quantum duel revisited

Quantum duel revisited

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J. Phys. A: Math. Theor. 45 (2012) 125304 (11pp) doi:10.1088/1751-8113/45/12/125304

Quantum duel revisited

Alexandre G M Schmidt1,3 and Milena M Paiva2

1 Departamento de Fısica, Instituto de Ciencias Exatas, Universidade Federal Fluminense,R. Des. Ellis Hermydio Figueira 783, Volta Redonda RJ, CEP 27215-350, Brazil2 Departamento de Engenharia de Producao, Escola de Engenharia Industrial Metalurgica,Universidade Federal Fluminense, Av. dos Trabalhadores 420, Volta Redonda RJ,CEP 27255-125, Brazil

E-mail: [email protected] and [email protected]

Received 24 October 2011, in final form 13 February 2012Published 6 March 2012Online at stacks.iop.org/JPhysA/45/125304

AbstractWe revisit the quantum two-person duel. In this problem, both Alice and Bobeach possess a spin-1/2 particle which models dead and alive states for eachplayer. We review the Abbott and Flitney result—now considering non-zeroα1 and α2 in order to decide if it is better for Alice to shoot or not the secondtime—and we also consider a duel where players do not necessarily start alive.This simple assumption allows us to explore several interesting special cases,namely how a dead player can win the duel shooting just once, or how can Bobrevive Alice after one shot, and the better strategy for Alice—being either aliveor in a superposition of alive and dead states—fighting a dead opponent.

PACS numbers: 03.65.−w, 03.65.Ge, 73.21.Fg.

(Some figures may appear in colour only in the online journal)

1. Introduction

Game theory deals with conflict situations between two or more so-called players, for instance,it can model a simple coin toss game, what dish to order for dinner (in the sense of diner’sdilemma), or a couple choosing to go to football game or to the theatre (battle of sexes). Inthe last decade, researchers put forward the idea of bringing quantum mechanics into thisscenario in order to create quantum games, i.e. games—or conflict problems—where actionscould be modelled by quantum operators defined in Hilbert space. For this reason, quantumlinear superposition, interference and entanglement can take place in these games.

Several classical games were represented in terms of quantum mechanics, as we can readin the very good review [1]. Meyer [2] introduced the PQ-game, a coin toss game where theplayer P has only the classical moves of flipping and not-flipping a coin, and his opponent Qhas these two moves as well as a quantum strategy of simultaneously flipping and not-flipping

3 Author to whom any correspondence should be addressed.

1751-8113/12/125304+11$33.00 © 2012 IOP Publishing Ltd Printed in the UK & the USA 1

http://dx.doi.org/10.1088/1751-8113/45/12/125304

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J. Phys. A: Math. Theor. 45 (2012) 125304 A G M Schmidt and M M Paiva

the coin. It is well known that Q can always win the game. Eisert, Wilkens and Lewenstein[3] introduced the quantum version of the famous prisoner dilemma, where two players, whocannot communicate with each other, must decide if they cooperate or defect. Experimentalrealization of this game was accomplished by Du et al [4]. Minority game [5], which isvery important in economy applications, also has a quantum version presented by Benjaminand Hayden [6, 7]; it is related to Bell inequalities [8] and was realized experimentally forthe case of four players by Schmid et al [9]. Quantum auctions [10], quantum gambling[11], quantum Monty–Hall game [12], quantum ultimate game [13], quantum Parrondo game[14, 15] and even quantum duels [16] were also studied. One can rightfully ask: what newscan quantum mechanics tell us about strategies? These quantum strategies—i.e. actions thathave no classical counter-part, such as choosing three doors in the Monty–Hall problem ortossing a coin to both heads and tails in PQ-game, or as we will present in this work, howa gunfighter who started a duel in a dead state can shoot and even win the duel with 100%probability—can provide us new ways of thinking about conflict situations and to furtherincrease our knowledge about quantum dynamics, both theoretically and experimentally, aswell as provide some insights about how to create new quantum algorithms.

