aa r X i v : . [ qu a n t - ph ] J u l Quantum Monty Hall problem under decoherence
Salman Khan, M. Ramzan and M. K. Khan
Department of Physics Quaid-i-Azam UniversityIslamabad 45320, Pakistan (Dated: November 23, 2018)We study the effect of decoherence on quantum Monty Hall problem under the influenceof amplitude damping, depolarizing and dephasing channels. It is shown that under theeffect of decoherence, there is a Nash equilibrium of the game in case of depolarizingchannel for Alice’s quantum strategy. Where as in case of dephasing noise, the game isnot influenced by the quantum channel. For amplitude damping channel, the Bob’s payoffsare found symmetrical with maximum at p = 0 . . However, itis worth-mentioning that in case of depolarizing channel, Bob’s classical strategy remainsalways dominant against any choice of Alice’s strategy.
Keywords: Quantum Monty Hall problem; decoherence; payoffs.
I. INTRODUCTION
In the recent past, much interest has been developed in the discipline of quantum information [1]that has led to the creation of quantum game theory [2]. Quantum game theory [3-9] has attracteda lot of attention during the last few years. The quantum Monty Hall problem [9] is an interestingexample in this realm. The quantum game theory has been shown to be experimentally feasiblethrough the application of a measurement-based protocol by Prevedel et al. [10]. They realized aquantum version of the Prisoner’s Dilemma game based on the entangled photonic cluster states.In quantum information processing, the main problem is to faithfully transmit unknown quan-tum states through a noisy quantum channel. When quantum information is sent through achannel, the carriers of the information interact with the channel and get entangled with its manydegrees of freedom. This gives rise to the phenomenon of decoherence on the state space of theinformation carriers. Quantum games in the presence of decoherence have produced interestingresults [13, 14]. Recently, we have studied the correlated noise effects in the field of quantum gametheory [15].In this paper, we study the effect of quantum decoherence on the quantum Monty Hall problem[9]. It is shown that under the effect of decoherence, a Nash equilibrium of the game exists in case ofdepolarizing channel. On the other hand, in case of amplitude damping channel, the Bob’s payoffsare found symmetrical with maximum at p = 0 . p correspondsto the decoherence parameter ranging from 0 to 1 . The lower and upper limits of p represents theundecohered and fully decohered cases respectively. It is also seen that the dephasing noise haveno influence on the game dynamics. However, it is worth-mentioning that in case of depolarizingchannel, Bob’s classical strategy becomes dominant against any choice of Alice’s strategy. II. QUANTUM MONTY HALL PROBLEM
The well-known classical Monty Hall problem was originally set in the context of a televisiongame show ”
Let’s make a deal ” hosted by Monty Hall. It is a two-person zero sum game involvinga prize (car) and three doors. In the classical version of this problem, the host of the show (Alice)hides the car behind one of the three closed doors. The guest (Bob) is asked to select one door outof the three closed doors. Alice then opens one of the two remaining doors to show that there isno prize behind it. Then Bob has the option either to stick with his current selection or to choosethe second closed door. This is the actual dilemma of the game. Classically, switching to the otherdoor increases the winning probability from one-third to two-third for Bob.A number of authors have contributed towards the quantization of Monty Hall