A parallel repetition theorem for entangled two-player one-round ...
A parallel repetition theorem for entangled two-player one-round games under product distributions Rahul Jain
Attila Pereszlényi
Penghui Yao
Centre for Quantum Technologies and Department of Computer Science National University of Singapore Singapore [email protected]
Centre for Quantum Technologies National University of Singapore Singapore [email protected]
Centre for Quantum Technologies National University of Singapore Singapore [email protected]
Abstract—We show a parallel repetition theorem for the entangled value ω ∗ (G) of any two-player one-round game G where the questions (x, y) ∈ X × Y to Alice and Bob are drawn from a product distribution on X × Y. We show that for the k-fold product Gk of the game G (which represents the game G played in parallel k times independently) k Ω log(|A|·|B|) ω ∗ Gk = 1 − (1 − ω ∗ (G))3 where A and B represent the sets from which the answers of Alice and Bob are drawn. The arguments we use are information theoretic and are broadly on similar lines as that of Raz [1] and Holenstein [2] for classical games. The additional quantum ingredients we need, to deal with entangled games, are inspired by the work of Jain, Radhakrishnan, and Sen [3], where quantum information theoretic arguments were used to achieve message compression in quantum communication protocols. Index Terms—parallel repetition theorem; two-player game; entangled value
I. I NTRODUCTION
and the famous Unique Games Conjecture. One of the most fundamental problems regarding this model is the so called parallel repetition question, which concerns the behavior of multiple copies of the game played in parallel. For the game G = (µ, X , Y, A, B, V ), its k-fold product is given by Gk = (µk , X k , Y k , Ak , B k , V k ), where V k (x, y, a, b) = 1 if and only if V (xi , yi , ai , bi ) = 1 for all i ∈ [k]. Namely, Alice and Bob play k copies of game G simultaneously, and they win iff they win all the copies. It is easily seen that ω(Gk ) ≥ ω(G)k for any game G. The equality of the two quantities, for all games, was conjectured by Ben-Or, Goldwasser, Kilian and Wigderson [4]. The conjecture was shown to be false by Fortnow [5]. However one could still expect that ω(Gk ) goes down exponentially in k (asymptotically). This is referred to as the parallel repetition (also known as the direct product) conjecture. This was shown to be indeed true in a seminal paper by Raz [1]. Raz showed that k ω Gk = (1 − (1 − ω(G))c )Ω( log(|A||B|) )
A two-player one-round game G is specified by finite sets X , Y, A, and B, a distribution µ over X × Y, and a predicate where c is a universal constant. This result, along with the V : X × Y × A × B → {0, 1}. It is played as follows. the PCP theorem had deep consequences for the theory of • The referee selects questions (x, y) ∈ X × Y according inapproximability [6], [7], [8]. A series of works later exhibited to distribution µ. improved results for general and specific games [2], [9], [10], • The referee sends x to Alice and y to Bob. Alice and Bob [11], [12]. are spatially separated, so they do not see each other’s In the quantum setting, it is natural to consider the so input. called entangled games where Alice and Bob are, in addition, • Alice chooses answer a ∈ A and sends it back to the allowed to share a quantum state before the games starts. referee. Bob chooses answer b ∈ B and sends it back to The questions and answers in the game remain classical. the referee. On receiving questions, Alice and Bob can generate their • The referee accepts if V (x, y, a, b) = 1 and otherwise answers by making quantum measurements on their shared rejects. Alice and Bob win the game if the referee accepts. entangled state. The value of the entangled version of the The value of the game G, denoted by ω(G), is defined to be the game G is denoted by ω ∗ (G). The study of entangled games maximum winning probability (averaged over the distribution is deeply related to the foundation of quantum mechanics and that of quantum entanglement. These games have been µ) achieved by Alice and Bob. These games have played an important and pivotal role used to give a novel interpretation to Bell inequalities, one in the study of the rich theory of inapproximability, leading of the most famous and useful methods for differentiating to the development of Probabilistically Checkable Proofs classical and quantum mechanics (e.g., by Clauser, Horne,
Shimony and Holt [13]). Recently these games have also been studied from cryptographic motivations such as in Refs. [14], [15], [16]. Analogously to the classical case, the study of the parallel repetition question in this setting may potentially have applications in quantum complexity theory. The parallel repetition conjecture has been shown to be true for several sub-classes of entangled games, starting with the so called XOR games by Cleve, Slofstra, Unger and Upadhyay [17], later generalized to unique games by Kempe, Regev and Toner [18] and very recently further generalized to projection games by Dinur, Steurer and Vidick [19] (following an analytical framework introduced by Dinur and Steurer in [20] to deal with classical projection games). For general games, Kempe and Vidick [21] (following a framework by Feige and Killian [22] for classical games) showed a parallel repetition theorem albeit with only a polynomial decay in k, in the value ω ∗ (Gk ). In a recent work, Chailloux and Scarpa [23] showed an exponential decay in ω ∗ (Gk ) using information theoretic arguments.
Our techniques
The arguments we use are information theoretic and are broadly on similar lines as that of Raz [1] and Holenstein [2] for classical games. The additional quantum ingredients we need, to deal with entangled games, are inspired by the work of Jain, Radhakrishnan, and Sen [3], where quantum information theoretic arguments were used to achieve message compression in quantum communication protocols. Given the k-fold game Gk , let us condition on success on a set C ⊆ [k] of coordinates. If the overall success in coordinates in C is already as small as we want, then we are done. Otherwise, we exhibit another coordinate j ∈ / C such that the success in the j-th coordinate, even when conditioning on success in the coordinates inside C, is bounded away from 1. Here we assume that ω ∗ (G) is bounded away from 1. This way the overall success keeps going down and becomes exponentially small in k, after we have identified Ω(k) such coordinates. To argue that the probability with which Alice and Bob win the j-th coordinate, conditioned on success in C, Theorem I.1 ([23]). For any game G = (µ, X , Y, A, B, V ), is bounded away from 1, we show that close to this success probability can be achieved for a single instance of the game where µ is the uniform distribution on X × Y, it holds that G. That is, given inputs (x0 , y 0 ), drawn from µ, for a single k Ω ( ) instance of G, Alice and Bob can embed (x0 , y 0 ) to the j-th ω ∗ Gk = 1 − (1 − ω ∗ (G))2 log(|A||B||X ||Y|) . coordinate of Gk , conditioned on success in C, and generate the rest of the state with good approximation. So, if the probability As a corollary, for a general distribution µ, of success in the j-th coordinate, conditioned on success in C, k is very close to 1, there is a strategy for G with probability of ∗ k ∗ 2 Ω Q4 log(Q) log(|A||B|) ω G = 1 − (1 − ω (G)) success strictly larger than ω ∗ (G), reaching a contradicting to the definition of ω ∗ (G). where Suppose the global state, conditioned on success in C, is of the form 1 np o X Q = max , |X | · |Y| . XY AB σ XY AB = µ ˜(x, y) |xyihxy| ⊗ |φxy ihφxy | minx,y:µ(x,y)6=0 µ(x, y) x∈X k ,y∈Y k ∗
k
Note that here ω (G ) depends on |X | · |Y| as well, in addition to |A| · |B| (as in Raz’s result). Also the value of Q can be arbitrarily large, depending on the distribution µ. Our result In this paper we consider the case when the distribution µ is product across X × Y. That is, there are distributions µX , µY on X , Y respectively such that ∀(x, y) ∈ X × Y : µ(x, y) = µX (x) · µY (y). We show the following.
