Local-Data-Hiding and Causal Inseparability:
Probing Indefinite Causal Structures with Cryptographic Primitives

Sahil Gopalkrishna Naik Department of Physics of Complex Systems, S. N. Bose National Center for Basic Sciences, Block JD, Sector III, Salt Lake, Kolkata 700106, India. Samrat Sen Department of Physics of Complex Systems, S. N. Bose National Center for Basic Sciences, Block JD, Sector III, Salt Lake, Kolkata 700106, India. Ram Krishna Patra Department of Physics of Complex Systems, S. N. Bose National Center for Basic Sciences, Block JD, Sector III, Salt Lake, Kolkata 700106, India. Ananya Chakraborty Department of Physics of Complex Systems, S. N. Bose National Center for Basic Sciences, Block JD, Sector III, Salt Lake, Kolkata 700106, India. Mir Alimuddin ICFO-Institut de Ciencies Fotoniques, The Barcelona Institute of Science and Technology, Av. Carl Friedrich Gauss 3, 08860 Castelldefels (Barcelona), Spain. Manik Banik Department of Physics of Complex Systems, S. N. Bose National Center for Basic Sciences, Block JD, Sector III, Salt Lake, Kolkata 700106, India. Pratik Ghosal Department of Physics of Complex Systems, S. N. Bose National Center for Basic Sciences, Block JD, Sector III, Salt Lake, Kolkata 700106, India.

Abstract

Formulation of physical theories typically assumes a definite causal structure – either static or dynamic – among the set of physical events. Recent studies, however, suggest the possibility of indefiniteness in causal structure, which emerges as a novel information primitive offering advantages in various protocols. In this work, we explore utilities of this new primitive in cryptographic applications. To this aim, we propose a task called local-data-hiding, where a referee distributes encrypted messages among distant parties in such a way that the parties individually remain completely ignorant about the messages, and thus try to decrypt their respective messages through mutual collaboration. As we demonstrate, agents embedded in an indefinite causal structure can outperform their counterparts operating in a definite causal background. Considering the bipartite local-bit-hiding (LBH) task, we establish a strict duality between its optimal success probability and the optimal violation of a causal inequality obtained from the guess-your-neighbour’s-input game. This, in turn, provides a way forward to obtain Tsirelson-type bounds for causal inequalities. Furthermore, similar to Peres’s separability criterion, we derive a necessary criterion for quantum processes to be useful in the LBH task. We then report an intriguing super-activation phenomenon, where two quantum processes, each individually not useful for the LBH task, become useful when used together. We also analyze the utility of causal indefiniteness arising in classical setups and show its advantages in multipartite variants of the local-data-hiding task. Along with establishing new cryptographic applications our study illuminates various unexplored aspects of causal indefiniteness, and welcomes further studies on this new information primitive.

I Introduction

Classical physics, as described by Newton’s mechanics and Maxwell’s electrodynamics, is a deterministic theory formulated under the assumption that events are embedded in a predefined and definite causal structure [1]. Quantum theory, on the other hand, is inherently probabilistic but still assumes a fixed and definite causal structure [2]. For instance, two local quantum operations, $\mathcal{E}$ and $\mathcal{F}$ , described by completely positive trace-preserving maps acting on their respective local Hilbert spaces [3], are always assumed to be in a definite causal order: they are either time-like separated, with $\mathcal{E}$ in the causal past of $\mathcal{F}$ or vice versa, or they are space-like separated. General relativity, being a deterministic theory, allows a dynamically evolving causal structure determined by the underlying spacetime geometry [4]. A crucial insight by L. Hardy suggests that in quantum gravity, the causal structure, which represents the dynamical degrees of freedom of gravity, should exhibit indefiniteness, similar to other dynamical degrees of freedom in ordinary quantum theory [5, 6, 7]. This subsequently motivates several approaches to study the notion of causal indefiniteness [8, 9, 10, 11].

Beyond its profound foundational implications, more recently, causal indefiniteness has found novel advantages in various information protocols. The standard formulation of information theory, as established in Shannon’s seminal 1948 work [12], employs classical systems to store, transmit, and process information. Its quantum generalization, the quantum Shannon theory, exploits nonclassical features of quantum states, such as coherence, entanglement etc, to obtain advantages in information processing [13, 14, 15, 16, 17]. However, in quantum Shannon theory, operations on quantum systems are assumed to occur in a definite causal order. Embedding them in an indefinite causal structure promises even further advantages in information processing. These include testing properties of quantum channels [18, 19], winning non-causal games [9], reducing quantum communication complexity [20], enhancing the precision of quantum metrology [21], achieving thermodynamic advantages [22, 23, 24], and improving classical and quantum information transmission rates through noisy quantum channels [25, 26, 27]. Moreover, in the context of quantum gravity, the emergence of indefinite causal order due to the spatial superposition of massive objects [28] and its resourcefulness for local implementation of nonlocal quantum operations on distributed quantum systems has also been studied [29]. Along with these theoretical proposals, the demonstration of an indefinite causal primitive, called the quantum SWITCH, and its several advantages have also been reported in different experiments [30, 31, 32, 33, 34, 35, 36].

In this article, we explore utilities of causal indefiniteness in cryptographic tasks. The use of nonclassicality in cryptographic protocols is well-established. While Shor’s algorithm poses a threat to the classical RSA cryptosystem [16], pioneering work by Bennett & Brassard demonstrated the feasibility of key distribution protocols using quantum systems that are secure against eavesdroppers equipped with quantum computing power [37]. This subsequently motivated several other key distribution protocols with varying degrees of security [38, 39, 40, 41], as well as various other cryptographic protocols [42, 43, 44, 45, 46]. Here, we introduce a cryptographic primitive called local-data-hiding (LDH). A referee encodes an $N$ -dit string into an $N$ -partite quantum state and distributes it among $N$ different parties, with each dit assigned to a specific party. The encoding ensures that each party individually has zero information about the encoded string. The objective of each party is to retrieve their assigned dit through mutual collaboration. The probability of success depends on the type of collaboration employed. For instance, if the parties perform local operations on their respective subsystems and communicate classically, their success probability is generally higher than if they do not communicate at all. However, if they use quantum communication, the success probability is generally even higher. To make our scenario compatible with the process matrix framework [9], we consider a restricted collaboration scenario: at a given run, each party can receive a system in their laboratory, implement an operation on it, and send it out of the laboratory, with each step occurring only once. This condition ensures that a party cannot receive any further communication in their laboratory once they have already communicated to the other parties. We show that under such a scenario, the success probability is generally higher when the parties share causally non-separable processes compared to when they share causally separable processes.

In the bipartite scenario, we focus on a specific encoding scheme in which two-bit strings are encoded into two-qubit Bell basis states; we call this the LBH-B task. We establish a strict duality between the success probability of the LBH-B task and the success probability of the Guess-Your-Neighbour’s-Input (GYNI) game, which is studied to obtain causal inequalities [47]. We show that a protocol with quantum processes yielding nontrivial success in one can be suitably adapted to achieve similar success in the other. This duality clearly establishes the advantage of causally inseparable processes over causally separable ones in the LBH-B task. Furthermore, this duality promises an efficient approach to addressing the question of obtaining the maximum quantum violation – the Tsirelson-like bound [48] – of causal inequalities. Despite the aforementioned duality, we find that not all causally inseparable processes yielding nontrivial success in GYNI game are advantageous in LBH-B task on their own. Naturally this raises the question which class of quantum processes are useful for LBH-B task. Here, we derive a Peres-like necessary criterion to characterise such processes¹¹1Recall that the celebrated positive-partial-transposition (PPT) condition turns out to be a necessary criterion for a bipartite quantum state $\rho_{AB}\in\mathcal{D}(\mathbb{C}^{d_{A}}\otimes\mathbb{C}^{d_{B}})$ to be separable [49]. As shown by Horodecki et al. this also turns out to be the sufficient criterion for the quantum systems associated with Hilbert spaces $\mathbb{C}^{2}\otimes\mathbb{C}^{2}$ , $\mathbb{C}^{2}\otimes\mathbb{C}^{3}$ , and $\mathbb{C}^{3}\otimes\mathbb{C}^{2}$ [50].. Nevertheless, we show that any process yielding nontrivial success in the GYNI game, when combined with an appropriately chosen bipartite state, achieves nontrivial success in the LBH-B task. This demonstrates an interesting super-activation phenomenon, where two processes, each of which individually fails to achieve nontrivial success in the LBH-B task, collectively become advantageous. Subsequently, we consider the LDH task in a tripartite scenario. We show that not only quantum processes but also classical processes with causal indefiniteness offer advantages. In fact, we demonstrate that tripartite classical processes exhibiting genuine causal inseparability generally outperform bicausal quantum processes.

The rest of the paper is organized as follows. In Section II, we briefly review the framework of the process matrix as a preliminary, defining causally separable and inseparable processes. We also discuss the GYNI game and its associated causal inequality, which are violated by causally inseparable processes. In Section III, we introduce our cryptographic primitive, the local-data-hiding task. In III.1, we particularly focus on the bipartite case of local bit hiding in Bell States. We discuss in detail different possible collaboration scenarios and their associated optimal success probability. Section IV establishes the duality between the LBH-B task and the GYNI game, demonstrating the advantage of causally inseparable processes. Furthermore, in Section IV.1, we derive the Peres-like criterion necessary for bipartite processes to show nontrivial advantages in LBH-B, and discuss the super-activation phenomenon in Section IV.2. In Section V, we analyse the advantage of classical causally inseparable processes in the LDH task within a tripartite setting. Finally, we conclude in Section VI.

II Preliminaries

Recently, several approaches have been proposed to study the notion of causal indefiniteness. For instance, L. Hardy has introduced the causaloid framework [5, 6], while Chiribella, D’Ariano, and Perinotti have developed the higher-order maps framework [8] (see also [10]). On the other hand, Oreshkov, Costa, and Brukner have presented the process matrix framework [9] (see also [51] and references therein). Here, we briefly review the process matrix framework and some necessary concepts relevant for the present work.

II.1 Process Matrix Framework

This particular framework is based on the fundamental premise that physics in local laboratories is described by standard quantum theory. The most general quantum operation applied by an agent (say, $X$ ) is described by a quantum instrument $\mathcal{I}_{X}\equiv\left\{\Lambda^{k}_{X}~{}|~{}\Lambda^{k}_{X}:\mathcal{L}(\mathcal{H}_{X_{I}})\mapsto\mathcal{L}(\mathcal{H}_{X_{O}})\right\}_{k=1}^{N}$ , where $\Lambda^{k}_{X}$ ’s are completely positive (CP) maps such that $\Lambda_{X}:=\sum_{k=1}^{N}\Lambda^{k}_{X}$ is a completely positive and trace-preserving (CPTP) map, also called a channel. Here, $\mathcal{L}(\mathcal{X})$ is the space of linear operators acting on $\mathcal{X}$ , with $\mathcal{H}_{X_{I}}$ ( $\mathcal{H}_{X_{O}}$ ) corresponding to the Hilbert spaces associated with the input (output) quantum system of the instrument $\mathcal{I}_{X}$ . When the instrument is fed with an input quantum state $\rho\in\mathcal{D}(\mathcal{H}_{X_{I}})$ , it yields a classical outcome $k\in\{1,\cdots,N\}$ and the state gets updated to $\Lambda^{k}_{X}(\rho)/\operatorname{Tr}[\Lambda^{k}_{X}(\rho)]$ , where $\operatorname{Tr}[\Lambda^{k}_{X}(\rho)]$ denotes the probability of observing the $k^{\text{th}}$ outcome. The Choi-Jamiołkowsky (CJ) isomorphism provides a convenient way of representing any linear map $\Lambda^{k}_{X}$ [52, 53]:

\displaystyle M^{k}_{X_{I}X_{O}}:=\textbf{id}\otimes\Lambda^{k}_{X}(\ket{\tilde{\phi}^{+}}\bra{\tilde{\phi}^{+}})\in\mathcal{L}(\mathcal{H}_{X_{I}}\otimes\mathcal{H}_{X_{O}}),

(1)

where $\ket{\tilde{\phi}^{+}}:=\sum_{i=1}^{d_{X_{I}}}\ket{i}\ket{i}\in\mathcal{H}^{\otimes 2}_{X_{I}}$ is the unnormalized maximally entangled state, and id denotes the identity channel. Denoting CJ of $\Lambda_{X}$ to be $\mathbb{M}_{X_{I}X_{O}}$ , the complete positivity and trace preserving conditions respectively are given by


$\displaystyle M^{k}_{X_{I}X_{O}}$	$\displaystyle\geq 0,~{}\forall~{}k~{}\in~{}~{}\{1,\cdots,N\}$	(2a)
${}_{X_{O}}\mathbb{M}_{X_{I}X_{O}}$	$\displaystyle:=\frac{1}{d_{X_{O}}}\left(\operatorname{Tr}_{X_{O}}[\mathbb{M}_{X_{I}X_{O}}]\right)\otimes\mathbb{I}_{X_{O}}$
	$\displaystyle=\frac{1}{d_{X_{O}}}\mathbb{I}_{X_{I}X_{O}},$	(2b)

where $\mathbb{I}$ denotes the identity operator. The bold letter symbol $\mathbb{M}_{X_{I}X_{O}}$ is used to denote the CJ operator of the CPTP map. We denote the sets


	$\displaystyle\mathcal{M}_{A}:=\{M_{A_{I}A_{O}}\|M_{A_{I}A_{O}}\geq 0\},$		(3a)
	$\displaystyle\mathcal{M}_{B}:=\{M_{B_{I}B_{O}}\|M_{B_{I}B_{O}}\geq 0\},$		(3b)

as the set of all CJ matrices of CP maps corresponding to Alice and Bob respectively. Without assuming any background causal structure between Alice’s and Bob’s actions, the most general statistics is given by a bi-linear functional


$\displaystyle P:\mathcal{M}_{A}\times\mathcal{M}_{B}$	$\displaystyle\mapsto[0,\infty),~{}s.t.$	(4a)
$\displaystyle P(\mathbb{M}_{A_{I}A_{O}},\mathbb{M}_{B_{I}B_{O}})$	$\displaystyle=1,~{}\forall~{}\mathbb{M}_{A_{I}A_{O}},\mathbb{M}_{B_{I}B_{O}}.$	(4b)

Any such bi-linear functional can be written as

	$\displaystyle P\left(M_{A_{I}A_{O}},M_{B_{I}B_{O}}\right)$
	$\displaystyle\hskip 28.45274pt=\operatorname{Tr}\left[W_{A_{I}A_{O}B_{I}B_{O}}\left(M_{A_{I}A_{O}}\otimes M_{B_{I}B_{O}}\right)\right],$		(5)

where $W_{A_{I}A_{O}B_{I}B_{O}}\in\mbox{Herm}(\mathcal{H}_{A_{I}}\otimes\mathcal{H}_{A_{O}}\otimes\mathcal{H}_{B_{I}}\otimes\mathcal{H}_{B_{O}})$ is a Hermitian operator. The requirement (4a) ensures $W_{A_{I}A_{O}B_{I}B_{O}}$ to be a positive-on-product-test (POPT) [54, 55, 56] (see also [57, 58, 59, 60]). Furthermore, the requirement

\displaystyle\left\{\!\begin{aligned} &\operatorname{Tr}[(\rho_{A_{I}^{\prime}B_{I}^{\prime}}\otimes W)(M_{A_{I}^{\prime}A_{I}A_{O}}\otimes M_{B_{I}^{\prime}B_{I}B_{O}})]\geq 0,\\ &~{}~{}\forall~{}M_{A_{I}^{\prime}A_{I}A_{O}}\geq 0,~{}M_{B_{I}^{\prime}B_{I}B_{O}}\geq 0,~{}\rho_{A_{I}^{\prime}B_{I}^{\prime}}\geq 0\end{aligned}\right\},

(6)

ensures $W_{A_{I}A_{O}B_{I}B_{O}}$ to be a positive operator, i.e., $W_{A_{I}A_{O}B_{I}B_{O}}\geq 0$ [61]. Such a positive operator satisfying the normalization condition (4b) is called a process matrix [9]. Mathematically, the normalization condition boils down to [62]:

\displaystyle\left\{\!\begin{aligned} {}_{A_{I}A_{O}B_{I}B_{O}}W&=\frac{1}{d_{A_{I}}d_{B_{I}}}\mathbb{I}_{A_{I}A_{O}B_{I}B_{O}},\\ {}_{A_{I}A_{O}}W=_{A_{I}A_{O}B_{O}}&W,~{}~{}_{B_{I}B_{O}}W=_{B_{I}B_{O}A_{O}}W,\\ W=_{A_{O}}W+&{}_{B_{O}}W-_{A_{O}B_{O}}W\end{aligned}\right\}.

