# Entanglement cost for steering assemblages

Thomas Cope<EMAIL_ADDRESS>Institut für Theoretische Physik,
Leibniz Universität Hannover, Appelstr. 2, 30167 Hannover, Germany

###### Abstract

In this paper we answer the question of how one can _quantify_ the amount of
entanglement necessary to create a given steering assemblage, when the
original state and measurements are unknown. To do this, we extend the
concepts of _entanglement cost_ and _entanglement of formation_ to steering
assemblages, and provide easy to calculate upper and lower bounds. We prove
that the entanglement of formation for assemblages is not generally continuous
and not a flat roof extension (in contrast to the state-based quantity); and
use numerical analysis to illustrate these properties. Finally, we discuss the
consequences of these results for assemblage-to-assemblage conversion and
suggest potential applications.

Quantum steering is a quickly growing field in quantum information research,
although its conception can be traced far back into quantum mechanics’
history. Originally explored by Schrödinger [1, 2] to understand the
consequences of the EPR paradox [3], it can be summarised as follows:
measurement on one part of an entangled state can cause the other part of the
state to be “steered” into a preferred decomposition. Although this cannot be
used to transmit information, this is nevertheless a surprising result,
described by Schrödinger himself as both “disconcerting” and “discomforting”
[2].

Interest in quantum steering was revived in 2007 in [4], where the modern
definition was formalised by defining it as the negation of a “local hidden
state” model; similarly to how entanglement is the negation of separable
states, and Bell nonlocality the negation of a “local hidden variable” model.
Steering is often considered “halfway” between entanglement, which doesn’t
consider measurements at all, and nonlocality, which is concerned _only_ with
the outcomes of measurements. Therefore, steering inherits properties from
both of its near neighbours. Steering is also operationally relevant to _one-
sided_ (or semi) Device Independent Quantum Key Distribution (DIQKD) [5, 6,
7], for which steering is the natural resource theory [8]. In this scenario,
one device is treated as a block box, while the other performs trusted quantum
processes. As characterising a device is both expensive in terms of knowledge
and resources, such one-sided protocols are useful between users of asymmetric
capital e.g. a bank and its many customers.
It is known that in order to exhibit the phenomenon of steering, one must
measure an entangled state using incompatible measurements. Thus, observing
the phenomenon of steering implicitly verifies entanglement of the underlying
state. In this paper the goal is to _quantify_ that verification, by asking
how much entanglement is required to prepare a system exhibiting steering.

##  I Assemblages and Entanglement

Mathematically, steering is described via assemblages: a set of subnormalised
substates $\\{\sigma_{n|r}\\}_{n,r}$ such that

$\sigma_{n|r}:=\mathrm{Tr}_{A}\left(M^{A}_{n|r}\otimes\mathbb{I}^{B}\rho^{AB}\right),$
(1)

where $\rho^{AB}$ is a bipartite quantum state, and $M_{n|r}$ is an element of
the Postive Operator Valued Measure (POVM) $\\{M_{n|r}\\}_{n}$ i.e.
$\sum_{n}M_{n|r}=\mathbb{I}^{A}$. Normally, we fix the number of measurements
Alice can perform, and the number of outputs she can have, so that
$r\in\mathcal{R}=\\{0,\ldots|\mathcal{R}|-1\\}$ and
$n\in\mathcal{N}=\\{0,\ldots|\mathcal{N}|-1\\}$. From this definition come the
natural constraints

$\sigma_{n|r}\geq 0\;\forall
n,r\;\;\;\;\sum_{n}\sigma_{n|r}=\mathrm{Tr}_{A}(\rho^{AB})\,=:\rho_{\boldsymbol{\sigma}}^{B},$
(2)

known as _positivity_ and _nosignalling_ respectively. From now on we will
write an assemblage succinctly as
$\boldsymbol{\sigma}:=(\sigma_{0|0},\ldots,\sigma_{(|\mathcal{N}|-1)|0},\ldots,\sigma_{0|(|\mathcal{R}|-1)},\ldots,\sigma_{(|\mathcal{N}|-1)|(|\mathcal{R}|-1)})$.
We will refer to the marginal state of this assemblage as
$\rho_{\boldsymbol{\sigma}}$. In this paper, we are limiting our discussion to
finite (though arbitrary dimension) Hilbert spaces. This will allow us to use
tricks such as Schmidt decomposition and purification.
In the case of bipartite systems, the mapping from all quantum states and
measurements to assemblages is surjective, but not injective; the entangled
Werner state

$W_{1/\sqrt{2}}=\frac{1}{\sqrt{2}}|\Phi^{+}\rangle\langle\Phi^{+}|+\left(1-\frac{1}{\sqrt{2}}\right)\frac{\mathbb{I}}{4}$
(3)

(where $\Phi^{+}=\left(|00\rangle+|11\rangle\right)\sqrt{2}$), along with
measurements

$\displaystyle M_{0|0}$ $\displaystyle=|0\rangle\langle 0|,$ $\displaystyle
M_{0|1}$ $\displaystyle=|+\rangle\langle+|,$ $\displaystyle M_{1|0}$
$\displaystyle=|1\rangle\langle 1|,$ $\displaystyle M_{1|1}$
$\displaystyle=|-\rangle\langle-|,$

creates the assemblage

$\displaystyle\sigma_{0|0}$
$\displaystyle=\frac{1}{2}\left(\mathbb{I}+\frac{1}{\sqrt{2}}\sigma_{z}\right),$
$\displaystyle\sigma_{0|1}$
$\displaystyle=\frac{1}{2}\left(\mathbb{I}+\frac{1}{\sqrt{2}}\sigma_{x}\right),$
$\displaystyle\sigma_{1|0}$
$\displaystyle=\frac{1}{2}\left(\mathbb{I}-\frac{1}{\sqrt{2}}\sigma_{z}\right),$
$\displaystyle\sigma_{1|1}$
$\displaystyle=\frac{1}{2}\left(\mathbb{I}-\frac{1}{\sqrt{2}}\sigma_{x}\right).$

However, so too does the separable state

$\rho^{\mathrm{Sep}}=1/4\sum_{i,j=0}^{1}|ij\rangle^{A}\langle
ij|\otimes\sigma_{ij}^{B},$ (4)

where
$\sigma_{ij}=\frac{1}{2}\left(\mathbb{I}+\frac{(-1)^{j}}{\sqrt{2}}\sigma_{x}+\frac{(-1)^{i}}{\sqrt{2}}\sigma_{z}\right)$,
using measurements

$\displaystyle M^{A}_{n|0}$ $\displaystyle=|n\rangle\langle
n|\otimes\mathbb{I},$ $\displaystyle M^{A}_{n|1}$
$\displaystyle=\mathbb{I}\otimes|n\rangle\langle n|.$

This example in particular highlights the fact that the same assemblage can
arise from states with a different quantity of entanglement. This leads to a
natural question: _how much entanglement is necessary to create a given
assemblage_ $\boldsymbol{\sigma}$? To answer this we define the _entanglement
cost for assemblages_ ,

$\displaystyle E_{\mathrm{CA}}(\boldsymbol{\sigma}):=\inf\bigg{\\{}$
$\displaystyle E\;\bigg{|}\;\forall\,\epsilon>0,\delta>0,\exists\,p,q,$
$\displaystyle\forall\,\mathbf{n}\in\mathcal{N}^{\times
q},\mathbf{r}\in\mathcal{R}^{\times
q},\,\exists\,M_{\boldsymbol{n}|\boldsymbol{r}}\geq 0,\,\Lambda$
$\displaystyle\sum_{\boldsymbol{a}}M_{\boldsymbol{n}|\boldsymbol{r}}=\mathbb{I}^{A},\;|E-\frac{p}{q}|\leq\delta\;\text{and}$
$\displaystyle\sum_{\boldsymbol{n},\boldsymbol{r}}\big{\|}\sigma_{\boldsymbol{n}|\boldsymbol{r}}-\mathrm{Tr}_{A}$
$\displaystyle\left[\left(M_{\boldsymbol{n}|\boldsymbol{r}}\otimes\mathbb{I}^{B}\right)\Lambda(|\Phi^{+}\rangle\langle\Phi^{+}|^{\otimes
p})\right]\big{\|}\leq\epsilon\bigg{\\}}.$ (5)

In this formula $\Lambda$ is a Local Operations and Classical Communications
(LOCC) distillation procedure, used to create a shared state between Alice and
Bob. Alice then measures her system (which may be of arbitrary finite size)
with a measurement whose operators are given by
$M_{\boldsymbol{n}|\boldsymbol{r}}$, with the goal of creating an
approximation to $q$ copies of the steering assemblage. We represent $q$
copies of the steering assemblage by a larger assemblage, in which Alice may
choose as input any $q$ tuple $\mathbf{r}\in\mathcal{R}^{\times q}$,
representing the inputs of each individual component. The possible substates
which can occur as a result of this choice consist of all combinations of
possible substates output by the individual smaller assemblages, in such a way
that the outputs are independent of each other. Thus a possible substate is
represented by

$\sigma_{\boldsymbol{n}|\boldsymbol{r}}=\sigma_{n_{1}|r_{1}}\otimes\sigma_{n_{2}|r_{2}}\otimes\ldots\otimes\sigma_{n_{q}|r_{q}}.$
(6)

