Separation of quantum, spatial quantum, and approximate quantum correlations

Quantum nonlocal correlations are generated by implementation of local quantum measurements on spatially separated quantum subsystems. Depending on the underlying mathematical model, various notions of sets of quantum correlations can be defined. In this paper we prove separations of such sets of quantum correlations. In particular, we show that the set of bipartite quantum correlations with four binary measurements per party becomes strictly smaller once we restrict the local Hilbert spaces to be finite dimensional, i.e., $\mathcal{C}_{q}^{(4, 4, 2,2)} = \mathcal{C}_{qs}^{(4, 4, 2,2)}$. We also prove non-closure of the set of bipartite quantum correlations with four ternary measurements per party, i.e., $\mathcal{C}_{qs}^{(4, 4, 3,3)} = \mathcal{C}_{qa}^{(4, 4, 3,3)}$.

= C (4,4,2,2) qs . We also prove non-closure of the set of bipartite quantum correlations with four ternary measurements per party, i.e., C

Introduction
Nonlocality is one of the most fascinating features of quantum physics, stating that spatially separated parties can generate correlations that cannot be generated in the local hidden variable model [10]. In a bipartite nonlocality scenarioà la Bell [3], it is assumed that two parties, Alice and Bob can apply a measurement of their choice, which we denote by labels s and t respectively, on their respective subsystems. The measurement outcomes a, b, in the most general setting, are probabilistic. Thus we denote by p(a, b|s, t) the probability of obtaining outputs a, b by Alice and Bob when they apply measurements s, t respectively. The set of all correlations p(a, b|s, t) that can be represented in the local hidden variable model is denoted by C L , and is called the set of local correlations. Bell nonlocality theorem [3] states that there are correlations in the quantum world that do not belong to C L .
Correlations in the quantum theory are obtained as follows. Corresponding to Alice and Bob's local subsystems there are Hilbert spaces H A and H B , and the state of their joint system is described by a unit vector |ψ ∈ H A ⊗ H B . Alice and Bob's measurement settings s, t correspond to projective measurements {P 1 By enlarging the local subspaces H A , H B and taking a purification we may assume that the shared state is pure. Moreover, by Naimark's dilation theorem we may assume that the local measurement operators are projective. We stick to these assumptions all over the paper.  Table 1: Summary of correlation sets the probability p(a, b|s, t) of obtaining outputs a, b is given by The set of quantum correlations p(a, b|s, t) that can be written as above is denoted by C qs and is called spatial quantum correlations. We emphasize that for correlations in C qs the local Hilbert spaces H A , H B may have infinite dimensions. Then to distinguish the case of finite dimensional Hilbert spaces, we denote by C q the set of correlations of the form (1) in which the local Hilbert spaces H A , H B have finite dimensions. Then by the definitions we have C q ⊆ C qs . We also note that by standard tricks local correlations can be realized in the quantum world as well. Indeed, local correlations have quantum representations in finite dimensions, so we have C L ⊆ C q .
We may define yet two other sets of correlations related to the quantum theory. It is not hard to verify that C L is closed. However, it is not clear that C q and C qs are closed. Thus we denote by C qa the closure of C q and call it the set of approximate quantum correlations. As correlations in C qs can be approximated by finite dimensional ones, we have C qs ⊆ C qa and that C qa is the closure of C qs as well [18].
