CONTENTS 10 quantumgames 3 10.1 Quantum Game 1: Bell 3 10.2 Quantum Game 2: CHSH 5 10.3 Quantum Game 3: GHZ 6 11 why quantum mechanics? 9 11.1 Recap 9 11.2 Why Quantum Mechanics? 10 11.2.1 Well, why not? 10 11.2.2 Why complex numbers? 10 12 11 13 postulate 3: measurement 13 14 quantum computation 15 14.1 Quantum Computation: Introduction 15 14.1.1 Quantum Entanglement 15 14.1.2 Qubits 16 14.2 Quantum Cryptography 16 1 10 QUANTUM GAMES 10.1 quantum game 1: bell Quantum Casino: Bell Consider the following shell game: There are three boxes (labeled left, middle, right (L,M,R for short)) within which is contained a single card that is either black or red. The house states that the cards are arranged such that: If you reveal the left and middle (LM) boxes, or the right and middle (RM) the cards are always the same color. If you reveal the left and right boxes (LR) the cards have opposite color. Clearly we can not achieve this by having three independent cards (there does not exist a jpd), so two house employee’s assist with the game. Your goal is to try to maximize the number of times that the above conditions are violated, while only revealing 2 (or 1) of the boxes. (You are not allowed to open all three.) The two employees (Alice and Bob) will try to maximize the conditions above. Payouts: The house and you are given an initial fixed amount of dollars (chips). P1: If the conditions above are met, you pay the house $1 (-$1 to your balance). P2: If the conditions above are not met, the house pays you $2 (+$2). P3: : If each query the same box, and different outcomes result, house pays you $1 1012 and goes out of business × (game over). Alice and Bob are not allowed to communicate during any single trial. They can use prearranged strategies. Each round will occur as follows: 1) Contestant querying Alice to reveal the first box. Alice will state the outcome of the chosen box (red/black). 2) The contestant then queries Bob to reveal the second box (or the same). Bob states the outcome of the chosen box. 3) The dealer determines whether the conditions are met and performs the appropriate payment. The contestant can end at any time, or until one of the parties runs out of money. Classical probability analysis Let’s calculate the expectation for the contestant to win and see if this is a fair game. We ignore the P3 outcome as that ends the game. We have the correlations LM¯ = RM = +1, LR = 1. i h i h i − Consider the case where results are all equiprobable (construct jpd). 3 4 quantum games L M R satisfied violated b b b LM,RM LR b b r LR,LM RM b r b LM,RM,LR b r r LR,RM LM r b b LR,RM LM r b r LM, RM ,LR r r b LR,LM RM r r r LM,RM LR As a potential prearranged strategy, the house can set the probability for (b,r, b) and (r, b,r) to 0 to gain an advantage as those outcomes always violate the conditions. Examining the total number of outcomes, we see that in any round, there is at least 1/3 probability that the conditions are violated, 2/3 probability that the conditions are met. Thus the expected gain for the contestant is, ($2)(1/3)+ (-$1)(2/3)=$0. This is a fair game as constructed. Quantum analysis: “House always wins." With classical resources (prearranged strategies) we see that this is a fair game. The question is, can the house tilt the odds in their favor? With quantum resources, yes! (Up to -$0.25 for the contestant.) Here’s how. As they are allowed to share resources in advance, but not allowed to communicate, they share an entangled pair of photons. Say they are in the state ψ = 1 [ + + ], where each have a qubit. | i √2 | −i−|− i They now choose the following prearranged strategy to change the odds. Alice: Contestant C chooses: Set polarization to Measure: If+return: If-return: L +30o red black M 0o red black R 30o red black − Bob: Contestant C chooses: Set polarization to Measure: If+return: If-return: L +30o black red M 0o black red R 30o black red − Let’s check the payouts: P3: First, let’s check that the house never goes out of business: If C chooses the same box, they are in the same basis and they always return the same result. P1: If one of the choices includes M, notice that the angle between Alice and Bob is 30o . This gives the probability, | | P(+ ) = P( +) = 3/8, P(rr or bb) = 3/4 (detailsnext page). − − P2: If C chooses LR, the bases are 60o apart. This gives the probability, | | P(+ ) = P(+ ) = 1/8, P(rr or bb) = 1/4. − − The expected gain for C is now: ($2)(1/4)+ (-$1)(3/4)=-$1/4 = -$0.25. The house has the advantage! 