}}|\leq 1$, we obtain

$\displaystyle\quantity|\
\Big{\langle}\mathcal{\tilde{A}}_{3,1}^{\prime}B_{3,1}\mathcal{\tilde{C}}_{3,1}^{\prime}\Big{\rangle}\
|\ $ $\displaystyle\leq\iint d\lambda_{1}\ d\lambda_{2}\ \mu(\lambda_{1})\
\mu(\lambda_{2})\ \big{|}\langle A_{3,1}\rangle_{\lambda_{1}}+\langle
A_{3,2}\rangle_{\lambda_{1}}\big{|}\ \big{|}\langle
C_{3,1}\rangle_{\lambda_{2}}+\langle C_{3,2}\rangle_{\lambda_{2}}+\langle
C_{3,3}\rangle_{\lambda_{2}}\big{|}$ (94)

The other terms can also be written in a similar manner as Eq. (94). Thus,
from Eq. (93), we obtain the following

$(\mathcal{B}_{3}^{\prime})_{bl}\leq\left(\iint d\lambda_{1}\ d\lambda_{2}\
\mu(\lambda_{1})\ \mu(\lambda_{2})\ \delta_{1}^{\prime}\
\delta_{2}^{\prime}\right)^{\frac{1}{2}}$ (95)

where
$\delta_{1}^{\prime}=\big{[}|\langle{A_{3,1}}\rangle_{\lambda_{1}}+\langle{A_{3,2}}\rangle_{\lambda_{1}}|+|\langle{A_{3,2}}\rangle_{\lambda_{1}}+\langle{A_{3,3}}\rangle_{\lambda_{1}}|+|\langle{A_{3,3}}\rangle_{\lambda_{1}}+\langle{A_{3,4}}|+|\langle{A_{3,4}}\rangle_{\lambda_{1}}-\langle{A_{3,1}}\rangle_{\lambda_{1}}|\big{]}$
and
$\delta_{2}^{\prime}=\big{[}|\langle{C_{3,1}}\rangle_{\lambda_{2}}+\langle{C_{3,2}}\rangle_{\lambda_{2}}+\langle{C_{3,3}}\rangle_{\lambda_{2}}|+|\langle{C_{3,1}}\rangle_{\lambda_{2}}+\langle{C_{3,2}}\rangle_{\lambda_{2}}-\langle{C_{3,3}}\rangle_{\lambda_{2}}|+|\langle{C_{3,1}}\rangle_{\lambda_{2}}-\langle{C_{3,2}}\rangle_{\lambda_{2}}+\langle{C_{3,3}}\rangle_{\lambda_{2}}|+|-\langle{C_{3,1}}\rangle_{\lambda_{2}}+\langle{C_{3,2}}\rangle_{\lambda_{2}}+\langle{C_{3,3}}\rangle_{\lambda_{2}}|\big{]}$.

Since all the observables are dichotomic with eigenvalues $\pm 1$, it is
straightforward to derive that $\delta_{1}^{\prime}\leq 6$ and
$\delta_{2}^{\prime}\leq 6$. Therefore, from Eq. (95), integrating over
$\lambda_{1}$ and $\lambda_{2}$ we obtain

$(\mathcal{B}_{3}^{\prime})_{bl}\leq 6$ (96)

###  A.2 Optimal quantum bound of the asymmetric bilocality inequality for
scenario:II

Let us consider a suitable positive semi-definite operator
$\gamma_{3}^{\prime}$ satisfying
$\langle\gamma_{3}^{\prime}\rangle=\beta_{3}^{\prime}-\left(\mathcal{B}_{3}^{\prime}\right)_{Q}$.
The existence of such operator is constructed by considering a set of
operators $M^{\prime}_{3,j}\ \forall j\in\\{1,2,3,4\\}$ such that

