Incompatible measurements in a class of general probabilistic theories

TL;DR

This paper explores measurement incompatibility in finite-dimensional general probabilistic theories, linking it to steering and nonlocality, and introduces tools to quantify and analyze incompatibility in various state spaces.

Contribution

It introduces the concept of incompatibility witnesses, characterizes the maximum incompatibility degree in GPTs, and relates these findings to classical and quantum channels.

Findings

Maximum incompatibility degree for quantum measurements identified

Incompatibility is characterized by properties of affine maps into polysimplices

Classical and quantum channels are closely related to polysimplices

Abstract

We study incompatibility of measurements and its relation to steering and nonlocality in a class of finite dimensional general probabilistic theories (GPT).The basic idea is to represent finite collections of measurements as affine maps of a state space into a polysimplex and show that incompatibility is characterized by properties of these maps. We introduce the notion of an incompatibility witness and show its relation to incompatibility degree. We find the maximum incompatibility degree attainable by pairs of two-outcome quantum measurements and characterize state spaces for which incompatibility degree attains maximal values possible in GPT. As examples, we study the spaces of classical and quantum channels and show their close relation to polysimplices. This relation explains the super-quantum non-classical effects that were observed on these spaces.

Equations

$$(V^{+}{)}^{\ast }:=\{v^{\ast }\in V^{\ast },⟨v^{\ast },v⟩\ge 0,\forall v\in V^{+}\}.$$

$$V^{+}{\otimes }_{min}{W}^{+}$$

$$V^{+}{\otimes }_{max}{W}^{+}$$

$$E(K):=\{f\in A(K),0\le f\le 1_K\}.$$

$$\Vert \psi {\Vert }_K=\inf \{a+b,\psi =ax-by,a,b\ge 0,x,y\in K\}.$$

$$f(\rho )=\mathrm{Tr}M\rho ,\rho \in \mathfrak{S}(\mathcal{H})$$

$$V({\mathcal{C}}_{A,A^{\prime }})\equiv \{X\in B_h({\mathcal{H}}_{A^{\prime }A}),{\mathrm{Tr}}_{A^{\prime }}X\in \mathbb{R}I\},$$

$${\mathcal{C}}_{A,A^{\prime }}$$

$$f_i(\Phi )=\mathrm{Tr}{M}_i(\Phi \otimes i{d}_R)({\rho }_{AR})$$

$${\Phi }_T(\sigma )=\sum _{i,j}⟨i,\sigma |i{⟩}_{A}T(j|i)|j⟩⟨j{|}_{A^{\prime }},\sigma \in \mathfrak{S}(\mathcal{H}).$$

$$K_A\otimes K_B$$

$$K_A\hat{\otimes }{K}_B$$

$$\forall f_A\in E(K_A),f_B\in E(K_B),⟨1_A\otimes 1_B,y⟩=1\}$$

$$K_A\otimes K_B\subseteq K_A\tilde{\otimes }{K}_B\subseteq K_A\hat{\otimes }{K}_B.$$

$$V(K_A\otimes K_B)$$

$$A(K_A\otimes K_B)$$

$$V(K_A\otimes K_B{)}^{+}$$

$$V(K_A\hat{\otimes }{K}_B{)}^{+}$$

$$A(K_A\otimes K_B{)}^{+}$$

$$A(K_A\hat{\otimes }{K}_B{)}^{+}$$

$${\Delta }_{n_A}\otimes {\Delta }_{n_B}={\Delta }_{n_A}\hat{\otimes }{\Delta }_{n_B}={\Delta }_{n_{AB}},$$

$${\mathfrak{S}}_A\tilde{\otimes }{\mathfrak{S}}_B={\mathfrak{S}}_{AB},$$

$$⟨1_{{\mathcal{C}}_{A,A^{\prime }}}\otimes f_B,\Phi ⟩=\mathrm{Tr}(I_{RA}\otimes M_{B^{\prime }})(i{d}_R\otimes \Phi )({\rho }_{RA}\otimes {\sigma }_B)$$

$${\sigma }_B\mapsto {\mathrm{Tr}}_{RA}(i{d}_R\otimes \Phi )({\rho }_{RA}\otimes {\sigma }_B)$$

$${\mathcal{C}}_{A,A^{\prime }}\tilde{\otimes }{\mathcal{C}}_{B,B^{\prime }}:={\mathcal{C}}_{AB,A^{\prime }{B}^{\prime }}^{caus}.$$

$$\mathcal{A}({\Delta }_{m_A},{\Delta }_{n_A})\hat{\otimes }\mathcal{A}({\Delta }_{m_B},{\Delta }_{n_B})\subset \mathcal{A}({\Delta }_{m_{AB}},{\Delta }_{n_{AB}}),$$

$$T_A\otimes T_B\in \mathcal{A}(K_A\otimes K_B,V(K_A^{\prime }\otimes K_B^{\prime }{)}^{+})$$

$$T_A\otimes T_B\in \mathcal{A}(K_A\hat{\otimes }{K}_B,V(K_A^{\prime }\hat{\otimes }{K}_B^{\prime }{)}^{+}).$$

$$⟨(T_A\otimes T_B)(y),f_A^{\prime }\otimes f_B^{\prime }⟩=⟨y,T_A^{\ast }(f_A^{\prime })\otimes (T_B{)}^{\ast }(f_B^{\prime })⟩,$$

$${\mathsf{S}}_{l_0,\dots ,l_k}:={\Delta }_{l_0}\times \cdots \times {\Delta }_{l_k},$$

$$F(x)=(f^0(x),\dots ,f^k(x));f^i={\mathsf{m}}^{i}F,\forall i$$

$${\mathsf{s}}_{n_0,\dots ,n_k}:=({\delta }_{n_0}^0,\dots ,{\delta }_{n_k}^k),n_i=0,\dots ,l_i,i=0\dots ,k,$$

$$\bar{\mathsf{s}}:=(u_0,\dots ,u_k)=\frac{1}{{\Pi }_i(l_i+1)}\sum _{n_0,\dots ,n_k}{\mathsf{s}}_{n_0,\dots ,n_k}$$

$$\mathsf{m}^{i}_{j}(\mathsf{s}_{n_{0},\dots,n_{k}})=\left\{\begin{array}[]{cc}1&\mbox{ if }n_{i}=j\\ 0&\mbox{ otherwise}\end{array}\right..$$

Full text

I Introduction

Incompatibility of measurements is one of the fundamental features of quantum mechanics. As a key ingredient in quantum information protocols, incompatibility and related non-classical effects such as Bell nonlocality and steering, are viewed as important resources in quantum information theory.

General probabilistic theories (GPT) form a framework for description of physical theories admitting probabilistic processes. Within this framework, the quantum theory is specified by several axioms Hardy (2001); Chiribella et al. (2011). Some of these axioms (causality, prefect distinguishability and local tomography) define a class of theories that can be studied within the setting of ordered vector spaces and their tensor products.

Many of the basic features that distinguish quantum mechanics from any classical theory are shared by a large class of GPT. It is a natural question what are the properties that characterize quantum mechanics. This question has been studied for many years, see e.g. Araki (1980); Birkhoff and von Neumann (1936); Mackey (1963), but recently the advance of quantum information theory led to a renewed interest in this topic, Barrett (2007); Busch and Stevens (2014); Barnum et al. (2006, 2007). To answer this question, it is important to understand the nature of the non-classical features and relations between them.

Such relations were already observed: it is well known that steering and nonlocality require both entanglement and incompatible observables. The relations between non-locality, steering and incompatibility were studied in Wolf et al. (2009); Busch and Stevens (2014); Banik et al. (2013); Quintino et al. (2014); Banik (2015), both for quantum theory and in GPT. On the other hand, there exist unsteerable entangled quantum states Wiseman et al. (2007), and incompatible sets of quantum measurements that cannot lead to violation of Bell inequalities Quintino et al. (2014). To understand these relations, the more general setting is useful because it allows one to recognize which non-classical manifestations are consequences of convexity and the tensor product structure, and which are inherently quantum.

In this contribution, we study incompatibility of measurements and its relation to steering and nonlocality in a class of finite dimensional GPT, using the tools of convex geometry. The basic idea is to represent finite collections of measurements as affine maps of a state space into a polysimplex (that is, a Cartesian product of simplices) and show that incompatibility is characterized by properties of these maps. A generalization of this idea was already used to describe incompatibility of channels Plávala (2017). We introduce the notion of an incompatibility witness and show its relation to incompatibility degree, defined in Heinosaari et al. (2014, 2016). We find the maximum incompatibility degree attainable by pairs of two-outcome quantum measurements, generalizing the results of Banik et al. (2013), and characterize state spaces for which incompatibility degree attains maximal values possible in GPT. This completes the results of Busch et al. (2013) and Jenčová and Plávala (2017), where maximal incompatibility of pairs of two-outcome measurements is considered. The representation of collections of measurements as maps enables us to tie incompatibility directly to steering and non-locality of states of composite systems. The concept of incompatibility witnesses helps to explain the relation of incompatibility degree and maximal violation of Bell inequalities, as well as the observed limitations of these relations.

Besides the classical and quantum state spaces, we study the spaces of classical and quantum channels. It was observed in Jenčová and Plávala (2017); Sedlák et al. (2016) that these spaces admit maximally incompatible measurements, which is known to be impossible in finite dimensional quantum theory Heinosaari et al. (2014). Moreover, it was shown that causal bipartite quantum channels can be used to obtain maximal violation of the CHSH inequality Beckman et al. (2001); Plávala and Ziman (2017), in fact, all kinds of non-signalling correlations Hoban and Sainz (2017). We prove these results as easy consequences of the fact that the set of classical channels is a retract of the set of quantum channels and is affinely isomorphic to a polysimplex.

The paper is organized as follows. In the next section we briefly describe the main components of GPT in our setting and present the main examples of spaces of classical and quantum states and channels. The structure of polysimplices is detailed in Section III. The relations to spaces of channels are also proved. Section IV is devoted to incompatibility of measurements, incompatibility witnesses and degree. The last section deals with steering and Bell nonlocality.

II General probabilistic theories

We present a brief overview of GPT in the finite dimensional setting, explain our overall assumptions and introduce the mathematical tools needed in the sequel. Let us remark that the GPT framework is much broader, see e.g. Barrett (2007); Chiribella et al. (2010); Janotta and Hinrichsen (2014); Barnum and Wilce (2012) for more details.

We first recall a few definitions and facts about convex sets and ordered vector spaces that will be needed below. For a subset $X\subseteq V$ of a finite dimensional vector space $V$, we denote by $co(X)$ the convex hull and $aff(X)$ the affine span of $X$ in $V$. For a convex subset $C\subseteq V$, $ri(C)$ denotes the relative interior of $C$ in $aff(C)$ and $\dim(C):=\dim(aff(V))$. The set of all affine maps (that is, preserving the convex structure) between convex sets $C_{1}$ and $C_{2}$ will be denoted by $\mathcal{A}(C_{1},C_{2})$.

For the purposes of this paper, an ordered vector space is a pair $(V,V^{+})$, where $V$ is a real finite dimensional vector space and $V^{+}$ is a closed convex cone satisfying $V^{+}\cap-V^{+}=\{0\}$ and $V=V^{+}-V^{+}$. This induces a partial order in $V$ as $v\leq w$ if $w-v\in V^{+}$. Let $V^{*}$ be the vector space dual with duality $\langle\cdot,\cdot\rangle$. The order dual of $(V,V^{+})$ is the ordered vector space $(V^{*},(V^{+})^{*})$, with the closed convex cone of positive functionals

[TABLE]

Note that we have $(V^{+})^{**}=V^{+}$. A linear map between ordered vector spaces is called positive if it preserves the positive cone. We say that the cone $V^{+}$ is weakly self-dual if it is affinely isomorphic to $(V^{+})^{*}$.

Let $(V,V^{+})$ and $(W,W^{+})$ be ordered vector spaces. There are two distinguished ways to define a positive cone in the tensor product $V\otimes W$:

[TABLE]

We have $V^{+}\otimes_{min}W^{+}\subseteq V^{+}\otimes_{max}W^{+}$. The elements of $V^{+}\otimes_{min}W^{+}$ are called separable.

II.1 States, effects and measurements

The framework of GPT is build on basic notions of states, representing preparation procedures of a given system, and effects, assigning to each state the corresponding probabilities of outcomes in yes/no experiments. The state spaces have a natural convex structure, expressing the possibility of forming probabilistic mixtures of states. The effects must respect this structure and are therefore represented by affine functions from the state space into the unit interval. Throughout this paper, we will assume that any state space is a compact convex subset $K$ of a finite dimensional real vector space. Moreover, we adopt the no restriction hypothesis, requiring that all affine functions $K\to[0,1]$ correspond to physical effects. For a discussion of these assumptions in GPT see Chiribella et al. (2010, 2011); Janotta and Lal (2013).

