“All-versus-nothing” proof of genuine tripartite steering and entanglement certification in the two-sided device-independent scenario

Abstract

We consider the task of certification of genuine entanglement of tripartite states. For this purpose, we first present an “all-versus-nothing” proof of genuine tripartite Einstein–Podolsky–Rosen (EPR) steering by demonstrating the non-existence of a hybrid local hidden state (LHS) model in the tripartite network as a motivation to our main result. A full logical contradiction of the predictions of the hybrid LHS model with quantum mechanical outcome statistics for any three-qubit generalized Greenberger–Horne–Zeilinger (GGHZ) states and pure W-class states is shown. Using logical contradiction, we can distinguish between the GGHZ and W-class state in two-sided device-independent (2SDI) steering scenario. We next formulate a 2SDI steering inequality which is a generalisation of the fine-grained steering inequality (FGI) derived in [1] for the tripartite scenario. We show that the maximum quantum violation of this tripartite FGI can be used to certify genuine entanglement of three-qubit pure states.

Appendices

Appendix A: All-versus-nothing proof of genuine steering of GGHZ states in 1SDI scenario

Here, we demonstrate that the existence of hybrid LHS models leads to the contradiction “$2=1$” in the 1SDI scenario for any pure state that belongs the generalized GHZ (Greenberger–Horne–Zeilinger) class (7). Let us recapitulate the form of the assemblage as described by hybrid LHS models in the 1SDI scenario. In this case, Alice being the untrusted party performs the measurement and the tripartite state imposes the constraints on the observed assemblage at the Bob–Charlie end. If the shared state is of the form (5), the assemblage has the following form [31]:

$$\begin{aligned} \sigma _{a|X_x}^{BC} = \sum _\lambda p_\lambda ^{A':BC} p(a|X_x,\lambda )\rho _\lambda ^{BC} + \sum _\mu p_\mu ^{B:A'C} \rho _\mu ^B \otimes \sigma _{a|X_x,\mu }^C + \sum _\nu p_\nu ^{A'B:C} \sigma _{a|X_x,\nu }^B \otimes \rho _\nu ^C. \end{aligned}$$

(A1)

The assemblage (A1) contains three terms. The first term is an unsteerable assemblage from Alice to Bob–Charlie. Bob–Charlie’s assemblage is dependent on Alice’s input, and output through the common variable $\lambda$. The second term is unsteerable from Alice to Bob but not necessarily from Alice to Charlie. In this case, Bob–Charlie’s assemblage is dependent on Alice’s input, output at Charlie’s end only, and the common hidden variable $\mu$. The third term is unsteerable from Alice to Charlie but not necessarily from Alice to Bob. In this case, Bob–Charlie’s assemblage is dependent on Alice’s input, output at Bob’s end only, and the common hidden variable $\nu$. When each element of Bob–Charlie’s assemblage can be written in the form (A1), then the assemblage does not demonstrate genuine EPR steering in 1SDI scenario but it may demonstrate steering. On the other hand, if the assemblage (1) can be written in the form ($\sum _\lambda P(\lambda )P(a|X_x,\lambda )\rho _\lambda ^B \otimes \rho _\lambda ^C$), then the assemblage is unsteerable and no signature of steering is present in the 1SDI scenario.

Now, Alice performs dichotomic projective measurements corresponding to the observables: $X_0 = \sigma _x$ and $X_1 = \sigma _y$ on her part of the shared GGHZ state (7). After Alice’s measurements, a total of four unnormalized conditional states $\sigma _{a|X_x}^{BC^{\text {GGHZ}}}$ (with a, x $\in \{0, 1\}$) are prepared at Bob–Charlie’s end as mentioned below.

