Randomness in nonlocal games between mistrustful players

# Randomness in nonlocal games between mistrustful players

Carl A. Miller National Institute of Standards and Technology, 100 Bureau Dr., Gaithersburg, MD 20890, USA,
Joint Center for Quantum Information and Computer Science, University of Maryland, College Park, MD 20742, USA
Yaoyun Shi Dept. of Electrical Engineering and Computer Science, University of Michigan, Ann Arbor, MI 48109, USA
July 25, 2019
###### Abstract

If two quantum players at a nonlocal game achieve a superclassical score, then their measurement outcomes must be at least partially random from the perspective of any third player. This is the basis for device-independent quantum cryptography. In this paper we address a related question: does a superclassical score at guarantee that one player has created randomness from the perspective of the other player? We show that for complete-support games, the answer is yes: even if the second player is given the first player’s input at the conclusion of the game, he cannot perfectly recover her output. Thus some amount of local randomness (i.e., randomness possessed by only one player) is always obtained when randomness is certified from nonlocal games with quantum strategies. This is in contrast to non-signaling game strategies, which may produce global randomness without any local randomness. We discuss potential implications for cryptographic protocols between mistrustful parties.

## I Introduction

When two quantum parties Alice and Bob play a nonlocal game and achieve a score that exceeds the best classical score , their outputs must be at least partially random. In other words, all Bell inequality violations certify the existence of randomness. This fact is at the center of protocols for device-independent quantum cryptography, where untrusted devices are used to perform cryptographic procedures. In particular, this notion of certification is the basis for device-independent randomness expansion, where a small random seed is converted into a much larger uniformly random output by repeating Bell violations col:2006; pam:2010; ColbeckK:2011; Vazirani:dice; pironio13; fehr13; CoudronVY:2013; CY:STOC; Miller:2016; Universal-spot; Dupuis:2016; Arnon:2016.

A natural question arises: is new randomness also generated by one player from the perspective of the other player? Specifically, if denotes Alice’s outputs, denotes the post-measurement state that Bob has at the conclusion of the game, and denotes all side information (including Alice’s input), is there a certified lower bound for the conditional entropy ? Besides helping us understand the nature of certified randomness, this particular kind of randomness (local randomness) has applications in mutually mistrustful cryptographic settings, where Alice and Bob are cooperating but have different interests.

Quantifying local randomness (i.e., randomness that is only known to one player) is challenging because many of the known tools do not apply. Lower bounds for the total randomness (i.e, randomness from the perspective of an outside adversary) have been computed as a function of the degree of the Bell violation (see Figure 2 in pam:2010) but are not directly useful for certifying local randomness. One of the central challenges is that we are measuring randomness from the perspective of an active, rather than passive, adversary: Bob’s guess at Alice’s output occurs after Bob has carried out his part of the strategy for . Current tools for device-independent randomness expansion are not designed to address the case where the adversary is a participant in the nonlocal game.

Does the generation of certified randomness always involve the generation of local certified randomness? The answer is not obvious: for example, in the non-signaling setting, Alice and Bob could share a PR-box111That is, the unique -part non-signaling resource whose input bits and output bits always satisfy which generates bit of certified randomness per use, but no new local randomness – Bob could perfectly guess Alice’s output from his own if he were given Alice’s input.

Motivated by the above, we prove the following result in this paper (see Theorem LABEL:robustthm for a formal statement).

###### Theorem 1 (Informal).

For any complete-support game222That is, a game in which each input pair occurs with nonzero probability. , there is a constant such that the following holds. Suppose Alice and Bob use a strategy for which achieves a score that is above the best classical score (with ). Then, at the conclusion of the strategy and given Alice’s input, Bob can guess her output with probability at most .

We note that similar problems have been studied in the literature in settings different from ours. There has been other work examining the scenario where a third party tries to guess Alice’s output after a game (e.g., Pawlowski:2010, Kempe:2011b, Acin:PRA:2016), and single-round games have appeared where Bob is sometimes given only Alice’s input, and asked to produce her output (e.g., Mancinska:2014, Vazirani:QKD:PRL, Vidick:2013). (We believe the novelty of our scenario in comparison to these papers is that we consider the randomness of Alice’s output after Bob has performed his part of a quantum strategy, and thus has potentially lost information due to measurement.) Two recent papers also address randomness between multiple players, under assumptions about imperfect storage Kaniewski:2016; Rib:2016.