where µ ˜ is a distribution, potentially different from µ because of the conditioning on success. (Here we further fix the questions and answers in C to specific values and do not specify them in σ XY AB .) In protocol P for the single instance of G, we let Alice and Bob start with the shared pure state X p ˜ Y˜ Y XX AB |ϕi = µ ˜(x, y) |xxyyi ⊗ |φxy i . x∈X k ,y∈Y k
Note that |ϕi is a purification of σ XY AB , where registers ˜ and Y˜ are identical to X and Y . We introduce these X Theorem I.2 (Main Result). For any game copies of the registers X and Y so that the marginal state in these registers remains a classical state and these registers G = (µ, X , Y, A, B, V ) can be viewed as classical registers, which is important in our arguments. where µ is a product distribution on X × Y, it holds that Using the chain rule for mutual we are able information, k ˜ B and I Yj : X XA ˜ ∗ k ∗ 3 Ω( log(|A||B|) ) to argue that both I X : Y Y are very j ω G = 1 − (1 − ω (G)) . small (close to 0), in |ϕi. This, obviously, is only possible Note that the uniform distribution on X × Y is a product when the distribution µ is product. In addition, the distribution distribution and our result has no dependence on the size of of the questions in the j-th coordinate, in |ϕi, remains close X × Y. Hence, our result implies and strengthens on the result to µ, in the `1 -distance. In protocol P, when Alice and Bob of Chailloux and Scarpa [23] (up to the exponent of 1−ω ∗ (G)). get questions x0 and y 0 , suppose they measure registers Xj
E and Yj , in |ϕi, and get x0j and yj0 . Let ϕx0j yj0 be the resulting state. If by luck it so happens that (x0 , y 0 ) = x0j , y j0 , then E they can measure the answer registers Aj and Bj , in ϕx0j yj0 , respectively, and send the answers to the referee. However, the probability that (x0 , y 0 ) = x0j , yj0 can be very small and they want to get this desired outcome with probability very close to 1. We next how this can be achieved. describe E Let ϕx0j be the resulting state obtained after we measure 0 register Xj (in |ϕi) and obtain outcome xj . The fact that E I Xj : Y Y˜ B is close to 0 implies that Bob’s side of ϕx0 j
x0j .
is mostly independent of By the unitary equivalence of purifications and Uhlmann’s theorem, there is a unitary transformation Ux0j that Alice apply to take the state can E 0 |ϕi quite close to the state ϕxj . Similarly, let us define E 0 ˜ being close to 0 implies that ϕyj and again I Yj : X XA E 0 Alice’s side of ϕyj is mostly independent of yj0 . Again, by Uhlmann’s theorem, there is a unitary transformation Uyj0 that Bob Ecan apply to take the state |ϕi quite close to the state 0 ϕyj . Interestingly, as was argued in [3], when Alice and Bob simultaneously U 0 and Uyj0 , they take |ϕi quite close apply E xj 0 0 to the state ϕxj yj ! This again requires the distribution of questions to be independent across Alice and Bob. Organization of the paper In Section II, we present some background on information theory, as well as some useful lemmas that we will need for our proof. In Section III, we prove our main result, Theorem I.2. II. P RELIMINARIES In this section we present some notations, definitions, facts, and lemmas that we will use later in our proof. Information theory
and trace preserving (CPTP) linear map from states to states. Readers can refer to [24], [25], [26] for more details. Definition II.1. For quantum states ρ and σ, the `1 -distance √ bedef tween them is given by kρ − σk1 , where kXk1 = Tr X † X is the sum of the singular values of X. We say that ρ is ε-close to σ if kρ − σk1 ≤ ε. Definition II.2. For quantum states ρ and σ, the fidelity def √ √ between them is given by F(ρ, σ) = ρ σ 1 . The following proposition states that the distance between two states can’t be increased by quantum operations. Proposition II.3 ([25], pages 406 and 414). For states ρ, σ, and quantum operation E(·), it holds that kE(ρ) − E(σ)k1 ≤ kρ − σk1 and F(E(ρ), E(σ)) ≥ F(ρ, σ). The following proposition relates the `1 -distance and the fidelity between two states. Proposition II.4 ([25], page 416). For quantum states ρ and σ, it holds that p 2(1 − F(ρ, σ)) ≤ kρ − σk1 ≤ 2 1 − F(ρ, σ)2 . For two pure states |φi and |ψi, we have q 2 k|φihφ| − |ψihψ|k1 = 1 − F(|φihφ| , |ψihψ|) p = 1 − |hφ|ψi|2 . Let ρAB be a bipartite quantum state in registers AB. We use the same symbol to represent a quantum register and the Hilbert space associated with it. We define def X def ρB = TrA ρAB = (hi| ⊗ 1B )ρAB (|ii ⊗ 1B )
For integer n ≥ 1, let [n] represent the set {1, 2, . . . , n}. i Let X and Y be finite sets and k be a natural number. Let k X be the set X × · · · × X , the cross product of X , k times. where {|ii}i is a basis for the Hilbert space A and 1B is the Let µ be a probability distribution on X . Let µ(x) represent identity matrix in space B. The state ρB is referred to as the the probability of x ∈ X according to µ. Let X be a random marginal state of ρAB in register B. variable distributed according to µ. We use the same symbol to represent a random variable and its distribution whenever it Definition II.5. We say that a pure state |ψi ∈ A ⊗ B is a is clear from the context. The expectation value of function f purification of some state ρ if TrA (|ψihψ|) = ρ. def P on X is defined as Ex←X [f (x)] = x∈X Pr[X = x] · f (x), Theorem II.6 (Uhlmann’s theorem). Given quantum states where x ← X means that x is drawn from the distribution ρ, σ, and a purification |ψi of ρ, it holds that F(ρ, σ) = of X. A quantum state (or just a state) ρ is a positive semi- max|φi |hφ|ψi|, where the maximum is taken over all purificadefinite matrix with trace equal to 1. It is pure if and only if tions of σ. the rank is 1. Let |ψi be a unit vector. With slight abuse of The entropy of a quantum state ρ (in register X) is defined notation, we use ψ to represent the state and also the density def as S(ρ) = −Trρ log ρ. We also let S (X)ρ represent S(ρ). The matrix |ψihψ|, associated with |ψi. A classical distribution µ between quantum states ρ and σ is defined can be viewed as a quantum state with diagonal entries µ(x) relative entropy def min-entropy and non-diagonal entries 0. For two quantum states ρ and σ, as S(ρkσ) = Trρ log ρ − Trρ log σ. The relative def ρ ⊗ σ represents the tensor product (Kronecker product) of ρ between them is defined as S∞ (ρkσ) = min λ : ρ ≤ 2λ σ . and σ. A quantum super-operator E(·) is a completely positive Since the logarithm is operator-monotone, S(ρkσ) ≤ S∞(ρkσ).
Let ρXY be a quantum state in space X ⊗ Y . The mutual information between registers X and Y is defined to be def
I(X : Y )ρ = S (X)ρ + S (Y )ρ − S (XY )ρ .
