Device-independent self-testing of unsharp measurements

Measurement plays a pivotal role in quantum theory which is in stark contrast to classical theory. The textbook version of a quantum measurement is modeled by a set of orthogonal projectors belonging to the Hilbert space. However, there exist more general measurements defined in terms of positive-operator-valued measures (POVMs) satisfying the completeness relation. Note that the POVMs can be defined in many different ways. However, in this paper, we are concerned about those POVMs which are noisy or unsharp variants of the projective measurements.

In a sharp projective measurement [1], the system collapses to one of the eigenstates of the measured observable, and the system is completely disturbed by the process of sharp measurement, so that, no residual coherence remains in the system. On the other hand, in the case of unsharp measurements that are characterized by the POVMs, the system is less disturbed compared to the sharp projective measurement. Since a projective measurement maximally disturbs the quantum system and hence extracts more information from the system compared to POVMs, one may think that a sharp measurement is more advantageous in information processing tasks. However, there exist certain information processing tasks where POVMs are proven to be more useful compared to sharp measurement.

Advantage of POVMs over sharp measurement has been explored in the context of quantum state discrimination [2, 3], randomness certification [4–7], quantum tomography [8], state estimation [9], quantum cryptography [10], information acquisition from a quantum source [11], quantum violation of certain Bell inequalities [12], and many more. There is one more advantage that is particularly relevant in the present work is the sequential sharing of various forms of quantum correlations [13–18]. For example, a sequential Bell test by multiple independent observers inevitably requires the prior observers to perform the unsharp quantum measurements.

In this work, we aim to provide device-independent (DI) self-testing of the unsharp measurements, which are noisy variants of the sharp projective measurements. Self-testing [19–23] is the strongest form of certification protocol where the devices are treated as black boxes. Also, the dimension of the quantum system is assumed to be finite but unbounded. In that, observed experimental statistics uniquely certify the state and measurement observables of an unknown dimension. DI self-testing is advantageous over the standard certification methods such as those based on quantum tomography. Essentially, a self-testing protocol requires optimal quantum violation of a suitable Bell's inequality [24]. For example, optimal violation of Clauser–Horne–Shimony–Holt (CHSH) inequality self-tests the maximally entangled state and mutually anticommuting local observables in any arbitrary dimension. Note that the DI certification encounters practical challenges arising from the requirement of a loophole-free Bell test. Such tests have recently been realized [25–29] enabling experimental demonstrations of DI certification of randomness [30, 31]. Of late, the DI certification is used as a resource for secure quantum key distribution [32–35], randomness certification [36–39], witnessing Hilbert space dimension [40–46] and for achieving advantages in communication complexity tasks [47].

We provide DI self-testing schemes to certify the unsharpness parameters through the quantum violation of two well-known Bell inequalities viz., the CHSH inequality [24] and the elegant Bell inequality [48]. Note that optimal quantum violations of such Bell inequalities can be obtained only for sharp measurement. Any value less than the optimal quantum value may originate due to various reasons. It may come from the unsharp measurements of local observables but may also come from the nonideal preparation of the state or the inappropriate choices of the local observables than the ones required for the optimal quantum violation. However, a more serious issue regarding DI self-testing of unsharp measurement through a Bell test arises due to Neimark's theorem. It states that every non-projective measurement can be modeled as a projective measurement in a larger Hilbert space by introducing suitable ancilla in a higher dimension. Since in the DI Bell test, there is no bound on the dimension, a stubborn physicist may argue that the sub-optimal quantum violation is arising due to the inappropriate choice of observables in a higher dimension and not due to the unsharpness of the measurement. Hence, to self-test unsharp measurements, one needs to introduce a protocol that certifies the state, the observables, and the unsharpness parameters without referring to the dimension of the system.

Against this backdrop, we demonstrate that the sequential Bell test has such potential where the sub-optimal sequential quantum violations of a Bell inequality by multiple independent observers on one side can enable such a self-testing. Such a scheme successfully bypasses the constraints that would arise from Naimark's theorem as the optimization of sequential Bell expression is performed without imposing any constraint on the dimension of Hilbert space. As far as our knowledge goes, the DI self-testing of non-projective measurements through the Bell test has not hitherto been demonstrated. However, semi-DI certification of non-projective measurements has been demonstrated by using the qubit system. In [49–51], the extremal qubit POVMs were experimentally certified based on the theoretical proposal [4]. Those experiments [49–52] do certify the non-projective character of measurement, but not how it relates to a specific target POVM. The semi-DI certification of qubit unsharp measurements (noisy variants of projective measurement) in the prepare-measure scenario has recently been proposed [53–57]. The proposal of [53] has been experimentally verified [52, 58, 59].

Specifically, in our self-testing protocol, two sequential sub-optimal quantum violations of a Bell inequality form an optimal pair powering the DI certifications of the state, the local observables, and the unsharpness parameter of one of the two parties. We also note here that all the previous works that demonstrated the sharing of various quantum correlations [16, 17, 56, 57,60–67], the dimension of the system was assumed. In contrast, throughout this work, we impose no bound on the dimension of the Hilbert space, and we consider that the measurement devices are black boxes. We first demonstrate that, at most, two independent sequential observers on one side can violate the CHSH inequality. We simultaneously maximize the quantum violations of CHSH inequality for two sequential observers on one side and demonstrate that there is a trade-off between the two sub-optimal quantum violations. This, in turn, certifies the state, and the observables for both the sequential observers and the unsharpness parameter of the first observer. Since in a practical implementation there remains imperfection, we show how a range of an unsharpness parameter can be self-tested in that scenario.

The protocol developed for the CHSH inequality is further extended to the case of elegant Bell inequality, where we demonstrate that, at most, two observers can share the quantum advantage when considering the inequality's local bound. However, we argue that the elegant Bell inequality has two classical bounds, the local and the preparation non-contextual bound, and the latter is smaller than the former. We show that if we consider the preparation non-contextual bound of the elegant Bell inequality, then at most three observers can share the quantum advantage. Further, we demonstrate that if the quantum advantage is extended to a third sequential observer, then the range of the values of the unsharpness parameter for the first observer can be more restricted, thereby self-testing a narrow range of the values of the unsharpness parameter in the sub-optimal scenario.

This paper is organized as follows. In section 2, we demonstrate the optimal quantum violation of CHSH inequality using an elegant sum-of-squares approach. In section 3, we explicitly show the sequential violation of CHSH inequality and self-testing of the unsharpness parameter. In section 4, we briefly introduce the notion of the preparation non-contextuality in an ontological model and provide the preparation non-contextual bound of elegant Bell inequality. In section 5, we extend the sharing of preparation contextuality by three sequential observers based on the quantum violation of elegant Bell inequality and demonstrate the self-testing of two unsharpness parameters along with the states and the observables. Finally, we discuss our results in section 6.

The CHSH scenario consists of two space-like separated parties (say, Alice and Bob) who share a common physical system. Alice (Bob) performs local measurements on her (his) subsystem upon receiving inputs $x\in \{1,2\} (y \in \{1,2\})$, and returns outputs $a \in \{-1,+1\}(b \in \{-1,+1\})$. Representing M_x and N_y as respective dichotomic observables of Alice and Bob, the CHSH form of Bell's inequality can be written as

$$\begin{equation} \mathcal{B}=\left(M_{1}+M_{2}\right) N_{1} +\left(M_{1}-M_{2}\right) N_{2}\leqslant 2. \end{equation} \tag{ 1 }$$

The optimal quantum value of the CHSH expression is $\left(\mathcal{B}\right)^\mathrm{opt}_{Q} = 2\sqrt{2}$, commonly known as Tirelson bound [68]. The optimal value can be obtained when the shared state is maximally entangled in a two-qubit system and the local qubit observables are mutually anticommuting. However, one does not need to impose the bound on the dimension to obtain the optimal value. Also, the measurement of Alice and Bob remains uncharacterized. In other words, the optimal value $\left(\mathcal{B}\right)^\mathrm{opt}_Q$ can be achieved if Alice and Bob perform dichotomic measurements on a maximally entangled state $\vert \psi\rangle_{AB} \in \mathcal{C}^{d}\otimes \ \mathcal{C}^{d}$ where $d\geqslant 2$ is arbitrary.

For our purpose, we provide a derivation of the $\left(\mathcal{B}\right)^\mathrm{opt}_{Q}$ without imposing a bound on the Hilbert space dimension by introducing an elegant sum-of-squares (SOS) approach. Let us assume that $(\mathcal{B})_{Q}\leqslant \Omega_{2}$, where Ω₂ is a positive quantity and the upper bound of $(\mathcal{B})_{Q}$. Equivalently, one can argue that there is a positive semi-definite operator $\eta \geqslant 0$, that can be expressed as

$$\begin{equation} \langle \eta\rangle_{Q}=\Omega_{2} -(\mathcal{B})_{Q} \end{equation} \tag{ 2 }$$

for a quantum state $|\psi\rangle_{AB}$. This can be proven by considering two suitable positive operators, L₁ and L₂, which are polynomial functions of M_x and N_y , so that

$$\begin{align} \eta =\frac{1}{2} \left( \omega_{1} L_1^\dagger L_1+\omega_{2} L_2^\dagger L_2\right). \end{align} \tag{ 3 }$$

For our purpose, we suitably choose L₁ and L₂ as

$$\begin{align} L_{1}|\psi\rangle_{AB}&=\frac{M_{1}+M_{2}}{\omega_{1}} |\psi\rangle_{AB} -N_{1} |\psi\rangle_{AB}\nonumber\\ L_{2}|\psi\rangle_{AB}&=\frac{M_{1}-M_{2}}{\omega_{1}} |\psi\rangle_{AB} -N_{2} |\psi\rangle_{AB} \end{align} \tag{ 4 }$$

where $\omega_{1} = ||\left(M_{1}+M_{2}\right)|\psi\rangle_{AB}||_{2}$ an $\omega_{2} = ||\left(M_{1}-M_{2}\right)|\psi\rangle_{AB}||_{2}$. Here $||.||_{2}$ is the Frobenius norm of a vector, $||\ \mathcal{O}\ ||_2 = \sqrt{Tr[\mathcal{O}^{\dagger}\mathcal{O} \rho]}$. Plugging equation (4) into equation (3) and by noting that $M_{x}^{\dagger} M_{x} = N\,_y^{\dagger} N_y = \mathbb{I}$, we get

$$\begin{align} (\mathcal{B})_{Q}= \left(\omega_{1}+\omega_{2}\right) -\langle \eta\rangle_{Q} . \end{align} \tag{ 5 }$$

Optimal value of $(\mathcal{B})_{Q}$ can be obtained when $\langle\eta\rangle_{Q} = 0$, i.e.

$$\begin{align} (\mathcal{B})_{Q}^\mathrm{opt} &=\underset{{}}{max}\left(\omega_{1}+\omega_{2}\right)\nonumber\\ &=\underset{{}}{max}\left(\sqrt{2+\langle \{M_1, M_2\}\rangle}+\sqrt{2-\langle \{M_1, M_2\}\rangle}\right). \end{align} \tag{ 6 }$$

Thus, the maximization requires $\{M_1, M_2\} = 0$ implying Alice's observables have to be anticommuting. In turn, we find the values $\omega_{1} = \omega_{2} = \sqrt{2}$, and consequently the optimal value $(\mathcal{B})_{Q}^\mathrm{opt} = 2\sqrt{2}$. The explicit conditions for the optimization are $L_{1}|\psi\rangle_{AB} = 0$ and $L_{2}|\psi\rangle_{AB} = 0$, i.e. $N_{1} = (M_{1}+M_{2})/\sqrt{2}$ and $N_{2} = (M_{1}-M_{2})/\sqrt{2}$. It can be easily checked that $\{N_1, N_2\} = 0$, i.e. Bob's observables are also anticommuting.