A more detailed treatment of this is given in the supplementary material. The
above definition is designed in mind that the entanglement cost for creating
multiple copies of an assemblage may be lower than that of a single
assemblage. Furthermore, we allow the freedom for Alice and Bob to create
“good approximations” of the assemblage, such that in the asymptotic limit
these approximations become exact. This is directly analogous to the
entanglement cost for quantum states,

$\displaystyle E_{\mathrm{C}}(\rho):=\inf\bigg{\\{}$ $\displaystyle
E\mid\forall\epsilon>0,\delta>0,\exists m,n,\Lambda,|E-\frac{p}{q}|\leq\delta$
$\displaystyle\text{ and }\|\rho^{\otimes
q}-\Lambda(|\Phi^{+}\rangle\langle\Phi^{+}|^{\otimes
p})\|\leq\epsilon\bigg{\\}},$ (7)

where $\Lambda$ is again a LOCC distillation procedure.
One key difference between the two definitions is that, when creating
assemblages, it may be that the measurements are only asymptotically correct,
instead of (or as well as) the state. We also address in the supplemental
material, why no post-processing term $\Lambda^{\prime}$ is included after the
measurement.
Calculation of $E_{\mathrm{C}}(\rho)$ has been shown to be NP-complete [12], a
property we conjecture carries over to $E_{\mathrm{CA}}$ due to the
definition. However, we do know its value for some special cases: when the
assemblage has a Local Hidden State (LHS) model of the form 111Normally this
condition is expressed as an integral over the hidden variable $\lambda$,
however, for fixed $n,r$ the form written here is entirely equivalent.

$\sigma^{\mathrm{LHS}}_{n|r}=\sum_{\lambda=1}^{d}D(n|\lambda,r)\sigma_{\lambda},$
(8)

where $D(n|\lambda,r)$ describes a possible deterministic assignment of
outputs to all inputs, with $d$ the total number of these, and
$\sigma_{\lambda}\geq 0$. Then the entanglement cost is

$E_{\mathrm{CA}}(\boldsymbol{\sigma}^{\mathrm{LHS}})=0.$ (9)

This follows from the fact that every LHS assemblage, also known as
_unsteerable_ assemblages, can be created from a separable state [13, 14],
which can be constructed explicitly via a semidefinite program [15]. It is
also worth noting that all assemblages created from separable states are
unsteerable. We can always upper bound
$E_{\mathrm{CA}}(\boldsymbol{\sigma})\leq S(\rho_{\boldsymbol{\sigma}})$ i.e.
by the von Neumann entropy of the marginal state
$S(\rho_{\boldsymbol{\sigma}}):=-\mathrm{Tr}(\rho_{\boldsymbol{\sigma}}\log\rho_{\boldsymbol{\sigma}})$.
This is because we have an explicit realisation of the assemblage which is
constructed from the marginal
$\rho_{\boldsymbol{\sigma}}=\sum_{i}\lambda^{i}|v^{i}\rangle\langle v^{i}|$
via

$\displaystyle\rho^{AB}$
$\displaystyle=\sum_{i}\sqrt{\lambda^{i}}|v^{i}\rangle^{A}|v^{i}\rangle^{B},$
$\displaystyle M_{n|r}$
$\displaystyle=\rho_{\boldsymbol{\sigma}}^{-1/2}\sigma_{n|r}^{T}\rho_{\boldsymbol{\sigma}}^{-1/2},$

where transposition is taken with respect to the basis
$\\{|v^{i}\rangle\\}^{i}$, and the pseudo-inverse is used if necessary.
Putting these into Eq. (1) will reproduce our desired assemblage.
Since for pure states
$E_{\mathrm{C}}(|\Phi\rangle^{AB}\langle\Phi|)=S(\mathrm{Tr}_{A}(|\Phi\rangle^{AB}\langle\Phi|)$,
we have that $E_{\mathrm{C}}(\rho^{AB})=S(\rho_{\boldsymbol{\sigma}})$. This
provides an upper bound to $E_{\mathrm{CA}}(\boldsymbol{\sigma})$. However, in
general it provides quite a bad lower bound; for our earlier example, which we
know has a value of $E_{\mathrm{CA}}=0$, it would give the upper bound $\leq
1$. We want to do better than this.
Instead of considering the entanglement cost directly, we propose another
measure: the _entanglement of formation for assemblages_

$\displaystyle E_{\mathrm{FA}}(\boldsymbol{\sigma}):=\inf\bigg{\\{}$
$\displaystyle\sum_{i}p^{i}S(\rho_{\boldsymbol{\sigma^{i}}})\;|\;\sum_{i}p^{i}\boldsymbol{\sigma}^{i}=\boldsymbol{\sigma},\;p^{i}\geq
0,$ $\displaystyle\sum_{i}p^{i}=1,\;\boldsymbol{\sigma}^{i}\text{ extremal in
}\mathcal{G}_{\mathcal{N}|\mathcal{R}}\bigg{\\}}$ (10)

where $\mathcal{G}_{\mathcal{N}|\mathcal{R}}$ is the set of all assemblages
with inputs in $\mathcal{R}$ and outcomes in $\mathcal{N}$. One can verify
from the relations in Eq. (2) this set is convex. The assemblages
$\boldsymbol{\sigma}^{i}$ are _extremal_ points; that is, they can only be
written as trivial convex combinations of themselves [16].
This is a direct analogy with the _entanglement of formation_ for quantum
states [9]

$\displaystyle E_{\mathrm{F}}(\rho):=\inf\bigg{\\{}$
$\displaystyle\sum_{i}p^{i}S(\mathrm{Tr}_{A}(|\phi^{i}\rangle\langle\phi^{i}|))\;|\;\sum_{i}p^{i}|\phi^{i}\rangle\langle\phi^{i}|=\rho,\;$
$\displaystyle p^{i}\geq 0,\;\sum_{i}p^{i}=1\bigg{\\}},$ (11)

which is an upper bound $E_{\mathrm{F}}(\rho)\geq E_{\mathrm{C}}(\rho)$.
Moreover, the stronger relation holds that
$\lim_{n\rightarrow\infty}E_{\mathrm{F}}(\rho^{\otimes
n})/n=E_{\mathrm{C}}(\rho)$ [10].
Both the original entanglement of formation and the entanglement of formation
for assemblages are examples of _convex roof extensions_ [17]. These consist
of a convex set $\mathcal{X}$ with a function $g(x)$ defined on the boundary
($\partial\mathcal{X}$). This is then extended to a function on the entire set
($G(x)$) by

$\displaystyle G(x):=$
$\displaystyle\inf\left\\{\sum_{i}g(x^{i})\mid\sum_{i}p^{i}x^{i}=x,\;\sum_{i}p^{i}=1,\;p^{i}\geq
0,\;x^{i}\in\partial\mathcal{X}\right\\}.$ (12)

The convex roof extension is equivalently defined as the largest convex
function $F$ on $\mathcal{X}$ such that $F(x)=g(x)$ when
$x\in\partial\mathcal{X}$. Another important result is that for the
entanglement of formation for quantum states (along with many other convex
roof extensions), the optimal decomposition satisfies $G(x)=g(x^{i}),\,\forall
i$.
Why do we choose to define the entanglement of formation for assemblages this
way? To see this, first we show that $S(\rho_{\boldsymbol{\sigma}})$
corresponds to the entanglement for extremal assemblages.∎

######  Theorem 1.

If $\boldsymbol{\sigma}$ is extremal, then any state
$\rho^{AB}\in\mathcal{S}_{\boldsymbol{\sigma}}$ must have
$E_{\mathrm{F}}(\rho^{AB})=S(\rho_{\boldsymbol{\sigma}})$.

If $\rho^{AB}$ is pure, i.e. $=|\phi\rangle\langle\phi|$, then this is true
straightaway, as
$E_{\mathrm{F}}(|\phi\rangle\langle\phi|)=E_{\mathrm{C}}(|\phi\rangle\langle\phi|)=S(\mathrm{Tr}_{A}|\phi\rangle\langle\phi|)=S(\rho_{\boldsymbol{\sigma}})$.
If the state is mixed, i.e.
$\rho^{AB}=\sum_{i}p^{i}|\phi^{i}\rangle\langle\phi^{i}|$, one can decompose
$\boldsymbol{\sigma}$ as

$\boldsymbol{\sigma}=\sum_{i}p^{i}\boldsymbol{\sigma}^{i}\;\;\;\sigma_{n|r}^{i}:=\mathrm{Tr}_{A}\left(M^{A}_{n|r}\otimes\mathbb{I}^{B}|\phi^{i}\rangle\langle\phi^{i}|\right).$
(13)

However, our assemblage is extremal, and thus
$\boldsymbol{\sigma}^{i}=\boldsymbol{\sigma},\;\forall i$. In particular, this
forces all $|\phi^{i}\rangle$ to have the marginal
$\rho_{\boldsymbol{\sigma}}$. Thus the entanglement of $|\phi^{i}\rangle$ is
$S(\rho_{\boldsymbol{\sigma}})$, and since this holds for all pure-state
decompositions of $\rho^{AB}$,
$E_{\mathrm{F}}(\rho^{AB})=S(\rho_{\boldsymbol{\sigma}})$.

