Arbitrarily many independent observers can share the nonlocality of a single maximally entangled qubit pair
Peter J. Brown Roger Colbeck
quant-ph



peter.brown@ens-lyon.fr
ENS Lyon, LIP, F-69342, Lyon Cedex
	07, France
Department of Mathematics, University of York, Heslington, York, YO10 5DD, United Kingdom

roger.colbeck@york.ac.uk
Department of Mathematics, University of York, Heslington, York, YO10 5DD, United Kingdom




  Alice and Bob each have half of a pair of entangled qubits. Bob measures his half and then passes his qubit to a second Bob who measures again and so on. The goal is to maximize the number of Bobs that can have an expected violation of the Clauser-Horne-Shimony-Holt (CHSH) Bell inequality with the single Alice. This scenario was introduced in [Phys. Rev. Lett. 114, 250401 (2015)] where the authors mentioned evidence that when the Bobs act independently and with unbiased inputs then at most two of them can expect to violate the CHSH inequality with Alice. Here we show that, contrary to this evidence, arbitrarily many independent Bobs can have an expected CHSH violation with the single Alice. Our proof is constructive and our measurement strategies can be generalized to work with a larger class of two-qubit states that includes all pure entangled two-qubit states. Since violation of a Bell inequality is necessary for device-independent tasks, our work represents a step towards an eventual understanding of the limitations on how much device-independent randomness can be robustly generated from a single pair of qubits.

Arbitrarily many independent observers can share the nonlocality of a single maximally entangled qubit pair
    Roger Colbeck
    26^th August 2020
===========================================================================================================





§ INTRODUCTION
Introduction.|Quantum theory predicts the possibility of measuring correlations that cannot be explained by standard notions of causality <cit.>. These `non-local' correlations are a crucial resource for device-independent tasks such as key distribution <cit.>, randomness expansion <cit.> and randomness amplification <cit.>. The idea is that because such correlations cannot be explained using local hidden variables, the individual outcomes are random and unpredictable to an adversary <cit.>. Several recent advances have enabled the first experimental demonstrations of device-independence <cit.>. However, further theoretical and experimental advances are needed before this can become a practical technology.

In this work we study fundamental limits of non-locality, asking whether a single pair of entangled qubits could generate a long sequence of non-local correlations. Such an approach could be useful for situations where a significant bottleneck lies in the state generation, such as in nitrogen vacancy based experiments <cit.>. We study a scenario in which a single Alice tries to establish non-local correlations with a sequence of Bobs who sequentially measure one half of an entangled qubit pair. An additional restriction we impose is that each Bob in the sequence can only send a single qubit (his post-measurement state) to the next. In particular, the classical information pertaining to measurement choices and outcomes of each Bob is not shared. It is in this sense that the Bobs act independently of one another.

This sequential scenario (see Fig. <ref> and Sec. <ref>) was introduced in <cit.>. There it was shown that by modifying the input distributions of the Bobs, so that one of the inputs is highly favoured, an unbounded number of Bobs could each have an expected CHSH violation <cit.> with the single Alice who measures once. However, the authors also mentioned numerical evidence suggesting that if the input distributions were not modified (each Bob chooses a binary input uniformly at random) then at most two Bobs would be able to have an expected CHSH violation with Alice.

By constructing an explicit measurement strategy, we show that, contrary to what was previously thought, there is no bound on the number of independent Bobs (with uniform inputs) that can have an expected violation of the CHSH inequality with Alice. We exhibit a class of initial two-qubit entangled states that are capable of achieving an unbounded number of violations, which includes all pure two-qubit entangled states. A previous work <cit.> claimed a proof that at most two unbiased Bobs could achieve an expected CHSH violation with a single Alice, in line with the numerical evidence mentioned in <cit.>. However, within <cit.> there is an implicit assumption that the sharpness of the two measurements that each Bob used is equal. 
Equal sharpness was also taken in <cit.> to show that at most two Bobs can violate any 2-outcome Bell inequality with Alice in this scenario when starting with a maximally entangled state. 
In the present paper Letter we show that these limitations can be overcome by using measurements with unequal sharpness, highlighting the advantage of considering the most general measurement strategies.




§ THE SEQUENTIAL CHSH SCENARIO
The sequential CHSH scenario.|We consider the following measurement scenario wherein a single party (Alice) attempts to share nonlocal correlations with n independent parties (1, …, n) using only a single maximally entangled qubit pair, one half of which is passed between the n parties (see Fig. <ref>). We denote the binary input and output of Alice by X and A respectively. For each k ∈ we denote the binary input and output of k by k and k respectively. 



To begin, a two-qubit state ρ_A1 is shared between Alice and 1. 1 then proceeds by choosing a uniformly random input, performing the corresponding measurement and recording the outcome. The post-measurement state is then sent to 2. Suppose 1 performed the measurement according to 1=y and received the outcome 1=b. The post-measurement state can be described by the Lüders rule[Using the Lüders rule is optimal in terms of the information retained in the post-measurement state|see Appendix <ref>the Supplemental Material.]

    ρ_A1↦(𝕀⊗√(by^(1))) ρ_A1(𝕀⊗√(by^(1))),

where by^(1) is the POVM effect corresponding to outcome b of 1's measurement for input y. As each Bob is assumed to act independently of all previous Bobs, 2 is ignorant of the values of 1 and 1. Thus, the state shared between Alice and 2 is averaged over the inputs and outputs of 1, i.e.

    ρ_A2 = 1/2∑_1,1(𝕀⊗√(11^(1))) ρ_A1(𝕀⊗√(11^(1))).
 
Repeating this process, we can compute the state shared between Alice and k.

