Infinitely Repeated Quantum Games and Strategic Efficiency

We consider the quantum prisoner’s dilemma Eisert:1998vv ; 1999PhRvL..82.1052M . Let $\ket{C},\ket{D}$ be two normalized orthogonal states that represent "Cooperative" and "Non-cooperative" states, respectively. Then a general state of the game is a complex linear combination, spanned by the basis vectors $\ket{CC},\ket{CD},\ket{DC},\ket{DD}$. Each player $i$ chooses a unitary strategy $U_{i}$ and the game is in a state

$\displaystyle\Ket{\psi}=\mathcal{J}^{\dagger}(U_{A}\otimes U_{B})\mathcal{J}\Ket{CC},$

(1)

where $\mathcal{J}$ gives some entanglement. In what follows we use

$\displaystyle\mathcal{J}=\exp{\left[i\frac{\theta}{2}Y\otimes Y\right]}=\cos{\frac{\theta}{2}}+i\sin{\frac{\theta}{2}}(Y\otimes Y),$

(2)

where $\theta$ represents entanglement and two states become maximally entangled at $\theta=\frac{\pi}{2}$. We assume a quantum strategy of a player $i$ has a representation

$\displaystyle U_{i}=\alpha_{i}I+i\left(\beta_{i}X+\gamma_{i}Y+\delta_{i}Z\right)\quad(\alpha_{i},\cdots,\delta_{i}\in\mathbb{R})$

(3)

where the coefficients obey $\alpha^{2}_{i}+\cdots+\delta_{i}^{2}=1$.

Alice receives payoff $\$_{A}=\Braket{\psi}{\hat{\$}_{A}}{\psi}$, where $\hat{\$}_{A}$ is her payoff matrix

$\displaystyle\hat{\$}_{A}=r\Ket{CC}\bra{CC}+p\ket{DD}\bra{DD}+t\ket{DC}\bra{DC}+s\ket{CD}\bra{CD}.$

(4)

Without loss of generality, we can always proceed with the discussion as $s=0$ if necessary. Taking into account of $U_{A}$ (3), the average payoff $\hat{\$}_{A}=\hat{\$}_{A}(x_{A},x_{B})$ of Alice is a function of $x_{i}=(\alpha_{i},\cdots,\delta_{i})$

$\displaystyle\begin{split}\$_{A}(x_{A},x_{B})&=r(\alpha_{A}\alpha_{B}-\delta_{A}\delta_{B}+\sin{\theta}(\beta_{A}\gamma_{B}+\gamma_{A}\beta_{B}))^{2}\\ &+p(\beta_{A}\beta_{B}-\gamma_{A}\gamma_{B}+\sin{\theta}(\alpha_{A}\delta_{B}+\delta_{A}\alpha_{B}))^{2}\\ &+t(\gamma_{A}\alpha_{B}+\beta_{A}\delta_{B}+\sin{\theta}(-\alpha_{A}\beta_{B}+\delta_{A}\gamma_{B}))^{2}\\ &+\cos^{2}{\theta}\left(r(\alpha_{A}\delta_{B}+\delta_{A}\delta_{B})^{2}+p(\beta_{A}\gamma_{B}+\gamma_{A}\beta_{B})^{2}\right.\\ &\left.+t(\beta_{A}\alpha_{B}-\gamma_{A}\delta_{B})^{2}\right).\end{split}$

(5)

Since the game is symmetric for two players, $\$_{A}(x_{A},x_{B})=\$_{B}(x_{B},x_{A})$ is satisfied.

Although our mixed strategies are probabilistic mixtures of the four Pauli operators $\{I,X,Y,Z\}$, we show the game is powerful enough to obtain stronger results compared with classical prisoner’s dilemma with mixed strategy.

The classical pure strategic prisoner’s dilemma allows players to choose one of $\{I,X\}$ and it becomes a mixed strategy game if the state is written as

$$\Ket{\psi}=\sum_{i\in\{CC,CD,DC,DD\}}p_{i}\ket{i},$$

(7)

where $p_{i}$ is non-negative and satisfies $\sum_{i}p_{i}=1$. Note that for a single round quantum game, a generic state is represented as

$$\ket{\psi}=\sum_{i\in\{CC,CD,DC,DD\}}c_{i}\ket{i},$$

(8)

where $c_{i}$ is a complex number and a state $\ket{i}$ is measured with the probability $|c_{i}|^{2}$. However as long as one considers a single round game, the quantum game is resemble the classical mixed strategy game, where a strategy is chosen stochastically with a given probability distribution.

