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Role of nonclassical temporal correlation in powering quantum random access codes

Subhankar Bera [email protected] S. N. Bose National Centre for Basic Sciences, Block JD, Sector III, Salt Lake, Kolkata 700106, India    Ananda G. Maity [email protected] S. N. Bose National Centre for Basic Sciences, Block JD, Sector III, Salt Lake, Kolkata 700106, India    Shiladitya Mal [email protected] Physics Division, National Center for Theoretical Sciences, Taipei 10617, Taiwan Department of Physics and Center for Quantum Frontiers of Research and Technology (QFort), National Cheng Kung University, Tainan 701, Taiwan    A. S. Majumdar [email protected] S. N. Bose National Centre for Basic Sciences, Block JD, Sector III, Salt Lake, Kolkata 700106, India
Abstract

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 n1n\mapsto 1 random access code in presence of shared randomness is equivalent to the violation of the corresponding temporal inequality. As a consequence of this connection we also prove that maximal success probability of n1n\mapsto 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 n1n\mapsto 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 raco1 ; raco2 ; raco3 ; Nayak99 ; Ambainis_SR . In a general nmn\mapsto m RAC, there are nn number of bits (say, x1,x2,,xnx_{1},x_{2},...,x_{n}) of message which the sender (say, Alice) sends to the receiver (say, Bob) by encoding the message in a mm bit string (where m<nm<n). In each round, Bob picks out a random number ii (where i{1,2,,n}i\in\{1,2,...,n\}) and tries to guess the ii-th bit (xix_{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 raco3 , or communication of classical bits assisted with a shared quantum stateerac ; srac1 ; srac2 ; Das21 . 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 holevo 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 mm-qubit can in general be depicted by an unit vector in 2m2^{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 nn 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 which otherwise were frozen. This motivates to formulate the task of RAC without contradicting Holevo’s result. Random access codes have been shown to possess wide range of applications such as quantum finite automata raco2 ; raco3 ; Nayak99 , network coding netcod1 ; netcod , locally decodable codes ldc1 ; ldc2 ; ldc3 , non-local gamesBridge , dimension witness dw ; dw2 ; dw3 ; dw4 , quantum communication complexity ccx ; Aaronson04 ; Gavinsky06 ; Buhrman01 ; Mar18 , randomness certification rangen , quantum cryptography qkd , studies of no-signaling resources Grudka , self-testing of quantum measurements Mohan , and so on.

However, there are also some surprising results such that although n1n\mapsto 1 quantum RAC with shared randomness (SR) exists Ambainis_SR , but it does not exist without SR for n4n\geq 4 netcod1 . A variant of this communication task was introduced by Spekkens et al exprac1 under a cryptographic constraint where preparation contextuality is shown to be the actual cause for achieving the quantum advantages (see also alok1 ). Recently an analytical way of finding maximal quantum bound for preparation non-contextuality inequality has been derived Sharma . Investigating the fundamental cause behind the quantum advantage of RAC is not only important from the foundational perspective, but may also escalate the way of finding new applications in information processing and communication tasks.

It is well known from the celebrated Bell theorem Bell'64 ; chsh'69 that correlations between spatially separated events are more restricted in classical theory than allowed in quantum theory. On the other hand, two fundamental no go theorems for time-like separated events involve non-contextual hidden variable models (NCVM) Bell'66 ; Kochen'67 and macro-realism LG'85 , compatible with classical theory. macro-realism (MR) was introduced by Leggett and Garg to probe quantumness of ‘macroscopic’ systems. MR is a conjunction of two assumptions, macro-realism per se and noninvasive measurability LG'85 ; LG'02 ; emary'14 ; yearsley . A variant of MR is known as non-invasive realist model where macro-realism per se is replaced by realism. Quantum theory violates consequences of both the models, i.e., KCBS inequality derivable from NCVM KCBS and Legget-Garg inequality (LGI) arising from MR LG'85 . In recent years, LGI and other temporal correlations have acquired considerable attention from both fundamental yearsley ; kofler'07 ; Naik'22 ; kofler'13 ; brukner'04 ; mal'16 ; mal'18 ; das'16 ; mal'19 ; fritz'10 ; budroni'13 ; rand'16 ; bm'14 ; brierley'19 ; usha'13 ; alok'17 ; das'18 ; ku'18 ; titas'18 ; alok2 ; urbasi1 as well as application perspectives knee'12 ; robens'15 ; knee'16 ; ku'19 ; shayan'19 ; spee'20 ; Maity21 . (For a detailed survey, see emary'14 ; yearsley ).

In the present work, we consider the framework of the prepare and measure scenario (see, for instance, Armin'21 ). We propose new inequalities for temporal correlations arising from sequential measurements and show that corresponding to every scenario of n1n\mapsto 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 Vertesi14 .) 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 n1n\geq 1 with unproven optimality Ambainis_SR . 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. rangen ; qkd ; randomness1 .

The plan of the paper is as follows. In section II, a preliminary discussion on n1n\mapsto 1 RAC is provided. In section III, with the aim of designing operational criteria for testing RAC, temporal inequalities have been proposed for each n1n\mapsto 1 RAC. In section IV, it has been shown that any non-zero quantum advantage of n1n\mapsto 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 n1n\mapsto 1 RAC in some details. Consider a prepare and measure scenario where Alice depending on the nn-bit input string given to her uniformly at random, implements a preparation procedure by encoding the string in a qubit state ρx1,,xn\rho_{x_{1},...,x_{n}}. The qubit is then sent to the measurement device held by Bob, who, upon receiving an input y{1,,n}y\in\{1,...,n\}, implements a binary outcome measurement ByB_{y} and reports the outcome β{0,1}\beta\in\{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

𝔽n1\displaystyle\mathbb{F}_{{n\mapsto 1}} =P(βy=y-th bit of Alice)\displaystyle=P(\beta_{y}=\text{y-th bit of Alice})
=1n2nx1,,xn,yP(βy=xyx1,,xn,y),\displaystyle=\frac{1}{n2^{n}}\sum_{x_{1},...,x_{n},y}P(\beta_{y}=x_{y}\mid x_{1},...,x_{n},y), (1)

where the coefficient 1n2n\frac{1}{n2^{n}} appears due to normalization. Below we discuss some particular examples of n1n\mapsto 1 RAC and their explicit forms.

212\mapsto 1 RAC: In a 212\mapsto 1 RAC, Alice has an input string consisting of two bits x1x2{00,01,10,11}x_{1}x_{2}\in\{00,01,10,11\} which she wants to send to Bob by encoding the bits in a qubit ρx1x2\rho_{x_{1}x_{2}}. Bob’s task is to guess one of the bits (which is again chosen randomly) reliably. Therefore, after receiving an input y{1,2}y\in\{1,2\}, he performs a binary outcome measurement ByB_{y} and reports the outcome β{0,1}\beta\in\{0,1\} as his output. In this scenario, following Eq. (II) the average probability of perfectly guessing the bit is given by

𝔽21=18x1,x2,yP(βy=xyx1,x2,y).\displaystyle\mathbb{F}_{{2\mapsto 1}}=\frac{1}{8}\sum_{x_{1},x_{2},y}P(\beta_{y}=x_{y}\mid x_{1},x_{2},y). (2)

Now consider the most general preparations and measurements as,

ρxx\displaystyle\rho_{xx} =12[𝕀+(1)xa1^.σ],ρxx¯=12[𝕀+(1)xa2^.σ],\displaystyle=\frac{1}{2}\left[\mathbb{I}+(-1)^{x}\hat{a_{1}}.\vec{\sigma}\right],~{}~{}\rho_{x\bar{x}}=\frac{1}{2}\left[\mathbb{I}+(-1)^{x}\hat{a_{2}}.\vec{\sigma}\right], (3)
B1\displaystyle B_{1} =12[𝕀+(1)b1b1^.σ],B2=12[𝕀+(1)b2b2^.σ].\displaystyle=\frac{1}{2}\left[\mathbb{I}+\left(-1\right)^{b_{1}}\hat{b_{1}}.\vec{\sigma}\right],~{}~{}B_{2}=\frac{1}{2}\left[\mathbb{I}+\left(-1\right)^{b_{2}}\hat{b_{2}}.\vec{\sigma}\right]. (4)

Here, ai^\hat{a_{i}}, bj^\hat{b_{j}} are the Bloch vectors denoting Alice’s and Bob’s measurement directions respectively and σ\vec{\sigma} are the Pauli matrices.

Clearly, ρ00+ρ11=𝕀\rho_{00}+\rho_{11}=\mathbb{I}, ρ01+ρ10=𝕀\rho_{01}+\rho_{10}=\mathbb{I}, Bj0+Bj1=𝕀B_{j}^{0}+B_{j}^{1}=\mathbb{I} for j={1,2}j=\{1,2\} and the notation BjbjB_{j}^{b_{j}} denotes the eigenstate corresponding to the outcome bjb_{j} of measurement BjB_{j}. The average success probability now can be written as

𝔽21\displaystyle\mathbb{F}_{2\mapsto 1} =18[Tr[ρ00B10+ρ00B20+ρ11B11+ρ11B21\displaystyle=\frac{1}{8}[\text{Tr}[\rho_{00}B^{0}_{1}+\rho_{00}B^{0}_{2}+\rho_{11}B^{1}_{1}+\rho_{11}B^{1}_{2}
+ρ01B10+ρ01B21+ρ10B11+ρ10B20]].\displaystyle~{}~{}~{}~{}+\rho_{01}B^{0}_{1}+\rho_{01}B^{1}_{2}+\rho_{10}B^{1}_{1}+\rho_{10}B^{0}_{2}]]. (5)

The maximum achievable value for the above quantity is 12(1+12)\frac{1}{2}(1+\frac{1}{2}) with classical strategy, whereas it can reach up to 12(1+12)\frac{1}{2}(1+\frac{1}{\sqrt{2}}) if quantum strategy is used Ambainis_SR .

313\mapsto 1 RAC: In a 313\mapsto 1 RAC, a three-bit input string x1x2x3x_{1}x_{2}x_{3} from the set {000,001,010,011,100,101,110,111}\{000,001,010,011,100,101,110,111\} is given to Alice uniformly at random. Alice then encodes the input string in a qubit ρx1x2x3\rho_{x_{1}x_{2}x_{3}} and sends it to Bob. Bob upon receiving an input y{1,2,3}y\in\{1,2,3\}, implements a binary outcome measurement ByB_{y} and reports the outcome β{0,1}\beta\in\{0,1\} as his output. Following Eq. (II), the average probability of winning can be calculated as

𝔽31=124x1,x2,x3,yP(βy=xyx1,x2,x3,y).\displaystyle\mathbb{F}_{3\mapsto 1}=\frac{1}{24}\sum_{x_{1},x_{2},x_{3},y}P(\beta_{y}=x_{y}\mid x_{1},x_{2},x_{3},y). (6)

Let us now consider most general preparations

ρxxx\displaystyle\rho_{xxx} =12[𝕀+(1)xa1^.σ],\displaystyle=\frac{1}{2}\left[\mathbb{I}+(-1)^{x}\hat{a_{1}}.\vec{\sigma}\right],
ρxxx¯\displaystyle\rho_{xx\bar{x}} =12[𝕀+(1)xa2^.σ],\displaystyle=\frac{1}{2}\left[\mathbb{I}+(-1)^{x}\hat{a_{2}}.\vec{\sigma}\right],
ρxx¯x\displaystyle\rho_{x\bar{x}x} =12[𝕀+(1)xa3^.σ],\displaystyle=\frac{1}{2}\left[\mathbb{I}+(-1)^{x}\hat{a_{3}}.\vec{\sigma}\right],
ρxx¯x¯\displaystyle\rho_{x\bar{x}\bar{x}} =12[𝕀+(1)xa4^.σ],\displaystyle=\frac{1}{2}\left[\mathbb{I}+(-1)^{x}\hat{a_{4}}.\vec{\sigma}\right], (7)

and measurements as,

B1\displaystyle B_{1} =12[𝕀+(1)b1b1^.σ],\displaystyle=\frac{1}{2}\left[\mathbb{I}+\left(-1\right)^{b_{1}}\hat{b_{1}}.\vec{\sigma}\right],
B2\displaystyle B_{2} =12[𝕀+(1)b2b2^.σ],\displaystyle=\frac{1}{2}\left[\mathbb{I}+\left(-1\right)^{b_{2}}\hat{b_{2}}.\vec{\sigma}\right],
B3\displaystyle B_{3} =12[𝕀+(1)b3b3^.σ].\displaystyle=\frac{1}{2}\left[\mathbb{I}+\left(-1\right)^{b_{3}}\hat{b_{3}}.\vec{\sigma}\right]. (8)

Clearly, ρ000+ρ111=𝕀\rho_{000}+\rho_{111}=\mathbb{I}, ρ001+ρ110=𝕀\rho_{001}+\rho_{110}=\mathbb{I}, ρ010+ρ101=𝕀\rho_{010}+\rho_{101}=\mathbb{I}, and ρ011+ρ100=𝕀\rho_{011}+\rho_{100}=\mathbb{I}. The average success probability can be written as

𝔽31\displaystyle\mathbb{F}_{3\mapsto 1} =124[Tr[ρ000(B10+B20+B30)+ρ001(B10+B20+B31)\displaystyle=\frac{1}{24}[\text{Tr}[\rho_{000}(B_{1}^{0}+B_{2}^{0}+B_{3}^{0})+\rho_{001}(B_{1}^{0}+B_{2}^{0}+B_{3}^{1})
+ρ010(B10+B21+B30)+ρ011(B10+B21+B31)\displaystyle~{}~{}~{}~{}+\rho_{010}(B_{1}^{0}+B_{2}^{1}+B_{3}^{0})+\rho_{011}(B_{1}^{0}+B_{2}^{1}+B_{3}^{1})
+ρ100(B11+B20+B30)+ρ101(B11+B20+B31)\displaystyle~{}~{}~{}~{}+\rho_{100}(B_{1}^{1}+B_{2}^{0}+B_{3}^{0})+\rho_{101}(B_{1}^{1}+B_{2}^{0}+B_{3}^{1})
+ρ110(B11+B21+B30)+ρ111(B11+B21+B31)]].\displaystyle~{}~{}~{}~{}+\rho_{110}(B_{1}^{1}+B_{2}^{1}+B_{3}^{0})+\rho_{111}(B_{1}^{1}+B_{2}^{1}+B_{3}^{1})]]. (9)

Here, the average success probability, 𝔽31\mathbb{F}_{3\mapsto 1} can be achieved up to 12(1+13)\frac{1}{2}(1+\frac{1}{3}) and 12(1+13)\frac{1}{2}(1+\frac{1}{\sqrt{3}}), by classical and quantum strategies, respectively Ambainis_SR .

