# Almost device-independent certification of GME states with minimal
measurements

Shubhayan Sarkar<EMAIL_ADDRESS>Center for Theoretical Physics,
Polish Academy of Sciences, Aleja Lotników 32/46, 02-668 Warsaw, Poland
Laboratoire d’Information Quantique, Université libre de Bruxelles (ULB), Av.
F. D. Roosevelt 50, 1050 Bruxelles, Belgium  Alexandre C. Orthey, Jr  Center
for Theoretical Physics, Polish Academy of Sciences, Aleja Lotników 32/46,
02-668 Warsaw, Poland  Gautam Sharma  Center for Theoretical Physics, Polish
Academy of Sciences, Aleja Lotników 32/46, 02-668 Warsaw, Poland  Remigiusz
Augusiak  Center for Theoretical Physics, Polish Academy of Sciences, Aleja
Lotników 32/46, 02-668 Warsaw, Poland

###### Abstract

Device-independent certification of quantum states allows the characterization
of quantum states present inside a device by making minimal physical
assumptions. A major problem in this regard is to certify quantum states using
minimal resources. In this work, we consider the multipartite quantum steering
scenario with an arbitrary number of parties but only one of which is trusted
in the sense that the measurements performed by the trusted party are known.
Consequently, the self-testing scheme is almost device-independent.
Importantly, all the parties can only perform two measurements each which is
the minimal number of measurements required to observe any form of quantum
nonlocality. Then, we propose steering inequalities that are maximally
violated by three major classes of genuinely multipartite entangled (GME)
states, one, graph states of arbitrary local dimension, two, Schmidt states of
arbitrary local dimension, and, three, $N$-qubit generalized W states. Using
the proposed inequalities, we then provide an almost device-independent
certification of the above GME states.

##  I Introduction

Device-independent (DI) quantum information refers to a framework in quantum
information theory where protocols and tasks are designed without relying on
detailed knowledge or assumptions about the devices used to implement them. In
other words, DI protocols aim to achieve certain quantum information
processing tasks without specifying the internal workings or characteristics
of the quantum devices involved. This approach is particularly valuable for
enhancing security and reliability in quantum communication and computation
tasks. By removing assumptions about the devices, DI protocols provide a
higher level of security against potential attacks and imperfections in the
devices themselves. One of the key resources in DI quantum information
processing is Bell nonlocality [1, 2], which serves as a fundamental
ingredient for DI tasks such as quantum cryptography [3], random number
generation [4], and state and measurement certification [5, 6, 4].

The strongest DI certification scheme is referred to as self-testing [7] (see
also Ref. [8]). It allows one to infer the form of the underlying quantum
system solely from the observed measurement outcomes, without needing to know
the details of how the measurements were performed. Recently, a lot of
attention has been devoted to finding schemes to self-test bipartite quantum
states [9, 10, 11, 12, 13] and multipartite quantum states [14, 15, 16, 17,
18, 19, 20, 21, 22]. In particular [20] provides a scheme to self-test any
pure entangled quantum state and [21] provides a scheme to self-test any
quantum state.

A major problem in this regard is to find the optimal self-testing scheme,
that is, using the minimal number of measurements to certify a particular
quantum state. As the self-testing schemes are based on quantum nonlocality,
each party in the experiment needs to perform at least two measurements. In
the bipartite regime, there are only a few schemes that utilize two
measurements per party [9, 10, 12] and even fewer in the multipartite scenario
[14, 18]. Moreover, in the multipartite scenario most of the schemes are
devoted to quantum states that are locally qubits. The main difficulty here
consists in designing a suitable Bell inequality that is maximally violated by
a given multipartite entangled quantum state. The problem becomes even more
complex if one wants to find a Bell inequality which involves the minimal
number of two measurements per observer.

To lower the complexity, one can introduce physically motivated assumptions
into the considered scenario, leading to what is known as semi-DI
certification schemes. One can for instance assume that one of the parties are
trusted and their measuring devices perform known measurements. Such an
assumption gives rise to scenarios which are typically referred to as one-
sided-device-independent (1SDI), and the key resource enabling certification
in this case is a weaker form of quantum nonlocality known as quantum steering
[23, 24, 25]. Recent investigations have explored the potential of such
scenarios for self-testing quantum states and measurements [26, 27, 28, 29,
30, 31, 32, 33]. For this work, we utilise the framework of multipartite
quantum steering introduced in [34].

