* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Classical vs. Quantum Correlations
Survey
Document related concepts
Quantum computing wikipedia , lookup
Density matrix wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
History of quantum field theory wikipedia , lookup
Interpretations of quantum mechanics wikipedia , lookup
Canonical quantization wikipedia , lookup
Quantum entanglement wikipedia , lookup
Quantum state wikipedia , lookup
Hidden variable theory wikipedia , lookup
EPR paradox wikipedia , lookup
Quantum key distribution wikipedia , lookup
Quantum electrodynamics wikipedia , lookup
Quantum teleportation wikipedia , lookup
Probability amplitude wikipedia , lookup
Transcript
Classical vs. Quantum Correlations Michael Lachenmaier, Marco Michel June 16, 2016 During this session we will explore the signicance of quantum information theory compared to classical information theory. The dierences will be visualized with the help of a simple two-player game. 1 Classical Correlations In this section the aim is to describe correlations using some reasonable assumptions. Later we will see that not all quantum correlations can be described using those assumptions. 1.1 LHV Ansatz Assumption 1 We assume that quantities have values even before they are measured. They might be unknown, i.e. hidden, and thus a probabilistic description may be needed. Assumption 2 The properties of one subsystem should not immediately depend on events taking place on very distant subsystems. Example 1 Let us assume we have a source observers Alice A B m∈N and Bob Both can choose out of S that emits pairs of particles to distant who each perform ±1 measurement devices, valued measurements. Ax and By {1, . . . , m}: ±1 Ax S 1 By ±1 with x, y ∈ p(a, b | x, y) We denote by the probability that b ∈ {±1} A measures outcome while B Here P we want to consider the product of their expectations, i.e. measures outcome a,b∈{±1} using devices Ax and By , a ∈ {±1} respectively. hAx By i := ab p(a, b | x, y). Ansatz 3 (LHV Ansatz) Let Ax , By : Ω −→ {±1} be random variables and P a probability measure. We make the following ansatz for the expectation of the product: Z hAx By i = Ax (ω)By (ω)dP (ω) (1) Ω Here, ω∈Ω plays the role of the hidden variable assumed in assumption 1 and locality from assumption 2 is expressed by the fact that not depend on y and x, Ax and By do respectively. This ansatz is called the local hidden variable (LHV) ansatz. 1.2 Bell Inequalities Denition 1 Let C ∈ Rm×m with entries Cxy := hAx By i that are empirically obtained expectation values for pairs of ±1 valued measurements. We denote by C ⊆ Rm×m the set of all such C for which there exists an LHV description as in n×n ansatz 3. An inequality F : R → R for C is called Bell inequality if it holds for all C ∈ C, i.e. if F (C) ≥ 0 ∀C ∈ C Remark 1 1. C 2. |Cxy | ≤ 1 is a closed convex polytope. are trivial Bell inequalities. Proposition 1 ρ ∈ B(HA ⊗ HB ) classically correlated (i.e. not entangled) and Mx : B({±1}) → B(HA ), My0 : B({±1}) → B(HB ) be POVMs for x, y ∈ {1, . . . , m}. m×m Let C ∈ R be dened by X Cxy := ab tr[ρMx (a) ⊗ My0 (b)] (x, y = 1, . . . , m), (2) Let a,b∈{±1} then C ∈ C. Proof. If we dene Ax := P a∈{±1} aMx (a), By := −1 ≤ Ax , By ≤ 1. 