We would also like to briefly mention another generalization of quantum correlations motivated by quantum field theory. Here, instead of assuming that Alice and Bob have their individual subsystems and their associated Hilbert spaces, it is assumed that there is a single Hilbert space to which the shared state |ψ belong. Then the measurement operators P t of the previous picture, these measurement operators commute. Then C qc contains quantum (finite or infinite dimensional) correlations. Moreover, it is known that C qc is closed [11]. Therefore, C qa ⊆ C qc . Putting all these together the following hierarchy of sets of correlations are obtained: To get a finer perspective of these sets let us assume that Alice's measurement label s takes n A values, and there are n B values t for Bob's measurement setting. Also let m A and m B be the number of possible values of a and b, outputs of Alice and Bob respectively. That is, for instance, Alice chooses s amongst n A possible choices and after measurement outputs a among m A possible values. Then we denote the set of such correlations p(a, b|s, t) in C * for * ∈ {L, q, qs, qa, qc} by C Observe that if any of n A , n B , m A , m B equals 1, all the sets C (n A ,n B ,m A ,m B ) * become trivial and collapse to the lower level C . This is because when, e.g., m A = 1, Alice's output is fixed. Also if n A = 1, Bob knows Alice's measurement setting and can guess her output. Thus in the following we always assume that n A , n B , m A , m B ≥ 2.
Bell's theorem [3] says that the first inclusion in (2) is strict, i.e., C L = C q . More precisely, by the example of the CHSH inequality C (2,2,2,2) L = C (2,2,2,2) q [4]. Then (simply by ignoring some measurement settings and some output values) we also have C Separation of other inclusions in (2) is started by the work of Slofstra [20] who showed that C qs = C qc . He then improved his result in [19] and showed that C qs = C qa . This separation is proven in [19] for the parameters (n A , n B , m A , m B ) = (185, 235, 8,2). This result then improved in [9] and [14] for the parameters (n A , n B , m A , m B ) = (5, 5, 2, 2). Later, Coladangelo [6] gave a quite simple proof of this separation based on the ideas of self-testing and entanglement embezzlement [21]. Coladangelo's proof of C qs = C qc although simpler, is for the parameters (n A , n B , m A , m B ) = (5, 6, 3, 3) and does not improve the parameters of any of the previous results.
Separation of C q and C qs was first conjectured in [15] and then proved by Coladangelo and Stark in [8]. Their proof is quite elementary and is based on the idea of self-testing of the so called tilted CHSH inequality [1]. The separation C q = C qs in [8] is proven for the parameters (n A , n B , m A , m B ) = (4, 5, 3, 3).
We also add that the separation of C qa and C qc was very recently established in the breakthrough work of Ji et al. [12] who also showed the refutation of Connes' embedding conjecture.
Our results: In this paper we give new proofs for both the separations C q = C qs and C qs = C qa . Here is our first result. This theorem is an improvement on the result of [8] that establishes this separation for (n A , n B , m A , m B ) = (4,5,3,3). We also, building on this result, show that which again is an improvement on [8], but is not comparable to Theorem 1.
Similarly to [8] our proof of this theorem is based on the self-testing of tilted CHSH correlations. That is, we assume that the parties are playing two copies of the tilted CHSH game to self-test a certain entangled state. But these two tests are not independent, and there are certain correlations among their outputs. This forces a particular structure on the shared entangled state. Then by studying the ranks of measurement operators we conclude that they cannot be finite, so that C q = C qs .
To emphasize on the main difference of our proof with that of [8], we note that the shared entangled state |ψ in [8] is taken to be ∞ i=0 α i |i |i up to a normalization factor for some 0 < α < 1. Then this state can be written as which up to local isometries equals |00 + α|11 ⊗ |ψ ′ for some auxiliary state |ψ ′ . So we may apply the tilted CHSH game to self-test the shared state |00 + α|11 . In another decomposition we may write which again up to local isometries equals |00 + α|11 ⊗ |ψ ′ |22 . Thus again we may apply the tilted CHSH game on the orthogonal subspace to |2 to self-test |00 + α|11 . Considering these two tilted CHSH games and the non-trivial correlations between them the main result of [8] is derived. Here our proof strategy departs from [8] by considering the starting shared state This state can be written as |00 + α|11 ⊗ |ψ ′ up to local isometries in two different ways without any extra summand. This would allow us to decrease the number of outputs from m A = m B = 3 to m A = m B = 2. However, after this modification, the argument of [8] based on the analysis of Schmidt coefficients of the shared state does not directly work. To overcome this difficulty, we further use self-testing properties of tilted CHSH correlations and build our arguments based on the ranks of the measurement operators. This would give us our first result C We note that C (2,n B ,2,2) q = C (2,n B ,2,2) qs = C (2,n B ,2,2) qa since any pair of projections (Alice's measurement operators) can be block-diagonalized with blocks of size at most 2. Thus, the first place that the separation of these sets is expected, and is conjectured in [15], is for the parameters (n A , n B , m A , m B ) = (3, 3, 2, 2). Thus our results are not optimal, yet they get closer to what is conjectured.