(7)

Often we will avoid the suffixes of Hilbert spaces to avoid cluttering of notation.

II.2 Causally (In)Separable Processes

The set of process matrices can be of two types: (i) causally separable and (ii) causally inseparable. For bipartite case causally separable quantum processes reads as

\displaystyle W^{Sep}:=p_{1}W^{A\prec B}+p_{2}W^{B\prec A}+p_{3}W^{B\nprec\nsucc A},

(8)

where $W^{A\prec B}~{}(W^{B\prec A})$ denotes a process where Alice (Bob) is in the causal past of Bob (Alice), $W^{B\nprec\nsucc A}$ represents a process with $A$ and $B$ being spacelike separated, and $\vec{p}=(p_{1},p_{2},p_{3})^{T}$ denotes a probability vector. Alternatively, a causally separable process can also be expressed as

\displaystyle W^{Sep}:=pW^{A\not\prec B}+(1-p)W^{B\not\prec A},

(9)

for $p\in[0,1]$ , where $W^{A\not\prec B}~{}(W^{B\not\prec A})$ denotes a process where communication from Alice (Bob) to Bob (Alice) is impossible. A causally separable process satisfies [62]

\displaystyle\left\{\!\begin{aligned} &W^{A\prec B}=_{B_{O}}W^{A\prec B},~{}_{B_{I}B_{O}}W^{A\prec B}=_{A_{O}B_{I}B_{O}}W^{A\prec B},\\ &W^{B\prec A}=_{A_{O}}W^{B\prec A},~{}_{A_{I}A_{O}}W^{B\prec A}=_{B_{O}A_{I}A_{O}}W^{B\prec A},\\ &\hskip 71.13188ptW^{B\nprec\nsucc A}=_{A_{O}B_{O}}W^{B\nprec\nsucc A}\end{aligned}\right\}.

(10)

The authors in [9], first reported an example of process that is not embedded in definite causal structure. The causal indefiniteness is established through a causal inequality, derived under three assumptions: (i) definite causal structure, (ii) free choice, (iii) closed laboratories. Violation of this inequality with the last two assumptions holding true, establishes indefiniteness of causal structure. Subsequently, a symmetric variant of the causal game - guess your neighbour’s input (GYNI) - has been studied [47], which we briefly recall below.

II.3 Guess-Your-Neighbour’s-Input

The simplest version of the game involves two distant players, Alice and Bob. Alice (Bob) tosses a random coin to generate a random bit $i_{1}~{}(i_{2})\in\{0,1\}$ . Each party aims to guess the coin state of the other party. Denoting their guesses as $a$ and $b$ respectively, the success probability reads as

\displaystyle P_{succ}^{\scalebox{0.6}{GYNI}}=\sum_{i_{1},i_{2}=0}^{1}\frac{1}{4}P(a=i_{2},b=i_{1}|i_{1},i_{2})

(11)

As it turns out for any causally separable process the success probability of GYNI is bounded by $1/2$ [47], leading to the causal inequality

\displaystyle P_{succ}^{\scalebox{0.6}{GYNI}}\leq 1/2.

(12)

Interestingly, their exist process matrices that lead to violation of this inequality, and thus establishes causal indefiniteness. An explicit such example, along with the Alice’s and Bob’s instruments are given by

\displaystyle\left\{\!\begin{aligned} &W^{Cyril}_{A_{I}A_{O}B_{I}B_{O}}=\frac{1}{4}\left[\mathbb{I}^{\otimes 4}+\frac{1}{\sqrt{2}}\left(\sigma^{3}\sigma^{3}\sigma^{3}\mathbb{I}+\sigma^{3}\mathbb{I}\sigma^{1}\sigma^{1}\right)\right],\\ &\mathcal{I}^{(0)}\equiv\left\{M^{0|0}_{X_{I}X_{O}}=0,~{}M^{1|0}_{X_{I}X_{O}}=2\ket{\phi^{+}}\bra{\phi^{+}}\right\},\\ &\mathcal{I}^{(1)}\equiv\left\{M^{0|1}_{X_{I}X_{O}}=(\ket{0}\bra{0})^{\otimes 2},~{}M^{1|1}_{X_{I}X_{O}}=(\ket{1}\bra{1})^{\otimes 2}\right\},\end{aligned}\right\},

(13)

where $X\in\{A,B\}$ , and $\sigma^{1}~{}\&~{}\sigma^{3}$ are qubit Pauli-X and Pauli-Z operators. While the strategy (13) yields a success $P_{succ}^{\scalebox{0.6}{GYNI}}=5/16(1+1/\sqrt{2})\approx 0.5335>1/2$ , numerical evidence suggests possibility of other quantum processes leading to higher success [47].

III Local Dit Hiding

In this section, we first formally define the local-dit-hiding (LDH) task (see Fig.1). The complete protocol can be divided into two phases – the hiding phase and the revealing phase.

Hiding Phase: A referee distributes an $N$ -dit message string $\mathbf{x}=x_{1}x_{2}~{}\cdots~{}x_{N}\in\{0,1,\cdots,d-1\}^{N}$ among $N$ players $\{\text{Alice}^{k}\}_{k=1}^{N}$ , each residing in separate laboratories. The distribution is done in a way that no player can reveal any information about the string $\mathbf{x}$ individually. We call this the local hiding condition. To satisfy this condition the referee may adopt the following strategy: they encode the messages into states of an $N$ -partite physical system. A referee availing quantum systems can choose states $\rho^{\mathbf{x}}_{A^{1}\cdots A^{N}}\in\mathcal{D}(\otimes_{k=1}^{N}\mathcal{H}_{A^{k}})$ , and accordingly sends the $k^{th}$ part of the state to the $k^{th}$ Alice. The hiding condition demands individual marginals to be independent of $\mathbf{x}$ , i.e.,

\displaystyle\rho^{\mathbf{x}}_{A^{k}}:=\operatorname{Tr}_{A^{1}\cdots A^{N}\setminus A^{k}}\left[\rho^{\mathbf{x}}_{A^{1}\cdots A^{N}}\right]=\rho_{A^{k}},~{}\forall~{}\mathbf{x},k,

(14)

where $\operatorname{Tr}_{A^{1}\cdots A^{N}\setminus A^{k}}(\cdot)$ denotes partial trace over all the subsystems except the $k^{th}$ one. Note that, $\rho_{A^{k}}$ in general can be different from $\rho_{A^{k^{\prime}}}$ for $k\neq k^{\prime}$ .

Refer to caption — Figure 1: Local-data-hiding task involving three parties. Referee encodes the stings ${\bf x}\equiv x_{1}x_{2}x_{3}\in\{0,\cdots,d-1\}^{\times 3}$ into tripartite quantum states $\rho^{\bf X}_{A_{1}A_{2}A_{3}}\in\mathcal{D}(\mathcal{H}_{A_{1}}\otimes\mathcal{H}_{A_{2}}\otimes\mathcal{H}_{A_{3}})$ and distributes the subsystems to the respective parties. Local marginals being independent of ${\bf x}$ ensure that none of the parties can reveal any information about ${\bf x}$ on their own. However, collaboration among themselves might be helpful to know their respective messages.

Revealing Phase: Each player aims to retrieve their respective messages through mutual collaboration, i.e., $k^{th}$ Alice wants to retrieve the dit value $x_{k}$ . Depending on the resources available, the collaboration among the parties can be of different types. Players performing any quantum operations on their respective local parts of the composite system and communicating classically with each other leads to the operational paradigm of local operation and classical communication (LOCC), which naturally appears in the resource theory of quantum entanglement [63]. On the other hand, replacing classical communication lines by quantum channels one obtains a stronger form of collaboration - local operation and quantum communication (LOQC). It is important to note here that, in the hiding phase the encoding states are not demanded to be mutually orthogonal. For non-orthogonal state encoding, even under LOQC collaboration, the players cannot know their respective messages perfectly. In other words, the perfect success demands the encoded states to be mutually orthogonal, i.e., $\operatorname{Tr}[\rho^{\mathbf{x}}\rho^{\mathbf{x}^{\prime}}]=\delta_{\mathbf{x},\mathbf{x}^{\prime}}$ . Notably, both in LOQC and in LOCC collaborations, the protocol goes in multi rounds [64]. At this stage, one may impose restriction on the rounds of communication. For instance, consider that at a given run of the task each player can receive a system in their laboratory, implement an operation on it, and send it out of the laboratory, with each step occurring only once. Thus a player cannot communicate with another player from whom they have received communication. Such a collaboration scenario is considered while developing the process matrix framework [9]. In this one-round collaboration setup, we particularly focus whether causally inseparable processes could be advantageous over causally separable processes in LDH tasks. To address this question, in the following we consider an explicit example of such a task.

III.1 Bipartite Bit-hiding in Bell States

Consider the simplest case of LDH with $d=2$ and $N=2$ . Referee encodes the strings $\mathbf{x}=x_{1}x_{2}\in\{0,1\}^{2}$ into maximally entangled basis of $\mathbb{C}^{2}\otimes\mathbb{C}^{2}$ system:

\displaystyle{\bf x}\mapsto\ket{\mathcal{B}^{\bf x}}_{AB}:=\frac{1}{\sqrt{2}}(\ket{0x_{1}}+(-1)^{x_{2}}\ket{1\bar{x}_{1}})_{AB},

(15)

where $\{\ket{0},\ket{1}\}$ represents the computational basis. Accordingly, the encoded states are distributed between Alice and Bob. Clearly the local hiding demand is satisfied,

\displaystyle\rho^{\mathbf{x}}_{A(B)}=\operatorname{Tr}_{B(A)}[\ket{\mathcal{B}^{\mathbf{x}}}_{AB}\bra{\mathcal{B}^{\mathbf{x}}}]=\mathbf{I}_{A(B)}/2,~{}\forall~{}\mathbf{x}.

(16)

Since Bell states are used for encoding, we call this task bipartite local bit hiding in Bell states (LBH-B). Later we will see how this encoding plays a crucial role to establish an intriguing result. At the revealing phase Alice and Bob have to guess the bit value $x_{1}$ and $x_{2}$ , respectively. Denoting their respective guesses ‘ $a$ ’ and ‘ $b$ ’, the success of the task reads as

\displaystyle P_{succ}^{\scalebox{0.6}{LBH-B}}=\sum_{x_{1},x_{2}=0}^{1}\frac{1}{4}P(a=x_{1},b=x_{2}|\mathcal{B}^{\bf x}_{AB}).

(17)

Since the local parts of the encoded states do not contain any information of ${\bf x}$ , without any collaboration a random guess by Alice of Bob will yield $P_{succ}^{\scalebox{0.6}{LBH-B}}=1/4$ . However, they can come up with a better strategy even without any collaboration.

Proposition 1.

Without any collaboration Alice and Bob can achieve the success $P_{succ}^{\scalebox{0.6}{LBH-B}}=1/2$ .

Proof.

Their protocol goes as follows: both the players performs $\sigma^{2}$ (i.e. Pauli-Y) measurement on their part of the encoded state received from the referee. Alice answers $a=0$ ( $a=1$ ) for ‘up’ (‘down’) outcome, while Bob answers $b=1$ ( $b=0$ ) for ‘up’ (‘down’) outcome. The claimed success probability follows from Table 1. ∎

$x_{1}x_{2}$	$\ket{\mathcal{B}^{\mathbf{x}}}$	Alice’s outcome	Bob’s outcome	$a$	$b$	Status
00	$\ket{\phi^{+}}$	up	down	0	0	success
00	$\ket{\phi^{+}}$	down	up	1	1	failure
01	$\ket{\phi^{-}}$	up	up	0	1	success
01	$\ket{\phi^{-}}$	down	down	1	0	failure
10	$\ket{\psi^{+}}$	up	up	0	1	failure
10	$\ket{\psi^{+}}$	down	down	1	0	success
11	$\ket{\psi^{-}}$	up	down	0	0	failure
11	$\ket{\psi^{-}}$	down	up	1	1	success

Table 1: Protocol for LBH-B task as discussed in Proposition 1. Success probability turns out to be

P_{succ}^{\scalebox{0.6}{LBH-B}}=1/2

Proposition 2.

Under LOCC collaboration $P_{succ}^{\scalebox{0.6}{LBH-B}}\leq 1/2$ .

Proof.

The proof follows a reductio ad absurdum argument. For that, we recall the task of local discrimination of Bell states [65]. Given one of the states, randomly chosen from two-qubit Bell basis, it is known that two distant parties cannot distinguish the state under LOCC. In fact their success probability is upper bounded by $1/2$ [66, 67]. A LOCC protocol yielding a success $P_{succ}^{\scalebox{0.6}{LBH-B}}>1/2$ will imply local distinguishability of Bell states with the same probability of success – a contradiction; and hence completes the proof. ∎

Proposition 3.

Under LOQC collaboration $P_{succ}^{\scalebox{0.6}{LBH-B}}=1$ .

Proof.

The proof is straight-forward. Alice sends her part of the encoded state to Bob through a perfect qubit channel; Bob performs a Bell basis measurement to retrieve both $x_{1}$ & $x_{2}$ , and classically communicates back $x_{1}$ to Alice. ∎

Notably, Proposition 3 holds true for any LDH task whenever the encoded states are mutually orthogonal. We will now consider the scenario of one-round communication. Within this setup, we start by establishing a bound on LBH-B success whenever the players are embedded in a definite causal structure.

Proposition 4.

In one-round collaboration scenario $P_{succ}^{\scalebox{0.6}{LBH-B}}\leq 1/2$ , whenever the players are embedded in a definite causal structure.

Proof.

(Intuitive argument) Assume that Alice is in the causal past of Bob. Thus communication from Alice to Bob is possible, but not in other direction, i.e., they can share a process of type $W^{A\prec B}_{A_{I}A_{O}B_{I}B_{O}}$ , along with the given encoded state $\mathcal{B}^{\bf x}_{AB}:=\ket{\mathcal{B}^{\bf x}}_{AB}\bra{\mathcal{B}^{\bf x}}$ . Marginal of the encoded state being independent of ${\bf x}$ , Alice cannot obtained any information about ${\bf x}$ from $W^{A\prec B}_{A_{I}A_{O}B_{I}B_{O}}\otimes\mathcal{B}^{\bf x}_{AB}$ . Therefore she can at best randomly guess the bit value of $x_{1}$ , while Bob can identify both $x_{1}$ and $x_{2}$ perfectly. Therefore success probability in this case is upper bounded by $1/2$ . Similar argument holds for the processes of types $W^{B\prec A}_{A_{I}A_{O}B_{I}B_{O}}$ , and also for $W^{B\nprec\nsucc A}_{A_{I}A_{O}B_{I}B_{O}}$ (see Proposition 1). Finally note that any causally separable process can be expressed as Eq.(40), and hence the claim follows from convexity. ∎

As we will see a more formal proof of Proposition 4 can be obtained as a consequence of one of our core results established in the next section (see Remark 1).

IV Advantage of causal inseparability in LBH-B task

Here we will show that Alice and Bob can obtain advantage in LBH-B task when they share causally inseparable processes (see Fig.2). To this aim we proceed to establish a generic connection between the success probabilities of two independent tasks - the GYNI game and the LBH-B task.

Theorem 1.

A success probability $P_{succ}^{\scalebox{0.6}{LBH-B}}=\mu$ in LBH-B task is achievable if and only if the same success is achievable in GYNI game, i.e., $P_{succ}^{\scalebox{0.6}{GYNI}}=\mu$ .

Proof.