$$\begin{aligned} \sigma _{0|X_0}^{BC^{\text {GGHZ}}}= & {} \frac{1}{2} \Big | \theta _+^0 \Big \rangle \Big \langle \theta _+^0 \Big | = p(1) \rho _1^{BC},\quad \quad \sigma _{1|X_0}^{BC^{\text {GGHZ}}} = \frac{1}{2} \Big | \theta _-^0 \Big \rangle \Big \langle \theta _-^0 \Big | = p(2) \rho _2^{BC}, \nonumber \\ \sigma _{0|X_1}^{BC^{\text {GGHZ}}}= & {} \frac{1}{2} \Big | \theta _-^1 \Big \rangle \Big \langle \theta _-^1 \Big | = p(3) \rho _3^{BC},\quad \quad \sigma _{1|X_1}^{BC^{\text {GGHZ}}} = \frac{1}{2} \Big | \theta _+^1 \Big \rangle \Big \langle \theta _+^1 \Big | = p(4) \rho _4^{BC}, \end{aligned}$$

(A2)

where $\Big | \theta _\pm ^0 \Big \rangle = \cos \theta \vert 00\rangle \pm \sin \theta \vert 11\rangle$ and $\Big | \theta _\pm ^1 \Big \rangle = \cos \theta \vert 00\rangle \pm i \, \sin \theta \vert 11\rangle$. Hence, a total of four different conditional states are produced on Bob–Charlie’s end, each of which are pure states. Since the conditional states are pure, the assemblage is not the convex combination of the three terms of Eq. (A1) but any one of the terms of Eq. (A1). We find that the dependence of Bob and Charlie’s assemblage on Alice’s input and output may come from the common hidden variable and it is not the case that only Bob’s or Charlie’s state changes by the Alice’s input and output. This is the feature of the first term of Eq. (A1). Hence, if the conditional states have hybrid LHS description, then there exists an assemblage $\{p(\lambda ) \rho _{\lambda }^{BC}\}$ such that

$$\begin{aligned} \sum _{\lambda } p(\lambda ) \rho _{\lambda }^{BC} = \rho _{\text {GGHZ}}^{BC} = \text {Tr}_{A} \Big [\big |\psi (\theta )_{\text {GGHZ}} \big \rangle \big \langle \psi (\theta )_{\text {GGHZ}} \big | \Big ]. \end{aligned}$$

(A3)

It is well known that a pure state cannot be expressed as a convex sum of other different states, i.e., a density matrix of pure state can only be expanded by itself. We can therefore, claim that the ensemble $\{p(\lambda ) \rho _{\lambda }^{BC}\}$ consists of four hybrid LHS: $\{p(1)\rho _1^{BC}, p(2)\rho _2^{BC}, p(3)\rho _3^{BC}, p(4)\rho _4^{BC}\}$ which reproduces the conditional states $\{\sigma _{a|X_x}^{BC^{\text {GGHZ}}}\}_{a, X_x}$ at Bob–Charlie’s end. Now, using Eq. (A3), we can write,

$$\begin{aligned} \sum _{\lambda = 1}^{4} p(\lambda ) \rho _{\lambda }^{BC} = \rho _{\text {GGHZ}}^{BC}. \end{aligned}$$

(A4)

Next, summing Eq. (A2) and then taking trace, the left-hand sides give $2 {\text {Tr}}[\rho ^{BC}_{\text {GGHZ}}] = 2$. Here we have used the fact: $\sum _{a=0}^{1} \sigma ^{BC^{\text {GGHZ}}}_{a|X_x} = \rho ^{BC}_{\text {GGHZ}}$ $\forall$ x. On the other hand, the right-hand sides give ${\text {Tr}}[\rho ^{BC}_{\text {GGHZ}}] = 1$ following Eq. (A4). Hence, this leads to a full contradiction of “$2=1$”. Similarly in 2SDI scenario, the existence of hybrid LHS model leads to a contradiction of “$2=1$” when the shared state is a three-qubit GGHZ state (7) as shown in Sect. 3.

