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# Self-testing of quantum states using symmetric local hidden state model

Debasis Mondal  Centre for Quantum Technologies, National University of
Singapore, 3 Science Drive 2, Singapore 117543  Dagomir Kaszlikowski  Centre
for Quantum Technologies, National University of Singapore, 3 Science Drive 2,
Singapore 117543  Department of Physics, National University of Singapore, 2
Science Drive 3, 117542 Singapore, Singapore

###### Abstract

We introduce a symmetric local hidden state $(slhs)$ model in a scenario,
where two spacially separated parties receive quantum states from an unknown
source. We derive an inequality based on the model. A completely new form of
nonlocality emerges from the resource theoretic point of view. The inequality
singles out a larger set of quantum correlated states in the higher
dimensional scenarios $(d>2$ X $2)$ than what is predicted by the existing
$lhs$ model, opening a new front for the experimentalists to test the accuracy
of the prediction. We propose an experiment to show the experimental violation
of the inequality in the two qubit scenario and perform the experiment on the
IBM quantum computer. However, the experimental method adopted for the two-
qubit scenario does not naturally generalize in the higher dimensional
scenarios and leaves the experimental verification of the claim open. We also
show that the maximal violation of the inequality can be used to self-test the
Bell state and measurement bases, leading to complete device-independence.

##  I Introduction

In the last few decades, we have explored different avenues to understand the
difference between the classical and the quantum world. Quantum correlation,
for instance, have been of great interest for its quantum advantages over the
classical in communication, cryptography and computation. For a bipartite
system, quantum correlation depends on the protocol—following which the
correlation is generated and the fact whether we trust the parties or not. A
trustless model is where parties do not need to trust each other, for example,
the local hidden variable $(lhv)$ model bell ; jfclauser . Entanglement, on
the other hand, is a measure of the quantum correlation between two trusted
parties. A seemingly new addition to the list is the local hidden state
$(lhs)$ model wittmann ; lhsintro ; sjjones , where only one of the parties is
trusted.

A quantum state is shared by two parties— Alice and Bob. The statement that
the state of Alice is correlated to the state of Bob but not the state of Bob
with that of Alice, sounds like: Alice is married to Bob but Bob is not nguyen
; uola . The violation of the present description of the local hidden state
$(lhs)$ model lhsintro ; sjjones ; wittmann  sometimes predicts exactly that.
There are states for which Alice can steer Bob but Bob cannot steer Alice.

Although, the steerability of a quantum state has been shown to be a resource
resource2 ; resource3  and strongly believed to be a signature of quantum
correlations, it cannot be considered as a measure of quantum correlations.
This is due to the fact that quantum correlations like its classical counter
part must be symmetric with respect to parties. Is it then possible to find an
alternative and improved local hidden state model for which such scenarios do
not arise?

The situation arises primarily because of the game considered in the
description of local hidden state model lhsintro ; sjjones . We consider two
parties— Alice and Bob. Alice prepares a bipartite state. A part of which she
sends to Bob and keeps the other part with her. Alice claims, she can steer
his state and his state is entangled with her. Bob does not trust Alice. Bob
comes up with a strategy to verify her claim by constructing the steering
inequality based on the $lhs$ model.

In the existing $lhs$ model lhsintro ; sjjones ; uola ; wittmann ; nguyen ;
resource2 ; resource3 , Bob assumes that the states, he is receiving from
Alice, have single system description and Alice can in principle prepare such
states from an ensemble of hidden states $\\{p(\lambda),\rho_{\lambda}\\}$
such that

$\rho_{a|\Pi}=\sum_{\lambda}p(\lambda)p(a|\Pi,\lambda)\rho_{\lambda},$ (1)

where $p(a|\Pi,\lambda)$ is Alice’s stochastic map to convince Bob and
$\lambda$ is a hidden variable such that $\sum_{\lambda}p(\lambda)=1$, $\Pi$
is an observable in which Bob asked Alice to perform measurements and $a$ is
the outcome. Violation of a steering inequality based on the $lhs$ model
implies the existence of $EPR$ non-locality.

As it can be observed, the game is asymmetric by construction. A symmetric
scenario would be, where Alice $(A)$ and Bob $(B)$ both receives quantum
states from an unknown source $(S)$. They do not trust the source. Therefore,
they would like to come up with an effective strategy to verify whether they
have $EPR$-correlations between their particles or not.

One way to represent the symmetric local hidden state $(slhs)$ model is of
course to express the bipartite state by local hidden states, i.e.,

$\rho_{AB}=\sum_{\lambda}p(\lambda)\rho_{\lambda}^{A}\otimes\rho_{\lambda}^{B},$
(2)

where $\\{p(\lambda),\rho_{\lambda}^{A}\otimes\rho_{\lambda}^{B}\\}$ is an
ensemble of hidden states like before. Alice and Bob can construct an
inequality based on this model and the level of their trust.

Assumption in the above model is that any bipartite state must be expressible
in terms of local hidden states. However, this is nothing but a separable
state. Any state, which cannot be expressed by the above model is nothing but
an entangled state and to witness an entangled state in the most effective
way, Alice and Bob need to trust each-other.

To better understand how the trust is enforced in the measures of
entanglement, it is noted that the entanglement measures depend on the
elements of the form of $\langle i^{a}i^{b}|\rho_{AB}|i^{a}i^{b}\rangle$,
where $a,b\in\\{0,1\\}\text{ for two qubit states}$. Therefore, measurements
and outcomes need to be communicated to each-other to measure such quantities
and the entanglement depends on the exact values of measurement outcomes. This
now raises a question: is it possible to witness quantum nonlocality based on
the local hidden state model in a trust-less manner?

The present description of the $lhs$ model is asymmetric and as a result, only
leads to the semi-device independence. It naturally creates a curiosity why
there should not be a device-independent or trustless protocol to observe EPR
nonlocality just like $lhv$ model? Moreover, the $lhs$ model is stronger than
the $lhv$ model. Therefore, less quantum states have local hidden state
description than local hidden variable description and more states violate the
$lhs$ model than the $lhv$ model nguyen . In this sense, nonlocality based on
the violation of $lhs$ model is more accurate description of the nonlocality
jones ; pati  and a new symmetric, trustless description of such nonlocality
should be better suited for device independent $QKD$ protocols than the
existing protocols based on the Bell nonlocality and the EPR steering bennet ;
device ; branciard .

It turns out that there is indeed another way to represent the $slhs$ model.
We use the $slhs$ model to self-test the state and the measurement bases,
making it only the second, yet better alternative to the Bell inequality for
device independent protocols and trust-less entanglement witness.

##  II Symmetric local hidden state model

In the $lhv$ model bell ; jfclauser , we consider the joint probability
distribution. We use the principle of locality and determinism to write the
joint distribution in terms of local probability distributions and hidden
variables. The locality in Bell’s formalism does not imply EPR locality. Non-
locality in a different form may still exist. It is just that the non-locality
is not reflected in its local probability distributions.