Note also that for the state $\rho_{AB} = \vert \psi_{AB}\rangle\langle \psi_{AB}\vert \in \mathcal{C}^d \otimes \mathcal{C}^d$, optimal violation requires $Tr[N_{1}\otimes N_{1}\rho_{AB}]$ $= Tr[N_{2}\otimes N_{2}\rho_{AB}] = 1$, which again confirms that ρ_AB has to be a pure state. Let us choose the state in Hilbert–Schmidt form as

$$\begin{equation} \rho_{AB}=\frac{1}{d^{2}}\Bigg[\mathbb{I}\otimes\mathbb{I}+\sum_{i=1}^{d^2-1}N_{i}\otimes N_{i}\Bigg] \end{equation} \tag{ 7 }$$

where $\{N_{i},N_{j}\} = 0$ and consequently $[N_{i}\otimes N_{i},N_{j}\otimes N_{j}] = 0$ for any arbitrary dimension d. For a density matrix ρ_AB , $Tr[\rho_{AB}] = 1$ has to be satisfied. This in turn provides $Tr[N_{1}] = Tr[N_{2}] = 0$. Also, $Tr[\rho_{AB}^{2}] = 1$ ensures that the observables in the summation in equation (7) contains full set of mutually anticommuting observables $\{N_{i}\otimes N_{i}\}$. Consequently, $Tr_{A}[\rho_{AB}] = Tr_{B}[\rho_{AB}] = \frac{\mathbb{I}}{d}$, i.e. partial trace of ρ_AB is maximally mixed state for both Alice and Bob. Thus, the optimal violation of the CHSH inequality is achieved for the maximally entangled state $\vert \psi_{AB}\rangle$. We thus derive the optimal quantum value $(\mathcal{B})_{Q}^\mathrm{opt}$ which uniquely certifies the state and observables. The entire derivation is done without assuming the dimension of the system.

The sequential Bell-CHSH test in the DI scenario is depicted in figure 1. There is only one Alice, who always performs sharp measurement and arbitrary k number of sequential Bobs (say, Bob_k ). Alice and Bob₁ share an entangled state $\rho_{AB_{1}}$. Our aim is to demonstrate the sharing of nonlocality by multiple sequential observers. Since a projective measurement inevitably disturbs the system maximally; hence, in the sequential Bell test, if the first Bob (Bob₁) performs a sharp measurement, the entanglement is lost after the measurement. In such a case, no residual entanglement remains for the second sequential observer (Bob₂); consequently, the Bell inequality cannot be violated. If both the sequential observers obtain the sub-optimal quantum violations, then the first observer must have to perform an unsharp measurement.

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In this work, the unsharp measurement corresponds to the noisy variant of projective measurements, i.e. the number of measurement operators is restricted to two. After performing the unsharp measurement, Bob₁ relays the post-measurement state to Bob₂, who performs an unsharp measurement intending to violate the Bell inequality. The chain runs up to arbitrary k^th Bob (Bob_k ) until the quantum violation of Bell's inequality is obtained. The k^th Bob may perform the sharp measurement. It is quite known that in the standard scenario at most, two Bobs can sequentially demonstrate nonlocality through the violation of CHSH inequality [13]. We stress again that, before our work all the studies that demonstrated the sharing of nonlocality, the dimension of the system was assumed.

We consider that Alice, upon receiving input $x = \{1,2\}$, always performs the projective measurements of observables A₁ and A₂. The CHSH form of Bell's inequality can be written as

$$\begin{equation} \mathcal{B}=\left(A_{1}+A_{2}\right) B_{1} +\left(A_{1}-A_{2}\right) B_{2} \leqslant 2. \end{equation} \tag{ 8 }$$

There are arbitrary k number of sequential Bobs (say, Bob_k ), who upon receiving input $y_{k}\in \{1,2\}$ perform measurements of B₁ and B₂ producing outputs $b_{k}\in \{-1,+1\}$. We demonstrate how sequential quantum violations of CHSH inequality by multiple independent Bobs enable the DI certification of the unsharpness parameter.

Now, if Bob_k 's instrument is represented by measurement operators $\{K_{b_{k}|y_{k}}\}$ then after $(k-1)^{th}$ Bob's measurement, the average state shared between Alice and Bob_k is

$$\begin{equation} \rho_{AB_{k}} = \frac{1}{2}\sum_{b_{k}\in \pm}\sum_{y_{k}=1}^{2} K_{b_{k}|y_{k}}^{\dagger}\ \rho_{AB_{(k-1)}}\ K_{b_{k}|y_{k}} \end{equation} \tag{ 9 }$$

where $\forall k$, $\sum_{b_{k} = \pm 1} K_{b_{k}|y_{k}}^{\dagger}K_{b_{k}|y_{k}} = \mathbb{I}$. Also, $K_{b_{k}|y_{k}} = \sqrt{E_{b_{k}|y_{k}}}$ where $E_{b_{k}|y_{k}}$ is the POVM. We consider that $\forall k$ and $\forall y_{k}$ the POVM [69, 70] is of the form

$$\begin{equation} E_{\pm|y_{k}}=\frac{1\pm\lambda_{k}}{2}\Pi_{y_k}^{+}+\frac{1\mp\lambda_{k}}{2}\Pi_{y_k}^{-} \end{equation} \tag{ 10 }$$

where $\{\Pi_{y_k}^{\pm}\}$ are the projectors corresponding to the observable $B_{y_{k}}$ satisfying $\Pi_{y_k}^{+}+\Pi_{y_k}^{-} = \mathbb{I}$, and λ_k is the unsharpness parameter for k^th Bob. The measurement operators can then be written as, $K{\pm|y_{k}} = \alpha_{k} \mathbb{I} \pm \beta_{k} B_{y_{k}}$ where

$$\begin{align} \alpha_{k} &= \frac{1}{2}\Bigg(\sqrt{\frac{(1+\lambda_{k})}{2}}+ \sqrt{\frac{(1-\lambda_{k})}{2}}\Bigg)\nonumber\\ \beta_{k} &= \frac{1}{2}\Bigg(\sqrt{\frac{(1+\lambda_{k})}{2}}-\sqrt{\frac{(1-\lambda_{k})}{2}}\Bigg), \end{align} \tag{ 11 }$$

satisfying $\alpha_{k}^{2} +\beta_{k}^{2} = 1/2$ with $\alpha_{k}\geqslant \beta_{k}$.

We derive the maximum quantum value of CHSH expression for Alice and Bob₁ as

$$\begin{align} (\mathcal{B}^{1})_Q=\lambda_{1}( \mathcal{B})_Q^\mathrm{opt} \end{align} \tag{ 12 }$$

which is independent of assuming the dimension. Here, λ₁ is the unsharpness parameter of Bob₁.

After his unsharp measurement, Bob₁ relays the average state to Bob₂. From equation (9), the reduced state for Alice and Bob₂ can be written as

$$\begin{align} \rho_{AB_{2}}&=\dfrac{1}{2} \sum_{b_1 \in \{+,-\}}\sum_{y_{1}=1}^{2} \left(\mathbb{I} \otimes K_{b_{1}|y_{1}}\right) \rho_{AB_{1}} \left(\mathbb{I} \otimes K_{b_{1}|y_{1}}\right)\nonumber\\ &= 2\alpha^2_{1} \rho_{AB_{1}} + \beta^2_{1} \sum_{y_{1}=1}^{2}(\mathbb{I}\otimes B_{y_{1}})\rho_{AB_{1}} (\mathbb{I}\otimes B_{y_{1}}) . \end{align} \tag{ 13 }$$

For the time being, let us consider that each sequential Bob measures same set of observables B₁ and B₂, i.e. $\forall k, \ B_{y_{k = 1}}\equiv B_{1}$ and $B_{y_{k = 2}}\equiv B_{2}$. By using equation (13), the maximum quantum value of CHSH expression for Alice and Bob₂ can be written as

$$\begin{align} (\mathcal{B}^{2})_Q&= \mathrm{max}\Big(Tr[\rho_{AB_{2}} \mathcal{B}]\Big)\nonumber\\ &=\mathrm{max}\Bigg(Tr\Big[\rho_{AB_{1}}\Big((A_{1}+A_{2})\tilde{B}_{1}+(A_{1}-A_{2})\tilde{B}_{2}\Big)\Big]\Bigg) \end{align} \tag{ 14 }$$

where we assume Bob₂ performs sharp measurement. We derive the explicit forms of $\tilde{B}_{1}$ and $\tilde{B}_{2}$ as

$$\begin{align} \tilde{B}_{1}&=(2\alpha_{1}^2+\beta_{1}^2)B_{1}+\beta_{1}^{2}B_{2}B_{1}B_{2}\nonumber\\ \tilde{B}_{2}&=(2\alpha_{1}^2+\beta_{1}^2)B_{2}+\beta_{1}^{2}B_{1}B_{2}B_{1}. \end{align} \tag{ 15 }$$

Note that, equation (14) has a similar form of CHSH expression as in equation (8), if $\tilde{B}_{1}$ and $\tilde{B}_{2}$ are considered to be effective observables of Bob₂. We can use the earlier SOS approach to obtain the maximum quantum value. However, $(\tilde{B}_{1})^{2}\neq \mathbb{I}$ and $(\tilde{B}_{2})^{2}\neq \mathbb{I}$, and hence they need to be properly normalized. By considering $\tilde{\omega}_{1} = ||\tilde{B}_{1}||$ and $\tilde{\omega}_{2} = ||\tilde{B}_{2}||$, and by using the SOS approach we have

$$\begin{align} (\mathcal{B}^{2})_{Q}= max\left(\omega_{1}\tilde{\omega}_{1} +\omega_{2}\tilde{\omega}_{2}\right). \end{align} \tag{ 16 }$$

As we already proved earlier, to obtain the optimal quantum value Alice's and Bob's observables have to be mutually anticommuting. Hence, for Bob's (unnormalized) observables $\tilde{B}_{1}$ and $\tilde{B}_{2}$ we require,

$$\begin{align} \{\tilde{B}_{1},\tilde{B}_{2}\}=4\alpha_{1}^2(\alpha_{1}^2+2\beta_{1}^{2}) \{B_{1},B_{2}\}+\beta_{1}^{4}\{B_{1},B_{2}\}^{3}=0 . \end{align} \tag{ 17 }$$

Since $\alpha_{1}\gt0$ and $\beta_{1}\geqslant 0$, equation (17) gives $\{B_{1},B_{2}\} = 0$. In other words, the observables of Bob₂ have to be anticommuting to obtain the maximum quantum value of the Bell expression $(\mathcal{B}^{2})_{Q}$. This, in turn, provides

$$\begin{align} \tilde{\omega}_{1}=\sqrt{(2\alpha_{1}^2)^{2}+(2\alpha_{1}^2+\beta_{1}^2)\beta_{1}^{2}\{B_{1},B_{2}\}^{2}}=2\alpha_{1}^2, \end{align} \tag{ 18 }$$

and $\tilde{\omega}_{2} = \tilde{\omega}_{1} = 2\alpha_{1}^2$. Note that the above result is obtained by considering that Bob₁ and Bob₂ perform the measurements on the same set of observables B₁ and B₂. In appendix A, we prove that to obtain the maximum quantum value $(\mathcal{B}^{2})_{Q}$ the choices of observables of Bob₂ have to be the same as Bob₁.

