Tripartite counterfactual quantum cryptography

We show how two distrustful parties, “Bob” and “Charlie,” can share a secret key with the help of a mutually trusted “Alice” counterfactually; that is, with no information-carrying particles traveling between any of the three.

Figure 1

The chained quantum Zeno effect (CQZE). Bob (Charlie) can block the channel by switching on Pockels cell PC, directing any photon in the channel (always horizontally polarized) towards detector $D_4$. The overall action on Alice's horizontally $\left(H\right)$ polarized photon is the following, $|\text{H}\rangle \to |\text{H}\rangle$ when the channel is not blocked, and $|\text{H}\rangle \to |\text{V}\rangle$ when the channel is blocked. The inner working of the two-cycle CQZE is explained in the main text. Counterfactuality is ensured because any photon going into the channel would either trigger $D_3$ for the case of Bob (Charlie) not blocking, or else trigger $D_4$ for the case of Bob (Charlie) blocking.

Figure 2

Protocol for tripartite counterfactual quantum cryptography. Bob (Charlie) randomly encodes a “0” (“1”) by not blocking his channel and a “1” (“0”) by blocking it. Bob (Charlie) can block his channel by switching on Pockels cell ${\mathrm{PC}}_B$ $\left({\mathrm{PC}}_C\right)$, which flips polarization, directing the photon towards $D_4$. Initially, Alice sends an $H$ photon from her photon source S towards beam splitter BS, which puts the photon in an equal superposition of being on path B (leading to Bob) and path C (leading to Charlie). If Bob and Charlie encode different bit values, the two parts of the photon superposition reflected back towards Alice's BS (from top and from right) will be, by the action of the two CQZEs, identically polarized. Constructive interference therefore takes place, resulting in Alice's $D_2$ clicking with certainty (provided the photon was not lost to $D_3$ or $D_4)$. If, however, Bob and Charlie encode the same bit value, the two parts of the photon superposition reflected back towards Alice's BS will be oppositely polarized. Interference does not take place. $D_1$ and $D_2$ are therefore equally likely to click. A click at $D_1$ uniquely corresponds to Bob and Charlie randomly agreeing in their bit choices. (Here, OC stands for optical circulator, which directs a photon exiting left towards $D_1$.)