problem [9, 11,12, 16] They have shown that quantum entanglement affects the payoffs of the players. In practiceno system can be completely isolated from its environment. Therefore, the interaction betweensystem and environment leads to the destruction of quantum coherence of the system. It producesan inevitable noise and results in the loss of information encoded in the system [17]. We proceedwith the quantization protocol of [9] and study the effect of decoherence introduced by the threeprototype channels such as amplitude damping, depolarizing and dephasing channels on the game’sdynamics.We consider that the game space is a 3-dimensional complex Hilbert space with orthonormalbasis | i , | i and | i . Such a 3-dimensional system in Hilbert space is called a qutrit. Alice and Bobstrategies are operators acting on their respective qutrits and are generally given by A = a ij and B = b ij . The open box operator O, which is an open box marking operator and not a measuringoperator, is a unitary operator that can be written as [9] O = X ijkl | ε ijk | | njk ih ljk | + X jl | mjj ih ljj | (1)where | ε ijk | = 1 , if i, j, k are all different and is 0 otherwise, m = ( j + l + 1) ∗ ( mod n = ( i + l ) ( mod O provides an option to Alice for opening anun-chosen box by Bob. The switching operator S of Bob can be written as S = X ijkl | ε ijl | | ilk ih ijk | + X ij | iij ih iij | (2)The second term in equation (2) ensures the unitarity of operator S and is irrelevant to themechanics of the game[9]. The Bob’s not-switching operator N is the identity operator I that actson the three-qutrit state. The total operator for the not-switching situation of the game can bewritten as U N = ( N sin γ ) O ( I ⊗ B ⊗ A ) (3)and if the Bob switches, the total operator becomes U S = ( S cos γ ) O ( I ⊗ B ⊗ A ) (4)where γ ∈ (cid:2) , π (cid:3) , γ = 0 , π correspond to the switching and not-switching choices of the Bobrespectively. All the summations in above equations are over the range 0 , ρ f = Φ ρ i = X k E k ρ i E † k (5)where the Kraus operators E i satisfy the following completeness relation X k E † k E k = I (6)The Kraus operators, in our case, for the game are constructed from single qutrit operators bytaking their tensor product over all n combination of π ( i ) indices E k = ⊗ π e π ( i ) (7)where n is the number of Kraus operators for a single qutrit channel. The single qutrit Krausoperators for the amplitude damping channel are given by [18] E = √ − p
00 0 √ − p , E = √ p
00 0 00 0 0 , E = √ p (8)The Kraus operator for a single qutrit for the dephasing channel are given by [18] E = p − p , E = √ p ω
00 0 ω , (9)where p corresponds to the decoherence parameter and ω = e iπ/ . The single qutrit Krausoperators for the depolarizing channel are given by [19] E = p − pI, E = r p Y, E = r p Z, E = r p Y , E = r p Y Z (10) E = r p Y Z, E = r p Y Z , E = r p Y Z , E = r p Z (11)where Y = , Z = ω
00 0 ω (12)We consider the following maximally entangled qutrit state, which is shared between Alice andBob | ψ i i = | i ⊗ √ | i + | i + | i ) (13)In equation (13) the state | i stands for the open door. The final state for the case when Bob doesnot switch to the other door becomes ρ fN = U N X k E k ρ i E † k ! U † N (14)For the case of switching to the other door, the game final state becomes ρ fS = U S X k E k ρ i E † k ! U † S (15)where ρ i = | ψ i i h ψ i | (16)Bob wins if he chooses the door behind which the prize is located. Hence the payoff of Bob is givenby h $ B i = h $ B i N + h $ B i S = X ijj (cid:16)(cid:0) ρ fN (cid:1) ijj + (cid:0) ρ fS (cid:1) ijj (cid:17) (17)The payoff of Alice is then given by h $ A i = 1 − h $ B i . III. CALCULATION, RESULTS AND DISCUSSIONS
