bibkey: arunachalam2017quantumhedging
authors: Srinivasan Arunachalam; Abel Molina; Vincent Russo
year: 2018
title: "Quantum hedging in two-round prover-verifier interactions"
doi: 10.4230/LIPIcs.TQC.2017.5
url: https://arxiv.org/abs/1310.7954v5
claim: "For a two-round game in which Bob returns one half of an entangled two-qubit state to Alice for a projective measurement, the paper characterizes when Bob can win at least one of n parallel repetitions with certainty, and suggests from numerical evidence that Bob cannot perfectly hedge more than n/2 of the games."
strata_touched:
  - D5/S3/Quantum/Information/HedgingMajorityThreshold
license: citation-only
triage: anchor

Quantum hedging in two-round prover-verifier interactions

Srinivasan Arunachalam, Abel Molina and Vincent Russo, arXiv:1310.7954v5 (quant-ph), 2017-03-12, first version 2013-10-29; published in TQC 2017, LIPIcs 73 (2018) 5:1–5:30. Quotations are from the v5 TeX source.

The game (Section 3):

Alice prepares the 2-qubit state ${\rho }_{\alpha }=u_{\alpha }{u}_{\alpha }^{\ast }\in \text{Density}(\text{X}\otimes \mathbb{Z})$ in registers $(\text{reg}X,\text{reg}Z)$ where

u_{\alpha} = \alpha \ket{00} + \sqrt{1 - \alpha^2} \ket{11} \in \X \otimes \Z,

Bob applies a channel $\Phi \in \text{Channel}(\text{X},\text{Y})$ to the contents of $\text{reg}X$. This results in a state $\sigma \in \text{Density}(\text{Y}\otimes \mathbb{Z})$, contained in registers $(\text{reg}Y,\text{reg}Z)$. Register $\text{reg}Y$ is sent back to Alice.

P_{1,\theta} = v_{\theta} v_{\theta}^*, ~P_{0,\theta} = \I - P_{1,\theta}, \nonumber \ v_{\theta} = \cos(\theta) \ket{00} + \sin(\theta) \ket{11} \in \Y \otimes \Z.

An outcome of 0'' or 1’’ denotes a losing or winning outcome for Bob, respectively.

Parallel repetition (Section 3):

\Phi_n \in \Channel(\X_1 \otimes \cdots \otimes \X_n, \Y_1 \otimes \cdots \otimes \Y_n).

The open question (Section 5):

However, it still remains open to determine the conditions under which Bob can always win at least $k$ out of $n$ repetitions for some $k>1$.

Based on our current numerical evidence, it is possible that Bob cannot perfectly hedge more than $k=n/2$ games.

Verified locator

• URL: https://arxiv.org/abs/1310.7954v5 (v5, 2017-03-12; source arxiv_v5.tex, md5 78b99144eee8f68c949b2319e53fee68): the game (l. 196–209), parallel repetition (l. 211–220), and the open question (l. 574).
• DOI: 10.4230/LIPIcs.TQC.2017.5 (TQC 2017). The published text was not read.