Ensemble Steering, Weak Self-Duality, and the Structure of Probabilistic Theories

Abstract

In any probabilistic theory, we say that a bipartite state ω on a composite system AB steers its marginal state ω ^B if, for any decomposition of ω ^B as a mixture ω ^B=∑ _i p _i β _i of states β _i on B, there exists an observable {a _i } on A such that the conditional states $\omega_{B|a_{i}}$ are exactly the states β _i . This is always so for pure bipartite states in quantum mechanics, a fact first observed by Schrödinger in 1935. Here, we show that, for weakly self-dual state spaces (those isomorphic, but perhaps not canonically isomorphic, to their dual spaces), the assumption that every state of a system A is steered by some bipartite state of a composite AA consisting of two copies of A, amounts to the homogeneity of the state cone. If the state space is actually self-dual, and not just weakly so, this implies (via the Koecher-Vinberg Theorem) that it is the self-adjoint part of a formally real Jordan algebra, and hence, quite close to being quantum mechanical.

Examples for Sect. 5

Example A.1

(An example in which there is a section that enables steering)

Consider the following state in the maximal tensor product of two state spaces with square base. We’ll view this as the state space of two two-outcome tests {a,a′} and {b,b′}, and also identify a,a′,b,b′ with the atomic effects in the dual cone. We’ll label each of the four vertices of the normalized state space by the two atomic effects it makes certain: ab,ab′,a′b,a′b′.

Writing, e.g. ab⊗a′b′ for a bipartite product state, the state:

$$ \omega= {\frac{1}{2}} \bigl(ab \otimes ab + a'b \otimes a'b \bigr) $$

(5)

has

$$ \hat{\omega}(a) = {\frac{1}{2}}ab, ~~ \hat{\omega} \bigl(a' \bigr) = {\frac{1}{2}}a'b , $$

(6)

so measuring {a,a′} on the first system gives the ensemble $\{{\frac{1}{2}}ab, {\frac{1}{2}}a'b\}$ for the second-system marginal $\omega^{B} = {\frac{1}{2}}(ab + a'b)$. ${{\rm Face}}(\omega^{B})$ is generated as nonnegative linear combinations of ab and a′b, and is thus a two-dimensional ordered subspace of the three-dimensional state space of the second system. Its unit interval is the square that is the convex hull of (0,0),ab/2,a′b/2,(ab+a′b)/2. The quotient of the first system space, A, by the kernel of the linear map $\hat{\omega}$, can be represented by setting up the state space to have 90^∘ opening angle between opposite extremal rays, and taking the orthogonal (90^∘) projection onto the plane normal to the ray b′, i.e. the projection along the ray generated by b′. This indeed gives a two-dimensional classical state space, i.e. one isomorphic to ${{\rm Face}}(\omega^{B})$. Moreover, there is an affine section σ of this quotient map into $A^{*}_{+}$, given by inverting the relations (6), thus allowing us to map the unit interval [0,ω ^B] in ${{\rm Face}}(\omega^{B})$ to the diagonal cross section, in the a,a′ plane, of the unit interval of $A^{*}_{+}$. σ∘π is indeed a positive projection on $A^{*}_{+}$, namely the orthogonal projection onto this plane.

For the state (5), the mentioned section is the only affine section over ${{\rm Face}}(\omega^{B})$. A slight modification, however, gives an example in which many affine sections exist.

Example A.2

(An example in which there are many affine sections over the image of the order unit)

Let

$$ \hat{\omega}= {\frac{1}{2}} \bigl( x \otimes ab + y \otimes a'b \bigr) $$

(7)

with x any state in the line segment [ab,ab′] and y any state in the segment [a′b,a′b′] respectively. This still gives a state steering for the same marginal $\omega^{B} = {\frac{1}{2}}(ab + a'b)$ as in the preceding example. If we choose x=ab,y=a′b′, that is,

$$ \hat{\omega}= {\frac {1}{2}} \bigl( ab \otimes ab + a'b' \otimes a'b \bigr) , $$

(8)

we can steer the marginal using either of the two observables {a,a′},{b,b′}. In this case, there are two distinct sections over ${{\rm Face}}(\omega^{B})$ that enable steering.

In the example of Eq. (8), the kernel of $\hat{\omega}$ is generated, not by b′ as before, but by a′b−ab′, and the natural way of attempting to represent the quotient map, namely by projection onto a subspace complementary to the kernel in the original space, if we project orthogonally in the natural geometry, projects onto the subspace spanned by ab,a′b′, and the order unit, giving a slice of the order interval that includes the diagonal of the square.

Although the kernel of this map is of course one-dimensional, its positive kernel, which we’ve claimed must be an exposed face of the state cone, is the trivial such exposed face: the zero subspace. This example shows that quotients whose positive kernel is trivial are not necessarily uninteresting or ill-behaved.

Example A.3

(A positive section of the order-quotient map is not necessary for steering)

Let $A \simeq {\mathbb{R}}^{4}$ be an abstract state space whose normalized states form a cube; its dual cone is a regular polyhedral cone in ${\mathbb{R}} ^{4}$ with octahedral base. The atomic effects (largest effects on extremal rays of the effect cone) are the vertices of such an octahedral base. The example has B≃R ³ a cone with regular hexagonal base; the center of the hexagon is a state having three distinct two-state extremal ensembles, each composed of a pair of opposite vertices of the hexagon (with weights 1/2 on each one). $\hat{\omega}^{AB}$ is chosen to take the six vertices of the octahedron of atomic effects in A ^∗ to the six vertices of the hexagon of states normalized to 1/2, in such a way that opposite vertex-pairs are mapped to opposite vertex-pairs. This is clearly positive and linear, and maps the order-unit (twice the center of the octahedron of atomic effects) to the center of the hexagon of normalized states (which is twice the center of the hexagon of states normalized to 1/2). ${{\rm Face}}(\hat{\omega}^{B})$ is the entire hexagonal cone in ${\mathbb{R}}^{3}$, so $\hat{\omega}$ is itself (technically, is a representative of) the quotient π. All three ensembles for ω ^B consist of extremal states, and because they have unique $\hat{\omega}$-preimages, a section of $\hat{\omega}$, we know that a section of π must take these, the vertices of a hexagon, to the vertices of the octahedron, matching opposites to opposites. But no affine map can do this. The affine span of the hexagonal points has affine dimension 2, while that of the octagonal ones has affine dimension 3.