# Self-testing of binary Pauli measurements requiring neither entanglement nor
any dimensional restriction

Ananda G. Maity<EMAIL_ADDRESS>S. N. Bose National Centre for Basic
Sciences, Block JD, Sector III, Salt Lake, Kolkata 700 106, India  Shiladitya
Mal<EMAIL_ADDRESS>Harish-Chandra Research Institute, HBNI,
Chhatnag Road, Jhunsi, Allahabad 211 019, India  Department of Physics and
Center for Quantum Frontiers of Research and Technology (QFort), National
Cheng Kung University, Tainan 701, Taiwan  Chellasamy Jebarathinam
<EMAIL_ADDRESS>Department of Physics and Center for Quantum Frontiers
of Research and Technology (QFort), National Cheng Kung University, Tainan
701, Taiwan  Center for Theoretical Physics, Polish Academy of Sciences, Aleja
Lotników 32/46, 02-668 Warsaw, Poland  A. S. Majumdar<EMAIL_ADDRESS>S.
N. Bose National Centre for Basic Sciences, Block JD, Sector III, Salt Lake,
Kolkata 700 106, India

###### Abstract

Characterization of quantum devices received from unknown providers is a
significant primary task for any quantum information processing protocol.
Self-testing protocols are designed for this purpose of certifying quantum
components from the observed statistics under a set of minimal assumptions.
Here we propose a self-testing protocol for certifying binary Pauli
measurements employing the violation of a Leggett-Garg inequality. The
scenario based on temporal correlations does not require entanglement, a
costly and fragile resource. Moreover, unlike previously proposed self-testing
protocols in the prepare and measure scenario, our approach requires neither
dimensional restrictions, nor other stringent assumptions on the type of
measurements. We further analyse the robustness of this hitherto unexplored
domain of self-testing of measurements.

##  I Introduction

It is hard to overemphasise the role of emerging quantum technology in recent
times. Various real life applications, such as quantum key distribution qkd ,
quantum sensing qsensing , quantum metrology metrology , quantum internet
internet , and machine learning mlearning  have been investigated with great
prospects. For such prospects to materialize in practice, it becomes utmost
important to design experiments which can test whether the quantum components
of the required devices function properly. To this end, various certification
or verification protocols have been employed, such as those based on
tomography and benchmarking bench , as well as self-testing protocols based
essentially on observed statistics my'98 ; my'04 .

Among several certification protocols generic quantum tomography is the most
powerful, but at the same time it is the most resource consuming. Randomized
benchmarking refers to a collection of methods that aim at reliably estimating
the figure of merit of the overlap between the physical quantum process and
it’s ideal counterpart. For both randomized benchmarking or quantum
tomography, trusted mesurements are required in order to certify quantum
devices. On the other hand, self-testing of quantum components based on
statistics collected from lesser number of measurements and minimal physical
assumptions, require lesser trust on the measurement devices and are
experimentally more efficient and less resource consumingbench .

Self-testing based on the Bell test was proposed initially in the context of
quantum cryptography my'98 ; my'04 , where it was shown that maximum violation
of the Bell-CHSH inequality Bell'64 ; chsh'69  implies that the underlying
quantum state is maximally entangled and the local measurements performed by
two distant parties are anticommuting popescu'92 . In such self-testing
protocols one can uniquely identify quantum states and measurements (upto some
local isometries) by observing extreme correlations in measurement statistics.
The underlying assumptions are the same as those required for implementation
of loophole free Bell tests hensen'15 ; knill'15 ; zeilinger'15 ;
weinfurter'17 . Since then all pure bipartite entangled states and certain
multipartite quantum states such as graph states have been self-tested Col ;
mayers2003  in this scenario. Self-testing of pure entangled bipartite states
has also been investigated Supic ; goswami ; Goh ; SBK20  employing Einstein-
Podolski-Rosen steering Wiseman . Self-testing using steering is weaker in the
sense that it requires more assumptions compared to that based on Bell tests,
but the former is easier to verify experimentally peng .

In this regard, it is worthwhile to mention that self-testing based on maximal
violation of some nonclassicality revealing criterion is very fragile as
idealised situations in the laboratory are rare to occur. Therefore, a self-
testing statement is useful for practical purposes if it has reasonable
robustness Scarani2 ; Yang ; Bamps ; Kani2 . For instance, there are robust
self-testing protocols for which given a certain level of violation of a Bell
inequality (but not necessarily maximal), nontrivial lower bound on the
fidelity between the initially unknown state and a given target state has been
shown. Robust self-testing of higher dimensional state has also been studied
Scarani3 ; supic2018 ; Zhang'18 . Apart from various special class of quantum
states, different types of measurements including maximally incompatible
observables popescu'92 , entangling Bell-measurement bancal'18 ; renou'18 ,
binary observables kani'17 , more than two dichotomic observables Mck ;
Bowles'18 , and non-maximally anticommuting observables Pironio  have also
been self-tested (See Ref.supic'19  for a review).

Apart from the above schemes of self-testing of measurements based on the use
of quantum correlations in entangled states, some certification methods for
self-testing have been proposed that do not require entanglement chen'11 ;
brunner'15 ; bourennane'16 ; Tavakoli . A pair of anticommuting Pauli
observables have been self-tested in the prepare and measure scenario
tavakoli'18 , employing the quantum advantage of $2\mapsto 1$ random access
code (RAC) ambainis'99 ; tavakoli'15 . Such approaches in the context of self-
testing of local states and measurements are more resource efficient and
experimentally feasible compared to schemes based on nonlocal quantum
correlations since they do not require entanglement, an expensive resource.
However, the assumption of an upper bound on the system-dimension has been
crucial for such approaches, and hence, the verifier needs to trust the
measurement devices more.

Another self-testing protocol has been constructed for three-dimensional
states and measurements kishor'19  based on contextuality of quantum theory
ks'67 , which requires certain restrictive assumptions such as the
compatibility between subsequent projective measurements supic'19 . Recently,
an extension of the scheme has been proposed based on the sum of sqaures
decomposition of a family of non-contextual inequalities kcbs , relaxing some
of the above restrictions saha'20 . Self-testing of arbitrary high-dimensional
local quantum systems using contextuality has also been proposed where the
assumption of projective measurement is necessary kishor'19(1) . The above
protocols pertain to local measurements in three and above dimensions. On the
other hand, among all measurements in quantum information processing, the
measurements with dichotomic settings and outcomes are the most widely used
ones. Hence, the design of resource efficient self-testing protocols for such
meaurements with minimal assumptions is necessary for practical purposes.

In the present work, we develop a scheme of self-testing binary Pauli
measurements in a hitherto unexplored scenario, i.e., employing
nonclassicality of temporal correlations exhibited via violation of the
Leggett-Garg inequality(LGI) LG'85 ; emary'14  and predictabilitykofler'13 ;
rand'16 , which does not require any entanglement. LGI has been employed to
probe macroscopic coherence and to study quantum to classical transition LG'85
; emary'14 ; LG'02 ; kofler'07 ; kofler'13 ; brukner'04 ; mal'16 ; mal'18 ;
mal'19 . Various aspects of temporal correlations have been investigated
fritz'10 ; budroni'13 ; rand'16 ; bm'14 ; brierley'19 ; usha'13 ; alok'17 ;
das'18 ; ku'18 ; titas'18  and experimentally realized knee'12 ; robens'15 ;
knee'16 ; ku'19 ; shayan'19 ; spee'20 . Here, we employ the violation of LGI,
and invoke a minimal set of assumptions which can be easily met in a real
experiment, in order to provide a self-testing statement for binary Pauli
measurements without any need to restrict the dimension of the Hilbert space
accessed by the quantum devices. We further perform the robustness analysis of
our self-testing protocol by deriving a lower bound on the fidelity of the
measured observables with that of ideal ones.