######  Theorem 2.

$E_{\mathrm{FA}}(\boldsymbol{\sigma})=\min\left\\{E_{\mathrm{F}}(\rho^{AB})\mid\rho^{AB}\in\mathcal{S}_{\boldsymbol{\sigma}}\right\\}$,
where $\mathcal{S}_{\boldsymbol{\sigma}}$ is the set of all states such that
$\exists$ POVMs with
$\sigma_{n|r}=\mathrm{Tr}_{A}\left[\left(M_{n|r}\otimes\mathbb{I}^{B}\right)\rho^{AB}\right]$.

First we prove that for all $\rho^{AB}\in\mathcal{S}_{\boldsymbol{\sigma}}$
there exists a decomposition
$\boldsymbol{\sigma}=\sum_{i}p^{i}\boldsymbol{\sigma}^{i}$ such that
$\sum_{i}p^{i}S(\rho_{\boldsymbol{\sigma}^{i}})=E_{\mathrm{F}}(\rho^{AB})$.
This is done simply by decomposing

$\boldsymbol{\sigma}=\sum_{i}p^{i}\boldsymbol{\sigma}^{i},\;\sigma_{n|r}^{i}:=\mathrm{Tr}_{A}\left(M^{A}_{n|r}\otimes\mathbb{I}^{B}|\phi^{i}\rangle\langle\phi^{i}|\right),$
(14)

where $|\phi^{i}\rangle$ are in the optimal decomposition of $\rho^{AB}$. This
gives us that

$E_{\mathrm{FA}}(\boldsymbol{\sigma})\leq\sum
p^{i}S(\mathrm{Tr}_{A}(|\phi^{i}\rangle\langle\phi^{i}|)=E_{\mathrm{F}}(\rho_{AB}).$
(15)

This holds even if $\boldsymbol{\sigma}^{i}$ are not extremal, since the von
Neumann entropy is concave. This implies that
$E_{\mathrm{FA}}(\boldsymbol{\sigma})\leq\min E_{\mathrm{F}}(\rho_{AB})$.

Conversely, suppose there exists a decomposition
$\boldsymbol{\sigma}=\sum_{i}p^{i}\boldsymbol{\sigma}^{i}$ such that
$\sum_{i}p^{i}S(\boldsymbol{\sigma}^{i})=E$.
Let us define the corresponding pure state and measurements creating
$\boldsymbol{\sigma}^{i}$ as $|\phi^{i}\rangle\langle\phi^{i}|$,
$M_{a|x}^{i}$. Then we can construct $\boldsymbol{\sigma}$ via the state and
measurements

$\displaystyle\rho^{A^{\prime}AB}$
$\displaystyle:=\sum_{i}|i\rangle^{A^{\prime}}\langle
i|\otimes|\phi^{i}\rangle\langle\phi^{i}|,$ $\displaystyle
M^{A^{\prime}A}_{n|r}$ $\displaystyle:=\sum_{i}|i\rangle^{A^{\prime}}\langle
i|\otimes M_{n|r}^{i}.$

$\rho^{A^{\prime}AB}$ is decomposable into pure states $|i\rangle\langle
i|\otimes|\phi^{i}\rangle^{AB}\langle\phi^{i}|$, whose entanglement (with
respect to the bipartition $AA^{\prime}|B$) is
$S(\mathrm{Tr}_{A}|\phi^{i}\rangle^{AB}\langle\phi^{i}|)=S(\boldsymbol{\sigma}^{i})$.
Thus $E_{\mathrm{F}}(\rho^{A^{\prime}AB})\leq E$, implying $\min
E_{\mathrm{F}}(\rho^{AB})\leq E_{\mathrm{FA}}(\boldsymbol{\sigma})$ (note
there is no restriction of the dimension of $A$).∎

Theorem 2 gives us a direct connection between
$E_{\mathrm{FA}}(\boldsymbol{\sigma})$ and the entanglement of formation for
quantum states. Since for all states $E_{\mathrm{C}}(\rho)\leq
E_{\mathrm{F}}(\rho)$, we see that it also gives an upper bound
$E_{\mathrm{CA}}(\boldsymbol{\sigma})\leq
E_{\mathrm{FA}}(\boldsymbol{\sigma})$. An important question is whether there
exists an assemblage which is optimally created from a state satisfies
$E_{\mathrm{F}}=E_{\mathrm{C}}$, but where
$E_{\mathrm{CA}}(\boldsymbol{\sigma})<E_{\mathrm{FA}}(\boldsymbol{\sigma})$.

For this measure we may state that $E_{\mathrm{FA}}(\boldsymbol{\sigma})=0$
iff $\boldsymbol{\sigma}$ is unsteerable. For an unsteerable assemblage
$\boldsymbol{\sigma}$, we may find the corresponding separable state by
semidefinite programming - decomposing this into pure states will then the
desired extremals $\boldsymbol{\sigma}^{i}$. Conversely, any steering
assemblage created from a separable state is unsteerable [15]. Furthermore,
one can always upper bound $E_{\mathrm{FA}}(\boldsymbol{\sigma})$ by use of
the _steering weight_ , a steering monotone calculable by semidefinite
programming. The steering weight is defined as [18]

$SW(\boldsymbol{\sigma}):=\min\\{\;p\;|\boldsymbol{\sigma}=p\boldsymbol{\gamma}+(1-p)\boldsymbol{\sigma}^{\mathrm{LHS}}\\},$
(16)

i.e. a convex combination of a steerable and unsteerable assemblage. This
gives the upper bound

$E_{\mathrm{FA}}(\boldsymbol{\sigma})\leq
pS(\rho_{\boldsymbol{\gamma}}):=E_{\mathrm{SW}}(\boldsymbol{\sigma}).$ (17)

We can also provide a lower bound for $E_{\mathrm{CA}}$ (and thus
$E_{\mathrm{FA}}$) in the form of the _distillable secret key_ ,
$K_{\mathrm{D}}$ [19, 20]. To see this, first note that any key distillation
protocol from $\boldsymbol{\sigma}$ is also a key distillation protocol from
$|\Phi^{+}\rangle$, since we may asymptotically create
$1/E_{\mathrm{CA}}(\boldsymbol{\sigma})$ copies of $\boldsymbol{\sigma}$ from
one copy of $|\Phi\rangle^{+}$. Therefore if
$K_{\mathrm{D}}/(E_{\mathrm{CA}})>1$ would represent a key distillation rate
of $|\Phi^{+}\rangle$ larger than 1. However, we know that
$K_{D}(|\Phi^{+}\rangle)=1$.
In order to lower bound this rate, we use a local entropic uncertainty
relation [21, 22], where Alice and Bob are able to make measurements on some
state, and Bob’s measurements are characterised. This results in a state-
independent bound

$K_{\mathrm{D}}\geq-\log_{2}c-H(N^{\prime}|N)-H(\tilde{N}^{\prime}|\tilde{N}).$
(18)

In this formula $N$ and $\tilde{N}$ are uncharacterised measurements performed
by Alice, whilst $N^{\prime},\tilde{N}^{\prime}$ are approximations of $N$ and
$\tilde{N}$ respectively, based on known POVMs
$\\{M^{B,N^{\prime}}_{n^{\prime}}\\}$,
$\\{M^{B,\tilde{N}^{\prime}}_{\tilde{n}^{\prime}}\\}$ performed by Bob. The
value $c$ serves as the _incompatibility_ of the two POVMs, and is defined as

$c=\max_{n^{\prime},\tilde{n}^{\prime}}\left\|\sqrt{M^{B,N^{\prime}}_{n^{\prime}}}\sqrt{M^{B,\tilde{N}^{\prime}}_{\tilde{n}^{\prime}}}\right\|^{2}_{\infty}.$
(19)

One can see that the security of this protocol comes from Bob being able to
perform exactly his measurements on a generally unknown entangled state, which
allows Bob to correlate two incompatible measurements to Alice. As has been
noted in e.g. [5, 23], this is well suited to steering scenarios as Bob can
characterise his system, whilst Alice cannot. To make the association with
steering clearer, we are associating with the distribution $N$ with $P_{N|r}$
i.e. the distribution over outcomes given Alice’s specific choice of $r$, and
likewise $\tilde{N}$ represents $P_{N|\tilde{r}}$. To respect this, we shall
express Bob’s POVM choices as $\\{M^{B}_{n^{\prime}|r}\\}$,
$\\{M^{B}_{\tilde{n}^{\prime}|\tilde{r}}\\}$.
Any choice of $\\{M^{B}_{n^{\prime}|r}\\}$,
$\\{M^{B}_{\tilde{n}^{\prime}|\tilde{r}}\\}$ provides a lower bound on
$K_{\mathrm{D}}(\boldsymbol{\sigma})$ although it might not be a very tight
one. In order to gain an intuition about what a sensible choice might be, let
us consider the role of $M^{B}_{n^{\prime}|r}$. Alice performs measurement $r$
and obtains outcome $n$ with probability $p_{n|r}=\mathrm{tr}(\sigma_{n|r})$.
As a result Bob holds the state
$\hat{\sigma}_{n|r}=\sigma_{n|r}/\mathrm{tr}(\sigma_{n|r})$. The aim of
$\\{M^{B}_{n^{\prime}|r}\\}$ is to determine which state Bob holds and give
$n^{\prime}=n$, given that he receives state $\hat{\sigma}_{n|r}$ with
probability $p_{n|r}$. This is a more generally studied task in quantum
information known as _quantum state discrimination_ [24]. For the case when
$|\mathcal{N}|=2$, the optimal measurement is known, while for
$|\mathcal{N}|>2$ the “pretty good measurement” [25, 26, 27] is a reasonable
choice. We also use the Fano inequality,

