The space of logically consistent classical processes without causal order

We describe an operational framework without global assumptions (other than logical consistency). Causal relations are defined as in the interventionists' approach to causality [48, 49]: Outputs can be correlated to inputs and inputs are manipulated freely (see figure 1). Defining causality based on free randomness is the converse approach to the one used in recent literature [24, 25]; there, free randomness is defined based on causal relations.

Definition 1. (Causality [35]). For two correlated random variables X and A, where X is an output and A is an input, i.e., A is chosen freely, we say that X is in the causal future of A, or equivalently, that A is in the causal past of X, denoted by $X\succeq A$ or $A\preceq X$. The negations of these relations are denoted by $\not\hspace{-2pt}{\succeq }$ and $\not\hspace{-2pt}{\preceq }$.

Consider N parties ${\{{S}_{j}\}}_{0\leqslant j\lt N}$, where party S_j has access to an input random variable A_j and generates an output random variable X_j. This allows us to causally order parties: if A_j is correlated to X_k, then S_j is in the causal past of S_k (${S}_{j}\preceq {S}_{k}$). To simplify the presentation, we write $\vec{X}=({X}_{0},...,{X}_{N-1})$ and likewise for $\vec{A}$, $\vec{O}$, and $\vec{I}$.

Definition 2. (Two-party predefined causal order). A two-party predefined causal order is a causal ordering of party S with input A, output X, and party T with input B, output Y, such that the distribution ${P}_{X,Y| A,B}$ can be written as a convex combination of one-way signaling distributions

$$\begin{eqnarray*}&&{P}_{X,Y| A,B}={{pP}}_{X| A,B,Y}{P}_{Y| B}+(1-p){P}_{X| A}{P}_{Y| A,B,X},\end{eqnarray*}$$

for some $0\leqslant p\leqslant 1$.

A definition for multi-party predefined causal order is given in [42]. Such a definition turns out to be more subtle since a party S_j in the causal past of some other parties ${\{{S}_{{\ell }}\}}_{L}$ can in principle influence everything in her causal future; in particular, S_j can influence the causal order of the parties ${\{{S}_{{\ell }}\}}_{L}$. We just state a lemma that follows from such a definition and that is sufficient to prove our claims.

Lemma 1. (Necesarry condition for predefined causal order). A necessary condition for a predefined causal order is that the probability distribution ${P}_{\vec{X}| \vec{I}}$ can be written as a convex combination

$$\begin{eqnarray*}&&{P}_{\vec{X}| \vec{I}}=\displaystyle \sum _{k}{p}_{k}{P}_{k},\end{eqnarray*}$$

with ${\sum }_{k}{p}_{k}=1$ and $\forall k\;:{p}_{k}\geqslant 0$, such that in every distribution P_k at least one party is not in the causal future of any other party, i.e.

$$\begin{eqnarray*}&&\forall k\exists {\rm{i}}\forall j\;\ne \;{\rm{i}}\;:\;{S}_{i}\;{\not\hspace{-2pt}{\succeq }}^{{P}_{k}}{S}_{j},\end{eqnarray*}$$

where ${\not\hspace{-2pt}{\succeq }}^{{P}_{k}}$ stands for the causal relation that is deduced from the distribution P_k.

In the framework without predefined causal order, each party S_j receives a random variable I_j from the environment E on which S_j can act. After the interaction with I_j, party S_j outputs a random variable O_j to the environment. Both random variables I_j and O_j are output random variables. The only input random variable a party has is A_j. The operation of S_j is a stochastic process mapping ${A}_{j},{I}_{j}$ to ${X}_{j},{O}_{j}$ (see figure 3). A stochastic process is a probability distribution over the range conditioned on the domain; in this case, the stochastic process of party S_j (which in the following will also be called the local operation of party S_j) is ${P}_{{X}_{j},{O}_{j}| {A}_{j},{I}_{j}}$. All parties are allowed to apply any possible operation described by probability theory. Furthermore, they are isolated from each other, which means that they can interact only through the environment. Because we do not make global assumptions (beyond logical consistency), the most general picture is that the random variables that are sent from the environment E to the parties are the result of a map on the random variables fed back by all parties to the same environment E (see figure 4). Such a composition of parties with the environment combines states and communication channels in one framework.

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A party S_j has access to the four random variables X_j, O_j, I_j, and A_j, where A_j is chosen freely. If we consider all parties together, we should get a probability distribution ${P}_{\vec{X},\vec{I},\vec{O}| \vec{A}}$. Furthermore, we ask the environment E to be a multi-linear functional of all local operations. The motivation for this is that linear combinations of local operations should carry through to the probabilities ${P}_{\vec{X},\vec{O},\vec{I}| \vec{A}}$. This brings us to a definition of logical consistency.

Definition 3. (Logical consistency). An environment E is called logically consistent if and only if it is a multi-linear positive map on any choice of local operations ${\{{P}_{{X}_{j},{O}_{j}| {A}_{j},{I}_{j}}\}}_{0\leqslant j\lt N}$ of all parties such that the composition of E with the local operations results in a probability distribution ${P}_{\vec{X},\vec{I},\vec{O}| \vec{A}}$.

The linearity and positivity conditions from definition 3 imply theorem 1, which states that the environment must be a stochastic process (conditional probability distribution).

Theorem 1. (Logical consistent environment as stochastic process). The environment E is a stochastic process ${P}_{\vec{I}| \vec{O}}$ that maps $\vec{O}$ to $\vec{I}$.