Here, we assume that both Alice and Bob is receiving quantum states. However,
quantum states provide more information than just probability distributions.
One can also extract information about certain quantities, which have no
classical counterparts unlike probability distributions. We consider a global
or joint property with no classical counterpart and using the principle of
locality and hidden states, express in terms of the local property of the
hidden states. In particular, we consider the transition amplitude or
probability to be the property, which has no classical counter part. With
these assumptions, all the off-diagonal elements of a bipartite density matrix
must be expressible by some unknown local hidden states $\rho_{A}^{\lambda}$
and $\rho_{B}^{\lambda}$ generated by the unknown source with a distribution
$p(\lambda)$. This is probably the most general version of the $slhs$ model
one could write. In a simple mathematical term, this implies,

$\langle
i^{a}j^{b}|\rho_{AB}|i^{a^{\prime}}j^{b^{\prime}}\rangle=\sum_{\lambda}p(\lambda)\langle
i^{a}|\rho^{\lambda}_{A}|i^{a^{\prime}}\rangle\langle
j^{b}|\rho^{\lambda}_{B}|j^{b^{\prime}}\rangle$ (3)

for any arbitrary bipartite state $\rho_{AB}$,
$\forall|{i}^{a(a^{\prime})}\rangle$ and $|{j}^{b(b^{\prime})}\rangle$. Here,
$i$, $j$ are the labels for bases and $a$, $a^{\prime}$, $b$ and $b^{\prime}$
are the labels for vectors in the particular basis. In this section, we
consider transition probability to express a relatively weaker version of our
hidden state model in the following form,

$\displaystyle
p\bigg{(}|{i}^{a-a^{\prime}}\rangle^{A},|{j}^{b-b^{\prime}}\rangle^{B}\bigg{)}=\sum_{\lambda}p(\lambda)p\bigg{(}|{i}^{a-a^{\prime}}\rangle^{A}\bigg{|}\lambda\bigg{)}$
$\displaystyle
p\bigg{(}|{j}^{b-b^{\prime}}\rangle^{B}\bigg{|}\lambda\bigg{)}.$ (4)

Here, $p({i}^{a-a^{\prime}})$ denotes the transition probability of the state
from the $a$ Eigenvector to the $a^{\prime}$ Eigenvector of the observable $i$
on Alice’s side and
$p(|{i}^{a-a^{\prime}}\rangle)=|\langle{i}^{a}|\rho|{i}^{a^{\prime}}\rangle|$.
Note however that $p(|{i}^{a-a^{\prime}}\rangle)$ does not really have
properties of a probability distribution function. One can indeed show that
the Eq. (3) implies the Eq. (4) (see the supplemental material supple ).

A clear distinction of the model from the local hidden variable $(lhv)$ model
is that the $lhv$ model with dichotomic observables can explain the
probability distributions generated by two classical coins. However, the same
distribution cannot be described by the $slhs$ model of two qubit systems. The
expression in Eq. (3) or in Eq. (4) is an attempt to describe a (property of
a) bipartite system in terms of that of the local hidden states.

##  III  SLHS inequality

In this section, we derive an inequality based on the model for a two-qubit
states. We consider the joint transition probability of Alice and Bob in the
Eigen bases of the observables $i$ and $j$,
$p(|{i}^{a-a^{\prime}}\rangle^{A},|{j}^{a-a^{\prime}}\rangle^{B})$ and express
it in terms of the $slhs$ model as

$\displaystyle\sum_{\lambda}p(\lambda)p\bigg{(}|{i}^{a-a^{\prime}}\rangle^{A}|\lambda\bigg{)}p\bigg{(}|{j}^{a-a^{\prime}}\rangle^{B}|\lambda\bigg{)}$
$\displaystyle\leq\sqrt{\sum_{\lambda^{\prime}}p(\lambda^{\prime})p^{2}\bigg{(}|i^{a-a^{\prime}}\rangle^{A}\bigg{|}\lambda^{\prime}\bigg{)}\sum_{\lambda}p(\lambda)p^{2}\bigg{(}|j^{a-a^{\prime}}\rangle^{B}\bigg{|}\lambda\bigg{)}}$
$\displaystyle=\sqrt{\overline{p^{2}\bigg{(}|i^{a-a^{\prime}}\rangle^{A}\bigg{)}}.\overline{p^{2}\bigg{(}|j^{a-a^{\prime}}\rangle^{B}\bigg{)}}}\leq\frac{1}{4}$
$\displaystyle\implies\underset{a\neq
a^{\prime}}{\sum_{a,a^{\prime},}}p(|{i}^{a-a^{\prime}}\rangle^{A},|{j}^{a-a^{\prime}}\rangle^{B})\leq\frac{1}{2},$
(5)

where in the first inequality, we use the Cauchy-Schwarz inequality, the
second inequality comes from the fact that for a probability distribution
$p(x)$ over a real random variable $x$ and a positive function of the random
variable $f(x)$, if the the function is bounded from above by $l$, the
$n^{th}$-moment of $f(x)$ is bounded by $l^{n}$. In this case, the function of
the random variable $\lambda$ is
$p\bigg{(}|i^{a-b}\rangle^{A}\bigg{|}\lambda\bigg{)}$ and is bounded by the
maximal value of the transition probability of a local state can possibly
take. For a two-qubit scenario, the dimension of the local state is two and
the maximum transition probability between two arbitrary orthonormal states
labeled by $a$, $b$ such that $a\neq b$ turns out to be $\frac{1}{2}$. Thus,
the bound becomes $\frac{1}{4}$. The inequality can also be extended for a
general $d\otimes d$ bipartite systems as shown later in Eq. (8).

It is well known that Bell inequality cannot be violated by generating random
measurement statistics locally. This makes the inequality trustless. It should
be noted here that the violation of the inequality in Eq. (5) ensures the
existence of the nonlocality in the state and this inequality as well cannot
be violated by generating random local measurement outcomes making it an
another alternative to the Bell inequality in terms of trustless entanglement
detection.

Figure 1: We plot (in red) the left hand side of the Eq. (5) with respect to
the parameters for the Werner state
$\rho_{w}^{\phi+}=p|\phi_{+}^{\alpha}\rangle\langle\phi_{+}^{\alpha}|+\frac{1-p}{4}I_{4}$,
where
$|\phi_{+}^{\alpha}\rangle=\sqrt{\alpha}|00\rangle+\sqrt{1-\alpha}|11\rangle$,
when both Alice and Bob measures the transition probabilities in the
$\sigma_{3}$ basis. The straight line parallel to the $\alpha$-axis represents
the bound $p>\frac{1}{2}$, for which the state is nonlocal.

As it can be observed from the fig. (1), quantum states indeed violate the
inequality. We will leave now the question on how Alice and Bob are going to
measure the quantity on the left hand side of the inequality in Eq. (5) for
the next section and instead, we focus on to ask a more deeper question: what
would happen if one considered quantum coherence instead of quantum transition
probabilities? One would expect to get similar violation of a bound based on
quantum coherence. More so due to the fact that quantum steering and quantum
coherence have a deep connection deba1 ; deba2 ; deba3 ; deba4 . However, it
is surprisingly not the case. One can follow similar steps to come up with a
bound, which turns out to be a trivial bound with no violation by the
bipartite entangled states.

Free will.— One can verify that the same Werner state as considered for the
plot in fig. (1), will not show violation for a similar inequality as

$\underset{a\neq
a^{\prime}}{\sum_{a,a^{\prime}}}p(|{i}^{a^{\prime}-a}\rangle^{A},|{j}^{a-a^{\prime}}\rangle^{B})\leq\frac{1}{2}.$
(6)

The inequality in Eq. (6) is violated by yet another Werner state
$\rho_{w}^{\psi+}=p|\psi_{+}^{\alpha}\rangle\langle\psi_{+}^{\alpha}|+\frac{1-p}{4}I_{4}$,
where
$|\psi_{+}^{\alpha}\rangle=\sqrt{\alpha}|01\rangle+\sqrt{1-\alpha}|10\rangle$.