Appendix B: All-versus-nothing proof of genuine steering of W-class states in 1SDI scenario

Here, we demonstrate that the existence of hybrid LHS models leads to the contradiction “$2=1$” in the 1SDI scenario for any pure state that belongs the W-class ($|\psi _w \rangle = c_0 \vert 001\rangle + c_1 \vert 010\rangle + \sqrt{1-c_0^2 - c_1^2}\vert 100\rangle$). Alice performs dichotomic projective measurements corresponding to the observables: $X_0 = \sigma _x$ and $X_1 = \sigma _y$. After Alice’s measurements, a total of four unnormalized conditional states $\sigma _{a|X_x}^{BC^{\text {W}}}$ (with a, x $\in \{0, 1\}$) are prepared at Bob–Charlie’s end as mentioned below.

$$\begin{aligned} \sigma _{0|X_0}^{BC^{\text {W}}}= & {} \frac{1}{2} \Big |w_+^0 \Big \rangle \Big \langle w_+^0 \Big | = p(1)\rho _{1}^{BC}, \quad \quad \sigma _{1|X_0}^{BC^{\text {W}}} = \frac{1}{2} \Big |w_-^0 \Big \rangle \Big \langle w_-^0 \Big | = p(2)\rho _{2}^{BC}, \nonumber \\ \sigma _{0|X_1}^{BC^{\text {W}}}= & {} \frac{1}{2} \Big |w_+^1 \Big \rangle \Big \langle w_+^1 \Big | = p(3)\rho _{3}^{BC}, \quad \quad \sigma _{1|X_1}^{BC^{\text {W}}} = \frac{1}{2} \Big |w_-^1 \Big \rangle \Big \langle w_-^1 \Big | = p(4)\rho _{4}^{BC}, \end{aligned}$$

(B1)

where $\Big |w_{\pm }^0\Big \rangle := \sqrt{1-c_0^2-c_1^2} \vert 00\rangle \pm c_0 \vert 01\rangle \pm c_1 \vert 10\rangle$ and $\Big |w_{\pm }^1\Big \rangle := \sqrt{1-c_0^2-c_1^2} \vert 00\rangle \pm i \, c_0 \vert 01\rangle \pm i \, c_1 \vert 10\rangle$. Hence, a total of four different conditional states are produced on Bob–Charlie’s end, each of which are pure states. Since the conditional states are pure, the assemblage is not the convex combination of the three terms of Eq. (A1) but any one of the terms of Eq. (A1). We find that the dependence of Bob and Charlie’s assemblage on Alice’s input and output may come from the common hidden variable and it is not the case that only Bob’s or Charlie’s state changes by the Alice’s input and output. This is the feature of the first term of Eq. (A1). Hence, if the conditional states have hybrid LHS description, then there exists an assemblage $\{p(\lambda ) \rho _{\lambda }^{BC}\}$ such that

$$\begin{aligned} \sum _{\lambda } p(\lambda ) \rho _{\lambda }^{BC} = \rho _{w}^{BC} = \text {Tr}_{A} \Big [\big |\psi _w \big \rangle \big \langle \psi _w \big | \Big ]. \end{aligned}$$

(B2)

It is well known that a pure state cannot be expressed as a convex sum of other different states, i.e., a density matrix of pure state can only be expanded by itself. We can, therefore, claim that the ensemble $\{p(\lambda ) \rho _{\lambda }^{BC}\}$ consists of four hybrid LHS: $\{p(1)\rho _1^{BC}, p(2)\rho _2^{BC}, p(3)\rho _3^{BC}, p(4)\rho _4^{BC}\}$ which reproduces the conditional states $\{\sigma _{a|X_x}^{BC^{\text {W}}}\}_{a, X_x}$ at Bob–Charlie’s end. Now, using Eq. (B2), we can write,

$$\begin{aligned} \sum _{\lambda = 1}^{4} p(\lambda ) \rho _{\lambda }^{BC} = \rho _{w}^{BC}. \end{aligned}$$

(B3)

Next, summing Eq. (B1) and then taking trace, the left-hand sides give $2 {\text {Tr}}[\rho ^{BC}_{w}] = 2$. Here we have used the fact: $\sum _{a=0}^{1} \sigma ^{BC^{w}}_{a|X_x} = \rho ^{BC}_{w}$ $\forall$ x. On the other hand, the right-hand sides give ${\text {Tr}}[\rho ^{BC}_{w}] = 1$ following Eq. (B3). Hence, this leads to a full contradiction of “$2=1$”.