It is easy to see that I(X : Y )ρ =P S ρXY ρX ⊗ ρY . If X is a classical register, namely ρ = x µ(x) |xihx| ⊗ ρx , where µ is a probability distribution over X, then I(X : Y )ρ = S (Y )ρ − S (Y |X)ρ ! X X =S µ(x)ρx − µ(x)S (ρx ) x
Fact II.11. The relative entropy is non-increasing when XY XY subsystems are considered. be quantum
Let ρX Xand σ XY XY
states, then S ρ σ ≥S ρ σ . The following fact is easily verified. Fact II.12. Let 0 < ε, ε0 < 1, 0 < c, µ and µ0 be probability distributions on a set X , and f : X → [0, c] be a function. If Ex←µ [f (x)] ≤ ε and kµ − µ0 k1 ≤ ε0 then Ex←µ0 [f (x)] ≤ ε + cε0 . Useful lemmas Here we state and prove some lemmas that we will use later.
x
AB
Lemma II.13. Let |ψi be a bipartite pure state with the marginal state on register B being ρ. Let a 0/1 outcome def measurement be performed on register A with outcome O. S(Y |X)ρ = E [S(ρx )] . x←µ Let Pr [O = 1] = q. Let the marginal states on register B Let ρXY Z be a quantum state with Y being a classical register. conditioned on O = 0 and O = 1 be ρ0 and ρ1 respectively. The mutual information between X and Z, conditioned on Y , Then, S∞ (ρ1 kρ) ≤ log 1q . is defined as Proof. It is easily seen that ρ = qρ1 + (1 − q)ρ0 . Hence h i def S∞ (ρ1 kρ) ≤ log 1q . I(X : Z |Y )ρ = E I(X : Z |Y = y)ρ y←Y The following lemma states that when the concerned mutual = S (X|Y )ρ + S (Z|Y )ρ − S (XZ|Y )ρ . information is small, then a measurement on Alice’s side can The following chain rule for mutual information follows easily be simulated by a unitary operation on Alice’s side. from the definitions, when Y is a classical register. Lemma II.14. Let µ be a probability distribution on X . Let Xp ˜ def XX AB I(X : Y Z)ρ = I(X : Y )ρ + I(X : Z |Y )ρ . |ϕi = µ(x) |xxi ⊗ |ψx i
where the conditional entropy is defined as
x∈X
We will need the following basic facts. Fact II.7. The relative entropy is jointly convex in its arguments. That is, for quantum states ρ, ρ1 , σ, and σ 1 , and p ∈ [0, 1],
S pρ + (1 − p)ρ1 pσ + (1 − p)σ 1
≤ p · S(ρkσ) + (1 − p) · S ρ1 σ 1 . We have the following chain rule for the relative-entropy. Fact II.8. Let
˜ be a joint pure state of Alice and Bob, where registers XXA are with Alice and register B is with Bob. Let I(X : B)ϕ ≤ def
ε and |ϕx i = |xxi ⊗ |ψx i. There exist unitary operators ˜ {Ux }x∈X acting on XXA such that √ E [k|ϕx ihϕx | − (Ux ⊗ 1B ) |ϕihϕ| (U∗x ⊗ 1B )k1 ] ≤ 4 ε. x←µ
Proof. Let us denote the reduced state of Bob in |ϕx i and |ϕi by def
ρx = TrA (|ψx ihψx |) ρ=
X
µ(x) |xihx| ⊗ ρx
x
Using Fact II.10, it holds that x←µ
ρ1 =
def
ρ = TrXXA (|ϕihϕ|) . ˜
ε ≥ I(X : B) = E [S(ρx kρ)] ≥ 1 − E [F(ρx , ρ)] .
and X
and
µ1 (x) |xihx| ⊗ ρ1x .
x←µ
By the unitary equivalence of purifications and Theorem II.6, there exists a Ux for each x ∈ X such that
x
| hϕx | (Ux ⊗ 1B ) |ϕi | = F(ρx , ρ).
It holds that
S ρ1 ρ = S µ1 µ + E 1 S ρ1x ρx . x←µ
Fact II.9. For quantum states ρXY , σ X , and τ Y , it holds that
S ρXY σ X ⊗ τ Y ≥ S ρXY ρX ⊗ ρY = I(X : Y )ρ . Fact II.10 ([26], [27]). For quantum states ρ and σ, it holds that p kρ − σk1 ≤ S(ρkσ) and 1 − F(ρ, σ) ≤ S(ρkσ) .
The lemma follows from the following calculation. E [k|ϕx ihϕx | − (Ux ⊗ 1B ) |ϕihϕ| (U∗x ⊗ 1B )k1 ] i hp 1 − | hϕx | (Ux ⊗ 1B ) |ϕi |2 (4) =2 E x←µ r 2 ≤ 2 1 − E [| hϕx | (Ux ⊗ 1B ) |ϕi |] (5) x←µ r 2 = 2 1 − E [F(ρx , ρ)] x←µ √ ≤ 4 ε.
x←µ
where Eq. (4) follows from Proposition √ II.4 and at Eq. (5) we used the concavity of the function 1 − α2 . The following is a generalization of the above lemma that states that when the concerned mutual informations are small then the simultaneous measurements on Alice’s and Bob’s side can be simulated by unitary operations on Alice’s and Bob’s side. It is a special case of a more general result in Ref. [3]. Lemma II.15 ([3]). Let µ be a probability distribution over X × Y. Let µX and µY be the marginals of µ on X and Y. Let X p ˜ Y˜ Y def XX AB |ϕi = µ(x, y) |xxyyi ⊗ |ψx,y i x∈X ,y∈Y
˜ be a joint pure state of Alice and Bob, where registers XXA belong to Alice and registers Y˜ Y B belong to Bob. Let ˜ I X : BY Y˜ ≤ ε and I Y : AX X ≤ ε. ϕ
ϕ
def
Let |ϕx,y i = |xxyyi ⊗ |ψx,y i. There exist unitary operators ˜ {Ux }x∈X on XXA and {Vy }y∈Y on Y˜ Y B such that h i
|ϕx,y ihϕx,y | − (Ux ⊗ Vy ) |ϕihϕ| U∗x ⊗ Vy∗ E 1 (x,y)←µ √ ≤ 8 ε + 2 kµ − µX ⊗ µY k1 .