414\mapsto 1 RAC: In a 414\mapsto 1 RAC, Alice has a four-bits input string x1x2x3x4x_{1}x_{2}x_{3}x_{4} 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}\{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 ρx1x2x3x4\rho_{x_{1}x_{2}x_{3}x_{4}} and sends to Bob. Bob upon receiving an input y{1,2,3,4}y\in\{1,2,3,4\}, implements a binary outcome measurement ByB_{y} and reports the outcome β{0,1}\beta\in\{0,1\} as his output. The average probability of winning is given by Eq. (II) as

𝔽41=164x1,x2,x3,x4,yP(βy=xyx1,x2,x3,x4,y).\displaystyle\mathbb{F}_{4\mapsto 1}=\frac{1}{64}\sum_{x_{1},x_{2},x_{3},x_{4},y}P(\beta_{y}=x_{y}\mid x_{1},x_{2},x_{3},x_{4},y). (10)

Let us now consider most general preparations,

ρxxxx\displaystyle\rho_{xxxx} =12[𝕀+(1)xa1^.σ],\displaystyle=\frac{1}{2}\left[\mathbb{I}+(-1)^{x}\hat{a_{1}}.\vec{\sigma}\right],
ρxxxx¯\displaystyle\rho_{xxx\bar{x}} =12[𝕀+(1)xa2^.σ],\displaystyle=\frac{1}{2}\left[\mathbb{I}+(-1)^{x}\hat{a_{2}}.\vec{\sigma}\right],
ρxxx¯x\displaystyle\rho_{xx\bar{x}x} =12[𝕀+(1)xa3^.σ],\displaystyle=\frac{1}{2}\left[\mathbb{I}+(-1)^{x}\hat{a_{3}}.\vec{\sigma}\right],
ρxxx¯x¯\displaystyle\rho_{xx\bar{x}\bar{x}} =12[𝕀+(1)xa4^.σ],\displaystyle=\frac{1}{2}\left[\mathbb{I}+(-1)^{x}\hat{a_{4}}.\vec{\sigma}\right],
ρxx¯xx\displaystyle\rho_{x\bar{x}xx} =12[𝕀+(1)xa5^.σ],\displaystyle=\frac{1}{2}\left[\mathbb{I}+(-1)^{x}\hat{a_{5}}.\vec{\sigma}\right],
ρxx¯xx¯\displaystyle\rho_{x\bar{x}x\bar{x}} =12[𝕀+(1)xa6^.σ],\displaystyle=\frac{1}{2}\left[\mathbb{I}+(-1)^{x}\hat{a_{6}}.\vec{\sigma}\right],
ρxx¯x¯x\displaystyle\rho_{x\bar{x}\bar{x}x} =12[𝕀+(1)xa7^.σ],\displaystyle=\frac{1}{2}\left[\mathbb{I}+(-1)^{x}\hat{a_{7}}.\vec{\sigma}\right],
ρxx¯x¯x¯\displaystyle\rho_{x\bar{x}\bar{x}\bar{x}} =12[𝕀+(1)xa8^.σ],\displaystyle=\frac{1}{2}\left[\mathbb{I}+(-1)^{x}\hat{a_{8}}.\vec{\sigma}\right], (11)

and measurements

B1=12[𝕀+(1)b1b1^.σ],\displaystyle B_{1}=\frac{1}{2}\left[\mathbb{I}+\left(-1\right)^{b_{1}}\hat{b_{1}}.\vec{\sigma}\right],
B2=12[𝕀+(1)b2b2^.σ],\displaystyle B_{2}=\frac{1}{2}\left[\mathbb{I}+\left(-1\right)^{b_{2}}\hat{b_{2}}.\vec{\sigma}\right],
B3=12[𝕀+(1)b3b3^.σ],\displaystyle B_{3}=\frac{1}{2}\left[\mathbb{I}+\left(-1\right)^{b_{3}}\hat{b_{3}}.\vec{\sigma}\right],
B4=12[𝕀+(1)b4b4^.σ].\displaystyle B_{4}=\frac{1}{2}\left[\mathbb{I}+\left(-1\right)^{b_{4}}\hat{b_{4}}.\vec{\sigma}\right]. (12)

Clearly, ρ0000+ρ1111=𝕀\rho_{0000}+\rho_{1111}=\mathbb{I}, ρ0001+ρ1110=𝕀\rho_{0001}+\rho_{1110}=\mathbb{I}, ρ0010+ρ1101=𝕀\rho_{0010}+\rho_{1101}=\mathbb{I}, ρ0011+ρ1100=𝕀\rho_{0011}+\rho_{1100}=\mathbb{I}, ρ0100+ρ1011=𝕀\rho_{0100}+\rho_{1011}=\mathbb{I}, ρ0101+ρ1010=𝕀\rho_{0101}+\rho_{1010}=\mathbb{I}, ρ0110+ρ1001=𝕀\rho_{0110}+\rho_{1001}=\mathbb{I}, and ρ0111+ρ1000=𝕀\rho_{0111}+\rho_{1000}=\mathbb{I}. Therefore the explicit form of the average success probability of 414\mapsto 1 RAC is calculated to be

𝔽41\displaystyle\mathbb{F}_{4\mapsto 1} =164[Tr[ρ0000(B10+B20+B30+B40)+ρ0001(B10+B20\displaystyle=\frac{1}{64}[\text{Tr}[\rho_{0000}(B^{0}_{1}+B^{0}_{2}+B^{0}_{3}+B^{0}_{4})+\rho_{0001}(B^{0}_{1}+B^{0}_{2}
+B30+B41)+ρ0010(B10+B20+B31+B40)+ρ0011(B10\displaystyle+B^{0}_{3}+B^{1}_{4})+\rho_{0010}(B^{0}_{1}+B^{0}_{2}+B^{1}_{3}+B^{0}_{4})+\rho_{0011}(B^{0}_{1}
+B20+B31+B41)+ρ0100(B10+B21+B30+B40)+ρ0101\displaystyle+B^{0}_{2}+B^{1}_{3}+B^{1}_{4})+\rho_{0100}(B^{0}_{1}+B^{1}_{2}+B^{0}_{3}+B^{0}_{4})+\rho_{0101}
(B10+B21+B30+B41)+ρ0110(B10+B21+B31+B40)\displaystyle(B^{0}_{1}+B^{1}_{2}+B^{0}_{3}+B^{1}_{4})+\rho_{0110}(B^{0}_{1}+B^{1}_{2}+B^{1}_{3}+B^{0}_{4})
+ρ0111(B10+B21+B31+B41)+ρ1000(B11+B20+B30\displaystyle+\rho_{0111}(B^{0}_{1}+B^{1}_{2}+B^{1}_{3}+B^{1}_{4})+\rho_{1000}(B^{1}_{1}+B^{0}_{2}+B^{0}_{3}
+B40)+ρ1001(B11+B20+B30+B41)+ρ1010(B11+B20\displaystyle+B^{0}_{4})+\rho_{1001}(B^{1}_{1}+B^{0}_{2}+B^{0}_{3}+B^{1}_{4})+\rho_{1010}(B^{1}_{1}+B^{0}_{2}
+B31+B40)+ρ1011(B11+B20+B31+B41)+ρ1100(B11\displaystyle+B^{1}_{3}+B^{0}_{4})+\rho_{1011}(B^{1}_{1}+B^{0}_{2}+B^{1}_{3}+B^{1}_{4})+\rho_{1100}(B^{1}_{1}
+B21+B30+B40)+ρ1101(B11+B21+B30+B41)+ρ1110\displaystyle+B^{1}_{2}+B^{0}_{3}+B^{0}_{4})+\rho_{1101}(B^{1}_{1}+B^{1}_{2}+B^{0}_{3}+B^{1}_{4})+\rho_{1110}
(B11+B21+B31+B40)+ρ1111(B11+B21+B31+B41)]].\displaystyle(B^{1}_{1}+B^{1}_{2}+B^{1}_{3}+B^{0}_{4})+\rho_{1111}(B^{1}_{1}+B^{1}_{2}+B^{1}_{3}+B^{1}_{4})]]. (13)

For 414\mapsto 1 RAC, the exact value of classical and quantum average success probabilities are 12(1+14)\frac{1}{2}(1+\frac{1}{4}) and 12(1+1+342)\frac{1}{2}(1+\frac{1+\sqrt{3}}{4\sqrt{2}}) respectively Ambainis_SR .

n1n\mapsto 1 RAC: In a n1n\mapsto 1 RAC, Alice has an nn-bits input string x1x2x3x4xnx_{1}x_{2}x_{3}x_{4}\dots x_{n} which is given to her uniformly at random. She then implements a preparation procedure by encoding this string in a qubit denoted by ρx1x2x3x4xn\rho_{x_{1}x_{2}x_{3}x_{4}\dots x_{n}} and sends it to Bob. Bob upon receiving an input y{1,2,3,,n}y\in\{1,2,3,\dots,n\}, implements a binary outcome measurement ByB_{y} and reports the outcome β{0,1}\beta\in\{0,1\} as his output. The average probability of winning is given by Eq. (II). For n1n\mapsto 1 RAC, there are 2n12^{n-1} preparations and nn measurements. Let us now consider the most general preparations,

ρxxxxx\displaystyle\rho_{xxx\dots xx} =12[𝕀+(1)xa^1.σ],\displaystyle=\frac{1}{2}\left[\mathbb{I}+(-1)^{x}\hat{a}_{1}.\vec{\sigma}\right],
ρxxxxx¯\displaystyle\rho_{xxx\dots x\bar{x}} =12[𝕀+(1)xa^2.σ],\displaystyle=\frac{1}{2}\left[\mathbb{I}+(-1)^{x}\hat{a}_{2}.\vec{\sigma}\right],
ρxxxx¯x\displaystyle\rho_{xxx\dots\bar{x}x} =12[𝕀+(1)xa^3.σ],\displaystyle=\frac{1}{2}\left[\mathbb{I}+(-1)^{x}\hat{a}_{3}.\vec{\sigma}\right],
\vdots
ρxxx¯x¯x¯\displaystyle\rho_{xx\bar{x}\dots\bar{x}\bar{x}} =12[𝕀+(1)xa^2n11.σ],\displaystyle=\frac{1}{2}\left[\mathbb{I}+(-1)^{x}\hat{a}_{2^{n-1}-1}.\vec{\sigma}\right],
ρxx¯x¯x¯x¯\displaystyle\rho_{x\bar{x}\bar{x}\dots\bar{x}\bar{x}} =12[𝕀+(1)xa^2n1.σ],\displaystyle=\frac{1}{2}\left[\mathbb{I}+(-1)^{x}\hat{a}_{2^{n-1}}.\vec{\sigma}\right], (14)

and measurements

B1=12[𝕀+(1)b1b^1.σ],\displaystyle B_{1}=\frac{1}{2}\left[\mathbb{I}+\left(-1\right)^{b_{1}}\hat{b}_{1}.\vec{\sigma}\right],
B2=12[𝕀+(1)b2b^2.σ],\displaystyle B_{2}=\frac{1}{2}\left[\mathbb{I}+\left(-1\right)^{b_{2}}\hat{b}_{2}.\vec{\sigma}\right],
B3=12[𝕀+(1)b3b^3.σ],\displaystyle B_{3}=\frac{1}{2}\left[\mathbb{I}+\left(-1\right)^{b_{3}}\hat{b}_{3}.\vec{\sigma}\right],
\vdots
Bn=12[𝕀+(1)bnb^n.σ].\displaystyle B_{n}=\frac{1}{2}\left[\mathbb{I}+\left(-1\right)^{b_{n}}\hat{b}_{n}.\vec{\sigma}\right].

Clearly, ρm+ρm¯=𝕀\rho_{m}+\rho_{\bar{m}}=\mathbb{I}, where, mm are the elements of the n-bit string set. The exact form of the average success probability of n1n\mapsto 1 RAC can be calculated as

𝔽n1\displaystyle\mathbb{F}_{n\mapsto 1} =1n2n[Tr[ρ0000(B10+B20++Bn10+Bn0)+\displaystyle=\frac{1}{n2^{n}}[\text{Tr}[\rho_{00\dots 00}(B^{0}_{1}+B^{0}_{2}+\dots+B^{0}_{n-1}+B^{0}_{n})+
ρ0001(B10+B20++Bn10+Bn1)+ρ0010(B10\displaystyle\rho_{00\dots 01}(B^{0}_{1}+B^{0}_{2}+\dots+B^{0}_{n-1}+B^{1}_{n})+\rho_{00\dots 10}(B^{0}_{1}
+B20++Bn11+Bn0)++ρ1110(B11+B21\displaystyle+B^{0}_{2}+\dots+B^{1}_{n-1}+B^{0}_{n})+\dots+\rho_{11\dots 10}(B^{1}_{1}+B^{1}_{2}
++Bn11+Bn0)+ρ1111(B11+B21++\displaystyle+\dots+B^{1}_{n-1}+B^{0}_{n})+\rho_{11\dots 11}(B^{1}_{1}+B^{1}_{2}+\dots+
Bn11+Bn1)]].\displaystyle B^{1}_{n-1}+B^{1}_{n})]]. (15)

The average success probability for n1n\mapsto 1 RAC (with SR) using best classical strategy is known to be 12(1+1n)\frac{1}{2}(1+\frac{1}{n}) Ambainis_SR .