$H(N^{\prime}|N)\leq H_{2}(P(\neq_{r}))+P(\neq_{r})\log_{2}(|\mathcal{N}|-1),$
(20)

where $P(\neq_{r})=P(N^{\prime}\neq N)$. This simplifies our lower bound to

$\displaystyle\max_{r,\tilde{r}}-\log_{2}c$
$\displaystyle-\left(H_{2}(P(\neq_{r}))+P(\neq_{r})\log_{2}(|\mathcal{N}|-1)\right)$
$\displaystyle-\left(H_{2}(P(\neq_{\tilde{r}}))+P(\neq_{\tilde{r}})\log_{2}(|\mathcal{N}|-1)\right).$
(21)

For $|\mathcal{N}|=2$, the Fano inequality is tight, and we use

$\displaystyle M^{B}_{0|r}$ $\displaystyle=\mathbb{P}_{+}(\sigma_{r}),$
$\displaystyle M^{B}_{1|r}$ $\displaystyle=\mathbb{P}_{-}(\sigma_{r}),$
$\displaystyle P(\neq_{r})=\frac{1}{2}-\frac{1}{2}\|\sigma_{r}\|_{1},$

where $\mathbb{P}_{+(-)}(\sigma_{r})$ is the projector onto the positive
(negative) eigenspaces of $\sigma_{r}:=\sigma_{0|r}-\sigma_{1|r}$. When
$|\mathcal{N}|>2$, we use

$\displaystyle M^{B}_{n^{\prime}|r}$
$\displaystyle=\rho_{\boldsymbol{\sigma}}^{-1/2}\sigma_{n^{\prime}|r}\rho_{\boldsymbol{\sigma}}^{-1/2},$
$\displaystyle P(\neq_{r})$
$\displaystyle=1-\sum_{n^{\prime}}\mathrm{Tr}\left[\left(\sigma_{n^{\prime}|r}\rho_{\boldsymbol{\sigma}}^{-1/2}\right)^{2}\right].$
(22)

Figure 1: Bounds for the entanglement cost of assemblages created by measuring
an entangled Werner state
$W_{p}=p|\Phi^{+}\rangle\langle\Phi^{+}|+(1-p)\mathbb{I}/4$, with $1/3<p\leq
1$, in the Pauli-X and Pauli-Z bases. Also plotted is the _Regularized
Relative Entropy of Entanglement_ with respect to PPT states
($E_{\mathrm{PPT}}^{\infty}$) for Werner states [28]. As
$E_{\mathrm{PPT}}^{\infty}$ is a lower bound to the entanglement cost, we can
see that Werner states _cannot_ be optimal for $p<0.85406$.

##  II Properties of $\mathbf{E_{\mathrm{FA}}}$ and Numerical Results

Currently we can only calculate bounds on
$E_{\mathrm{FA}}(\boldsymbol{\sigma})$; either by the steering weight or by
approximation of extremal assemblages by a finite set, as can be done with
computations of the entanglement of formation for states [29]. Nevertheless,
we can say something about its properties:

  * •

it is not a continuous function;

  * •

the optimal decomposition, in general, does not consist of
$\boldsymbol{\sigma}^{i}$ with equal $S(\rho_{\boldsymbol{\sigma}^{i}})$.

The second is in direct contrast to the case for quantum states, a property
which was used to prove the remarkable formula for the 2-qubit entanglement of
formation [30].
To prove these two statements, we turn to the specific case of
$|\mathcal{N}|=|\mathcal{R}|=2$ and $\mathrm{dim}(\mathcal{H}_{B})=2$. For
this set, extremal assemblages take the form

$\displaystyle\sigma_{n|r}=p_{n|r}|\phi_{n|r}\rangle\langle\phi_{n|r}|\;\forall
n,\,r\;\;$
$\displaystyle\\{|\phi_{0|0}\rangle,|\phi_{1|0}\rangle\\}\neq\\{|\phi_{0|1}\rangle,|\phi_{1|1}\rangle\\}.$
(23)

This is best seen by considering the “slice” of the Bloch sphere considering
the points $\\{|\phi_{n|r}\rangle\\}$; we see there is only one unique set of
weights such that the nosignalling condition holds.

Figure 2: An example of an extremal assemblage: the weights of the
subnormalised states are given by the relative length to the crossing point,
which represents the marginal $\rho_{\boldsymbol{\sigma}}$. Only one crossing
of the two lines is possible, and the individual points cannot be decomposed;
these two properties makes the assemblage extremal.

There are also extremal assemblages of the form

$\sigma_{n|r}=\delta_{nn_{r}}|\phi\rangle\langle\phi|.$ (24)

which correspond to embeddings of $|\mathcal{N}|=1,\,|\mathcal{R}|=2$ into the
space.
The noncontinuity of $E_{\mathrm{FA}}$ can seen by considering the assemblage

$\displaystyle\sigma^{\epsilon}_{0|0}$
$\displaystyle=\frac{1}{2}|0\rangle\langle 0|,$
$\displaystyle\sigma^{\epsilon}_{1|0}$
$\displaystyle=\frac{1}{2}|\phi_{+}^{\epsilon}\rangle\langle\phi_{+}^{\epsilon}|,$
$\displaystyle\sigma^{\epsilon}_{0|1}$
$\displaystyle=\frac{1}{2}|1\rangle\langle 1|,$
$\displaystyle\sigma^{\epsilon}_{1|1}$
$\displaystyle=\frac{1}{2}|\phi_{-}^{\epsilon}\rangle\langle\phi_{-}^{\epsilon}|,$

where

$\displaystyle|\phi_{+}^{\epsilon}\rangle$
$\displaystyle=\epsilon|0\rangle+\sqrt{1-\epsilon^{2}}|1\rangle,$
$\displaystyle|\phi_{-}^{\epsilon}\rangle$
$\displaystyle=\epsilon|1\rangle-\sqrt{1-\epsilon^{2}}|0\rangle.$

As long as $\epsilon>0$, this is an extremal assemblage, and thus
$E_{\mathrm{FA}}(\boldsymbol{\sigma}^{\epsilon})=S(\mathbb{I}/2)=1$. However,
as soon as $\epsilon=0$, the assemblage is decomposable into
$\boldsymbol{\sigma}^{0}=1/2(\boldsymbol{\sigma}^{a}+\boldsymbol{\sigma}^{b})$,
where

$\displaystyle\sigma^{a}_{0|0}$ $\displaystyle=|0\rangle\langle
0|=\sigma^{a}_{0|1}$ $\displaystyle\sigma^{a}_{1|0}$
$\displaystyle=0=\sigma^{a}_{1|1}$ $\displaystyle\sigma^{b}_{1|0}$
$\displaystyle=|1\rangle\langle 1|=\sigma^{b}_{1|1}$
$\displaystyle\sigma^{b}_{0|0}$ $\displaystyle=0=\sigma^{b}_{0|1}$

and thus $E_{\mathrm{FA}}(\boldsymbol{\sigma}^{0})=0$. This kind of
noncontinuity is to be expected: the same example can be used to show the
noncontinuity of the steering weight.

In order to prove that extremal points with equal $E_{\mathrm{FA}}$ do not
form the optimal decomposition, we first introduce a parametrisation of
$|\mathcal{N}|=|\mathcal{R}|=\mathrm{dim}(\mathcal{H}_{B})=2$ assemblages.
Using the Bloch vector notation, we can represent the marginal
$\rho_{\boldsymbol{\sigma}}$ and the matrix
$\sigma_{r}:=\sigma_{0|r}-\sigma_{1|r}$ as

$\displaystyle\rho_{\boldsymbol{\sigma}}$
$\displaystyle=1/2\left(\begin{array}[]{cc}1+z&x-iy\\\
x+iy&1-z\end{array}\right),$ $\displaystyle\sigma_{r}$
$\displaystyle=1/2\left(\begin{array}[]{cc}t_{r}+z_{r}&x_{r}-iy_{r}\\\
x_{r}+iy_{r}&t_{r}-z_{r}\end{array}\right).$

In this representation we represent the marginal as a Bloch vector
$\mathbf{x}_{\sigma}$, and each measurement as a four-vector
$\tilde{\boldsymbol{x}}_{r}=(t_{r},x_{r},y_{r},z_{r})$ which may be separated
into a trace component, $t_{r}$, and a Pauli component,
$\mathbf{x}_{r}=(x_{r},y_{r},z_{r})$. This takes care of the nosignalling
condition. Extremal assemblages in this parametrization satisfy

$\displaystyle\frac{|\mathbf{x}+\mathbf{x}_{r}|+|\mathbf{x}-\mathbf{x}_{r}|}{2}$
$\displaystyle=1,$
$\displaystyle\frac{|\mathbf{x}+\mathbf{x}_{r}|-|\mathbf{x}-\mathbf{x}_{r}|}{2}$
$\displaystyle=t_{r}$

where $\mathbf{x}_{0}\neq\mathbf{x}_{1}$. In the degenerate case,
$\mathbf{x}=\pm\mathbf{x}_{0}=\pm\mathbf{x}_{1}$ and $|\mathbf{x}|=1$.

