Optimal quantum preparation contextuality in n -bit parity-oblivious multiplexing task
aa r X i v : . [ qu a n t - ph ] J un Optimal quantum preparation contextuality in n -bit parity-oblivious multiplexing task Shouvik Ghorai ∗ and A. K. Pan † Indian Institute of Science Education and Research Kolkata, Mohanpur, Nadia 741246, India and National Institute of Technology Patna, Ashok Rajhpath, Patna 800005, India
In [ PRL, 102, 010401 (2009)], Spekkens et al. have shown that quantum preparation contextu-ality can power the parity-oblivious multiplexing (POM) task. The bound on the optimal successprobability of n -bit POM task performed with the classical resources was shown to be the same asin a preparation non-contextual theory. This non-contextual bound is violated if the task is per-formed with quantum resources. While in 2-bit POM task the optimal quantum success probabilityis achieved, in 3-bit case optimality was left as an open question. In this paper, we show thatthe quantum success probability of a n -bit POM task is solely dependent on a suitable 2 n − × n Bell’s inequality and optimal violation of it optimizes the success probability of the said POM task.Further, we discuss how the degree of quantum preparation contextuality restricts the amount ofquantum violations of Bell’s inequalities, and consequently the success probability of a POM task.
I. INTRODUCTION
By demonstrating an ingenious gedanken experimentEinstein, Podoloski and Rosen had remarked[1] that thequantum mechanical description of nature by using ψ function is inherently incomplete. The ontological mod-els of an operational quantum theory seek to provide a‘complete specification’ of the state of a quantum systemso that the individual measured values of any dynami-cal variable are predicted by an appropriate set of onticstates (usually denoted as λ ’s). Studies on this issue haveresulted in spectacular discoveries about the constraintsthat need to be imposed on the ontological models inorder to be consistent with the statistics of quantum me-chanics (QM). Bell’s theorem [2] is the first which pro-vides a constraint that an ontological model has to benonlocal. Shortly after the Bell’s theorem, Kochen andSpecker (KS) [3, 4] demonstrated an inconsistency be-tween QM and the non-contextual ontological models.In a non-contextual ontological model the individualmeasured values of an observable that occur for an appro-priate set of λ s is irrespective of the way it is being mea-sured. Let an observable b A be commuting with b B and b C ,with b B and b C being non-commuting. Then the assump-tion of non-contextuality asserts that the value occurringin a measurement b A is independent of, whether the mea-surement is performed with b B or b C . KS theorem [3, 4]demonstrates that such a non-contextual assignment ofvalues is impossible for all possible set of measurementsfor d ≥
3. The original KS proof was demonstrated us-ing 117 projectors for qutrit system. Later, simpler ver-sions of it using lower number of projectors have beenprovided [5]. Apart from the KS-type all-versus-nothingproof, inequality based proofs have also been provided[6]. Note that KS-type proof has a limited applicabilitybecause it uses an additional assumption of outcome de- ∗ [email protected] † [email protected] terminism for sharp measurement along with the usualmeasurement non-contextuality assumption. Also, it isnot applicable to any arbitrary operational theory, ratheris specific to quantum theory. The traditional notion ofnon-contextuality was further generalized by Spekkens[8]for any arbitrary operational theory and extended theformulation to the transformation and preparation non-contextuality. The present paper concerns the notionof preparation non-contextuality of an ontological modeland its quantum violation.Before proceeding further let us recapitulate theessence of an ontological model reproducing the quan-tum statistics [7]. Given a preparation procedure P ∈ P and a measurement procedures M ∈ M , an opera-tional theory assigns probability p ( k | P, M ) of obtaininga particular outcome k ∈ K M . Here M is the set ofmeasurement procedures and P is the set of prepara-tion procedures. In QM, a preparation procedure pro-duces a density matrix ρ and measurement procedure(in general described by a suitable POVM E k ) providesthe probability of a particular outcome k is given by p ( k | P, M ) =
