Entangled states cannot be classically simulated in generalized Bell experiments with quantum inputs

In the bipartite case, we define via the triple $(\rho ^{\mathrm {AB}}, \{\mathcal {A}^{\mathrm {A}}_{a|s}\},\{\mathcal {B}^{\mathrm {B}}_{b|t}\})$ the scenario to be simulated, where ρ^AB is a bipartite entangled state shared between Alice and Bob, while $\{ \mathcal {A}^{\mathrm {A}}_{a|s}\}$ and $\{ \mathcal {B}^{\mathrm {B}}_{b|t}\}$ are positive operator-valued measure (POVM) elements [1] describing the respective measurements to be made on their subsystems; each POVM, labeled by settings s or t, has outcomes a or b. The simulation task is defined as follows. Alice and Bob are given beforehand classical descriptions of the triple specifying the scenario and have access to shared randomness. In each round of the simulation, they each receive a classical description of their input s or t but do not know the other party's input. Their task is to produce—possibly after some rounds of classical communication with each other—classical outputs (a,b) such that they reproduce exactly the joint conditional probability distribution

$$\begin{equation} P_\rho(a,b|s,t) = {\rm tr} \left [ \left ( \mathcal{A}^{\mathrm{A}}_{a|s} \otimes \mathcal{B}^{\mathrm{B}}_{b|t} \right ) \rho^{\mathrm{AB}} \right ] \end{equation} \tag{ 1 }$$

as predicted by quantum mechanics.

It is well known that some quantum correlations can violate Bell inequalities, and as such cannot be simulated using only shared randomness. The canonical no-go result is represented by the Clauser–Horne–Shimony–Holt (CHSH) [30] Bell scenario, where each of (s,t,a,b) is binary. In this scenario, the CHSH inequality defines constraints that have to be satisfied by Bell-local correlations [7, 8, 31]; these constraints can be maximally violated with judicious choices of measurements $\{ \mathcal {A}^{\mathrm {A}}_{a|s} \}$, $\{ \mathcal {B}^{\mathrm {B}}_{b|t}\}$ when Alice and Bob share, for instance, a two-qubit singlet state ρ^AB = |Ψ⁻〉〈Ψ⁻|, with $\vert \Psi ^-\rangle = \frac {1}{\sqrt {2}}(\vert 01\rangle -\vert 10\rangle )$. Note that only entangled states ρ^AB can generate Bell non-local correlations; entanglement is however not a sufficient resource, as some entangled states can only generate Bell-local correlations when measured one copy at a time [21, 22].

When any amount of communication is allowed, a simple strategy to achieve the simulation task is the following. Alice communicates the input that she receives to Bob, and then they produce outputs in accordance with some pre-established strategy via shared randomness. However, there may be more efficient strategies when more inputs are considered: for instance, the simulation of all projective measurements on the singlet state can be achieved using only 1 bit of communication in addition to shared randomness [11].

Let us now consider a variant inspired by Buscemi's work [29]. The scenario of a QI simulation task is defined via the five-tuple $(\rho ^{\mathrm {AB}},\{\vert \varphi _{s}\rangle ^{\mathrm {A'}}\},\{\vert \psi _{t}\rangle ^{\mathrm {B'}}\},\{ \mathcal {A}^{\mathrm {A\!'A}}_{a} \},\{ \mathcal {B}^{\mathrm {BB'}}_{b}\})$ where the superscripts identify the relevant subsystems (see figure 1); the sets {|φ_s〉^A'},{|ψ_t〉^B'} specify the QI states provided to Alice and Bob, while $\{ \mathcal {A}^{\mathrm {A\!'A}}_{a} \}$ and $\{ \mathcal {B}^{\mathrm {BB'}}_{b}\}$ are POVM elements acting, respectively, on the systems A'A and BB'. As with the standard simulation task, Alice and Bob are given beforehand classical descriptions of the five-tuple specifying the scenario and are assumed to have access to shared randomness. However, instead of a classical description of the inputs s and t, in each round of the simulation task, Alice and Bob now receive respectively from external state preparation devices the quantum states |φ_s〉^A' and |ψ_t〉^B' to be measured jointly with ρ^AB. The goal here is again for Alice and Bob to produce, possibly with the help of classical communication, outputs a and b such that they reproduce exactly the conditional probability distribution