The proof is divided into two parts:

(i)

only if part: $P_{succ}^{\scalebox{0.6}{LBH-B}}=\mu$ ensures a protocol for GYNI game yielding success probability $P_{succ}^{\scalebox{0.6}{GYNI}}=\mu$ .
(ii)

if part: $P_{succ}^{\scalebox{0.6}{GYNI}}=\mu$ ensures a protocol for LBH-B task yielding success probability $P_{succ}^{\scalebox{0.6}{LBH-B}}=\mu$ .

only if part:
Given the encoded states $\{\mathcal{B}^{\bf x}_{AB}\}$ , let the process matrix $W_{A_{I}A_{O}B_{I}B_{O}}$ yields a success $P_{succ}^{\scalebox{0.6}{LBH-B}}=\mu$ with Alice and Bob applying the quantum instruments $\mathcal{I_{A}}=\{M^{a}_{AA_{I}A_{O}}\}_{a=0}^{1}$ and $\mathcal{I_{B}}=\{M^{b}_{BB_{I}B_{O}}\}_{b=0}^{1}$ , respectively. Thus we have,

	$\displaystyle P_{succ}^{\scalebox{0.6}{LBH-B}}=\frac{1}{4}\sum_{x_{1},x_{2}=0}^{1}p(a=x_{1},b=x_{2}\|\mathcal{B}^{x_{1}x_{2}}_{AB})=\mu,~{}~{}\mbox{with},$		(18)
	$\displaystyle p(a,b\|\mathcal{B}^{x_{1}x_{2}}_{AB}):=\operatorname{Tr}[(\mathcal{B}^{x_{1}x_{2}}_{AB}\otimes W)(M^{a}_{AA_{I}A_{O}}\otimes M^{b}_{BB_{I}B_{O}})].$

For playing the $GYNI$ game, let Alice and Bob share the process Matrix $W^{\prime}=W_{A_{I}A_{O}B_{I}B_{O}}\otimes\mathcal{B}^{00}_{AB}$ . Based on their coin states $i_{1},i_{2}\in\{0,1\}$ , Alice and Bob respectively perform quantum instruments

	$\displaystyle\mathcal{I_{A}}^{(i_{1})}$	$\displaystyle:=\left\{\mathcal{Z}^{i_{1}}_{A}\left(M^{a}_{AA_{I}A_{O}}\right)\right\}_{a=0}^{1}\equiv\left\{Z^{i_{1}}_{A}M^{a}_{AA_{I}A_{O}}Z^{i_{1}}_{A}\right\}_{a=0}^{1},$
	$\displaystyle\mathcal{I_{B}}^{(i_{2})}$	$\displaystyle=\left\{\mathcal{X}^{i_{2}}_{B}\left(M^{b\|i_{2}}_{BB_{I}B_{O}}\right)\right\}_{b=0}^{1}\equiv\left\{X^{i_{2}}_{B}M^{b}_{BB_{I}B_{O}}X^{i_{2}}_{B}\right\}_{b=0}^{1},$

where $Z$ & $X$ are qubit Pauli gates and $\{M^{a}_{AA_{I}A_{O}}\}_{a=0}^{1}$ & $\{M^{b}_{BB_{I}B_{O}}\}_{b=0}^{1}$ are the instruments used in LBH-B task. The success probability of GYNI game, therefore, reads as

	$\displaystyle P_{succ}^{\scalebox{0.6}{GYNI}}=\sum_{i_{1},i_{2}=0}^{1}\frac{1}{4}p(a=i_{2},b=i_{1}\|i_{1},i_{2})$
	$\displaystyle=\frac{1}{4}\sum_{i_{1},i_{2}=0}^{1}\operatorname{Tr}\left[\left(\mathcal{B}^{00}_{AB}\otimes W\right)\left(M^{a=i_{2}\|i_{1}}_{AA_{I}A_{O}}\otimes M^{b=i_{1}\|i_{2}}_{BB_{I}B_{O}}\right)\right]$
	$\displaystyle=\frac{1}{4}\sum_{i_{1},i_{2}=0}^{1}\operatorname{Tr}\left[\left(\mathcal{B}^{00}_{AB}\otimes W\right)\left(\mathcal{Z}^{i_{1}}_{A}\left(M^{a=i_{2}}_{AA_{I}A_{O}}\right)\otimes\mathcal{X}^{i_{2}}_{B}\left(M^{b=i_{1}}_{BB_{I}B_{O}}\right)\right)\right]$
	$\displaystyle=\frac{1}{4}\sum_{i_{1},i_{2}=0}^{1}\operatorname{Tr}\left[\left(\left(\mathcal{Z}^{i_{1}}\otimes\mathcal{X}^{i_{2}}\left(\mathcal{B}^{00}\right)\right)\otimes W\right)\left(M^{a=i_{2}}_{AA_{I}A_{O}}\otimes M^{b=i_{1}}_{BB_{I}B_{O}}\right)\right]$
	$\displaystyle=\frac{1}{4}\sum_{i_{1},i_{2}=0}^{1}\operatorname{Tr}\left[\left(\mathcal{B}^{i_{2}i_{1}}_{AB}\otimes W\right)\left(M^{a=i_{2}}_{AA_{I}A_{O}}\otimes M^{b=i_{1}}_{BB_{I}B_{O}}\right)\right]$
	$\displaystyle=\mu=P_{succ}^{\scalebox{0.6}{LBH-B}},~{}~{}[\mbox{using~{}Eq}.(\ref{lbh})].$		(19)

This completes the only if part of the claim.

if part:
Given $x_{1}$ and $x_{2}$ being the respective coin states of Alice and Bob, let the process matrix $W^{\prime}_{A_{I}A_{O}B_{I}B_{O}}$ yields a success $P_{succ}^{\scalebox{0.6}{GYNI}}=\mu$ , with Alice and Bob performing quantum instruments $\mathcal{I_{A}}^{(x_{1})}=\{M^{a|x_{1}}_{A_{I}A_{O}}\}_{a=0}^{1}$ and $\mathcal{I_{B}}^{(x_{2})}=\{M^{b|x_{2}}_{B_{I}B_{O}}\}_{b=0}^{1}$ , respectively. Thus we have,

	$\displaystyle P_{succ}^{\scalebox{0.6}{GYNI}}=\frac{1}{4}\sum_{x_{1},x_{2}=0}^{1}p(a=x_{2},b=x_{1}\|x_{1},x_{2})=\mu,~{}~{}\mbox{with},$		(20)
	$\displaystyle p(a,b\|x_{1},x_{2}):=\operatorname{Tr}\left[\left(M^{a\|x_{1}}_{A_{I}A_{O}}\otimes M^{b\|x_{2}}_{B_{I}B_{O}}\right)W^{\prime}\right].$

To perform the LBH-B task, Alice and Bob share the Process Matrix $W^{\prime}_{A_{I}A_{O}B_{I}B_{O}}\otimes\mathcal{B}^{00}_{A^{\prime}B^{\prime}}$ . Now, given the encoded state $\mathcal{B}^{x_{1}x_{2}}_{AB}$ , Alice and Bob apply the following unitary operation on parts of their local systems

\displaystyle U_{AA^{\prime}}=U_{BB^{\prime}}=\frac{1}{\sqrt{2}}\begin{pmatrix}~{}~{}1&~{}~{}0&~{}~{}0&~{}~{}1\\ ~{}~{}0&~{}~{}1&~{}~{}1&~{}~{}0\\ ~{}~{}1&~{}~{}0&~{}~{}0&-1\\ ~{}~{}0&~{}~{}1&-1&~{}~{}0\\ \end{pmatrix},

(21)

which results in

	$\displaystyle\mathcal{U}_{AA^{\prime}}\otimes\mathcal{U}_{BB^{\prime}}\left(W^{\prime}_{A_{I}A_{O}B_{I}B_{O}}\otimes\mathcal{B}^{00}_{A^{\prime}B^{\prime}}\otimes\mathcal{B}^{x_{1}x_{2}}_{AB}\right)$
	$\displaystyle\hskip 28.45274pt=W^{\prime}_{A_{I}A_{O}B_{I}B_{O}}\otimes\mathcal{B}^{x_{1}0}_{A^{\prime}B^{\prime}}\otimes\mathcal{B}^{x_{2}x_{1}}_{AB},$		(22)

where $\mathcal{U}(\rho):=U\rho U^{\dagger}$ . On $A$ and $A^{\prime}$ sub-parts of the evolved process Alice performs computational basis measurement (i.e. the Pauli- $\sigma^{3}$ measurement), resulting into outcomes $u,u^{\prime}\in\{0,1\}$ , where $0~{}(1)$ corresponds to ‘up’ (‘down’) outcome. Bob also does the same resulting into outcomes $v,v^{\prime}\in\{0,1\}$ . Clearly, due to correlation of the state, we have

\displaystyle u\oplus v=x_{2},~{}~{}\&~{}~{}u^{\prime}\oplus v^{\prime}=x_{1}.

(23)

Therefore, guessing the value of $v^{\prime}$ and $u$ respectively by Alice and Bob with the probability $\mu$ will ensure the same success in LBH-B task. At this point the process $W^{\prime}_{A_{I}A_{O}B_{I}B_{O}}$ proves to be helpful in this task, which can be accordingly chosen looking into its advantage in GYNI game. Rest of the protocol mimics GYNI strategy with $u$ and $v^{\prime}$ being the inputs of Alice and Bob, respectively. Denoting $a^{\prime}$ and $b^{\prime}$ as the output of the GYNI strategy the final guess in LBH-B task by Alice and Bob are respectively,

\displaystyle a=a^{\prime}\oplus u^{\prime},~{}~{}\&~{}~{}b=b^{\prime}\oplus v.

(24)

On the composite process $W^{\prime}_{A_{I}A_{O}B_{I}B_{O}}\otimes\mathcal{B}^{00}_{A^{\prime}B^{\prime}}\otimes\mathcal{B}^{x_{1}x_{2}}_{AB}$ , the effective instruments $\{M^{a}_{AA^{\prime}A_{I}A_{O}}\}_{a=0}^{1}$ for Alice and $\{M^{b}_{BB^{\prime}B_{I}B_{O}}\}_{b=0}^{1}$ for Bob are respectively given by

	$\displaystyle\sum_{u,u^{\prime},a^{\prime}=0}^{1}\delta_{a,a^{\prime}\oplus u^{\prime}}\mathcal{U}_{AA^{\prime}}\otimes\textbf{id}_{A_{I}A_{O}}(\ket{uu^{\prime}}_{AA^{\prime}}\bra{uu^{\prime}}\otimes M^{a^{\prime}\|u}_{A_{I}A_{O}}),$
	$\displaystyle\sum_{v,v^{\prime},b^{\prime}=0}^{1}\delta_{b,b^{\prime}\oplus v}\mathcal{U}_{BB^{\prime}}\otimes\textbf{id}_{B_{I}B_{O}}(\ket{vv^{\prime}}_{BB^{\prime}}\bra{vv^{\prime}}\otimes M^{b^{\prime}\|v^{\prime}}_{B_{I}B_{O}}),$

where, $\{M^{a^{\prime}|u}_{A_{I}A_{O}}\}$ and $\{M^{b^{\prime}|v^{\prime}}_{B_{I}B_{O}}\}$ are the instruments used by Alice and Bob in GYNI game. The success probability of LBH-B task with the aforesaid protocol turns out to be

	$\displaystyle P_{succ}^{\scalebox{0.6}{LBH-B}}=\sum_{x_{1},x_{2}=0}^{1}\frac{1}{4}P(a=x_{1},b=x_{2}\|\mathcal{B}^{x_{1}x_{2}}_{AB})$
	$\displaystyle=\sum_{x_{1},x_{2}=0}^{1}\frac{1}{4}\operatorname{Tr}\left[\left(W^{\prime}\otimes\mathcal{B}^{00}\otimes\mathcal{B}^{x_{1}x_{2}}\right)\left(M^{a=x_{1}}_{AA^{\prime}A_{I}A_{O}}\otimes M^{b=x_{2}}_{BB^{\prime}B_{I}B_{O}}\right)\right]$
	$\displaystyle=\sum_{\begin{subarray}{c}x_{1},x_{2},u,u^{\prime},\\ a^{\prime},v,v^{\prime},b^{\prime}=0\end{subarray}}^{1}\frac{1}{4}\delta_{a=x_{1},a^{\prime}\oplus u^{\prime}}\delta_{b=x_{2},b^{\prime}\oplus v}\operatorname{Tr}\left[\left(W^{\prime}\otimes\mathcal{B}^{x_{1}0}_{A^{\prime}B^{\prime}}\otimes\mathcal{B}^{x_{2}x_{1}}_{AB}\right)\right.$
	$\displaystyle\left.\hskip 56.9055pt\left(\ket{uvu^{\prime}v^{\prime}}_{ABA^{\prime}B^{\prime}}\bra{uvu^{\prime}v^{\prime}}\otimes M^{a^{\prime}\|u}_{A_{I}A_{O}}\otimes M^{b^{\prime}\|v^{\prime}}_{B_{I}B_{O}}\right)\right]$
	$\displaystyle=\sum_{\begin{subarray}{c}x_{1},x_{2},u,u^{\prime},\\ a^{\prime},v,v^{\prime},b^{\prime}=0\end{subarray}}^{1}\frac{1}{16}\delta_{x_{1},a^{\prime}\oplus u^{\prime}}\delta_{x_{2},b^{\prime}\oplus v}\delta_{x_{2},u\oplus v}\delta_{x_{1},u^{\prime}\oplus v^{\prime}}$
	$\displaystyle\hskip 56.9055pt\operatorname{Tr}\left[(W^{\prime}\left(M^{a^{\prime}\|u}_{A_{I}A_{O}}\otimes M^{b^{\prime}\|v^{\prime}}_{B_{I}B_{O}}\right)\right]$
	$\displaystyle=\sum_{\begin{subarray}{c}x_{1},x_{2},u,u^{\prime},\\ a^{\prime},v,v^{\prime},b^{\prime}=0\end{subarray}}^{1}\frac{1}{16}\delta_{x_{1},a^{\prime}\oplus u^{\prime}}\delta_{x_{2},b^{\prime}\oplus v}\delta_{x_{2},u\oplus v}\delta_{x_{1},u^{\prime}\oplus v^{\prime}}P(a^{\prime},b^{\prime}\|u,v^{\prime})$
	$\displaystyle=\sum_{\begin{subarray}{c}x_{1},x_{2},u,u^{\prime},\\ v,v^{\prime}=0\end{subarray}}^{1}\frac{1}{16}\delta_{x_{2},u\oplus v}\delta_{x_{1},u^{\prime}\oplus v^{\prime}}P(x_{1}\oplus u^{\prime},x_{2}\oplus v\|u,v^{\prime})$
	$\displaystyle=\sum_{x_{1},x_{2},u,v^{\prime}=0}^{1}\frac{1}{16}P(x_{1}\oplus v^{\prime}\oplus x_{1},x_{2}\oplus u\oplus x_{2}\|u,v^{\prime})$
	$\displaystyle=\sum_{x_{1},x_{2}}\frac{1}{4}\sum_{u,v^{\prime}}\frac{1}{4}P(v^{\prime},u\|u,v^{\prime})=\sum_{x_{1},x_{2}}\frac{1}{4}\mu=\mu~{}\left[\mbox{using eq.}(\ref{gyni})\right].$		(25)

This completes the if part of the claim, and hence the Theorem is proved. ∎

Remark 1.

Theorem 1 provides a necessary condition on process matrices to be useful in LBH-B task (i.e. providing a success $>1/2$ ). The success probability of GYNI game with a causally separable process is known to be upper bounded by $1/2$ [47]. Therefore success of LBH-B task with causally separable process will also be upper bounded by $1/2$ . Otherwise, according to Theorem 1, a causally separable process $W^{sep}$ yielding $P_{succ}^{\scalebox{0.6}{LBH-B}}>1/2$ will imply that the causally separable process $W^{sep}\otimes\phi^{+}$ will result in $P_{succ}^{\scalebox{0.6}{GYNI}}>1/2$ , contradicting the result of [47]. This formally proofs Proposition 4.

Tsirelson bound for causal inequalities: Theorem 1 also implies that the optimal success probabilities of the GYNI game and the LHB-B task are same when Alice and Bob share valid processes. While a recent result by Liu & Chiribella provides an algorithm to obtain a nontrivial upper bound for the GYNI game, however it is not yet known whether the bound obtained by them is tight or not [48]. Our LBH-B task could play a crucial role here, as finding the optimal success in LBH-B task will derive a Tsirelson kind of bound for the GYNI game. At this point, it is important to note the crucial difference between GYNI game and LBH-B task. Finding the optimal success in GYNI game involves optimization over the set of bipartite processes shared by Alice and Bob, and over the pairs of two-outcome instruments for each player. On the other hand, in the LBH-B task the optimization involves a single two-outcome instrument for each player along with the set of bipartite processes. This seemingly reduces the complexity of the optimization problem, indicating an alternative efficient approach to find the Tsirelson bound for causal inequalities. We believe this could be a promising direction for further exploration. By considering higher input GYNI game and the corresponding generalization of LBH-B task, a generalization of Theorem 1 can be obtained. For the sake of completeness, we discuss the proof in Appendix A. In the following subsection we proceed to obtain another necessary condition on the process matrices to be useful in LBH-B task.