T r [ ρE k ], which is the Born rule. In anontological model of QM, it is assumed that whenever ρ is prepared by a specific preparation procedure P ∈ P aprobability distribution µ P ( λ | ρ ) in the ontic space is pre-pared, satisfying R Λ µ P ( λ | ρ ) dλ = 1 where λ ∈ Λ and Λis the ontic state space. The probability of obtaining anoutcome k is given by a response function ξ M ( k | λ, E k )satisfying P k ξ M ( k | λ, E k ) = 1 where a measurementoperator E k is realized through a particular measure-ment procedure M ∈ M . A viable ontological modelshould reproduce the Born rule, i.e., ∀ ρ , ∀ E k and ∀ k , R Λ µ P ( λ | ρ ) ξ M ( k | λ, E k ) dλ = T r [ ρE k ].According to the generalized notion of non-contextuality proposed by Spekkens [8], an ontologicalmodel of an operational theory can be assumed tobe non-contextual if two experimental proceduresare operationally equivalent, then they have equiv-alent representations in the ontological model. Iftwo measurement procedures M and M ′ producessame observable statistics for all possible prepara-tions then the measurements M and M ′ belong tothe equivalent class. An ontological model of QM isassumed to be measurement non-contextual if ∀ P : p ( k | P, M ) = p ( k | P, M ′ ) ⇒ ξ M ( k | λ, E k ) = ξ M ′ ( k | λ, E k )is satisfied. KS non-contextuality assumes the afore-mentioned measurement non-contextuality along withthe outcome determinism for the sharp measure-ment. Similarly, an ontological model of QM can beconsidered to be preparation non-contextual one if ∀ M : p ( k | P, M ) = p ( k | P ′ , M ) ⇒ µ P ( λ | ρ ) = µ P ′ ( λ | ρ )is satisfied where P and P ′ are two distinct prepa-ration procedures but in the same equivalent class.In an ontological model of QM, the preparation non-contextuality implies the outcome determinism for sharpmeasurements and also implies KS non-contextuality[8]. Then any KS proof can be considered as a proof ofpreparation contextuality but converse does not hold. Inthis sense, preparation non-contextuality is a strongernotion than traditional KS non-contextuality [9]. Veryrecently, it is also shown [10] that any ontological modelsatisfying both the assumptions of preparation and themeasurement non-contextuality cannot reproduce allquantum statistics, even if the assumption of outcomedeterminism for sharp measurement is dropped. Exper-imental test of such an universal non-contextuality hasalso been provided which are free from idealized assump-tions of noiseless measurements and exact operationalequivalences [11].The failure of non-contextuality is a signature of non-classicality which is of foundational importance. It wouldthen be interesting if this non-classical feature of contex-tuality can be used as a resource for providing advantagein various information processing and computation tasks,similar to the spirit of the violations of Bell inequali-ties which have been shown to be a resource for device-independent key distribution [12] and certified random-ness [13]. In an interesting work Spekkens et al. [14]have first demonstrated how quantum preparation con-textuality can power a communication game termed asparity-oblivious multiplexing (POM) task.The essence of a n -bit POM task can be encapsu-lated as follows. Alice has a n -bit string x chosen uni-formly at random from { , } n . Bob can choose any bit y ∈ { , , ..., n } and recover the bit x y with a probability.The condition of the task is, Bob’s output must be thebit b = x y , i.e., the y th bit of Alice’s input