$\langle\gamma_{3}^{\prime}\rangle=\sum\limits_{j=1}^{4}\dfrac{\sqrt{\omega_{3,j}}}{2}\quantity|M^{\prime}_{3,j}\ket{\psi}|^{2}$
(97)

where $\omega_{3,j}\geq 0$ and
$\omega_{3,j}=\omega^{A}_{3,j}\cdot\omega^{C}_{3,j}$. We choose
$M^{\prime}_{3,j}$ and the quantity $\omega_{3,j}$ as the following way

$\displaystyle\quantity|M^{\prime}_{3,j}\ket{\psi}|=\sqrt{\quantity|\quantity(\frac{\mathcal{\tilde{A}}_{3,j}^{\prime}}{\omega_{3,j}^{A}}\otimes\frac{\mathcal{\tilde{C}}_{3,j}^{\prime}}{\omega_{3,j}^{C}})\ket{\psi}|}-\sqrt{|B_{3,j}\ket{\psi}|}\
\ \ \forall j\in\\{1,2,3,4\\}$ (98) $\displaystyle\hskip
28.45274pt\omega^{A}_{3,j}=||\mathcal{\tilde{A}}_{3,j}^{\prime}\ket{\psi}||_{2}\
;\ \omega^{C}_{3,j}=||\mathcal{\tilde{C}}_{3,j}^{\prime}\ket{\psi}||_{2}$ (99)

Putting $\quantity|M^{\prime}_{3,j}\ket{\psi}|$ and $\omega_{3,j}^{k}$ from
Eq. (98) into Eq. (97) and, by using the inequalities given in footnotes (3)
and (4), we obtain the quantum optimal value as follows

$\displaystyle\left(\mathcal{B}_{3}^{\prime}\right)_{Q}^{opt}=\max\left[\prod_{k=A,C}\left(4\sum_{j=1}^{4}\left(\omega^{k}_{3,j}\right)^{2}\right)\right]^{\frac{1}{4}}\
;\ \ \ \ \text{with the optimality condition: }M^{\prime}_{3,j}\ \ket{\psi}=0\
\ \forall j\in\\{1,2,3,4\\}$ (100)

In the following we evaluate $\sum_{j=1}^{4}\left(\omega^{A}_{3,j}\right)^{2}$
and $\sum_{j=1}^{4}\left(\omega^{C}_{3,j}\right)^{2}$ from Eq. (99)
separately.

#### Evaluation of $\sum_{j=1}^{4}\left(\omega^{A}_{3,j}\right)^{2}$:

From Eq (99) we obtain

$\displaystyle\sum\limits_{j=1}^{4}\bigg{(}\omega^{A}_{4,j}\bigg{)}^{2}$
$\displaystyle=$
$\displaystyle\bra{\psi}(8+\\{A_{4,2},(A_{4,1}+A_{4,3})\\}+\\{A_{4,4},(A_{4,3}-A_{4,1})\\})\ket{\psi}$
(101)

Note that in the above Eq. (101) the quantities $A_{4,2}$ and $A_{4,4}$
appeared independently with $\\{A_{4,2},(A_{4,1}+A_{4,3})\\}$ and
$\\{A_{4,4},(A_{4,3}-A_{4,1})\\}$ respectively. Thus, we can always define
$A_{4,2}$ and $A_{4,4}$ independently. Hence, without loss of generality, we
chose $A_{4,2}=(A_{4,3}+A_{4,1})/\nu_{1}$ and
$A_{4,4}=(A_{4,3}-A_{4,1})/\nu_{2}$, where $\nu_{1}=||\ (A_{4,3}+A_{4,1})\
||_{2}$ and $\nu_{2}=||\ (A_{4,3}-A_{4,1})\ ||_{2}$. therefore, the above Eq.
(101) reduces to

$\sum\limits_{j=1}^{4}\bigg{(}\omega^{A}_{4,j}\bigg{)}^{2}=8+2\Bigg{[}\sqrt{4+2\sqrt{4-\langle\\{A_{4,1},A_{4,3}\\}\rangle}}\Bigg{]}\
\leq\ 4\left(2+\sqrt{2}\right)$ (102)