$$\begin{equation} P_\rho(a,b \, | \, \vert\varphi_{s}\rangle,\vert\psi_{t}\rangle) = \, {\rm tr} \big [ \! (\mathcal{A}^{\mathrm{A\!'A}}_a \! \otimes \mathcal{B}^{\mathrm{BB'}}_b) ( {|\varphi_{s}\rangle\!\langle\varphi_{s}|}^{\mathrm{A\!'}} \! \otimes \rho^{\mathrm{AB}} \! \otimes {|\psi_{t}\rangle\!\langle\psi_{t}|}^{\mathrm{B'}}) \! \big ] \end{equation}\noindent \tag{ 2 }$$

as predicted by quantum mechanics.

The correlations (2) can also be written in a similar form as (1), namely

$$\begin{equation} P_\rho(a,b \, | \, \vert\varphi_{s}\rangle,\vert\psi_{t}\rangle) = {\rm tr} \left [ \left ( \mathcal{A}^{\mathrm{A}}_{a|\vert\varphi_{s}\rangle} \otimes \mathcal{B}^{\mathrm{B}}_{b|\vert\psi_{t}\rangle} \right ) \rho^{\mathrm{AB}} \right ], \end{equation}\noindent \tag{ 3 }$$

where the operators

$$\begin{equation} \mathcal{A}^{\mathrm{A}}_{a|\vert\varphi_{s}\rangle} = ^{\mathrm{A\!'}}\!\langle\varphi_{s}\vert \!\otimes\! \mathbbm{1} \cdot \mathcal{A}^{\mathrm{A\!'A}}_a \cdot \vert\varphi_{s}\rangle^{\!\mathrm{A\!'}} \otimes\! \mathbbm{1}, \qquad \mathcal{B}^{\mathrm{B}}_{b|\vert\psi_{t}\rangle} = \mathbbm{1} \!\otimes \, ^{\mathrm{B'}}\!\!\langle\psi_{t}\vert \cdot \mathcal{B}^{\mathrm{BB'}}_b \cdot \mathbbm{1} \!\otimes\! \vert\psi_{t}\rangle^{\!\mathrm{B'}} \end{equation}\noindent \tag{ 4 }$$

describe Alice and Bob's effective POVMs acting on ρ^AB, for each input state |φ_s〉,|ψ_t〉. Here, in contrast to standard simulation tasks (1), Alice and Bob do not know a priori which effective POVMs $\{\mathcal {A}^{\mathrm {A}}_{a|\vert \varphi _{s}\rangle }\}_a$ and $\{\mathcal {B}^{\mathrm {B}}_{b|\vert \psi _{t}\rangle }\}_b$ they should simulate, as they do not know (and may not be able to determine with certainty) the classical indices s, t of their QIs |φ_s〉,|ψ_t〉: these states are chosen randomly and are prepared by external devices they do not control⁴.

Note, however, that when all of Alice's (respectively Bob's) input states {|φ_s〉} ({|ψ_t〉}) are orthogonal to one another, Alice (Bob) can perfectly distinguish between them; then the QI simulation task simply reduces to the standard one. As such, there is a distinction between the two tasks only when the QI states are non-orthogonal and therefore non-distinguishable. Note that the non-orthogonality of the set of QI states only implies, according to quantum theory, that the received state cannot be determined with certainty. If the states in {|φ_s〉}_s are non-orthogonal but still linearly independent, they can be unambiguously discriminated with a non-zero probability of getting a conclusive answer [33].

We consider a state ρ^AB of dimension d_A × d_B, and prescribe the following. Alice and Bob make local measurements $\{ \mathcal {A}^{\mathrm {A\!'A}}_{a} \},\{ \mathcal {B}^{\mathrm {BB'}}_{b}\}$ with binary outcomes a,b, where the outcome a or b = 1 corresponds to the successful projection onto the maximally entangled state $\vert \Phi ^{\!+}_d\rangle = \sum _{k=0}^{d-1} \vert kk\rangle / \sqrt {d}$. Thus