string x . Inother words, Alice and Bob try to optimize the proba-bility p ( b = x y ) with the constraint that no informationabout any parity of x can be transmitted to Bob.In [14], it is shown that a n -bit POM task performedwith classical resources is constrained by an inequal-ity. Interestingly, the same inequality can be obtainedin any ontological model satisfying preparation non-contextuality. For the case of 2-bit POM task they haveachieved the optimal quantum success probability, whichis recently reaffirmed [15] through the Cirelson bound.For 3-bit POM task, Spekkens et al. [14] provided thequantum advantage over the non-contextual POM taskbut the question pertaining to its optimality is left as an open problem. In this paper, we first show that thequantum success probability provided in [14] is indeedthe optimal one for the 3 − bit POM task. In order todemonstrate this we first prove that the optimal successprobability of 3 − bit POM task in QM is solely dependenton the optimal quantum violation of the elegant Bell in-equality proposed by Gisin [16]. We then generalize ourapproach to show that one can find a suitable 2 n − × n elegant Bell’s inequality for n -bit POM task and optimalviolation of it provides the optimal success probabilityof the POM task. Further, we discuss how the degreeof quantum preparation contextuality puts constraint onthe quantum violations of Bell’s inequalities and conse-quently on the quantum success probability of a POMtask. II. THE POM TASK AND PREPARATIONNON-CONTEXTUALITY
Following [14], we define a parity set P n = { x | x ∈{ , } n , P r x r ≥ } with r ∈ { , , ..., n } . The crypto-graphic constraint here is the following. For any s ∈ P n ,no information about s.x = ⊕ r s r x r (s-parity) is to betransmitted to Bob, where ⊕ is sum modulo 2. For ex-ample, when n = 2 the set is P = { } , so no informationabout x ⊕ x can be transmitted by Alice.The maximum probability of success in a classical n -bit POM task is ( n + 1) / n , because only those encodingof x which does not provide any information about parityare those which encodes a single bit. While the explicitproof can be found in [14], a simple trick can saturate thebound as follows. Assume that Alice always encodes thefirst bit (pre-discussed between Alice and Bob) and sendsto Bob. If y = 1, occurring with probability 1 /n , Bobcan predict the outcome with certainty and for y = 1,occurring with probability of ( n − /n , he at best guessesthe bit with probability 1 /
2. Hence the total probabilityof success is 1 /n + ( n − / n = ( n + 1) / n . Since y is chosen uniformly it is irrelevant which bit is encoded.This does not disclose the parity information to Bob.Let in an operational theory, Alice encodes her n -bitstring of x prepared by a procedure P x . Next, after re-ceiving the particle , for every y ∈ { , , ..., n } , Bob per-forms a two-outcome measurement M y and reports out-come b as his output. Then the probability of success isgiven by p ( b = x y ) = 12 n n X x,y p ( b = x y | P x , M y ) . (1)The parity-obliviousness condition in the operationaltheory guarantees that there is no outcome of any mea-surement for which the probabilities for s-parity 0 ands-parity 1 are different. Mathematically, ∀ s ∀ M ∀ k X x | x.s =0 p ( P x | k, M ) = X x | x.s =1 p ( P x | k, M ) . (2)For a preparation non-contextual ontological model, itis proved [14] that the success probability in n -bit POMtask satisfies the bound p ( b = x y ) pnc ≤ (cid:18) n (cid:19) . (3)In order to derive this bound it is proved that in apreparation non-contextual models, parity-obliviousnessat the operational level implies equivalent represen-tation in the ontological model, so that, ∀ M : P x | x.s =0 µ ( λ | P x , M ) = P x | x.s =1 µ ( λ | P x , M ) is satisfied. Thisis true even if Bob can perfectly determine the ontic state λ . In quantum POM task, Alice encodes her n -bit stringof x into pure quantum states ρ x = | ψ x ih ψ x | , pre-pared by a procedure P x . After receiving the particle,Bob performs a two-outcome measurement M y for ev-ery y ∈ { , , ..., n } and reports outcome b as his out-put. Spekkens et al. [14] have proved that the opti-mal quantum success probability for 2-bit POM task is p optQ = (1 / / √ > p ( b = x y ) pnc = 3 / − bit quantum POM task they provided a success prob-ability p Q = (1 / / √