The maximum value of
$\sum\limits_{j=1}^{4}\bigg{(}\omega^{A}_{4,j}\bigg{)}^{2}$ is then achieved
when $\\{A_{4,1},A_{4,3}\\}=0$ which automatically implies
$\nu_{1}=\nu_{2}=\sqrt{2}$. Therefore, for the optimal quantum violation, the
linear constraints on Alice’s observables are given as

$\displaystyle A_{4,1}-\sqrt{2}A_{4,2}+A_{4,3}=0\ ;\ \ \ \
-A_{4,1}+A_{4,3}-\sqrt{2}A_{4,4}=0$ (103)

#### Evaluation of $\sum_{j=1}^{3}\left(\omega^{C}_{3,j}\right)^{2}$:

By Eq. (99), a straightforward calculation leads to

$\sum_{j=1}^{4}\left(\omega^{C}_{3,j}\right)^{2}\leq 12$ (104)

with the constraints on Charlie’s observables are given by

$\displaystyle\\{C_{3,1},C_{3,2}\\}=\\{C_{3,2},C_{3,3}\\}=\\{C_{3,1},C_{3,3}\\}=0$
(105)

Finally, we obtain the optimal quantum value from Eqs. (100), (102) and (104)
as

$\displaystyle\left(\mathcal{\mathcal{B}}_{3}^{\prime}\right)_{Q}^{opt}=4\left[3\left(2+\sqrt{2}\right)\right]^{\frac{1}{4}}$
(106)

###  A.3 The state and observables for the optimal quantum violation of
$(\mathcal{B}^{\prime})_{3}$

We further obtain relationships between the observables of all the parties for
achieving the optimal quantum violation in terms of the anti-commuting
relations. The anti-commutation relations for Charlie’s observables are
already given in Eq. (38). From Eqs. (35), we obtain the following anti-
commuting relations for Alice’s observables as

$\displaystyle\\{A_{3,1},A_{3,2}\\}=\\{A_{3,2},A_{3,3}\\}=\\{A_{3,3},A_{3,4}\\}=-\\{A_{3,1},A_{3,4}\\}=\sqrt{2}\
\mathbb{I}_{2}\ ;\ \ \ \\{A_{3,1},A_{3,3}\\}=\\{A_{3,2},A_{3,4}\\}=0$ (107)

By using the above relations between the observables given by Eqs. (105) and
(107) on the observables, one can always construct a set of observables for
Alice and Charlie in the Hilbert space dimension $\mathcal{H}^{d},\ \ \forall
d\geq 2$.

Next, we recall the optimization condition obtained in the SOS method from the
Eq. (100) to find the constraints on Bob’s observable. The specific condition
$M^{\prime}_{3,j}\ket{\psi}=0\ \ \forall j\in\\{1,2,3\\}$ implies the
following

$\displaystyle
B_{3,j}=\frac{\mathcal{\tilde{A}}_{3,j}^{\prime}}{\omega_{3,j}^{A}}\otimes\frac{\mathcal{\tilde{C}}_{3,j}^{\prime}}{\omega_{3,j}^{C}}\
\ \forall j\in\\{1,2,3,4\\}$ (108)

We explicitly construct a set of observables for the Hilbert space dimension
$\mathcal{H}^{2}$ are the following.