Using equations (6) and (18), from equation (16) we can then write

$$\begin{align*} (\mathcal{B}^{2})_{Q} = 2\alpha_{1}^2 \ max\Big(\omega_{1}+\omega_{2}\Big)=2\alpha_{1}^2(\mathcal{B})_{Q}^\mathrm{opt}. \end{align*}$$

Putting the value of α₁, we write $(\mathcal{B}^{2})_Q$ in terms of unsharpness parameter λ₁ of Bob₁ as

$$\begin{equation} (\mathcal{B}^{2})_Q=\dfrac{1}{2}\Big(1+\sqrt{1-\lambda_{1}^2}\Big)(\mathcal{B})_Q^\mathrm{opt} . \end{equation} \tag{ 19 }$$

Hence, we have derived the maximum quantum values of Bell expressions for two sequential Bobs where $(\mathcal{B})_Q^\mathrm{opt}$ is common, but they differ by the coefficients, which are solely dependent only on λ₁. Using equation (12), the quantum value of the CHSH expression $(\mathcal{B}^{2})_Q$ in equation (19) can be written in terms of $(\mathcal{B}^{1})_Q$ as

$$\begin{equation} (\mathcal{B}^{2})_Q=\sqrt{2}\left(1+\sqrt{1-\left(\frac{ (\mathcal{B}^{1})_Q}{(\mathcal{B})_{Q}^\mathrm{opt}}\right)^{2}}\right) . \end{equation} \tag{ 20 }$$

This implies that if $(\mathcal{B}^{1})_Q$ increases, then $(\mathcal{B}^{2})_Q$ decreases, i.e. the more the Bob₁ disturbs the system, the more he gains the information, and consequently the quantum value $(\mathcal{B}^{2})_Q$ of Bob₂ decreases. Hence, there is a trade-off between $(\mathcal{B}^{2})_Q$ and $(\mathcal{B}^{1})_Q$, which eventually form an optimal pair demonstrating the certification of unsharpness parameter.

In figure 2, we plot the optimal trade-off characteristics between $(\mathcal{B}^{1})_Q$ and $(\mathcal{B}^{2})_Q$. The green line corresponds to the maximum classical value of the CHSH expression where Bob₂ can get a maximum value independent of Bob₁, i.e. there is no trade-off in classical theory. The blue curve exhibits the trade-off between quantum values of $(\mathcal{B}^{2})_Q$ and $(\mathcal{B}^{1})_Q$ where each point on it certifies a unique value of unsharpness parameter λ₁. For example, when $(\mathcal{B}^{2})_Q = (\mathcal{B}^{1})_Q$, the value of $\lambda_1 = 4/5 = 0.80$ is certified, as shown in the figure by brown dot.

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Note that in our protocol we consider Alice always performs sharp measurements. Before making the self-testing statements of our protocol, let us examine whether our protocol certifies sharp measurements of Alice. As mentioned earlier that the blue curve in figure 2 represents the optimal trade-off between $(\mathcal{B}^{2})_Q$ and $(\mathcal{B}^{1})_Q$. If Alice performs unsharp measurement then the trade-off curve will always be below the blue curve. For instance, $(\mathcal{B}^{2})_Q = (\mathcal{B}^{1})_Q = 8\sqrt{2}/5$ (represented by the brown dot over the blue line) can never be reached unless Alice performs the sharp measurement of her anti-commuting observables. The same argument holds for any point over the blue curve in figure 2. We are now in the position to make the DI self-testing statements of our protocol.

3.1. DI self-testing statements

3.2. Robust certification of unsharpness parameter

Note, however, that the experimental implementation of any protocol inevitably introduces noise and imperfections. We provide an argument to demonstrate how our certification protocol is robust to the noise. In the real experimental scenario, the maximum values of CHSH expressions $(\mathcal{B}^{2})_Q$ and $(\mathcal{B}^{1})_Q$ may not be achieved, and hence unique certification of λ₁ may not be accurate. In such a case, we can certify the range within which λ₁ can belong.

The quantum advantage for Bob₁ requires $(\mathcal{B}^{1})_Q \gt 2$, which fixes the $(\lambda_{1})_\mathrm{min} = 1/\sqrt{2}\approx0.707$ as $(\mathcal{B})_{Q}^\mathrm{opt} = 2\sqrt{2}$. Thus, any value of $\lambda_{1} \in [1/\sqrt{2},1]$ provides quantum advantage for Bob₁. However, this range has to be further restricted if the nonlocality is extended to Bob₂. To obtain advantage for Bob₂, two sequential Bob requires $(\mathcal{B}^{1})_Q, (\mathcal{B}^{2})_Q\gt2$ and in turn necessitates $\lambda_{2} = 1$. From equation (19), we get

$$\begin{align} \lambda_{1} < \sqrt{1-\Bigg(\frac{2(\mathcal{B}^{2})_{Q}}{(\mathcal{B})_{Q}^\mathrm{opt}}-1\Bigg)^{2}} \end{align} \tag{ 21 }$$

which in turn fixes the upper bound of $(\lambda_{1})_\mathrm{max} = \sqrt{2(\sqrt{2}-1)}\approx 0.912$. Hence, when both Bob₁ and Bob₂ get quantum advantage, the interval $0.707\lt\lambda_{1}\lt 0.912$ is certified.

Now, as an example, let us consider an experiment in which we attempt to certify the desired value of $\lambda_{1} = 0.74$ corresponding to the optimal pair $\{(\mathcal{B}^{1})_Q,(\mathcal{B}^{2})_Q\}$. In such a scenario, we need the value of $\{(\mathcal{B}^{1})_Q,(\mathcal{B}^{2})_Q\}\approx\{2.093,2.365\}$. However, due to imperfections, accurate quantum values may not be obtained. Instead, experimentalist gets $\{(\mathcal{B}^{1})_Q,(\mathcal{B}^{2})_Q\}\approx\{2.05,2.34\}$. In such a case, we cannot obtain the desired value of λ₁ but will deduce a range of λ₁ within which it has to be confined. From equations (12) and (21), we can calculate the range of λ₁ as $0.724\lt\lambda_{1}\lt0.755$. Thus, depending upon the observed quantum values, the range of λ₁ can be confined, i.e. the more perfect the experimental determination of $\{(\mathcal{B}^{1})_Q,(\mathcal{B}^{2})_Q\}$, the certified range of λ₁ becomes narrower.

3.3. Sharing of nonlocality by the third Bob

We examine whether the sharing of nonlocality can be extended to the third Bob (Bob₃). If so, we can certify two unsharpness parameters. Also, the range of λ₁ can be made more restrictive if the third Bob can share the nonlocality. By using equation (9), the maximum quantum value of CHSH expression $(\mathcal{B}^{3})_Q = \mathrm{max}(Tr[\rho_{AB_{3}} (\mathcal{B})])$ between Alice and Bob₃ can be calculated where $\rho_{AB_{3}}$ is the average state shared between Alice and Bob₃. Then,

$$\begin{align} (\mathcal{B}^{3})_Q&=\mathrm{max}\left(Tr\left[\Big(2\alpha^2_{2} \rho_{AB_{2}} + \beta^2_{2} \sum_{y_2=1}^{2}(\mathbb{I}\otimes B_{y_2})\rho_{AB_{2}} (\mathbb{I}\otimes B_{y_2})\Big) \mathcal{B}\right]\right) \end{align} \tag{ 22 }$$

which can be re-written in a similar form as in equation (14), is given by

$$\begin{align} (\mathcal{B}^{3})_Q=\mathrm{max}\Bigg(Tr\Big[\rho_{AB_{1}}\Big((A_{1}+A_{2})\tilde{\tilde{B}}_{1}+(A_{1}-A_{2})\tilde{\tilde{B}}_{2}\Big)\Big]\Bigg). \end{align} \tag{ 23 }$$

Here, $\tilde{\tilde{B}}_{1}$ and $\tilde{\tilde{B}}_{2}$ represent effective observables of Bob₃ are derived as

$$\begin{align} \tilde{\tilde{B}}_{i}&=(4\alpha^2_{1}\alpha_2^{2}+2\beta_{1}^{2}\beta_2^{2})B_{i}+(2\alpha_2^{2}\beta_{1}^{2}+2\alpha_{1}^{2}\beta_{2}^{2})(B_{i}+B_{j}B_{i}B_{j}) +\beta_{1}^{2}\beta_2^{2}(B_{j}B_{i}B_{j}+B_{i}B_{j}B_{i}B_{j}B_{i}) \end{align} \tag{ 24 }$$

with $i(j)_{i\neq j} \in \{1,2\}$. Again $(\tilde{\tilde{B}}_{i})^{2}\neq \mathbb{I}$ and hence $\tilde{\tilde{B}}_{i}$'s need to be properly normalized. Following the earlier argument, we can use the aforesaid SOS approach to obtain the optimal value of $(\mathcal{B}^{3})_{Q}$. By considering $\tilde{\tilde{\omega}}_{1} = ||\tilde{\tilde{B}}_{1}||$ and $\tilde{\tilde{\omega}}_{2} = ||\tilde{\tilde{B}}_{2}||$, and by using the SOS approach we obtain

$$\begin{align} (\mathcal{B}^{3})_{Q}= \mathrm{max}\left(\omega_{1}\tilde{\tilde{\omega}}_{1} +\omega_{2}\tilde{\tilde{\omega}}_{2}\right) . \end{align} \tag{ 25 }$$

Note that, to obtain the maximum quantum value, $\tilde{\tilde{B}}_{1}$ and $\tilde{\tilde{B}}_{2}$ have to be mutually anticommuting, i.e.

$$\begin{align*} &\{\tilde{\tilde{B}}_{1},\tilde{\tilde{B}}_{2}\}=\Big(8(4 \alpha^2_{1} \alpha^4_{2} \beta^2_{1} +2\alpha^4_{1} (\alpha^4_{2} + 2 \alpha^2_{2} \beta^2_{1}) -3 \beta^4_{1} \beta^4_{1})\Big) \{B_{1},B_{2}\} \\ &\qquad\qquad\quad+\Big(4 (\alpha_{2}^{4} \beta_{1}^{4} + 2 \alpha_{2}^{2} \beta_{1}^{2} (2\alpha_{1}^{2} + \beta_{1}^{2}) \beta_{2}^{2} + \alpha_{1}^{2} (\alpha_{1}^{2} + 2\beta_{1}^{2}) \beta_{2}^{4})\Big)\{B_{1},B_{2}\}^{3} \\ &\qquad\qquad\quad+(\beta_{1}^{2}\beta_2^{2})^{2}\{B_{1},B_{2}\}^{5}=0 \end{align*}$$

which provides $\{B_{1},B_{2}\} = 0$. Then, Bob₃ also requires the anticommuting observables to obtain the maximum quantum value. We then calculate

$$\begin{align} \tilde{\tilde{\omega}}_{1}&=\Bigg(16\alpha^4_{1}\alpha_2^{4}+\Big(4\alpha_2^{4}\beta_{1}^{4}+2\alpha_{1}^{2}\alpha_2^{2}\beta_{1}^{2}(\alpha_2^{2}+2\beta_{2}^{2})+\alpha^4_{1}(2\alpha_{2}^{2}\beta_{2}^{2}+\beta_{2}^{4})\Big) \nonumber\\ &\quad\times\{B_{1},B_{2}\}^{2}+\Big(2\alpha_2^{2}\beta_{1}^{4}\beta_{2}^{2}+\beta_{1}^{2}(2\alpha_1^{2}+\beta_{1}^2)\beta_{2}^4\Big)\{B_{1},B_{2}\}^{4}\Bigg)^{1/2} . \end{align} \tag{ 26 }$$

Since, $\{B_1,B_2\} = 0$, we obtain $\tilde{\tilde{\omega}}_{1} = 4\alpha^2_{1}\alpha_2^{2}$, and similarly we find $\tilde{\tilde{\omega}}_{2} = \tilde{\tilde{\omega}}_{1} = 4\alpha^2_{1}\alpha_2^{2}$.