In addition to the above, we prove a structural theorem for quantum strategies that allow perfect guessing by Bob. Not only do such strategies not achieve Bell inequalities, but they are also essentially classical in the following sense. Let denote the quantum systems possessed by Alice and Bob, respectively

###### Theorem 2 (Informal).

Suppose that Alice’s and Bob’s strategy is such that if the game is played and then Bob is given Alice’s input, he can perfectly guess her output. Then, there is an isometry mapping Bob’s system to such that Bob’s strategy for involves only , and all of Alice’s observables commute with the reduced state on .

(See Theorem 5 and Corollary 7 for a formal statement.) Thus, in the case of perfect guessing, the strategy is equivalent to one in which Alice’s measurements have no effect on the shared state.

### i.1 Structure of the paper

We begin with the case of perfect guessing. We formalize the concept of an essentially classical strategy, using a definition of equivalence between strategies which is similar to definitions used in results on quantum rigidity. We then give the proof of Theorem 2. It is known that two sets of mutually commuting measurements on a finite-dimensional space can be expressed as the pullback of bipartite measurements. This fact is used along with matrix algebra arguments to show the necessary splitting of Bob’s system into .

Then we proceed with the proof of Theorem 1. It has been observed by previous work (e.g., Mancinska:2014, vidickthesis) that if a measurement on a system from bipartite state are highly predictable via measurements on , then the measurement does not disturb the reduced state by much: . In this paper we give a simplified proof of that fact (Proposition 11). The interesting consequence for our purpose is that if Alice’s measurements are highly predictable to Bob, then Alice can copy out her measurement outcomes in advance, thus making her strategy approximately classical. We take this a step further, and show that if Bob first performs his own measurement on the resulting classical-quantum correlation is also approximately preserved by Alice’s measurements (which is not necessarily true of the original entangled state ). This is sufficient to show that an approximately-guessable strategy yields an approximately classical strategy.

The subtleties in the proof are in establishing the error terms that arise when Alice copies out multiple measures from her side of the state. We note that the proof crucially requires that the game has complete support. An interesting further avenue is to explore how local randomness may break down if the condition is not satisfied.

In section LABEL:sec:disc we discuss the implications of our result.

## Ii Preliminaries

For any finite-dimensional Hilbert space , let denote the vector space of linear automorphisms of . For any , we let denote . If is a subspace of , let denote orthogonal projection onto .

Throughout this paper we fix four disjoint finite sets , which denote, respectively, the first player’s input alphabet, the second player’s input alphabet, the first player’s output alphabet, and the second player’s output alphabet. A -player (input-output) correlation is a vector of nonnegative reals, indexed by , satisfying for all pairs , and satisfying the condition that the quantities

 pxa:=∑ypxyab,pyb:=∑xpxyab (1)

are independent of and , respectively (no-signaling).

A -player game is a pair where

 q:A×B→[0,1] (2)

is a probability distribution and

 H:A×B×X×Y→[0,1] (3)

is a function. If for all and , the game is said to have a complete support. The expected score associated to such a game for a -player correlation is

 ∑a,b,x,yq(a,b)H(a,b,x,y)pxyab. (4)

We will extend notation by writing , , and (if ).

A -player strategy is a -tuple

 Γ = (D,E,{{Rxa}x}a,{{Syb}y}b,γ) (5)

such that are finite dimensional Hilbert spaces, is a family of -valued positive operator valued measures (POVMs) on (indexed by ), is a family of -valued positive operator valued measures on , and is a density operator on . The second player states of are defined by

 ρxyab := TrD[√Rxa⊗Sybγ√Rxa⊗Syb] (6)

(These states are, more explicitly, the subnormalized states of Bob’s system that arise after both Alice and Bob have performed their measurements.) Define by the same expression with replaced by the identity operator. (These represent the pre-measurement states of the second-player.) Define for any .

We say that the strategy achieves the -player correlation if for all . If a -player correlation can be achieved by a -player strategy then we say that it is a quantum correlation.

If is a convex combination of product distributions (i.e., distributions of the form where and ) then we say that is a classical correlation. Note that if the underlying state of a quantum strategy is separable (i.e., it is a convex combination of bipartite product states) then the correlation it achieves is classical. The maximum expected score that can be achieved for a game by a classical correlation is denoted .