Using the above, we get the bound of Eq. (3) from the calculation that is on the bottom of this page. Equation (1) follows from the triangle inequality, the second term in Eq. (2) is because Ux doesn’t change the `1 -distance, and the first term in Eq. (2) follows from Proposition II.3 with the superoperator that corresponds to measuring Y in the standard basis and storing the outcome in a new register. The lemma follows from the following calculation. h i
|ϕx,y ihϕx,y | − (Ux ⊗ Vy ) |ϕihϕ| U∗x ⊗ Vy∗ E 1 (x,y)←µ
= E |xyihxy| ⊗ |ϕx,y ihϕx,y |
(x,y)←µ
∗ ∗ − |xyihxy| ⊗ (Ux ⊗ Vy ) |ϕihϕ| Ux ⊗ Vy
1
≤ E [|xyihxy| ⊗ |ϕx,y ihϕx,y |]
(x,y)←µ − E |xyihxy| (x,y)←µX ⊗µY
⊗ (Ux ⊗ Vy ) |ϕihϕ| U∗x ⊗ Vy∗
1
+ E |xyihxy|
(x,y)←µ X ⊗µY ⊗ (Ux ⊗ Vy ) |ϕihϕ| U∗x ⊗ Vy∗ − E |xyihxy| (x,y)←µ
⊗ (Ux ⊗ Vy ) |ϕihϕ| U∗x ⊗ Vy∗
1 √ ≤ 8 ε + 2 kµ − µX ⊗ µY k1
Proof. Let |ϕx i be the state obtained when we measure register X in |ϕi and obtain x. Similarly let |ϕy i be the state obtained when we measure register Y in |ϕi and obtain y. By Lemma II.14, there exist unitary operators {Ux }x∈X and {Vy }y∈Y such that where the first inequality follows from the triangle inequality √ ∗ E [k|ϕx ihϕx | − (Ux ⊗ 1B ) |ϕihϕ| (Ux ⊗ 1B )k1 ] ≤ 4 ε and at the last inequality we used Eq. (3) and Fact II.12. x←µX
III. P ROOF OF THE MAIN RESULT
and E
y←µY
h i √
|ϕy ihϕy | − (1A ⊗ Vy ) |ϕihϕ| 1A ⊗ Vy∗ ≤ 4 ε. 1
Let a game G = (µ, X , Y, A, B, V ) be given. We assume that the distribution µ = µX ⊗ µY is product across X and
∗ ∗
E [|xyihxy| ⊗ |ϕx,y ihϕx,y |] − E |xyihxy| ⊗ (Ux ⊗ Vy ) |ϕihϕ| Ux ⊗ Vy
(x,y)←µ (x,y)←µX ⊗µY 1
∗
≤ E [|xyihxy| ⊗ |ϕx,y ihϕx,y |] − E [|xyihxy| ⊗ (Ux ⊗ 1B ) |ϕy ihϕy | (Ux ⊗ 1B )]
(x,y)←µ (x,y)←µX ⊗µY 1
∗ ∗ ∗
+ E |xyihxy| ⊗ (Ux ⊗ 1B ) |ϕy ihϕy | (Ux ⊗ 1B ) − |xyihxy| ⊗ (Ux ⊗ Vy ) |ϕihϕ| Ux ⊗ Vy (x,y)←µX ⊗µY 1
∗ ≤ E [|xihx| ⊗ |ϕx ihϕx | − |xihx| ⊗ (Ux ⊗ 1B ) |ϕihϕ| (Ux ⊗ 1B )] x←µX 1
∗ + E |xyihxy| ⊗ |ϕ ihϕ | − |xyihxy| ⊗ (1 ⊗ V ) |ϕihϕ| 1 ⊗ V y y A y A y
xy←µX ⊗µY
(1)
(2)
1
E [k|ϕx ihϕx | − (Ux ⊗ 1B ) |ϕihϕ| (U∗x ⊗ 1B )k1 ] h i + E |ϕy ihϕy | − (1A ⊗ Vy ) |ϕihϕ| 1A ⊗ Vy∗ 1 y←µY √ ≤8 ε =
x←µX
(3)
Y. Before the game starts, Alice and Bob share a pure state Now, 0 0 on the registers AEA BEB , where A and B are used to store the answers for Alice and Bob, respectively. After getting the − log q + |C| · log(|A| · |B|)
h ˜ ˜ i inputs, Alice and Bob perform unitary operations independently ˜ ¯Y˜ ¯XY EA EB XC¯YC¯XY EA EB X C C ≥ E S ϕ
θ ∞ aC bC and then they measure registers A and B. The outcomes of aC bC ←ϕAC BC
h ˜ ˜ i the measurements are sent to the referee. Now, let’s consider ˜ ¯Y˜ ¯XY EA EB XC¯YC¯XY EA EB X C C k k k ≥ E S ϕ
θ aC bC the game G . Let x = x1 . . . xk ∈ X , y = y1 . . . yk ∈ Y , aC bC ←ϕAC BC k
h ˜ ˜ i a = a1 . . . ak ∈ Ak , and b = bQ 1 . . . bk ∈ B . To make notations ˜ ¯Y˜ ¯XY EA EB XC¯YC¯XY EA EB X C C ≥ E S ϕ
θ short, we denote µ(x, y) = µ(x , y ) and V (x, y, a, b) = x y x y a b C C i i C C C C i xC yC aC bC Q ←ϕXC YC AC BC i V (xi , yi , ai , bi ), whenever it is clear from the context. Let C ⊆ [k] and let C¯ represent its complement in [k]. Let xC represent the substring of x corresponding to the indices in C. where the last inequality follows from Fact II.8. (Similarly, we will use yC , aC , bC .) Let’s define def
|θi =
Xp
˜ Y˜ Y XX
µ(x, y) |xxyyi
x,y
⊗
X
AC BC
|aC bC i
EA EB
⊗ |γxyaC bC i
aC bC def
def
0 0 where EA = EA AC¯, EB = EB BC¯, and aC bC |aC bC i ⊗ |γxyaC bC i is the shared state after Alice and Bob performed their unitary operations corresponding to questions x and y. (Note that |γxyaC bC i is unnormalized.) Consider the state
P
Xp ˜ Y˜ Y def 1 XX |ϕi = √ µ(x, y) |xxyyi q x,y X A B E E ⊗ |aC bC i C C ⊗ |γxyaC bC i A B
For each i ∈ [k], let us define a binary random variable Ti ∈ {0, 1}, which indicates success in the i-th repetition. That is, Ti = V (Xi , Yi , Ai , Bi ). Our main theorem will follow from the following lemma. Lemma III.2. Let 0.1 > δ1 , δ2 , δ3 > 0 such that δ3 = δ2 + def δ1 · log(|A| · |B|). Let k 0 = bδ1 kc. For any quantum strategy k for the k-fold game G , there exists a set {i1 , . . . , ik0 }, such that for each 1 ≤ r ≤ k 0 − 1, either h i Pr T (r) = 1 ≤ 2−δ2 k or h i p Pr Tir+1 = 1 T (r) = 1 ≤ ω ∗ (G) + 12 10δ3
aC bC :V (xC ,yC ,aC ,bC )=1
where normalizer q is the probability of success on C.
E
r Y
Tij .
j=1
Lemma III.1. xC yC aC bC ←ϕXC YC AC BC
def
where T (r) =
i h ˜ ˜ X Y XY E E ˜ ˜ S ϕxCC¯yCC¯aC bC A B θxXCC¯yYCC¯XY EA EB
≤ − log q + |C| · log(|A| · |B|) .
Proof. In the following, we assume that 1 ≤ r < k 0 . However, the same argument also works when r = 0, i.e., for identifying the first coordinate, which we skip for the sake of avoiding repetition. Suppose that we have already identified r coordinates i1 , . . . , ir satisfying that
Proof. Note that, by Lemma II.13,
˜ ˜
˜ ˜ S∞ ϕXC¯YC¯XY EA EB θXC¯YC¯XY EA EB ≤ − log q.