Then, the maximum quantum value $(\mathcal{B}^{3})_Q$ in equation (25) becomes

$$\begin{align*} (\mathcal{B}^{3})_{Q}= 4\alpha^2_{1}\alpha_2^{2} \ max\Big(\omega_{1}+\omega_{2}\Big)=4\alpha^2_{1}\alpha_2^{2}(\mathcal{B})_{Q}^\mathrm{opt} . \end{align*}$$

Putting the values of α₁ and α₂, we write $(\mathcal{B}^{3})_Q$ in terms of unsharpness parameters of Bob₁ and Bob₂ as

$$\begin{equation} (\mathcal{B}^{3})_Q=\dfrac{1}{4}\Big(1+\sqrt{1-\lambda_{1}^2}\Big)\Big(1+\sqrt{1-\lambda_{2}^2}\Big)(\mathcal{B})_Q^\mathrm{opt} . \end{equation} \tag{ 27 }$$

Now, to exhibit the quantum violation for three sequential Bobs, we have to show that $(\mathcal{B}^{1})_Q,(\mathcal{B}^{2})_Q,(\mathcal{B}^{3})_Q\gt2$. For $(\mathcal{B}^{1})_Q\gt2$, the lower bound on λ₁ is $\lambda_{1} \gt 1/\sqrt{2}$. From equation (19), we can calculate the lower bound of λ₂ for $(\mathcal{B}^{2})_Q\gt2$ is

$$\begin{align} \lambda_2 > \frac{4}{(\mathcal{B})_{Q}^\mathrm{opt}\Big(1+\sqrt{1-\Big(\frac{2}{(\mathcal{B})_{Q}^\mathrm{opt}}}\Big)^2}= \dfrac{2}{\sqrt{2}+1} \approx 0.828 . \end{align} \tag{ 28 }$$

By considering that Bob₁ and Bob₂ implement their unsharp measurements with minimum required values of unsharpness parameters at their respective sites, and by substituting the values of $(\lambda_{1})_\mathrm{min}\approx0.707$ and $(\lambda_{2})_\mathrm{min}\approx0.828$ in equation (27), we get $(\mathcal{B}^{3})_Q = 1.89$. This means Alice and Bob₃ cannot violate CHSH inequality when both Bob₁ and Bob₂ violate it.

We extend the sequential sharing of nonlocality by using another well-known Bell inequality known as Gisin's elegant Bell inequality [48]. The elegant Bell expression can be written as,

$$\begin{align} \mathbb{E}=(A_{1}+A_{2}+A_{3}-A_{4})\otimes B_{1}+(A_{1}+A_{2}-A_{3}+A_{4}) \otimes B_{2}+(A_{1}-A_{2}+A_{3}+A_{4})\otimes B_{3}. \end{align} \tag{ 29 }$$

whose local bound is $(\mathbb{E})_{l} \leqslant 6$. We show that the elegant Bell expression has two bounds, the local bound (which can be considered as trivial preparation non-contextual bound) and a nontrivial preparation non-contextual bound when there exists a relational constraint between the observables of Alice and Bob. We show that the non-trivial preparation non-contextual bound $(\mathbb{E})_{pnc} \leqslant 4$.

Before proceeding further, we briefly introduce the notion of preparation non-contextuality in an ontological model of quantum theory. Consider that a preparation procedure P prepares a density matrix ρ, and a measurement procedure M realizes the measurement of a POVM E_k . Quantum theory predicts the probability of obtaining a particular outcome k is $p(k|P, M) = Tr[\rho E_{k}]$—the Born rule. In an ontological model of quantum theory, preparation procedures assign a probability distribution $\mu_{P}(\lambda|\rho)$ on ontic states $\lambda \in \Lambda$ where Λ is the ontic state space. Given the measurement procedure, the ontic state λ assigns a response function $\xi_{M}(k|\lambda,E_{k})$. A viable ontological model must reproduce the Born rule, i.e. $\forall k, \rho, E_{k}:\ \ p(k|P, M) = \int_{\Lambda} \mu_{P}(\lambda|\rho) \xi_{M}(k|\lambda,E_{k})d\lambda$.

The dependencies of P and M do not appear if the ontological model is preparation and measurement non-contextual, respectively. Two preparation procedures P and Pʹ are said to be operationally equivalent if they can not be distinguished by any measurement, implying, $\forall k,M:\hspace{.1cm}p(k|P, M) = p(k|P^{^{\prime}},M)$. In quantum theory, such preparation procedures are realized by the density matrix ρ. Such equivalence in operational theory can be reflected in the ontic state level, assuming preparation non-contextuality [71]. An ontological model of an operational quantum theory is considered to be preparation non-contextual if two preparation procedures P and Pʹ prepare the same density matrix ρ, and no measurement can operationally distinguish the context by which ρ is prepared, i.e. $\forall k, M :\hspace{.1cm} p(k|P,M) = p(k|P^{^{\prime}},M) \Rightarrow \forall \lambda : \mu(\lambda|\rho,P) = \mu(\lambda|\rho,P^{^{\prime}})$, implying two ontic state distributions are equivalent irrespective of the contexts P and Pʹ [71–75].

Let us intuitively understand the notion of preparation non-contextuality in the CHSH scenario. Consider that Alice and Bob share an entangled state ρ_AB and Alice measures A₁ and A₂. Alice's measurements on her local system produces density matrices $\rho_{A_{1}}$ and $\rho_{A_{2}}$ on Bob's side corresponding to measurement contexts A₁ and A₂, respectively. The non-signaling condition demands that $\rho_{A_{1}}$ and $\rho_{A_{2}}$ cannot be distinguishable by any measurement of Bob, i.e. $\rho_{A_{1}} = \rho_{A_{2}}\equiv \sigma$. Equivalently, in an ontological model we assume that $\mu(\lambda|\sigma, A_{1}) = \mu(\lambda|\sigma, A_{2})$, i.e. the distribution of ontic states are preparation non-contextual. Intuitively, preparation non-contextuality implies the locality assumption in the Bell scenario. In other words, every probability distribution that violates a Bell inequality can also be regarded as proof of preparation contextuality as proved in [76]. Here, we provide a modified version of the proof.

For this, by using Bayes' theorem we write the joint probability distribution in the ontological model as

$$\begin{equation} p(a, b|A_{i}, B_{j}) = \sum_{\lambda}p(a|A_{i}, B_{j})p(\lambda|a,A_{i})p(b|B_{j}, \lambda). \end{equation} \tag{ 30 }$$

Now, the no-signaling condition implies the marginal probability of Alice's side is independent of Bob's input and hence we can write

$$\begin{equation} p(a, b|A_{i}, B_{j}) = \sum_{\lambda}p(a|A_{i})p(\lambda|a,A_{i})p(b|B_{j}, \lambda). \end{equation} \tag{ 31 }$$

Using Bayes' theorem we can write $p(a|A_{i})p(\lambda|a,A_{i}) = \mu(\lambda|A_{i})p(a|\lambda,A_{i})$ where we specifically denoted the probability distribution $p(\lambda|A_{i})$ as $\mu(\lambda|A_{i})$.

From equation (31), we then obtain

$$\begin{equation} p(a, b|A_{i}, B_{j}) = \sum_{\lambda}\mu(\lambda|A_{i})p(a|\lambda,A_{i})p(b|B_{j}, \lambda). \end{equation} \tag{ 32 }$$

Due to the assignment of the same ontic-state distribution for the different preparation procedures on Bob's side, the assumption of preparation non-contextuality is enforced. Then the preparation non-contextual assumption for Bob's preparation reads as $\mu(\lambda|\rho,{A_{1}}) = \mu(\lambda|\rho,{A_{2}})\equiv\mu(\lambda)$, which in turn provides

$$\begin{equation} p(a, b|A_{i}, B_{j}) = \sum_{\lambda}\mu(\lambda)p(a|\lambda,A_{i})p(b| \lambda, B_{j}), \end{equation} \tag{ 33 }$$

which is the desired factorizability condition commonly derived for a local hidden variable model. Therefore, we can argue that whenever in an ontological model $p(a, b|A_{i},B_{j})$ satisfies preparation non-contextuality, this, in turn, satisfies locality in a hidden variable model.

Here we go one step further. The above case involving the CHSH scenario is a trivial one. We introduce a nontrivial form of preparation non-contextuality in the Bell experiment involving more than two inputs. Consider a Bell experiment where Alice receives four inputs and performs the measurements of four observables A₁, A₂, A₃ and A₄. Bob receives three inputs and performs the measurement of three observables B₁ , B₂ and B₃. In quantum theory, when Alice and Bob share an entangled state ρ_AB

$$\begin{align} \rho_{a|A_{i}}+\rho_{a\oplus 1|A_{i}}=\rho_{a|A_{i^{^{\prime}}}}+\rho_{a\oplus 1|A_{i^{^{\prime}}}}\equiv \sigma \end{align} \tag{ 34 }$$

where $i,i^{^{\prime}} = 1,2,3,4$ with $i\neq i^{^{\prime}}$, and $\rho_{a|A_{i}} = Tr_{A}\left[\rho_{AB}\Pi_{A_{i}}\otimes \mathbb{I}\right]$. This is within the premise of preparation non-contextuality in an ontological model as the distribution of ontic states is assumed to be equivalent for those two preparation procedures i and iʹ. As argued above, such a trivial preparation non-contextuality can be attributed to the locality in a Bell experiment. In that case, the local bound of the elegant Bell expression $(\mathbb{E})_{l}\leqslant6$.

We introduce a nontrivial form of preparation non-contextuality in an ontological model of quantum theory by imposing an additional relational constraint on Alice's measurement observables. For this, we consider that the joint probability satisfies

$$\begin{align} \forall b, j, &P(a, b|A_{1},B_{j})+\sum\limits_{i=2}^{4} P(a\oplus 1, b|A_{i},B_{j}) =P(a\oplus 1, b|A_{1},B_{j})+\sum\limits_{i=2}^{4} P(a, b|A_{i},B_{j}) \end{align} \tag{ 35 }$$

which in quantum theory implies that

$$\begin{align} &\rho_{a\oplus 1|A_{1}}+\rho_{a |A_{2}}+\rho_{a|A_{3}}+\rho_{a|A_{4}} = \rho_{a|A_{1}}+\rho_{a\oplus 1 |A_{2}}+\rho_{a\oplus 1|A_{3}}+\rho_{a\oplus 1|A_{4}}. \end{align} \tag{ 36 }$$

Since $\rho_{a|A_{i}} = Tr_{A}\left[\rho_{AB}\Pi_{A_{i}} \otimes \mathbb{I}\right]$ with $\Pi_{A_{i}} = \left(\mathbb{I}+A_{i}\right)/2$, the above equation (36) implies that $A_{1}-A_{2}-A_{3}-A_{4} = 0$. Along with the equivalence in equation (34) a non-trivial constraint equation (36) is also imposed on Alice's preparation procedures. In such a case, the local bound reduces to the preparation non-contextual bound $(\mathbb{E})_{l}\leqslant 4$ as $A_{1} = A_{2}+A_{3}+A_{4}$. Thus, the quantum violation of it provides a weaker notion of nonlocality, which we call nontrivial preparation contextuality. It is well-known that steering is a weaker form of nonlocality, but the relation between steering and nontrivial preparation contextuality needs to be properly explored. While we plan to do it in another occasion, interested reader may see a relevant work [76]. However, for our present purpose it is not directly relevant and hence we skip such discussion.