In this section, we present the results of our calculations based on the three prototype channelssuch as amplitude damping, depolarizing and dephasing channels parametrized by the decoherenceparameter p . A. Amplitude damping channel
By using equations (8, 13 and 16), the Bob’s payoff for not-switching case can be written as h $ B i N = −
13 ( A ( − − p ) + A ( − p − p )+( − p )( A p + ( a b + a b ) a ∗ b ∗ + ( a b + a b − a b p ) a ∗ b ∗ + ( a b + a b − a b p ) a ∗ b ∗ +( a b + a b ) a ∗ b ∗ + ( a b + a b − a b p ) a ∗ b ∗ +( a b + a b − a b p ) a ∗ b ∗ +( a b + a b ) a ∗ b ∗ + ( a b + a b − a b p ) a ∗ b ∗ +( a b + a b − a b p ) a ∗ b ∗ ) sin γ (18)where the coefficients A i are given in appendix A.For the case when Bob switches to the other door, the payoff can be obtained by using equations(8, 15 and 16) which reads h $ B i S = −
13 (( − − p ) B + ( − p ) pB + ( − p ) B +( − a b a ∗ − a b a ∗ ) b ∗ − ( a b a ∗ b + a b a ∗ ) b ∗ − ( a b a ∗ + a b a ∗ ) b ∗ − ( a b a ∗ + a b a ∗ ) b ∗ − ( a b a ∗ + a b a ∗ ) b ∗ − ( a b a ∗ + a b a ∗ ) b ∗ )+( − p )( a ∗ (( a b + a b ) b ∗ + ( a b + a b ) b ∗ )+ a a ∗ ( b b ∗ + b b ∗ ) + a a ∗ ( b b ∗ + b b ∗ )+ a ∗ (( a b + a b ) b ∗ + ( a b + a b ) b ∗ )+ a ∗ (( a b + a b ) b ∗ + ( a b + a b ) b ∗ )+ a a ∗ ( b b ∗ + b b ∗ ) + a a ∗ ( b b ∗ + b b ∗ )+ a ( a ∗ ( b b ∗ + b b ∗ ) + a ∗ ( b b ∗ + b b ∗ )))) cos γ (19)Where B i ,are given in appendix A. The total payoff of Bob is the sum of equations (18 and 19).To analyze our results, we consider that let Bob has access to a classical strategy only (i.e., B = I ), therefore, he can select any door with equal probability. Then Bob’s total payoff becomes h $ B i = −
13 ( − ( | a | + | a | + | a | + | a | )( − p ) +(2 | a | + | a | + | a | + | a | + | a | + | a | + | a | )( − p ) p + ( | a | + | a | )( − − p )) cos γ − / | a | ( − − p ) + ( | a | + | a | )( − p − p )+( | a | + | a | + | a | + | a | )( − p + p )) sin γ (20)Now, Alice can make the game fair if she uses an operator whose every diagonal element has anabsolute value of √ and every off-diagonal element has an absolute value of . One such SU(3)operator is H = √ −
12 3 − i √ √ i √ √ − − i √ √ − i √
78 5+ i √ (21)The total payoff for Bob by using the above operator is obtained as h $ B i = 16 (cid:0) (3 − − p ) p ) cos γ + (3 + 2( − p ) p ) sin γ (cid:1) (22)It is easy to check that by setting p = 0 , in equation (22), the Bob’s payoff reduces to the resultsobtained in ref. [9]. For p = 0 .
5, the probability of Bob to win increases to 58 .
33% if he changesthe current selection and switch to the other door. This result for the Bob’s payoff is differentboth from the classical result (66%) and quantum mechanical result (50%) without decoherence.If Bob sticks to his current selection, his winning probability is 41 . . The dependence of Bob’s payoff on p for both switching and not-switchingcases is shown in figure 1. However, instead of classical strategy (identity operator) if Bob hasaccess to the quantum strategy too, that is, if Bob uses any one of the following operators M = M = (23)and then switches, his payoff increases. The dependence of Bob’s payoff for B = M i is shown infigure 3.Now consider the situation where Alice is restricted to a classical strategy, that is, Alice operates A = I . Then, in the undecohered case, Bob wins ($ B = 1) by using B = I, if he does not switch.However, under decoherence, Bob’s winning probability reduces to two-third for p = 0 . p = 0 . . Similarly,for p = 0 Bob wins if he uses M i ( M or M ) and then switches. Where as in the presence ofnoise, Bob’s winning probability reduces to 83 .