##  II Description of the scenario

We start with a brief description of the Leggett-Garg (LG) test (Fig.-1) which
enables self-testing of Pauli measurements whenever extremal correlations of
LGI are observed. In a LG test, a single system is measured sequentially at
different instants of time in order to obtain temporal correlations. Two
sequential binary measurements are performed (by Alice and Bob, respectively)
on an identical initial state prepared by the experimenter in every run of the
experiment. In contrast to the sequential measurement scenario of self-testing
using contextuality kishor'19 , here subsequent measurements are not required
to commute with each other.

Alice and Bob have two choices of binary measurements, say,
$\\{A_{1},A_{2}\\}$ and $\\{B_{1},B_{2}\\}$ to perform in each run. The
probability of obtaining outcome $a_{i}$ and $b_{j}$ is denoted by
$P(a_{i},b_{j}\mid A_{i},B_{j})$, when Alice measures $A_{i}$ at time $t_{m}$
and Bob measures $B_{j}$ at some later instant $t_{m+1}$, with
$i,j\in\\{1,2\\}$ and $a_{i},b_{j}\in\\{0,1\\}$. Let us denote,
$\mathcal{P}_{a_{i}|A_{i}}$, $\mathcal{P}_{b_{j}|B_{j}}$ as projectors so that
$\sum_{a_{i}}\mathcal{P}_{a_{i}|A_{i}}=\mathbb{I},\sum_{b_{j}}\mathcal{P}_{b_{j}|B_{j}}=\mathbb{I}$.
The two-time joint probability can be obtained using Bayes’ rule as,

$\displaystyle P(a_{i},b_{j}\mid A_{i},B_{j})=P(a_{i}\mid A_{i})P(b_{j}\mid
a_{i},A_{i},B_{j})$
$\displaystyle=\text{Tr}\left[\mathcal{P}_{a_{i}|A_{i}}\rho_{in}\right]\text{Tr}\left[\mathcal{P}_{b_{j}|B_{j}}\frac{\mathcal{P}_{a_{i}|A_{i}}\rho_{in}\mathcal{P}_{a_{i}|A_{i}}^{\dagger}}{\text{Tr}\left[\mathcal{P}_{a_{i}|A_{i}}\rho_{in}\mathcal{P}_{a_{i}|A_{i}}^{\dagger}\right]}\right].$
(1)

The two-time correlation is defined as,

$\mathcal{C}_{ij}=\sum_{a_{i},b_{j}}(-1)^{a_{i}\oplus b_{j}}P(a_{i},b_{j}\mid
A_{i},B_{j}),$ (2)

where $\oplus$ denotes addition modulo $2$. The four-term LGI in terms of the
above correlators is given by,

$\mathcal{K}_{4}=C_{11}+C_{21}+C_{22}-C_{12}\leq 2.$ (3)

Figure 1: Sequential-measurement set-up. The joint probability distribution
$P(a_{i},b_{j}\mid A_{i},B_{j})$, i.e., the probability of obtaining outcome
$a_{i}$ and $b_{j}$ when Alice measures $A_{i}$ in an arbitrary input state at
some instant $t_{m}$, and Bob performs measurement $B_{j}$ at some later
instant $t_{m+1}$, respectively, is observed to obtain the violation of LGI
(3).

Any model compatible with classical theory predicts that the maximal value of
the above expression is $2$, whereas the quantum theory can violate this
inequality up to $2\sqrt{2}$. Suppose, in a single run, Alice performs a
projective qubit measurement, $A_{1}\equiv\hat{a_{1}}.\vec{\sigma}$ (to make
the preliminary discussion simple, but it can be any general measurement too)
on an arbitrary input state, $\rho_{in}$ at time $t_{1}$, and Bob performs a
qubit measurement, $B_{1}\equiv\hat{b_{1}}.\vec{\sigma}$ at some later time
$t_{2}$, where $\hat{a_{1}}$ and $\hat{b_{1}}$ are the Bloch vectors denoting
Alice’s and Bob’s measurement directions respectively and $\vec{\sigma}$ is
the vector of Pauli matrices. Then the maximum quantum violation, $2\sqrt{2}$,
can be achieved by the following measurement settings,

$\displaystyle A_{1}^{\text{ideal}}=\sigma_{z},$ $\displaystyle
A_{2}^{\text{ideal}}=\sigma_{x},$ $\displaystyle
B_{1}^{\text{ideal}}=\frac{\sigma_{x}+\sigma_{z}}{\sqrt{2}},$ $\displaystyle
B_{2}^{\text{ideal}}=\frac{\sigma_{x}-\sigma_{z}}{\sqrt{2}},$ (4)

upto some local unitary. In the context of self-testing, all the four
measurements are a priori unknown to Alice and Bob. Our goal is to certify
above measurements from the observed statistics, under some suitable and
minimal assumptions.

##  III Derivation of Leggett-Garg inequality

LGI has been earlier derived using the assumptions of non-invasive measurement
and realism brukner'04 ; mal'16 . However, those assumptions are at
ontological level, and hence, cannot be verified individually. Rather a
conjunction of them can be verified in an experiment. On the other hand, in
rand'16 , LGI was derived from some operational assumptions called
predictability and no signalling in time (NSIT) kofler'13 . Subsequently, the
NSIT condition has been experimentally observed on a variety of input states
expnsit . Here we present the essential features of the LGI derivation rand'16
. Let us first state the two assumptions in precise mathematical form.

Predictability : A model is said to be predictable if the joint statistics
$P(a_{i},b_{j}\mid A_{i},B_{j})\in\\{0,1\\}~{}\forall a_{i},b_{j},A_{i},B_{j}$
wiseman'12 .
NSIT : NSIT is defined by the condition that measurement statistics is not
influenced by the earlier measurements. Mathematically, $P(b_{j}\mid
B_{j})=P(b_{j}\mid A_{i},B_{j})~{}\forall A_{i},B_{j},b_{j}$.
In order to derive LGI from the above two assumptions, our aim is to show that

$\displaystyle NSIT\wedge Predictability\Rightarrow LGI.$ (5)

Suppose $\lambda$ denotes some underlying variable at the ontological level,
averaging over which we obtain joint probabilities observed in an experiment,
i.e.,

$\displaystyle P(a_{i},b_{j}\mid A_{i},B_{j})=\int_{\lambda}d\lambda
p(\lambda)P(a_{i},b_{j}\mid A_{i},B_{j},\lambda).$

LGI follows in a straightforward way when the joint probability at the
ontological level gets factorised, or in other words, we have to show that
predictability together with NSIT leads to

$\displaystyle P(a_{i},b_{j}\mid A_{i},B_{j},\lambda)=P(a_{i}\mid
A_{i},\lambda)P(b_{j}\mid B_{j},\lambda).$ (6)

As further conditioning does not change the deterministic probability
distribution, we have from the predictability, $P(a_{i},b_{j}\mid
A_{i},B_{j},\lambda)=P(a_{i},b_{j}\mid A_{i},B_{j})$. Using Bayes’ rule one
has

$P(a_{i},b_{j}\mid A_{i},B_{j})=P(a_{i}\mid A_{i},B_{j},b_{j})P(b_{j}\mid
A_{i},B_{j}).$

The condition of NSIT implies $P(b_{j}\mid A_{i},B_{j})=P(b_{j}\mid B_{j})$.
Also, as physically reasonable and broadly accepted, a later measurement
cannot influence the past measurement result, and hence $P(a_{i}\mid
A_{i},B_{j},b_{j})=P(a_{i}\mid A_{i})$. With the above implications, one can
construct a theory at the ontological level where $P(a_{i}\mid
A_{i},\lambda)=P(a_{i}\mid A_{i})$, $P(b_{j}\mid B_{j},\lambda)=P(b_{j}\mid
B_{j})$, which lead to Eq. (6). Now, a straightforward calculation leads us to
derive the four-term LGI (3) which is bounded by $2$.