$\displaystyle A_{3,1}$ $\displaystyle=$ $\displaystyle
r\sigma_{x}+\sqrt{1-r^{2}}\ \sigma_{z}\ \ ;\ \
A_{3,2}=t\sigma_{x}+\sqrt{1-t^{2}}\ \sigma_{z}\ \ ;\ A_{3,3}=t\
\sigma_{x}-\sqrt{1-t^{2}}\ \sigma_{z}\ \ ;\ \ A_{3,4}=r\
\sigma_{x}-\sqrt{1-r^{2}}\ \sigma_{z}\ \ $ $\displaystyle C_{3,1}$
$\displaystyle=$ $\displaystyle\sigma_{x}\ \ ;\ \ C_{3,2}=\sigma_{y}\ \ ;\ \
C_{3,3}=\sigma_{z}\ \ \ \ \ \ [\text{where $r=\frac{1}{2}\sqrt{2-\sqrt{2}}$
and $t=\frac{1}{2}\sqrt{2+\sqrt{2}}$.}]$ (109)

Bob’s observables can be constructed from Eq. (108). Note that employing the
above-mentioned observables, we find that the quantum optimal value
$\left(\mathcal{B}_{3}^{\prime}\right)_{Q}^{opt}=6$ is achieved when two
maximally entangled two-qubit states are shared between Alice-Bob and Bob-
Charlie.

##  Appendix B Detailed calculation for the asymmetric trilcality network
scenario-II

###  B.1 Trilocal bound for the asymmetric Trilocality network scenario-II

$\mathcal{T}_{3}^{\prime}=\sum_{j=1}^{3}|\mathcal{J}_{3,j}^{\prime}|^{\frac{1}{3}}\leq(\mathcal{T}_{3}^{\prime})_{tl}\
\ \ \text{with}\ \ \
\mathcal{J}_{3,j}^{\prime}=\Big{\langle}\mathscr{\tilde{A}}_{3,j}^{\prime}B_{3,j}\mathscr{\tilde{C}}_{3,j}^{\prime}\mathscr{\tilde{D}}_{3,j}^{\prime}\Big{\rangle}$
(110)

where the quantities $\mathscr{\tilde{A}}_{3,j}^{\prime}$,
$\mathscr{\tilde{C}}_{3,j}^{\prime}$ and $\mathscr{\tilde{D}}_{3,j}^{\prime}$
are defined in Sec. (VI) of the main text. Now, invoking the reproducibility
condition given by Eq. (II) and Since
$\quantity|\braket{B_{3,1}}_{\lambda_{1},\lambda_{2},\lambda_{3}}|\leq 1$, we
obtain

$\displaystyle\mathcal{J}_{3,1}^{\prime}$ $\displaystyle=$
$\displaystyle\iiint d\lambda_{1}d\lambda_{2}d\lambda_{3}\
\mu(\lambda_{1})\mu(\lambda_{2})\mu(\lambda_{3})\ \Big{|}\langle
A_{3,1}\rangle_{\lambda_{1}}+\langle A_{3,2}\rangle_{\lambda_{1}}+\langle
A_{3,3}\rangle_{\lambda_{1}}-\langle A_{3,4}\rangle_{\lambda_{1}}\Big{|}\
\Big{|}\langle C_{3,1}\rangle_{\lambda_{2}}+\langle
C_{3,2}\rangle_{\lambda_{2}}\Big{|}\ \Big{|}\langle
D_{3,1}\rangle_{\lambda_{3}}+\langle D_{3,2}\rangle_{\lambda_{3}}+\langle
D_{3,3}\rangle_{\lambda_{3}}\Big{|}$

The terms $|\mathcal{J}_{3,2}^{\prime}|$, and $|\mathcal{J}_{3,3}^{\prime}|$
given by Eq. (110) can also be written in a similar manner as Eq. (B.1). Then,
we obtain the following

