The strategic form of quantum prisoners' dilemma

TL;DR

This paper explores the quantum version of the prisoners' dilemma, analyzing how entanglement and measurement basis affect the game's strategic form and equilibrium outcomes.

Contribution

It introduces a generalized quantization scheme for PD, identifying entanglement thresholds that preserve or alter the game's strategic structure.

Findings

Existence of entanglement cutoffs where game behavior changes

Quantum PD can behave like chicken game or classical PD depending on parameters

Measurement basis influences the strategic form and Nash equilibria

Abstract

In its normal form prisoners' dilemma (PD) is represented by a payoff matrix showing players strategies and payoffs. To obtain distinguishing trait and strategic form of PD certain constraints are imposed on the elements of its payoff matrix. We quantize PD by generalized quantization scheme to analyze its strategic behavior in quantum domain. The game starts with general entangled state of the form $\left}\psi\right\rangle =\cos\frac{\xi}% {2}\left|00\right\rangle +i\sin\frac{\xi}{2}\left|11\right\rangle $ and the measurement for payoffs is performed in entangled and product bases. We show that for both measurements there exist respective cutoff values of entanglement of initial quantum state up to which strategic form of game remains intact. Beyond these cutoffs the quantized PD behaves like chicken game up to another cutoff value. For the measurement in entangled basis the dilemma is resolved for\ $\sin\xi>\frac{1}{7}$ with $Q\otimes Q$ as a NE but the quantized game behaves like PD when $\sin\xi>\frac{1}{3}$; whereas in the range $\frac{1}{7}<\sin\xi<\frac{1}{3}$ it behaves like chicken game (CG)\ with $Q\otimes Q$ as a NE. For the measurement in product basis the quantized PD behaves like classical PD for $\sin^{2}\frac{\xi}{2}<\frac{1}{3}$ with $D\otimes D$ as a NE. In region $\frac{1}{3}<\sin^{2}\frac{\xi}{2}% <\frac{3}{7}$ the quantized PD behaves like classical CG with $C\otimes D$ and $D\otimes C$ as NE.