One sided indeterminism alone is not a useful resource to simulate any nonlocal correlation

Abstract

Determinism, no signaling and measurement independence are some of the constraints required for framing Bell inequality. Any model simulating nonlocal correlations must either individually or jointly give up these constraints. Recently Hall (Phys Review A, 84:022102, 2011) derived different forms of Bell inequalities under the assumption of individual or joint relaxation of those constraints on both (i.e., two) the sides of a bipartite system. In this work, we have investigated whether one sided relaxation can also be a useful resource for simulating nonlocal correlations or not. We have derived Bell-type inequalities under the assumption of joint relaxation of these constraints only by one party of a bipartite system. Interestingly, we found that any amount of randomness in correlations of one party in absence of signaling between two parties is incapable of showing any sort of Bell–CHSH violation, whereas signaling and measurement dependence individually can simulate any nonlocal correlations. We have also completed the proof of a recent conjecture due to Hall (Phys. Rev. A 82:062117, 2010; Phys. Rev. A 84:022102, 2011) for one-sided relaxation scenario only.

Appendix

1.1 Appendix 1: Proof of the Theorem for $M_2 = 0$

The technique of the proof is similar to that given in [7]. The relaxed Bell inequality given in Eqs.(13) and (14) for $M_2 = 0$ is proved here. The outcomes of Alice and Bob are labeled by $\pm 1$. Let the joint measurement outcomes be $(+,\,+),\,(+,\,-),\,(-,\,+)$ and $(-,\,-)$.

Let $p(+,\,+|x,\,y,\,\lambda )\,=\,c$; $p^{(1)}(+|x,\,y,\,\lambda )\,=\,m$ and $p^{(2)}(+|x,\,y,\,\lambda )\,=\,n$. Hence, if $\langle xy \rangle _{\lambda }$ denotes the average product of the measurement outcomes for a fixed value of $\lambda$, then $\langle xy \rangle _{\lambda }\,=\,1+4c-2(m+n)$ [7]. For positivity of joint probabilities, $\max \{0,\,m+n-1\}\,\le c\,\le \,\min \{m,\,n\}$. Thus,

$$\begin{aligned} 2|m+n-1|\,-\,1\,\le \,\langle xy \rangle _{\lambda }\,\le \,1-2|m-n|. \end{aligned}$$

(20)

Suitable choices of $c$ give upper and lower bounds.

Let, $p_1\,\equiv \,p(*|x,\,y),\,p_2\,\equiv \,p(*|x,\,y'),\,p_3\,\equiv \,p(*|x',\,y)$ and $p_4\,\equiv \,p(*|x',\,y')$;

$$\begin{aligned} E({\lambda }):=\,\langle xy \rangle _{\lambda }\,+\,\langle xy' \rangle _{\lambda }\,+\,\langle x'y\rangle _{\lambda }\,-\,\langle x'y' \rangle _{\lambda }, \end{aligned}$$

then,

$$\begin{aligned} E({\lambda })\,\le \,4\,-\,2J, \end{aligned}$$

(21)

where

$$\begin{aligned} J=\,|m_1\,-\,n_1|\,+\,|m_2\,-\,n_2|\,+\,|m_3\,-\,n_3|\,+\,|m_4\,+\,n_4\,-\,1|. \end{aligned}$$

(22)

For suitable choices of $c_1,\,c_2,\,c_3$ and $c_4$ the upper bound is attainable.

As Alice’s correlations abide by no signaling principle, $m_1\,=\,m_2$ and $m_3\,=\,m_4$. Therefore,

$$\begin{aligned} J=\,|m_1\,-\,n_1|\,+\,|m_1\,-\,n_2|\,+\,|m_3\,-\,n_3|\,+\,|m_3\,+\,n_4\,-\,1|. \end{aligned}$$

(23)

Due to determinism on Alice’s part, $m_1,\,m_3\in \{0,\,1\}$. The indeterminism and signaling constraints of the theorem on Bob’s outcomes imply $n_j\in [0,\,I]\,\bigcup \,[1-I,\,1]\quad \forall \,\text { j }=1,2,3,4$ and $|n_1\,-\,n_3|\,\le \,S$; $|n_2\,-\,n_4|\,\le \,S$.