Proof. The environment is a multi-linear positive map ${ \mathcal E }$ on the probabilities (we omit the arguments for the sake of presentation)

$$\begin{eqnarray*}&&{p}_{j}:= {P}_{{X}_{j},{O}_{j}| {A}_{j},{I}_{j}}({x}_{j},{o}_{j},{a}_{j},{i}_{j})\end{eqnarray*}$$

that party S_j outputs o_j to the environment and generates x_j conditioned on the setting a_j and on ${I}_{j}={i}_{j}$. Therefore, we write

$$\begin{eqnarray*}&&{P}_{\vec{X},\vec{I},\vec{O}| \vec{A}}(\vec{x},\vec{i},\vec{o},\vec{a})={ \mathcal E }({p}_{0},...,{p}_{N-1}).\end{eqnarray*}$$

Since ${ \mathcal E }$ is a multi-linear positive map and since it depends on $\vec{O}$ and $\vec{I}$ only, the above probability can be written as

$$\begin{eqnarray}&&{P}_{\vec{X},\vec{I},\vec{O}| \vec{A}}(\vec{x},\vec{i},\vec{o},\vec{a})=E(\vec{o},\vec{i}){p}_{0}\cdots {p}_{N-1},\end{eqnarray} \tag{ 1 }$$

where $E(\vec{o},\vec{i})$ is a number. This number must be non-negative, as otherwise the above expression (1) is not a probability. By fixing $\vec{A}=\vec{a}$ and by summing over $\vec{x}$, we get

$$\begin{eqnarray*}{P}_{\vec{I},\vec{O}| \vec{A}=\vec{a}}(\vec{i},\vec{o}) & = & \displaystyle \sum _{\vec{x}}E(\vec{o},\vec{i}){p}_{0}\cdots {p}_{N-1}\\ & = & E(\vec{o},\vec{i})\displaystyle \sum _{\vec{x}}{p}_{0}\cdots {p}_{N-1}\\ & = & E(\vec{o},\vec{i}){{p}^{\prime }}_{0}\cdots {{p}^{\prime }}_{N-1},\end{eqnarray*}$$

where

$$\begin{eqnarray*}&&p{{}^{\prime }}_{j}:= {P}_{{O}_{j}| {I}_{j},{A}_{j}={a}_{j}}({o}_{j},{i}_{j}).\end{eqnarray*}$$

Let us fix the local operations ${p}_{j}^{\prime }$ of all parties to be

$$\begin{eqnarray*}{P}_{{O}_{j}| {I}_{j},{A}_{j}={a}_{j}}({o}_{j},{i}_{j})=\left\{\begin{array}{ll}1 & {o}_{j}=0,\\ 0 & \mathrm{otherwise}.\end{array}\right.\end{eqnarray*}$$

From the total-probability condition we obtain

$$\begin{eqnarray*}&&\displaystyle \sum _{\vec{o},\vec{i}}E(\vec{o},\vec{i}){p}_{0}^{\prime }\cdots p{{}^{\prime }}_{N-1}=\displaystyle \sum _{\vec{i}}E(\vec{0},\vec{i})=1.\end{eqnarray*}$$

By repeating this calculation for different choices of local operations where the parties deterministically output a value, we get

$$\begin{eqnarray*}&&\forall \vec{o}\;:\;\displaystyle \sum _{\vec{i}}E(\vec{o},\vec{i})=1.\end{eqnarray*}$$

Therefore, E is a stochastic process ${P}_{\vec{I}| \vec{O}}$. □

The following corollary follows from theorem 1.

Corollary 1. A logical consistent environment ${P}_{\vec{I}| \vec{O}}$ fulfills the property that under any choice of the local operations ${\{{P}_{{X}_{j},{O}_{j}| {A}_{j},{I}_{j}}\}}_{0\leqslant j\lt N}$ of all parties, the expression ${P}_{\vec{I}| \vec{O}}{\prod }_{j=0}^{N-1}{P}_{{X}_{j},{O}_{j}| {A}_{j},{I}_{j}}$ form a conditional probability distribution ${P}_{\vec{X},\vec{I},\vec{O}| \vec{A}}$.

Note that not every conditional distribution ${P}_{\vec{I}| \vec{O}}$ is logically consistent. Some stochastic processes lead to grandfather-paradox-type [50] inconsistencies. Consider the following two extreme examples of such inconsistencies. We describe the examples in the single-party scenario as depicted in figure 5 and where O, I, X, and A are binary random variables.

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Example 1. Let the environment as well as the party S forward the random variable, i.e., the operation of the environment is

$$\begin{eqnarray*}{P}_{I| O}(i,o)=\left\{\begin{array}{ll}1 & i=o,\\ 0 & \mathrm{otherwise},\end{array}\right.\end{eqnarray*}$$

and the operation of the party S is

$$\begin{eqnarray*}{P}_{X,O| A,I}(x,o,a,i)=\left\{\begin{array}{ll}1 & o=i=x,\\ 0 & \mathrm{otherwise}.\end{array}\right.\end{eqnarray*}$$

Since the environment E and the party S forward the random variable, we have $\mathrm{Pr}(O=I)=1$. However, it is unclear what value the probability ${P}_{O}(0)$ should take. This is also known as the causal-loop paradox.

Example 2. We alter the local operation of party S to negate the binary random variable

$$\begin{eqnarray*}{P}_{X,O| A,I}(x,o,a,i)=\left\{\begin{array}{ll}1 & o=i\oplus 1=x,\\ 0 & \mathrm{otherwise}.\end{array}\right.\end{eqnarray*}$$

Now, we are faced with the grandfather paradox: if party S receives i = 1 from the environment, then she sends the value o = 0 to the environment. But in that case, she should receive i = 0 and not i = 1.

Let $\{{q}_{0},{q}_{1},...\}$ be the sample space of a random variable Q with the probability measure P_Q.