Our entanglement quantity (on extremal states) is then calculated by

$\displaystyle E_{\mathrm{FA}}(\boldsymbol{\sigma})=E(\mathbf{x}):=$
$\displaystyle\frac{1+|\mathbf{x}|}{2}\log\left(\frac{1+|\mathbf{x}|}{2}\right)$
$\displaystyle+$
$\displaystyle\frac{1-|\mathbf{x}|}{2}\log\left(\frac{1-|\mathbf{x}|}{2}\right).$
(25)

(This is the same formula as the von Neumann entropy for qubits). We will
focus on assemblages with the condition

$\displaystyle\sigma_{0|0}$ $\displaystyle=|0\rangle\langle 0|/2,$
$\displaystyle\sigma_{1|0}$ $\displaystyle=|1\rangle\langle 1|/2.$

Any decomposition of this into extremal points $\boldsymbol{\sigma}^{i}$ must
have $\sigma^{i}_{0|0}\propto|0\rangle\langle 0|$,
$\sigma^{i}_{1|0}\propto|1\rangle\langle 1|$, since these are only valid
convex decompositions of rank-1 operators. These extremal assemblages must
therefore have $\mathbf{x}^{i}_{0}=(0,0,1)$ and $\mathbf{x}^{i}=(0,0,t_{0})$.
Therefore the set of extremal states with constant $E$ have the same value of
$|t_{0}|$. Furthermore the parameters of $\tilde{\boldsymbol{x}}_{1}^{i}$ must
satisfy

$\displaystyle(x^{i}_{1})^{2}+(y^{i}_{1})^{2}=(1-(z^{i}_{1})^{2})(1-(t^{i}_{0})^{2}),$
$\displaystyle t^{i}_{1}$ $\displaystyle=z^{i}_{1}t^{i}_{0}.$ (26)

In our restricted subspace, we can therefore describe states by $t_{0}$ and
$\tilde{\boldsymbol{x}}_{1}$.
Let us consider an assemblages $\boldsymbol{\sigma}$ with $t_{0}=0$,
$\tilde{\boldsymbol{x}}_{1}=(0,x_{1},y_{1},z_{1})$. For which values of $T\geq
0$ can it be decomposed into extremal points all with $|t_{0}^{i}|=T$? We need
to weight contributions of $t_{0}^{i}=T$ and $t_{0}^{i}=-T$ equally, and we
can always set $t_{1}=0$ by choosing with equal weight points
$t^{i_{1}}_{0}=T,\,\tilde{\boldsymbol{x}}^{i_{1}}=(t^{i}_{0}z_{1}^{i},x^{i}_{1},y^{i}_{1},z^{i}_{1})$
and
$t^{i_{2}}_{0}=-T,\,\tilde{\boldsymbol{x}}^{i_{2}}=(-t^{i}_{0}z^{i},x^{i}_{1},y^{i}_{1},z^{i}_{1})$.
Thus from Eq. (26) we can decompose $\boldsymbol{\sigma}$ into extremal
assemblages with equal $T$ (and thus equal $E$) iff

$\frac{(x_{1})^{2}+(y_{1})^{2}}{1-T^{2}}+(z_{1})^{2}\leq 1.$ (27)

As $E$ is monotonically decreasing with the value of $T$, that means the
optimal equal-value decomposition is given by the largest $T$ such that the
above equation holds. Eq. (27) for a fixed $T$ defines a $z$-axis aligned
ellipsoid of principal radius 1, and therefore the largest $T$ for which the
equation holds is when $(x_{1},y_{1},z_{1})$ lies on the ellipsoid’s surface.
We name this value of $T$ as $T_{\boldsymbol{\sigma}}$, and find that
$T_{\boldsymbol{\sigma}}=\sqrt{(1-x_{1}^{2}-y_{1}^{2}-z_{1}^{2})/(1-z_{1}^{2})}$,
or $T_{\boldsymbol{\sigma}}=1$ if $z_{1}=1$. The optimal equal-value
decomposition is then given by
$E(T_{\boldsymbol{\sigma}}):=E((0,0,T_{\boldsymbol{\sigma}}))$.
We can compare the value of this decomposition to that of another explicit
decomposition: that provided by the steering weight. This is done in Fig. 3.
For clarity, we have done this only for _rebit_ assemblages, where $y_{1}=0$.
Some further properties of rebit assemblages are given in the supplementary
information. From this one can see a clear region, close to the extremal
assemblages, in which the steering weight performs better. Thus for these
assemblages $E_{\mathrm{FA}}(\boldsymbol{\sigma})\lneq
E(T_{\boldsymbol{\sigma}})$, which we have already shown to be the lowest
valued decomposition for which $E_{\mathrm{FA}}(\boldsymbol{\sigma}^{i})$ are
all equal.

Figure 3: Comparison of decompositions for $t_{0}=0$,
$\tilde{\boldsymbol{x}}_{1}=(0,x_{1},0,z_{1})$ assemblages. Inside the red
region equal-value decompositions give a better upper bound, outside the
region the steering weight gives the better bound.

From this argument we are forced to admit the optimal solution cannot consist
of equal-valued extremal assemblages; but one could conjecture that by
augmenting this strategy with the steering weight decomposition, the optimum
is always attained. However, we can again show this is not the case. Fig. 4
shows the result of a numerical optimisation in which an assemblage
$\boldsymbol{\sigma}$ is optimally decomposed into two further assemblages
$\boldsymbol{\sigma}^{a},\,\boldsymbol{\sigma}^{b}$, which are assigned the
value of their equal weight decompositions $E(T_{\boldsymbol{\sigma}^{a}})$,
$E(T_{\boldsymbol{\sigma}^{b}})$. This allows for both the steering weight and
equal-weight decompositions. The results of the optimisation show that there
is a region with decompositions of a lower value than both
$E(T_{\boldsymbol{\sigma}})$ and $E_{\mathrm{SW}}(\boldsymbol{\sigma})$. These
decompositions correspond to the mixing of an unsteering and steering
assemblage, where the steering assemblage has higher weight than in the
steering-weight achieving decomposition.

Figure 4: Results of numerical decomposition optimisation for $t_{0}=0$,
$\tilde{\boldsymbol{x}}_{1}=(0,x_{1},0,z_{1})$ assemblages. We find there is a
region where neither the equal-value nor steering weight decompositions are
optimal.

##  III Conclusions and Discussions

In this work, we have defined two new quantities for studying the connection
between entanglement and steering, by quantifying the amount of entanglement
required to create an assemblage. We have shown that both quantities vanish if
the considered assemblage is unsteerable. Both of these values can be upper
bounded via the steering weight, which requires the solution of just one
semidefinite problem. We also provide a lower bound taken from secret key
distillation, which requires only the computation of certain Schatten norms.
However, this lower bound could be further optimised with the use of
semidefinite programming. The entanglement cost for assemblages,
$E_{\mathrm{CA}}$, has a natural role as a “proof of prior entanglement” in
semi-device-independent systems; the ability to repeatedly generate the
assemblage $\boldsymbol{\sigma}$ requires that Alice’s device must share
entanglement with Bob with a rate of at least
$E_{\mathrm{CA}}(\boldsymbol{\sigma})$ per assemblage. This may find
application in semi-device-independent protocols in testing rounds, for which
an analysis of the relation under a finite number of runs would be vital.

Of these two quantities, we believe the entanglement of formation for
assemblages to be the easier to work with, as one can apply the results of
convex roof extension theory, and exploit its connection to the entanglement
of formation for quantum states (shown in Theorem 2). One important problem
therefore is to determine the class of assemblages for which
$E_{\mathrm{CA}}=E_{\mathrm{FA}}$; the analogous equality holds true for a
wide class of quantum states, and therefore we expect the same to be true for
assemblages. We do not expect it to hold in general, however. In fact, we even
expect that the regularisation,
$\lim_{k\rightarrow\infty}E_{\mathrm{FA}}(\boldsymbol{\sigma}^{\times
k})/k\neq E_{\mathrm{CA}}(\boldsymbol{\sigma})$. This is because the
corresponding result for quantum states [10],
$\lim_{k\rightarrow\infty}E_{\mathrm{F}}(\rho^{\otimes
k})/k=E_{\mathrm{C}}(\rho)$, relies on the continuity properties of
$E_{\mathrm{F}}$, which we have shown do not transfer to $E_{\mathrm{FA}}$.
Intuitively, an example of nonequality would follow from an optimal
distillation protocol in which the _measurement_ choices are only
asymptotically exact; since our definition $E_{\mathrm{FA}}$ requires there
exact measurements which create exactly the assemblage from a given state.
Another important question is how to calculate $E_{\mathrm{FA}}$ for general
assemblages: a big result here would be to classify for which assemblages the
steering weight gives the optimal value, since this is easily calculable. This
question is a direct analogy to asking when the “fully entangled fraction” [9]
of a state equals its entanglement of formation.
One can also imagine performing a similar analysis as is done here, but taking
_measurement incompatibility_ as the primary resource; asking what are the
most compatible measurements that can create a given assemblage. It is already
a known result [13, 14] that commuting measurements cannot create a steerable
assemblage, and vice-versa, that any unsteerable assemblage can be created
with commuting measurements. Some work in this direction has already been done
[31, 32].
A natural consequence of this work is that we obtain a _total ordering_ on
steering assemblages. If $E_{\mathrm{CA}}(\boldsymbol{\sigma}^{a})\leq
E_{\mathrm{CA}}(\boldsymbol{\sigma}^{b})$, then sufficient entanglement to
create $\boldsymbol{\sigma}^{b}$ is sufficient to create
$\boldsymbol{\sigma}^{a}$. The relation can be used to give a necessity
condition about conversion of assemblages via 1-way LOCC (the set of mappings
which do not increase steering [8]): if $\boldsymbol{\sigma}^{a}$ can be
created by 1-way LOCC from $\boldsymbol{\sigma}^{b}$ then we must have
$E_{\mathrm{CA}}(\boldsymbol{\sigma}^{a})\leq
E_{\mathrm{CA}}(\boldsymbol{\sigma}^{b})$. This is because creating first
$\boldsymbol{\sigma}^{b}$ from our entanglement, then transforming it to
$\boldsymbol{\sigma}^{a}$, gives us a valid sufficient entanglement for
$\boldsymbol{\sigma}^{a}$. We note here that such relations have been studied
in detail for general resource conversion under Local Operations and Shared
Randomness (LOSR) in [33].
Finally, this work could be extended to continuous variable steering, which is
also an active field of research [4, 34, 35, 36].

