May 06, 2026
Identifying causal order from restricted projective data is generally nontrivial. When two quantum players interact only through an unobserved environment, the available local measurement statistics are typically not tomographically complete, so the underlying process cannot in general be reconstructed exactly from the observed distribution. As a result, causal direction can be statistically identifiable in some cases but fundamentally indistinguishable in others.
In this work, we determine necessary and sufficient conditions for deciding when an observed distribution is compatible with a memoryless sequential quantum process in a fixed direction. We show that directional conditional-independence structure and the positivity criterion based on the pseudo-density matrix, as developed in recent work by Liu, Qiu, Dahlsten, and Vedral, are not sufficient by themselves. The missing ingredient is an additional algebraic consistency requirement, and together these conditions yield a complete criterion for membership in the memoryless sequential class.
We then specialize to the two-qubit Pauli setting, where the problem remains non-tomographic but becomes explicitly tractable. In this regime, we characterize when the two sequential directions are statistically indistinguishable, and we show by example that positivity alone does not exclude more general memoryful strategies, whereas the additional algebraic consistency requirement does.
Keywords: Causal order identification; quantum processes; restricted projective measurements; pseudo-density matrix; algebraic consistency criterion
Causality has become a central topic across a wide range of disciplines, from statistics and machine learning to quantum information science [1]–[5]. In the quantum setting, recent work has shown that causal structure is not only a conceptual issue but also an operational resource, with applications to channel discrimination, communication, metrology, and thermodynamic tasks [6]–[10]. Among the basic questions in this area, one of the most fundamental is the problem of determining temporal order: from observed data alone, can one decide which subsystem influences which, and when is such a decision impossible?
This problem is relatively transparent when one can prepare a sufficiently rich set of observations to reconstruct the underlying process matrix [3]–[5], [11]. In that case, the task reduces to the characterization of process matrices compatible with a given causal order. In the earlier work [11], we treated in detail the complementary regime in which the available observations are tomographically complete and hence identify the process matrix itself. In that full-tomography regime, the order-identification problem becomes essentially a process-level one, and for memoryless sequential structures it is shown that the problem is effectively reduced to the corresponding Markovian property. By contrast, when the available observations are not tomographically complete, the process matrix cannot be reconstructed exactly, and the temporal-order problem becomes substantially more subtle. Instead of process-level reconstruction, one must ask a distribution-level question: given only the observed statistics, can the underlying process belong to a given causal class, and if not, why not?
A natural approach to this restricted-observation problem uses state-like representations of temporal correlations. In particular, the pseudo-density matrix (PDM) framework and related spatiotemporal formalisms provide a unified way to represent temporal and spatial correlations by Hermitian operators [12]–[16]. Recent work by Liu, Qiu, Dahlsten, and Vedral used the PDM formalism to extract causal information from restricted measurement data. More precisely, the observed correlations determine, through the PDM construction, a channel-like operator associated with a candidate temporal evolution; positivity of this TP-CP-map-like operator is then a necessary compatibility condition for the corresponding causal structure, and its failure provides a witness of causal incompatibility. In addition, time asymmetry in the PDM-based description can provide information about temporal order [17]–[19]. Related works have developed state-over-time, quantum Bayesian, and operator-representation frameworks for spatiotemporal quantum correlations [20]–[22]. These works provide useful formal languages for representing temporal correlations and relating forward and reverse descriptions, but they do not by themselves give the memoryless-sequential membership criterion. Further compatibility-based approaches have studied whether observed measurement statistics or bipartite states admit spatial, temporal, or causal explanations [23], [24]. These developments strongly suggest that restricted projective observations may still carry substantial information about causal order. At the same time, they also indicate a basic limitation: positivity of reconstructed objects alone does not provide a complete characterization of memoryless sequential dynamics.
The purpose of the present paper is to resolve this issue for memoryless sequential quantum processes under restricted projective observations. Our starting point is that the usual ingredients suggested by non-signaling structure and by positivity of reconstructed state-like objects are not sufficient by themselves. What is missing is an additional algebraic consistency requirement that captures whether the observed statistics are generated by a single underlying channel. We show that, under projective state reduction, this additional condition completes the characterization of the memoryless sequential class. The contributions of the present paper can be summarized as follows.
First, we establish a general distribution-level criterion for deciding whether restricted projective data are compatible with a memoryless sequential quantum process in a fixed direction. This result gives a complete characterization of memoryless sequential dynamics under restricted observations and identifies the precise role of the additional algebraic consistency condition introduced in this work.
Second, we specialize the general theory to the two-qubit Pauli setting, which remains non-tomographic but becomes explicitly tractable. In this regime, we characterize when the two sequential directions are statistically indistinguishable and derive explicit conditions for representative channel families.
Third, we clarify the observable hierarchy underlying these results by comparing positivity-based criteria with the membership criterion obtained here. This comparison shows that positivity alone does not exclude more general memoryful strategies and that the algebraic consistency condition constitutes a genuinely new ingredient.
Finally, we formulate the reverse-direction membership problem in the same qubit-Pauli regime. This problem should be understood as the unrestricted reverse-direction question, whereas the order-indistinguishability analysis studied above addresses the same question under the additional assumption that the observed distribution is already known to arise from a forward memoryless sequential process.
The remainder of this paper is organized as follows. Section 3 introduces the general process-matrix framework and the statistical task. Sections 4 and 5 develop the state-reduction framework and its specialization to projective observations. Section 6 derives the general fixed-direction membership criterion for memoryless sequential dynamics. Section 7 specializes these results to the two-qubit Pauli setting and analyzes order-indistinguishability. Section 9 compares the resulting observable hierarchies. Finally, Section 8 formulates the unrestricted reverse-direction membership problem in the same qubit-Pauli regime.
We briefly summarize the logic leading to the main result, Theorem 6. Because restricted projective data are not tomographically complete, the problem is treated at the level of distributional membership rather than process reconstruction. The pseudo-density-matrix formalism provides a natural state-like description of the observed temporal correlations, but PDM-based positivity alone is not sufficient; the missing ingredient is the algebraic consistency condition introduced later. The strategy of the paper can be summarized as follows.
We first pass from process-level descriptions to distribution-level objects. Using the pseudo-density-matrix formalism together with compatible marginals, we associate to the observed distribution a state-like object that captures the relevant temporal correlations at the distribution level. Under projective state reduction, this PDM-based description also leads to an effective channel-like object, namely a candidate Choi matrix reconstructed from the observed statistics. This is the role of the general state-reduction framework developed in Sections 4 and 5.
We then identify the basic necessary conditions for memoryless sequentiality. At the distribution level, every memoryless sequential strategy must satisfy a directional Markovian condition expressing the appropriate one-way non-signaling structure. In addition, the PDM/reconstruction formalism yields a candidate Choi matrix, whose positivity is another necessary condition for compatibility with a memoryless sequential process.
However, neither directional Markovianity nor PDM-based positivity is sufficient. The genuinely new ingredient of the present paper is an additional algebraic consistency condition, later formalized as condition (C1). Its role is to enforce that the observed correlations arise from a single underlying channel acting consistently on the post-measurement states, rather than from different effective maps selected by hidden memory. At an operational level, condition (C1) links single-site and two-site statistics and excludes precisely those distributions that satisfy the standard positivity and non-signaling requirements but nevertheless fail to admit a representation by memoryless sequential dynamics.
The main theorem of the paper, Theorem 6, shows that these three ingredients together are complete. More precisely, under the assumptions used for the reconstruction procedure, an observed distribution is compatible with a memoryless sequential process in a fixed direction if and only if it satisfies the directional Markovian condition, the positivity of the reconstructed Choi matrix, and the algebraic consistency condition (C1). Moreover, whenever these conditions hold, the reconstructed pair actually reproduces the original observed distribution.
Once the fixed-direction membership criterion has been established, the next natural question is whether a distribution already known to arise from a forward memoryless sequential process can also be explained in the opposite direction. This is the order-indistinguishability problem. Because this problem starts from a distribution with additional structure already identified, it is more tractable than the respective-direction membership problem and is therefore analyzed first.
Finally, we specialize these questions to the two-qubit Pauli setting. Although that regime remains non-tomographic, the abstract conditions can there be converted into explicit inequalities for the parameters of the channel and the initial state. This yields concrete analyses both of the order-indistinguishability and the respective-direction membership problem, together with the corresponding observable hierarchy comparisons.
It is worth emphasizing the conceptual role of condition (C1). This condition is not an additional assumption on the measurement scheme and is not a reformulation of PDM positivity, Choi positivity, or directional non-signaling. Rather, it is a distribution-level consistency requirement expressing the compatibility of the observed data with a single physical channel. In this sense, condition (C1) is the genuinely new contribution of the present work: it is the minimal additional observable requirement needed to distinguish memoryless sequential dynamics from more general processes with hidden memory.
The later sections make this point precise in two complementary ways. On the one hand, Proposition 5 shows that condition (C1) is automatically satisfied by every memoryless sequential strategy. On the other hand, the two-qubit example in Section 9 shows that positivity-based criteria alone do not exclude more general memoryful strategies, whereas condition (C1) does. Thus the proof of the main theorem should be read not simply as a reconstruction argument based on the PDM formalism, but as the identification of the missing observable principle that closes the gap between positivity and genuine memoryless sequentiality.
We consider two isolated quantum players, Alice (player 1) and Bob (player 2), who interact only through an external environment controlled by a third party, Charlie. The aim of this subsection is to formalize Charlie’s strategy by introducing a process matrix \(W_S\). In each experimental run, Alice receives a quantum input system \(\mathcal{H}_{I,1}\) from Charlie and returns a quantum output system \(\mathcal{H}_{O,1}\) to Charlie, while Bob analogously receives \(\mathcal{H}_{I,2}\) and returns \(\mathcal{H}_{O,2}\). We denote the corresponding dimensions by \(d_{I,1}, d_{O,1}, d_{I,2}, d_{O,2}\). During the experiment Alice and Bob are not allowed to communicate; after all runs are completed, they may share their classical records to form empirical statistics. Our goal is to infer structural information about Charlie’s “strategy” purely from the joint input–output statistics observed by Alice and Bob.
Alice’s most general admissible operation in one run is a quantum instrument \(\{\Gamma_{1,z_1}\}_{z_1\in\mathcal{Z}_1}\), where each \(\Gamma_{1,z_1}\) is completely positive (CP) and trace-nonincreasing, and \(\sum_{z_1}\Gamma_{1,z_1}\) is trace preserving (TP). The classical outcome \(z_1\) is recorded by Alice. Bob’s operation is similarly described by an instrument \(\{\Gamma_{2,z_2}\}_{z_2\in\mathcal{Z}_2}\) with classical outcome \(z_2\). We write \(C[\Gamma]\) for the unnormalized Choi matrix of a CP map \(\Gamma\), using Choi’s convention [25]: \[\operatorname{Tr}_{I}\bigl[(\rho^{\mathsf T}\otimes I_{O})\,C[\Gamma]\bigr] = \Gamma(\rho), \label{eq:choi-convention}\tag{1}\] where \({\mathsf T}\) denotes transpose in a fixed input basis, and the partial trace is over the input space. Equivalently, \[C[\Gamma]=\sum_{a,b}|a\rangle\langle b|\otimes \Gamma(|a\rangle\langle b|).\] This is the convention used throughout the paper. It differs from Jamiołkowski’s original convention [26] by a partial transpose on the input system; accordingly, whenever a partial transpose is needed below, it is written explicitly as \(T_1\) or \(T_2\). For a linear map \(\Gamma:{\cal B}({\cal H}_{I,i})\to{\cal B}({\cal H}_{O,i})\), we denote by \(\Gamma^\dagger:{\cal B}({\cal H}_{O,i})\to{\cal B}({\cal H}_{I,i})\) its adjoint supermap with respect to the Hilbert–Schmidt inner product, i.e., \[\mathrm{Tr}\!\bigl[A\,\Gamma(B)\bigr] = \mathrm{Tr}\!\bigl[\Gamma^\dagger(A)\,B\bigr] \label{eq:adjoint-supermap}\tag{2}\] for all Hermitian operators \(A\in{\cal B}({\cal H}_{O,i})\) and \(B\in{\cal B}({\cal H}_{I,i})\). In particular, \(\Gamma^\dagger(I)\) represents the corresponding effect operator on the input space.
We now formalize Charlie’s strategy by introducing a process matrix \(W_S\). Charlie does not access Alice’s and Bob’s laboratories directly; instead, he mediates their interaction by preparing inputs and processing outputs across the global interface \(\mathcal{H}_{I,1}\otimes\mathcal{H}_{O,1}\otimes\mathcal{H}_{I,2}\otimes\mathcal{H}_{O,2}\) [3]–[5]. We model Charlie’s strategy \(S\) by a bipartite process matrix \(W_S\ge 0\) acting on this interface. When Alice and Bob perform their local instruments \(\{\Gamma_{1,z_1}\}_{z_1\in\mathcal{Z}_1}\) and \(\{\Gamma_{2,z_2}\}_{z_2\in\mathcal{Z}_2}\), respectively, the joint distribution of their outcomes \(z_1\) and \(z_2\) is given by the generalized Born rule \[P_{Z_1,Z_2}(z_1,z_2) = \mathrm{Tr}\Bigl[ W_S\bigl(C[\Gamma_{1,z_1}]\otimes C[\Gamma_{2,z_2}]\bigr) \Bigr]. \label{eq:GBR}\tag{3}\] Thus the observed joint distribution is determined by Charlie’s process matrix \(W_S\). Valid process matrices satisfy the standard linear “no-signalling-in-time” constraints ensuring that 3 defines a normalized probability distribution for all TP choices \(\sum_{z_i}\Gamma_{i,z_i}\) [4], [5]. In what follows, however, we only use 3 together with basic positivity and normalization properties, and we do not need the full structural characterization of admissible process matrices.
Fixing the local operational capabilities, namely the families of instruments accessible to Alice and Bob, Charlie induces a statistical model \[\mathcal{M}:\quad S\longmapsto \mathcal{M}[S] := \bigl\{ P_{Z_1,Z_2}(z_1,z_2) \bigr\}_{(z_1,z_2)\in\mathcal{Z}_1\times\mathcal{Z}_2}. \label{eq:model-map}\tag{4}\]
Our primary question is:
(Q)Given only the observed distribution \(\mathcal{M}[S]\), what aspects of Charlie’s strategy—in particular, its causal direction and the presence/absence of memory—can Alice and Bob determine, and when is the causal direction fundamentally indistinguishable?
When Alice and Bob have sufficiently rich local operations, for example tomographically complete sets of instruments as in [11], the map \(\mathcal{M}\) can become injective on broad strategy classes, so that reconstruction-based classification is possible. The present paper addresses the complementary regime in which the available operations are genuinely restricted. We focus on observation schemes based on state reduction, with particular emphasis on projective state reduction. In such settings, exact process reconstruction is generally impossible, and the problem becomes one of nontrivial set-membership at the distribution level. Our main results therefore provide complete characterizations for memoryless sequential strategies and for the order-indistinguishable region in the minimal nontrivial two-qubit setting.
In addition to the general process-matrix description in Section 3, we will use several definite-order strategy classes that capture parallel vs. sequential wiring and the presence/absence of memory. We adopt the definitions in [11] and record them here for completeness. Throughout this subsection, \(W_S\) denotes a valid bipartite process matrix on \({\cal H}_{I,1}\otimes{\cal H}_{O,1}\otimes{\cal H}_{I,2}\otimes{\cal H}_{O,2}\).
(i) General class. \({\cal S}_G\) denotes the set of all physically valid bipartite process matrices satisfying the standard no-signalling-in-time constraints as in [11]. In general, this class may also contain processes that are not of definite order.
(ii) Individual (product) strategy. \(S\in{\cal S}_I\) if there exist states \(\rho_1\) on \({\cal H}_{I,1}\) and \(\rho_2\) on \({\cal H}_{I,2}\) such that \[W_S=\rho_1\otimes\rho_2\otimes I_{O,1}\otimes I_{O,2}.\] Operationally, Charlie’s inputs are independent and do not depend on either output (See Fig. 1).
(iii) Parallel strategy (quantum memory allowed). \(S\in{\cal S}_P\) if there exists a (possibly entangled) state \(\rho_{12}\) on \({\cal H}_{I,1}\otimes{\cal H}_{I,2}\) such that \[W_S=\rho_{12}\otimes I_{O,1}\otimes I_{O,2}.\] This class is called quantum parallel in [11]. We keep the legacy notation \({\cal S}_P\) used in the remainder of this manuscript (See Fig. 2).
(iv) Sequential (memoryless) strategy. \(S\in{\cal S}_{N,1\to 2}\) if there exist an input state \(\rho_1\) on \({\cal H}_{I,1}\) and a CPTP map \(\Lambda_{1\to2}:{\cal B}({\cal H}_{O,1})\to{\cal B}({\cal H}_{I,2})\) such that \[W_S=\rho_1\otimes C[\Lambda_{1\to2}]\otimes I_{O,2},\] where \(C[\Lambda]\) is the (unnormalized) Choi matrix in the convention of 1 (See Fig. 3). The opposite direction class \({\cal S}_{N,2\to 1}\) is defined analogously.
(v) Sequential strategy with quantum memory. \(S\in{\cal S}_{Q,1\to 2}\) if Charlie may keep a quantum memory that mediates the map from Alice’s output to Bob’s input; as presented in [11], an equivalent process-matrix characterization satisfies \[\begin{align} (\mathrm{Tr}_{(O,2)}W_{S})\otimes \rho_{mix,(O,2)} &= W_{S}\tag{5},\\ (\mathrm{Tr}_{(O,1),(I,2),(O,2)}W_{S})\otimes I_{(O,1)} &= \mathrm{Tr}_{(I,2),(O,2)}W_{S}. \tag{6} \end{align}\] See Fig. 4. The opposite direction class \({\cal S}_{Q,2\to 1}\) is defined analogously.
The hierarchy of the above classes with the definite-order is summarized as \[\begin{align} \begin{array}{ccccc} {\cal S}_{N,1\to 2} & \supset & {\cal S}_{I} & \subset & {\cal S}_{N,2\to 1} \\ \cap & & \cap & & \cap \\ {\cal S}_{Q,1\to 2} & \supset & {\cal S}_{P} & \subset & {\cal S}_{Q,2\to 1} \end{array} \label{HI1} \end{align}\tag{7}\] Here, \({\cal S}_{I}\) is the smallest class in this hierarchy, \({\cal S}_{P}\) is the parallel class with arbitrary bipartite input state, and \({\cal S}_{N,1\to 2}\) and \({\cal S}_{Q,1\to 2}\) (and their reversed counterparts) describe sequential strategies without and with quantum memory, respectively. See [11] for the corresponding process-matrix characterizations.
In addition to these classes, one may also consider subclasses of \({\cal S}_{Q,1\to 2}\), \({\cal S}_{Q,2\to 1}\), or \({\cal S}_{Q}\) in which Charlie’s memory is restricted to be classical rather than quantum. While such classes are well defined at the process level, they are not well suited to the distribution-level approach developed in the present paper, and we therefore do not consider them here.
For example, in the class \({\cal S}_{Q}\), the subclass in which Charlie’s memory is classical corresponds simply to separable states on \({\cal H}_{I,1} \otimes {\cal H}_{I,2}\). The analysis of such classical-memory structures requires techniques that are conceptually different from those employed in this work.
The final goal of this paper is a sharp characterization of memoryless sequential dynamics under restricted projective observations. Before imposing that projective-state-reduction assumption, however, it is useful to formulate the problem at the more general level of state reduction associated with randomized measurements. This section develops that general framework. Its role is to identify what can already be said at the distribution level without yet restricting the post-measurement dynamics to the projective case, and to prepare the structures that will later be specialized to projective state reduction. In this way, the subsequent projective analysis can be understood as a refinement of the general state-reduction framework introduced here.
We consider the case when the operations by Alice and Bob are given as a randomized choice of measurements. In this case, their information can be described by two random variables, the random variable \(Y\in {\cal Y}\) describing the choice of measurement, and the random variable \(X\in {\cal X}\) describing the measurement outcome. Hence, two players, Alice (1st player) and Bob (2nd player), have their respective random variables \(X_1,Y_1\) and \(X_2,Y_2\), respectively.
In this scenario, Alice and Bob independently decide their choice \(Y_1\) and \(Y_2\) according to the uniform distribution independently, i.e., \(Y_1\perp Y_2\). We denote the state reduction with the measurement choice \(Y_1=y_1\in \overline{\cal Y}_1\) in Alice’s side by \(\{\Gamma_{x_1|y_1}^1\}_{x_1 \in {\cal X}_{y_1}}\), where the state \(\rho\) is changed to \(\Gamma_{x_1|y_1}^1(\rho)/\mathrm{Tr}\Gamma_{x_1|y_1}(\rho)\) with the measurement outcome \(x_1\). Here, \(\Gamma_{x_1|y_1}^1\) is a CP map from the input system \({\cal H}_{(I,1)}\) to the output system \({\cal H}_{(O,1)}\), and \(\sum_{x_1 \in {\cal X}_{y_1}}\Gamma_{x_1|y_1}^1\) is a TP-CP map. Similarly, we define \(\{\Gamma_{x_2|y_2}^2\}_{x_2 \in {\cal X}_{y_2}}\) for \(y_2\in \overline{\cal Y}_2\).
Hence, Alice’s and Bob’s operations are described as \(\{\Gamma_{1,(x_1,y_1)}\}\) and \(\{\Gamma_{2,(x_2,y_2)}\}\) under the formalism give in Section 3, respectively, where \(\Gamma_{i,(x_i,y_i)}:= \frac{1}{|\overline{\cal Y}_i|}\Gamma_{x_i|y_i}^i\) for \(i=1,2\). Using 3 , we define the joint distribution \(P_{X_1,Y_1,X_2,Y_2}\), which will be obtained by repeating several experiments. This joint distribution gives the concrete form of the map \({\cal M}\) defined in 4 . Due to the condition \(\Gamma_{i,(x_i,y_i)}= \frac{1}{|\overline{\cal Y}_i|}\Gamma_{x_i|y_i}^i\), the obtained distribution belongs to the following set \[\begin{align} {\cal M}[{\cal S}_G] \subset {\cal P}_{{\cal U}}:= \big\{P_{X_1,Y_1,X_2,Y_2}\big| P_{Y_1,Y_2}(y_1,y_2)=\frac{1}{|\overline{\cal Y}_1||\overline{\cal Y}_2|} ,\quad \forall y_1\in \overline{\cal Y}_1,y_2 \in \overline{\cal Y}_2\big\}. \end{align}\]
Generally, the set \(\{ C[\Gamma_{1,(x_1,y_1)}]\otimes C[\Gamma_{2,(x_2,y_2)}] \}\) is not tomographically complete on \({\cal H}_{(I,1)} \otimes {\cal H}_{(O,1)} \otimes{\cal H}_{(I,2)}\otimes{\cal H}_{(O,2)}\). In this case, we cannot recover the process matrix from the obtained distribution \(P_{X_1,Y_1,X_2,Y_2}\). Therefore, there is a process matrix \(W_S\) that cannot be classified in the sense of 7 from the obtained distribution \(P_{X_1,Y_1,X_2,Y_2}\), i.e., the map \({\cal M}\) defined in 4 is not one-to-one.
Checking of whether \(S\) belongs to \({\cal S}_{C|1\to 2}\) or \({\cal S}_{C}\) is difficult even when the process matrix can be reconstructed. Hence, we focus on the hierarchies given in 7 , i.e., we do not discuss the classes \({\cal S}_{C|1\to 2}\), \({\cal S}_{C|2\to 1}\), or \({\cal S}_{C}\). Although the hierarchies given in 7 yield the hierarchies: \[\begin{align} \begin{array}{ccccc} {\cal M}[{\cal S}_{N,1\to 2}] & \supset & {\cal M}[{\cal S}_{I}] &\subset & {\cal M}[{\cal S}_{N,2\to 1}] \\ \cap & &\cap & & \cap \\ {\cal M}[{\cal S}_{Q,1\to 2}] & \supset & {\cal M}[{\cal S}_{P}] &\subset & {\cal M}[{\cal S}_{Q,2\to 1}] , \end{array}\label{HI8} \end{align}\tag{8}\] there is a possibility that the intersection \({\cal M}[{\cal S}_{N,1\to 2}] \cap {\cal M}[{\cal S}_{N,2\to 1}]\) is strictly larger than \({\cal M}[{\cal S}_{I}]\). In the following, we discuss how to characterize the sets in the hierarchies given in 8 .
For the characterization, we have the following theorem.
Theorem 1. When Charlie’s strategy \(S\) belongs to \({\cal S}_{Q,1\to2}\) or \({\cal S}_{N,1\to2}\), the distribution \({\cal M}[S]\) satisfies the Markovian chain \(X_1-Y_1-Y_2\). When \(S \in {\cal S}_{P}\), it satisfies both \(X_1-Y_1-Y_2\) and \(Y_1-Y_2-X_2\). When \(S \in {\cal S}_{I}\), it satisfies the Markovian chain \(X_1-Y_1-Y_2-X_2\).
Except for \({\cal S}_{I}\), the above conditions come from the non-signaling condition. Since the classes \({\cal S}_{N|1\to 2}\) and \({\cal S}_{Q|1\to 2}\) are characterized by the same Markovian chains \(X_1-Y_1-Y_2\), the hierarchy \({\cal S}_{N|1\to 2}\subset {\cal S}_{Q|1\to 2}\) cannot be distinguished by using the Markov chain. In the next section, we study whether there exist other conditions to detect the above hierarchies under a special example.
We now introduce the formalism of pseudo-density matrix (PDM)[12], [16], [17] as a state-like representation of the observed distribution \(P_{X_1,Y_1,X_2,Y_2}\). Its role in the present paper is to describe the compatible bipartite and marginal states associated with the observed data. To formulate this representation, we first fix a family of auxiliary Hermitian operators built from the state-reduction maps introduced in the previous section. These operators will serve to extract the relevant moments of \(P_{X_1,Y_1,X_2,Y_2}\) and hence to define compatibility with a PDM.
To encode the observed statistics in a convenient linear form, we introduce, for each measurement setting, a family of independent centered observables. These observables will later be used to define the moments appearing in the pseudo-density-matrix description.
We assume that \({\cal H}_{(I,i)}={\cal H}_{(O,i)}\) and denote their common dimension by \(d_i\). We also enlarge the set of measurement choices to \(\overline{\cal Y}_i={\cal Y}_i\cup\{0\}\) for \(i=1,2\), where the additional label \(0\) represents the trivial choice of doing nothing. For this choice, we define \(\Gamma_{0|0}^i\) to be the identity map on the density matrices on \({\cal H}_{(I,i)}\).
Fix \(y_i\in{\cal Y}_i\). Since the outcome probabilities for a fixed measurement \(y_i\) satisfy one normalization constraint, only \(|{\cal X}_{y_i}|-1\) independent linear combinations are needed. We therefore choose an auxiliary index set \({\cal Z}_{y_i}\) with \(|{\cal Z}_{y_i}|=|{\cal X}_{y_i}|-1\). For each \(z_i\in{\cal Z}_{y_i}\), we choose real coefficients \(\{g_{x_i|y_i,z_i}\}_{x_i\in{\cal X}_{y_i}}\) and define a Hermitian operator \(G_{y_i,z_i}\) by \[G_{y_i,z_i} = \sum_{x_i\in{\cal X}_{y_i}} g_{x_i|y_i,z_i}\,\Gamma_{x_i|y_i}^{i,\dagger}(I).\] We require that the family \(\{G_{y_i,z_i}\}_{z_i\in{\cal Z}_{y_i}}\cup\{I\}\) be linearly independent and that each \(G_{y_i,z_i}\) be traceless: \(\mathrm{Tr}G_{y_i,z_i}=0\). Thus, the identity operator \(I\) accounts for normalization, while the operators \(G_{y_i,z_i}\) describe the nontrivial independent components of the measurement statistics.
