1303.6831 (Min Li et al.)

Qunatum Parrondo's games constructed by quantum random walk    [PDF]

Min Li, Yong-Sheng Zhang, Guang-Can Guo

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 $U_{A}(\alpha_{A},\beta_{A},\gamma_{A})$ and $U_{B}(\alpha_{B},\beta_{B},\gamma_{B})$, 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 $\beta_{A}=45^{\circ},\gamma_{A}=0,\alpha_{B}=0,\beta_{B}=88^{\circ}$, 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 $\alpha_{A}=0,\beta_{A}=45^{\circ},\alpha_{B}=0,\beta_{B}=88^{\circ}$ 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.

View original: http://arxiv.org/abs/1303.6831