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Nonclassical temporal correlation powers quantum random access codes

We explore the fundamental origin of the quantum advantage behind random access code. We propose new temporal inequalities compatible with noninvasive-realist models and show that any non-zero quantum advantage of n (cid:55)→ 1 random access code in presence of shared randomness is equivalent to the violation of the corresponding temporal inequality. As a consequence of this con-nection we also prove that maximal success probability of n (cid:55)→ 1 random access code can be obtained when the maximal violation of the corresponding inequality is achieved. We further show that any non-zero quantum advantage of n (cid:55)→ 1 random access code, or in other words, any non-zero violation of the corresponding temporal inequality can certify genuine randomness.


I. INTRODUCTION
Random access codes (RAC) are among the most fundamental and powerful communication scenarios where a sender encodes a string of message into fewer bits and sends it to the receiver who then aims to recover any of the initial bits with some good probability of success [1][2][3][4][5]. In a general n → m RAC, there are n number of bits (say x 0 , x 1 , ..., x n−1 ) of message which the sender (say, Alice) sends to the receiver (say, Bob) by encoding the message in a m bit string (where m < n). In each round, Bob picks out a random number i (where i ∈ {0, 1, ..., n − 1}) and tries to guess the i-th bit (x i ) of Alice. The probability with which Bob can decode the message can be increased if instead of a classical system, Alice sends a quantum system, e.g., by using either quantum communication [3], or communication of classical bits assisted with a shared quantum state [6][7][8][9]. Such quantum random access codes have been introduced and developed for qubit systems as well as higher dimensional quantum systems.
Historically, RAC was first proposed in order to show the enormous information carrying capabilities of a quantum system compared to a classical system of the same dimension. Although the well-known Holevo bound [10] stipulates that by using a qubit system one cannot transmit more information reliably to the receiver than one classical bit, this does not solely portray the information carrying capabilities of a quantum system. A m-qubit can in general be depicted by an unit vector in 2 m dimensional complex Hilbert space revealing the possibility of encoding classical information with exponentially fewer qubits. In general, Bob may not need to know the information of all n bits together, but rather may choose to extract some bits of classical information out of the encoding depending on some task and therefore, may explore some degrees of freedom hand, Noninvasive measurability implies that 'it is possible, in principle, to determine which of the states the system is in, without affecting the state itself or the system's subsequent evolution'. Quantum theory violates consequences of both the models, i.e., KCBS inequality derivable from NCVM [40] and Legget-Garg inequality (LGI) arising from MR [36]. In recent years, LGI and other temporal correlations have acquired considerable attention from both fundamental [39,[41][42][43][44][45][46][47][48][49][50][51][52][53][54][55][56][57][58][59][60] as well as application perspectives [61][62][63][64][65][66][67]. (For a detailed survey, see [38,39]).
In the present work, we propose new inequalities for temporal correlations arising from sequential measurements and show that corresponding to every scenario of n → 1 RAC with SR there exists such an temporal inequality. Any quantum advantage of these communication tasks are implied by an associated violation of the derived temporal inequalities. These inequalities are asymmetric with respect to number of measurements. (For space-like separated correlations, asymmetric Bell inequalities were introduced to disprove the Peres conjecture [68].) Moreover, as an immediate consequence of our result, maximal violation of these temporal inequalities provide the maximal success probability of the corresponding RAC with SR. In general, maximal success probability of such RAC is derived numerically for n ≥ 1 with unproven optimality [5]. Moving on, we find an important application of our scheme in a cryptographically primitive task, viz. randomness generation. We show that any non-zero quantum advantage of RAC can be used for certifying randomness, while all previously proposed protocols for randomness generation based on RAC do not generate genuine randomness for any arbitrary success probability. [26,69,70].
The plan of the paper is as follows. In section II, a preliminary discussion on n → 1 RAC is provided. In section III, with the aim of designing operational criteria for testing RAC, temporal inequalities have been proposed for each n → 1 RAC. In section IV, it has been shown that any non-zero quantum advantage of n → 1 RAC can be used to generate genuine randomness. Finally, section V is reserved for discussion on the results obtained in this paper along with some future directions.