One can derive another inequality based on these two inequalities in Eq. (5)
and (6) just by summing both the inequalities. The new inequality is although
not tight, to show the violation, one does not need to invoke the assumption
of free will. Both the Werner states for $p>\frac{1}{\sqrt{2}}$, violates the
inequality as given below,

$\underset{a\neq a^{\prime},b\neq
b^{\prime}}{\sum_{a,a^{\prime},b,b^{\prime}=0}^{1}}p(|{i}^{a^{\prime}-a}\rangle^{A},|{j}^{b-b^{\prime}}\rangle^{B})\leq
1.$ (7)

For a $d\otimes d$ bipartite system, an equivalent inequality corresponding to
the inequality in Eq. (6) will be

$\underset{a\neq
a^{\prime}}{\sum_{a,a^{\prime}=0}^{d-1}}p(|{i}^{a^{\prime}-a}\rangle^{A},|{j}^{a-a^{\prime}}\rangle^{B})\leq\frac{d-1}{d}.$
(8)

whereas, an equivalent inequality to that given in Eq. (7) will be

$\underset{a\neq a^{\prime},b\neq
b^{\prime}}{\sum_{a,a^{\prime},b,b^{\prime}=0}^{d-1}}p(|{i}^{a^{\prime}-a}\rangle^{A},|{j}^{b-b^{\prime}}\rangle^{B})\leq\bigg{(}d-1\bigg{)}^{2},$
(9)

The inequality in Eq. (8) is tighter than that in Eq. (9). The two
inequalities above do not follow directly from the two-qubit scenario. For a
detailed analysis, see supplemental material supple .

Let us now focus on to compare the inequalities with that of the existing
bounds. For example, it is well-known that a generalized isotropic state
belonging to the Hilbert space $\mathcal{H}_{d}\otimes\mathcal{H}_{d}$ of the
form of

$\rho_{AB}=(1-p)\frac{I_{d^{2}}}{d^{2}}+\frac{p}{d}\sum_{i,j}|ii\rangle\langle
jj|$ (10)

is steerable for $p>\frac{\sum_{r=2}^{d}\frac{1}{r}}{d-1}$ and entangled for
$p>\frac{1}{d+1}$ nguyen . A similar bound on $p$ can be derived from our
inequalities and it turns out to be $p>\frac{1}{d}$. Therefore, although our
model predicts the same bound as the existing LHS model for a two-qubit Werner
state, in the higher dimensions, our model predicts tighter bounds on $p$ than
the existing model and opens a new avenue to test the new boundary of quantum
correlations.

##  IV  Experimental realization

So far, $EPR$ steering has been observed in several experiments wittmann ;
saunders ; eberle ; smith ; bennet1 ; tischler ; ksun ; yxiao . We propose an
experimental setup to show the violation of our inequality in this section.
Before we proceed further on how to realize an experiment for quantum
violation of the inequality just like the theoretical violation, let us first
focus on to understand what does this quantity on the left hand side of Eq.
(5) represent? One could write any transition probability of the state $\rho$
from $|i^{a}\rangle$ to $|i^{b}\rangle$ as $p(|i^{a-b}\rangle)=|\langle
i^{a}|\rho|i^{b}\rangle|$, where $i$ is the label depicting the eigenbasis of
an observable and $a$ and $b$ are labels for the eigenvectors in the basis.
For $\sigma_{3}$ observable, it turns out to be

$p(|z^{0-1}\rangle)=p_{01}=|\textbf{Tr}(\rho|0\rangle\langle 1|)|.$ (11)

Here, we provide a method to measure the quantity by measuring a overlap
between two states. D. K. L. Oi ${\it et.al.}$ in (oi1, ) gave the first
proposal to measure various linear and nonlinear functions of density matrices
in the interferometry directly. Later, the method was used in oi2  to measure
various overlaps as shown in the Fig. (2).

Figure 2:  We provide a schematic diagram to generate entangled states and
perform measurements in different bases. Here, $+$ denotes the $CNOT$ gate,
$H$ stands for the Hadamard gate and $z$ stands for the measurement detector.
To verify the quantum violation of the inequality in Eq. (5), one does not
need to measure the overlaps directly. Instead, one can use the identities as
given in equations in Eq. (12) and Eq. (13) in the two qubit scenario.  Figure
3:  We implemented the circuit in Fig (2) on the IBM quantum computer 5q-ibmq-
vigo. It turns out that the entangled states generated by the IBM quantum
computer maximally violates the inequality in Eq. (5) or (13) for
$i=j=\sigma_{z}$. Therefore, to measure the four terms of the Eq. (12) or
(13), first we added the Pauli-x gate (Pauli-x is not available on the
platform. Therefore, we implemented it by using the other two Pauli gates.)
and then Pauli-y gate before both the detectors. We performed two runs of the
experiment with 8192 shots each—one for each of the gates before the
detectors. The probability of getting $|00\rangle$ (or $|11\rangle$) in the
above figures is therefore, actually corresponds to the probability of getting
the state $|++\rangle$ (or $|--\rangle$) when the Pauli-x is placed before the
detectors. Thus, the entangled states produced by the IBM quantum computer
violate the Eq. (6) and the left hand side reaches up to $~{}0.92$.

We consider a state $\rho$ and measure its overlap with
$|+\rangle=\frac{|0\rangle+|1\rangle}{\sqrt{2}}$ and
$|i+\rangle=\frac{|0\rangle+i|1\rangle}{\sqrt{2}}$. We consider the value to
be $p_{+}$ and $p_{i+}$ respectively. We can also measure the overlap with the
state $|0\rangle$ and $|1\rangle$, which we consider to be $p_{0}$ and
$p_{1}$. We define $a_{01}=\text{Tr}(\rho|0\rangle\langle 1|)$ and
$a_{10}=\text{Tr}(\rho|1\rangle\langle 0|)$. One can easily show that
$p_{+}=\frac{1}{2}(p_{0}+p_{1}+a_{01}+a_{10})$ and
$p_{i+}=\frac{1}{2}(p_{0}+p_{1}-ia_{01}+ia_{10})$. Since, we know the values
of $p_{+}$, $p_{i+}$, $p_{0}$ and $p_{1}$ from the measurements of overlaps
with the corresponding states, we should be able to calculate the values of
$a_{01}=a_{10}^{*}$ and $p_{01}$ by solving the two equations.