The main objective of this paper is to study the two-person quantum duel, first proposedin [16] by Flitney and Abbott. The authors analysed the quantum duel considering the playersAlice and Bob both having a spin-1/2 particle which represents their states alive and dead. Theplayers’ objective in this game is to flip the spin of his/her rival, maintaining his/her own in thealive state. However, the operator, which represents the action of shooting in the opponent, inorder to be unitary, must contain the possibility of reviving or resurrecting him/her; this is anextra ingredient of this game if we compare it to the classical counterpart. The game setup isas follows: we consider an initial state |ψ〉 = |alice〉 ⊗ |bob〉, and then the players shoot eachother, first Alice, then Bob and so on. The state |0〉 represents a dead player, |1〉 an alive playerand the action of shooting means, in Hilbert space, to apply a certain operator to the qubit |ψ〉.Since probabilities are calculated only at the end of the duel, a dead player is able to shoot,and a living player can revive the other. We also consider that players act only according totheir self-interest.

Our main original results are the analysis of this two-person duel considering the initialstate to be a superposition of |0〉 and |1〉, which was not done in [16], as well as a correctionof a previous result where it was stated that when Alice was not a good shooter and she wasfacing a good one, then she should have wasted her second bullet to prevent Bob’s revival.We show that wasting the second shot is not always a good strategy for Alice. The outline forour paper is as follows: in section 2, we review the problem of quantum duel; in section 3, weanalyse the duel when the initial state is not |11〉; in section 4, we briefly discuss an exampleof a quantum truel; and in section 5, we conclude the work.

2. Quantum duel

The quantum duel can be formulated using a two-level system, where the players Alice andBob fight with each other. The states |0〉 and |1〉 represent dead and alive, respectively. Thus,the state of our system is written as the product

|ψ〉 = |alice〉 ⊗ |bob〉, (1)

and in this four-dimensional Hilbert space, one has the basis {|11〉, |10〉, |01〉, |00〉}. The gunduel proceeds with the players shooting: first Alice, then Bob and so on. The initial state is|11〉, i.e. both players begin alive, however, in the next section we explore the duel when one orboth players begin in a superposition of dead and alive states. In order to calculate the payoff,

2


second time,

|ψ2〉 =[

F1e−iα2 cos

(θ2

2

)+ iF2 e−iβ2 sin

(θ2

2

)]|11〉 + F3|10〉

+[

iF1 eiβ2 sin

(θ2

2

)+ F2 eiα2 cos

(θ2

2

)]|01〉, (8)

where we define the factors

F1 = e−i(2α1+α2 ) cos2

(θ1

2

)cos

(θ2

2

)− sin2

(θ1

2

)(9)

and

F2 = iei(β2−α1 ) cos

(θ1

2

)sin

(θ2

2

), (10)

and

F3 = ieiβ1 sin

(θ1

2

)cos

(θ1

2

) [e−i(α1+α2 ) cos

(θ2

2

)+ eiα1

]. (11)

Now we analyse the probability of Alice winning the duel, namely we calculate

|〈10|ψ2〉|2 = a(1 − a)[1 + b + 2√

b cos(2α1 + α2)], (12)

where the parameters a = cos2(θ1/2) and b = cos2(θ2/2). In order to write the payofffunction, we need to calculate also the probability of Bob being alive,

|〈01|ψ2〉|2 = (1 − b)[ab(1 + a) + (1 − a2) + 2ab√

a cos(α1 + 2α2) − 2a(1 − a)√

b

× cos(2α1 + α2) − 2√

ab(1 − a) cos(α1 − α2)]. (13)

Using equations (12) and (13), we can write the function that defines Alice’s payoff,usually represented by 〈$A〉,

〈$A〉(a, b) = 12 (1 + |〈10|ψ2〉|2 − |〈01|ψ2〉|2 − |〈00|ψ2〉|2); (14)

substituting the above equations (12), (13) and |〈00|ψ2〉|2 = 0 into (14), we obtain Alice’spayoff after two rounds,

〈$A〉(a, b, α1, α2) = 12 {a(3 − 2a) − ab(2 + a − b − ab)

+ 2(1 − a)√

ab[(1 − b) cos(α1 − α2)

+√a(2 − b) cos(2α1 + α2)] − 2a3/2b(1 − b) cos(α1 + 2α2)}. (15)

The first interesting result we obtain occurs when both α1 = α2 = 0; Alice’s payoff becomesminimum when a = b = 0, that is, when both players are excellent shooters Bob alwayswins the duel since 〈$A〉(0, 0, 0, 0) = 0; the maximum can be unity if a = 1/2 and b = 1. Ifα1 = α2, then Alice’s payoff becomes equal to unity if a = 1/2, b = 1 and α2 = 2.0944, andthe minimum occurs for a = 1, b = 1/2 and α2 = 0.