Definition 4. (States, operations, evolution, and composition). We represent a state corresponding to a random variable P_Q as the probability vector

$$\begin{eqnarray*}&&{\vec{P}}_{Q}={({P}_{Q}({q}_{0}),{P}_{Q}({q}_{1}),...)}^{T}.\end{eqnarray*}$$

A stochastic process ${P}_{R| Q}$ from Q to a random variable R with sample space $\{{r}_{0},{r}_{1},...\}$ describes an operation and is modeled by the stochastic matrix

$$\begin{eqnarray*}{\hat{P}}_{R| Q}=\left(\begin{array}{ccc}{P}_{R| Q}({r}_{0},{q}_{0}) & {P}_{R| Q}({r}_{0},{q}_{1}) & ...\\ {P}_{R| Q}({r}_{1},{q}_{0}) & {P}_{R| Q}({r}_{1},{q}_{1}) & ...\\ \vdots & \vdots & \ddots \end{array}\right).\end{eqnarray*}$$

The result P_R of evolving the random variable P_Q through the operation ${P}_{R| Q}$ is given by the matrix multiplication

$$\begin{eqnarray*}&&{\vec{P}}_{R}={\hat{P}}_{R| Q}{\vec{P}}_{Q}.\end{eqnarray*}$$

Finally, vectors and matrices are composed in parallel using the Kronecker product $\otimes$.

For example, by this definition, the output of a stochastic process ${P}_{R| {Q}_{0},{Q}_{1}}$ taking two inputs and producing one output is expressed by

$$\begin{eqnarray*}&&{\hat{P}}_{R| {Q}_{0},{Q}_{1}}({\vec{P}}_{{Q}_{0}}\otimes {\vec{P}}_{{Q}_{1}}).\end{eqnarray*}$$

We derive the conditions on the environment E (stochastic process) such that it is logically consistent. For simplicity, we start with the single-party scenario as depicted in figure 5; the party is denoted by S and the environment by E. We can further simplify our picture by fixing the value of A to a and by summing over X:

$$\begin{eqnarray*}&&\displaystyle \sum _{x}{P}_{X=x,O| A=a,I}={P}_{O| I}.\end{eqnarray*}$$

The stochastic process of the environment E is ${P}_{I| O}$. For now, let us assume that S performs a deterministic operation ${D}_{O| I}$. This assumption is dropped later. The operation applied by S can be written as a function

$$\begin{eqnarray*}&&o=f(i),\end{eqnarray*}$$

where i is a deterministic input value. By embedding f into the process of E, we get

$$\begin{eqnarray*}&&{P}_{I| O}(i,f(i)).\end{eqnarray*}$$

This can be interpreted as a probability measure of party S receiving the value i from the environment E:

$$\begin{eqnarray*}&&Q(i)={P}_{I| O}(i,f(i)).\end{eqnarray*}$$

For $Q(i)$ to represent a probability measure, the values of Q for every deterministic value i must be non-negative and have to sum up to 1:

$$\begin{eqnarray*}&&\forall i:\;Q(i)\geqslant 0,\\ &&\quad \displaystyle \sum _{i}Q(i)=1.\end{eqnarray*}$$

We express both conditions in the matrix picture. Non-negativity is achieved whenever all entries of the matrix ${\hat{P}}_{I| O}$ are non-negative. The total-probability conditions are formulated in the following way. The value $f(i)$ that is fed into the environment E is

$$\begin{eqnarray*}&&{\hat{D}}_{O| I=\vec{i}}={\hat{D}}_{O| I}\vec{i}.\end{eqnarray*}$$

The matrix ${\hat{P}}_{I| O}$ fixed to providing the state $\vec{i}$ to the party S is

$$\begin{eqnarray*}&&{\hat{P}}_{I=\vec{i}| O}={\vec{i}}^{T}{\hat{P}}_{I| O}.\end{eqnarray*}$$

Therefore, the probability of party S observing i is

$$\begin{eqnarray*}&&Q(i)={\vec{i}}^{T}{\hat{P}}_{I| O}{\hat{D}}_{O| I}\vec{i},\end{eqnarray*}$$

and the law of total probability requires

$$\begin{eqnarray*}&&\mathrm{Tr}({\hat{P}}_{I| O}{\hat{D}}_{O| I})=1.\end{eqnarray*}$$

This condition remains the same if we relax the input to a stochastic input and the operation of S to a stochastic process ${P}_{O| I}$. The reason for this is that any stochastic input can be written as a convex combination of deterministic inputs, and any stochastic process can be written as a convex combination of deterministic operations. Therefore, the logical-consistency requirement asks the environment E to be restricted to those processes $\hat{E}$ where, under any choice of the local operation ${P}_{O| I}$ of party S, the law of total probability

$$\begin{eqnarray}&&\mathrm{Tr}(\hat{E}{\hat{P}}_{O| I})=1\end{eqnarray} \tag{ 2 }$$

and the non-negativity condition

$$\begin{eqnarray}&&\forall i,j\;:\;{\hat{E}}_{i,j}\geqslant 0\end{eqnarray} \tag{ 3 }$$

hold. Because a stochastic process can be written as a convex mixture of deterministic operations, it is sufficient to ask for

$$\begin{eqnarray*}&&\forall \hat{D}\in { \mathcal D }\;:\;\mathrm{Tr}(\hat{E}\hat{D})=1\;\\ &&\qquad \forall i,j\;:\;{\hat{E}}_{i,j}\geqslant 0\end{eqnarray*}$$

for every operation $\hat{D}$ from the set ${ \mathcal D }$ of all deterministic operations. Thanks to linearity, we can straightforwardly extend these requirements to multiple parties, and arrive at theorems 2 and 3.

Theorem 2. (Total probability). The law that the sum of the probabilities over the exclusive states the parties receive is 1 is satisfied if and only if

$$\begin{eqnarray*}&&\forall {\hat{D}}_{0},{\hat{D}}_{1},...\in { \mathcal D }\;:\;\mathrm{Tr}(\hat{E}({\hat{D}}_{0}\otimes {\hat{D}}_{1}\otimes \cdots ))=1,\end{eqnarray*}$$

where ${\hat{D}}_{j}$ represents a deterministic operation of party S_j.