IV.1 Usefulness of Quantum Processes: A Peres-like Criterion

Our Theorem 1 establishes that any bipartite process yielding non-trivial success in the GYNI game, when combined with an additional maximally entangled state, achieves similar success in the LBH-B task. It is now natural to ask whether all such causally inseparable processes are advantageous in the LBH-B task on their own. Interestingly, we answer this question in negative. In the following we show that a given process matrix needs to be of some particular form to be useful in LBH-B task.

Theorem 2.

A bipartite process matrix, $W_{A_{I}A_{O}B_{I}B_{O}}$ yielding a nontrivial success in the LBH-B task (i.e., $P_{succ}^{\scalebox{0.6}{LBH-B}}>1/2$ ) must have negative partial transposition across $(A_{I}A_{O})|(B_{I}B_{O})$ bipartition.

Proof.

Consider Alice and Bob share a bipartite process $W_{A_{I}A_{O}B_{I}B_{O}}$ for the LBH-B task. Given the encoded state $\mathcal{B}^{\textbf{x}}_{AB}:=\ket{\mathcal{B}^{\textbf{x}}}_{AB}\bra{\mathcal{B}^{\textbf{x}}}$ , Alice and Bob respectively implement two-outcome instruments $\{\mathcal{N}^{a}_{AA_{I}\to A_{O}}\}_{a=0}^{1}$ and $\{\mathcal{M}^{b}_{BB_{I}\to B_{O}}\}_{b=0}^{1}$ on their respective shares of the joint process $\mathcal{B}^{\textbf{x}}_{AB}\otimes W_{A_{I}A_{O}B_{I}B_{O}}$ , and give the classical outcomes $a$ and $b$ as their respective guesses for $x_{1}$ and $x_{2}$ . The success probability is then given by


	$\displaystyle P_{succ}^{\scalebox{0.6}{LBH-B}}=\frac{1}{4}\sum_{x_{1},x_{2}=0}^{1}p(a=x_{1},b=x_{2}\|\mathcal{B}^{x_{1}x_{2}}_{AB})$		(26a)
	$\displaystyle=\operatorname{Tr}\left[(\mathcal{B}^{\textbf{x}}_{AB}\otimes W_{A_{I}A_{O}B_{I}B_{O}})\left\{\textbf{id}_{AA_{I}}\otimes\mathcal{N}^{a=x_{1}}_{A^{\prime}A^{\prime}_{I}\to A_{O}}\otimes\right.\right.$
	$\displaystyle\left.\left.\textbf{id}_{BB_{I}}\otimes\mathcal{M}^{b=x_{2}}_{B^{\prime}B^{\prime}_{I}\to B_{O}}(\tilde{\phi}^{+}_{AA_{I}A^{\prime}A^{\prime}_{I}}\otimes\tilde{\phi}^{+}_{BB_{I}B^{\prime}B^{\prime}_{I}})\right\}\right].$		(26b)

Here, $p(a,b|\mathcal{B}^{\textbf{x}}_{AB}):=p(a=x_{1},b=x_{2}|\mathcal{B}^{x_{1}x_{2}}_{AB})$ is the probability of getting outcomes $a=x_{1}$ and $b=x_{2}$ , with $\tilde{\phi}^{+}:=\ket{\tilde{\phi}^{+}}\bra{\tilde{\phi}^{+}}$ . Accordingly we have

	$\displaystyle p(a,b\|\mathcal{B}^{\textbf{x}}_{AB})=\operatorname{Tr}\left[\left\{\textbf{id}_{ABA_{I}B_{I}}\otimes\mathcal{N}^{()a}_{A_{O}\to A^{\prime}A^{\prime}_{I}}\otimes\mathcal{M}^{()b}_{B_{O}\to B^{\prime}B^{\prime}_{I}}\right.\right.$
	$\displaystyle\left.\left.\left(\mathcal{B}^{\textbf{x}}_{AB}\otimes W_{A_{I}A_{O}B_{I}B_{O}}\right)\right\}(\tilde{\phi}^{+}_{AA_{I}A^{\prime}A^{\prime}_{I}}\otimes\tilde{\phi}^{+}_{BB_{I}B^{\prime}B^{\prime}_{I}})\right],$		(27)

where $\mathcal{N}^{(*)}$ and $\mathcal{M}^{(*)}$ are the dual maps²²2Recall that, a map $\Lambda^{(*)}_{D\to C}$ is called dual to $\Lambda_{C\to D}$ if $\operatorname{Tr}_{D}[\rho_{D}\{\Lambda_{C\to D}(\sigma_{C})\}]=\operatorname{Tr}_{C}[\{\Lambda^{(*)}_{D\to C}(\rho_{D})\}\sigma_{C}]$ for all $\rho_{D}\in\mathcal{D}(\mathcal{H}_{D})~{}\&~{}\sigma_{C}\in\mathcal{D}(\mathcal{H}_{C})$ . of $\mathcal{N}$ and $\mathcal{M}$ , respectively. Defining, $\tilde{W}^{a,b}_{A_{I}A^{\prime}A^{\prime}_{I}B_{I}B^{\prime}B^{\prime}_{I}}:=\textbf{id}_{A_{I}B_{I}}\otimes\mathcal{N}^{(*)a}_{A_{O}\to A^{\prime}A^{\prime}_{I}}\otimes\mathcal{M}^{(*)b}_{B_{O}\to B^{\prime}B^{\prime}_{I}}(W_{A_{I}A_{O}B_{I}B_{O}})$ , we obtain


	$\displaystyle p(a,b\|\mathcal{B}^{\textbf{x}}_{AB})=\operatorname{Tr}\left[(\mathcal{B}^{\textbf{x}}_{AB}\otimes\tilde{W}^{a,b})(\tilde{\phi}^{+}_{AA^{\prime}}\otimes\tilde{\phi}^{+}_{A_{I}A^{\prime}_{I}}\otimes\tilde{\phi}^{+}_{BB^{\prime}}\otimes\tilde{\phi}^{+}_{B_{I}B^{\prime}_{I}})\right]$
	$\displaystyle=\operatorname{Tr}_{A,A^{\prime},B,B^{\prime}}\left[(\mathcal{B}^{\textbf{x}}_{AB}\otimes\chi^{a,b}_{A^{\prime}B^{\prime}})(\tilde{\phi}^{+}_{AA^{\prime}}\otimes\tilde{\phi}^{+}_{BB^{\prime}})\right]$
	$\displaystyle=\operatorname{Tr}_{A,B}\left[\mathcal{B}^{\textbf{x}}_{AB}\Pi^{a,b}_{AB}\right],$		(28a)
	$\displaystyle\Pi^{a,b}_{AB}:=\operatorname{Tr}_{A^{\prime},B^{\prime}}[(\mathbb{I}_{AB}\otimes\chi^{a,b}_{A^{\prime}B^{\prime}})(\tilde{\phi}^{+}_{AA^{\prime}}\otimes\tilde{\phi}^{+}_{BB^{\prime}})],$		(28b)
	$\displaystyle\chi^{a,b}_{A^{\prime}B^{\prime}}:=\operatorname{Tr}_{A_{I},A^{\prime}_{I},B_{I},B^{\prime}_{I}}\left[\tilde{W}^{a,b}(\mathbb{I}_{A^{\prime}B^{\prime}}\otimes\tilde{\phi}^{+}_{A_{I}A^{\prime}_{I}}\otimes\tilde{\phi}^{+}_{B_{I}B^{\prime}_{I}})\right]$		(28c)

Eq.(28a) tells that, any protocol followed by Alice and Bob on the joint process $\mathcal{B}^{\textbf{x}}_{AB}\otimes W_{A_{I}A_{O}B_{I}B_{O}}$ boils down to performing a POVM $\{\Pi^{a,b}_{AB}\}_{a,b=0}^{1}$ on $\mathcal{B}^{\textbf{x}}_{AB}$ . Consider now, $W_{A_{I}A_{O}B_{I}B_{O}}$ is a PPT (positive partial transpose) operator across $A_{I}A_{O}|B_{I}B_{o}$ bipartition, i.e., $W^{\mathrm{T}_{B_{I}B_{O}}}_{A_{I}A_{O}B_{I}B_{O}}\geq 0$ . Furthermore, $\mathcal{N}^{a}$ and $\mathcal{M}^{b}$ being local CP maps on Alice’s and Bob’s parts, their corresponding dual maps $\mathcal{N}^{(\star)a}$ and $\mathcal{M}^{(\star)b}$ are also CP on the respective parts. This ensures

\displaystyle\left(\tilde{W}^{a,b}_{A_{I}A^{\prime}A^{\prime}_{I}B_{I}B^{\prime}B^{\prime}_{I}}\right)^{\mathrm{T}_{B_{I}B^{\prime}B^{\prime}_{I}}}\geq 0,~{}\forall~{}a,b\in\{0,1\}.

(29)

Eq.(28c) and Eq.(29) together imply $\chi^{a,b}_{A^{\prime}B^{\prime}}\in\mbox{PPT}(\mathbb{C}^{2}_{A^{\prime}}\otimes\mathbb{C}^{2}_{B^{\prime}})$ , and in-fact due to Peres-Horodecki criteria [49, 50], $\chi^{a,b}_{A^{\prime}B^{\prime}}\in\mbox{Sep}(\mathbb{C}^{2}_{A^{\prime}}\otimes\mathbb{C}^{2}_{B^{\prime}})$ . Now, Eq.(28b) ensures $\Pi^{a,b}_{AB}\in\mbox{Sep}(\mathbb{C}^{2}_{A}\otimes\mathbb{C}^{2}_{B})$ . Therefore, Eqs.(26a) & (28a) imply

\displaystyle P_{succ}^{\scalebox{0.6}{LBH-B}}=\frac{1}{4}\sum_{x_{1},x_{2}}\operatorname{Tr}_{A,B}\left[\mathcal{B}^{x_{1}x_{2}}_{AB}\Pi^{a=x_{1},b=x_{2}}_{AB}\right],

(30)

which can be thought of as the success probability of distinguishing two-qubit Bell Basis $\{\ket{\phi^{\pm}},\ket{\psi^{\pm}}\}$ under separable measurement $\{\Pi^{a,b}\}_{ab}$ . Recalling the result from [67], we know this success probability is upper bounded by $1/2$ . This concludes the proof. ∎

Note that, Theorem 2 provides only a necessary criterion on bipartite processes to be useful in LBH-B task. The fact that it is not a sufficient criterion can be seen from the example of the no-signaling process $W_{A_{I}A_{O}B_{I}B_{O}}:=\ket{\phi^{+}}_{A_{I}B_{I}}\bra{\phi^{+}}\otimes\mathbb{I}_{A_{O}B_{O}}$ . This particular process is NPT (negative-partial-transpose) across $A_{I}A_{O}|B_{I}B_{O}$ bipartition, but according to Proposition 4 it does not provide a nontrivial success in LBH-B task.

Theorem 2 also indicates that not all causally inseparable processes are advantageous in LBH-B task. For instance, consider the process $W^{Cyril}_{A_{I}A_{O}B_{I}B_{O}}$ in Eq. (20), which is known to be advantageous in GYNI task. However, it turns out that $(W^{Cyril}_{A_{I}A_{O}B_{I}B_{O}})^{\mathrm{T}_{B_{I}B_{O}}}\geq 0$ , implying $W^{Cyril}_{A_{I}A_{O}B_{I}B_{O}}$ to be PPT across $A_{I}A_{O}|B_{I}B_{O}$ bipartition. Moreover, it admits a fully separable decomposition across $A_{I}|A_{O}|B_{I}|B_{O}$ partition³³3A positive operator $\mathcal{O}_{YZ}\in\mathcal{L}_{+}(\mathcal{H}_{Y}\otimes\mathcal{H}_{Z})$ is called separable across $Y|Z$ cut if it allows a decomposition of the form $\mathcal{O}_{YZ}=\sum_{i}\mathcal{O}^{i}_{Y}\otimes\mathcal{O}^{i}_{Z}$ , where $\forall~{}i,~{}\mathcal{O}^{i}_{Y}\in\mathcal{L}_{+}(\mathcal{H}_{Y})~{}\&~{}\mathcal{O}^{i}_{Z}\in\mathcal{L}_{+}(\mathcal{H}_{Z})$ .:

	$\displaystyle W^{Cyril}_{A_{I}A_{O}B_{I}B_{O}}=\frac{1}{4}\left[\mathbb{I}^{\otimes 4}+\frac{1}{\sqrt{2}}\left(\sigma^{3}\sigma^{3}\sigma^{3}\mathbb{I}+\sigma^{3}\mathbb{I}\sigma^{1}\sigma^{1}\right)\right]_{A_{I}A_{O}B_{I}B_{O}}$
	$\displaystyle~{}~{}~{}~{}~{}=\frac{1}{2}\left[P^{z}_{+}P^{z}_{+}P^{\alpha}_{+}P^{x}_{+}+P^{z}_{-}P^{z}_{+}P^{\alpha}_{-}P^{x}_{+}+P^{z}_{+}P^{z}_{+}P^{\beta}_{+}P^{x}_{-}\right.$
	$\displaystyle\hskip 28.45274pt+P^{z}_{-}P^{z}_{+}P^{\beta}_{-}P^{x}_{-}+P^{z}_{+}P^{z}_{-}P^{\beta}_{-}P^{x}_{+}+P^{z}_{-}P^{z}_{-}P^{\beta}_{+}P^{x}_{+}$
	$\displaystyle\left.\hskip 28.45274pt+P^{z}_{+}P^{z}_{-}P^{\alpha}_{-}P^{x}_{-}+P^{z}_{-}P^{z}_{-}P^{\alpha}_{+}P^{x}_{-}\right]_{A_{I}A_{O}B_{I}B_{O}}~{};$		(31)
	$\displaystyle\mbox{where},~{}~{}P^{z}_{\pm}:=\frac{1}{2}\left(\mathbb{I}\pm\sigma^{3}\right),~{}~{}~{}P^{\alpha}_{\pm}:=\frac{1}{2}\left(\mathbb{I}\pm\frac{1}{\sqrt{2}}\left(\sigma^{3}+\sigma^{1}\right)\right),~{}$
	$\displaystyle\hskip 27.03003ptP^{x}_{\pm}:=\frac{1}{2}\left(\mathbb{I}\pm\sigma^{1}\right),~{}~{}~{}P^{\beta}_{\pm}:=\frac{1}{2}\left(\mathbb{I}\pm\frac{1}{\sqrt{2}}\left(\sigma^{3}-\sigma^{1}\right)\right).$

While Proposition 4 excludes all causally separable bipartite processes to be useful in LBH-B task, Theorem 2 excludes processes that are PPT in Alice vs Bob bipartition. In fact, a larger class of bipartite processes can be excluded for the task in question.

Corollary 1.

Any bipartite process matrix $W$ will not provide a nontrivial success in LBH-B task if it can be obtained through probabilistic mixture of two other processes $W^{\prime}~{}\&~{}W^{\prime\prime}$ , where $W^{\prime}$ is causally separable and $W^{\prime\prime}$ is PPT in Alice vs Bob bipartition.

Proof simply follows from Proposition 4 and Theorem 2 due to the fact that success probability is a linear function of processes. A pictorial depiction of this corollary is shown in Fig.3. We now proceed to present an intriguing super-activation phenomenon involving process matrices.

IV.2 Super-Activation Phenomenon

Super-activation and super-additivity represent two intriguing phenomena observed while processing information with quantum systems. These denote the phenomena where collective utility of multiple quantum resources exceeds the sum of their individual contributions, and formally can be defined as follows.

Definition 1.

Given two quantum resources $\mathcal{R}_{1}$ and $\mathcal{R}_{2}$ , super-additivity generally implies $\mathcal{K}(\mathcal{R}_{1}\otimes\mathcal{R}_{2})>\mathcal{K}(\mathcal{R}_{1})+\mathcal{K}(\mathcal{R}_{2})$ , where $\mathcal{K}$ denotes some utility function from the set of resources ( $\mathbf{R}$ ) to real numbers, i.e., $\mathcal{K}:\mathbf{R}\mapsto\mathbb{R}_{\geq}$ .