2.3. Alice changes strategy

The well-known result of Abbott and Flitney [16] claims that if Alice has a poormarksmanship—and Bob has an intermediary one—it would be better for her to avoid shootingthe second time, because she could revive Bob. In order to analyse that argumentation, let uscalculate |ψ ′

2〉 = (BABAAB)|11〉. In order to do so, we simply apply (3) to (6) and obtain

|ψ ′2〉 = √

a e−iα1 [b(1 + e−2iα2 ) − 1]|11〉 + i eiβ1√

1 − a|10〉+ 2i

√ab(1 − b) ei(β2−α1 ) cos(α2)|01〉; (16)

4


Figure 1. Contour plot of⟨$air

A

⟩as a function of a, b, α2, given by equation (19). Observe that it

has a minimum equal to zero and a maximum equal to unity, and they depend on all variables.The colour scale begins from zero (blue or rightmost surface) and ends at unity (red or leftmostsurface); the intermediate colours represent (from right to left) 0.25, 0.50 and 0.75.

the referred authors analyse only the case where α1 = α2 = 0, and we propose to show thatthe present strategy cannot be good because it depends on particular values of α1 and α2. Theprobability for Alice remaining alive is written as

|〈10|ψ ′2〉|2 = 1 − a, (17)

as can be straightforwardly calculated. However, the second amplitude, namely the one forBob alive, has a misprint in [16],

|〈01|ψ ′2〉|2 = 2ab(1 − b)(1 + cos 2α2); (18)

we claim that the correct amplitude is this one, with (1 + cos 2α2) instead of equation (13) of[16]. We checked it performing the same calculation using Mathematica 8.0.1. Now the newpayoff function is given by 1

2 (1 + |〈10|ψ ′2〉|2 − |〈01|ψ ′

2〉|2 − |〈00|ψ ′2〉|2), namely

〈$airA 〉(a, b, α2) = 1 − a

2− 2ab(1 − b) cos2 α2. (19)

In figure 1, we present a contour plot of⟨$air

A

⟩as a function of (a, b, α2). The surface (red) on

the left is the one where the function reaches the maximum, and we observe that a high payoff(orange surface for instance) can be obtained if Alice has a � 0.50, depending on the α2 value.For instance, if α2 = π/2, then her payoff is independent of b and reduces to

⟨$air

A

⟩ = 1 − a/2,even if she is a poor shooter, e.g. a = 4/5, her payoff can be as good as 0.60. On the otherhand, if α2 = 0 or ±π , then

⟨$air

A

⟩has a maximum equal to 0.60 (when b = 0 or b = 1) and a

minimum of 0.20 for b = 0.50.In fact, the authors of [16] calculate and plot the difference between strategies,

⟨$air

A

⟩−〈$A〉,and this is a function of both α1 and α2, and as such can have several minima and maxima.The idea is to analyse this function in order to let Alice choose, considering the specificparameters a and b, if she will shoot or not the second time. The function to be analysed is⟨$diff

A

⟩ = ⟨$air

A

⟩ − 〈$A〉, given by (19)–(14). The first special case is the one studied in [16], i.e.for α1 = α2 = 0 the difference between payoffs has a maximum equal to unity if a = b = 0,and a minimum equal to −0.378 when a = b = 0.61. So, if both players are excellent

5


Figure 2. Left: contour plot of⟨$diff

A

⟩ = ⟨$air

A

⟩ − 〈$A〉, as a function of a, b, α, where α1 = α2 = α.Observe that when this function becomes negative, it is a better strategy for Alice to shoot asecond time. On the other hand, the red surface (frontmost first surface) shows the range of values(a, b, α) for which such a difference approaches the maximum. The intermediate colours represent0.15 (third surface) and 0.48 (second surface). Right: plot of the same function for α = π . Themaximum takes place for a = b = 0.

