The critical detection efficiency for closing the detection loophole of some modified Bell inequalities

Abstract

As we know, the violations of Bell inequalities reveal nonlocality. On the other hand, loopholes in any Bell tests can cause the issues in the interpretation of the above conclusion, since an apparent violation of Bell inequality may not correspond to a real violation of local realism. The detection loophole, as an important example, arises when the overall detection efficiency is not larger than a certain threshold value. With the detection loophole, the above-mentioned effect can be observed in many experiments where the partial detected events can cause apparent Bell violations, while the entire ensemble cannot violate them. However, the real violations of Bell inequalities are necessary for device independence quantum information processing tasks. So, it is crucial to investigate the critical detection efficiency for closing the detection loophole of Bell inequalities. Here, by considering some novel Bell inequalities (Mironowicz and Pawłowski in Phys Rev A 88:032319, 2013), we give the critical detection efficiency for closing the detection loophole of these Bell inequalities. Furthermore, we prove the tightness of these detection efficiency bounds. That is, if detection efficiency is not larger than the critical one, we construct local hidden variable models to reproduce these violations. To sum up, if the detection efficiency of experiment exceeds the critical one derived here, the detection loophole is eliminated.

5. Appendix

1.1 5.1 The Proof of Lemma 1

Proof

Based on Ref. [32], it is obvious that $p(\varLambda _{A_{j}}|\varLambda _{A_{j}B_{k}})=p(\varLambda _{B_{k}}|\varLambda _{A_{j}B_{k}})=1$.

For $j'\ne j$, $p(\varLambda _{A_{j'}}|\varLambda _{A_{j}B_{k}})$ is given by

$$\begin{aligned} \begin{aligned} p(\varLambda _{A_{j'}}|\varLambda _{A_{j}B_{k}})&=\frac{p(\varLambda _{A_{j'}A_{j}}|\varLambda _{B_{k}})}{p(\varLambda _{A_{j}}|\varLambda _{B_{k}})}\\&=\frac{p(\varLambda _{A_{j'}}|\varLambda _{B_{k}})+p(\varLambda _{A_{j}}|\varLambda _{B_{k}})-p(\varLambda _{A_{j'}}\cup \varLambda _{A_{j}}|\varLambda _{B_{k}})}{p(\varLambda _{A_{j}}|\varLambda _{B_{k}})}\\&\ge \frac{2\eta -1}{\eta }, \end{aligned} \end{aligned}$$

(46)

where we use the definition of detection efficiency of each party and Eq. (8). So, $\eta \le p(\varLambda _{A_{j'}}|\varLambda _{B_{k}}), \eta \le p(\varLambda _{A_{j}}|\varLambda _{B_{k}})$.

Furthermore, when $j'\ne j$ and $k'\ne k$, $p(\varLambda _{A_{j'}B_{k'}}|\varLambda _{A_{j}B_{k}})$ can be deduced by

$$\begin{aligned} \begin{aligned}&p(\varLambda _{A_{j'}B_{k'}}|\varLambda _{A_{j}B_{k}})\\&=p(\varLambda _{A_{j'}}|\varLambda _{A_{j}B_{k}})+p(\varLambda _{B_{k'}}|\varLambda _{A_{j}B_{k}}) -p(\varLambda _{A_{j'}}\cup \varLambda _{B_{k'}}|\varLambda _{A_{j}B_{k}})\\&\ge 3-\frac{2}{\eta }, \end{aligned} \end{aligned}$$

(47)

where the first line holds according to the fact $\varLambda _{A_{j'}B_{k'}}=\varLambda _{A_{j'}}\cap \varLambda _{B_{k'}}$ and $p(x y)=p(x)+p(y)-p(x\cup y)$, the second line holds based on Eq. (46) and $p(\varLambda _{A_{j'}}\cup \varLambda _{B_{k'}}|\varLambda _{A_{j}B_{k}})\le 1$.