Definition 2.

Super-activation is a kind of strong version of super-additivity phenomenon, where $\mathcal{K}(\mathcal{R}_{1}\otimes\mathcal{R}_{2})>0$ , even when $\mathcal{K}(\mathcal{R}_{1})=0=\mathcal{K}(\mathcal{R}_{2})$ .

For instance, two quantum channels (the resources), each individually possessing zero quantum capacity (the utility function), can collectively facilitate reliable transmission of quantum information at a nonzero rate when used in combination [68] (see also [69, 70, 71]). Similar observations extend to other utility functions of quantum channels, such as the private capacity [72] and the zero-error quantum capacity [73]. Turning attention to the inherent characteristics of quantum states as static resources, it has been demonstrated that all bipartite quantum states with NPT become distillable through the utilization of PPT entangled states which are known to be the bound entangled states [74]. The presence of bound NPT states, a longstanding open question in entanglement theory [75], would substantiate the phenomenon of super-activation for distillable entanglement. The seminal quantum superdense coding protocol [14] can also be interpreted as a super-additivity phenomenon. In this case, the static resource of quantum entanglement, which, in isolation, is ineffective for communication, effectively doubles the classical communication power of a quantum channel.

Naturally one can ask the question: is there an analogous feature of super-activation for causal indefiniteness, if it exists at all? At this point one can consider different kind of utility functions, such as the amount of violation of some causal inequality. Here we consider the success probability of LBH-B task as the utility function. Accordingly, the question of super-activation for causal indefiniteness can be formulated as follows:

Q.

Can there be two quantum processes $W$ and $W^{\prime}$ , neither providing any advantage in LBH-B task, yields a nontrivial success while their composition $W\otimes W^{\prime}$ is considered?

Before addressing this question, a careful analysis is required to determine whether the composite object $W\otimes W^{\prime}$ represents a bona fide quantum process matrix. As pointed out by Jia & Sakharwade [76], $W\otimes W^{\prime}$ violates the normalization condition of probabilities, leading to paradoxes when both $W$ and $W^{\prime}$ are causally indefinite processes (see also [77]). However, the composition represents a valid quantum process when one of them is a no-signaling process, namely a bipartite quantum state, and the other is any general quantum process. In fact, the existence of such a composition is required to prove the positivity of a generic quantum process matrix [9]. Particularly, Eq. (4a) ensures $W_{A_{I}A_{O}B_{I}B_{O}}\in\mathcal{L}(\mathcal{H}_{A_{I}}\otimes\mathcal{H}_{A_{O}}\otimes\mathcal{H}_{B_{I}}\otimes\mathcal{H}_{B_{O}})$ , to be to POPT, whereas its positivity in ensured in Eq. (6), demanding existence of the composite process $W_{A_{I}A_{O}B_{I}B_{O}}\otimes\rho_{AB}$ . Thus, the question of super-activation of causal indefiniteness still makes sense, and we provide an affirmative answer to the question ${\bf Q}$ .

SA.

Consider $W=W^{Cyril}_{A_{I}A_{O}B_{I}B_{O}}$ and $W^{\prime}=\phi^{+}_{AB}$ . While Proposition 4 bounds success probability of LBH-B task for $W^{\prime}$ to be $1/2$ , Theorem 2 imposes the same bound for $W$ . In both cases, the success $1/2$ can be achieved simply by following the protocol stated in Proposition 1. On the other hand, using the protocol stated in Eq.(13), a success $5/16(1+1/\sqrt{2})>1/2$ can be achieved in GYNI game with the process $W$ . Therefore, following the protocol discussed in the ‘if part’ proof of Theorem 1, we can obtain the success $5/16(1+1/\sqrt{2})$ in the LBH-B task with the composite process $W\otimes W^{\prime}$ . This establishes the super-activation of causal indefiniteness, answering ${\bf Q}$ affirmatively.

Notably, the pair $(W,W^{\prime})\equiv(W^{Cyril},\phi^{+})$ is not the only instance of process-pair exhibiting such super-activation phenomenon – here $W^{Cyril}$ can be replaced by any process $W_{A_{I}A_{O}B_{I}B_{O}}\in\mbox{ConvHull}(\textbf{W}_{CS},\textbf{W}_{PPT})$ and yielding nontrivial advantage in GYNI game. An interesting question is which other no-signaling processes (i.e, bipartite quantum states) can be used as $W^{\prime}_{AB}$ to activate causal indefiniteness of such $W_{A_{I}A_{O}B_{I}B_{O}}$ ’s. A partial answer follows from Theorem 2. Any PPT state $\rho^{PPT}_{AB}$ cannot be used for the purpose as $W_{A_{I}A_{O}B_{I}B_{O}}\otimes\rho^{PPT}_{AB}$ is PPT across $AA_{I}A_{O}|BB_{I}B_{O}$ whenever $W_{A_{I}A_{O}B_{I}B_{O}}$ is PPT across $A_{I}A_{O}|B_{I}B_{O}$ . In general, it would be interesting to see which NPT states will be useful for this purpose. Particularly, finding the range of parameter $\lambda$ in $W^{\prime}=\lambda\phi^{+}+(1-\lambda)\mathbb{I}/4$ that activate $W^{Cyril}$ in LBH-B task seems to be a simpler question to address.

At this point, the results in [78, 79] are worth mentioning. In [78], the authors introduce the notion of causal and causally separable quantum processes. While the causal processes never violates any causal inequality, the causally separable processes allow a canonical decomposition (see Theorem 2.2 in [78]). In a sense, they have analogy with the notions of Bell-local and separable (non-entangled) quantum states. The authors also provide example of a tripartite quantum process that is causal but not causally separable. They also show example of tripartite causally separable processes that become non-causal when extended by supplying the parties with entangled ancillas. This exhibits a kind of ‘causal activation’ phenomenon. In [79], the authors provide example of bipartite causally nonseparable processes that allow causal model, and they also show evidence of ‘causal activation’ phenomenon where combination of two causal process becomes non-causal.

IV.3 Bell-states Encoding & Super-Activation

In this subsection we analyse the critical role of Bell basis used by the referee in LBH-B task during the encoding step. In particular we will show that this encoding is crucial to establish the super-activation phenomenon. To see that here we present the same LBH task but with a different encoding strategy. Referee encodes the string ${\bf x}\equiv x_{1}x_{2}\in\{0,1\}^{\times 2}$ into the states

	$\displaystyle{\bf x}\mapsto\rho^{\bf x}_{\bf AB}=\rho^{\bf x}_{A_{1}A_{2}B_{1}B_{2}}:=\rho^{x_{1}}_{A_{1}B_{1}}\otimes\rho^{x_{2}}_{A_{2}B_{2}},$		(32)
	$\displaystyle\mbox{where},~{}~{}\rho^{0}:=\frac{1}{2}(\ket{00}\bra{00}+\ket{11}\bra{11}),$
	$\displaystyle\hskip 32.72049pt\rho^{1}:=\frac{1}{2}(\ket{01}\bra{01}+\ket{10}\bra{10}),$

and distributes the $A_{1}$ & $A_{2}$ subsystems to Alice and $B_{1}$ & $B_{2}$ subsystems to Bob. The local hiding condition is satisfied as

\displaystyle\rho^{\bf x}_{{\bf A}({\bf B})}:=\operatorname{Tr}_{{\bf B}({\bf A})}\left[\rho^{\bf x}_{\bf AB}\right]=(\mathbb{I}/2)^{\otimes 2},~{}\forall~{}{\bf x}.

(33)

Since separable states are used for encoding we call it the LBH-Sep task. We now proceed to see the status of Propositions (1)-(4) in this case.

Proposition 1^′.

Without any collaboration Alice and Bob can achieve the success $P_{succ}^{\scalebox{0.6}{LBH-Sep}}=1/4$ .

Proof.

The success probability for Alice in guessing $x_{1}$ can be $1/2$ only. Similar is the case for Bob while guessing $x_{2}$ . Since $x_{1}$ and $x_{2}$ are respectively encoded in the states of $A_{1}B_{1}$ and $A_{2}B_{2}$ systems independently, the success probability that both are guessing correctly is $1/4$ . ∎

Proposition 2^′.

Under LOCC collaboration $P_{succ}^{\scalebox{0.6}{LBH-Sep}}=1$ .

Proof.

Alice (Bob) performs computational basis measurement on $A_{1}$ & $A_{2}$ ( $B_{1}$ & $B_{2}$ ) subsystems. Alice communicates the outcome on $A_{2}$ subsystem to Bob, and Bob communicates the outcome on $B_{1}$ subsystem to Alice. ∎

Proposition 4^′.

Within single-direction communication setup $P_{succ}^{\scalebox{0.6}{LBH-Sep}}\leq 1/2$ , whenever Alice and Bob are embedded in definite causal structure.

Proof.

Using unidirectional communication one of the players can guess perfectly, whereas the other player’s guess is completely random. ∎

Proposition 5.

A process matrix yielding a success $P^{\scalebox{0.6}{GYNI}}_{succ}=\mu$ in GYNI game yields the same success in LBH-Sep task.

Proof.

The encoding in this case is such that Eq.(61) holds where $u^{\prime}$ & $u$ are the outcome of computational basis measurement on $A_{1}$ & $A_{2}$ subsystems respectively, and $v^{\prime}$ & $v$ are the computational basis outcome on $B_{1}$ & $B_{2}$ subsystems respectively. Therefore, the GYNI strategy with $u$ and $v^{\prime}$ being the respective input of Alice and Bob ensure the same success in LBH-Sep. ∎

Important to note that the scope of super-activation phenomenon ${\bf A}$ does not arise in this case.

V Advantage of classical causal-indefinite processes in LDH task

Assuming quantum theory to be valid locally, relaxation of global time order between multiple parties led to the formalism of Process Matrices that accommodates the notion of causal indefiniteness [9]. Notably, the authors in [9] have shown impossibility of bipartite causally inseparable processes in classical case, conjecturing the same to hold in the multipartite setting as well. However, quite surprisingly the authors in [80, 81] prove the above conjecture to be false, implying causal indefiniteness to be a feature not inherent to quantum theory only. In this section we will analyse whether such causally indefinite classical process could be advantageous in multipartite LDH task. Before presenting our findings we briefly recall the relevant framework first.

V.1 Causal Indefiniteness in Classical Setup

The state cone ( $\Omega^{n}_{+}$ ) and normalised state space ( $\Omega^{n}$ ) of an $n$ level classical system is described as


$\displaystyle\Omega^{n}_{+}$	$\displaystyle:=\left\{\vec{p}~{}\|~{}\vec{p}\in\mathbb{R}^{n}~{},p_{i}\geq 0~{}\forall~{}i\right\},$	(34a)
$\displaystyle\Omega^{n}$	$\displaystyle:=\left\{\vec{p}~{}\|~{}\vec{p}\in\mathbb{R}^{n}~{},p_{i}\geq 0~{}\forall~{}i~{},\&~{}\sum_{i=0}^{n-1}p_{i}=1\right\}.$	(34b)

Pure state of $\Omega^{n}$ are $\vec{l}:=\{\delta_{il}\}_{i=0}^{n-1}$ for $l\in\{0,\cdots,n-1\}$ . Later, sometime we will denote $\vec{l}\equiv l$ simply. The most general operation that an agent (say X) can apply on a classical system is described by a classical instrument

\displaystyle\mathcal{I}^{c}_{X}\equiv\left\{S^{k}_{X}|S^{k}_{X}:\Omega^{I_{X}}_{+}\mapsto\Omega^{O_{X}}_{+}\right\}_{k=1}^{N},

(35)

where $S^{k}_{X}$ are positive linear maps mapping the state cone of the input $I_{X}$ level classical system to the state cone of the output $O_{X}$ level classical system, with $I_{X},O_{X}<\infty$ . Moreover, $S^{k}_{X}$ ’s sum up to a stochastic map $\mathbb{S}_{X}$ , i.e.,

\displaystyle\mathbb{S}_{X}:=\sum_{k=1}^{N}S^{k}_{X},~{}s.t.~{}\mathbb{S}_{X}(\Omega^{I_{X}})\subseteq\Omega^{O_{X}}.

(36)

The stochasticity condition is analogous to the trace preserving condition in the quantum case. Let us consider the case involving two parties say Alice and Bob with


	$\displaystyle\mathcal{S}_{A}:=\{S_{A}~{}\|~{}S_{A}:\Omega^{I_{A}}_{+}\mapsto\Omega^{O_{A}}_{+}\},$		(37a)
	$\displaystyle\mathcal{S}_{B}:=\{S_{B}~{}\|~{}S_{B}:\Omega^{I_{B}}_{+}\mapsto\Omega^{O_{B}}_{+}\},$		(37b)

denoting the sets of all state-cone preserving maps for Alice and Bob, respectively. Any such linear map $S:\Omega^{n}_{+}\mapsto\Omega^{m}_{+}$ can be represented as an $m\times n$ real matrix, which can be uniquely specified by it’s action on pure states $\{l\}_{l=0}^{n-1}$ of $\Omega^{n}$ . Without assuming any background causal structure among Alice’s and Bob’s actions, the most general statistics observed is given by a bi-linear functional,


	$\displaystyle P:\mathcal{S}_{A}\times\mathcal{S}_{B}\mapsto[0,\infty),$		(38a)
	$\displaystyle P(\mathbb{S}_{A},\mathbb{S}_{B})=1,~{}\forall~{}\mathbb{S}_{A},\mathbb{S}_{B}.$		(38b)

Any such bi-linear functional reads as a Trace-rule over a stochastic map $\mathbb{E}_{AB}$ [81], i.e.,


	$\displaystyle P(S_{A},S_{B})=\operatorname{Tr}\left[\mathbb{E}_{AB}\left(S_{A}\otimes S_{B}\right)\right],$		(39a)
	$\displaystyle\operatorname{Tr}[\mathbb{E}_{AB}(\mathbb{S}_{A}\otimes\mathbb{S}_{B})]=1,~{}\forall~{}\mathbb{S}_{A},\mathbb{S}_{B}$		(39b)
	$\displaystyle\mathbb{E}_{AB}(\Omega^{O_{A}O_{B}})\subseteq\Omega^{I_{A}I_{B}}.$		(39c)

Such a $\mathbb{E}_{AB}$ is termed as logically-consistent-classical process (LCCP). As shown in [81] (see also [9]) all bipartite LCCPs are causally definite, i.e.,

\displaystyle\mathbb{E}_{AB}=p_{1}\mathbb{E}_{AB}^{A\prec B}+p_{2}\mathbb{E}_{AB}^{B\prec A}+p_{3}\mathbb{E}_{AB}^{B\nprec\nsucc A},

(40)

where $\mathbb{E}_{AB}^{A\prec B}~{}(\mathbb{E}_{AB}^{B\prec A})$ denotes a process where Alice (Bob) is in the causal past of Bob (Alice), and $\mathbb{E}_{AB}^{B\nprec\nsucc A}$ represents a process with $A$ and $B$ being spacelike separated; with $\vec{p}=(p_{1},p_{2},p_{3})^{T}$ being a probability vector. However, for multipartite case there are classical processes which do not admit a notion of causal ordering among the parties. For instance, consider the classical process,


	$\displaystyle\mathbb{E}^{BW}_{ABC}(o_{A},o_{B},o_{C})=(i_{A},i_{B},i_{C}),~{}{\mbox{with}}$		(41a)
	$\displaystyle(i_{A},i_{B},i_{C})\equiv\begin{cases}(o_{C},o_{A},o_{B}),\\ \hskip 28.45274pt\mbox{if~{}maj}(o_{A},o_{B},o_{C})=0,\\ (\bar{o}_{B},\bar{o}_{C},\bar{o}_{A}),\\ \hskip 28.45274pt\mbox{if~{}maj}(o_{A},o_{B},o_{C})=1,\end{cases}$		(41b)

where $o_{A},o_{B},o_{C},i_{A},i_{B},i_{C}\in\{0,1\},\bar{0}=1$ and $\bar{1}=0$ (see Fig.4). As shown by the the authors in [81], the LCCP $\mathbb{E}^{BW}$ violates a tripartite causal inequality, establishing that causal indefiniteness is no longer an artifact of quantum processes. At this point, one might ask whether advantage in LDH task stems from indefinite quantum processes only or is it a general trait of causal indefiniteness. We answer this question in affirmative by providing a tripartite variant of LDH task wherein $\mathbb{E}^{BW}$ provides a nontrivial advantage.