Considering repeated quantum games is a way to make quantum games much more meaningful, since transition of quantum states is different from that of classical states. To begin with, we first address single round quantum prisoner’s dilemma and equilibria of the game.

A generic repeated quantum game is summarized as follows.

Now let us consider the repeated quantum prisoner’s dilemma. Let $\ket{\psi^{t}}\in\mathcal{H}_{A}\otimes\mathcal{H}_{B}$ be a state at the $t$-th round. Let $\{\tau_{i}\}_{i=1,2,\cdots}$ be a set of positive integers. We assume measurement of quantum states is done at the end of the $t_{i}=\sum_{k=1}^{i}\tau_{k}$ round for each $i$. For example, $\tau_{i}=1$ for all $i$ means we measure quantum states every round. Suppose $V^{\tau_{i-1}+1},\cdots,V^{\tau_{i}}$ are operators of Alice, then her strategy is $U^{t_{i}}_{A}=V^{t_{i-1}+1}_{A}\otimes\cdots\otimes V^{t_{i-1}+\tau_{i}}_{A}$ and the game is in a state

$$\ket{\psi^{t_{i}}}=\mathcal{J}^{\dagger}(U^{t_{i}}_{A}\otimes U^{t_{i}}_{B})\mathcal{J}\Ket{\psi^{t_{i-1}}}$$

(105)

which should be measured at the end of $t_{i}$ round. The payoff could depend on $\tau_{i}$:

$$\$^{t_{i}}_{A}=f(\tau_{i})\bra{\psi^{t_{i}}}\hat{\$}_{A}\Ket{\psi^{t_{i}}},$$

(106)

where $f(\tau_{i})$ is a certain function of $\tau_{i}$. In repeated games, the total payoff is written with discount factor $\delta\in(0,1)$ and in our case it can be introduced in such a way that

$$\$_{A}=\sum_{i}\delta^{t_{i}}\$^{t_{i}}_{A}.$$

(107)

An interpretation of the discount factor is that the importance of the future payoffs decreases with time. The measurement periods $\{\tau_{i}\}$ can be defined randomly or predeterminedly. When game’s state is measured every round, the probability of monitoring signals is classical (see (15)). In order to enjoy full quantum games, a game should evolve with unitary operation.

Since this work is the first study on infinitely repeated quantum games, we focus on the most fundamental case and investigate a role of entanglement. We will work on a generic case elsewhere. For this purpose, in what follows, we put $\tau_{i}=1$ for all $i$ and $f(1)=1$. Currently, we are considering only the prisoner’s dilemma, but there are frequent cases when the game is repeated, for example in negotiation games. Negotiations usually involve thinking not only about the pie on the table today, but also about the pie after tomorrow. It is natural that interest may accrue and costs may change during negotiations. The function $f$ represents the change in cost that can occur during the period between the finalization of one state and the transition to the next state. Negotiation games are the basic repeated games in contract theory. The formulation of contract theory as quantum games has been recently given in Ikeda_cat2021 . Further developing that model as a game theory is a very interesting problem.

Then the game evolves in such a way that

$\displaystyle\Ket{\psi^{t+1}}=\mathcal{J}^{\dagger}(U_{A}^{t}\otimes U_{B}^{t})\mathcal{J}\Ket{\psi^{t}},$

(108)

where $U_{A}^{t}$ is her quantum strategy for the $t$th round. Then her average payoff is

$$\$^{t}_{A}=\Braket{\psi^{t}}{\hat{\$}_{A}}{\psi^{t}}.$$

(109)

We define a trigger strategy as follows.

We show that Trigger 1 is an equilibrium of the repeated quantum prisoner’s dilemma.

Fig.4 and 5 present the relation between $\delta_{\inf}$ and entanglement. The bigger $\theta$ becomes, the more profit the players can receive, therefore the discount factor $\delta$ should be sufficiently big in order to maintain the cooperative relation when their states are strongly entangled.

Deviation 1 means that Alice (Bob) deviates the cooperation and chooses $X$ or $Y$. Deviation 2 means that Alice (Bob) chooses $I$ or $Z$ when they should choose $X$ or $Y$ with the equal probability. According to Proposition 2.6, it can happen when $\sin^{2}\theta>\frac{2(p-s)}{t-r+p-s}$. Deviation 3 means that Alice (Bob) chooses $I$ or $Z$ when Alice should play $X$ ($Y$) and Bob should play $Y$ $(X)$, respectively. It can happen at $\sin^{2}\theta>\frac{p-s}{t-r+p-s}$. This is because that if an opponent chooses $Y$ $(X)$, choosing $Z$ $(I)$ against $X$ $(Y)$ can yield more profit.