Theorem 3. (Non-negative probabilities). The law that the probability of the parties observing a state is non-negative is satisfied if and only if

$$\begin{eqnarray*}&&\forall i,j\;:\;{\hat{E}}_{i,j}\geqslant 0.\end{eqnarray*}$$

The ingredients of the framework by Oreshkov et al [31] are process matrices and local operations—described by matrices as well. All the matrices are completely positive trace-preserving quantum maps in the Choi–Jamiołkowski [51, 52] picture. In the classical limit, the matrices become diagonal in the computational basis [31, 35]. In the single-party scenario, the process matrix W is a map from the Hilbert space ${{ \mathcal H }}_{O}$ to the Hilbert space ${{ \mathcal H }}_{I}$. The party's local operation A then again is a map from the Hilbert space ${{ \mathcal H }}_{I}$ to the Hilbert space ${{ \mathcal H }}_{O}$. The conditions a process matrix W in a single-party scenario has to fulfill [31] are

$$\begin{eqnarray}&&\forall A\in { \mathcal M }\;:\;\mathrm{Tr}(\mathrm{WA})=1,\end{eqnarray} \tag{ 4 }$$

$$\begin{eqnarray}&&W\geqslant 0,\end{eqnarray} \tag{ 5 }$$

where ${ \mathcal M }$ is the set of all completely positive trace-preserving maps from the space ${{ \mathcal H }}_{I}$ to the space ${{ \mathcal H }}_{O}$. Intuitively, the condition given by equation (4) 'short-circuits' both maps and enforces the probabilities of the outcomes to sum up to 1.

Theorem 4. (Equivalence). The quantum framework given by equations (4) and (5) in the classical limit is equivalent to the description of classical correlations without predefined causal order given by equations (2) and (3).

Proof. The process matrix W in the quantum framework corresponds to the stochastic process of the environment E in our framework, and the local operations correspond to the stochastic process of the parties. We show a bijection between process matrices and stochastic processes of the environment, and between local operations and stochastic processes of the parties.

A stochastic matrix $\hat{E}$, representing the environment E in our framework, can be translated into the quantum framework by

$$\begin{eqnarray*}&&{W}_{\hat{E}}=\displaystyle \sum _{k}| k\rangle {\langle k| }_{{{ \mathcal H }}_{O}}\otimes d\left(\hat{E}| k\rangle \displaystyle \sum _{{\ell }}{\langle {\ell }| }_{{{ \mathcal H }}_{I}}\right),\end{eqnarray*}$$

where $| k\rangle$ and $| {\ell }\rangle$ are computational-basis states of the same dimension as $\hat{E}$, and where the subscripts denote the respective Hilbert spaces. This completely positive trace-preserving map (expressed in the Choi–Jamiołkowski picture) acts in the same way as the stochastic matrix $\hat{E}$: the state $| k\rangle$ is mapped to $\hat{E}| k\rangle$. The function $d(\rho )$ takes the matrix ρ and cancels all off-diagonal terms, i.e.

$$\begin{eqnarray*}&&d(\rho )=\displaystyle \sum _{m}| m\rangle \langle m| \rho | m\rangle \langle m| .\end{eqnarray*}$$

We can rewrite ${W}_{\hat{E}}$ as

$$\begin{eqnarray*}&&{W}_{\hat{E}}=\displaystyle \sum _{k}| k\rangle {\langle k| }_{{{ \mathcal H }}_{O}}\otimes \displaystyle \sum _{m}| m\rangle \langle m| \hat{E}| k\rangle {\langle m| }_{{{ \mathcal H }}_{I}}.\end{eqnarray*}$$

Analogously, the stochastic matrix ${\hat{P}}_{O| I}$ of the party can be translated into the quantum framework and becomes

$$\begin{eqnarray*}&&{A}_{{\hat{P}}_{O| I}}=\displaystyle \sum _{{k}^{\prime },{m}^{\prime }}| {m}^{\prime }\rangle \langle {m}^{\prime }| {\hat{P}}_{O| I}| {k}^{\prime }\rangle {\langle {m}^{\prime }| }_{{{ \mathcal H }}_{O}}\otimes | {k}^{\prime }\rangle {\langle {k}^{\prime }| }_{{{ \mathcal H }}_{I}}.\end{eqnarray*}$$

The reverse direction of the bijection follows from the description above.

Now, we show that the conditions (4) and (5) in a single-party scenario on a process matrix W coïncide with the conditions (2) and (3) in our framework. The non-negativity condition (5) forces the probabilities of the outputs of W to be non-negative; the same holds for the condition (3) in our framework. That the condition (4) coincides with the condition (2) is shown below. Forcing W and A to be diagonal in the computational basis gives

$$\begin{eqnarray*}\mathrm{Tr}(\mathrm{WA}) & = & \displaystyle \sum _{i,j}\langle i,j| {WA}| i,j\rangle \\ & = & \displaystyle \sum _{i,j}\langle i,j| W| i,j\rangle \langle i,j| A| i,j\rangle .\end{eqnarray*}$$

Substituting W with ${W}_{\hat{E}}$ and A with ${A}_{{\hat{P}}_{O| I}}$ yields

$$\begin{eqnarray*}&&\displaystyle \sum _{i,j,m,k,{m}^{\prime },{k}^{\prime }}\langle i| k\rangle \langle k| i\rangle \langle j| m\rangle \langle m| \hat{E}| k\rangle \langle m| j\rangle \langle i| {m}^{\prime }\rangle \langle {m}^{\prime }| {\hat{P}}_{O| I}| {k}^{\prime }\rangle \langle {m}^{\prime }| i\rangle \langle j| {k}^{\prime }\rangle \langle {k}^{\prime }| j\rangle \\ &&\qquad =\displaystyle \sum _{i,j}\langle j| \hat{E}| i\rangle \langle i| {\hat{P}}_{O| I}| j\rangle \\ &&\qquad =\mathrm{Tr}(\hat{E}{\hat{P}}_{O| I}),\end{eqnarray*}$$

which proves the claim. The multi-party case follows through linearity.□

Convex polytopes can be represented in two different ways: the H-representation is a list of half-spaces where the intersection is the polytope, and the V-representation is a list of the extremal points of the polytope. Algorithms like the double-description method [53, 54] enumerate all extremal points of the polytope given the H-representation. We used cdd+ [55] for vertex enumeration. The inverse problem is solved by its dual: a convex-hull algorithm.