Note that the quantum value of $(\mathbb{E})_{Q}$ has also to be calculated by considering this constraint. Below we show that the optimal quantum value $(\mathbb{E})_{Q}^\mathrm{opt} = 4\sqrt{3}$ satisfies the constraint $A_{1} = A_{2}+A_{3}+A_{4}$.

We derive the optimal quantum value $(\mathbb{E})_{Q}^\mathrm{opt}$ without using the dimension of the system. For this, we again use the SOS approach developed in section 2. We define a positive semidefinite operator $\langle\chi\rangle_Q \geqslant 0$ so that $\langle\chi\rangle_Q +\mathbb{E} = \Omega_3$ where Ω₃ is a positive quantity. By considering suitable positive operators L₁, L₂ and L₃ we can write

$$\begin{align} \chi =\frac{1}{2} \left( \omega_{1} L_1^\dagger L_1+\omega_{2} L_2^\dagger L_2+\omega_{3} L_3^\dagger L_3\right) \end{align} \tag{ 37 }$$

where ω₁, ω₂ and ω₃ are positive numbers that will be determined soon. For our purpose, we choose

$$\begin{align} L_{1}|\psi\rangle_{AB}=\frac{A_{1}+A_{2}+A_{3}-A_{4}}{\omega_{1}} |\psi\rangle_{AB} -B_{1} |\psi\rangle_{AB}\nonumber\\ L_{2}|\psi\rangle_{AB}=\frac{A_{1}+A_{2}-A_{3}+A_{4}}{\omega_{2}} |\psi\rangle_{AB} -B_{2} |\psi\rangle_{AB}\nonumber\\ L_{3}|\psi\rangle_{AB}=\frac{A_{1}-A_{2}+A_{3}+A_{4}}{\omega_{3}} |\psi\rangle_{AB} -B_{3} |\psi\rangle_{AB} \end{align} \tag{ 38 }$$

where $\omega_{1} = ||(A_{1}+A_{2}+A_{3}-A_{4})|\psi\rangle_{AB}||_{2}, \omega_{2} = ||(A_{1}+A_{2}-A_{3}+A_{4})|\psi\rangle_{AB}||_{2}$ and $\omega_{3} = ||(A_{1}-A_{2}+A_{3}+A_{4})|\psi\rangle_{AB}||_{2}$. Substituting equation (38) in equation (37), we get

$$\begin{equation} \langle\chi\rangle_Q=-(\mathbb{E})_Q+\sum_{i=1}^{3}\omega_{i} . \end{equation} \tag{ 39 }$$

Hence, the optimal value $(\mathbb{E})_Q^\mathrm{opt}$ is obtained if $\langle\chi\rangle_Q = 0$, i.e.

$$\begin{equation} (\mathbb{E})_Q^\mathrm{opt}=max(\omega_{1}+\omega_{2}+\omega_{3}) \end{equation} \tag{ 40 }$$

where

$$\begin{align} \omega_1= \sqrt{4+\langle\{A_{1},(A_{2}+A_{3}-A_{4})\}+\{A_{2},(A_{3}-A_{4})\}-\{A_{3},A_{4}\}\rangle} \end{align} \tag{ 41 }$$

and similarly for ω₂ and ω₃.

We use the concavity inequality $\sum_{i = 1}^{n}\omega_{i}\leqslant \sqrt{n\sum_{i = 1}^{n}(\omega_{i})^{2}}$ where the equality holds when ω_i 's are equal to each other. We can then write

$$\begin{equation} (\mathbb{E})_{Q}\leqslant max\Big(\sqrt{3(\omega_{1}^{2}+\omega_{2}^{2}+\omega_{3}^{2})}\Big). \end{equation} \tag{ 42 }$$

Now, by using the expressions of ω₁, ω₂ and ω₃ we can write

$$\begin{align} \omega_{1}^{2}+\omega_{2}^{2}+\omega_{3}^{2}={_{AB}\langle \psi\vert}(12+\delta)\vert \psi\rangle_{AB} \end{align} \tag{ 43 }$$

where

$$\begin{align} \delta=(\{A_{1},(A_{2}+A_{3}+A_{4})\}-\{A_{2},(A_{3}+A_{4})\}-\{A_{3},A_{4}\}). \end{align} \tag{ 44 }$$

To maximize δ, we consider that there exist a state $\vert \psi^{^{\prime}}\rangle$ which can be written as $\vert \psi^{^{\prime}}\rangle = (A_{1}-A_{2}-A_{3}-A_{4})\vert \psi\rangle_{AB}$ such that $\vert \psi\rangle_{AB}\neq 0$. We can show that $\langle{\psi^{^{\prime}}}|{\psi^{^{\prime}}}\rangle = 4-\langle\delta\rangle$. By rearranging we can write $\langle\delta\rangle = 4-\langle{\psi^{^{\prime}}}|{\psi^{^{\prime}}}\rangle$. Clearly, the maximum value $\langle\delta\rangle_\mathrm{max} = 4$ is obtained when $\langle{\psi^{^{\prime}}}|{\psi^{^{\prime}}}\rangle = 0$. Since, $\vert \psi\rangle_{AB}\neq 0$, we find the condition of maximization

$$\begin{align} A_{1}-A_{2}-A_{3}-A_{4}=0. \end{align} \tag{ 45 }$$

Since, $\langle\delta\rangle_\mathrm{max} = 4$, we find $\mathrm{max}(\omega_{1}^{2}+\omega_{2}^{2}+\omega_{3}^{2}) = 16$. From equation (42) we can then write $(\mathbb{E})_{Q}\leqslant4\sqrt{3}$.

Using equation (45), a few steps of calculations gives the following relations that need to be satisfied by Alice's observables given by

$$\begin{align} \{A_{1},A_{2}\}=\{A_{1},A_{3}\}=\{A_{1},A_{4}\}=\frac{2}{3}, \end{align} \tag{ 46 }$$

$$\begin{align} \{A_{2},A_{3}\}=\{A_{2},A_{4}\}=\{A_{3},A_{4}\}=-\frac{2}{3}. \end{align} \tag{ 47 }$$

Putting equations (46) and (47) in equation (41), we get $\omega_{1} = 4/\sqrt{3}$. A similar calculation gives us $\omega_{2} = \omega_{3} = 4/\sqrt{3}$. Since, $\omega_{1} = \omega_{2} = \omega_{3}$, the optimal quantum value of elegant Bell expression is $(\mathbb{E})_{Q}^\mathrm{opt} = 4\sqrt{3}$. Note here again that we have not imposed any bound on the dimension of the system throughout the derivation.

As mentioned, the optimal quantum value of $\mathbb{E}$ for any given dimension is obtained when $\langle\chi\rangle_Q = 0$, implying that

$$\begin{align} \forall i\in [3],\ L_{i}\vert \psi\rangle_{AB}=0. \end{align} \tag{ 48 }$$

Further, using the conditions in equation (38), Bobs observables B₁, B₂, and B₃ can be written as

$$\begin{align} B_{1}&=\frac{\sqrt{3}(A_{1}+A_{2}+A_{3}-A_{4})}{4}, \nonumber\\ B_{2}&=\frac{\sqrt{3}(A_{1}+A_{2}-A_{3}+A_{4})}{4}, \nonumber\\ B_{3}&=\frac{\sqrt{3}(A_{1}-A_{2}+A_{3}+A_{4})}{4}. \end{align} \tag{ 49 }$$

It is straightforward to show that Bob's observables satisfy,

$$\begin{align} \{B_{1},B_{2}\} = \frac{\sqrt{3}}{4}\{(A_{1}+A_{2}+A_{3}-A_{4}),(A_{1}+A_{2}-A_{3}+A_{4})\} =\frac{\sqrt{3}}{4}\left(\{A_{1},A_{2}\}+\{A_{3},A_{4}\}\right). \end{align} \tag{ 50 }$$

Putting the values of Alice's anticommuting relations from equations (46) and (47), we get $\{B_{1},B_{2}\} = 0$. Similarly, it can also be shown that $\{B_{1},B_{3}\} = 0$ or $\{B_{2},B_{3}\} = 0$. Thus, Bobs observables $B_{1},B_{2}$ and B₃ have to be mutually anticommuting to obtain the optimal quantum value $(\mathbb{E})_{Q}^\mathrm{opt} = 4\sqrt{3}$. Following the argument developed for CHSH, it can be proved that the shared state between Alice and Bob has to be a maximally entangled state as in equation (7).

5.1. Certification of multiple unsharpness parameters

Similar to the sequential CHSH scenario, for elegant Bell inequality, the average state shared between Alice and Bob_k if Alice and Bob$_{(k-1)}$ shares a maximally entangled state and each Bob performs a dichotomic POVM measurement in the sequential scheme is

$$\begin{align} \rho_{AB_{k}} = \frac{1}{3}\sum_{b_{k}\in \pm}\sum_{y_{k}=1}^{3} K_{b_{k}|y_{k}}\rho_{AB_{(k-1)}}K_{b_{k}|y_{k}} \end{align} \tag{ 51 }$$

where $\{K_{b_{k}|y_{k}}\}$ are the Kraus operators satisfying $\sum_{b_{k}} K_{b_{k}|y_{k}}^{\dagger}K_{b_{k}|y_{k}} = \mathbb{I}$ and $K_{b_{k}|y_{k}} = \sqrt{E_{b_{k}|y_{k}} }$.