III Temporal inequalities associated with the Random access codes

We would now like to present a temporal inequality corresponding to each n1n\mapsto 1 RAC. To derive such temporal inequalities, we use the assumptions of realism and noninvasive measurability. The term ‘realism’ implies “at any instant, irrespective of any measurement, a system is definitely in any one of the available states such that all its observable properties have definite values”. On the other hand, the term ‘noninvasive measurability’ assures 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”. It can be shown that under these two assumptions the joint probability distribution get factorized at the ontological level LG'85 ; emary'14 ; yearsley ; kofler'13 . Mathematically,

P(ai,bjAi,Bj)=λρ(λ)P(aiAi,λ)P(bjBj,λ)P(a_{i},b_{j}\mid A_{i},B_{j})=\int_{\lambda}\rho(\lambda)P(a_{i}\mid A_{i},\lambda)P(b_{j}\mid B_{j},\lambda) (16)

with λ\lambda being the hidden variable. Based on this macro-realistic definition of classicality, later we derive temporal inequalities corresponding to each n1n\mapsto 1 RAC and establish that violation of such inequalities implies non-classical temporal correlation. The underlying experimental setup for both the scenarios are the same, as will be clear from the following description.

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. If Alice’s measurement is thought of as playing the role of preparation, then it is not very difficult to realize the similarity between RACs and macrorealistic inequalities. This observation is further substantiated by formulating the relevant inequalities which build the connection quantitatively. Suppose the measurements performed by Alice and the corresponding outcomes are denoted by AiA_{i} and aia_{i} respectively. Similarly, the measurements performed by Bob are denoted by BjB_{j} with corresponding outcome bjb_{j}. All the measurements performed by both the parties are considered to be dichotomic i.e., ai,bj{0,1}a_{i},b_{j}\in\{0,1\}.

Initially, the state on which Alice performs her measurement is denoted by ρin\rho_{in}. The correlation between Alice’s and Bob’s measurement outcome depends on ρin\rho_{in}. In the present analysis we took a maximally mixed state, i.e., ρin=𝕀/2\rho_{in}=\mathbb{I}/2, for reasons that will be clear later. It may be noted here that other states may be chosen for which the same maximum violation for our temporal inequalities can be achieved using different Bloch vectors (ai^\hat{a_{i}}, bj^\hat{b_{j}}). However, even if the directions of Alice’s and Bob’s measurements are changed, the maximum violation will remain the same.

Let the probability of obtaining outcome aia_{i} and bjb_{j} be denoted by P(ai,bjAi,Bj)P(a_{i},b_{j}\mid A_{i},B_{j}), when Alice measures AiA_{i} at time tit_{i} and Bob measures BjB_{j} at some later instant tjt_{j} respectively and ai,bj{0,1}a_{i},b_{j}\in\{0,1\}. Let us denote, AiaiA_{i}^{a_{i}}, BjbjB_{j}^{b_{j}} as projectors so that aiAiai=𝕀,bjBjbj=𝕀\sum_{a_{i}}A_{i}^{a_{i}}=\mathbb{I},\sum_{b_{j}}B_{j}^{b_{j}}=\mathbb{I}. Now, following the standard procedure, the joint probability distribution can be obtained using Bayes’ rule as,

P(ai,bjAi,Bj)=P(aiAi)P(bjai,Ai,Bj)\displaystyle P(a_{i},b_{j}\mid A_{i},B_{j})=P(a_{i}\mid A_{i})P(b_{j}\mid a_{i},A_{i},B_{j})
=Tr[Aiaiρin]Tr[BjbjAiaiρinAiaiTr[AiaiρinAiai]].\displaystyle=\text{Tr}\left[A_{i}^{a_{i}}\rho_{in}\right]\text{Tr}\left[B_{j}^{b_{j}}\frac{A_{i}^{a_{i}}\rho_{in}A_{i}^{a_{i}\dagger}}{\text{Tr}\left[A_{i}^{a_{i}}\rho_{in}A_{i}^{a_{i}\dagger}\right]}\right]. (17)

With this joint probability distribution, the two-time correlation is defined as,

Cij=ai,bj(1)aibjP(ai,bjAi,Bj),C_{ij}=\sum_{a_{i},b_{j}}(-1)^{a_{i}\oplus b_{j}}P(a_{i},b_{j}\mid A_{i},B_{j}), (18)

where \oplus denotes addition modulo 2. In the subsections below we present the temporal inequalities corresponding to cases of 212\mapsto 1, 313\mapsto 1 and 414\mapsto 1 RACs. These inequalities are derived with a close look at the success probabilities of the corresponding RAC games. The general form of the temporal inequality for n1n\mapsto 1 RAC, i.e., 𝒦n1\mathcal{K}_{n\mapsto 1} is provided in the Appendix-(A).

As mentioned earlier, in our temporal scenario, the initially prepared state is considered to be 𝕀/2\mathbb{I}/2. Since we are interested in finding the maximum violation of the temporal inequality 𝒦n1\mathcal{K}_{n\mapsto 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

Aiai=12[𝕀+(1)aiai^.σ],A_{i}^{a_{i}}=\frac{1}{2}\left[\mathbb{I}+(-1)^{a_{i}}\hat{a_{i}}.\vec{\sigma}\right], (19)

where aia_{i} represents the outcome corresponding the measurement AiA_{i} and a^i\hat{a}_{i} represents the direction along which the AiA_{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

Bjbj=12[𝕀+(1)bjbj^.σ],B_{j}^{b_{j}}=\frac{1}{2}\left[\mathbb{I}+(-1)^{b_{j}}\hat{b_{j}}.\vec{\sigma}\right], (20)

where bjb_{j} represents the outcome corresponding to measurement BjB_{j} and b^j\hat{b}_{j} represents the direction along which BjB_{j} measurement is being performed. In the subsections below we state the explicit form of the quantum strategy for which maximum quantum violation of the temporal inequality 𝒦n1\mathcal{K}_{n\mapsto 1} RAC is achieved.

III.1 Temporal inequality for 212\mapsto 1 RAC

Now, to derive a temporal inequality corresponding to the 212\mapsto 1 RAC, let us first assume that Alice and Bob have two choices of binary measurements, say, {A1,A2}\{A_{1},A_{2}\} and {B1,B2}\{B_{1},B_{2}\} to perform in each run and ρin=𝕀2\rho_{in}=\frac{\mathbb{I}}{2}. Let us now consider the following quantity in terms of the above correlators as,

𝒦21=C11+C21+C12C22.\mathcal{K}_{2\mapsto 1}=C_{11}+C_{21}+C_{12}-C_{22}. (21)

Following Eq. (18), we calculate the C11C_{11} term explicitly:

C11=\displaystyle C_{11}= P(0,0|A1,B1)+P(1,1|A1,B1)\displaystyle P(0,0|A_{1},B_{1})+P(1,1|A_{1},B_{1})-
P(0,1|A1,B1)P(1,0|A1,B1)\displaystyle P(0,1|A_{1},B_{1})-P(1,0|A_{1},B_{1})
=\displaystyle= 12Tr[A10B10+A11B11A10B11A11B10]\displaystyle\frac{1}{2}\text{Tr}[A_{1}^{0}B_{1}^{0}+A_{1}^{1}B_{1}^{1}-A_{1}^{0}B_{1}^{1}-A_{1}^{1}B_{1}^{0}]
=\displaystyle= 12Tr[A10B10+A11B11A10(𝕀B10)A11(𝕀B11)]\displaystyle\frac{1}{2}\text{Tr}[A_{1}^{0}B_{1}^{0}+A_{1}^{1}B_{1}^{1}-A_{1}^{0}(\mathbb{I}-B_{1}^{0})-A_{1}^{1}(\mathbb{I}-B_{1}^{1})]
=\displaystyle= Tr[A10B10+A11B11]1.\displaystyle\text{Tr}[A^{0}_{1}B_{1}^{0}+A_{1}^{1}B_{1}^{1}]-1.

where, AiaiA_{i}^{a_{i}} represents the eigenstate corresponding to the outcome ai{0,1}a_{i}\in\{0,1\} of the measurement AiA_{i} and similarly for Bob. Here, the second equality can be derived using Eq.(III), 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., Ai0+Ai1=𝕀A_{i}^{0}+A_{i}^{1}=\mathbb{I} and Bj0+Bj1=𝕀B_{j}^{0}+B_{j}^{1}=\mathbb{I} for all i,ji,j.

Similarly, the other terms can be evaluated as

C12\displaystyle C_{12} =Tr[A10B20+A11B21]1,\displaystyle=\text{Tr}[A_{1}^{0}B^{0}_{2}+A_{1}^{1}B^{1}_{2}]-1,
C21\displaystyle C_{21} =Tr[A20B10+A21B11]1,\displaystyle=\text{Tr}[A_{2}^{0}B^{0}_{1}+A_{2}^{1}B^{1}_{1}]-1,
C22\displaystyle C_{22} =Tr[A21B20+A20B21]+1.\displaystyle=-\text{Tr}[A_{2}^{1}B^{0}_{2}+A_{2}^{0}B^{1}_{2}]+1.

Hence, the expression for 𝒦21\mathcal{K}_{2\mapsto 1} becomes,

𝒦21\displaystyle\mathcal{K}_{2\mapsto 1} =C11+C21+C12C22\displaystyle=C_{11}+C_{21}+C_{12}-C_{22}
=Tr[A10B10+A11B11+A10B20+A11B21+A20B10\displaystyle=\text{Tr}[A^{0}_{1}B_{1}^{0}+A_{1}^{1}B_{1}^{1}+A_{1}^{0}B^{0}_{2}+A_{1}^{1}B^{1}_{2}+A_{2}^{0}B^{0}_{1}
+A21B11+A21B20+A20B21]4.\displaystyle~{}~{}~{}~{}~{}+A_{2}^{1}B^{1}_{1}+A_{2}^{1}B^{0}_{2}+A_{2}^{0}B^{1}_{2}]-4. (22)

It may be noted here that the eigenstate of the measurements performed by Alice AiaiA_{i}^{a_{i}} is the same as that of the preparations considered in Eq.(3). Now, comparing the above equation with Eq.(II), one obtains

𝒦21=8(𝔽2112).\mathcal{K}_{2\mapsto 1}=8(\mathbb{F}_{2\mapsto 1}-\frac{1}{2}). (23)

Conversely, 𝔽21=12+18𝒦21\mathbb{F}_{2\mapsto 1}=\frac{1}{2}+\frac{1}{8}\mathcal{K}_{2\mapsto 1}.

It can be shown that the maximum quantum violation of the temporal inequality 𝒦21\mathcal{K}_{2\mapsto 1} corresponding to 212\mapsto 1 RAC can be achieved up to 2.8282.828 for the following sets of measurements:

a^1\displaystyle\hat{a}_{1} =12(1,1,0),\displaystyle=\frac{1}{\sqrt{2}}(1,1,0),
a^2\displaystyle\hat{a}_{2} =12(1,1,0),\displaystyle=\frac{1}{\sqrt{2}}(1,-1,0), (24)

and

b^1\displaystyle\hat{b}_{1} =(1,0,0),\displaystyle=(1,0,0),
b^2\displaystyle\hat{b}_{2} =(0,1,0).\displaystyle=(0,1,0). (25)

One can see that the strategy to reach the maximum violation of the temporal inequality 𝒦21\mathcal{K}_{2\mapsto 1} with initially prepared state 𝕀/2\mathbb{I}/2 is the same with that of the quantum strategy for which the maximum success probability for 212\mapsto 1 RAC is achieved.

Note further, a noninvasive-realist bound for the term 𝒦21\mathcal{K}_{2\mapsto 1} was derived to be 22 (See, Appendix-(A.1)). One can see from this relation that whenever the value of the term 𝒦21\mathcal{K}_{2\mapsto 1} falls below 22, the success probability of the 212\mapsto 1 random access code also falls below 34\frac{3}{4} which is the maximum probability of success with classical strategy. Moreover, the maximum success probability of 212\mapsto 1 RAC reaches 12(1+12)\frac{1}{2}(1+\frac{1}{\sqrt{2}}) whenever the maximal qubit strategy of 𝒦21\mathcal{K}_{2\mapsto 1} reaches 222\sqrt{2}, and vice versa. Therefore, any quantum advantage of 212\mapsto 1 RAC implies a violation of the corresponding macro-realist model.