We are concerned with the expected CHSH value between Alice and each of the Bobs independently. Given a conditional probability distribution p_AB^(k)|XY^(k) on the binary random variables A, B^(k), X and Y^(k), the CHSH value is defined as 

    k :=     2[ p(A=B^(k)|00) + p(A=B^(k)|01)  
    
     +     p(A=B^(k)|10) + p(A≠ B^(k)|11) - 2].

When Alice and k behave classically (with shared randomness) or do not share entanglement then k≤2, which is the CHSH inequality.

The aim of this work is to explain how, given n∈, we can define a sequence of pairs of POVMs for 1, …, n such that k > 2 for all k = 1,…,n. That is, by initially sharing two qubits in an appropriate entangled state, there is no bound on the number of independent Bobs that can expect to violate the CHSH inequality with a single Alice.



§ THE MEASUREMENT STRATEGY
The measurement strategy.|This uses two-outcome POVMs, {E,𝕀-E}, where E has the form E=12(𝕀+γσ_r). Here r∈^3 with r=1, γ∈ [0,1] is the sharpness of the measurement, and σ_r=r_1σ_1+r_2σ_2+r_3σ_3, where {σ_1,σ_2,σ_3} are the Pauli operators. Since there are only two outcomes, a single effect is enough to define the POVM. Note that if γ=1 the measurement is sharp (projective), and the post-measurement state is unentangled, while if γ=0, the measurement ignores the state and is equivalent to an unbiased coin toss.

Our measurement strategy for nonlocality sharing between Alice and n Bobs is fully specified by n + 1 parameters. In this strategy Alice's POVMs are defined by the effects

    A_0|0   := 12(𝕀 + cos(θ) σ_3 + sin(θ) σ_1)  and
    
    A_0|1   := 12(𝕀 + cos(θ) σ_3 - sin(θ) σ_1)

for some θ∈(0, π/4]. For each k = 1, …, n, k's POVMs are defined by the effects

    B_0|0^(k)   := 12(𝕀+ σ_3)  and
    
    B_0|1^(k)   := 12(𝕀 +γ_k σ_1).

For these measurements and with an initial state ρ_A1 = Ψ with |Ψ⟩ = 1√(2)(|00⟩ + |11⟩), the expected CHSH value between Alice and k is given by

    k = 2^2-k(γ_k sin(θ) + cos(θ) ∏_j=1^k-1(1+ √(1-γ_j^2))).

(see Appendix <ref>the Supplemental Material). 



§ AN UNBOUNDED NUMBER OF VIOLATIONS
An unbounded number of violations.|Our aim is to choose the values of γ_k and θ to achieve k>2 for arbitrarily many values of k. Using (<ref>), we require

    γ_k > 2^k-1 - cos(θ) ∏_j=1^k-1(1 + √(1-γ_j^2))/sin(θ).

Our strategy is as follows. Let ϵ > 0, γ_1(θ)=(1+ϵ)1-cos(θ)sin(θ) and for k∈{2,…,n} recursively set

    γ_k(θ)=(1+ϵ)2^k-1-cos(θ)P_k/sin(θ)   if γ_k-1(θ)∈(0,1)
    ∞   otherwise,

where P_k=∏_j=1^k-1(1 + √(1-γ_j^2(θ))), and γ_k(θ)=∞ indicates that violation of k>2 is not possible for this k.










The following lemma shows that the sharpness parameters of such a measurement strategy (when they are finite) are strictly increasing, i.e., 0 < γ_1(θ) < γ_2(θ) < …. 

	Let θ∈ (0, π/4] and ϵ > 0. Then the sequence (γ_k(θ))_k ∈ as defined by (<ref>) 
	with γ_1(θ) = (1+ϵ)1-cos(θ)sin(θ), is positive and increasing. Moreover, the subsequence consisting of all finite terms is a strictly increasing sequence. 


In addition to the sequence being monotonically increasing, each term in the sequence also has a vanishing limit as θ approaches 0. 

	Let ϵ > 0 and let (γ_n(θ))_n ∈ be the sequence defined by (<ref>) 
		with γ_1(θ) = (1+ϵ)1-cos(θ)sin(θ). Then for any n ∈ there exists some θ_n ∈ (0,π/4] such that for all k ≤ n  and θ∈ (0, θ_n), 
    γ_k(θ) < 1
    		..
    		
    		 Moreover, we have 
        lim_θ→ 0^+γ_n(θ) = 0
                      for all n ∈.
                    
        
        Using the two lemmas above, the proofs of which are given in Appendices <ref> and <ref> the Supplemental Material, we can now state and prove our main result.
        
        	For each n ∈, there exists a sequence (γ_k(θ))_k=1^n and a θ_n ∈ (0, π/4] such that k > 2 for all k = 1,…,n.
        	
        	By Lemma <ref>, we know that there exists some θ_n ∈ (0, π/4] such that γ_n(θ_n) < 1. By Lemma <ref>, we then also have that 0< γ_1(θ_n) < γ_2(θ_n) < … < γ_n(θ_n) < 1.  As such, this value of θ_n defines a sequence of n valid sharpness parameters which, by construction, each satisfy the conditions (<ref>) and thus each achieve an expected CHSH value greater than 2. 
        	
        
        The above theorem shows that an unbounded number of independent Bobs are able to violate the CHSH inequality with a single Alice, by sequentially measuring one half of a maximally entangled pair. 
        
        	To increase the number n of Bobs violating the CHSH inequality, we require smaller values of θ_n ∈ (0, π/4]. By upper and lower bounding γ_n by polynomials of θ, we find evidence to suggest that θ_n must decrease double-exponentially fast with n. A more detailed discussion of this is given in  Appendix <ref> the Supplemental Material. Our findings agree with the numerics presented in Fig. <ref> where we plot θ (log-scale) against the number of CHSH violations possible.
        