3% for p = 0 . p for the case when both Alice and Bob use classicalstrategies is shown in figure 2. For Alice to make the game fair (i.e. A = H ), the maximum valueof Bob’s payoff occurs if he uses either of M i and then switches. In conclusion, the Bob’s payoffsare found symmetrical with maximum at p = 0 . . B. Depolarizing channel
In case of depolarizing channel, the Bob’s payoff for not-switching case can be found by usingequations (14 and 11) as h $ B i N = − C − (8 − p ) (( a b + a b ) a ∗ b ∗ + ( a b + a b ) a ∗ b ∗ +( a b + a b ) a ∗ b ∗ + ( a b + a b ) a ∗ b ∗ + ( a b + a b ) a ∗ b ∗ +( a b + a b ) a ∗ b ∗ + ( a b + a b ) a ∗ b ∗ +( a b + a b ) a ∗ b ∗ + ( a b + a b ) a ∗ b ∗ )) sin γ (24)where C is given in appendix A. Similarly, using equations (11, 15 and 16), the Bob’s payoff forthe switching case becomes h $ B i S = 1192 ((48 p − p ) D + (64 − p + 54 p ) D +(8 − p ) (( a b + a b ) a ∗ b ∗ + ( a b + a b ) a ∗ b ∗ +( a b + a b ) a ∗ b ∗ + ( a b + a b ) a ∗ b ∗ +( a b + a b ) a ∗ b ∗ + ( a b + a b ) a ∗ b ∗ +( a b + a b ) a ∗ b ∗ + ( a b + a b ) a ∗ b ∗ +( a b + a b ) a ∗ b ∗ + ( a b + a b ) a ∗ b ∗ +( a b + a b ) a ∗ b ∗ + ( a b + a b ) a ∗ b ∗ +( a b + a b ) a ∗ b ∗ + ( a b + a b ) a ∗ b ∗ +( a b + a b ) a ∗ b ∗ + ( a b + a b ) a ∗ b ∗ +( a b + a b ) a ∗ b ∗ + ( a b + a b ) a ∗ b ∗ )) cos γ (25)where D i are given in appendix A.Bob’s total payoff is the sum of equations (24 and 25). To analyze the effect of decoherence onBob’s payoff we consider for example that Bob plays the classical strategy B = I, then Bob’s totalpayoff becomes h $ B i = 1 / − p + 54 p )( | a | + | a | + | a | + | a | + | a | + | a | ) + (48 p − p )(2 | a | + | a | + | a | + | a | + 2 | a | + | a | + | a | + | a | + 2 | a | ) cos γ − (( − p + 27 p )( | a | + | a | + | a | + | a | + | a | + | a | )+( −
64 + 96 p − p )( | a | + | a | + | a | )) sin γ ) (26)The game becomes fair if Alice uses the operator H as given in equation (21). Bob’s total payoffthen becomes h $ B i = 1128 ((64 + 3(16 − p ) p ) cos γ + (64 + 3 p ( −
16 + 9 p )) sin γ ) (27)We can see that in the absence of decoherence i.e. p = 0, our results equation (27) reduces tothe results of ref. [9]. However, in the presence of decoherence i.e. for p = 0 .
5, Bob’s winningprobability increases to 63 .
47% instead of 50% if he switches to the other door and if he sticks0to his current selection, the winning probability is just 36 . . On the other hand, for p = 0 . A = H and B = I in the presence ofdecoherence for both switching and not-switching cases are shown in figure 4.Bob’s payoffs in the presence of decoherence for B = A = I, are shown in figure 5. We seefrom figure 5 that Bob can win with a two-third probability for p = 0 . p = 0 situation, where Bob loses ( h $ B i = 0) in thecase of switching to the other door. Thus, it gives rise to the Nash equilibrium of the game underdecoherence. Further more, If Bob uses M i and then switches, his winning probability varies from1 to as p varies from 0 to 1. C. Dephasing Channel
Bob’s payoff for not-switching case in the presence of dephasing noise can be written, by usingequations (14, 16) and 9), as h $ B i N = 16 (2 E + 2(( a b + a b ) a ∗ b ∗ + ( a b + a b ) a ∗ b ∗ +( a b + a b ) a ∗ b ∗ + ( a b + a b ) a ∗ b ∗ +( a b + a b ) a ∗ b ∗ + ( a b + a b ) a ∗ b ∗ +( a b + a b ) a ∗ b ∗ + ( a b + a b ) a ∗ b ∗ +( a b + a b ) a ∗ b ∗ ) + 3( − p ) p (( a b + a b ) a ∗ b ∗ +( a b + a b ) a ∗ b ∗ + ( a b + a b ) a ∗ b ∗ +( a b + a b ) a ∗ b ∗ + ( a b + a b ) a ∗ b ∗ +( a b + a b ) a ∗ b ∗ + ( a b + a b ) a ∗ b ∗ +( a b + a b ) a b + ( a b + a b ) a ∗ b ∗ )+ √ ip ( − p )( − ( a b − a b ) a ∗ b ∗ + ( a b − a b ) a ∗ b ∗ − ( a b − a b ) a ∗ b ∗ − ( a b − a b ) a ∗ b ∗ +( a b − a b ) a ∗ b ∗ − ( a b − a b ) a ∗ b ∗ − ( a b − a b ) a ∗ b ∗ +( a b − a b ) a ∗ b ∗ − ( a b − a b a ∗ b ∗ )) sin γ (28)where E is given in appendix A. However, if Bob switches to the other door his payoff becomes1 h $ B i S = 16 (2 F + 3(1 + √ i ) p F + (3 − √ i ) p F +(2 + 2( − √ i ) p ) F + (2 − √ i ) p ) F ) cos γ (29)Where F i are given in appendix A. It is important to note here that Bob’s payoffs for not-switchingand switching cases are written in general form. Bob’s total payoff reduces to the result of ref.