V.2 Tripartite LDH (T-LDH)

A Referee encodes the strings ${\bf x}\equiv\mathbf{x_{1}}\mathbf{x_{2}}\mathbf{x_{3}}\equiv x_{1}x^{\prime}_{1}x_{2}x^{\prime}_{2}x_{3}x^{\prime}_{3}\in\{0,1\}^{\times 6}$ into

\displaystyle\rho_{ABC}^{\mathbf{x}}=(\mathcal{B}_{AC}^{\mathbf{x_{1}}})^{\otimes 2}\otimes(\mathcal{B}_{BA}^{\mathbf{x_{2}}})^{\otimes 2}\otimes(\mathcal{B}_{CB}^{\mathbf{x_{3}}})^{\otimes 2}.

(42)

and distributes respective subsystems to Alice, bob, and Charlie. The hiding condition is satisfied as

\displaystyle\rho^{\bf x}_{\mathcal{K}}:=\operatorname{Tr}_{\bar{\mathcal{K}}}\rho^{\bf x}_{ABC}=\left(\mathbb{I}/2\right)^{\otimes 4},~{}~{}\forall~{}{\bf x},

(43)

for all $\mathcal{K}\in\{A,B,C\}$ , where $\bar{A}:=BC$ and etc. Each player guesses a two bit string and accordingly will be given some payoff. Their guesses are correct if they have some definitive information about the given messages. For instance, Alice’s guess $a_{1}a^{\prime}_{1}$ could be correct in two ways: (i) she perfectly predicts the given string $x_{1}x^{\prime}_{1}$ , (ii) she perfectly eliminates one of the strings not given to her. Let us define sets


$\displaystyle\$^{\bf y}$	$\displaystyle\equiv\$^{yy^{\prime}}:=\{0y\bar{y}^{\prime},0\bar{y}y^{\prime},0\bar{y}\bar{y}^{\prime},1yy^{\prime}\},$	(44a)
$\displaystyle\pounds^{\bf x}$	$\displaystyle\equiv\pounds^{\bf x_{1}x_{2}x_{3}}:=\$^{\bf x_{1}}\times\$^{\bf x_{2}}\times\$^{\bf x_{3}}.$	(44b)

Accordingly, the winning condition reads as

	$\displaystyle\left\{\!\begin{aligned} &\left({\bf a}:=a_{0}a_{1}a^{\prime}_{1}\in\$^{\bf x_{1}}\right)~{}\land~{}\left({\bf b}:=b_{0}b_{1}b^{\prime}_{1}\in\$^{\bf x_{2}}\right)\\ &\hskip 56.9055pt\land~{}\left({\bf c}:=c_{0}c_{1}c^{\prime}_{1}\in\$^{\bf x_{3}}\right)\end{aligned}\right\},$
	$\displaystyle\hskip 28.45274pt\mbox{or~{}equivalently},~{}~{}{\bf g}\equiv{\bf abc}\in\pounds^{\bf x}.$		(45)

The first bit of a player’s guess, i.e., $a_{0}/b_{0}/c_{0}$ denotes whether they chooses to identify the string given to them or chooses to eliminate it.

V.3 T-LDH Success Under Different Collaboration Scenarios

In this subsection we analyse the success probability of T-LDH task under different collaboration scenarios. We start by considering the no-collaboration case.

Proposition 6.

Without any collaboration Alice, Bob and Charlie can achieve the success $P_{succ}^{\scalebox{0.6}{T-LDH}}=27/64$ .

Proof.

The only way Alice can learn something definitive about her string is through communication from Charlie. Similarly, Bob and Charlie need communication from Alice and Bob, respectively. Without any such communication, the best Alice can do is to answer $a_{0}=0$ and guess a value for $a_{1}a^{\prime}_{1}$ , which leads to a success $3/4$ . A similar strategy followed by Bob and Charlie yields an overall success $P_{succ}^{\scalebox{0.6}{T-LDH}}=(3/4)^{3}=27/64$ . ∎

However, unlike the LBH-B task, LOCC collaboration turns out to be advantageous in this case.

Proposition 7.

Under LOCC collaboration $P_{succ}^{\scalebox{0.6}{T-LDH}}=1$ .

Proof.

Recall that two-qubit Bell basis shared between two distant parties cannot be perfectly distinguished by LOCC [65]. However, according to the result in [82], given two copies of the states, they can be perfectly distinguished under LOCC. The protocol goes as follows: both the players perform $Z$ -basis measurement on their parts of the first copy and $X$ -basis measurement on second. One of the players communicate the results to the other player, who can accordingly identify the given Bell state. This ensures a perfect success of T-LDH task under LOCC. ∎

However, the protocol in Proposition 7 demands multi-round communication among the players. For instance, let Alice first communicate her results to Bob implying Alice’s measurement event to be in the causal past of Bob’s guess. Similarly, Bob to Charlie communication demands Bob’s measurement event to be in the causal past of Charlie’s guess. Finally, Charlie to Alice communication demands Charlie’s measurement event to be in the causal past of Alice’s guess. Since, in the single-opening setup communication entering into a local laboratory must happen before any communication going outside it, therefore the above protocol cannot be implemented within this setup. Therefore, the question of optimal success of T-LDH is worth exploring in single-opening scenario. However, likewise the notions of genuine and non-genuine entanglement on multipartite case [83], the notion of causal indefiniteness can also have different manifestations when more than two parties are involved. Before proceeding further, here we first recall the definition of bi-causal /genuine quantum process.

Definition 3.

A multipartite quantum process $W$ is called bi-causal if it allows a convex decomposition $W=\sum_{i}p_{i}W_{i}$ , where each $W_{i}$ are causally separable across some bipartition. Processes that are not bi-causal are termed as genuine causal inseparable.

Recalling Eq.(9), a tripartite bi-causal process $W$ can always be written as

	$\displaystyle W=$	$\displaystyle p_{1}W^{A\not\prec BC}+p_{2}W^{B\not\prec AC}+p_{3}W^{C\not\prec AB}$
		$\displaystyle+p_{4}W^{BC\not\prec A}+p_{5}W^{AC\not\prec B}+p_{6}W^{AB\not\prec C},$		(46)

with $\vec{p}:=\{p_{i}\}_{i=1}^{6}$ denoting a probability vector. Here, the term $W^{A\not\prec BC}$ denotes a process where Alice cannot communicate neither to Bob nor to Charlie, whereas in process $W^{BC\not\prec A}$ neither Bob nor Charlie can communicate to Alice. The other terms carry similar meanings. Importantly, in the process $W^{A\not\prec BC}/W^{BC\not\prec A}$ causal inseparability could be present between Bob and Charlie. Our next result shows that success of T-LDH is non-trivially bounded for any such bi-causal process.

Proposition 8.

Under bi-causal collaboration $P_{succ}^{\scalebox{0.6}{T-LDH}}\leq 3/4$ .

Proof.

In a process of type $W^{A\not\prec BC}$ communication from Alice is not possible to Bob as well as to Charlie. Thus Bob’s success is bounded by $3/4$ (see Proposition 6). On the other hand, in a process of type $W^{BC\not\prec A}$ Alice’s success is bounded by $3/4$ as neither Bob nor Charlie can communicate to Alice. Similar arguments hold for all other terms in Eq.(V.3), and hence the claim follows from convexity. To achieve the bound, they can share a definite order process $W^{A\prec B\prec C}$ where Alice is in the causal past of Bob, who is in the causal past of Charlie. Using the strategy discussed in Proposition 7, Bob’s and Charlie’s guesses will be perfect whereas Alice’s success is bounded by $3/4$ . This completes the proof. ∎

Naturally, the question arises whether a genuine inseparable causal process could be advantageous over the bi-causal processes. In the following section, we show this is indeed possible, even with a classical indefinite process.

V.4 Nontrivial Success in T-LDH with LCCP

Given the LCCP $\mathbb{E}^{BW}\in$ , the players can obtain a nontrivial success in T-LDH task. In the encoded state given to the players, two Bell states are shared between each pair of the players. Of course, the identity of the Bell state is not known to the individual parties. Given an encoded state, Alice performs $Z$ -basis measurement on her part of one of the the Bell state shared with Bob, and performs $X$ -basis measurement on her part of the the Bell state shared with Bob. Similarly, $Z$ and $X$ measurements are performed on the parts of Bell states shared with Charlie. Bob and Charlie follow a similar protocol. Outcome of all these different measurements can be compactly expressed as $G^{(H)}_{K}\in\{0,1\}$ – outcome of $K$ -basis measurement performed by the player $G$ on her part of the Bell state shared with the player $H$ ; with $K\in\{Z,X\}$ and $G,H\in\{A,B,C\}$ . Given the encoded state $\rho^{\mathbf{x_{1}}\mathbf{x_{2}}\mathbf{x_{3}}}_{ABC}$ , we have


$\displaystyle A^{(C)}_{Z}\oplus C^{(A)}_{Z}$	$\displaystyle=x_{1},~{}~{}A^{(C)}_{X}\oplus C^{(A)}_{X}=x^{\prime}_{1},$	(47a)
$\displaystyle B^{(A)}_{Z}\oplus A^{(B)}_{Z}$	$\displaystyle=x_{2},~{}~{}B^{(A)}_{X}\oplus A^{(B)}_{X}=x^{\prime}_{2},$	(47b)
$\displaystyle C^{(B)}_{Z}\oplus B^{(C)}_{Z}$	$\displaystyle=x_{3},~{}~{}C^{(B)}_{X}\oplus B^{(C)}_{X}=x^{\prime}_{3}.$	(47c)

For their local measurement outcomes the players respectively evaluate a bit values

\displaystyle o_{A}=A^{(B)}_{Z}A^{(B)}_{X},~{}~{}o_{B}=B^{(C)}_{Z}B^{(C)}_{X},~{}~{}o_{C}=C^{(A)}_{Z}C^{(A)}_{X},

(48)

and send them to the environment $\mathbb{E}^{BW}$ , which on the other hand returns back the bits $i_{A}$ , $i_{B}$ , and $i_{C}$ to Alice, Bob, and Charlie. The guesses in T-LDH task for Alice, Bob and Charlie are given by


$\displaystyle a_{0}$	$\displaystyle=i_{A},~{}~{}~{}a_{1}=\overline{A^{(C)}_{Z}},~{}~{}~{}a^{\prime}_{1}=\overline{A^{(C)}_{X}}$	(49a)
$\displaystyle b_{0}$	$\displaystyle=i_{B},~{}~{}~{}b_{1}=\overline{B^{(A)}_{Z}},~{}~{}~{}b^{\prime}_{1}=\overline{B^{(A)}_{X}}$	(49b)
$\displaystyle c_{0}$	$\displaystyle=i_{C},~{}~{}~{}a_{1}=\overline{C^{(B)}_{Z}},~{}~{}~{}c^{\prime}_{1}=\overline{C^{(B)}_{X}}$	(49c)

maj	$(o_{A},o_{B},o_{C})$	$(i_{A},i_{B},i_{C})$	$\bf a$	$\bf b$	$\bf c$
0	$(0,0,0)$	$(0,0,0)$	$0\neg(00)$	$0\neg(00)$	$0\neg(00)$
	$(0,0,1)$	$(1,0,0)$	$100$	$0\neg(00)$	$0\neg(00)$
	$(1,0,0)$	$(0,1,0)$	$0\neg(00)$	$100$	$0\neg(00)$
	$(0,1,0)$	$(0,0,1)$	$0\neg(00)$	$0\neg(00)$	$100$
1	$(0,1,1)$	$(0,0,1)$	${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}000}$	$0\neg(00)$	$100$
	$(1,0,1)$	$(1,0,0)$	$100$	${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}000}$	$0\neg(00)$
	$(1,1,0)$	$(0,1,0)$	$0\neg(00)$	$100$	${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}000}$
	$(1,1,1)$	$(0,0,0)$	${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}000}$	${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}000}$	${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}000}$

Table 2: Input

{\bf x}=\mathbf{0}

. For the case “maj

(o_{A},o_{B},o_{C})=0

”, all three players guess correctly. However, for the case “maj

(o_{A},o_{B},o_{C})=1

”, at-least one of players’ guess is not correct (marked in red). Here,

\neg(00)

indicates any string not equal to

00

i.e.

01/10/11

The success probability for ${\bf x}=\mathbf{0}\equiv 000000$ , turns out to be

	$\displaystyle P^{\scalebox{0.6}{T-LDH}}_{succ}({\bf x}=\mathbf{0})=\sum_{o_{A},o_{B},o_{C}}\sum_{{\bf g}\in\pounds^{\bf 0}}p(o_{A}o_{B}o_{C}{\bf g}\|{\bf x}=\mathbf{0})$
$\displaystyle=$	$\displaystyle\sum_{maj(o_{A},o_{B},o_{C})=0}\sum_{{\bf g}\in\pounds^{\bf 0}}p(o_{A}o_{B}o_{C}\|{\bf x}=\mathbf{0})p({\bf g}\|{\bf x}=\mathbf{0}o_{A}o_{B}o_{C})~{}+$
	$\displaystyle\sum_{maj(o_{A},o_{B},o_{C})=1}\sum_{{\bf g}\in\pounds^{\bf 0}}p(o_{A}o_{B}o_{C}\|{\bf x}=\mathbf{0})p({\bf g}\|{\bf x}=\mathbf{0}o_{A}o_{B}o_{C}).$	(50)

As we can see from Table 2, for the case “maj $(o_{A},o_{B},o_{C})=1$ ”, atleast one of the players violates the winning condition (45), i.e.,

\displaystyle\sum_{{\bf g}\in\pounds^{\bf 0}}p({\bf g}|{\bf x}=\mathbf{0}o_{A}o_{B}o_{C})=0.

(51)

However, for the case “maj $(o_{A},o_{B},o_{C})=0$ ”, all the players satisfy the winning condition (45), i.e.,

\displaystyle\sum_{{\bf g}\in\pounds^{\bf 0}}p({\bf g}|{\bf x}=\mathbf{0}o_{A}o_{B}o_{C})=1.

(52)

Consequently, Eq.(50) becomes

	$\displaystyle P^{\scalebox{0.6}{T-LDH}}_{succ}({\bf x}=\mathbf{0})$	$\displaystyle=\sum_{maj(o_{A},o_{B},o_{C})=0}p(o_{A}o_{B}o_{C}\|{\bf x}=\mathbf{0})$
		$\displaystyle=\left[\frac{3^{3}}{4^{3}}+3\times\frac{3^{2}}{4^{3}}\right]=\frac{27}{32}\approx 0.84>\frac{3}{4}~{}.$		(53)

Similarly, it can be shown that $P^{\scalebox{0.6}{T-LDH}}_{succ}({\bf x})=27/32,~{}\forall~{}{\bf x}\in\{0,1\}^{\times 6}$ , leading to $P^{\scalebox{0.6}{T-LDH}}_{succ}=27/32$ . Therefore, the classical causally indefinite process $\mathbb{E}^{EW}$ exhibits nontrivial advantage over the quantum bi-causal processes in T-LDH task. In fact, the success establishes the genuine multipartite nature of causal indefiniteness. Note that in the above mentioned protocol all the players are efficiently able to communicate the required information by effectively implementing the maj $(o_{A},o_{B},o_{C})=0$ loop in Fig.(4) with a high probability. One can say that effectively clockwise communication is happening between the players. This clockwise communication is also in some sense necessary, as the encoding states also have this ”clockwise” property (see Eq.(42)) i.e. Alice needs help from Charlie, Bob needs help from Alice and Charlie needs help from Bob . In Appendix B we discuss an interesting variant of the T-LDH task where the referee does not reveal whether they have done a clockwise or anticlockwise encoding but rather encodes this information in the distributed state itself. Interestingly, we show that even though the three players beforehand do not know whether the referee has encoded in a clockwise or anticlockwise fashion the process $\mathbb{E}^{BW}$ still can provide a advantage by effectively using both branches in Fig.(4). Which is impossible to do by a definite ordered process.

VI Conclusion and Outlook

In this work, we have analysed several aspects of causal indefiniteness from a cryptographic perspective. To this aim, we have introduced a novel cryptographic primitive called Local-Data-Hiding (LDH). We have shown that under the restricted collaboration scenario where each party’s laboratory opens only once — during which they can receive a system, operate on it, and send it out of their laboratory — parties sharing causally inseparable processes generally have a higher payoff in revealing the encoded data compared to their counterparts operating in a definite causal background. Along this line, we have obtained several interesting results. Below we discuss comprehensively our specific results and their nontrivial consequences.