A state space $K$ determines a pair of dual ordered vector spaces as follows. Let $A(K):=\mathcal{A}(K,\mathbb{R})$ and $A(K)^{+}:=\mathcal{A}(K,\mathbb{R}^{+})$. Let also $1_{K}$ be the constant map $1_{K}(x)\equiv 1$. Then $(A(K),A(K)^{+})$ is an ordered vector space. The function $1_{K}$ is an interior element in $A(K)^{+}$ and hence is an order unit: for any $f\in A(K)$ we have $-t1_{K}\leq f\leq t1_{K}$ for some $t>0$. The set of effects is thus given by

[TABLE]

For $x\in K$, the evaluation map $f\mapsto f(x)$ defines a linear functional on $A(K)$ that is clearly positive and unital: $1_{K}(x)=1$. The converse is also true Alfsen (1971), so that $K$ can be identified with the set of positive unital functionals, or states, on $A(K)$. With this identification, $K$ is a base of the dual cone $(A(K)^{+})^{*}$, in the sense that each $0\neq\varphi\in(A(K)^{+})^{*}$ can be expressed in a unique way as a multiple of some element in $K$. We therefore have $(A(K)^{+})^{*}\equiv V(K)^{+}:=\cup_{\lambda\geq 0}\lambda K$ and $A(K)^{*}\equiv V(K):=V(K)^{+}-V(K)^{+}$. These identifications will be used throughout. For $\psi\in V(K)$, let

[TABLE]

Then $\|\cdot\|_{K}$ is a norm in $V(K)$, called the base norm. It is the dual of the supremum norm $\|\cdot\|_{max}$ in $A(K)$.

Similarly as for two-outcome measurements, any measurement on a system with state space $K$ is fully described by its outcome statistics in each state. A measurement with $n+1$ outcomes is therefore identified with a map $f\in\mathcal{A}(K,\Delta_{n})$, where $\Delta_{n}$ is the $n$-dimensional simplex of probability measures over $\{0,\dots,n\}$. The measurement $f$ is determined by $n+1$ effects $f_{0},\dots,f_{n}\in E(K)$, satisfying $\sum_{i}f_{i}=1_{K}$, given as $f_{i}(x)=f(x)(i)$. As before, we assume that each element of $\mathcal{A}(K,\Delta_{n})$ describes a valid measurement.

Example 1*.*

(Classical state spaces.) The state space of a classical system is an $m$-dimensional simplex $\Delta_{m}$. We have $A(\Delta_{m})\simeq V(\Delta_{m})=\mathbb{R}^{m+1}$, with $V(\Delta_{m})^{+}\simeq A(\Delta_{m})^{+}$ the positive cone of vectors with nonnegative entries and $E(\Delta_{m})$ is the set of vectors with entries in $[0,1]$. The base norm in this case is the $l_{1}$-norm in $\mathbb{R}^{m+1}$. Measurements $f\in\mathcal{A}(\Delta_{m},\Delta_{n})$ are classical channels and can be identified with $(m+1)\times(n+1)$ stochastic matrices $\{T(j|i)\}_{i,j}$, where $T(\cdot|i)\in\Delta_{n}$, $i=0,\dots,m$ are determined by the values of $f$ on the vertices of $\Delta_{m}$.

Example 2*.*

(Quantum state spaces.) A quantum state space is the set of density operators $\mathfrak{S}(\mathcal{H})$ on a finite dimensional Hilbert space $\mathcal{H}$. We will sometimes use labels $\mathcal{H}_{A}$, $\mathcal{H}_{B}$, etc. for the Hilbert spaces, then we use the notations $d_{A}:=\dim(\mathcal{H}_{A})$, $\mathfrak{S}_{A}:=\mathfrak{S}(\mathcal{H}_{A})$. Any $f\in A(\mathfrak{S}(\mathcal{H}))$ has the form

[TABLE]

for some $M\in B_{h}(\mathcal{H})$, the set of Hermitian operators on $\mathcal{H}$. In this way, we have $A(\mathfrak{S}(\mathcal{H}))\simeq V(\mathfrak{S}(\mathcal{H}))=B_{h}(\mathcal{H})$, $A(\mathfrak{S}(\mathcal{H}))^{+}\simeq V(\mathfrak{S}(\mathcal{H}))^{+}=B(\mathcal{H})^{+}$, the cone of positive operators on $\mathcal{H}$, $1_{\mathfrak{S}(\mathcal{H})}=I$, the identity operator and $E(\mathfrak{S}(\mathcal{H}))\simeq E(\mathcal{H})$, the set of quantum effects. The base norm $\|\cdot\|_{\mathfrak{S}(\mathcal{H})}$ is the trace norm $\|X\|_{1}=\mathrm{Tr}\,|X|$. The measurements are given by positive operator valued measures (POVMs) on $\mathcal{H}$, that is, tuples of effects $M_{0},\dots,M_{n}\in B(\mathcal{H})^{+}$, $\sum_{i}M_{i}=I$.

Example 3*.*

(Spaces of quantum channels.) Let $\mathcal{H}_{A},\mathcal{H}_{A^{\prime}}$ be finite dimensional Hilbert spaces. We will denote by $\mathcal{C}_{A,A^{\prime}}$ the set of all quantum channels $\mathcal{H}_{A}\to\mathcal{H}_{A^{\prime}}$, that is, all completely positive and trace preserving linear maps $B(\mathcal{H}_{A})\to B(\mathcal{H}_{A^{\prime}})$. We now describe the corresponding cones and measurements for $\mathcal{C}_{A,A^{\prime}}$, see Jenčová (2012) for details.

By the Choi representation, $\mathcal{C}_{A,A^{\prime}}$ is isomorphic to a compact convex subset of the quantum state space $\mathfrak{S}_{A^{\prime}A}$. Using this isomorphism, $V(\mathcal{C}_{A,A^{\prime}})$ can be identified with the subspace

[TABLE]

where $\mathrm{Tr}_{A^{\prime}}$ is the partial trace over $\mathcal{H}_{A^{\prime}}$. We then have

[TABLE]

Consequently, $A(\mathcal{C}_{A,A^{\prime}})$ is a quotient of $B_{h}(\mathcal{H}_{A^{\prime}A})$ and $A(\mathcal{C}_{A,A^{\prime}})^{+}$ is the set of equivalence classes containing some positive element. The base norm $\|\cdot\|_{\mathcal{C}_{A,A^{\prime}}}$ is identified with the diamond norm $\|\cdot\|_{\diamond}$ Jenčová (2014). One can also show that any measurement $f\in\mathcal{A}(\mathcal{C}_{A,A^{\prime}},\Delta_{n})$ has the form

[TABLE]

for some POVM $M_{0},\dots,M_{n}$ on $\mathcal{H}_{A^{\prime}R}$ and some state $\rho\in\mathfrak{S}_{AR}$ where $\mathcal{H}_{R}$ is an ancilla, $d_{R}\leq d_{A}$, but the representation in this form is not unique, see also Ziman (2008). In particular, the unit effect $1_{\mathcal{C}_{A,A^{\prime}}}$ is obtained from any state $\rho_{AR}$ and the trivial measurement $M_{0}=I_{A^{\prime}R}$.

Example 4*.*

(Spaces of classical channels.) The set of classical channels $\mathcal{A}(\Delta_{m},\Delta_{n})$ is isomorphic to a subset of $\mathcal{C}_{A,A^{\prime}}$, with $d_{A}=m+1$, $d_{A^{\prime}}=n+1$. Such an isomorphism is obtained by fixing orthonormal bases $|i\rangle_{A}$ of $\mathcal{H}_{A}$ and $|j\rangle_{A^{\prime}}$ of $\mathcal{H}_{A^{\prime}}$ and putting for any stochastic matrix $T\in\mathcal{A}(\Delta_{m},\Delta_{n})$,

[TABLE]

Quantum channels of this form are called classical-to-classical, or c-c channels. The cones and measurements for this state space will be identified later (cf. Proposition 3).

II.2 Composition of state spaces: tensor products

Let $K_{A}$ and $K_{B}$ be state spaces, corresponding to two systems in a GPT. To describe the state space of the joint system, we need the notion of a tensor product of state spaces. Let the composite state space be denoted by $K_{A}\widetilde{\otimes}K_{B}$. Assuming the local tomography axiom Hardy (2001); Chiribella et al. (2010, 2011), $K_{A}\widetilde{\otimes}K_{B}$ is a subset of the tensor product $V(K_{A})\otimes V(K_{B})$ such that

1. (a)

$x_{A}\otimes x_{B}\in K_{A}\widetilde{\otimes}K_{B}$ for all $x_{A}\in K_{A}$, $x_{B}\in K_{B}$, 2. (b)

$f_{A}\otimes f_{B}\in E(K_{A}\widetilde{\otimes}K_{B})$ for all $f_{A}\in E(K_{A}),f_{B}\in E(K_{B})$ 3. (c)

$1_{K_{A}\widetilde{\otimes}K_{B}}=1_{A}\otimes 1_{B}$, here $1_{A}:=1_{K_{A}}$, $1_{B}:=1_{K_{B}}$.

This is based on the requirement that for the composite system, all product states and all product effects are valid. These conditions determine the minimal and the maximal tensor product of state spaces. Let

[TABLE]

Note that both sets satisfy the conditions for a composite state space and we always have

[TABLE]

The states in $K_{A}\otimes K_{B}$ are called separable, all other states in $K_{A}\widetilde{\otimes}K_{B}$ are called entangled. The particular form of the composite state space is specified by the theory in question, see the examples below. In terms of the related spaces and cones, we have

[TABLE]

this follows easily from the definitions and duality relations.

Example 5*.*

For classical state spaces, we have

[TABLE]

$n_{AB}:=n_{A}n_{B}+n_{A}+n_{B}$, is the set of probability measures on $\{0,\dots,n_{A}\}\times\{0,\dots,n_{B}\}$. In fact, we have $K\otimes\Delta_{n}=K\widehat{\otimes}\Delta_{n}$ for any state space $K$ and this property characterizes the simplices in a general infinite-dimensional setting, see Namioka and Phelps (1969).

Example 6*.*

For quantum state spaces, we have

[TABLE]

with the usual tensor product of Hilbert spaces $\mathcal{H}_{AB}=\mathcal{H}_{A}\otimes\mathcal{H}_{B}$. Here the minimal tensor product $\mathfrak{S}_{A}\otimes\mathfrak{S}_{B}$ is the subset of separable states and the maximal tensor product $\mathfrak{S}_{A}\widehat{\otimes}\mathfrak{S}_{B}$ is the set of entanglement witnesses (with unit trace).

Example 7*.*

For spaces of quantum channels, let $\Phi\in\mathcal{C}_{A,A^{\prime}}\widetilde{\otimes}\mathcal{C}_{B,B^{\prime}}$. It is natural to require that $\Phi\in\mathcal{C}_{AB,A^{\prime}B^{\prime}}$, so that $\Phi$ is a bipartite quantum channel. By the condition (b), each product of effects is a valid effect, in particular, $1_{\mathcal{C}_{A,A^{\prime}}}\otimes f_{B}$ is a valid effect for any $f_{B}\in E(\mathcal{C}_{B,B^{\prime}})$. By Example 3, $1_{\mathcal{C}_{A,A^{\prime}}}$ is obtained from any state $\rho_{RA}$ and the identity $I_{RA^{\prime}}$. Let $f_{B}$ be given by $\sigma_{B}$ and an effect $M_{B^{\prime}}$, then

[TABLE]

and this expression does not depend on $\rho_{RA}$. It follows that

[TABLE]

defines a channel in $\mathcal{C}_{B,B^{\prime}}$ that does not depend on $\rho$; similarly, we obtain a channel in $\mathcal{C}_{A,A^{\prime}}$ by applying the unit effect on the second part. Channels with this property are called causal or no-signalling bipartite channels, see Beckman et al. (2001), the set of all such channels is denoted by $\mathcal{C}^{caus}_{AB,AB^{\prime}}$. We define the composite state space as

[TABLE]

The minimal tensor product is the set of local bipartite channels, which are convex combinations of channels prepared by each party separately. The maximal tensor product is strictly larger than $\mathcal{C}^{caus}_{AB,A^{\prime}B^{\prime}}$, since its elements are not necessarily completely positive.

Example 8*.*

It is clear that

[TABLE]

where $\Delta_{m_{AB}}=\Delta_{m_{A}m_{B}+m_{A}+m_{B}}=\Delta_{m_{A}}\otimes\Delta_{m_{B}}$ and similarly for $\Delta_{n_{AB}}$. We will see later (Section V.5) that the maximal tensor product is the set of all classical bipartite causal channels, characterized by the no-signalling conditions (16) and (17).

II.3 Channels and positive maps

Channels in GPT describe transformations of the systems allowed in the theory and are represented by affine maps between state spaces. Although all affine maps between simplices are classical channels, we do not assume in general that all elements in $\mathcal{A}(K,K^{\prime})$ for state spaces $K$ and $K^{\prime}$ correspond to valid channels. For the spaces of quantum states and channels, it is required that the maps have completely positive extensions. Completely positive maps $B(\mathcal{H}_{A^{\prime}A})\to B(\mathcal{H}_{B^{\prime}B})$ that map $\mathcal{C}_{AA^{\prime}}$ into $\mathcal{C}_{BB^{\prime}}$ are called quantum supermaps Chiribella et al. (2008) or quantum combs Chiribella et al. (2009) and belong to a hierarchy describing quantum networks.