[9] when we set the decoherence parameter p = 0 in the general relation. Further more, wheneither Bob or Alice is restricted to a classical strategy the total Bob’s payoff becomes independentof decoherence parameter p . The same is true for the case of when Alice and Bob use quantumstrategies. Therefore, the dephasing noise does not influence the game. IV. CONCLUSIONS
We study the quantum Monty Hall problem under the influence of amplitude damping, de-polarizing and dephasing channels. A Nash equilibrium of the game exists under the effect ofdecoherence against Alice’s quantum strategy in the case of depolarizing channel. It is also seenthat the dephasing noise does not influence the game in contrary to the depolarizing and amplitudedamping channels. It is worth-mentioning that for amplitude damping and depolarizing channels,Bob’s classical strategy is superior over any choice of Alice’s strategy.
Appendix A
The coefficients A i in equation (18) are given below, A = | a | | b | + | a | | b | + | a | | b | A = | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | A = | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | (30)2The coefficients B i in equation (19) are given below, B = | a | ( | b | + | b | ) + | a | ( | b | + | b | ) + | a | ( | b | + | b | ) B = | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | ( | b | + | b | )+ | a | ( | b | + | b | ) + | a | | b | + | a | | b | + | a | | b | + | a | | b | B = | a | | b | + | a | | b | + | a | | b | + | a | ( | b | + | b | ) + | a | | b | + | a | | b | + | a | | b | + | a | ( | b | + | b | ) + | a | | b | + | a | | b | (31)The coefficient C in equation (24) is given below, C = | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | ) + ( − p + 27 p )( | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | (32)The coefficients D i in equation (25) are given below,3 D = | a | | b | + | a | | b | + | a | | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | D = | a | | b | + | a | | b | + | a | | b | ]+ | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | (33)The coefficient E in equation (28) is given below, E = | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | (34)4The coefficients F i in equation (29) are given below, F = | a | | b | + | a | | b | + | a | | b | + | a | ( | b | + | b | ) + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | ( | b | + | b | ) + | a | | b | + | a | | b | + | a | | b | + | a | | b | + | a | | b | F = ( a b a ∗ b ∗ + a b a ∗ b ∗ + a b a ∗ b ∗ + a b a ∗ b ∗ + a b a ∗ b ∗ + a b a ∗ b ∗ + a b a ∗ b ∗ + a b a ∗ b ∗ + a b a ∗ b ∗ + a b a ∗ b ∗ + a b a ∗ b ∗ + a b a ∗ b ∗ + a b a ∗ b ∗ + a b a ∗ b ∗ + a b a ∗ b ∗ + a b a ∗ b ∗ + a b a ∗ b ∗ + a b a ∗ b ∗ ) F = a a ∗ ( b b ∗ + b b ∗ ) + a a ∗ ( b b ∗ + b b ∗ )+ a a ∗ ( b b ∗ + b b ∗ ) + a a ∗ ( b b ∗ + b b ∗ )+ a a ∗ ( b b ∗ + b b ∗ ) + a a ∗ ( b b ∗ + b b ∗ )+ a a ∗ ( b b ∗ + b b ∗ ) + a a ∗ ( b b ∗ + b b ∗ )+ a a ∗ ( b b ∗ + b b ∗ ) (35) F = a a ∗ ( b b ∗ + b b ∗ ) + a a ∗ ( b b ∗ + b b ∗ )+ a a ∗ ( b b ∗ + b b ∗ ) + a a ∗ ( b b ∗ + b b ∗ )+ a a ∗ ( b b ∗ + b b ∗ ) + a a ∗ ( b b ∗ + b b ∗ )+ a a ∗ ( b b ∗ + b b ∗ ) + a a ∗ ( b b ∗ + b b ∗ )+ a a ( b b ∗ + b b ∗ ) F = a a ∗ ( b b ∗ + b b ∗ ) + a a ∗ ( b b ∗ + b b ∗ )+ a a ∗ ( b b ∗ + b b ∗ ) + a a ∗ ( b b ∗ + b b ∗ )+ a a ∗ ( b b ∗ + b b ∗ ) + a a ∗ ( b b ∗ + b b ∗ )+ a a ∗ ( b b ∗ + b b ∗ ) + a a ∗ ( b b ∗ + b b ∗ )+ a a ∗ ( b b ∗ + b b ∗ ) (36)5 [1] M. A. Nielson, I. L. Chuang, Quantum Computation and Quantum Information (Cambridge: Cam-bridge University Press), (2000).[2] Meyer, D. A., Phys. Rev. Lett.