For later convenience, we collect these operators into a single indexed family \[\{G_{t_i}\}_{t_i\in{\cal T}_i} := \{I\}\cup \{G_{y_i,z_i}\}_{y_i\in{\cal Y}_i,\;z_i\in{\cal Z}_{y_i}},\] where the index \(0\in{\cal T}_i\) is reserved for the identity operator, so that \(G_0=I\). For \(t_i\neq0\), we identify \(t_i\) with the pair \((y(t_i),z(t_i))\), and we abbreviate \(g_{x_i|y(t_i),z(t_i)}\) to \(g_{x_i|t_i}\).
Intuitively, for each measurement setting \(y_i\), we do not work directly with the raw outcome labels \(x_i\). Instead, we replace them by a set of \(|{\cal X}_{y_i}|-1\) independent contrast variables. The corresponding operators \(G_{y_i,z_i}\) play the role of a basis for these contrasts and will allow us to express the observed distribution through linear moments.
Although the joint distribution \(P_{X_1,Y_1,X_2,Y_2}\) is defined in 3 , the conditional distribution \(P_{X_1,X_2|Y_1,Y_2}\) is given as \[\begin{align} P_{X_1,X_2|Y_1,Y_2}(x_1,x_2|y_1,y_2):= |\overline{\cal Y}|^2 P_{X_1,Y_1,X_2,Y_2}(x_1,y_1,x_2,y_2). \end{align}\] We call a Hermitian matrix \(R_{1,2}\) with trace \(1\) on \({\cal H}_{(I,1)} \otimes {\cal H}_{(I,2)}\) a pseudo density matrix (PDM).
Definition 1. We say that a PDM \(R_{1,2}\) is compatible with the distribution \(P_{X_1,Y_1,X_2,Y_2}\) when the relation \[\begin{align} \mathrm{Tr}(G_{t_1}\otimes G_{t_2})R_{1,2}= \langle G_{t_1}\otimes G_{t_2}\rangle_{P_{X_1,Y_1,X_2,Y_2}}\label{VBSA} \end{align}\qquad{(1)}\] holds for \(t_1 \in {\cal T}_1\), \(t_2 \in {\cal T}_2\), where \[\begin{align} \langle G_{t_1}\otimes G_{t_2}\rangle_{P_{X_1,Y_1,X_2,Y_2}} := \sum_{x_1,x_2} g_{x_1|t_1} g_{x_2|t_2} P_{X_1,X_2|Y_1,Y_2}(x_1,x_2|y(t_1),y(t_2)).\label{BAS} \end{align}\qquad{(2)}\] We say that a density matrix \(\rho_1\) on \({\cal H}_{(I,1)}\) is compatible with the distribution \(P_{X_1,Y_1,X_2,Y_2}\) when the relation \[\begin{align} \mathrm{Tr}G_{t_1} \rho_1= \langle G_{t_1}\otimes I \rangle_{P_{X_1,Y_1,X_2,Y_2}}\label{VBSA1} \end{align}\qquad{(3)}\] holds for \(t_1 \in {\cal T}_1\). The compatibility of \(\rho_2\) on \({\cal H}_{(I,2)}\) is defined in the same way.
Lemma 1. For \(P_{X_1,Y_1,X_2,Y_2} \in {\cal M}[{\cal S}_G]\), there exist a density matrix \(\rho_1\) on \({\cal H}_{(I,1)}\) and a density matrix \(\rho_2\) on \({\cal H}_{(I,2)}\) that are compatible with the distribution \(P_{X_1,Y_1,X_2,Y_2}\).
Proof. Let \(\rho_1\) be Alice’s input state when Bob makes nothing, and \(\rho_2\) be Bob’s input state when Alice makes nothing. Then, \(\rho_1\) satisfies the condition ?? . Hence, the density matrix \(\rho_1\) on \({\cal H}_{(I,1)}\) is compatible with the distribution \(P_{X_1,Y_1,X_2,Y_2}\). Similarly, the density matrix \(\rho_2\) on \({\cal H}_{(I,2)}\) is compatible with the distribution \(P_{X_1,Y_1,X_2,Y_2}\). ◻
Any parallel strategy is completely specified by a bipartite initial state \(\rho\) on \({\cal H}_{(I,1)}\otimes {\cal H}_{(I,2)}\), since Charlie introduces no sequential dependence between Alice and Bob. We denote the corresponding parallel strategy by \(S_P(\rho)\). For such a strategy, the state \(\rho\) itself is compatible with the observed distribution \({\cal M}[S_P(\rho)]\). The following relation shows that the converse also holds: a distribution belongs to \({\cal M}[{\cal S}_P]\) if and only if it satisfies the two non-signalling conditions characteristic of parallel strategies and admits a compatible bipartite density matrix; \[\begin{align} {\cal M}[{\cal S}_{P}] = \left\{ P_{X_1,Y_1,X_2,Y_2} \in {\cal P}_{\cal U} \left| \begin{array}{l} P_{X_1,Y_1,X_2,Y_2}\;satisfiesX_1-Y_1-Y_2 \;and\;Y_1-Y_2-X_2,\\ there exists a bipartite density matrix\rhoon{\cal H}_{(I,1)}\otimes{\cal H}_{(I,2)}\\ that is compatible withP_{X_1,Y_1,X_2,Y_2}. \end{array} \right. \right\}. \label{ZL1} \end{align}\tag{9}\] Hence, the existence of a compatible density matrix is essential.
An individual strategy is precisely a parallel strategy whose initial bipartite state is a product state. Accordingly, within the characterization of parallel strategies in 9 , the additional Markovian condition \(X_1-Y_1-Y_2-X_2\) singles out the individual class, since for fixed measurement choices it expresses the independence of the two observed outcomes \(X_1\) and \(X_2\). Moreover, for any distribution in \({\cal M}[{\cal S}_G]\), Lemma 1 provides compatible local density matrices \(\rho_1\) on \({\cal H}_{(I,1)}\) and \(\rho_2\) on \({\cal H}_{(I,2)}\); under the condition \(X_1-Y_1-Y_2-X_2\), the product state \(\rho_1\otimes\rho_2\) is itself compatible with the distribution. Therefore, \[\begin{align} {\cal M}[{\cal S}_{I}] &= \left\{P_{X_1,Y_1,X_2,Y_2} \in{\cal P}_{\cal U} \left| \begin{array}{l} P_{X_1,Y_1,X_2,Y_2}satisfiesX_1-Y_1-Y_2-X_2,\\ there exists a compatible density matrix \rhowith P_{X_1,Y_1,X_2,Y_2} \end{array} \right. \right\} \notag\\ &= \{P_{X_1,Y_1,X_2,Y_2} \in {\cal M}[{\cal S}_G] \mid P_{X_1,Y_1,X_2,Y_2}satisfiesX_1-Y_1-Y_2-X_2 \}. \label{CWSI} \end{align}\tag{10}\]
Equation 10 should be compared with the parallel characterization 9 . There, the class \({\cal M}[{\cal S}_P]\) is described by the two directional non-signalling conditions \(X_1-Y_1-Y_2\) and \(Y_1-Y_2-X_2\), together with the existence of a compatible bipartite density matrix. The present condition 10 strengthens this by requiring that the compatible bipartite state be effectively a product state. At the distribution level, this product structure is reflected exactly by the additional Markovian condition \(X_1-Y_1-Y_2-X_2\), which expresses the conditional independence of the two local outcomes once the measurement choices are fixed. In this sense, the individual class is obtained from the parallel class by imposing the extra independence structure that removes all nontrivial input correlations between Alice and Bob.
We now specialize the general state-reduction framework to the case of projective state reduction. The purpose of this section is to record what can already be derived in this more concrete setting by combining projective measurements with the pseudo-density-matrix formalism. In particular, we obtain compatibility and reconstruction results for memoryless sequential strategies, and clarify how these results can be used to study order-indistinguishability at the distribution level. These observations will serve as the basis for the sharper characterization developed in the next section.
To make more detailed analysis, we focus on the case with projective state reductions. For an element \(y_i \in {\cal Y}_i\), we choose a projection-valued measure \(\{E_{x_i|y_i}^i\}_{x_i \in {\cal X}_{y_i}}\), where \(E_{x_i|y_i}^i\) is a projection for \(x_i\in {\cal X}_{y_i}\). Then, we choose the measurement process \(\{\Gamma_{x_i|y_i}^i\}_{x_i \in {\cal X}_{y_i}}\) in the formalism of Section 3 as \(\Gamma_{x_i|y_i}^i(\rho):= E_{x_i|y_i}^i\rho E_{x_i|y_i}^i\). Hence, we have \(\mathrm{Tr}_{(O,i)} C[\Gamma_{x_i|y_i}^i] =\Gamma_{x_i|y_i}^{i,\dagger}(I) =E_{x_i|y_i}^i\).
We study the class \({\cal S}_{N|1\to 2}\). When Charlie’s strategy \(S\) belongs to the classes \({\cal S}_{N|1\to 2}\), the strategy is written as the pair of Alice’s initial state \(\rho_1\) before the measurement by Alice and the Choi matrix \(C_{1\to 2}\) of the TP-CP map that Charlie applies to Alice’s output state. In the following, we denote this strategy by \(S_{1\to 2}(\rho_1,C_{1\to 2})\). The aim of this section is the development of a method to identify whether the distribution \({\cal M}[S_{1\to 2}(\rho_1,C_{1\to 2})]\) belongs to \({\cal M}[{\cal S}_{N,1\to 2}]\cap {\cal M}[{\cal S}_{N,2\to 1}]\).
Definition 2. We say that a pair of a Hermitian matrix \(C_{1\to 2}\) on \({\cal H}_{(I,1)}\otimes {\cal H}_{(I,2)}\) and a density matrix \(\rho_1\) on \({\cal H}_{(I,1)}\) is compatible with \(P_{X_1,Y_1,X_2,Y_2}\) via a pseudo density matrix \(R_{1,2}\) when \(R_{1,2}\) is compatible with \(P_{X_1,Y_1,X_2,Y_2}\) and \[\begin{align} \rho_1=\mathrm{Tr}_2 R_{1,2}, \quad \mathrm{Tr}_2 {C}_{1\to 2}=I, \quad R_{1,2}= C_{1\to 2}^{T_1} \circ (\rho_{1} \otimes I)\label{XS} , \end{align}\qquad{(4)}\] where \(T_i\) expresses the partial transpose of the \(i\)-th system.
When \(\rho_1\) is invertible, the second condition in ?? is implied by the other two conditions, because they yield \(\rho_1 = (\mathrm{Tr}_2 C_{1\to2}) \circ \rho_1\).
Similarly, we say that a pair of a Hermitian matrix \(C_{2\to 1}\) on \({\cal H}_{(I,1)}\otimes {\cal H}_{(I,2)}\) and a density matrix \(\rho_2\) on \({\cal H}_{(I,2)}\) is compatible with \(P_{X_1,Y_1,X_2,Y_2}\) via a pseudo density matrix \(R_{1,2}\) when \(R_{1,2}\) is compatible with \(P_{X_1,Y_1,X_2,Y_2}\) and \[\begin{align} \rho_2=\mathrm{Tr}_1 R_{1,2}, \quad \mathrm{Tr}_1 {C}_{2\to 1}=I, \quad R_{1,2}= C_{2\to 1}^{T_2} \circ (I \otimes \rho_{2} )\label{XS2} . \end{align}\tag{11}\] In particular, we omit the word “via a pseudo density matrix \(R_{1,2}\)” in the above definitions.
Lemma 2. A pair of a Hermitian matrix \(C_{1\to 2}\) on \({\cal H}_{(I,1)}\otimes {\cal H}_{(I,2)}\) and a density matrix \(\rho_1\) on \({\cal H}_{(I,1)}\) is compatible with \(P_{X_1,Y_1,X_2,Y_2}\) if and only if the relations \[\begin{align} \mathrm{Tr}_2 {C}_{1\to 2}&=I \label{MGH} \\ \langle G_{t_1}\otimes G_{t_2} \rangle_{P_{X_1,Y_1,X_2,Y_2}} &=\mathrm{Tr} \big({C}_{1\to 2}^{T_1} \circ\big( \rho_1 \otimes I \big)\big) \big( G_{t_1} \otimes G_{t_2}\big) \label{FN3Y} \end{align}\] {#eq: sublabel=eq:MGH,eq:FN3Y} hold for \(t_1,t_2 \in {\cal T}\).
Proof. Assume that a pair of a Hermitian matrix \(C_{1\to 2}\) on \({\cal H}_{(I,1)}\otimes {\cal H}_{(I,2)}\) and a density matrix \(\rho_1\) on \({\cal H}_{(I,1)}\) is compatible with \(P_{X_1,Y_1,X_2,Y_2}\) via \(R_{1,2}\). The combination of ?? and the third equation of ?? yields ?? . Assume the relations ?? and ?? . We set \(R_{1,2}\) to be \(\big({C}_{1\to 2}^{T_1} \circ\big( \rho_1 \otimes I \big)\big)\). The relation ?? guarantees that \(R_{1,2}\) is compatible with \(P_{X_1,Y_1,X_2,Y_2}\), which implies the third equation of ?? . The combination of the ?? and the third equation of ?? implies the first equation of ?? . ◻
Theorem 2.
Alice’s initial state \(\rho_1\) is compatible with \({\cal M}[S_{1\to 2}(\rho_1,C_{1\to 2})]\).
Proof. The initial state \(\rho_1\) on Alice’s system satisfies \[\begin{align} P_{X_1|Y_1,Y_2}(x_1|y_1,0) = \mathrm{Tr}\rho_1 \Gamma_{x_1|y_1}^{1,\dagger}(I) = \mathrm{Tr}\rho_1 E_{x_1|y_1}^1 \end{align}\] for \(x_1,y_1\). Since the distribution \({\cal M}[S_{1\to 2}(\rho_1,C_{1\to 2})]\) satisfies the Markovian chain \(X_1-Y_1-Y_2\), \(\rho_1\) is compatible with \({\cal M}[S_{1\to 2}(\rho_1,C_{1\to 2})]\). ◻
To generalize [17], [19], we introduce the following condition for \(\{G_t\}_{t \in {\cal T}}\) and \(\{E_y\}_{y\in {\cal Y}}\):
For any \(t_i \in {\cal T}_i\setminus \{0\}\), \(G_{t_i}^2\) is a constant times of the identity operator. Also, \(E_{y(t_i)}\) is the spectral decomposition of \(G_{t_i}\), i.e., \(|{\cal X}_{y(t_i)}|=2\) and any two distinct elements \(x_i,x_i'\) in \({\cal X}_{y(t_i)}\) satisfy the condition \(g_{x_i|y(t_i),z(t_i)}\neq g_{x_i'|y(t_i),z(t_i)} \neq 0\).
Pauli matrices on the \(n\)-qubit system satisfies the above condition.
Lemma 3. Assume the condition (A2).
We have \[\begin{align} \langle G_{t_1}\otimes G_{t_2} \rangle_{{\cal M}[S_{1\to 2}(\rho_1,C_{1\to 2})]} =\mathrm{Tr} \big({C}_{1\to 2}^{T_1} \circ\big( \rho_1 \otimes I \big)\big) \big( G_{t_1} \otimes G_{t_2}\big) \label{FN3} \end{align}\qquad{(5)}\] for \(t_1,t_2 \in {\cal T}\).
Proof. In this proof, we denote the joint distribution \({\cal M}[S_{1\to 2}(\rho_1,C_{1\to 2})]\) by \(P_{X_1,Y_1,X_2,Y_2}(x_1,y_1,x_2,y_2)\). Due to the condition (A2), for \(t (\neq 0)\in {\cal T}\), the set \({\cal X}_{y(t)}\) is composed of two elements. For simplicity, we assume that \({\cal X}_{y(t)} =\{0,1\}\) and \(g_{x|t}=(-1)^x \kappa_{t}\), where \(G_{t}^2=\kappa_{t}^2 I\). Hence, we have \(G_{t}= \kappa_t E_{0|y(t)} -\kappa_t E_{0|y(t)}\). Since \[\begin{align} &\sum_{x_1} g_{x_1|t} E_{x_1|y(t)} \rho_{1} E_{x_1|y(t)} \notag \\ =& \kappa_{t} E_{0|y(t)} \rho_{1} E_{0|y(t)} -\kappa_{t} E_{1|y(t)} \rho_{1} E_{1|y(t)} \notag \\ =& \frac{1}{2} \Big(\kappa_t E_{0|y(t)} \rho_{1} E_{0|y(t)} -\kappa_t E_{0|y(t)} \rho_{1} E_{1|y(t)} +\kappa_t E_{1|y(t)} \rho_{1} E_{0|y(t)} -\kappa_t E_{1|y(t)} \rho_{1} E_{1|y(t)} \Big)\notag \\ &+\frac{1}{2} \Big(\kappa_t E_{0|y(t)} \rho_{1} E_{0|y(t)} +\kappa_t E_{0|y(t)} \rho_{1} E_{1|y(t)} -\kappa_t E_{1|y(t)} \rho_{1} E_{0|y(t)} -\kappa_t E_{1|y(t)} \rho_{1} E_{1|y(t)} \Big)\notag \\ =& \frac{1}{2} \Big(E_{0|y(t)} \rho_{1} G_{t} +E_{1|y(t)} \rho_{1} G_{t} +G_{t} \rho_{1} E_{0|y(t)} +G_{t} \rho_{1} E_{0|y(t)} \Big) = \rho_1 \circ G_{t}, \end{align}\] we have \[\begin{align} &\langle G_{t_1}\otimes G_{t_2} \rangle_{{\cal M}[S_{1\to 2}(\rho_1,C_{1\to 2})]} \notag \\ =& \sum_{x_1,x_2\in {\cal X}} g_{x_1|t_1} g_{x_2|t_2} P_{X_1,X_2|Y_1,Y_2}(x_1,x_2|y(t_1),y(t_2)) \notag \\ =& \sum_{x_1,x_2}g_{x_1|t_1}g_{x_2|t_2} \mathrm{Tr}_2 \Big(\mathrm{Tr}_1 \Big({C}_{1\to 2}^{T_1} (E_{x_1|y(t_1)}\rho_{1}E_{x_1|y(t_1)} \otimes I) E_{x_2|y(t_2)}\Big) \notag \\ =&\mathrm{Tr} {C}_{1\to 2}^{T_1} \Big( \Big(\sum_{x_1} g_{x_1|t_1} E_{x_1|y(t_1)} \rho_{1} E_{x_1|y(t_1)}\Big) \otimes G_{y(t_2)}\Big) \notag \\ =&\mathrm{Tr} {C}_{1\to 2}^{T_1} \big( (\rho_1 \circ G_{t_1}) \otimes G_{t_2}\big) =\mathrm{Tr} {C}_{1\to 2}^{T_1} \big( \rho_1 \otimes I \big) \circ \big( G_{t_1} \otimes G_{t_2}\big) =\mathrm{Tr} {C}_{1\to 2}^{T_1} \circ\big( \rho_1 \otimes I \big) \big( G_{t_1} \otimes G_{t_2}\big) .\label{FN3B} \end{align}\tag{12}\] ◻
Lemma 3 implies that the PDM \({C}_{1\to 2}^{T_1} \circ\big( \rho_1 \otimes I \big)\) is compatible with \({\cal M}[S_{1\to 2}(\rho_1,C_{1\to 2})]\). Hence, we obtain the following theorem.
Theorem 3. Assume the condition (A2).
Then, the pair of a Hermitian matrix \(C_{1\to 2}\) on \({\cal H}_{(I,1)}\otimes {\cal H}_{(I,2)}\) and a density matrix \(\rho_1\) on \({\cal H}_{(I,1)}\) is compatible with the distribution \({\cal M}[S_{1\to 2}(\rho_1,C_{1\to 2})]\).
Then, Theorem 3 implies the following corollary.
Corollary 1. Assume the condition (A2). We have \[\begin{align} &{\cal M}[{\cal S}_{N,1\to 2}]\notag \\ =& \left \{P_{X_1,Y_1,X_2,Y_2} \in {\cal P}_{\cal U} \left| \begin{array}{l} There exists a compatible pair of\rho_1\ge 0 and C_{1\to 2}\ge 0 withP_{X_1,Y_1,X_2,Y_2} \\ such thatP_{X_1,Y_1,X_2,Y_2}={\cal M}[S_{1\to 2}(\rho_1, C_{1\to 2})]. \end{array} \right.\right\} \label{FFT1}. \end{align}\qquad{(6)}\]
Proof. The relation \(\supset\) is trivial. The relation \(\subset\) follows from Theorem 3. ◻
This is still not a direct distribution-level test, because it is formulated through the existence of a generating compatible pair. Its role is only to prepare the sharper criterion given later.
Lemma 4. Assume the condition (A2). Assume that a pair of a Hermitian matrix \(C_{1\to 2}\) on \({\cal H}_{(I,1)}\otimes {\cal H}_{(I,2)}\) and a density matrix \(\rho_1\) on \({\cal H}_{(I,1)}\) is compatible with \(P_{X_1,Y_1,X_2,Y_2}\). Then, the PDM \({C}_{1\to 2}^{T_1} \circ\big( \rho_1 \otimes I \big)\) is compatible with \({\cal M}[{\cal S}_{1\to 2}(\rho_1,C_{1\to 2})]\). That is, the distribution \(P_{X_1,Y_1,X_2,Y_2}\) satisfies
\[\begin{align} \langle G_{t_1}\otimes G_{t_2} \rangle_{P_{X_1,Y_1,X_2,Y_2}} =\mathrm{Tr}\big({C}_{1\to 2}^{T_1} \circ\big( \rho_1 \otimes I \big)\big) \big( G_{t_1} \otimes G_{t_2}\big) =\langle G_{t_1}\otimes G_{t_2} \rangle_{{\cal M}[S_{1\to 2}(\rho_1,C_{1\to 2})]} \label{FN334} \end{align}\qquad{(7)}\] for \(t_1,t_2 \in {\cal T}\).
Proof. It is sufficient to show ?? . The first equation of ?? follows from ?? . The second equation of ?? follows from ?? of Lemma 3. ◻
Lemma 5. Assume the condition (A2). Assume that a pair of a Hermitian matrix \(C_{1\to 2}\) on \({\cal H}_{(I,1)}\otimes {\cal H}_{(I,2)}\) and a density matrix \(\rho_1\) on \({\cal H}_{(I,1)}\) is compatible with \(P_{X_1,Y_1,X_2,Y_2}\in {\cal M}[{\cal S}_G]\). When the distributions \(P_{X_1,Y_1,X_2,Y_2}\) and \({\cal M}[S_{1\to 2}(\rho_1, C_{1\to 2})]\) satisfy the Markovian conditions \(X_1-Y_1-Y_2\) and \(Y_1-Y_2-X_2\), we have \({\cal M}[S_{1\to 2}(\rho_1, C_{1\to 2})]=P_{X_1,Y_1,X_2,Y_2}\).
Proof. In this proof, we denote the distribution \({\cal M}[S_{1\to 2}(\rho_1, {C}_{1\to 2})\) by \(\tilde{P}_{X_1,Y_1,X_2,Y_2}\). It is sufficient to show the relation \[\begin{align} \tilde{P}_{X_1,X_2|Y_1=y_1,Y_2=y_2}= P_{X_1,X_2|Y_1=y_1,Y_2=y_2} \end{align}\] for \(y_1,y_2 \in {\cal Y}\). To show the above relation, it is sufficient to show that these two conditional distributions have the same expectation of the three random variables \(g_{X_1|y_1,z_1}\), \(g_{X_2|y_2,z_2}\), \(g_{X_1|y_1,z_1} g_{X_2|y_2,z_2}\) for \(z_1\in {\cal Z}_{y_1}\), \(z_2\in {\cal Z}_{y_2}\) because these random variables span the linear function space on \({\cal X}_{y_1}\times {\cal X}_{y_2}\) with the constant function.
Since their expectations of the variable \(g_{X_1|y_1,z_1} g_{X_2|y_2,z_2}\) is given as \(\langle G_{y_1,z_1}\otimes G_{y_1,z_1} \rangle_{\tilde{P}_{X_1,Y_1,X_2,Y_2}}\) and \(\langle G_{y_1,z_1}\otimes G_{y_2,z_2} \rangle_{{P}_{X_1,Y_1,X_2,Y_2}}\), ?? guarantees that these two expectations take the same value. The Markovian chain condition \(X_1-Y_1-Y_2\) guarantees that their expectations of the variable \(g_{X_1|y_1,z_1}\) is given as \(\langle G_{y_1,z_1}\otimes I \rangle_{\tilde{P}_{X_1,Y_1,X_2,Y_2}}\) and \(\langle G_{y_1,z_1}\otimes I \rangle_{{P}_{X_1,Y_1,X_2,Y_2}}\). Then, ?? guarantees that these two expectations take the same value. Also, in the same way, we can show that their expectations of the variable \(g_{X_2|y_2,z_2}\) take the same value. ◻
Lemma 6. Assume the condition (A2). Given \(S \in {\cal S}_{N|1\to 2}\), we assume that a pair of a Hermitian matrix \(C_{2\to 1}\) on \({\cal H}_{(I,1)}\otimes {\cal H}_{(I,2)}\) and a density matrix \(\rho_2\) on \({\cal H}_{(I,2)}\) is compatible with \({\cal M}[S]\). The following conditions are equivalent.
The distributions \({\cal M}[S]\) and \({\cal M}[S_{2\to 1}(\rho_2, C_{2\to 1})]\) satisfy the Markovian conditions \(X_1-Y_1-Y_2\) and \(Y_1-Y_2-X_2\).
\({\cal M}[S_{2\to 1}(\rho_2, C_{2\to 1})]={\cal M}[S]\).
Proof. Lemma 5 guarantees that the condition (B1) implies the condition (B2). The opposite direction can be shown as follows. When the relation \({\cal M}[S_{2\to 1}(\rho_2, C_{2\to 1})] ={\cal M}[S]\) holds, the distribution \({\cal M}[S_{2\to 1}(\rho_2, C_{2\to 1})] ={\cal M}[S]\) satisfies the Markovian conditions \(X_1-Y_1-Y_2\) and \(Y_1-Y_2-X_2\) because \({\cal M}[S]\) satisfies the Markovian condition \(X_1-Y_1-Y_2\) and \({\cal M}[S_{2\to 1}(\rho_2, C_{2\to 1})]\) satisfies the Markovian condition \(Y_1-Y_2-X_2\). ◻
The above conditions do not guarantee the condition \({C}_{2\to 1}\ge 0\). Adding this condition, we have the following corollary of Lemma 6.