shooters, Alice should waste her second shot in order to win the duel. When both playershave poor marksmanship, the best strategy for Alice is not firing into the air. On the otherhand, the second special case comes from α1 = α2 = α, where the difference

⟨$diff

A

⟩has a

maximum equal to unity for a = b = 0 and arbitrary α, and a minimum equal to −0.28 whena = 0.625, b = 1, α = 2.09. We present a contour plot of

⟨$diff

A

⟩in figure 2 where we can

observe, in red, the surface where this difference is close to unity. The third special case, namelyα = α1 = α2, has the same maximum equal to unity and negative minimum equal to −0.28;however, they take place for a = b = 0, α = −2.74 and a = 0.625, b = 1, α = −0.78,respectively.

3. Dead players’ duel

Let us explore what would happen if both players could start the duel in a superposition ofalive and dead states, namely

|ψi〉 = (cos ω1|0〉 + eiφ1 sin ω1|1〉) ⊗ (cos ω2|0〉 + eiφ2 sin ω2|1〉), (20)

where both ω1 and ω2 ∈ [0, 2π) and we denote the vector states produced after n rounds by∣∣ψgn⟩. The previous analysis can be obtained taking ω1 = ω2 = π/2 and φ1 = φ2 = 0, and

again we use label ‘1’ for Alice and ‘2’ for Bob.One could rightfully ask: what could be meant by a player starting the game in a dead

state? In the quantum mechanical framework, we do not need to feel uneasy if a player beginsthe duel, or the conflict, in a dead state. The quantum duel models a conflict between twoplayers using quantum resources; it is not supposed to model a real classical duel. In this sense,Alice could in principle, unwittingly, flip back Bob’s qubit, and we use the term ‘revive’ torefer to this simple effect. We can make an analogy with the Schrodinger cat paradox, since ineach and every round of this quantum duel, both players are in a superposition of ‘alive’ and‘dead’ states, just like Schrodinger’s cat in the famous gedanken experiment—as long as oneremembers that we are dealing with quantum resources such as spin and/or photon polarization

6


states and also that this quantum game ends only when one takes the final measurement in thefinal round.

3.1. Round I

Applying (2) and then (3) to (20), we obtain the duel state after the first round,∣∣ψg

1

⟩ = ei(φ1−α2 ) cos

(θ2

2

)sin ω1

[G1 + iei(φ2−β2 ) sin

(θ2

2

)cos ω1 sin ω2

]|11〉 + eiφ1 G2

× sin ω1|10〉 +[

iG1 ei(β2+φ1) sin

(θ2

2

)sin ω1

+ ei(α2+φ2) cos

(θ2

2

)cos ω1 sin ω2

]|01〉 + cos ω1 cos ω2|00〉; (21)

observe that now we have the |00〉 state corresponding to dead players and the factors G1 andG2 are defined as

G1 = i e−iβ1 sin

(θ1

2

)cos ω2 + ei(φ2−α1 ) cos

(θ1

2

)sin ω2,

and

G2 = eiα1 cos

(θ1

2

)cos ω2 + i ei(φ2+β1 ) sin

(θ1

2

)sin ω2.

Now, in order to write down the payoff functions, we need to calculate the probabilities forfinal states alive–alive, alive–dead, dead–alive and dead–dead. The first one written in termsof the a and b parameters reads∣∣⟨11

∣∣ψg1

⟩∣∣2 = b[−a cos(2ω2) + cos2 ω2 +√

a(1 − a)(2a − 1) sin(2ω2) sin(φ2 + β1 − α1)]

× sin2 ω1; (22)

we clearly see that if both players start the duel in the dead state, ω1 = ω2 = 0, then there isno chance at all that both revive after one round. On the other hand, if the initial state is |10〉,namely if ω1 = π/2 and ω2 = 0, then

∣∣⟨11∣∣ψg

1

⟩∣∣2 = (1 − a)b, which can be made equal tounity if Alice is an excellent shooter (a = 0) and Bob a very poor shooter (b = 1).