Pr[Ti1 = 1] ≤ ω ∗ (G) + 12
p 10δ3
and
Let p(aC , bC ) be the probability of obtaining (aC , bC ) when measuring registers (AC , BC ) in |ϕi. Consider,
h i p Pr Tij+1 = 1 T (j) = 1 ≤ ω ∗ (G) + 12 10δ3
h ˜ ˜ i ˜ ¯Y˜ ¯XY EA EB X Y XY EA EB X C C S∞ ϕaCC¯bCC¯
θ aC bC ←ϕAC BC
h ˜ ˜ ˜ ¯Y˜ ¯XY EA EB X Y XY EA EB X C C ≤ E S∞ ϕaCC¯bCC¯
ϕ aC bC ←ϕAC BC
i ˜ ˜
˜ ˜ + S∞ ϕXC¯YC¯XY EA EB θXC¯YC¯XY EA EB
for 1 ≤ j ≤ r − 1. If Pr T (r) = 1 ≤ 2−δ2 k then we are done, so from now on, we assume that Pr T (r) = 1 > 2−δ2 k . def def Let C = {i1 , . . . , ir }. To simplify notations, let A˜ = def def ˜ C¯XEA , B ˜ = Y˜C¯Y EB , and Ri = XC YC X
Attila Pereszlényi
Penghui Yao
Centre for Quantum Technologies and Department of Computer Science National University of Singapore Singapore [email protected]
Centre for Quantum Technologies National University of Singapore Singapore [email protected]
Centre for Quantum Technologies National University of Singapore Singapore [email protected]
Abstract—We show a parallel repetition theorem for the entangled value ω ∗ (G) of any two-player one-round game G where the questions (x, y) ∈ X × Y to Alice and Bob are drawn from a product distribution on X × Y. We show that for the k-fold product Gk of the game G (which represents the game G played in parallel k times independently) k Ω log(|A|·|B|) ω ∗ Gk = 1 − (1 − ω ∗ (G))3 where A and B represent the sets from which the answers of Alice and Bob are drawn. The arguments we use are information theoretic and are broadly on similar lines as that of Raz [1] and Holenstein [2] for classical games. The additional quantum ingredients we need, to deal with entangled games, are inspired by the work of Jain, Radhakrishnan, and Sen [3], where quantum information theoretic arguments were used to achieve message compression in quantum communication protocols. Index Terms—parallel repetition theorem; two-player game; entangled value
I. I NTRODUCTION
and the famous Unique Games Conjecture. One of the most fundamental problems regarding this model is the so called parallel repetition question, which concerns the behavior of multiple copies of the game played in parallel. For the game G = (µ, X , Y, A, B, V ), its k-fold product is given by Gk = (µk , X k , Y k , Ak , B k , V k ), where V k (x, y, a, b) = 1 if and only if V (xi , yi , ai , bi ) = 1 for all i ∈ [k]. Namely, Alice and Bob play k copies of game G simultaneously, and they win iff they win all the copies. It is easily seen that ω(Gk ) ≥ ω(G)k for any game G. The equality of the two quantities, for all games, was conjectured by Ben-Or, Goldwasser, Kilian and Wigderson [4]. The conjecture was shown to be false by Fortnow [5]. However one could still expect that ω(Gk ) goes down exponentially in k (asymptotically). This is referred to as the parallel repetition (also known as the direct product) conjecture. This was shown to be indeed true in a seminal paper by Raz [1]. Raz showed that k ω Gk = (1 − (1 − ω(G))c )Ω( log(|A||B|) )
A two-player one-round game G is specified by finite sets X , Y, A, and B, a distribution µ over X × Y, and a predicate where c is a universal constant. This result, along with the V : X × Y × A × B → {0, 1}. It is played as follows. the PCP theorem had deep consequences for the theory of • The referee selects questions (x, y) ∈ X × Y according inapproximability [6], [7], [8]. A series of works later exhibited to distribution µ. improved results for general and specific games [2], [9], [10], • The referee sends x to Alice and y to Bob. Alice and Bob [11], [12]. are spatially separated, so they do not see each other’s In the quantum setting, it is natural to consider the so input. called entangled games where Alice and Bob are, in addition, • Alice chooses answer a ∈ A and sends it back to the allowed to share a quantum state before the games starts. referee. Bob chooses answer b ∈ B and sends it back to The questions and answers in the game remain classical. the referee. On receiving questions, Alice and Bob can generate their • The referee accepts if V (x, y, a, b) = 1 and otherwise answers by making quantum measurements on their shared rejects. Alice and Bob win the game if the referee accepts. entangled state. The value of the entangled version of the The value of the game G, denoted by ω(G), is defined to be the game G is denoted by ω ∗ (G). The study of entangled games maximum winning probability (averaged over the distribution is deeply related to the foundation of quantum mechanics and that of quantum entanglement. These games have been µ) achieved by Alice and Bob. These games have played an important and pivotal role used to give a novel interpretation to Bell inequalities, one in the study of the rich theory of inapproximability, leading of the most famous and useful methods for differentiating to the development of Probabilistically Checkable Proofs classical and quantum mechanics (e.g., by Clauser, Horne,
Shimony and Holt [13]). Recently these games have also been studied from cryptographic motivations such as in Refs. [14], [15], [16]. Analogously to the classical case, the study of the parallel repetition question in this setting may potentially have applications in quantum complexity theory. The parallel repetition conjecture has been shown to be true for several sub-classes of entangled games, starting with the so called XOR games by Cleve, Slofstra, Unger and Upadhyay [17], later generalized to unique games by Kempe, Regev and Toner [18] and very recently further generalized to projection games by Dinur, Steurer and Vidick [19] (following an analytical framework introduced by Dinur and Steurer in [20] to deal with classical projection games). For general games, Kempe and Vidick [21] (following a framework by Feige and Killian [22] for classical games) showed a parallel repetition theorem albeit with only a polynomial decay in k, in the value ω ∗ (Gk ). In a recent work, Chailloux and Scarpa [23] showed an exponential decay in ω ∗ (Gk ) using information theoretic arguments.
Our techniques
The arguments we use are information theoretic and are broadly on similar lines as that of Raz [1] and Holenstein [2] for classical games. The additional quantum ingredients we need, to deal with entangled games, are inspired by the work of Jain, Radhakrishnan, and Sen [3], where quantum information theoretic arguments were used to achieve message compression in quantum communication protocols. Given the k-fold game Gk , let us condition on success on a set C ⊆ [k] of coordinates. If the overall success in coordinates in C is already as small as we want, then we are done. Otherwise, we exhibit another coordinate j ∈ / C such that the success in the j-th coordinate, even when conditioning on success in the coordinates inside C, is bounded away from 1. Here we assume that ω ∗ (G) is bounded away from 1. This way the overall success keeps going down and becomes exponentially small in k, after we have identified Ω(k) such coordinates. To argue that the probability with which Alice and Bob win the j-th coordinate, conditioned on success in C, Theorem I.1 ([23]). For any game G = (µ, X , Y, A, B, V ), is bounded away from 1, we show that close to this success probability can be achieved for a single instance of the game where µ is the uniform distribution on X × Y, it holds that G. That is, given inputs (x0 , y 0 ), drawn from µ, for a single k Ω ( ) instance of G, Alice and Bob can embed (x0 , y 0 ) to the j-th ω ∗ Gk = 1 − (1 − ω ∗ (G))2 log(|A||B||X ||Y|) . coordinate of Gk , conditioned on success in C, and generate the rest of the state with good approximation. So, if the probability As a corollary, for a general distribution µ, of success in the j-th coordinate, conditioned on success in C, k is very close to 1, there is a strategy for G with probability of ∗ k ∗ 2 Ω Q4 log(Q) log(|A||B|) ω G = 1 − (1 − ω (G)) success strictly larger than ω ∗ (G), reaching a contradicting to the definition of ω ∗ (G). where Suppose the global state, conditioned on success in C, is of the form 1 np o X Q = max , |X | · |Y| . XY AB σ XY AB = µ ˜(x, y) |xyihxy| ⊗ |φxy ihφxy | minx,y:µ(x,y)6=0 µ(x, y) x∈X k ,y∈Y k ∗
k
Note that here ω (G ) depends on |X | · |Y| as well, in addition to |A| · |B| (as in Raz’s result). Also the value of Q can be arbitrarily large, depending on the distribution µ. Our result In this paper we consider the case when the distribution µ is product across X × Y. That is, there are distributions µX , µY on X , Y respectively such that ∀(x, y) ∈ X × Y : µ(x, y) = µX (x) · µY (y). We show the following.