Finally, we calculate $\delta _{I_{0}}$, without loss of generality, we assume that $j=1, k=2$,

$$\begin{aligned}&\begin{aligned} p(\varLambda _{0}|\varLambda _{A_{1}B_{2}})&=p(\varLambda _{A_{1}A_{2}B_{2}B_{3}}|\varLambda _{A_{1}B_{2}})\\&=p(\varLambda _{A_{2}B_{3}}\cap \varLambda _{A_{1}B_{2}}|\varLambda _{A_{1}B_{2}})\\&=p(\varLambda _{A_{2}B_{3}}|\varLambda _{A_{1}B_{2}})+p(\varLambda _{A_{1}B_{2}}|\varLambda _{A_{1}B_{2}})-p(\varLambda _{A_{2}B_{3}}\cup \varLambda _{A_{1}B_{2}}|\varLambda _{A_{1}B_{2}})\\&\ge p(\varLambda _{A_{2}B_{3}}|\varLambda _{A_{1}B_{2}})\\&\ge 3-\frac{2}{\eta }, \end{aligned}\nonumber \\ \end{aligned}$$

(48)

where the forth line holds since $p(\varLambda _{A_{1}B_{2}}|\varLambda _{A_{1}B_{2}})=1, p(\varLambda _{A_{2}B_{3}}\cup \varLambda _{A_{1}B_{2}}|\varLambda _{A_{1}B_{2}})\le 1.$ The fifth line holds based on Eq. (47). So, we get

$$\begin{aligned} \delta _{I_{0}}\ge 3-\frac{2}{\eta }. \end{aligned}$$

(49)

$\square$

1.2 5.2 The Proof of Lemma 2

Proof

Obviously, $\varLambda _{0}\subset \varLambda _{A_{j}B_{k}}$, so we define $\overline{\varLambda _{0}}=\varLambda _{A_{j}B_{k}}\setminus \varLambda _{0}$, which satisfies that

$$\begin{aligned}&\varLambda _{0}\cap \overline{\varLambda _{0}}=\emptyset , \end{aligned}$$

(50)

$$\begin{aligned}&\varLambda _{0}\cup \overline{\varLambda _{0}}=\varLambda _{A_{j}B_{k}}. \end{aligned}$$

(51)

Next, we have

$$\begin{aligned}&|E(A_{j}B_{k}|\varLambda _{A_{j}B_{k}})-\delta _{I_{0}}E(A_{j}B_{k}|\varLambda _{0})|\nonumber \\&\quad = |p(\overline{\varLambda _{0}}|\varLambda _{A_{j}B_{k}})E(A_{j}B_{k}|\overline{\varLambda _{0}}) +p(\varLambda _{0}|\varLambda _{A_{j}B_{k}})E(A_{j}B_{k}|\varLambda _{0}) -\delta _{I_{0}}E(A_{j}B_{k}|\varLambda _{0})|\nonumber \\&\quad \le |p(\overline{\varLambda _{0}}|\varLambda _{A_{j}B_{k}})E(A_{j}B_{k}|\overline{\varLambda _{0}})| +|p(\varLambda _{0}|\varLambda _{A_{j}B_{k}})E(A_{j}B_{k}|\varLambda _{0}) -\delta _{I_{0}}E(A_{j}B_{k}|\varLambda _{0})|\nonumber \\&\quad \le p(\overline{\varLambda _{0}}|\varLambda _{A_{j}B_{k}})E(|A_{j}B_{k}||\overline{\varLambda _{0}}) +(p(\varLambda _{0}|\varLambda _{A_{j}B_{k}})-\delta _{I_{0}})E(|A_{j}B_{k}||\varLambda _{0})\nonumber \\&\quad \le 1-\delta _{I_{0}}, \end{aligned}$$

(52)

where $E(|A_{j}B_{k}|)=\sum _{a,b}|ab|p(a,b|j,k)$.