Our second result is regarding the separation C qs = C qa .  This result is an improvement over [6] that shows this separation for larger parameters (n A , n B , m A , m B ) = (5, 6, 3, 3), but is not comparable to [9,14].
Our proof idea of this result is similar to [6] and is based on self-testing and entanglement embezzlement. The only difference with [6] is that in order to self-test the maximally entangled state of Schmidt rank 3, instead of the protocol of [5], we use the protocol of [17] which allows to self-test this state with question sets of size 2 and answer sets of size 3.
As mentioned above, our proof of C is based on the recent result of [17]. Sarkar et al. in this paper show that any maximally entangled state of Schmidt rank d can be self-tested via the Bell inequality of [16]. We describe this Bell inequality in details in the following section. We also present a proof of this self-testing property in Appendix A that is shorter than the original proof in [17] and may enlighten our understanding of the aforementioned Bell inequality.
In the next section we describe two self-testing protocols that will be used in the proofs of our main results. In the following two sections we will prove Theorem 1 and Theorem 2.

Self-testing protocols
Recall that the CHSH inequality states that for binary observables A s , B t , s, t ∈ {0, 1}, with ±1 values we have where AB is the expectation value of the observable AB. The CHSH inequality belongs to a family of such inequalities called the tilted CHSH inequalities [1], which given a parameter β ∈ [0, 2] states that Observe that for β = 0 we recover the standard CHSH inequality. Moreover, similar to the CHSH inequality, the tilted CHSH inequality is violated in the quantum regime. Indeed, there are quantum binary observables A s , B t , s, t ∈ {0, 1} and an entangled state |φ such that Moreover, 8 + 2β 2 is the maximum value that can be gained by a quantum strategy.
Let us describe the observables and the shared state that give (3). Given β ∈ [0, 2] define θ, µ ∈ [0, π/4], and α ∈ [0, 1] by tan(µ) = sin(2θ) = 4 − β 4 4 + β 2 , α = tan(θ). Let be a two-qubit state. Also let σ z and σ x be the Pauli operators and define Observe that σ are also binary observables. Next let x . Then with the shared state |Φ (α) in (4) and these four observables we obtain (3). Surprisingly, the above shared state and observables represent essentially the unique strategy that achieves (3). This is called the self-testing property of the tilted CHSH inequality and is proven in [22,2]. Theorem 3. (Self-testing of tilted CHSH [22,2]) Let |ψ AB ∈ H A ⊗ H B be a bipartite entangled state and A s , B t , s, t ∈ {0, 1} be binary observables (with A 2 s = I, Here |Φ (α) is given in (4) and σ are defined in (5). Moreover, A s , B t are the restrictions of A s , B t to the invariant subspaces H A and H B respectively, and I A ′ , I B ′ are the identity operators acting on H A ′ and H B ′ respectively.
There are some remarks in line.