The probability of Alice winning the duel, eliminating Bob, is calculated straightforwardly,

∣∣⟨10∣∣ψg

1

⟩∣∣2 = [a cos(2ω2) + sin2 ω2 + 1

2

√a(1 − a) sin(2ω2) sin(α1 − β1 − φ2)

]sin2 ω1;

(23)

when both players have ω1 = ω2 = π/4, it reaches a maximum of 3/8 for a = 1/2 andα1 − β1 − φ2 = (n + 1/2)π .

It is interesting to investigate the payoff function 〈$A〉 after the first round—we do notpresent the cumbersome probability

∣∣⟨01∣∣ψg

1

⟩∣∣2, despite it can be easily calculated using (21).

Even if Alice starts dead, ω1 = 0, her payoff, given by 〈$A〉 = 12 (1 − b) sin2 ω2, can reach

a maximum of 1/2, if Bob is an excellent shooter b = 0 and he was alive in the beginningω2 = π/2. Surprisingly Bob can reduce Alice’s payoff—considering a one-shot duel—if heis a poor shooter and/or starts the duel in a superposition of alive–dead states ω2 ≈ 0.

3.2. Round II

The full expression of both payoff and vector state after the second round is very cumbersome.So let us first analyse the special case where Alice starts the duel in a dead state, i.e. ω1 = 0.

7


Figure 3. Contour plot of Alice’s payoff function when she begins dead ω1 = 0 and Bob beginsalive ω2 = π/2, as a function of (a, b, z), where z = cos(α1 + 2α2). The lower surfaces representhigher payoff according to the colour scale. The intermediate colours represent (from highest tolowest surfaces) 0.2, 0.4, 0.6 and 0.8.

The probabilities we use to write down the payoff function are given as follows. Alice aliveafter the second round and Bob dead:∣∣⟨10

∣∣ψg2

⟩∣∣2ω1=0 = (1 − a)(1 − b) sin2 ω2; (24)

the inverse situation, Alice dead and Bob alive:∣∣⟨01∣∣ψg

2

⟩∣∣2ω1=0 = {a(1 − b)2 + b[b − 2

√a(1 − b) cos(α1 + 2α2)]} sin2 ω2; (25)

and both players dead:∣∣⟨00∣∣ψg

2

⟩∣∣2ω1=0 = cos2 ω2. (26)

From equation (24), we observe that Alice can be revived and win the duel if both playershave a = b = 0 and Bob begins alive ω2 = π/2. Using the above equations (24)—(26) andsubstituting into (14) we obtain a function 〈$A〉 = 〈$A〉(a, b, cos[α1 + 2α2]). In figure 3, wepresent a contour plot of it for the whole ranges of a, b, as well as for cos(α1 + 2α2). Thelower surface (in red) is the one for higher payoff. It is clear that Alice has a good chance ofwinning the duel, as long as she and Bob have a � 0.5 and b � 0.3, respectively.

In the special case, both players begin in a superposition of states,

|ψi〉 = 12 (|00〉 + eiφ1 |10〉 + eiφ2 |01〉 + ei(φ1+φ2)|11〉), (27)

namely when ω1 = ω2 = π/4; we can ask: what is maximum payoff Alice can achieve? Thispayoff function in particular has several variables, so in order to find maximum/minimum, wemake α1 = α2 = φ1 = φ2 = β2 = 0, and to allow Alice to control at least one parameter, letus take β1 = β. Now the payoff can be 0.625 at maximum, and β values are very importantbecause, as we can infer from figure 4, a slight shift in this parameter can turn a good payoffinto a small one.

8


Figure 4. Both contour plots show Alice’s payoff when the initial state is given by equation (27),with α1 = α2 = φ1 = φ2 = β2 = 0 and β1 = β. The minimum is equal to 0.125for a = 0.808, b = 0.277, as we can see in the left figure which is a plot of equation (27)for β = 1.02. The maximum is equal to 0.625 and takes place at the same values of a and b but forβ = −2.12, as we can observe in the right figure. The intermediate colours represent 0.250, 0.375and 0.500.