Here, we derive the polytope of classical processes without predefined causal order. This polytope is represented by the dashed–dotted lines in figure 2. A projection of the polytope for three parties and binary inputs/outputs onto a plane is given in figure 6.

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We start with the polytope for one party (see figure 5) with a binary input and a binary output. In this case, a process is described by a square matrix of dimension 2. The most general process of the environment E is

$$\begin{eqnarray*}\hat{E}={\hat{P}}_{I| O}=\left(\begin{array}{cc}{w}_{0} & {w}_{1}\\ {w}_{2} & {w}_{3}\end{array}\right),\end{eqnarray*}$$

consisting of four variables. The deterministic operations party S can apply are

$$\begin{eqnarray*}{\hat{D}}_{0} & = & \left(\begin{array}{cc}1 & 1\\ 0 & 0\end{array}\right),\qquad {\hat{D}}_{1}=\left(\begin{array}{cc}0 & 0\\ 1 & 1\end{array}\right),\\ {\hat{D}}_{2} & = & \left(\begin{array}{cc}1 & 0\\ 0 & 1\end{array}\right),\qquad {\hat{D}}_{3}=\left(\begin{array}{cc}0 & 1\\ 1 & 0\end{array}\right),\end{eqnarray*}$$

where ${\hat{D}}_{0}$, ${\hat{D}}_{1}$ produce a constant 0, 1, respectively, and where the matrix ${\hat{D}}_{2}$ is the identity and ${\hat{D}}_{3}$ the negation. The equalities

$$\begin{eqnarray}&&\mathrm{Tr}(\hat{E}{\hat{D}}_{0})=1,\end{eqnarray} \tag{ 6 }$$

$$\begin{eqnarray}&&\mathrm{Tr}(\hat{E}{\hat{D}}_{1})=1,\end{eqnarray} \tag{ 7 }$$

$$\begin{eqnarray}&&\mathrm{Tr}(\hat{E}{\hat{D}}_{2})=1,\end{eqnarray} \tag{ 8 }$$

enforce

$$\begin{eqnarray*}&&\mathrm{Tr}(\hat{E}{\hat{D}}_{3})=1.\end{eqnarray*}$$

This is shown as follows:

$$\begin{eqnarray*} & & \mathrm{Tr}(\hat{E}{\hat{D}}_{0})+\mathrm{Tr}(\hat{E}{\hat{D}}_{1})+\mathrm{Tr}(\hat{E}{\hat{D}}_{2}) & = & ({w}_{0}+{w}_{2})+({w}_{1}+{w}_{3})+({w}_{0}+{w}_{3})\\ & & & = & 2({w}_{0}+{w}_{3})+{w}_{1}+{w}_{2}\\ & & & = & 2\mathrm{Tr}(\hat{E}{\hat{D}}_{2})+\mathrm{Tr}(\hat{E}{\hat{D}}_{3}).\end{eqnarray*}$$

By eliminating three variables using the total-probability conditions (6)–(8) from above, we get

$$\begin{eqnarray*}{\hat{P}}_{I| O}=\left(\begin{array}{cc}{w}_{0} & {w}_{0}\\ 1-{w}_{0} & 1-{w}_{0}\end{array}\right)\end{eqnarray*}$$

with the non-negativity conditions

$$\begin{eqnarray*}{w}_{0} & \geqslant & 0,\\ 1-{w}_{0} & \geqslant & 0.\end{eqnarray*}$$

This solution set is a one-dimensional polytope with the extremal points 0 and 1. All solutions describe a state. This implies that all correlations that can be obtained in this framework with a single party and binary input and output, can also be obtained in a framework without feedback, i.e., these correlations can be obtained causally (see figure 7).

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In the two-party case with a binary input and a binary output for each party, the process $\hat{E}={\hat{P}}_{{I}_{0},{I}_{1}| {O}_{0},{O}_{1}}$ of the environment is described by a square matrix of dimension 2². The conditions are

$$\begin{eqnarray}\forall i,j\in \{0,1,2\}\;:\;\mathrm{Tr}(\hat{E}({\hat{D}}_{i}\otimes {\hat{D}}_{j})) & = & 1,\\ \forall i,j\;:\;{\hat{E}}_{i,j} & \geqslant & 0.\end{eqnarray} \tag{ 9 }$$

With a similar argument as above, one can show that the operation ${\hat{D}}_{3}$ does not need to be considered for either party. The matrix $\hat{E}$ consists of 4² unknowns, out of which 3² are eliminated by the total-probability conditions given by equation (9). Thus, we are left with seven unknowns, forming a seven-dimensional polytope with 16 inequalities.

The resulting V-representation of the polytope consists of 12 extremal points, all of which represent deterministic processes:

$$\begin{eqnarray*}{\hat{E}}_{0} & = & \left(\begin{array}{cccc}1 & 1 & 1 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{array}\right),\qquad {\hat{E}}_{1}=\left(\begin{array}{cccc}0 & 0 & 0 & 0\\ 1 & 1 & 1 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{array}\right),\\ {\hat{E}}_{2} & = & \left(\begin{array}{cccc}0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 1 & 1 & 1 & 1\\ 0 & 0 & 0 & 0\end{array}\right),\qquad {\hat{E}}_{3}=\left(\begin{array}{cccc}0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 1 & 1 & 1 & 1\end{array}\right),\\ {\hat{E}}_{4} & = & \left(\begin{array}{cccc}1 & 1 & 0 & 0\\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{array}\right),\qquad {\hat{E}}_{5}=\left(\begin{array}{cccc}0 & 0 & 1 & 1\\ 1 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{array}\right),\\ {\hat{E}}_{6} & = & \left(\begin{array}{cccc}0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 1 & 1 & 0 & 0\\ 0 & 0 & 1 & 1\end{array}\right),\qquad {\hat{E}}_{7}=\left(\begin{array}{cccc}0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 1\\ 1 & 1 & 0 & 0\end{array}\right),\\ {\hat{E}}_{8} & = & \left(\begin{array}{cccc}1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 1\\ 0 & 0 & 0 & 0\end{array}\right),\qquad {\hat{E}}_{9}=\left(\begin{array}{cccc}0 & 1 & 0 & 1\\ 0 & 0 & 0 & 0\\ 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\end{array}\right),\\ {\hat{E}}_{10} & = & \left(\begin{array}{cccc}0 & 0 & 0 & 0\\ 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 1\end{array}\right),\qquad {\hat{E}}_{11}=\left(\begin{array}{cccc}0 & 0 & 0 & 0\\ 0 & 1 & 0 & 1\\ 0 & 0 & 0 & 0\\ 1 & 0 & 1 & 0\end{array}\right).\end{eqnarray*}$$