10.2 quantum game 2: chsh 5 Quantum analysis 2: Let’s examine the details of calculating the probabilities. The general form of the state takes this form in one of the three bases (L,M,R). 1 ψ = [ + + ] | i √2 | −i−|− i We want to find P(rr or bb), which corresponds to P(+ or +), − − 1 P(M+, L ) = M + L ( M+ M M M+ ) 2 − 2 |h |h −| | i| −i−| −i| i | 2 1 2 1 2 1 2 o 1 3 3 = M + M+ L M = L M = cos 30 = = 2 |h || ih −|| −i| 2 |h −| −i| 2 2 4 8 where the figure shows the projection. When squared, we see this probability is 3/8. It is not hard to convince yourself that P( M, +L) also equals 3/8. − Thus, P(rr or bb) = 3/4. For this case, we want to find P(rborbr), which corresponds to P(+ + or ), − − 1 P(L+, R+) = L + R + ( L+ L L L+ ) 2 2 |h |h | | i| −i−| −i| i | 2 1 2 1 2 o 1 3 3 = R + L = cos 30 = = 2 |h | −i| 2 2 4 8 Again, putting this altogether, the expected gain for C is: ($2)(1/4)+ (-$1)(3/4)= -$1/4 = -$0.25. 10.2 quantum game 2: chsh In this game we will represent inputs and outputs in binary form. In this game Alice and Bob are isolated from each other and each are given a random bit, Alice receives x = 0,1 and Bob receives y = 0, 1, and their goal is to provide outputs, a and b, that maximize the following relation, a b = xy. ⊕ You may recognize this as the defining relation for a PR box. If they satisfy this condition, they win $1 and if not, they lose $ 3. (We will see that for classical resources, this is a fair game.) Classical analysis The probability table representing the above relation is, (x, y) (a, b) (0,0) (0,1) (1,0) (1,1) | → 1 (0,0) 1 0 0 s 2 2 1 (0,1) 1 0 0 2 2 1 (1,0) 1 0 0 2 2 1 1 (1,1) 0 0 2 2 Thus, a simple (pre-established) strategy that Alice and Bob could employ is to simply always return 0. In this case, 3 they win the game 4 of the times. It can be proven that this is the maximal possible success probability in this game utilizing classical resources. The expected gain in this scenario is, 3 1 $ = ($1)( )+( $3)( ) = 0. h i 4 − 4 Thus, with only classical resources, this is a fair game. 6 quantum games The quantum edge We now consider that Alice and Bob have access to quantum recourses, mainly entangled states. Thus, their prear- ranged strategy is as follows: They each obtain a photon in the entangled state ψ = 1 [ 0 0 + 1 1 ]. • | i √2 | i| i | i| i If Alice receives input x = 0, she simply measures her qubit. If she receives x = 1 • ˆ π x = 1, she applies the unitary U( 4 ) to her qubit and then measures it. She outputs the outcome of her measurement. y = 0 If Bob receives y = 0 he applies Uˆ ( π ) and measures, for y = 1 he applies θ = • 8 θ x 0 Uˆ ( π ) − 8 θ x The figure shows the effect of Uˆ (θ) on the H state. This algorithm yields the y = 1 highest possible winning probability with quantum resources. It is left as an exercise to compute the maximum winning probability. A little algebra will show that the maximum classical winning strategy corresponds to S = 2, the boundary of the local polytope, while the maximal quantum probability corresponds to S = 2√2. And of course, if Alice and Bob could share PR boxes, then they could win this game every time. 10.3 quantum game 3: mermin square 1 We will preform the following two player game in class. There are two players, Alice and Bob (A & B) along with a referee (R). The procedure for the game is as follows: There is a 3 3 board upon which Alice and Bob will be asked to fill with either 0, 1. They are not allowed to × communicate once the round has begun. 1. The referee picks a random row, r 1,2,3 and a random column c 1,2,3 .