CS/Ph120 Homework 4 Solutions

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1 CS/Ph10 Homework 4 Solutions November 3, 016 Problem 1: Robustness of GHZ and W states, part Solution: Due to Bolton Bailey a For the GHZ state, we have T r N GHZ N GHZ N = 1 0 N 1 0 N N 1 1 N 1 So therefore, the rank is r GHZ =. For the W state, we have N 1 W N = N W N N 1 1 N So we get T r N W N W N = N 1 N W N 1 W N N 0 N 1 0 N 1 And so the rank is r W =. b The minimum purity for a density matrix ρ on a d-dimensional vector space is 1. To see that this d can be attained, consider ρ = 1 d I d This matrix is positive semidefinite as a postive multiple of the identiy, and it has trace 1, since the I d has trace d. Note that And so ρ = 1 d I d = 1 d I d T rρ = 1 d d = 1 d 1

2 To see it is impossible to have a density matrix of this dimension with a smaller purity, note that T rρ = i ρ i 1 i d To see it is impossible to have a density matrix of this dimension with a smaller purity, let ρ = i p i ψ i ψ i And so therefore ρ = i,j p i p j ψ i ψ i ψ j ψ j By linearity of the trace T rρ = T r p i p j ψ i ψ i ψ j ψ j i,j T rρ = p i p j T r ψ i ψ i ψ j ψ j i,j By invariance of trace under cyclic permutations T rρ = i,j p i p j T r ψ j ψ i ψ i ψ j T rρ = i,j p i p j ψ j ψ i ψ i ψ j T rρ = i,j p i p j ψ i ψ j Now, we separate the sum into the cases where i = j and i j T rρ = i p i ψ i ψ i + i j p i p j ψ i ψ j Since ψ i ψ i = 1, we get T rρ = i p i + i j p i p j ψ i ψ j Now, since the p i sum to 1, the minimum value of the sum of the p i is 1 by the Cauchy-Schwarz d inequality. The minimum value of ψ i ψ j is 0 since the norm squared is nonnegative. T rρ 1 d As we claimed. The maximum value of the purity of a density matrix on d dimensions is 1. To see that this can be attained, consider ρ = 0 0

3 Which is a pure density matrix, and satisfies T rρ = T r = T r = 1 To see that this is maximal, recall that we have shown for arbitrary ρ = p i ψ i ψ i i That T rρ = i,j p i p j ψ i ψ j And since 0 ψ i ψ j 1, and p i, p j are positive T rρ = i,j p i p j ψ i ψ j i,j p i p j 1 1 = 1 So 1 is the maximum purity. c As a state gets more entangled, we expect the purity to decrease. We think of a state being more entangled as the partial state being more heavily correlated with the other half of the bipartite state. Thus, if we trace out the other state, the partial state will be more mixed. d Again, we have T r N GHZ N GHZ N = 1 0 N 1 0 N N 1 1 N 1 And the purity of this state is T r T r N GHZ N GHZ N 1 = T r 0 N 1 0 N N 1 1 N = T r 0 N 1 0 N N 1 1 N 1 4 = 1 So in the limit as N, the purity of this state is 1 e Evaluating the purity T r T r N W N W N N 1 = T r W N 1 W N N 0 N 1 0 N 1 N N 1 = T r W N 1 W N 1 + N N 1 1 = + N N = N N + N 1 0 N 1 0 N 1 N 3

4 So in the limit as N, the purity of this state is 1. Since this value is higher than that for the GHZ states, we can conclude the W states are more robust, as they remain mostly pure even when a bit is traced out. f We now repeat the analysis tracing out n qubits instead of 1, which we will indicate by T r B, Again, we have T r B GHZ N GHZ N = 1 0 N n 0 N n + 1 N n 1 N n And the purity of this state is T r T r B GHZ N GHZ N 1 = T r 0 N n 0 N n + 1 N n 1 N n 4 1 = T r 0 N n 0 N n + 1 N n 1 N n 4 = 1 So no matter how many bits are traced out of the GHZ density matrix, the purity is 1 and so in the limit as N, the purity is 1. Evaluating the purity for the W density matrix T r T r B W N W N N n = T r N W N 1 W N 1 + n N 0 N n 0 N n N n n = T r W N n W N n + 0 N n 0 N n N N N n n = + N N = N Nn + n N So in the limit as N, the purity of this state still goes to 1 although slower for larger N. Since this value is higher than that for the GHZ states, we can again conclude the W states are more robust, as they remain mostly pure even when many bits are traced out. Problem : Dimension of a purifying system Solution: Due to De Huang a Let r be the Schmidt rank of Ψ AB, then D r. On the other hand, given ρ A = , 4