where µ ˜ is a distribution, potentially different from µ because of the conditioning on success. (Here we further fix the questions and answers in C to specific values and do not specify them in σ XY AB .) In protocol P for the single instance of G, we let Alice and Bob start with the shared pure state X p ˜ Y˜ Y XX AB |ϕi = µ ˜(x, y) |xxyyi ⊗ |φxy i . x∈X k ,y∈Y k
Note that |ϕi is a purification of σ XY AB , where registers ˜ and Y˜ are identical to X and Y . We introduce these X Theorem I.2 (Main Result). For any game copies of the registers X and Y so that the marginal state in these registers remains a classical state and these registers G = (µ, X , Y, A, B, V ) can be viewed as classical registers, which is important in our arguments. where µ is a product distribution on X × Y, it holds that Using the chain rule for mutual we are able information, k ˜ B and I Yj : X XA ˜ ∗ k ∗ 3 Ω( log(|A||B|) ) to argue that both I X : Y Y are very j ω G = 1 − (1 − ω (G)) . small (close to 0), in |ϕi. This, obviously, is only possible Note that the uniform distribution on X × Y is a product when the distribution µ is product. In addition, the distribution distribution and our result has no dependence on the size of of the questions in the j-th coordinate, in |ϕi, remains close X × Y. Hence, our result implies and strengthens on the result to µ, in the `1 -distance. In protocol P, when Alice and Bob of Chailloux and Scarpa [23] (up to the exponent of 1−ω ∗ (G)). get questions x0 and y 0 , suppose they measure registers Xj
E and Yj , in |ϕi, and get x0j and yj0 . Let ϕx0j yj0 be the resulting state. If by luck it so happens that (x0 , y 0 ) = x0j , y j0 , then E they can measure the answer registers Aj and Bj , in ϕx0j yj0 , respectively, and send the answers to the referee. However, the probability that (x0 , y 0 ) = x0j , yj0 can be very small and they want to get this desired outcome with probability very close to 1. We next how this can be achieved. describe E Let ϕx0j be the resulting state obtained after we measure 0 register Xj (in |ϕi) and obtain outcome xj . The fact that E I Xj : Y Y˜ B is close to 0 implies that Bob’s side of ϕx0 j
x0j .
is mostly independent of By the unitary equivalence of purifications and Uhlmann’s theorem, there is a unitary transformation Ux0j that Alice apply to take the state can E 0 |ϕi quite close to the state ϕxj . Similarly, let us define E 0 ˜ being close to 0 implies that ϕyj and again I Yj : X XA E 0 Alice’s side of ϕyj is mostly independent of yj0 . Again, by Uhlmann’s theorem, there is a unitary transformation Uyj0 that Bob Ecan apply to take the state |ϕi quite close to the state 0 ϕyj . Interestingly, as was argued in [3], when Alice and Bob simultaneously U 0 and Uyj0 , they take |ϕi quite close apply E xj 0 0 to the state ϕxj yj ! This again requires the distribution of questions to be independent across Alice and Bob. Organization of the paper In Section II, we present some background on information theory, as well as some useful lemmas that we will need for our proof. In Section III, we prove our main result, Theorem I.2. II. P RELIMINARIES In this section we present some notations, definitions, facts, and lemmas that we will use later in our proof. Information theory
and trace preserving (CPTP) linear map from states to states. Readers can refer to [24], [25], [26] for more details. Definition II.1. For quantum states ρ and σ, the `1 -distance √ bedef tween them is given by kρ − σk1 , where kXk1 = Tr X † X is the sum of the singular values of X. We say that ρ is ε-close to σ if kρ − σk1 ≤ ε. Definition II.2. For quantum states ρ and σ, the fidelity def √ √ between them is given by F(ρ, σ) = ρ σ 1 . The following proposition states that the distance between two states can’t be increased by quantum operations. Proposition II.3 ([25], pages 406 and 414). For states ρ, σ, and quantum operation E(·), it holds that kE(ρ) − E(σ)k1 ≤ kρ − σk1 and F(E(ρ), E(σ)) ≥ F(ρ, σ). The following proposition relates the `1 -distance and the fidelity between two states. Proposition II.4 ([25], page 416). For quantum states ρ and σ, it holds that p 2(1 − F(ρ, σ)) ≤ kρ − σk1 ≤ 2 1 − F(ρ, σ)2 . For two pure states |φi and |ψi, we have q 2 k|φihφ| − |ψihψ|k1 = 1 − F(|φihφ| , |ψihψ|) p = 1 − |hφ|ψi|2 . Let ρAB be a bipartite quantum state in registers AB. We use the same symbol to represent a quantum register and the Hilbert space associated with it. We define def X def ρB = TrA ρAB = (hi| ⊗ 1B )ρAB (|ii ⊗ 1B )
For integer n ≥ 1, let [n] represent the set {1, 2, . . . , n}. i Let X and Y be finite sets and k be a natural number. Let k X be the set X × · · · × X , the cross product of X , k times. where {|ii}i is a basis for the Hilbert space A and 1B is the Let µ be a probability distribution on X . Let µ(x) represent identity matrix in space B. The state ρB is referred to as the the probability of x ∈ X according to µ. Let X be a random marginal state of ρAB in register B. variable distributed according to µ. We use the same symbol to represent a random variable and its distribution whenever it Definition II.5. We say that a pure state |ψi ∈ A ⊗ B is a is clear from the context. The expectation value of function f purification of some state ρ if TrA (|ψihψ|) = ρ. def P on X is defined as Ex←X [f (x)] = x∈X Pr[X = x] · f (x), Theorem II.6 (Uhlmann’s theorem). Given quantum states where x ← X means that x is drawn from the distribution ρ, σ, and a purification |ψi of ρ, it holds that F(ρ, σ) = of X. A quantum state (or just a state) ρ is a positive semi- max|φi |hφ|ψi|, where the maximum is taken over all purificadefinite matrix with trace equal to 1. It is pure if and only if tions of σ. the rank is 1. Let |ψi be a unit vector. With slight abuse of The entropy of a quantum state ρ (in register X) is defined notation, we use ψ to represent the state and also the density def as S(ρ) = −Trρ log ρ. We also let S (X)ρ represent S(ρ). The matrix |ψihψ|, associated with |ψi. A classical distribution µ between quantum states ρ and σ is defined can be viewed as a quantum state with diagonal entries µ(x) relative entropy def min-entropy and non-diagonal entries 0. For two quantum states ρ and σ, as S(ρkσ) = Trρ log ρ − Trρ log σ. The relative def ρ ⊗ σ represents the tensor product (Kronecker product) of ρ between them is defined as S∞ (ρkσ) = min λ : ρ ≤ 2λ σ . and σ. A quantum super-operator E(·) is a completely positive Since the logarithm is operator-monotone, S(ρkσ) ≤ S∞(ρkσ).
Let ρXY be a quantum state in space X ⊗ Y . The mutual information between registers X and Y is defined to be def
I(X : Y )ρ = S (X)ρ + S (Y )ρ − S (XY )ρ .