The quantum value of elegant Bell expression due to the unsharp measurement of Bob₁ irrespective of the dimension is calculated as

$$\begin{align} (\mathbb{E}^{1})_Q=\lambda_{1}( \mathbb{E})_Q^\mathrm{opt}. \end{align} \tag{ 52 }$$

Let us assume that every sequential Bob measures same set of observables B₁, B₂ and B₃ i.e. $\forall k, \ B_{y_k = 1}\equiv B_{1}$ and so on. The average post-measurement state after Bob₁'s unsharp measurement can be written as

$$\begin{align} \rho_{AB_{2}}&=\dfrac{1}{3} \sum_{b_{1}\in \{+,-\}}^{}\sum_{y_{1}=1}^{3} \left(\mathbb{I} \otimes K_{b_{1}|y_{1}}\right) \rho_{AB_{1}} \left(\mathbb{I} \otimes K_{b_{1}|y_{1}}\right) =\dfrac{1}{3} \left[ 6\alpha^2_{1} \rho_{AB_{1}} +2 \beta^2_{1} \sum_{y_{1}=1}^{3}(\mathbb{I}\otimes B_{y_{1}})\rho_{AB_{1}}(\mathbb{I}\otimes B_{y_{1}})\right] . \end{align} \tag{ 53 }$$

Using $\rho_{AB_{2}}$ in equation (53), the maximum quantum value of elegant Bell expression between Alice and Bob₂ is given by

$$\begin{align} (\mathbb{E}^{2})_Q &= \mathrm{max}\Big(Tr[\rho_{AB_{2}} \mathbb{E}]\Big)\nonumber\\ &=\mathrm{max}\Big(Tr\Big[\rho_{AB_{1}}\Big((A_{1}+A_{2}+A_{3}-A_{4})B_{1}^{^{\prime}}+(A_{1}+A_{2} -A_{3}+A_{4})B_{2}^{^{\prime}}+(A_{1}-A_{2}+A_{3}+A_{4})B_{3}^{^{\prime}}\Big)\Big]\Big). \end{align} \tag{ 54 }$$

We derive $B_{1}^{^{\prime}}$, $B_{2}^{^{\prime}}$ and $B_{3}^{^{\prime}}$ as

$$\begin{align} B_{1}^{^{\prime}}&=2\left(\alpha_{1}^2+\frac{\beta_{1}^2}{3}\right)B_{1}+\frac{2}{3}\beta_{1}^2\left(B_{2}B_{1}B_{2}+B_{3}B_{1}B_{3}\right),\nonumber\\ B_{2}^{^{\prime}}&=2\left(\alpha_{1}^2+\frac{\beta_{1}^2}{3}\right)B_{2}+\frac{2}{3}\beta_{1}^2\left(B_{1}B_{2}B_{1}+B_{3}B_{2}B_{3}\right),\nonumber\\ B_{3}^{^{\prime}}&=2\left(\alpha_{1}^2+\frac{\beta_{1}^2}{3}\right)B_{3}+\frac{2}{3}\beta_{1}^2\left(B_{1}B_{3}B_{1}+B_{2}B_{3}B_{2}\right). \end{align} \tag{ 55 }$$

Equation (54) has a complete resemblance with elegant Bell inequality in equation (29) with the effective observables of Bob are $B_{1}^{^{\prime}}$, $B_{2}^{^{\prime}}$ and $B_{3}^{^{\prime}}$. However, $(B_{i}^{^{\prime}})^{2}\neq \mathbb{I}$ with $i\in [3]$ and hence they need to be normalized. By assuming, $\omega_{i}^{^{\prime}} = ||B_{i}^{^{\prime}}||$ and using the SOS approach we get

$$\begin{align} (\mathbb{E}^{2})_{Q}= max \left(\omega_{1}\omega_{1}^{^{\prime}}+\omega_{2}\omega_{2}^{^{\prime}}+\omega_{3}\omega_{3}^{^{\prime}}\right). \end{align} \tag{ 56 }$$

Optimization of quantum value of $(\mathbb{E}^{2})_{Q}$ demands Bob's observables $B_{1}^{^{\prime}}$, $B_{2}^{^{\prime}}$ and $B_{3}^{^{\prime}}$ have to be mutually anticommuting. It is explicitly shown in appendix B that the anticommutation $\{B_{i}^{^{\prime}},B_{j}^{^{\prime}}\}$ can be written in terms of $\{B_{i},B_{j}\}$ where $i(j) \in 1,2,3$ with i ≠ j. It is proved that $\{B_{i}^{^{\prime}},B_{j}^{^{\prime}}\} = 0$ implying $\{B_{i},B_{j}\} = 0$ as the parameters α₁ and β₁ are positive.

In other words, Bob₂ also requires the anticommuting observables to obtain the maximum quantum value of $(\mathbb{E}^{2})_{Q}$. We then calculate

$$\begin{align} \omega_{1}^{^{\prime}}&=\Bigg((2\alpha_{1}^2+\frac{2}{3}\beta_{1}^2)^{2}+(\frac{2}{3}\beta_{1}^2)^{2}\Big(4+B_{2}\{B_{1},B_{2}\}B_{3}\{B_{1},B_{3}\} -B_{2}\{B_{1},B_{2}\}B_{1}-B_{1}B_{3}\{B_{1},B_{3}\}+B_{3}\{B_{1},B_{3}\}\nonumber\\ &\quad \times B_{2}\{B_{1},B_{2}\}-B_{3}\{B_{1}B_{3}\}B_{1}-B_{1}B_{2}\{B_{1},B_{2}\}\Big)+(2\alpha_{1}^2+\frac{2}{3}\beta_{1}^2)\frac{2}{3}\beta_{1}^2 \Big(\{B_{1},B_{2}\}^{2}+\{B_{1},B_{3}\}^{2}-4\Big)\Bigg)^{1/2}. \end{align} \tag{ 57 }$$

Using $\{B_{i},B_{j}\} = 0$, we get $\omega_{1}^{^{\prime}} = 2(\alpha_{1}^2-\frac{\beta_{1}^2}{3})$. Similarly, we find $\omega_{2}^{^{\prime}} = \omega_{3}^{^{\prime}} = \omega_{1}^{^{\prime}}$.

Putting the values of $\omega_{1}^{^{\prime}}$, $\omega_{2}^{^{\prime}}$ and $\omega_{3}^{^{\prime}}$, the maximum quantum value of the elegant Bell expression $(\mathbb{E}^{2})_Q$ can be obtained from equation (56) as

$$\begin{align*} (\mathbb{E}^{2})_Q&=2\Bigg(\alpha_{1}^2-\frac{\beta_{1}^2}{3}\Bigg)(\mathbb{E})_Q^\mathrm{opt}. \end{align*}$$

By inserting the values of α₁ and β₁, we have $(\mathbb{E}^{2})_Q$ in terms of unsharpness parameter as

$$\begin{equation} (\mathbb{E}^{2})_Q=\dfrac{1}{3}\Big(1+2\sqrt{1-\lambda_{1}^2}\Big)(\mathbb{E})_Q^\mathrm{opt}. \end{equation} \tag{ 58 }$$

Note that, the maximum values of $(\mathbb{E}^{1})_Q$ and $(\mathbb{E}^{2})_Q$ in equations (52) and (58) are dependent only on the unsharpness parameter λ₁ as $(\mathbb{E})^\mathrm{opt}_Q$ is common. Writing $(\mathbb{E}^{2})_Q$ in terms of $(\mathbb{E}^{1})_Q$ we get

$$\begin{equation} (\mathbb{E}^{2})_Q=\frac{4}{\sqrt{3}}\left(1+2\sqrt{1-\Bigg(\frac{ (\mathbb{E}^{1})_Q}{(\mathbb{E})^\mathrm{opt}_Q}}\Bigg)^{2}\right) . \end{equation} \tag{ 59 }$$

It can be seen for equation (59) that if Bob₁ extracts more information then the quantum value of $(\mathbb{E}^{1})_Q$ increases and consequently decreases the value of $(\mathbb{E}^{2})_Q$ and hence there is a trade-off between the values of $(\mathbb{E}^{2})_Q$ and $(\mathbb{E}^{1})_Q$. Figure 3 represents the optimal trade-off relation written in equation (59) where the green curve shows the quantum values where each point on its surface certifies a unique value of unsharpness parameter λ₁. For $(\mathbb{E}^{1})_Q = (\mathbb{E}^{2})_Q = 12(4+\sqrt{3})/13\approx5.291$, both the Bob₁ and Bob₂ gets equal advantage as shown by the blue point on the surface of the curve which uniquely certifies the value of sharpness parameter $\lambda_1 = (\sqrt{57+24\sqrt{3}})/13\approx0.763$.

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Similarly, the quantum value of elegant Bell expression between Alice and Bob₃ can be calculated as $(\mathbb{E}^{3})_Q = Tr[\rho_{AB_{3}} (\mathbb{E})]$ where $\rho_{AB_{3}}$ is the average state shared between Alice and Bob₃. Then, the elegant Bell expression between Alice and Bob₃ can be re-written as

$$\begin{align} (\mathbb{E}^{3})_Q&=Tr\Big[\rho_{AB_{1}}\Big((A_{1}+A_{2}+A_{3}-A_{4})B_{1}^{^{\prime\prime}}+(A_{1}+A_{2} -A_{3}+A_{4})B_{2}^{^{\prime\prime}}+(A_{1}-A_{2}+A_{3}+A_{4})B_{3}^{^{\prime\prime}}\Big)\Big] \end{align} \tag{ 60 }$$

where $B^{^{\prime\prime}}_{1}$, $B^{^{\prime\prime}}_{2}$ and $B^{^{\prime\prime}}_{3}$ are explicitly defined in appendix C.

Now, we can use the SOS approach as mentioned earlier to obtain the optimal value of $(\mathbb{E}^{3})_{Q}$. But, $(B^{^{\prime\prime}}_{i})^{2}\neq \mathbb{I}$ and hence needs to be normalised. By considering $\omega^{^{\prime\prime}}_{1} = ||B^{^{\prime\prime}}_{1}||$, $\omega^{^{\prime\prime}}_{2} = ||B^{^{\prime\prime}}_{2}||$ and $\omega^{^{\prime\prime}}_{3} = ||B^{^{\prime\prime}}_{3}||$ and by using the SOS approach we obtain

$$\begin{align} (\mathbb{E}^{3})_{Q}= max\left(\omega_{1}\omega^{^{\prime\prime}}_{1} +\omega_{2}\omega^{^{\prime\prime}}_{2}++\omega_{3}\omega^{^{\prime\prime}}_{3}\right). \end{align} \tag{ 61 }$$

Note that, $B^{^{\prime\prime}}_{i}$ and $B^{^{\prime\prime}}_{j}$ with $i(j)_{i\neq j} \in \{1,2,3\}$ can again be proved to be mutually anticommuting. That is, Bob₃ requires the anticommuting observables to obtain the optimal quantum value. Using the anticommutation relation we then calculate $\omega^{^{\prime\prime}}_{1} = 4(\alpha_{1}^2-\frac{\beta_{1}^2}{3})(\alpha_{2}^2-\frac{\beta_{2}^2}{3})$. Similarly, we find $\omega^{^{\prime\prime}}_{2} = \omega^{^{\prime\prime}}_{3} = \omega^{^{\prime\prime}}_{1}$.

The quantum value for Alice and Bob₃ can be written as

$$\begin{align} (\mathbb{E}^{3})_{Q} &= 4\Bigg(\alpha_{1}^2-\frac{\beta_{1}^2}{3}\Bigg)\Bigg(\alpha_{2}^2-\frac{\beta_{2}^2}{3}\Bigg) \ max\Big(\omega_{1}+\omega_{2}+\omega_{3}\Big) \end{align} \tag{ 62 }$$

$$\begin{align} &=4\left(\alpha_{1}^2-\frac{\beta_{1}^2}{3}\right)\left(\alpha_{2}^2-\frac{\beta_{2}^2}{3}\right)(\mathbb{E})_{Q}^\mathrm{opt}. \end{align} \tag{ 63 }$$

Writing $(\mathbb{E}^{3})_{Q}$ in terms of the unsharpness parameters λ₁ and λ₂ we get

$$\begin{equation} (\mathbb{E}^{3})_Q=\dfrac{1}{9}\Big(1+2\sqrt{1-\lambda_{1}^2}\Big)\Big(1+2\sqrt{1-\lambda_{2}^2}\Big)(\mathbb{E})_Q^\mathrm{opt} \end{equation} \tag{ 64 }$$

which can be generalised for Alice and any arbitrary k^th Bob (Bob_k ) as

$$\begin{equation} (\mathbb{E}^{k})_Q=\dfrac{1}{3^{k}}\prod_{i=1}^{k-1}\sqrt{3}\Big(1+2\sqrt{1-\lambda_{i}^2}\Big)(\mathbb{E})_Q^\mathrm{opt}. \end{equation} \tag{ 65 }$$

Further, $(\mathbb{E}^{3})_Q$ can be simplified in term of $(\mathbb{E}^{1})_Q$, $(\mathbb{E}^{2})_Q$ as

$$\begin{equation} (\mathbb{E}^{3})_Q=\dfrac{1}{9}(1+\Delta_{1})\Bigg(1+\sqrt{\frac{4+4\Delta_{1}-\Delta_{2}}{1+\Delta_{1}}}\Bigg) \end{equation} \tag{ 66 }$$

where $\Delta_{1} = 2\sqrt{1-\Big(\frac{(\mathbb{E}^{1})_{Q}}{(\mathbb{E})^\mathrm{opt}_Q}\Big)^{2}}$ and $\Delta_{2} = \dfrac{\sqrt{3}(\mathbb{E}^{1})_Q}{4}$.