Corollary 2. Assume the condition (A2). We have \[\begin{align} &{\cal M}[{\cal S}_{N,1\to 2}] \cap {\cal M}[{\cal S}_{N,2\to 1}] \\ =& \left\{{\cal M}[S]\in {\cal M}[{\cal S}_{N,1\to 2}]\left| \begin{array}{l} There exists a compatible pair of a Hermitian matrix C_{2\to 1}\ge 0 on {\cal H}_{(I,1)}\otimes {\cal H}_{(I,2)} \\ and a density matrix \rho_2 on {\cal H}_{(I,2)}with{\cal M}[S]such that{\cal M}[S]={\cal M}[S_{2\to 1}(\rho_2, C_{2\to 1})] \end{array} \right.\right \} \notag\\ =& \left\{{\cal M}[S]\in {\cal M}[{\cal S}_{N,1\to 2}]\left| \begin{array}{l} There exists a compatible pair of a Hermitian matrix C_{2\to 1}\ge 0 on {\cal H}_{(I,1)}\otimes {\cal H}_{(I,2)} \\ and a density matrix \rho_2 on {\cal H}_{(I,2)}with{\cal M}[S] \\ such that{\cal M}[S_{2\to 1}(\rho_2, C_{2\to 1})]and {\cal M}[S]satisfy X_1-Y_1-Y_2,Y_1-Y_2-X_2. \end{array} \right.\right\} . \end{align}\]
Corollary 1 is useful for identifying whether a strategy \(S \in {\cal S}_{N,1\to 2}\) satisfies the condition \({\cal M}[S] \in {\cal M}[{\cal S}_{N,1\to 2}]\cap {\cal M}[{\cal S}_{N,2\to 1}]\). This characterization of \({\cal M}[{\cal S}_{N,1\to 2}]\cap {\cal M}[{\cal S}_{N,2\to 1}]\) can be used for making its concrete description in Section 7. When the observed distribution belongs to the set \({\cal M}[{\cal S}_{N,1\to 2}] \cap {\cal M}[{\cal S}_{N,2\to 1}]\), we cannot distinguish which order, \(1\to 2\) or \(2 \to 1\), is correct. The characterization of this set clarifies the limitation of the method of projective state reduction.
We next explain how the preceding analysis extends beyond the memoryless case. Once the projective-state-reduction setting has been fixed, strategies with memory can still be related to the memoryless framework in a structurally simple way. For classical memory, the observed distribution is obtained by probabilistically mixing memoryless sequential strategies. For quantum memory, the same idea survives after enlarging the input space so as to include Charlie’s memory system.
In the class \({\cal S}_{C,1\to 2}\), Charlie has a classical memory system \(M\). Conditioned on the hidden memory value \(M=m\), Charlie chooses an initial state \(\rho_{1,m}\ge 0\) and a Choi matrix \(C_{1\to 2,m}\ge 0\). Thus, for each fixed value of \(m\), the corresponding branch is an ordinary memoryless sequential strategy. Since the value of \(M\) is not observed by Alice and Bob, the actual observed distribution is obtained only after averaging over \(m\). Hence \({\cal M}[{\cal S}_{C,1\to 2}]\) is given by the convex hull of \({\cal M}[{\cal S}_{N,1\to 2}]\) characterized in ?? .
It is important, however, not to confuse this branchwise positivity with the positivity of the reconstructed Choi matrix obtained from the coarse-grained observed distribution. Even though every branch satisfies \(C_{1\to 2,m}\ge 0\), the reconstructed matrix \(\hat{C}_{1\to 2}[P]\) associated with the averaged distribution \(P\) need not be positive, because the hidden classical memory variable \(M\) is not accessible at the distribution level. In particular, as shown later in Example 4 in Section 9, positivity of \(\hat{C}_{1\to 2}[P]\) can already fail in the classical-memory subclass.
Next, we study the class \({\cal S}_{Q,1\to 2}\). In this case, without loss of generality, we may assume that Charlie’s quantum memory system \({\cal H}_{M,1}\) has the same dimension as \({\cal H}_{(I,1)}\). Then Alice’s projective measurement can be regarded as \(\{E_{x_1\mid y_1}^1\otimes I\}_{x_1\in{\cal X}_{y_1}}\) on the enlarged system \({\cal H}_{(I,1)}\otimes{\cal H}_{(M,1)}\). In this sense, a quantum-memory strategy can be viewed as an ordinary sequential strategy on a larger input space. Hence, the analysis of Section 5.1 applies to \({\cal S}_{Q,1\to 2}\) after this enlargement of the input space.
In the above analysis, the compatible pair cannot be uniquely determined. For the uniqueness, we assume the following condition.
The set \(\{G_{y_i,z_i}\}_{y_i\in {\cal Y}_i, z_i\in {\cal Z}_{y_i}}\cup \{I\}\) forms a basis of \({\cal B}({\cal H}_{(I,i)})\).
Under the above condition, \({\cal T}_i\) is given as \(\{0,1,\ldots, d_i^2-1\}\).
Example 1. In a typical case, we set \({\cal Y}_1={\cal Y}_2=\{1, \ldots, d+1\}\), and \({\cal X}_y=\{1, \ldots, d\}\) and \({\cal Z}_y=\{1, \ldots, d-1\}\) for \(y \in {\cal Y}\). Then, we choose a rank-one projection-valued measure \(\{E_{x|y}\}_{x \in {\cal X}_y}\). When \(d\) is a prime power, there are \(d+1\) mutually unbiased bases. When \(d+1\) rank-one projection-valued measures \(\{E_{x|y}\}_{x \in {\cal X}_y}\) constructed from \(d+1\) mutually unbiased bases, the condition (A1) holds.
To describe the unique compatible elements, we employ the dual basis \(\{H_t\}_{t \in {\cal T}}\) of \(\{ G_t\}_{t \in {\cal T}}\) as \[\begin{align} \mathrm{Tr}G_{t'}H_t=\delta_{t',t}.\label{dualB} \end{align}\tag{13}\] Thus, \(H_0=\frac{1}{d}I\). Then, the unique compatible PDM with \(P_{X_1,Y_1,X_2,Y_2}\) is given as \[\begin{align} {\cal R}[P_{X_1,Y_1,X_2,Y_2}]:= \sum_{t_1,t_2\in {\cal T}} \langle G_{t_1}\otimes G_{t_2}\rangle_{P_{X_1,Y_1,X_2,Y_2}} H_{t_1}\otimes H_{t_2} \end{align}\] because the unique compatiblility with \(P_{X_1,Y_1,X_2,Y_2}\) is guaranteed by the relation \[\begin{align} \mathrm{Tr}(G_{t_1}\otimes G_{t_2}){\cal R}[P_{X_1,Y_1,X_2,Y_2}]= \langle G_{t_1}\otimes G_{t_2}\rangle_{P_{X_1,Y_1,X_2,Y_2}}\label{VBSA2} \end{align}\tag{14}\] for \(t_1,t_2 \in {\cal T}\). Also, the matrices \(\rho_1[P_{X_1,Y_1,X_2,Y_2}]\) and \(\rho_2[P_{X_1,Y_1,X_2,Y_2}]\) defined below are compatible with \(P_{X_1,Y_1,X_2,Y_2}\); \[\begin{align} \rho_1[P_{X_1,Y_1,X_2,Y_2}]:=\mathrm{Tr}_2 {\cal R}[P_{X_1,Y_1,X_2,Y_2}],\quad \rho_2[P_{X_1,Y_1,X_2,Y_2}]:=\mathrm{Tr}_1 {\cal R}[P_{X_1,Y_1,X_2,Y_2}]. \label{HJK1} \end{align}\tag{15}\] For \(P_{X_1,Y_1,X_2,Y_2} \in {\cal M}[{\cal S}_G]\), \(\rho_1[P_{X_1,Y_1,X_2,Y_2}]\) is the unique compatible matrix with \(P_{X_1,Y_1,X_2,Y_2}\). Hence, Lemma 1 implies the following lemma.
Lemma 7. For \(P_{X_1,Y_1,X_2,Y_2} \in {\cal M}[{\cal S}_G]\), \(\rho_1[P_{X_1,Y_1,X_2,Y_2}]\) is the unique compatible density matrix with \(P_{X_1,Y_1,X_2,Y_2}\).
Further, when \(\rho_1[P_{X_1,Y_1,X_2,Y_2}]>0\), the Hermitian matrix \(\tilde{C}_{1\to 2}[P_{X_1,Y_1,X_2,Y_2}]\) on \({\cal H}_{(I,1)}\otimes {\cal H}_{(I,2)}\) satisfying the following condition is uniquely determined; The pair of the Hermitian matrix \(C_{1\to 2}\) and a density matrix \(\rho_1[P_{X_1,Y_1,X_2,Y_2}]\) is compatible with \(P_{X_1,Y_1,X_2,Y_2}\), i.e., \[\begin{align} {\cal R}[P_{X_1,Y_1,X_2,Y_2}] = \tilde{C}_{1\to 2}[P_{X_1,Y_1,X_2,Y_2}]^{T_1}\circ (\rho_{1}[P_{X_1,Y_1,X_2,Y_2}] \otimes I)\label{XSA} . \end{align}\tag{16}\] Similarly, when \(\rho_2[P_{X_1,Y_1,X_2,Y_2}]>0\), we define \(\tilde{C}_{2\to 1}[P_{X_1,Y_1,X_2,Y_2}]\) as \[\begin{align} {\cal R}[P_{X_1,Y_1,X_2,Y_2}] = \tilde{C}_{2\to 1}[P_{X_1,Y_1,X_2,Y_2}]^{T_2}\circ (I \otimes \rho_{2}[P_{X_1,Y_1,X_2,Y_2}] ).\label{XSA2} \end{align}\tag{17}\]
When the conditions (A1) and (A2) holds, Theorem 2 guarantees \[\begin{align} \rho_1[{\cal M}[S_{1\to 2}(\rho_1,C_{1\to 2})]]=\rho_1. \end{align}\] In addition, when \(\rho_1>0\), Theorem 3 guarantees the relation \[\begin{align} \tilde{C}_{1\to 2}[{\cal M}[S_{1\to 2}(\rho_1,C_{1\to 2})]] =C_{1\to 2}. \end{align}\] That is, the pair \((\rho_1[{\cal M}[S_{1\to 2}(\rho_1,C_{1\to 2})]]\), \(\tilde{C}_{1\to 2}[{\cal M}[S_{1\to 2}(\rho_1,C_{1\to 2})]])\) recovers the joint distribution \({\cal M}[S_{1\to 2}(\rho_1,C_{1\to 2})]\). This statement is a generalization of [17], [19].
Next, we characterize parallel strategies. When the condition (A1) holds, for a given distribution \(P_{X_1,Y_1,X_2,Y_2}\) we have the unique compatible PDM \({\cal R}[P_{X_1,Y_1,X_2,Y_2}]\), and it equals the true density matrix \(\rho\) on \({\cal H}_{I,1} \otimes {\cal H}_{I,2}\), which implies the relation \({\cal R}[P_{X_1,Y_1,X_2,Y_2}] \ge 0\). Since there exists a one-to-one correspondence between the PDM \({\cal R}[P_{X_1,Y_1,X_2,Y_2}]\) and the joint distribution \(P_{X_1,Y_1,X_2,Y_2}\), we have the following relation under the condition (A1); \[\begin{align} {\cal M}[{\cal S}_{P}] = \{P \mid {\cal R}[P_{X_1,Y_1,X_2,Y_2}] \ge 0 \}. \label{FG1VS} \end{align}\tag{18}\] This characterization is closely related to the PDM viewpoint of Liu et al. [16], according to which spatial correlations are represented by ordinary density matrices while temporal correlations are encoded only in a coarse-grained manner. In the present restricted-observation framework, under (A1), this yields the distribution-level criterion 18 . In addition, since the unique compatible PDM \({\cal R}[P_{X_1,Y_1,X_2,Y_2}]\) uniquely determines the joint distribution \(P_{X_1,Y_1,X_2,Y_2}\), the relation \({\cal R}[P_{X_1,Y_1,X_2,Y_2}] \ge 0\) implies the non-signaling Markovian conditions of both directions \(Y_1-Y_2-X_2\) and \(X_1-Y_1-Y_2\).
In this section, we derive the necessary-and-sufficient characterization of memoryless sequential dynamics under restricted projective observations. We first show that, under projective state reduction, the observed distribution allows reconstruction of the effective Choi matrix. We then use this reconstruction to complete the characterization of \({\cal M}[{\cal S}_{N,1\to 2}]\).
The analysis of adaptive strategies without memory in Section 5.1 characterized the set \({\cal M}[{\cal S}_{N,1\to 2}]\) in Corollary 1. However, in order to decide whether a given joint distribution \(P_{X_1,Y_1,X_2,Y_2}\) belongs to \({\cal M}[{\cal S}_{N,1\to 2}]\), this characterization is still limited in two respects. First, it assumes the strong condition (A2) for \(\{G_{t_i}\}_{t_i \in {\cal T}_i}\) and \(\{E_{y_i}\}_{y_i\in {\cal Y}_i}\). Second, Corollary 1 does not yet provide a direct membership criterion in terms of the observed distribution itself. To obtain a more concrete characterization, we now replace (A2) by the following rank-one assumption, under which the effective Choi matrix can be reconstructed directly from \(P_{X_1,Y_1,X_2,Y_2}\).
Each operator \(E_{x_i|y_i}^i\) has rank one for any \(y_i \in {\cal Y}_i\) and \(x_i \in {\cal X}_{y_i}\). That is, \(E_{x_i|y_i}^i\) is written as \(|x_i,y_i\rangle\langle x_i,y_i|\) by using a vector \(|x_i,y_i\rangle\). [27]
Then, we will derive a more useful characterization of the set \({\cal M}[{\cal S}_{N,1\to 2}]\). For this aim, we describe the conditional distribution \(P_{X_2|X_1,Y_1,Y_2}\) as follows \[\begin{align} P_{X_2|X_1,Y_1,Y_2}(x_2|x_1,y_1,y_2) :=\frac{P_{X_1,X_2|Y_1,Y_2}(x_1,x_2|y_1,y_2) }{ \sum_{x_2'}P_{X_1,X_2|Y_1,Y_2}(x_1,x_2'|y_1,y_2)}. \end{align}\]
For example, Example 1 satisfies the condition (A3). We compare the cases satisfying conditions (A1) and (A2) with those satisfying conditions (A1) and (A3) under the \(n\)-qubit system. In the \(n\)-qubit system, the size \(|{\cal Y}|\) in Example 1 with the \(n\)-qubit system is \(2^n-1\). Next, we discuss the case satisfying the conditions (A1) and (A2) under the \(n\)-qubit system. In the \(n\)-qubit system, the Pauli matrices are given as \(\{\sigma_{s_1}\otimes \cdots \otimes \sigma_{s_n}\}_{ s_j \in \{0,1,2,3\}}\). When the identity matrix is excluded, the number of Pauli matrices is \(4^n-1\). We choose the set \({\cal Y}\) to be the set of Pauli matrices by excluding the identity matrix. Then, the projective measurement is decided to be the spectral decomposition of the corresponding Pauli matrix. This case satisfies the conditions (A1) and (A2). In this case, the size \(|{\cal Y}|\) is \(4^n-1\). That is, the latter case has a much larger size \(|{\cal Y}|\), i.e., the latter case requires so many types of experimental settings. In fact, the condition (A2) has the requirement \(|{\cal X}_{y(t)}|=2\), which considerably increases the size \(|{\cal Y}|\).
Under the condition (A3), we define \(\hat{C}_{1\to 2}[P_{X_1,Y_1,X_2,Y_2}]\) as \[\begin{align} &\hat{C}_{1\to 2}[P_{X_1,Y_1,X_2,Y_2}]\notag\\ :=& \sum_{t_1\neq 0, t_2} \sum_{x_1,x_2} g_{x_1|t_1} g_{x_2|t_2} P_{X_2|X_1,Y_1,Y_2}(x_2|x_1,y(t_1),y(t_2)) H_{t_1}^T \otimes H_{t_2}\notag\\ &+ \sum_{t_2} \Big(\sum_{x_2} g_{x_2|t_2} P_{X_2|Y_1,Y_2}(x_2|0,y(t_2)) \notag\\ &-\sum_{t_1,t_1'\neq 0} \sum_{x_1} g_{x_1|t_1} P_{X_1|Y_1}(x_1|y(t_1)) h_{t_1,t_1'} \sum_{x_1',x_2} g_{x_1'|t_1'} g_{x_2|t_2} P_{X_2|X_1,Y_1,Y_2}(x_2|x_1',y(t_1'),y(t_2)) \Big) I \otimes H_{t_2},\label{VBTR} \end{align}\tag{19}\] where \(h_{t,t'}\) is defined as \[\begin{align} h_{t,t'}:=\mathrm{Tr}H_{t} H_{t'}. \end{align}\] The significance of this definition is that, for distributions generated by memoryless sequential strategies, the matrix \(\hat{C}_{1\to 2}[P_{X_1,Y_1,X_2,Y_2}]\) recovers the underlying Choi matrix exactly.
Theorem 4. Assume the conditions (A1) and (A3). Then, \(\hat{C}_{1\to 2}[{\cal M}[S_{1\to 2}(\rho_1,C_{1\to 2})]]\) equals the Choi matrix \(C_{1\to 2}\).
The combination of Theorems 2 and 4 guarantees that the pair \((\rho_1[{\cal M}[S_{1\to 2}(\rho_1,C_{1\to 2})]]\), \(\hat{C}_{1\to 2}[{\cal M}[S_{1\to 2}(\rho_1,C_{1\to 2})]])\) recovers the joint distribution \({\cal M}[S_{1\to 2}(\rho_1,C_{1\to 2})]\).
Proof. In this proof, we denote the distribution \({\cal M}[S_{1\to 2}(\rho_1,C_{1\to 2})\) by \(\hat{P}_{X_1,Y_1,X_2,Y_2}\). Theorem 2 guarantees that \(\rho_1[{\cal M}[S_{1\to 2}(\rho_1,C_{1\to 2})]]=\rho_1\). In order to show \(\hat{C}_{1\to 2}[{\cal M}[S_{1\to 2}(\rho_1,C_{1\to 2})]]=C_{1\to 2}\), it is sufficient to show \[\begin{align} \mathrm{Tr}(G_{t_1} \otimes G_{t_2} ) \hat{C}_{1\to 2}^{T_1} = \mathrm{Tr}(G_{t_1} \otimes G_{t_2} ) C_{1\to 2}^{T_1}\label{VX7} \end{align}\tag{20}\] for \(t_1,t_2 \in {\cal T}\).
For this aim, we denote the channel corresponding to \(C_{1\to 2}\) by \(\Lambda\). Therefore, for \(y_1,y_2 \in {\cal Y}\), and \(x_1 \in {\cal X}_{y_2}\), \(x_2 \in {\cal X}_{y_2}\), the conditional distribution \(\hat{P}_{X_2|X_1,Y_1,Y_2}(x_2|x_1,y_1,y_2)\) is given as \[\begin{align} &\hat{P}_{X_2|X_1,Y_1,Y_2}(x_2|x_1,y_1,y_2) \notag \\ =& \langle x_2,y_2| \Lambda_{1\to 2} (|x_1,y_1\rangle\langle x_1,y_1|)|x_2,y_2\rangle \label{VBI1}. \end{align}\tag{21}\] Therefore, for \(t_1,t_2 \in {\cal T}\setminus \{0\}\), we have \[\begin{align} & \mathrm{Tr}(G_{t_1} \otimes G_{t_2} ) \hat{C}_{1\to 2}^{T_1} \notag\\ =&\sum_{x_1,x_2} g_{x_1|t_1} g_{x_2|t_2} \hat{P}_{X_2|X_1,Y_1,Y_2}(x_2|x_1,y(t_1),y(t_2)) \notag\\ =&\sum_{x_1}g_{x_1|t_1} \mathrm{Tr}G_{t_2}\Lambda_{1\to 2}( |x_1,y(t_1)\rangle \langle x_1,y(t_1)|) \notag\\ =& \sum_{x_1}g_{x_1|t_1} \mathrm{Tr} ((|x_1,y(t_1)\rangle \langle x_1,y(t_1)|) \otimes G_{t_2} ) C_{1\to 2}^{T_1} \notag\\ =& \mathrm{Tr}(G_{t_1} \otimes G_{t_2} ) C_{1\to 2}^{T_1}.\label{VX1} \end{align}\tag{22}\] When \(t_2\) is \(0\), we replace \(y(t_2)\) and \(g_{x_2|t_2}\) by an arbitrary element \(y \in {\cal Y}\) and \(1\), respectively, which guarantees the relation 22 . Thus, we obtain 20 when \(t_1 \neq 0\).
Since \(\rho_1=\frac{1}{d}I +\sum_{t_1\neq 0}\langle G_{t_1} \rangle H_{t_1}\), for \(t_2 \in {\cal T}\setminus \{0\}\), we have \[\begin{align} &\sum_{x_2} g_{x_2|t_2} \hat{P}_{X_2|Y_1,Y_2}(x_2|0,y(t_2)) =\mathrm{Tr}(\rho_1 \otimes G_{t_2} ) C_{1\to 2}^{T_1}\notag \\ =& \mathrm{Tr}\Big(\Big(\frac{1}{d}I + \sum_{t_1\neq 0} \langle G_{t_1} \rangle H_{t_1} \Big) \otimes G_{t_2} \Big) C_{1\to 2}^{T_1}\notag \\ =& \mathrm{Tr}\Big( \Big(\frac{1}{d}I + \sum_{t_1,t_1'\neq 0} \langle G_{t_1} \rangle h_{t_1,t_1'} G_{t_1'} \Big) \otimes G_{t_2} \Big) C_{1\to 2}^{T_1}\notag \\ =& \mathrm{Tr}\Big(\frac{1}{d}I \otimes G_{t_2} \Big) C_{1\to 2}^{T_1} +\sum_{t_1,t_1'\neq 0} \langle G_{t_1} \rangle h_{t_1,t_1'} \mathrm{Tr}(G_{t_1'} \otimes G_{t_2} ) C_{1\to 2}^{T_1}\notag \\ =& \mathrm{Tr}\Big(\frac{1}{d}I \otimes G_{t_2} \Big) C_{1\to 2}^{T_1} +\sum_{t_1,t_1'\neq 0} \langle G_{t_1} \rangle h_{t_1,t_1'} \sum_{x_1,x_2} g_{x_1|t_1'} g_{x_2|t_2} \hat{P}_{X_2|X_1,Y_1,Y_2}(x_2|x_1,y(t_1'),y(t_2)) \notag \\ =& \mathrm{Tr}\Big(\frac{1}{d}I \otimes G_{t_2} \Big) C_{1\to 2}^{T_1} +\sum_{t_1,t_1'\neq 0} \sum_{x_1} g_{x_1|t_1} \hat{P}_{X_1|Y_1}(x_1|y(t_1)) h_{t_1,t_1'} \sum_{x_1',x_2} g_{x_1'|t_1'} g_{x_2|t_2} \hat{P}_{X_2|X_1,Y_1,Y_2}(x_2|x_1',y(t_1'),y(t_2)) , \end{align}\] which implies \[\begin{align} & \mathrm{Tr}(I \otimes G_{t_2} ) C_{1\to 2}^{T_1} \notag \\ =& d\sum_{x_2} g_{x_2|t_2} \hat{P}_{X_2|Y_1,Y_2}(x_2|0,y(t_2)) -d\sum_{t_1,t_1'\neq 0} \sum_{x_1} g_{x_1|t_1} \hat{P}_{X_1|Y_1}(x_1|y(t_1)) \delta_{t_1,t_1'} \sum_{x_1',x_2} g_{x_1'|t_1'} g_{x_2|t_2} \hat{P}_{X_2|X_1,Y_1,Y_2}(x_2|x_1',y(t_1'),y(t_2))\notag \\ =& \mathrm{Tr}(I \otimes G_{t_2} ) \hat{C}_{1\to 2}^{T_1}. \label{VH7} \end{align}\tag{23}\] When \(t_2\) is \(0\), we replace \(y(t_2)\) and \(g_{x_2|t_2}\) by an arbitrary element \(y \in {\cal Y}\) and \(1\), respectively, which guarantees the relation 23 . Hence, we obtain 20 when \(t_1 = 0\). ◻
Theorem 4 shows that, under (A1) and (A3), the reconstructed matrix \(\hat{C}_{1\to 2}[P_{X_1,Y_1,X_2,Y_2}]\) coincides with the true Choi matrix for memoryless sequential strategies. Together with the directional Markovianity established earlier, this immediately yields the following necessary conditions on \(P_{X_1,Y_1,X_2,Y_2}\).
Corollary 3. Assume the conditions (A1) and (A3). For \(P_{X_1,Y_1,X_2,Y_2}\in {\cal M}[{\cal S}_{N|1\to 2}]\), the relations \(X_1-Y_1-Y_2\) and \(\hat{C}_{1\to 2}[P_{X_1,Y_1,X_2,Y_2}]\ge 0\) hold.
Corollary 3 therefore gives basic necessary conditions for membership in \({\cal M}[{\cal S}_{N,1\to 2}]\): the directional Markovian condition \(X_1-Y_1-Y_2\) and the positivity of the reconstructed Choi matrix \(\hat{C}_{1\to 2}[P]\).
The previous subsection showed that, under projective state reduction, memoryless sequential distributions satisfy directional Markovianity together with positivity of the reconstructed Choi matrix. However, these conditions are still not sufficient to characterize \({\cal M}[{\cal S}_{N,1\to 2}]\) completely. What is still missing is a consistency requirement ensuring that the observed correlations arise from a single underlying channel. We therefore introduce an additional algebraic condition for the joint distribution \(P_{X_1,Y_1,X_2,Y_2}\).
For every \(y_1\in{\cal Y}\) and every \(x_1\in{\cal X}_{y_1}\), the conditional probability \(P_{X_1\mid Y_1}(x_1\mid y_1)\) is strictly positive.
The condition (C0) holds. The relation \[\begin{align} &\sum_{x_1} P_{X_2|X_1,Y_1,Y_2}(x_2|x_1,y_1,y_2) \notag\\ =& d P_{X_2|Y_1,Y_2}(x_2|0,y_2) -d\sum_{t_1,t_1'\neq 0} \sum_{x_1} g_{x_1|t_1} P_{X_1|Y_1}(x_1|y(t_1)) h_{t_1,t_1'} \sum_{x_1'} g_{x_1'|t_1'} P_{X_2|X_1,Y_1,Y_2}(x_2|x_1',y(t_1'),y_2) \label{NM5} \end{align}\tag{24}\] holds for any \(y_1,y_2 \in {\cal Y}, x_2 \in {\cal X}_{y_2}\).
Here it is important to distinguish the role of the assumptions (A1), (A2), and (A3) from that of the new condition introduced above. The assumptions (A1)–(A3) specify the operator and measurement structure under which the reconstruction method is carried out. By contrast, the condition (C1) is not an additional assumption on the measurement setting, but a consistency requirement imposed directly on the observed distribution \(P_{X_1,Y_1,X_2,Y_2}\).
Condition (C1) should be understood as a compatibility requirement linking single-site and two-site statistics at the distribution level. Operationally, it enforces that the observed joint probabilities can be generated by a single physical channel acting consistently on the post-measurement states produced by projective state reduction, rather than by different effective maps conditioned on hidden memory.
Importantly, (C1) does not introduce any additional physical assumption beyond memorylessness. Instead, it formalizes the fact that, under restricted projective measurements, Markovian conditional-independence relations and positivity constraints alone do not guarantee that all observed correlations originate from one underlying channel. Condition (C1) precisely excludes those distributions that satisfy all standard non-signaling and positivity requirements but nevertheless fail to admit a channel-level representation compatible with sequential, memoryless dynamics.
As shown below, (C1) is automatically satisfied by all memoryless sequential strategies, while it is generically violated by strategies involving classical or quantum memory, even when these strategies reproduce identical marginal and pairwise statistics. This makes (C1) the minimal algebraic condition required to distinguish genuinely memoryless dynamics from more general processes under restricted observational access.