It is easy to see that I(X : Y )ρ =P S ρXY ρX ⊗ ρY . If X is a classical register, namely ρ = x µ(x) |xihx| ⊗ ρx , where µ is a probability distribution over X, then I(X : Y )ρ = S (Y )ρ − S (Y |X)ρ ! X X =S µ(x)ρx − µ(x)S (ρx ) x
Fact II.11. The relative entropy is non-increasing when XY XY subsystems are considered. be quantum
Let ρX Xand σ XY XY
states, then S ρ σ ≥S ρ σ . The following fact is easily verified. Fact II.12. Let 0 < ε, ε0 < 1, 0 < c, µ and µ0 be probability distributions on a set X , and f : X → [0, c] be a function. If Ex←µ [f (x)] ≤ ε and kµ − µ0 k1 ≤ ε0 then Ex←µ0 [f (x)] ≤ ε + cε0 . Useful lemmas Here we state and prove some lemmas that we will use later.
x
AB
Lemma II.13. Let |ψi be a bipartite pure state with the marginal state on register B being ρ. Let a 0/1 outcome def measurement be performed on register A with outcome O. S(Y |X)ρ = E [S(ρx )] . x←µ Let Pr [O = 1] = q. Let the marginal states on register B Let ρXY Z be a quantum state with Y being a classical register. conditioned on O = 0 and O = 1 be ρ0 and ρ1 respectively. The mutual information between X and Z, conditioned on Y , Then, S∞ (ρ1 kρ) ≤ log 1q . is defined as Proof. It is easily seen that ρ = qρ1 + (1 − q)ρ0 . Hence h i def S∞ (ρ1 kρ) ≤ log 1q . I(X : Z |Y )ρ = E I(X : Z |Y = y)ρ y←Y The following lemma states that when the concerned mutual = S (X|Y )ρ + S (Z|Y )ρ − S (XZ|Y )ρ . information is small, then a measurement on Alice’s side can The following chain rule for mutual information follows easily be simulated by a unitary operation on Alice’s side. from the definitions, when Y is a classical register. Lemma II.14. Let µ be a probability distribution on X . Let Xp ˜ def XX AB I(X : Y Z)ρ = I(X : Y )ρ + I(X : Z |Y )ρ . |ϕi = µ(x) |xxi ⊗ |ψx i
where the conditional entropy is defined as
x∈X
We will need the following basic facts. Fact II.7. The relative entropy is jointly convex in its arguments. That is, for quantum states ρ, ρ1 , σ, and σ 1 , and p ∈ [0, 1],
S pρ + (1 − p)ρ1 pσ + (1 − p)σ 1
≤ p · S(ρkσ) + (1 − p) · S ρ1 σ 1 . We have the following chain rule for the relative-entropy. Fact II.8. Let
˜ be a joint pure state of Alice and Bob, where registers XXA are with Alice and register B is with Bob. Let I(X : B)ϕ ≤ def
ε and |ϕx i = |xxi ⊗ |ψx i. There exist unitary operators ˜ {Ux }x∈X acting on XXA such that √ E [k|ϕx ihϕx | − (Ux ⊗ 1B ) |ϕihϕ| (U∗x ⊗ 1B )k1 ] ≤ 4 ε. x←µ
Proof. Let us denote the reduced state of Bob in |ϕx i and |ϕi by def
ρx = TrA (|ψx ihψx |) ρ=
X
µ(x) |xihx| ⊗ ρx
x
Using Fact II.10, it holds that x←µ
ρ1 =
def
ρ = TrXXA (|ϕihϕ|) . ˜
ε ≥ I(X : B) = E [S(ρx kρ)] ≥ 1 − E [F(ρx , ρ)] .
and X
and
µ1 (x) |xihx| ⊗ ρ1x .
x←µ
By the unitary equivalence of purifications and Theorem II.6, there exists a Ux for each x ∈ X such that
x
| hϕx | (Ux ⊗ 1B ) |ϕi | = F(ρx , ρ).
It holds that
S ρ1 ρ = S µ1 µ + E 1 S ρ1x ρx . x←µ
Fact II.9. For quantum states ρXY , σ X , and τ Y , it holds that
S ρXY σ X ⊗ τ Y ≥ S ρXY ρX ⊗ ρY = I(X : Y )ρ . Fact II.10 ([26], [27]). For quantum states ρ and σ, it holds that p kρ − σk1 ≤ S(ρkσ) and 1 − F(ρ, σ) ≤ S(ρkσ) .
The lemma follows from the following calculation. E [k|ϕx ihϕx | − (Ux ⊗ 1B ) |ϕihϕ| (U∗x ⊗ 1B )k1 ] i hp 1 − | hϕx | (Ux ⊗ 1B ) |ϕi |2 (4) =2 E x←µ r 2 ≤ 2 1 − E [| hϕx | (Ux ⊗ 1B ) |ϕi |] (5) x←µ r 2 = 2 1 − E [F(ρx , ρ)] x←µ √ ≤ 4 ε.
x←µ
where Eq. (4) follows from Proposition √ II.4 and at Eq. (5) we used the concavity of the function 1 − α2 . The following is a generalization of the above lemma that states that when the concerned mutual informations are small then the simultaneous measurements on Alice’s and Bob’s side can be simulated by unitary operations on Alice’s and Bob’s side. It is a special case of a more general result in Ref. [3]. Lemma II.15 ([3]). Let µ be a probability distribution over X × Y. Let µX and µY be the marginals of µ on X and Y. Let X p ˜ Y˜ Y def XX AB |ϕi = µ(x, y) |xxyyi ⊗ |ψx,y i x∈X ,y∈Y
˜ be a joint pure state of Alice and Bob, where registers XXA belong to Alice and registers Y˜ Y B belong to Bob. Let ˜ I X : BY Y˜ ≤ ε and I Y : AX X ≤ ε. ϕ
ϕ
def
Let |ϕx,y i = |xxyyi ⊗ |ψx,y i. There exist unitary operators ˜ {Ux }x∈X on XXA and {Vy }y∈Y on Y˜ Y B such that h i
|ϕx,y ihϕx,y | − (Ux ⊗ Vy ) |ϕihϕ| U∗x ⊗ Vy∗ E 1 (x,y)←µ √ ≤ 8 ε + 2 kµ − µX ⊗ µY k1 .