5 we have r = ranktr B Ψ Ψ AB = rankρ A =. Therefore D, the minimum value of D is no less than. In particular, if we let B be the space of single qubitd =, and take Ψ AB = 1 0 A 0 B + 3 A 1 B, then we have tr B Ψ Ψ AB = ρ A. Thus the minimum of D is. b We can rewrite ρ A as ρ A = Notice that 1, 1 + 3, are orthogonal to each other, we have Agian we have rankρ A = 3. D ranktr B Ψ Ψ AB = rankρ A = 3. In particular, if we let B be a 3 dimensional qudit space, and take Ψ AB = A 0 B A 1 B A B, then we have tr B Ψ Ψ AB = ρ A. Thus the minimum of D is 3. c For a general ρ A, let r be the rank of ρ A. Consider the eigenvalue decomposition of ρ A, r 1 ρ A = λ i φ i φ i, where φ i, i = 0, 1,..., r 1, are normalized orthogonal eigen states, λ i > 0, i = 0, 1,..., r 1, and r 1 λ i = 1. If tr B Ψ Ψ AB = ρ A, then D Schmidt rank Ψ AB = ranktr B Ψ Ψ AB = rankρ A = r. In particular, let B be a r dimensional qudit space with standard baisi { i, i = 0, 1,..., r 1}, and take r 1 Ψ AB = λi φ i A i B, then we have tr B Ψ Ψ AB = ρ A, and thus the minimum of D is r. d Still, consider the eigenvalue decomposition of ρ A with rank r, r 1 ρ A = λ i φ i φ i, 5

6 where { φ i, i = 0, 1,..., d 1} is a orthogonal basis of the qudit space A, λ i > 0, i = 0, 1,..., r 1, and r 1 λ i = 1. Also consider the eigenvalue decomposition of σ AB, r 1 σ AB = s k Ψ k Ψ k AB, k=0 where r = rankσ AB, and s k > 0, k = 0, 1,..., r 1. i First we find the lowest attainable rank of σ AB. For each Ψ k AB, we have ranktr B Ψ k Ψ k AB = Schmidt rank Ψ k AB m, since the dimension of B system is m Then we have ranktr B σ AB = rank r 1 s k tr B Ψ k Ψ k AB r 1 rank tr B Ψ k Ψ k AB r m. k=0 But since tr B σ AB = ρ A, we have k=0 ranktr B σ AB = rankρ A = r, and thus we get r m r, i.e. r r m, since r is an integer. Let p = r, q = r mp, then we can take m s k = λ mk+i, Ψ k AB = 1 sk λmk+i φ mk+i A i B, k = 1,,..., p 1, q 1 s p = λ mp+i, Ψ p AB = 1 q 1 λmp+i φ mp+i A i B, if q > 0. sp It s easy to check that Ψ k, k = 0, 1,..., p, are orthogonal to each other, and that tr B σ AB = ρ A, r = rankσ AB = p + signq 1 = p + r m r m = r m. In this case the lower-bound is achieved. Thus the minimum attainable rank of σ AB is r m. ii Next we find the highest attainable purity of σ AB, i.e. the maximum of r 1 trσab = s k. We now should find out the constraints on s = s 0, s,..., s rnotice the we now have no upper bound on r, even though there is indeed one as r m r. Let Ω be the 6 k=0

7 feasible set for s, and that s be an optimal solution that achieves the maximum of attainable purity. The first constraints on s are s k 0, k = 0, 1,..., r, r 1 s k = 1, since σ AB is a density matrix. Before we go further, we give a Lemma: if a b 0, c d 0, a + b = c + d = e, a c, then a + b c + d. Proof: Now we inductively definde k=0 a + b c + d = a ca + c + b db + d 0. = a ca + c + c ae a c = a ca + c e g 0 = max s Ω s 0, g k = max s Ω,s i =g i,,...,k 1 s k, k = 1,,..., r 1, then using the lemma, we can always make s k = g k, k = 0, 1,..., r 1it s not a hard proof, and we skip it here. We should find out what g k are. Let s assume that in the expression of σ AB we give above, where r 1 j=0 r 1 Ψ k AB = j=0 α k i,j φ i A j B, α k i,j = Ψ k Ψ k = 1, k = 0, 1,..., r 1, r 1 j=0 α k i,jα l i,j = Ψ l Ψ k = 0, k l, since Ψ k AB are normalized and orthogonal to each other. The condition tr B σ AB = ρ A gives that r 1 λ i = s k αi,j k, i = 0, 1,..., r 1. k=0 Now define matrices A k C r m as j=0 A k i,j = s k a k i,j, k = 0, 1,..., r 1, then all the constraints above can be summarized as λ 0 r 1 A k A k = Λ = λ 1.. k=0., λ r 1 7