However, one does not need to measure these local overlaps directly. Let us
consider two states $|i^{a}\rangle$ and $|i^{a^{\prime}}\rangle$ on the
Alice’s side and similarly, two states $|j^{a}\rangle$ and
$|j^{a^{\prime}}\rangle$ $(a\neq a^{\prime}$ and for two qubit scenario
$a,a^{\prime}\in\\{1,1\\})$ on the Bob’s side. Let us also form x-plus and
x-minus states as
$|i^{x\pm}\rangle=\frac{1}{\sqrt{2}}(|i^{a}\rangle\pm|i^{a^{\prime}}\rangle)$.
Similarly, y-plus and y-minus states are defined as
$|i^{y\pm}\rangle=\frac{1}{\sqrt{2}}(|i^{a}\rangle\pm
i|i^{a^{\prime}}\rangle)$. One can easily show that the left hand sides of the
inequalities in Eq. (5) and 6 turn out to be

$\displaystyle\underset{a\neq
a^{\prime}}{\sum_{a,a^{\prime},}}p(|{i}^{a-a^{\prime}}\rangle^{A},|{j}^{a-a^{\prime}}\rangle^{B})$
$\displaystyle=$ $\displaystyle\langle
i^{x+}j^{x+}|\rho|i^{x+}j^{x+}\rangle+\langle
i^{x-}j^{x-}|\rho|i^{x-}j^{x-}\rangle-\langle
i^{y+}j^{y+}|\rho|i^{y+}j^{y+}\rangle-\langle
i^{y-}j^{y-}|\rho|i^{y-}j^{y-}\rangle$ (12) and $\displaystyle\underset{a\neq
a^{\prime}}{\sum_{a,a^{\prime},}}p(|{i}^{a-a^{\prime}}\rangle^{A},|{j}^{a^{\prime}-a}\rangle^{B})$
$\displaystyle=$ $\displaystyle\langle
i^{x+}j^{x+}|\rho|i^{x+}j^{x+}\rangle+\langle
i^{x-}j^{x-}|\rho|i^{x-}j^{x-}\rangle+\langle
i^{y+}j^{y+}|\rho|i^{y+}j^{y+}\rangle$ (13) $\displaystyle+$
$\displaystyle\langle i^{y-}j^{y-}|\rho|i^{y-}j^{y-}\rangle-1.$

We use these identities for the two-qubit scenario to show the violation of
the symmetric local hidden state model by the quantum mechanical states in the
IBM quantum computer as depicted in the Fig (3). However, showing the
violation in the higher dimensions is a nontrivial task and at the same time,
is of utmost importance for verifying our prediction.

##  V Self-testing

One of the ab initio motivations behind the wanderlust in the dark for an
alternative description of $lhs$ model is of course to find a larger set of
self-testable states and measurements, which can be used in various quantum
information theoretic protocols. A state $|\psi_{AB}\rangle$ and the
measurement basis $|\phi^{a}\rangle$ $(a=0,1,2,....,$ a is the outcome of a
measurement in the $\phi$ basis and ranges upto the dimension of the
corresponding system$)$ are self-testable by the correlation measured via
joint transition probability
$p(|\phi^{a-a^{\prime}}\rangle^{A},|\phi^{b-b^{\prime}}\rangle^{B})$ if all
states $\rho_{AB}$ and measurements $|\widetilde{\phi}^{a}\rangle^{A}$
compatible with
$p(|\phi^{a-a^{\prime}}\rangle^{A},|\phi^{b-b^{\prime}}\rangle^{B})$ or
$\langle\phi^{a}_{A}\phi^{b}_{B}|\rho_{AB}|\phi^{a^{\prime}}_{A}\phi^{b^{\prime}}_{B}\rangle$,
turns out to be $|\psi_{AB}\rangle$ and the measurement $|\phi^{a}\rangle$.

To self-test one of the Bell states and measurement bases, we start with the
assumptions that the transition amplitudes of an arbitrary state $\rho_{AB}$
in the measurement bases $|\phi^{a}\rangle^{A}$ and $|\xi^{a}\rangle^{A}$ take
the maximal possible value of $\frac{1}{2}$, i.e.,

$\displaystyle\langle\phi^{0}_{A}\phi^{0}_{B}|\rho_{AB}|\phi^{1}_{A}\phi^{1}_{B}\rangle$
$\displaystyle=$ $\displaystyle\frac{1}{2},\hskip
2.84544pt\langle\xi^{0}_{A}\xi^{0}_{B}|\rho_{AB}|\xi^{1}_{A}\xi^{1}_{B}\rangle=\frac{1}{2}$
$\displaystyle\langle\phi^{0}_{A}\phi^{1}_{B}|\rho_{AB}|\phi^{1}_{A}\phi^{0}_{B}\rangle$
$\displaystyle=$ $\displaystyle 0,$ (14)

where $|\phi^{0}_{A(B)}\rangle$ and $|\phi^{1}_{A(B)}\rangle$ are two
orthonormal states of Alice(Bob).

We define
$|\xi^{0}_{A(B)}\rangle=\frac{1}{\sqrt{2}}(|\phi^{0}_{A(B)}\rangle+|\phi^{1}_{A(B)}\rangle)$
and
$|\xi^{1}_{A(B)}\rangle=\frac{1}{\sqrt{2}}(|\phi^{0}_{A(B)}\rangle-|\phi^{1}_{A(B)}\rangle)$.
We have the freedom to express the state $\rho_{AB}$ in the
$|\phi^{i}_{A(B)}\rangle$ basis as $\rho_{AB}=\underset{i,k,\in
S_{d_{A}}}{\sum_{j,l\in
S_{d_{B}}}}\alpha_{ij}^{kl}|\phi^{i}_{A}\phi^{j}_{B}\rangle\langle\phi^{k}_{A}\phi^{l}_{B}|$
(without loss of generality), where $S_{d}=\\{0,1,2,....,d\\}$. Here, $d$ is
the dimension of the system. Hermiticity condition of the density matrix
implies $\alpha^{kl}_{ij}=\alpha^{ij*}_{kl}$ and the normalization condition
implies $\sum_{ij}\alpha^{ij}_{ij}=1$. First two equations and the Hermiticity
condition imply that $\alpha_{11}^{00}=\alpha_{00}^{11}=\frac{1}{2}$ and
$\alpha_{01}^{10}=\alpha_{10}^{01}=0$. The third equation together with the
Hermiticity condition implies
$\underset{k,l}{\sum_{i,j=0}^{1}}(-1)^{k+l}\alpha_{ij}^{kl}=2$ and
$\underset{k,l}{\sum_{i,j=0}^{1}}(-1)^{i+j}\alpha_{ij}^{kl}=2.$ Adding both
the equations and putting the values of $\alpha_{11}^{00}=\alpha_{00}^{11}$
and $\alpha_{01}^{10}=\alpha_{10}^{01}$ we get,

$\alpha^{00}_{00}-\alpha^{01}_{01}-\alpha^{10}_{10}+\alpha^{11}_{11}=1.$ (15)

Subtracting the Eq. (15) from the normalization condition, we get

$2\alpha^{01}_{01}+2\alpha^{10}_{10}+\sum_{i,j\neq 0,1}\alpha_{ij}^{ij}=0.$
(16)

The fact that all the diagonal elements of a density matrix are non-negative
and the Eq. (16) imply $\alpha_{ij}^{ij}=0$ $\forall i,j$ except
$\alpha_{00}^{00}$ and $\alpha_{11}^{11}$ and we get,

$\alpha_{00}^{00}+\alpha_{11}^{11}=1.$ (17)

Using the Cauchy-Schwartz inequality, one can show that

$\langle\phi^{k}\phi^{l}|\rho|\phi^{i}\phi^{j}\rangle|^{2}\leq|\langle\phi^{i}\phi^{j}|\rho|\phi^{i}\phi^{j}\rangle||\langle\phi^{k}\phi^{l}|\rho|\phi^{k}\phi^{l}\rangle|.$
(18)