We define condition (C2) by exchanging the roles of \({\cal H}_{I,1},{\cal H}_{O,1}\) and \({\cal H}_{I,2},{\cal H}_{O,2}\). We compare the non-signaling condition \(Y_1-Y_2-X_2\) of the opposite direction, which is written as \[\begin{align} \sum_{x_1}P_{X_1|Y_1,Y_2}(x_1|y_1,y_2) P_{X_2|X_1,Y_1,Y_2}(x_2|x_1,y_1,y_2) = P_{X_2|Y_1,Y_2}(x_2|0,y_2) \label{NVO} \end{align}\tag{25}\] for any \(y_1,y_2 \in {\cal Y}, x_2 \in {\cal X}_{y_2}\). The following theorem mentions that any element \(P_{X_1,Y_1,X_2,Y_2}\in{\cal M}[ {\cal S}_{N,1\to 2}]\) satisfies (C1) while it does not satisfy 25 in general.
Proposition 5. Assume the conditions (A1) and (A3). For Charlie’s strategy \(S\in {\cal S}_{N,1\to 2}\), the distribution \({\cal M}[S]\) satisfies the condition (C1).
Proof. We employ the same notation as the proof of Theorem 4. For \(y_1\in {\cal Y}\setminus \{0\}\),
we have \[\begin{align} & \mathrm{Tr}(I \otimes G_{y_2,z_2} ) C_{1\to 2}^{T_1} =\mathrm{Tr}G_{y_2,z_2} \Lambda_{1\to 2}( I) = \sum_{x_1} \mathrm{Tr}G_{y_2,z_2}\Lambda_{1\to 2}( |x_1,y_1\rangle \langle x_1,y_1|)\notag\\ \stackrel{(a)}{=} & \sum_{x_1} \sum_{x_2} g_{x_2|y_2,z_2} P_{X_2|X_1,Y_1,Y_2}(x_2|x_1,y_1,y_2), \tag{26} \\ & \mathrm{Tr}(I \otimes I ) C_{1\to 2}^{T_1} =\mathrm{Tr} \Lambda_{1\to 2}( I) = \sum_{x_1} \mathrm{Tr}\Lambda_{1\to 2}( |x_1,y_1\rangle \langle x_1,y_1|)\notag\\ \stackrel{(b)}{=} & \sum_{x_1} \sum_{x_2} P_{X_2|X_1,Y_1,Y_2}(x_2|x_1,y_1,y_2) \tag{27} \end{align}\] for any \(y_2 \in {\cal Y}\) and \(z_2 \in {\cal Z}_{y_2}\), where \((a)\) and \((b)\) follow from 21 . The combination of 23 with \(t_2=(y_2,z_2)\) and 26 implies \[\begin{align} &\sum_{x_1} \sum_{x_2} g_{x_2|y_2,z_2} P_{X_2|X_1,Y_1,Y_2}(x_2|x_1,y_1,y_2) \notag \\ =&d\sum_{x_2} g_{x_2|y_2,z_2} P_{X_2|Y_1,Y_2}(x_2|0,y_2) -d\sum_{t_1,t_1'\neq 0} \sum_{x_1} g_{x_1|t_1} P_{X_1|Y_1}(x_1|y(t_1)) h_{t_1,t_1'} \sum_{x_1',x_2} g_{x_1'|t_1'} g_{x_2|y_2,z_2} P_{X_2|X_1,Y_1,Y_2}(x_2|x_1',y(t_1'),y_2).\label{MV1} \end{align}\tag{28}\] Also, the combination of 23 with \(t_2=0\) and 27 implies \[\begin{align} &\sum_{x_1} \sum_{x_2} P_{X_2|X_1,Y_1,Y_2}(x_2|x_1,y_1,y_2) \notag \\ =&d\sum_{x_2} P_{X_2|Y_1,Y_2}(x_2|0,y_2) -d\sum_{t_1,t_1'\neq 0} \sum_{x_1} g_{x_1|t_1} P_{X_1|Y_1}(x_1|y(t_1)) h_{t_1,t_1'} \sum_{x_1',x_2} g_{x_1'|t_1'} P_{X_2|X_1,Y_1,Y_2}(x_2|x_1',y(t_1'),y_2).\label{MV2} \end{align}\tag{29}\]
We choose are real numbers \(f_{x,y,z}\) such that \(\sum_{z \in {\cal Z}_{y}'}f_{z,y,x}g_{x'|y,z}=\delta_{x,x'}\). Then, for \(x_2\), we have \[\begin{align} &\sum_{x_1} P_{X_2|X_1,Y_1,Y_2}(x_2|x_1,y_1,y_2) \notag\\ \stackrel{(a)}{=} & \sum_{z_2\in {\cal Z}_{y_2}'}f_{x_2,y_2,z_2} \sum_{x_1} \sum_{x_2'} g_{x_2'|y_2,z_2} P_{X_2|X_1,Y_1,Y_2}(x_2'|x_1,y_1,y_2) \notag\\ \stackrel{(b)}{=} & \sum_{z_2\in {\cal Z}_{y_2}'} f_{x_2,y_2,z_2} \Big( d\sum_{x_2'} g_{x_2'|y_2,z_2} P_{X_2|Y_1,Y_2}(x_2'|0,y_2) \notag\\ &-d\sum_{t_1,t_1'\neq 0} \sum_{x_1} g_{x_1|t_1} P_{X_1|Y_1}(x_1|y(t_1)) h_{t_1,t_1'} \sum_{x_1',x_2'} g_{x_1'|t_1'} g_{x_2'|y_2,z_2} P_{X_2|X_1,Y_1,Y_2}(x_2'|x_1',y(t_1'),y_2)\Big)\notag \\ \stackrel{(c)}{=} & d P_{X_2|Y_1,Y_2}(x_2|0,y_2) -d\sum_{t_1,t_1'\neq 0} \sum_{x_1} g_{x_1|t_1} P_{X_1|Y_1}(x_1|y(t_1)) h_{t_1,t_1'} \sum_{x_1'} g_{x_1'|t_1'} P_{X_2|X_1,Y_1,Y_2}(x_2|x_1',y(t_1'),y_2) ,\label{SK1} \end{align}\tag{30}\] where \((a)\) and \((c)\) follow from the relation \(\sum_{z \in {\cal Z}_{y}'}f_{z,y,x}g_{x'|y,z}=\delta_{x,x'}\), and \((b)\) follows from the combination of 28 and 29 . The relation 30 implies the condition (C1). ◻
We are now in a position to state the main result of this paper. The previous subsection showed that any distribution generated by a memoryless sequential strategy necessarily satisfies the directional Markovian condition \(X_1-Y_1-Y_2\) together with the positivity of the reconstructed Choi matrix \(\hat{C}_{1\to 2}[P]\). Proposition 5 further showed that such distributions also satisfy the additional algebraic condition (C1). The following theorem shows that, under (A1) and (A3), these three conditions are not only necessary but also sufficient, and hence give a complete characterization of \({\cal M}[{\cal S}_{N,1\to 2}]\).
Theorem 6. Assume the conditions (A1), (A3), and (C0). For a distribution \(P \in {\cal M}[{\cal S}_G]\), the following conditions are equivalent:
\(P \in {\cal M}[{\cal S}_{N,1\to 2}]\).
\(P\) satisfies the Markovian condition \(X_1-Y_1-Y_2\), the condition (C1), and the positivity condition \[\hat{C}_{1\to 2}[P]\ge 0.\]
Equivalently, \[\begin{align} {\cal M}[{\cal S}_{N,1\to 2}] = \{P \in {\cal M}[{\cal S}_G]\mid P \text{ satisfies } X_1-Y_1-Y_2,\;(C1),\; \hat{C}_{1\to 2}[P]\ge 0 \}. \label{FG1} \end{align}\qquad{(8)}\] Moreover, if \(P \in {\cal M}[{\cal S}_G]\) satisfies \(X_1-Y_1-Y_2\) and (C1), then the reconstructed pair \(\bigl(\rho_1[P],\hat{C}_{1\to 2}[P]\bigr)\) reproduces the distribution \(P\) in the sense that \[\begin{align} P_{X_1,Y_1,X_2,Y_2} = {\cal M}\!\left[ S_{1\to 2}\!\left( \rho_1[P_{X_1,Y_1,X_2,Y_2}], \hat{C}_{1\to 2}[P_{X_1,Y_1,X_2,Y_2}] \right) \right].\label{EE54} \end{align}\qquad{(9)}\] In particular, if \(\hat{C}_{1\to 2}[P]\ge 0\), then this reconstructed pair defines a memoryless sequential strategy in \({\cal S}_{N,1\to 2}\).
Theorem 6 is the point at which the reconstruction formalism becomes a complete membership criterion. It shows that, in the present restricted projective setting, neither the directional Markovian condition nor the positivity of the reconstructed Choi matrix is sufficient by itself: the missing ingredient is precisely the distribution-level consistency condition (C1). Hence, under (A1) and (A3), membership in \({\cal M}[{\cal S}_{N,1\to 2}]\) can be decided exactly from the observed distribution.
The second part of the theorem is equally important. Whenever these conditions are satisfied, the reconstructed pair \((\rho_1[P],\hat{C}_{1\to 2}[P])\) is not merely a formal witness but actually reproduces the original distribution. This point will be made concrete in Subsection 9.2, where two complementary types of examples are exhibited inside \({\cal M}[{\cal S}_{Q,1\to 2}]\): one in which the positivity condition fails while the directional Markovian condition and (C1) still hold, and another in which (C1) fails while the directional Markovian condition and the positivity condition still hold. Taken together, those examples show that the two requirements in Theorem 6 detect genuinely different obstructions, and this is exactly why both are needed for the characterization of the memoryless sequential class in addition to the non-signaling Markovian condition.
Proof. Since Lemma 7 guarantees \[\begin{align} & \{P \in {\cal M}[{\cal S}_G]| PsatisfiesX_1-Y_1-Y_2, (C1), \rho_1[P] \ge 0, \hat{C}_{1\to 2}[P] \ge 0 \}\notag\\ =& \{P \in {\cal M}[{\cal S}_G]| PsatisfiesX_1-Y_1-Y_2, (C1), \hat{C}_{1\to 2}[P] \ge 0 \}, \end{align}\] it is sufficient to show the relation \[\begin{align} {\cal M}[{\cal S}_{N,1\to 2}]= \{P \in {\cal M}[{\cal S}_G]| PsatisfiesX_1-Y_1-Y_2, (C1), \rho_1[P] \ge 0, \hat{C}_{1\to 2}[P] \ge 0 \}\label{FG1B}. \end{align}\tag{31}\] Since the relation \(\subset\) in 31 follows from Theorems 2, 4, and Proposition 5, it is sufficient to show the relation \(\supset\) in 31 . For this aim, it is sufficient to show the relation \[\begin{align} P_{X_1,Y_1,X_2,Y_2}= {\cal M}[S_{1\to 2}(\rho_1[P_{X_1,Y_1,X_2,Y_2}], C_{1\to 2}[P_{X_1,Y_1,X_2,Y_2}])] \label{DJ2} \end{align}\tag{32}\] for an element \(P_{X_1,Y_1,X_2,Y_2} \in \{P \in {\cal M}[{\cal S}_G]| P satisfies X_1-Y_1-Y_2, (C1), \rho_1[P] \ge 0, \hat{C}_{1\to 2}[P] \ge 0 \}\).
Since \(P_{X_1,Y_1,X_2,Y_2} \in {\cal M}[{\cal S}_G]\) satisfies the Markovian condition \(X_1-Y_1-Y_2\), we have \[\begin{align} P_{X_1,X_2|Y_1,Y_2}(x_1,x_2|y_1,y_2) = P_{X_1|Y_1}(x_1|y_1) P_{X_2|X_1,Y_1,Y_2}(x_1,x_2|y_1,y_2). \end{align}\] Hence, it is sufficient for 32 to show that \[\begin{align} P_{X_1|Y_1}(x_1|y_1) &=\mathrm{Tr}|x_1,y_1\rangle \langle x_1,y_1|\rho_1[P_{X_1,Y_1,X_2,Y_2}] \tag{33} \\ P_{X_2|X_1,Y_1,Y_2}(x_2|x_1,y_1,y_2) &=\mathrm{Tr} (|x_1,y_1\rangle \langle x_1,y_1|\otimes |x_2,y_2\rangle \langle x_2,y_2|) \hat{C}_{1\to 2}[P_{X_1,Y_1,X_2,Y_2}]^{T_1} \tag{34} \end{align}\] for \(y_1,y_2\in{\cal Y}\), \(x_2\in{\cal X}_{y_2}\), and \(x_1\in{\cal X}_{y_1}\). Since \(E_{x|y}\) is written as a linear sum of \(\{G_{y,z}\}_{z \in {\cal Z}_{y}'}\), there are real numbers \(f_{x,y,z}\) such that \(\sum_{z \in {\cal Z}_{y}'}f_{z,y,x}g_{x'|y,z}=\delta_{x,x'}\). Hence, \(|x_1,y_1\rangle \langle x_1,y_1| =\sum_{z_1 \in {\cal Z}_{y_1}}f_{z_1,y_1,x_1} G_{y_1,z_1}\). Thus, we have \[\begin{align} P_{X_1|Y_1}(x_1|y_1) \stackrel{(a)}{=} & P_{X_1|Y_1,Y_2}(x_1|y_1,0) =\sum_{z_1 \in {\cal Z}_{y_1}}f_{z_1,y_1,x_1} \langle G_{y_1,z_1}\otimes I\rangle_{P_{X_1,Y_1,X_2,Y_2}} \notag\\ =&\sum_{z_1 \in {\cal Z}_{y_1}}f_{z_1,y_1,x_1} \mathrm{Tr}G_{y_1,z_1}\rho_1[P_{X_1,Y_1,X_2,Y_2}] =\mathrm{Tr}|x_1,y_1\rangle \langle x_1,y_1|\rho_1[P_{X_1,Y_1,X_2,Y_2}], \end{align}\] which implies 33 . Here, Step \((a)\) follows from the Markovian condition \(X_1-Y_1-Y_2\) of \(P_{X_1,Y_1,X_2,Y_2} \in {\cal M}[{\cal S}_G]\).
The definition of \(\hat{C}_{1\to 2}[P_{X_1,Y_1,X_2,Y_2}]\) implies \[\begin{align} &\mathrm{Tr}(G_{t_1}\otimes G_{t_2}) \hat{C}_{1\to 2}[P_{X_1,Y_1,X_2,Y_2}]^{T_1}\notag\\ =& \sum_{x_1,x_2} g_{x_1|t_1} g_{x_2|t_2} P_{X_2|X_1,Y_1,Y_2}(x_2|x_1,y(t_1),y(t_2)) , \tag{35}\\ &\mathrm{Tr}(I \otimes G_{t_2}) \hat{C}_{1\to 2}[P_{X_1,Y_1,X_2,Y_2}]^{T_1}\notag\\ =& d \Big(\sum_{x_2} g_{x_2|t_2} P_{X_2|Y_1,Y_2}(x_2|0,y(t_2)) \notag\\ &-\sum_{t_1,t_1'\neq 0} \sum_{x_1} g_{x_1|t_1} P_{X_1|Y_1}(x_1|y(t_1)) h_{t_1,t_1'} \sum_{x_1',x_2} g_{x_1'|t_1'} g_{x_2|t_2} P_{X_2|X_1,Y_1,Y_2}(x_2|x_1',y(t_1'),y(t_2)) \Big)\tag{36} \end{align}\] for \(t_1 \in {\cal T}\setminus \{0\}\) and \(t_2 \in {\cal T}\). The combination of the condition (C1) and 36 implies \[\begin{align} \mathrm{Tr}(I \otimes G_{t_2}) \hat{C}_{1\to 2}[P_{X_1,Y_1,X_2,Y_2}]^{T_1} =\sum_{x_2} g_{x_2|t_2} \sum_{x_1} P_{X_2|X_1,Y_1,Y_2}(x_2|x_1,y_1,y_2). \label{XK3} \end{align}\tag{37}\] For \(z_1\in {\cal Z}_{y_1}'\) and \(z_2\in {\cal Z}_{y_2}'\), the combination of 35 and 37 with \(t_1=(y_1,z_1)\) and \(t_2=(y_2,z_2)\) implies \[\begin{align} & \sum_{x_1,x_2} g_{x_1|y_1,z_1} g_{x_2|y_2,z_2} \mathrm{Tr} (|x_1,y_1\rangle \langle x_1,y_1|\otimes |x_2,y_2\rangle \langle x_2,y_2|) \hat{C}_{1\to 2}[P_{X_1,Y_1,X_2,Y_2}]^{T_1}\notag\\ =& \sum_{x_1,x_2} g_{x_1|y_1,z_1} g_{x_2|y_1,z_1} P_{X_2|X_1,Y_1,Y_2}(x_2|x_1,y_1,y_2) \label{XK4}. \end{align}\tag{38}\] Thus, \[\begin{align} & \mathrm{Tr} (|x_1,y_1\rangle \langle x_1,y_1|\otimes |x_2,y_2\rangle \langle x_2,y_2|) \hat{C}_{1\to 2}[P_{X_1,Y_1,X_2,Y_2}]^{T_1}\notag\\ =& \sum_{z_1,z_2} f_{x_1,y_1,z_1} f_{x_2,y_2,z_2} \sum_{x_1',x_2'} g_{x_1'|y_1,z_1} g_{x_2'|y_2,z_2} \mathrm{Tr} (|x_1',y_1\rangle \langle x_1',y_1|\otimes |x_2',y_2\rangle \langle x_2',y_2|) \hat{C}_{1\to 2}[P_{X_1,Y_1,X_2,Y_2}]^{T_1}\notag\\ =& \sum_{z_1,z_2} f_{x_1,y_1,z_1} f_{x_2,y_2,z_2} \sum_{x_1',x_2'} g_{x_1'|y_1,z_1} g_{x_2'|y_1,z_1} P_{X_2|X_1,Y_1,Y_2}(x_2'|x_1',y_1,y_2) \notag\\ =& P_{X_2|X_1,Y_1,Y_2}(x_2|x_1,y_1,y_2) \label{XK5}. \end{align}\tag{39}\] Hence, we obtain 34 . ◻
By summarizing 9 , 10 , Theorems 1, and 4, the conditions for the classes in the hierarchies 7 are characterized as follows. \[\begin{align} \begin{array}{ccccc} X_1-Y_1-Y_2~\&~ \hat{C}_{1\to2}\ge0~\&~(C1) & \supset &X_1-Y_1-Y_2-X_2 &\subset & Y_1-Y_2-X_2~\&~ \hat{C}_{2\to1}\ge0 ~\&~(C2) \\ \cap & &\cap & & \cap \\ X_1-Y_1-Y_2 & \supset &X_1-Y_1-Y_2 ~\& ~Y_1-Y_2-X_2 ~\&~ R_{1,2}\ge 0 &\subset & Y_1-Y_2-X_2 \end{array}\label{HI16} \end{align}\tag{40}\]
However, the conditions given in 40 show only sufficient conditions for the conditions in 7 .
Remark 7. The full-support assumption (C0)* is imposed only to ensure that the conditional probabilities appearing in (C1) are well defined. When (C0) fails, the observed distribution necessarily assigns zero probability to some rank-one projective outcome. Under (A3), this means that the relevant state is supported on the orthogonal complement of the corresponding projector, so that the analysis may be reformulated from the outset on the smaller support subspace. After this compression of the input space and removal of the impossible outcomes, the same reconstruction argument applies to the reduced model. Thus Theorem 6 should be understood as the nondegenerate (full-support) version of the general support-restricted statement.*
We now specialize the general framework to the minimal nontrivial setting in which the order-indistinguishability problem can be analyzed explicitly. The aim of this section is not to solve the full reverse-direction membership problem for arbitrary observed distributions. Instead, we restrict attention to distributions already known to arise from a forward memoryless sequential strategy. Starting from a forward memoryless sequential strategy \(S=S_{1\to 2}(\rho_1,C_{1\to 2}) \in {\cal S}_{N,1\to 2}\), we ask whether the resulting observed distribution \({\cal M}[S]\) can also be explained by a memoryless sequential strategy of the opposite direction. Equivalently, we ask when \({\cal M}[S]\) belongs to the intersection \({\cal M}[{\cal S}_{N,1\to 2}] \cap {\cal M}[{\cal S}_{N,2\to 1}]\). This is the sense in which we use the term order-indistinguishability in the present section.
Similar to [18], as a special case of Section 5, we consider the case in which \({\cal H}_{I,1}\), \({\cal H}_{O,1}\), \({\cal H}_{I,2}\), and \({\cal H}_{O,2}\) are qubit systems, and the measurements are given by projective measurements of the Pauli matrices \[\begin{align} \sigma_0&:= \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right),~ \sigma_1:= \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right), \\ \sigma_2&:= \left( \begin{array}{cc} 0 & -\sqrt{-1} \\ \sqrt{-1} & 0 \end{array} \right),~ \sigma_3:= \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right). \end{align}\] That is, the state reduction on \({\cal H}_{(I,i)}\) to \({\cal H}_{(O,i)}\) is given by the projection hypothesis for Pauli measurements.
Both Alice and Bob independently choose one of these Pauli measurements at random. Hence, \(Y_1\) and \(Y_2\) take values in \(\{0,1,2,3\}\) and are uniformly distributed, while \(X_1\) and \(X_2\) record the corresponding outcomes. When \(\sigma_0\) is chosen, there is no nontrivial measurement outcome. Thus, the pair \((X_i,Y_i)\) takes one of seven possible values for each \(i=1,2\).
This measurement scheme is genuinely restricted and is not tomographically complete. In the present qubit-Pauli setting, the family \(\{C[\Gamma_{1,z_1}]\otimes C[\Gamma_{2,z_2}]\}_{z_1,z_2}\) contains at most \(49\) linearly independent elements. On the other hand, Appendix 12 shows that the relevant quotient space has dimension \(88\). Hence these tensors cannot span the quotient space, and exact process-matrix reconstruction is impossible in this setting. Accordingly, the underlying process matrix cannot in general be reconstructed from the observed distribution. The role of the present two-qubit analysis is therefore not process-level reconstruction, but the explicit study of order distinguishability directly at the level of the observed statistics.
At the same time, this Pauli setting satisfies the assumptions (A1), (A2), and (A3). In particular, the reconstructed quantities \(\hat{C}_{1\to 2}\) and \(\tilde{C}_{1\to 2}\) coincide, so that they can be used as concrete distribution-level tools in the analysis below.
We now turn to the central question of this section. Starting from a forward memoryless sequential strategy \(S=S_{1\to 2}(\rho_1,C_{1\to 2}) \in {\cal S}_{N,1\to 2}\), we ask whether the resulting observed distribution \({\cal M}[S]\) can also be explained by a memoryless sequential strategy of the opposite direction. Equivalently, we ask when \({\cal M}[S]\) belongs to the intersection \({\cal M}[{\cal S}_{N,1\to 2}] \cap {\cal M}[{\cal S}_{N,2\to 1}]\).
By Corollary 2, an element \({\cal M}[S]\in {\cal M}[{\cal S}_{N,1\to 2}]\) belongs to this intersection if and only if the following three conditions hold:
\({\cal M}[S]\) satisfies \(Y_1-Y_2-X_2\).
\({\cal M}[S_{2\to 1}(\rho_2[{\cal M}[S]], \tilde{C}_{2\to 1}[{\cal M}[S]])]\) satisfies \(X_1-Y_1-Y_2\).
\(\tilde{C}_{2\to 1}[{\cal M}[S]]\ge 0\).
The first two conditions express directional non-signaling/Markov structure, whereas the third is the complete-positivity requirement for the reconstructed reverse-direction channel. Among them, (D3) is the hardest condition to check explicitly.
For this reason, we proceed in two stages. We first analyze a general qubit channel in a family-independent manner, which reveals the structural form of (D1)–(D3) as a function of the rank parameter \(t\). We then specialize to representative channel families—Pauli, phase-damping, and depolarizing channels—for which the indistinguishable region can be described by explicit inequalities.
We first study a general qubit channel before specializing to particular families. The point of this subsection is to translate the abstract reverse-compatibility conditions (D1)–(D3) into explicit constraints on the parameters of a forward memoryless sequential strategy \(S=S_{1\to 2}(\rho_1,C_{1\to 2})\). This family-independent analysis isolates the geometric structure behind the order-indistinguishable region and explains why the later channel-specific formulas take the form they do.
In this section, we employ the formulas \[\begin{align} I \circ \sigma_i=\sigma_i, \quad \sigma_j \circ \sigma_i=\delta_{i,j} I \label{CZER} \end{align}\tag{41}\] for \(i,j=1,2,3\). We parameterize a forward memoryless sequential strategy \(S=S_{1\to 2}(\rho_1, C_{1\to 2})\in {\cal S}_{N,1\to 2}\) as follows. The initial state \(\rho_1\) is written as \[\begin{align} \rho_1=\frac{1}{2}(I+ \sum_{i=1}^3 c^1_i \sigma_i) \end{align}\] with the condition \(\sum_{i=1}^3 (c^1_i)^2\le 1\). The Choi matrix \(C_{1\to 2}\) of a qubit TP-CP map \(\Lambda_{1\to 2}\) can be written as [25] \[\begin{align} C_{1\to 2}^{T_1}= I \otimes \rho_2+ \sum_{j=1}^t \frac{\lambda_j}{2} \alpha_j \otimes \beta_j,\label{HJ8} \end{align}\tag{42}\] where \(\alpha_j\) and \(\beta_j\) are traceless matrices, \(\alpha_1,\alpha_2,\alpha_3\) are orthogonal to each other, \(\beta_1,\beta_2,\beta_3\) are orthogonal to each other, \(\|\alpha_j\|=\|\beta_j\|=1\), and \(1\ge \lambda_j> 0\). Here, the qubit state \(\rho_2\) appearing in 42 can be written in Bloch form as \[\begin{align} \rho_2=\frac{1}{2}(I+ \sum_{i=1}^3 c^2_i \sigma_i). \end{align}\] The rank parameter \(t\in\{0,1,2,3\}\) will determine the form of the feasible region below.
We begin with condition (D1), which asks whether the observed distribution of the forward strategy is compatible with the non-signaling structure required from the reverse direction. The following lemma shows that this already imposes strong restrictions on the input Bloch vector \(\rho_1\), and that the form of these restrictions depends only on the rank parameter \(t\in\{0,1,2,3\}\).
Lemma 8. \(S_{1\to 2}(\rho_1, C_{1\to 2})\) satisfies the condition (D1) if and only if the pair satisfies one of the following four cases depending on \(t\):
\(t=3\): \(c^1_i=0\) for \(i=1,2,3\).
\(t=2\): at least two \(i\) satisfy \(c^1_i=0\); for instance, if \(c^1_2=c^1_3=0\), then \(c^1_1\mathrm{Tr}\alpha_1\sigma_1 =c^1_1 \mathrm{Tr}\alpha_2\sigma_1=0\).
\(t=1\): at least one \(i\) satisfies \(c^1_i=0\); for instance, if \(c^1_3=0\), then \(c^1_1 \mathrm{Tr}\alpha_1\sigma_1 =c^1_2 \mathrm{Tr}\alpha_1\sigma_2=0\).
\(t=0\): no further condition is required.
Proof deferred to Appendix 11.2.
Once (D1) is imposed, the reconstructed quantities simplify considerably. In particular, the next lemma identifies the reconstructed marginal state \(\rho_2[{\cal M}[S]]\) and the reconstructed bipartite operator \({\cal R}[{\cal M}[S]]\) explicitly. It also translates condition (D2) into the corresponding restrictions on the output Bloch vector \(\rho_2\). Thus, Lemma 9 should be read as the companion to Lemma 8: the first lemma resolves the input-side restriction coming from (D1), while the second resolves the output-side restriction coming from (D2).
Lemma 9. Assume that \(S=S_{1\to 2}(\rho_1, C_{1\to 2})\) satisfies the condition (D1).