The first equation holds based on $\varLambda _{A_{j}B_{k}}=\varLambda _{0}+\overline{\varLambda _{0}}$, $E(A_{j}B_{k}$ $|\varLambda _{A_{j}B_{k}})=p(\overline{\varLambda _{0}}|\varLambda _{A_{j}B_{k}})$ $E(A_{j}B_{k}|\overline{\varLambda _{0}}) +p(\varLambda _{0}|\varLambda _{A_{k}B_{k}})E(A_{j}B_{k}|\varLambda _{0})$, and the first inequality holds based on $|x+y-z|\le |x|+|y-z|$. The second inequality holds based on $|E(xy)|\le E(|xy|)$. The last inequality holds because that $E(|A_{j}B_{k}||\overline{\varLambda _{0}})=1$ and $E(|A_{j}B_{k}||\varLambda _{0})=1$ due to measurement outcomes $a,b\in \{+1,-1\}$. $\square$

1.3 5.3 The Proof of (31)

Proof

Based on $\delta _{I_{1}}=\min _{(j,k)}p(\varLambda _{0}|\varLambda _{A_{j}B_{k}})$ for $j, k \in \{1,2,3\}$, without loss of generality, we assume that $j=1, k=2$,

$$\begin{aligned} \begin{aligned} p(\varLambda _{0}|\varLambda _{A_{1}B_{2}})&=p(\varLambda _{A_{2}B_{3}}\cap \varLambda _{A_{3}B_{1}}\cap \varLambda _{A_{1}B_{2}}|\varLambda _{A_{1}B_{2}})\\&=p(\varLambda _{A_{1}B_{2}}|\varLambda _{A_{1}B_{2}})p(\varLambda _{A_{2}B_{3}}\cap \varLambda _{A_{3}B_{1}}|\varLambda _{A_{1}B_{2}})\\&=p(\varLambda _{A_{2}B_{3}}\cap \varLambda _{A_{3}B_{1}}|\varLambda _{A_{1}B_{2}}),\\ \end{aligned} \end{aligned}$$

(53)

where $\varLambda _{0}=\varLambda _{A_{1}}\cap \varLambda _{A_{2}}\cap \varLambda _{A_{3}}\cap \varLambda _{B_{1}}\cap \varLambda _{B_{2}}\cap \varLambda _{B_{3}}=\varLambda _{A_{1}B_{2}}\cap \varLambda _{A_{2}B_{3}}\cap \varLambda _{A_{3}B_{1}}$, $p(\varLambda _{A_{1}B_{2}}|\varLambda _{A_{1}B_{2}})=1$.

Furthermore, $P(\varLambda _{0}|\varLambda _{A_{1}B_{2}})$ can be deduced by

$$\begin{aligned}&\begin{aligned} p(\varLambda _{0}|\varLambda _{A_{1}B_{2}})&=p(\varLambda _{A_{2}B_{3}}\cap \varLambda _{A_{3}B_{1}}|\varLambda _{A_{1}B_{2}})\\&=p(\varLambda _{A_{2}B_{3}}|\varLambda _{A_{1}B_{2}})+p(\varLambda _{A_{3}B_{1}}|\varLambda _{A_{1}B_{2}}) -p(\varLambda _{A_{2}B_{3}}\cup \varLambda _{A_{3}B_{1}}|\varLambda _{A_{1}B_{2}})\\&\ge p(\varLambda _{A_{2}B_{3}}|\varLambda _{A_{1}B_{2}})+p(\varLambda _{A_{3}B_{1}}|\varLambda _{A_{1}B_{2}})-1\\&\ge 5-\frac{4}{\eta }, \end{aligned}\nonumber \\ \end{aligned}$$

(54)

where the second line holds because $\varLambda _{A_{2}B_{3}}\cap \varLambda _{A_{3}B_{1}}=\varLambda _{A_{2}B_{3}}+\varLambda _{A_{3}B_{1}} -\varLambda _{A_{2}B_{3}}\cup \varLambda _{A_{3}B_{1}}$, the third line holds since $p(\varLambda _{A_{2}B_{3}}\cup \varLambda _{A_{3}B_{1}}|\varLambda _{A_{1}B_{2}})\le 1$, and then the last inequality holds based on the relation (46).

Furthermore, $p(\varLambda _{0}|\varLambda _{A_{j}B_{k}})\ge 5-\frac{4}{\eta }$. So, we get

$$\begin{aligned} \delta _{I_{1}}=\min _{(j,k)}p(\varLambda _{0}|\varLambda _{A_{j}B_{k}})\ge 5-\frac{4}{\eta }. \end{aligned}$$

(55)

$\square$