Appendix C: All-versus-nothing proof of genuine steering of W-class states in 2SDI scenario

Here, we demonstrate that the existence of hybrid LHS models leads to the contradiction “$4=1$” in the 2SDI scenario for any pure state that belongs the W-class ($|\psi _w \rangle = c_0 \vert 001\rangle + c_1 \vert 010\rangle + \sqrt{1-c_0^2 - c_1^2}\vert 100\rangle$). Alice performs dichotomic projective measurements corresponding to the observables: $X_0 = \sigma _x$ and $X_1 = \sigma _y$. On the other hand, Bob performs dichotomic projective measurements corresponding to the observables: $Y_0 = \frac{\sigma _x + \sigma _z}{\sqrt{2}}$ and $Y_1 = \frac{\sigma _y + \sigma _z}{\sqrt{2}}$. After Alice’s and Bob’s measurements, a total of sixteen unnormalized conditional states $\sigma _{a,b|X_x,Y_y}^{C^{\text {W}}}$ (with a, b, x, y $\in \{0, 1\}$) are prepared at Charlie’s end as mentioned below.

$$\begin{aligned} \sigma _{0, 0|X_0, Y_0}^{C^{\text {W}}}= & {} \frac{N_{w_1}^{00100}}{8} \Big | \psi _{w_1} \Big \rangle ^{00100} \Big \langle \psi _{w_1} \Big | = p(1) \rho _1^{C},\quad \quad \sigma _{0,1|X_0, Y_0}^{C^{\text {W}}} = \frac{N_{w_1}^{11001}}{8} \Big | \psi _{w_1} \Big \rangle ^{11001} \Big \langle \psi _{w_1} \Big | = p(2) \rho _2^{C}, \nonumber \\ \sigma _{1, 0|X_0, Y_0}^{C^{\text {W}}}= & {} \frac{N_{w_1}^{00110}}{8} \Big | \psi _{w_1} \Big \rangle ^{00110} \Big \langle \psi _{w_1} \Big | = p(3) \rho _3^{C},\quad \quad \sigma _{11|X_0,Y_0}^{C^{\text {W}}} = \frac{N_{w_1}^{11011}}{8} \Big | \psi _{w_1} \Big \rangle ^{11011} \Big \langle \psi _{w_1} \Big | = p(4) \rho _4^{C}, \nonumber \\ \sigma _{0, 0|X_0, Y_1}^{C^{\text {W}}}= & {} \frac{N_{w_2}^{01100}}{8} \Big | \psi _{w_2} \Big \rangle ^{01100} \Big \langle \psi _{w_2} \Big | = p(5) \rho _5^{C},\quad \quad \sigma _{0,1|X_0, Y_1}^{C^{\text {W}}} = \frac{N_{w_2}^{10001}}{8} \Big | \psi _{w_2} \Big \rangle ^{10001} \Big \langle \psi _{w_2} \Big | = p(6) \rho _6^{C}, \nonumber \\ \sigma _{1, 0|X_0, Y_1}^{C^{\text {W}}}= & {} \frac{N_{w_2}^{01110}}{8} \Big | \psi _{w_2} \Big \rangle ^{01110} \Big \langle \psi _{w_2} \Big | = p(7) \rho _7^{C},\quad \quad \sigma _{11|X_0,Y_1}^{C^{\text {W}}} = \frac{N_{w_2}^{10011}}{8} \Big | \psi _{w_2} \Big \rangle ^{10011} \Big \langle \psi _{w_2} \Big | = p(8) \rho _8^{C}, \nonumber \\ \sigma _{0, 0|X_1, Y_0}^{C^{\text {W}}}= & {} \frac{N_{w_3}^{00110}}{8} \Big | \psi _{w_3} \Big \rangle ^{00110} \Big \langle \psi _{w_3} \Big | = p(9) \rho _9^{C},\quad \quad \sigma _{0,1|X_1, Y_0}^{C^{\text {W}}} = \frac{N_{w_3}^{11011}}{8} \Big | \psi _{w_3} \Big \rangle ^{11011} \Big \langle \psi _{w_3} \Big | = p(10) \rho _{10}^{C}, \nonumber \\ \sigma _{1, 0|X_1, Y_0}^{C^{\text {W}}}= & {} \frac{N_{w_3}^{00100}}{8} \Big | \psi _{w_3} \Big \rangle ^{00100} \Big \langle \psi _{w_3} \Big | = p(11) \rho _{11}^{C},\quad \quad \sigma _{11|X_0,Y_0}^{C^{\text {W}}} = \frac{N_{w_3}^{11001}}{8} \Big | \psi _{w_3} \Big \rangle ^{11001} \Big \langle \psi _{w_3} \Big | = p(12) \rho _{12}^{C}, \nonumber \\ \sigma _{0, 0|X_1, Y_1}^{C^{\text {W}}}= & {} \frac{N_{w_4}^{01100}}{8} \Big | \psi _{w_4} \Big \rangle ^{01100} \Big \langle \psi _{w_4} \Big | = p(13) \rho _{13}^{C},\quad \quad \sigma _{0,1|X_1, Y_1}^{C^{\text {W}}} = \frac{N_{w_4}^{10011}}{8} \Big | \psi _{w_4} \Big \rangle ^{10011} \Big \langle \psi _{w_4} \Big | = p(14) \rho _{14}^{C}, \nonumber \\ \sigma _{1, 0|X_1, Y_1}^{C^{\text {W}}}= & {} \frac{N_{w_4}^{01110}}{8} \Big | \psi _{w_4} \Big \rangle ^{01110} \Big \langle \psi _{w_4} \Big | = p(15) \rho _{15}^{C},\quad \quad \sigma _{11|X_1,Y_1}^{C^{\text {W}}} = \frac{N_{w_4}^{10001}}{8} \Big | \psi _{w_4} \Big \rangle ^{10001} \Big \langle \psi _{w_4} \Big | = p(16) \rho _{16}^{C}, \end{aligned}$$