To prove the inequality (18), we start with
$|\langle\phi^{k}\phi^{l}|\rho|\phi^{i}\phi^{j}\rangle|^{2}=\langle\phi^{k}\phi^{l}|\rho|\phi^{i}\phi^{j}\rangle\langle\phi^{i}\phi^{j}|\rho|\phi^{k}\phi^{l}\rangle=\text{Tr}(|\phi^{i}\phi^{j}\rangle\langle\phi^{i}\phi^{j}|\rho|\phi^{k}\phi^{l}\rangle\langle\phi^{k}\phi^{l}|\rho)=\text{Tr}(A\rho
B\rho)$, where we consider
$A=|\phi^{i}\phi^{j}\rangle\langle\phi^{i}\phi^{j}|$ and
$|\phi^{k}\phi^{l}\rangle\langle\phi^{k}\phi^{l}=B$. Now using the Cauchy-
Schwarz inequality, one can show that $\text{Tr}(A\rho
B\rho)\leq\sqrt{\text{Tr}(\rho A\rho A)\text{Tr}(\rho B\rho B)}$. This is
nothing but the inequality (18), which in turn implies,

$|\alpha^{kl}_{ij}|^{2}\leq|\alpha_{ij}^{ij}||\alpha_{kl}^{kl}|$ (19)

We use the inequality to show that for all vanishing diagonal elements, i.e.,
$\alpha_{ij}^{ij}=0$, all the off-diagonal elements are zero, i.e.,
$\alpha_{ij}^{kl}=\alpha_{kl}^{ij}=0\forall i,j,k,l$.

Using the same inequality, one can also show that
$\alpha_{00}^{00}\alpha_{11}^{11}\geq\frac{1}{4}$. From the bound together
with the Eq. (17), one can show that

$(\alpha^{00}_{00}-\alpha^{11}_{11})^{2}\\\
=\alpha_{00}^{00^{2}}+\alpha_{11}^{11^{2}}-2\alpha_{00}^{00}\alpha_{11}^{11}=1-4\alpha_{00}^{00}\alpha_{11}^{11}\leq
0.$ (20)

Since, $0\leq(\alpha_{00}^{00}-\alpha_{11}^{11})^{2}\leq 0$ implies
$\alpha_{00}^{00}=\alpha_{11}^{11}$, we get
$\alpha_{00}^{00}=\alpha_{11}^{11}=\frac{1}{2}$ from eq. (17). Using the fact
that $\alpha_{01}^{01}=0$ in the above inequality, one can easily show that
$|\alpha_{01}^{ij}|^{2}\leq 0$. This implies
$\alpha_{01}^{ij}=\alpha^{01}_{ij}=0$ $\forall i,j.$ Therefore, we are left
with terms
$\alpha_{11}^{00}=\alpha_{00}^{11}=\alpha_{00}^{00}=\alpha_{11}^{11}=\frac{1}{2}$
implying that the state $\rho_{AB}$ is the Bell state $|\phi^{+}\rangle$ and
the corresponding measurement bases are $z$ and $x$.

The next natural question to ask is possibly: Is it possible to extend the
formalism of self-testing here for arbitrary states? A general proof for
arbitrary two qubit entangled states of the form of
$|\psi\rangle=\sqrt{\alpha}|00\rangle+\sqrt{1-\alpha}e^{i\phi}|11\rangle$ can
indeed be extended from the simplified proof given here.

##  VI Resource theory

We put forward a resource theory of this new form of nonlocality in this
section. Just like other resources resource1 ; resource2 ; resource3 ;
resource4  in quantum information theory, the first element we need to dig out
is the free operations, i.e., operations under which a state belonging to
$slhs$ model remains in $slhs$. Here we show that like entanglement or Bell
nonlocality EPR ; Schrodinger ; Schrodinger1 ; bell , this new form of
nonlocality also cannot be created by local completely positive trace
preserving operations $(lcptp)$ i.e.,

######  Theorem 1.

a state $\rho_{ab}$ belonging to the $slhs$ model remains in the $slhs$ model
under $lcptp$ operations $(\mathcal{LCPTP})$.

We lay down the proof of the theorem in the supplemental material supple .

Once the free operations are found, the next logical step is to introduce an
axiomatic approach to define the measures of the nonlocality. Finding the
measure of a resource is a difficult task and even more difficult without
knowing the conditions, it must satisfy. In this regard, we lay down a set of
axioms, which must be satisfied by the measure $\mathcal{N}$ of the new form
of nonlocality—$(a)$ $\mathcal{N}(\rho_{AB})=0$ $\forall\rho_{AB}\in slhs$,
$(b)$
$\sum_{k}p_{k}\mathcal{N}(\mathcal{L}_{k}(\rho_{AB}))\leq\mathcal{N}(\rho_{AB})$
$\forall\mathcal{L}_{k}\in\mathcal{LCPTP}$, such that
$\mathcal{L}_{k}(.)=\frac{r_{k}^{A}\otimes r_{k}^{B}(.)r_{k}^{A\dagger}\otimes
r_{k}^{B\dagger}}{p_{k}}$, where $p_{k}=\textbf{Tr}(r_{k}^{A}\otimes
r_{k}^{B}(.)r_{k}^{A\dagger}\otimes r_{k}^{B\dagger})$ and bipartite states
$\rho_{AB}$. Additionally, it must not increase under classical mixing of
bipartite states $\rho_{AB}^{i}$, i.e., $(c)$
$\mathcal{N}(\rho_{AB})\leq\sum_{i}p_{i}\mathcal{N}(\rho_{AB}^{i})$, such that
$\rho_{AB}=\sum_{i}p_{i}\rho_{AB}^{i}$ and $\sum_{i}p_{i}=1$.

In the supplemental material, we show that $|\langle ab|\rho_{AB}|cd\rangle|$
satisfies the axioms $(b)$ and $(c)$ for any bipartite state $\rho_{AB}$. By
some numerical adjustments, we turn the quantity into a measure of the new
form of non-locality.

##  VII Conclision

There was a formidable belief that the only way to describe the $lhs$ game is
by giving up the freedom whether to trust or not to trust a party and
retaining it for the other, thereby introducing an asymmetry in the game. In
the $lhs$ game described earlier, Alice is considered to be an un-trusted
party. Beauty of science is that it allows one to look at each and everything
with suspicion, doubts and skepticism. Our skepticism about the game led to an
alternative description of the $lhs$ model, which is symmetric, trust-less and
can be used in self-testing the states and measurements. Therefore, it
provides a better and operationally more efficient alternative to the Bell
inequality for the self-testing and other device-independent protocols device
; branciard ; bennet . The symmetric form of $lhs$ model reveals more
nonlocality than the existing $lhs$ or $lhv$ models (particularly in higher
dimensions $d>2\otimes 2$) opening a new frontier for the experimentalists to
test the new boundary of quantum correlation. In the two qubit scenario, an
experimental violation of the inequalities can easily be shown and we
performed an experiment in this scenario on the IBM quantum computer.

Bell nonlocality and $EPR$ nonlocality have already been verified
experimentally tittel1 ; tittel2 ; nature ; panexp1 ; zhang ; zhao ; weihs ;
optcomm ; deng ; cavalcanti ; steinlechner ; hanchen ; bennet ; wittmann ;
smith ; saunders ; sun ; armstrong ; cmli  under different scenarios and
assumptions in the last four decades. It will be really interesting to show
the experimental violation of the inequalities given in this letter. Moreover,
our study is limited to the bipartite scenario. Extending the formalism in the
multipartite scenarios will be another interesting direction to explore.

Acknowledgements—DM and DK are supported by the National Research Foundation,
Prime Minister’s Office, Singapore and the Ministry of Education, Singapore
under the Research Centres of Excellence programme.