Despite some structural similarities, the LG-test is essentially different
from the Bell-test clemente , and also has some loopholes which are different
from that of the Bell-test wilde . The approach adopted in this paper for the
derivation of LGI is based on the conjunction of Predictability and NSIT. In
the present scenario it is ensured that NSIT condition is satisfied, as we see
in the next section. This means that if LGI is violated, then predictability
must have to be violated. Thus, the violation of predictability naturally
guarantees non-classicality which is required for the purpose of self-testing.

##  IV Self-testing of measurements using Leggett-Garg inequality

The above derivation of LGI with the assumptions of predictability and NSIT
helps us to devise a self-testing protocol for binary Pauli measurements (II)
whenever extremal non-classical temporal correlations are observed. We
formulate the self-testing protocol in such a way that the NSIT condition is
satisfied so that maximal violation of the LGI will imply the violation of the
predictability condition, and hence, no classical strategy is able to
reproduce the statistics. We now make a minimal assumption which is again very
natural in the sequential measurement scenario, since otherwise, a classical
model may simulate the quantum violation of LGI.

Assumption :  _The measurement device of Alice acts only on the input state
prepared by the experimenter, and the measurement device of Bob acts only on
the state produced by Alice’s measurement, with both returning only the
respective post-measurement states._

A similar assumption was also considered in Ref.saha'20  for the purpose of
self-testing of local three dimensional measurements using the sum-of-squares
decomposition of a family of non-contextual inequalities kcbs . However, in
our approach using LGI we are interested here in the self-testing of binary
qubit measurements. Employing the maximum violation of LGI along with NSIT
condition we can self-test the binary qubit measurements given by Eq. (II)
upto some local unitaries. Our self-testing protocol does not depend on the
input state. However, for the sake of a formal proof of self-testing, we first
consider a general qubit state (in Lemma 1). Then, making use of this proof,
we extend our result for states in any dimension (Lemma 2). Our formal proof
of self-testing is thus accomplished in the following two steps (lemma 1 and
2), and finally we present the proof of isometry in Theorem 1.

Lemma 1. _The maximum violation of LGI (i.e,
$\mathcal{K}_{4}^{\text{max}}=2\sqrt{2}$) implies implementation of the qubit
measurement observables given by equation (II) upto some local unitaries,
satisfying the NSIT condition. _

###### Proof.

Maximum quantum violation of LGI can only be achieved if the measurements are
taken to be projective . We maximize the LGI considering the most general
positive operator valued measure (POVM) with two outcomes. Maximizing the
four-time LGI expression numerically over all the POVM parameters, it is found
that the maximum value, $2\sqrt{2}$ can only be achieved if the measurements
are taken to be projective (the details of the proof are given in Appendix-A).
Hence, without any loss of generality, we restrict ourselves to projective
measurements only.
To obtain the two-time correlation $\mathcal{C}_{ij}$ we have to calculate
joint probabilities $P(a_{i},b_{j}\mid A_{i},B_{j})$ given in Eq.(II).
Considering the general qubit input state,
$\rho_{in}=\frac{\mathbb{I}+\hat{n}.\vec{\sigma}}{2}$, the first term of
Eq.(II) can be simplified to
$\text{Tr}\left[\mathcal{P}_{a_{i}|A_{i}}\rho_{in}\right]=\frac{1}{2}(1+(-1)^{a_{i}}\hat{a_{i}}.\hat{n})$.
The post-measurement state after obtaining outcome $a_{i}$ is given by,
$\frac{\mathcal{P}_{a_{i}|A_{i}}\rho_{in}\mathcal{P}_{a_{i}|A_{i}}^{\dagger}}{\text{Tr}\left[\mathcal{P}_{a_{i}|A_{i}}\rho_{in}\mathcal{P}_{a_{i}|A_{i}}^{\dagger}\right]}=\frac{1}{2}\left[\mathbb{I}+\left(-1\right)^{a_{i}}\hat{a_{i}}.\vec{\sigma}\right]$.
Now, the term,
$\text{Tr}\left[\mathcal{P}_{b_{j}|B_{j}}\frac{\mathcal{P}_{a_{i}|A_{i}}\rho_{in}\mathcal{P}_{a_{i}|A_{i}}^{\dagger}}{\text{Tr}\left[\mathcal{P}_{a_{i}|A_{i}}\rho_{in}\mathcal{P}_{a_{i}|A_{i}}^{\dagger}\right]}\right]$
can be simplified to
$\frac{1}{2}\left[1+(-1)^{a_{i}+b_{j}}\hat{a_{i}}.\hat{b_{j}}\right]$. So, the
joint probability distribution of getting outcomes $a_{i}$ and $b_{j}$ when
measurements $A_{i}$ and $B_{j}$ are performed respectively, is given by,

$P(a_{i},b_{j}\mid
A_{i},B_{j})=\frac{1}{4}(1+(-1)^{a_{i}}\hat{a_{i}}.\hat{n})(1+(-1)^{a_{i}+b_{j}}\hat{a_{i}}.\hat{b_{j}}).$
(7)

After simplification, we get, $\mathcal{C}_{ij}=\hat{a_{i}}.\hat{b_{j}}.$
Then, the four-time LGI in terms of four correlators will be,

$\displaystyle\mathcal{K}_{4}$
$\displaystyle=\hat{a_{1}}.\hat{b_{1}}+\hat{a_{2}}.\hat{b_{1}}+\hat{a_{2}}.\hat{b_{2}}-\hat{a_{1}}.\hat{b_{2}}\leq
2.$ (8)

It follows that the maximum value of $\mathcal{K}_{4}$ is $2\sqrt{2}$ when all
$|\hat{a_{i}}.\hat{b_{j}}|=\frac{1}{\sqrt{2}}$ fritz'10 , which can be
obtained for the settings described in Eq. (4) or their local unitary
rotation.
Here, one can verify that the NSIT condition is also satisfied in case of
maximum LGI violation when $|\hat{a_{i}}.\hat{b_{j}}|=\frac{1}{\sqrt{2}}$. The
NSIT condition on the probabilities, $P(b_{j}\mid A_{i},B_{j})=P(b_{j}\mid
B_{j})$ $\forall b_{j},A_{i},B_{j}$, implies
$(-1)^{a_{1}+b_{1}}\hat{a_{1}}.\hat{b_{1}}=(-1)^{a_{2}+b_{1}}\hat{a_{2}}.\hat{b_{1}}=(-1)^{a_{1}+b_{2}}\hat{a_{1}}.\hat{b_{2}}=(-1)^{a_{2}+b_{2}}\hat{a_{2}}.\hat{b_{2}}$.
Thus clearly, when maximum violation of $\mathcal{K}_{4}$ is observed along
with the NSIT condition, it is obvious that the predictability condition does
not hold anymore. ∎

We now show how maximal violation of LGI can also be used to self-test the
Pauli measurements even if the measurement operators act on a higher
dimensional Hilbert space. Let the initial state be $\rho_{in}$ in an
arbitrary dimensional Hilbert space, and the measurements $A_{i},B_{j}$ both
act on this space.