8 tra k A k = s k, k = 0, 1,..., r 1; tra k A l = 0, k l. We first give a lemma without proof: if m r, then max trp P 1,P Ω 1 MP = m σ i M, M C r r, i=1 where Ω = {P C r m : P P = I m }, and σ i M denote the i th large singular value of M. Assume that λ 0 λ 1... λ r > 0. Then for our case, the lemma above reduces to max trp P 1,P Ω 1 ΛP = λ m. Consider the reduced eigenvalue decomposition of A 0 A k 0, A 0 A k 0 = QΣQ, where Q Ω, and Σ C m m is a positive diagonal matrix. Then since we have r 1 A 0 A k 0 A k A k = Λ, = Σ = Q A 0 A k 0Q Q ΛQ, k=0 s 0 = tra 0 A k 0 = trσ trq ΛQ max trp P 1,P Ω 1 ΛP = λ m, thus g0 λ m. Recall that in di we provided an example in which s 0 = λ m, therefore we indeed have g0 = λ m. Let p = r, q = r mp. Then similarly we can prove that m g k = λ mk+i, k = 0, 1,..., p 1, and if q > 0, g p = q 1 λ mp+i. Now using the previous result, we can always make s k = g k = q 1 λ mk+i, k = 0, 1,..., p 1; s p = gp = λ mp+i, if q > 0, and therefore the highest attainable purity of σ AB is p 1 p 1 q 1, s k + signqs p = λ mk+i + signq λ mp+i k=0 k=0 where signq = 1, if q > 0; signq = 0, if q = 0. Indeed the example in di, s k = λ mk+i, Ψ k AB = 1 sk λmk+i φ mk+i A i B, k = 1,,..., p 1, q 1 s p = λ mp+i, Ψ p AB = 1 q 1 λmp+i φ mp+i A i B, if q > 0, sp achieves this maximum of attainable purity.

9 Problem 3: Secret sharing among three people Solution: Due to Bolton Bailey a We compute the reduced density matrix ρ A Ψ Ψ = 1 0 A 0 B 0 C + 1 b 1 A 1 B 1 C 1 0 A 0 B 0 t C + 1 b 1 A 1 B 1 t C Ψ Ψ = ABC + 1 b ABC + 1 b ABC ABC + Ψ Ψ = A BC + 1 b 0 1 A BC + 1 b 1 0 A BC A B T r BC Ψ Ψ = A A = 1 I So ρ A is the maximally mixed state, no matter the value of b. Thus, A alone gains no information about the secret. By the symmetry of the state Ψ, we see that the other density matrices are the same T r AC Ψ Ψ = T r AB Ψ Ψ = 1 I ρ A = ρ B = ρ C = 1 I So none of the three can recover the secret on their own. b We compute the two-party reduced density matrix ρ BC, from part b Ψ Ψ = A BC + 1 b 0 1 A BC + 1 b 1 0 A BC A B So T r A Ψ Ψ = A A So ρ BC is the same mixed state, no matter the value of b. Thus, B and C together gain no information about the secret. By the symmetry of the state Ψ, we see that the other two-party density matrices are again the same T r B Ψ Ψ = T r C Ψ Ψ = A A So these pairs also do not have any information about the secret. c Consider the following LOCC protocol. Alice, Bob and Charlie all apply local Hadamard transformations to their qubit. This yields the state H H H = 1 [ b ] = 1 [ b ] 9