II. PRELIMINARIES
Let us now discuss the idea of n → 1 RAC in some details. Consider a prepare and measure scenario where Alice depending on the n-bit input string given to her uniformly at random, implements a preparation procedure by encoding the string in a qubit state ρ x 0 ,...,x n−1 . The qubit is then sent to the measurement device held by Bob, who, upon receiving an input y ∈ {0, ..., n − 1}, implements a binary outcome measurement B y and reports the outcome β ∈ {0, 1} as his output. Bob wins the game if he can perfectly guess the encoded bit sent by Alice. The average probability of winning is given by F n →1 = P(β y = y-th bit of Alice) = 1 n.2 n ∑ x 0 ,...,x n−1 ,y P(β y = x y | x 0 , ..., x n−1 , y), (1) where the coefficient 1 n.2 n appears due to normalization. Below we discuss some particular examples of n → 1 RAC and their explicit forms.
2 → 1 RAC: In a 2 → 1 RAC, Alice has an input string consisting of two bits x 0 x 1 ∈ {00, 01, 10, 11} which she wants to send to Bob by encoding the bits in a qubit ρ x 0 x 1 . Bob's task is to guess one of the bits (which is again chosen randomly) reliably. Therefore, after receiving an input y ∈ {1, 2}, he performs a binary outcome measurement B y and reports the outcome β ∈ {0, 1} as his output. In this scenario, following Eq. (1) the average probability of perfectly guessing the bit is given by Now consider the most general preparations and measurements as, Clearly, ρ 00 + ρ 11 = I, ρ 01 + ρ 10 = I, B 0 j + B 1 j = I for j = {1, 2} and the notation B b j j denotes the eigenstate corresponding to the outcome b j of measurement B j . The average success probability now can be written as The maximum achievable value for the above quantity is 1 2 (1 + 1 2 ) with classical strategy, whereas it can reach up to 1 ) if quantum strategy is used [5]. 3 → 1 RAC: In a 3 → 1 RAC, a three-bit input string x 0 x 1 x 2 from the set {000, 001, 010, 011, 100, 101, 110, 111} is given to Alice uniformly at random. Alice then encodes the input string in a qubit ρ x 0 x 1 x 2 and sends it to Bob. Bob upon receiving an input y ∈ {1, 2, 3}, implements a binary outcome measurement B y and reports the outcome β ∈ {0, 1} as his output. Following Eq. (1), the average probability of winning can be calculated as Let us now consider most general preparations and measurements as, Clearly, ρ 000 + ρ 111 = I, ρ 001 + ρ 110 = I, ρ 010 + ρ 101 = I, ρ 011 + ρ 100 = I. The average success probability can be written as Here, the average success probability, F 3 →1 can be achieved up to 1 2 (1 + 1 3 ) and 1 2 (1 + 1 √ 3 ), by classical and quantum strategies, respectively [5]. 4 → 1 RAC: In a 4 → 1 RAC, Alice has a four-bits input string x 0 x 1 x 2 x 3 which is given to her uniformly at random from the set {0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111}. She then implements a preparation procedure by encoding in a qubit ρ x 0 x 1 x 2 X 3 and sends to Bob. Bob upon receiving an input y ∈ {1, 2, 3, 4}, implements a binary outcome measurement B y and reports the outcome β ∈ {0, 1} as his output. The average probability of winning is given by Eq. (1) as Let us now consider most general preparations, and measurements Clearly, ρ 0000 + ρ 1111 = I, ρ 0001 + ρ 1110 = I, ρ 0010 + ρ 1101 = I, ρ 0011 + ρ 1100 = I, ρ 0100 + ρ 1011 = I, ρ 0101 + ρ 1010 = I, ρ 0110 + ρ 1001 = I, ρ 0111 + ρ 1000 = I. Therefore the explicit form of the average success probability of 4 → 1 RAC is calculated to be For 4 → 1 RAC, the exact value of classical and quantum average success probabilities are 1 2 (1 + 1 4 ) and 1 2 (1 + ) respectively [5]. n → 1 RAC: In a n → 1 RAC, Alice has an n-bits input string x 0 x 1 x 2 x 3 . . . x n−1 which is given to her uniformly at random. She then implements a preparation procedure by encoding this string in a qubit denoted by ρ x 0 x 1 x 2 x 3 ...x n−1 and sends it to Bob. Bob upon receiving an input y ∈ {1, 2, 3, . . . , n}, implements a binary outcome measurement B y and reports the outcome β ∈ {0, 1} as his output. The average probability of winning is given by Eq. (1). For n → 1 RAC, there are 2 n−1 preparations and n measurements. Let us now consider the most general preparations, and measurements Clearly, ρ m + ρm = I, where, m are the elements of the n-bit string set. The exact form of the average success probability of n → 1 RAC can be calculated as The average success probability for n → 1 RAC (with SR) using best classical strategy is known to be 1 2 (1 + 1 n ) [5].