Lemma 2. _The maximum violation of LGI (i.e,
$\mathcal{K}_{4}^{\text{max}}=2\sqrt{2}$) implies implementation of the block
diagonal measurement, i.e.,
$A_{1}=\oplus_{i}\sigma_{z}^{i},A_{2}=\oplus_{i}\sigma_{x}^{i},B_{1}=\oplus_{j}(\sigma_{x}^{j}+\sigma_{z}^{j})/\sqrt{2},B_{2}=\oplus_{j}(\sigma_{x}^{j}-\sigma_{z}^{j})/\sqrt{2}$._

###### Proof.

Suppose $\left\\{\mathcal{P}_{a_{i}|A_{i}}\right\\}$ and
$\left\\{\mathcal{P}_{b_{j}|B_{j}}\right\\}$ are two dichotomic measurements
which act on an arbitrary dimensional Hilbert space. Then, according to
Jordan’s lemma Kani2 ,
$\mathcal{P}_{a_{i}|A_{i}}=\oplus_{m}\mathcal{P}_{a_{i}|A_{i}}^{m}$ and
$\mathcal{P}_{b_{j}|B_{j}}=\oplus_{n}\mathcal{P}_{b_{j}|B_{j}}^{n}$, where
$\mathcal{P}_{a_{i}|A_{i}}^{m}$ and $\mathcal{P}_{b_{j}|B_{j}}^{n}$ are
projectors on $\mathcal{H}_{d}$ with $d\leq 2$ (here, in view of our proof of
lemma 1, we again stick to projective measurements without any loss of
generality). Therefore, one has

$\displaystyle P(a_{i},b_{j}\mid A_{i},B_{j})=$
$\displaystyle\sum_{m,n}p_{m}\text{Tr}\left[\mathcal{P}_{a_{i}|A_{i}}^{m}\rho_{m}\right]\text{Tr}\left[\mathcal{P}_{b_{j}|B_{j}}^{n}\frac{\mathcal{P}_{a_{i}|A_{i}}^{m}\rho_{m}\mathcal{P}_{a_{i}|A_{i}}^{m\dagger}}{\text{Tr}\left[\mathcal{P}_{a_{i}|A_{i}}^{m}\rho_{m}\mathcal{P}_{a_{i}|A_{i}}^{m\dagger}\right]}\right]$

where $p_{m}=\text{Tr}(\rho_{in}\Pi_{m})$, with
$\Pi_{m}=\sum_{a_{i}}\mathcal{P}^{m}_{a_{i}|A_{i}}$, and
$\rho_{m}=(\Pi_{m}\rho_{in}\Pi_{m})/p_{m}$, which is at most a qubit state.
with $\rho=\oplus_{m}p_{m}\rho_{m}$, where
$\rho_{m}=\frac{\mathbb{I}+\hat{r}.\vec{\sigma}}{2}$ acts on $\mathcal{H}_{d}$
with $d\leq 2$. Now, the operators of the above equation acting on
$\mathcal{H}_{d}$ with $d=2$ implies that

$\displaystyle P(a_{i},b_{j}\mid A_{i},B_{j})=$
$\displaystyle\sum_{m}p_{m}\frac{1}{4}(1+(-1)^{a^{m}_{i}}\hat{a^{m}_{i}}.\hat{r})(1+(-1)^{a_{i}+b_{j}}\hat{a^{m}_{i}}\cdot\hat{b^{n}_{j}}).$

From the above equation it follows that,

$\displaystyle C_{ij}=\sum_{m,n}p_{m}\hat{a^{m}_{i}}\cdot\hat{b^{n}_{j}}$ (9)

The above expectation implies $C_{11}+C_{21}+C_{22}-C_{12}=2\sqrt{2}$ if and
only if
$\hat{a_{1}}^{m}=\hat{z},\hat{a_{2}}^{m}=\hat{x},\hat{b_{1}}^{n}=(\hat{z}+\hat{x})/\sqrt{2},\hat{b_{2}}^{n}=(\hat{z}-\hat{x})/\sqrt{2}$
and $\sum_{m}p_{m}=1$. This is achieved if and only if
$A_{1}=\oplus_{m}\sigma_{z}^{m},A_{2}=\oplus_{m}\sigma_{x}^{m},B_{1}=\oplus_{n}(\sigma_{x}^{n}+\sigma_{z}^{n})/\sqrt{2},B_{2}=\oplus_{n}(\sigma_{x}^{n}-\sigma_{z}^{n})/\sqrt{2}$
and $\rho_{in}=\oplus_{m} \rho_{m}$. ∎

The above two lemmas enables us to present the following theorem.

Theorem 1. _If $\mathcal{K}_{4}^{\text{max}}=2\sqrt{2}$ is observed in LG-test
under Assumption 1, with the measurements of Alice, $A_{i}$ acting on $H_{d}$,
producing the post measurement states
$\left\\{\frac{\mathcal{P}_{a_{i}|A_{i}}\rho_{in}\mathcal{P}_{a_{i}|A_{i}}^{\dagger}}{\text{Tr}\left[\mathcal{P}_{a_{i}|A_{i}}\rho_{in}\mathcal{P}_{a_{i}|A_{i}}^{\dagger}\right]}\right\\}_{a}$,
and the measurements of Bob $B_{j}$ acting on these post measurement states,
then there exists an isometry
$\Phi:\mathcal{H}^{d}\rightarrow\mathcal{C}^{2}\otimes\mathcal{H}^{d}$ such
that_

$\displaystyle\Phi\left(B_{j}\frac{\mathcal{P}_{a_{i}|A_{i}}\rho_{in}\mathcal{P}_{a_{i}|A_{i}}^{\dagger}}{\text{Tr}\left[\mathcal{P}_{a_{i}|A_{i}}\rho_{in}\mathcal{P}_{a_{i}|A_{i}}^{\dagger}\right]}\right)\Phi^{\dagger}$
$\displaystyle=$ $\displaystyle
B^{\texttt{ideal}}_{j}\left|\psi^{\texttt{ideal}}_{a|A_{i}}\right\rangle\left\langle\psi^{\texttt{ideal}}_{a|A_{i}}\right|\otimes\left|\texttt{junk}\right\rangle\left\langle\texttt{junk}\right|$

_where $\left|\psi^{\texttt{ideal}}_{a|A_{i}}\right\rangle$ are the
eigenstates of Alice’s ideal measurements and $B^{\texttt{ideal}}_{j}$ are
Bob’s ideal measurements given by Eq. (4) respectively, and
$\left|\texttt{junk}\right\rangle$ is a junk state acting on
$\mathcal{H}^{d}$._

###### Proof.

The details proof are given in the Appendix-B. Choosing the eigenbasis of
$A_{1}$ as the computational basis, from lemma $1$ and $2$, it follows that
the post measurement states of Alice can be written as
$\frac{\mathcal{P}_{a_{i}|A_{i}}\rho_{in}\mathcal{P}_{a_{i}|A_{i}}^{\dagger}}{\text{Tr}\left[\mathcal{P}_{a_{i}|A_{i}}\rho_{in}\mathcal{P}_{a_{i}|A_{i}}^{\dagger}\right]}=\oplus_{m}p_{m}\left|\psi^{m}_{a|A_{i}}\right\rangle\left\langle\psi^{m}_{a|A_{1}}\right|$,
with $\left|\psi^{m}_{0|A_{1}}\right\rangle=\left|2m\right\rangle$ and
$\left|\psi^{m}_{1|A_{1}}\right\rangle=\left|2m+1\right\rangle$. We append an
ancilla qubit prepared in the state $\left|0\right\rangle$ and look for an
isometry $\Phi$ such that

$\displaystyle\Phi\left(B_{j}\frac{\mathcal{P}_{a_{i}|A_{i}}\rho_{in}\mathcal{P}_{a_{i}|A_{i}}^{\dagger}}{\text{Tr}\left[\mathcal{P}_{a_{i}|A_{i}}\rho_{in}\mathcal{P}_{a_{i}|A_{i}}^{\dagger}\right]}\otimes\left|0\right\rangle\left\langle
0\right|\right)\Phi^{\dagger}$ $\displaystyle=$ $\displaystyle
B^{\texttt{ideal}}_{j}\left|\psi^{\texttt{ideal}}_{a|A_{i}}\right\rangle\left\langle\psi^{\texttt{ideal}}_{a|A_{i}}\right|\otimes\left|\texttt{junk}\right\rangle\left\langle\texttt{junk}\right|$

This can be achieved for $\Phi$ defined by the map

$\displaystyle\Phi\left|2m,0\right\rangle$
$\displaystyle\rightarrow\left|2m,0\right\rangle$
$\displaystyle\Phi\left|2m+1,0\right\rangle$
$\displaystyle\rightarrow\left|2m,1\right\rangle$