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##  VIII Appendix

## Stronger SLHS model

A stronger $slhs$ model would be just to express the complex off-diagonal
terms corresponding to the transition probabilities of a bipartite state in
terms of the complex off-diagonal terms of hidden states,

$\langle
i^{a}i^{b}|\rho_{AB}|i^{a^{\prime}}i^{b^{\prime}}\rangle=\sum_{\lambda}p(\lambda)\langle
i^{a}|\rho^{\lambda}_{A}|i^{a^{\prime}}\rangle\langle
i^{b}|\rho^{\lambda}_{B}|i^{b^{\prime}}\rangle,$ (A1)

where $\rho^{\lambda}_{A}$ and $\rho^{\lambda}_{B}$ are hidden states of Alice
and Bob respectively with a distribution $p(\lambda)$ prepared by the unknown
source to cheat them.

The inequality in Eq. (A1) is stronger than the Eq. (4) of the article is due
to the fact that the former implies the later but not the other way around. To
prove it, we start with,

$\displaystyle|\langle
i^{a}i^{b}|\rho_{AB}|i^{a^{\prime}}i^{b^{\prime}}\rangle|$ $\displaystyle\leq$
$\displaystyle\sum_{\lambda}p(\lambda)|\langle
i^{a}|\rho^{\lambda}_{A}|i^{a^{\prime}}\rangle||\langle
i^{b}|\rho^{\lambda}_{B}|i^{b^{\prime}}\rangle|,$

where we use the Eq. (A1) and the triangle inequality. The inequality above
implies that there must exist a set of hidden states $\sigma_{A}^{\lambda}$
and $\sigma_{B}^{\lambda}$ with a distribution $p^{\prime}(\lambda)$ such that

$\displaystyle|\langle
i^{a}i^{b}|\rho_{AB}|i^{a^{\prime}}i^{b^{\prime}}\rangle|=\sum_{\lambda}p^{\prime}(\lambda)|\langle
i^{a}|\sigma^{\lambda}_{A}|i^{a^{\prime}}\rangle||\langle
i^{b}|\sigma^{\lambda}_{B}|i^{b^{\prime}}\rangle|,$

which is nothing but the Eq. (4) of the letter.

## Proof of Eq. 8.

In the symmetric local hidden state model section of the main article, we have
given two inequalities in Eq. (8) and (9) for general bipartite scenario. The
inequalities do not follow directly from the proof given in Eq. (5) for the
two-qubit scenario. Here we provide the general proof. We start with the joint
transition probability as

$\displaystyle p$
$\displaystyle(|i^{a-a^{\prime}}\rangle^{A},|j^{a-a^{\prime}}\rangle^{B})=\sum_{\lambda}p(\lambda)p\bigg{(}|{i}^{a-a^{\prime}}\rangle^{A}|\lambda\bigg{)}p\bigg{(}|{j}^{a-a^{\prime}}\rangle^{B}|\lambda\bigg{)}$
(A2) $\displaystyle\leq$
$\displaystyle\sqrt{\sum_{\lambda^{\prime}}p(\lambda^{\prime})p^{2}\bigg{(}|i^{a-a^{\prime}}\rangle^{A}\bigg{|}\lambda^{\prime}\bigg{)}\sum_{\lambda}p(\lambda)p^{2}\bigg{(}|j^{a-a^{\prime}}\rangle^{B}\bigg{|}\lambda\bigg{)}}$
$\displaystyle=$
$\displaystyle\sqrt{\overline{p^{2}\bigg{(}|i^{a-a^{\prime}}\rangle^{A}\bigg{)}}.\overline{p^{2}\bigg{(}|j^{a-a^{\prime}}\rangle^{B}\bigg{)}}}.$

Here, we used the fact that the transition probability
$p(|i^{a-a^{\prime}}\rangle)$ is bounded from above by $\frac{1}{2}$ in the
derivation of the two-qubit inequality in Eq. (5). In the higher dimensions,
where $a$ and $a^{\prime}$ have more than two values $0$ and $1$,
$p(|i^{0-1}\rangle)=\frac{1}{2}$ implies
$p(|i^{1-2}\rangle)=p(|i^{0-2}\rangle)=0$. Therefore, for
$p(|i^{a-a^{\prime}}\rangle=\frac{1}{2})$, the maximal value cannot be reached
for local dimensions more than two. The fact that $\sum_{t}\max_{x}f(t,x)$
will not necessarily be greater than $\max_{x}\sum_{t}f(t,x)$ for any $f(t,x)$
and we are supposed to find the maximum value of a quantity of the form of
$\max_{x}\sum_{t}f(t,x)$, implies we need to take an alternative route. We
need to calculate the maximum of
$\sum_{a,a^{\prime}}\sqrt{\overline{p^{2}\bigg{(}|i^{a-a^{\prime}}\rangle^{A}\bigg{)}}.\overline{p^{2}\bigg{(}|j^{a-a^{\prime}}\rangle^{B}\bigg{)}}}$
for arbitrary states $\rho_{A}$ and $\sigma_{B}$. One can show that the maxima
can be found when $p(|i^{a-a^{\prime}}\rangle^{A})=\frac{1}{d}\text{$\forall
a$ and $a^{\prime}$}$. Therefore,

$\displaystyle\max\sum_{a,a^{\prime}}\sqrt{\overline{p^{2}\bigg{(}|i^{a-a^{\prime}}\rangle^{A}\bigg{)}}.\overline{p^{2}\bigg{(}|j^{a-a^{\prime}}\rangle^{B}\bigg{)}}}=\frac{d-1}{d}.$
(A3)

This proves the inequality in Eq. (8) in the article.

## Proof of Theorem 1.

We consider a state $\rho_{AB}$ belongs to the set of $slhs$ states, such that

$p\bigg{(}|i^{a-a^{\prime}}\rangle,|j^{a-a^{\prime}}\rangle\bigg{)}=|\langle
i^{a}j^{a}|\rho_{AB}|i^{a^{\prime}}j^{a^{\prime}}\rangle|\overset{SLHS}{=}\sum_{\lambda}p(\lambda)|\langle
i^{a}|\rho_{A}^{\lambda}|i^{a^{\prime}}\rangle||\langle
j^{a}|\rho_{B}^{\lambda}|j^{a^{\prime}}\rangle|,$ (A4)

for any basis labeled by $i$ and a particular vector from the basis labeled by
$a$, such that $a\neq a^{\prime}$. The state $\rho_{AB}$, under local $cptp$
operation, transforms to the state $\rho^{\prime}_{AB}$, such that
$\rho^{\prime}_{AB}=\mathcal{L}(\rho_{AB})=\sum_{k}r_{k}^{A}\otimes
r_{k}^{B}\rho_{AB}r_{k}^{A\dagger}\otimes r_{k}^{B\dagger}$. We define
$\mathcal{L}_{k}(\rho_{AB})=\frac{r_{k}^{A}\otimes
r_{k}^{B}\rho_{AB}r_{k}^{A\dagger}\otimes
r_{k}^{B\dagger}}{\text{Tr}(r_{k}^{A}\otimes
r_{k}^{B}\rho_{AB}r_{k}^{A\dagger}\otimes r_{k}^{B\dagger})}$,
$p_{k}=\text{Tr}(r_{k}^{A}\otimes r_{k}^{B}\rho_{AB}r_{k}^{A\dagger}\otimes
r_{k}^{B\dagger})$, such that $\sum_{k}p_{k}=1$ and
$\sum_{k}r_{k}^{\dagger}r_{k}\leq I$ for both the systems. Here, we show that
the state $\rho^{\prime}_{AB}$ also belongs to the set of $SLHS$ states. We
start with the quantity $\langle
i^{a}i^{a}|\rho^{\prime}_{AB}|i^{a^{\prime}}i^{a^{\prime}}\rangle$ such that
$a\neq a^{\prime}$