10 So if b = 0, the state is now And if b = 1, the state is now Now, Alice, Bob, and Charlie measure in the standard basis. We note that these three measurements are equivalent to a measurement in the computational in the product space. Thus, if b = 0, the only possible measurements for Alice, Bob, and, Charlie are 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0. If b = 1, the only possible measurements for Alice, Bob, and, Charlie are 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1 Thus, to determine b, Bob and CHarlie send the bits from their measurements to Alice and Alice computes the parity of the three measurements, which is then equal to b. Problem 4: Non-local boxes Solution: Due to De Huang a U For any input x, y A 1 A, we have pa, b x, y = = 1. PR If x, y 1, 1, pa, b x, y = p0, 0 x, y + p1, 1 x, y = = 1, if x, y = 1, 1, pa, b x, y = p1, 0 x, y + p0, 1 x, y = = 1. CH If x, y 1, 1, pa, b x, y = 1 π cos + 1 π cos + 1 π sin + 1 π sin = 1, if x, y = 1, 1, pa, b x, y = 1 π sin + 1 π sin + 1 π cos + 1 π cos = 1. 10

11 SIG For any input x, y A 1 A, we have pa, b x, y = py, x x, y = 1. b Let pa, x, y = b {0,1} pa, b x, y denote the marginal probability of the first output being a given the input x, y, and p, b x, y = pa, b x, y a {0,1} the marginal probability of the second output being b given the input x, y. U It s non-signaling, because pa, x, 0 = pa, x, 1 = 1, x {0, 1}, a {0, 1}, PR It s non-signaling, because p, b 0, y = p, b 1, y = 1, y {0, 1}, b {0, 1}. pa, x, y = pa, a x y x, y = 1, x, y {0, 1}, a {0, 1}, p, b x, y = pb x y, b x, y = 1, x, y {0, 1}, b {0, 1}. CH It s non-signaling, because given any input x, y, we always have p0, 0 x, y = p1, 1 x, y = x y cos π + x y sin π, and thus p1, 0 x, y = p0, 1 x, y = x y cos π + x y sin π, pa, x, y = pa, 0 x, y + pa, 1 x, y = 1, x, y {0, 1}, a {0, 1}, p, b x, y = p0, b x, y + p1, b x, y = 1, x, y {0, 1}, b {0, 1}. SIG It s not non-signaling, because p1, 0, 0 = 0, p1, 0, 1 = p1, 0 0, 1 = 1, p1, 0, 0 p1, 0, 1. 11

12 c We always have p win = x,y {0,1} px, y a b=x y pa, b x, y = 1 4 p0, 0 0, 0 + p1, 1 0, p0, 0 0, 1 + p1, 1 0, p0, 0 1, 0 + p1, 1 1, p0, 1 1, 1 + p1, 0 1, 1. 4 We just need to specify each probability for each cases. U Everything is 1 4, thus p win = = 1. PR p win = = 1. CH SIG p win = cos π + 1 cos π 4 = cos π = p win = 1 4 p0, 0 0, 0 = 1 4. d U Can. Let ρ AB = I I, A 0 0 = B 0 0 = 0 0, A 1 0 = B 1 0 = 1 1, A 0 1 = B 0 1 = + +, A 1 1 = B 1 1 =. It s easy to check that tr A a x Byρ b 1 AB =, x, y, a, b {0, 1}. 4 PR Can not. We will prove by contradiction. For any POVM {A 0 x, A 1 x} a, x {0, 1}, {B 0 y, B 1 y} b, y {0, 1}, and any density matrix ρ AB, consider the folloing formula A 0 0 A 1 0 B 0 0 B A 0 0 A 1 0 B 0 1 B A 0 1 A 1 1 B 0 0 B 1 0 A 0 1 A 1 1 B 0 1 B 1 1 = M 0 N 0 + M 0 N 1 + M 1 N 0 M 1 N 1, where M i = A 0 i A 1 i I, N i = I B 0 i B 1 i, i {0, 1}. It s easy to check that I 4 M i I 4, I 4 N i I 4,, i = {0, 1}, M i N j = N j M i, i, j {0, 1}. 1