          We also investigate the behaviour of the sequence of violations of the CHSH inequality. For the strategy described in this  paper Letter, we can bound k< 2^2-k(γ_k(θ_n) θ_n + 2^k-1)<  2 + 2^2-kθ_n for k=1,…,n, where we have used 1+ √(1-γ_j^2(θ_n))<2 for each j=1,…,n and γ_k(θ_n)<1. Therefore, the sizes of the CHSH violations must decrease as least as fast as θ_n.
        
        	The analysis of this section was restricted to the setting where Alice and 1 initially share a maximally entangled qubit pair. However, the measurement strategy presented can be readily extended to a larger class of two qubit states. In particular, if Alice and 1 initially share any pure two-qubit entangled state then it is also possible to define a measurement strategy that gives rise to arbitrarily many CHSH violations in the scenario under consideration. The larger class of states and their respective measurement strategies are discussed in  Appendix <ref> the Supplemental Material.
        
        
        § DISCUSSION
         Discussion.|
        In this work we have shown that it is possible for arbitrarily many independent Bobs to violate the CHSH inequality with a single Alice using only a single maximally entangled qubit pair and while making uniformly distributed inputs, answering the problem posed in <cit.> in the negative and overturning a claim in <cit.> due to an implicit assumption made there. In addition, we showed that the same can be achieved if the parties initially share any pure, entangled two-qubit state. We also provided bounds on the size of the violations achievable with our strategy, giving evidence that they decrease double-exponentially fast with the number of Bobs.
        
        The magnitude of the violations suggests that our measurement strategies, whilst able to achieve an unbounded number of violations, are not suited to device-independent tasks like randomness expansion in which the quantity of randomness certifiable increases with the size of the CHSH violation.  However, it is known that by using the tilted CHSH inequality <cit.> and allowing one party to choose from an exponentially large number of measurements, an unbounded number of random bits can (in principle, but not in a robust way) be certified from a single pair of entangled qubits in a device-independent manner <cit.>. More recently, in <cit.> it was shown that more than two bits of local randomness can be robustly certified by performing sequential measurements on a two-qubit system and with each party choosing from at most three measurements. Further work is needed to understand the limitations on robust device-independent randomness expansion from a single pair of entangled qubits.
        
        The scenario of the present  paper  Letter  has also been studied for the related tasks of steering <cit.> and entanglement witnessing <cit.>.  Since the presence of non-locality implies the possibility of steering and the presence of entanglement, our work shows that it is also possible to achieve both of these with arbitrarily many independent Bobs using only a single pair of entangled qubits. This goes against some of the results in <cit.>, and suggests that it is worth rethinking others that study the space of such correlations <cit.> or look at other Bell inequalities <cit.> or more parties <cit.> after removing the assumption that both measurements used by each party have the same sharpness. 
        
        Several interesting questions remain open. Firstly, can we fully characterise the set of two-qubit states that allow for an unbounded number of CHSH violations? Here, we gave a sufficient condition for two-qubit state to achieve unbounded violations. In the standard case of a single Bob the set of states for which a violation of the CHSH inequality is possible has been fully characterised <cit.>. It would be interesting to know whether the conditions presented in  Theorem <ref> the Supplemental Material  are both necessary and sufficient for unbounded violations.
        
        Second, in the scenario analysed here we only permit a single qubit to be transmitted between subsequent Bobs. We could also consider the setting in which we additionally allow the Bobs to share classical information, i.e., the inputs and outputs of previous Bobs. Such a setting would open up the possibility of the Bobs using an adaptive strategy. In <cit.> some steps were made in this direction but the authors' strategy required that the inputs and outputs of the Bobs were also sent to Alice before she made her measurement. Whilst our work here implies that it is also possible in the classically-assisted setting to achieve unbounded violations, it would be interesting to know if the classical communication could be used to produce larger violations and more noise-tolerant strategies. 
        
        Finally, in a different direction, it would be interesting to explore the scenario where we also consider multiple Alices. In particular, how many expected pairwise violations could a sequence of Alices and Bobs achieve from a single pair of qubits if they act independently?
        
         Acknowledgements.|
        	PB thanks Mirjam Weilenmann for feedback on an earlier version of the manuscript. This work was supported by EPSRC's Quantum Communications Hub (grant numbers EP/M013472/1 and EP/T001011/1), an EPSRC First Grant (grant number EP/P016588/1) and the French National Research Agency via Project No. ANR-18-CE47-0011 (ACOM). The majority of this work was carried out whilst PB was at the University of York.
        
        
        
        
        
        
        
        
        
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          10.1103/PhysRevA.101.042340journaljournalPhysical Review A volume101, pages042340 (year2020)NoStop
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          authorauthorG. Foletto, authorL. Calderaro,
          authorA. Tavakoli, authorM. Schiavon, authorF. Picciariello, authorA. Cabello, authorP. Villoresi,  and authorG. Vallone, titletitleExperimental demonstration of sustained entanglement and nonlocality
          after sequential measurements, 
          10.1103/PhysRevApplied.13.044008journaljournalPhysical Review Applied volume13, pages044008 (year2020)NoStop
        
        
        § OPTIMALITY OF THE LÜDERS RULE
        
        The following lemma shows that use of the Lüders rule is optimal in terms of the information retained in the post-measurement state.
        
          Let {E_i}_i=1^s be a POVM on . Any instrument {_i}_i=1^s that implements this POVM can be constructed by performing the Lüders measurement followed by a quantum channel that depends on the outcome. 
        
          Let () be the set of density operators on . If {K_i^j}_j are the Kraus operators representing the channel _i, then, when acting on a state ρ∈(), the post-measurement state on obtaining outcome i is proportional to _i(ρ)=∑_jK_i^jρ(K_i^j)^†.
        