∎

##  V Robust self-testing of measurements

Robustness of self-testing of measurements is quantified as how the actual
observables which are to be self-tested differ from the ideal ones. Hence,
characterization of the fidelities between the real measurements and the ideal
measurements is warranted. Let us first perform the robustness analysis of the
measurements on Alice’s side i.e,
$\left\\{\mathcal{P}_{a_{i}|A_{i}}\right\\}$. Given an arbitrary set of
measurements $\left\\{\mathcal{P}_{a_{i}|A_{i}}\right\\}$, the average
fidelity with the ideal measurements are,

$S(\left\\{\mathcal{P}_{a_{i}|A_{i}}\right\\})=\text{max}_{\Lambda}\sum_{i,a_{i}}F(\mathcal{P}_{a_{i}|A_{i}}^{\text{ideal}},\Lambda[\mathcal{P}_{a_{i}|A_{i}}])/4.$
(10)

Here $\Lambda$ is a quantum channel and the fidelities are defined as usual,
$F(\rho,\sigma)=\text{Tr}(\sqrt{\sqrt{\rho}\sigma\sqrt{\rho}})$. The
fidelities between the real and the ideal measurements,
$F(\mathcal{P}_{a_{i}|A_{i}}^{\text{ideal}},\Lambda[\mathcal{P}_{a_{i}|A_{i}}])$
simplify to
$\text{Tr}(\Lambda\left[\mathcal{P}_{a_{i}|A_{i}}\right]\mathcal{P}_{a_{i}|A_{i}}^{\text{ideal}})$.
The lower bound on the smallest possible value of the average fidelity $S$,
given a particular violation of the four-time LGI $\mathcal{K}_{4}$, can be
found by minimizing over all sets of four measurements of
$\mathcal{P}_{a_{i}|A_{i}}$,

$\mathcal{F}(\mathcal{K}_{4})=\text{min}_{\mathcal{P}_{a_{i}|A_{i}}}S(\\{\mathcal{P}_{a_{i}|A_{i}}\\}).$
(11)

In order to a lower bound for $\mathcal{F}$ as a function of LGI-violation
$\mathcal{K}_{4}$, we use the operator inequality approach given in Refs.
Kani2 ; tavakoli'18 . Rewriting $\mathcal{K}_{4}$ in terms of an operator
$W_{ia_{i}}$ and $\mathcal{P}_{a_{i}|A_{i}}$, where
$W_{ia_{i}}=\frac{1}{2}[(-1)^{a_{i}}(\mathcal{P}_{B_{1}}+(-1)^{i-1}\mathcal{P}_{B_{2}})]$,
$\mathcal{P}_{B_{i}}\equiv\mathcal{P}_{0|B_{i}}-\mathcal{P}_{1|B_{i}}$, we
have,

$\mathcal{K}_{4}=\sum_{i,a_{i}}\text{Tr}[W_{ia_{i}}\mathcal{P}_{a_{i}|A_{i}}].$
(12)

Let us now define another operator
$K_{ia_{i}}(\mathcal{P}_{B_{1}},\mathcal{P}_{B_{2}})\equiv\Lambda^{\dagger}(\mathcal{P}_{B_{1}},\mathcal{P}_{B_{2}})[\mathcal{P}_{a_{i}|A_{i}}^{\text{ideal}}]$.
Considering an operator inequality of the form,

$K_{ia_{i}}(\mathcal{P}_{B_{1}},\mathcal{P}_{B_{2}})\geq
sW_{ia_{i}}+\mu_{ia_{i}}(\mathcal{P}_{B_{1}},\mathcal{P}_{B_{2}})\mathbb{I}$
(13)

with $s$ and $\mu_{ia_{i}}(\mathcal{P}_{B_{1}},\mathcal{P}_{B_{2}})$ being
real coeficients, the average fidelity $S$ (and hence $\mathcal{F}$) can be
lower bounded as,

$\displaystyle\mathcal{F}\geq S$ $\displaystyle\geq$
$\displaystyle\frac{1}{4}\sum_{i,a_{i}}F(\mathcal{P}_{a_{i}|A_{i}}^{\text{ideal}},\Lambda\left[\mathcal{P}_{a_{i}|A_{i}}\right]$
(14) $\displaystyle=$
$\displaystyle\frac{1}{4}\sum_{i,a_{i}}Tr[K_{ia_{i}}\mathcal{P}_{a_{i}|A_{i}}]$
$\displaystyle\geq$
$\displaystyle\frac{s}{4}\sum_{i,a_{i}}Tr[W_{ia_{i}}\mathcal{P}_{a_{i}|A_{i}}]+\frac{1}{4}\sum_{i,a_{i}}\mu_{ia_{i}}$
$\displaystyle=$ $\displaystyle\frac{s}{4}\mathcal{K}_{4}+\mu$

where $\mu\equiv
1/4~{}min_{\mathcal{P}_{B_{1}},\mathcal{P}_{B_{2}}}\sum_{i,a_{i}}\mu_{ia_{i}}(\mathcal{P}_{B_{1}},\mathcal{P}_{B_{2}})$.

Since we can use Jordan’s lemma to write the observables of Alice in a block-
diagonal form with block of size at most $2\times 2$, it suffices for our
purpose to focus on each block as in Ref. Kani2  to find the constants $s$ and
$\mu$ for lower bounding the fidelity. The goal of the robustness analysis is
to obtain a self-testing bound as tight as possible. Following Ref.
tavakoli'18 , adoption of the dephasing channel $\Lambda$, as an extraction
map suffices to achieve our aim of finding an optimal self-testing bound.

$\Lambda_{\theta}(\rho)=\frac{1+\xi(\theta)}{2}\rho+\frac{1-\xi(\theta)}{2}\Gamma(\theta)\rho\Gamma(\theta);$
(15)

where $\xi(\theta)\in[-1,1]$ and

$\displaystyle\Gamma(\theta)$
$\displaystyle=\left\\{\begin{array}[]{lr}\sigma_{z}&\quad\text{for}\quad\theta\in\left[0,\frac{\pi}{4}\right]\\\
\sigma_{x}&\quad\text{for}\quad\theta\in\left(\frac{\pi}{4},\frac{\pi}{2}\right]\\\
\end{array}\right.$ (18)

For the interval $0\leq\theta\leq\pi/4$

$\displaystyle K_{10}$ $\displaystyle=$
$\displaystyle\Lambda^{\dagger}[\mathcal{P}_{0|A_{1}}]$ $\displaystyle=$
$\displaystyle\frac{1+\xi(\theta)}{2}\left[\frac{\mathbb{I}+\sigma_{z}}{2}\right]+\frac{1-\xi(\theta)}{2}\left[\sigma_{z}\frac{\mathbb{I}+\sigma_{z}}{2}\sigma_{z}\right]$
$\displaystyle=$ $\displaystyle\frac{\mathbb{I}+\sigma_{z}}{2},$

and similarly $K_{20}=\frac{\mathbb{I}+\xi(\theta)\sigma_{x}}{2}$ ,
$K_{21}=\frac{\mathbb{I}-\xi(\theta)\sigma_{x}}{2}$,
$K_{11}=\frac{\mathbb{I}-\sigma_{z}}{2}$.
whereas, for the interval $\pi/4\leq\theta\leq\pi/2$,

$K_{10}=\frac{\mathbb{I}+\xi(\theta)\sigma_{z}}{2}$,
$K_{20}=\frac{\mathbb{I}+\sigma_{x}}{2}$,

$K_{11}=\frac{\mathbb{I}-\xi(\theta)\sigma_{z}}{2}$,
$K_{21}=\frac{\mathbb{I}-\sigma_{x}}{2}$.