Optimal trade-off between quantum bound of elegant Bell inequality of Bob₁, Bob₂ and Bob₃ are given by equation (66) and plotted in figure 4. The brown cube represents the preparation non-contextual bound showing no trade-off. The three-dimensional semi-paraboloid over the cube represents the trade-off between quantum bound. For sharp measurement of Bob₃ with $\lambda_{3} = 1$, each point on the surface of the semi paraboloid in figure 4 uniquely certifies λ₁ and λ₂. The black point on the surface of the semi paraboloid uniquely certifies $\lambda_{1} = 0.644$ and $\lambda_{2} = 0.763$ for $(\mathbb{E}^{1})_Q = (\mathbb{E}^{2})_Q = (\mathbb{E}^{3})_Q = 4.462$.

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5.2. DI self-testing statement in sequential scenario

5.3. Robust certification of unsharpness parameters

Similar to the sequential CHSH scenario, for sub-optimal violation of elegant Bell expression, the certifiable range of λ₁ from equations (52) and (58) can be calculated as

$$\begin{align} \frac{(\mathbb{E}^{1})_Q}{(\mathbb{E})^{opt }_Q}<\lambda_{1}<\sqrt{1-\frac{1}{4}\Bigg(\frac{3(\mathbb{E}^{2})_Q}{\lambda_{2}(\mathbb{E})^{opt }_Q}-1\Bigg)^{2}}. \end{align} \tag{ 67 }$$

Equation (67) shows that for the optimal quantum value of $(\mathbb{E})_Q^\mathrm{opt}$, the lower bound of λ₁ depends only on the maximum quantum value of $(\mathbb{E}^{1})_Q$. Since, $(\mathbb{E})_Q^\mathrm{opt}$ is $4\sqrt{3}$, for violating the elegant Bell inequality Bob₁ requires $(\mathbb{E}^{1})_Q\gt4$, which fixes lower bound of unsharpness parameter as $(\lambda_{1})_\mathrm{min} \approx 1/\sqrt{3}\approx 0.57$. Any value of $\lambda_{1} \in [1/\sqrt{3},1]$ therefore provides the quantum advantage for Bob₁.

The upper bound of unsharpness parameter λ₁ is a function of λ₂ and the maximum quantum value of $(\mathbb{E}^{2})_Q$. To get advantage for Bob₂, two sequential Bob requires $(\mathbb{E}^{1})_Q, (\mathbb{E}^{2})_Q\gt4$. Considering Bob₂ performs sharp measurement ($\lambda_{2} = 1$), the upper bound on the unsharpness parameter i.e. $(\lambda_{1})_\mathrm{max}$ from equation (67) is calculated as $(\lambda_{1})_\mathrm{max} = \sqrt{\sqrt{3}/2}\approx 0.93$. Thus, when both Bob₁ and Bob₂ get quantum advantage, the interval $0.57\lt\lambda_{1}\lt 0.93$ can be certified.

Now, the range of λ₁ becomes more restricted if the quantum advantage is further extended to Bob₃ which demands $(\mathbb{E}^{1})_Q, (\mathbb{E}^{2})_Q, \text{and}\ (\mathbb{E}^{3})_Q\gt4$. In such a case, from equations (52) and (64), the interval of λ₁ is narrower down to $0.57 \lt \lambda_{1} \lt 0.77$ which requires a more efficient experimental realization.

Again, if sequential Bobs up to Bob₃ get the quantum advantage, the range of λ₂ can be calculated from equations (58) and (64) as

$$\begin{align} \dfrac{3(\mathbb{E}^{2})_Q}{\xi_{1}(\mathbb{E})_Q^\mathrm{opt}}<\lambda_2<\sqrt{1-\dfrac{1}{4}\Bigg(\dfrac{9(\mathbb{E}^{3})_Q}{\xi_{1}}-1\Bigg)^{2}} \end{align} \tag{ 68 }$$

where $\xi_{1} = \Big(1+2\sqrt{1-\lambda_{1}^{2}}\Big)$. Equation (68) shows that both the upper and lower bound of λ₂ are dependent on λ₁. Thus, the range $0.65 \lt \lambda_{2} \lt 0.87$ can be certified. Further, from equation (64) the lower bound of $(\lambda_{3})_\mathrm{min}$ is calculated as,

$$\begin{equation} (\lambda_3)_\mathrm{min}=\dfrac{9(\mathbb{E}^{3})_Q}{\xi_{1} \Big(1+2\sqrt{1-\lambda_{2}^{2}}\Big)(\mathbb{E})_Q^\mathrm{opt}}\approx 0.78 . \end{equation} \tag{ 69 }$$

Now, considering Bob₁, Bob₂ and Bob₃ implement their unsharp measurements with lower critical values of unsharpness parameters at their respective sites, from equation (65) we get $(\mathbb{E}^{4})_Q = 3.84$, i.e. Alice and Bob₄ cannot violate elegant Bell inequality, thereby will not provide any quantum advantage over preparation non-contextual bound.

The precise control of quantum devices plays a crucial role in the development of quantum technologies. Therefore, developing elegant protocols for certifying quantum devices is indispensable in quantum information theory. In this work, we provided DI self-testing of unsharp instruments based on two Bell inequalities, viz., the CHSH inequality and the elegant Bell inequality. Note that the optimal quantum violation of a Bell inequality uniquely self-tests the state and the observables. We discussed that such a violation does not certify the post-measurement states, and hence self-testing the unsharp instrument is not possible in that way. We demonstrated that the sequential Bell test by multiple independent observers on one side of the Bell experiment enables the self-testing of post-measurement states, which, in turn, uniquely self-tests the unsharp instruments. The optimal quantum violation of CHSH inequality can be achieved only for sharp measurement, and for unsharp measurement, one obtains the sub-optimal value. Notably, the sub-optimal quantum value may also arise due to the nonideal preparation of the state or the inappropriate choices of the local observables. However, as explicitly discussed in [52], the primary criticism regarding DI self-testing of unsharp measurement through a Bell test arises from Naimark's theorem, which states that any non-projective measurement can be viewed as a projective measurement in a higher-dimensional Hilbert space. Since there is no bound on dimension in DI self-testing, one can always argue that the sub-optimal quantum violation arises from the inappropriate choice of observables in higher-dimension and not from the unsharp measurement.

By providing the dimension-independent optimization of the sequential Bell test, we provided a scheme for self-testing of unsharp instruments. As mentioned, the sequential sharing of nonlocality by multiple independent observers plays a crucial role in our work. Crucially, such a sharing of quantum correlation without assuming the dimension of the system has not hitherto been discussed. Here, we impose no bound on the dimension of the system, and the quantum devices are taken as black boxes. We introduced an elegant SOS approach enabling us to derive the maximum quantum values in the sequential Bell test. We note here that the semi-DI certification of the unsharp instruments in a sequential prepare-measure scenario was demonstrated in [53] by using a qubit system. In [53], the authors leave DI self-testing of an unsharp instrument as an open question which is now provided in our work.

We first considered the sequential sharing of CHSH nonlocality where Alice always performs sharp measurement and an arbitrary k number of sequential Bobs (Bob_k ) who perform unsharp measurements on their local sub-system. If the first observer (Bob₁) performs a sharp projective measurement, the entanglement between Alice and Bob₁ will be lost, and there is no chance that Alice and Bob₂ will violate the CHSH inequality. On the other hand, if Bob₁ performs an unsharp measurement, a sufficient residual entanglement may remain to exhibit the violation of CHSH inequality between Alice and Bob₂. In such a case, we jointly maximized the sub-optimal quantum advantages for both sequential observers and demonstrated that there exists a trade-off relation between the two sequential quantum violations. The sub-optimal quantum violations form an optimal pair which eventually self-tests the unsharp instrument of Bob₁ along with the entangled state and the observables of Alice, Bob₁ and Bob₂. In the CHSH scenario, at most, two sequential Bobs can share the nonlocality, and hence only one unsharpness parameter can be self-tested.

We extend our treatment to the sharing of preparation contextuality based on the elegant Bell inequality. We demonstrated that the preparation contextuality can be shared up to a maximum of three sequential Bobs. We jointly maximized the sub-optimal sequential quantum values by using the SOS approach without assuming the dimension of the system. There is a trade-off between the three sub-optimal quantum values for Bob₁, Bob₂, and Bob₃ violating the elegant Bell inequality. Three sub-optimal quantum values form an optimal triple which in turn self-tests the unsharp instruments of Bob₁ and Bob₂. In the process of maximization of the sub-optimal values, the entangled state and the observables of Alice and of Bob₁, Bob₂, and Bob₃ are self-tested.

We note that due to unavoidable losses and imperfection in the actual experimental scenarios, the unique certification of the unsharpness parameter is not possible. In such a case, we provided an analysis of robust certification so that a range of the unsharpness parameter can be certified. The more perfect the actual experiment, one can achieve more accurate the certification of the unsharpness parameters.

We conclude by proposing the following future direction and potential applications. Our work can be further generalized to self-test an arbitrary number of unsharp instruments based on the quantum violation of a family of preparation non-contextual inequalities proposed in [74]. It can be interesting to study the sequential sharing of preparation contextuality by multiple numbers of observers on one side of the Bell experiment. Using the SOS approach one can simultaneously maximize the sub-optimal quantum values corresponding to the independent sequential observers. This can be an interesting line of future study. Our work has immediate application in generating a higher amount of DI-certified randomness. In a recent work, based on the sequential sharing of nonlocality by using the two-qubit entangled state, the generation of a higher amount of randomness was proposed [5], which is experimentally tested in [77]. The self-testing protocol provided here can be used to generate a higher amount of DI-certified randomness. This calls for further study.

The Authors acknownledge Sumit Mukherjee for helpful discusssions. PR acknowledges the support from the research grant DST/ICPS/QuEST/2019/Theme-1/Q42. AKP acknowledges the support from the research Grant MTR/2021/000908.

No new data were created or analysed in this study.

Here we show that the set of observables require for achieving the maximum quantum violation for Bob₁ and Bob₂ are the same. Let Bob₁ performs measurement of B₁ and B₂ on his local subsystems upon receiving input $y_{1}\in \{1,2\}$ on the entangled state $\rho_{AB_{1}}$, and Bob₂ upon receiving input $y_{2}\in \{1,2\}$ performs measurements of the observables B₃ and B₄ on a state $\rho_{AB_{2}}$ producing outputs $b_{2}\in \{-1,+1\}$. Putting $\rho_{AB_{2}}$ from equation (13), we can explicitly write the CHSH expression between Alice and Bob₂ as

$$\begin{align} (\mathcal{B}^{2})_Q&=Tr\Bigg[\rho_{AB_{1}}\left(\Big(A_{1}+A_{2}\Big) \bar{B}+\Big((A_{1}-A_{2})\bar{\bar{B}}\right)\Bigg] \end{align} \tag{ A1 }$$

where $\bar{B} = 2\alpha^2_{1}B_{3}+\beta^2_{1}\Big(B_{1}B_{3}B_{1}+B_{2}B_{3}B_{2}\Big)$ and $\bar{\bar{B}} = 2\alpha^2_{1}B_{4}+\beta^2_{1}\Big(B_{1}B_{4}B_{1}+B_{2}B_{4}B_{2}\Big)$.