###### Acknowledgements.

This work was supported, in part, by the DFG through SFB 1227 (DQ-mat), the
RTG 1991, and funded by the Deutsche Forschungsgemeinschaft (DFG, German
Research Foundation) under Germany’s Excellence Strategy – EXC-2123 Quantum
Frontiers – 390837967. The author would like to thank Tobias Osborne, Reinhard
Werner and Ramona Wolf for useful discussions.

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##  Appendix A Copies and Distances of Steering Assemblages

In this section we explain in more detail what it means to have multiple
copies of an assemblage, and what distance we are implicitly using when
talking about continuity.
We consider an assemblage as an ordered tuple of subnormalised density
matrices
$\boldsymbol{\sigma}:=(\sigma_{0|0},\ldots,\sigma_{(|\mathcal{N}|-1)|0},\ldots,\sigma_{0|(|\mathcal{R}|-1)},\ldots,\sigma_{(|\mathcal{N}|-1)|(|\mathcal{R}|-1)})$
which satisfy the nosignalling condition,
$\sum_{n}\sigma_{n|r}=\rho_{\boldsymbol{\sigma}}$. What does it mean to have
two copies of an assemblage? Intuitively, an assemblage is a system in which
by Alice choosing an input $r$, she can probabilistically steer Bob’s state
into the collection of substates $\\{\sigma_{n|r}\\}_{n}$. Thus, with two
copies of an assemblage, Alice should be able to choose an $r_{1}$ and
$r_{2}$, which independently steer Bob’s first state into
$\\{\sigma_{n_{1}|r_{1}}\\}_{n_{1}}$ and his second into
$\\{\sigma_{n_{2}|r_{2}}\\}_{n_{2}}$. From this idea, we define two copies of
an assemblage as a larger assemblage, written as the ordered tuple

$\displaystyle\boldsymbol{\sigma}^{\times 2}:=($
$\displaystyle\sigma_{11|11}\ldots\sigma_{(|\mathcal{N}|-1)1|(|\mathcal{R}|-1)1}\ldots$
(28) $\displaystyle\ldots\,$
$\displaystyle\sigma_{1(|\mathcal{N}|-1)|1(|\mathcal{R}|-1)}\ldots\sigma_{(|\mathcal{N}|-1)(|\mathcal{N}|-1)|(|\mathcal{R}|-1)(|\mathcal{R}|-1)}),$

where
$\sigma_{n_{1}n_{2}|r_{1}r_{2}}:=\sigma_{n_{1}|r_{1}}\otimes\sigma_{n_{2}|r_{2}}\geq
0$. Furthermore, we have that

$\sum_{n_{1},n_{2}}\sigma_{n_{1}n_{2}|r_{1}r_{2}}=\left(\sum_{n_{1}}\sigma_{n_{1}|r_{1}}\right)\otimes\left(\sum_{n_{2}}\sigma_{n_{2}|r_{2}}\right)=\rho_{\boldsymbol{\sigma}}\otimes\rho_{\boldsymbol{\sigma}}.$
(29)

So $\rho_{\boldsymbol{\sigma}^{\times 2}}=\rho_{\boldsymbol{\sigma}}^{\otimes
2}$.

Suppose we have a state and measurements $\rho^{AB}$ and $M^{A}_{n|r}$ such
that
$\sigma_{n|r}=\mathrm{Tr}_{A}\left(M^{A}_{n|r}\otimes\mathbb{I}^{B}\rho^{AB}\right)$.
We then see that

$\displaystyle\mathrm{Tr}_{A_{1}A_{2}}\left(M^{A_{1}}_{n_{1}|r_{1}}\otimes
M^{A_{2}}_{n_{2}|r_{2}}\otimes\mathbb{I}^{B_{1}B_{2}}\rho^{A_{1}B_{1}}\otimes\rho^{A_{2}B_{2}}\right)=$
$\displaystyle\mathrm{Tr}_{A_{1}}\left(M^{A_{1}}_{n_{1}|r_{1}}\otimes\mathbb{I}^{B_{1}}\rho^{A_{1}B_{1}}\right)\otimes\mathrm{Tr}_{A_{2}}\left(M^{A_{2}}_{n_{2}|r_{2}}\otimes\mathbb{I}^{B_{2}}\rho^{A_{2}B_{2}}\right)=$
$\displaystyle\sigma_{n_{1}|r_{1}}\otimes\sigma_{n_{2}|r_{2}}=:\sigma_{n_{1}n_{2}|r_{1}r_{2}}.$

This fits with our intuition that having two copies of the state and
performing separately the correct measurements, we should be able to create
two copies of the assemblage. The concept described above then generalises to
$\times k$. We can describe an element of the assemblage
$\boldsymbol{\sigma}^{\times k}$ succinctly using $\sigma_{\bf{n}|\bf{r}}$,
where $\bf{n},\bf{r}$ are two length-$k$ vectors, where the $i$th entry
represents an output or input respectively for the $i$th copy of the
assemblage.

For the concept of continuity, note that we are describing our assemblages
within the more general set of a Cartesian product of finite operators.
Therefore, we choose our norm for this space as

$\|\boldsymbol{\sigma}\|=\sum_{n,r}\|\sigma_{n|r}\|,$ (30)

where $\|\sigma_{n|r}\|$ is a norm for finite operators e.g. Hilbert-Schmidt.
We can check the three norm properties:

$\|\boldsymbol{\sigma}\|=0\Rightarrow\|\sigma_{n|r}\|=0\Rightarrow\sigma_{n|r}=0\Rightarrow\boldsymbol{\sigma}=\mathbf{0},$
$\displaystyle\|\alpha\boldsymbol{\sigma}\|=\sum_{n,r}\|\alpha\sigma_{n|r}\|=|\alpha|\sum_{n,r}\|\sigma_{n|r}\|=|\alpha|\|\boldsymbol{\sigma}\|$

and

$\displaystyle\|\boldsymbol{\sigma}+\boldsymbol{\gamma}\|$
$\displaystyle=\sum_{n,r}\|\sigma_{n|r}+\gamma_{n|r}\|$
$\displaystyle\leq\sum_{n,r}\left(\|\sigma_{n|r}\|+\|\gamma_{n|r}\|\right)$
$\displaystyle=\|\boldsymbol{\sigma}\|+\|\boldsymbol{\gamma}\|.$

Also note that

$\displaystyle\|\boldsymbol{\sigma}^{\times 2}\|$
$\displaystyle=\sum_{n_{1},n_{2},r_{1},r_{2}}\|\sigma_{n_{1}n_{2}|r_{1}r_{2}}\|$
$\displaystyle=\sum_{n_{1},n_{2},r_{1},r_{2}}\|\sigma_{n_{1}|r_{1}}\otimes\sigma_{n_{2}|r_{2}}\|$
$\displaystyle=\sum_{n_{1},n_{2},r_{1},r_{2}}\|\sigma_{n_{1}|r_{1}}\|\|\sigma_{n_{2}|r_{2}}\|$
$\displaystyle=\left(\sum_{n_{1},r_{1}}\|\sigma_{n_{1}|r_{1}}\|\right)\left(\sum_{n_{2},r_{2}}\|\sigma_{n_{2}|r_{2}}\|\right)$
$\displaystyle=\|\boldsymbol{\sigma}\|^{2}.$

where we have assumed a norm is chosen such that $\|A\otimes B\|=\|A\|\|B\|$
such as Hilbert-Schmidt. We use the metric induced by this norm for
discussions of continuity.