$\displaystyle\langle
i^{a}j^{a}|\rho^{\prime}_{AB}|i^{a^{\prime}}j^{a^{\prime}}\rangle$
$\displaystyle=$ $\displaystyle\sum_{k}\langle i^{a}j^{a}|r_{k}^{A}\otimes
r_{k}^{B}\rho_{AB}r_{k}^{A\dagger}\otimes
r_{k}^{B\dagger}|i^{a^{\prime}}j^{a^{\prime}}\rangle=\sum_{k}\text{Tr}\Big{(}r_{k}^{A}\otimes
r_{k}^{B}\rho_{AB}r_{k}^{A\dagger}\otimes
r_{k}^{B\dagger}|i^{a^{\prime}}j^{a^{\prime}}\rangle\langle
i^{a}j^{a}|\Big{)}$ (A5) $\displaystyle=$
$\displaystyle\sum_{k}\text{Tr}\Big{(}\rho_{AB}A_{k}\otimes
B_{k}\Big{)}=\sum_{k,m,n,p,q}\langle mn|\rho_{AB}|pq\rangle\langle
pq|A_{k}\otimes B_{k}|mn\rangle,$ where
$A_{k}=r_{k}^{A\dagger}|i^{a^{\prime}}\rangle\langle i^{a}|r_{k}^{A}$ and
$B_{k}=r_{k}^{B\dagger}|j^{a^{\prime}}\rangle\langle j^{a}|r_{k}^{B}$
$\displaystyle=$ $\displaystyle\sum_{k,m,n,p,q,\lambda}\langle
p|A_{k}|m\rangle\langle q|B_{k}|n\rangle p(\lambda)\langle
m|\rho_{A}^{\lambda}|p\rangle\langle
n|\rho_{B}^{\lambda}|q\rangle=\sum_{k,p,q,\lambda}p(\lambda)\langle
p|A_{k}\rho^{\lambda}_{A}|p\rangle\langle q|B_{k}\rho_{B}^{\lambda}|q\rangle$
$\displaystyle=$ $\displaystyle\sum_{k,\lambda}p(\lambda)\langle
i^{a}|r_{k}^{A}\rho_{A}^{\lambda}r_{k}^{A\dagger}|i^{a^{\prime}}\rangle\langle
j^{a}|r_{k}^{A}\rho_{B}^{\lambda}r_{k}^{B\dagger}|j^{a^{\prime}}\rangle.$

This proof is for the stronger version of our $slhs$ model. To prove it for
the weaker version, we proceed further with the following,

$\displaystyle\Big{|}\langle
i^{a}j^{a}|\rho^{\prime}_{AB}|i^{a^{\prime}}j^{a^{\prime}}\rangle\Big{|}$
$\displaystyle=$ $\displaystyle\Big{|}\sum_{k,\lambda}p(\lambda)\langle
i^{a}|r_{k}^{A}\rho_{A}^{\lambda}r_{k}^{A\dagger}|i^{a^{\prime}}\rangle\langle
j^{a}|r_{k}^{A}\rho_{B}^{\lambda}r_{k}^{B\dagger}|j^{a^{\prime}}\rangle\Big{|}\leq\sum_{k,\lambda}p(\lambda)\Big{|}\langle
i^{a}|r_{k}^{A}\rho_{A}^{\lambda}r_{k}^{A\dagger}|i^{a^{\prime}}\rangle\langle
j^{a}|r_{k}^{A}\rho_{B}^{\lambda}r_{k}^{B\dagger}|j^{a^{\prime}}\rangle\Big{|}$
(A6) $\displaystyle=$ $\displaystyle\sum_{k,\lambda}p(\lambda)\Big{|}\langle
i^{a}|r_{k}^{A}\rho_{A}^{\lambda}r_{k}^{A\dagger}|i^{a^{\prime}}\rangle\Big{|}\Big{|}\langle
j^{a}|r_{k}^{A}\rho_{B}^{\lambda}r_{k}^{B\dagger}|j^{a^{\prime}}\rangle\Big{|}$

which implies that the state $\rho_{AB}^{\prime}$ also belongs to the set of
$slhs$. In the equation above in Eq. (A5), in the second line, we use the fact
that $r_{k}^{A(B)}|i^{a^{\prime}}\rangle\langle
i^{a}|r_{k}^{A(B)\dagger}=A_{k}(B_{k})$, in the third line, we use the fact
that the state $\rho_{AB}\in slhs$, i.e., the Eq. (A1). In the first
inequality in Eq. (A6), we use the triangle inequality for complex numbers.

##  IX Proof of Axioms.

We consider $\mathcal{N}(\rho_{AB})=\underset{a\neq a^{\prime},b\neq
b^{\prime}}{\sum_{a,a^{\prime},b,b^{\prime}}}\max\Bigg{\\{}\max_{i,j}|\langle
i^{a}j^{b}|\sum_{i}p_{i}\rho_{AB}^{i}|i^{a^{\prime}}j^{b^{\prime}}\rangle|-\frac{1}{d^{2}},0\Bigg{\\}}$
to be a measure of the new form of $EPR$ correlation, where $d$ is the
dimension of the subsystems. To prove the axiom (b), we start with the fact
that

$\displaystyle|\langle
ab|\mathcal{L}_{k}^{AB}\rho_{AB}|cd\rangle|=\frac{|\sum_{n,m,p,q}\langle
mn|\rho_{AB}|pq\rangle\langle pq|A_{k}\otimes B_{k}|mn\rangle|}{p_{k}},$

where $A_{k}=r^{A\dagger}_{k}|c\rangle\langle a|r_{k}^{A}$ and
$B_{k}=r^{B\dagger}_{k}|d\rangle\langle b|r_{k}^{B}$. Therefore,

$\displaystyle\sum_{k}p_{k}|\langle
ab|\mathcal{L}_{k}^{AB}(\rho_{AB})|cd\rangle|$ $\displaystyle\leq$
$\displaystyle\sum_{n,m,p,q,k}|\langle mn|\rho_{AB}|pq\rangle\langle
pq|A_{k}\otimes B_{k}|mn\rangle|$ (A8) $\displaystyle=$
$\displaystyle\sum_{n,m,p,q,k}|\langle mn|\rho_{AB}|pq\rangle||\langle
p|A_{k}|m\rangle||\langle q|B_{k}|n\rangle|,$

where in the inequality, we use the triangle inequality. Now, we start with
the quantity $|\langle p|A_{k}|m\rangle|$ as

$\displaystyle\sum_{k}|\langle p|A_{k}|m\rangle|=\sum_{k}|\langle
p|r^{A\dagger}_{k}|c\rangle\langle a|r_{k}^{A}|m\rangle|$
$\displaystyle=\sum_{k}|[r^{A\dagger}_{k}]_{cp}[r^{A}_{k}]_{ma}|$
$\displaystyle=\sum_{k}|[r^{A\dagger}_{k}]_{cp}[r^{A}_{k}]_{pa}|+\underset{p\neq
m}{\sum_{k}}|[r^{A\dagger}_{k}]_{cp}[r^{A}_{k}]_{ma}|,$ (A9)

where the first term is zero due to the fact that
$\sum_{k}r^{\dagger}_{k}r_{k}=I_{d}$ and off-diagonal terms of an identity
matrix are all zero. We start with the second term as