## Iii Perfect Guessing

We first address the case of perfect guessing — that is, the case when the second-player states that remain after the game is played are perfectly distinguishable by Bob. It turns out that this condition will imply some strong structural conditions on the strategy used by Alice and Bob, and it will imply in particular that Alice’s and Bob’s score at the game cannot be better than that of any classical strategy.

### iii.1 Congruent strategies

It is necessary to identify pairs of strategies that are essentially the same from an operational standpoint. We use a definition that is similar to definitions from quantum self-testing (e.g., Definition 2.13 in McKagueBQP_published).

A unitary embedding from a -player strategy

 Γ = (D,E,{{Rxa}x}a,{{Syb}y}b,γ) (7)

to another -player strategy

 ¯¯¯¯Γ = (¯¯¯¯¯D,¯¯¯¯E,{{¯¯¯¯Rxa}x}a,{{¯¯¯¯Syb}y}b,¯¯¯γ) (8)

is a pair of unitary embeddings and such that , , and .

Additionally, if is such that , and for all , then we will call the strategy given by

 (D1,E,{{Gxa}a}x,{{Syb}y}b,TrD2γ) (9)

a partial trace of . We can similarly define a partial trace on the second subspace if it is a tensor product space.

We will say that two strategies and are congruent if there exists a sequence of strategies such that for each , either is a partial trace of , or vice versa, or there is a unitary embedding of into , or vice versa. This is an equivalence relation. Intuitively, two strategies are congruent if one can be constructed from the other by adding or dropping irrelevant information. Note that if two strategies are congruent then they achieve the same correlation.

### iii.2 Essentially classical strategies

We are ready to define the key concept in this section and to state formally our main theorem.

###### Definition 3.

A quantum strategy (5) is said to be essentially classical if it is congruent to one where commutes with for all and .

Note that if the above condition holds, then applying the measurement map

 X ↦ ∑x√GxaX√Gxa (10)

to the system leaves the state of unchanged.

We are interested in strategies after the application of which Bob can predict Alice’s output given her input. This is formalized as follows. If are positive semidefinite operators on some finite dimensional Hilbert space , then we say that is perfectly distinguishable if and have orthogonal support for any . This is equivalent to the condition that there exists a projective measurement on which perfectly identifies the state from the set .

###### Definition 4.

A quantum strategy (5) allows perfect guessing (by Bob) if for any , is perfectly distinguishable.

###### Theorem 5 (Main Theorem).

If a strategy for a complete-support game allows perfect guessing, then it is essentially classical.

(We note that the converse of the statement is not true. This is because even in a classical strategy, Alice’s output may depend on some local randomness, which Bob cannot perfectly predict.)

Before giving the proof of this result, we note the following proposition, which taken together with Theorem 5 implies that any strategy that permits perfect guessing yields a classical correlation.

###### Proposition 6.

The correlation achieved by an essentially classical strategy must be classical.

###### Proof.

We need only to consider the case that commutes with for all . For each , let , and let be the nondestructive measurement defined by

 Φa(T) = ∑x∈X|x⟩⟨x|⊗√RxaT√Rxa. (11)

Note that by the commutativity assumption, such operation leaves the state of unchanged.

Since the measurements do not disturb the state of , Alice can copy out all of her measurement outcomes in advance. Without loss of generality, assume . Let be the state that arises from applying the superoperators , in order, to . For any , the reduced state is precisely the same as the result of taking the state , applying the measurement to , and recording the result in . Alice and Bob can therefore generate the correlation from the marginal state alone (if Alice possesses and Bob possesses ). Since this state is classical on Alice’s side, and therefore separable, the result follows.

###### Corollary 7.

If a strategy for a complete-support game allows perfect guessing, the correlation achieved must be classical.

### iii.3 Proving Theorem 5

The proof will proceed as follows. First, we show that Alice’s measurements induce projective measurements on Bob’s system. Next, we argue that commutes with Bob’s own measurement for any . This allows us to isometrically decompose Bob’s system into two subsystems , such that acts trivially on , while alone can be used to predict given . The latter property allows us to arrive at the conclusion that commutes with .

We will need the following lemma, which is well-known and commonly attributed to Tsirelson. We will only sketch the proof, and more details can be found in Appendix A of Doherty:2008. The lemma asserts that for families of positive semidefinite operators on a finite-dimensional space , commutativity (i.e., the condition that commute for and ) implies bipartiteness (i.e., the condition that and can be obtained as pullbacks via a map of operators on and , respectively).