3) but left open the questionof optimality of it.Recently, Chailloux et al. [17] have shown that for evenPOM task optimal success probability is (1 / / √ n ).However, they first proved that POM task can be shownto be equivalent to an another game in some condi-tions and then optimize the success probability of thatgame. By taking a different approach, Banik et al. [18]obtained the quantum optimal success probability of the2 − bit POM task through the Tsirelson bound [19] ofCHSH inequality [20]. Instead of two-outcome mea-surement, Hameedi et al. [21] have derived the non-contextual bound for m -outcome scenario is given by p ( b = x y ) pnc ≤ ( n + m − / ( nm ). However, they havenumerically optimized the quantum success probabilityof POM task for n = 2 and m = 3 ... n − bit POM taskfor dichotomic outcomes. Interestingly, the success prob-ability can be shown to be solely linked to the 2 n − × n elegant Bell’s inequality [16] which obviously reduces tothe CHSH inequality for 2-bit POM task. Further, byusing an interesting technique we analytically optimize2 n − × n elegant Bell’s inequality which in turn providesthe optimal quantum success probability ( p optQ ) of n -bitPOM task. In order to showing this, let us first providean explicit derivation of p optQ for 3-bit POM task whichwill help the reader to understand the optimization of p Q for n -bit quantum POM task. III. 3-BIT POM TASK AND OPTIMALQUANTUM SUCCESS PROBABILITY
For 3-bit POM task Alice chooses her bit x randomlyfrom { , } . We rewrite all the possible x as an ordered set D = (000 , , , , , , , P = { , , , } . If weconsider the case when s = 110, the bits having s − parity0 are { } and the bits have s − parity1 are { } . From Eq. (2), the parity-obliviousness in a non-contextual ontological model canthen be ensured if ∀ M and ∀ kp ( P | k, M ) + p ( P | k, M ) + p ( P | k, M )+ p ( P | k, M ) = p ( P | k, M ) + p ( P | k, M )+ p ( P | k, M ) + p ( P | k, M ) . (4)Similar parity-oblivious conditions can be found for ev-ery other element of P .Let us consider an entangled quantum state ρ AB = | ψ AB ih ψ AB | where | ψ AB i ∈ C ⊗ C . Alice randomly per-forms one of the four projective measurements { P A i , I − P A i } where i = 1 , , , x into eightpure qubits as ρ x , are given by12 ρ = T r h ( P A ⊗ I ) ρ AB i (5a)12 ρ = T r h ( I − P A ⊗ I ) ρ AB i (5b)12 ρ = T r h ( P A ⊗ I ) ρ AB i (5c)12 ρ = T r h ( I − P A ⊗ I ) ρ AB i (5d)12 ρ = T r h ( P A ⊗ I ) ρ AB i (5e)12 ρ = T r h ( I − P A ⊗ I ) ρ AB i (5f)12 ρ = T r h ( P A ⊗ I ) ρ AB i (5g)12 ρ = T r h ( I − P A ⊗ I ) ρ AB i . (5h)After receiving the information from Alice, Bob per-forms three projective measurements { P B y , I − P B y } with y = 1 , , ρ + ρ = ρ + ρ = ρ + ρ = ρ + ρ = I / ( ρ + ρ + ρ + ρ ) = ( ρ + ρ + ρ + ρ ).Spekkens et al. [14] has obtained a quantum successprobability p Q = (1 / (cid:0) / √ (cid:1) of the 3-bit POM task.Given an entangled state | Ψ AB i = ( | i + | i ) / √
2, if thefollowing choices of observables in Alice’s end are made,so that, A = ( σ x + σ y + σ z ) / √ A = ( σ x + σ y − σ z ) / √ A = ( σ x − σ y + σ z ) / √ A = ( − σ x + σ y + σ z ) / √ B = σ x , B = − σ y and B = σ z , thenthe above bound can be achieved. Similarly, one maychose another entangled state for which a different set ofobservables is required to obtain that bound. The ques-tion is whether the above quantum success probability isoptimal.In this paper, we first prove that quantum successprobability for 3 − bit POM task obtained by Spekkens et al. [14] is indeed the optimal one. This is shown throughthe optimal violation of the elegant Bell inequality [16].In order to showing this, let us explicitly write down thequantum success probability for 3-bit POM task by usingEq.