$$\begin{equation} \mathcal{A}^{\mathrm{A\!'A}}_1 = {|\Phi^{\!+}_{d_{\mathrm{A}}}\rangle\!\langle\Phi^{\!+}_{d_{\mathrm{A}}}|}, \mathcal{A}^{\mathrm{A\!'A}}_0 = \mathbbm{1} - \mathcal{A}^{\mathrm{A\!'A}}_1, \mathcal{B}^{\mathrm{BB'}}_1 = {|\Phi^{\!+}_{d_{\mathrm{B}}}\rangle\!\langle\Phi^{\!+}_{d_{\mathrm{B}}}|}, \mathcal{B}^{\mathrm{BB'}}_0 = \mathbbm{1} - \mathcal{B}^{\mathrm{BB'}}_1. \end{equation} \tag{ 5 }$$

The input states are chosen to have the same dimensions as the respective subsystem of A and B in ρ^AB, $\vert \varphi _{s}\rangle \in \mathbbm {C}^{d_{\mathrm {A}}}$, $\vert \psi _{t}\rangle \in \mathbbm {C}^{d_{\mathrm {B}}}$, and constructed such that the corresponding density matrices {|φ_s〉〈φ_s|}, {|ψ_t〉〈ψ_t|} span the space of linear operators acting on $\mathbbm {C}^{d_{\mathrm {A}}}$, $\mathbbm {C}^{d_{\mathrm {B}}}$, as is also done in the proof and example of [29]. Such sets can always be constructed using the minimal number d²_A,d²_B of elements, with s = 1,...,d²_A and t = 1,...,d²_B. Together with ρ^AB, these elements define completely our QI simulation task.

The effective POVMs applied to ρ^AB are then described (from equation (4)) as

$$\begin{equation} \mathcal{A}^{\mathrm{A}}_{1|\vert\varphi_{s}\rangle} \! = \frac{ {|\varphi_{s}\rangle\!\langle\varphi_{s}|}^{\!\top}}{d_{\mathrm{A}}}, \mathcal{A}^{\mathrm{A}}_{0|\vert\varphi_{s}\rangle} \! = \mathbbm{1} - \frac{ {|\varphi_{s}\rangle\!\langle\varphi_{s}|}^{\!\top}}{d_{\mathrm{A}}}, \end{equation} \tag{ 6 }$$

$$\begin{equation} \mathcal{B}^{\mathrm{B}}_{1|\vert\psi_{t}\rangle} \! = \frac{ {|\psi_{t}\rangle\!\langle\psi_{t}|}^{\!\top}}{d_{\mathrm{B}}}, \mathcal{B}^{\mathrm{B}}_{0|\vert\psi_{t}\rangle} \! = \mathbbm{1} - \frac{ {|\psi_{t}\rangle\!\langle\psi_{t}|}^{\!\top}}{d_{\mathrm{B}}}, \end{equation} \tag{ 7 }$$

where $^{\top}$ indicates transposition in the computational bases {|k〉} of $\mathbbm {C}^{d_{\mathrm {A}}}$ and $\mathbbm {C}^{d_{\mathrm {B}}}$.

3.1.1. Canonical QI simulation task of the singlet state

As a concrete example, consider the case where Alice and Bob share the singlet state ρ^AB = |Ψ⁻〉〈Ψ⁻|. The input states can, for instance, be chosen as the vertices of a regular tetrahedron on the surface of the Bloch sphere. Using $\vec {\sigma } = (\sigma _1, \sigma _2, \sigma _3)$ the vector of Pauli matrices, and $\vec {v}_1 = (1,1,1)/\sqrt {3}$, $\vec {v}_2 = (1,-1,-1) /\sqrt {3}$, $\vec {v}_3 = (-1,1,-1) /\sqrt {3}$, $\vec {v}_4 = (-1,-1,1) /\sqrt {3}$, we define, for s,t = 1,...,4,

$$\begin{equation} {|\varphi_{s}\rangle\!\langle\varphi_{s}|} = \frac{\mathbbm{1} + \vec{v}_{s} \cdot \vec{\sigma}}{2}, \quad {|\psi_{t}\rangle\!\langle\psi_{t}|} = \frac{\mathbbm{1} + \vec{v}_t \cdot \vec{\sigma}}{2}. \end{equation} \tag{ 8 }$$