Any $T\in\mathcal{A}(K,V(K^{\prime}))$ extends uniquely to a linear map $T:V(K)\to V(K^{\prime})$ and $\mathcal{A}(K,V(K^{\prime}))$ has the structure of a real vector space. The subset $\mathcal{A}(K,V(K^{\prime})^{+})\subseteq\mathcal{A}(K,V(K^{\prime}))$ is a closed convex cone of elements whose extensions are positive maps. With this cone, $\mathcal{A}(K,V(K^{\prime}))$ is an ordered vector space.

Let $T_{A}\in\mathcal{A}(K_{A},V(K_{A}^{\prime})^{+})$ and $T_{B}\in\mathcal{A}(K_{B},V(K_{B}^{\prime})^{+})$. It is easy to see that $T_{A}\otimes T_{B}$ is positive with respect to both the maximal and minimal tensor product cones, that is,

[TABLE]

and

[TABLE]

Indeed, the first inclusion follows from the definition of the minimal tensor product and the second one from

[TABLE]

for all $y\in K_{A}\widehat{\otimes}K_{B}$, $f_{A}^{\prime}\in A(K_{A}^{\prime})^{+}$ and $f_{B}^{\prime}\in A(K_{B}^{\prime})^{+}$, here $T^{*}$ denotes the adjoint of the linear extension of $T$. We say that $T_{A}$ is entanglement breaking (ETB) if for any state space $K_{B}$, we have $T_{A}\otimes id_{K_{B}}\in\mathcal{A}(K_{A}\widehat{\otimes}K_{B},V(K_{A}^{\prime}\otimes K_{B})^{+})$. The set of all ETB maps will be denoted by $\mathcal{A}_{sep}(K_{A},V(K_{A}^{\prime})^{+})$, it is a closed convex subcone in $\mathcal{A}(K_{A},V(K_{A}^{\prime})^{+})$.

There is a well known relation between linear maps and tensor products of vector spaces, with respect to which the positive maps correspond to elements of the maximal tensor product and ETB maps to elements of the minimal one. Details on these relations, as well as the proofs of the following results, are given in Appendix A.

Proposition 1**.**

Let $T\in\mathcal{A}(K,V(K^{\prime})^{+})$. Then $T$ is ETB if and only if $T$ factorizes through a simplex: there are a simplex $\Delta_{n}$ and maps $T_{0}\in\mathcal{A}(K,V(\Delta_{n})^{+})$ and $T_{1}\in\mathcal{A}(\Delta_{n},V(K^{\prime})^{+})$ such that $T=T_{1}T_{0}$. If $T$ is a channel, $T_{0}$ and $T_{1}$ may be chosen to be channels as well.

It is clear that any constant map factorizes through the 1-dimensional simplex $\Delta_{0}$ and hence must be ETB.

We now look at the dual spaces and cones. For $T\in\mathcal{A}(K,V(K))$, let $\mathrm{Tr}\,T$ denote the usual trace of its linear extension. It is not difficult to see that the dual space of $\mathcal{A}(K,V(K^{\prime}))$ can be identified with $\mathcal{A}(K^{\prime},V(K))$, with duality $\langle S,T\rangle=\mathrm{Tr}\,ST$.

Proposition 2**.**

The dual cone to $\mathcal{A}(K,V(K^{\prime})^{+})$ is $\mathcal{A}_{sep}(K^{\prime},V(K)^{+})$.

III Polysimplices and their structure

Let $k,l_{0},\dots,l_{k}\in\mathbb{N}$. A polysimplex is a Cartesian product of simplices

[TABLE]

with pointwise defined convex structure. This is a compact convex set, more precisely a convex polytope. Elements of $\mathsf{S}_{l_{0},\dots,l_{k}}$ represent states of a device determined by a set of inputs indexed by $0,\dots,k$, each of which has an allowed set of outputs $0,\dots,l_{i}$. Such devices were introduced by Popescu and Rohrlich Popescu and Rohrlich (1994) as toy theories exhibiting super-quantum correlations. In the framework of GPT, theories with state spaces of this form were studied in Barrett (2007); Janotta and Lal (2013).

If $l_{0}=\dots=l_{k}=n$, the polysimplex will be denoted by $\Delta_{n}^{k+1}$. The $(k+1)$-hypercube $\Delta_{1}^{k+1}$ will be denoted by $\square_{k+1}$. If $l_{0},\dots,l_{k}\in\mathbb{N}$ are assumed fixed, we often drop the multiindex $l_{0},\dots,l_{k}$ and denote the polysimplex by $\mathsf{S}$.

Let $f^{0},\dots,f^{k}$ be a collection of measurements on $K$, such that $f^{i}\in\mathcal{A}(K,\Delta_{l_{i}})$. By definition of the Cartesian product, such collections correspond precisely to elements of $\mathcal{A}(K,\mathsf{S}_{l_{0},\dots,l_{k}})$. Explicitly, the relations between $f^{i}\in\mathcal{A}(K,\Delta_{l_{i}})$, $i=0,\dots,k$ and $F=(f^{0},\dots,f^{k})\in\mathcal{A}(K,\mathsf{S}_{l_{0},\dots,l_{k}})$ are given by

[TABLE]

where $\mathsf{m}^{i}:\mathsf{S}_{l_{0},\dots,l_{k}}\to\Delta_{l_{i}}$ is the projection onto the $i$-th component. Since our main results are based on the relation of properties of such maps to incompatibility, it will be necessary to describe the structure of the polysimplices and related spaces and cones.

The vertices of $\mathsf{S}=\mathsf{S}_{l_{0},\dots,l_{k}}$ are the $(k+1)$-tuples

[TABLE]

where $\delta^{i}_{n_{i}}$ denotes the $n_{i}$-th vertex of the $i$-th simplex. If $u_{i}\in\Delta_{l_{i}}$ is the uniform probability distribution for all $i$, then

[TABLE]

is the barycenter of $\mathsf{S}$. Further, note that each projection $\mathsf{m}^{i}$ is a measurement on $\mathsf{S}$, with effects determined by

[TABLE]

Since all faces of $\mathsf{S}$ have the form $F_{0}\times\dots\times F_{k}$, where $F_{i}$ is a face of $\Delta_{l_{i}}$, it is clear that the maximal faces are precisely the null spaces of $\mathsf{m}^{i}_{j}$.

It will be convenient to fix a pair of dual bases of the spaces $A(\mathsf{S})$ and $V(\mathsf{S})$, such that the basis of $A(\mathsf{S})$ consists of the effects $1_{\mathsf{S}}$ and $\mathsf{m}^{i}_{j}$. Since $\sum_{j}\mathsf{m}^{i}_{j}=1_{\mathsf{S}}$ for all $i$, we will fix a linearly independent subset. For the dual basis, we need to describe the edges of $\mathsf{S}$. Since the edges are 1-dimensional faces, they have the form

[TABLE]

where $E_{i}$ is an edge of $\Delta_{l_{i}}$. We see that the vertices adjacent to a vertex $\mathsf{s}_{n_{0},\dots,n_{k}}$ are those that differ from $\mathsf{s}_{n_{0},\dots,n_{k}}$ in exactly one index, that is, the vertices

[TABLE]

Pick the vertex $\mathsf{s}_{l_{0},\dots,l_{k}}$ and let

[TABLE]

denote the vectors given by the adjacent edges.

Lemma 1**.**

1. (i)

The extreme rays of the cone $A(\mathsf{S})^{+}$ are generated by the effects $\mathsf{m}^{i}_{j}$, $i=0,\dots,k$, $j=0,\dots,l_{i}$. 2. (ii)

The effects

[TABLE]

form a basis of the vector space $A(\mathsf{S})$. 3. (iii)

The elements

[TABLE]

form a basis of the vector space $V(\mathsf{S})$, dual to (3).

Proof.

Since the null spaces $(\mathsf{m}^{i}_{j})^{-1}(0)$ are exactly the maximal faces of $\mathsf{S}$, these effects generate the extreme rays of $A(\mathsf{S})^{+}$. Further, let $f_{1},f_{2},\dots$ denote the elements in (3) and $x_{1},x_{2},\dots$ the elements of (4). It is easy to see that $\langle f_{i},x_{j}\rangle=\delta_{ij}$, so that both sets are linearly independent. The statements (ii) and (iii) now follow from the fact that $\dim(A(\mathsf{S}))=\sum_{i=0}^{k}l_{i}+1$.

∎

The basis elements (4) are visualized in Fig. 1. With respect to this basis, the vertices are expressed as

[TABLE]

Remark 1*.*

We can get another pair of dual bases using any vertex $\mathsf{s}_{n_{0},\dots,n_{k}}$ and its adjacent edges for a basis of $V(\mathsf{S})$ and

[TABLE]

for the dual basis of $A(\mathsf{S})$.

Example 9*.*

(The square.) The simplest example is the square $\square_{2}=\Delta_{1}\times\Delta_{1}$. The vertices $\mathsf{s}_{0,0},\mathsf{s}_{0,1},\mathsf{s}_{1,0},\mathsf{s}_{1,1}$ satisfy the relation $\mathsf{s}_{0,0}+\mathsf{s}_{1,1}=\mathsf{s}_{0,1}+\mathsf{s}_{1,0}$. This state space is also called the gbit or square-bit, see Popescu and Rohrlich (1994); Barrett (2007); Busch et al. (2013). The dual cone $A(\square_{2})^{+}$ is generated by four effects $\mathsf{m}^{0}_{0},\mathsf{m}^{0}_{1},\mathsf{m}^{1}_{0},\mathsf{m}^{1}_{1}$. Since we have $\mathsf{m}^{0}_{0}+\mathsf{m}^{0}_{1}=1_{\square_{2}}=\mathsf{m}^{1}_{0}+\mathsf{m}^{1}_{1}$, these effects again form a square, so that $V(\square_{2})^{+}$ is weakly self-dual. Note that the square is the only polysimplex with this property. (This follows from the fact that the extreme rays of $V(\mathsf{S})^{+}$ are generated by vertices of $\mathsf{S}$, whereas the extreme rays of $A(\mathsf{S})^{+}$ correspond to maximal faces of $\mathsf{S}$. Therefore, $V(\mathsf{S})^{+}\simeq A(\mathsf{S})^{+}$ implies $\Pi_{i=0}^{k}(l_{i}+1)=\sum_{i=0}^{k}(l_{i}+1)$, this holds only for the square).

The following relation of polysimplices and spaces of classical channels is immediate (see also Examples 1 and 4).

Proposition 3**.**

We have $\Delta_{n}^{k+1}\simeq\mathcal{A}(\Delta_{k},\Delta_{n})$. More generally, any polysimplex $\mathsf{S}_{l_{0},\dots,l_{k}}$ is isomorphic to a face of $\mathcal{A}(\Delta_{k},\Delta_{n})$, with $n\geq\max_{i}l_{i}$. This isomorphism is given by

[TABLE]

There is also a relation of polysimplices and spaces of quantum channels. To describe this relation, we will need the following notion.

Let $K$ and $K^{\prime}$ be state spaces. A map $R\in\mathcal{A}(K,K^{\prime})$ such that there is a map $S\in\mathcal{A}(K^{\prime},K)$ with $RS=id_{K^{\prime}}$ is called a retraction. The map $S:K^{\prime}\to K$ is then called a section. For any retraction-section pair $(R,S)$, the map $P=SR$ is a projection on $K$ onto the range of $S$, that is, an affine idempotent map $K\to K$ such that $P(K)=S(K^{\prime})$. Moreover, any map in $\mathcal{A}(K^{\prime},C)$ for a convex set $C$ has an extension to a map in $\mathcal{A}(K,C)$.

Proposition 4**.**

There exists a retraction-section pair $R\in\mathcal{A}(C_{A,A^{\prime}},\Delta^{d_{A}}_{d_{A^{\prime}}-1})$ and $S\in\mathcal{A}(\Delta^{d_{A}}_{d_{A^{\prime}}-1},C_{A,A^{\prime}})$, determined by

[TABLE]

and

[TABLE]

Proof.

Since $\mathsf{m}^{i}_{j}R(\Phi)\geq 0$ for all $i,j$ and $\sum_{j}\mathsf{m}^{i}_{j}R(\Phi)=\mathrm{Tr}\,\Phi(|i\rangle\langle i|)=1$ for all $i$, $R$ is a well defined element in $\mathcal{A}(C_{A,A^{\prime}},\Delta^{d_{A}}_{d_{A^{\prime}}-1})$. For each $s$, $S(s)$ is a c-c channel and is therefore completely positive. It is quite clear that

[TABLE]

so that $RS=id$.

∎

Remark 2*.*

The above Proposition implies that there is a projection of $\mathcal{C}_{A,A^{\prime}}$ onto a set of c-c channels and that any map in $\mathcal{A}(\Delta^{d_{A}}_{d_{A^{\prime}}-1},C)$ can be extended to a map in $\mathcal{A}(\mathcal{C}_{A,A^{\prime}},C)$, for any convex set $C$. The consequences of this fact will become clear later on.

IV Incompatibility of measurements

Let $K$ be a state space and let $f^{i}\in\mathcal{A}(K,\Delta_{l_{i}})$ be a measurement with values in $\{0,\dots,l_{i}\}$, $i=0,\dots,k$. We say that $f^{0},\dots,f^{k}$ are compatible if they are the marginals of a single joint measurement with values in $\{0,\dots,l_{0}\}\times\dots\times\{0,\dots,l_{k}\}$. For an exposition of incompatibility in our setting, see Heinosaari et al. (2016).