$(\mathcal{T}_{3}^{\prime})_{bl}\leq\left(\iiint
d\lambda_{1}d\lambda_{2}d\lambda_{3}\
\mu(\lambda_{1})\mu(\lambda_{2})\mu(\lambda_{3})\ \
\eta_{1}^{\prime}\eta_{2}^{\prime}\eta_{3}^{\prime}\right)^{\frac{1}{3}}$
(112)

where
$\eta_{1}^{\prime}=\big{[}|\langle{A_{3,1}}\rangle_{\lambda_{1}}+\langle{A_{3,2}}\rangle_{\lambda_{1}}+\langle{A_{3,3}}\rangle_{\lambda_{1}}-\langle{A_{3,4}}\rangle_{\lambda_{1}}|+|\langle{A_{3,1}}\rangle_{\lambda_{1}}+\langle{A_{3,2}}\rangle_{\lambda_{1}}-\langle{A_{3,3}}\rangle_{\lambda_{1}}+\langle{A_{3,4}}\rangle_{\lambda_{1}}|+|\langle{A_{3,1}}\rangle_{\lambda_{1}}-\langle{A_{3,2}}\rangle_{\lambda_{1}}+\langle{A_{3,3}}\rangle_{\lambda_{1}}+\langle{A_{3,4}}\rangle_{\lambda_{1}}|\big{]}$,
$\eta_{2}^{\prime}=\big{[}|\langle{C_{3,1}}\rangle_{\lambda_{2}}+\langle{C_{3,2}}\rangle_{\lambda_{2}}|+|\langle{C_{3,2}}\rangle_{\lambda_{2}}+\langle{C_{3,3}}\rangle_{\lambda_{2}}|+|\langle{C_{3,3}}\rangle_{\lambda_{2}}-\langle{C_{3,1}}\rangle_{\lambda_{2}}|\big{]}$,
and $\eta_{3}^{\prime}=\big{[}|\langle D_{3,1}\rangle_{\lambda_{3}}+\langle
D_{3,2}\rangle_{\lambda_{3}}+\langle D_{3,3}\rangle_{\lambda_{3}}|+|\langle
D_{3,1}\rangle_{\lambda_{3}}+\langle D_{3,2}\rangle_{\lambda_{3}}-\langle
D_{3,3}\rangle_{\lambda_{3}}|+|\langle D_{3,1}\rangle_{\lambda_{3}}-\langle
D_{3,2}\rangle_{\lambda_{3}}-\langle D_{3,3}\rangle_{\lambda_{3}}|\big{]}$.

Since all the observables are dichotomic with eigenvalues $\pm 1$, it is
straightforward to derive that $\eta_{1}^{\prime}\leq 6$,
$\eta_{2}^{\prime}\leq 4$, and $\eta_{3}^{\prime}\leq 5$. Therefore, from Eq.
(112), integrating over $\lambda_{1}$ and $\lambda_{2}$ we obtain

$(\mathcal{T}_{3})_{bl}\leq 2(15)^{\frac{1}{3}}\approx 4.93$ (113)

###  B.2 Optimal quantum bound of the asymmetric trilocality inequality for
scenario-II

To derive the optimal quantum bound of $\mathcal{T}_{3}^{\prime}$ without
assuming the dimension of the system, we again invoke the SOS approach
discussed in the preceding Sec. II.1. Following the similar argument presented
earlier in the Sec. II.1, we first show that there exists a positive semi-
definite operator
$\langle\Gamma_{3}^{\prime}\rangle=\zeta_{3}^{\prime}-\quantity(\mathcal{T}_{3}^{\prime})_{Q}$.
The existence of such operator can be proved by considering a set of operators
$L_{3,j}^{\prime}\ \forall j\in\\{1,2,3\\}$ such that

$\langle\Gamma_{3}^{\prime}\rangle=\sum\limits_{j=1}^{4}\dfrac{\quantity(\omega_{3,j})^{\frac{1}{3}}}{2}\quantity|L_{3,j}^{\prime}\ket{\psi}|^{2}$
(114)

where $\omega_{3,j}\geq 0$ and
$\omega_{3,j}=\omega^{A}_{3,j}\cdot\omega^{C}_{3,j}\cdot\omega^{D}_{3,j}$. We
choose $L^{\prime}_{3,j}$ and the quantity $\omega_{3,j}$ as the following way