13 Now define the we have i {0,1} XY = trxy ρ AB, X = XX, M i N j = N j M i, i, j {0, 1}, α i M i + β j N j 0, α 0, α 1, β 0, β 1. j {0,1} Then by direct calculation 1, we have tr M 0 N 0 + M 0 N 1 + M 1 N 0 M 1 N 1 ρ AB = M 0 N 0 + M 0 N 1 + M 1 N 0 M 1 N 1 = 1 M 0 + M1 + N0 + N M 0 + M1 + N0 + N M 0 N 0 + M 1 N 1 + 1M 0 N 1 M 1 N M 1 N 0 + M 0 + N 1 + 1M 1 + N 1 M 0 N 0 1 I4 + I 4 + I 4 + I 4 =. However, if there exist such a quantum strategy that can implement PR box, than it s easy to check that tr M 0 N 0 + M 0 N 1 + M 1 N 0 M 1 N 1 ρ AB = tr A 0 0 A 1 0 B 0 0 B 1 0ρ AB + tr A 0 0 A 1 0 B 0 1 B 1 1ρ AB = 4, + tr A 0 1 A 1 1 B 0 0 B 1 0ρ AB tr A 0 1 A 1 1 B 0 1 B 1 1ρ AB which violates the upper bound we obtain above. This contradiction implies that we can not use quantum strategy to implement PR box. CH Can. Let ρ AB = EP R EP R AB = AB = AB, A 0 0 = 0 0, A 1 0 = 1 1, A 0 1 = + +, A 1 1 =, B.S. Cirel son, Quantum generalizations of Bell s inequality. Letters in Mathematical Physics 4,

14 where B 0 0 = φ 0 φ 0, B 1 0 = φ 1 φ 1, B 0 1 = ψ 0 ψ 0, B 1 1 = ψ 1 ψ 1, φ 0 = cos π 0 + sin π 1, φ 1 = sin π 0 + cos π 1, It s easy to check that ψ 0 = cos π 0 sin π 1, ψ 1 = sin π 0 + cos π 1. tr A a x Byρ b 1 π AB = cos, ifa b = x y, tr A a x Byρ b 1 π AB = sin, ifa b x y. SIG Can not. We will show this by contradiction. Assume that there is such a strategy, then tr A 0 0 B 0 0ρ AB = tr A 1 1 B 1 1ρ AB = 1, tr A 1 0 B 0 1ρ AB = tr A 0 1 B 1 0ρ AB = 1. In my HW3 problem, I have shown a lemma that if X Y, Z 0, then trxz try Z. We will use this lemma here again. Since {A a x} a and {B b y} b are POVMs for all x, y, we have I B 0 0 0, I B 1 0 0, A 0 0 0, A 1 0 0, A A 1 0 = I, = A 0 0 I A 0 0 B 0 0 0, A 1 0 I A 1 0 B 0 1 0, = I 4 = I I = A 0 0 I + A 1 0 I A 0 0 B A 1 0 B 0 1 0, then using the lemma, we have 1 = tri 4 ρ AB tr A 0 0 B 0 0+A 1 0 B 0 1ρ AB = tr A 0 0 B 0 0ρ AB +tr A 1 0 B 0 1ρ AB =. This contradiction implies that we can not find such a quantum strategy. e Consider an non-signaling extension {q,, x, y, z} of the PR box. Using non-signaling condition, we have qa, b, c x, y, z = qa, b, c x, y, z, c, z, x, y, x, y {0, 1}, therefore we can define p c z = qa, b, c 0, 0, z 0, c, z, then we have p c z = qa, b, c x, y, z, x, y {0, 1}. 14

15 Also we can check that for all z, p c, z = c qa, b, c 0, 0, z = c pa, b 0, 0 = 1, thus {p c z} z is a family of well defined distributions. Now using the properties of PR box, we have q0, 1, c x, y, z = p0, 1 x, y = 0, z, if x, y 1, 1, c q1, 0, c x, y, z = p1, 0 x, y = 0, z, if x, y 1, 1, c q0, 0, c x, y, z = p0, 0 x, y = 0, z, if x, y = 1, 1, c q1, 1, c x, y, z = p1, 1 x, y = 0, z. if x, y = 1, 1, c Since all probabilities are non negative, we have q0, 1, c x, y, z = 0 = p0, 1 x, yp c z, c, z, if x, y 1, 1, q1, 0, c x, y, z = 0 = p1, 0 x, yp c z, c, z, if x, y 1, 1, q0, 0, c x, y, z = 0 = p0, 0 x, yp c z, c, z, if x, y = 1, 1, q1, 1, c x, y, z = 0 = p1, 1 x, yp c z, c, z, if x, y = 1, 1. Then using the non-signaling condition for fixing two inputs and outputs, we have q0, 0, c 0, 0, z = q0, 0, c 0, 1, z = q1, 0, c 1, 1, z = q1, 1, c 1, 0, z = q1, 1, c 0, 0, z, c, z, and since we also have qa, b, c 0, 0, z = q0, 0, c 0, 0, z + q1, 1, c 0, 0, z = p c z, c, z, thus Similarly we can also prove that q0, 0, c 0, 0, z = 1 p c z = p0, 0 0, 0p c, z, c, z, q1, 1, c 0, 0, z = 1 p c z = p1, 1 0, 0p c, z, c, z. q0, 0, c x, y, z = 1 p c z = p0, 0 x, yp c, z, c, z, if x, y 1, 1, q1, 1, c x, y, z = 1 p c z = p1, 1 x, yp c, z, c, z, if x, y 1, 1, q0, 1, c x, y, z = 1 p c z = p0, 1 x, yp c, z, c, z, if x, y = 1, 1, q1, 0, c x, y, z = 1 p c z = p1, 0 x, yp c, z, c, z, if x, y = 1, 1. Therefore the extension {q,, x, y, z} is in a product form. 15