###### Lemma 8.

Let be positive semidefinite operators on a finite-dimensional Hilbert space such that for all . Then, there exists a unitary embedding and and positive semidefinite operators on and on such that and for all .

Proof sketch. Via the theory of von Neumann algebras, there exists an isomorphism

 V ≅ ⨁ℓVℓ⊗Wℓ (12)

under which

 Mj ≅ ⨁ℓMℓj⊗I, (13) Nk ≅ ⨁ℓI⊗Nℓk. (14)

Let , and let .

###### Proof of Theorem 5.

Express as in (5). Without loss of generality, we may assume that . By the assumption that allows perfect guessing, for any , the second-player states must be perfectly distinguishable (since otherwise the post-measurement states would not be). Therefore, we can find projective measurements on such that

 QxaρQxa=ρxa. (15)

Note that for any fixed , if Alice and Bob were to prepare the state and Alice were to measure with and Bob were to measure with , their outcomes would be the same.

We have that the states

 ρxyab = √SybQxaρQxa√Syb (16) ρx′yab = √SybQx′aρQx′a√Syb (17)

have orthogonal support for any . Since , we have for some . Therefore,

 ⟨√SybcQxa√Syb,√SybcQx′a√Syb⟩ = 0, (18)

which implies, using the cyclicity of the trace function,

 ∥∥QxaSybQx′a∥∥2 = 0 (19)

Therefore, the measurements and commute for any . (This is clear from writing out the matrix in block form under the subspaces determined by the projections .)

By Lemma 8, we can find a unitary embedding and POVMs on such that and . With

 ¯¯¯γ=(ID⊗i)γ(ID⊗i∗), (20)

the strategy embeds into the strategy

 Γ′:=(D,E1⊗E2,{{Rxa}x}a,{¯¯¯¯Syb⊗IE2}y}b,¯¯¯γ).

For any fixed , the state is such that applying the measurement to the system and the measurement to the system always yields the same outcome. In particular, if we let

 τxa = TrE2(¯¯¯¯Qxa¯¯¯γ), (21)

then will always be equal to if and equal to otherwise. Therefore commutes with the operators , and thus also with their sum .

Thus if we trace out the strategy over the system , we obtain a strategy (congruent to the original strategy ) in which Alice’s measurement operators commute with the shared state. ∎

## Iv Approximate Guessing

Next we address the case where the second-player states are not necessarily perfectly distinguishable as varies, but are approximately distinguishable. (Thus, if Bob were given Alice’s input after the game was played and asked to guess her output, he could do so with probability close to .) We begin by quantizing “approximate” distinguishability.

###### Definition 9.

Let denote a finite set of positive semidefinite operators on a finite dimensional Hilbert space . Then, let

 Dist{ρi} = max∑iTr(Tiρi), (22)

where the maximum is taken over all POVMs on .

Note that if , and each is nonzero, then this quantity has the following interpretation: if Alice gives Bob a state from the set at random according to the distribution , then is the optimal probability that Bob can correctly guess the state. This quantity is well-studied (see, e.g., Spehner:2014_published).

When we discussed perfect distinguishability, we made use of measurements that commuted with a given state. In the current section we will need an approximate version of such commutativity, and thus we make the following definition.

###### Definition 10.

Let denote a completely positive trace-preserving map over a finite-dimensional Hilbert space . Let denote a density operator on . Then we say that is -commutative with if

 ∥Φ(β)−β∥1 ≤ ϵ. (23)

Note that this relation obeys a natural triangle inequality: if is -commutative with , and is -commutative with , then

 ∥Φ2(Φ1(β))−β∥1 ≤ ∥Φ2(Φ1(β))−Φ2(β)∥1+∥Φ2(β)−β∥1 ≤ ∥Φ1(β)−β∥1+ϵ2 ≤ ϵ1+ϵ2.

The following known proposition will be an important building block. We give a proof that is a significant simplification of a method from Lemma 29 in vidickthesis. (See also Lemma 2 in Mancinska:2014 for a related result.)

###### Proposition 11.

Let be a density operator and a projective measurement on such that the induced states satisfy

 Dist{ΛBi} = 1−δ. (24)

Then, the superoperator is -commutative with .

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