We can, without loss of generality, represent the two measurements in the x-z
plane as,

$\displaystyle\mathcal{P}_{B_{1}}$ $\displaystyle=$
$\displaystyle\cos(\theta)\sigma_{z}+\sin(\theta)\sigma_{x}$
$\displaystyle\mathcal{P}_{B_{2}}$ $\displaystyle=$
$\displaystyle\cos(\theta)\sigma_{z}-\sin(\theta)\sigma_{x}$ (19)

and therefore, $W_{10}=\cos(\theta)\sigma_{z}$,
$W_{20}=\sin(\theta)\sigma_{x}$, and $W_{21}=-\sin(\theta)\sigma_{x}$,
$W_{11}=-\cos(\theta)\sigma_{z}$.

We analyze the operator inequality in the interval $0\leq\theta\leq\pi/4$ and
$\pi/4\leq\theta\leq\pi/2$ separately. One can see that the effective number
of inequalities may be reduced since there are symmetries in the expression of
$K_{ia_{i}}$ and $W_{ia_{i}}$, and hence, without loss of generality we can
choose $\mu_{o}\equiv\mu_{20}=\mu_{21}$ and $\mu_{e}\equiv\mu_{10}=\mu_{11}$.
After simplication we get,

$\displaystyle\mu_{e}$ $\displaystyle\leq$ $\displaystyle 1-s\cos(\theta),$
$\displaystyle\mu_{e}$ $\displaystyle\leq$ $\displaystyle s\cos(\theta).$ (20)

and,

$\displaystyle\mu_{o}$ $\displaystyle\leq$
$\displaystyle\frac{1}{2}+\frac{1}{2}\xi(\theta)-s\sin(\theta),$
$\displaystyle\mu_{o}$ $\displaystyle\leq$
$\displaystyle\frac{1}{2}-\frac{1}{2}\xi(\theta)+s\sin(\theta).$ (21)

In order to obtain the strongest bound, we have to choose the largest value of
$\mu_{o}$ and $\mu_{e}$ consistant with their respective constraints, i.e., in
the first interval,

$\displaystyle\mu_{e}$
$\displaystyle=\text{min}\left\\{1-s\cos(\theta),s\cos(\theta)\right\\},$
$\displaystyle\mu_{o}$
$\displaystyle=\text{min}\left\\{\frac{1}{2}+\frac{1}{2}\xi(\theta)-s\sin(\theta),\frac{1}{2}-\frac{1}{2}\xi(\theta)+s\sin(\theta)\right\\}.$
(22)

A similar procedure in the second interval leads to,

$\displaystyle\mu_{o}$
$\displaystyle=\text{min}\left\\{1-s\sin(\theta),s\sin(\theta)\right\\},$
$\displaystyle\mu_{e}$
$\displaystyle=\text{min}\left\\{\frac{1}{2}+\frac{1}{2}\xi(\theta)-s\cos(\theta),\frac{1}{2}-\frac{1}{2}\xi(\theta)+s\cos(\theta)\right\\}.$
(23)

The expressions in the two intervals are related to each other by the
transfomations $\mu_{o}\leftrightarrow\mu_{e}$ and
$\sin(\theta)\leftrightarrow\cos(\theta)$.

Hence, the lower bound in fidelity becomes,

$\mathcal{F}(\mathcal{K}_{4})\geq\frac{s}{4}\mathcal{K}_{4}+\text{min}_{\mathcal{P}_{B_{1}},\mathcal{P}_{B_{2}}}\mu(\mathcal{P}_{B_{1}},\mathcal{P}_{B_{2}}).$
(24)

where $\mu(\mathcal{P}_{B_{1}},\mathcal{P}_{B_{2}})=(\mu_{e}+\mu_{o})/2$. To
compute this value, we fix $s=\frac{1+\sqrt{2}}{2}$, and choosing the
dephasing function as $\xi(\theta)=\text{min}\\{1,2s\sin(\theta)\\}$ in the
interval $\theta\in[0,\pi/4]$, and
$\xi(\theta)=\text{min}\\{1,2s\cos(\theta)\\}$ in the interval
$\theta\in(\pi/4,\pi/2]$. After simplification, we get,
$\mu=\frac{2-\sqrt{2}}{4}$ which gives the lower bound,

$\mathcal{F}(\mathcal{K}_{4})\geq\frac{(1+\sqrt{2})}{8}\mathcal{K}_{4}+\frac{2-\sqrt{2}}{4}.$
(25)

This provides robust self-testing of Alice’s measurements. Clearly, the
maximal quantum violation of the LGI, i.e., $\mathcal{K}_{4}=2\sqrt{2}$
implies $\mathcal{F}(\mathcal{K}_{4})=1$, which suggests that the measurements
must be the ideal ones. For $\mathcal{K}_{4}=2$,
$\mathcal{F}(\mathcal{K}_{4})\geq 3/4$. This bound can be obtained by
$A_{1}=A_{2}=B_{1}=B_{2}=\sigma_{z}$. Therefore, we see that our bound is
optimal.

We now quantify the average fidelity of the measurements on Bob’s side with
respect to the ideal ones:
$\mathcal{S}^{\prime}(\\{\mathcal{P}_{b_{i}|B_{i}}\\})=\text{max}_{\Lambda}\sum_{i,b_{i}}F((\mathcal{P}_{b_{i}|B_{i}})^{\text{ideal}}\Lambda[\mathcal{P}_{b_{i}|B_{i}}])/4$,
where $\Lambda$ must be a unital channel. Now, define,

$\mathcal{F}^{\prime}(\mathcal{K}_{4})=\text{min}_{\\{\mathcal{P}_{b_{i}|B_{i}}\\}}\mathcal{S}^{\prime}(\\{\mathcal{P}_{b_{i}|B_{i}}\\}).$
(26)

First, we rewrite
$\mathcal{K}_{4}=\sum_{i,b_{i}}Tr[\mathcal{P}_{b_{i}|B_{i}}Z_{ib_{i}}]$ where
$Z_{ib_{i}}=\frac{1}{2}(-1)^{b_{i}}[\mathcal{P}_{A_{1}}+(-1)^{i-1}\mathcal{P}_{A_{2}}]$
and $\mathcal{P}_{A_{i}}\equiv\mathcal{P}_{0|A_{i}}-\mathcal{P}_{1|A_{i}}$.

Let us take an operator inequality of the form,

$K_{ib_{i}}(\\{\mathcal{P}_{A_{1}},\mathcal{P}_{A_{2}}\\})\geq
sZ_{ib_{i}}+\mu_{ib_{i}}(\\{\mathcal{P}_{A_{1}},\mathcal{P}_{A_{2}}\\})\mathbb{I},$
(27)

with $K_{ib_{i}}=\Lambda^{\dagger}[\mathcal{P}_{b_{i}|B_{i}}^{\text{ideal}}]$.
Similar to the previous case,

$\mathcal{F}^{\prime}(\mathcal{K}_{4})\geq\text{min}_{\mathcal{P}_{A_{1}},\mathcal{P}_{A_{2}}}\frac{1}{4}\sum_{i,b_{i}}Tr[K_{ib_{i}}\mathcal{P}_{b_{i}|B_{i}}].$
(28)

Using the same map and the same technique, we have
$\xi(\theta)=\text{min}\\{1,2s\sin(\theta)\\}$ in interval
$\theta\in[0,\pi/4]$ and $\xi(\theta)=\text{min}\\{1,2s\cos(\theta)\\}$
$\theta\in(\pi/4,\pi/2]$. If we repeat the same procedure as in Alice’s
measurements, we get the bound,
$\mathcal{F}^{\prime}(\mathcal{K}_{4})\geq\frac{(1+\sqrt{2})}{8}\mathcal{K}_{4}+\frac{2-\sqrt{2}}{4}$.
This provides robust self-testing of Bob’s measurements which is again
optimal.