III.2 Temporal inequality for 313\mapsto 1 RAC

To derive a temporal inequality analogous to 313\mapsto 1 RAC, we need to consider four measurements on Alice’s side say {A1,A2,A3,A4}\{A_{1},A_{2},A_{3},A_{4}\} and three measurements on Bob’s side say {B1,B2,B3}\{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, ρin=𝕀2\rho_{\text{in}}=\frac{\mathbb{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 (18) as,

𝒦31\displaystyle\mathcal{K}_{3\mapsto 1} =C11+C12+C13+C22+C21C23+C31C32\displaystyle=C_{11}+C_{12}+C_{13}+C_{22}+C_{21}-C_{23}+C_{31}-C_{32}
+C33+C41C42C43.\displaystyle~{}~{}~{}~{}~{}+C_{33}+C_{41}-C_{42}-C_{43}. (26)

Following Eq. (18) and after some straightforward calculations, one can obtain the general form of the correlators as

Cij\displaystyle C_{ij} =(1)ai[Tr[AiaiBj0+Aiai¯Bj1]1].\displaystyle=(-1)^{a_{i}}[\text{Tr}[A_{i}^{a_{i}}B_{j}^{0}+A_{i}^{\bar{a_{i}}}B_{j}^{1}]-1]. (27)

Here, ii and jj denotes Alice’s and Bob’s measurement indices, respectively. The outcome of Alice’s measurement AiaiA_{i}^{a_{i}} are denoted as aia_{i} and ai¯\bar{a_{i}} represents the complement of aia_{i}.

Therefore, the term 𝒦31\mathcal{K}_{3\mapsto 1} becomes,

𝒦31=j=13i=14(1)ai[Tr[AiaiBj0+Aiai¯Bj1]1]\displaystyle\mathcal{K}_{3\mapsto 1}=\sum_{j=1}^{3}\sum_{i=1}^{4}(-1)^{a_{i}}[\text{Tr}[A_{i}^{a_{i}}B_{j}^{0}+A_{i}^{\bar{a_{i}}}B_{j}^{1}]-1]
=j=13i=14(1)aiTr[AiaiBj0+Aiai¯Bj1]12.\displaystyle=\sum_{j=1}^{3}\sum_{i=1}^{4}(-1)^{a_{i}}\text{Tr}[A_{i}^{a_{i}}B_{j}^{0}+A_{i}^{\bar{a_{i}}}B_{j}^{1}]-12. (28)

Taking the eigenstate of Alice’s measurement Aiai,i{1,2,3,4}A_{i}^{a_{i}},~{}i\in\{1,2,3,4\} as that of the preparations for 313\mapsto 1 RAC, i.e., Eq.(II) we obtain the success probability of 313\mapsto 1 RAC to be

𝒦31=24𝔽3112\displaystyle\mathcal{K}_{3\mapsto 1}=24\mathbb{F}_{3\mapsto 1}-12 (29)

or equivalently, 𝔽31=12+124𝒦31\mathbb{F}_{3\mapsto 1}=\frac{1}{2}+\frac{1}{24}\mathcal{K}_{3\mapsto 1}.

It can be shown that the maximum violation of the temporal inequality 𝒦31\mathcal{K}_{3\mapsto 1} corresponding to the 313\mapsto 1 RAC can be achieved up to 6.9286.928 for the following measurement settings:

a^1\displaystyle\hat{a}_{1} =13(1,1,1),\displaystyle=\frac{1}{\sqrt{3}}(1,1,1),
a^2\displaystyle\hat{a}_{2} =13(1,1,1),\displaystyle=\frac{1}{\sqrt{3}}(1,1,-1),
a^3\displaystyle\hat{a}_{3} =13(1,1,1),\displaystyle=\frac{1}{\sqrt{3}}(1,-1,1),
a^4\displaystyle\hat{a}_{4} =13(1,1,1).\displaystyle=\frac{1}{\sqrt{3}}(1,-1,-1). (30)

and

b^1\displaystyle\hat{b}_{1} =(1,0,0),\displaystyle=(1,0,0),
b^2\displaystyle\hat{b}_{2} =(0,1,0),\displaystyle=(0,1,0),
b^3\displaystyle\hat{b}_{3} =(0,0,1).\displaystyle=(0,0,1). (31)

This is again the same strategy for which maximum quantum success probability of 313\mapsto 1 RAC is achieved.

A noninvasive-realist bound for the above quantity 𝒦31\mathcal{K}_{3\mapsto 1} is derived in Appendix-(A.2) to be 44. One can see that when 𝒦31=4\mathcal{K}_{3\mapsto 1}=4, the success probability 𝔽31\mathbb{F}_{3\mapsto 1} reaches 23\frac{2}{3} which is the best classical strategy to win the 313\mapsto 1 RAC. On the other hand, with the maximal qubit strategy, 𝒦31\mathcal{K}_{3\mapsto 1} can however achieve value up to 6.9286.928, and for this the success probability 𝔽31\mathbb{F}_{3\mapsto 1} to win 313\mapsto 1 RAC reaches up to 12(1+13)\frac{1}{2}(1+\frac{1}{\sqrt{3}}). This is again the maximum success probability of winning 313\mapsto 1 RAC using quantum strategy. Therefore, any violation of 313\mapsto 1 RAC again does not possess any macro-realist description.

III.3 Temporal inequality for 414\mapsto 1 RAC

To derive a temporal inequality corresponding to 414\mapsto 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, ρin=𝕀2\rho_{\text{in}}=\frac{\mathbb{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,

𝒦41\displaystyle\mathcal{K}_{4\mapsto 1} =C11+C12+C13+C14+C21+C22+C23C24\displaystyle=C_{11}+C_{12}+C_{13}+C_{14}+C_{21}+C_{22}+C_{23}-C_{24}
+C31+C32C33+C34+C41+C42C43C44\displaystyle+C_{31}+C_{32}-C_{33}+C_{34}+C_{41}+C_{42}-C_{43}-C_{44}
+C51C52+C53+C54+C61C62+C63C64\displaystyle+C_{51}-C_{52}+C_{53}+C_{54}+C_{61}-C_{62}+C_{63}-C_{64}
+C71C72C73+C74+C81C82C83C84.\displaystyle+C_{71}-C_{72}-C_{73}+C_{74}+C_{81}-C_{82}-C_{83}-C_{84}. (32)

Now, the explicit form of the correlators CijC_{ij} can be evaluated directly from Eq.(27). Therefore, the term 𝒦41\mathcal{K}_{4\mapsto 1} reduces to

𝒦41=j=14i=18(1)ai[Tr[AiaiBj0+Aiai¯Bj1]1]\displaystyle\mathcal{K}_{4\mapsto 1}=\sum_{j=1}^{4}\sum_{i=1}^{8}(-1)^{a_{i}}[\text{Tr}[A_{i}^{a_{i}}B_{j}^{0}+A_{i}^{\bar{a_{i}}}B_{j}^{1}]-1]
=j=14i=18(1)aiTr[AiaiBj0+Aiai¯Bj1]32\displaystyle=\sum_{j=1}^{4}\sum_{i=1}^{8}(-1)^{a_{i}}\text{Tr}[A_{i}^{a_{i}}B_{j}^{0}+A_{i}^{\bar{a_{i}}}B_{j}^{1}]-32 (33)

with the symbols having usual meanings. Now comparing the eigenstate of Alice’s measurement Aiai,i{1,2,,8}A_{i}^{a_{i}},~{}i\in\{1,2,...,8\} with that of the preparations for the 414\mapsto 1 i.e., Eq. (II), we obtain

𝒦41=64(𝔽4112).\mathcal{K}_{4\mapsto 1}=64(\mathbb{F}_{4\mapsto 1}-\frac{1}{2}). (34)

Equivalently, 𝔽41=12+164𝒦41\mathbb{F}_{4\mapsto 1}=\frac{1}{2}+\frac{1}{64}\mathcal{K}_{4\mapsto 1}.

The maximum quantum violation of the temporal inequality 𝒦41\mathcal{K}_{4\mapsto 1} can be achieved up to 15.45415.454 for the following measurement settings:

a^1\displaystyle\hat{a}_{1} =16(1,1,2),\displaystyle=\frac{1}{\sqrt{6}}(1,1,2),
a^2\displaystyle\hat{a}_{2} =16(1,1,2),\displaystyle=\frac{1}{\sqrt{6}}(1,1,-2),
a^3\displaystyle\hat{a}_{3} =16(1,1,2),\displaystyle=\frac{1}{\sqrt{6}}(1,-1,2),
a^4\displaystyle\hat{a}_{4} =16(1,1,2),\displaystyle=\frac{1}{\sqrt{6}}(1,-1,-2),
a^5\displaystyle\hat{a}_{5} =16(3,3,0),\displaystyle=\frac{1}{\sqrt{6}}(\sqrt{3},\sqrt{3},0),
a^6\displaystyle\hat{a}_{6} =16(3,3,0),\displaystyle=\frac{1}{\sqrt{6}}(\sqrt{3},-\sqrt{3},0),
a^7\displaystyle\hat{a}_{7} =16(3,3,0),\displaystyle=\frac{1}{\sqrt{6}}(\sqrt{3},\sqrt{3},0),
a^8\displaystyle\hat{a}_{8} =16(3,3,0).\displaystyle=\frac{1}{\sqrt{6}}(\sqrt{3},-\sqrt{3},0). (35)

and

b^1\displaystyle\hat{b}_{1} =(1,0,0),\displaystyle=(1,0,0),
b^2\displaystyle\hat{b}_{2} =(0,1,0),\displaystyle=(0,1,0),
b^3\displaystyle\hat{b}_{3} =(0,0,1),\displaystyle=(0,0,1),
b^4\displaystyle\hat{b}_{4} =(0,0,1).\displaystyle=(0,0,1). (36)

This is again the same strategy for which maximum success probability of 414\mapsto 1 RAC is achieved.

A noninvasive-realist bound for 𝒦41\mathcal{K}_{4\mapsto 1} is derived in Appendix-(A.3) to be 88. In addition, the best classical strategy to win the 414\mapsto 1 RAC is 58\frac{5}{8}. One can see from the above relation that whenever the 𝒦41\mathcal{K}_{4\mapsto 1} rises above 88 there is no macro-realist model, and only in this case the quantum advantage of 414\mapsto 1 RAC can be obtained. The term 𝒦41\mathcal{K}_{4\mapsto 1} can reach up to 15.45415.454 with maximal qubit strategy which also matches with the maximum average success probability, 𝔽41=0.741\mathbb{F}_{4\mapsto 1}=0.741.

III.4 Temporal inequality for n1n\mapsto 1 RAC

Let us now derive a temporal inequality corresponding to n1n\mapsto 1 RAC. To do so we need to consider 2n12^{n-1} measurements on Alice’s side say {A1,A2,A3,,A2n1}\{A_{1},A_{2},A_{3},\dots,A_{2^{n-1}}\} and n measurements on Bob’s side say {B1,B2,,Bn}\{B_{1},B_{2},\dots,B_{n}\}. At first, Alice performs one out of the 2n12^{n-1} dichotomic measurements on an initially prepared input state, ρin=𝕀2\rho_{\text{in}}=\frac{\mathbb{I}}{2} and then, Bob performs one out of the nn possible measurements on the post measurement state of Alice. Finally, they evaluate the quantity 𝒦n1\mathcal{K}_{n\mapsto 1}, represented in terms of the correlators (18) as,

𝒦n1\displaystyle\mathcal{K}_{n\mapsto 1} =C11+C12+C13++C1n+C21+C22+\displaystyle=C_{11}+C_{12}+C_{13}+\dots+C_{1n}+C_{21}+C_{22}+
C23+C2n+C31+C32+C33++C3n\displaystyle C_{23}+\dots-C_{2n}+C_{31}+C_{32}+C_{33}+\dots+C_{3n}
++C(2n11)1C(2n11)2C(2n11)3\displaystyle+\dots+C_{{(2^{n-1}-1)}1}-C_{{(2^{n-1}-1)}2}-C_{{(2^{n-1}-1)}3}
++C(2n11)n++C2n11C2n12\displaystyle+\dots+C_{{(2^{n-1}-1)}n}+\dots+C_{2^{n-1}1}-C_{2^{n-1}2}
C2n13C2n1n.\displaystyle-C_{2^{n-1}3}-\dots-C_{2^{n-1}n}. (37)

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.(27).

Therefore, the term 𝒦n1\mathcal{K}_{n\mapsto 1} becomes,

𝒦n1=j=1ni=12n1(1)ai[Tr[AiaiBj0+Aiai¯Bj1]1]\displaystyle\mathcal{K}_{n\mapsto 1}=\sum_{j=1}^{n}\sum_{i=1}^{2^{n-1}}(-1)^{a_{i}}[\text{Tr}[A_{i}^{a_{i}}B_{j}^{0}+A_{i}^{\bar{a_{i}}}B_{j}^{1}]-1]
=j=1ni=12n1(1)aiTr[AiaiBj0+Aiai¯Bj1]n2n1.\displaystyle=\sum_{j=1}^{n}\sum_{i=1}^{2^{n-1}}(-1)^{a_{i}}\text{Tr}[A_{i}^{a_{i}}B_{j}^{0}+A_{i}^{\bar{a_{i}}}B_{j}^{1}]-n2^{n-1}. (38)

Now if we take the eigenstate of Alice’s measurement Aiai,i{1,2,,2n1}A_{i}^{a_{i}},~{}i\in\{1,2,\dots,2^{n-1}\} as that of the preparations for n1n\mapsto 1 RAC, i.e., Eq.(II), we obtain the success probability of n1n\mapsto 1 RAC to be

𝒦n1=n2n𝔽n1n2n1\displaystyle\mathcal{K}_{n\mapsto 1}=n2^{n}\mathbb{F}_{n\mapsto 1}-n2^{n-1} (39)

or equivalently, 𝔽n1=12+1n2n𝒦n1\mathbb{F}_{n\mapsto 1}=\frac{1}{2}+\frac{1}{n2^{n}}\mathcal{K}_{n\mapsto 1}.