(C1)

where $\Big | \psi _{w_1}^{abcde} \Big \rangle = \dfrac{\sqrt{2+(-1)^a\sqrt{2}}c_0\vert 1\rangle +(-1)^b\sqrt{2+(-1)^c\sqrt{2}}c_1\vert 0\rangle +(-1)^d\sqrt{2+(-1)^e\sqrt{2}}\sqrt{1-c_0^2-c_1^2}\vert 0\rangle }{\sqrt{N_{w_1}^{abcde}}}$,

$\Big | \psi _{w_2}^{abcde} \Big \rangle = \dfrac{\sqrt{2+(-1)^a\sqrt{2}}c_0\vert 1\rangle +(-1)^b \iota \sqrt{2+(-1)^c\sqrt{2}}c_1\vert 0\rangle +(-1)^d\sqrt{2+(-1)^e\sqrt{2}}\sqrt{1-c_0^2-c_1^2}\vert 0\rangle }{\sqrt{N_{w_2}^{abcde}}}$,

$\Big | \psi _{w_3}^{abcde} \Big \rangle = \dfrac{\sqrt{2+(-1)^a\sqrt{2}}c_0\vert 1\rangle +(-1)^b\sqrt{2+(-1)^c\sqrt{2}}c_1\vert 0\rangle +(-1)^d \iota \sqrt{2+(-1)^e\sqrt{2}}\sqrt{1-c_0^2-c_1^2}\vert 0\rangle }{\sqrt{N_{w_3}^{abcde}}}$,

$\Big | \psi _{w_4}^{abcde} \Big \rangle = \dfrac{\sqrt{2+(-1)^a\sqrt{2}}c_0\vert 1\rangle +(-1)^b \iota (\sqrt{2+(-1)^c\sqrt{2}}c_1\vert 0\rangle +(-1)^d\sqrt{2+(-1)^e\sqrt{2}}\sqrt{1-c_0^2-c_1^2}\vert 0\rangle )}{\sqrt{N_{w_4}^{abcde}}}$.