$\displaystyle\underset{p\neq
m}{\sum_{k}}|[r^{A\dagger}_{k}]_{cp}[r^{A}_{k}]_{ma}|\leq\sum_{k}|[r^{A\dagger}_{k}]_{cp}[r^{A}_{k}]_{ma}|$
(A10) $\displaystyle\leq$
$\displaystyle\sqrt{\sum_{k}|[r_{k}^{A\dagger}]_{cp}|^{2}\sum_{l}|[r_{l}^{A}]_{ma}|^{2}}.$

Now, we start with the first term under the square-root, i.e.,

$\displaystyle\sum_{k}|[r_{k}^{A\dagger}]_{cp}|^{2}$ $\displaystyle=$
$\displaystyle\sum_{k}\langle c|r^{A\dagger}_{k}|p\rangle\langle
p|r_{k}^{A}|c\rangle.$ (A11)

We know $\text{Tr}(\sum_{k}r_{k}^{A}\rho_{0}r_{k}^{A\dagger})=1$ and
$\text{Tr}(r_{k}^{A}\rho_{0}r_{k}^{A\dagger})=p_{k}$ such that
$\sum_{k}p_{k}=1$ for any arbitrary state $\rho_{0}$. We consider
$\rho_{0}=|c\rangle\langle c|$. Therefore, we get

$\displaystyle p_{k}$ $\displaystyle=$
$\displaystyle\text{Tr}(r_{k}^{A}\rho_{0}r_{k}^{A\dagger})=\sum_{m}\langle
m|r_{k}^{A}|c\rangle\langle c|r_{k}^{A\dagger}|m\rangle$ or,
$\displaystyle\langle m|r_{k}^{A}|c\rangle\langle
c|r_{k}^{A\dagger}|m\rangle\leq p_{k},$ (A12)

where $|m\rangle$ is an arbitrary state and forms a complete basis and the
inequality comes from the fact that each term within the summation is
positive. Now, we start with the same quantity again to show that

$\displaystyle p_{k}$ $\displaystyle=$ $\displaystyle\sum_{m}\langle
m|r_{k}^{A}|c\rangle\langle c|r_{k}^{A\dagger}|m\rangle$ (A13)
$\displaystyle=$ $\displaystyle\sum_{m}\langle
m|r_{k}^{A\dagger}r_{k}^{A}|c\rangle\langle c||m\rangle=\langle
c|r_{k}^{A\dagger}r_{k}^{A}|c\rangle.$

Thus, from Eq. (A12) and (A13), we get

$\displaystyle|\langle m|r_{k}^{A}|c\rangle|^{2}\leq p_{k}=\delta_{mc}\langle
m|r_{k}^{A\dagger}r_{k}^{A}|c\rangle$ (A14)

for any states $|m\rangle$ and $|c\rangle$. Now, from Eq. (A14) and (A11), we
get

$\displaystyle\sum_{k}|[r_{k}^{A\dagger}]_{cp}|^{2}\leq\delta_{cp}.$ (A15)

Using Eq. (A8), (A9), (A10) and (A15), we get

$\displaystyle\sum_{k}p_{k}|\langle
i^{a}i^{b}|\frac{\mathcal{L}_{k}^{AB}(\rho_{AB})}{p_{k}}|i^{c}i^{d}\rangle|\leq|\langle
i^{a}i^{b}|\rho_{AB}|i^{c}i^{d}\rangle|,$ (A16)

which proves the axiom (b).

Now, we turn to the axiom (c). We consider a bipartite state $\rho_{AB}$ such
that $\rho_{AB}=\sum_{i}p_{i}\rho_{AB}^{i}$, where $\sum_{i}p_{i}=1$. We start
with the following

$\displaystyle|\langle
ab|\sum_{i}p_{i}\rho_{AB}^{i}|cd\rangle|=|\sum_{i}p_{i}\langle
ab|\rho_{AB}^{i}|cd\rangle|$ $\displaystyle\leq\sum_{i}|p_{i}\langle
ab|\rho_{AB}^{i}|cd\rangle|=\sum_{i}p_{i}|\langle ab|\rho_{i}|cd\rangle|.$
(A17)

Thus, we prove that
$\mathcal{N}(\rho_{AB})\leq\sum_{i}p_{i}\mathcal{N}(\rho_{AB}^{i})$.

## No separable state can violate the inequalities

We consider the transition probability of a bipartite, two-qubit state from
$|0\rangle$ to $|1\rangle$ on Alice and as well as Bob’s side as
$p(|{z}^{0-1}\rangle^{A},|{z}^{0-1}\rangle^{B})$, where $|z^{0}\rangle$ or
$|z^{1}\rangle$ stands for the states $|0\rangle$ or $|1\rangle$ respectively.
For a two-qubit separable state $\rho_{AB}$, the transition probability is
given by

$\displaystyle p(|{z}^{0-1}\rangle^{A},|{z}^{0-1}\rangle^{B})=\langle
00|\rho_{AB}|11\rangle\langle 11|\rho_{AB}|00\rangle$
$\displaystyle=\sum_{\lambda,\lambda^{\prime}}p_{\lambda}p_{\lambda^{\prime}}\rho^{A}_{01}(\lambda)\rho^{B}_{01}(\lambda)\rho^{A}_{10}(\lambda^{\prime})\rho^{B}_{10}(\lambda^{\prime})$
$\displaystyle\leq\sum_{\lambda}p_{\lambda}|\rho^{A}_{01}(\lambda)|^{2}\sum_{\lambda^{\prime}}p_{\lambda^{\prime}}|\rho^{B}_{01}(\lambda^{\prime})|\leq\frac{1}{4}.$
(A18)

Therefore, two terms $p(|{z}^{0-1}\rangle^{A},|{z}^{0-1}\rangle^{B})$ and
$p(|{z}^{1-0}\rangle^{A},|{z}^{1-0}\rangle^{B})$ can at most add upto
$\frac{1}{2}$ and cannot go beyond that value. The arguments here can be
extended even for the general bipartite separable states. Therefore, separable
states cannot violate the inequality given in Eq. (5)-(9) in the main article.
Does the same hold for the existing one-sided LHS model as well? We show that
the answer to this question to be affirmative.

To prove it, we start with the steering game, where Alice prepares a two-qubit
bipartite state. She sends one part of it to Bob and keeps another part with
her. She claims that his state is entangled with her and she can steer his
state. Bob does not believe Alice. He asks Alice to perform measurements in
certain bases. After Alice’s measurements in each basis, Bob measures certain
quantities on his system and he claims that the (conditional) states can in
principle be explained by hidden state model as laid down in Eq. (1). The
overall state of Alice and Bob can be written as (Bob’s assumption)

$\rho_{AB}=\sum_{a,\theta}p(a,\theta)|a\rangle\langle
a|\otimes\rho_{a|\Pi(\theta)}.$ (A19)

Now on this state, if Alice and Bob really performs the measurements as
prescribed in the article, one can easily show following the previous
arguments for separable states that such states cannot violate the inequality
based on the symmetric hidden state model.