It can be checked that when 𝒦n1=2n1\mathcal{K}_{n\mapsto 1}=2^{n-1}, the success probability 𝔽n1\mathbb{F}_{n\mapsto 1} reaches 12(1+1n)\frac{1}{2}(1+\frac{1}{n}) which is the best classical strategy to win the n1n\mapsto 1 RAC. In Appendix-(A.4) we derive the noninvasive-realist bound for the temporal inequality corresponding to the n1n\mapsto 1 RAC which turns out to be 2n12^{n-1}. Now, from the relation between the average success probability 𝔽n1\mathbb{F}_{n\mapsto 1} and corresponding temporal inequality 𝒦n1\mathcal{K}_{n\mapsto 1}, we can conclude that any non-zero quantum advantage of general n1n\mapsto 1 RAC necessitates non-classical temporal correlation. On the basis of the above investigation, following the main idea stated at the beginning of this section, we clearly summarize our first main result, given bellow.

\bullet Result 1:- For every n1n\mapsto 1 RAC with SR, there exists a temporal inequality where Alice, the first observer has 2n12^{n-1} measurement settings and Bob who measures later has nn measurement settings. The maximum success probability of each n1n\mapsto 1 RAC with best classical strategy is related to the maximum noninvasive-realist bound, and any non-zero quantum advantage of a n1n\mapsto 1 RAC translates to the violation of the corresponding temporal inequality.

Moreover, based on our analysis for the cases of 212\mapsto 1, 313\mapsto 1 and 414\mapsto 1 RACs, we can further make the following conjecture. If the maximum success probability of n1n\mapsto 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\mathbb{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 signaling in time (NSIT) and predictability kofler'13 ; rand'16 ; Maity'21 . The NSIT condition states that the measurement statistics are not influenced by the earlier measurements, or mathematically, P(bj|Bj)=P(bj|Ai,Bj)Ai,Bj,bjP(b_{j}|B_{j})=P(b_{j}|A_{i},B_{j})~{}\forall A_{i},B_{j},b_{j} kofler'13 . On the other hand a model is said to be predictable if P(ai,bj|Ai,Bj){0,1}ai,bj,Ai,BjP(a_{i},b_{j}|A_{i},B_{j})\in\{0,1\}~{}\forall a_{i},b_{j},A_{i},B_{j} eric .

Now, if λ\lambda 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\lambda, i.e., p(ai,bj|Ai,Bj)=λ𝑑λp(λ)p(ai,bj|Ai,Bj,λ)p(a_{i},b_{j}|A_{i},B_{j})=\int_{\lambda}d\lambda p(\lambda)p(a_{i},b_{j}|A_{i},B_{j},\lambda). Note that a crucial step to derive the temporal inequality is to show that the probability distribution at ontological level gets factorised, i.e.,

p(ai,bj|Ai,Bj,λ)=p(ai|Ai,λ)p(bj|Bj,λ)p(a_{i},b_{j}|A_{i},B_{j},\lambda)=p(a_{i}|A_{i},\lambda)p(b_{j}|B_{j},\lambda) (40)

Now, using predictability one can write p(ai,bj|Ai,Bj,λ)=p(ai,bj|Ai,Bj)p(a_{i},b_{j}|A_{i},B_{j},\lambda)=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 distribution p(ai,bj|Ai,Bj)=p(ai|Ai,Bj,bj)p(bj|Ai,Bj)p(a_{i},b_{j}|A_{i},B_{j})=p(a_{i}|A_{i},B_{j},b_{j})p(b_{j}|A_{i},B_{j}). Now, from the NSIT conditions one has p(bj|Ai,Bj)=p(bj|Bj)p(b_{j}|A_{i},B_{j})=p(b_{j}|B_{j}). Also, from a physically reasonable perspective, it is broadly accepted that a later measurement cannot influence the past measurement result, and hence p(ai|Ai,Bj,bj)=p(ai|Ai)p(a_{i}|A_{i},B_{j},b_{j})=p(a_{i}|A_{i}). Since, at the ontological level p(ai|Ai,λ)=p(ai|Ai)p(a_{i}|A_{i},\lambda)=p(a_{i}|A_{i}) and p(bj|Bj,λ)=p(bj|Bj)p(b_{j}|B_{j},\lambda)=p(b_{j}|B_{j}), the probability distribution can be written in a factorized form p(ai,bj|Ai,Bj,λ)=p(ai|Ai,λ)p(bj|Bj,λ)p(a_{i},b_{j}|A_{i},B_{j},\lambda)=p(a_{i}|A_{i},\lambda)p(b_{j}|B_{j},\lambda). Therefore,

NSITpredictabilityfactorizability\text{NSIT}\wedge\text{predictability}\implies\text{factorizability} (41)

or

¬factorizabilityNSIT¬predictability.\neg~{}\text{factorizability}\wedge\text{NSIT}\implies\neg~{}\text{predictability}. (42)

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)H_{\infty}(X) which captures the associated randomness that a particular distribution XX contains. Therefore, any probability distribution P(ai,bj|Ai,Bj)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 min-entropy . For some distribution P(ai,bj|Ai,Bj)P(a_{i},b_{j}|A_{i},B_{j}), the min-entropy is defined as

H(ai,bj|Ai,Bj)\displaystyle H_{\infty}(a_{i},b_{j}|A_{i},B_{j}) =log2[maxai,bjP(ai,bj|Ai,Bj)]\displaystyle=-\log_{2}[\text{max}_{a_{i},b_{j}}P(a_{i},b_{j}|A_{i},B_{j})]
=minai,bj[log2[P(ai,bj|Ai,Bj)]].\displaystyle=\text{min}_{a_{i},b_{j}}[-\log_{2}[P(a_{i},b_{j}|A_{i},B_{j})]]. (43)

To calculate the randomness associated with the 𝒦n1\mathcal{K}_{n\mapsto 1}, we need to find the maximum probability distribution P(ai,bj|Ai,Bj)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 randomness .

P(ai,bj|Ai,Bj)\displaystyle P^{*}(a_{i},b_{j}|A_{i},B_{j}) =maxP(ai,bj|Ai,Bj)\displaystyle=\text{max}P(a_{i},b_{j}|A_{i},B_{j})
constraints to𝒦n1=𝒦n1MR+ϵ,\displaystyle\text{constraints to}~{}\mathcal{K}_{n\mapsto 1}=\mathcal{K}^{MR}_{n\mapsto 1}+\epsilon,
P(ai,bj|Ai,Bj)0,\displaystyle P(a_{i},b_{j}|A_{i},B_{j})\geq 0,
ai,bjP(ai,bj|Ai,Bj)=1Ai,Bj,\displaystyle\sum_{a_{i},b_{j}}P(a_{i},b_{j}|A_{i},B_{j})=1~{}~{}\forall A_{i},B_{j},
andP(ai,bj|Ai,Bj)satisfy NSIT\displaystyle\text{and}~{}P(a_{i},b_{j}|A_{i},B_{j})~{}\text{satisfy NSIT} (44)

where P(ai,bj|Ai,Bj)P^{*}(a_{i},b_{j}|A_{i},B_{j}) denotes the maximized value of P(ai,bj|Ai,Bj)P(a_{i},b_{j}|A_{i},B_{j}) and 𝒦n1MR\mathcal{K}^{MR}_{n\mapsto 1} is the MR bound of the temporal inequality 𝒦n1\mathcal{K}_{n\mapsto 1} corresponding to n1n\mapsto 1 RAC. We use linear programming to solve this optimization problem. The parameters α\alpha and β\beta are chosen in such a way that the inequality maintain its linear form. By putting some boundary conditions, α\alpha and β\beta are calculated for each 𝒦n1\mathcal{K}_{n\mapsto 1} so that P(ai,bj|Ai,Bj)α𝒦n1+βP^{*}(a_{i},b_{j}|A_{i},B_{j})\leq\alpha\mathcal{K}_{n\mapsto 1}+\beta. Here, it may be noted that for any particular n1n\mapsto 1 RAC, if α\alpha and β\beta depend on ai,bj,Ai,Bja_{i},b_{j},A_{i},B_{j}, then assuming P(ai,bj|Ai,Bj)α𝒦n1+βP^{*}(a_{i},b_{j}|A_{i},B_{j})\leq\alpha\mathcal{K}_{n\mapsto 1}+\beta is not consistent, since we chose α\alpha and β\beta to be some constant so that the linear form of the inequality can be maintained. However, in case of a different n1n\mapsto 1 RAC, this α\alpha and β\beta in general depends on the inequality corresponding to 𝒦n1\mathcal{K}_{n\mapsto 1} as well as on P(ai,bj|Ai,Bj)P^{*}(a_{i},b_{j}|A_{i},B_{j}). Therefore, in general it may depend on ai,bj,Ai,Bja_{i},b_{j},A_{i},B_{j}. A more detailed discussion on the choice of α\alpha and β\beta (for the Bell-scenario) is provided explicitly in Ref.randomness .

Note for example, in the case of 212\mapsto 1 RAC, the MR or classical bound of 𝒦21\mathcal{K}_{2\mapsto 1} is 22. Now, 𝒦21=2+ϵ\mathcal{K}_{2\mapsto 1}=2+\epsilon (with ϵ>0\epsilon>0) implies a non-zero violation of the inequality 𝒦21\mathcal{K}_{2\mapsto 1}. Therefore, optimization of P(ai,bj|Ai,Bj)P(a_{i},b_{j}|A_{i},B_{j}) under the constraints 𝒦n1=𝒦n1MR+ϵ\mathcal{K}_{n\mapsto 1}=\mathcal{K}^{MR}_{n\mapsto 1}+\epsilon enables one to determine the maximum value of P(ai,bj|Ai,Bj)P(a_{i},b_{j}|A_{i},B_{j}) for a certain amount of violation ϵ\epsilon. The other constraints P(ai,bj|Ai,Bj)0P(a_{i},b_{j}|A_{i},B_{j})\geq 0 and ai,bjP(ai,bj|Ai,Bj)=1Ai,Bj\sum_{a_{i},b_{j}}P(a_{i},b_{j}|A_{i},B_{j})=1~{}~{}\forall A_{i},B_{j} can be easily understood from the properties of a valid probability distribution. The last constraint that P(ai,bj|Ai,Bj)satisfies NSITP(a_{i},b_{j}|A_{i},B_{j})~{}\text{satisfies NSIT} implies that the probability distribution P(ai,bj|Ai,Bj)P(a_{i},b_{j}|A_{i},B_{j}) satisfies the NSIT condition given by P(bj|Bj)=P(bj|Ai,Bj)Ai,Bj,bjP(b_{j}|B_{j})=P(b_{j}|A_{i},B_{j})~{}\forall A_{i},B_{j},b_{j}.

We are now interested to obtain a lower bound on the min-entropy H(ai,bj|Ai,Bj)H_{\infty}(a_{i},b_{j}|A_{i},B_{j}) as a function of 𝒦n1\mathcal{K}_{n\mapsto 1}, i.e., we need to derive an inequality of the form H(ai,bj|Ai,Bj)f(𝒦n1)H_{\infty}(a_{i},b_{j}|A_{i},B_{j})\geq f(\mathcal{K}_{n\mapsto 1}). Below we provide a general lower bound of H(ai,bj|Ai,Bj)H_{\infty}(a_{i},b_{j}|A_{i},B_{j}) from the perspective of no-signalling in time conditions. Using linear programming and imposing NSIT conditions, it can be shown that solving Eq.(IV) one can obtain P(ai,bj|Ai,Bj)α𝒦n1+βP^{*}(a_{i},b_{j}|A_{i},B_{j})\leq\alpha\mathcal{K}_{n\mapsto 1}+\beta, where α\alpha and β\beta in general may depend on ai,bj,Ai,Bja_{i},b_{j},A_{i},B_{j}.

For 212\mapsto 1 RAC, in the case of the classical strategy, a deterministic point can achieve P(ai,bj|Ai,Bj)P(a_{i},b_{j}|A_{i},B_{j}) upto 11, i.e., P(ai,bj|Ai,Bj)1P^{*}(a_{i},b_{j}|A_{i},B_{j})\leq 1 when 𝒦21=2\mathcal{K}_{2\mapsto 1}=2, and for the no signaling (in time) box (which is equivalent to the Popescu-Rorlich box for spatial correlation), P(ai,bj|Ai,Bj)=1/2P(a_{i},b_{j}|A_{i},B_{j})=1/2 i.e., when 𝒦21=4\mathcal{K}_{2\mapsto 1}=4, P(ai,bj|Ai,Bj)1/2P^{*}(a_{i},b_{j}|A_{i},B_{j})\leq 1/2 . Therefore, analyzing the above inequalities one can obtain the values of α\alpha and β\beta to be 1/4-1/4 and 3/23/2, respectively randomness ; rand'16 . Hence,

P(ai,bj|Ai,Bj)32𝒦214or\displaystyle P^{*}(a_{i},b_{j}|A_{i},B_{j})\leq\frac{3}{2}-\frac{\mathcal{K}_{2\mapsto 1}}{4}~{}~{}\text{or}
H(ai,bj|Ai,Bj)log2[32𝒦214].\displaystyle H_{\infty}(a_{i},b_{j}|A_{i},B_{j})\geq-\log_{2}[\frac{3}{2}-\frac{\mathcal{K}_{2\mapsto 1}}{4}]. (45)

In fig.(1), a graphical representation of the lower bound of H(ai,bj|Ai,Bj)H_{\infty}(a_{i},b_{j}|A_{i},B_{j}) corresponding to 212\mapsto 1 RAC is provided.