III. TEMPORAL INEQUALITIES ASSOCIATED WITH THE RANDOM ACCESS CODES
We would now like to present a temporal inequality corresponding to each n → 1 RAC. Let us first give a brief description of the setup for the temporal scenario. In a temporal scenario, temporal correlations are obtained by measuring a single system sequentially at different instants of time. In each run of the experiment, two sequential measurements are performed on an identically prepared initial state. We assume that the first measurement is performed by Alice whereas the second one is performed by Bob. Suppose the measurements performed by Alice and the corresponding outcomes are denoted by A i and a i respectively. Similarly, the measurements performed by Bob are denoted by B j with corresponding outcome b j . All the measurements performed by both the parties are considered to be dichotomic i.e., a i , b j ∈ {0, 1}. Initially, the state on which Alice performs her measurement is denoted by ρ in .
Probability of obtaining outcome a i and b j is denoted by P(a i , b j | A i , B j ), when Alice measures A i at time t i and Bob measures B j at some later instant t j respectively following the standard procedure, the joint probability distribution can be obtained using Bayes' rule as, With this joint probability distribution, the two-time correlation is defined as, where ⊕ denotes addition modulo 2. In the subsections below we present the temporal inequalities corresponding to cases of 2 → 1, 3 → 1 and 4 → 1 RACs. The general form of the temporal inequality for n → 1 RAC is provided in the Appendix-(A).
A. Temporal inequality for 2 → 1 RAC Now, in order to derive a temporal inequality corresponding to the 2 → 1 RAC, let us first assume that Alice and Bob have two choices of binary measurements, say, {A 1 , A 2 } and {B 1 , B 2 } to perform in each run and ρ in = I 2 . Let us now consider the following quantity in terms of the above correlators as, Following Eq. (17), we calculate the C 11 term explicitly: where, A a i i represents the eigenstate corresponding to the outcome a i ∈ {0, 1} of the measurement A i and similarly for Bob. Here, the second equality can be derived using Eq.(16), i.e., by evaluating all the probability terms explicitly, and the third equality follows from the fact that two eigenstate corresponding to the same dichotomic measurement add up to unity, i.e., A 0 i + A 1 i = I and B 0 j + B 1 j = I for all i, j. Similarly, the other terms can be evaluated as Hence, the expression for K 2 →1 becomes, It may be noted here that the eigenstate of the measurements performed by Alice A a i i are the same as that of the preparations considered in Eq.(3). Now, comparing above equation with Eq.(5), one obtains Conversely, F 2 →1 = 1 2 + 1 8 K 2 →1 . In Appendix-(A 1), a noninvasive-realist bound for the term K 2 →1 has been derived to be 2. One can see from the above relation that whenever the value of the term K 2 →1 falls below 2, the success probability of 2 → 1 random access code also falls below 3 4 which is the maximum probability of success with classical strategy. Nonetheless, the maximum success probability of 2 → 1 RAC reaches 1 2 (1 + 1 √ 2 ) whenever the maximal qubit strategy of K 2 →1 reaches 2 √ 2 and vice versa. The quantum strategy for obtaining the maximum violation of the corresponding temporal inequality is presented in Appendix-(B). Therefore, any quantum advantage of 2 → 1 RAC implies violation of the corresponding macro-realist model.

B. Temporal inequality for 3 → 1 RAC
In order to derive a temporal inequality analogous to 3 → 1 RAC, we need to consider four measurements on Alice's side say {A 1 , A 2 , A 3 , A 4 } and three measurements on Bob's side say {B 1 , B 2 , B 3 }. In each run of the experiment, Alice performs one out of the four dichotomic measurements on an initially prepared input state, ρ in = I 2 , and then Bob implements one out of the three possible measurements on the post-measurement state of Alice. In this way let us define the following quantity in terms of the correlators (17) as, Following Eq. (17) and after some straightforward calculations, one can obtain the general form of the correlators as Here, i and j denotes Alice's and Bob's measurement indices, respectively. The outcome of Alice's measurement A a i i are denoted as a i andā i represents the complement of a i . Therefore, the term K 3 →1 becomes, Taking the eigenstate of Alice's measurement A a i i , i ∈ {1, 2, 3, 4} as that of the preparations for 3 → 1 RAC, i.e., Eq. (7) we get the success probability of 3 → 1 RAC to be or equivalently, F 3 →1 = 1 2 + 1 24 K 3 →1 . A noninvasive-realist bound for the above quantity K 3 →1 has been derived in Appendix-(A 2) to be 4. One can see that when K 3 →1 = 4, the success probability F 3 →1 reaches 2 3 which is the best classical strategy to win the 3 → 1 RAC. On the other hand, with the maximal qubit strategy (see Appendix-(B)), K 3 →1 can however achieve value up to 6.928, and for this the success probability F 3 →1 to win 3 → 1 RAC reaches up to 1 2 (1 + 1

