Let us consider the three-person Prisoner’s Dilemma that was studied in the quantum domain (via the EWL scheme) by Du et al. [
15]. In terms of matrices the game is defined as follows:
Here, players 1 and 2 choose between the rows and the columns, respectively, whereas player 3 chooses between the matrices. We recall that the only Nash equilibrium in (
27) is a profile consisting of the players’ second strategies. Thus, the most reasonable result of the game is
\((1,1,1)\). Similar to the best-known 2-person Prisoner’s Dilemma, the players would increase their payoffs if at least two of them played their first strategies. However, the first strategy cannot be played by a rational player since for each profile of the opponents’ strategies this strategy always yields a worse payoff than the second strategy. In what follows, we apply scheme (
22)–(
24) to game (
27). According to the reasoning used immediately before Example
1, we identify each player’s strategies in game (
27) with local operators
\(U_{0}\) and
\(U_{1}\). Moreover, let us assume that player
\(i\),
\(i=1,2,3\) acts on the system of
\(i\)th and
\((i+3)\)th qubit. As a result, scheme (
22)–(
24) comes down to one defined on
\((\mathbb {C}^{2})^{\otimes 3} \otimes (\mathbb {C}^2)^{\otimes 3}\) with the positive operator
$$\begin{aligned} H = \left( {\mathbb {1}}^{\otimes 3} - |111\rangle \langle 111|\right) \otimes |000\rangle \langle 000| + |111\rangle \langle 111| \otimes |\varPsi \rangle \langle \varPsi |, \end{aligned}$$
(28)
the player
\(i\)’s strategy set
$$\begin{aligned} \left\{ P^{(i)}_{0}\otimes U^{(i+3)}_{0}, P^{(i)}_{0}\otimes U^{(i+3)}_{1}, P^{(i)}_{1}\otimes U^{(i+3)}_{0}, P^{(i)}_{1}\otimes U^{(i+3)}_{1}\right\} , \end{aligned}$$
(29)
and the triple of payoff operators
$$\begin{aligned} (M_{1}, M_{2}, M_{3})= & {} {\mathbb {1}}^{\otimes 3}\otimes \bigl [(3,3,3)|000\rangle \langle 000| + (2,2,5)|001\rangle \langle 001| \nonumber \\&+ (2,5,2)|010\rangle \langle 010|+ (0,4,4)|011\rangle \langle 011| + (5,2,2)|100\rangle \langle 100| \nonumber \\&+ (4,0,4)|101\rangle \langle 101|+ (4,4,0)|110\rangle \langle 110| + (1,1,1)|111\rangle \langle 111|\bigr ].\nonumber \\ \end{aligned}$$
(30)
Let us fix now the players’ joint strategy
\(|\varPsi \rangle \) as:
$$\begin{aligned} |\varPsi \rangle = \frac{1}{2}\left( |001\rangle + |010\rangle + |100\rangle + |111\rangle \right) \end{aligned}$$
(31)
and determine the resulting players’ payoffs that correspond to profiles
$$\begin{aligned} \bigotimes ^{3}_{k=1}P^{(k)}_{j_{k}}\otimes \bigotimes ^6_{k=4}U^{(k)}_{j_{k}},~j_{k} \in \{0,1\}. \end{aligned}$$
(32)
Note that for fixed
\(\bigotimes ^6_{k=4} U^{(k)}_{j_{k}}\), the value
$$\begin{aligned} \mathrm{tr}{\left[ \left( \bigotimes ^{3}_{k=1}P^{(k)}_{j_{k}} \otimes \bigotimes ^6_{k=4} U^{(k)}_{j_{k}}\right) H \left( \bigotimes ^{3}_{k=1}P^{(k)}_{j_{k}} \otimes \bigotimes ^6_{k=4} U^{(k)}_{j_{k}}\right) M_{i}\right] },~i=1,2,3 \end{aligned}$$
(33)
is the same for each
\(\bigotimes ^{3}_{k=1}P^{(k)}_{j_{k}} \ne |111\rangle \langle 111|\). Therefore, the problem of determining all the 64 payoff profiles actually reduces to determining
\(64 - 6\cdot 8 = 16\) of them. For example,
$$\begin{aligned}&\left( P^{(1)}_{1}\otimes P^{(2)}_{0} \otimes P^{(3)}_{0} \otimes U^{(4)}_{0}\otimes U^{(5)}_{0} \otimes U^{(6)}_{1}\right) \nonumber \\&\quad H\left( P^{(1)}_{1}\otimes P^{(2)}_{0} \otimes P^{(3)}_{0} \otimes U^{(4)}_{0}\otimes U^{(5)}_{0} \otimes U^{(6)}_{1}\right) \nonumber \\&\quad = |100\rangle \langle 100|\otimes \left( U^{(4)}_{0} \otimes U^{(5)}_{0} \otimes U^{(6)}_{1}\right) |000\rangle \langle 000|\left( U^{(4)}_{0} \otimes U^{(5)}_{0} \otimes U^{(6)}_{1}\right) \nonumber \\&\quad = |100\rangle \langle 100| \otimes |001\rangle \langle 001|. \end{aligned}$$