The joint measurement is described by a map $g\in\mathcal{A}(K,\Delta_{L})$ with $L:=\Pi_{i}(l_{i}+1)-1$ (note that $\Delta_{L}\simeq\bigotimes_{i}\Delta_{l_{i}}$) and can be parametrized as

[TABLE]

where $g_{n_{0},\dots,n_{k}}\in E(K)$ and $\delta_{n_{0},\dots,n_{k}}$ is the probability measure concentrated at $(n_{0},\dots,n_{k})$. We then have

[TABLE]

for the effects of $f^{i}$. Let $J\in\mathcal{A}(\Delta_{L},\mathsf{S})$ be determined by $J(\delta_{n_{0},\dots,n_{k}})=\mathsf{s}_{n_{0},\dots,n_{k}}$. Then it is easy to see that (6) can be written as $f^{i}_{j}=\mathsf{m}^{i}_{j}Jg$. In other words, $f^{0},\dots,f^{k}$ is compatible if and only if the corresponding $F\in\mathcal{A}(K,\mathsf{S})$ satisfies

[TABLE]

The following observation is simple but important.

Theorem 1**.**

The measurements $f^{0},\dots,f^{k}$ are compatible if and only if the corresponding channel $F$ is entanglement breaking.

Proof.

If $f^{0},\dots,f^{k}$ are compatible, then $F$ is ETB by (7) and Proposition 1. Conversely, let $F$ be ETB. Proposition 1 implies that there is some simplex $\Delta_{n}$ and channels $g^{\prime}\in\mathcal{A}(K,\Delta_{n})$, $T\in\mathcal{A}(\Delta_{n},\mathsf{S})$ such that $F=Tg^{\prime}$. The channel $T$ corresponds to a collection of measurements $t^{i}:=\mathsf{m}^{i}T\in\mathcal{A}(\Delta_{n},\Delta_{l_{i}})$. Since all measurements on a simplex are compatible, there is some $h\in\mathcal{A}(\Delta_{n},\Delta_{L})$ such that $T=Jh$. Putting $g=hg^{\prime}$ finishes the proof.

∎

Remark 3*.*

The above characterization of incompatible measurements as non-ETB channels suggests that these channels should be admissible in the GPT in question, which also means that we need to include the polysimplices into the theory. For quantum theory this might seem strange, since the polysimplices are certainly not quantum state spaces. On the other hand, the retraction-section pairs of Proposition 4 allow us to include maps in $\mathcal{A}(\mathfrak{S}(\mathcal{H}),\mathsf{S})$ into the larger setting of quantum networks. If $F\in\mathcal{A}(\mathfrak{S}(\mathcal{H}),\Delta_{n}^{k+1})$ is a collection of quantum measurements and $S$ is the section of Proposition 4, then $SF$ is a map from states into quantum channels. Using the Choi representation, one can see that this map is also completely positive, hence a quantum comb, Chiribella et al. (2009). Moreover, since $RS=id$, $SF$ is ETB iff $F$ is. One should be aware that ”entanglement breaking” has a different meaning here than for usual cp maps: $F$ is compatible iff $(SF\otimes id)(\rho)$ is a local bipartite channel for any bipartite state $\rho$.

IV.1 Incompatibility witnesses

Let $F=(f^{0},\dots,f^{k})\in\mathcal{A}(K,\mathsf{S})$ be a collection of measurements. By Proposition 2, $F$ is non-ETB if and only if there is some $W\in\mathcal{A}(\mathsf{S},V(K)^{+})$ such that $\mathrm{Tr}\,FW<0$. Such a $W$ will be called an incompatibility witness. As can be seen from Proposition 1, this notion has a close relation to entanglement witnesses.

Any $W\in\mathcal{A}(\mathsf{S},V(K))$ is determined by the images of the vertices of $\mathsf{S}$. The elements

[TABLE]

will be called the vertices of $W$ (although not all of these points must be vertices of the image $W(\mathsf{S})$). The map $W$ is positive if and only if all its vertices are in $V(K)^{+}$. The image of the barycenter of $\mathsf{S}$, $\bar{w}:=W(\bar{\mathsf{s}})$, will be called the barycenter of $W$. We say that $W$ is degenerate if $\dim(W(\mathsf{S}))<\dim(\mathsf{S})$. A description of the cones $\mathcal{A}(\mathsf{S},V(K)^{+})$ and $\mathcal{A}_{sep}(\mathsf{S},V(K)^{+})$ can be found in Appendix B.

It is clear that an ETB map cannot be an incompatibility witness. As we shall see (Fig. 2 below), not all non-ETB maps are witnesses. For a characterization of witnesses, we will need the following notion. Let $W,\tilde{W}\in\mathcal{A}(\mathsf{S},V(K)^{+})$. We say that $\tilde{W}$ is a translation of $W$ in the direction $v\in V(K)$ if $\tilde{W}=W+L_{v}$ where $L_{v}$ is the constant map $L_{v}(s)\equiv v$. Equivalently, the vertices of $\tilde{W}$ satisfy $\tilde{w}_{n_{0},\dots,n_{k}}=w_{n_{0},\dots,n_{k}}+v$ for all $n_{0},\dots,n_{k}$. If $v$ is such that $\langle 1_{K},v\rangle=0$, we say that $\tilde{W}$ is a translation of $W$ along $K$.

Theorem 2**.**

A map $W\in\mathcal{A}(\mathsf{S},V(K)^{+})$ is an incompatibility witness if and only if no translation of $W$ along $K$ is ETB.

Proof.

Note that $\mathcal{A}(K,\mathsf{S})$ is a compact convex subset of $\mathcal{A}(K,V(\mathsf{S})^{+})$. We need to describe the generated space and cone. For shorter notations, let us denote $\mathcal{V}:=V(\mathcal{A}(K,\mathsf{S}))$ and $\mathcal{V}^{+}:=V(\mathcal{A}(K,\mathsf{S}))^{+}$. We have

[TABLE]

and $\mathcal{V}^{+}=\mathcal{A}(K,V(\mathsf{S})^{+})\cap\mathcal{V}$. Let $\mathcal{V}^{\perp}$ be the annihilator of $\mathcal{V}$ in the dual space $\mathcal{A}(\mathsf{S},V(K))$, then it is not difficult to see that

[TABLE]

Since $int(\mathcal{V}^{+})\neq\emptyset$ (for example, any constant map of $K$ onto $s\in ri(\mathsf{S})$ is in $int(\mathcal{V}^{+})$), Krein’s theorem Naimark (1959) implies that any positive functional on $(\mathcal{V},\mathcal{V}^{+})$ extends to an element in the dual cone $\mathcal{A}(K,V(\mathsf{S})^{+})^{*}=\mathcal{A}_{sep}(\mathsf{S},V(K)^{+})$. If $W\in\mathcal{A}(\mathsf{S},V(K)^{+})$ is not a witness, then $F\mapsto\mathrm{Tr}\,FW$ extends to a positive functional on $(\mathcal{V},\mathcal{V}^{+})$, so that there is some $\tilde{W}\in\mathcal{A}_{sep}(\mathsf{S},V(K)^{+})$ such that

[TABLE]

Hence $W-\tilde{W}\in\mathcal{V}^{\perp}$, so that $\tilde{W}$ is an ETB translation of $W$ along $K$. Conversely, assume that $W$ is a witness. Let $F\in\mathcal{A}(K,\mathsf{S})$ be such that $\mathrm{Tr}\,FW<0$, then for any translation $\tilde{W}$ of $W$ along $K$, we have $\mathrm{Tr}\,F\tilde{W}=\mathrm{Tr}\,FW<0$. It follows that $\tilde{W}$ is a witness as well and cannot be ETB.

∎

We will find another characterization of incompatibility witnesses for two-outcome measurements later (Corollary 3).

IV.1.1 Extremal and non-ETB elements in $\mathcal{A}(\square_{2},V(K)^{+})$

For detection of incompatibility, it suffices to use witnesses that are extremal in the cone $\mathcal{A}(\mathsf{S},V(K)^{+})$. More precisely, for an ordered vector space $(V,V^{+})$, we say that and element $v\in V^{+}$ is extremal if it is nonzero and lies on an extreme ray of $V^{+}$. Alternatively, $v$ is extremal if $v\neq 0$ and $v^{\prime}\leq v$ for any $v^{\prime}\in V^{+}$ implies that $v^{\prime}=tv$ for some $t\geq 0$. A description of extremal elements in $\mathcal{A}(\mathsf{S},V(K)^{+})$ will be also useful in the next section.

So far, we can do this in the simplest case when $\mathsf{S}=\square_{2}$. Any $W\in\mathcal{A}(\square_{2},V(K)^{+})$ is given by four vertices $w_{i,j}\in V(K)^{+}$, $i,j=0,1$, satisfying

[TABLE]

Proposition 5**.**

Let $W\in\mathcal{A}(\square_{2},V(K)^{+})$ have vertices $w_{i,j}$, $i,j=0,1$ and barycenter $\bar{w}\neq 0$. Let $F_{i,j}$ denote the face of $V(K)^{+}$ generated by $w_{i,j}$ and let $L_{i,j}=F_{i,j}-F_{i,j}$ be the generated subspace. Then $W$ is extremal if and only if

[TABLE]

Proof.

Assume that the two conditions are fulfilled and let $W^{\prime}\leq W$, with vertices $w_{i,j}^{\prime}$ and barycenter $\bar{w}^{\prime}$. Then clearly $w_{i,j}^{\prime}\in F_{i,j}$. By (10) we must have $\bar{w}^{\prime}=t\bar{w}$ for some $t\in[0,1]$, but then $w^{\prime}_{i,j}=tw_{i,j}$ for all $i,j$ by (9). It follows that $W$ is extremal.

Conversely, let us denote the subspace on the LHS of (10) by $L$ and assume that there is some $\psi\neq t\bar{w}$ in $L$. Then there are some $\psi_{i,j}\in L_{i,j}$ such that $\psi=\psi_{0,0}+\psi_{1,1}=\psi_{0,1}+\psi_{1,0}$. By definition of $L_{i,j}$, there is some $u>0$ such that $w_{i,j}^{\pm}:=\tfrac{1}{2}w_{i,j}\pm u\psi_{i,j}\in V(K)^{+}$. Obviously, $w_{i,j}^{+}$ and $w_{i,1}^{-}$ are vertices of some $W^{+}$ and $W^{-}$, which are not multiples of $W$, and we have $W=W^{+}+W^{-}$. It follows that $W$ is not extremal.

If (9) is not true, then there are some $\eta_{i,j}\in L_{i,j}$ such that not all of them are 0 and $\eta_{0,0}+\eta_{1,1}=\eta_{0,1}+\eta_{1,0}=0$. We may then proceed as above to show that $W$ is not extremal.

∎

We are now interested in extremal elements that are non-ETB. We start with a simple observation.

Lemma 2**.**

Let $W\in\mathcal{A}(\square_{2},V(K)^{+})$. If $W$ is degenerate, then it is ETB.

Proof.

$W$ is degenerate iff $\dim(W(\square_{2}))\leq 1$. If the dimension is 0, then $W$ is constant, hence clearly ETB. Assume that the dimension is 1, then all vertices $w_{i,j}$ of $W$ lie on a segment. It is easy to see that we may find a decomposition as in Proposition B.1 using multiples of the endpoints of the segment, so that $W$ is ETB.

∎

Corollary 1**.**

Assume that $W$ is non-degenerate and each vertex is extremal in $V(K)^{+}$. Then $W$ is non-ETB and extremal in $\mathcal{A}(\square_{2},V(K)^{+})$.

Proof.

Any ETB map which is extremal in $\mathcal{A}(\square_{2},V(K)^{+})$ has the form $\mathsf{m}^{i}_{j}(\cdot)\phi$ for some $i,j\in\{0,1\}$ and $\phi$ an extremal element in $V(K)^{+}$. Such a map is clearly degenerate. It is therefore enough to show that $W$ is extremal. Since $w_{i,j}$ are extremal in $V(K)^{+}$, $\dim(L_{i,j})=1$ for all $i,j$. If (9) or (10) is not satisfied, then it is easy to see that $\dim(W(\square_{2})=1$. Since $W$ is non-degenerate, this is impossible.

∎

Corollary 2**.**

Let $\dim(K)=2$ and assume that $W\in\mathcal{A}(\square_{2},V(K)^{+})$ is non-ETB. Then $W$ is extremal if and only if all its vertices are extremal in $V(K)^{+}$.

Proof.