          Using the Lüders update, the post measurement state on receiving outcome i is proportional to √(E_i)ρ√(E_i).  We now consider the channel with Kraus operators {J_i^j}_j∪{𝕀-Π_E_i}, where Π_E_i is the projector onto the support of E_i, J_i^j=K_i^jE_i^-1/2 and the inverse is interpreted as the pseudo-inverse (so that E_i^-1/2E_iE_i^-1/2=Π_E_i).  These form a valid quantum channel since
          
            ∑_j(J^j_i)^† J^j_i+𝕀-Π_E_i=∑_jE_i^-1/2(K_i^j)^† K_i^jE_i^-1/2+𝕀-Π_E_i=E_i^-1/2E_iE_i^-1/2+𝕀-Π_E_i=𝕀 ,
        
        where we have used ∑_j(K_i^j)^† K_i^j=E_i which follows because _i has corresponding POVM element E_i. Furthermore, the concatenation of the Lüders measurement with this channel corresponds to the map
        
            ρ↦(𝕀-Π_E_i)√(E_i)ρ√(E_i)(𝕀-Π_E_i)+∑_jJ^j_i√(E_i)ρ√(E_i)(J^j_i)^†=∑_jK_i^jΠ_E_iρΠ_E_i(K_i^j)^†=∑_jK_i^jρ(K_i^j)^†=_i(ρ) ,
        
        where we have used K_i^jΠ_E_i=K_i^j for all j, which follows from Corollary <ref> noting that Π_E_i≥Π_(K_i^j)^† K_i^j.
        In other words, using the Lüders rule followed by the channel with Kraus operators {J^j_i}_j∪{𝕀-Π_E_i} is equivalent to the relevant instrument channel.
        
        Let K be a linear map from _A to _B and Π_K^† K be the projector onto the support of K^† K. Then KΠ_K^† K=K.
        
          We first suppose K is of the form ∑_id_i|i⟩_B⟨i|_A, where {|i⟩_A} ({|i⟩_B}) is an orthonormal basis for _A (_B).  We have K^† K=∑_i|d_i|^2|i⟩_A⟨i|_A and Π_K^† K=∑_i:d_i≠0|i⟩_A⟨i|_A. Thus, KΠ_K^† K
          =K.
        
        More generally, let K=UDV be the singular value decomposition of K, where U:_B→_B and V:_A→_A are unitary and D:_A→_B is non-zero only on the diagonal.  We have K^† K=V^† D^† DV and hence Π_K^† K=V^†Π_D^† DV. Thus, KΠ_K^† K=UDVV^†Π_D^† DV=UDV=K, using the first part of the proof.
        
        Let K be a linear map from _A to _B and Π be a projector on _A such that Π≥Π_K^† K. Then KΠ=K.
        
        
        
        § DERIVATION OF CHSH EXPRESSION
        
        
        In this section we will derive the CHSH value for a more general strategy, recovering the expression  in Equation (<ref>) from the main text  as a special case.  The notation here and basis for the strategy is inspired by <cit.>. Given a state ρ_AB of two qubits, we define T(ρ_AB) to be the 3× 3 matrix whose entries are given by T_i,j(ρ_AB) = ρ_AB(σ_i ⊗σ_j). Considering this matrix for the initial state ρ_AB^(1) shared between Alice and 1, we define λ_0, λ_1 be the two largest eigenvalues of T(ρ_AB^(1))T^T(ρ_AB^(1)) and let c_0, c_1 be the corresponding orthonormal eigenvectors. We also define b_i = T^T(ρ_AB^(1))c_iT^T(ρ_AB^(1))c_i. Note that for two unit-length vectors a, b∈^3 we have ρ(σ_a⊗σ_b) = (a, T(ρ) b) where (· , ·) is the standard Euclidean inner product for the real vector space ^3.
        
        Let the measurement strategy of Alice and k be defined by the effects
        
            A_0|0   = 12(𝕀 + cos(θ) σ_c_0 + sin(θ)σ_c_1), 
            
            A_0|1   = 12(𝕀 + cos(θ) σ_c_0 - sin(θ)σ_c_1), 
            
            B_0|0^(k)   = 12(𝕀 + σ_b_0), 
            
            B_0|1^(k)   = 12(𝕀 + γ_k  σ_b_1).
        
        Defining the expectation operatorsA_x = A_0|x - A_1|x and B_y = B_0|y - B_1|y, the CHSH value of Alice and k can be written 
        
            k   = ρ_AB^(k)((A_0 + A_1)⊗ B_0^(k)) + ρ_AB^(k)((A_0 - A_1)⊗ B_1^(k)), 
               = 2 cos(θ) (c_0, T(ρ_AB^(k)) b_0) + 2 γ_k sin(θ) (c_1, T(ρ_AB^(k)) b_1) .
        
        For the case k=1 this gives
        
            1   = 2 cos(θ) (c_0, T(ρ_AB^(1)) b_0) + 2 γ_1 sin(θ) (c_1, T(ρ_AB^(1)) b_1) , 
               = 2 cos(θ) (c_0, T(ρ_AB^(1)) T^T(ρ_AB^(1)) c_0)/T^T(ρ_AB^(1)) c_0 + 2 γ_1 sin(θ) (c_1, T(ρ_AB^(1)) T^T(ρ_AB^(1)) c_1)/T^T(ρ_AB^(1)) c_1 , 
               = 2 cos(θ) T^T(ρ_AB^(1)) c_0 + 2 γ_1 sin(θ)T^T(ρ_AB^(1)) c_1, 
               = 2 cos(θ) √(λ_0) + 2 γ_1 sin(θ)√(λ_1).
        