205 (2007)[11] Zander, Claudia., et al , Annals of the Brazilian Academy of Sciences
417 (2006)[12] . D’Ariano, G.M., et al , Quantum Information and Computation
355 (2002)[13] Chen, L. K., Ang, H., Kiang, D., Kwek, L. C., Lo, C. F., Phys. Lett. A Figures Captions
Figure 1. Bob’s payoff is plotted as a function of decoherence parameter p when Alice plays H and Bobplays I for amplitude damping channel . The solid line represents the Bob’s payoff for γ = 0 and thedashed line represents the Bob’s payoff for γ = π . Figure 2. Bob’s payoff is plotted as a function of decoherence parameter p, when both Alice and Bobplay I for amplitude damping channel . The solid line represents the Bob’s payoff for γ = 0 and thedashed line represents the Bob’s payoff for γ = π . Figure 3. Bob’s payoff is plotted as a function of decoherence parameter p when Alice plays H and Bobplays M or M for amplitude damping channel . The solid line represents the Bob’s payoff for γ = 0and the dashed line represents the Bob’s payoff for γ = π . Figure 4. Bob’s payoff is plotted as a function of decoherence parameter p when Alice plays H and Bobplays I for depolarizing channel . The solid line represents the Bob’s payoff for γ = 0 and the dashedline represents the Bob’s payoff for γ = π . Figure 5. Bob’s payoff is plotted as a function of decoherence parameter p when both Alice and Bobplay I for depolarizing channel . The solid line represents the Bob’s payoff for γ = 0 and the dashedline represents the Bob’s payoff for γ = π . Figure 6. Bob’s payoff is plotted as a function of decoherence parameter p when Alice plays H and Bobplays M or M for depolarizing channel . The solid line represents the Bob’s payoff for γ = 0 and thedashed line represents the Bob’s payoff for γ = π . P a y o ff p NSW SW FIG. 1: Bob’s payoff is plotted as a function of decoherence parameter p when Alice plays H and Bob plays I for amplitude damping channel . The solid line represents the Bob’s payoff for γ = 0 and the dashed linerepresents the Bob’s payoff for γ = π2
Figure 1. Bob’s payoff is plotted as a function of decoherence parameter p when Alice plays H and Bobplays I for amplitude damping channel . The solid line represents the Bob’s payoff for γ = 0 and thedashed line represents the Bob’s payoff for γ = π . Figure 2. Bob’s payoff is plotted as a function of decoherence parameter p, when both Alice and Bobplay I for amplitude damping channel . The solid line represents the Bob’s payoff for γ = 0 and thedashed line represents the Bob’s payoff for γ = π . Figure 3. Bob’s payoff is plotted as a function of decoherence parameter p when Alice plays H and Bobplays M or M for amplitude damping channel . The solid line represents the Bob’s payoff for γ = 0and the dashed line represents the Bob’s payoff for γ = π . Figure 4. Bob’s payoff is plotted as a function of decoherence parameter p when Alice plays H and Bobplays I for depolarizing channel . The solid line represents the Bob’s payoff for γ = 0 and the dashedline represents the Bob’s payoff for γ = π . Figure 5. Bob’s payoff is plotted as a function of decoherence parameter p when both Alice and Bobplay I for depolarizing channel . The solid line represents the Bob’s payoff for γ = 0 and the dashedline represents the Bob’s payoff for γ = π . Figure 6. Bob’s payoff is plotted as a function of decoherence parameter p when Alice plays H and Bobplays M or M for depolarizing channel . The solid line represents the Bob’s payoff for γ = 0 and thedashed line represents the Bob’s payoff for γ = π . P a y o ff p NSW SW FIG. 1: Bob’s payoff is plotted as a function of decoherence parameter p when Alice plays H and Bob plays I for amplitude damping channel . The solid line represents the Bob’s payoff for γ = 0 and the dashed linerepresents the Bob’s payoff for γ = π2 . P a y o ff p NSW SW FIG. 2: Bob’s payoff is plotted as a function of decoherence parameter p, when both Alice and Bob play I , for amplitude damping channel . The solid line represents the Bob’s payoff for γ = 0 and the dashed linerepresents the Bob’s payoff for γ = π2