The resulting correlations are then

$$\begin{equation} P_{\vert\Psi^-\rangle}(a,b \, | \, \vert\varphi_{s}\rangle,\vert\psi_{t}\rangle) = \left\{ \! \begin{array}{cl} \! \displaystyle\frac{2-(a+b)}{4} & \!\! \mathrm{ if } ~s=t, \\ \! \displaystyle\frac{7-5a-5b+4ab}{12} & ~\mathrm{otherwise}. \end{array} \right. \end{equation} \tag{ 9 }$$

This means that Alice and Bob must never output a = b = 1 when their input states are identical, but must otherwise produce this combination of outputs with some non-zero probability. Recall that Alice and Bob only have access to their respective quantum states |φ_s〉 and |ψ_t〉 and not their classical labels s and t. Since these input states are not linearly independent, they cannot be unambiguously distinguished from one another [33], i.e. Alice and Bob are bound to make mistakes if they try to guess the classical label s and t in some rounds of the simulation; thus the task of reproducing the correlations in equation (9) cannot be achieved perfectly if Alice and Bob only make use of shared randomness and classical communication.

The no-go result given in the previous paragraph is in fact not specific to pure two-qubit maximally entangled states, but rather is a general feature of all entangled states, as we shall demonstrate by constructing an explicit Bell-like inequality in the following theorem and proof.

Theorem 1. For any entangled quantum state ρ of dimension dA × dB, and sets of inputs states {|φs〉}d2As=1, {|ψt〉}d2Bt=1 whose corresponding density matrices are informationally complete (i.e. span the space of linear operators acting on $\mathbbm {C}^{d_{\mathrm {A}}}$, $\mathbbm {C}^{d_{\mathrm {B}}}$, respectively), there exist real coefficients βρst defining a Bell-like inequality

$$\begin{equation} I^\rho(P) := \sum_{st} \ \beta^\rho_{st} \ P(1,1 \, | \, \vert\varphi_{s}\rangle,\vert\psi_{t}\rangle) \ \geq \ 0 \end{equation} \tag{ 10 }$$

which is satisfied by all correlations obtainable from LOCC, but is violated by the quantum correlation P_ρ(a,b | |φ_s〉,|ψ_t〉) obtained from the entangled state ρ, when Alice and Bob perform projections onto a maximally entangled state as in (5).

Corollary. Any entangled state can generate correlations that cannot be simulated by classical resources in a QI simulation task.

Proof of theorem 1. We use the fact that for any entangled state ρ, there exists a positive (but not completely positive) map Λ, such that $\mathbbm {1} \otimes \Lambda (\rho )$ is not positive [34]; it has at least one eigenstate |ξ〉 with negative eigenvalue λ < 0. Let us denote by Λ* the (positive) dual map of Λ (defined such that tr[MΛ*(N)] = tr[Λ(M)N] for all linear operators M, N), and let us decompose the Hermitian operator $\mathbbm {1} \otimes \Lambda ^{\!*} ({|\xi \rangle \!\langle \xi |})$ as follows:

$$\begin{equation} \mathbbm{1} \otimes \Lambda^{\!*} ({|\xi\rangle\!\langle\xi|}) = \sum_{st} \ \beta^\rho_{st} \ {|\varphi_{s}\rangle\!\langle\varphi_{s}|}^{\!\top\!} \! \otimes \! {|\psi_{t}\rangle\!\langle\psi_{t}|}^{\!\top\!} , \end{equation} \tag{ 11 }$$

where β^ρ_st are real coefficients. Note that such a decomposition is always possible⁵ since, by assumption, the Hermitian operators $\left \{ {|\varphi _{s}\rangle \!\langle \varphi _{s}|}^{\!\top \!} \right \}_{s=1}^{d_{\mathrm {A}}^2}$, $\left \{{|\psi _{t}\rangle \!\langle \psi _{t}|}^{\!\top \!} \right \}_{t=1}^{d_{\mathrm {B}}^2}$ used in theorem 1 form a basis for Hermitian operators acting, respectively, on $\mathbbm {C}^{d_{\mathrm {A}}},\mathbbm {C}^{d_{\mathrm {B}}}$. The coefficients β^ρ_st introduced in (11) are used to define the linear combination I^ρ(P) in (10).