        To derive the general expression we will show how to relate (c_0, T(ρ_AB^(k)) b_0) to (c_0, T(ρ_AB^(1)) b_0) and (c_1, T(ρ_AB^(k)) b_1) to (c_1, T(ρ_AB^(1)) b_1). 
        
        Let ρ_AB^(k-1) be the state shared between Alice and k-1 prior to k-1's measurement (and where Alice has not measured). Using the Lüders update rule the state sent to k is
        
            ρ_AB^(k)   = 1/2∑_b,y(𝕀⊗√(B_b|y^(k))) ρ_AB^(k-1)(𝕀⊗√(B_b|y^(k)))  
               = (𝕀⊗12(𝕀 + σ_b_0)) ρ_AB^(k-1)(𝕀⊗12(𝕀 + σ_b_0)) + (𝕀⊗12(𝕀 - σ_b_0)) ρ_AB^(k-1)(𝕀⊗12(𝕀 - σ_b_0)) 
                + (𝕀⊗√(12(𝕀 + γ_k  σ_b_1))) ρ_AB^(k-1)(𝕀⊗√(12(𝕀 + γ_k  σ_b_1))) + (𝕀⊗√(12(𝕀 - γ_k  σ_b_1))) ρ_AB^(k-1)(𝕀⊗√(12(𝕀 - γ_k  σ_b_1)))  
               =2 + √(1-γ_k-1^2)/4ρ_AB^(k-1) + 1/4 (𝕀⊗σ_b_0) ρ_AB^(k-1) (𝕀⊗σ_b_0) + 1-√(1-γ_k-1^2)/4 (𝕀⊗σ_b_1) ρ_AB^(k-1) (𝕀⊗σ_b_1).
        
        The final line follows from a direct calculation using the identity
        
            √(12(𝕀±γ_k  σ_b_1)) = (√(1+γ_k) + √(1-γ_k)) 𝕀± (√(1+γ_k) - √(1-γ_k)) σ_b_1/2√(2).
        
        Now consider the quantity (c_0, T(ρ_AB^(k)) b_0). Using (<ref>) we may write this as 
        
            (c_0, T(ρ_AB^(k)) b_0)    = 2 + √(1-γ_k-1^2)/4 (c_0, T(ρ_AB^(k-1)) b_0) + (𝕀⊗σ_b_0) ρ_AB^(k-1) (𝕀⊗σ_b_0)(σ_c_0⊗σ_b_0)/4
               +(1-√(1-γ_k-1^2))(𝕀⊗σ_b_1) ρ_AB^(k-1) (𝕀⊗σ_b_1)(σ_c_0⊗σ_b_0)/4.
        
        Using the cyclicity of the trace this may be written more succinctly as 
        
            (c_0, T(ρ_AB^(k)) b_0)    = 2 + √(1-γ_k-1^2)/4 (c_0, T(ρ_AB^(k-1)) b_0) + ρ_AB^(k-1)(c_0⊗σ_b_0σ_b_0σ_b_0)/4
               +(1-√(1-γ_k-1^2))ρ_AB^(k-1)(c_0⊗σ_b_1σ_b_0σ_b_1)/4.
        
        We then use the identity σ_aσ_b = (a,b)𝕀 + iσ_a×b to simplify the matrix products in the above expressions. In particular we have
        σ_b_0σ_b_0σ_b_0=σ_b_0 and
        σ_b_1σ_b_0σ_b_1=-σ_b_0.
        Inserting this into the above expression we get
        
            (c_0, T(ρ_AB^(k)) b_0)    = (2 + √(1-γ_k-1^2)/4 + 1/4 - 1-√(1-γ_k-1^2)/4)(c_0, T(ρ_AB^(k-1)) b_0) 
               = (1 + √(1-γ_k-1^2)/2)(c_0, T(ρ_AB^(k-1)) b_0).
        
        Performing an analogous calculation for (c_1, T(ρ_AB^(k)) b_1) yields
        
            (c_1, T(ρ_AB^(k)) b_1)    = (2 + √(1-γ_k-1^2)/4 - 1/4 + 1-√(1-γ_k-1^2)/4)(c_1, T(ρ_AB^(k-1)) b_1) 
               = 1/2(c_1, T(ρ_AB^(k-1)) b_1).
        
        By recursion these give
        
            (c_0, T(ρ_AB^(k)) b_0)    = 2^1-k(c_0, T(ρ_AB^(1)) b_0)∏_j=1^k-1(1+√(1-γ_j^2)) 
            
            (c_1, T(ρ_AB^(k)) b_1)    = 2^1-k(c_1, T(ρ_AB^(1)) b_1) .
        
        Inserting these into the CHSH expression (<ref>) and noting that √(λ_i) = (c_i, T(ρ_AB^(1)) b_i) we obtain
        
            k = 2^2-k(γ_k √(λ_1)sin(θ) + √(λ_0)cos(θ)∏_j=1^k-1(1+ √(1-γ_j^2))).
        
        Note that for the maximally entangled state we have λ_0 = λ_1 = 1 and so in this case k reduces to the CHSH expression (<ref>) from the main text.
        
        
        
        § PROOF OF LEMMA <REF>
        
        
        § PROOF OF LEMMA 1
        
        We prove this for the more general sequence (cf. the expression given in  Section <ref> of  the main text),
        
            γ_k(θ) := 
            (1+ϵ)2^k-1 - cos(θ) ∏_j=1^k-1(1 + √(1-γ_j^2(θ)))/√(λ_1)sin(θ)       if  0 < γ_k-1(θ) < 1 
            ∞   otherwise ,
        
        with ϵ>0, λ_1>0 and γ_1(θ) = (1+ϵ)1-cos(θ)/√(λ_1)sin(θ).  The lemma as stated in the main text corresponds to the case where λ_1 = 1.
        