By writing equation (A1) in terms of the unsharpness parameter and rearranging, the CHSH expression can be written as

$$\begin{align} (\mathcal{B}^{2})_Q&=Tr\Bigg[\rho_{AB_{1}}\Bigg(\gamma_{1}\Big((A_{1}+A_{2}) B_{3}+(A_{1}-A_{2})B_{4}\Big)\nonumber \\ &\qquad\quad +\dfrac{\gamma_{1}^{^{\prime}}}{2}\Big((A_{1}+A_{2}) (B_{1}B_{3}B_{1}+B_{2}B_{3}B_{2})+(A_{1}-A_{2})(B_{1}B_{4}B_{1}+B_{2}B_{4}B_{2}\Big)\Bigg)\Bigg] \end{align} \tag{ A2 }$$

where $\gamma_{1} = \dfrac{1}{2}({1+\sqrt{1-\lambda_{1}^{2}}})$ and $\gamma_{1}^{^{\prime}} = \dfrac{1}{2}({1-\sqrt{1-\lambda_{1}^{2}}})$.

The first term in equation (A2) has the form of CHSH inequality where B₃ and B₄ can be proved as mutually anticommuting by the SOS approach. Since $\gamma_{1} \gt \gamma_{1}^{^{\prime}}$, we can then write

$$\begin{align} &(\mathcal{B}^{2})_Q\leqslant\gamma_{1}\ max\Big(||(A_{1}+A_{2})|| +||(A_{1}-A_{2})||\Big) +\dfrac{\gamma_{1}^{^{\prime}}}{2}\Big((A_{1}+A_{2}) (B_{1}B_{3}B_{1}+B_{2}B_{3}B_{2})+(A_{1}-A_{2})(B_{1}B_{4}B_{1}+B_{2}B_{4}B_{2})\Big). \end{align} \tag{ A3 }$$

It is simple to show that $\{(B_{1}B_{3}B_{1}+B_{2}B_{3}B_{2}),(B_{1}B_{4}B_{1}+B_{2}B_{4}B_{2})\} = 0$ which requires $B_{1} = B_{3}$ and $B_{2} = B_{4}$, and in turn $\{B_{1},B_{2}\} = 0$.

To derive optimal quantum value of the elegant Bell expression written in equation (54) we require the (unnormalized) observables to be $B_{1}^{^{\prime}}$, $B_{2}^{^{\prime}}$ and $B_{3}^{^{\prime}}$ mutually anticommuting. We show that $\{B_{i}^{^{\prime}}, B_{j}^{^{\prime}}\} = 0$ implies $\{B_{i}, B_{j}\} = 0$ where $i,j = 1,2,3$ with i ≠ j.

$$\begin{align} &\{B_{1}^{^{\prime}},B_{2}^{^{\prime}}\} \end{align} \tag{ B1 }$$

$$\begin{align} &=4\Bigg(\alpha_{1}^2+\frac{\beta_{1}^2}{3}\Bigg)^{2}\{B_{1},B_{2}\}+\frac{4}{3}\beta_{1}^2\Bigg(\alpha_{1}^2+\frac{\beta_{1}^2}{3}\Bigg) \Bigg(B_{1}B_{3}\{B_{2},B_{3}\}+B_{2}B_{3}\{B_{1},B_{3}\}+B_{3}\{B_{1},B_{3}\}B_{2}+B_{3}\{B_{2},B_{3}\}B_{1}\Bigg)\nonumber\\ &\quad+\Bigg(\frac{2}{3}\beta_{1}^2\Bigg)^{2}\Bigg(\{B_{1},B_{2}\}^{3}-3\{B_{1},B_{2}\}-B_{1}B_{3}\{B_{2},B_{3}\} -B_{2}\{B_{1},B_{2}\}B_{2}-B_{1}\{B_{1},B_{2}\}B_{1}-B_{3}\{B_{2},B_{3}\}B_{1}\nonumber\\ & \quad-B_{3}\{B_{1},B_{3}\}B_{2} -B_{2}B_{3}\{B_{1},B_{3}\}+B_{2}\{B_{1},B_{2}\}B_{3}\{B_{2},B_{3}\}+B_{3}\{B_{2},B_{3}\}B_{2}\{B_{1},B_{2}\}+B_{3}\{B_{1},B_{3}\}B_{1}\{B_{1},B_{2}\}\nonumber\\ & \quad +B_{1}\{B_{1},B_{2}\}B_{3}\{B_{1},B_{3}\}+B_{3}\{B_{1},B_{2}\}B_{3}\Bigg) \end{align} \tag{ B2 }$$

The co-efficients $\alpha_{1}\gt0$, $\beta_{1}\geqslant 0$. So, the above expression can only be zero when $\{B_{i},B_{j}\} = 0$ where $i(j) \in 1,2,3$.

To derive the optimal quantum value of $(\mathbb{E}^{2})_Q$ in the equation (54) we use SOS approach where $\omega_{i}^{^{\prime}} = ||B_{i}^{^{\prime}}||$ with $i \in \{1,2,3\}$ where $||.||_{2}$ is the Euclidean norm of a vector. The optimal quantum value of $(\mathbb{E}^{2})_Q$ can be obtained as

$$\begin{align} (\mathbb{E}^{2})_Q^\mathrm{opt} ={max}\left(\omega_{1}\omega^{^{\prime}}+\omega_{2}\omega_{2}^{^{\prime}}+\omega_{3}\omega_{3}^{^{\prime}}\right)=2\Bigg(\alpha_{1}^2-\frac{\beta_{1}^2}{3}\Bigg)(\mathbb{E})_Q^\mathrm{opt} \end{align} \tag{ B3 }$$

The average reduced state for Alice and Bob₃ is calculated by using equation (51) as

$$\begin{align} \rho_{AB_{3}}&=\dfrac{1}{3} \sum_{b_{2}\in \{+,-\}}^{}\sum_{y_{2}=1}^{3} \left(\mathbb{I} \otimes K_{b_{2}|y_{2}}\right) \rho_{AB_{2}} \left(\mathbb{I} \otimes K_{b_{2}|y_{2}}\right)\nonumber\\ &= 4\Big(\alpha_{1}^2\alpha_{2}^2+\frac{\beta_{1}^2\beta_{2}^2}{3}\Big) \rho_{AB_{1}} +\frac{4}{3}\Big(\alpha_{1}^2\beta_{2}^2+\beta_{1}^2\alpha_{2}^2\Big)\sum_{{y_{2}}=1}^{3}(\mathbb{I}\otimes B_{y_{2}})\rho_{AB_{1}} (\mathbb{I}\otimes B_{y_2})+\frac{4\beta_{1}^2\beta_{2}^2}{9}\nonumber\\ &\quad\times\Big( B_{1}B_{2}\rho_{AB_{1}}B_{2} B_{1}+B_{2}B_{1}\rho_{AB_{1}}B_{1} B_{2} +B_{1}B_{3}\rho_{AB_{1}}B_{3} B_{1}+B_{3}B_{1}\rho_{AB_{1}}B_{1} B_{3}+B_{2}B_{3}\rho_{AB_{1}}B_{3} B_{2}+B_{3}B_{2}\rho_{AB_{1}}B_{2} B_{3}\Big) \end{align} \tag{ C1 }$$

The quantum value of elegant Bell expression between Alice and Bob₃ from equation (C1) is given by

$$\begin{align*} (\mathbb{E}^{3})_Q =max\left(Tr\left[\rho_{AB_{1}}\Big((A_{1}+A_{2}+A_{3}-A_{4})B_{1}^{^{\prime\prime}}+(A_{1}+A_{2}-A_{3}+A_{4})B_{2}^{^{\prime\prime}}+(A_{1}-A_{2}+A_{3}+A_{4})B_{3}^{^{\prime\prime}}\Big)\right]\right) \end{align*}$$

where

$$\begin{align} B_{1}^{^{\prime\prime}}&=\left(4\alpha_{1}^2\alpha_{2}^2+\frac{4\beta_{1}^2\beta_{2}^2}{3}+\frac{4\alpha_{1}^2\beta_{2}^2}{3}+\frac{4\beta_{1}^2\alpha_{2}^2}{3}\right)B_{1}+\left(\frac{4\alpha_{1}^2\beta_{2}^2}{3}+\frac{4\beta_{1}^2\alpha_{2}^2}{3}+\frac{4\beta_{1}^2\beta_{2}^2}{9}\right)(B_{2}B_{1}B_{2}+B_{3}B_{1}B_{3})\nonumber\\ &\quad+\frac{4\beta_{1}^2\beta_{2}^2}{9}(B_{1}B_{2}B_{1}B_{2}B_{1}+B_{1}B_{3}B_{1}B_{3}B_{1}+B_{3}B_{2}B_{1}B_{2}B_{3}+B_{2}B_{3}B_{1}B_{3}B_{2}), \end{align} \tag{ C2 }$$

$$\begin{align} B_{2}^{^{\prime\prime}}&=\left(4\alpha_{1}^2\alpha_{2}^2+\frac{4\beta_{1}^2\beta_{2}^2}{3}+\frac{4\alpha_{1}^2\beta_{2}^2}{3}+\frac{4\beta_{1}^2\alpha_{2}^2}{3}\right)B_{2}+\left(\frac{4\alpha_{1}^2\beta_{2}^2}{3}+\frac{4\beta_{1}^2\alpha_{2}^2}{3}+\frac{4\beta_{1}^2\beta_{2}^2}{9}\right)(B_{1}B_{2}B_{1}+B_{3}B_{2}B_{3})\nonumber\\ &\quad+\frac{4\beta_{1}^2\beta_{2}^2}{9}(B_{2}B_{1}B_{2}B_{1}B_{2}+B_{2}B_{3}B_{2}B_{3}B_{2}+B_{3}B_{1}B_{2}B_{1}B_{3}+B_{1}B_{3}B_{2}B_{3}B_{1}), \end{align} \tag{ C3 }$$

$$\begin{align} B_{3}^{^{\prime\prime}}&=\left(4\alpha_{1}^2\alpha_{2}^2+\frac{4\beta_{1}^2\beta_{2}^2}{3}+\frac{4\alpha_{1}^2\beta_{2}^2}{3}+\frac{4\beta_{1}^2\alpha_{2}^2}{3}\right)B_{3}+\left(\frac{4\alpha_{1}^2\beta_{2}^2}{3}+\frac{4\beta_{1}^2\alpha_{2}^2}{3}+\frac{4\beta_{1}^2\beta_{2}^2}{9}\right)(B_{1}B_{3}B_{1}+B_{2}B_{3}B_{2})\nonumber\\ &\quad+\frac{4\beta_{1}^2\beta_{2}^2}{9}(B_{3}B_{1}B_{3}B_{1}B_{3}+B_{3}B_{2}B_{3}B_{2}B_{3}+B_{2}B_{1}B_{3}B_{1}B_{2}+B_{1}B_{2}B_{3}B_{2}B_{1}). \end{align} \tag{ C4 }$$

It can also be proved that $\{B_{i}^{^{\prime\prime}},B_{j}^{^{\prime\prime}}\} = 0$ with $i,j = 1,2,3$ provided $\{B_{i},B_{j}\} = 0$.