To maximize $E({\lambda }),\,J$ must be minimized. There are four possible cases corresponding to the four possible values of $(m_1,\,m_3)$: $(0,\,0),\,(1,\,0),\,(0,\,1) ,\,\text {and},\, (1,\,1)$. Now for the first case, i.e., for $m_1=0$ and $m_3 = 0,\,J$ defined in Eq. (23) becomes

$$\begin{aligned} J=\,|n_1|\,+\,|n_2|\,+\,|n_3|\,+\,|1\,-\,n_4|. \end{aligned}$$

(24)

Since $n_1,\,n_2,\,n_3$ and $(1-n_4)$ all are positive, hence

$$\begin{aligned} J&= \,n_1\,+\,n_3\,+\,n_2\,-\,n_4\,+\,1\nonumber \\&\ge 1-S_{1\rightarrow 2}+n_1+n_3\nonumber \\&\ge 1-S_{1\rightarrow 2} \end{aligned}$$

(25)

where we have used the constraints $-S_{1\rightarrow 2}\le n_2-n_4 \le S_{1\rightarrow 2}$. Therefore, $J\ge 1-S_{1\rightarrow 2}$. Equality is obtained when $n_1 = 0,\,n_3 =0$ and $n_2-n_4 = -S_{1\rightarrow 2}$. Similarly, for the other three cases $(m_1,\,m_3)$: $(1,\,0),\,(0,\,1)\text { and }\, (1,\,1),\,J \ge 1-S_{1\rightarrow 2}$ where the equality is obtained for the three cases when $\{n_1 = 0,\,n_3 =0$ and $n_2-n_4 = -S_{1\rightarrow 2}\},\,\{n_2 = 1,\,n_4 = 1$ and $n_3-n_1 = -S_{1\rightarrow 2}\}$ and $\{n_1 = 0,\,n_3 =1$ and $n_4-n_2 = -S_{1\rightarrow 2}\}$, respectively. Hence in any case

$$\begin{aligned} J\ge 1-S_{1\rightarrow 2}. \end{aligned}$$

(26)

Now, let $S_{1\rightarrow 2}<1-2I_2$, then the maximum value of $S_{1\rightarrow 2}$ will give the minimum value of the lower bound of $J$ in the Eq. (26). The maximum value that $S_{1\rightarrow 2}$ can take for $S_{1\rightarrow 2}<1-2I_2$ is $I_2$(by definition). Hence, for $S_{1\rightarrow 2}<1-2I_2$ Eq. (26) gets modified as

$$\begin{aligned} J\ge 1-I_2 \end{aligned}$$

(27)

and equality is obtained when $S_{1\rightarrow 2} = I_2$. The above relation implies [Via Eq.(21)] the tight bound $E({\lambda })\,\le \,2\,+\,2I_2,$ where $E({\lambda })$ achieve maximum value $2\,+\,2I_2$, for the following cases $\{m_j = 0, n_1=0 , n_3=0, n_4=I,n_2 = 0 (j=1,2,3,4))\},\,\{m_1 = m_2 = 0, m_3 = m_4 = 1, n_2 = 0 , n_4 = 0, n_3 = I, n_1 = 0\},\,\{m_1 = m_2 = 1, m_3 = m_4 = 0, n_2 = 1 , n_4 = 1, n_3 = I, n_1 = 0\}$ and $\{m_j= 1, n_1 = 1 , n_3 = 1, n_2 = I, n_4 = 0(j=1,2,3,4)\}$. Finally, let $S_{1\rightarrow 2} \ge 1-2I_2$, then the value of at least one pair of marginal probabilities (($n_1,\,n_3$) and/or ($n_2,\,n_4$)) must shift across the gap between the subintervals $[0,I]$ and $[1-I,1]$. Now, the constraint $S_{1\rightarrow 2} \ge 1-2I_2$, will not improve the lower bound $1-S_{1\rightarrow 2}$ of $J$ given in Eq. (26), hence $J\ge 1-S_{1\rightarrow 2}$, implying [Via Eq. (21)] the tight bound $E({\lambda })\,\le \,2\,+\,2S_{1\rightarrow 2},$.

1.2 Appendix 2: Proof of Theorem for $M_2 \ne 0$

We start the derivation of the bounds in the theorem [given in (13) and (14)] by defining

$$\begin{aligned} T(\lambda )&= P_{xy}(\lambda )\langle xy \rangle + P_{xy'}(\lambda )\langle xy' \rangle \nonumber \\&+ P_{x'y}(\lambda )\langle x'y\rangle - P_{x'y'}(\lambda )\langle x'y'\rangle \end{aligned}$$

(28)

where $p_{xy} \!= p(a,b|x,y,\lambda ),\,p_{xy'} \!= p(a,b|x,y',\lambda )$, etc., and $P_{xy} = p(\lambda |x,y),\,P_{xy'} =p(\lambda |x,y')$, etc.

Now by Eq. (20) the above relation takes the form

$$\begin{aligned} T(\lambda )\le P_{xy}(\lambda ) + P_{xy'}(\lambda ) + P_{x'y}(\lambda ) + P_{x'y'}(\lambda ) - 2J(\lambda ) \end{aligned}$$

(29)

where

$$\begin{aligned} J(\lambda )&= P_{xy}(\lambda )|m_{1} - n_{1}| + P_{xy'}(\lambda ) |m_2 - n_2| \nonumber \\&+ P_{x'y}(\lambda )|m_3 - n_3| + P_{x'y'}(\lambda )|m_4 + n_4 -1|. \end{aligned}$$

(30)

Here $m_j,\,n_j$ and $c_j$ all depend on the underlying variable $\lambda$.

