Playing distributed two-party quantum games on quantum networks

Abstract

This paper investigates quantum games between two remote players on quantum networks. We propose two schemes for distributed remote quantum games: the client–server scheme based on states transmission between nodes of the network and the peer-to-peer scheme devised upon remote quantum operations. Following these schemes, we construct two designs of the distributed prisoners’ dilemma game on quantum entangling networks, where concrete methods are employed for teleportation and nonlocal two-qubits unitary gates, respectively. It seems to us that the requirement for playing distributed quantum games on networks is still an open problem. We explore this problem by comparing and characterizing the two schemes from the viewpoints of network structures, quantum and classical operations, experimental realization and simplification.

Appendices

Appendix A: Game process of the example based on the client–server scheme

• 1. The referee of the game applies quantum gate $J(\gamma )=\exp (-i\gamma \sigma _y\otimes \sigma _y )$ on the qubits CA1 and CB1. This operation can transform the tensor product state $|0_{CA1}\rangle \otimes |0_{CB1}\rangle$ into an entangled state

  $$\begin{aligned} |\psi _{in}\rangle =J(\gamma ) \left( |0_{CA1}\rangle \otimes |0_{CB1} \rangle \right) = \cos \frac{\gamma }{2} |0_{CA1}0_{CB1}\rangle + i \sin \frac{\gamma }{2} |1_{CA1}1_{CB1}\rangle .\nonumber \\ \end{aligned}$$

  (1)

• 2. The referee teleports quantum state $|\psi _{in}\rangle$ to the players with the help of two ancillary EPR pairs, $|\psi _{A1}\rangle =\frac{1}{\sqrt{2}} \left( |0_{A}0_{A1}\rangle +|1_{A}1_{A1}\rangle \right)$ and $|\psi _{B1}\rangle =\frac{1}{\sqrt{2}} \left( |0_{B}0_{B1}\rangle +|1_{B}1_{B1}\rangle \right)$. The referee conducts CNOT gates on qubit pairs (CA1, A1) and (CB1, B1) and then performs projective measurements on them. Through classical channels, the measurement results (l, k) and (m, n) are sent to Alice and Bob, respectively. With these results, Alice performs ${\sigma _z}^k_A{\sigma _x}^l_A$ on qubit A and Bob performs ${\sigma _z}^n_B{\sigma _x}^m_B$ on qubit B. After the teleportation process, the system will be in state

  $$\begin{aligned} |\psi _{CT}\rangle= & {} |\psi _{T}\rangle \otimes |k_{CA1}l_{A1}\rangle \otimes | m_{CB1}n_{B1}\rangle \nonumber \\= & {} \left( \cos \frac{\gamma }{2} |0_A0_B\rangle + i \sin \frac{\gamma }{2} |1_A1_B\rangle \right) \otimes |k_{CA1}l_{A1}\rangle \otimes |m_{CB1}n_{B1}\rangle . \end{aligned}$$

  (2)

  where k, l, m, $n=0$, 1.

• 3. According to their strategies, Alice and Bob apply local operators $U_A$ and $U_B$ to qubits A and B, respectively. Corresponding to the local quantum prisoners’ dilemma game, the strategy space of players is restricted to the 2-parameter set of unitary $2\times 2$ matrices, $U_A=U_P (\theta _A,\phi _A )$ and $U_B=U_P (\theta _B,\phi _B )$, where

  $$\begin{aligned} U_P(\theta , \phi ) = \left[ \begin{array}{cc} e^{i\phi }\cos \frac{\theta }{2} &{} \sin \frac{\theta }{2} \\ -\sin \frac{\theta }{2} &{} e^{-i\phi }\cos \frac{\theta }{2} \\ \end{array} \right] . \end{aligned}$$

  (3)

  Hence the final state of qubits A and B in this step can be denoted as follows

  $$\begin{aligned} |\psi _{S}\rangle= & {} (U_A \otimes U_B) |\psi _T\rangle \nonumber \\= & {} a|0_A0_B\rangle +b|0_A1_B\rangle +c|1_A0_B\rangle +d|1_A1_B\rangle , \end{aligned}$$

  (4)

  where

  $$\begin{aligned} a=&e^{i(\phi _A+\phi _B)}\cos \frac{\gamma }{2} \cos \frac{\theta _A}{2} \cos \frac{\theta _B}{2} +i \sin \frac{\gamma }{2} \sin \frac{\theta _A}{2} \sin \frac{\theta _B}{2}, \nonumber \\ b=&-e^{i\phi _A}\cos \frac{\gamma }{2} \cos \frac{\theta _A}{2} \sin \frac{\theta _B}{2} +i e^{-i\phi _B} \sin \frac{\gamma }{2} \sin \frac{\theta _A}{2} \cos \frac{\theta _B}{2}, \nonumber \\ c=&-e^{i\phi _B}\cos \frac{\gamma }{2} \sin \frac{\theta _A}{2} \cos \frac{\theta _B}{2} +i e^{-i\phi _A} \sin \frac{\gamma }{2} \cos \frac{\theta _A}{2} \sin \frac{\theta _B}{2}, \nonumber \\ d=&\cos \frac{\gamma }{2} \sin \frac{\theta _A}{2} \sin \frac{\theta _B}{2} +i e^{-i(\phi _A+\phi _B)}\sin \frac{\gamma }{2} \cos \frac{\theta _A}{2} \cos \frac{\theta _B}{2}. \end{aligned}$$

  (5)