In order to provide an operational analysis of robustness, let us choose a
dephasing channel,
$\Lambda_{\theta}(\rho)=\frac{1+\xi(\theta)}{2}\rho+\frac{1-\xi(\theta)}{2}\sigma_{z}\rho\sigma_{z}$.
It keeps the measurement, $A_{1}=\sigma_{z}$ as it is and dephase the
measurement, $A_{2}=\sigma_{x}$ to $\xi(\theta)\sigma_{x}$. Let us consider
Bob’s measurements as, $B_{1}=\cos(\phi)\sigma_{z}+\sin(\phi)\sigma_{x}$ and
$B_{2}=\cos(\phi)\sigma_{z}-\sin(\phi)\sigma_{x}.$ Taking
$\xi(\theta)=\tan(\phi)$, we get,

$\mathcal{K}_{4}=2\sqrt{1+\tan^{2}(\phi)}~{}~{}~{}~{}~{}~{}\text{and}~{}~{}~{}~{}~{}~{}\mathcal{F}=\frac{1}{4}(3+\tan(\phi)).$

The above expressions provide a parametric curve as a function of
$\phi\in[0,\pi/4]$, which is the dashed (blue) line in Fig.-2. The straight
(red) line represents the lower bound of fidelity as given by Eq.(25). A
straightforward calculation shows that the robustness analysis for Bob’s
measurements($\mathcal{F}^{\prime}$) produces a matching parametric curve.

Figure 2: The average fidelity $\mathcal{F}$ ($\mathcal{F^{{}^{\prime}}}$),
for Alice’s (Bob’s) measurement as a function of $\mathcal{K}_{4}$ is plotted.
The red (solid) line represents the lower bound of average fidelity obtained
analytically from Eq. (25). The blue (dotted) line is the average fidelity
under dephasing noise.

##  VI Conclusions

With the rapid development of quantum technologies, it is important to first
characterize and certify various quantum devices. Among various certification
protocols, self-testing protocols are designed for the purpose of certifying
quantum components from the observed statistics under a set of minimal
assumptions. Self-testing is regarded to be more resource-efficient, and
requires lesser trust on measurement devices. Previously, self-testing of
measurements has been performed mainly employing nonlocal spatial quantum
correlations my'04 ; Col ; supic2018 ; supic'19 ; bancal'18 ; renou'18 , as
exhibited by the violation of Bell-type inequalities, which require
entanglement, a costly resource. Schemes of self-testing measurements without
entanglement proposed earlier require either dimensional restrictions
tavakoli'18 ; tavakoli'15 , or other stringent assumptions like compatibility
and projectivity of the measurements kishor'19 ; kishor'19(1)  in case of
approaches based on contextuality, though it may be possible to relax some of
these assumptions for local measurements in three and higher dimensions
saha'20 .

In this work our purpose is to self-test binary measurements with dichotomic
settings and outcomes without using entangled states. To this end we have
exploited another fundamental property of quantum mechanics, viz., temporal
quantum correlations exhibited by violation of Leggett-Garg inequalities
together with violation of predictability LG'85 ; emary'14 ; kofler'13 ;
brukner'04 ; rand'16 . We have presented a scheme of self-testing of binary
Pauli measurements through LGI violation using a minimal assumption that is
inevitable in the present context. We have shown how the maximum quantum
violation of the four-term LGI along with the NSIT condition can be used to
provide a self-testing statement of binary measurements without entanglement,
which requires neither compatibility or projectivity, nor any dimensional
restriction. Moreover, we have formulated the robustness bound of our self-
testing protocol, and analysed it with an example of dephasing noise.

Before concluding, we would like to highlight certain salient features of our
analysis. First, it is evident from our analysis that the proposed self-
testing protocol for binary qubit measurements does not depend on the input
state. Secondly, though in the present work we have not discussed self-testing
of quantum states, once the measurement devices are characterized through our
approach of self-testing through the LGI framework, any subsequent
certification of quantum states using such devices may reduce essentially to
the task of decoherence control. Finally, the NSIT condition used here is
experimentally implementable, as shown recently, for a variety of states
expnsit . Experimental viability of our protocol is ensured in the backdrop of
several recent LG tests knee'12 ; robens'15 ; knee'16 ; expnsit ; ku'19 ;
shayan'19 ; spee'20 .

###### Acknowledgements.

AGM would like to thank Debarshi Das and Debashis Saha for helpful comments.
SM acknowledges the Ministry of Science and Technology in Taiwan (Grant no.
110-2811-M-006 -501). CJ thanks Huan-Yu Ku for discussions and acknowledges
the financial support from the Ministry of Science and Technology, Taiwan
(Grant No. MOST 108-2811-M-006-516) and the Foundation for Polish Science
through the First Team project (No First TEAM/2017-4/31). ASM acknowledges
support from the project DST/ICPS/QuEST/Q98 from the Department of Science and
Technology, India.

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##  Appendix A Maximal violation of the LGI with POVMs.

Here we obtain the maximal of the Leggett-Garg inequality for POVMs busch'91 ;
das'18  which are a set of positive operators that add to identity, i. e.,
$E\equiv\\{E_{i}|\sum_{i}E_{i}=\mathbb{I},0<E_{i}\leq\mathbb{I}\\}$. Each of
the operators $E_{i}$, called effect operator determines the probability
$\text{Tr}[\rho E_{i}]$ of obtaining the $i^{\text{th}}$ outcome when applied
on the state $\rho$. The most general POVM with two outcomes are defined by
two parameters,– sharpness parameter and biasedness parameter. Let us consider
that $\gamma_{a},\gamma_{b}$ are the biasedness parameter and $\lambda_{i}$
and $\mu_{j}$ are the sharpness parameters for Alice’s $i$-th and Bob’s $j$-th
measurements, respectively. The most general effect operators with two
outcomes can be written as,

$\displaystyle
E^{\lambda_{i}}_{a_{i}}=\lambda_{i}\mathcal{P}_{a_{i}|A_{i}}+(1\pm\gamma_{a}-\lambda_{i})\frac{\mathbb{I}_{2}}{2},$
$\displaystyle
E^{\mu_{j}}_{b_{j}}=\mu_{j}\mathcal{P}_{b_{j}|B_{j}}+(1\pm\gamma_{b}-\mu_{j})\frac{\mathbb{I}_{2}}{2}$
(29)

The post measurement state can be derived using generalized von Neumann-Lüders
transformation rule,
$\dfrac{\sqrt{E^{\lambda_{i}}_{a_{i}}}\rho\sqrt{E^{\lambda_{i}}_{a_{i}}}}{\text{Tr}[E^{\lambda_{i}}_{a_{i}}\rho]}$.
Now, in order to derive the LGI we shall follow the same scenario and
procedure as discussed in the main text. The two-time joint probability is
obtained using Baye’s rule,

$\displaystyle P(a_{i},b_{j}\mid A_{i},B_{j})=P(a_{i}\mid A_{i})P(b_{j}\mid
a_{i},A_{i},B_{j})$ (30) $\displaystyle=$ $\displaystyle
Tr\left[E^{\lambda_{i}}_{a_{i}}\rho_{in}\right]Tr\left[E^{\mu_{j}}_{b_{j}}\dfrac{\sqrt{E^{\lambda_{i}}_{a_{i}}}\rho\sqrt{E^{\lambda_{i}}_{a_{i}}}}{\text{Tr}[E^{\lambda_{i}}_{a_{i}}\rho]}\right].$

The two-time correlation is defined as,

$\displaystyle\mathcal{C}_{ij}=\sum_{a_{i},b_{j}}(-1)^{a_{i}\oplus
b_{j}}P(a_{i},b_{j}\mid A_{i},B_{j}),$ (31)

where $\oplus$ denotes addition modulo $2$. The four-term LGI in terms of the
above correlators is given by,

$\mathcal{K}_{4}=C_{11}+C_{21}+C_{22}-C_{12}\leq 2$ (32)

For a general input state
$\rho_{in}=\frac{\mathbb{I}+\hat{n}.\vec{\sigma}}{2}$, the first term
(probability obtained by Alice) reduces to,

$Tr\left[E^{\lambda_{i}}_{a_{i}}\rho_{in}\right]=\frac{1}{2}((1\pm\gamma_{a})+(-1)^{a_{i}}\lambda_{i}\hat{a_{i}}.\hat{n})$

Similarly, one can calcultate
$Tr\left[E^{\mu_{j}}_{b_{j}}\dfrac{\sqrt{E^{\lambda_{i}}_{a_{i}}}\rho\sqrt{E^{\lambda_{i}}_{a_{i}}}}{\text{Tr}[E^{\lambda_{i}}_{a_{i}}\rho]}\right]$
One can now calculate $\mathcal{K}_{4}$ and maximize it numerically. It can be
found that the maximum quantum value of $\mathcal{K}_{4}$ is $2\sqrt{2}$ when
all $|\hat{a_{i}}.\hat{b_{j}}|=\frac{1}{\sqrt{2}}$ along with
$\lambda_{i}=\mu_{j}=1$ and $\gamma_{a}=\gamma_{b}=0$. Hence, the maximum
value of LGI can only be achieved if Alice and Bob perform sharp-projective
measurements.