(i) We have \[\begin{align} \rho_{2}[{\cal M}[S]]&= \rho_2 \label{DSJK2}\\ {\cal R}[{\cal M}[S]]&= \rho_1 \otimes \rho_2 + \sum_{j=1}^t \frac{\lambda_j}{4} \alpha_j \otimes \beta_j . \label{DSJK} \end{align}\] {#eq: sublabel=eq:DSJK2,eq:DSJK}
(ii) \(S\) satisfies the condition (D2) if and only if the pair \((\rho_1, C_{1\to 2})\) satisfies one of the following four cases:
\(1\)
: *$t=3$: $c^2_i=0$ for $i=1,2,3$.*
\(2\)
: *$t=2$: at least two $i$ satisfy $c^2_i=0$; for instance, if
$c^2_2=c^2_3=0$, then
$c^2_1\mathrm{Tr}\alpha_1\sigma_1 =c^2_1 \mathrm{Tr}\alpha_2\sigma_1=0$.*
\(3\)
: *$t=1$: at least one $i$ satisfies $c^2_i=0$; for instance, if
$c^2_3=0$, then
$c^2_1 \mathrm{Tr}\alpha_1\sigma_1 =c^2_2 \mathrm{Tr}\alpha_1\sigma_2=0$.*
\(4\)
: *$t=0$: no further condition is required.*
Proof deferred to Appendix 11.3.
After the directional constraints (D1) and (D2) have been reduced to explicit conditions on \(\rho_1\) and \(\rho_2\), the remaining task is positivity. We first examine the positivity of the original forward Choi matrix \(C_{1\to 2}\) itself. The following lemma gives the corresponding positivity region under the constraints already imposed by (D1) and (D2).
Lemma 10. Assume that \(S\) satisfies the conditions (D1) and (D2). The relation \(C_{1\to 2}\ge 0\) is equivalent to the following condition.
\(t=3\). There are two local-unitary normal forms.
The positivity condition is equivalent to \(1\ge \max(-\lambda_1 +\lambda_2+\lambda_3, \lambda_1 -\lambda_2+\lambda_3, \lambda_1 +\lambda_2-\lambda_3)\).
The positivity condition is equivalent to \(1\ge \lambda_1+\lambda_2+\lambda_3\).
\(t=2\): \(1\ge (\lambda_1+\lambda_2)^2+(c^2_1)^2\).
\(t=1\): \(1 \ge \lambda_1^2+(c^2_1)^2+(c^2_2)^2\).
\(t=0\): no further condition is imposed.
Proof deferred to Appendix 11.4.
The final step is to translate condition (D3), namely the complete positivity of the reconstructed reverse-direction channel, into explicit inequalities. The next lemma first writes the reconstructed reverse-direction Choi matrix \(\tilde{C}_{2\to 1}[{\cal M}[S]]\) in closed form and then gives the positivity criterion for this matrix. This is the key lemma for the order-indistinguishability problem, because (D3) is the only genuinely channel-level constraint among (D1)–(D3).
Lemma 11. Assume that \(S\) satisfies the conditions (D1) and (D2).
(i) The following relation holds: \[\begin{align} \tilde{C}_{2\to 1}[{\cal M}[S]]^{T_2} = \rho_1 \otimes I + \sum_{j=1}^t \frac{\lambda_j}{2} \alpha_j \otimes \beta_j \label{VNI1} \end{align}\qquad{(10)}\]
(ii) The relation \(\tilde{C}_{2\to 1}[{\cal M}[S]]\ge 0\) is equivalent to the following condition.
\(1\)
: *$t=3$. There are two local-unitary normal forms.*
\(A\)
: *The positivity condition is equivalent to
$1\ge \max(-\lambda_1 +\lambda_2+\lambda_3,
\lambda_1 -\lambda_2+\lambda_3,
\lambda_1 +\lambda_2-\lambda_3)$.*
\(B\)
: *The positivity condition is equivalent to
$1\ge \lambda_1+\lambda_2+\lambda_3$.*
\(2\)
: *$t=2$: $1\ge (\lambda_1+\lambda_2)^2+(c^1_1)^2$.*
\(3\)
: *$t=1$: $1 \ge \lambda_1^2+(c^1_1)^2+(c^1_2)^2$.*
\(4\)
: *$t=0$: no further condition is imposed.*
Proof deferred to Appendix 11.4.
Taken together, Lemmas 8–11 reduce the reverse-compatibility problem for a general forward qubit strategy \(S_{1\to 2}(\rho_1,C_{1\to 2})\) to explicit constraints on a small number of geometric parameters. In this sense, the general qubit analysis isolates the common structure behind the order-indistinguishable region before any channel family is specified. We now specialize to representative families—Pauli, phase-damping, and depolarizing channels—for which these abstract constraints become concrete and easily interpretable inequalities.
We first consider the full Pauli family before passing to the more transparent subfamilies treated later. The point of this subsection is to make the reverse-compatibility conditions (D1)–(D3) explicit in a canonical and analytically tractable channel family. In particular, this family already contains the depolarizing channel as a special case.
We consider the Pauli channel \[\begin{align} \Lambda(\rho):= \sum_{j=0}^3 p_j \sigma_j \rho \sigma_j . \end{align}\] In particular, the depolarizing channel \[\Lambda_\lambda(\rho):= (1-\lambda)\rho+\lambda \frac{I}{2}\] is obtained as the special case \(p_0=1-\frac{3}{4}\lambda\) and \(p_j=\frac{1}{4}\lambda\) for \(j=1,2,3\). Its Choi matrix can be written as \[\begin{align} C_{1\to 2}^{T_1} = I \otimes \frac{I}{2} +\sum_{j=1}^3 \frac{\lambda_j}{2}\sigma_j\otimes\sigma_j, \label{HJ51} \end{align}\tag{43}\] where \(\lambda_j:=2p_0+2p_j-1\) for \(j=1,3\) and \(\lambda_2:=-2p_0-2p_2+1\).
We fix the initial state in the form \[\rho_1=\frac{1}{2}(I+\kappa\sigma_1).\]
Lemma 12. For the Pauli channel with Choi matrix 43 and initial state \(\rho_1=\frac{1}{2}(I+\kappa\sigma_1)\), the reverse-compatibility conditions (D1)–(D3) take the following form:
(i) (D1) holds if and only if \[\kappa\lambda_1=0,\] that is, \(\lambda_1=0\) or \(\kappa=0\).
(ii) (D2) holds if and only if \[\lambda_1=0, \qquad\text{or}\qquad \kappa=0, \qquad\text{or}\qquad \kappa=\pm1 \;\text{ and }\;\lambda_1<1.\]
(iii) For \(|\kappa|<1\), condition (D3) is equivalent to \[\begin{align} 1+ \frac{\kappa(\lambda_1^2 -1)}{\kappa^2\lambda_1^2-1} +\frac{ \lambda_1 (\kappa^2 -1)}{\kappa^2\lambda_1^2-1} &\ge 0,\label{CBQ1}\\ 1- \frac{\kappa(\lambda_1^2 -1)}{\kappa^2\lambda_1^2-1} -\frac{ \lambda_1 (\kappa^2 -1)}{\kappa^2\lambda_1^2-1} &\ge 0,\label{CBQ2}\\ \left(\frac{(\kappa^2 \lambda_1 -1)( \lambda_1 -1)}{\kappa^2\lambda_1^2-1}\right)^2 = \left(1- \frac{ \lambda_1 (\kappa^2 -1)}{\kappa^2\lambda_1^2-1} \right)^2 & \ge \left(\frac{\kappa(\lambda_1^2 -1)}{\kappa^2\lambda_1^2-1}\right)^2 +(\lambda_2+\lambda_3)^2 , \label{CBV1}\\ \left(\frac{(\kappa^2 \lambda_1 -1)( \lambda_1 +1)}{\kappa^2\lambda_1^2-1}\right)^2 = \left(1+ \frac{ \lambda_1 (\kappa^2 -1)}{\kappa^2\lambda_1^2-1} \right)^2 & \ge \left(\frac{\kappa(\lambda_1^2 -1)}{\kappa^2\lambda_1^2-1}\right)^2 +(\lambda_2-\lambda_3)^2 . \label{CBV2} \end{align}\] {#eq: sublabel=eq:CBQ1,eq:CBQ2,eq:CBV1,eq:CBV2}
(iv) For \(|\kappa|=1\), condition (D3) is equivalent to \[\lambda_2=\lambda_3=0.\]
Proof deferred to Appendix 11.5.
Lemma 12 gives the generic Pauli-family form of the reverse-compatibility conditions. The first two conditions constrain the directional parameters \((\kappa,\lambda_1)\), whereas the third is the complete-positivity condition for the reconstructed reverse-direction channel. Thus, the Pauli family already exhibits the basic geometry of the order-indistinguishable region in a concrete form. The next subsections extract especially transparent representative examples from this family, namely the phase-damping and depolarizing channels.
We now consider a particularly simple representative of the Pauli family, namely the phase-damping channel in the \(\sigma_1\) basis. This example is useful because, unlike the general Pauli case, the directional conditions (D1) and (D2) collapse to the same simple restriction on the input state, so that the remaining nontrivial issue is the reverse-direction complete-positivity condition (D3).
We fix the initial state to be \(\rho_1=\frac{1}{2}(I+\kappa\sigma_1)\). We consider the phase-damping channel with respect to the \(\sigma_1\) basis, whose Choi matrix \(C_{1\to 2}\) satisfies the condition \(\lambda_1=1\) under the form 42 . Then the directional conditions simplify as follows:
(D1) holds if and only if \(\kappa=0\).
(D2) holds if and only if \(\kappa=0\).
Thus, in this channel family, the only remaining nontrivial constraint is the CP-feasibility condition (D3), which is characterized by the following lemma.
Lemma 13. The condition (D3) holds if and only if the relations \(\kappa=\pm 1\) and \(\lambda_2=\lambda_3=0\) or the relations \(|\kappa|<1\), \(\lambda_2=-\lambda_3\), and \(|\lambda_3|<1\) hold.
Proof deferred to Appendix 11.6.
This lemma determines the reverse-direction positivity condition (D3) in the present channel family. Since, as noted above, the directional conditions (D1) and (D2) are both equivalent to \(\kappa=0\), the full reverse-compatibility conditions (D1)–(D3) can hold simultaneously only when \(\kappa=0\). Thus, in the \(\sigma_1\)-basis phase-damping family, the order-indistinguishable region is governed entirely by the directional constraints, while the above lemma shows that the CP-feasibility condition (D3) is consistent with that same restriction.
We next consider the phase-damping channel in the \(\sigma_3\) basis, under the same fixed initial state \(\rho_1=\frac{1}{2}(I+\kappa\sigma_1)\). Compared with the previous subsection, the main point is that the directional conditions (D1) and (D2) remain unchanged, whereas the reverse-direction positivity condition (D3) takes a different form.
More precisely, we consider the phase-damping channel whose Choi matrix \(C_{1\to 2}\) satisfies the condition \(\lambda_3=1\) under the form 42 . For simplicity, we also assume \(\lambda_2=-\lambda_1\). Since the conditions (D1) and (D2) do not depend on \(\lambda_3\), they have exactly the same form as in the previous subsection. Thus, the only new issue in the present family is the CP-feasibility condition (D3), which is characterized by the following lemma.
Lemma 14. The condition (D3) holds if and only if the relation \(\kappa=0\) or the relations \(\lambda_1=\pm 1\) and \(\kappa^2<1\) hold.
Proof deferred to Appendix 11.7.
This lemma determines the reverse-direction positivity condition (D3) in the present channel family. Since, as noted above, the directional conditions (D1) and (D2) are both equivalent to \(\kappa=0\), the three conditions (D1)–(D3) can hold simultaneously only when \(\kappa=0\). Thus, in the \(\sigma_3\)-basis phase-damping family, the order-indistinguishable region is characterized simply by the vanishing of the input Bloch parameter \(\kappa\).
We next consider the depolarizing channel, which provides the cleanest analytic description of the order-indistinguishable region in the main text. Recall that, for a forward memoryless sequential strategy \(S=S_{1\to 2}(\rho_1,C_{1\to 2})\), the observed distribution \({\cal M}[S]\) is order-indistinguishable if and only if it satisfies all three reverse-compatibility conditions (D1), (D2), and (D3). In the depolarizing family, these conditions can be written explicitly, and their combination yields a sharp boundary in the \((\lambda,\kappa)\)-plane.
Assume that the initial state is \[\rho_1=\frac{1}{2}(I+\kappa \sigma_1), \qquad \kappa\in[-1,1].\] When the channel corresponds to the depolarizing Choi matrix 43 , we have \[\lambda_j=1-\lambda \quad \text{for } j=1,3, \qquad \lambda_2=-(1-\lambda).\] The directional conditions are then given by:
(D1) holds if and only if \(\lambda=1\) or \(\kappa=0\).
(D2) holds if and only if \(\lambda=1\), or \(\kappa=0\), or \(\kappa=\pm 1\) and \(\lambda>0\).
The remaining condition (D3) takes the following explicit form.
Lemma 15. The condition (D3) is equivalent to the following cases:
\(\lambda = 0\): \(|\kappa|<1\).
\(\lambda = 1\): \(-1\le \kappa \le 1\).
\(0<\lambda < 1\): \[\begin{align} |\kappa| \le \sqrt{\frac{4-3\lambda}{(3-2\lambda)(2\lambda^2-5\lambda+4)}} . \label{BNE} \end{align}\qquad{(11)}\]
Proof deferred to Appendix 11.8.
Therefore, in the depolarizing family, the three conditions (D1), (D2), and (D3) are satisfied simultaneously if and only if \(\lambda=1\) or \(\kappa=0\). Indeed, when \(\lambda=1\), all three conditions hold for every \(\kappa\in[-1,1]\). For \(0\le \lambda<1\), condition (D1) already forces \(\kappa=0\), and then both (D2) and (D3) are automatically satisfied. Thus, in this family the order-indistinguishable region is described exactly by the union of the completely depolarizing line \(\lambda=1\) and the unbiased input line \(\kappa=0\).
In this section, we consider the membership problem for \({\cal M}[{\cal S}_{N,2\to 1}]\) in the qubit-Pauli setting. Unlike Section 7, which treated the reverse-direction question only on the restricted domain \({\cal M}[{\cal S}_{N,1\to 2}]\), we now allow an arbitrary observed distribution as input. That is, given \(P \in {\cal M}[{\cal S}_G]\), we ask whether \(P\) belongs to \({\cal M}[{\cal S}_{N,2\to 1}]\). Thus, the present section addresses the full reverse-direction membership problem, whereas Section 7 addressed its restriction to distributions already known to lie in \({\cal M}[{\cal S}_{N,1\to 2}]\), namely the order-indistinguishability problem.
This problem should be distinguished from the reverse-compatibility analysis of Section 7.1. There, starting from a strategy \(S=S_{1\to 2}(\rho_1,C_{1\to 2})\in{\cal S}_{N,1\to 2}\), we studied whether the resulting distribution \({\cal M}[S]\) also belongs to \({\cal M}[{\cal S}_{N,2\to 1}]\). Because that problem already assumes membership in \({\cal M}[{\cal S}_{N,1\to 2}]\), the formulation based on Corollary 2 and the explicit conditions (D1)–(D3) is the more effective route there. By contrast, the present problem is a genuine membership test for \({\cal M}[{\cal S}_{N,2\to 1}]\) itself.
At the general level, this membership problem is governed by Theorem 6, with the roles of the two directions exchanged. Namely, in order to decide whether a distribution belongs to \({\cal M}[{\cal S}_{N,2\to 1}]\), one needs the reverse-direction analogue of the directional Markovian condition, the corresponding positivity condition for the reconstructed Choi matrix, and the algebraic consistency condition (C2), which is obtained from (C1) by exchanging the roles of the two parties. In the qubit-Pauli setting, these general conditions can be written in an explicit distribution-level form.
We therefore proceed in two steps. First, we record the qubit-Pauli specialization of the relevant algebraic consistency condition in the symmetric conditional model used below. Second, we combine this specialized form with the directional Markovian and positivity conditions to obtain a concrete membership criterion for \({\cal M}[{\cal S}_{N,2\to 1}]\).
We first recall the explicit two-qubit Pauli form of condition (C1). Although the present section is devoted to the membership problem for \({\cal M}[{\cal S}_{N,2\to 1}]\), it is convenient to begin with the original condition (C1), because the corresponding reverse-direction condition (C2) is obtained simply by exchanging the roles of the two parties. The following proposition is the two-qubit Pauli specialization of condition (C1) under the symmetric conditional model introduced below.
Proposition 8. In the two-qubit case, the relation 24 in condition (C1)* is simplified as \[\begin{align} \sum_{x_1\in \{0,1\}} P_{X_2|X_1,Y_1,Y_2}(x_2|x_1,y_1,y_2) &= 2 P_{X_2|Y_1,Y_2}(x_2|0,y_2) \notag\\ &\quad -\sum_{y_1'\in \{1,2,3\}} \sum_{x_1\in \{0,1\}} (-1)^{x_1} P_{X_1|Y_1}(x_1|y_1') \sum_{x_1'\in \{0,1\}} (-1)^{x_1'} P_{X_2|X_1,Y_1,Y_2}(x_2|x_1',y_1',y_2) \notag\\ &= 2 P_{X_2|Y_1,Y_2}(x_2|0,y_2) -\sum_{y_1'\in \{1,2,3\}} \langle \sigma_{y_1'}\otimes I\rangle \sum_{x_1'\in \{0,1\}} (-1)^{x_1'} P_{X_2|X_1,Y_1,Y_2}(x_2|x_1',y_1',y_2) \label{C1-2q} \end{align}\tag{44}\] for any \(y_1,y_2 \in \{1,2,3\}\) and \(x_2 \in \{0,1\}\).*
Assume further that, for parameters \(\lambda_1,\lambda_2,\lambda_3 \in [0,1]\), the conditional distribution \(P_{X_2|X_1,Y_1,Y_2}\) has the symmetry \[\begin{align} P_{X_2|X_1,Y_1,Y_2}(x_2|x_1,y_1,y_2) = \left\{ \begin{array}{ll} \frac{1}{2} &wheny_1 \neq y_2, \\[1mm] \frac{1}{2}(1-\lambda_{y_1})+\lambda_{y_1} \delta_{x_2,x_1} &wheny_1 = y_2, \end{array} \right. \label{C1-symmetry} \end{align}\qquad{(12)}\] and define \(\eta_1,\eta_2,\eta_3,\zeta_1,\zeta_2,\zeta_3 \in [-1,1]\) by \[\begin{align} P_{X_1|Y_1}(x_1|y_1)&= \frac{1}{2}(1-\eta_{y_1})+\eta_{y_1} x_1, \label{eta-def}\\ P_{X_2|Y_1,Y_2}(x_2|0,y_2)&= \frac{1}{2}(1-\zeta_{y_2})+\zeta_{y_2} x_2. \label{zeta-def} \end{align}\] {#eq: sublabel=eq:eta-def,eq:zeta-def} Then:
(i) condition (C1)* implies \[\begin{align} \zeta_{y_2}=0 \qquad\text{for all }y_2\in\{1,2,3\}; \label{zeta-zero} \end{align}\tag{45}\] *
(ii) if, in addition, the non-signaling condition \(Y_1-Y_2-X_2\) holds, then \[\begin{align} \eta_{y_1}\lambda_{y_1}=0 \qquad\text{for all }y_1\in\{1,2,3\}. \label{etalambda-zero} \end{align}\qquad{(13)}\]
Proof. The two-qubit simplification 44 follows directly from 24 by setting \(d=2\) and using the Pauli-measurement outcomes \(X_i\in\{0,1\}\), \(Y_i\in\{1,2,3\}\). The second equality in 44 is just the identity \[\langle \sigma_{y_1'}\otimes I\rangle = \sum_{x_1\in\{0,1\}}(-1)^{x_1}P_{X_1|Y_1}(x_1|y_1').\]
We now analyze 44 under the symmetric model ?? –?? .
Step 1: the case \(y_1\neq y_2\). When \(y_1\neq y_2\), the condition 44 becomes \[\begin{align} 1 &= (1-\zeta_{y_2})+2\zeta_{y_2} x_2 + \sum_{y_1'\in \{1,2,3\}} \eta_{y_1'} \sum_{x_1'\in \{0,1\}} (-1)^{x_1'} P_{X_2|X_1,Y_1,Y_2}(x_2|x_1',y_1',y_2) \notag\\ &= (1-\zeta_{y_2})+2\zeta_{y_2} x_2 +\eta_{y_2} \sum_{x_1'\in \{0,1\}} (-1)^{x_1'} P_{X_2|X_1,Y_1,Y_2}(x_2|x_1',y_2,y_2) \notag\\ &= (1-\zeta_{y_2})+2\zeta_{y_2} x_2-\eta_{y_2}\lambda_{y_2}. \label{C1-case-neq} \end{align}\tag{46}\] The left-hand side is independent of \(x_2\). Hence the coefficient of \(x_2\) on the right-hand side must vanish, which implies \[\zeta_{y_2}=0.\] Since \(y_2\) is arbitrary, this proves 45 .
Step 2: the case \(y_1=y_2\). Under the same model, we have \[\begin{align} P_{X_2,X_1|Y_1,Y_2}(x_2,x_1|y_1,y_2) =\frac{1}{2} \qquad\text{when }y_1\neq y_2, \label{PXY-neq} \end{align}\tag{47}\] whereas for \(y_1=y_2\), \[\begin{align} P_{X_2,X_1|Y_1,Y_2}(x_2,x_1|y_1,y_2) = \Bigl(\frac{1}{2}(1-\eta_{y_1})+\eta_{y_1}x_1\Bigr) \Bigl(\frac{1}{2}(1-\lambda_{y_1})+\lambda_{y_1}\delta_{x_2,x_1}\Bigr). \label{PXY-eq} \end{align}\tag{48}\] Therefore, \[\begin{align} P_{X_2|Y_1,Y_2}(x_2|y_1,y_2) &= \frac{1}{2}(1-\eta_{y_1}) +\eta_{y_1} \Bigl(\frac{1}{2}(1-\lambda_{y_1})+\lambda_{y_1}\delta_{x_2,1}\Bigr) \notag\\ &= \frac{1}{2} +\eta_{y_1}\lambda_{y_1}\Bigl(-\frac{1}{2}+\delta_{x_2,1}\Bigr). \label{PX2-cond} \end{align}\tag{49}\]
Now impose the non-signaling condition \(Y_1-Y_2-X_2\). Since 45 implies \[P_{X_2|Y_1,Y_2}(x_2|0,y_2)=\frac{1}{2},\] the non-signaling condition gives \[\begin{align} \frac{1}{2}(1-\zeta_{y_2})+\zeta_{y_2} x_2 = \frac{1}{2} = \frac{1}{2}+\eta_{y_1}\lambda_{y_1}\Bigl(-\frac{1}{2}+\delta_{x_2,1}\Bigr). \end{align}\] Hence \[\eta_{y_1}\lambda_{y_1}=0 \qquad\text{for all }y_1\in\{1,2,3\},\] which proves ?? . ◻
Proposition 8 gives the explicit qubit-Pauli form of the algebraic consistency condition in the symmetric conditional model ?? . By exchanging the roles of the two parties, one obtains the corresponding reverse-direction condition required for the membership problem of \({\cal M}[{\cal S}_{N,2\to 1}]\).
We now combine the above algebraic characterization with the other structural conditions. In the qubit-Pauli setting, the assumptions (A1), (A2), and (A3) hold. Hence, by Theorem 6 with the roles of the two directions exchanged, membership in \({\cal M}[{\cal S}_{N,2\to 1}]\) is governed by three ingredients: the reverse-direction Markovian condition, the positivity of the reconstructed Choi matrix for the direction \(2\to1\), and the reverse-direction analogue of the algebraic consistency condition.
The proposition above provides a concrete qubit-Pauli expression for the algebraic part. Therefore, under the symmetric conditional model ?? , the membership problem for \({\cal M}[{\cal S}_{N,2\to 1}]\) is reduced to an explicit check of these three conditions.
Theorem 9. Assume the qubit-Pauli setting of Section 7.1 and the symmetric conditional model ?? . Let \(P_{X_1,Y_1,X_2,Y_2}\in {\cal M}[{\cal S}_G]\) be an observed distribution. If
(i) the reverse-direction Markovian condition \(Y_1-Y_2-X_2\) holds,
(ii) the reconstructed Choi matrix for the direction \(2\to1\) satisfies \(\hat{C}_{2\to1}[P_{X_1,Y_1,X_2,Y_2}] \ge 0\),
(iii) and the reverse-direction analogue of Proposition 8 holds (equivalently, the qubit-Pauli form of condition (C2)* obtained by exchanging the labels \(1\) and \(2\)),*
then the distribution belongs to \({\cal M}[{\cal S}_{N,2\to 1}]\). Conversely, every element of \({\cal M}[{\cal S}_{N,2\to 1}]\) satisfies these three conditions.
Proof. This is exactly Theorem 6 with the two directions exchanged, specialized to the present qubit-Pauli setting. Condition (i) is the directional Markovian condition for the direction \(2\to1\). Condition (ii) is the corresponding positivity requirement for the reconstructed Choi matrix. Condition (iii) is the algebraic consistency condition needed in addition to Markovianity and positivity; in the qubit-Pauli setting, its explicit form is the reverse-direction counterpart of Proposition 8. Hence the statement follows. ◻
Theorem 9 clarifies the relation between the present section and the distinguishability analysis of Section 7.1. The previous section was optimized for the special problem in which the observed distribution is already known to come from a forward memoryless sequential strategy, and for that reason the formulation based on Corollary 2 and the explicit reverse-compatibility conditions (D1)–(D3) is more effective there. By contrast, the present section deals with the genuine membership problem for \({\cal M}[{\cal S}_{N,2\to 1}]\) itself. In that setting, the formulation descending from Theorem 6 is the appropriate one, and in the qubit-Pauli case it becomes fully explicit through the above proposition and theorem.
Under the same setting as the previous section, we next examine how the parallel class \({\cal M}[{\cal S}_{P}]\) and the memoryless sequential class \({\cal M}[{\cal S}_{N,1\to 2}]\) sit inside the quantum-memory class \({\cal M}[{\cal S}_{Q,1\to 2}]\). The first comparison concerns the inclusion \({\cal M}[{\cal S}_{P}]\subset {\cal M}[{\cal S}_{Q,1\to 2}]\) and asks how far positivity of reconstructed objects can detect the parallel class. The second concerns \({\cal M}[{\cal S}_{N,1\to 2}]\subset {\cal M}[{\cal S}_{Q,1\to 2}]\) and asks what additional structure is needed to distinguish genuinely memoryless sequential dynamics from more general quantum-memory strategies.
We begin with the hierarchy \({\cal M}[{\cal S}_{P}]\subset {\cal M}[{\cal S}_{Q,1\to 2}]\). Due to 18 , within the class \({\cal M}[{\cal S}_{Q,1\to 2}]\), the parallel subclass \({\cal M}[{\cal S}_{P}]\) is distinguished at the distribution level by the positivity of the reconstructed pseudo-density matrix \({\cal R}[{\cal M}[S]]\), which automatically implies the additional reverse-direction non-signaling condition \(Y_1-Y_2-X_2\). Since the characterization of the non-signaling conditions (D1) and (D2), are simpler than the characterization of the positivity condition \({\cal R}[{\cal M}[S]]\ge 0\), we discuss this condition under the assumptions (D1) and (D2) as follows.
Lemma 16. Assume that \(S \in {\cal S}_{Q,1\to 2}\) satisfies the conditions (D1) and (D2). The relation \({\cal R}[{\cal M}[S]]\ge 0\) is equivalent to the following condition.
\(t=3\). There are two local-unitary normal forms.