Remark 1. In the statement of the theorem we assume that U, V are not only isometries, but also invertible, i.e., they are onto. Indeed, we claim that for such isometries U, V , their images U H A and V H B are always of the form C 2 ⊗ H ′ A ′ and C 2 ⊗ H ′ B ′ respectively, so that we may identify H A ′ and H B ′ in the statement of the theorem by H ′ A ′ and H ′ B ′ respectively. To prove this claim, we compute where we used V † V = I, i.e., V is an isometry, and the cyclic property of trace. This means that U sends H A ′ = supp tr B |ψ ψ| to the support of the partial trace |Φ (α) ⊗ |ψ ′ which of course is of the form C 2 ⊗ H ′ A ′ . The proof for V is similar. Remark 2. Suppose that |ψ in the statement of the theorem has finite Schmidt rank. Then by definition Sch-rank(|ψ ) = dim H A = dim H B , where Sch-rank(|ψ ) is the Schmidt-rank of |ψ . On the other hand, Then using the fact that U, V are invertible and by comparing dimensions, Sch-rank(|ψ Remark 3. We note that the operators A s , B t , s, t ∈ {0, 1} may act on larger spaces than H A = supp(tr B |ψ ψ|) and H B = supp(tr A |ψ ψ|), and the fact that the tilted CHSH inequality is maximally violated does not give any information about the action of these operators on the complementary subspaces. Because of this, in the statement of the theorem we had to replace A s , B t with their restricted versions A s , B t . By Theorem 3 any entangled state of Schmidt rank 2 can be self-tested. It is then shown in [7] that in fact any entangled state of finite Schmidt rank can be self-tested. In the special case of the maximally entangled state of Schmidt rank d, it is shown that it can be self-tested via a correlation in C 3,4,d,d q [5]. However, one would expect to get self-testing protocols for maximally entangled states with smaller question and answer sets, particularly with n A = n B = 2 and m A = m B = d. This problem was resolved recently in [17]. To state this result we need to develop some notations.
Binary observables are usually represented by unitary operators A with eigenvalues ±1 (equivalently with A 2 = I). Indeed, to a binary projective measurement {P 0 , P 1 } one can associate the binary observable A = P 0 − P 1 . Conversely, having such A, the operators of the associated projective measurement are given by P 0 = (I + A)/2 and P 1 = (I − A)/2. We can apply the same idea to represent d-valued measurements, i.e., measurements with outcomes in {0, 1, . . . , d − 1}.
where ω = e 2πi/d is the d-th root of unity. Observe that A is a unitary and A d = I. Conversely, eigen-decomposition of any unitary A with A d = I corresponds to a d-valued projective measurement with measurement operators: As a result, similar to the binary case, any Bell inequality with d-valued measurements, can be written in terms of unitary operators A i , B j with A d i = B d j = 1, and their powers. We call such unitaries d-valued observables.
We now state our desired Bell inequality which following [17] we call the SATWAP Bell inequality [16].
Then the SATWAP Bell operator is given by where as before ω = 2 2πi/d , andr k is the complex conjugate of r k given byr k = r d−k . It is known that the maximum of O d in the local hidden variable model, i.e., in C , equals 2 cot(π/4d) − cot(3π/4d) − 4 /2. Also, for any entangled state |ψ we have Moreover, there is essentially a unique strategy to saturate the above inequality.
Theorem 4. (Self-testing of SATWAP [17]) i=0 |i |i be the maximally entangled state of Schmidt rank d. Let |i . Then letting |ψ = |Φ d and We notice that the unitary operators A s , B t given in equations (7)-(10) differ from those of [17]. However, this is not hard to verify that they are indeed equivalent under local unitaries. In fact, this simple representation of the optimal strategy in part (i) of the theorem would help us to attain a simpler proof of the self-testing property in part (ii). We give a proof of this theorem in Appendix A.

Quantum correlations in finite vs infinite dimensions
In this section we prove our first main result stated in Theorem 1. Let us recall that C is the set of correlations p(a, b|s, t), with s, t taking n A , n B values and a, b taking m A , m B values respectively, such that it has a representation of the form (1) is defined similarly except that there is no restriction on the dimension of local spaces.