• (i)  
  To show that inequality (10) holds true for all correlations P(a,b | |φ_s〉,|ψ_t〉) that can be obtained from LOCC operations, we shall now recall that all classes of LOCC operations, distinguished by the amount and/or the number of rounds of communication [4], are included in the set of the so-called separable operations [4, 37–39]. As a result, the set of correlations that can be obtained by performing arbitrary LOCC operations on the input states |φ_s〉,|ψ_t〉 (LOCC correlations, for short) is included in the set of correlations obtainable from separable measurements on |φ_s〉,|ψ_t〉. Our strategy is to prove that inequality (10) holds for all separable measurements, which in turn, implies that it holds also for all LOCC correlations.Let then P_SEP(a,b | |φ_s〉,|ψ_t〉) be the correlations obtained from some separable measurements on |φ_s〉,|ψ_t〉. In our QI simulation task, such correlations take the form of

  $$\begin{equation} P_{\mathrm{SEP}}(a,b \, | \, \vert\varphi_{s}\rangle,\vert\psi_{t}\rangle) = \mathrm{tr}[ {|\varphi_{s}\rangle\!\langle\varphi_{s}|} \!\otimes\! {|\psi_{t}\rangle\!\langle\psi_{t}|}\, \Pi_{ab} ], \end{equation} \tag{ 12 }$$

  where the Π_ab are some POVM elements corresponding to outcomes a and b. The separability constraint implies that for each a and b, these can be decomposed as $\Pi _{ab} = \sum _k \Pi _{ab}^{\mathrm {A},k} \otimes \Pi _{ab}^{\mathrm {B},k}$, where the Π^A,k_ab and Π^B,k_ab are positive operators [4, 37–39]. Thus

  $$\begin{eqnarray} &&\fl I^\rho(P_{\mathrm{SEP}}) = \sum_{st} \ \beta^\rho_{st} \ P_{\mathrm{SEP}}(1,1 \, | \, \vert\varphi_{s}\rangle,\vert\psi_{t}\rangle) = \sum_k \sum_{st} \ \beta^\rho_{st} \ \mathrm{tr}[ {|\varphi_{s}\rangle\!\langle\varphi_{s}|} \!\otimes\! {|\psi_{t}\rangle\!\langle\psi_{t}|} \, \Pi_{11}^{\mathrm{A},k} \otimes \Pi_{11}^{\mathrm{B},k} ] \nonumber \\ &&\hspace{-25pt}= \sum_k \ \mathrm{tr}[ \mathbbm{1} \!\otimes\! \Lambda^{\!*} ({|\xi\rangle\!\langle\xi|}) (\Pi_{11}^{\mathrm{A},k} \!\otimes\! \Pi_{11}^{\mathrm{B},k})^{\top} ] \!=\! \sum_k \ \mathrm{tr}[ {|\xi\rangle\!\langle\xi|} (\Pi_{11}^{\mathrm{A},k})^{\!\top} \!\!\otimes\! \Lambda\big((\Pi_{11}^{\mathrm{B},k})^{\!\top}\big) ] \ {\!\geq\!} \ 0, \nonumber\\ \end{eqnarray} \tag{ 13 }$$

  where the last inequality is due to the fact that for each k, $(\Pi_{11}^{\mathrm{A},k})^{\!\top} \!\!\otimes\! \Lambda\big((\Pi_{11}^{\mathrm{B},k})^{\!\top}\big)$ is a positive operator. This proves that inequality (10) indeed holds for separable measurements on |φ_s〉,|ψ_t〉. Since, as recalled above, correlations obtained using LOCC are a subset of those obtained from separable measurements [4, 37–39], I^ρ(P_LOCC) ⩾ 0 holds also for all LOCC correlations P_LOCC(a,b | |φ_s〉,|ψ_t〉).
• (ii)  
  On the other hand, the quantum correlations P_ρ(a,b | |φ_s〉,|ψ_t〉) can be computed using (3) and the effective POVM elements of (6) and (7). One then obtains