Hence, a total of sixteen different conditional states are produced on Charlie’s end, each of which are pure states. Since the conditional states are pure, the assemblage is not the convex combination of the three terms of Eq. (6) but any one of the terms of Eq. (6). We find that the dependence of Charlie’s assemblage on Alice and Bob’s input and output may come from the common hidden variable and it is not the case that either Alice or Bob can change Charlie’s state through their input and output choices. This is the feature of the third term of the Eq. (6). Hence, if the conditional states have hybrid LHS description, then there exists an assemblage $\{p(\nu ) \rho _{\nu }^{C}\}$ such that

$$\begin{aligned} \sum _{\nu } p(\nu ) \rho _{\nu }^{C} = \rho _{w}^{C} = \text {Tr}_{AB} \Big [\big |\psi _w \big \rangle \big \langle \psi _w \big | \Big ]. \end{aligned}$$

(C2)

It is well known that a pure state cannot be expressed as a convex sum of other different states, i.e., a density matrix of pure state can only be expanded by itself. We can therefore, claim that the ensemble $\{p(\nu ) \rho _{\nu }^{C}\}$ consists of sixteen hybrid LHS: $\{p(1)\rho _1^{C}, p(2)\rho _2^{C}, p(3)\rho _3^{C}, p(4)\rho _4^{C}, p(5)\rho _5^{C}$, $p(6)\rho _6^{C}, p(7)\rho _7^{C}, p(8)\rho _8^{C}, p(9)\rho _9^{C}, p(10)\rho _{10}^{C}, p(11)\rho _{11}^{C}, p(12)\rho _{12}^{C}, p(13)\rho _{13}^{C}, p(14)\rho _{14}^{C},$ $p(15)\rho _{15}^{C}, p(16)\rho _{16}^{C}\}$ which reproduces the conditional states $\{\sigma _{a, b|X_x, Y_y}^{C}\}_{a, b, X_x, Y_y}$ at Charlie’s end. Now, using Eq. (C2) we can write,

$$\begin{aligned} \sum _{\nu = 1}^{16} p(\nu ) \rho _{\nu }^{C} = \rho _{w}^{C}. \end{aligned}$$

(C3)

Next, summing Eq. (C1) and then taking trace, the left-hand sides give $4 {\text {Tr}}[\rho ^{C}_w] = 4$. Here we have used the fact: $\sum _{a,b = 0}^{1} \sigma ^{C}_{a, b|X_x, Y_y} = \rho ^{C}_{w}$ $\forall$ x, y. On the other hand, the right-hand sides give ${\text {Tr}}[\rho ^{C}_{w}] = 1$ following Eq. (C3). Hence, this leads to a full contradiction of “$4=1$”.

Appendix D: All-versus-nothing proof of steering of pure bi-separable states in 1SDI and 2SDI scenario

We first demonstrate that the existence of a hybrid LHS model leads to no contradiction in the 1SDI scenario for bi-separable state, but the existence of LHS models may lead to a contradiction. Consider a state $\psi$ of the form (35). In particular, consider the following bi-separable state:

$$\begin{aligned} \psi _{bs} = (\cos \theta _1 \vert 00\rangle _{AB}+\sin \theta _1 \vert 11\rangle _{AB}) \otimes (\cos \theta _2 \vert 0\rangle _C+\sin \theta _2 \vert 1\rangle _C). \end{aligned}$$

(D1)

Alice performs dichotomic projective measurements corresponding to the observables: $X_0 = \sigma _x$ and $X_1 = \sigma _y$ on her part of the shared bi-separable state (D1). After Alice’s measurements, a total of four unnormalized conditional states $\sigma _{a|X_x}^{BC^{\text {bs}}}$ (with a, x $\in \{0, 1\}$) are prepared at Bob–Charlie’s end as mentioned below.