        First we note that γ_1(θ) > 0 for all θ∈ (0,π/4] and ϵ>0. Furthermore, γ_1(θ)<1 whenever 0<θ < arctan(2 (1+ϵ) √(λ_1)(1+ϵ)^2 - λ_1), so the sequence may admit additional finite terms.
        
        If a term in the sequence is infinite then so are all subsequent terms. Suppose then that the k^th term of the sequence is finite. This implies that 0 < γ_j(θ) < 1 for all j = 1, …, k-1 and in particular we have 1 < 1+√(1- γ_k-1^2(θ)) < 2. The bound 1+√(1- γ_k-1^2(θ)) < 2 then implies that γ_k(θ)/γ_k-1(θ) > 2 and so the finite terms of the sequence are strictly increasing and the sequence as a whole is a positive, increasing sequence.
        
        
        
        § PROOF OF LEMMA <REF>
        
        
        § PROOF OF LEMMA 2
        
        Again, we prove this lemma for the more general sequence (cf. the expression given in  Section <ref> of  the main text),
        
            γ_k(θ) := 
            (1+ϵ)2^k-1 - cos(θ) ∏_j=1^k-1(1 + √(1-γ_j^2(θ)))/√(λ_1)sin(θ)       if  0 < γ_k-1(θ) < 1 
            ∞   otherwise ,
        
        with ϵ>0, λ_1>0 and γ_1(θ) = (1+ϵ)1-cos(θ)/√(λ_1)sin(θ).  The lemma as stated in the main text corresponds to the case where λ_1 = 1.
        
        We use the inequalities √(1-x^2)≥ 1 - x^2, for x ∈ [0,1]; cos(θ) ≥ 1-θ^2/2, for θ∈ (0, π/4]; and sin(θ)≥θ/2, for θ∈ (0, π/4]. Applying these to γ_k(θ) we have that when γ_k(θ) is finite,
        
            γ_k(θ)    ≤ (1+ϵ) 2^k-1 - (1-θ^2/2)∏_j=1^k-1(2 - γ_j^2(θ))/√(λ_1)θ/2
               = (1+ϵ)2^k 1 - (1-θ^2/2)∏_j=1^k-1(1 - γ_j^2(θ)/2)/√(λ_1)θ.
        
        Now define a new sequence p_k(θ) based on this upper bound where
        
            p_k(θ) = 
            (1+ϵ)2^k 1 - (1-θ^2/2)∏_j=1^k-1(1 - p_j^2(θ)/2)/√(λ_1)θ       if  0 < p_k-1(θ) < 1 
            ∞   otherwise ,
        
        and p_1(θ) = (1+ϵ) θ/√(λ_1). Since the right hand side of (<ref>) is increasing with γ_j(θ) for any j<k, we have that p_k(θ) ≥γ_k(θ). Therefore, γ_k(θ) is finite whenever p_k(θ) is finite. 
        
        
        
        We now show that there always exists a θ_k ∈ (0, π/4] such that 0 < p_k(θ) < 1 for all θ∈ (0, θ_k) and that lim_θ→ 0^+ p_k(θ) = 0.
        
        Firstly, note that 0 < p_1(θ) < 1 for all θ∈ (0, √(λ_1)1+ϵ) and that p_1(θ) is an odd polynomial in θ. Therefore on the interval (0, √(λ_1)1+ϵ), p_2(θ) is an odd polynomial. As p_2(θ) > p_1(θ) > 0 and p_2(θ) has no constant term we know that lim_θ→ 0^+ p_2(θ) = 0 and therefore, as p_2(θ) is a continuous positive function on the interval, there exists some θ_2 ∈ (0, √(λ_1)1+ϵ) such that 0 < p_2(θ) < 1 for all θ∈ (0, θ_2). We now proceed by induction to show that there exists some θ_k>0 such that p_k(θ) is an odd polynomial on the interval (0, θ_k) and 0 < p_k(θ) < 1. 
        
        Suppose there exists a θ_k-1 such that on the interval (0, θ_k-1) all p_j(θ) for j = 1,…, k-1 are odd polynomials in θ and 0 < p_j(θ) < 1. If this is the case then on the interval (0, θ_k-1)p_k(θ) is finite and the numerator in (<ref>) is an even polynomial in θ with no constant term. Cancelling the θ in the denominator, p_k(θ) is an odd polynomial in θ. In particular, this implies that lim_θ→ 0^+ p_k(θ) = 0 and also, because p_k(θ) > 0, there exists some θ_k ∈ (0, θ_k-1) such that 0 < p_k(θ) < 1 for all θ∈ (0, θ_k). It follows by induction that for any n ∈ we can find a θ_n∈(0,π/4] such that 0 < p_1(θ) < … < p_n(θ) < 1 for all θ∈ (0, θ_n). 
        
        
        Finally, as for each n ∈ we have 0 <γ_n(θ) ≤ p_n(θ) it follows that there exists a θ_n ∈ (0, π/4] such that 0 < γ_1(θ) < … < γ_n(θ) < 1 for all θ∈ (0, θ_n) and that lim_θ→ 0^+γ_n(θ) = 0.
        
        
        
        
        § UNBOUNDED VIOLATIONS FOR A LARGER SET OF TWO-QUBIT STATES
        
        Following the same proof strategy as presented in the main text, in order to observe k > 2 for the more general measurement strategy (<ref>) we require
        
            γ_k > 2^k-1 - √(λ_0)cos(θ) ∏_j=1^k-1(1 + √(1-γ_j^2))/√(λ_1)sin(θ).
        