$\bullet$ Duality between the payoff in the LDH task and the success of the GYNI game.– For the bipartite case, considering the LDH task with maximally entangled states encoding we have shown that the success in LDH is same as that of the GYNI game. In other words, our Theorems 1 & 3 establish a duality between the success probabilities of two apparently different tasks. Recall that, duality often plays important roles in both mathematics and physics. By bridging seemingly distinct concepts, it provides alternative avenues to answer questions that are difficult or impossible to address in one field, whereas becoming easier to tackle in their dual formulations. For instance, the seminal AdS/CFT duality has provided valuable insights into quantum gravity problems [84]. The duality relation established in our work, in the same spirit, holds promise for exploring new insights regarding the nature of indefinite causal structures. In particular, obtaining optimal violation of causal inequalities admissible by quantum processes (analogous to Tsirelson bound for quantum nonlocal correlations) is a central question in this field. Recently, Liu and Chiribella have proposed an algorithmic approach to address this question [48]. For a class of causal inequalities ( the single trigger causal inequalities) while their approach yields tight bounds, for others it provides nontrivial upper bounds only. Notably, in causal games (such as the GYNI game) finding the optimal success requires optimization over multiple quantum instruments for each parties as well as the shared processes. Likewise, the best success in LDH also demands optimization over the shared quantum processes. However, in this case, the optimization involves only a single instrument for each of the parties. This significantly reduces the complexity in optimization, and in turn, promises a more efficient approach to obtain Tsirelson bounds for causal inequalities due to our established duality.

$\bullet$ Peres-like criterion for bipartite quantum processes.– Considering a bipartite local-bit-hiding task with Bell states encoding (LBH-B) we have obtained a necessary criterion for bipartite quantum processes to be useful in this task. Specifically, we have shown that bipartite processes $W_{A_{I}A_{O}B_{I}B_{O}}$ that are positive under partial transpose (PPT) across the $A_{I}A_{O}|B_{I}B_{O}$ partition tunrs out to be not useful in LBH-B (Theorem 2). This suggests that a process violating a causal inequality need not be advantageous in the LBH-B task. Consequently, in Corollary 1 we have shown that any bipartite process lying within the convex hull of causally separable processes and the PPT processes does not yield any non-trivial advantage in the LBH-B task. This in a sense classifies quantum processes from an operational perspective. Processes yielding nontrivial success in LBH-B task, i.e., $P^{\scalebox{0.6}{LBH-B}}\geq 1/2$ , do not belong to $\mbox{ConvHull}(\textbf{W}_{CS},\textbf{W}_{PPT})$ [Fig.3]. However, it remains an open question whether there are processes outside $\mbox{ConvHull}(\textbf{W}_{CS},\textbf{W}_{PPT})$ that do not yield nontrivial success in LBH-B task. Furthermore, classifying quantum processes depending on similar such task could bring more insight regarding their structure and is left here as an interesting direction for future research.

$\bullet$ Super-activation of causal indefiniteness.– We have reported an intriguing super-activation phenomenon involving quantum processes. Particularly, an entangled state shared between Alice and Bob, being a no-signalling resource by its own does not provide a nontrivial success in LBH-B task. On the other hand, a process lying within the set $\tilde{\bf W}\equiv\mbox{ConvHull}(\textbf{W}_{CS},\textbf{W}_{PPT})\setminus\textbf{W}_{CS}$ is also not useful for this task by its own. However, as shown in only if part of our Theorem 1, a process $W\in\tilde{\bf W}$ violating GYNI inequality will become useful in LBH-B task when combined with a two-qubit maximally entangled states. An explicit example is the process $W^{Cyril}$ of Eq.(13). In other words, causal indefiniteness of this process gets activated in assistance with another no-signalling and hence causally separable process, together yielding nontrivial success in LBH-task. It remains an open question whether all processes in $\tilde{\bf W}$ together with some quantum state will depict such super-activation property. Furthermore, it could be also interesting to explore other possibilities of such super-activation phenomenon involving multipartite processes.

$\bullet$ Advantage of causally inseparable classical processes.– Considering the tripartite version of LDH task (T-LDH), we have shown that the advantage of causal indefiniteness in LDH task is not exclusive to the quantum nature of process matrices, rather it persists in classical processes as well. Our T-LDH task demonstrates that certain tripartite classical processes can outperform bicausal quantum processes (Section V.4), establishing efficacy of genuine causal indefiniteness. Moreover, our FT-LDH task demonstrates how it is possible for the three players to communicate their respective messages in a clockwise or anti-clockwise fashion based on their will without giving rise to casual loops (Appendix B).

Quantum cryptography is one of the first applications to demonstrate the significant utility of quantum properties at the single-quantum level, and it has subsequently played a major role in advancing research in quantum information science [85]. Apart from its applications, several cryptographic protocols have played a nontrivial role in the axiomatic derivation of Hilbert space quantum theory [86, 87, 88]. In the present work we analyze the new information primitive of causal indefiniteness from cryptographic perspective. While we have reported several intriguing results, our work also suggests exciting avenues for future research. Firstly, exploring the role of causal indefiniteness in other cryptographic protocols could reveal new advantages and applications, potentially leading to more secure and efficient systems. Furthermore, investigating the structural implications of causal indefiniteness from cryptographic perspective could provide deeper insights into the classification and characterization of quantum and classical processes. Understanding how causal structures influence the operational capabilities of processes may lead to new theoretical developments and practical applications. Finally, extending our duality results to broader classes of causal inequalities could enhance our understanding of the fundamental limits of both quantum and classical cryptographic systems, offering new perspectives on the interplay between causal structures and information processing.

Acknowledgment: We heartily thank Ananda Maity and Ognyan Oreshkov for useful suggestions on earlier version of the manuscript. SGN acknowledges support from the CSIR project 09/0575(15951)/2022-EMR-I. MA acknowledges the funding supported by the European Union (Project QURES- GA No. 101153001). MB acknowledges funding from the National Mission in Interdisciplinary Cyber-Physical systems from the Department of Science and Technology through the I-HUB Quantum Technology Foundation (Grant no: I-HUB/PDF/2021-22/008).

Appendix A Bipartite Local Dit Hiding in Maximally Entangled States

In the main manuscript, we observed a strict duality between the success probability of the bipartite LBH-B task and the success probability of the GYNI game. In this section, we will extend this concept of theorem 1 with Local Dit Hiding in higher dimensional Maximally Entangled States(LDH-ME) and GYNI with dit inputs.

LDH-ME: Referee encodes the string $\mathbf{x}=x_{1}~{}x_{2}\in\{0,1,\cdots,d-1\}^{2}$ in bipartite maximally entangles states as follows:


$\displaystyle\mathbf{x}$	$\displaystyle\to\ket{\mathcal{B}^{\mathbf{x}}}_{AB}:=\frac{1}{\sqrt{d}}\sum_{k=0}^{d-1}\omega^{x_{2}k}\ket{k}_{A}\otimes\ket{k\oplus_{d}x_{1}}_{B},$
	$\displaystyle\hskip 28.45274pt=\left(Z^{x_{2}}_{A}\otimes X^{x_{1}}_{B}\right)\frac{1}{\sqrt{d}}\sum_{k=0}^{d-1}\ket{k}_{A}\otimes\ket{k}_{B},$	(54a)
	$\displaystyle~{}Z_{A}\ket{k}:=e^{\frac{2\pi ik}{d}}\ket{k},~{}~{}\&~{}~{}X_{A}\ket{k}:=\ket{k\oplus_{d}1}$	(54b)

with $\omega=e^{\frac{2\pi i}{d}}$ and $\oplus_{d}$ representing modulo $d$ addition. The local marginals of Alice and Bob are the maximally mixed states $\mathbf{I}/d$ for every encoded state, and thus the hiding condition is satisfied. The success probability of LDH-ME task is given by

\displaystyle P_{succ}^{\scalebox{0.6}{LDH-ME}}:=\sum_{x_{1},x_{2}=0}^{d-1}\frac{1}{d^{2}}P(a=x_{1},b=x_{2}|\mathcal{B}^{x_{1}x_{2}}_{AB}).

(55)

GYNI-d: Alice (Bob) tosses a random $d$ sided coin to generate a random dit $i_{1}~{}(i_{2})\in\{0,1,\cdots,d-1\}$ . Each party aims to guess the coin state of the other party. Denoting their guesses as $a$ and $b$ respectively, the success probability reads as

\displaystyle P_{succ}^{\scalebox{0.6}{GYNI-d}}=\sum_{i_{1},i_{2}=0}^{d-1}\frac{1}{d^{2}}P(a=i_{2},b=i_{1}|i_{1},i_{2}).

(56)

The optimal winning probabilities for GYNI-d with an indefinite causal ordered process are unknown. However, the duality established in Theorem 1 extends to this higher dimensional case.

Theorem 3.

A success probability $P_{succ}^{\scalebox{0.6}{LDH-ME}}=\mu$ in LDH-ME task is achievable if and only if the same success is achievable in GYNI-d game, i.e., $P_{succ}^{\scalebox{0.6}{GYNI-d}}=\mu$ .

Proof.

As before the proof is done in two parts:

(i)

only if part: $P_{succ}^{\scalebox{0.6}{LDH-ME}}=\mu$ ensures a protocol for GYNI-d game yielding success probability $P_{succ}^{\scalebox{0.6}{GYNI-d}}=\mu$ .
(ii)

if part: $P_{succ}^{\scalebox{0.6}{GYNI-d}}=\mu$ ensures a protocol for LDH-ME task yielding success probability $P_{succ}^{\scalebox{0.6}{LDH-ME}}=\mu$ .

only if part:
Given the encoded states $\{\mathcal{B}^{\bf x}_{AB}\}$ , let the process matrix $W_{A_{I}A_{O}B_{I}B_{O}}$ yields a success $P_{succ}^{\scalebox{0.6}{LDH-ME}}=\mu$ with Alice and Bob applying the quantum instruments $\mathcal{I_{A}}=\{M^{a}_{AA_{I}A_{O}}\}_{a=0}^{d-1}$ and $\mathcal{I_{B}}=\{M^{b}_{BB_{I}B_{O}}\}_{b=0}^{d-1}$ , respectively. Thus we have,

	$\displaystyle P_{succ}^{\scalebox{0.6}{LDH-ME}}=\frac{1}{d^{2}}\sum_{x_{1},x_{2}=0}^{d-1}p(a=x_{1},b=x_{2}\|\mathcal{B}^{x_{1}x_{2}}_{AB})=\mu,~{}~{}\mbox{with},$		(57)
	$\displaystyle p(a,b\|\mathcal{B}^{x_{1}x_{2}}_{AB}):=\operatorname{Tr}[(\mathcal{B}^{x_{1}x_{2}}_{AB}\otimes W)(M^{a}_{AA_{I}A_{O}}\otimes M^{b}_{BB_{I}B_{O}})].$

For playing the GYNI-d game, let Alice and Bob share the process Matrix $W^{\prime}=W_{A_{I}A_{O}B_{I}B_{O}}\otimes\mathcal{B}^{00}_{AB}$ . Based on their coin states $i_{1},i_{2}\in\{0,1,\cdots,d-1\}$ , Alice and Bob respectively perform quantum instruments

	$\displaystyle\mathcal{I_{A}}^{(i_{1})}$	$\displaystyle:=\left\{\mathcal{Z}^{i_{1}}_{A}\left(M^{a}_{AA_{I}A_{O}}\right)\right\}_{a=0}^{d-1}\equiv\left\{Z^{i_{1}}_{A}M^{a}_{AA_{I}A_{O}}Z^{i_{1}}_{A}\right\}_{a=0}^{d-1},$
	$\displaystyle\mathcal{I_{B}}^{(i_{2})}$	$\displaystyle=\left\{\mathcal{X}^{i_{2}}_{B}\left(M^{b\|i_{2}}_{BB_{I}B_{O}}\right)\right\}_{b=0}^{d-1}\equiv\left\{X^{i_{2}}_{B}M^{b}_{BB_{I}B_{O}}X^{i_{2}}_{B}\right\}_{b=0}^{d-1},$

where $Z$ & $X$ are Pauli gates on $\mathbb{C}^{d}$ as defined in eq.(54b) and $\{M^{a}_{AA_{I}A_{O}}\}_{a=0}^{d-1}$ & $\{M^{b}_{BB_{I}B_{O}}\}_{b=0}^{d-1}$ are the instruments used in LDH-ME task. With this protocol the success probability of GYNI-d game becomes

	$\displaystyle P_{succ}^{\scalebox{0.6}{GYNI-d}}=\sum_{i_{1},i_{2}=0}^{d-1}\frac{1}{d^{2}}p(a=i_{2},b=i_{1}\|i_{1},i_{2})$
	$\displaystyle=\frac{1}{d^{2}}\sum_{i_{1},i_{2}=0}^{d-1}\operatorname{Tr}\left[\left(\mathcal{B}^{00}_{AB}\otimes W\right)\left(M^{a=i_{2}\|i_{1}}_{AA_{I}A_{O}}\otimes M^{b=i_{1}\|i_{2}}_{BB_{I}B_{O}}\right)\right]$
	$\displaystyle=\frac{1}{d^{2}}\sum_{i_{1},i_{2}=0}^{d-1}\operatorname{Tr}\left[\left(\mathcal{B}^{i_{2}i_{1}}_{AB}\otimes W\right)\left(M^{a=i_{2}}_{AA_{I}A_{O}}\otimes M^{b=i_{1}}_{BB_{I}B_{O}}\right)\right]$
	$\displaystyle=\mu=P_{succ}^{\scalebox{0.6}{LDH-ME}}.~{}~{}[\mbox{using~{}Eq}.(\ref{lbhd})].$

This completes the only if part of the claim.

if part:
Given $x_{1}$ and $x_{2}$ being the respective coin states of Alice and Bob, let the process matrix $W^{\prime}_{A_{I}A_{O}B_{I}B_{O}}$ yields a success $P_{succ}^{\scalebox{0.6}{GYNI}}=\mu$ , with Alice and Bob performing quantum instruments $\mathcal{I_{A}}^{(x_{1})}=\{M^{a|x_{1}}_{A_{I}A_{O}}\}_{a=0}^{d-1}$ and $\mathcal{I_{B}}^{(x_{2})}=\{M^{b|x_{2}}_{B_{I}B_{O}}\}_{b=0}^{d-1}$ , respectively. Thus we have,

	$\displaystyle P_{succ}^{\scalebox{0.6}{GYNI}}=\frac{1}{d^{2}}\sum_{x_{1},x_{2}=0}^{d-1}p(a=x_{2},b=x_{1}\|x_{1},x_{2})=\mu,~{}~{}\mbox{with},$		(58)
	$\displaystyle p(a,b\|x_{1},x_{2}):=\operatorname{Tr}\left[\left(M^{a\|x_{1}}_{A_{I}A_{O}}\otimes M^{b\|x_{2}}_{B_{I}B_{O}}\right)W^{\prime}\right].$

To perform the LDH-ME task, Alice and Bob share the Process Matrix $W^{\prime}_{A_{I}A_{O}B_{I}B_{O}}\otimes\mathcal{B}^{00}_{A^{\prime}B^{\prime}}$ . Now, given the encoded state $\mathcal{B}^{x_{1}x_{2}}_{AB}$ , Alice and Bob apply the following Controlled-Shift(CS) unitary operation on parts of their local systems

\displaystyle CS_{AA^{\prime}}\ket{m}_{A}\ket{n}_{A^{\prime}}=\ket{m}_{A}\ket{n\oplus_{d}m}_{A^{\prime}}

(59)

They follow this with a discrete Fourier transformation, $F\ket{k}=\frac{1}{\sqrt{d}}\sum_{q=0}^{d-1}\omega^{qk}\ket{q}$ , on their respective unprimed parts, which results in

	$\displaystyle\mathcal{F}_{A^{\prime}}\circ\mathcal{CS}_{AA^{\prime}}\otimes\mathcal{F}_{B^{\prime}}\circ\mathcal{CS}_{BB^{\prime}}\left(W^{\prime}_{A_{I}A_{O}B_{I}B_{O}}\otimes\mathcal{B}^{00}_{A^{\prime}B^{\prime}}\otimes\mathcal{B}^{x_{1}x_{2}}_{AB}\right)$
	$\displaystyle\hskip 28.45274pt=W^{\prime}_{A_{I}A_{O}B_{I}B_{O}}\otimes\mathcal{B}^{x_{1}0}_{A^{\prime}B^{\prime}}\otimes\mathcal{B}^{x_{1}x_{2}}_{AB},$		(60)

where $\mathcal{F}$ and $\mathcal{CS}$ denote the linear maps corresponding to the unitary operations $F$ and $CS$ respectively. After this Alice performs computational basis measurement on $A$ and $A^{\prime}$ , resulting in outcomes $u,u^{\prime}\in\{0,1,\cdots,d-1\}$ . Similarly, Bob obtains the outcomes $v,v^{\prime}\in\{0,1,\cdots,d-1\}$ . Clearly, due to the correlation of the state, we have

\displaystyle u\oplus_{d}v=x_{2},~{}~{}\&~{}~{}u^{\prime}\oplus_{d}v^{\prime}=x_{1}.