16 f Let c X 3 = {0, 1}, z A 3 = {0}. Let q0, 0, 0 x, y, z = q1, 1, 0 x, y, z = 1, 4 q0, 0, 1 x, y, z = q1, 1, 1 x, y, z = 1 cos π sin π, q1, 0, 1 x, y, z = q0, 1, 1 x, y, z = 1 4 sin π, q1, 0, 0 x, y, z = q0, 1, 0 x, y, z = 1 4 sin π, q0, 0, 0 x, y, z = q1, 1, 0 x, y, z = 1 4 sin π, if x, y 1, 1, q0, 0, 1 x, y, z = q1, 1, 1 x, y, z = 1 4 sin π, q1, 0, 1 x, y, z = q0, 1, 1 x, y, z = 1 cos π, sin π, q1, 0, 0 x, y, z = q0, 1, 0 x, y, z = 1 4, if x, y = 1, 1. It s easy to check that this {q,, x, y, z} defines a tripartite non local box. Also, using the fact that 1 + cos π sin π = π cos, and noticing that z is always 0, we can check that qa, b, c x, y, 0 = qa, b, c x, y, 0, a, x, y, y, b,c b,c qa, b, c x, y, 0 = qa, b, c x, y, 0, b, x, y, x, a,c a,c qa, b, c x, y, 0 = qa, b, c x, y, 0, a, x, y, x, y, a,b a,b qa, b, c x, y, 0 = qa, b, c x, y, 0, a, c, x, y, y, b b qa, b, c x, y, 0 = qa, b, c x, y, 0, b, c, x, y, x, a a therefore it s non-signaling. Moreover, if x, y 1, 1 we have q0, 0, c x, y, z = 1 cos π q1, 0, c x, y, z = 1 sin π if x, y = 1, 1 we have q0, 0, c x, y, z = 1 sin π = p0, 0 x, y, = p1, 0 x, y, = p0, 0 x, y, 16 q1, 1, c x, y, z = 1 cos π q0, 1, c x, y, z = 1 sin π q1, 1, c x, y, z = 1 sin π = p1, 1 x, y, = p0, 1 x, y, = p1, 1 x, y,

17 q1, 0, c x, y, z = 1 π cos = p1, 0 x, y, q0, 1, c x, y, z = 1 cos π Therefore this {q,, x, y, z} is a non-signaling extension of CH box. However, if this {q,, x, y, z} has a product form, then we have = p0, 1 x, y. q0, 0, 0 0, 0, z = p0, 0 0, 0p 0 z = p 0 z = q0, 0, 0 1, 1, z = p0, 0 1, 1p 0 z = p 0 z = q0, 0, 0 0, 0, z p0, 0 0, 0 = q0, 0, 0 1, 1, z p0, 0 1, 1 1 cos π, = 1. This contraction implies that this {q,, x, y, z} is a non-product, non-signaling extension of CH box. g Given that each pair is chosen uniformly and x, y, z is generated uniformly, the success probability is Pr win = 1 1 pa, b, c x, y, z + pa, b, c x, y, z + pa, b, c x, y, z 3 x,y,z {0,1} a b=x y a c=x z b c=y z = 1 pa, b, c x, y, z + pa, b, c x, y, z + pa, b, c x, y, z 4 = 1 4 x,y,z {0,1} x,y,z {0,1} a b=x y a c=x z I 1 x, y, z + I x, y, z + I 3 x, y, z, b c=y z where I i x, y, z, i = 1,, 3 denote the success probability under the condition that the input is x, y, z and the i th pair is chosen. Here the first, second and third pair means A, B, A, C and B, C respectively. The following table gives occurrence number of each term pa, b, c x, y, z in the summation I 1 x, y, z + I x, y, z + I 3 x, y, z. xyz abc