Refer to caption
Figure 1: Min entropy H(ai,bj|Ai,Bj)H_{\infty}(a_{i},b_{j}|A_{i},B_{j}) is plotted with 𝒦21\mathcal{K}_{2\mapsto 1}

A similar approach based on NSIT conditions can also be applied to other n1n\mapsto 1 RAC. For the case of 313\mapsto 1 RAC, the no-signalling polytope can achieve the value of 𝒦31\mathcal{K}_{3\mapsto 1} up to 1212 and for 414\mapsto 1 the value of 𝒦41\mathcal{K}_{4\mapsto 1} up to 3232. Therefore, solving linear equations one can obtain P(ai,bj|Ai,Bj)54𝒦3116P^{*}(a_{i},b_{j}|A_{i},B_{j})\leq\frac{5}{4}-\frac{\mathcal{K}_{3\mapsto 1}}{16} and P(ai,bj|Ai,Bj)76𝒦4148P^{*}(a_{i},b_{j}|A_{i},B_{j})\leq\frac{7}{6}-\frac{\mathcal{K}_{4\mapsto 1}}{48} for 313\mapsto 1 and 414\mapsto 1 RAC, respectively. The lower bound of H(ai,bj|Ai,Bj)H_{\infty}(a_{i},b_{j}|A_{i},B_{j}) corresponding to 𝒦31\mathcal{K}_{3\mapsto 1} and 𝒦41\mathcal{K}_{4\mapsto 1} RAC are plotted in fig.(2) and fig.(3) respectively. Here, it may be pertinent to mention that if instead of P(ai,bj|Ai,Bj)P^{*}(a_{i},b_{j}|A_{i},B_{j}), randomness is certified from P(bj|Ai,Bj,ai)P^{*}(b_{j}|A_{i},B_{j},a_{i}), then one obtains the maximum value of H(ai,bj|Ai,Bj)H_{\infty}(a_{i},b_{j}|A_{i},B_{j}) to be 11 for any n1n\mapsto 1 RAC, because P(bj|Ai,Bj,ai)P^{*}(b_{j}|A_{i},B_{j},a_{i}) is 1/21/2 in this case irrespective of the value of nn. Below, we summarize our key findings for the generation of genuine randomness based on our protocol.

Refer to caption
Figure 2: Min entropy H(ai,bj|Ai,Bj)H_{\infty}(a_{i},b_{j}|A_{i},B_{j}) is plotted with 𝒦31\mathcal{K}_{3\mapsto 1}
Refer to caption
Figure 3: Min entropy H(ai,bj|Ai,Bj)H_{\infty}(a_{i},b_{j}|A_{i},B_{j}) is plotted with 𝒦41\mathcal{K}_{4\mapsto 1}

\bullet Result 2:- Any violation of the noninvasive-realist model of the temporal inequalities with initially prepared input state 𝕀/2\mathbb{I}/2, or equivalently, any non-zero quantum advantage of n1n\mapsto 1 RAC with SR can be exploited 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 n1n\mapsto 1 RAC qkd ; randomness1 .

V Conclusions

Despite the fact that the 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 showed that any non-zero quantum advantage of RAC with shared randomness necessarily violates a noninvasive-realist model. We proposed temporal inequalities corresponding to each n1n\mapsto 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 n1n\mapsto 1 RAC with best classical strategy is connected to the maximum noninvasive-realist bound. Moreover, any non-zero quantum advantage of a n1n\mapsto 1 RAC was equivalent to the violation of the corresponding temporal inequality.

Next, using an alternative derivation of the noninvasive-realist model, we showed 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 the arbitrary quantum advantage of RAC rangen ; qkd ; randomness1 . Before concluding a few remarks are in order. The maximum success probability using quantum strategy for a general n1n\mapsto 1 RAC is hard to compute for large nn, 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 urbasi1 , and hence, the generation of genuine randomness without entanglement based on our protocol might be exemplary for practical purposes.

VI Acknowledgements

SB and ASM acknowledge support of the Project No. DST/ICPS/QuEST/2018/98 of the Department of Science and Technology, Government of India. AGM acknowledges fruitful discussions with Debarshi Das and Bihalan Bhattacharya of SNBNCBS, Kolkata. SM acknowledges the support from the Ministry of Science and Technology, Taiwan (Grant No. MOST 111-2119-M-008-002).

Appendix A Derivation of the classical bound for the temporal inequality using macrorealism

A.1 Ontic model of 212\mapsto 1 RAC:

The average of a quantum operator in the Heisenberg picture can be written as an average over a set of hidden variables λ\lambda. The role of the initial state ρ(λ)\rho(\lambda) 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

At=𝑑λAt(λ)ρ(λ)\langle A_{t}\rangle=\int d\lambda A_{t}(\lambda)\rho(\lambda) (46)

where A(λt{}_{t}(\lambda) is the value taken by the observable on the hidden variable λ\lambda. The correlation between two observables AtiA_{t_{i}} (measured at some time tit_{i}), BtjB_{t_{j}} (measured at some later time tjt_{j}) is given by

AtiBtj=𝑑λAti(λ)Btj(λ)ρ(λAti)\langle A_{t_{i}}B_{t_{j}}\rangle=\int d\lambda A_{t_{i}}(\lambda)B_{t_{j}}(\lambda)\rho(\lambda\mid A_{t_{i}}) (47)

In general, we can always ignore the effect of final measurement due to noninvasive measurability (NIM). NIM can be defined as ρ(λAti,Btj,)\rho(\lambda\mid A_{t_{i}},B_{t_{j}},\dots) = ρ(λ)\rho(\lambda), i.e., a measurement does not change the distribution of λ\lambda. In Eq.(47), ρ(λ)\rho(\lambda) not depends on observable BtjB_{t_{j}} due to observable BtjB_{t_{j}} being measured after the measurement of observable AtiA_{t_{i}}.

Let us take Alice’s preparations to be eigenstates of observables {A1,A2}\{A_{1},A_{2}\} and Bob’s measurements to be {B1,B2}\{B_{1},B_{2}\}. Let, measurement A be carried out at some time tit_{i}, and measurement B be carried out at some later time tjt_{j}. Now, imposing the conditions of realism and NIM, we obtain

A1B1+A2B1\displaystyle\langle A_{1}B_{1}\rangle+\langle A_{2}B_{1}\rangle
=𝑑λ[A1(λ)B1(λ)ρ(λA1)+A2(λ)B1(λ)ρ(λA2)]\displaystyle=\int d\lambda[A_{1}(\lambda)B_{1}(\lambda)\rho(\lambda\mid A_{1})+A_{2}(\lambda)B_{1}(\lambda)\rho(\lambda\mid A_{2})]
=𝑑λA1(λ)B1(λ)[1A2(λ)B2(λ)]ρ(λA1)\displaystyle=\int d\lambda A_{1}(\lambda)B_{1}(\lambda)[1\mp A_{2}(\lambda)B_{2}(\lambda)]\rho(\lambda\mid A_{1})
+𝑑λA2(λ)B1(λ)[1±A1(λ)B2(λ)]ρ(λA2).\displaystyle+\int d\lambda A_{2}(\lambda)B_{1}(\lambda)[1\pm A_{1}(\lambda)B_{2}(\lambda)]\rho(\lambda\mid A_{2}). (48)

Taking the modulus on both sides and using the triangle inequality we obtain,

|A1B1+A2B1|\displaystyle|\langle A_{1}B_{1}\rangle+\langle A_{2}B_{1}\rangle|
2±[dλA1(λ)B2(λ)ρ(λA1)\displaystyle\leq 2\pm[\int d\lambda A_{1}(\lambda)B_{2}(\lambda)\rho(\lambda\mid A_{1})
dλA2(λ)B2(λ)ρ(λA2)].\displaystyle-\int d\lambda A_{2}(\lambda)B_{2}(\lambda)\rho(\lambda\mid A_{2})]. (49)

Now invoking NIM, we have,

|A1B1+A2B1|\displaystyle|\langle A_{1}B_{1}\rangle+\langle A_{2}B_{1}\rangle|
[A1B2A2B2]2.\displaystyle\mp[\langle A_{1}B_{2}\rangle-\langle A_{2}B_{2}\rangle]\leq 2.
or,\displaystyle\text{or},
𝒦212.\displaystyle\mathcal{K}_{2\mapsto 1}\leq 2. (50)

This is the four term Leggett-Garg inequality.

A.2 Ontic model of 313\mapsto 1 RAC :

Let us take Alice’s preparations to be eigenstates of {A1,A2,A3,A4}\{A_{1},A_{2},A_{3},A_{4}\} and Bob’s measurements to be {B1,B2,B3}\{B_{1},B_{2},B_{3}\}. Now, following similar steps as in the derivation of the Bell inequality, we obtain the sum of four correlations,

A1B1+A4B1+A2B1+A3B1\displaystyle\langle A_{1}B_{1}\rangle+\langle A_{4}B_{1}\rangle+\langle A_{2}B_{1}\rangle+\langle A_{3}B_{1}\rangle
=dλ[A1(λ)B1(λ)ρ(λA1)+A4(λ)B1(λ)ρ(λA4)+\displaystyle=\int d\lambda[A_{1}(\lambda)B_{1}(\lambda)\rho(\lambda\mid A_{1})+A_{4}(\lambda)B_{1}(\lambda)\rho(\lambda\mid A_{4})+
A2(λ)B1(λ)ρ(λA2)+A3(λ)B1(λ)ρ(λA3)]\displaystyle A_{2}(\lambda)B_{1}(\lambda)\rho(\lambda\mid A_{2})+A_{3}(\lambda)B_{1}(\lambda)\rho(\lambda\mid A_{3})]
=𝑑λA1(λ)B1(λ)[1A4(λ)(B2(λ)+B3(λ))]ρ(λA1)\displaystyle=\int d\lambda A_{1}(\lambda)B_{1}(\lambda)[1\mp A_{4}(\lambda)(B_{2}(\lambda)+B_{3}(\lambda))]\rho(\lambda\mid A_{1})
+𝑑λA4(λ)B1(λ)[1±A1(λ)(B2(λ)+B3(λ))]ρ(λA4)\displaystyle+\int d\lambda A_{4}(\lambda)B_{1}(\lambda)[1\pm A_{1}(\lambda)(B_{2}(\lambda)+B_{3}(\lambda))]\rho(\lambda\mid A_{4})
+𝑑λA2(λ)B1(λ)[1A3(λ)(B2(λ)B3(λ))]ρ(λA2)\displaystyle+\int d\lambda A_{2}(\lambda)B_{1}(\lambda)[1\mp A_{3}(\lambda)(B_{2}(\lambda)-B_{3}(\lambda))]\rho(\lambda\mid A_{2})
+𝑑λA3(λ)B1(λ)[1±A2(λ)(B2(λ)B3(λ))]ρ(λA3)\displaystyle+\int d\lambda A_{3}(\lambda)B_{1}(\lambda)[1\pm A_{2}(\lambda)(B_{2}(\lambda)-B_{3}(\lambda))]\rho(\lambda\mid A_{3}) (51)

Now, we take the modulus of both sides and use the triangle inequality to obtain,

|A1B1+A4B1+A2B1+A3B1|\displaystyle|\langle A_{1}B_{1}\rangle+\langle A_{4}B_{1}\rangle+\langle A_{2}B_{1}\rangle+\langle A_{3}B_{1}\rangle|
4±[dλA1(λ)(B2(λ)+B3(λ))ρ(λA1)\displaystyle\leq 4\pm[\int d\lambda A_{1}(\lambda)(B_{2}(\lambda)+B_{3}(\lambda))\rho(\lambda\mid A_{1})
𝑑λA4(λ)(B2(λ)+B3(λ))ρ(λA4)\displaystyle-\int d\lambda A_{4}(\lambda)(B_{2}(\lambda)+B_{3}(\lambda))\rho(\lambda\mid A_{4})
+𝑑λA2(λ)(B2(λ)B3(λ))ρ(λA2)\displaystyle+\int d\lambda A_{2}(\lambda)(B_{2}(\lambda)-B_{3}(\lambda))\rho(\lambda\mid A_{2})
dλA3(λ)(B2(λ)B3(λ))ρ(λA3)].\displaystyle-\int d\lambda A_{3}(\lambda)(B_{2}(\lambda)-B_{3}(\lambda))\rho(\lambda\mid A_{3})]. (52)

Invoking NIM, we have,

|A1B1+A4B1+A2B1+A3B1|\displaystyle|\langle A_{1}B_{1}\rangle+\langle A_{4}B_{1}\rangle+\langle A_{2}B_{1}\rangle+\langle A_{3}B_{1}\rangle|
[A1(B2+B3)A4(B2+B3)+A2(B2B3)\displaystyle\mp[\langle A_{1}(B_{2}+B_{3})\rangle-\langle A_{4}(B_{2}+B_{3})\rangle+\langle A_{2}(B_{2}-B_{3})\rangle
A3(B2B3)]4.\displaystyle-\langle A_{3}(B_{2}-B_{3})]\leq 4.
or,\displaystyle\text{or},
𝒦314.\displaystyle\mathcal{K}_{3\mapsto 1}\leq 4. (53)

A.3 Ontic model of 414\mapsto 1 RAC :

Similarly, for 414\mapsto 1 RAC, here Alice has 8 preparations to be eigenstates of {A1,A2,,A8}\{A_{1},A_{2},...,A_{8}\} and Bob has 4 measurements to be {B1,B2,B3,B4}\{B_{1},B_{2},B_{3},B_{4}\}. Now using a similar procedure presented for deriving the classical bound corresponding to 212\mapsto 1 RAC, we can obtain