Using the above, we get the bound of Eq. (3) from the calculation that is on the bottom of this page. Equation (1) follows from the triangle inequality, the second term in Eq. (2) is because Ux doesn’t change the `1 -distance, and the first term in Eq. (2) follows from Proposition II.3 with the superoperator that corresponds to measuring Y in the standard basis and storing the outcome in a new register. The lemma follows from the following calculation. h i
|ϕx,y ihϕx,y | − (Ux ⊗ Vy ) |ϕihϕ| U∗x ⊗ Vy∗ E 1 (x,y)←µ
= E |xyihxy| ⊗ |ϕx,y ihϕx,y |
(x,y)←µ
∗ ∗ − |xyihxy| ⊗ (Ux ⊗ Vy ) |ϕihϕ| Ux ⊗ Vy
1
≤ E [|xyihxy| ⊗ |ϕx,y ihϕx,y |]
(x,y)←µ − E |xyihxy| (x,y)←µX ⊗µY
⊗ (Ux ⊗ Vy ) |ϕihϕ| U∗x ⊗ Vy∗
1
+ E |xyihxy|
(x,y)←µ X ⊗µY ⊗ (Ux ⊗ Vy ) |ϕihϕ| U∗x ⊗ Vy∗ − E |xyihxy| (x,y)←µ
⊗ (Ux ⊗ Vy ) |ϕihϕ| U∗x ⊗ Vy∗
1 √ ≤ 8 ε + 2 kµ − µX ⊗ µY k1
Proof. Let |ϕx i be the state obtained when we measure register X in |ϕi and obtain x. Similarly let |ϕy i be the state obtained when we measure register Y in |ϕi and obtain y. By Lemma II.14, there exist unitary operators {Ux }x∈X and {Vy }y∈Y such that where the first inequality follows from the triangle inequality √ ∗ E [k|ϕx ihϕx | − (Ux ⊗ 1B ) |ϕihϕ| (Ux ⊗ 1B )k1 ] ≤ 4 ε and at the last inequality we used Eq. (3) and Fact II.12. x←µX
III. P ROOF OF THE MAIN RESULT
and E
y←µY
h i √
|ϕy ihϕy | − (1A ⊗ Vy ) |ϕihϕ| 1A ⊗ Vy∗ ≤ 4 ε. 1
Let a game G = (µ, X , Y, A, B, V ) be given. We assume that the distribution µ = µX ⊗ µY is product across X and
∗ ∗
E [|xyihxy| ⊗ |ϕx,y ihϕx,y |] − E |xyihxy| ⊗ (Ux ⊗ Vy ) |ϕihϕ| Ux ⊗ Vy
(x,y)←µ (x,y)←µX ⊗µY 1
∗
≤ E [|xyihxy| ⊗ |ϕx,y ihϕx,y |] − E [|xyihxy| ⊗ (Ux ⊗ 1B ) |ϕy ihϕy | (Ux ⊗ 1B )]
(x,y)←µ (x,y)←µX ⊗µY 1
∗ ∗ ∗
+ E |xyihxy| ⊗ (Ux ⊗ 1B ) |ϕy ihϕy | (Ux ⊗ 1B ) − |xyihxy| ⊗ (Ux ⊗ Vy ) |ϕihϕ| Ux ⊗ Vy (x,y)←µX ⊗µY 1
∗ ≤ E [|xihx| ⊗ |ϕx ihϕx | − |xihx| ⊗ (Ux ⊗ 1B ) |ϕihϕ| (Ux ⊗ 1B )] x←µX 1
∗ + E |xyihxy| ⊗ |ϕ ihϕ | − |xyihxy| ⊗ (1 ⊗ V ) |ϕihϕ| 1 ⊗ V y y A y A y
xy←µX ⊗µY
(1)
(2)
1
E [k|ϕx ihϕx | − (Ux ⊗ 1B ) |ϕihϕ| (U∗x ⊗ 1B )k1 ] h i + E |ϕy ihϕy | − (1A ⊗ Vy ) |ϕihϕ| 1A ⊗ Vy∗ 1 y←µY √ ≤8 ε =
x←µX
(3)
Y. Before the game starts, Alice and Bob share a pure state Now, 0 0 on the registers AEA BEB , where A and B are used to store the answers for Alice and Bob, respectively. After getting the − log q + |C| · log(|A| · |B|)
h ˜ ˜ i inputs, Alice and Bob perform unitary operations independently ˜ ¯Y˜ ¯XY EA EB XC¯YC¯XY EA EB X C C ≥ E S ϕ
θ ∞ aC bC and then they measure registers A and B. The outcomes of aC bC ←ϕAC BC
h ˜ ˜ i the measurements are sent to the referee. Now, let’s consider ˜ ¯Y˜ ¯XY EA EB XC¯YC¯XY EA EB X C C k k k ≥ E S ϕ
θ aC bC the game G . Let x = x1 . . . xk ∈ X , y = y1 . . . yk ∈ Y , aC bC ←ϕAC BC k
h ˜ ˜ i a = a1 . . . ak ∈ Ak , and b = bQ 1 . . . bk ∈ B . To make notations ˜ ¯Y˜ ¯XY EA EB XC¯YC¯XY EA EB X C C ≥ E S ϕ
θ short, we denote µ(x, y) = µ(x , y ) and V (x, y, a, b) = x y x y a b C C i i C C C C i xC yC aC bC Q ←ϕXC YC AC BC i V (xi , yi , ai , bi ), whenever it is clear from the context. Let C ⊆ [k] and let C¯ represent its complement in [k]. Let xC represent the substring of x corresponding to the indices in C. where the last inequality follows from Fact II.8. (Similarly, we will use yC , aC , bC .) Let’s define def
|θi =
Xp
˜ Y˜ Y XX
µ(x, y) |xxyyi
x,y
⊗
X
AC BC
|aC bC i
EA EB
⊗ |γxyaC bC i
aC bC def
def
0 0 where EA = EA AC¯, EB = EB BC¯, and aC bC |aC bC i ⊗ |γxyaC bC i is the shared state after Alice and Bob performed their unitary operations corresponding to questions x and y. (Note that |γxyaC bC i is unnormalized.) Consider the state
P
Xp ˜ Y˜ Y def 1 XX |ϕi = √ µ(x, y) |xxyyi q x,y X A B E E ⊗ |aC bC i C C ⊗ |γxyaC bC i A B
For each i ∈ [k], let us define a binary random variable Ti ∈ {0, 1}, which indicates success in the i-th repetition. That is, Ti = V (Xi , Yi , Ai , Bi ). Our main theorem will follow from the following lemma. Lemma III.2. Let 0.1 > δ1 , δ2 , δ3 > 0 such that δ3 = δ2 + def δ1 · log(|A| · |B|). Let k 0 = bδ1 kc. For any quantum strategy k for the k-fold game G , there exists a set {i1 , . . . , ik0 }, such that for each 1 ≤ r ≤ k 0 − 1, either h i Pr T (r) = 1 ≤ 2−δ2 k or h i p Pr Tir+1 = 1 T (r) = 1 ≤ ω ∗ (G) + 12 10δ3
aC bC :V (xC ,yC ,aC ,bC )=1
where normalizer q is the probability of success on C.
E
r Y
Tij .
j=1
Lemma III.1. xC yC aC bC ←ϕXC YC AC BC
def
where T (r) =
i h ˜ ˜ X Y XY E E ˜ ˜ S ϕxCC¯yCC¯aC bC A B θxXCC¯yYCC¯XY EA EB
≤ − log q + |C| · log(|A| · |B|) .
Proof. In the following, we assume that 1 ≤ r < k 0 . However, the same argument also works when r = 0, i.e., for identifying the first coordinate, which we skip for the sake of avoiding repetition. Suppose that we have already identified r coordinates i1 , . . . , ir satisfying that
Proof. Note that, by Lemma II.13,
˜ ˜
˜ ˜ S∞ ϕXC¯YC¯XY EA EB θXC¯YC¯XY EA EB ≤ − log q.
Pr[Ti1 = 1] ≤ ω ∗ (G) + 12
p 10δ3
and
Let p(aC , bC ) be the probability of obtaining (aC , bC ) when measuring registers (AC , BC ) in |ϕi. Consider,
h i p Pr Tij+1 = 1 T (j) = 1 ≤ ω ∗ (G) + 12 10δ3
h ˜ ˜ i ˜ ¯Y˜ ¯XY EA EB X Y XY EA EB X C C S∞ ϕaCC¯bCC¯
θ aC bC ←ϕAC BC
h ˜ ˜ ˜ ¯Y˜ ¯XY EA EB X Y XY EA EB X C C ≤ E S∞ ϕaCC¯bCC¯
ϕ aC bC ←ϕAC BC
i ˜ ˜
˜ ˜ + S∞ ϕXC¯YC¯XY EA EB θXC¯YC¯XY EA EB
for 1 ≤ j ≤ r − 1. If Pr T (r) = 1 ≤ 2−δ2 k then we are done, so from now on, we assume that Pr T (r) = 1 > 2−δ2 k . def def Let C = {i1 , . . . , ir }. To simplify notations, let A˜ = def def ˜ C¯XEA , B ˜ = Y˜C¯Y EB , and Ri = XC YC X