(34)
Then,
$$\begin{aligned} \mathrm{tr}{\left( |100\rangle \langle 100| \otimes |001\rangle \langle 001|M_{i}\right) } = {\left\{ \begin{array}{ll}2 &{}\hbox {if }i\in \{1,2\} \\ 5 &{}\hbox {if }i=3.\end{array}\right. } \end{aligned}$$
(35)
Hence, we obtain the same payoffs if
\(P^{(1)}_{1}\otimes P^{(2)}_{0} \otimes P^{(3)}_{0}\) is replaced by
\(P^{(1)}_{j_{1}}\otimes P^{(2)}_{j_{2}} \otimes P^{(3)}_{j_{3}} \ne |111\rangle \langle 111|\). For case
\(P^{(1)}_{1}\otimes P^{(2)}_{1} \otimes P^{(3)}_{1}\), we have
$$\begin{aligned}&\left( P^{(1)}_{1}\otimes P^{(2)}_{1} \otimes P^{(3)}_{1} \otimes U^{(4)}_{0}\otimes U^{(5)}_{0} \otimes U^{(6)}_{1}\right) \nonumber \\&\quad H\left( P^{(1)}_{1}\otimes P^{(2)}_{1} \otimes P^{(3)}_{1} \otimes U^{(4)}_{0}\otimes U^{(5)}_{0} \otimes U^{(6)}_{1}\right) \nonumber \\&\quad = |111\rangle \langle 111| \otimes \left( U^{(4)}_{0}\otimes U^{(5)}_{0} \otimes U^{(6)}_{1}\right) |\varPsi \rangle \langle \varPsi |\left( U^{(4)}_{0}\otimes U^{(5)}_{0} \otimes U^{(6)}_{1}\right) \nonumber \\&\quad = |111\rangle \langle 111|\otimes \left( |\varPsi '\rangle \langle \varPsi '|\right) \end{aligned}$$
(36)
where
\(|\varPsi '\rangle = (|000\rangle + |011\rangle + |101\rangle + |110\rangle )/2\). State (
36) implies the payoff
$$\begin{aligned} \mathrm{tr}{\left[ |111\rangle \langle 111|\otimes \left( U^{(4)}_{0}\otimes U^{(5)}_{0} \otimes U^{(6)}_{1}\right) |\varPsi \rangle \langle \varPsi |\left( U^{(4)}_{0}\otimes U^{(5)}_{0} \otimes U^{(6)}_{1}\right) M_{i}\right] } = \frac{11}{4}\nonumber \\ \end{aligned}$$
(37)
for each
\(i=1,2,3\). Having determined the payoffs associated with each strategy profile, we can describe the game given by scheme (
28)–(
30) with the use of four matrices
We see from the matrix representation that there are two types of pure Nash equilibria. The first one corresponds to the unique equilibrium in game (
27), and it is generated by profiles
$$\begin{aligned} \bigotimes ^3_{k=1}P^{(k)}_{j_{k}} \otimes \bigotimes ^{6}_{k=4} U^{(k)}_{1}~~\text{ where }~~(j_{1},j_{2},j_{3}) \in \{(0,0,0), (1,0,0), (0,1,0), (0,0,1)\}. \end{aligned}$$
(38)
Each profile of (
38) is a Nash equilibrium since each player’s unilateral deviation from the equilibrium strategy yields the payoff 0 or 1. It also follows from the construction of (
22)–(
24). Namely, if a player cannot cause the joint strategy
\(|\varPsi \rangle \) to be played by changing her own strategy, the equilibrium analysis is restricted to studying the local operations on state
\(|000\rangle \). That, in turn, coincides with the problem of finding Nash equilibria in game (
27), and
\( \bigotimes ^{6}_{k=4}U^{(k)}_{1}\) is just the counterpart of the profile of the players’ second strategies that forms the unique equilibrium in (
27). However, in contrast to (
27), the quantum game has another equilibrium given by profile
$$\begin{aligned} \bigotimes ^3_{k=1}P^{(k)}_{1} \otimes \bigotimes ^6_{k=4}U^{(k)}_{1}. \end{aligned}$$
(39)
Indeed, player 1 suffers a loss of at least 1/4 by unilateral deviation from strategy
\(P^{(1)}_{1}\otimes U^{(4)}_{1}\) and the same occurs in the case of players 2 and 3. Profile (
39) is more profitable than (
38) since it implies 11/4 for each player instead of 1. Thus, the players gain by making use of the joint strategy
\(|\varPsi \rangle \), i.e., by playing
\(\bigotimes ^3_{k=1}P^{(k)}_{1}\).