(1) is given by p Q = 124 h p (0 | ρ , P B ) + p (0 | ρ , P B ) + p (0 | ρ , P B )+ p (0 | ρ , P B ) + p (0 | ρ , P B ) + p (0 | ρ , I − P B )+ p (0 | ρ , P B ) + p (1 | ρ , I − P B ) + p (0 | ρ , P B )+ p (1 | ρ , I − P B ) + p (0 | ρ , P B ) + p (0 | ρ , P B )+ p (0 | ρ , P B ) + p (1 | ρ , I − P B ) + p (1 | ρ , I − P B )+ p (1 | ρ , I − P B ) + p (0 | ρ , P B ) + p (1 | ρ , I − P B )+ p (1 | ρ , I − P B ) + p (1 | ρ , I − P B ) + p (0 | ρ , P B )+ p (1 | ρ , I − P B ) + p (1 | ρ , I − P B )+ p (1 | ρ , I − P B ) i . (6)Further simplification and rearrangements provide thefollowing form is given by p Q = 12 + hB i
24 (7)where B is the elegant Bell expression[16] is given by B =( A + A + A − A ) ⊗ B +( A + A − A + A ) ⊗ B (8)+( A − A + A + A ) ⊗ B The detailed calculation to derive Eq.(7) from Eq.(6) isshown in the Appendix A.We have thus shown that the optimality of p Q for 3-bit POM task requires the optimal violation of elegantBell inequality. For this, by following [22], we define γ = 4 √ I − B . Since A † i A i = I = B † y B y , γ can bedecomposed as γ = ( √ / P i =1 M † i M i where M i ’s arelinear combination of A i ’s and B y ’s M = ( B + B + B ) / √ − A M = ( B + B − B ) / √ − A M = ( B − B + B ) / √ − A M = ( − B + B + B ) / √ − A . (9)Since γ is positive semi-definite, we have hB i opt =4 √
3. This in turn optimize the success probability givenby Eq.(7), so that, p optQ = (1 / / √ − bitPOM task. Thus, p optQ of 3 − bit POM task is achievedthrough the optimal quantum violation of the elegantBell’s inequality.It is interesting to note here that hB i can be saturatedif the choice of B y ’s can be made in the following way, sothat A + A + A − A = (4 / √ B , A + A − A + A =(4 / √ B and A − A + A + A = (4 / √ B . Then, hB i = (4 / √ P y =1 h B y ⊗ B y i provides hB i opt = 4 √ h B y ⊗ B y i isequal to 1. The important question is whether such achoice of the observables and the state can be found.In fact, the choice made by in [14] satisfies the aboverequirements.Note that, the algebraic maximum of Eq.(8) is 12which may be obtained for a post-quantum theory (PRbox is an example for the case of Bell-CHSH expression)providing the maximum violation of parity-obliviousnesscondition.The above calculation is performed by assuming ele-ment s = 110 from the set P . One may take any of theother three elements of P to find the optimal successprobability. However, p optQ will remain same for any ofsuch cases. We now proceed to generalize the approachfor n -bit POM task. IV. n -BIT POM TASK AND OPTIMALSUCCESS PROBABILITY IN QM For n -bit POM task Alice chooses her bit x δ ran-domly from { , } n with δ ∈ { , ... n } . The relevantordered set D n can be written as D n = ( x δ | x i ⊕ x j =111 ...
11 and i + j = 2 n + 1) and i ∈ { , , ... n − } . Here, x = 00 ... , x = 00 ... , .... , and so on. The parity setis defined as P n = { x δ | x δ ∈ { , } n , P r x δr ≥ } . Wechoose x s = 1100 ...
00 and fix the s-parity 0 and s-parity1 sets.Let us consider a suitable entangled state ρ AB = | ψ AB i h ψ AB | with | ψ AB i ∈ C d ⊗ C d . Alice performs one ofthe 2 n − projective measurements { P A i , I − P A i } where i ∈ { , , ... n − } to encode her n -bits into 2 n pure quan-tum states are given by12 ρ x i = tr A [( P A i ⊗ I ) ρ AB ] (10a)12 ρ x j = tr A [( I − P A i ⊗ I ) ρ AB ]; (10b)with i + j = 2 n + 1.We define Bob’s measurements as M y = ( M iy , when b = x iy M jy , when b = x jy (11) M i ( j ) y = ( P B y , when x i ( j ) y = 0 I − P B y , when x i ( j ) y = 1 (12)The quantum success probability can then be written as p Q = 12 n n n X y =1 2 n − X i =1 p ( b = x iy | ρ x i , M iy )+ p ( b = x jy | ρ x j , M jy )= 12 n n n X y =1 2 n − X i =1 tr [ ρ x i M iy ] + tr [ ρ x j M jy ] (13)Since ∀ i, j , x i ⊕ x j = 111 ...