$$\begin{aligned} \sigma _{0|X_0}^{BC^{\text {bs}}}= & {} \frac{1}{2} \Big | \theta _{++}^0 \Big \rangle \Big \langle \theta _{++}^0 \Big | = p(1) \sigma _{0|X_0,1}\otimes \rho _1^{C},\quad \quad \sigma _{1|X_0}^{BC^{\text {bs}}} = \frac{1}{2} \Big | \theta _{-+}^0 \Big \rangle \Big \langle \theta _{-+}^0 \Big | = p(1)\sigma _{1|X_0,1}\otimes \rho _1^{C}, \nonumber \\ \sigma _{0|X_1}^{BC^{\text {bs}}}= & {} \frac{1}{2} \Big | \theta _{-+}^1 \Big \rangle \Big \langle \theta _{-+}^1 \Big | = p(2) \sigma _{0|X_1,2}\otimes \rho _2^{C},\quad \quad \sigma _{1|X_1}^{BC^{\text {bs}}} = \frac{1}{2} \Big | \theta _{++}^1 \Big \rangle \Big \langle \theta _{++}^1 \Big | = p(2) \sigma _{1|X_1,2} \otimes \rho _2^{C}, \end{aligned}$$

(D2)

where $\Big | \theta _{\pm ,\pm }^0 \Big \rangle = (\cos \theta _1 \vert 0\rangle \pm \sin \theta _1 \vert 1\rangle ) \otimes (\cos \theta _2 \vert 0\rangle \pm \sin \theta _2 \vert 1\rangle )$ and $\Big | \theta _{\pm ,\pm }^1 \Big \rangle = (\cos \theta _1 \vert 0\rangle \pm \iota \sin \theta _1 \vert 1\rangle ) \otimes (\cos \theta _2 \vert 0\rangle \pm \sin \theta _2 \vert 1\rangle )$. Hence, a total of two distinguishable conditional states are produced at Bob–Charlie’s end, all of which are pure states. Since the conditional states are pure, the assemblage is not a convex combination of the three terms of Eq. (A1) but any one of the terms of Eq. (A1). We find that the dependence of Bob and Charlie’s assemblage on Alice’s input and output may come from the common hidden variable and it is the case that only Bob’s state changes by the Alice’s input and output. This is the feature of third term of Eq. (A1). The common variable for the conditional state $\sigma _{1|X_0}^{BC^{bs}}$ is same as that of $\sigma _{0|X_0}^{BC^{bs}}$. So, using a hybrid LHS model we cannot distinguish between them. Similarly, the states $\sigma _{0|X_1}^{BC^{bs}}$ and $\sigma _{1|X_1}^{BC^{bs}}$ are same according to the hybrid LHS model. Now, if the conditional states have hybrid LHS description, then there exists an assemblage $\{p(\nu ) \rho _\nu ^B \otimes \rho _{\nu }^{C}\}$ such that

$$\begin{aligned} \sum _{\nu } p(\nu ) \rho _{\nu }^{B} \otimes \rho _{\nu }^{C} = \rho _{\text {bs}}^{BC} = \text {Tr}_{A} \Big [\big |\psi _{\text {bs}} \big \rangle \big \langle \psi _{\text {bs}} \big | \Big ]. \end{aligned}$$

(D3)

So, according to the hybrid LHS model, the ensemble $\{p(\nu ) \rho _{\nu }^{B} \otimes \rho _{\nu }^{C}\}$ consists of four conditional states out of which two are distinguishable. These two reproduce the conditional states $\{\sigma _{a|X_x}^{BC^{\text {bs}}}\}_{a, X_x}$ at Bob–Charlie’s end. Now, using Eq. (D3) we can write,

$$\begin{aligned} \sum _{\nu = 1}^{2} p(\nu ) \rho _{\nu }^{B} \otimes \rho _{\nu }^{C} = \rho _{\text {bs}}^{BC}. \end{aligned}$$

(D4)

Next, summing Eq. (D2) and then taking trace, both the left-hand and right-hand sides give $2 {\text {Tr}}[\rho ^{BC}_{\text {GGHZ}}] = 2$. Here, we have used the fact: $\sum _{a=0}^{1} \sigma ^{BC^{\text {bs}}}_{a|X_x} = \rho ^{BC}_{\text {bs}}$ $\forall$ x and ${\text {Tr}}[\rho ^{BC}_{\text {bs}}] = 1$ following Eq. (D4). Hence, in this case there is no contradiction.

• 1. Remark-1 Note that in case of an LHS model, there will be four distinct conditional states and it leads to the contradiction $``2 = 1''$ in the 1SDI scenario for the bi-separable states of the form (D1) when Alice performs dichotomic projective measurements corresponding to the observables: $X_0 = \sigma _x$, $X_1 = \sigma _y$. This demonstrates steering in such states.