##  Appendix B Details of proof of Theorem 1.

Appending an ancilla qubit prepared in the state $\left|0\right\rangle$, let
us show that there exists an isometry $\Phi$ defined by the map,

$\displaystyle\Phi\left|2m,0\right\rangle$
$\displaystyle\rightarrow\left|2m,0\right\rangle$
$\displaystyle\Phi\left|2m+1,0\right\rangle$
$\displaystyle\rightarrow\left|2m,1\right\rangle$ (33)

such that

$\displaystyle\Phi\left(B_{j}\frac{\mathcal{P}_{a_{i}|A_{i}}\rho_{in}\mathcal{P}_{a_{i}|A_{i}}^{\dagger}}{\text{Tr}\left[\mathcal{P}_{a_{i}|A_{i}}\rho_{in}\mathcal{P}_{a_{i}|A_{i}}^{\dagger}\right]}\otimes\left|0\right\rangle\left\langle
0\right|\right)\Phi^{\dagger}$ $\displaystyle=$ $\displaystyle
B^{\texttt{ideal}}_{j}\left|\psi^{\texttt{ideal}}_{a|A_{i}}\right\rangle\left\langle\psi^{\texttt{ideal}}_{a|A_{i}}\right|\otimes\left|\texttt{junk}\right\rangle\left\langle\texttt{junk}\right|$

holds.
We present here the calculation for one term (say,
$\Phi\left(B_{1}\frac{\mathcal{P}_{a_{1}|A_{1}}\rho_{in}\mathcal{P}_{a_{1}|A_{1}}^{\dagger}}{\text{Tr}\left[\mathcal{P}_{a_{1}|A_{1}}\rho_{in}\mathcal{P}_{a_{1}|A_{1}}^{\dagger}\right]}\otimes\left|0\right\rangle\left\langle
0\right|\right)\Phi^{\dagger}$) explicitly. Other terms can be calculated in a
similar fashion.
Choosing the eigenbasis of $A_{1}$ as the computational basis, from lemma $1$
and $2$, it follows that the post-measurement states of Alice can be written
as
$\frac{\mathcal{P}_{a_{i}|A_{i}}\rho_{in}\mathcal{P}_{a_{i}|A_{i}}^{\dagger}}{\text{Tr}\left[\mathcal{P}_{a_{i}|A_{i}}\rho_{in}\mathcal{P}_{a_{i}|A_{i}}^{\dagger}\right]}=\oplus_{m}p_{m}\left|\psi^{m}_{a|A_{1}}\right\rangle\left\langle\psi^{m}_{a|A_{1}}\right|$,
with $\left|\psi^{m}_{0|A_{1}}\right\rangle=\left|2m\right\rangle$ and
$\left|\psi^{m}_{1|A_{1}}\right\rangle=\left|2m+1\right\rangle$. The
measurement $B_{1}$ on Bob’s side is
$\oplus_{n}(\sigma_{x}^{n}+\sigma_{z}^{n})/\sqrt{2}$. Expanding the sum for
$a=0$ leads to

$\displaystyle\oplus_{n}\frac{\sigma_{x}^{n}+\sigma_{z}^{n}}{\sqrt{2}}\oplus_{m}p_{m}\left|2m\right\rangle\left\langle
2m\right|$ $\displaystyle=\bigg{(}\frac{|0\rangle\langle 0|-|1\rangle\langle
1|+|0\rangle\langle 1|+|1\rangle\langle 0|}{\sqrt{2}}+\frac{|2\rangle\langle
2|-|3\rangle\langle 3|+|2\rangle\langle 3|+|3\rangle\langle
2|}{\sqrt{2}}+\cdots$ $\displaystyle+\frac{|2m\rangle\langle
2m|-|2m+1\rangle\langle 2m+1|+|2m\rangle\langle 2m+1|+|2m+1\rangle\langle
2m|}{\sqrt{2}}\bigg{)}\cdot\left(p_{0}\ket{0}\bra{0}+p_{1}\ket{2}\bra{2}+\cdots+p_{m}\ket{2m}\bra{2m}\right)$
$\displaystyle=\bigg{(}p_{0}\frac{\ket{0}+\ket{1}}{\sqrt{2}}\bra{0}+p_{1}\frac{\ket{2}+\ket{3}}{\sqrt{2}}\bra{2}+\cdots+p_{m}\frac{\ket{2m}+\ket{2m+1}}{\sqrt{2}}\bra{2m}\bigg{)}$
(34)

Now, for the map defined above,

$\displaystyle\Phi\left(\oplus_{n}\frac{\sigma_{x}^{n}+\sigma_{z}^{n}}{\sqrt{2}}\oplus_{m}p_{m}\left|2m\right\rangle\left\langle
2m\right|\otimes\ket{0}\bra{0}\right)\Phi^{\dagger}$
$\displaystyle=\frac{\ket{0}+\ket{1}}{\sqrt{2}}\bra{0}\otimes\oplus_{m}p_{m}\left|2m\right\rangle\left\langle
2m\right|$
$\displaystyle=B^{\texttt{ideal}}_{1}\left|\psi^{\texttt{ideal}}_{0|A_{1}}\right\rangle\left\langle\psi^{\texttt{ideal}}_{0|A_{1}}\right|\otimes\left|\texttt{junk}\right\rangle\left\langle\texttt{junk}\right|$
(35)

with
$\left|\texttt{junk}\right\rangle\left\langle\texttt{junk}\right|=\sum_{m}p_{m}\left|2m\right\rangle\left\langle
2m\right|$.
Next, for $a=1$, we have

$\displaystyle\oplus_{n}\frac{\sigma_{x}^{n}+\sigma_{z}^{n}}{\sqrt{2}}\oplus_{m}p_{m}\left|2m+1\right\rangle\left\langle
2m+1\right|$
$\displaystyle=\bigg{(}p_{0}\frac{\ket{0}-\ket{1}}{\sqrt{2}}\bra{0}+p_{1}\frac{\ket{2}-\ket{3}}{\sqrt{2}}\bra{2}+\cdots+p_{m}\frac{\ket{2m}-\ket{2m+1}}{\sqrt{2}}\bra{2m}\bigg{)}$
(36)

Hence, for the map defined above,

$\displaystyle\Phi\left(\oplus_{n}\frac{\sigma_{x}^{n}+\sigma_{z}^{n}}{\sqrt{2}}\oplus_{m}p_{m}\left|2m+1\right\rangle\left\langle
2m+1\right|\otimes\ket{0}\bra{0}\right)\Phi^{\dagger}$
$\displaystyle=\frac{\ket{0}-\ket{1}}{\sqrt{2}}\bra{0}\otimes\oplus_{m}p_{m}\left|2m\right\rangle\left\langle
2m\right|$
$\displaystyle=B^{\texttt{ideal}}_{1}\left|\psi^{\texttt{ideal}}_{1|A_{1}}\right\rangle\left\langle\psi^{\texttt{ideal}}_{1|A_{1}}\right|\otimes\left|\texttt{junk}\right\rangle\left\langle\texttt{junk}\right|.$
(37)