Figure 5. Left: Alice’s payoff, equation (28), when the initial state is given by equation (20), withω1 = π/2, ω2 = 0 and α1 = α2 = 0. When Alice is a very poor shooter, e.g. a = 1, her payoffbecomes equal to 1. Right: the same function equation (28) when b = 0.38. The payoff reaches aminimum equal to zero if a is also equal to 0.38.

Finally, another interesting special case occurs when Alice (ω1 = π/2) faces an initiallydead opponent, that is, ω2 = 0. For simplicity, let us consider α1 = α2 = 0; then we obtainthe payoff

〈$A〉(a, b) = 12 {1 + a2(1 +

√b)2(2 − b) + b2 − a(1 + 4

√b + b − 2b3/2)

+ 2√

a(√

b − a)[b3/2 + b −√

b − 1]}, (28)

which becomes constant and equal to unity if a = 1, i.e. 〈$A〉(1, b) = 1. On the otherhand, if Alice is an excellent shooter, her payoff increases quadratically with b, namely〈$A〉(0, b) = (1 + b2)/2. We plot (28) in figure 5 to show that her payoff can vary betweenzero and unity, being minimum when both a = b = 0.38.

9


4. Quantum truels

In this section, we briefly discuss what would happen in a three-player duel [16] when theplayers are allowed to begin in a superposition of dead and alive states. An interesting setup isa conflict of Alice and both Bob and Charles (BC team), and in order to balance this one-shottruel, let Alice begin alive, Bob in a superposition of states and Charles in a dead state,

|ψi〉 = |1〉 ⊗ (cos ω2|0〉 + eiφ2 sin ω2|1〉) ⊗ |0〉, (29)

where the ket |ψ〉 = |alice〉 ⊗ |bob〉 ⊗ |charles〉 and our basis has now eight states. In thiscase, Alice fires at Bob, Bob targets Charles (in order to try to flip back his partner’s spin) andthen Charles shoots at Alice; the payoff of the BC team, i.e. 〈$BC〉 = 〈$B〉 + 〈$C〉, is given by

〈$BC〉 = [8b2 + (13 − 24c)b − 24(1 − c)][−1 + (2a − 1) cos(2ω2)]

24, (30)

which is maximum, i.e. 〈$BC〉 = 2, when a = b = c = 0 and ω2 = nπ (which means thatBob should start in a completely alive state in order to maximize the above quantity); itsminimum is equal to zero and the parameters are the same except for ω2 = (n + 1/2)π—Bobstarts the truel in a dead state—where n is an integer and c = cos2(θ3/2). In other words, theminimum payoff is achieved if Bob starts completely dead and uses his only bullet to try torevive Charles, i.e. this is the worst strategy for team BC if Bob is dead. On the other hand, ifBob instead of trying to revive his partner Charles fires at Alice, then the expression for teamBC payoff becomes simpler,

〈$BC〉 = (2 − b)[1 + (1 − 2a) cos(2ω2)]

8, (31)

which has a maximum equal to 1/2 for a = b = 0, c = 1 and ω2 = nπ , and a minimumequal to zero for a = 1, b = c = 0 and ω2 = nπ . In this analysis, we considered the specialcase where all parameters αi, βi, φi are equal to zero, and the utilities for Alice are one-thirdif all three players survive, one-half if she survives in a pair and one if the shooter is ableto eliminate her rivals. For the BC team, the utilities are as follows: one for both surviving,one-half when all three survive, one-third for surviving alone (Bob or Charles) and one-fourthwhen Alice is alive with only one member of team BC.

A comprehensive analysis of the truel using both superposition of dead and alive statesfor the three players and using entangled states will be presented in a forthcoming paper.

5. Conclusions

We studied the quantum duel game and carried out a complete analysis of the Alice payoffas a function of the parameters α1 and α2. In a previous paper [16], the authors introduced astrategy—considering the possibility of one player to shoot poorly and the other to not—forAlice not to shoot the second time, because she could have the chance to revive Bob. However,a careful analysis shows that this strategy is not always effective, since the final payoff dependson α1 and α2. Interesting results were obtained considering the possibilities of players to startthe duel in a superposition of dead and alive states. We presented some scenarios wherethe initial states were linear superpositions of the form (20) and analysed how Alice couldmaximize her payoff for three cases: when the initial state is |10〉, |01〉 and the one given byequation (27).