Let us start by introducing the shared entangled state to be used for generating the target nonlocal correlation. Let H Z be the separable Hilbert space with orthonormal basis where 0 < α < 1 is an arbitrary constant, |i| is the absolute value of integer i and C is a normalization factor |Ψ can be written as 1 √ 1+α 2 |00 + α|11 ⊗ |ψ ′ up to local isometries in two different ways, one of which by pairing the basis states as {|i Z : i ∈ Z} = ∪ j∈Z |2j Z , |2j + 1 Z and the other one by pairing them as {|i Z : i ∈ Z} = ∪ j∈Z |2j Z , |2j − 1 Z . To make this more precise let us introduce two isometries given by and Then observe that for some |ψ ′ 0 , |ψ ′ 2 ∈ H Z ⊗ H Z . Thus using |Ψ as the shared state, the tilted CHSH game can be played in two different ways. To this end let us define the observables A s , B t , s, t ∈ {0, 1, 2, 3} according to Table 2. Then A s , B t for s, t ∈ {0, 1} generate the tilted CHSH correlation and A s , B t for s, t ∈ {2, 3} generate another copy of this correlation.    (1) . Then the correlation generated by the shared state (11) and observables in Table 2 is given by By the definitions we have p ∈ C t , s, t ∈ {0, 1, 2, 3} acting on H A and H B respectively, such that where p(a, b|s, t) is given by (14). As mentioned before, p r (a, b|s, t) = p(a, b|s + r, t + r) for s, t ∈ {0, 1} form two copies of the tilted CHSH correlation for r ∈ {0, 2}. Then by Theorem 3 there are Hilbert spaces H A ′ r , H B ′ r , r ∈ {0, 2}, and invertible isometries U r : and Here H A ⊆ H A and H B ⊆ H B are supports of tr B |φ φ| and tr A |φ φ| respectively, and |φ ′ r ∈ H A ′ r ⊗ H B ′ r is some bipartite state.
The above equations show that P s and Q t , s, t ∈ {0, 1, 2, 3}, leave the subspaces H A and H B invariant, respectively. Thus by restricting everything to these subspaces, we may assume with no loss of generality that tr A |φ φ| and tr B |φ φ| are invertible and H A = H A and H B = H B .

Let us define
Then by the definitions of σ Thus M is an observable. Moreover, is the projection on the eigenspace of M with eigenvalue (−1) b . We also have Let us compute where |Ψ is given by (11), the operator A s , B t , s, t ∈ {0, 1, 2, 3} are given in Table 2 and Now by the definitions of B 2 , B 3 and the isometry W 2 we have We also note that Putting these together, and using the particular form of |Ψ we find that Therefore, taking the partial trace of both sides over the first subsystem we find that As a result, based on Remark 2 we have Next, by comparing the dimensions, using (17) and dim H B ′ 0 = dim H B ′ 2 = 1 2 dim H B , we find that equality holds in the above inclusion. Equivalently, we obtain On the other hand, following similar computations as in (18) we have where we used the fact that the supports of both A (0) 0 and D (0) contain |0 Z ∈ H Z . This is in contradiction with (19). We are done.