√ 3
). This is again the maximum success probability of winning 3 → 1 RAC using quantum strategy. Therefore, any violation of 3 → 1 RAC again does not possess any macro-realist description.

C. Temporal inequality for 4 → 1 RAC
In order to derive a temporal inequality corresponding to 4 → 1 RAC, Alice and Bob need to perform eight and four measurements respectively in their respective parts. In each run of the experiment Alice performs one out of the eight dichotomic measurements on an initially prepared input state, ρ in = I 2 , and Bob performs one out of the four dichotomic measurements on the post measurement state of Alice. Let us now consider the following quantity, consisting of thirty-two correlators given by, Now, the explicit form of the correlators C ij can be evaluated directly from Eq. (22). Therefore, the term K 4 →1 reduces to with the symbols having usual meanings. Now comparing the eigenstate of Alice's measurement A a i i , i ∈ {1, 2, ..., 8} with that of the preparations for the 4 → 1 i.e., Eq. (11), we get Equivalently, F 4 →1 = 1 2 + 1 64 K 4 →1 . A noninvasive-realist bound for K 4 →1 has been derived in Appendix-(A 3) to be 8. Also, the best classical strategy to win the 4 → 1 RAC is 5 8 . One can see from the above relation that whenever the K 4 →1 rises above 8 there is no macro-realist model, and only in this case the quantum advantage of 4 → 1 RAC can be obtained. The term K 4 →1 can reach up to 15.454 with maximal qubit strategy (see Appendix-(B)) which also matches with the maximum average success probability, F 4 →1 = 0.741.
It might be noted here that some of the correlators will contain negative sign. This is because Alice's and Bob's measurements are anti-correlated for those particular terms. Now, the explicit form of the correlators can be written from Eq. (22). Therefore, the term K n →1 becomes, Now if we take the eigenstate of Alice's measurement A a i i , i ∈ {1, 2, . . . , 2 n−1 } as that of the preparations for n → 1 RAC, i.e., Eq.(14), we get the success probability of n → 1 RAC to be K n →1 = n2 n F n →1 − n2 n−1 (30) or equivalently, F n →1 = 1 2 + 1 n2 n K n →1 . It can be checked that when K n →1 = 2 n−1 , the success probability F n →1 reaches 1 2 (1 + 1 n ) which is the best classical strategy to win the n → 1 RAC. In Appendix-(A 4) we derive the noninvasive-realist bound for the temporal inequality corresponding to the n → 1 RAC which turns out to be 2 n−1 . Now, from the relation between the average success probability F n →1 and corresponding temporal inequality K n →1 , we can conclude that any non-zero quantum advantage of general n → 1 RAC necessitates non-classical temporal correlation.
• Result 1:-For every n → 1 RAC with SR, there exists a temporal inequality where Alice, the first observer has 2 n−1 measurement settings and Bob who measures later has n measurement settings. The maximum success probability of each n → 1 RAC with best classical strategy is related to the maximum noninvasiverealist bound, and any non-zero quantum advantage of a n → 1 RAC translates to the violation of the corresponding temporal inequality.
Moreover, based on our analysis for the cases of 2 → 1, 3 → 1 and 4 → 1 RACs, we can further make the following conjecture. If the maximum success probability of n → 1 RAC occurs with a set of encoding states for Alice and decoding measurements for Bob, then maximal violation of the corresponding temporal inequality is obtained with Alice measuring observables whose eigenstates are exactly the encoded states and Bob's measurements are decoding observables with the initially prepared input state I/2.

