Self-testing with untrusted random number generators

TL;DR

Surprisingly, it is shown that all pure bipartite partially entangled states can be self-tested provided that the random number generator obeys a residual randomness constraint strictly weaker than the independence assumption.

Abstract

Self-testing--the attractive possibility to infer the underlying physics of a quantum device in a black-box scenario--has gained increased traction in recent years, with applications to device-independent quantum information processing. Thus far, self-testing has been done under the assumption that the settings for the requisite Bell test are chosen freely and independently of the device tested in the experiment. That is, the random number generator used to generate the settings has been assumed to have no correlations with the device tested. Here, we extend self-testing protocols beyond the independence assumption. Surprisingly, we show that all pure bipartite partially entangled states can be self-tested provided that the random number generator obeys a residual randomness constraint strictly weaker than the independence assumption. This in itself provides a semi-device-independent certification of independence between the randomness source and the device.

Figures (2)

• Figure 1: Bell scenario with untrusted sources. The sources $S$ and $T$ are chosen to pick the setting from the input space $X$ and $Y$, which produce outputs from spaces $A$ and $B$ with no explicit dependence on the sources. The hidden variable $\Lambda$ influences $S, T, A, B$. We can understand $S$ and $T$ as the output spaces of two additional parties in the Bell experiments with no inputs. We consider classical-quantum states $\rho = \sum_{st} |st \rangle \langle st | \otimes \rho_{st}$, which produce probabilities $p(stab|xy) = \langle A_{a|x} \otimes B_{b|y} \rangle_{st} = \tr (\rho_{st} A_{a|x} \otimes B_{b|y})$ satisfying $p(st|abxy) > 0$. The protocols only have access to the observed probabilities, which have the form $p(stab|st)$.
• Figure 2: Description of the self-testing protocol for the state $\ket{\psi} = 1/6 \ket{00} + 1/8 \ket{11} + 1/6 \ket{22} + 1/6\ket{33} + 1/8\ket{44} + 1/4 \ket{55}$. Take the inhomogeneous covering tree rooted at $0$ with edges $(0,1), (0,4), (0,5), (1,2)$ and $(1,3)$. Each party has one measurement with six outcomes $A_0$ and $B_0$. For each edge $(0_i, 1_i)$, say $(0, 4)$, each party has one dichotomic measurement $A_{i,1}$ and $B_{i, 1}$ and the virtual dichotomic measurement $A_{i,0} = (A_{0|0}, A_{4|0})$ and $B_{i,0} = (B_{0|0}, B_{4,|0})$ that is used for a Hardy self-test of the substate $|\psi_i \rangle = c_0|00\rangle + c_4 |44 \rangle$. Namely, the distribution $p_i(ab|xy)$ for these measurements satisfies $p_i(01|01) = p_i(10|10) = p_i(00|11) = 0$ and $p_i(00|00) + w_i p_i(11|00) = p_i q(w_i)$. Here, $p_i = (c_0^2 + c_4^2)$ and $w_i$ is the parameter that self-tests the state with angle $\operatorname{arctan}(c_4/c_0)$.

Theorems & Definitions (10)

• Lemma 1
• proof
• Proposition 2
• proof
• Lemma 3
• Proposition 4
• proof
• Proposition 5
• proof
• Theorem 6