In the following, we use $A={S}_{0}$ and $B={S}_{1}$. The first four processes ${\hat{E}}_{0},{\hat{E}}_{1},{\hat{E}}_{2},{\hat{E}}_{3}$ represent the four constants $(0,0),(0,1),(1,0),(1,1)$ as inputs to the parties A and B. The next four processes represent a constant input to party A (processes ${\hat{E}}_{4}$ and ${\hat{E}}_{5}$ produce the constant 0, and the other two processes produce the constant 1 ) and a channel from party A to party B; the processes ${\hat{E}}_{4}$ and ${\hat{E}}_{6}$ describe the identity channel, and ${\hat{E}}_{5}$ and ${\hat{E}}_{7}$ describe the bit-flip channel. The last four processes are analogous, with a channel from B to A and where party B receives a constant. All these 12 processes act deterministically on bits for two parties where at least one party receives a constant (see figure 8). Therefore, every such channel can be simulated in a causal fashion. This result generalized to higher dimensions was already shown by taking the classical limit ofthe framework for quantum correlations without predefined causal order [31].

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The process of the environment E in a three party setup with binary inputs and outputs is described by a square matrix $\hat{E}={\hat{P}}_{{I}_{0}{I}_{1}{I}_{2}| {O}_{0}{O}_{1}{O}_{2}}$ of dimension 2³. The matrix $\hat{E}$ consists of 4³ variables, out of which 3³ can be eliminated with the total-probability conditions

$$\begin{eqnarray}&&\forall i,j,k\in \{0,1,2\}\;:\mathrm{Tr}(\hat{E}({\hat{D}}_{i}\otimes {\hat{D}}_{j}\otimes {\hat{D}}_{k}))=1,\end{eqnarray} \tag{ 10 }$$

resulting in a 37 -dimensional polytope with 4³ linear constraints (non-negative probabilities):

$$\begin{eqnarray}&&\forall i,j\;:\;{\hat{E}}_{i,j}\geqslant 0.\end{eqnarray} \tag{ 11 }$$

Solving this polytope yields ${710}^{\prime }760$ extremal points. Only 744 extremal points out of these ${710}^{\prime }760$ are deterministic, i.e., consist of 0–1 values; the remaining extremal points are so-called proper mixtures of logically inconsistent processes. Such proper mixtures are not convex combinations of deterministic extremal points inside the polytope, but are convex combinations of deterministic points where some lie outside of the polytope—any process from outside of the polytope leads to logical inconsistencies. Interestingly, this smaller polytope (hence, also the polytope described by the equations (10) and (11)) consists of processes that cannot be simulated using a predefined causal order, i.e., processes where no party receives a constant, implying that every party causally succeeds some other party. The 744 deterministic extremal points are discussed in section 4 along with the general polytope restricted to the deterministic extremal points.

We describe the polytope for logically consistent classical processes without predefined causal order in the general case. Let n be the number of parties and let d be the dimension of the states entering and leaving every laboratory. This leaves us with a ${d}^{n}\times {d}^{n}$ stochastic matrix $\hat{E}$ describing the environment. Every party can perform an operation that is a convex mixture of all d^d deterministic operations. The set of all deterministic operations is denoted by ${ \mathcal D }$. For every party, under any choice of deterministic operation $D\in { \mathcal D }$, the trace of the environment $\hat{E}$ multiplied with the local operations is constrained to give 1 (see theorem 2). However—as in the binary-input/output case above—, some of these constraints are redundant.

Theorem 5. (Sufficient set for total-probability conditions). The total-probability conditions to this family of operations

$$\begin{eqnarray*}{\hat{D}}_{i,j}=\left(\begin{array}{ccccccccc}1 & 1 & 1 & ... & 1 & 0 & 1 & ... & 1\\ 0 & 0 & 0 & ... & 0 & 0 & 0 & ... & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & ... & 0 & 0 & 0 & ... & 0\\ 0 & 0 & 0 & ... & 0 & 1 & 0 & ... & 0\\ 0 & 0 & 0 & ... & 0 & 0 & 0 & ... & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & ... & 0 & 0 & 0 & ... & 0\end{array}\right),\end{eqnarray*}$$

where j is output for input i and 0 otherwise, imply the total-probability conditions for all remaining deterministic operations of the same dimension, i.e.

$$\begin{eqnarray*}&&\quad \forall {i}_{0},{j}_{0},{i}_{1},{j}_{j},...,{i}_{n-1},{j}_{n-1}\geqslant 0:\mathrm{Tr}(\hat{E}({\hat{D}}_{{i}_{0},{j}_{0}}\otimes ...\otimes \;{\hat{D}}_{{i}_{n-1},{j}_{n-1}}))=1\\ &&\Longrightarrow \\ &&\quad \forall {\hat{D}}_{0},...,{\hat{D}}_{n-1}\in { \mathcal D }:\mathrm{Tr}(\hat{E}({\hat{D}}_{0}\otimes ...\otimes \;{\hat{D}}_{n-1}))=1.\end{eqnarray*}$$