IV. CERTIFYING TRUE RANDOMNESS
For the purpose of certifying randomness we describe here an alternative derivation of temporal inequalities, instead of the one based on realism and noninvasive measurability. This alternative derivation was proposed based on some operational assumptions which can be tested in a real experiment. In this alternative derivation the pertaining assumptions are no signalling in time (NSIT) and predictability [43,60,71]. The NSIT condition states that the measurement statistics are not influenced by the earlier measurements, or mathematically, [43]. On the other hand a model is said to be predictable if P(a [72]. Now, if λ denotes some classical variable at the ontological level, then in order to predict the experimental results at the operational level one needs to integrate over all λ, i.e., p(a i , b j |A i , B j ) = λ dλp(λ)p(a i , b j |A i , B j , λ). Note that a crucial step to derive the temporal inequality is to show that the probability distribution at ontological level gets factorised, i.e., Now, using predictability one can write p(a i , b j |A i , B j , λ) = p(a i , b j |A i , B j ) as further conditioning does not change the deterministic probability distribution. Using Bayes' rule one can write the probability . Also, from a physically reasonable perspective, it is broadly accepted that a later measurement cannot influence the past measurement result, and hence Therefore, NSIT ∧ predictability =⇒ factorizability (32) or ¬ factorizability ∧ NSIT =⇒ ¬ predictability. (33) Hence, if we consider a set of probability distributions which satisfies the NSIT conditions but does not fulfill the conditions for factorizability, then it is sure that predictability must be violated. In other words, if for a set of probability distributions the NSIT conditions hold and the temporal inequality is violated simultaneously, then predictability must not hold. Let us quantify this randomness by min-entropy H ∞ (X) which captures the associated randomness that a particular distributionTherefore, any probability distribution P(a i , b j |A i , B j ) will not be predictable and hence some genuine randomness must be associated with that probability distribution. X contains [73]. For some distribution P(a i , b j |A i , B j ), the min-entropy is defined as In order to calculate the randomness associated with the K n →1 , we need to find the maximum probability distribution P(a i , b j |A i , B j ) corresponding to some violation of this inequality. In other words, we need to solve the following optimization problem [74].
We are now interested to obtain a lower bound on the min-entropy H ∞ (a i , b j |A i , B j ) as a function of K n →1 , i.e., we need to derive an inequality of the form H ∞ (a i , b j |A i , B j ) ≥ f (K n →1 ). Below we provide a general lower bound of H ∞ (a i , b j |A i , B j ) from the perspective of no-signalling in time conditions. Imposing NSIT conditions one can obtain P(a i , b j |A i , B j ) ≥ αK n →1 + β where α and β in general depends on a i , b j , A i , B j .
For 2 → 1 RAC, in case of classical strategy, a deterministic point can achieve P(a i , b j |A i , B j ) upto 1, i.e., P * (a i , b j |A i , B j ) ≤ 1 when K 2 →1 = 2, and for the no signalling (in time) box (which is equivalent to Popescu-Rorlich box for spatial correlation) i.e., when K 2 →1 = 4, P * (a i , b j |A i , B j ) ≤ 1/2 . Therefore, analyzing the above inequalities one can obtain the value of α and β to be −1/4 and 3/2, respectively [60]. Hence, In fig.(1), a graphical representation of lower bound of H ∞ (a i , b j |A i , B j ) corresponding to 2 → 1 RAC is provided.
A similar approach based on NSIT conditions can also be applied to other n → 1 RAC. For the case of 3 → 1 RAC, the no-signalling polytope can achieve the value of K 3 →1 up to 12 and for 4 → 1 the value of K 4 →1 up to 32. Therefore, solving linear equations one can obtain P * (a i , b j |A i , B j ) ≤ 5 4 − K 3 →1 16 and P * (a i , b j |A i , B j ) ≤ 7 6 − K 4 →1 48 for 3 → 1 and 4 → 1 RAC, respectively. The lower bound of H ∞ (a i , b j |A i , B j ) corresponding to K 3 →1 and K 4 →1 RAC are plotted in fig.(2) and fig.(3) respectively. Here, it may be pertinent to mention that if instead of P * (a i , b j |A i , B j ), randomness is certified from P * (b j |A i , B j , a i ), then one obtains the maximum value of H ∞ (a i , b j |A i , B j ) to be 1 for any n → 1 RAC, because P * (b j |A i , B j , a i ) is 1/2 in this case irrespective of the value of n. Below, we summarize our key findings for the generation of genuine randomness based on our protocol.
• Result 2:-Any violation of the noninvasive-realist model of the temporal inequalities with initially prepared input state I/2, or equivalently, any non-zero quantum advantage of n → 1 RAC with SR can be ex-ploited to generate genuine randomness. It might be noted here that previous results to generate genuine randomness exploiting RAC do not guaranty genuine randomness for any non-zero quantum advantage of n → 1 RAC [69,70].