  $$\begin{eqnarray} &&\fl I^\rho(P_\rho) \! = \sum_{st} \ \beta^\rho_{st} \ P_\rho(1,1 \, | \, \vert\varphi_{s}\rangle,\vert\psi_{t}\rangle) = \sum_{st} \ \beta^\rho_{st} \ \mathrm{tr}[\mathcal{A}^{\mathrm{A}}_{1|\vert\varphi_{s}\rangle} \otimes \mathcal{B}^{\mathrm{B}}_{1|\vert\psi_{t}\rangle} \rho] \nonumber \\ &&\hspace{-38pt} = \sum_{st} \ \frac{\beta^\rho_{st}}{d_{\mathrm{A}} d_{\mathrm{B}}} \ \mathrm{tr}\left[ {|\varphi_{s}\rangle\!\langle\varphi_{s}|}^{\!\top\!} \!\otimes\! {|\psi_{t}\rangle\!\langle\psi_{t}|}^{\!\top\!} \,\rho\right] = \frac{\mathrm{tr}[\mathbbm{1} \!\otimes\! \Lambda^{\!*} ({|\xi\rangle\!\langle\xi|}) \rho]}{d_{\mathrm{A}} d_{\mathrm{B}}} = \frac{{\langle\xi|\mathbbm{1} \!\otimes\! \Lambda (\rho)|\xi\rangle}}{d_{\mathrm{A}} d_{\mathrm{B}}} \nonumber\\ &&\hspace{-38pt}= \frac{\lambda}{d_{\mathrm{A}} d_{\mathrm{B}}} < 0 \end{eqnarray} \tag{ 14 }$$

  violating inequality (10), and thus concluding the proof.   □

Before providing explicit examples of such Bell-like inequalities (10), we remark that they have, in general, two interesting properties.

First, let us emphasize that these inequalities always use sets of input states that render unambiguous quantum state discrimination impossible. As seen in section 2, QI simulation tasks offer richer scenarios specifically because they allow sets of non-orthogonal QI states. As also noted before, non-orthogonality by itself does not rule out the possibility to perform unambiguous state discrimination (USD) [33]. Now, if USD was possible, Alice (Bob) could learn the classical label s (respectively t) of her state |φ_s〉 (|ψ_t〉) with non-zero probability. Then, for all s and t, both Alice and Bob would obtain conclusive results in some rounds of the simulation task. Alice and Bob could then coordinate to output a = b = 1 only when (s,t) is known and β_st is negative, filtering out non-negative contributions to the inequality. However, this strategy cannot be used against inequalities constructed in theorem 1: informationally complete sets of d²_A,d²_B rank-1 projectors in dimension d_A, d_B have corresponding states |φ_s〉, |ψ_t〉 linearly dependent on $\mathbbm {C}^{d_{\mathrm {A}}}$, $\mathbbm {C}^{d_{\mathrm {B}}}$, rendering USD impossible [33].

We also note that the Bell-like inequality (10) can be used to certify the entanglement of any state that violates it (and in particular the entangled state ρ for which it is constructed). It corresponds indeed to a measurement-device-independent entanglement witness (MDI-EW), as defined in [40], in a scenario with (trusted) QIs. Our proof here made use of positive but not completely positive maps, but can equivalently be based on the existence of entanglement witnesses [34], as in the proof of [40]. Our present theorem 1 thus implies that the MDI-EWs constructed in [40] can be used to distinguish entangled states not only from separable states, but also from arbitrary LOCC resources.

When Alice and Bob share the state ρ₂, and use the inputs given in (8) corresponding to the measurements specified in equations (6) and (7), their correlations are a mixture

$$\begin{equation} \fl P_{\rho_2}(a,b \, | \, \vert\varphi_{s}\rangle,\vert\psi_{t}\rangle) = \, v \ P_{\vert\Psi^-\rangle}(a,b \, | \, \vert\varphi_{s}\rangle,\vert\psi_{t}\rangle) + \, (1\!-\!v) \ P_0(a,b \, | \, \vert\varphi_{s}\rangle,\vert\psi_{t}\rangle) \end{equation} \tag{ 17 }$$

of the singlet correlations given in (9) and noise P₀(a,b | |φ_s〉,|ψ_t〉) = (3 − 2a)(3 − 2b)/16.