From the statement of the theorem it is clear that the restrictions are,

$$\begin{aligned}&m_j\in \{0,1\} , \,\, n_j \in [0\,,\,I_2] \cup [1-I_2\,,\,1]\end{aligned}$$

(31)

$$\begin{aligned}&m_1 = m_2,\, m_3 = m_4 \end{aligned}$$

(32)

$$\begin{aligned}&|n_1 - n_3| \text {,}\, |n_2 - n_4| \le S_{1\rightarrow 2} \end{aligned}$$

(33)

and

$$\begin{aligned}&\int |P_{xy}(\lambda ) - P_{xy'}(\lambda )|d\lambda , \int |P_{x'y}(\lambda ) - P_{x'y'}(\lambda )|d\lambda \le M_2 ;\nonumber \\&\qquad \qquad P_{xy}(\lambda ) = P_{x'y}(\lambda ) \, \text { and } \, P_{xy'}(\lambda )= P_{x'y'}(\lambda ). \end{aligned}$$

(34)

Hence $J(\lambda )$ takes the form,

$$\begin{aligned} J(\lambda )&= P_{xy}(\lambda )|m_{1} - n_{1}| + P_{xy'}(\lambda ) |m_1 - n_2| \nonumber \\&\quad + P_{xy}(\lambda )|m_3 - n_3| + P_{xy'}(\lambda )|m_3 + n_4 -1|. \end{aligned}$$

(35)

The left hand side of the Eq. (13) is obtained by integrating both sides of Eq. (28) over $\lambda$ and using Eq. (29), hence

$$\begin{aligned}&\langle xy \rangle + \langle xy' \rangle + \langle x'y \rangle - \langle x'y' \rangle \nonumber \\&\quad = \int T(\lambda )d\lambda \le 4 - 2 \int J(\lambda )d\lambda . \end{aligned}$$

(36)

In order to get the maximum value of the left hand side of Eq. (36), we have to minimize the integral of the positive quantity $J(\lambda )$ under the restriction given in the statement of the theorem. There are four possible cases corresponding to the four possible values of $(m_1,\,m_3)$: $(0,\,0),\,(1,\,0),\,(0,\,1) ,\, \text { and } \, (1,\,1)$. Similarly(as in the $M_2 = 0$ case), one trivially has

$$\begin{aligned} J(\lambda )\ge (1-S_{1\rightarrow 2})\min \{P_{xy},\,P_{xy'}\}. \end{aligned}$$

(37)

Then Eq. (36) becomes [by using (37)],

$$\begin{aligned}&\langle xy \rangle + \langle xy' \rangle + \langle x'y \rangle - \langle x'y' \rangle \nonumber \\&\le 4 - 2(1-S_{1\rightarrow 2})\int \min \{P_{xy},\,P_{xy'}\} d\lambda . \end{aligned}$$

(38)

Now,

$$\begin{aligned} \min \{P_{xy},\,P_{xy'}\}\,=\,\frac{1}{2}( P_{xy} + P_{xy'}) - \frac{1}{2}|P_{xy} - P_{xy'}|. \end{aligned}$$

(39)

Hence,

$$\begin{aligned} \int \min \{P_{xy},P_{xy'}\} d\lambda \, \ge \, \max \left\{ 0,1 - \frac{M_2}{2}\right\} . \end{aligned}$$

(40)

Then, one obtains the bound [using Eqs. (38) and (40)],

$$\begin{aligned}&\qquad \qquad \langle xy \rangle + \langle xy' \rangle + \langle x'y \rangle \nonumber \\&- \langle x'y' \rangle \le 4 - 2(1-S_{1\rightarrow 2})\max \left\{ 0,1 - \frac{M_2}{2}\right\} \end{aligned}$$

(41)

Now, just as in the previous case (i.e for $M_2 = 0$), it is clear that

$$\begin{aligned} B(I_2, S_{1\rightarrow 2}, M_2)&= 4 - (1 - I_2)(2 - M_2),\,\text { for } \, S_{1\rightarrow 2} < 1 - 2I_2 \nonumber \\&\text { and } \, M_2 < 2 \,(\text {tight upper bound}) \nonumber \\&= 4 - (1 - S_{1\rightarrow 2})(2 - M_2),\,\text { for } \, S_{1\rightarrow 2} \ge 1 - 2I_2 \nonumber \\&\text { and } \, M_2 < 2 \,(\text {tight upper bound}) \nonumber \\&= 4, \qquad \qquad \qquad \qquad \qquad \text { otherwise}. \end{aligned}$$

(42)

It still remain to show that the bound in Eq. (40) is tight. For this we consider a model with two variables $\lambda _1$ and $\lambda _2$ in Table 1 similar to that given in [10]. It is clear from the table that $\int \min \{P_{xy},P_{xy'}\} d\lambda \, = \, \max \{0,1 - \frac{M_2}{2}\}$.                                                                         $\square$

Table 1 A class of deterministic no signaling model