Contextuality-based quantum conferencing

Abstract

Nonlocality inequalities for multi-party systems act as contextuality inequalities for single qudit systems of suitable dimensions (Heywood and Redhead in Found Phys 13(5):481–499, 1983; Abramsky and Brandenburger in New J Phys 13(11):113036, 2011). In this paper, we propose the procedure for adaptation of nonlocality-based quantum conferencing protocols (NQCPs) to contextuality-based QCPs (CQCPs). Unlike the NQCPs, the CQCPs do not involve nonlocal states. As an illustration of the procedure, we present a QCP based on Mermin’s contextuality inequality. As a significant improvement, we propose a QCP based on CHSH contextuality inequality involving only four-dimensional states irrespective of the number of parties sharing the key. The key generation rate of the latter is twice that of the former. Although CQCPs allow for an eavesdropping attack which has no analog in NQCPs, a way out of this attack is demonstrated. Finally, we examine the feasibility of experimental implementation of these protocols with orbital angular momentum states.

Appendices

A Mermin’s nonlocality inequality as contextuality inequality

The interrelation between the two-party nonlocality and single-qudit contextuality, shown in Sect. 3.1 for CHSH inequality, continues to hold even for multi-party nonlocality inequalities. We illustrate it through the example of Mermin’s inequality. For that, we briefly recapitulate Mermin’s inequality for an N-party system.

Mermin’s inequality distinguishes the states admitting completely factorisable local hidden variable model from nonlocal states.

Consider a pair of dichotomic observables $\{A_{j}, A'_{j} \}$ in the space of the $j^\mathrm{th}$ party ($j\in \{1, \cdots , N\}$). The inequality is as follows [43]:

$$\begin{aligned} {{\mathcal {M}}}_N&=\frac{1}{2i}\Bigg \vert \bigg \langle \prod _{j=1}^{N} (A_j+i A'_j)-\prod _{j=1}^{N}(A_j-i A'_j)\bigg \rangle \Bigg \vert \le c\nonumber \\ c&= 2^{N/2},\quad N=\mathrm{even},\nonumber \\&= 2^{\frac{N-1}{2}},\quad N= \mathrm{odd}; N \ge 3. \end{aligned}$$

(39)

The observables corresponding to different parties commute and $[A_k,A'_k] \ne 0$.

For the special case of an N qubit system, inequality (39) gets maximally violated by the GHZ state,

$$\begin{aligned} |\phi _N\rangle =\dfrac{1}{\sqrt{2}}\bigg (|0\rangle ^{\otimes N}+i|1\rangle ^{\otimes N}\bigg ), \end{aligned}$$

(40)

for the following choice of the observables:

$$\begin{aligned} A_k =X_k;~A'_k=Y_k. \end{aligned}$$

(41)

It is pertinent to illustrate the contextual behaviour of $|\phi _N\rangle$ to substantiate our claim. For that purpose, we consider the following two contexts: Context 1 ($Z_1 Z_2\cdots Z_{N-1}Z_N$): Suppose the outcome of $Z_1$ is $+1(-1)$. Then, the outcomes of observables $Z_2, Z_3, \cdots , Z_N$ will be $+1(-1)$ with unit probability. Context 2 ($Y_1Z_2\cdots Z_N$): Let the outcome of the measurement of $Y_1$ be once again, +1. Following it, the measurement of observable $Z_2$ will yield $+1$ or $-1$ with equal probability. Thereafter, the measurement of $Z_3, \cdots , Z_N$ will definitely yield +1. Thus, the outcome of $Z_2$ depends on the set of commuting observables it is measured with, i.e. it depends on the context. Exactly in the same manner as in Sect. 3.1, the observables and the state in the $2^N$ dimensional Hilbert space can be identified.

$$\begin{aligned} |\phi _N\rangle \mapsto |\Phi \rangle =\dfrac{1}{\sqrt{2}}(|0\rangle +i |2^N-1\rangle ) \end{aligned}$$

(42)

The equivalent observables $\{{\mathbb {Z}}_1,\cdots , {\mathbb {Z}}_N,{\mathbb {Y}}_1 \}$ can be obtained by following the prescription in point (2) and (3) of Sect. 2, and they satisfy the same commutation relations as $\{{Z}_1,\cdots , {Z}_N, {Y}_1 \}$ because of operator isomorphism. Thus, Mermin’s inequality probes contextuality in a qudit of appropriate higher dimension and can be referred to as Mermin’s contextuality inequality.