Entangled Werner states have non-positive partial transposes [41]; hence, the map Λ introduced in our construction above can simply be taken to be the transposition (which is self-dual). The four eigenvalues of the partial transpose $\rho_2^{\top_{\!\!B}}$ of ρ2 are $\{ \frac {1-3v}{4}, \frac {1+v}{4}, \frac {1+v}{4}, \frac {1+v}{4} \}$, with the eigenvalue $\lambda = \frac {1-3v}{4}$ corresponding to the eigenvector $\vert\xi_2\rangle = \vert\Phi_2^+\rangle $. $\rho_2^{\top_{\!\!B}}$ thus has a negative eigenvalue λ < 0 if and only if $v > \frac {1}{3}$, which indeed corresponds to the necessary and sufficient condition for ρ2 to be entangled. We solve (11) to obtain the βρ2st, with which we compute the value of the inequality (10) for the correlations (17):

$$\begin{equation} \beta^{\rho_2}_{st} = \frac{6 \delta_{st} - 1}{8}, \quad I^{\rho_2}(P_{\rho_2}) = \frac{1-3v}{16} \end{equation} \tag{ 18 }$$

making use of the Kronecker delta δ_st. One obtains a violation I^ρ₂(P_ρ2) < 0 of the Bell-like inequality (10) whenever $v > \frac {1}{3}$, i.e. for all entangled two-qubit Werner states, while all LOCC correlations satisfy I^ρ₂(P_LOCC) ⩾ 0: entangled two-qubit Werner states correlations (17) cannot be simulated by classical resources in QI scenarios.

Note that the Bell-like inequality thus obtained is exactly the same as the first MDI-EW derived in [40] to certify the entanglement of two-qubit Werner states.

The two-qubit example above generalizes readily to the d × d dimensional case. To simplify our computations, we consider input states $\left \{ \vert \varphi _{s}\rangle \right \}_{s=1}^{d^2}$ satisfying the requirement that

$$\begin{equation} \left| {\langle\varphi_{s'}|\varphi_{s}\rangle}\right|^2 = \frac{\delta_{s'\!s} \, d + 1}{d+1} \end{equation} \tag{ 19 }$$

and the same inputs for Bob, i.e. |ψ_t〉 = |φ_t〉. Condition (19) is satisfied by the QI states of equation (8) for d = 2; sets of such states are given, mostly numerically, in [42] for d ⩽ 67, and are conjectured to exist for all dimensions [42]. Using the identity

$$\begin{equation} {\rm tr}(F\cdot A\otimes B)={\rm tr}(A\cdot B) \end{equation} \tag{ 20 }$$

for any d × d matrices A and B, one can show that for ρ_d and the measurements specified in equations (6) and (7), one obtains, from equation (3),

$$\begin{equation} P_{\rho_d}(a,b \, | \, \vert\varphi_{s}\rangle,\vert\psi_{t}\rangle)\!=\!\frac{(d^2\!\!-\!\!1)^{3-\!a\!-\!b}+(\!-1)^{a\!+\!b} (1\!-\!d^2 \delta_{st})v}{d^4(d^2\!\!-\!\!1)}. \end{equation} \tag{ 21 }$$

The positive map Λ can again be taken to be the transposition. To obtain βρdst from (11), we note that for every entangled ρd, i.e. for $v > \frac {1}{d+1}$, its partial transpose

$$\begin{equation} \rho_d^{\top_{\!\!\mathrm{B}}} = \left[ 1 + \frac{v}{d-1} \right] \frac{\mathbbm{1}}{d^2} - \frac{v}{d-1} {|\Phi_d^+\rangle\!\langle\Phi_d^+|} \end{equation} \tag{ 22 }$$

has a negative eigenvalue $\lambda =\frac {1-v(d+1)}{d^2}$, corresponding to the eigenstate |ξ_d〉 = |Φ⁺_d〉. Solving equation (11) with ${|\xi _d\rangle \!\langle \xi _d|}^{\!\top _{\!\!\mathrm {B}}} = \frac {F}{d}$, we obtain

$$\begin{equation} \beta^{\rho_d}_{st} = \frac{d(d+1)\delta_{st} - 1}{d^3} \end{equation} \tag{ 23 }$$

which gives $I^{\rho _d}(P_{\rho _d}) = \frac {\lambda }{d^2}=\frac {1-v(d+1)}{d^4}{<}0$. Hence, one obtains a violation I^ρ_d(P_ρd) < 0 of the Bell-like inequality (10) whenever $v > \frac {1}{d+1}$, i.e. for all entangled Werner states.