2 P b∈{±1} bMy0 (b), then Assume ρ Cxy (ω) P (ω) pω ρA ⊗ ρB . We h i h i X (ω) (ω) = tr[ρAx ⊗ By ] = pω tr ρA Ax tr ρB By has convex combination ρ= ω calculate: ω h i (ω) The last term gives a LHV description as in eq. (1) with Ax (ω) = tr ρA Ax h i (ω) and By (ω) = tr ρB By on some discrete probability space. Since eq. (2) is just the expected value, the previous proposition yields that Bell inequalities hold in particular for any unentangled quantum state. Next we will state and prove a non-trivial Bell inequality. Theorem 2 (Causer-Horne-Shimony-Holt (CHSH) inequality) Every LHV theory for the description of suring devices A1 , A2 , B1 , B2 ±1 valued measurements with mea- (as in ansatz 3) satises: |hA1 B1 i + hA1 B2 i + hA2 B1 i − hA2 B2 i| ≤ 2 (3) Proof. By the denition of the expected values, we can rewrite the left-hand side as Z A1 (ω) (B1 (ω) + B2 (ω)) + A2 (ω) (B1 (ω) − B2 (ω)) dP (ω) . Ω If we x ω ∈ Ω, we can distinguish two cases: B1 (ω) − B2 (ω) = 0 B1 (ω) + B2 (ω) ∈ {±2}. 1. B1 (ω) = B2 (ω), 2. B1 (ω) 6= B2 (ω): Since there are only values ±1 possible, this implies B1 (ω) = −B2 (ω), i.e. B1 (ω) + B2 (ω) = 0 and B1 (ω) − B2 (ω) ∈ {±2}. i.e. and Since both A1 (ω) and A2 (ω) are in {±1}, in both cases, the integrand is in {±2} which yields, using the fact that P is a probability measure, the desired inequality. We will later see an example of how entangled states can violate this inequality. 2 The CHSH game in this section we will rst look at classical strategies for the CHSH game. Later we will also consider quantum strategies. 3 qA A qB B qA rB rA p(rA , rB | qA , qB ) qB C rA rB Figure 1: structure of a nonlocal game with two players, questions answers qA , qB and rA , rB 2.1 Denition Denition 2 (Nonlocal Game with two Players) A nonlocal game with two players (henceforth only nonlocal game) consists of players Alice, A, and Bob, B, and a referee Charlie, C. All communication in the game is only between any player and the referee, however the players are allowed to choose a strategy beforehand. The referee chooses a question at random for each of the two players. Every player then responds with an answer (dependent on the question she got). After receiving both answers, the referee determines whether both players won or lost the game. It is therefore not possible for one player to win and for the other one to lose. See also g. 1. Denition 3 (CHSH game) The CHSH game is a nonlocal game with two players. The referee Charlie qA , qB for Alice and Bob, respectively, uniformly {0, 1}. The answers of Alice and Bob, rA , rB , respectively, must 0 or 1. The players win if the following condition is met: chooses questions from the set be either rA ⊕ rB = qA ∧ qB . Here, ⊕ (4) denotes the exclusive or which is equivalent to addition modulo The following table shows the winning conditions: rA ⊕ rB 0 0 0 1 (qA , qB ) (0, 0) (0, 1) (1, 0) (1, 1) 4 2. 2.2 Classical Strategies Proposition 3 Using a classical strategy, i.e. Alice and Bob are only allowed to return classically correlated states, the maximal probability for Alice and Bob to win the game is 3/4, i.e. P (W IN ) ≤ 3/4. This bound is sharp. Proof. For the sake of contradiction, let us assume that there is a strategy P (W IN ) > 3/4. Since there are only 4 possible question-pairs, this means that P (W IN ) = 1, i.e. they would win every time. Let rA , rB : {0, 1} −→ {0, 1} be the strategies used by Bob and Alice, for Alice and Bob with accordingly. We write the winning conditions as a system of equations: rA (0) ⊕ rB (0) rA (0) ⊕ rB (1) rA (1) ⊕ rB (0) rA (1) ⊕ rB (1) 2 Adding those four equations modulo = = = = 0 0 0 1 0 = 1, gives a contradiction. Even when using probabilistic strategies rather than deterministic ones, the expected probability to win will never be greater than with a deterministic So far we have proven that there cannot be a strategy P (W IN ) > 3/4, thus P (W IN ) ≤ 3/4 must hold. Next we will give a strategy with P (W IN ) = 3/4: Suppose, Alice and Bob will always return 0 as an answer. one. Then in three out of four cases, they will win the game. Since the cases occur with equal probability (are chosen uniformly), this strategy wins with probability 3/4. Relation between the CHSH Game and the CHSH Inequality We will look into the relation between the CHSH game and the CHSH inequality from the previous section. {±1} rather than swers r̃A and r̃B {0, 1}, If we assume that Alice and Bob return values we get new winning conditions: Call the new an- for Alice and Bob, respectively. In the case the winning condition is that the sum of to saying that both are either product of r̃A and condition is that r̃B rA is one. and rB 0 or 1. rA and (qA , qB ) = (1, 1), are dierent, i.e. 5 is zero which is equivalent This is equivalent to saying that the Similarly, for minus one. We summarize: rB (qA , qB ) 6= (1, 1) the product of the winning r̃A and r̃B is (qA , qB ) rA ⊕ rB (0, 0) 0 0 (0, 1) (1, 0) 0 (1, 1) 1 r̃A r̃B 1 1 1 −1 A0 , A1 for Alice and B0 , B1 for Bob. Alice uses A0 if qA = 0 and A1 if qA = 1. Similarly for Bob. We are now in the situation of example 1, i.e. we can make a LHV ansatz. For x, y ∈ {0, 1} We now consider measuring devices consider the expectation of the product X hAx By i = a,b∈{±1} a=b For (x, y) 6= (1, 1), X p(a, b | x, y) − p(a, b | x, y) (5) a,b∈{±1} a6=b eq. (5) gives the probability of winning minus the probability of losing, while for (x, y) = (1, 1), eq. (5) gives the probability of losing minus the probability of winning. Since all devices are chosen with equal probability 1/4, we get the total probability of winning minus the probability of losing by the following formula: P (W IN ) − P (LOSE) = 1 [hA0 B0 i + hA0 B1 i + hA1 B0 i − hA1 B1 i] 4 (6) We can now apply the CHSH inequality to eq. (6) to obtain 1 2 P (W IN ) ≤ 3/4, P (W IN ) − P (LOSE) ≤ Using P (W IN ) = 1 − P (LOSE), we obtain (7) the same bound as in proposition 3. 2.3 Quantum strategies However, if Alice and Bob share an entangled two-qubit system which is 1 initialized in the Bell state |ψi = √ (|00i + |11i), then they can gain a 2 2 probability of cos (π/8) ≈ 0.85 if they use following strategy: For a given angle θ ∈ [0, 2π), dene: |φ0 (θ)i = cos(θ) |0i + sin(θ) |1i (8) |φ1 (θ)i = − sin(θ) |0i + cos(θ) |1i (9) The strategy is now: 6 1. Alice takes the rst qubit and Bob takes the second qubit from the quantum system. 2. If Alice receives x=0 then she measures her qubit with respect to the basis {|φ0 (0)i , |φ1 (0)i} (10) and if she receives the question 1, she will measure her qubit with respect to the basis {|φ0 (π/4)i , |φ1 (π/4)i} (11) 3. Bob uses a similar strategy, except that he measures with respect to the basis {|φ0 (π/8)i , |φ1 (π/8)i} or {|φ0 (−π/8)i , |φ1 (−π/8)i} (12) depending on whether his question was 0 or 1. 4. Alice and Bob output the values obtained as their answers a and b. Each of these matrices is a rank-one projection matrix, so the measurements Alice and Bob are making are examples of projective measurements. 1 T Given our particular choice of |ψi, we have hψ| X ⊗ Y |ψi = tr(X Y ) for 2 b a arbitrary matrices X and Y . Thus, as each of the matrices Xs and Yt is real and symmetric, the probability that Alice and Bob answer (s, t) with (a, b) is 1 tr(Xsa Ytb )). Is is now routine to check that in every case, the correct answer 2 2 is given with probability cos (π/8) and the incorrect answer with probability 2 sin (π/8). 