##  Appendix B Creation of Steering Assemblages

The entanglement cost we consider is the following,

$\displaystyle E_{\mathrm{CA}}(\boldsymbol{\sigma}):=\inf\bigg{\\{}$
$\displaystyle E\mid\forall\epsilon>0,\delta>0,\exists p,q,$
$\displaystyle\forall\mathbf{n}\in\mathcal{N}^{\times
q},\mathbf{x}\in\mathcal{X}^{\times q},$ $\displaystyle\exists
M_{\boldsymbol{n}|\boldsymbol{r}}\geq
0,\sum_{\boldsymbol{a}}M_{\boldsymbol{n}|\boldsymbol{r}}=\mathbb{I}^{A},\Lambda\;\text{and}$
$\displaystyle\sum_{\boldsymbol{n},\boldsymbol{r}}\big{\|}\sigma_{\boldsymbol{n}|\boldsymbol{r}}-\mathrm{Tr}_{A}$
$\displaystyle\left[\left(M_{\boldsymbol{n}|\boldsymbol{r}}\otimes\mathbb{I}_{B}\right)\Lambda(|\Phi^{+}\rangle\langle\Phi+|^{\otimes
p})\right]\big{\|}\leq\epsilon\bigg{\\}}.$ (31)

Similar to the entanglement cost for quantum states, we allow for the
possibility of a cheaper cost when preparing many copies of the assemblage,
and also allow for protocols which are only asymptotically perfect in creating
the assemblage. Note that these imperfections may be in the state, or in the
measurements. Although the state before measurement can be quite general
(reflected by e.g. the $\mathbb{I}^{A}$, denoting an arbitrary sized system
for Alice before measurement) it should result in an assemblage with the same
input/output set as $\sigma_{\boldsymbol{n}|\boldsymbol{r}}$ and the same
Hilbert space dimension, in order for the distance comparison to be valid.

A reasonable question to ask is whether this definition should allow for some
LOCC operations _after_ the measurement; which itself is an LOCC operation. We
believe that is not necessary, and that all operations possible after the
measurement can be absorbed into $\Lambda$, or the definition of the
measurement operators. We will argue this below.

Another point of consideration is, if LOCC operations are allowed after the
measurement operator, are all LOCC operations allowed? Here we argue that the
answer to this second question is no. Let us suppose that once an assemblage
is created, it is distributed to two parties (Alexa and Boris) who use it for
a semi-device independent protocol. Before Alexa has chosen an input $r$, they
can ensure that no signalling can enter or leave their laboratories.
Therefore, we do not allow distillation procedures in which Alice’s system
signals information about $n$ or $r$, after the input is chosen and output
determined (done by $M_{n|r}$). The device may only be prepared to locally
react to input choices and output results, explained by the _wirings_ given
below. We now show that all other allowed operations can be absorbed into
$\Lambda$, or the definition of the measurement operators.

  * •

_Sharing of randomness for joint operations and deciding which operations to
be done._
All of this can be shared and communicated during the LOCC protocol $\Lambda$,
before the measurement takes place.

  * •

_Tracing out of extra systems, and operations independent of $n,r$ and
$\rho_{\mathbf{\sigma}}$_.
Since these either commute with the measurement operators, or do not require
the assemblage to exist when performing them, these can also be moved into the
$\Lambda$ protocol before the measurements.

  * •

_Bob performs an operation on his state, and communicates the result to
Alice._
Since Alice’s measurements are performed locally, they commute with Bob’s
local operations, which can take place beforehand. So this is also absorbed
into $\Lambda$.

  * •

_Alice rewires her possible inputs._
Given Alice has access to input choices in $\mathcal{R}$, a rewiring of inputs
corresponds to a conditional probability distribution
$p(r|\Omega,r^{\prime})$, with $r^{\prime}$ in a (potentially) different input
choice set $\mathcal{R}^{\prime}$. Alice will then choose an input
$\mathcal{R^{\prime}}$, which will, along with the outcome of some random
variables $\Omega$, determine which choice is inputted into the original
assemblage. As discussed, any generation of $\Omega$ can be absorbed into the
$\Lambda$ beforehand; meanwhile, a rewiring can be absorbed into the
measurements by rewriting
$M_{n|r^{\prime}}=\sum_{r}p(r|\Omega,r^{\prime})M_{n|r}$. This is again
positive and moreover we have that
$\sum_{n}M_{n|r^{\prime}}=\sum_{n,r}p(r|\Omega,r^{\prime})M_{n|r}=\sum_{r}p(r|\Omega,r^{\prime})\sum_{n}M_{n|r}=\sum_{r}p(r|\Omega,r^{\prime})\mathbb{I}^{A_{p}}$.
Since we are considering LOCCs and not more general stochastic LOCCs (SLOCCs),
this wiring must be deterministic (normalised implying
$\sum_{r}p(r|\Omega,r^{\prime}=1)$) and so we see that $r^{\prime}$ remains a
valid measurement.

  * •

_Alice rewires her possible outputs._
Similarly, Alice may decide to implement a conditional probability
distribution $p(n^{\prime}|\Omega,r^{\prime},r,n)$, which receives output $n$
from the original assemblage and maps it to $n^{\prime}$, an output in a
(possibly different) output set $\mathcal{N}^{\prime}$. We may in a similar
manner define
$M_{n^{\prime}|r}=\sum_{n}p(n^{\prime}|\Omega,r^{\prime},r,n)M_{n|r}$. This
again gives us positive operators, where
$\sum_{n^{\prime}}M_{n^{\prime}|r}=\sum_{n}\sum_{n^{\prime}}p(n^{\prime}|\Omega,r^{\prime},r,n)M_{n|r}=\sum_{n}M_{n|r}=\mathbb{I}^{A_{p}}$,
where we have again used the deterministic nature of the wirings. Thus, we
have incorporated our wiring into the new measurement $\\{M_{n|r}\\}$.

##  Appendix C Rebit Assemblages

In this appendix we talk shortly about, when calculating the $E_{\mathrm{CA}}$
for an assemblage with $\sigma_{0|0}=|0\rangle\langle 0|/2$,
$\sigma_{1|0}=|1\rangle\langle 1|/2$, how one can ensure that the matrix
$\sigma_{1}=\sigma_{0|1}-\sigma_{1|1}$ can always be taken to be real. To see
this, we note that the unitary
$U_{\theta}=\cos{\theta}\sigma_{x}+\sin{\theta}\sigma_{y}$ (here
$\sigma_{x},\sigma_{y}$ are the Pauli matrices) transforms the matrix
$\sigma_{1}$ as

$U_{\theta}\sigma_{1}U^{\dagger}_{\theta}=1/2\left(\begin{array}[]{cc}t_{1}+z_{1}&e^{-2i\theta}(x_{1}-iy_{1})\\\
e^{2i\theta}(x_{1}+iy_{1})&t_{1}-z_{1}\end{array}\right).$ (32)

Thus, by choosing $\theta=\mathrm{Arg}(x_{1}-iy_{1})/2$, this transformation
takes it to a real matrix with $x_{1}^{\prime}=\sqrt{x_{1}^{2}+y_{1}^{2}}$ and
$y_{1}^{\prime}=0$.
To see that this transformation can be realised, consider what happens if Bob
performs this unitary on his local system. Since we constrained
$\sigma_{0|0}=|0\rangle\langle 0|/2$ and $\sigma_{1|0}=|1\rangle\langle 1|/2$,
these are left invariant by the unitary, as is the marginal
$\rho_{\boldsymbol{\sigma}}=\sigma_{0|0}+\sigma_{1|0}$, since it is diagonal
in the Pauli-$Z$ basis. However,
$\sigma_{0|1}\rightarrow\sigma^{\prime}_{0|1}=U_{\theta}\sigma_{0|1}U^{\dagger}_{\theta}$
and
$\sigma_{1|0}\rightarrow\sigma^{\prime}_{1|1}=U_{\theta}\sigma_{1|1}U^{\dagger}_{\theta}$
. Therefore,

$\sigma^{\prime}_{1}=\sigma^{\prime}_{0|1}-\sigma^{\prime}_{1|1}=U_{\theta}(\sigma_{0|1}-\sigma_{1|1})U^{\dagger}_{\theta}=U_{\theta}\sigma_{1}U^{\dagger}_{\theta},$
(33)

and so we see that the transformation in Eq. (32) is a physically realisable
one. We also note that such a unitary commutes with Alice’s measurement (since
they take place on different subsytems), and any state used to create the
first assemblage is locally unitarily equivalent to any to create the second
assemblage. Since entanglement measures are invariant under local unitaries,
we can conclude that such local unitaries do not affect the value of
$E_{\mathrm{CA}}/E_{\mathrm{FA}}$.
We also make the comment here that the optimal decomposition of a rebit
assemblage must be into extremal rebit assemblages. Let us assume the optimal
decomposition contains the extremal point $\boldsymbol{\sigma}^{i}$ with
weight $p^{i}$, whose parameters are $t^{i}_{0}$ and
$\tilde{\boldsymbol{x}}^{i}_{1}=(t^{i}_{0}z^{i}_{1},x_{1}^{i},y_{1}^{i},z_{1}^{i}),\;y_{1}^{i}\neq
0$. As our assemblage is invariant under the transformation $y_{1}\rightarrow-
y_{1}$, so too must our optimal decomposition, and therefore it must contain
the assemblage $\boldsymbol{\sigma}^{j}$, also with weight $p_{i}$, whose
parameters are $t^{j}_{0}=k^{i}_{0}$ and
$\tilde{\boldsymbol{x}}^{j}_{1}=(t^{i}_{0}z^{i}_{1},x_{1}^{i},-y_{1}^{i},z_{1}^{i})$.
Together these two points contribute, with weight $2p_{i}$, the assemblage
with $t_{0}=t^{i}_{0}$,
$\tilde{\boldsymbol{x}}_{1}=(t^{i}_{0}z^{i}_{1},x_{1}^{i},0,z_{1}^{i})$.
However, there exists an extremal point (which we call
$\boldsymbol{\sigma}^{i^{\prime}}$) with $|t^{i^{\prime}}_{0}|>|t^{i}_{0}|$
and
$\tilde{\boldsymbol{x}}^{i^{\prime}}=(t^{i^{\prime}}_{0}z_{1}^{i},x_{1}^{i},0,z_{1}^{i})$
(this follows from Eq. (25) in the main body). Furthermore, the point
$\boldsymbol{\sigma}^{j^{\prime}}$ with
$t^{j^{\prime}}_{0}=-t^{i^{\prime}}_{0}$ and
$\tilde{\boldsymbol{x}}^{j^{\prime}}=(-t^{i^{\prime}}_{0}z_{1}^{i},x_{1}^{i},0,z_{1}^{i})$
is also extremal. By mixing together these two points in the ratio
$(1+t_{0}^{i})/(2t^{i^{\prime}})\boldsymbol{\sigma}^{i^{\prime}}+(1-t_{0}^{i})/(2t^{i^{\prime}})\boldsymbol{\sigma}^{j^{\prime}}$
we find we create the same assemblage as
$1/2\boldsymbol{\sigma}^{i}+1/2\boldsymbol{\sigma}^{j}$, and can therefore
replace $\boldsymbol{\sigma}^{i},\boldsymbol{\sigma}^{j}$ in the decomposition
with $\boldsymbol{\sigma}^{i^{\prime}},\boldsymbol{\sigma}^{j^{\prime}}$.