10


Acknowledgments

The authors gratefully acknowledge FAPERJ (Fundacao Carlos Chagas Filho de Amparoa Pesquisa do Estado do Rio de Janeiro, grant E-26/110.192/2011) and CNPq (ConselhoNacional de Desenvolvimento Cientıfico e Tecnologico, grant 480825/2009-2) for partialfinancial support.

Appendix.

As we stated in section 2 there is a misprint in equation (13) of [16] and the correct expressionfor |〈01|ψ ′

2〉|2 is given by our equation (18). In order to show that, let us calculate in detail thestate after the second round when Alice wastes her second shot, namely |ψ ′

2〉, which can beobtained applying (3) to (6) and using 〈ab|cd〉 = δacδbd , where δi j is the Kronecker delta; weobtain

|ψ ′2〉 = e−i(α1+α2 ) cos

(θ1

2

)cos

(θ2

2

)[e−iα2 cos (θ2/2)|11〉 + i eiβ2 sin (θ2/2)|01〉]

+ i ei(β2−α1 ) cos

(θ1

2

)sin

(θ2

2

)[eiα2 cos (θ2/2)|01〉 + i e−iβ2 sin (θ2/2)|11〉]

+ i eiβ1 sin

(θ1

2

)|10〉; (A.1)

after substituting a = cos2(θ1/2) and b = cos2(θ2/2), one obtains (16). The amplitude 〈01|ψ ′2〉

can be easily calculated using (16), since the only non-vanishing scalar product comes fromthe last term,

〈01|ψ ′2〉 = 2i

√ab(1 − b) ei(β2−α1 ) cos(α2), (A.2)

and finally, squaring this amplitude, we obtain (18).

References

[1] Guo H, Zhang J and Koehler G J 2008 Decis. Support Syst. 46 318[2] Meyer D A 1999 Phys. Rev. Lett. 82 1052[3] Eisert J, Wilkens M and Lewenstein M 1999 Phys. Rev. Lett. 83 3077[4] Du J, Li H, Xu X, Shi M, Wu J, Zhou X and Han R 2002 Phys. Rev. Lett. 88 137902[5] Challet D and Zhang Y-C 1997 Physica A 246 407[6] Benjamin S C and Hayden P M 2001 Phys. Rev. A 64 030301[7] Satinover J B and Sornette D 2007 Eur. Phys. J. B 60 369[8] Flitney A P, Schlosshauer M, Schmid C, Laskowski W and Hollenberg L C L 2009 Phys. Lett. A 373 521[9] Schmid C, Flitney A P, Wieczorek W, Kiesel N, Weinfurter H and Hollenberg L C L 2010 New J.

Phys. 12 063031[10] Hogg T, Harsha P and Chen K-Y 2007 Int. J. Quantum Inform. 5 751[11] Goldenberg L, Vaidman L and Wiesner S 1999 Phys. Rev. Lett. 82 3356[12] Flitney A P and Abbott D 2002 Phys. Rev. A 65 062318[13] Vilela Mendes R 2005 Quantum Inform. Proc. 4 1[14] Parrondo J M R, Harmer G P and Abbott D 2000 Phys. Rev. Lett. 85 5226[15] Flitney A P and Abbott D 2003 Physica A 324 152[16] Flitney A P and Abbott D 2004 J. Opt. B 6 S860

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http://dx.doi.org/10.1016/j.dss.2008.07.001

http://dx.doi.org/10.1103/PhysRevLett.82.1052



http://dx.doi.org/10.1016/S0378-4371(97)00419-6

http://dx.doi.org/10.1103/PhysRevA.64.030301

http://dx.doi.org/10.1140/epjb/e2007-00353-6

http://dx.doi.org/10.1016/j.physleta.2008.12.003

http://dx.doi.org/10.1088/1367-2630/12/6/063031

http://dx.doi.org/10.1142/S0219749907003183


http://dx.doi.org/10.1103/PhysRevA.65.062318

http://dx.doi.org/10.1007/s11128-005-3192-7


http://dx.doi.org/10.1016/S0378-4371(02)01909-X

http://dx.doi.org/10.1088/1464-4266/6/8/036

quantum duel revisited

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