A1B1+A5B1+A1B3+A2B3+A2B1+A6B1\displaystyle\langle A_{1}B_{1}\rangle+\langle A_{5}B_{1}\rangle+\langle A_{1}B_{3}\rangle+\langle A_{2}B_{3}\rangle+\langle A_{2}B_{1}\rangle+\langle A_{6}B_{1}\rangle
+A3B1+A7B1+A4B1+A8B1+A5B3\displaystyle+\langle A_{3}B_{1}\rangle+\langle A_{7}B_{1}\rangle+\langle A_{4}B_{1}\rangle+\langle A_{8}B_{1}\rangle+\langle A_{5}B_{3}\rangle
+A6B3A4B3A7B3A8B3A3B3\displaystyle+\langle A_{6}B_{3}\rangle-\langle A_{4}B_{3}\rangle-\langle A_{7}B_{3}\rangle-\langle A_{8}B_{3}\rangle-\langle A_{3}B_{3}\rangle
=dλ[A1(λ)B1(λ)ρ(λA1)+A5(λ)B1(λ)ρ(λA5)+\displaystyle=\int d\lambda[A_{1}(\lambda)B_{1}(\lambda)\rho(\lambda\mid A_{1})+A_{5}(\lambda)B_{1}(\lambda)\rho(\lambda\mid A_{5})+
A1(λ)B3(λ)ρ(λA1)+A2(λ)B3(λ)ρ(λA2)+\displaystyle A_{1}(\lambda)B_{3}(\lambda)\rho(\lambda\mid A_{1})+A_{2}(\lambda)B_{3}(\lambda)\rho(\lambda\mid A_{2})+
A2(λ)B1(λ)ρ(λA2)+A6(λ)B1(λ)ρ(λA6)+\displaystyle A_{2}(\lambda)B_{1}(\lambda)\rho(\lambda\mid A_{2})+A_{6}(\lambda)B_{1}(\lambda)\rho(\lambda\mid A_{6})+
A3(λ)B1(λ)ρ(λA3)+A7(λ)B1(λ)ρ(λA7)+\displaystyle A_{3}(\lambda)B_{1}(\lambda)\rho(\lambda\mid A_{3})+A_{7}(\lambda)B_{1}(\lambda)\rho(\lambda\mid A_{7})+
A4(λ)B1(λ)ρ(λA4)+A8(λ)B1(λ)ρ(λA8)+\displaystyle A_{4}(\lambda)B_{1}(\lambda)\rho(\lambda\mid A_{4})+A_{8}(\lambda)B_{1}(\lambda)\rho(\lambda\mid A_{8})+
A5(λ)B3(λ)ρ(λA5)+A6(λ)B3(λ)ρ(λA6)\displaystyle A_{5}(\lambda)B_{3}(\lambda)\rho(\lambda\mid A_{5})+A_{6}(\lambda)B_{3}(\lambda)\rho(\lambda\mid A_{6})-
A4(λ)B3(λ)ρ(λA4)A7(λ)B3(λ)ρ(λA7)\displaystyle A_{4}(\lambda)B_{3}(\lambda)\rho(\lambda\mid A_{4})-A_{7}(\lambda)B_{3}(\lambda)\rho(\lambda\mid A_{7})-
A8(λ)B3(λ)ρ(λA8)A3(λ)B3(λ)]ρ(λA3)]\displaystyle A_{8}(\lambda)B_{3}(\lambda)\rho(\lambda\mid A_{8})-A_{3}(\lambda)B_{3}(\lambda)]\rho(\lambda\mid A_{3})]
=𝑑λA1(λ)B1(λ)[1A5(λ)B2(λ)]ρ(λA1)\displaystyle=\int d\lambda A_{1}(\lambda)B_{1}(\lambda)[1\mp A_{5}(\lambda)B_{2}(\lambda)]\rho(\lambda\mid A_{1})
+𝑑λA5(λ)B1(λ)[1±A1(λ)B2(λ)]ρ(λA5)\displaystyle+\int d\lambda A_{5}(\lambda)B_{1}(\lambda)[1\pm A_{1}(\lambda)B_{2}(\lambda)]\rho(\lambda\mid A_{5})
+𝑑λA1(λ)B3(λ)[1A2(λ)B4(λ)]ρ(λA1)\displaystyle+\int d\lambda A_{1}(\lambda)B_{3}(\lambda)[1\mp A_{2}(\lambda)B_{4}(\lambda)]\rho(\lambda\mid A_{1})
+𝑑λA2(λ)B3(λ)[1±A1(λ)B4(λ)]ρ(λA2)\displaystyle+\int d\lambda A_{2}(\lambda)B_{3}(\lambda)[1\pm A_{1}(\lambda)B_{4}(\lambda)]\rho(\lambda\mid A_{2})
+𝑑λA2(λ)B1(λ)[1A6(λ)B2(λ)]ρ(λA2)\displaystyle+\int d\lambda A_{2}(\lambda)B_{1}(\lambda)[1\mp A_{6}(\lambda)B_{2}(\lambda)]\rho(\lambda\mid A_{2})
+𝑑λA6(λ)B1(λ)[1±A2(λ)B2(λ)]ρ(λA6)\displaystyle+\int d\lambda A_{6}(\lambda)B_{1}(\lambda)[1\pm A_{2}(\lambda)B_{2}(\lambda)]\rho(\lambda\mid A_{6})
+𝑑λA3(λ)B1(λ)[1A7(λ)B2(λ)]ρ(λA3)\displaystyle+\int d\lambda A_{3}(\lambda)B_{1}(\lambda)[1\mp A_{7}(\lambda)B_{2}(\lambda)]\rho(\lambda\mid A_{3})
+𝑑λA7(λ)B1(λ)[1±A3(λ)B2(λ)]ρ(λA7)\displaystyle+\int d\lambda A_{7}(\lambda)B_{1}(\lambda)[1\pm A_{3}(\lambda)B_{2}(\lambda)]\rho(\lambda\mid A_{7})
+𝑑λA4(λ)B1(λ)[1A8(λ)B2(λ)]ρ(λA4)\displaystyle+\int d\lambda A_{4}(\lambda)B_{1}(\lambda)[1\mp A_{8}(\lambda)B_{2}(\lambda)]\rho(\lambda\mid A_{4})
+𝑑λA8(λ)B1(λ)[1±A4(λ)B2(λ)]ρ(λA8)\displaystyle+\int d\lambda A_{8}(\lambda)B_{1}(\lambda)[1\pm A_{4}(\lambda)B_{2}(\lambda)]\rho(\lambda\mid A_{8})
+𝑑λA5(λ)B3(λ)[1A6(λ)B4(λ)]ρ(λA5)\displaystyle+\int d\lambda A_{5}(\lambda)B_{3}(\lambda)[1\mp A_{6}(\lambda)B_{4}(\lambda)]\rho(\lambda\mid A_{5})
+𝑑λA6(λ)B3(λ)[1±A5(λ)B4(λ)]ρ(λA6)\displaystyle+\int d\lambda A_{6}(\lambda)B_{3}(\lambda)[1\pm A_{5}(\lambda)B_{4}(\lambda)]\rho(\lambda\mid A_{6})
𝑑λA4(λ)B3(λ)[1A7(λ)B4(λ)]ρ(λA4)\displaystyle-\int d\lambda A_{4}(\lambda)B_{3}(\lambda)[1\mp A_{7}(\lambda)B_{4}(\lambda)]\rho(\lambda\mid A_{4})
𝑑λA7(λ)B3(λ)[1±A4(λ)B4(λ)]ρ(λA7)\displaystyle-\int d\lambda A_{7}(\lambda)B_{3}(\lambda)[1\pm A_{4}(\lambda)B_{4}(\lambda)]\rho(\lambda\mid A_{7})
𝑑λA8(λ)B3(λ)[1A3(λ)B4(λ)]ρ(λA8)\displaystyle-\int d\lambda A_{8}(\lambda)B_{3}(\lambda)[1\mp A_{3}(\lambda)B_{4}(\lambda)]\rho(\lambda\mid A_{8})
𝑑λA3(λ)B3(λ)[1±A8(λ)B4(λ)]ρ(λA3)\displaystyle-\int d\lambda A_{3}(\lambda)B_{3}(\lambda)[1\pm A_{8}(\lambda)B_{4}(\lambda)]\rho(\lambda\mid A_{3}) (54)

Taking the modulus on both sides and using the triangle inequality we obtain,

|A1B1+A5B1+A1B3+A2B3+A2B1+\displaystyle|\langle A_{1}B_{1}\rangle+\langle A_{5}B_{1}\rangle+\langle A_{1}B_{3}\rangle+\langle A_{2}B_{3}\rangle+\langle A_{2}B_{1}\rangle+
A6B1+A3B1+A7B1+A4B1+A8B1+\displaystyle\langle A_{6}B_{1}\rangle+\langle A_{3}B_{1}\rangle+\langle A_{7}B_{1}\rangle+\langle A_{4}B_{1}\rangle+\langle A_{8}B_{1}\rangle+
A5B3+A6B3A4B3A7B3A8B3\displaystyle\langle A_{5}B_{3}\rangle+\langle A_{6}B_{3}\rangle-\langle A_{4}B_{3}\rangle-\langle A_{7}B_{3}\rangle-\langle A_{8}B_{3}\rangle
A3B3|8±[dλA1(λ)B2(λ)ρ(λA1)\displaystyle-\langle A_{3}B_{3}\rangle|\leq 8\pm[\int d\lambda A_{1}(\lambda)B_{2}(\lambda)\rho(\lambda\mid A_{1})
𝑑λA5(λ)B2(λ)ρ(λA5)+𝑑λA1(λ)B4(λ)ρ(λA1)\displaystyle-\int d\lambda A_{5}(\lambda)B_{2}(\lambda)\rho(\lambda\mid A_{5})+\int d\lambda A_{1}(\lambda)B_{4}(\lambda)\rho(\lambda\mid A_{1})
𝑑λA2(λ)B4(λ)ρ(λA2)+𝑑λA2(λ)B2(λ)ρ(λA2)\displaystyle-\int d\lambda A_{2}(\lambda)B_{4}(\lambda)\rho(\lambda\mid A_{2})+\int d\lambda A_{2}(\lambda)B_{2}(\lambda)\rho(\lambda\mid A_{2})
𝑑λA6(λ)B2(λ)ρ(λA6)+𝑑λA3(λ)B2(λ)ρ(λA3)\displaystyle-\int d\lambda A_{6}(\lambda)B_{2}(\lambda)\rho(\lambda\mid A_{6})+\int d\lambda A_{3}(\lambda)B_{2}(\lambda)\rho(\lambda\mid A_{3})
𝑑λA7(λ)B2(λ)ρ(λA7)+𝑑λA4(λ)B2(λ)ρ(λA4)\displaystyle-\int d\lambda A_{7}(\lambda)B_{2}(\lambda)\rho(\lambda\mid A_{7})+\int d\lambda A_{4}(\lambda)B_{2}(\lambda)\rho(\lambda\mid A_{4})
𝑑λA8(λ)B2(λ)ρ(λA8)+𝑑λA5(λ)B4(λ)ρ(λA5)\displaystyle-\int d\lambda A_{8}(\lambda)B_{2}(\lambda)\rho(\lambda\mid A_{8})+\int d\lambda A_{5}(\lambda)B_{4}(\lambda)\rho(\lambda\mid A_{5})
𝑑λA6(λ)B4(λ)ρ(λA6)+𝑑λA7(λ)B4(λ)ρ(λA4)\displaystyle-\int d\lambda A_{6}(\lambda)B_{4}(\lambda)\rho(\lambda\mid A_{6})+\int d\lambda A_{7}(\lambda)B_{4}(\lambda)\rho(\lambda\mid A_{4})
𝑑λA4(λ)B4(λ)ρ(λA7)+𝑑λA3(λ)B4(λ)ρ(λA8)\displaystyle-\int d\lambda A_{4}(\lambda)B_{4}(\lambda)\rho(\lambda\mid A_{7})+\int d\lambda A_{3}(\lambda)B_{4}(\lambda)\rho(\lambda\mid A_{8})
dλA8(λ)B4(λ)ρ(λA3)].\displaystyle-\int d\lambda A_{8}(\lambda)B_{4}(\lambda)\rho(\lambda\mid A_{3})]. (55)

Invoking NIM, we have,

|A1B1+A5B1+A1B3+A2B3+A2B1+\displaystyle|\langle A_{1}B_{1}\rangle+\langle A_{5}B_{1}\rangle+\langle A_{1}B_{3}\rangle+\langle A_{2}B_{3}\rangle+\langle A_{2}B_{1}\rangle+
A6B1+A3B1+A7B1+A4B1+A8B1+\displaystyle\langle A_{6}B_{1}\rangle+\langle A_{3}B_{1}\rangle+\langle A_{7}B_{1}\rangle+\langle A_{4}B_{1}\rangle+\langle A_{8}B_{1}\rangle+
A5B3+A6B3A4B3A7B3A8B3\displaystyle\langle A_{5}B_{3}\rangle+\langle A_{6}B_{3}\rangle-\langle A_{4}B_{3}\rangle-\langle A_{7}B_{3}\rangle-\langle A_{8}B_{3}\rangle
A3B3|[A1B2A5B2+A1B4A2B4\displaystyle-\langle A_{3}B_{3}\rangle|\mp[\langle A_{1}B_{2}\rangle-\langle A_{5}B_{2}\rangle+\langle A_{1}B_{4}\rangle-\langle A_{2}B_{4}\rangle
+A2B2A6B2+A3B2A7B2+A4B2\displaystyle+\langle A_{2}B_{2}\rangle-\langle A_{6}B_{2}\rangle+\langle A_{3}B_{2}\rangle-\langle A_{7}B_{2}\rangle+\langle A_{4}B_{2}\rangle-
A8B2+A5B4A6B4A4B4+A7B4\displaystyle\langle A_{8}B_{2}\rangle+\langle A_{5}B_{4}\rangle-\langle A_{6}B_{4}\rangle-\langle A_{4}B_{4}\rangle+\langle A_{7}B_{4}\rangle-
A8B4+A3B4]8.\displaystyle\langle A_{8}B_{4}\rangle+\langle A_{3}B_{4}\rangle]\leq 8.
or,\displaystyle\text{or},
𝒦418.\displaystyle\mathcal{K}_{4\mapsto 1}\leq 8. (56)