• 4. Players teleport the states of qubits A and B back to the local qubits of the referee, CA2 and CB2. The procedure is similar to what we introduced in step 2, and the ancillary EPR pairs are now $|\psi _{A2}\rangle = \frac{1}{\sqrt{2}} ( |0_{CA2}0_{A2}\rangle + |1_{CA2}1_{A2}\rangle )$ and $|\psi _{B2}\rangle = \frac{1}{\sqrt{2}} ( |0_{CB2}0_{B2}\rangle + |1_{CB2}1_{B2}\rangle )$, respectively. Qubits CA2 and CB2 are turned into the following state:

  $$\begin{aligned} |\psi _{T2}\rangle =&a|0_{CA2}0_{CB2}\rangle +b|0_{CA2}1_{CB2}\rangle +c|1_{CA2}0_{CB2}\rangle +d|1_{CA2}1_{CB2}\rangle . \end{aligned}$$

  (6)

• 5. The referee applies the conjugate operator of $J(\gamma )$ to the entangled qubits CA2 and CB2 as the postoperation, which leads to the final state

  $$\begin{aligned} |\psi _{fin}\rangle= & {} \cos \frac{\gamma }{2} ( a|0_{CA2}0_{CB2}\rangle +b|0_{CA2}1_{CB2}\rangle +c|1_{CA2}0_{CB2}\rangle +d|1_{CA2}1_{CB2}\rangle ) \nonumber \\&- i \sin \frac{\gamma }{2}( a|1_{CA2}1_{CB2}\rangle -b|1_{CA2}0_{CB2}\rangle -c|0_{CA2}1_{CB2}\rangle +d|0_{CA2}1_{CB2}\rangle ).\nonumber \\ \end{aligned}$$

  (7)

• 6. By measuring $|\psi _{fin}\rangle$, the referee calculates payoffs of players from the measurement result according to the payoff matrix of the game.

Appendix B: Game process of the example based on the peer-to-peer scheme

• 1. With an ancillary EPR pair of state $|\psi _1\rangle$, Alice and Bob conduct a nonlocal operation $J(\gamma )=\exp \left( -i\gamma {\sigma _y}_A \otimes {\sigma _y}_B \right)$ on the tensor product state $|0_A\rangle \otimes |0_B\rangle$ of qubits A and B. In this step, Alice should successively perform on her local qubits two controlled unitary operators, $U_{YA1}=I_A \otimes |0\rangle _{A1}\langle 0|_{A1} + {\sigma _y}_A \otimes |1\rangle _{A1}\langle 1|_{A1}$ and $U_{XA1}=I_{A1} \otimes |0\rangle _{A}\langle 0|_{A} + {\sigma _x}_{A1} \otimes |1\rangle _{A}\langle 1|_{A}$. On the other hand, Bob firstly applies the controlled unitary operator $U_{YB1}=I_B \otimes |0\rangle _{B1}\langle 0|_{B1} + {\sigma _y}_B \otimes |1\rangle _{B1}\langle 1|_{B1}$ on qubits B and B1 and then conducts the transformation $U_S(\gamma )=i\sin {\frac{\gamma }{2}}\sigma _y+i\cos {\frac{\gamma }{2}}\sigma _z$ on qubit B1. Subsequently, he measures qubit B1 with orthogonal bases ${|f_{B1}\rangle }$, $f=0,1$ and tells Alice the measurement result f via a classical channel. With operations ${\sigma _x}_A^f \otimes {\sigma _x}_B^f$, players can turn $|\psi _{N1}\rangle$ into state

  $$\begin{aligned} |\psi _{Cin}\rangle =|\psi _{in}\rangle \otimes |0_{A1}f_{B1}\rangle = \left( \cos \frac{\gamma }{2}|0_A0_B\rangle + i \sin \frac{\gamma }{2}|1_A1_B\rangle \right) \otimes |0_{A1}f_{B1}\rangle .\nonumber \\ \end{aligned}$$

  (8)

• 2. Alice and Bob apply strategy operators $U_A$ and $U_B$ to local qubits A and B, respectively, with $U_A=U_P (\theta _A,\phi _A )$ and $U_B=U_P (\theta _B,\phi _B )$. The final state of qubits A and B can be denoted by $|\psi _{S}\rangle$ in Eq. (4).

• 3. Once more, with the ancillary EPR pair provided by the quantum network, Alice and Bob apply $J^\dag (\gamma )=\exp \left( i\gamma {\sigma _y}_A \otimes {\sigma _y}_B \right)$ to the entangled qubits A and B through a nonlocal quantum gate. Here the process is similar to the first step, although the controlled $\sigma _x$ gate on qubit A1 is now replaced by a Hadamard gate followed by measurement on A2 and a $\sigma _z^g$ gate on B2 performed according to the measurement result g. After the process, the state of qubits A and B becomes

  $$\begin{aligned} |\psi _{fin}\rangle= & {} \cos \frac{\gamma }{2} ( a|0_A0_B\rangle +b|0_A1_B\rangle +c|1_A0_B\rangle +d|1_A1_B\rangle ) \nonumber \\&- i \sin \frac{\gamma }{2}( a|0_A0_B\rangle -b|0_A1_B\rangle -c|1_A0_B\rangle +d|1_A1_B\rangle ). \end{aligned}$$

  (9)

• 4. Alice and Bob measure their qubit A and B in the orthogonal bases $|0_A\rangle$, $|1_A\rangle$ and $|0_B\rangle$, $|1_B\rangle$, respectively. They send the results i and j to the referee of the game.

• 5. The referee calculates players’ payoffs from the measurement results according to the payoff matrix of the game.