Proof. We restrict ourselves to the single-party scenario—the multi-party case follows through linearity. Let ${\vec{v}}_{i}$ be the d-dimensional vector with a 1-entry at position i and 0 's everywhere else. We can write a d-dimensional matrix ${\hat{D}}_{i,j}$ as

$$\begin{eqnarray*}&&{\hat{D}}_{i,j}=\left(\displaystyle \sum _{m\ne i}{\vec{v}}_{0}{\vec{v}}_{m}^{T}\right)+{\vec{v}}_{j}{\vec{v}}_{i}^{T}.\end{eqnarray*}$$

A general deterministic matrix $D\in { \mathcal D }$ of the same dimension, where k is mapped to a_k, is expressed as

$$\begin{eqnarray*}&&D=\displaystyle \sum _{k}{\vec{v}}_{{a}_{k}}{\vec{v}}_{k}^{T}.\end{eqnarray*}$$

On the one hand, using the antecedent above, we have

$$\begin{eqnarray}&&\mathrm{Tr}\left(\hat{E}\displaystyle \sum _{{\rm{k}}:{{\rm{a}}}_{k}\ne 0}{\hat{D}}_{k,{a}_{k}}\right)=\displaystyle \sum _{{\rm{k}}:{{\rm{a}}}_{k}\ne 0}\mathrm{Tr}(\hat{E}{\hat{D}}_{k,{a}_{k}})={\ell }\end{eqnarray} \tag{ 12 }$$

with ${\ell }=| \{k| {a}_{k}\;\ne \;0\}|$. On the other hand, we can rewrite ${\sum }_{k:{a}_{k}\ne 0}{\hat{D}}_{k,{a}_{k}}$ as

$$\begin{eqnarray*}\displaystyle \sum _{k:{a}_{k}\ne 0}{\hat{D}}_{k,{a}_{k}} & = & \displaystyle \sum _{k:{a}_{k}\ne 0}\left(\displaystyle \sum _{m\ne k}{\vec{v}}_{0}{\vec{v}}_{m}^{T}\right)+{\vec{v}}_{{a}_{k}}{\vec{v}}_{k}^{T}\\ & = & \displaystyle \sum _{k:{a}_{k}\ne 0}{\vec{v}}_{{a}_{k}}{\vec{v}}_{k}^{T}+\displaystyle \sum _{k:{a}_{k}=0}{\vec{v}}_{0}{\vec{v}}_{k}^{T}-\displaystyle \sum _{k:{a}_{k}=0}{\vec{v}}_{0}{\vec{v}}_{k}^{T}+\displaystyle \sum _{k:{a}_{k}\ne 0}\displaystyle \sum _{m\ne k}{\vec{v}}_{0}{\vec{v}}_{m}^{T}\\ & = & D+\displaystyle \sum _{k:{a}_{k}\ne 0}\displaystyle \sum _{m\ne k}{\vec{v}}_{0}{\vec{v}}_{m}^{T}-\displaystyle \sum _{k:{a}_{k}=0}{\vec{v}}_{0}{\vec{v}}_{k}^{T}\\ & = & D+{\ell }\displaystyle \sum _{k:{a}_{k}=0}{\vec{v}}_{0}{\vec{v}}_{k}^{T}+({\ell }-1)\displaystyle \sum _{k:{a}_{k}\ne 0}{\vec{v}}_{0}{\vec{v}}_{k}^{T}-\displaystyle \sum _{k:{a}_{k}=0}{\vec{v}}_{0}{\vec{v}}_{k}^{T}\\ & = & D+({\ell }-1)\displaystyle \sum _{k}{\vec{v}}_{0}{\vec{v}}_{k}^{T}\\ & = & D+({\ell }-1){\hat{D}}_{\mathrm{0,0}}.\end{eqnarray*}$$

Therefore

$$\begin{eqnarray*}\mathrm{Tr}\left(\hat{E}\displaystyle \sum _{{\rm{k}}:{{\rm{a}}}_{k}\ne 0}{\hat{D}}_{k,{a}_{k}}\right) & = & \mathrm{Tr}(\hat{E}{\rm{D}})+({\ell }-1)\mathrm{Tr}(\hat{E}{\hat{D}}_{\mathrm{0,0}})\\ & = & \mathrm{Tr}(\hat{E}{\rm{D}})+{\ell }-1,\end{eqnarray*}$$

which, with the identity (12), implies

$$\begin{eqnarray*}&&\mathrm{Tr}(\hat{E}{\rm{D}})=1.\end{eqnarray*}$$

□

The family $\{{\hat{D}}_{i,j}| i,j\in I\}$ of deterministic operations with the set $I=\{0,...,n-1\}$ has size $d(d-1)+1$.

Theorem 6. (Polytope). The H-representation of the polytope of logically consistent classical processes without predefined causal order is

$$\begin{eqnarray*}\forall {\hat{D}}_{0},{\hat{D}}_{1},...,{\hat{D}}_{n-1}\in {\{{\hat{D}}_{i,j}\}}_{I\times I}:\mathrm{Tr}(\hat{E}({\hat{D}}_{0}\otimes {\hat{D}}_{1}\otimes ...\otimes \;{\hat{D}}_{n-1})) & = & 1,\\ \forall i,j\;:\;{\hat{E}}_{i,j} & \geqslant & 0.\end{eqnarray*}$$

The polytope has ${d}^{2n}$ facets and dimension

$$\begin{eqnarray*}&&{d}^{2n}-{(d(d-1)+1)}^{n},\end{eqnarray*}$$

which is exponential in the number of parties.

We discuss the deterministic-extrema polytope in the setting of three parties and binary inputs and outputs. To simplify the presentation, we use $A={S}_{0}$, $B={S}_{1}$, $C={S}_{2}$, ${O}_{A}={O}_{0}$, ${O}_{B}={O}_{1}$, ${O}_{C}={O}_{2}$, ${I}_{A}={I}_{0}$, ${I}_{B}={I}_{1}$, and ${I}_{C}={I}_{2}$. As described in section 3.4, this polytope has 744 extremal points. They can be characterized as follows.