111 we have x iy ⊕ x jy = 1.Then, while x iy = 0 we can write tr [ ρ x i M iy ]+ tr [ ρ x j M jy ] = tr [ ρ x j ] + tr [( ρ x i − ρ x j ) P B y ], and while x jy = 0 we have tr [ ρ x i M iy ] + tr [ ρ x j M jy ] = tr [ ρ x i ] − tr [( ρ x i − ρ x j ) P B y ].Hence, if x iy = 0 the term tr [ ρ x j ] exists, while x iy = 1 , then tr [ ρ x i ] exists. So, Eq.(13) can be written as p Q = 12 n n n X y =1 2 n − X i =1 ( − x iy tr [( ρ x i − ρ x j ) P B y ] + tr [ ρ x ( i.xiy + j.xjy ) ]= 2 n − n n n + 12 n n n X y =1 2 n − X i =1 ( − x iy h (2 P A i − I ) ⊗ P B y i = 12 + 12 n n n X y =1 2 n − X i =1 ( − x iy h A i ⊗ B y i (14)Then the success probability of n -bit POM task is de-pendent on the 2 n − × n elegant Bell expression B n = n X y =1 2 n − X i =1 ( − x iy A i ⊗ B y (15)In order to optimize B n we define γ n = 2 n − √ n I − B n .By considering A † i A i = I = B † y B y , γ n can be written inthe following way γ n = √ n P n − i =1 M † i M i where M i = P y ( − x iy B y √ n − A i . Since γ n ≥ B n ≤ n − √ n I (16)It is then straightforward to see from Eq.(14) thatthe optimal quantum success probability for n − bit POMtask is p optQ = 12 (cid:16) √ n (cid:17) . (17)Thus, p optQ ≥ p optpnc for n -bit POM task. Thequestion remains whether such an amount of successprobability can be achieved for any n if Alice usesqubit system for encoding her input into pure states.Clearly, if the choices of observables is found for which(2 n − / √ n ) P n − i =1 ( − x iy A i = B y is satisfied then wehave hB n i = P ny =1 h B y ⊗ B y i which may provide B optn =2 n − √ n provided each of the h B y ⊗ B y i = 1. We havealready shown that for n = 2 and 3 such choices of ob-servables are available for qubit system. However, for n > n/ ] for n − bit POMtask. V. SUMMARY AND DISCUSSIONS
We studied how the quantum preparation contextu-ality provides advantage in a POM task. The successprobability of the n -bit POM task is shown to exceedthe non-contextual bound if performed with quantum re-sources. Spekkens et al. [14] have provided the optimalquantum success probability of 2-bit POM task whichis reaffirmed [18] through the Cirelson bound of CHSHinequality. The p Q of 3-bit POM task is shown [14] tobe larger than non-contextual bound but optimality of itwas left as an open question.By using an interesting approach, we showed that thesuccess probability of a n -bit POM task can be solelydependent on the quantum violation of 2 n − × n Bell’sinequality. Thus, the derivation of p optQ of n − bit POMtask reduces to the optimization of the relevant Bell ex-pression. For n = 2, the Bell inequality is the CHSHone and for n = 3 we have the elegant Bell’s inequality[16]. By using an interesting technique [22], we first opti-mize the elegant Bell expression arising from 3-bit POMtask and further generalized it for n -bit case. The opti-mal quantum value of 2 n − × n elegant Bell’s expressionis 2 n − √ n which in turn provides the optimal successprobability p optQ = (1 / / √ n ) for n -bit POM task.Note that for n = 2 , p optQ can be obtained even if Alicechooses pure qubit states for encoding her bits. But, for n > p optQ for n -bit POM task in AppendixC.Note that, the success probability can be unity ifthe value of relevant Bell expression reaches to its al-gebraic maximum. However, such amount of violationof Bell’s inequality may be obtained in a post-quantumtheory which then implies the highest degree of prepara-tion contextuality. In such a case, the overlap betweenthe respective probability distributions µ ( λ | ρ x | x.s =0 ) and µ ( λ | ρ x | x.s =1 ) corresponding to s-parity 0 and s-parity 1requires to be maximum. In QM, the maximum successprobability is (1 / / √ p optQ ≥ p ( b = x i ) pnc for any n > p optQ decreases with the increment of the number of bit n .The effect of preparation contextuality is then prominenthere. The condition of parity-obliviousness produce twomixed states in Bob’s side and such preparation proce-dures fix the relevant Bell’s inequality. Then, the overlapbetween µ ( λ | ρ x | x.s =0 ) and µ ( λ | ρ x | x.s =1 ) in the ontic spaceΛ for the case of 2 − bit POM task is larger than 3 − bitcase. It is then straightforward to understand that for n − bit POM tasks, both µ ( λ | ρ x | x.s =0 ) and µ ( λ | ρ x | x.s =1 )contains distributions corresponding to 2 n − pure states,so that, every pure state in ρ x | x.s =0 is very much close toa pure state in ρ x | x.s =1 yielding the distributions of onticstates for the mixed state indistinguishable (i.e., prepa-ration noncontextual) in the ontic space which therebyproviding the lowest success probability. Thus, optimal quantum preparation contextuality limits the amount ofviolation of Bell’s inequality and fixes the maximum suc-cess probability of the POM task. Acknowledgments
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Appendix A
Explicit derivation of Eq.(7) in the main text is shown. The quantum success probability given by Eq.(6) can berearranged as p Q = 124 (cid:16) T r [ ρ ] + T r [ ρ ] + T r [ ρ ] + 2 T r [ ρ ] + 2 T r [ ρ ] + 2 T r [ ρ ] + 3 T r [ ρ ]+ T r (cid:2) ( ρ − ρ ) P B (cid:3) + T r (cid:2) ( ρ − ρ ) P B (cid:3) + T r (cid:2) ( ρ − ρ ) P B (cid:3) − T r (cid:2) ( ρ − ρ ) P B (cid:3) + T r (cid:2) ( ρ − ρ ) P B (cid:3) + T r (cid:2) ( ρ − ρ ) P B (cid:3) − T r (cid:2) ( ρ − ρ ) P B (cid:3) + T r (cid:2) ( ρ − ρ ) P B (cid:3) + T r (cid:2) ( ρ − ρ ) P B (cid:3) − T r (cid:2) ( ρ − ρ ) P B (cid:3) + T r (cid:2) ( ρ − ρ ) P B (cid:3) + T r (cid:2) ( ρ − ρ ) P B (cid:3)(cid:17) (A1)Since ρ = 2 T r h ( P A ⊗ I ) ρ AB i and ρ = 2 T r h ( I − P A ⊗ I ) ρ AB i , we can write T r [( ρ − ρ ) P B ] =2 T r h (2 P A − I ) ⊗ P B ) i , T r [( ρ + ρ )] = 2 T r h I ⊗ I i = 2 and T r [( ρ − ρ )] = 2 T r h (2 P A − I ) ⊗ I i . Similarly, writing other terms in Eq.