18 Using this table and the condition that pa, b, c x, y, z = 1, x, y, z, we can easily see that a,b,c{0,1} Pr win = p0, 0, 0 0, 0, 0 + p0, 0, 0 1, 0, 0 + p0, 0, 0 0, 1, 0 + p0, 0, 0 0, 0, 1 + p1, 1, 1 0, 0, 0 + p1, 1, 1 1, 0, 0 + p1, 1, 1 0, 1, 0 + p1, 1, 1 0, 0, 1 p1, 0, 0 0, 1, 1 p0, 1, 0 1, 0, 1 p0, 0, 1 1, 1, 0 p0, 0, 0 1, 1, 1 p0, 1, 1 0, 1, 1 p1, 0, 1 1, 0, 1 p1, 1, 0 1, 1, 0 p1, 1, 1 1, 1, p0, 0, 0 0, 0, 0 + p0, 0, 0 1, 0, 0 + p0, 0, 0 0, 1, 0 + p0, 0, 0 0, 0, 1 + p1, 1, 1 0, 0, 0 + p1, 1, 1 1, 0, 0 + p1, 1, 1 0, 1, 0 + p1, 1, 1 0, 0, 1 p0, 0, 0 1, 1, 1 p1, 1, 1 1, 1, 1. Notice that whichever pair of players is chosen, a, b, c is a valid answer to the input x, y, z if and only if a 1, b 1, c 1 is a valid answer to x, y, z. Therefore, given a tripartite box {p,, x, y, z}, if we define a new tripartite {q,, x, y, z} box like qa, b, c x, y, z = 1 pa, b, c x, y, z + pa 1, b 1, c 1 x, y, z, a, b, c, x, y, z, then it s easy to check that we will have Pr win p = Pr win q, that is to say such transformation preserves the success probability. Also if {p,, x, y, z} is non-signaling, then for any a, x, we have b, qa, b, c x, y, z = 1 = 1 b, b, = b, pa, b, c x, y, z + pa 1, b 1, c 1 x, y, z pa, b, c x, y, z + pa 1, b 1, c 1 x, y, z qa, b, c x, y, z, y, z, y z, and this is also true for any b, y or any c, z. Also for any a, b, x, y, we have qa, b, c x, y, z = 1 = 1 = pa, b, c x, y, z + pa 1, b 1, c 1 x, y, z pa, b, c x, y, z + pa 1, b 1, c 1 x, y, z qa, b, c x, y, z, z, z, 1