3 Tsirelson bound One might wonder if it is possible to do even better with another quantum strategy for the CHSH game. The answer is "no" due to Tsirelson's bound: Theorem 4 (Tsirelson bound for the CHSH inequality) β̂ := A1 ⊗ (B1 + B2 ) + A2 ⊗ (B1 − B2 ) with Ai , Bj density operators ρ ∈ B(HA ⊗ HB ) : √ tr[ρβ̂] ≤ 2 2 For as above and for all (13) √ That is, quantum theory violates the CHSH inequality at most by a factor 2. 2 2 2 Moreover, there exists a pure state on C ⊗C and observables Ax , Ay ∈ B(C ) with eigenvalues ±1, s.t. equality holds in this equation. 7 i ∈ {1, 2}. Since these are ane −1 ≤ Ai ≤ 1, the supremum and inmum is attained for some extreme point for which spec(Ai ) = {±1} and 2 thus Ai = 1. The same holds for Bj , j ∈ {1, 2}. Using this property, direct Proof. Consider the map Ai 7→ tr[ρβ̂] for functionals over the closed convex set computations leads to β̂ 2 = 41 ⊗ 1 + [A2 , A1 ] ⊗ [B1 , B2 ]. (14) We exploit this via positivity of the variance and obtain tr[ρβ̂]2 ≤ tr[ρβ̂ 2 ] = 4 + tr[ρ[A2 , A1 ] ⊗ [B1 , B2 ]] (15) ≤ 4 + ||[A2 , A1 ] ⊗ [B1 , B2 ]|| ≤ 8 (16) where the last inequality uses that ||[A1 , A2 ]|| ≤ ||A1 A2 || + ||A2 A1√|| ≤ 2||A1 ||||A2 || = 2 and similarly for the B's. So we end up with tr[ρβ̂] ≤ 2 2 as claimed. In order to prove that equality can be achieved, assume that ρ = |ψi hψ| where |ψi is an eigenvector of β̂ with eigenvalue ν . Then equality holds in (15). A1 = B1 = σ1 and A2 = B2 = σ2 Pauli matrices, so that β̂ 2 = 41 + 4σ3 ⊗ σ3 has eigenvalues 0 and 8. Hence, ν can be chosen such that tr[ρβ̂]2 = ν 2 = 8. Now take Remark 2 We have previously seen that β̂ can be interpreted as the total probability of winning minus the probability of losing for the CHSH game. So for the previously used strategy we archive equality in Tsirelson's bound: √ P (LOSE)) = 4 × (cos(π/8)2 − sin(π/8)2 ) = 2 2. 4 × (P (W IN ) − Remark 3 [A2 , A1 ] = 0 or [B1 , B2 ] = hCHSHi ≤ 2. Clearly, the noncom- Note that if the local observables are commuting, i.e., 0, then we recover the classical bound mutativity of observables in quantum mechanics plays a crucial role in the quantum advantage, yielding a winning probability of 85% compared to the classical 75% in the CHSH game. Remark 4 The violation of CHSH by a factor of for the rst time in the early 80's. √ 2 has been veried experimentally This was done using down conversion in a non-linear crystal, which produces entangled pairs of photons, whose polarization degrees of freedom violate CHSH. A more recent experiment [3] tested the CHSH Bell inequality on photon pairs in maximally entangled states of polarization in which a value 2.8276 ± 0.00082 8 was observed. The argumentation can be generalized to more than two observables: hAx By i =: Cx y, x, y ∈ {1, ..., m} as before, γ ∈ Rm×m and dene Consider ||γ||LHV X γxy ax by := sup a,b∈{±1}m (17) x,y ||γ||quantum X γxy tr[ρAx ⊗ By ] := sup ρ,{Ax By } (18) x,y where ρ is density operator and −1 ≤ Ax , By ≤ 1. The Trirelson bound for the CHSH inequality then reads: ||γ||quantum √ ν(γ) := = 2 ||γ||LHV for 1 1 γ= 1 −1 (19) Theorem 5 (General Cirelson bounds) √ 2×2 1. γ∈R ⇒ ν(γ) ≤ 2. γ ∈ Rm×m , m ∈ N ⇒ ν(γ) ≤ KG < 2 π √ 2ln(1+ 2) ≈ 1.782 Remark 5 1. means that the choice of coecients in CHSH is optimal 2. is a non-trivial statement whose proof is based on a deep result of Grothendieck. KG is called Grothendieck's constant, which is unknown ν(γ) over all γ ∈ Rm×m and all m ∈ N but equal to the supremum of References [1] M. Wolf, Quantum eects, Lecture Notes, 2014. [2] J. Watrous, Quantum Computation - Lecture 20: Bell inequalities and nonlocality, Lecture Notes, 2006. [3] Hou Shun Poh, et al., Approaching Tsirelsons Bound in a Photon Pair Experiment, Physical Review Letters 115(18), 2015. 9