Figure 1. Bob’s payoff is plotted as a function of decoherence parameter p when Alice plays H and Bobplays I for amplitude damping channel . The solid line represents the Bob’s payoff for γ = 0 and thedashed line represents the Bob’s payoff for γ = π . Figure 2. Bob’s payoff is plotted as a function of decoherence parameter p, when both Alice and Bobplay I for amplitude damping channel . The solid line represents the Bob’s payoff for γ = 0 and thedashed line represents the Bob’s payoff for γ = π . Figure 3. Bob’s payoff is plotted as a function of decoherence parameter p when Alice plays H and Bobplays M or M for amplitude damping channel . The solid line represents the Bob’s payoff for γ = 0and the dashed line represents the Bob’s payoff for γ = π . Figure 4. Bob’s payoff is plotted as a function of decoherence parameter p when Alice plays H and Bobplays I for depolarizing channel . The solid line represents the Bob’s payoff for γ = 0 and the dashedline represents the Bob’s payoff for γ = π . Figure 5. Bob’s payoff is plotted as a function of decoherence parameter p when both Alice and Bobplay I for depolarizing channel . The solid line represents the Bob’s payoff for γ = 0 and the dashedline represents the Bob’s payoff for γ = π . Figure 6. Bob’s payoff is plotted as a function of decoherence parameter p when Alice plays H and Bobplays M or M for depolarizing channel . The solid line represents the Bob’s payoff for γ = 0 and thedashed line represents the Bob’s payoff for γ = π . P a y o ff p NSW SW FIG. 1: Bob’s payoff is plotted as a function of decoherence parameter p when Alice plays H and Bob plays I for amplitude damping channel . The solid line represents the Bob’s payoff for γ = 0 and the dashed linerepresents the Bob’s payoff for γ = π2 . P a y o ff p NSW SW FIG. 2: Bob’s payoff is plotted as a function of decoherence parameter p, when both Alice and Bob play I , for amplitude damping channel . The solid line represents the Bob’s payoff for γ = 0 and the dashed linerepresents the Bob’s payoff for γ = π2 . P a y o ff p NSW SW FIG. 3: Bob’s payoff is plotted as a function of decoherence parameter p when Alice plays H and Bob plays M or M for amplitude damping channel . The solid line represents the Bob’s payoff for γ = 0 and thedashed line represents the Bob’s payoff for γ = π . P a y o ff p NSW SW FIG. 4: Bob’s payoff is plotted as a function of decoherence parameter p when Alice plays H and Bob plays I for depolarizing channel . The solid line represents the Bob’s payoff for γ = 0 and the dashed line representsthe Bob’s payoff for γ = π . P a y o ff p NSW SW FIG. 5: Bob’s payoff is plotted as a function of decoherence parameter p when both Alice and Bob play I for depolarizing channel . The solid line represents the Bob’s payoff for γ = 0 and the dashed line representsthe Bob’s payoff for γ = π . P a y o ff p NSW SW FIG. 6: Bob’s payoff is plotted as a function of decoherence parameter p when Alice plays H and Bob plays M or M for depolarizing channel . The solid line represents the Bob’s payoff for γ = 0 and the dashed linerepresents the Bob’s payoff for γ = π2