19 and this is true for any a, c, x, z or any b, c, y, z. Therefor {q,, x, y, z} is also nonsignaling. Thus this transformation transformation also preserves the non-signaling property. Out of this reason, from now on we can alwasy assume that pa, b, c x, y, z = pa 1, b 1, c 1 x, y, z, a, b, c, x, y, z, otherwise we can do this transformation to {p,, x, y, z} to make it so. Now we continue our task of finding the upper bound of success probability, Pr win p0, 0, 0 0, 0, 0 + p0, 0, 0 1, 0, 0 + p0, 0, 0 0, 1, 0 + p0, 0, 0 0, 0, 1 + p1, 1, 1 0, 0, 0 + p1, 1, 1 1, 0, 0 + p1, 1, 1 0, 1, 0 + p1, 1, 1 0, 0, 1 p0, 0, 0 1, 1, 1 p1, 1, 1 1, 1, 1 = p0, 0, 0 0, 0, 0 + 4p0, 0, 0 1, 0, 0 + 4p0, 0, 0 0, 1, 0 + 4p0, 0, 0 0, 0, 1 4p0, 0, 0 1, 1, 1 = p0, 0, 0 0, 0, 0 + p0, 0, 0 1, 0, 0 + p0, 0, 0 0, 1, 0 + p0, 0, 0 0, 0, 1 p0, 0, 0 1, 1, 1, where we have used the condition. Next we will have to do some painful calculation. Using condition and non-signaling condition for fixing two inputs, we can show that p0, 0, 1 0, 1, 1 + p0, 1, 0 0, 1, 1 = p0, 0, 1 1, 1, 1 + p0, 1, 0 1, 1, 1, p0, 0, 1 1, 0, 1 + p1, 0, 0 1, 0, 1 = p0, 0, 1 1, 1, 1 + p1, 0, 0 1, 1, 1, p0, 1, 0 1, 1, 0 + p1, 0, 0 1, 1, 0 = p0, 1, 0 1, 1, 1 + p1, 0, 0 1, 1, 1, = p0, 0, 1 0, 1, 1 + p0, 1, 0 0, 1, 1 + p0, 0, 1 1, 0, 1 + p1, 0, 0 1, 0, 1 + p0, 1, 0 1, 1, 0 + p1, 0, 0 1, 1, 0 = p0, 0, 1 1, 1, 1 + p0, 1, 0 1, 1, 1 + p1, 0, 0 1, 1, 1 = 1 p0, 0, 0 1, 1, 1. On the other hand, still using condition and non-signaling condition for fixing two inputs, we can check that p0, 0, 1 1, 0, 0 + p0, 1, 0 1, 0, 0 + p1, 0, 0 1, 0, 0 + p0, 0, 1 0, 1, 0 + p0, 1, 0 0, 1, 0 + p1, 0, 0 0, 1, 0 + p0, 0, 1 0, 0, 1 + p0, 1, 0 0, 0, 1 + p1, 0, 0 0, 0, 1 = p0, 0, 1 0, 1, 1 + p0, 1, 0 0, 1, 1 + p0, 0, 1 1, 0, 1 + p1, 0, 0 1, 0, 1 + p0, 1, 0 1, 1, 0 + p1, 0, 0 1, 1, 0 + p1, 0, 0 0, 1, 1 + p0, 1, 0 1, 0, 1 + p0, 0, 1 1, 1, 0 = 1 p0, 0, 0 1, 1, 1 + p1, 0, 0 0, 1, 1 + p0, 1, 0 1, 0, 1 + p0, 0, 1 1, 1, 0 19

20 Notice that condition gives thus p0, 0, 0 1, 0, 0 + p0, 0, 0 0, 1, 0 + p0, 0, 0 0, 0, 1 + p0, 0, 1 1, 0, 0 + p0, 1, 0 1, 0, 0 + p1, 0, 0 1, 0, 0 + p0, 0, 1 0, 1, 0 + p0, 1, 0 0, 1, 0 + p1, 0, 0 0, 1, 0 + p0, 0, 1 0, 0, 1 + p0, 1, 0 0, 0, 1 + p1, 0, 0 0, 0, 1 3, p0, 0, 0 1, 0, 0 + p0, 0, 0 0, 1, 0 + p0, 0, 0 0, 0, p0, 0, 0 1, 1, 1 + p1, 0, 0 0, 1, 1 + p0, 1, 0 1, 0, 1 + p0, 0, 1 1, 1, 0 3, = p0, 0, 0 1, 0, 0 + p0, 0, 0 0, 1, 0 + p0, 0, 0 0, 0, 1 p0, 0, 0 1, 1, 1 1 p1, 0, 0 0, 1, 1 p0, 1, 0 1, 0, 1 p0, 0, 1 1, 1, 0 1. Finally we have Pr win p0, 0, 0 0, 0, 0 + p0, 0, 0 1, 0, 0 + p0, 0, 0 0, 1, 0 + p0, 0, 0 0, 0, 1 6 p0, 0, 0 1, 1, p0, 0, 0 0, 0, = 3 4. Here we have used the condition, so p0, 0, 0 0, 0, 0 1. Now we have proved that 3 4 is a upper bound of the success probability. Indeed if we take p0, 0, 0 x, y, z = p1, 1, 1 x, y, z = 1, x, y, z, and all unspecified probabilities being 0, then we can check that this {p,, x, y, z} is a non-signaling tripartite nonlocal box, and we have Pr win = p0, 0, 0 0, 0, 0 + p0, 0, 0 1, 0, 0 + p0, 0, 0 0, 1, 0 + p0, 0, 0 0, 0, 1 = p1, 1, 1 0, 0, 0 + p1, 1, 1 1, 0, 0 + p1, 1, 1 0, 1, 0 + p1, 1, 1 0, 0, 1 p0, 0, 0 1, 1, 1 p1, 1, 1 1, 1, 1 Therefore the optimum success probability achieved by any non-signaling tripartite box in this game is