In the definition of the nonlocal correlation p(a, b|s, t) in the statement of Theorem 5 we could exchange the operators A 2 , A 3 with operators B 2 , B 3 in Table 2. In that case we again get two copies of the tilted CHSH correlation (but with roles of the two players being exchanged in one of them), and can follow similar steps as in the above proof to show that this new correlation does not belong to C (4,4,2,2) q . Indeed, we get a slightly simpler proof since in this case we would not need to introduce a new operator as in (16). However, our particular choice of the correlation p(a, b|s, t) would help us to get yet another separation of C q and C qs . Proof. We first describe a correlation q(a, b|s, t) in C (3,4,3,2) qs . Let the local Hilbert spaces be H Z as before and |Ψ ∈ H Z ⊗ H Z be as in (11). Next we need to introduce three projective measurements A (a) s : a ∈ {0, 1, 2} , s ∈ { * , 1, 3}, and four (binary) observables B t , t ∈ {0, 1, 2, 3}. Here for convenience the inputs of the first player are indexed by s in { * , 1, 3} instead of {0, 1, 2}. We let B t , t ∈ {0, 1, 2, 3} be as before given by Table 2. We also let A s for s ∈ {1, 3} be binary observables with values in {0, 1}, i.e., we assume that A for a ∈ {0, 1} be given according to the binary observables A 1 , A 3 in Table 2. Finally, we define A (a) * : a ∈ {0, 1, 2} by These give the correlation q(a, b|s, t) = Ψ| A (a) which belongs to C We claim that q does not belong to C (3,4,3,2) q . The main point behind the proof is that the binary observables A 0 , A 2 given in Table 2 can be written in terms of A (a) * , a ∈ {0, 1, 2}. In fact, using the definitions of the isometries W 0 , W 2 we have This means that the correlation p ∈ C t |Ψ = q(0, b| * , t) + q(2, b| * , t).
To make the above argument more precise, suppose that there are finite dimensional Hilbert spaces H A , H B , |φ ∈ H A ⊗ H B , local measurements P (a) s : a ∈ {0, 1, 2} , s ∈ { * , 1, 3}, and binary observable Q t , t ∈ {0, 1, 2, 3} such that q(a, b|s, t) = φ| P (a) Since A  s . Also we may define binary measurements P 0 , P 1 based on (20) and (21): Then it is not hard to verify that the correlation generated by the shared state |φ and binary observables P s , Q t , s, t ∈ {0, 1, 2, 3} in finite dimensions equals p. However, in Theorem 5 we showed that p does not belong to C (4,4,2,2) q . This is a contradiction.

Non-closure of the set of quantum correlations
In this section we prove our second main result (Theorem 2) that is the separation of C . To prove this separation, we introduce a sequence of correlations p n (a, b|s, t) : n ≥ 1 in C (4,4,3,3) q that converge to a correlation p * (a, b|s, t). By definition p * (a, b|s, t) belongs to C . This gives our main result.
In constructing p * (a, b|s, t) and proving that it does not belong to C qs we follow similar steps as in [6]. The only difference is that instead of using the protocol of [5] to self-test the maximally entangled state of Schmidt rank 3, we use Theorem 4. This would enable us to reduce the number of inputs in the target nonlocal correlation.
Following [6] our construction of p * (a, b|s, t) is based on entanglement embezzlement [21,13]. Let |ψ , |φ be two bipartite states that are not equivalent up to local isometries, i.e., they have different multisets of Schmidt coefficients. Then for any other bipartite state |χ , the states |ψ ⊗ |χ and |φ ⊗ |χ are still inequivalent up to local isometries. Nevertheless, by choosing an appropriate state |χ with increasing Schmidt rank, |ψ ⊗ |χ and |φ ⊗ |χ become approximately equivalent up to local isometries with an arbitrarily small error. This is called entanglement embezzlement, which besides self-testing are the main ingredients of our separation theorem in this section. Let be the maximally entangled state in C 3 ⊗ C 3 . Also let Then let H E n = H F n = C 3 ⊗n and define |χ n ∈ H E n ⊗ H F n by where C n is a normalization factor. Now let be the shared state in the nonlocal correlation p n (a, b|s, t) in C (4,4,3,3) q . Since |Φ 3 is a part of this shared state, the two players can generation the SATWAP correlation by measuring this part. More precisely, following part (i) of Theorem 4 we let A 0 , A 1 and B 0 , B 1 be 3-valued observables be given by the first two rows of Table 3.