We will use the notation of Proposition 5. Assume that $W$ is extremal and that, say, $w_{0,1}=0$. Then $w_{0,0}+w_{1,1}=w_{0,1}=2\bar{w}$ and $F_{0,0},F_{1,1}\subseteq F_{0,1}$. By (10), we must have $L_{0,1}=\mathbb{R}\bar{w}$, so that $\phi$ must be extremal in $V(K)^{+}$. Consequently, both $w_{0,0}$ and $w_{1,1}$ are multiples of $\bar{w}$, but then $W$ is degenerate and hence ETB. It follows that all vertices must be nonzero. Then it follows from (9), (10) by dimension counting that we must have $\dim(L_{i,j})=1$, so that all vertices are extremal in $V(K)^{+}$. The converse is Corollary 1. ∎

Example 10*.*

(The square) Since $\dim(\square_{2})=2$, all non-ETB extremal maps must have extremal vertices. Therefore, $w_{0,0}$ and $w_{1,1}$ must be some multiples of opposite vertices, similarly $w_{0,1}$ and $w_{1,0}$ must be multiples of the other pair of opposite vertices. Applying effects $\mathsf{m}^{i}_{j}$ to the equality (8), we see that all coefficients must be the same. It follows that $W$ is (a multiple of) an automorphism of $\square_{2}$. Hence there are 8 extremal rays in $\mathcal{A}(\square_{2},V(\square_{2})^{+})$ that are non-ETB. It is easy to see that elements in these rays are witnesses, since they have no nontrivial translations along $\square_{2}$.

Example 11*.*

(Quantum state spaces) Let $W\in\mathcal{A}(\square_{2},B(\mathcal{H})^{+})$ be extremal and let $\rho=\bar{w}$ be the barycenter. Let $P=\mathrm{supp}(\rho)$ be the support projection of $\rho$. Let $0\leq E_{i,j}\leq P$ be effects such that $\frac{1}{2}w_{i,j}=\rho^{1/2}E_{i,j}\rho^{1/2}$. We will show that all $E_{i,j}$ are projections. Indeed, let $M$ be an effect majorized by both $E_{0,0}$ and $I-E_{0,0}$. Then $\sigma:=\rho^{1/2}M\rho^{1/2}\leq w_{0,0},w_{1,1}$, so that $\sigma\in L_{0,0}\cap L_{1,1}=\{0\}$. Since $M\leq P$, it follows that $M=0$, so that $E_{0,0}\wedge(I-E_{0,0})=0$ and this implies that $E_{0,0}$ is a projection. Then $E_{1,1}=P-E_{0,0}$ is a projection as well, orthogonal to $E_{0,0}$. Similarly for $E_{0,1}$ and $E_{1,0}$.

Let $A\in B_{h}(P\mathcal{H})$ be such that $A$ commutes with both $E_{0,0}$ and $E_{0,1}$. Then $\rho^{1/2}AE_{i,j}\rho^{1/2}\in L_{i,j}$ and $\rho^{1/2}A\rho^{1/2}\in(L_{0,0}+L_{1,1})\cap(L_{0,1}+L_{1,0})$. By (10), this implies $\rho^{1/2}A\rho^{1/2}=t\rho$ for some $t\in\mathbb{R}$, so that $A=tP$. Hence any element commuting with both $E_{0,0}$ and $E_{1,1}$ must be a multiple of $P$. But the commutant of two projections is always nontrivial, unless one of the following two cases occurs:

1. (a)

$P$ is rank one and all $E_{i,j}$ are either $P$ or 0. Then $\rho$ is rank one and $W=\mathsf{m}^{i}_{j}(\cdot)\rho$ for some $i,j\in\{0,1\}$, these are precisely the ETB extremal maps. 2. (b)

$P$ is rank 2 and $E_{0,0}$ and $E_{0,1}$ are rank one non-commuting projections. In that case, $\rho$ is rank 2 and all $w_{i,j}$ are rank 1 operators.

It follows that a non-ETB map $W\in\mathcal{A}(\square_{2},B(\mathcal{H})^{+})$ is extremal if and only if all its vertices are extremal.

IV.2 Incompatibility degree

Incompatibility degree of a collection of measurements can be defined as the least amount of noise that has to be added to obtain a compatible collection. Following Heinosaari et al. (2016), the noise will have the form of coin-toss measurements, see Busch et al. (2013); Cavalcanti and Skrzypczyk (2016); Wolf et al. (2009) for related definitions.

For $p\in\Delta_{l}$, a coin-toss measurement is defined as a constant map $f_{p}(x)\equiv p$. It is immediate that $f_{p}$ is compatible with any $g\in\mathcal{A}(K,\Delta_{l})$ (since the map $(g,f_{p})$ factorizes through $\Delta_{l}\times\Delta_{0}\simeq\Delta_{l}$). Let us again fix $l_{0},\dots,l_{k}\in\mathbb{N}$ and let $p^{i}\in\Delta_{l_{i}}$, $0=1,\dots,k$. Then the channel given by the collection of coin-tosses $(f_{p^{0}},\dots,f_{p^{k}})$ is the constant map $F_{s}(x)\equiv s:=(p^{1},\dots,p^{k})\in\mathsf{S}$.

For $F\in\mathcal{A}(K,\mathsf{S})$ and $s\in\mathsf{S}$, we define the incompatibility degree as

[TABLE]

We also define

[TABLE]

We first show that $ID(F)$ is attained at an interior point of $\mathsf{S}$.

Lemma 3**.**

$ID(F)=\inf_{s\in ri(\mathsf{S})}ID_{s}(F)$.

Proof.

Let $s_{0}\in\partial\mathsf{S}$ and let $s_{1}\in ri(\mathsf{S})$, then $s_{t}:=ts_{1}+(1-t)s_{0}\in ri(\mathsf{S})$ for all $t\in(0,1]$. Put $\mu:=\frac{\lambda(1-t)}{1-\lambda t}$. We have $F_{s_{t}}=tF_{s_{1}}+(1-t)F_{s_{0}}$ and

[TABLE]

Assume that $ID_{s_{0}}(F)=\mu$, then $(1-\mu)F+\mu F_{s_{0}}$ is ETB, so that $(1-\lambda)F+\lambda F_{s_{t}}$ is ETB as well. It follows that

[TABLE]

Letting $t\to 0$ implies the result.

∎

The next result shows that the incompatibility degree can be obtained using incompatibility witnesses. Note that the minimum $q_{s}(F)$ below is attained at an extremal element in $\mathcal{A}(\mathsf{S},V(K)^{+})$.

For pairs of two-outcome measurements, the following expression for incompatibility degree is related to the dual linear program of Plávala (2016); Jenčová and Plávala (2017). In the quantum case, similar results using SDP were obtained in Wolf et al. (2009).

Proposition 6**.**

Let $s\in ri(\mathsf{S})$. Let us denote

[TABLE]

and for $F\in\mathcal{A}(K,\mathsf{S})$,

[TABLE]

Then

[TABLE]

For the proof, we will need the following lemma.

Lemma 4**.**

Let $(V,V^{+})$ be an ordered vector space and $u\in V$ an order unit. Let $\mathfrak{S}(V,V^{+},u)=\{\sigma\in(V^{+})^{*},\ \langle\sigma,u\rangle=1\}$. Then for $v\in V$,

[TABLE]

Proof.

Let $t_{0}$ denote the infimum on the LHS and $s_{0}$ the supremum on the RHS. Let $t\in\mathbb{R}$ be such that $v+tu\in V^{+}$. Then for any $\sigma\in\mathfrak{S}(V,V^{+},u)$, we have

[TABLE]

so that $-\langle\sigma,w\rangle\leq t$. It follows that $s_{0}\leq t_{0}$. Conversely, note that

[TABLE]

for all $\sigma\in\mathfrak{S}(V,V^{+},u)$, hence for all elements in $(V^{+})^{*}$, this implies $t_{0}\leq s_{0}$.

∎

Proof of Proposition 6.

Note that $F_{s}=1_{K}(\cdot)s$ is an interior element in the cone $\mathcal{A}_{sep}(K,V(\mathsf{S})^{+})$, hence an order unit and we have $\mathrm{Tr}\,F_{s}W=\langle 1_{K},W(s)\rangle$. Clearly, if $q_{s}(F)>0$ then $F$ is compatible. Otherwise, by the above Lemma, $-q_{s}(F)$ is the smallest $t\geq 0$ such that $F+tF_{s}$ is ETB, so that $ID_{s}(F)=\frac{-q_{s}(F)}{1-q_{s}(F)}$.

∎

We next use the pair of dual bases in (3) and (4) to find a suitable expression for $\mathrm{Tr}\,FW$. To shorten the notations, put $w^{i}_{j}:=w_{l_{0},\dots,l_{i-1},j,l_{i+1},\dots,l_{k}}$, note that $w^{i}_{l_{i}}=w_{l_{0},\dots,l_{k}}$ for all $i$. Then

[TABLE]

We have

[TABLE]

Using this, we obtain another characterization of witnesses for two-outcome measurements. Recall that $\mathsf{e}^{0}_{0}$,…,$\mathsf{e}^{k}_{0}$ are edges adjacent to the vertex $\mathsf{s}_{1,\dots,1}$.

Corollary 3**.**

Let $W\in\mathcal{A}(\square_{k+1},V(K)^{+})$. Then $W$ is a witness if and only if

[TABLE]

Proof.

We may assume that $\langle 1_{K},\bar{w}\rangle=1$, so that $W\in\mathcal{W}_{\bar{\mathsf{s}}}$. Put $\mu^{i}_{j}:=\langle 1_{K},w^{i}_{j}\rangle$. Then for any $F\in\mathcal{A}(K,\square_{k+1})$, we have using (12)

[TABLE]

$W$ is a witness if and only if

[TABLE]

The last equality follows from the fact that $\max_{f\in E(K)}\langle f,\psi\rangle=\frac{1}{2}(\|\psi\|_{K}+\langle 1_{K},\psi\rangle)$ for all $\psi\in V(K)$.

∎

Observe that by the above proof, the maximal value of $ID_{\bar{\mathsf{s}}}$ attainable by $k+1$ two-outcome measurements on a state space $K$ is obtained from the maximal value of $\sum_{i=0}^{k}\|W(\mathsf{e}^{i}_{0})\|_{K}$ over all maps $W\in\mathcal{A}(\square_{k+1},V(K)^{+})$ with $\bar{w}\in K$. For $k=1$, the maximal value of $ID_{\bar{\mathsf{s}}}$ for quantum state spaces was obtained in Banik et al. (2013). We prove this result by our method and show that $ID$ attains the same value.

Corollary 4**.**

For a quantum state space $\mathfrak{S}=\mathfrak{S}(\mathcal{H})$, we have

[TABLE]

Proof.

Let $F\in\mathcal{A}(\mathfrak{S},\square_{2})$ be incompatible and let $W\in\mathcal{W}_{\bar{\mathsf{s}}}$ be such that $\mathrm{Tr}\,FW=q_{\bar{\mathsf{s}}}(F)$. We may assume that $W$ is extremal in $\mathcal{A}(\square_{2},B(\mathcal{H})^{+})$, so by Example 11, there are some unit vectors $x_{i,j}\in\mathcal{H}$ and a rank 2 density operator $\rho$ with support projection $P$ such that

[TABLE]

and the vertices of $W$ satisfy $\frac{1}{2}w_{i,j}=\rho^{1/2}|x_{i,j}\rangle\langle x_{i,j}|\rho^{1/2}$. By Hölder’s inequality, we have

[TABLE]

Similarly,

[TABLE]

Since $\langle x_{0,1},x_{1,0}\rangle=0$, we have $c^{2}+d^{2}=1$ and hence $c+d\leq\sqrt{2}$. By the proof of Corollary 3, it follows that

[TABLE]

On the other hand, let $W_{0}$ have vertices $\rho_{i,j}=|x_{i,j}\rangle\langle x_{i,j}|$, with unit vectors $x_{i,j}$ such that $|\langle x_{0,1}|x_{1,1}\rangle|=|\langle x_{1,0}|x_{1,1}\rangle|=\sqrt{2}/2$. In this case, $W(s)\in\mathfrak{S}$ for any $s\in\square_{2}$. Let $F_{0}=(f^{0},f^{1})$ be determined by the effects

[TABLE]

Then we have

[TABLE]

It follows that $\max_{F}ID_{\bar{\mathsf{s}}}(F)=1-\frac{1}{\sqrt{2}}$, this corresponds to the results obtained in Banik et al. (2013). Note further that the witness $W_{0}\in\mathcal{W}_{s}$ for any $s\in ri(\square_{2})$. We obtain

[TABLE]

It follows that

[TABLE]

so that $ID_{\bar{\mathsf{s}}}(F_{0})=ID(F_{0})$. We then have

[TABLE]

∎

IV.3 Maximally incompatible measurements

Let $\mathsf{S}$ be any polysimplex and let $F=(f^{0},\dots,f^{k})\in\mathcal{A}(K,\mathsf{S})$ be a collection of measurements. It is well known that

[TABLE]

for any $s\in\mathsf{S}$ (see e.g. Heinosaari et al. (2016)): the joint measurement for $\tfrac{1}{k+1}F+\tfrac{k}{k+1}F_{s}$ can be defined by choosing one of the measurements uniformly at random and replace all other measurements by the corresponding coin tosses. If $ID(F)=\frac{k}{k+1}$, we say that $F$ is maximally incompatible. We now give a general characterization of maximal incompatibility.

Theorem 3**.**

Let $F\in\mathcal{A}(K,\mathsf{S})$. Then the following are equivalent.

1. (i)

$F$* is maximally incompatible.* 2. (ii)

$ID_{s}(F)=\frac{k}{k+1}$* for all $s\in\mathsf{S}$.* 3. (iii)

$ID_{s}(F)=\frac{k}{k+1}$* for some $s\in ri(\mathsf{S})$.* 4. (iv)

There is some $W\in\mathcal{A}(\mathsf{S},K)$ such that $\mathrm{Tr}\,FW=-k$. 5. (v)

There is some $W\in\mathcal{A}(\mathsf{S},K)$ such that

[TABLE]

Proof.

