When Are Popescu-Rohrlich Boxes and Random Access Codes Equivalent?

We study a problem of interconvertibility of two supraquantum resources: one is the so-called Popescu-Rohrlich (PR) box, which violates Clauser-Horne-Shimony-Holt inequality up to the maximal algebraic bound, and the second is the so-called random access code (RAC). The latter is a functionality that enables Bob (receiver) to choose one of two bits of Alice. It is known that a PR box supplemented with one bit of communication can be used to simulate a RAC. We ask the converse question: to what extent can a RAC can simulate a PR box? To this end, we introduce a “racbox”: a box such that when it is supplemented with one bit of communication it offers a RAC. As said, a PR box can simulate a racbox. The question we raise is whether any racbox can simulate a PR box. We show that a nonsignaling racbox, indeed, can simulate a PR box; hence, these two resources are equivalent. We also provide an example of a signaling racbox that cannot simulate a PR box. We give a resource inequality between racboxes and PR boxes and show that it is saturated.

Figure 1

(a) PR box. (b) RAC. (c) Racbox acts as RAC, provided that the input $y^{\prime }$ is equal to $a$. Thus, in particular, if the output $a$ is sent to Bob and he inputs it to $y^{\prime }$ (as depicted by dashed line), then $b=x_y$. (d) Nonsignaling racbox satisfies $b=x_y\oplus a\oplus y^{\prime }$.

Figure 2

(a) Simulation of a nonsignaling racbox with a PR box. (b) Simulation of a PR box with a nonsignaling racbox. We set inputs as $x_0=0$, $x_1=x$, $y^{\prime }=0$, while leaving $y$ and the outputs $a$ and $b$ unchanged. This simulation precisely cancels the actions of CNOT gate in the previous one so that we get a PR box again.

Figure 3

A protocol for achieving resource inequality (7). The bit to be transmitted is denoted by $z$. $\mathcal{E}$ is the erasure channel: with probability $\epsilon =p(y=1)$ the message is lost, whereas with probability $1-\epsilon$ the message is delivered intact. The receiver knows which is the case. The inputs $x$, $y$ and the outputs $a$, $b$ satisfy Eq. (2).