Qunatum Parrondo's games constructed by quantum random walk

Abstract

We construct a Parrondo's game using discrete time quantum walks. Two lossing games are represented by two different coin operators. By mixing the two coin operators $U_A(\hat{I}{\pm }_A,{\hat{I}}_A^2,{\hat{I}}_A^3)$ and $U_B(\hat{I}{\pm }_B,{\hat{I}}_B^2,{\hat{I}}_B^3)$, we may win the game. Here we mix the two games in position instead of time. With a number of selections of the parameters, we can win the game with sequences ABB, ABBB, \emph{et al}. If we set ${\hat{I}}_A^2=45^{\circ },{\hat{I}}_A^3=0,\hat{I}{\pm }_B=0,{\hat{I}}_B^2=88^{\circ }$, we find the game 1\emph{}with {\normalsize $U_A^S=U^S(-51^{\circ },45^{\circ },0)$, $U_B^S=U^S(0,88^{\circ },-16^{\circ })$ will win and get the most profit.}If we set $\hat{I}{\pm }_A=0,{\hat{I}}_A^2=45^{\circ },\hat{I}{\pm }_B=0,{\hat{I}}_B^2=88^{\circ }$ and{\normalsize{} the game 2 with $U_A^S=U^S(0,45^{\circ },-51^{\circ })$, $U_B^S=U^S(0,88^{\circ },-67^{\circ })$, will win most. And}game 1\emph{}{\normalsize is equivalent to the}game\emph{}2\emph{}with the changes of sequences and steps. But at a large enough steps, the game will loss at last.