B Illustration of eavesdropping

To illustrate the point that how Eve obtains information about the key without disturbing the violation of contextuality inequality, we consider the QCP based on Mermin’s contextuality inequality. Consider an example of three parties, Bob$_1$, Bob$_2$ and Bob$_3$. We choose following set of observables:

$$\begin{aligned}&{\mathbb {Z}}_1=|0\rangle \langle 0|+|1\rangle \langle 1|+|2\rangle \langle 2|+|3\rangle \langle 3|\nonumber \\&~~~~~-\Big (|4\rangle \langle 4|+|5\rangle \langle 5|+|6\rangle \langle 6|+|7\rangle \langle 7|\Big )\nonumber \\&{\mathbb {Z}}_2=|0\rangle \langle 0|+|1\rangle \langle 1|+|4\rangle \langle 4|+|5\rangle \langle 5|\nonumber \\&~~~~~-\Big (|2\rangle \langle 2|+|3\rangle \langle 3|+|6\rangle \langle 6|+|7\rangle \langle 7|\Big )\nonumber \\&{\mathbb {Z}}_3=|0\rangle \langle 0|+|2\rangle \langle 2|+|4\rangle \langle 4|+|6\rangle \langle 6|\nonumber \\&~~~~~~-\Big (|1\rangle \langle 1|+|3\rangle \langle 3|+|5\rangle \langle 5|+|7\rangle \langle 7|\Big )\nonumber \\&{\mathbb {X}}_1=|0^+\rangle \langle 0^+|+|1^+\rangle \langle 1^+|+|2^+\rangle \langle 2^+|+|3^+\rangle \langle 3^+|\nonumber \\&~~~~~-\Big (|0^-\rangle \langle 0^-|+|1^-\rangle \langle 1^-|+|2^-\rangle \langle 2^-|+|3^-\rangle \langle 3^-|\Big ), \end{aligned}$$

(43)

where the symbols $|l^{\pm }\rangle$ are defined as,

$$\begin{aligned} |l^\pm \rangle =\frac{1}{\sqrt{2}}(|l\rangle \pm |l+4\rangle );~~~ l \in \{0, 1, 2, 3\}. \end{aligned}$$

(44)

Consider the following two situations, one in which there is no eavesdropping, the other in which Eve is present.

• 1. Case I: Following the steps of Mermin’s contextuality-based QCP, let Bob$_1$ prepares a state $|\Psi _M'\rangle =|0\rangle$ for the +1 outcome of ${\mathbb {Z}}_1$ on the reference state $|\Psi _M\rangle$. He sends this state $|\Psi _M'\rangle$ to Bob$_2$. If Bob$_2$ measures, say, ${\mathbb {Z}}_2$ on this state, he obtains the outcome +1 with unit probability. Thereafter, he sends the post-measurement state to Bob$_3$. Bob$_3$ measures, say, ${\mathbb {Z}}_3$, he is bound to get outcome +1 with unit probability.

• 2. Case II: Let there be an eavesdropping between Bob$_2$ and Bob$_3$. Eve measures an observable, $\mathbb {X_1}$. She obtains $\pm 1$ with equal probability and the state collapses to $|\Phi ^{\pm }\rangle =\frac{1}{\sqrt{2}}(|0\rangle \pm |4\rangle )$. She sends the post-measurement state to Bob$_3$. Bob$_3$ measures an observable ${\mathbb {Z}}_3$, as in the previous case, and obtains outcome +1 with unit probability. Thus, the measurement of Eve does not affect the outcome of Bob$_3$’s measurement. Due to this, Bob$_3$ will never detect presence of Eve. In this way, Eve can obtain full information about the key by choosing appropriate observable without being detected.