A.4 Ontic model of n1n\mapsto 1 RAC :

Here Alice has 2n12^{n-1} preparations which are the eigenstates corresponding to {A1,A2,,A2n1}\{A_{1},A_{2},...,A_{2^{n-1}}\} and Bob has n measurements, {B1,B2,B3,,Bn}\{B_{1},B_{2},B_{3},...,B_{n}\}. Following Eq.(III.4)\eqref{kn} the term 𝒦n1\mathcal{K}_{n\mapsto 1} can be explicitly written as,

𝒦n1=A1B1+A1B2+A1B3+A1B4++\displaystyle\mathcal{K}_{n\mapsto 1}=\langle A_{1}B_{1}\rangle+\langle A_{1}B_{2}\rangle+\langle A_{1}B_{3}\rangle+\langle A_{1}B_{4}\rangle+\dots+
A1Bn+A2B1+A2B2+A2B3+A2B4A2Bn\displaystyle\langle A_{1}B_{n}\rangle+\langle A_{2}B_{1}\rangle+\langle A_{2}B_{2}\rangle+\langle A_{2}B_{3}\rangle+\langle A_{2}B_{4}\rangle-\langle A_{2}B_{n}\rangle
+A3B1+A3B2+A3B3+A3B4+A3Bn\displaystyle+\langle A_{3}B_{1}\rangle+\langle A_{3}B_{2}\rangle+\langle A_{3}B_{3}\rangle+\langle A_{3}B_{4}\rangle+\langle A_{3}B_{n}\rangle
++A2n11B1A2n11B2A2n11B3\displaystyle+\dots+\langle A_{2^{n-1}-1}B_{1}\rangle-\langle A_{2^{n-1}-1}B_{2}\rangle-\langle A_{2^{n-1}-1}B_{3}\rangle
+A2n11Bn+A2n1B1A2n1B2\displaystyle-\dots+\langle A_{2^{n-1}-1}B_{n}\rangle+\langle A_{2^{n-1}}B_{1}\rangle-\langle A_{2^{n-1}}B_{2}\rangle
A2n1B3A2n1Bn3A2n1Bn1\displaystyle-\langle A_{2^{n-1}}B_{3}\rangle-\dots-\langle A_{2^{n-1}}B_{n-3}\rangle-\langle A_{2^{n-1}}B_{n-1}\rangle
A2n1Bn.\displaystyle-\langle A_{2^{n-1}}B_{n}\rangle. (57)

Now, we can obtain the classical bound for n1n\mapsto 1 RAC if we adopt a similar procedure as presented for deriving the classical bound corresponding to 212\mapsto 1 RAC. Let us now derive the classical bound explicitly when nn is even.

A1B1+A1B3++A2B1+A2B3++\displaystyle\langle A_{1}B_{1}\rangle+\langle A_{1}B_{3}\rangle+\dots+\langle A_{2}B_{1}\rangle+\langle A_{2}B_{3}\rangle+\dots+
A3B1+A3B3++A2n11B1\displaystyle\langle A_{3}B_{1}\rangle+\langle A_{3}B_{3}\rangle+\dots+\langle A_{2^{n-1}-1}B_{1}\rangle-
A2n11B3+A2n1B1A2n1B3\displaystyle\langle A_{2^{n-1}-1}B_{3}\rangle-\dots+\langle A_{2^{n-1}}B_{1}\rangle-\langle A_{2^{n-1}}B_{3}\rangle-\dots
A2n1Bn3A2n1Bn1\displaystyle-\langle A_{2^{n-1}}B_{n-3}\rangle-\langle A_{2^{n-1}}B_{n-1}\rangle
=dλ[A1(λ)B1(λ)ρ(λA1)+A1(λ)B3(λ)ρ(λA1)\displaystyle=\int d\lambda[A_{1}(\lambda)B_{1}(\lambda)\rho(\lambda\mid A_{1})+A_{1}(\lambda)B_{3}(\lambda)\rho(\lambda\mid A_{1})
++A2(λ)B1(λ)ρ(λA2)+A2(λ)B3(λ)ρ(λA2)+\displaystyle+\dots+A_{2}(\lambda)B_{1}(\lambda)\rho(\lambda\mid A_{2})+A_{2}(\lambda)B_{3}(\lambda)\rho(\lambda\mid A_{2})+
+A3(λ)B1(λ)ρ(λA3)+A3(λ)B3(λ)ρ(λA3)\displaystyle\dots+A_{3}(\lambda)B_{1}(\lambda)\rho(\lambda\mid A_{3})+A_{3}(\lambda)B_{3}(\lambda)\rho(\lambda\mid A_{3})
++A2n11(λ)B1(λ)ρ(λA2n11)\displaystyle+\dots+A_{2^{n-1}-1}(\lambda)B_{1}(\lambda)\rho(\lambda\mid A_{2^{n-1}-1})
A2n11(λ)B3(λ)ρ(λA2n11)+\displaystyle-A_{2^{n-1}-1}(\lambda)B_{3}(\lambda)\rho(\lambda\mid A_{2^{n-1}-1})-\dots+
A2n1(λ)B1(λ)ρ(λA2n1)A2n1(λ)B3(λ)ρ(λA2n1)\displaystyle A_{2^{n-1}}(\lambda)B_{1}(\lambda)\rho(\lambda\mid A_{2^{n-1}})-A_{2^{n-1}}(\lambda)B_{3}(\lambda)\rho(\lambda\mid A_{2^{n-1}})
A2n1(λ)Bn3(λ)ρ(λA2n1)\displaystyle-\dots-A_{2^{n-1}}(\lambda)B_{n-3}(\lambda)\rho(\lambda\mid A_{2^{n-1}})
A2n1(λ)Bn1(λ)ρ(λA2n1)]\displaystyle-A_{2^{n-1}}(\lambda)B_{n-1}(\lambda)\rho(\lambda\mid A_{2^{n-1}})]
=𝑑λA1(λ)B1(λ)[1A2n1(λ)B2(λ)]ρ(λA2n1)\displaystyle=\int d\lambda A_{1}(\lambda)B_{1}(\lambda)[1\mp A_{2^{n-1}}(\lambda)B_{2}(\lambda)]\rho(\lambda\mid A_{2^{n-1}})
+𝑑λA2(λ)B1(λ)[1A2n11(λ)B2(λ)]ρ(λA2n11)\displaystyle+\int d\lambda A_{2}(\lambda)B_{1}(\lambda)[1\mp A_{2^{n-1}-1}(\lambda)B_{2}(\lambda)]\rho(\lambda\mid A_{2^{n-1}-1})
+𝑑λA1(λ)Bn1(λ)[1A2(λ)Bn(λ)]ρ(λA1)\displaystyle+\int d\lambda A_{1}(\lambda)B_{n-1}(\lambda)[1\mp A_{2}(\lambda)B_{n}(\lambda)]\rho(\lambda\mid A_{1})
+𝑑λA2(λ)Bn1(λ)[1±A1(λ)Bn(λ)]ρ(λA2)\displaystyle+\int d\lambda A_{2}(\lambda)B_{n-1}(\lambda)[1\pm A_{1}(\lambda)B_{n}(\lambda)]\rho(\lambda\mid A_{2})
++dλA2n11(λ)B1(λ)[1±\displaystyle+\dots+\int d\lambda A_{2^{n-1}-1}(\lambda)B_{1}(\lambda)[1\pm
A2(λ)B2(λ)]ρ(λA2n11)\displaystyle A_{2}(\lambda)B_{2}(\lambda)]\rho(\lambda\mid A_{2^{n-1}-1})
+dλA2n1(λ)B1(λ)[1±\displaystyle+\int d\lambda A_{2^{n-1}}(\lambda)B_{1}(\lambda)[1\pm
A1(λ)B2(λ)]ρ(λA2n1)\displaystyle A_{1}(\lambda)B_{2}(\lambda)]\rho(\lambda\mid A_{2^{n-1}})
+dλA2n11(λ)Bn1(λ)[1\displaystyle+\dots-\int d\lambda A_{2^{n-1}-1}(\lambda)B_{n-1}(\lambda)[1\mp
A2n1(λ)Bn(λ)]ρ(λA2n11)\displaystyle A_{2^{n-1}}(\lambda)B_{n}(\lambda)]\rho(\lambda\mid A_{2^{n-1}-1})
dλA2n1(λ)Bn1(λ)[1±\displaystyle-\int d\lambda A_{2^{n-1}}(\lambda)B_{n-1}(\lambda)[1\pm
A2n11(λ)Bn(λ)]ρ(λA2n1)\displaystyle A_{2^{n-1}-1}(\lambda)B_{n}(\lambda)]\rho(\lambda\mid A_{2^{n-1}})

Taking the modulus on both sides and using the triangle inequality we obtain,

|A1B1+A1B3++A2B1+A2B3++\displaystyle|\langle A_{1}B_{1}\rangle+\langle A_{1}B_{3}\rangle+\dots+\langle A_{2}B_{1}\rangle+\langle A_{2}B_{3}\rangle+\dots+
A3B1+A3B3++A2n11B1\displaystyle\langle A_{3}B_{1}\rangle+\langle A_{3}B_{3}\rangle+\dots+\langle A_{2^{n-1}-1}B_{1}\rangle-
A2n11B3+A2n1B1A2n1B3\displaystyle\langle A_{2^{n-1}-1}B_{3}\rangle-\dots+\langle A_{2^{n-1}}B_{1}\rangle-\langle A_{2^{n-1}}B_{3}\rangle
A2n1Bn3A2n1Bn1|2n1\displaystyle-\dots-\langle A_{2^{n-1}}B_{n-3}\rangle-\langle A_{2^{n-1}}B_{n-1}\rangle|\leq 2^{n-1}
±[dλA1(λ)B2(λ)ρ(λA1)\displaystyle\pm[\int d\lambda A_{1}(\lambda)B_{2}(\lambda)\rho(\lambda\mid A_{1})
+𝑑λA2(λ)B2(λ)ρ(λA2)+\displaystyle+\int d\lambda A_{2}(\lambda)B_{2}(\lambda)\rho(\lambda\mid A_{2})+\dots
+𝑑λA1(λ)Bn(λ)ρ(λA1)\displaystyle+\int d\lambda A_{1}(\lambda)B_{n}(\lambda)\rho(\lambda\mid A_{1})
𝑑λA2(λ)Bn(λ)ρ(λA2)+\displaystyle-\int d\lambda A_{2}(\lambda)B_{n}(\lambda)\rho(\lambda\mid A_{2})+\dots
𝑑λA2n11(λ)B2(λ)ρ(λA2n11)\displaystyle-\int d\lambda A_{2^{n-1}-1}(\lambda)B_{2}(\lambda)\rho(\lambda\mid A_{2^{n-1}-1})
𝑑λA2n1(λ)B2(λ)ρ(λA2n1)\displaystyle-\int d\lambda A_{2^{n-1}}(\lambda)B_{2}(\lambda)\rho(\lambda\mid A_{2^{n-1}})-\dots
𝑑λA2n11(λ)Bn(λ)ρ(λA2n11)\displaystyle-\int d\lambda A_{2^{n-1}-1}(\lambda)B_{n}(\lambda)\rho(\lambda\mid A_{2^{n-1}-1})
dλA2n1(λ)Bn(λ)ρ(λA2n1)].\displaystyle-\int d\lambda A_{2^{n-1}}(\lambda)B_{n}(\lambda)\rho(\lambda\mid A_{2^{n-1}})]. (59)

Invoking NIM, we have,

|A1B1+A1B3++A2B1+A2B3++\displaystyle|\langle A_{1}B_{1}\rangle+\langle A_{1}B_{3}\rangle+\dots+\langle A_{2}B_{1}\rangle+\langle A_{2}B_{3}\rangle+\dots+
A3B1+A3B3++A2n11B1\displaystyle\langle A_{3}B_{1}\rangle+\langle A_{3}B_{3}\rangle+\dots+\langle A_{2^{n-1}-1}B_{1}\rangle-
A2n11B3+A2n1B1A2n1B3\displaystyle\langle A_{2^{n-1}-1}B_{3}\rangle-\dots+\langle A_{2^{n-1}}B_{1}\rangle-\langle A_{2^{n-1}}B_{3}\rangle
A2n1Bn3A2n1Bn1|[A1B2+\displaystyle-\langle A_{2^{n-1}}B_{n-3}\rangle-\langle A_{2^{n-1}}B_{n-1}\rangle|\mp[\langle A_{1}B_{2}\rangle+
A1B4++A1BnA2B2\displaystyle\langle A_{1}B_{4}\rangle+\dots+\langle A_{1}B_{n}\rangle-\langle A_{2}B_{2}\rangle
+A2B4+A2Bn+A3B2+A3B4\displaystyle+\langle A_{2}B_{4}\rangle+\dots-\langle A_{2}B_{n}\rangle+\langle A_{3}B_{2}\rangle+\langle A_{3}B_{4}\rangle
++A3Bn+A2n11B2+\displaystyle+\dots+\langle A_{3}B_{n}\rangle+\dots-\langle A_{2^{n-1}-1}B_{2}\rangle-\dots+
A2n11BnA2n1B2A2n1Bn]2n1.\displaystyle\langle A_{2^{n-1}-1}B_{n}\rangle-\langle A_{2^{n-1}}B_{2}\rangle-\dots-\langle A_{2^{n-1}}B_{n}\rangle]\leq 2^{n-1}.
or,\displaystyle\text{or},
𝒦n12n1.\displaystyle\mathcal{K}_{n\mapsto 1}\leq 2^{n-1}. (60)

Similarly, one can also obtain the same classical bound for 𝒦n1\mathcal{K}_{n\mapsto 1} when nn is odd.

References