Now observe that
Therefore, where α = 1 √ 2 . Thus Γ n ⊗Λ n |ψ n approximately contains |Φ (α) in its first register. Therefore, the two players may generate the tilted CHSH correlation by locally measuring it. Thus we define the measurements A (2) t for s, t ∈ {2, 3} according to the last two rows of Table 3. We should explain that the measurement operators of the tilted CHSH correlation are binary, yet here A s , B t are 3-valued measurements. Nevertheless, as is clear from (23), the state Γ n ⊗ Λ n |ψ n is locally orthogonal to both span{|2 } ⊗ H E n and span{|2 } ⊗ H F n . Thus we may implement, e.g., Pauli measurements on the first registers of Γ n ⊗ Λ n |ψ n . This gives us A t , for s, t ∈ {2, 3} and a, b ∈ {0, 1} as in Table 3. Then we let The shared state |ψ n given in (22) and the measurements given in Table 3 give us a correlation p n (a, b|s, t) ∈ C (4,4,3,3) q .
Since C (4,4,3,3) q is bounded, the sequence {p n (a, b|s, t) : n ≥ 1} has a limit p * (a, b|s, t) which by definition belongs to C (4,4,3,3) qa . In the following we identify some crucial properties of this limiting correlation.

Theorem 7.
There is no q(a, b|s, t) in C Proof. Suppose that q(a, b|s, t) ∈ C (4,4,3,3) qs satisfies the aforementioned properties. Suppose that q(a, b|s, t) is obtained by the shared state |φ ∈ H A ⊗ H B and 3-valued measurements P s , Q t , s, t ∈ {0, 1, 2, 3}: q(a, b|s, t) = φ|P (a) is the projection on the eigenspace of P s with eigenvalue ω a , where ω = e 2πi/3 , and Q (b) t is defined similarly. Using Theorem 4, property (i) implies that there are invertible isometries U 0 : and where H A and H B are supports of tr B |φ φ| and tr A |φ φ| respectively, Q 0 is the restriction of Q 0 to H B , and |φ ′ 0 ∈ H A ′ 0 ⊗ H B ′ 0 . Similarly, using Theorem 3, property (ii) implies that there are invertible isometries and and P 2 is the restriction of P 2 to H A . Next, property (iii) implies that This, in particular, gives Therefore, As a result, |11 ⊗ |φ ′ 0 and |11 ⊗ |φ ′ 1 are equivalent up to local isometries. Equivalently, letting S j , j ∈ {0, 1}, be the multiset of the Schmidt coefficients of |φ ′ j we find that On the other hand, comparing (24) and (25) we find that |Φ 3 ⊗ |φ ′ 0 and 1 √ 1+α 2 (|00 + α|11 ) ⊗ |φ ′ 1 are equivalent up to local isometries, and have the same multisets of Schmidt coefficients. That is, where by xS we mean xS = {xs : s ∈ S}. Then taking the supremum of both sides and using (26) we find that which is a contradiction since sup S 1 = 0. We are done.
The main result of [6] is more general than a separation of C qs and C qa . Indeed, in [6] a nonlocal game is introduced that has the following property: in order to win the game with probability ǫ-close to optimal, the Schmidt-rank of the shared entangled state must be 2 Ω(ǫ −1/8 ) . The main tool of [6] for proving this result is the so call stability of the self-testing protocols under noise. We know that the self-testing property of the tilted CHSH correlation is stable under noise. However, this is not known for the SATWAP correlation. Thus, it is not clear if Theorem 7 can be generalized to a result similar to that of [6].
Next we have where we usedr 2 k ω k + r 2 k = r 2 k ω −k +r 2 k = 0. Therefore, This means that for any bipartite state |ψ we have This equation encourages us to prove the following lemma which will be used later.