(61)

Therefore, guessing the value of $v^{\prime}$ and $u$ respectively by Alice and Bob with the probability $\mu$ will ensure the same success in the LDH-ME task. At this point the process $W^{\prime}_{A_{I}A_{O}B_{I}B_{O}}$ proves to be helpful in this task, which can be accordingly chosen looking into its advantage in GYNI-d game. The rest of the protocol mimics the GYNI-d strategy with $u$ and $v^{\prime}$ being the inputs of Alice and Bob, respectively. Denoting $a^{\prime}$ and $b^{\prime}$ as the output of the GYNI strategy the final guess in the LDH-ME task by Alice and Bob are respectively,

\displaystyle a=a^{\prime}\oplus_{d}u^{\prime},~{}~{}\&~{}~{}b=b^{\prime}\oplus_{d}v.

(62)

On the composite process $W^{\prime}_{A_{I}A_{O}B_{I}B_{O}}\otimes\mathcal{B}^{00}_{A^{\prime}B^{\prime}}\otimes\mathcal{B}^{x_{1}x_{2}}_{AB}$ , the effective instruments $\{M^{a}_{AA^{\prime}A_{I}A_{O}}\}_{a=0}^{d-1}$ and $\{M^{b}_{BB^{\prime}B_{I}B_{O}}\}_{b=0}^{d-1}$ are respectively given by

	$\displaystyle\sum_{u,u^{\prime},a^{\prime}=0}^{d-1}\delta_{a,a^{\prime}\oplus_{d}u^{\prime}}\mathcal{F}_{A^{\prime}}\circ\mathcal{CS}_{AA^{\prime}}\otimes\textbf{id}_{A_{I}A_{O}}(\ket{uu^{\prime}}_{AA^{\prime}}\bra{uu^{\prime}}\otimes M^{a^{\prime}\|u}_{A_{I}A_{O}}),$
	$\displaystyle\sum_{v,v^{\prime},b^{\prime}=0}^{d-1}\delta_{b,b^{\prime}\oplus_{d}v}\mathcal{F}_{B^{\prime}}\circ\mathcal{CS}_{BB^{\prime}}\otimes\textbf{id}_{B_{I}B_{O}}(\ket{vv^{\prime}}_{BB^{\prime}}\bra{vv^{\prime}}\otimes M^{b^{\prime}\|v^{\prime}}_{B_{I}B_{O}}),$

where $\{M^{a^{\prime}|u}_{A_{I}A_{O}}\}$ and $\{M^{b^{\prime}|v^{\prime}}_{B_{I}B_{O}}\}$ are the instruments used by Alice and Bob in GYNI-d game. The success probability of the LDH-ME task with the aforesaid protocol becomes

	$\displaystyle P_{succ}^{\scalebox{0.6}{LBH-B}}=\sum_{x_{1},x_{2}=0}^{1}\frac{1}{d^{2}}P(a=x_{1},b=x_{2}\|\mathcal{B}^{x_{1}x_{2}}_{AB})$
	$\displaystyle=\sum_{x_{1},x_{2}=0}^{d-1}\frac{1}{d^{2}}\operatorname{Tr}\left[\left(W^{\prime}\otimes\mathcal{B}^{00}\otimes\mathcal{B}^{x_{1}x_{2}}\right)\left(M^{a=x_{1}}_{AA^{\prime}A_{I}A_{O}}\otimes M^{b=x_{2}}_{BB^{\prime}B_{I}B_{O}}\right)\right]$
	$\displaystyle=\sum_{x_{1},x_{2}}\frac{1}{d^{2}}\sum_{u,v^{\prime}}\frac{1}{d^{2}}P(v^{\prime},u\|u,v^{\prime})$
	$\displaystyle=\sum_{x_{1},x_{2}}\frac{1}{d^{2}}\mu=\mu.~{}~{}[\mbox{using~{}Eq}.(\ref{gynid})].$		(63)

This completes the if part of the claim, and hence the Theorem is proved. ∎

Appendix B Flagged T-LDH

In this flagged version of T-LDH task (FT-LDH) referee encodes the strings ${\bf x}={\bf x_{1}x_{2}x_{3}}$ into

$\displaystyle\rho_{AA^{\prime}BB^{\prime}CC^{\prime}}^{\bf x}$	$\displaystyle=\frac{1}{2}\left[\ket{000}\bra{000}_{A^{\prime}B^{\prime}C^{\prime}}\right.$
	$\displaystyle\hskip 28.45274pt\left.\otimes(\mathcal{B}_{AC}^{\mathbf{x_{1}}})^{\otimes 2}\otimes(\mathcal{B}_{BA}^{\mathbf{x_{2}}})^{\otimes 2}\otimes(\mathcal{B}_{CB}^{\mathbf{x_{3}}})^{\otimes 2}\right]$
	$\displaystyle~{}~{}+\frac{1}{2}\left[\ket{111}\bra{111}_{A^{\prime}B^{\prime}C^{\prime}}\right.$
	$\displaystyle\hskip 28.45274pt\left.\otimes(\mathcal{B}_{AB}^{\mathbf{x_{1}}})^{\otimes 2}\otimes(\mathcal{B}_{BC}^{\mathbf{x_{2}}})^{\otimes 2}\otimes(\mathcal{B}_{CA}^{\mathbf{x_{3}}})^{\otimes 2}\right].$	(64)

Winning condition for FT-LDH remains same as of Eq.(45). All the players perform $Z$ basis measurement on the flagged state (primed systems). If they obtain outcome ‘ $0$ ’, they follow the strategy of T-LDH with $\mathbb{E}^{BW}$ . Otherwise, Eqs.(47) get modified as


$\displaystyle A^{(B)}_{Z}\oplus B^{(A)}_{Z}$	$\displaystyle=x_{1},~{}~{}A^{(B)}_{X}\oplus B^{(A)}_{X}=x^{\prime}_{1},$	(65a)
$\displaystyle B^{(C)}_{Z}\oplus C^{(B)}_{Z}$	$\displaystyle=x_{2},~{}~{}B^{(C)}_{X}\oplus C^{(B)}_{X}=x^{\prime}_{2},$	(65b)
$\displaystyle C^{(A)}_{Z}\oplus A^{(C)}_{Z}$	$\displaystyle=x_{3},~{}~{}C^{(A)}_{X}\oplus A^{(C)}_{X}=x^{\prime}_{3}.$	(65c)

In this case the players encode as

\displaystyle\bar{o}_{A}=A^{(C)}_{Z}A^{(C)}_{X},~{}\bar{o}_{B}=B^{(A)}_{Z}B^{(A)}_{X},~{}\bar{o}_{C}=C^{(B)}_{Z}C^{(B)}_{X}.

(66)

And their guesses are


$\displaystyle a_{0}$	$\displaystyle=i_{A},~{}a_{1}=\overline{A^{(B)}_{Z}},~{}a^{\prime}_{1}=\overline{A^{(B)}_{X}}$	(67a)
$\displaystyle b_{0}$	$\displaystyle=i_{B},~{}b_{1}=\overline{B^{(C)}_{Z}},~{}b^{\prime}_{1}=\overline{B^{(C)}_{X}}$	(67b)
$\displaystyle c_{0}$	$\displaystyle=i_{C},~{}a_{1}=\overline{C^{(A)}_{Z}},~{}c^{\prime}_{1}=\overline{C^{(A)}_{X}}$	(67c)

From symmetry of $\mathbb{E}^{BW}$ , it follows that $P^{\scalebox{0.6}{FT-LDH}}_{succ}\approx 0.84$ . While $P^{\scalebox{0.6}{T-LDH}}_{succ}=3/4$ can be achieved in definite causal structure, it is not the case for FT-LDH task. To see this consider the case $A\prec B\prec C$ .
(i) if outcome on flagged state is ‘ $0$ ’, then they can ensure a success $3/4$ : Alice and Bob can respectively help Bob and Charlie to guess their respective messages correctly.
(ii) for ‘ $1$ ’ outcome on flagged state, a success of $3^{2}/4^{2}$ can be ensured. While Alice can help Charlie only, Alice and Bob have to guess theire respective messages.

Thus on an average the success becomes

\displaystyle P^{\scalebox{0.6}{FT-LDH}}_{succ}=\frac{1}{2}\left(\frac{3}{4}+\frac{3^{2}}{4^{2}}\right)=\frac{21}{32}<\frac{3}{4}.

(68)

This demonstrates that sharing $\mathbb{E}^{BW}$ allows the players to effectively communicate in clockwise or anticlockwise fashion by suitably modifying their protocols of T-LDH task. However any causally ordered process would fail miserably to do so. Like proposition 8, in this case too obtaining a nontrivial bound for bi-causal quantum processes is not straightforward.

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	$\displaystyle P_{succ}^{\scalebox{0.6}{LBH-B}}=\sum_{x_{1},x_{2}=0}^{1}\frac{1}{4}P(a=x_{1},b=x_{2}\|\mathcal{B}^{x_{1}x_{2}}_{AB})$
	$\displaystyle=\sum_{x_{1},x_{2}=0}^{1}\frac{1}{4}\operatorname{Tr}\left[\left(W^{\prime}\otimes\mathcal{B}^{00}\otimes\mathcal{B}^{x_{1}x_{2}}\right)\left(M^{a=x_{1}}_{AA^{\prime}A_{I}A_{O}}\otimes M^{b=x_{2}}_{BB^{\prime}B_{I}B_{O}}\right)\right]$
	$\displaystyle=\sum_{\begin{subarray}{c}x_{1},x_{2},u,u^{\prime},\\ a^{\prime},v,v^{\prime},b^{\prime}=0\end{subarray}}^{1}\frac{1}{4}\delta_{a=x_{1},a^{\prime}\oplus u^{\prime}}\delta_{b=x_{2},b^{\prime}\oplus v}\operatorname{Tr}\left[\left(W^{\prime}\otimes\mathcal{B}^{x_{1}0}_{A^{\prime}B^{\prime}}\otimes\mathcal{B}^{x_{2}x_{1}}_{AB}\right)\right.$
	$\displaystyle\left.\hskip 56.9055pt\left(\ket{uvu^{\prime}v^{\prime}}_{ABA^{\prime}B^{\prime}}\bra{uvu^{\prime}v^{\prime}}\otimes M^{a^{\prime}\|u}_{A_{I}A_{O}}\otimes M^{b^{\prime}\|v^{\prime}}_{B_{I}B_{O}}\right)\right]$
	$\displaystyle=\sum_{\begin{subarray}{c}x_{1},x_{2},u,u^{\prime},\\ a^{\prime},v,v^{\prime},b^{\prime}=0\end{subarray}}^{1}\frac{1}{16}\delta_{x_{1},a^{\prime}\oplus u^{\prime}}\delta_{x_{2},b^{\prime}\oplus v}\delta_{x_{2},u\oplus v}\delta_{x_{1},u^{\prime}\oplus v^{\prime}}$
	$\displaystyle\hskip 56.9055pt\operatorname{Tr}\left[(W^{\prime}\left(M^{a^{\prime}\|u}_{A_{I}A_{O}}\otimes M^{b^{\prime}\|v^{\prime}}_{B_{I}B_{O}}\right)\right]$
	$\displaystyle=\sum_{\begin{subarray}{c}x_{1},x_{2},u,u^{\prime},\\ a^{\prime},v,v^{\prime},b^{\prime}=0\end{subarray}}^{1}\frac{1}{16}\delta_{x_{1},a^{\prime}\oplus u^{\prime}}\delta_{x_{2},b^{\prime}\oplus v}\delta_{x_{2},u\oplus v}\delta_{x_{1},u^{\prime}\oplus v^{\prime}}P(a^{\prime},b^{\prime}\|u,v^{\prime})$
	$\displaystyle=\sum_{\begin{subarray}{c}x_{1},x_{2},u,u^{\prime},\\ v,v^{\prime}=0\end{subarray}}^{1}\frac{1}{16}\delta_{x_{2},u\oplus v}\delta_{x_{1},u^{\prime}\oplus v^{\prime}}P(x_{1}\oplus u^{\prime},x_{2}\oplus v\|u,v^{\prime})$
	$\displaystyle=\sum_{x_{1},x_{2},u,v^{\prime}=0}^{1}\frac{1}{16}P(x_{1}\oplus v^{\prime}\oplus x_{1},x_{2}\oplus u\oplus x_{2}\|u,v^{\prime})$
	$\displaystyle=\sum_{x_{1},x_{2}}\frac{1}{4}\sum_{u,v^{\prime}}\frac{1}{4}P(v^{\prime},u\|u,v^{\prime})=\sum_{x_{1},x_{2}}\frac{1}{4}\mu=\mu~{}\left[\mbox{using eq.}(\ref{gyni})\right].$		(25)

	$\displaystyle P^{\scalebox{0.6}{T-LDH}}_{succ}({\bf x}=\mathbf{0})=\sum_{o_{A},o_{B},o_{C}}\sum_{{\bf g}\in\pounds^{\bf 0}}p(o_{A}o_{B}o_{C}{\bf g}\|{\bf x}=\mathbf{0})$
$\displaystyle=$	$\displaystyle\sum_{maj(o_{A},o_{B},o_{C})=0}\sum_{{\bf g}\in\pounds^{\bf 0}}p(o_{A}o_{B}o_{C}\|{\bf x}=\mathbf{0})p({\bf g}\|{\bf x}=\mathbf{0}o_{A}o_{B}o_{C})~{}+$
	$\displaystyle\sum_{maj(o_{A},o_{B},o_{C})=1}\sum_{{\bf g}\in\pounds^{\bf 0}}p(o_{A}o_{B}o_{C}\|{\bf x}=\mathbf{0})p({\bf g}\|{\bf x}=\mathbf{0}o_{A}o_{B}o_{C}).$	(50)

Local-Data-Hiding and Causal Inseparability: Probing Indefinite Causal Structures with Cryptographic Primitives

Abstract

I Introduction

II Preliminaries

II.1 Process Matrix Framework

II.2 Causally (In)Separable Processes

II.3 Guess-Your-Neighbour’s-Input

III Local Dit Hiding

III.1 Bipartite Bit-hiding in Bell States

Proposition 1.

Proof.

Proposition 2.

Proof.

Proposition 3.

Proof.

Proposition 4.

Proof.

IV Advantage of causal inseparability in LBH-B task

Theorem 1.

Proof.

Remark 1.

IV.1 Usefulness of Quantum Processes: A Peres-like Criterion

Theorem 2.

Proof.

Corollary 1.

IV.2 Super-Activation Phenomenon

Definition 1.

Definition 2.

IV.3 Bell-states Encoding & Super-Activation

Proposition 1′.

Proof.

Proposition 2′.

Proof.

Proposition 4′.

Proof.

Proposition 5.

Proof.

V Advantage of classical causal-indefinite processes in LDH task

V.1 Causal Indefiniteness in Classical Setup

V.2 Tripartite LDH (T-LDH)

V.3 T-LDH Success Under Different Collaboration Scenarios

Proposition 6.

Proof.

Proposition 7.

Proof.

Definition 3.

Proposition 8.

Proof.

V.4 Nontrivial Success in T-LDH with LCCP

VI Conclusion and Outlook

Appendix A Bipartite Local Dit Hiding in Maximally Entangled States

Theorem 3.

Proof.

Appendix B Flagged T-LDH

References

Local-Data-Hiding and Causal Inseparability:
Probing Indefinite Causal Structures with Cryptographic Primitives

Proposition 1^′.

Proposition 2^′.

Proposition 4^′.