The condition is equivalent to \(1\ge \lambda_1+\lambda_2+\lambda_3\).
The condition is equivalent to \(1\ge \max(-\lambda_1 +\lambda_2+\lambda_3, \lambda_1 -\lambda_2+\lambda_3, \lambda_1 +\lambda_2-\lambda_3)\).
\(t=2\): \[\begin{align} (1-(c_1^1)^2)(1-(c_1^2)^2) \ge (\lambda_1+\lambda_2)^2. \label{BP1T} \end{align}\qquad{(14)}\]
\(t=1\): \[\begin{align} \bigl(1-((c_1^1)^2+(c_2^1)^2)\bigr) \bigl(1-((c_1^2)^2+(c_2^2)^2)\bigr) \ge (\lambda_1)^2. \label{BP2E} \end{align}\qquad{(15)}\]
\(t=0\): no further condition is imposed.
Proof deferred to Appendix 11.9.
Lemma 16 should be compared with Lemmas 10 and 11, which describe the positivity regions of the forward and reverse reconstructed Choi matrices. Under the directional assumptions (D1) and (D2), the three positivity notions \[\tilde{C}_{1\to 2}[{\cal M}[S]]\ge 0,\qquad \tilde{C}_{2\to 1}[{\cal M}[S]]\ge 0,\qquad {\cal R}[{\cal M}[S]]\ge 0\] do not coincide. In the cases \(t=2\) and \(t=1\), the condition \({\cal R}[{\cal M}[S]]\ge 0\) is stronger than reconstructed-Choi positivity, whereas in the case \(t=3\) the relative strength depends on the local-unitary normal form. Thus, positivity of the reconstructed pseudo-density matrix detects a generally stricter observable hierarchy than reconstructed-Choi positivity alone, but not in a uniform way across all channel geometries.
This comparison clarifies why positivity by itself does not close the hierarchy problem. Even after the directional conditions have been imposed, different positivity tests lead to different observable boundaries. To distinguish the memoryless sequential class from the full quantum-memory class, one therefore needs an additional ingredient beyond positivity.
We now turn to the hierarchy \({\cal M}[{\cal S}_{N,1\to 2}] \subset {\cal M}[{\cal S}_{Q,1\to 2}]\). Every strategy in \({\cal S}_{Q,1\to 2}\) already satisfies the non-signaling condition \(X_1-Y_1-Y_2\), so this directional condition alone cannot separate the memoryless sequential class from the larger quantum-memory class. Proposition 5 shows that the positivity condition \(\hat{C}_{1 \to 2}[P]\ge 0\) and the additional algebraic condition (C1) are satisfied by all memoryless sequential strategies. Here, both conditions are useful to distinguish an element of \({\cal M}[{\cal S}_{N,1\to 2}]\) from elements in \({\cal M}[{\cal S}_{Q,1\to 2}]\).
First, we show an example illustrating the usefulness of the positivity condition \(\hat{C}_{1 \to 2}[P]\ge 0\).
Example 2. Charlie prepares a maximally entangled state \(|\Phi^+\rangle := \frac{1}{\sqrt2}(|00\rangle+|11\rangle)\) on Alice’s input system and an internal memory qubit \(M\). He sends Alice’s half to Alice and keeps the memory qubit \(M\). After Alice performs her projective Pauli measurement, Charlie discards Alice’s output system and sends the memory qubit \(M\) to Bob’s input system.
Lemma 17. The strategy given by Example 2 satisfies the condition (C1)* and does not satisfy the positivity condition \(\hat{C}_{1 \to 2}[P]\ge 0\).*
Proof deferred to Appendix 11.10.
Next, we present an example for the usefulness of the condition (C1).
Example 3. Charlie prepares the classical memory \({\cal X}=\{0,1\}\) with the uniform distribution. He inputs the state \(|(-1)^X,3\rangle\) to Alice’s input system. Later, Charlie applies the unitary \(\sigma_X\) to Alice’s output system, then applies the depolarizing channel \(D_{1/2}(\rho):=\frac{1}{2}\,\rho+\frac{1}{2} \rho_{mix}\), where \(\rho_{mix}:=\frac{I}{2}\), and sends the result to Bob’s input system.
Lemma 18. The strategy given by Example 3 satisfies the positivity condition \(\hat{C}_{1\to 2}[P]\ge 0\), and does not satisfy the condition (C1).
Proof deferred to Appendix 11.11.
These two examples play complementary roles. Example 2 shows that the positivity condition \(\hat{C}_{1\to2}[P]\ge0\) is genuinely useful, since it can already exclude certain strategies in \({\cal M}[{\cal S}_{Q,1\to2}]\). By contrast, Example 3 shows that positivity alone is not sufficient: even when \(\hat{C}_{1\to2}[P]\ge0\) holds, the algebraic consistency condition (C1) may still fail. Taken together, the two examples make clear that the positivity condition and (C1) detect different obstructions, and that both are needed for the distribution-level characterization of \({\cal M}[{\cal S}_{N,1\to2}]\).
Example 2 showed that, when Charlie’s memory is quantum, the positivity condition \(\hat{C}_{1\to 2}[P]\ge 0\) may fail. This failure of positivity is not restricted to the quantum-memory case. It can already occur when Charlie’s memory is purely classical. The following example provides such a classical-memory strategy. Unlike Example 2, however, the present example does not isolate positivity alone: in this case, the observed distribution violates not only the positivity condition \(\hat{C}_{1\to 2}[P]\ge 0\), but also condition (C1). Thus, the example should be understood only as showing that the positivity condition can already fail in the classical-memory subclass.
Example 4. Charlie prepares the classical memory \({\cal X}=\{0,1\}\). He chooses \(X=0\) with probability \(\lambda \in (0,1)\), and chooses \(X=1\) with probability \(1-\lambda\). He inputs the state \(|(-1)^X,3\rangle\) to Alice’s input system. Later, Charlie applies the unitary \(\sigma_{X}\) to Alice’s output system and sends the result to Bob’s input system. Here, to satisfy the condition (C0), the condition \(\lambda \neq 0,1\) is needed.
Lemma 19. The strategy given by Example 4 satisfies neither condition (C1)* nor the positivity condition \(\hat{C}_{1\to 2}[P]\ge 0\).*
Proof deferred to Appendix 11.12.
We studied the statistical identification of causal structure in a restricted projective-observation regime, where Alice and Bob have access only to local measurement data and full process tomography is impossible. Our main focus was the memoryless sequential scenario, in which Charlie mediates a definite order without retaining memory, and the central question was whether this structure can be recognized directly from the observed distribution.
Our first main result is a complete distribution-level characterization of \({\cal M}[{\cal S}_{N,1\to 2}]\) under the assumptions (A1) and (A3). We showed that the directional Markovian condition \(X_1-Y_1-Y_2\) and the positivity of the reconstructed Choi matrix \(\hat{C}_{1\to2}[P]\) are not sufficient by themselves. The missing ingredient is the additional algebraic consistency condition (C1). Theorem 6 shows that these three conditions are together necessary and sufficient, and moreover that the reconstructed pair \((\rho_1[P],\hat{C}_{1\to2}[P])\) reproduces the original distribution. In this sense, the restricted projective problem can still be solved exactly at the level of observed statistics, even though exact process-matrix reconstruction is unavailable.
A central conceptual point of the paper is that condition (C1) is genuinely new. It is not a reformulation of positivity, nor does it follow from the usual non-signaling or directional Markovian constraints. Rather, it expresses the consistency requirement that the observed joint statistics arise from a single underlying channel acting on the post-measurement states. This clarifies why restricted projective observations require more than the standard combination of Markovianity and complete positivity. The example constructed in Section 9 makes this point explicit by showing that a strategy may satisfy the relevant positivity requirements and still fail to belong to the memoryless sequential class because (C1) is violated.
Our second main contribution is the explicit two-qubit Pauli analysis. Although this setting is not tomographically complete, it is analytically tractable and already captures the essential obstruction to order identification. For a forward memoryless sequential strategy \(S_{1\to2}(\rho_1,C_{1\to2})\), we translated the reverse-compatibility problem into the concrete conditions (D1)–(D3), and we analyzed these conditions both for general qubit channels and for representative channel families such as Pauli, phase-damping, and depolarizing channels. This yields a closed-form description of the order-indistinguishable region in the minimal nontrivial dimension.
Our third contribution is the clarification of observable hierarchies. We showed that positivity of reconstructed objects is not, by itself, enough to distinguish the relevant operational classes. In particular, the comparison between reconstructed-Choi positivity and pseudo-density-matrix positivity reveals that different positivity tests lead to different observable boundaries. To separate the memoryless sequential class from the broader quantum-memory class, an additional distribution-level condition is indispensable, and this is precisely the role played by (C1).
Finally, we formulated the reverse-direction membership problem for \({\cal M}[{\cal S}_{N,2\to 1}]\) in the same qubit-Pauli regime. This should be viewed as the full version of the reverse-direction question: in Section 7 we considered the same question only for distributions already known to belong to \({\cal M}[{\cal S}_{N,1\to 2}]\), which led to the order-indistinguishability problem, whereas here we treated arbitrary observed distributions. By applying the reverse-direction analogue of Theorem 6, we obtained a concrete membership criterion in terms of the reverse Markovian condition, the positivity of the reconstructed reverse-direction Choi matrix, and the corresponding algebraic consistency condition.
Taken together, these results turn the qualitative limitation of causal-order identification under restricted projective observations into an explicit set-membership problem. They show that even in a genuinely non-tomographic regime, one can still obtain sharp and operationally checkable criteria for memorylessness, causal compatibility, and order-indistinguishability. This provides a concrete statistical framework for causal inference in quantum network scenarios where experimental control is limited to fixed projective measurements.
The author thanks Dr Baichu Yu for valuable discussions. The author was supported in part by the General R&D Projects of 1+1+1 CUHK-CUHK(SZ)-GDST Joint Collaboration Fund (Grant No. GRDP2025-022), the Guangdong Provincial Quantum Science Strategic Initiative (Grant No. GDZX2505003), and the Shenzhen International Quantum Academy (Grant No. SIQA2025KFKT07). Large language model tools were used as auxiliary aids in preparing this manuscript, including assistance with exposition, literature search, and exploratory discussions of possible approaches to the research problem. The manuscript was written under the author’s direction, and the author is solely responsible for all mathematical content, proofs, references, and conclusions.
The following lemma provides the basic positivity criterion used in the following appendices.
Lemma 20. The relation \(I \otimes I +s_1 \sigma_1 \otimes I +s_2 I \otimes \sigma_1 +\sum_{j=1}^3 \tau_j \sigma_j \otimes \sigma_j \ge 0\) holds if and only if the relations \[\begin{align} 1+\tau_1+s_1+s_2 &\ge 0 \label{AP1}\\ 1-\tau_1-s_1+s_2 &\ge 0 \label{AP2}\\ (1-\tau_1)^2 &\ge (s_1-s_2)^2+(\tau_2+\tau_3)^2 \label{BP1}\\ (1+\tau_1)^2 &\ge (s_1+s_2)^2+(\tau_2-\tau_3)^2 .\label{BP2} \end{align}\] {#eq: sublabel=eq:AP1,eq:AP2,eq:BP1,eq:BP2} hold.
Proof. Applying Hadamard matrix, we exchange the bases of \(\sigma_1\) and \(\sigma_3\). After this conversion, we make the matrix representation of \(I \otimes I +s_1 \sigma_1 \otimes I +s_2 I \otimes \sigma_1 +\sum_{j=1}^3 \tau_j \sigma_j \otimes \sigma_j\) based on the basis \(|0,0\rangle, |1,0\rangle,|0,1\rangle,|1,1\rangle\) as \[\begin{align} \left( \begin{array}{cccc} 1+\tau_1+s_1+s_2 & 0 & 0 &\tau_3-\tau_2 \\ 0& 1-\tau_1-s_1+s_2 & \tau_3+\tau_2 &0 \\ 0& \tau_3+\tau_2 & 1-\tau_1+s_1-s_2 & 0 \\ \tau_3-\tau_2 & 0& 0 & 1+\tau_1-s_1-s_2 \end{array} \right) \end{align}\] The positivity of the determinant of the components with basis \(|0,0\rangle, |1,1\rangle\) means ?? and \[\begin{align} (1+\tau_1+s_1+s_2) (1+\tau_1-s_1-s_2) \ge (\tau_2-\tau_3)^2 , \end{align}\] which is equivalent to ?? . The positivity of the determinant of the components with basis \(|1,0\rangle, |0,1\rangle\) means ?? and \[\begin{align} (1-\tau_1-s_1+s_2) (1-\tau_1+s_1-s_2) \ge (\tau_2+\tau_3)^2 , \end{align}\] which is equivalent to ?? . ◻
Proof. Alice’s averaged output state is \(\frac{1}{2}(I+ c^1_{Y_1} \sigma_{Y_1})\) for when \(Y_1=1,2,3\), and is \(\frac{1}{2}(I+ \sum_{j=1}^3 c^1_j \sigma_j)\) for \(Y_1=0\). The condition (D1) is equivalent to the condition \(\Lambda_{1\to 2}(\frac{1}{2}(I+ c^1_{i} \sigma_{i}))= \Lambda_{1\to 2}(\frac{1}{2}(I+ \sum_{y=1}^3 c^1_y \sigma_y))\) for \(i=1,2,3\), which means that \[\begin{align} \rho_2+\sum_{j=1}^t \frac{c^1_i \lambda_j}{4}\mathrm{Tr}(\alpha_j \sigma_{i}) \beta_j = \rho_2+\sum_{j=1}^t \sum_{y=1}^3 \frac{c^1_y \lambda_j}{4} \mathrm{Tr}(\alpha_j \sigma_{y}) \beta_j \label{BF0} \end{align}\tag{50}\] for \(y_1=1,2,3\). Since \(\lambda_j>0\), the above condition is equivalent to the condition \[\begin{align} c^1_i \mathrm{Tr}(\alpha_j \sigma_{i}) = \sum_{y=1}^3 c^1_y \mathrm{Tr}(\alpha_j \sigma_{y}) \label{BF1} \end{align}\tag{51}\] for \(i=1,2,3\) and \(j\le t\). This condition is equivalent to \[\begin{align} c^1_i \mathrm{Tr}(\alpha_j \sigma_{i}) =0 \label{BF2} \end{align}\tag{52}\] for \(i=1,2,3\) and \(j\le t\).
Case (1). \(t=3\). Since \(\alpha_1,\alpha_2,\alpha_3\) are linearly independent, the matrix \((\mathrm{Tr}(\alpha_j \sigma_{i}))_{j,i}\) is invertible. The condition (D1), i.e., the condition 51 is equivalent to the condition \(c^1_i=0\) for \(i=1,2,3\).
Case (2). \(t=2\). Since \(\alpha_1,\alpha_2\) are linearly independent, At most, two \(i\) satisfies \((\mathrm{Tr}(\alpha_1 \sigma_{i}) ,\mathrm{Tr}(\alpha_2 \sigma_{i}) ) \neq (0,0)\). For simplicity, we assume that \((\mathrm{Tr}(\alpha_1 \sigma_{i}) ,\mathrm{Tr}(\alpha_2 \sigma_{i}) ) \neq (0,0)\) for \(i=2,3\). Then, the condition (D1), i.e., the condition 51 is equivalent to the condition that \(c^1_2=c^1_3=0\) and \(c^1_1\mathrm{Tr}\alpha_1\sigma_1 =c^1_1 \mathrm{Tr}\alpha_2\sigma_1=0\). That is, at least, two \(i\) satisfies \(c^1_i=0\).
Case (3).
\(t=1\). At most, one \(i\) satisfies \(\mathrm{Tr}(\alpha_1 \sigma_{i}) \neq 0\). For simplicity, we assume that \(\mathrm{Tr}(\alpha_1 \sigma_{3}) \neq 0\). Then, the condition (D1), i.e., the condition 51 is equivalent to the condition that \(c^1_3=0\) and \(c^1_1 \mathrm{Tr}\alpha_1\sigma_1 =c^1_2 \mathrm{Tr}\alpha_1\sigma_2=0\). That is, at least, one \(i\) satisfies \(c^1_i=0\).
Case (4).
\(t=0\). The condition (D1) always holds because the relation 50 always holds. That is, no condition is required. ◻
Proof. First, we show (i). The conditions given in Lemma 8 and the formula 41 yield that \((\lambda_j \beta_j)\circ \rho_1= \frac{\lambda_j}{2} \beta_j\), which implies ?? . Then, ?? implies ?? by using 15 .
Next, we proceed to (ii). We denote \(\tilde{C}_{2\to 1}[{\cal M}[S]]\) as \[\begin{align} \tilde{C}_{2\to 1}[{\cal M}[S]]^{T_2} =& \tilde{\rho}_1 \otimes I + \sum_{j=1}^t \frac{\tilde{\lambda}_j}{2} \tilde{x}_j \otimes \tilde{y}_j , \end{align}\] where \(I\), \(\tilde{y}_1,\tilde{y}_2,\tilde{y}_3\) are orthogonal to each other, and \(\tilde{x}_1,\tilde{x}_2,\tilde{x}_3\) are orthogonal to each other. We assume that \(\|\tilde{x}_j\|=\|\tilde{y}_j\|=1\) for \(j=1,2,3\). Since \(\tilde{C}_{2\to 1}[{\cal M}[S]] \circ (I\otimes \rho_2) ={\cal R}[{\cal M}[S]]\), we have \[\begin{align} {\cal R}[{\cal M}[S]]= \rho_1 \otimes \rho_2 + \sum_{j=1}^t \frac{\lambda_j}{4} \alpha_j \otimes \beta_j = \tilde{\rho}_1 \otimes \rho_2 + \sum_{j=1}^t \frac{\tilde{\lambda}_j}{2} \tilde{x}_j \otimes (\tilde{y}_j \circ \rho_2). \end{align}\] Since \(\mathrm{Tr}_2 {\cal R}[{\cal M}[S]]=\rho_1\), we have \(\rho_1=\tilde{\rho}_1\). Since \(\mathrm{Tr}_1 {\cal R}[{\cal M}[S]]=\rho_2\), we have \[\begin{align} \rho_2= \rho_2 + \sum_{j=1}^t \frac{\tilde{\lambda}_j}{2} (\mathrm{Tr}\tilde{x}_j) (\tilde{y}_j \circ \rho_2).\label{XHJ} \end{align}\tag{53}\] Since \(\rho_2\), \(\{\tilde{y}_j \circ \rho_2\}_{j=1}^3\) are linearly independent, 53 implies \(\frac{\tilde{\lambda}_j}{2} \mathrm{Tr}\tilde{x}_j =0\) for \(j=1,2,3\).
Applying the conditions in Lemma 8 to the pair \((\rho_2,\tilde{C}_{2\to 1}[{\cal M}[S]])\), we find that the condition (D2) implies \(\sum_{j=1}^t \frac{\tilde{\lambda}_j}{2} \tilde{x}_j\otimes (\tilde{y}_j \circ \rho_2)= \sum_{j=1}^t \frac{\tilde{\lambda}_j}{4} \tilde{x}_j \otimes\tilde{y}_j\). When the condition (D2) holds, changing the indexes of \(\tilde{x}_j,\tilde{y}_j\), we have \(\lambda_j=\tilde{\lambda}_j\), \(\alpha_j=\tilde{x}_j\), \(\alpha_j=\tilde{y}_j\) for \(j=1,2,3\). Then, we obtain the conditions (1),(2),(3),(4).
Conversely, we assume that the conditions (1),(2),(3),(4) hold. The relation ?? implies \(\lambda_j \beta_j\circ \rho_2=\lambda_j \beta_j\) for \(j=1,2,3\), which yields the relation \[\begin{align} \tilde{C}_{2\to 1}[{\cal M}[S]]^{T_2} =& \rho_1 \otimes I + \sum_{j=1}^t \frac{\lambda_j}{2} \alpha_j \otimes \beta_j . \label{HU5} \end{align}\tag{54}\] The relation 54 and Lemma 8 guarantees the condition (D2). ◻
Proof of Lemma 10. Throughout the proof, we use the structural restrictions obtained in Lemma 9 under the assumptions (D1) and (D2), and we repeatedly reduce the positivity problem to Lemma 20 by choosing convenient local-unitary normal forms.
Lemma 9 guarantees that \(\rho_1=I/2\). Applying a unitary on the first system, we can transform \(\alpha_1,\alpha_2,\alpha_3\) into either \(\sigma_1,\sigma_2,\sigma_3\) or \(\sigma_1,-\sigma_2,\sigma_3\). Similarly, applying a unitary on the second system, we can transform \(\beta_1,\beta_2,\beta_3\) into either \(\sigma_1,\sigma_2,\sigma_3\) or \(\sigma_1,-\sigma_2,\sigma_3\). Hence there are two local-unitary normal forms.
Case (A). Then \(C_{1\to 2}\) is unitarily equivalent to \[\frac{1}{2}\bigl( I\otimes I +\lambda_1\sigma_1\otimes\sigma_1 -\lambda_2\sigma_2\otimes\sigma_2 +\lambda_3\sigma_3\otimes\sigma_3 \bigr).\] Since ?? and ?? in Lemma 20 are automatic in this case, Lemma 20 implies that \(C_{1\to 2}\ge0\) is equivalent to \[(1-\lambda_1)^2\ge(-\lambda_2+\lambda_3)^2, \qquad (1+\lambda_1)^2\ge(-\lambda_2-\lambda_3)^2.\] Because \(\lambda_j>0\), this is equivalent to \[1\ge \max(-\lambda_1+\lambda_2+\lambda_3,\, \lambda_1-\lambda_2+\lambda_3,\, \lambda_1+\lambda_2-\lambda_3).\]
Case (B). Then \(C_{1\to 2}\) is unitarily equivalent to \[\frac{1}{2}\bigl( I\otimes I +\lambda_1\sigma_1\otimes\sigma_1 +\lambda_2\sigma_2\otimes\sigma_2 +\lambda_3\sigma_3\otimes\sigma_3 \bigr).\] Again ?? and ?? are automatic, and Lemma 20 yields \[(1-\lambda_1)^2\ge(\lambda_2+\lambda_3)^2, \qquad (1+\lambda_1)^2\ge(\lambda_2-\lambda_3)^2.\] Since \(\lambda_j>0\), this is equivalent to \[1\ge \lambda_1+\lambda_2+\lambda_3.\]
Lemma 9 guarantees that \[C_{1\to 2} = \frac{1}{2} I\otimes (I+c_1^2\sigma_1)^T +\sum_{j=1}^2\frac{\lambda_j}{2}\alpha_j\otimes\beta_j^T.\] Assume first that \(c_1^2\neq0\). Since \(c_1^2\mathrm{Tr}\beta_1\sigma_1=c_1^2\mathrm{Tr}\beta_2\sigma_1=0\), we can, by a unitary on the second system fixing \(\sigma_1\), transform \(\beta_1,\beta_2\) into \(\pm\sigma_2,\sigma_3\). By a unitary on the first system, we can simultaneously transform \(\alpha_1,\alpha_2\) into \(\pm\sigma_2,\sigma_3\) with the same sign choice. Thus \(C_{1\to 2}\) is unitarily equivalent to \[\frac{1}{2}\bigl( I\otimes I +c_1^2 I\otimes \sigma_1 -\lambda_1\sigma_2\otimes\sigma_2 +\lambda_2\sigma_3\otimes\sigma_3 \bigr).\] Applying Lemma 20, we obtain \[1\ge (\lambda_1+\lambda_2)^2+(c_1^2)^2.\] If \(c_1^2=0\), the \(t=3\) discussion shows that \(1\ge\lambda_1+\lambda_2\), which is equivalent to the same inequality. This proves Case (2).
Lemma 9 yields \[C_{1\to 2} = \frac{1}{2} I\otimes (I+c_1^2\sigma_1+c_2^2\sigma_2) +\frac{\lambda_1}{2}\alpha_1\otimes\beta_1^T.\] If \(c_1^2\neq0\) and \(c_2^2\neq0\), then \(c_1^2\mathrm{Tr}\beta_1\sigma_1=c_2^2\mathrm{Tr}\beta_1\sigma_2=0\), so \(\beta_1=\pm\sigma_3\). A unitary on the first system transforms \(\alpha_1\) to \(\sigma_3\). Hence \(C_{1\to 2}\) is unitarily equivalent to \[\frac{1}{2}\bigl( I\otimes I +c_1^2 I\otimes \sigma_1 +c_2^2 I\otimes \sigma_2 \pm \lambda_1 \sigma_3\otimes\sigma_3 \bigr),\] and Lemma 20 implies \[1\ge \lambda_1^2+(c_1^2)^2+(c_2^2)^2.\] If one of \(c_1^2,c_2^2\) vanishes, the \(t=2\) discussion reduces again to the same inequality. This proves Case (3).
In this case \[C_{1\to 2}=I\otimes \rho_2\ge0,\] so no further condition is required. ◻
Proof of Lemma 11. Statement (i) was already obtained in the proof of Lemma 9, namely \[\tilde{C}_{2\to1}[{\cal M}[S]]^{T_2} = \rho_1\otimes I+\sum_{j=1}^t\frac{\lambda_j}{2}\alpha_j\otimes\beta_j. }\]
We now prove (ii). The point is that the positivity analysis for \(\tilde{C}_{2\to1}[{\cal M}[S]]\) is exactly parallel to the positivity analysis for \(C_{1\to2}\) in Lemma 10. Indeed, once ?? is available, the only difference is that the Bloch parameters entering the local state now come from \(\rho_1\) rather than \(\rho_2\). Therefore the same local-unitary reductions and the same applications of Lemma 20 go through verbatim, with \(c_i^2\) replaced by \(c_i^1\).
More explicitly:
Under (D1) and (D2), Lemma 8 and Lemma 9 imply \(\rho_1=I/2\) and \(\rho_2=I/2\). Hence ?? has exactly the same two local-unitary normal forms as in Lemma 10, namely \[\frac{1}{2}\bigl( I\otimes I +\lambda_1\sigma_1\otimes\sigma_1 -\lambda_2\sigma_2\otimes\sigma_2 +\lambda_3\sigma_3\otimes\sigma_3 \bigr)\] and \[\frac{1}{2}\bigl( I\otimes I +\lambda_1\sigma_1\otimes\sigma_1 +\lambda_2\sigma_2\otimes\sigma_2 +\lambda_3\sigma_3\otimes\sigma_3 \bigr).\] Applying Lemma 20 exactly as in the proof of Lemma 10, we obtain the two conditions \[1\ge \max(-\lambda_1+\lambda_2+\lambda_3,\, \lambda_1-\lambda_2+\lambda_3,\, \lambda_1+\lambda_2-\lambda_3)\] and \[1\ge \lambda_1+\lambda_2+\lambda_3.\]
Now ?? has the same form as the \(t=2\) case of Lemma 10, except that the local Bloch parameter comes from \(\rho_1\). Thus the same unitary reduction gives a matrix unitarily equivalent to \[\frac{1}{2}\bigl( I\otimes I +c_1^1 \sigma_1\otimes I -\lambda_1\sigma_2\otimes\sigma_2 +\lambda_2\sigma_3\otimes\sigma_3 \bigr),\] and Lemma 20 yields \[1\ge (\lambda_1+\lambda_2)^2+(c_1^1)^2.\]
Similarly, ?? is reduced to a matrix unitarily equivalent to \[\frac{1}{2}\bigl( I\otimes I +c_1^1 \sigma_1\otimes I +c_2^1 \sigma_2\otimes I \pm \lambda_1\sigma_3\otimes\sigma_3 \bigr),\] and Lemma 20 yields \[1\ge \lambda_1^2+(c_1^1)^2+(c_2^1)^2.\]
In this case ?? becomes \[\tilde{C}_{2\to1}[{\cal M}[S]]^{T_2} = \rho_1\otimes I,\] which is always positive.