V. CONCLUSIONS
Despite the fact that random access code has enormous applications in information processing and communication tasks, the fundamental cause behind this advantage was hitherto not well understood. Here we have shown that any non-zero quantum advantage of RAC with shared randomness necessarily violates a noninvasive-realist model. We have proposed new temporal inequalities corresponding to each n → 1 RAC with SR using the assumption of Realism and Noninvasive measurability. We have then established the fact that the maximum success probability of each n → 1 RAC with best classical strategy is connected to the maximum noninvasive-realist bound. Moreover, any nonzero quantum advantage of a n → 1 RAC is equivalent to the violation of the corresponding temporal inequality.
Next, using an alternative derivation of the noninvasive-realist model, we have shown that any non-zero advantage of RAC can be used to certify genuine randomness. This is particularly significant as all the previously proposed protocols based on RAC do not exhibit genuine randomness for arbitrary quantum advantage of RAC [26,69,70]. Before concluding a few remarks are in order. The maximum success probability using quantum strategy for a general n → 1 RAC is hard to compute for large n, and numerical strategies may be needed to tackle this problem. Finally, it may be reemphasized that our proposed protocol based on LGI violation is amenable for experimental realization [59], and hence, generation of genuine randomness without entanglement based on our protocol might be exemplary for practical purposes. The average of a quantum operator in the Heisenberg picture can be written as an average over a set of hidden variables λ. The role of the initial state ρ(λ) is to provide a probability distribution on the set of hidden variables, called the ontic state. The average of an observable A (measurement carried out at time t can be written as where A t (λ) is the value taken by the observable on the hidden variable λ. The correlation between two observables A t i (measured at some time t i ), B t j (measured at some later time t j ) is given by In general, we can always ignore the effect of final measurement due to NIM (Non-invasive measurability). NIM can be defined as ρ(λ | A t i , B t j , . . . ) = ρ(λ), i.e., a measurement does not change the distribution of λ. In equation (B2), ρ(λ) not depends on observable B t j due to observable B t j being measured after the measurement of observable A t i . Let us take Alice's preparations to be eigenstates of observables {A 1 , A 2 } and Bob's measurements to be {B 1 , B 2 }. Let, measurement A be carried out at some time t i , and measurement B be carried out at some later time t j . Now, imposing the conditions of realism and NIM, we obtain Taking modulus on both sides and using the triangle inequality we get, Now invoking NIM, we have, This is the four term Leggett-Garg inequality.

Ontic model of 3 → 1 RAC :
Let us take Alice's preparations to be eigenstates of {A 1 , A 2 , A 3 , A 4 } and Bob's measurements to be {B 1 , B 2 , B 3 }. Now, following similar steps as in the derivation of the Bell inequality, we obtain sum of four correlations, Now, we take the modulus of both sides and use the triangle inequality to get, (A7) Invoking NIM, we have,

Ontic model of 4 → 1 RAC :
Similarly, for 4 → 1 RAC, here Alice has 8 preparations to be eigenstates of {A 1 , A 2 , ..., A 8 } and Bob has 4 measurements to be {B 1 , B 2 , B 3 , B 4 }. Now, similar procedure presented for deriving classical bound corresponding to 2 → 1 RAC, we can obtain Taking modulus on both sides and using the triangle inequality we get, Invoking NIM, we have,
Similarly, one can also obtain the same classical bound for K n →1 when n is odd.

Appendix B: Quantum strategy to achieve the maximum violation of the temporal inequalities
In our temporal scenario, the initially prepared state is considered to be I/2. Since we are interested to find the maximum violation of the temporal inequality K n →1 , without loss of generality we can stick to projective measurements only. Consider the general form of the measurements on Alice's side to be where a i represents the outcome corresponding the measurement A i andâ i represents the direction along which the A i measurement is being performed. On the other hand the general form of the measurements performed on Bob's side can be considered to be where b j represents the outcome corresponding to measurement B j andb j represents the direction along which