$\displaystyle\quantity|L^{\prime}_{3,j}\ket{\psi}|=\quantity|\quantity(\frac{\mathscr{\tilde{A}}_{3,j}^{\prime}}{\omega_{3,j}^{A}}\otimes\frac{\mathscr{\tilde{C}}_{3,j}^{\prime}}{\omega_{3,j}^{C}}\otimes\frac{\mathscr{\tilde{D}}_{3,j}^{\prime}}{\omega_{3,j}^{C}})\ket{\psi}|^{\frac{1}{3}}-\quantity|B_{3,j}\ket{\psi}|^{\frac{1}{3}}\
\ \ \forall j\in\\{1,2,3\\}$ (115) $\displaystyle\hskip
28.45274pt\omega^{A}_{3,j}=||\mathscr{\tilde{A}}_{3,j}^{\prime}\ket{\psi}||_{2}\
;\ \omega^{C}_{3,j}=||\mathscr{\tilde{C}}_{3,j}^{\prime}\ket{\psi}||_{2}\ ;\
\omega^{D}_{3,j}=||\mathscr{\tilde{D}}_{3,j}^{\prime}\ket{\psi}||_{2}$ (116)

Where $||\cdot||_{2}$ denotes the Frobenious norm given by $||\ \mathcal{O}\ \
||_{2}=\sqrt{\bra{\psi}\mathcal{O}^{\dagger}\mathcal{O}\ket{\psi}}$.

Putting $L_{3,j}^{\prime}$ and $\omega_{3,j}$ from Eqs. (115) and (116) into
Eq. (114), and by using the inequalities given in footnotes (3) and (4), we
obtain the quantum optimal value as follows

$\displaystyle\left(\mathcal{T}_{3}^{\prime}\right)_{Q}^{opt}=\max\left[\prod_{k=A,C,D}\left(3\sum_{j=1}^{3}\left(\omega^{k}_{3,j}\right)^{2}\right)\right]^{\frac{1}{6}}\
;\ \ \ \ \text{with the optimality condition: }L^{\prime}_{3,j}\ \ket{\psi}=0\
\ \forall j\in\\{1,2,3,4\\}$ (117)

Note that from Eqs. (36) and (37)
$\max\sum_{j=1}^{3}\left(\omega^{A}_{3,j}\right)^{2}=16$ and
$\max\sum_{j=1}^{3}\left(\omega^{C}_{3,j}\right)^{2}=9$, respectively. We
evaluate $\sum_{j=1}^{3}\left(\omega^{D}_{3,j}\right)^{2}$ as follows

$\sum_{j=1}^{3}\quantity(\omega^{D}_{3,j})^{2}=\bra{\psi}(9+\\{D_{3,1},(D_{3,2}-D_{3,3})\\}+\\{D_{3,2},D_{3,3}\\})\ket{\psi}=\bra{\psi}(9+3\mathbb{I}-(D_{3,1}-D_{3,2}+D_{3,3})^{2})\ket{\psi}\leq
12\\\ $ (118)

The above Eq. (118) provides maximum value when

$D_{3,1}-D_{3,2}+D_{3,3}=0$ (119)

By placing the value of $\sum_{j=1}^{3}\quantity(\omega^{A}_{3,j})^{2}$,
$\sum_{j=1}^{3}\quantity(\omega^{C}_{3,j})^{2}$, and
$\sum_{j=1}^{3}\quantity(\omega^{D}_{3,j})^{2}$ in Eq. (117) we obtain the
optimal quantum bound as

$\displaystyle\left(\mathcal{\mathcal{T}}_{3}^{\prime}\right)_{Q}^{opt}=6$
(120)

It is important to remark here that the optimal quantum value
$\left(\mathcal{T}_{3}\right)_{Q}^{opt}=6$ is evaluated without specifying the
dimension of both the system and observables. The optimal value fixes the
states, and the observables are the following.