(i) $\implies$ (ii) follows from the definition of $ID(F)$ and the fact that $ID_{s}(F)\leq\frac{k}{k+1}$ for all $s$, (ii) $\implies$ (iii) is trivial. Assume (iii), then by Proposition 6 there is some $W\in\mathcal{W}_{s}$ such that $\mathrm{Tr}\,FW=q_{s}(F)=-k$. For any $n_{0},\dots,n_{k}$, choose the pair of dual bases of $V(\mathsf{S})$ and $A(\mathsf{S})$ by fixing the vertex $\mathsf{s}_{n_{0},\dots,n_{k}}$ as in Remark 1, then exactly as in (12), we obtain

[TABLE]

This implies that $\langle 1_{K},w_{n_{0},\dots,n_{k}}\rangle\geq 1$ for all $n_{0},\dots,n_{k}$. On the other hand, since $W(s)\in K$ is a convex combination of all $w_{n_{0},\dots,n_{k}}$ with nonzero coefficients, we must have $\langle 1_{K},w_{n_{0},\dots,n_{k}}\rangle=1$, hence (iv) holds. Further, if $W$ is as in (iv), the inequality in the above computation must be an equality, so that $\langle f^{i}_{j},w_{n_{0},\dots,n_{i-1},j,n_{i+1},\dots,n_{k}}\rangle=0$ for all $i$ and $j$, hence (v) holds.

Assume (v), then $W(s)\in K$ for any $s\in\mathsf{S}$ and $\mathrm{Tr}\,FW=-k$ by (12). It follows that $ID_{s}(F)$ is maximal for all $s\in ri(\mathsf{S})$. By Lemma 3, this implies (i).

∎

Maximal incompatibility has a nice geometric interpretation for two-outcome measurements. For $k=1$ the following results were essentially proved in Jenčová and Plávala (2017). Recall the definition of retraction-section pairs in Section III.

Corollary 5**.**

Let $F\in\mathcal{A}(K,\square_{k+1})$. Then $F$ is maximally incompatible if and only if $F$ is a retraction.

Proof.

Assume that $F$ is a retraction and let $S:\mathsf{S}\to K$ be the corresponding section. Let $U$ be the automorphism of $\mathsf{S}$ given as $U(\mathsf{s}_{n_{0},\dots,n_{k}})=\mathsf{s}_{1-n_{0},\dots,1-n_{k}}$ and put $W=SU$. Then $W\in\mathcal{A}(\mathsf{S},K)$ and we have

[TABLE]

Hence $F$ is maximally incompatible. Conversely, assume that $F$ is maximally incompatible and let $W$ be the witness as in Thm. 3 (v). Observe that then

[TABLE]

it follows that $FW=U$. Putting $S=WU$ we obtain $FS=id_{\mathsf{S}}$, so that $F$ is a retraction.

∎

Corollary 6**.**

There exist $k+1$ maximally incompatible two-outcome measurements on $K$ if and only if there exists a projection $K\to K$ whose range is affinely isomorphic to $\square_{k+1}$.

Proof.

If $F$ is maximally incompatible, then by Corollary 5 there is a section $S\in\mathcal{A}(\square_{k+1},K)$ such that $FS=id$. It follows that $P:=SF$ is a projection $K\to K$ such that $PS=S$ and $FP=F$, hence the restriction of $F$ to the range of $P$ is an isomorphism onto $\square_{k+1}$ whose inverse is $S$.

Conversely, assume that $P:K\to K$ is such a projection and let $U:P(K)\to\square_{k+1}$ be the isomorphism onto the cube. Then $F=UP$ is obviously a retraction, since then $FU^{-1}=UPU^{-1}=UU^{-1}=id_{\square_{k+1}}$.

∎

Example 12*.*

(Maximal incompatibility in $\mathsf{S}$) We will show that any collection of effects $\mathsf{m}^{0}_{n_{0}},\dots,\mathsf{m}^{k}_{n_{k}}$ with $n_{i}\in\{0,\dots,l_{i}\}$ is maximally incompatible. Indeed, let $F\in\mathcal{A}(\mathsf{S},\square_{k+1})$ be the corresponding channel and let $S\in\mathcal{A}(\square_{k+1},\mathsf{S})$ be determined by the collection of measurements $t^{0},\dots,t^{k}$, given by

[TABLE]

for some $n_{i}^{\prime}\neq n_{i}$. Then $FS=id_{\square_{k+1}}$, so that $F$ is a retraction. Note that this also implies that the projections $\mathsf{m}^{0},\dots,\mathsf{m}^{k}$ of $\mathsf{S}$ are maximally incompatible as well, since they determine the identity map $id_{\mathsf{S}}$ and if $W$ is a witness as in Theorem 3 (iv), then $\mathrm{Tr}\,id_{\mathsf{S}}FW=\mathrm{Tr}\,FW=-k$ and $FW\in\mathcal{A}(\mathsf{S},\mathsf{S})$.

We next show that maximally incompatible measurements exist in the space of quantum channels, cf. Sedlák et al. (2016); Jenčová and Plávala (2017). This result is a simple consequence of Proposition 4 and Example 12, see also Remark 2.

Corollary 7**.**

There exists a maximally incompatible collection of $d_{A}$ two-outcome measurements on $\mathcal{C}_{A,A^{\prime}}$.

Proof.

Let $R\in\mathcal{A}(\mathcal{C}_{A,A^{\prime}},\Delta^{d_{A}}_{d_{A^{\prime}}-1})$ be the retraction as in Proposition 4 and let $F\in\mathcal{A}(\Delta^{d_{A}}_{d_{A^{\prime}}-1},\square_{d_{A}})$ be as in Example 12. Then $F$ is a retraction, so that $FR\in\mathcal{A}(\mathcal{C}_{A,A^{\prime}},\square_{d_{A}})$ is a retraction as well.

∎

Remark 4*.*

As in the above proof, any retraction $K\to\mathsf{S}$ is maximally incompatible. On the other hand, let $F^{\prime}=(f^{0},\dots,f^{k})$ be such that the collection $G$ of two-outcome measurements determined by the effects $f^{0}_{n_{0}},\dots,f^{k}_{n_{k}}$ is maximally incompatible. Then $F^{\prime}$ is maximally incompatible as well. Indeed, with the notation of Example 12, $G=FF^{\prime}$. Let $W^{\prime}$ be the witness $\square_{k+1}\to K$ such that $\mathrm{Tr}\,GW^{\prime}=-k$, then $W^{\prime}F:\mathsf{S}\to K$ is a witness demonstrating maximal incompatibility of $F^{\prime}$. Since the other effects are not involved, $F^{\prime}$ is not necessarily a retraction, so that Corollary 5 cannot be extended to all polysimplices.

V Steering and nonlocality

Quantum steering refers to the property of entangled quantum states which allows one to ”steer” the state of one component by choosing suitable measurements on the other Schrödinger (1936). A rigorous operational definition was given in Wiseman et al. (2007) and can be easily rephrased in the setting of GPT.

V.1 Steering in GPT

Let $K_{A},K_{B}$ be state spaces and let $y\in K_{A}\widetilde{\otimes}K_{B}$ be a joint state. If a measurement $f_{A}\in\mathcal{A}(K_{A},\Delta_{n})$ is applied on system $A$, then $y$ is mapped onto some element $(f_{A}\otimes id_{B})(y)\in\Delta_{n}\otimes K_{B}$. It means that there are some states $x_{j}\in K_{B}$ and a probability measure $p\in\Delta_{n}$ such that $(f_{A}\otimes id_{B})(y)=\sum_{j=0}^{n}p(j)\delta_{j}\otimes x_{j}$. This has the interpretation that with probability $p(j)$, the outcome $j$ is observed on $A$ and the state of $B$ turns into $x_{j}$. The collection $\{p(j),x_{j}\}$ of states and probabilities is called an ensemble. If $B$ has no information about the outcome, the state of $B$ is just the average state $\sum_{j}p(j)x_{j}=(1_{A}\otimes id_{B})(y)=:y_{B}$.

Assume now that an observer on the system $A$ can choose from a collection of measurements $f^{i}_{A}\in\mathcal{A}(K_{A},\Delta_{l_{i}})$. Then we obtain a set of ensembles $\{p(j|i),x_{j|i}\}$ with a common average state $y_{B}$. Such a set is called an assemblage. According to Wiseman et al. (2007), an assemblage does not demonstrate steering if there is a (finite) set $\Lambda$ of ”classical messages” distributed according to a probability measure $q$, corresponding elements $\{x_{\lambda}\in K_{B},\ \lambda\in\Lambda\}$ and conditional probabilities $q(j|i,\lambda)$ such that

[TABLE]

In this case, the assemblage can be explained by a local hidden state model, see Wiseman et al. (2007) for more details. The next result shows that steering can be conveniently expressed in terms of the minimal and maximal tensor products of compact convex sets.

Theorem 4**.**

Let $K$ be a state space and $\mathsf{S}$ a polysimplex. Let $\beta\in\mathsf{S}\widehat{\otimes}K$.

1. (i)

There is an assemblage $\{p(j|i),x_{j|i},j=0,\dots,l_{i},i=0,\dots,k\}$ of elements in $K$ with average state $\sum_{j}p(j|i)x_{j|i}=x\in K$, such that

[TABLE] 2. (ii)

The assemblage in (i) does not demonstrate steering if and only if $\beta$ is separable.

Moreover, any element of the form (14) is in $\mathsf{S}\widehat{\otimes}K$.

Proof.

Using the basis (4), we have

[TABLE]

for some $\phi,\phi^{i}_{j}\in V(K)$. By definition of $\mathsf{S}\widehat{\otimes}K$, we must have $\langle\beta,\mathsf{m}^{i}_{j}\otimes f\rangle\geq 0$ for all $i,j$ and $f\in E(K)$. For $j\neq l_{i}$ this is true if and only if $\phi^{i}_{j}\in V(K)^{+}$. We also have for all $i$ and $f\in E(K)$

[TABLE]

hence $\sum_{j}\phi^{i}_{j}\leq\phi\in V(K)^{+}$ and $\langle\beta,1_{\mathsf{S}}\otimes 1_{K}\rangle=\langle\phi,1_{K}\rangle=1$. Put $\phi^{i}_{l_{i}}:=\phi-\sum_{j=0}^{l_{i}-1}\phi^{i}_{j}$ and $p(j|i):=\langle 1_{K},\phi^{i}_{j}\rangle$, $x_{j|i}:=p(j|i)^{-1}\phi^{i}_{j}$ (if $p(j|i)>0$, otherwise $x(j|i)$ can be anything) for all $i$ and $j$. Then $\{p(j|i),x_{j|i}\}$ is an assemblage with average state $x:=\phi$. This proves (i).

For (ii), assume that (13) holds. Then $x=\sum_{\lambda}q(\lambda)x_{\lambda}$ and we have

[TABLE]

where $s_{\lambda}:=\mathsf{s}_{l_{0},\dots,l_{k}}+\sum_{i=0}^{k}\sum_{j=0}^{l_{i}-1}q(j|i,\lambda)\mathsf{e}^{i}_{j}\in\mathsf{S}$, this follows from $\langle\mathsf{m}^{i}_{j},s_{\lambda}\rangle=q(j|i,\lambda)\geq 0$ and $\langle 1_{\mathsf{S}},s_{\lambda}\rangle=\langle 1_{\mathsf{S}},\mathsf{s}_{l_{0},\dots,l_{k}}\rangle=1$. Hence $\beta$ is separable. Conversely, let

[TABLE]

for some $\alpha_{n_{0},\dots,n_{k}}\in V(K)^{+}$. Put $\Lambda:=\{(n_{0},\dots,n_{k}),n_{i}=0,\dots,l_{i}\}$, $q(n_{0},\dots,n_{k}):=\langle 1_{K},\alpha_{n_{0}\dots n_{k}}\rangle$, $x_{n_{0},\dots,n_{k}}:=q(n_{0},\dots,n_{k})^{-1}\alpha_{n_{0}\dots n_{k}}$, and

[TABLE]

Then for all $i$ and $j$,

[TABLE]

This proves (ii). The last statement follows from the proof of (i).

∎

In view of the preceding theorem, any element in $\mathsf{S}\widehat{\otimes}K$ will be called an assemblage. The common average state $x$ will be called the barycenter of $\beta$.

It is already known that there is no steering if $y$ is separable or the measurements $f^{i}_{A}$ are compatible. This can be seen immediately from Theorem 4. Let $F_{A}=(f^{0}_{A},\dots,f^{k}_{A})$, then $\beta=(F_{A}\otimes id_{B})(y)\in\mathsf{S}\widehat{\otimes}K_{B}$ is the associated assemblage. If $y\in K_{A}\otimes K_{B}$, then $\beta$ must be separable and hence does not demonstrate steering. If the measurements are compatible, then $\beta$ is separable by Theorem 1.