• 2. Remark-2 The existence of LHS or hybrid LHS models lead to NO contradiction in the 1SDI scenario for bi-separable states of the form (31).

• 3. Remark-3 The characteristics of the pure bi-separable states in which Alice and Charlie are entangled are same as the state of the form (D1). This means no contradiction occurs for hybrid LHS models but the existence of LHS models lead to a contradiction.

• 4. Remark-4 Following similar reasoning and using a hybrid LHS model of the form (6), it can be shown that the existence of LHS models lead to no contradiction in the 2SDI scenario for bi-separable states, but the existence of LHS models may lead to a contradiction.

Appendix E

Proposition 1

For any two genuinely entangled three-qubit pure states (W-class pure states or GHZ-class pure states), FGI (26) does not give maximum quantum violation ($= 4$) for the same set of measurement settings by the two untrusted parties.

Proof

We use the following numerical strategy to show that no two pure genuinely entangled states can give rise to the maximum violation of FGI (26) for the same set of measurement settings by the untrusted parties:

Numerical strategy: The precision is set at the 6th decimal place for the numerical calculations. Following steps are evaluated for $10^6$ randomly generated states: (i) State parameters are chosen randomly in the allowed range. There are three state parameters for the pure w-class state (E1) (taking normalization into account) and 5 state parameters for the pure GHZ-class (E2). (ii) We then numerically maximize the FGI (26) over the measurement parameters of the untrusted parties (Alice and Bob). Charlie’s measurements are as usual $\sigma _x$ and $\sigma _y$. FGI is maximally violated numerically if the violation is more than or equal to 3.99. Note that if we keep the same precision for maximum violation i.e. FGI is maximally violated if the violation is above 3.99999 then that are stricter conditions and are already a part of our observations with aforementioned relaxed conditions.

Examining equality of measurement parameters: All the states that maximally violates the FGI, their state parameters and the measurement parameters are printed out in distinct row in a file. Using sort filename | uniq -c command that outputs the measurement parameters (for one state, eight measurement parameters are in one line and task is to examine whether any two or more lines in the files are same) in ascending order along with their repeated values. If the row is not repeated, it outputs 1 otherwise the number of time it is repeated.

• Pure W-class state: A general pure w-class state has the following form:

  $$\begin{aligned} \vert \psi _w\rangle = \sqrt{a}\vert 001\rangle + \sqrt{b}\vert 010\rangle + \sqrt{c}\vert 100\rangle +\sqrt{d}\vert 000\rangle \end{aligned}$$

  (E1)

  We observed that out of $10^6$ randomly generated states, 44001 states maximally violates the FGI but no two states have the same set of measurement parameters for untrusted side.

• Pure GHZ-class state: A general pure GHZ-class state has the following form:

  $$\begin{aligned} \vert \psi _{GHZ}^{C}\rangle : \cos \delta \vert 000\rangle + \sin \delta e^{\iota \phi } \vert \phi _A\phi _B\phi _C\rangle \end{aligned}$$

  (E2)

  where, $\vert \phi _A\rangle = \cos \alpha \vert 0\rangle + \sin \alpha \vert 1\rangle$, $\vert \phi _B\rangle = \cos \beta \vert 0\rangle + \sin \beta \vert 1\rangle$ and $\vert \phi _C\rangle = \cos \gamma \vert 0\rangle + \sin \gamma \vert 1\rangle$. The above state is a GGHZ state (7) for $\delta = \theta$, $\phi = 0$, $\alpha = \beta = \gamma = \frac{\pi }{2}$.

  We observed that only 6879 states out of $10^6$ maximally violates the FGI but no two states have the same set of measurement parameters for the untrusted sides.

• Both w-class and GHZ-class pure state: Even if we take both GHZ-class and w-class pure states together, we found no two states have the same set of measurement parameters for the untrusted sides. *The datasets generated during and/or analysed during the current study are available from the corresponding author on reasonable request.

$\square$