Assume that $\hat{E}$, when the parties locally apply the identity operation, maps $(0,0,0)$ to $(0,0,0)$, i.e., $(0,0,0)$ is a fixed-point. Then, any other extremal point ${\hat{E}}^{\prime }$ is obtained by the local operations identity ${\hat{D}}_{3}$ and bit-flip ${\hat{D}}_{4}$, where we embed these local operations into the environment. Let ${\hat{L}}_{i,j,k}$ be the local operation of the three parties

$$\begin{eqnarray*}&&{\hat{L}}_{i,j,k}={\hat{D}}_{4}^{i}\otimes {\hat{D}}_{4}^{j}\otimes {\hat{D}}_{4}^{k},\end{eqnarray*}$$

i.e., party A performs the identity if i = 0 and the bit-flip operation if i = 1—the other parties' local operations are defined in the same way. The extremal point ${\hat{E}}^{\prime }$ can be described as

$$\begin{eqnarray*}&&{\hat{E}}^{\prime }={\hat{L}}_{i,j,k}\hat{E}{\hat{L}}_{i,j,k},\end{eqnarray*}$$

where, as described above, the operations are embedded into the environment (see figure 9). Logical consistency of the environment ${\hat{E}}^{\prime }$ follows because we started with a logically consistent $\hat{E}$ and the operations act on single parties. The process ${\hat{E}}^{\prime }$ maps ($i,j,k$) to $(i,j,k)$. Thus, starting with $\hat{E}$, for any choice of $i,j,k$, we obtain a different extremal point. There are ${2}^{3}-1$ alternative extremal points that can be constructed in this fashion. From this we conclude that $744/8=93$ extremal points are such, that $(0,0,0)$ is a fixed-point under locally applying the identity. We restrict our analysis to these 93 extremal points; all others can be obtained by the above construction. The following analysis is structured depending on the number of parties that receive a constant from the environment.

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There exists only one extremal point where $(0,0,0)$ is mapped to $(0,0,0)$ under applying identity locally, and where every party receives a constant: the constant $(0,0,0)$.

Assume exactly two parties receive a constant, which leaves us with three possibilities of choosing them. Fix these parties to be A and B. The third party C receives a value that depends on the operation of A or of B or of both. Thus, we are in the case $A\preceq C$ or $B\preceq C$. Inevitably, the constant must be $(0,0);$ otherwise the fixed-point $(0,0,0)$ is not recovered. Party C receives a value that depends on the value fed back by A and B; there exist ${2}^{3}-1=7$ such functions where we have excluded the constant and all operations where C receives a value different from 0 on inputs $(0,0)$ to the environment from A and B. Therefore, under all permutations of the parties, 21 extremal points give a constant to two parties and have the fixed-point $(0,0,0)$ when the identity is applied locally.

In a next step, assume that exactly one party receives a constant. This assumption, again, allows for three different setups, as we can choose which party receives a constant. Without loss of generality, let A be this party, i.e., $A\preceq B$ and $A\preceq C;$ the constant must be 0 again in order to comply with the requirement of the fixed-point. Now, we are left with several possibilities on how B and C depend on A and on each other. As a first case, we assume that B and C do not depend on each other, but depend on A only ($A\preceq B$ and $A\preceq C$). This dependency cannot be different from the identity channels from A to B and from A to C; the alternative would be the bit-flip channel that would not reproduce the desired fixed-point $(0,0,0)$. This gives us  three different extremal points under all permutations of the parties. Another possibility on the dependencies is that B depends on A, and C depends on B, i.e., $A\preceq B\preceq C$, and the interchange of parties B and C. The channels—by following the same reasoning above—again must be the identity channels: this gives us six extremal points. Now, we look at the case where B depends on A and where C depends on both, A and B, i.e., $A\preceq B$, $A\preceq C$, and $B\preceq C$, and any permutation of the parties. There are six permutations. The constant that A receives must be 0, party B must depend trivially on A (the identity channel) and party C can depend in  five different ways on A and B: these are all 2 -to- 1 -bit functions where $(0,0)$ is mapped to 0 (2³) minus the constant and minus the dependencies on A only and on B only. In total, there are $6\times 5=30$ such extremal points. We are left with the last scenario: B depends on A and on C, and C depends on A and on B, i.e., $A\preceq B$, $C\preceq B$, $A\preceq C$, and $B\preceq C$. The constant, as above, is 0. Given the random variable O_A fed to the environment by A, the environment can either describe a channel from B to C ($B\preceq C$) or describe a channel from C to B ($C\preceq B$); any other channel would lead to a causal loop. The direction of the channel must differ under different values o_A fed-back by A, as otherwise B and C would not mutually depend on each other. We have two possibilities on the direction given the value fed-back by A is ${O}_{A}=0$. Assume the direction to be $B\preceq C$. For the case ${O}_{A}=0$, the channels from A to B and from B to C are the identity channels in order to comply with the fixed point $(0,0,0)$. In the other case ${O}_{A}=1$, the direction of the channel between B and C is in the reverse direction compared to ${O}_{A}=0$, i.e., $C\preceq B$. Then, because of ${O}_{A}\;\ne \;0$, the random variables I_B and I_C are not forced to be $(0,0);$ there exist two channels from A to C and another two channels from B to C. Therefore, we are left with 3 × 2 × 4 = 24 possibilities. An overview over these setups is given in figure 10.

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The last setup (see figure 10(g)) where no party receives a constant builds a family of  eight extremal points. All extremal points are equivalent up to relabelling of the inputs to and outputs from the environment. One such extremal point is

$$\begin{eqnarray*}{\hat{E}}_{\mathrm{det}1}=\left(\begin{array}{cccccccc}1 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\end{array}\right).\end{eqnarray*}$$

The behaviour of this extremal point is

$$\begin{eqnarray}&&{I}_{A}={\bar{O}}_{B}{O}_{C},\quad {I}_{B}={O}_{A}{\bar{O}}_{C},\quad {I}_{C}={\bar{O}}_{A}{O}_{B},\end{eqnarray} \tag{ 13 }$$

where $\bar{x}$ is the negation of x. This solution is depicted in figure 11. Its location in the polytope is shown in figure 6.

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