As we have $E(\mathbf{x}^{i^{\prime}})<E(\mathbf{x}^{i})$ and
$E(\mathbf{x}^{j^{\prime}})<E(\mathbf{x}^{j})$ (due to the monotonicity of
$E$) our original decomposition was not optimal – this is a contradiction.
Thus, we can conclude that $y_{1}^{i}=0$ for all extremal
$\boldsymbol{\sigma}^{i}$ in the optimal decomposition.

##  Appendix D Numerical Algorithm

In this section we explain briefly how the numerical optimisation algorithm
works, and how it incorporates the equal value decomposition and steering
weight into the optimisation.
The algorithm uses MATLAB’s _fmincon_ function to split an assemblage
$\boldsymbol{\sigma}=p\boldsymbol{\sigma}^{a}+(1-p)\boldsymbol{\sigma}^{b}$,
in order to minimise the quantity
$pE(T_{\boldsymbol{\sigma}^{a}})+(1-p)E(T_{\boldsymbol{\sigma}^{b}})$. By
setting $p=1$ and $\boldsymbol{\sigma}^{a}=\boldsymbol{\sigma}$, we see the
equal value decomposition is considered by this optimisation.
In order to prove that this optimisation can also achieve the steering weight
decomposition, we need to show that for an assemblage in our restricted
subspace with $t_{0}=0=t_{1}$, there exists a decomposition
$\boldsymbol{\sigma}=(1-SW(\boldsymbol{\sigma}))\boldsymbol{\sigma}^{\mathrm{LHS}}+SW(\boldsymbol{\sigma})\boldsymbol{\gamma}$,
where $t^{\mathrm{LHS}}_{0}=t^{\mathrm{LHS}}_{1}=0$. To do this, we first
rewrite the steering weight in an equivalent form

$1-SW(\boldsymbol{\sigma})=\max\\{\mathrm{Tr}(\boldsymbol{\sigma}^{\mathrm{LHS}})\mid\boldsymbol{\sigma}=\boldsymbol{\sigma}^{\mathrm{LHS}}+\boldsymbol{\gamma}\\},$
(34)

where now $\boldsymbol{\sigma}^{\mathrm{LHS}},\boldsymbol{\gamma}$ are
subnormalised assemblages, i.e.
$\mathrm{Tr}(\rho_{\boldsymbol{\sigma}^{\mathrm{LHS}}}),\mathrm{Tr}(\rho_{\boldsymbol{\gamma}})\leq
1$.
Next, we note that due to our restricted subspace, we are limited to
assemblages with $\sigma_{0|0}\propto|0\rangle\langle 0|$,
$\sigma_{1|0}\propto|1\rangle\langle 1|$. This means our unsteerable
assemblage must be a convex combination of the following four assemblages:

$\displaystyle\sigma^{1}_{0|0}$ $\displaystyle=|0\rangle\langle
0|=\sigma^{1}_{0|1}$ $\displaystyle\sigma^{1}_{1|0}$
$\displaystyle=0=\sigma^{1}_{1|1},$ $\displaystyle\sigma^{2}_{0|0}$
$\displaystyle=|0\rangle\langle 0|=\sigma^{2}_{1|1}$
$\displaystyle\sigma^{2}_{1|0}$ $\displaystyle=0=\sigma^{2}_{0|1},$
$\displaystyle\sigma^{3}_{1|0}$ $\displaystyle=|1\rangle\langle
1|=\sigma^{3}_{0|1}$ $\displaystyle\sigma^{3}_{1|0}$
$\displaystyle=0=\sigma^{3}_{1|1},$ $\displaystyle\sigma^{4}_{1|0}$
$\displaystyle=|1\rangle\langle 1|=\sigma^{4}_{1|1}$
$\displaystyle\sigma^{4}_{1|0}$ $\displaystyle=0=\sigma^{4}_{0|1}.$

Therefore any $\boldsymbol{\sigma}^{\mathrm{LHS}}$ must be of the form
$\boldsymbol{\sigma}^{\mathrm{LHS}}=\sum_{i}\alpha^{i}\boldsymbol{\sigma}^{i}$,
$\alpha^{i}\geq 0$, $\sum_{i}\alpha^{i}\leq 1$. In particular,

$\displaystyle t^{\mathrm{LHS}}_{0}$
$\displaystyle=\alpha^{1}+\alpha^{2}-\alpha^{3}-\alpha^{4},$ $\displaystyle
t^{\mathrm{LHS}}_{1}$
$\displaystyle=\alpha^{1}+\alpha^{3}-\alpha^{2}-\alpha^{4}.$

The steering weight problem is now equivalent to $\max\sum_{i}\alpha^{i}$ such
that

$\displaystyle(1/2-\alpha^{1}-\alpha^{2})|0\rangle\langle 0|\geq 0,$
$\displaystyle(1/2-\alpha^{3}-\alpha^{4})|1\rangle\langle 1|\geq 0,$
$\displaystyle\sigma_{0|1}-\alpha^{1}|0\rangle\langle
0|-\alpha^{3}|1\rangle\langle 1|\geq 0,$
$\displaystyle\sigma_{1|1}-\alpha^{2}|0\rangle\langle
0|-\alpha^{4}|1\rangle\langle 1|\geq 0.$

Using the fact that a 2x2 matrix is positive semidefinite iff its trace and
determinant are nonnegative, and that our original assemblage can be
explicitly expressed in our parameterisation as

$\displaystyle\sigma_{0|1}$
$\displaystyle=1/4\left(\begin{array}[]{cc}1+z_{1}&x_{1}-iy_{1}\\\
x_{1}+iy_{1}&1-z_{1}\end{array}\right)$ $\displaystyle\sigma_{1|1}$
$\displaystyle=1/4\left(\begin{array}[]{cc}1-z_{1}&-x_{1}+iy_{1}\\\
-x_{1}-iy_{1}&1+z_{1}\end{array}\right)$

We can rewrite the optimization as $\max\sum_{i}\alpha^{i}$ such that

$\displaystyle(1/2-\alpha^{1}-\alpha^{2})\geq 0,$
$\displaystyle(1/2-\alpha^{3}-\alpha^{4})\geq 0,$
$\displaystyle(1/2-\alpha^{1}-\alpha^{3})\geq 0,$
$\displaystyle(1/2-\alpha^{2}-\alpha^{4})\geq 0,$
$\displaystyle(1+z_{1}-4\alpha^{1})(1-z_{1}-4\alpha^{3})-x_{1}^{2}-y_{1}^{2}\geq
0,$
$\displaystyle(1+z_{1}-4\alpha^{4})(1-z_{1}-4\alpha^{2})-x_{1}^{2}-y_{1}^{2}\geq
0.$

We can see that this problem is invariant under the transformations
$\alpha^{1}\leftrightarrow\alpha^{4}$, $\alpha^{2}\leftrightarrow\alpha^{3}$.
Thus, if there exists an optimal solution
$\alpha^{1}=\beta^{1},\,\alpha^{2}=\beta^{2},\,\alpha^{3}=\beta^{3},\,\alpha^{4}=\beta^{4}$,
then
$\alpha^{1}=\beta^{4},\,\alpha^{2}=\beta^{3},\,\alpha^{3}=\beta^{2},\,\alpha^{4}=\beta^{1}$
is also an optimal solution. Furthermore, an equal mixture of the two
$\alpha^{1}=\alpha^{4}=(\beta^{1}+\beta^{4})/2,\,\alpha^{2}=\alpha^{3}=(\beta^{2}+\beta^{3})/2$
is also an optimal solution. Moreover, it is an optimal solution with
$t^{\mathrm{LHS}}_{0}=t^{\mathrm{LHS}}_{1}=0$. This means that it is eligible
to be considered in our MATLAB optimisation, and consequently our optimisation
will achieve a value less than or equal to
$E_{\mathrm{SW}}(\boldsymbol{\sigma})$.