        We can define a sequence (γ_i(θ))_i analogous to that  of (<ref>) from the main text, i.e., for some fixed ϵ > 0 we have
        
            γ_k(θ) =  (1+ϵ) 2^k-1 - √(λ_0)cos(θ) ∏_j=1^k-1(1 + √(1-γ_j^2))/√(λ_1)sin(θ)   if γ_j∈ (0,1) for allj∈{1,…,k-1}
            ∞   otherwise.
        
        By  Lemma <ref> Lemma 1  we have that for any non-zero λ_0, λ_1, γ_k(θ) is a strictly increasing sequence when finite. However, in the more general setting  Lemma <ref> Lemma 2  holds only when λ_0 = 1. From this, the proof of unbounded violations ( Theorem <ref> Theorem 1  from the main text) can be replicated for the measurement strategy (<ref>) when λ_0 = 1 and λ_1 > 0. We therefore arrive at the following, more general, theorem. 
        
        
        	Let ρ_AB^(1) be an entangled two-qubit state and let λ_0, λ_1 be the two largest eigenvalues of the matrix T(ρ_AB^(1))T^T(ρ_AB^(1)). If λ_0 = 1 and λ_1 > 0, then for any n ∈, there exists a sequence (γ_i)_i and a θ∈ (0, π/4] such that the measurement strategy (<ref>) achieves
        	
            k > 2,
        
        	for all k = 1,…,n. 
        
        
        As an example, for two-qubit states of the form
        
            [   α   0   0   β;   0   0   0   0;   0   0   0   0; β^*   0   0 1-α ]
        
        where α∈[0,1] and |β|≤√(α(1-α)), we have λ_0=1 and λ_1=4|β|^2. Moreover, we have σ_b_0=σ_3 and σ_b_1=σ_1 and so their measurement strategy coincides with the strategy presented in the main text.[For this family of states the second and third largest eigenvalues coincide, i.e., λ_1 = λ_2, and so one could take σ_b_1 = σ_2 instead.] By the Schmidt-decomposition we can always write a pure two-qubit entangled state as |ψ⟩ = cos(φ) |00⟩ + sin(φ) |11⟩ for φ∈ (0, π/4]. As such, all pure two-qubit entangled states may be written in the form (<ref>) for some suitably chosen basis. Furthermore, we have λ_1 = sin^2(2 φ) which means that λ_1 > 0 for all φ∈(0, π/4].
        
        
        	By initially sharing any pure, entangled two-qubit state, the number of Bobs that can violate the CHSH inequality with a single Alice is unbounded.
        
        	There are also some mixed states that satisfy the conditions λ_0 = 1 and λ_1 >0. If the condition λ_0 = 1 holds then there exists some vector c∈^3 of unit length such that ρ(σ_ c⊗σ_ c) = ± 1. If c_+ and c_- are the projectors onto the eigenstates of σ_ c, then any mixture of ρ with c_+⊗c_+ or c_-⊗c_- will produce a state that still satisfies λ_0 = 1.
        
        
        § BOUNDS ON THE SEQUENCE (Θ_N)
        
        Suppose θ is chosen small enough such that the k^th term in the sequence (γ_k)_k ∈ is finite, i.e.
        
            γ_k(θ) = (1+ϵ)2^k-1 - cos(θ)∏_j=1^k-1(1+ √(1-γ_j^2(θ)))/sin(θ).
        
        Using the inequalities √(1-x^2)≥ 1 - x^2 for x ∈ [0,1], cos(θ) ≥ 1 - θ^2/2 for θ∈, and sin(θ) ≥θ/2 for θ∈ [0,π/2] we arrive at the upper bound
        
            γ_k(θ) ≤ (1+ϵ) 2^k1- (1- θ^2/2)∏_j=1^k-1( 1 - γ_j^2(θ)/2) /θ.
        
        Similarly, using the inequalities √(1-x^2)≤ 1 - x^2/2 for x ∈ [0,1], cos(θ) ≤ 1 - θ^2/4 for θ∈ [0,π/4], and sin(θ) ≤θ for θ∈ [0,π/2] we find the lower bound
        
            γ_k(θ) ≥ (1+ϵ) 2^k-11- (1-θ^2/4)∏_j=1^k-1( 1 - γ_j^2(θ)/4 ) /θ.
        
        We now define two sequences p_k^↑(θ) and p_k^↓(θ) based on the upper and lower bounds respectively:
        
            p^↑_k(θ) := (1+ϵ) 2^k1- (1- θ^2/2)∏_j=1^k-1( 1 - p^↑_j(θ)^2/2) /θ,
        
            p^↓_k(θ) := (1+ϵ) 2^k-11- (1-θ^2/4)∏_j=1^k-1( 1 - p^↓_j(θ)^2/4 ) /θ,
        
        with p^↑_1(θ) = (1+ϵ)  θ and p^↓_1(θ) = (1+ϵ)θ/4. This gives us two sequences that satisfy p^↓_k(θ) ≤γ_k(θ) ≤ p_k^↑(θ) whenever p_j^↑(θ) < 1 for all j < k. For all k, p_k^↑(θ) and p_k^↓(θ) are polynomials in θ with no constant term.  We proceed to estimate the growth of γ_k(θ) by bounding the growth of the coefficients of θ. Let c^↑_k and c^↓_k be the coefficient of θ in p^↑_k(θ) and p_k^↓(θ) respectively. Then,
        
            c^↑_k = (1+ϵ)2^k-1(1+∑_j=1^k-1 (c^↑_j)^2 )
        
        and
        
            c^↓_k = (1+ϵ)2^k-3(1 + ∑_j=1^k-1 (c^↓_j)^2 )
        
        with c^↑_1 = (1+ϵ) and c^↓_1 = (1+ϵ)/4. Unfortunately, no general method exists for solving nonlinear recurrence relations of higher orders. A simpler lower bounding sequence for c^↓_k can be found by noting ∑_j=1^k-1 c_j^2 ≥ c_k-1^2. Using this we get a sequence
        
            _k =(1+ϵ)2^k-3 (_k-1)^2 .
        