In general, not all assemblages in $\mathsf{S}\widehat{\otimes}K_{B}$ are obtained from some collection $F_{A}$ and a bipartite state $y\in K_{A}\widetilde{\otimes}K_{B}$. If this is the case, we say that the state space $K_{B}$ admits steering. Note that quantum state spaces satisfy this condition. Somewhat stronger conditions were studied in Barnum et al. (2013) and their relations to homogeneity and weak self-duality of the state spaces were found.

V.2 Steering degree

Let $\beta\in\mathsf{S}\widehat{\otimes}K$ be an assemblage with barycenter $x$. A steering degree can be defined similarly as incompatibility degree, as the smallest amount of noise that has to be added to $\beta$ to obtain a separable element. For the noise, we use assemblages of the form $s\otimes x$ for $s\in\mathsf{S}$. This is a separable assemblage with $p(j|i)=\mathsf{m}^{i}_{j}(s)$ and $x_{j|i}=x$ for all $i,j$. We put

[TABLE]

and

[TABLE]

Observe that for any $F_{A}\in\mathcal{A}(K_{A},\mathsf{S})$ and $y\in K_{A}\widetilde{\otimes}K_{B}$, we have

[TABLE]

To see this, note that the barycenter of $(F_{A}\otimes id_{B})(y)$ is the marginal $y_{B}$ and $s\otimes y_{B}=(F_{s}\otimes id_{B})(y)$. We therefore have

[TABLE]

and this is separable if $(1-\lambda)F_{A}+\lambda F_{s}$ is ETB. The possibility of attaining equality depends on the properties of $K_{A}$ and the form of composite state spaces in the GPT. Assume that $V(K_{A})^{+}$ is weakly self-dual, so that there is an affine isomorphism $\Psi:A(K_{A})^{+}\to V(K_{A})^{+}$. With the notations of Appendix A, we have $(id\otimes\Psi)(\chi_{K_{A}})\in K_{A}\widehat{\otimes}K_{A}$, see Lemma A.1. We are now prepared to state the following result.

Theorem 5**.**

Assume that there is an isomorphism $\Psi:A(K_{A})^{+}\to V(K_{A})^{+}$ such that $(id\otimes\Psi)(\chi_{K_{A}})\in K_{A}\widetilde{\otimes}K_{A}$. Then for any polysimplex $\mathsf{S}$, $F_{A}\in\mathcal{A}(K,\mathsf{S})$ and $s\in\mathsf{S}$, we have

[TABLE]

Proof.

By (15), the supremum on the left is never larger than $ID_{s}(F_{A})$. Put $y=(id\otimes\Psi)(\chi_{K_{A}})$ and let $F_{\lambda}:=(1-\lambda)F_{A}+\lambda F_{s}$. Then $(F_{\lambda}\otimes id)(y)$ is separable if and only if $(F_{\lambda}\otimes id)(\chi_{K_{A}})$ is separable, which by Proposition A.1 (ii) means that $F_{\lambda}$ is ETB. It follows that $SD_{s}((F_{A}\otimes id_{A})(y))\geq ID_{s}(F_{A})$.

∎

The conditions in the previous theorem are fulfilled in quantum state spaces, where $y$ is a pure maximally entangled state. In this case, this result was proved in (Cavalcanti and Skrzypczyk, 2016, pp. 8-9).

Remark 5*.*

Similarly as for incompatibility, we may define steering witnesses and their relation to steering degree, maximal steering degree, etc. The witnesses will now be elements in $A(\mathsf{S}\otimes K)^{+}$. We will not investigate this here, only remark that since the assemblages generate all of the positive cone $V(\mathsf{S}\widehat{\otimes}K)^{+}$, any non-separable element in $A(\mathsf{S}\otimes K)^{+}$ is a witness.

V.3 Nonlocality and Bell’s inequalities

Let $f^{i}_{A}\in\mathcal{A}(K_{A},\Delta_{l^{A}_{i}})$, $i=0,\dots,k_{A}$ and $f^{i}_{B}\in\mathcal{A}(K_{B},\Delta_{l^{B}_{i}})$, $i=0,\dots,k_{B}$, and let $y\in K_{A}\widetilde{\otimes}K_{B}$. If a measurement $f^{i_{A}}_{A}$ is chosen for $A$ and $f^{i_{B}}_{B}$ for $B$, then the result is a pair $(j_{A},j_{B})$ with probability

[TABLE]

These conditional probabilities satisfy the no-signalling properties

[TABLE]

where $p_{A}(j_{A}|i_{A}):=\langle(f^{i_{A}}_{A})_{j_{A}},y_{A}\rangle$, $p_{B}(j_{B}|i_{B}):=\langle(f^{i_{B}}_{B})_{j_{B}},y_{B}\rangle$. Following Wiseman et al. (2007), we say that the state $y$ is Bell local if for all measurements $f^{i}_{A}$ and $f^{i}_{B}$, these probabilities admit a local hidden variable (LHV) model, that is, there is a probability distribution $q$ on a set $\Lambda$ and conditional probabilities $q_{A}(j_{A}|i_{A},\lambda)$ and $q_{A}(j_{B}|i_{B},\lambda)$ such that

[TABLE]

Let $F_{A}=(f_{A}^{0},\dots,f_{A}^{k_{A}})$ and let $\mathsf{S}_{A}$ be the related polysimplex, similarly define $F_{B}$ and $\mathsf{S}_{B}$. Then $\gamma:=(F_{A}\otimes F_{B})(y)\in\mathsf{S}_{A}\widehat{\otimes}\mathsf{S}_{B}$ and

[TABLE]

It can be seen by putting $K_{B}=\mathsf{S}_{B}$ in Theorem 4 that the elements $\gamma\in\mathsf{S}_{A}\widehat{\otimes}\mathsf{S}_{B}$ are characterized by the property that $p(j_{A},j_{B}|i_{A},i_{B}):=(\mathsf{m}^{i_{A}}_{j_{A}}\otimes\mathsf{m}^{i_{B}}_{j_{B}})(\gamma)$ are no-signalling conditional probabilities and (18) describes precisely the separable elements. The tensor product $\mathsf{S}_{A}\widehat{\otimes}\mathsf{S}_{B}$ is therefore called the no-signalling polytope and $\mathsf{S}_{A}\otimes\mathsf{S}_{B}$ the local polytope.

The steering witnesses in this case (see Remark 5) will be called Bell witnesses. These are precisely the elements of $A(\mathsf{S}_{A}\otimes\mathsf{S}_{B})^{+}$ that are not separable. With some normalization, there is a finite number of extremal Bell witnesses $\mu_{1},\dots,\mu_{N}$ that completely determine the local polytope: if $\gamma\in\mathsf{S}_{A}\widehat{\otimes}\mathsf{S}_{B}$, then $\gamma$ is local if and only if

[TABLE]

These are the Bell inequalities. Lemma A.1 (iv) shows that, similarly as in the case of incompatibility, the Bell witnesses correspond to affine maps of the polysimplex $\mathsf{S}_{A}$ into a positive cone, this time it is the cone $A(\mathsf{S}_{B})^{+}$. All Bell inequalities are given by extremal non-ETB elements in $\mathcal{A}(\mathsf{S}_{A},A(\mathsf{S}_{B})^{+})$.

Example 13*.*

(The CHSH inequality) Assume that there is a pair of two-outcome measurements on both sides, so that $\mathsf{S}_{A}=\mathsf{S}_{B}=\square_{2}$. Since $V(\square_{2})^{+}\simeq A(\square_{2})^{+}$, we see by Example 10 that all extremal witnesses are precisely (multiples of) the isomorphisms $\Psi_{i,j,k}\in\mathcal{A}(\square_{2},A(\square_{2})^{+})$ that map the extreme points of $\square_{2}$ to the four effects $\mathsf{m}^{i}_{j}$:

[TABLE]

Let $\mu_{i,j,k}$ be the witness corresponding to $\Psi_{i,j,k}$. Using the basis elements (3) and (4), we get

[TABLE]

Let $F_{A}=(f^{0}_{A},f^{1}_{A})\in\mathcal{A}(K_{A},\square_{2})$, $F_{B}=(f^{0}_{B},f^{1}_{B})\in\mathcal{A}(K_{B},\square_{2})$ and put

[TABLE]

then we can see that

[TABLE]

where $\mathbb{B}=\langle a_{1}\otimes(b_{1}+b_{2})+a_{2}\otimes(b_{1}-b_{2}),y\rangle$, so that (19) becomes the CHSH inequality.

V.4 Bell inequalities and the incompatibility degree

The maximal value of $\mathbb{B}$ in Example 13 that can be attained by two-outcome measurements is called the CHSH bound. It is well known that the outcome probabilities satisfy the LHV model (18) if and only if $\mathbb{B}\leq 2$ and we always have $\mathbb{B}\leq 4$. In quantum state spaces, the Tsirelson bound holds: $\mathbb{B}\leq 2\sqrt{2}$. It was observed in Wolf et al. (2009) that the incompatibility degree for pairs of quantum effects is connected to this bound.

The relation of incompatibility degree and CHSH bound in GPT was proved in Busch and Stevens (2014). We include a proof in our setting.

Theorem 6**.**

Let $\mathsf{S}_{A},\mathsf{S}_{B}$ be polysimplices. Let $F_{A}\in\mathcal{A}(K_{A},\mathsf{S}_{A})$, $F_{B}\in\mathcal{A}(K_{B},\mathsf{S}_{B})$, $y\in K_{A}\widetilde{\otimes}K_{B}$ and assume that $F_{A}$ is incompatible. Then for any $\mu\in A(\mathsf{S}_{A}\otimes\mathsf{S}_{B})^{+}$ and $s\in ri(\mathsf{S}_{A})$, we have

[TABLE]

If $K_{A}$ admits steering and $\mathsf{S}_{A}=\square_{2}$, then there is some state space $K_{B}$, $F_{B}\in(K_{B},\square_{2})$ and $y\in K_{A}\widetilde{\otimes}K_{B}$ such that

[TABLE]

for the witness $\mu_{i,j,k}$ as in example 13.

Proof.

By Lemma A.1 (iv), there are some $T\in\mathcal{A}(K_{B}^{\dagger},V(K_{A})^{+})$ and $M\in\mathcal{A}(\mathsf{S}_{A},V(\mathsf{S}_{B}^{\dagger})^{+})$ such that $y=(T\otimes id)(\chi_{K_{B}^{\dagger}})$ and $\mu=(M^{*}\otimes id)(\chi_{\mathsf{S}_{B}})$. Then by using Lemma A.1 (iii),

[TABLE]

Put $W:=TF_{B}^{*}M$, then $W\in\mathcal{A}(\mathsf{S}_{A},V(K_{A}^{+}))$. Moreover,

[TABLE]

where $s^{\prime}:=F_{B}T^{*}(1_{K_{A}})$. It is easy to see that $T^{*}(1_{K_{A}})\in K_{B}$, so that $s^{\prime}\in\mathsf{S}_{B}$. It follows that

[TABLE]

We have $t^{-1}W\in\mathcal{W}_{s}$, so that

[TABLE]

The final inequality follows by the assumption that $F_{A}$ is incompatible, so that $q_{s}(F)<0$.

Assume now that $K_{A}$ admits steering. Let $W\in\mathcal{A}(\square_{2},V(K_{A})^{+})$ be a witness in $\mathcal{W}_{\bar{\mathsf{s}}}$ such that $\mathrm{Tr}\,F_{A}W=q_{\bar{\mathsf{s}}}(F_{A})$. In view of the above proof, it is enough to show that $W=TF_{B}^{*}M$ for suitable $T$, $F_{B}$ and $M$. So let $M=\Psi_{i,j,k}$ be the isomorphism as in Example 13. Put $\beta:=(WM^{-1}\otimes id)(\chi_{\square_{2}^{\dagger}})$, then clearly $\beta\in V(K_{A}\widehat{\otimes}\square_{2})^{+}$ and

[TABLE]

Hence $\frac{1}{2}\beta$ is an assemblage. Since $K_{A}$ admits steering, there is some state space $K_{B}$, $F_{B}\in\mathcal{A}(K_{B},\square_{2})$ and $y\in K_{A}\widetilde{\otimes}K_{B}$ such that

[TABLE]

It follows that $\frac{1}{2}W=TF_{B}^{*}M$, this finishes the proof.

∎

Note that the crucial part of the proof of the equality in the above theorem is that $V(\square_{2})^{+}$ is weakly self-dual, so we may chose $M$ to be an isomorphism. This is not true for any other $\mathsf{S}_{A}$ and $\mathsf{S}_{B}$. As we have seen in the proof, the Bell scenario provides incompatibility witnesses only of the form $W=TF_{B}^{*}M$, that is, factorizing through some $A(\mathsf{S}_{B})^{+}$, which can be weaker for detection of some types of incompatibility. This seems to be the reason why already for three quantum effects, incompatibility in some cases cannot be detected by violation of Bell inequalities. This was observed in Quintino et al. (2014) in the case of qubit states, but the above arguments suggest that such effect exist in any non-classical theory in our class of GPT.