###  B.3 The state and observables for the optimal quantum violation of
$\mathcal{T}^{\prime}_{3}$

We further obtain relationships between the observables of all the parties for
achieving the optimal quantum violation. It follows from the earlier derived
results (Eqs. 35 and 38) and Eq. (119) the following anti-commuting relations
of the observables for all the parties.

$\displaystyle\\{A_{3,1},A_{3,2}\\}=\\{A_{3,1},A_{3,3}\\}=\\{A_{3,1},A_{3,4}\\}=\frac{2}{3}\mathbb{I}_{d};\
\\{A_{3,2},A_{3,3}\\}=\\{A_{3,2},A_{3,4}\\}=\\{A_{3,3},A_{3,4}\\}=-\frac{2}{3}\mathbb{I}_{d}$
(121)
$\displaystyle\\{C_{3,1},C_{3,2}\\}=\\{C_{3,2},C_{3,3}\\}=-\\{C_{3,1},C_{3,3}\\}=\\{D_{3,1},C_{3,2}\\}=\\{D_{3,2},D_{3,3}\\}=-\\{D_{3,1},D_{3,3}\\}=\mathbb{I}_{d}$
(122)

By using the above relations between the observables given by Eqs. (121-122)
on the observables, one can always construct a set of observables for Alice
and Charlie in the Hilbert space dimension $\mathcal{H}^{d},\ \ \forall d\geq
2$.

Next, we recall the optimization condition obtained in the SOS method from the
Eq. (117) to find the constraints on Bob’s observable. The specific condition
$L^{\prime}_{3,j}\ket{\psi}=0\ \ \forall j\in\\{1,2,3\\}$ implies the
following

$\displaystyle
B_{3,j}=\frac{\mathscr{\tilde{A}}_{3,j}^{\prime}}{\omega_{3,j}^{A}}\otimes\frac{\mathscr{\tilde{C}}_{3,j}^{\prime}}{\omega_{3,j}^{C}}\otimes\frac{\mathscr{\tilde{D}}_{3,j}^{\prime}}{\omega_{3,j}^{C}}$
(123)

We explicitly construct a set of observables for the Hilbert space dimension
$\mathcal{H}^{2}$ are the following.

$\displaystyle A_{3,1}$ $\displaystyle=$
$\displaystyle\frac{\sigma_{x}+\sigma_{y}+\sigma_{z}}{\sqrt{3}}\ ;\
A_{3,2}=\frac{\sigma_{x}+\sigma_{y}-\sigma_{z}}{\sqrt{3}};\
A_{3,3}=\frac{\sigma_{x}-\sigma_{y}+\sigma_{z}}{\sqrt{3}};\
A_{3,4}=\frac{-\sigma_{x}+\sigma_{y}+\sigma_{z}}{\sqrt{3}}\ ;\
C_{3,1}=\sigma_{z}$ $\displaystyle C_{3,2}$ $\displaystyle=$
$\displaystyle\left(\frac{\sqrt{3}}{2}\sigma_{x}+\frac{\sigma_{z}}{2}\right);\
C_{3,3}=\left(\frac{\sqrt{3}}{2}\sigma_{x}-\frac{\sigma_{z}}{2}\right)\ ;\
D_{3,3}=-\sigma_{z}\ ;\
D_{3,1}=\left(\frac{-\sqrt{3}}{2}\sigma_{x}+\frac{\sigma_{z}}{2}\right);\
D_{3,2}=\left(-\frac{\sqrt{3}}{2}\sigma_{x}-\frac{\sigma_{z}}{2}\right)$ (124)

Note that Bobs observables can be constructed from Eq. (123). Now, employing
the above-mentioned observables, we find that the quantum optimal value
$\left(\mathcal{T}_{3}^{\prime}\right)_{Q}^{opt}=6$ is achieved when three
maximally entangled two-qubit states are shared between Alice-Bob, Charlie-Bob
and Diana-Bob.

## References

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