        By direct computation we can compute c^↓_4, a polynomial in ϵ with positive co-efficients, and with constant term 
        larger than 4.  We choose to start the sequence _k at _4=4≤ c^↓_4. Note that in principle we could start the sequence at _1=(1+ϵ)/4, but starting at the fourth term, which is always larger than 1, gives a tighter bound. (In addition, the choice of the bound of 4 enables a neater final expression).
        Now consider an upper bound for c_k^↑, using ∑_j=1^k-1 c_j^2 ≤ (k-1) c_k-1^2 we have 
        
            c_k^↑   = (1+ϵ)2^k-1(1 + ∑_j=1^k-1  (c^↑_j)^2 )
               ≤ (1+ϵ)2^k-1 k (c^↑_k-1)^2  
               ≤ (1+ϵ) 2^2k - 1  (c^↑_k-1)^2,
        
        where on the second line we used k ≤ 2^k. Using this upper bound we define the new sequence 
        
            _k =(1+ϵ)2^2k-1 (_k-1)^2,
        
        with _1 = (1+ϵ).
        The following two lemmas give us a closed form for _k and _k.
        
        	Let k ∈ with k ≥ k_0 and t, c ∈. Then the sequence
        	
            b_k := 2 b_k-1 + tk + c
        
        	admits the closed form solution
        	
            b_k=2^k-1(b_k_0+(2+k_0)t+c)-(k+2)t-c .
        
        		Rearranging, we find the relation 
        		
            b_k - 2 b_k-1 = t k + c
        
        		and through a telescoping procedure we can write 
        		
            b_k-2^k-k_0b_k_0   = ∑_r=0^k-k_0-1 2^r(b_k-r-2b_k-r-1) 
               = ∑_r=0^k-k_0-1 2^r (t(k-r)+c).
        
        		Evaluating the summation we find
        		
            b_k-2^k-k_0b_k_0   =2^k-k_0((2+k_0)t+c)-(k+2)t-c
        
        		and so 
        		
            b_k    = 2^k-k_0(b_k_0+(2+k_0)t+c)-(k+2)t-c.
        
        	Let ϵ>0 and k ∈ with k ≥ 1. Let 
        	
            _k    =  (1+ϵ)2^k-3 (_k-1)^2
            		and_k    = (1+ϵ)2^2k-1 (_k-1)^2
        
        	define two sequences with _4=4 and _1 = (1+ϵ). Then for k≥4 we have
        	
            _k    = (1+ϵ)^2^k-4-1 2^5×2^k-4-k+1and for k≥1 we have_k    = (1+ϵ)^2^k-1 2^5× 2^k-1 - 2 k - 3 .
        
        		We derive a closed-form expression for the sequence
        		
            d_k = (1+ϵ)2^tk-a d_k-1^2,
         
        		for t,a,k ∈ and k≥ k_0. Taking logarithms of both sides we have 
        		
            log(d_k) = (tk-a) + log(1+ϵ) + 2 log(d_k-1).
        
        		Defining a new sequence via the replacements b_k = log(d_k) and c = log(1+ϵ) - a, we may apply Lemma <ref> to get
        		
            b_k   =2^k-k_0(b_k_0+(2+k_0)t+c)-(k+2)t-c.
        
        	Equivalently,
        	
            d_k   =2^2^k-k_0(log(d_k_0)+(2+k_0)t+c)-(k+2)t-c
               =2^2^k-k_0(log(d_k_0)+(2+k_0)t+log(1+ϵ)-a)-(k+2)t-log(1+ϵ)+a
        
        	To recover _k we first set t=1, a=3, k_0=4 and d_4=4 to obtain
        	
            _k   =2^2^k-4(5+log(1+ϵ))-(k+2)-log(1+ϵ)+3
               = 2^(2^k-4-1)(log(1+ϵ)2^5×2^k-4-k+1
               = (1+ϵ)^2^k-4-1 2^5×2^k-4+1-k.
        
        	Setting t=2, a=1, k_0=1 and d_1=(1+ϵ) we find
        	
            _k    = 2^2^k-1(log(1+ϵ)+log(1+ϵ) + 5) - 2(k+2) - log(1+ϵ) + 1
               = 2^2^k-1(2log(1+ϵ)+5)-(2k+3)-log(1+ϵ)
               = 2^(2^k-1)log(1+ϵ)+5× 2^k-1-2k-3
               = (1+ϵ)^2^k-1 2^5× 2^k-1-2k-3
        
        Hence both _k and _k exhibit double-exponential growth with k.  Relating this back to γ_n(θ), from the proof  of Lemma <ref> in Section <ref> we know that there exists some θ_n ∈ (0, π/4) such that on the interval (0, θ_n) we have 0 < p_k^↓(θ) ≤γ_k(θ) ≤ p_k^↑(θ) for all k < n. Furthermore, from the two previous lemmas we have established that there exist polynomials d_n^↓θ + O(θ^3) and d_n^↑θ + O(θ^3) such that
        
            d_n^↓θ+O(θ^3)≤γ_n(θ)≤ d_n^↑θ+O(θ^3)
        
        on this interval. The two inequalities together imply that γ_n(θ) grows double-exponentially fast in n for sufficiently small θ, which suggests that we may need θ to be double exponentially small to observe γ_n(θ) ≤ 1. However, because we have not ruled out that the higher order terms remain significant when θ_n is double-exponentially small in n, our argument is not conclusive, although a double-exponental scaling agrees with the numerics presented in  Fig. <ref> the main text.