Indeed, if we let |J ℓ = Z ℓ |J , then {|J 0 , . . . , |J d−1 } forms an orthonormal basis and Using this we find that B d 1 = I. By similar calculations we obtain where T denotes transposition with respect to the computational basis. We compute Therefore, Proof of (ii). By assumptions we know that (27) Thus since partial traces of |ψ are invertible, we have C s,k C s,ℓ = C s,k+ℓ . This means that Next by a similar argument (using A d s = I), we find that Observe that C † s,k = C s,d−k = C d−k s,1 = C −k s,1 = C −1 s,k . Thus C s,k is unitary. We compute Therefore, since both C 0,k and √ 2r k B −k 0 are unitary, is self-adjoint. D k is also unitary since it is a multiplication of unitary operators. Thus, D k is a unitary with eigenvalues ±1. Therefore, for some orthogonal projection P k , and we have By a similar calculation we have We now use the fact that C s,k+ℓ = C s,k C s,ℓ . By (30) this gives Subtracting this equation from its adjoint, we find that B ℓ 0 P k B −ℓ 0 P ℓ + P ℓ B ℓ 0 P k B −ℓ 0 = 0. This means that for any |v ∈ supp(P ℓ ) we have 2 v|B ℓ 0 P k B −k 0 |v = 0. Then, using the fact that B ℓ 0 P k B −k 0 is a projection, we find that B ℓ 0 P k B −k 0 |v = 0 for all |v ∈ supp(P ℓ ). As a result, and B ℓ 0 P k B −ℓ 0 + P ℓ = P k+ℓ . Then by a simple induction we arrive at Indeed, P k is the summation of k projections B ℓ 0 P 1 B −ℓ 0 , 0 ≤ ℓ ≤ k − 1 that are mutually orthogonal. Moreover, since C d 0,1 = C 0,1 C 0,d−1 = I we have Let us define By the above equations, the orthogonal projections B ℓ 0 P 1 B −ℓ 0 , 0 ≤ ℓ ≤ d − 1 form an eigendecomposition of F . Thus, F is a unitary operator and F d = I. Next, by a simple calculation we find that B 0 F k B −1 0 = ω k F k and As a result, eigenspaces of B 0 are isomorphic. In fact, letting H (k) ⊆ H B , 0 ≤ k ≤ d − 1, to be the eigenspace of B 0 with eigenvalue ω k , we have H B = k H (k) and that Let H B ′ = H (0) and define V : V is well-defined since H B = k H (k) and by (33) we have F −k H (k) = H (0) = H B ′ . Moreover, since F is unitary, V is an invertible isometry. Now, for any |w 0 ∈ H B ′ = H (0) by (33) there exists |w k ∈ H (k) such that |w 0 = F −k |w k . Then by the definition of V we have This means that as desired. Next, taking |w 0 , |w k as before, we have where in the third line we use F |w k ∈ H (k+1) . Therefore, where X : C d → C d is defined by X|i = |i + 1 (mod d) .
Now recall that by (32), P 1 is the projection on the eigenspace of F with eigenvalue 1. Then using V F V † = X ⊗ I B ′ we find that |i . Next, using this in (28) and (29) we obtain as desired. Also using (30) and (31) we have and To characterize the state |ψ and operators A 0 , A 1 we once again use (27). We compute C s,k tr A |ψ ψ| C † s,k = tr A I A ⊗ C s,k |ψ ψ| I A ⊗ C † s,k = tr A A −k s ⊗ I B |ψ ψ| A k s ⊗ I B = tr A |ψ ψ| .
Thus tr A |ψ ψ| commutes with C s,k , s ∈ {0, 1}, 1 ≤ k ≤ d − 1, and with any operator in the algebra generated by them. On the other hand, it is not hard to verify that the algebra generated by the operators in (34) and (35) equals the space of all operators of the form Q ⊗ I B ′ . 3 As a result, V tr A |ψ ψ| V † commutes with any operator of the form Q ⊗ I B ′ . This means that tr A I A ⊗ V |ψ ψ|I A ⊗ V † = V tr A |ψ ψ| V † = 1 d I ⊗ ρ ′ B ′ , for some density operator ρ ′ acting on H B ′ . Therefore, there exists an invertible isometry U : where |ψ ′ A ′ B ′ is a purification of ρ ′ . Using this in (27) we find that where T is the transpose with respect to the computation basis of C d . Equivalently, we have We are done.