(A1), we get the following expression of the success probability in QM is givenby p Q = 124 (cid:16) T r [ ρ + ρ ] + T r [ ρ + ρ ] + T r [ ρ + ρ ] + T r [ ρ ] + T r [ ρ ] + T r [ ρ ] + 3 T r [ ρ ] (cid:17) + 112 h(cid:10) P A − I ⊗ P B (cid:11) + (cid:10) P A − I ⊗ P B (cid:11) + (cid:10) P A − I ⊗ P B (cid:11) − (cid:10) P A − I ⊗ P B (cid:11) + (cid:10) P A − I ⊗ P B (cid:11) + (cid:10) P A − I ⊗ P B (cid:11) − (cid:10) P A − I ⊗ P B (cid:11) + (cid:10) P A − I ⊗ P B (cid:11) + (cid:10) P A − I ⊗ P B (cid:11) − (cid:10) P A − I ⊗ P B (cid:11) + (cid:10) P A − I ⊗ P B (cid:11) + (cid:10) P A − I ⊗ P B (cid:11)i (A2)So, the success probability can be written as p Q = 12 + 124 h(cid:10) A ⊗ B (cid:11) + (cid:10) A ⊗ B (cid:11) + (cid:10) A ⊗ B (cid:11) − (cid:10) A ⊗ B (cid:11) + (cid:10) A ⊗ B (cid:11) + (cid:10) A ⊗ B (cid:11) − (cid:10) A ⊗ B (cid:11) + (cid:10) A ⊗ B (cid:11) + (cid:10) A ⊗ B (cid:11) − (cid:10) A ⊗ B (cid:11) + (cid:10) A ⊗ B (cid:11) + (cid:10) A ⊗ B (cid:11)i (A3)= 12 + hB i
24 (A4)where B = 2 P B − I and A = 2 P A − I are used. Eq.(A4) is the Eq.(7) in the main text. Appendix B
In this section, we provide explicit derivation of the optimal quantum success probability of 4-bit POMtask. Similar to 3 − bit case, let us define an ordered set D , where all possible x ’s are written as D =(0000 , , , , , , , , , , , , , , , D = ( x , x , ...x | x i + x j = 1111 and i + j = 17). We have the following parity set P = { , , , , , , , , } and for our purpose we take s = 1100. For s -parity 0 set, Aliceencodes her inputs in the following pure states are given by12 ρ = T r h ( P A ⊗ I ) ρ i ; 12 ρ = T r h ( I − P A ⊗ I ) ρ i (B1a)12 ρ = T r h ( P A ⊗ I ) ρ i ; 12 ρ = T r h ( I − P A ⊗ I ) ρ i (B1b)12 ρ = T r h ( P A ⊗ I ) ρ i ; 12 ρ = T r h ( I − P A ⊗ I ) ρ i (B1c)12 ρ = T r h ( P A ⊗ I ) ρ i ; 12 ρ = T r h ( I − P A ⊗ I ) ρ i . (B1d)and similar encoding for the s -parity 1 set. The quantum success probability can be calculated as p Q = 12 + hB i
64 (B2)where B = ( A + A + A + A − A + A + A + A ) ⊗ B + ( A + A + A − A + A + A − A − A ) ⊗ B +( A + A − A + A + A − A + A − A ) ⊗ B + ( A − A + A + A + A − A − A + A ) ⊗ B If we define,14 ( A + A + A + A − A + A + A + A ) ⊗ I = B ⊗ I
14 ( A + A + A − A + A + A − A − A ) ⊗ I = B ⊗ I (B3)14 ( A + A − A + A + A − A + A − A ) ⊗ I = B ⊗ I
14 ( A − A + A + A + A − A − A + A ) ⊗ I = B ⊗ I then B = 4 X y =1 B y ⊗ B y (B4)It is possible to find a choice of observables and states so that each of the h B y ⊗ B y i is 1. In such a case, B = 16providing the desired optimal probability p optQ = (1 + 1 / /
2. A choice observables and the state are the following. B = σ x ⊗ σ x , B = σ x ⊗ σ y , B = σ x ⊗ σ z and B = σ y ⊗ I and A = 12 ( σ x ⊗ σ x + σ x ⊗ σ y + σ x ⊗ σ z + σ y ⊗ I ) A = 12 ( σ x ⊗ σ x + σ x ⊗ σ y + σ x ⊗ σ z − σ y ⊗ I ) A = 12 ( σ x ⊗ σ x + σ x ⊗ σ y − σ x ⊗ σ z + σ y ⊗ I ) A = 12 ( σ x ⊗ σ x − σ x ⊗ σ y + σ x ⊗ σ z + σ y ⊗ I ) A = 12 ( − σ x ⊗ σ x + σ x ⊗ σ y + σ x ⊗ σ z + σ y ⊗ I ) A = 12 ( σ x ⊗ σ x + σ x ⊗ σ y − σ x ⊗ σ z − σ y ⊗ I ) A = 12 ( σ x ⊗ σ x − σ x ⊗ σ y + σ x ⊗ σ z − σ y ⊗ I ) A = 12 ( σ x ⊗ σ x − σ x ⊗ σ y − σ x ⊗ σ z + σ y ⊗ I ) (B5)For an entangled state | ψ i AB = 12 ( | i + | i + | i + | i ), the requirement of P y =1 B y ⊗ B y = 4 can beachieved. Appendix C