This proves statement (ii). ◻
We consider the Pauli channel \[\begin{align} \Lambda(\rho):= \sum_{j=0}^3 p_j \sigma_j \rho \sigma_j . \end{align}\] Its Choi matrix \(C_{1\to 2}\) is given as \[\begin{align} C_{1\to 2}^{T_1} = I \otimes \frac{I}{2} + \sum_{j=1}^3 \frac{\lambda_j}{2} \sigma_j \otimes \sigma_j, \label{HJ51A} \end{align}\tag{55}\] where \(\lambda_j:=2p_0+2p_j-1\) for \(j=1,3\) and \(\lambda_2:=-2p_0-2p_2+1\). Assume that the initial state is \[\rho_1 =\frac{1}{2}(I+ \kappa \sigma_1).\]
We first compute \({\cal R}[{\cal M}[S]]\). A direct calculation gives \[\begin{align} &{\cal R}[{\cal M}[S]] = \rho_1 \otimes \frac{I}{2} +\frac{\kappa}{2}\frac{\lambda_1}{2} I \otimes \sigma_1 + \frac{1}{2}\sum_{j=1}^3 \frac{\lambda_j}{2}\sigma_j \otimes \sigma_j \notag\\ =& \frac{I}{2} \otimes \Bigl(\frac{I}{2} +\kappa \frac{\lambda_1}{2} \sigma_1\Bigr) +\frac{\kappa}{2} \sigma_1 \otimes \frac{ I}{2} + \frac{1}{2}\sum_{j=1}^3 \frac{\lambda_j}{2}\sigma_j \otimes \sigma_j \notag\\ =& \frac{I}{2} \otimes \Bigl(\frac{I}{2} +\kappa \frac{\lambda_1}{2} \sigma_1\Bigr) +\frac{\kappa}{2} \sigma_1 \otimes \frac{ I}{2} + \frac{\lambda_1}{4}\sigma_1 \otimes \sigma_1 + \frac{1}{2}\sum_{j=2}^3 \frac{\lambda_j}{2}\sigma_j \otimes \sigma_j \notag\\ =& \frac{I}{2} \otimes \Bigl(\frac{I}{2} +\kappa \frac{\lambda_1}{2} \sigma_1\Bigr) +\frac{1}{2}\sigma_1 \otimes \Bigl(\frac{\kappa}{2} I +\frac{\lambda_1}{2} \sigma_1 \Bigr) + \frac{1}{2}\sum_{j=2}^3 \frac{\lambda_j}{2}\sigma_j \otimes \sigma_j . \label{HJ58A} \end{align}\tag{56}\] Hence, \[\begin{align} \rho_2[{\cal M}[S]] = \frac{I}{2} +\kappa \frac{\lambda_1}{2}\sigma_1 . \label{HJ58B} \end{align}\tag{57}\]
Since \[(a I+b\sigma_1)\circ (c I+d\sigma_1) = (ac+bd) I+(ad+bc)\sigma_1,\] for \(|\kappa|<1\) we have \[\begin{align} \tilde{C}_{2\to 1}[{\cal M}[S]]^{T_2} =& \frac{I}{2} \otimes I +\frac{1}{2}\sigma_1\otimes \Big( \frac{\kappa(\lambda_1^2 -1)}{\kappa^2\lambda_1^2-1}I +\frac{ \lambda_1 (\kappa^2 -1)}{\kappa^2\lambda_1^2-1} \sigma_1\Big) \notag\\ & + \sum_{j=2}^3 \frac{\lambda_j}{2}\sigma_j \otimes \sigma_j \notag\\ =& \frac{I}{2}\otimes I +\frac{\kappa(\lambda_1^2 -1)}{2(\kappa^2\lambda_1^2-1)} \sigma_1 \otimes I + \frac{ \lambda_1 (\kappa^2 -1)}{2(\kappa^2\lambda_1^2-1)} \sigma_1\otimes \sigma_1 \notag\\ & + \sum_{j=2}^3 \frac{\lambda_j}{2} \sigma_j \otimes \sigma_j . \label{HJ53A} \end{align}\tag{58}\] When \(|\kappa|=1\), we have \[\begin{align} \tilde{C}_{2\to 1}[{\cal M}[S]]^{T_2} =& \frac{I}{2} \otimes I +\frac{\kappa}{2}\sigma_1\otimes I + \sum_{j=2}^3 \frac{\lambda_j}{2}\sigma_j \otimes \sigma_j \notag\\ =& \frac{I}{2}\otimes I +\frac{\kappa}{2}\sigma_1 \otimes I + \sum_{j=2}^3 \frac{\lambda_j}{2} \sigma_j \otimes \sigma_j . \label{HJ5GA} \end{align}\tag{59}\]
We can now read off the directional conditions. The condition (D1) holds if and only if \(\kappa \lambda_1=0\), which is equivalent to the condition \(\lambda_1=0\) or \(\kappa=0\). This proves (i).
The condition (D2) holds if and only if \(\kappa \lambda_1=0\) or \[\frac{ \lambda_1 (\kappa^2 -1)}{2(\kappa^2\lambda_1^2-1)} =0,\] which is equivalent to the condition \(\lambda_1=0\) or \(\kappa=0\), or \(\kappa=\pm 1\) and \(\lambda_1<1\). This proves (ii).
We next prove (iii), namely the characterization of (D3) in the interior case \(|\kappa|<1\). Applying Lemma 20 to 58 , we find that condition (D3) holds if and only if \[\begin{align} 1+ \frac{\kappa(\lambda_1^2 -1)}{\kappa^2\lambda_1^2-1} +\frac{ \lambda_1 (\kappa^2 -1)}{\kappa^2\lambda_1^2-1} &\ge 0,\tag{60}\\ 1- \frac{\kappa(\lambda_1^2 -1)}{\kappa^2\lambda_1^2-1} -\frac{ \lambda_1 (\kappa^2 -1)}{\kappa^2\lambda_1^2-1} &\ge 0, \tag{61}\\ \left(\frac{(\kappa^2 \lambda_1 -1)( \lambda_1 -1)}{\kappa^2\lambda_1^2-1}\right)^2 = \left(1- \frac{ \lambda_1 (\kappa^2 -1)}{\kappa^2\lambda_1^2-1} \right)^2 & \ge \left(\frac{\kappa(\lambda_1^2 -1)}{\kappa^2\lambda_1^2-1}\right)^2 +(\lambda_2+\lambda_3)^2 , \tag{62}\\ \left(\frac{(\kappa^2 \lambda_1 -1)( \lambda_1 +1)}{\kappa^2\lambda_1^2-1}\right)^2 = \left(1+ \frac{ \lambda_1 (\kappa^2 -1)}{\kappa^2\lambda_1^2-1} \right)^2 & \ge \left(\frac{\kappa(\lambda_1^2 -1)}{\kappa^2\lambda_1^2-1}\right)^2 +(\lambda_2-\lambda_3)^2 . \tag{63} \end{align}\]
Since \(\kappa^2\lambda_1^2-1 <0\), the conditions 60 and 61 are simplified as \[\begin{align} \kappa^2\lambda_1^2-1+ \kappa(\lambda_1^2 -1) + \lambda_1 (\kappa^2 -1) &\le 0,\notag\\ \kappa^2\lambda_1^2-1- \kappa(\lambda_1^2 -1) - \lambda_1 (\kappa^2 -1) &\le 0, \end{align}\] which are equivalent to \[\begin{align} (1+\kappa)(1+\lambda_1)(\kappa\lambda_1 - 1)&\le 0,\tag{64}\\ (1-\kappa)(1-\lambda_1)(\kappa\lambda_1 - 1)&\le 0.\tag{65} \end{align}\] Since \(1+\kappa, 1+\lambda_1,1-\kappa, 1-\lambda_1\ge 0\), the above conditions are equivalent to \[\begin{align} \kappa \le 1/\lambda_1 . \label{ABQC} \end{align}\tag{66}\] But this condition always holds. Hence, for \(|\kappa|<1\), condition (D3) is equivalent exactly to 60 –68 . This proves (iii).
Finally, we treat the boundary case \(|\kappa|=1\) and prove (iv). In this case, since the conditions ?? and ?? in Lemma 20 hold automatically for 59 , Lemma 20 guarantees that condition (D3) holds if and only if \[\begin{align} 1\ge 1 + \max \bigl((\lambda_2+\lambda_3)^2,(\lambda_2-\lambda_3)^2\bigr), \label{CBV2Y} \end{align}\tag{67}\] which is equivalent to \(\lambda_2=\lambda_3=0\). This proves (iv).
We fix the initial state to be \(\rho_1 =\frac{1}{2}(I+ \kappa \sigma_1)\). We consider the phase damping channel with respect to the \(\sigma_1\)-basis, whose Choi matrix \(C_{1\to 2}\) satisfies the condition \(\lambda_1=1\) under the form 42 . Then, the condition (D1) holds if and only if \(\kappa=0\). The condition (D2) holds if and only if \(\kappa=0\). We discuss the condition (D3).
Lemma 21. The condition (D3) holds if and only if the relations \(\kappa=\pm 1\) and \(\lambda_2=\lambda_3=0\) or the relations \(|\kappa|<1\), \(\lambda_2=-\lambda_3\), and \(|\lambda_3|<1\) hold.
Proof. When the condition (D3) holds, \(\kappa\) can take the value \(\pm 1\) when \(\lambda_2=\lambda_3=0\) due to 66 . Hence, we consider the condition (D3) under the condition \(\kappa^2 <1\). In this case, the condition (D3) is equivalent to 62 and 68 , which are simplified as \[\begin{align} 0 & \ge(\lambda_2+\lambda_3)^2 \\ 4=(\frac{2(\kappa^2 -1)}{\kappa^2-1 })^2 & \ge (\lambda_2-\lambda_3)^2 .\label{CBV2} \end{align}\tag{68}\] Thus, \(\lambda_2=-\lambda_3\) and \(|\lambda_3|<1\). ◻
Next, under the same the fixed initial state \(\rho_1 =\frac{1}{2}(I+ \kappa \sigma_1)\), we consider the phase damping channel with respect to the \(\sigma_3\)-basis, whose Choi matrix \(C_{1\to 2}\) satisfies the condition \(\lambda_3=1\) under the form 42 . Additionally, we assume \(\lambda_2=-\lambda_1\) for simplicity. Then, the conditions (D1) and (D2) have the same form as the previous section because these two condition does not depend on \(\lambda_3\). We discuss the condition (D3).
Lemma 22. The condition (D3) holds if and only if the relation \(\kappa=0\) or the relations \(\lambda_1=\pm 1\) and \(\kappa^2<1\) hold.
Proof. When the condition (D3) holds, \(\kappa\) cannot take the value \(\pm 1\) due to 66 . Hence, we consider the condition (D3) under the condition \(\kappa^2 <1\). In this case, the condition (D3) is equivalent to 62 and 68 , which are simplified as \[\begin{align} (\frac{(\kappa^2 \lambda_1 -1)( \lambda_1 -1)}{\kappa^2\lambda_1^2-1 })^2 =(1- \frac{ \lambda_1 (\kappa^2 -1)}{\kappa^2\lambda_1^2-1 } )^2 & \ge (\frac{\kappa(\lambda_1^2 -1)}{\kappa^2\lambda_1^2-1 })^2 +(1-\lambda_1)^2 \tag{69}\\ (\frac{(\kappa^2 \lambda_1 -1)( \lambda_1 +1)}{\kappa^2\lambda_1^2-1 })^2 =(1+ \frac{ \lambda_1 (\kappa^2 -1)}{\kappa^2\lambda_1^2-1 } )^2 & \ge (\frac{\kappa(\lambda_1^2 -1)}{\kappa^2\lambda_1^2-1 })^2 +(1+\lambda_1)^2 .\tag{70} \end{align}\]
First, we discuss the condition 69 by breaking down into the following two cases:
Case 1: \(\lambda_1 = 1\). If \(\lambda_1 = 1\), then \(1-\lambda_1 = 0\). The inequality 69 becomes: \[\left(\frac{0}{\kappa^2-1}\right)^2 \ge \left(\frac{0}{\kappa^2-1}\right)^2 + 0^2,\] which is true for all \(\kappa\) as long as \(\kappa^2 <1\).
Case 2: \(\lambda_1 \ne 1\). Multiplying by \((\kappa^2\lambda_1^2-1)^2/(1-\lambda_1)^2\) in 69 , we have \[(\kappa^2\lambda_1+1)^2 \ge \kappa^2(\lambda_1+1)^2 + (\kappa^2\lambda_1^2-1)^2\] Expanding both sides, we have \[0 \ge \kappa^4\lambda_1^4 - \kappa^4\lambda_1^2 - \kappa^2\lambda_1^2 + \kappa^2.\] Factoring out common terms, we obtain \[0 \ge \kappa^2(\kappa^2\lambda_1^2 - 1)(\lambda_1^2 - 1),\] which implies \(\kappa=0\) or \(\lambda_1=-1\) due to the conditions \(\lambda \neq 1\) and \(\kappa^2 <1\).
Next, we discuss the condition 70 by breaking down into the following two cases:
Case 1: \(\lambda_1 = -1\). If \(\lambda_1 = -1\), then \(1+\lambda_1 = 0\). The inequality 70 becomes: \[\left(\frac{0}{\kappa^2-1}\right)^2 \ge \left(\frac{0}{\kappa^2-1}\right)^2 + 0^2,\] which is true for all \(\kappa\) as long as \(\kappa^2 <1\).
Case 2: \(\lambda_1 \ne -1\). Multiplying by \((\kappa^2\lambda_1^2-1)^2/(1+\lambda_1)^2\) in 70 , we have \[(\kappa^2\lambda_1-1)^2 \ge \kappa^2(\lambda_1-1)^2 + (\kappa^2\lambda_1^2-1)^2\] Expanding both sides, we have \[0 \ge \kappa^4\lambda_1^4 - \kappa^4\lambda_1^2 - \kappa^2\lambda_1^2 + \kappa^2.\] Factoring out common terms: \[0 \ge \kappa^2(\lambda_1^2 - 1)(\kappa^2\lambda_1^2 - 1),\] which implies \(\kappa=0\) or \(\lambda_1=1\) due to the conditions \(\lambda \neq -1\) and \(\kappa^2 <1\).
In summary, under the condition \(\kappa^2 <1\), the conditions 69 and 70 are equivalent to the relation \(\kappa=0\) or \(\lambda_1=\pm 1\). ◻
Proof. When \(\kappa^2=1\) the condition (D3) is equivalent with the condition \(\lambda_2=\lambda_3=0\), which implies \(\lambda =1\). The statement of this lemma allows \(\kappa^2\) takes \(1\) only when \(\lambda=1\). Hence, the statement of this lemma is shown when \(\kappa=1\). In the following, we discuss the case when \(\kappa^2<1\), which converts the condition (D3) to the conditions 62 and 68 .
Analysis for 62 : We show that the condition 62 always holds. This condition is simplified as \[\begin{align} (1- \frac{ \lambda_1 (\kappa^2 -1)}{\kappa^2\lambda_1^2-1 } )^2 & \ge (\frac{\kappa(\lambda_1^2 -1)}{\kappa^2\lambda_1^2-1 })^2 \end{align}\] Since \(\kappa^2\lambda_1^2-1 <0\), it is converted to \[\begin{align} |(\lambda_1 - 1)(\kappa^2\lambda_1 + 1)| =|\kappa^2\lambda_1^2-1 - \lambda_1 (\kappa^2 -1)| & \ge |\kappa(\lambda_1^2 -1)|. \end{align}\] When \(\kappa\ge 0\), it is converted to \[\begin{align} (1-\lambda_1)(\kappa^2\lambda_1 + 1) \ge -\kappa(\lambda_1^2 -1)= \kappa(1-\lambda_1)(1+\lambda_1). \end{align}\] That is, \[\begin{align} (1-\kappa )(1-\lambda_1 )(1-\kappa\lambda_1)\ge 0.\label{GH1} \end{align}\tag{71}\] When \(\kappa\le 0\), it is converted to \[\begin{align} (1-\lambda_1)(\kappa^2\lambda_1 + 1) \ge \kappa(\lambda_1^2 -1)=-\kappa(1-\lambda_1)(1+\lambda_1). \end{align}\] That is, \[\begin{align} (1+\kappa )(1-\lambda_1)(1+\kappa\lambda_1 )\ge 0. \label{GH2} \end{align}\tag{72}\] However, the conditions 71 and 72 always hold.
Analysis for 68 : The condition 68 is simplified as \[\begin{align} (\frac{(\kappa^2 (1-\lambda)-1)(2-\lambda)}{\kappa^2(1-\lambda)^2-1 })^2 & \ge (\frac{\kappa((1-\lambda)^2 -1)}{\kappa^2(1-\lambda)^2-1 })^2 +4(1-\lambda)^2 ,\label{CBF2} \end{align}\tag{73}\] which is equivalent to \[\begin{align} \left(\frac{(\kappa^2 (1-\lambda)-1)(2-\lambda)}{\kappa^2(1-\lambda)^2-1}\right)^2 - \left(\frac{\kappa((1-\lambda)^2 -1)}{\kappa^2(1-\lambda)^2-1}\right)^2 \ge 4(1-\lambda)^2.\label{CBF8} \end{align}\tag{74}\] Hence, we solve the inequality 74 with respect to \(\kappa\) where \(0 \le \lambda \le 1\). To solve the inequality with respect to \(\kappa\), we first simplify the expression. This inequality is defined for all \(\kappa\) such that the denominator is not zero, i.e., \(\kappa^2(1-\lambda)^2-1 \neq 0\).
We apply the formula \(A^2-B^2 = (A-B)(A+B)\) when \(A = \frac{(\kappa^2 (1-\lambda)-1)(2-\lambda)}{\kappa^2(1-\lambda)^2-1}\), \(B = \frac{\kappa((1-\lambda)^2 -1)}{\kappa^2(1-\lambda)^2-1} = \frac{\kappa(\lambda^2-2\lambda)}{\kappa^2(1-\lambda)^2-1} = \frac{-\kappa\lambda(2-\lambda)}{\kappa^2(1-\lambda)^2-1}\). Since \(A-B = \frac{(2-\lambda)}{\kappa^2(1-\lambda)^2-1} \left( \kappa^2(1-\lambda)-1 + \kappa\lambda \right)\), \(A+B = \frac{(2-\lambda)}{\kappa^2(1-\lambda)^2-1} \left( \kappa^2(1-\lambda)-1 - \kappa\lambda \right)\), the LHS of 74 equals \[\frac{(2-\lambda)^2}{(\kappa^2(1-\lambda)^2-1)^2} \left( \kappa^2(1-\lambda)-1 + \kappa\lambda \right) \left( \kappa^2(1-\lambda)-1 - \kappa\lambda \right).\] Let’s simplify the product of the two parentheses: \((\kappa^2(1-\lambda)-1)^2 - (\kappa\lambda)^2 = ((\kappa-1)(\kappa(1-\lambda)+1))((\kappa+1)(\kappa(1-\lambda)-1)) = (\kappa^2-1)(\kappa^2(1-\lambda)^2-1)\). This simplification is a key step. The product becomes: \[\frac{(2-\lambda)^2 (\kappa^2-1)(\kappa^2(1-\lambda)^2-1)}{(\kappa^2(1-\lambda)^2-1)^2} = \frac{(2-\lambda)^2(\kappa^2-1)}{\kappa^2(1-\lambda)^2-1}.\] Thus, the inequality 74 simplifies to: \[\frac{(2-\lambda)^2(\kappa^2-1)}{\kappa^2(1-\lambda)^2-1} \ge 4(1-\lambda)^2.\] We can rearrange this to: \[\frac{(2-\lambda)^2(\kappa^2-1) - 4(1-\lambda)^2(\kappa^2(1-\lambda)^2-1)}{\kappa^2(1-\lambda)^2-1} \ge 0\]
Let’s analyze the numerator: \[\begin{align} N =& (\kappa^2-1)(2-\lambda)^2 - 4(1-\lambda)^2(\kappa^2(1-\lambda)^2-1) \\ =& \kappa^2((2-\lambda)^2 - 4(1-\lambda)^4) - ((2-\lambda)^2 - 4(1-\lambda)^2) \\ =& \kappa^2\lambda(3-2\lambda)(2\lambda^2-5\lambda+4) - \lambda(4-3\lambda) \\ =& \lambda \left[ \kappa^2(3-2\lambda)(2\lambda^2-5\lambda+4) - (4-3\lambda) \right]. \end{align}\]
So the inequality is: \[\frac{\lambda \left[ \kappa^2(3-2\lambda)(2\lambda^2-5\lambda+4) - (4-3\lambda) \right]}{\kappa^2(1-\lambda)^2-1} \ge 0.\] We now analyze this inequality for different values of \(\lambda \in [0, 1]\).
Case 1: \(\lambda=0\). The inequality becomes \(\frac{0}{\kappa^2-1} \ge 0\), which is \(0 \ge 0\). This is true for all \(\kappa\) for which the expression is defined, which is \(\kappa^2-1 \neq 0\). Thus, for \(\lambda=0\), the condition 68 holds for any \(\kappa\in (-1,1)\).
Case 2: \(\lambda=1\). The numerator is \(1[\kappa^2(1)(1) - 1] = \kappa^2-1\). The denominator is \(\kappa^2(0)^2-1 = -1\). The inequality is \(\frac{\kappa^2-1}{-1} \ge 0\), which implies \(\kappa^2-1 \le 0\), or \(\kappa^2 \le 1\). the condition 68 holds for any \(\kappa\in (-1,1)\).
Case 3: \(0 < \lambda < 1\). Since \(\lambda>0\), we can divide by it. The inequality becomes: \[\frac{\kappa^2(3-2\lambda)(2\lambda^2-5\lambda+4) - (4-3\lambda)}{\kappa^2(1-\lambda)^2-1} \ge 0\] For \(0 < \lambda < 1\), the terms \((3-2\lambda)\), \((2\lambda^2-5\lambda+4)\), and \((4-3\lambda)\) are all positive. Since \(\kappa^2<1\), we have \(\kappa^2(1-\lambda)^2-1<0\). Thus, 74 is equivalent to \[\begin{align} |\kappa| \le \sqrt{\frac{4-3\lambda}{(3-2\lambda)(2\lambda^2-5\lambda+4)}}. \label{BNF} \end{align}\tag{75}\] ◻
Proof. Since \(S\) satisfies the conditions (D1) and (D2), the conditions in Lemmas 8 and 9 hold. We employ the form ?? of \({\cal R}[{\cal M}[S]]\).
Case (1). Lemmas 8 and 9 guarantee that \(\rho_1=\rho_2=I/2\). Applying a unitary in the first system we can convert \(\alpha_1\), \(\alpha_2\), and \(\alpha_3\) to \(\sigma_1\), \(\sigma_2\), and \(\sigma_3\) or \(\sigma_1\), \(-\sigma_2\), and \(\sigma_3\). Also, applying a unitary in the second system we can convert \(\beta_1\), \(\beta_2\), and \(\beta_3\) to \(\sigma_1\), \(\sigma_2\), and \(\sigma_3\) or \(\sigma_1\), \(-\sigma_2\), and \(\sigma_3\). Therefore, we have two cases (A) and (B) in Case (1). In case (A), \({\cal R}[{\cal M}[S]]\) is unitary equivalent to \(\frac{1}{2} (I\otimes I + \lambda_1 \sigma_1 \otimes \sigma_1 + \lambda_2 \sigma_2 \otimes \sigma_2 + \lambda_3 \sigma_3 \otimes \sigma_3\). Thus, Lemma 20 guarantees that the relation \(\tilde{C}_{2\to 1}[{\cal M}[S]]\ge 0\) is equivalent to \[\begin{align} (1-\lambda_1)^2 \ge (\lambda_2 +\lambda_3)^2,\quad (1+\lambda_1)^2 \ge (\lambda_2 -\lambda_3)^2. \end{align}\] Since \(\lambda_j>0\), the above condition is equivalent to \(1\ge \lambda_1+\lambda_2+\lambda_3\).
In case (B), \({\cal R}[{\cal M}[S]]\) is unitary equivalent to \(\frac{1}{2} (I\otimes I + \lambda_1 \sigma_1 \otimes \sigma_1 - \lambda_2 \sigma_2 \otimes \sigma_2 + \lambda_3 \sigma_3 \otimes \sigma_3\). Thus, Lemma 20 guarantees that the relation \(C_{1\to 2}\ge 0\) is equivalent to \[\begin{align} (1-\lambda_1)^2 \ge (-\lambda_2 +\lambda_3)^2,\quad (1+\lambda_1)^2 \ge (-\lambda_2 -\lambda_3)^2. \end{align}\] Since \(\lambda_j>0\), the above condition is equivalent to the condition \(1\ge \max(-\lambda_1 +\lambda_2+\lambda_3, \lambda_1 -\lambda_2+\lambda_3, \lambda_1 +\lambda_2-\lambda_3)\).
Case (2) Lemmas 8 and 9 guarantee that \[\begin{align} {\cal R}[{\cal M}[S]] = \frac{1}{4}(I+c_1^1 \sigma_1 ) \otimes (I+c_1^2 \sigma_1 ) + \sum_{j=1}^2 \frac{\lambda_j}{4} \alpha_j \otimes \beta_j . \end{align}\]
When \(c^2_1 \neq 0\), since \(c^2_1\mathrm{Tr}\beta_1\sigma_1 =c^2_1 \mathrm{Tr}\beta_2\sigma_1=0\), applying a unitary in the second system we can convert \(\beta_1\), \(\beta_2\) to \(\pm \sigma_2\), \(\sigma_3\) while \(\sigma_1\) is fixed. When \(c^2_1 = 0\), we can convert \(\beta_1\), \(\beta_2\) to \(\pm \sigma_2\), \(\sigma_3\) without fixing \(\sigma_1\). Also, when \(c^1_1 \neq 0\), since \(c^1_1\mathrm{Tr}\alpha_1\sigma_1 =c^1_1 \mathrm{Tr}\alpha_2\sigma_1=0\), applying a unitary in the first system we can convert \(\alpha_1\), \(\alpha_2\) to \(\pm \sigma_2\), \(\sigma_3\), where the choice of \(\pm\) is not necessarily the same as the above. When \(c^1_1 = 0\), we can convert \(\alpha_1\), \(\alpha_2\) to \(\pm \sigma_2\), \(\sigma_3\) without fixing \(\sigma_1\).
Then, \(4{\cal R}[{\cal M}[S]]\) is unitary equivalent to \(I\otimes I +c_1^1 \sigma_1 \otimes I +c_1^2 I \otimes \sigma_1 + c_1^1 c_1^2 \sigma_1 \otimes \sigma_1 \pm \lambda_1 \sigma_2 \otimes \sigma_2 + \lambda_2 \sigma_3 \otimes \sigma_3\). Lemma 20 guarantees that the relation \({\cal R}[{\cal M}[S]]\ge 0\) is equivalent to \[\begin{align} (1-c_1^1 c_1^2)^2 &\ge (c_1^1-c_1^2)^2+(\pm\lambda_1+\lambda_2)^2 \tag{76}\\ (1+c_1^1 c_1^2)^2 &\ge (c_1^1+c_1^2)^2+(\pm\lambda_1-\lambda_2)^2 .\tag{77} \end{align}\] These two conditions are equivalent to \[\begin{align} (1-(c_1^1)^2)(1-(c_1^2)^2) &\ge (\pm\lambda_1+\lambda_2)^2 \tag{78}\\ (1-(c_1^1)^2)(1-(c_1^2)^2) &\ge (\pm\lambda_1-\lambda_2)^2 .\tag{79} \end{align}\] The pair of these two conditions are equivalent to ?? .