V.5 Nonlocality in spaces of quantum channels

It was proved in Beckman et al. (2001) that one can obtain probabilities maximally violating the CHSH inequality, that is, attaining the value $\mathbb{B}=4$, by using causal bipartite quantum channels. In fact, it was shown recently in Hoban and Sainz (2017) that one can obtain all no-signalling probabilities in this way. In these works, the GPT setting was not used, but nevertheless it was shown that any element of the no-signalling polytope can be obtained by applying sets of channel measurements to both parts of an element of $\mathcal{C}^{caus}_{AB,A^{\prime}B^{\prime}}$. Maximal violation of the CHSH inequality in spaces of quantum channels was also proved using GPT in Plávala and Ziman (2017). Note that the channel used in Beckman et al. (2001); Plávala and Ziman (2017) was a bipartite c-c channel.

The aim of the present paragraph is to remark that this feature of quantum channels is immediate from Proposition 4. Indeed, the isomorphism $\Delta_{n}^{k+1}\simeq\mathcal{A}(\Delta_{k},\Delta_{n})$ also implies that

[TABLE]

In this way, any no-signalling polytope is isomorphic to a face in the space of causal classical bipartite channels and the local polytope corresponds to the local channels in this face (see also Example 7).

Let now $(R_{A},S_{A})$, $(R_{B},S_{B})$ be the retraction-section pairs as in Proposition 4 and let $\gamma\in\Delta^{k_{A}+1}_{n_{A}}\widehat{\otimes}\Delta^{k_{B}+1}_{n_{B}}$ be a collection of no-signalling conditional probabilities. Then $\gamma$ corresponds to a classical causal channel $T_{\gamma}:\Delta_{k_{AB}}\to\Delta_{n_{AB}}$ (cf. the notation in Example 5), given by $T_{\gamma}(j_{A},j_{B}|i_{A},i_{B})=(\mathsf{m}^{i_{A}}_{j_{A}}\otimes\mathsf{m}^{i_{B}}_{j_{B}})(\gamma)$. Put $\Phi:=\Phi_{T_{\gamma}}$ as in (1), choosing product bases in $\mathcal{H}_{A}\otimes\mathcal{H}_{B}$ and $\mathcal{H}_{A^{\prime}}\otimes\mathcal{H}_{B^{\prime}}$. Then $\Phi=(S_{A}\otimes S_{B})(\gamma)\in\mathcal{C}_{A,A^{\prime}}\widehat{\otimes}\mathcal{C}_{B,B^{\prime}}$, but since $\Phi$ is completely positive, we have $\Phi\in\mathcal{C}^{caus}_{AB,A^{\prime}B^{\prime}}=\mathcal{C}_{A,A^{\prime}}\widetilde{\otimes}\mathcal{C}_{B,B^{\prime}}$. Clearly,

[TABLE]

We have proved the following result (cf. Hoban and Sainz (2017)).

Theorem 7**.**

For any collection of no-signalling conditional probabilities $\gamma\in\Delta^{k_{A}+1}_{n_{A}}\widehat{\otimes}\Delta^{k_{B}+1}_{n_{B}}$, there is a causal bipartite quantum channel $\Phi\in\mathcal{C}^{caus}_{AB,A^{\prime}B^{\prime}}$, with $d_{A}=k_{A}+1$, $d_{A^{\prime}}=n_{A}+1$ and $d_{B}=k_{B}+1$, $d_{B^{\prime}}=n_{B}+1$, and collections of measurements $R_{A}\in\mathcal{A}(\mathcal{C}_{A,A^{\prime}},\Delta^{k_{A}+1}_{n_{A}})$ and $R_{B}\in\mathcal{A}(\mathcal{C}_{B,B^{\prime}},\Delta^{k_{B}+1}_{n_{B}})$ such that $(R_{A}\otimes R_{B})(\Phi)=\gamma$.

Example 14*.*

Let $\mathsf{S}_{A}=\mathsf{S}_{B}=\square_{2}$ and choose

[TABLE]

We obtain the same c-c bipartite channel and sets of measurements attaining maximal CHSH violation as in Beckman et al. (2001); Plávala and Ziman (2017).

VI Conclusion and further questions

We have studied incompatibility of measurements in a family of convex finite dimensional GPTs by representing collections of measurements as affine maps into a polysimplex. We have shown how properties of these maps (like being ETB or a retraction) are tied to incompatibility. We introduced incompatibility witnesses and used them to characterize incompatibility degree. Our results suggest that incompatibility is closely related to the geometry of polysimplices, for example, the maximal incompatibility degree attainable for a given state space $K$ can be obtained by considering positive maps on $K$ that factorize through the polysimplex of the given shape.

Our setting allows us to study the relations of incompatibility to steering and non-locality through incompatibility witnesses. We have shown that the Bell scenario provides incompatibility witnesses of a restricted type, more precisely, factorizing through the dual of a polysimplex. In general, the family of such witnesses is strictly smaller than the set of all witnesses and is therefore weaker for the detection of incompatibility. This explains the existence of incompatible collections of measurements that do not violate Bell inequalities and suggests that this feature is not specific for quantum theory, but is common in GPT.

There is a number of questions left for further research. For example, the incompatibility degree attainable by more general collections of quantum measurement can be investigated using witnesses as in Corollary 4. For this, a characterization of extremal maps of a polysimplex into $B(\mathcal{H})^{+}$ would be useful. For the study of relations between incompatibility and non-locality, one could describe the witnesses that can be obtained from Bell inequalities as in Theorem 6. It is an interesting question to what extend are these witnesses weaker and how it depends on the theory in question. It might be also worthwhile to study collections of quantum measurements and their incompatibility, as well as steering and Bell nonlocality, within the framework of quantum networks as suggested in Remark 3.

Maps into a polysimplex can be used to describe collections of measurements only up to a fixed size and number of outcomes. The isomorphism with (faces of) classical channels (Proposition 3) suggests that it might be possible do include all collections of measurements as affine maps into the space of more general Markov kernels.

Appendix A Positive and ETB maps

We list some well known results on the cones of positive maps. Let $K$ be a compact convex set. Let $x_{0},\dots,x_{n}$ be a basis of $V(K)$ and let $e_{0},\dots,e_{n}\in A(K)$ be the dual basis. Put

[TABLE]

Let $f\in A(K)$, then $f=\sum_{i}\langle f,x_{i}\rangle e_{i}$, therefore we have

[TABLE]

Consequently, $\chi_{K}$ does not depend on the choice of the basis.

Lemma A.1**.**

Let $x^{\prime}\in ri(K)$ and put $K^{\dagger}:=\{f\in A(K)^{+},f(x^{\prime})=1\}$. Then

1. (i)

$K^{\dagger}$* is a compact convex set and we have $V(K^{\dagger})^{+}=A(K)^{+}$, $A(K^{\dagger})^{+}=V(K)^{+}$ and $1_{K\dagger}=x^{\prime}$.* 2. (ii)

$\chi_{K}\in K\widehat{\otimes}K^{\dagger}$. 3. (iii)

Let $T\in\mathcal{A}(K,V(K^{\prime}))$, then $(T\otimes id)(\chi_{K})=(id\otimes T^{*})(\chi_{K^{\prime}})$. 4. (iv)

For any $\xi\in V(K^{\prime}\widehat{\otimes}K^{\dagger})^{+}$, there is a unique $T\in\mathcal{A}(K,V(K^{\prime})^{+})$ such that $(T\otimes id)(\chi_{K})=\xi$, determined by

[TABLE]

Proof.

Since $x^{\prime}\in ri(K)$, $f(x^{\prime})=0$ for $f\in A(K)^{+}$ implies that $f=0$, so that $K^{\dagger}$ is a base of $A(K)^{+}$. This proves (i). The statement (ii) follows immediately from (A.2) and (i). For (iii), let $f^{\prime}\in A(K^{\prime})$ and $y\in K$, then by (A.2), we have

[TABLE]

For (iv), it is clear that (A.3) determines an element $T\in\mathcal{A}(K,V(K^{\prime})^{+})$ and $(T\otimes id)(\chi_{K})=\xi$ holds by (A.2).

∎

We now have the following characterizations of ETB maps.

Proposition A.1**.**

Let $T\in\mathcal{A}(K,V(K^{\prime})^{+})$. The following are equivalent.

1. (i)

$T$* is ETB.* 2. (ii)

$(T\otimes id)(\chi_{K})$* is separable.* 3. (iii)

$T$* factorizes through a simplex: there are a simplex $\Delta_{n}$ and maps $T_{0}\in\mathcal{A}(K,V(\Delta_{n})^{+})$ and $T_{1}\in\mathcal{A}(\Delta_{n},V(K^{\prime})^{+})$ such that $T=T_{1}T_{0}$.*

If $T$ is a channel, $T_{0}$ and $T_{1}$ in (iii) may be chosen to be channels as well.

Proof.

(i) $\implies$ (ii) is clear. Assume (ii), then there are some $\phi_{j}\in V(K^{\prime})^{+}$ and $f_{j}\in A(K)^{+}$ such that $(T\otimes id)(\chi_{K})=\sum_{j=0}^{n}\phi_{j}\otimes f_{j}$. By (A.3), we have for $y\in K$, $g\in A(K^{\prime})$,

[TABLE]

so that $T=\sum_{j}f_{j}(\cdot)\phi_{j}$. Let $\delta_{j}$ be the extreme points of $\Delta_{n}$ and put $T_{0}=\sum_{j}f_{j}(\cdot)\delta_{j}$, $T_{1}(\delta_{j})=\phi_{j}$, then $T=T_{1}T_{0}$. (iii) $\implies$ (i) since any $T_{0}\in\mathcal{A}(K,V(\Delta_{n})^{+})$ is ETB, see Example 5.

Assume that $T$ is an ETB channel, so that $T=\sum_{j}f_{j}(\cdot)\phi_{j}$ as above. Let $c_{j}:=\langle\phi_{j},1_{K^{\prime}}\rangle$. We may assume $c_{j}>0$, otherwise we may replace $\Delta_{n}$ by a smaller simplex. Put $\tilde{f}_{j}:=c_{j}f_{j}$ and $\tilde{\phi}_{j}:=c_{j}^{-1}\phi_{j}$, then the corresponding maps $\tilde{T}_{0}$ and $\tilde{T}_{1}$ are channels such that $T=\tilde{T}_{1}\tilde{T}_{0}$.

∎

We next prove Proposition 2. Let $x_{0},\dots,x_{n}\in K$ be a basis of $V(K)$ and $e_{0},\dots,e_{n}$ the dual basis, then

[TABLE]

Proof of Proposition 2.

Let $S\in\mathcal{A}_{sep}(K^{\prime},V(K))$, so that $S=\sum_{j}f^{\prime}_{j}(\cdot)\phi_{j}$ for some $f^{\prime}_{j}\in A(K^{\prime})^{+}$ and $\phi_{j}\in V(K)^{+}$. Then $\mathrm{Tr}\,TS=\sum_{j}\langle T(\phi_{j}),f^{\prime}_{j}\rangle\geq 0$. Conversely, assume that for all $T\in\mathcal{A}(K,V(K^{\prime})^{+})$

[TABLE]

By Lemma A.1 (iv), we have $(S^{*}\otimes id)\chi_{K^{\dagger}}\in A(K\widehat{\otimes}K^{\dagger})^{+}$, so it must be separable. By Proposition A.1, $S^{*}$, and hence also $S$, is ETB.

∎

Appendix B The cones $\mathcal{A}(\mathsf{S},V(K)^{+})$ and $\mathcal{A}_{sep}(\mathsf{S},V(K)^{+})$

We describe the cone of positive maps $\mathcal{A}(\mathsf{S},V(K)^{+})$ and characterize the ETB ones.

Proposition B.1**.**

The elements $w_{n_{0},\dots,n_{k}}\in V(K)^{+}$, $n_{i}=0,\dots,l_{i}$, $i=0,\dots,k$ are vertices of some $W\in\mathcal{A}(\mathsf{S},V(K)^{+})$ if and only if they satisfy

[TABLE]

for all $n_{0},\dots,n_{k}$, $n_{0}^{\prime},\dots,n_{k}^{\prime}$ and $i$. Moreover, $W$ is ETB if and only if there are some $\psi^{i}_{j}\in V(K)^{+}$, $j=0,\dots,l_{i}$, $i=0,\dots,k$, such that

[TABLE]

Proof.

Let $w_{n_{)},\dots,n_{k}}$ be vertices of $W\in\mathcal{A}(\mathsf{S},V(K)^{+})$. We have

[TABLE]

hence (B.4) must hold. Conversely, assume $w_{n_{0},\dots,n_{k}}$ satisfy (B.4) and put

[TABLE]

This determines a map $W\in\mathcal{A}(\mathsf{S},V(K))$. By (5), we have $W(\mathsf{s}_{n_{0},\dots,n_{k}})=w_{l_{0},\dots,l_{k}}+\sum_{i=0}^{k}W(\mathsf{e}^{i}_{n_{i}})$. Using repeatedly the relations (B.4), we get $W(\mathsf{s}_{n_{0},\dots,n_{k}})=w_{n_{0},\dots,n_{k}}$. For the second statement, note that since the effects $\mathsf{m}^{i}_{j}$ generate $A(\mathsf{S})^{+}$, $W$ is ETB if and only if there are $\psi^{i}_{j}\in V(K)^{+}$ such that

[TABLE]

Applying this to the vertices of $\mathsf{S}$, we obtain the statement.

∎

Acknowledgements.

This research was supported by the grants VEGA 2/0069/16 and APVV-16-0073.