Case (3). Lemma 9 guarantees that \[\begin{align} 4{\cal R}[{\cal M}[S]] = (I+c_1^1 \sigma_1+c_2^1 \sigma_2 ) \otimes (I+c_1^2 \sigma_1+c_2^2 \sigma_2 ) + \lambda_1 \alpha_1 \otimes \beta_1 \end{align}\] Since \(\mathrm{Tr}(c_1^1 \sigma_1+c_2^1 \sigma_2)\alpha_1= \mathrm{Tr}(c_1^2 \sigma_1+c_2^2 \sigma_2)\alpha_2=0\), \(4{\cal R}[{\cal M}[S]]\) is unitary equivalent to \[\begin{align} (I+\sqrt{(c_1^1)^2+(c_2^1)^2} \sigma_1 ) \otimes (I+\sqrt{(c_1^2)^2+(c_2^2)^2} \sigma_2 ) + \lambda_1 \sigma_3 \otimes \sigma_3. \end{align}\] Lemma 20 guarantees that the relation \(4{\cal R}[{\cal M}[S]]\ge 0\) is equivalent to the pair of \[\begin{align} \big(1-\sqrt{(c_1^1)^2+(c_2^1)^2} \sqrt{(c_1^2)^2+(c_2^2)^2}\big)^2 &\ge \big(\sqrt{(c_1^1)^2+(c_2^1)^2}-\sqrt{(c_1^2)^2+(c_2^2)^2}\big)^2+ (\lambda_1)^2 \tag{80}\\ \big(1+\sqrt{(c_1^1)^2+(c_2^1)^2} \sqrt{(c_1^2)^2+(c_2^2)^2}\big)^2 &\ge \big(\sqrt{(c_1^1)^2+(c_2^1)^2}+\sqrt{(c_1^2)^2+(c_2^2)^2}\big)^2 +(\lambda_1)^2 .\tag{81} \end{align}\] These two conditions are equivalent to \[\begin{align} \big(1-((c_1^1)^2+(c_2^1)^2)\big) \big(1- ((c_1^2)^2+(c_2^2)^2)\big) \ge (\lambda_1)^2 \label{BP1ED}. \end{align}\tag{82}\]
Case (4). We have \({\cal R}[{\cal M}[S]]= \rho_1 \otimes \rho_2 \ge 0\). ◻
Proof. Conditioned on Alice’s measurement choice \(Y_1=y_1\) and outcome \(X_1=x_1\), the memory qubit is projected onto the complex-conjugate state \(|x_1,y_1\rangle^*\), where \(^*\) expresses the complex conjugate. Hence Bob receives the state \(|x_1,y_1\rangle^*\langle x_1,y_1|^* =(|x_1,y_1\rangle\langle x_1,y_1|)^T\) for the same conditioning. In the qubit-Pauli setting, this coincides with the action of the transpose map \(\rho\mapsto \rho^{\mathsf T}\) on the post-measurement state \(|x_1,y_1\rangle\langle x_1,y_1|\). Therefore, under the reconstruction formula of Theorem 4, the matrix \(\hat{C}_{1\to2}[P]\) is exactly the Choi matrix of the transpose map. With our Choi convention, this matrix is the swap operator \(\sum_{m,n=0}^1 |m\rangle\langle n| \otimes |n\rangle\langle m|\), whose eigenvalues are \(1,1,1,-1\). Hence \(\hat{C}_{1\to2}[P]\) is not positive. ◻
Proof. We verify condition (C1) by using the two-qubit Pauli form given in Proposition 8. Conditioned on Alice’s measurement choice \(Y_1=y_1\) and outcome \(X_1=x_1\), the memory qubit is projected onto the complex-conjugate state \(|x_1,y_1\rangle^{*}\). Hence, when Bob measures the Pauli observable \(\sigma_{y_2}\), the conditional distribution is \[\begin{align} P_{X_2\mid X_1,Y_1,Y_2}(x_2\mid x_1,y_1,y_2) = \langle x_2,y_2| \left(|x_1,y_1\rangle^{*}\!\langle x_1,y_1|^{*}\right) |x_2,y_2\rangle. \label{ExB-cond} \end{align}\tag{83}\] Since complex conjugation in the computational basis acts trivially on the \(\sigma_1\)- and \(\sigma_3\)-eigenbases and exchanges the two \(\sigma_2\)-eigenstates, the right-hand side of 83 is symmetric under the replacement \(x_1\mapsto 1-x_1\) up to relabeling of the \(\sigma_2\)-outcomes. As a consequence, for every \(y_1',y_2\in\{1,2,3\}\) and \(x_2\in\{0,1\}\), we have \[\begin{align} \sum_{x_1'\in\{0,1\}}(-1)^{x_1'} P_{X_2\mid X_1,Y_1,Y_2}(x_2\mid x_1',y_1',y_2)=0. \label{ExB-alt} \end{align}\tag{84}\] Moreover, Alice’s local statistics are unbiased for every Pauli choice, so \[\begin{align} P_{X_1\mid Y_1}(x_1\mid y_1)=\frac{1}{2}, \qquad \eta_{y_1} := \sum_{x_1\in\{0,1\}}(-1)^{x_1}P_{X_1\mid Y_1}(x_1\mid y_1) =0 \label{ExB-eta} \end{align}\tag{85}\] for all \(y_1\in\{1,2,3\}\). Hence the correction term in Proposition 8, \[\begin{align} \sum_{y_1'\in\{1,2,3\}} \eta_{y_1'} \sum_{x_1'\in\{0,1\}}(-1)^{x_1'} P_{X_2\mid X_1,Y_1,Y_2}(x_2\mid x_1',y_1',y_2), \end{align}\] vanishes identically. Therefore the right-hand side of 44 reduces to \(2P_{X_2\mid Y_1,Y_2}(x_2\mid 0,y_2)\). On the other hand, summing 83 over \(x_1\in\{0,1\}\) gives \[\begin{align} \sum_{x_1\in\{0,1\}} P_{X_2\mid X_1,Y_1,Y_2}(x_2\mid x_1,y_1,y_2) = 2P_{X_2\mid Y_1,Y_2}(x_2\mid 0,y_2), \end{align}\] because the two conditional input states to Bob are the two eigenstates of the same Pauli observable (or their complex conjugates), whose average is the maximally mixed state. Thus the left-hand side and the right-hand side of 44 coincide for all \(y_1,y_2\in\{1,2,3\}\) and \(x_2\in\{0,1\}\). Hence condition (C1) holds. ◻
Proof. We first rewrite the reconstruction formula 19 in the present two-qubit Pauli setting. Here we have \[G_0=I,\qquad G_j=\sigma_j,\qquad H_0=\frac{I}{2},\qquad H_j=\frac{\sigma_j}{2}\qquad (j=1,2,3),\] and the coefficients are \(g_{x,j}=(-1)^x\) for \(x\in\{0,1\}\). Since in Example 3 the classical memory \(X\) is chosen uniformly, Alice’s local distribution is unbiased for every Pauli choice, namely \[P_{X_1\mid Y_1}(0\mid y_1)=P_{X_1\mid Y_1}(1\mid y_1)=\frac{1}{2} \qquad (y_1=1,2,3).\] Hence the correction term in 19 vanishes, and the reconstructed matrix reduces to \[\begin{align} \hat{C}_{1\to2}[P] = \frac{1}{2}\,I\otimes I +\frac{1}{2}\sum_{j=1}^3 b_j\,I\otimes \sigma_j +\frac{1}{4}\sum_{i,j=1}^3 c_{ij}\,\sigma_i^{T}\otimes \sigma_j, \label{Ex1prime-recon} \end{align}\tag{86}\] where \[\begin{align} b_j &:= \sum_{x_2\in\{0,1\}}(-1)^{x_2} P_{X_2\mid Y_1,Y_2}(x_2\mid 0,j), \tag{87} \\ c_{ij} &:= \sum_{x_1,x_2\in\{0,1\}}(-1)^{x_1+x_2} P_{X_2\mid X_1,Y_1,Y_2}(x_2\mid x_1,i,j). \tag{88} \end{align}\]
We now compute these coefficients. When Alice does nothing, Bob receives the state \[D_{1/2}(|0,3\rangle\langle 0,3|) = \frac{1}{2}|0,3\rangle\langle 0,3|+\frac{1}{4} I,\] so \[b_1=b_2=0,\qquad b_3=\frac{1}{2}.\] Next, conditioned on Alice’s measurement, the relevant conditional states at Bob are as follows.
If \(Y_1=1\), then the two values of the memory variable \(X\) lead to the same post-unitary state, so conditioned on \(X_1=x_1\) Bob receives \(D_{1/2}(|x_1,1\rangle\langle x_1,1|)\). Hence \[P_{X_2\mid X_1,Y_1,Y_2}(x_2\mid x_1,1,1) = \frac{3}{4}\,\delta_{x_2,x_1}+\frac{1}{4}\,(1-\delta_{x_2,x_1}),\] while for \(Y_2=2,3\) the distribution is unbiased. Therefore \[c_{11}=1,\qquad c_{12}=c_{13}=0.\]
If \(Y_1=2\), then averaging over \(X\in\{0,1\}\) yields the maximally mixed state at Bob, so \[c_{21}=c_{22}=c_{23}=0.\]
If \(Y_1=3\), then conditioned on Alice’s outcome one has \(X=X_1\), and the subsequent unitary \(\sigma_X\) always maps Alice’s output to \(|0,3\rangle\). Thus Bob receives the fixed state \(D_{1/2}(|0,3\rangle\langle 0,3|)\), independently of \(X_1\), and hence \[c_{31}=c_{32}=c_{33}=0.\]
Substituting these coefficients into 86 , and noting that \(\sigma_1^{T}=\sigma_1\), we obtain \[\begin{align} \hat{C}_{1\to2}[P] = \frac{1}{2}\Bigl( I\otimes I +\frac{1}{2}\,I\otimes \sigma_3 +\frac{1}{2}\,\sigma_1\otimes \sigma_1 \Bigr). \label{Ex1prime-C} \end{align}\tag{89}\]
In the computational basis \(\{|00\rangle,|01\rangle,|10\rangle,|11\rangle\}\), the matrix 89 is written as \[\begin{align} \hat{C}_{1\to2}[P] = \begin{pmatrix} \frac{3}{4} & 0 & 0 & \frac{1}{4}\\ 0 & \frac{1}{4} & \frac{1}{4} & 0\\ 0 & \frac{1}{4} & \frac{3}{4} & 0\\ \frac{1}{4} & 0 & 0 & \frac{1}{4} \end{pmatrix}. \label{Ex1prime-C-matrix} \end{align}\tag{90}\] Its eigenvalues are \[\begin{align} \frac{1}{2}-\frac{\sqrt2}{4},\quad \frac{1}{2}-\frac{\sqrt2}{4},\quad \frac{1}{2}+\frac{\sqrt2}{4},\quad \frac{1}{2}+\frac{\sqrt2}{4}, \end{align}\] all of which are strictly positive. Hence, \(\hat{C}_{1\to2}[P]\ge 0\). ◻
Proof. By Proposition 8, in the two-qubit Pauli setting, condition (C1) is equivalent to \[\begin{align} \sum_{x_1\in\{0,1\}} P_{X_2\mid X_1,Y_1,Y_2}(x_2\mid x_1,y_1,y_2) &= 2P_{X_2\mid Y_1,Y_2}(x_2\mid0,y_2) \notag\\ &\quad -\sum_{y_1'\in\{1,2,3\}}\eta_{y_1'} \sum_{x_1'\in\{0,1\}}(-1)^{x_1'} P_{X_2\mid X_1,Y_1,Y_2}(x_2\mid x_1',y_1',y_2), \label{eq:C1-2q-Ex1prime} \end{align}\tag{91}\] for all \(y_1,y_2\in\{1,2,3\}\) and \(x_2\in\{0,1\}\), where \(\eta_{y_1'} := \sum_{x_1\in\{0,1\}}(-1)^{x_1}P_{X_1\mid Y_1}(x_1\mid y_1')\).
We show that 91 fails for \((y_1,y_2,x_2)=(1,3,0)\). Since the classical memory \(X\) is chosen uniformly, Alice’s local statistics are unbiased for every Pauli choice, and hence \[\begin{align} \eta_{1}=\eta_{2}=\eta_{3}=0. \label{Ex1prime-eta} \end{align}\tag{92}\] Therefore the correction term in 91 vanishes identically.
Next, when Alice measures \(Y_1=1\) and Bob measures \(Y_2=3\), Bob receives a depolarized \(\sigma_1\)-eigenstate. Hence Bob’s \(\sigma_3\)-measurement is unbiased, so \[P_{X_2\mid X_1,Y_1,Y_2}(0\mid x_1,1,3)=\frac{1}{2} \qquad (x_1=0,1).\] Therefore the left-hand side of 91 equals \[\begin{align} \sum_{x_1\in\{0,1\}} P_{X_2\mid X_1,Y_1,Y_2}(0\mid x_1,1,3) = 1. \label{Ex1prime-LHS} \end{align}\tag{93}\]
On the other hand, when Alice does nothing (\(Y_1=0\)), Bob receives the depolarized state \[D_{1/2}(|0,3\rangle\langle 0,3|) = \frac{1}{2}|0,3\rangle\langle 0,3|+\frac{1}{4} I.\] Hence \[\begin{align} P_{X_2\mid Y_1,Y_2}(0\mid 0,3)=\frac{3}{4}. \label{Ex1prime-Bobmarginal} \end{align}\tag{94}\] Since the correction term vanishes by 92 , the right-hand side of 91 becomes \[\begin{align} 2P_{X_2\mid Y_1,Y_2}(0\mid 0,3)=\frac{3}{2}. \label{Ex1prime-RHS} \end{align}\tag{95}\] Comparing 93 and 95 , we obtain \(1\neq \frac{3}{2}\). Thus 91 is violated, and therefore Example 3 does not satisfy (C1). ◻
Proof. We employ the notation in Subsection 11.11. That is, we use 91 as a condition equivalent to the condition (C1). We show that the relation 91 fails for \((y_1,y_2,x_2)=(1,3,0)\). The important point is that, for this choice of \((y_1,y_2,x_2)\), the relevant probabilities are independent of the mixing parameter \(\lambda\). Indeed, when \(Y_1=1\), Alice’s post-measurement state is a \(\sigma_1\)-eigenstate, and after Charlie applies \(\sigma_X\), Bob still receives a \(\sigma_1\)-eigenstate. Hence Bob’s \(\sigma_3\)-measurement is unbiased, regardless of the value of \(X\) and therefore regardless of \(\lambda\).
Thus, \[P_{X_2\mid X_1,Y_1,Y_2}(0\mid x_1,1,3)=\frac{1}{2} \qquad (x_1=0,1),\] and the left-hand side is \[\sum_{x_1\in\{0,1\}} P_{X_2\mid X_1,Y_1,Y_2}(0\mid x_1,1,3)=1.\]
Next, when Alice does nothing (\(Y_1=0\)), Charlie sends \(\sigma_X|(-1)^X,3\rangle\) to Bob. But \[\sigma_X|(-1)^X,3\rangle = |0,3\rangle \qquad (X=0,1),\] so Bob always receives the state \(|0,3\rangle\), again independently of \(\lambda\). Hence \[P_{X_2\mid Y_1,Y_2}(0\mid 0,3)=1,\] and therefore the first term on the right-hand side equals \[2P_{X_2\mid Y_1,Y_2}(0\mid 0,3)=2.\]
Finally, for every \(y_1'\in\{1,2,3\}\), the correction term vanishes: \[\sum_{x_1'\in\{0,1\}}(-1)^{x_1'} P_{X_2\mid X_1,Y_1,Y_2}(0\mid x_1',y_1',3)=0.\] Indeed, for \(y_1'=1,2\), Bob’s \(\sigma_3\)-measurement is unbiased, so the two probabilities are both \(1/2\). For \(y_1'=3\), Bob always obtains \(x_2=0\), independently of \(x_1'\), so the two terms are both \(1\) and their alternating sum is \(1-1=0\). Therefore the right-hand side is \(2\), whereas the left-hand side is \(1\). This contradiction shows that condition (C1) fails for the strategy of Example 4. ◻
Proof. Recall that, in Example 4, Charlie uses a hidden classical variable \(X\in\{0,1\}\), chooses \(X=0\) with probability \(\lambda\in(0,1)\) and \(X=1\) with probability \(1-\lambda\), inputs the state \(|(-1)^X,3\rangle\) to Alice, and then applies the unitary \(\sigma_X\) to Alice’s output before sending the system to Bob.
We first compute the local statistics on Alice’s side. For \(Y_1=1,2\), Alice’s outcome is unbiased: \[P_{X_1\mid Y_1}(0\mid y_1)=P_{X_1\mid Y_1}(1\mid y_1)=\frac{1}{2} \qquad (y_1=1,2).\] For \(Y_1=3\), we have \[P_{X_1\mid Y_1}(0\mid 3)=\lambda, \qquad P_{X_1\mid Y_1}(1\mid 3)=1-\lambda.\] Hence the reconstructed initial state is \[\rho_1[P] = \frac{1}{2}\bigl(I+(2\lambda-1)\sigma_3\bigr).\]
Next, when Alice does nothing (\(Y_1=0\)), Bob always receives the state \(|0,3\rangle\). Therefore, \[\rho_2[P] = |0,3\rangle\langle 0,3| = \frac{1}{2}(I+\sigma_3).\]
We now determine the conditional distribution \(P_{X_2\mid X_1,Y_1,Y_2}\). From the definition of Example 4, Bob’s conditional statistics are as follows:
If \(Y_1=1\), then \[Y_2=1:\;x_2=x_1 \;\text{with probability }1, \qquad Y_2=2,3:\;\text{uniform}.\]
If \(Y_1=2\), then \[Y_2=2:\;x_2=x_1 \;\text{with probability }\lambda, \quad x_2=1-x_1 \;\text{with probability }1-\lambda,\] and for \(Y_2=1,3\) the distribution is uniform.
If \(Y_1=3\), then \[Y_2=3:\;x_2=0 \;\text{with probability }1, \qquad Y_2=1,2:\;\text{uniform}.\]
Substituting these statistics into the qubit-Pauli reconstruction formula for \(\hat{C}_{1\to2}[P]\), with \[H_0=\frac{I}{2}, \qquad H_j=\frac{\sigma_j}{2} \quad (j=1,2,3),\] we obtain \[\hat{C}_{1\to2}[P] = I\otimes \frac{I+\sigma_3}{2} +\frac{1}{2}\,\sigma_1\otimes\sigma_1 -\frac{2\lambda-1}{2}\,\sigma_2\otimes\sigma_2.\]
In the computational basis \(\{|00\rangle,|01\rangle,|10\rangle,|11\rangle\}\), this matrix is \[\hat{C}_{1\to2}[P] = \begin{pmatrix} 1 & 0 & 0 & \lambda\\ 0 & 0 & 1-\lambda & 0\\ 0 & 1-\lambda & 1 & 0\\ \lambda & 0 & 0 & 0 \end{pmatrix}.\] This matrix decomposes into two \(2\times2\) blocks, so its eigenvalues are \[\frac{1\pm \sqrt{1+4\lambda^2}}{2}, \qquad \frac{1\pm \sqrt{1+4(1-\lambda)^2}}{2}.\] Since \(\lambda\in(0,1)\), we have \[\sqrt{1+4\lambda^2}>1, \qquad \sqrt{1+4(1-\lambda)^2}>1,\] and therefore \[\frac{1-\sqrt{1+4\lambda^2}}{2}<0, \qquad \frac{1-\sqrt{1+4(1-\lambda)^2}}{2}<0.\] Thus \(\hat{C}_{1\to2}[P]\) has negative eigenvalues, and hence \[\hat{C}_{1\to2}[P]\not\ge 0.\]
We conclude that the strategy of Example 4 satisfies neither condition (C1) nor the positivity condition \(\hat{C}_{1\to2}[P]\ge 0\). ◻
In this appendix, we justify the counting argument used in Section 7.1. The point is that, even before imposing any special structure on Charlie’s strategy beyond physical validity, the generalized Born rule depends on the process matrix only through its class in a certain quotient space. Thus, exact process-matrix reconstruction requires the accessible tensors \(C[\Gamma_{1,z_1}]\otimes C[\Gamma_{2,z_2}]\) to span that quotient space. We explain this reduction and evaluate the corresponding dimension in the two-qubit case.
Let \({\cal B}\) denote the real vector space of Hermitian matrices on \[{\cal H}_{I,1}\otimes{\cal H}_{O,1}\otimes {\cal H}_{I,2}\otimes{\cal H}_{O,2}.\] For Alice’s and Bob’s local operations \(\Gamma_{1,z_1}\) and \(\Gamma_{2,z_2}\), the observed joint distribution is given by the generalized Born rule \[\label{APP-GBR} P_{Z_1,Z_2}(z_1,z_2) = \mathrm{Tr}\!\bigl( W_S\, C[\Gamma_{1,z_1}]\otimes C[\Gamma_{2,z_2}] \bigr),\tag{96}\] where \(W_S\) is Charlie’s process matrix and \(C[\Gamma]\) is the (unnormalized) Choi matrix of \(\Gamma\).
When a strategy \(S\) belongs to the most general physical class \({\cal S}_{G}\), its process matrix \(W_S\) satisfies the standard linear no-signalling-in-time constraints [6] \[\begin{align} (\mathrm{Tr}_{(I,1),(O,1),(O,2)}W_S)\otimes \rho_{mix,(O,2)} &= \mathrm{Tr}_{(I,1),(O,1)}W_S, \tag{97}\\ (\mathrm{Tr}_{(O,1),(I,2),(O,2)}W_S)\otimes \rho_{mix,(O,1)} &= \mathrm{Tr}_{(I,2),(O,2)}W_S, \tag{98}\\ W_S = (\mathrm{Tr}_{(O,1)}W_S)\otimes \rho_{mix,(O,1)} +(\mathrm{Tr}_{(O,2)}W_S)\otimes \rho_{mix,(O,2)} &\notag\\ \qquad -(\mathrm{Tr}_{(O,1),(O,2)}W_S)\otimes \rho_{mix,(O,1),(O,2)}. \tag{99} \end{align}\] Here, \(\rho_{mix}\) denotes the maximally mixed state on the indicated space.
These conditions can be rewritten in the dual form \[\begin{align} \mathrm{Tr}W_S \Bigl[ (\mathrm{Tr}_{(I,1),(O,1),(O,2)}A)\otimes \rho_{mix,(I,1),(O,1),(O,2)} - \mathrm{Tr}_{(I,1),(O,1)}A \otimes \rho_{mix,(I,1),(O,1)} \Bigr] &=0, \tag{100}\\ \mathrm{Tr}W_S \Bigl[ (\mathrm{Tr}_{(O,1),(I,2),(O,2)}A)\otimes \rho_{mix,(O,1),(I,2),(O,2)} - \mathrm{Tr}_{(I,2),(O,2)}A \otimes \rho_{mix,(I,2),(O,2)} \Bigr] &=0, \tag{101}\\ \mathrm{Tr}W_S \Bigl[ A +(\mathrm{Tr}_{(O,1),(O,2)}A)\otimes \rho_{mix,(O,1)}\otimes \rho_{mix,(O,2)} -(\mathrm{Tr}_{(O,1)}A)\otimes \rho_{mix,(O,1)} -(\mathrm{Tr}_{(O,2)}A)\otimes \rho_{mix,(O,2)} \Bigr] &=0 \tag{102} \end{align}\] for every Hermitian matrix \(A\in{\cal B}\).
Motivated by these identities, define the following subspaces of \({\cal B}\): \[\begin{align} {\cal V}_1 := \Bigl\{ \sum_j \rho_{mix,(I,1),(O,1)}\otimes A_{j,(I,2)}\otimes B_{j,(O,2)} \;\Big|\; \mathrm{Tr}B_{j,(O,2)}=0 \Bigr\}, \\ {\cal V}_2 := \Bigl\{ \sum_j A_{j,(I,1)}\otimes B_{j,(O,1)}\otimes \rho_{mix,(I,2),(O,2)} \;\Big|\; \mathrm{Tr}B_{j,(O,1)}=0 \Bigr\}, \\ {\cal V}_3 := \Bigl\{ \sum_j A_{j,(I,1)}\otimes B_{j,(O,1)}\otimes A_{j,(I,2)}\otimes B_{j,(O,2)} \;\Big|\; \mathrm{Tr}B_{j,(O,1)}=\mathrm{Tr}B_{j,(O,2)}=0 \Bigr\}. \end{align}\] By construction, 100 –102 imply that \[\mathrm{Tr}W_S A=0 \qquad \forall A\in {\cal V}_1\oplus{\cal V}_2\oplus{\cal V}_3.\] Hence, in the generalized Born rule 96 , only the equivalence class of \(C[\Gamma_{1,z_1}]\otimes C[\Gamma_{2,z_2}]\) modulo \({\cal V}_1\oplus{\cal V}_2\oplus{\cal V}_3\) can influence the observed statistics. Therefore, exact reconstruction of the process matrix requires the accessible family \[\bigl\{ C[\Gamma_{1,z_1}]\otimes C[\Gamma_{2,z_2}] \bigr\}_{z_1,z_2}\] to span the quotient space \[{\cal B}/({\cal V}_1\oplus{\cal V}_2\oplus{\cal V}_3).\]
A direct dimension count gives \[\begin{align} \dim {\cal V}_1 &= d_{I,2}^2(d_{O,2}^2-1),\\ \dim {\cal V}_2 &= d_{I,1}^2(d_{O,1}^2-1),\\ \dim {\cal V}_3 &= d_{I,1}^2 d_{I,2}^2 (d_{O,1}^2-1)(d_{O,2}^2-1), \end{align}\] while \[\dim {\cal B} = d_{I,1}^2 d_{O,1}^2 d_{I,2}^2 d_{O,2}^2.\] Consequently, \[\begin{align} \dim {\cal B}/({\cal V}_1\oplus{\cal V}_2\oplus{\cal V}_3) &= d_{I,1}^2 d_{I,2}^2 (d_{O,1}^2+d_{O,2}^2-1) -d_{I,2}^2(d_{O,2}^2-1) -d_{I,1}^2(d_{O,1}^2-1). \label{APP-quotient-dim} \end{align}\tag{103}\]
We now specialize to the two-qubit case \[d_{I,1}=d_{O,1}=d_{I,2}=d_{O,2}=2.\] Then 103 becomes \[\begin{align} \dim {\cal B}/({\cal V}_1\oplus{\cal V}_2\oplus{\cal V}_3) = 4\cdot 4\cdot(4+4-1)-4\cdot(4-1)-4\cdot(4-1) =88. \label{APP-quotient-dim-qubit} \end{align}\tag{104}\]
This is the minimum number of linearly independent quotient-space directions that any tomographically complete reconstruction scheme must access in the two-qubit setting.
In Section 7.1, however, we consider the restricted qubit-Pauli measurement scheme. There, for each player \(i=1,2\), the pair \((X_i,Y_i)\) takes only seven possible values. Hence the family \(\bigl\{ C[\Gamma_{1,z_1}]\otimes C[\Gamma_{2,z_2}] \bigr\}_{z_1,z_2}\) contains at most \(7\times 7 = 49\) elements. Since \(49<88\), these tensors cannot span the quotient space \({\cal B}/({\cal V}_1\oplus{\cal V}_2\oplus{\cal V}_3)\). Therefore, exact reconstruction of the underlying process matrix is impossible in the qubit-Pauli setting considered in Section 7.1.