We consider here a much more complicated problem of certifying genuinely
multipartite entangled (GME) states shared among an arbitrary number of
parties and arbitrary local dimension using minimal number of measurements,
that is, two. To simplify the problem we make the aforementioned assumption
that only one party is trusted, thus, making our scheme almost device-
independent. Furthermore, we restrict the parties to only choose two
measurements each. We then construct steering inequalities that are maximally
violated by three major classes of GME states. We then prove that maximal
violation of our inequalities can be used for certification of the
corresponding states. Since, among the arbitrary number of parties only one is
trusted, the presented results below are almost DI certificates of the quantum
state. The first family of states we can certify in this way is any connected
graph state of arbitrary local dimension. Graph states are highly useful in
quantum information tasks [35, 36, 37, 38]. The only other known schemes to
self-test graph states are restricted to local dimension two [15, 14] and
three [19]. The second class of states we consider here are the Schmidt states
of arbitrary local dimension, for which a self-testing method was provided in
Ref. [17]. The latter, however, requires more measurements per observer and
thus is not optimal as far as physical implementations are concerned.
Interestingly, our scheme can also be utilized to certify two measurements
with the untrusted parties to be mutually unbiased using almost no
entanglement across multiple parties. Finally, we self-test the generalized
$W$ state (see below for a definition) shared among an arbitrary number of
parties but only with local dimension two. While self-testing methods for the
standard $N$-qubit $W$ state were already introduced in Refs. [16, 17], here,
by making a mild assumption that only one party is trusted, we can
significantly generalize these results to a multi-parameter class of genuinely
entangled states.

Before proceeding to the results, let us introduce the relevant notions
required for this work.

##  II Preliminaries

We consider a one-sided DI (1SDI) scenario consisting of $N$-parties, namely
Alice and $N-1$ Bobs, whom we refer to as $A,B_{1},\ldots,B_{N-1}$. All the
parties are located in space-like separated laboratories and share a quantum
state $\rho_{A\mathbf{B}}$, where we denote $\mathbf{B}\equiv B_{1}\ldots
B_{N-1}$. In this sense, each one of the $N$ parties has a device to measure
their share of $\rho_{A\mathbf{B}}$. Since this is a 1SDI scenario, we assume
that Alice’s measurement device is trusted, i.e., her measurements are known.
Yet, we do not assume that her device performs full tomography over her share
of $\rho_{A\mathbf{B}}$. On the other side, the measurement devices that
belong to the $N-1$ Bobs are untrusted and we treat them as “black boxes”. The
scenario we just described is also referred to as local steering [39], in
which the untrusted parties perform local measurements on their respective
states to steer the state owned by the trusted party. In this work, we
consider that the trusted party performs the measurements given by the
$d-$dimensional generalization of the Pauli $Z$ and $X$ measurement

$\displaystyle Z_{d}=\sum_{i=0}^{d-1}\omega^{i}|i\rangle\\!\langle i|,\qquad
X_{d}=\sum_{i=0}^{d-1}|i+1\rangle\\!\langle i|$ (1)

which for ease of notation will be denoted as $Z$ and $X$ respectively. Fig. 1
contains an illustration of the scenario considered here.

We denote the inputs of all the parties $A,B_{1},\ldots,B_{N-1}$ with
$x,y_{1},\ldots,y_{N-1}$ respectively and the corresponding outcomes with
$a,b_{1},\ldots,b_{N-1}$. For convenience, we chose the notation
$\mathbf{y}\equiv y_{1},\ldots,y_{N-1}$ and $\mathbf{b}\equiv
b_{1},\ldots,b_{N-1}$. Thus, the correlations in this scenario are captured by
the set of probability distributions
$\vec{p}=\\{p(a,\mathbf{b}|x,\mathbf{{y}})\\}$, where
$p(a,\mathbf{b}|x,\mathbf{y})$ tells us the probability of getting outcomes
$a$ and $\mathbf{b}$ given that the measurements $x$ and $\mathbf{y}$ have
been performed. According to Born’s rule, the latter is expressed as

$\displaystyle
p(a,\mathbf{b}|x,\mathbf{y})=\mathrm{Tr}\left[\rho_{A\mathbf{B}}\left(M_{a|x}\otimes\bigotimes_{i=1}^{N-1}N^{(i)}_{b_{i}|y_{i}}\right)\right],$
(2)

where $M_{x}=\\{M_{a|x}\\}_{a}$,
$N_{y_{i}}^{(i)}=\\{N^{(i)}_{b_{i}|y_{i}}\\}_{b_{i}}$ are the measurement
operators of Alice and Bobs $B_{i}$, respectively. We recall that the
measurement operators $M_{a|x}$ and $N^{(i)}_{b_{i}|y_{i}}$ are positive semi-
definite and sum to identity on their corresponding Hilbert spaces.

Figure 1: Depiction of the local steering scenario that we consider. A GME
state is shared among $N$ parties: one Alice and $N-1$ Bobs. While Alice
performs measurements $Z$ and $X$, all the other Bobs perform unknown
measurements.

In our multipartite 1SDI scenario, the presence of local quantum steering is
demonstrated if the probability distribution $p(a,\mathbf{b}|x,\mathbf{y})$
can not be described by a local hidden state (LHS) model [40, 39], i.e., if it
can not be written as

$\displaystyle
p(a,\mathbf{b}|x,\mathbf{y})=\sum_{\lambda}p(\lambda)\mathrm{Tr}\left[M_{a|x}\rho_{A}^{\lambda}\right]p(\mathbf{b}|\mathbf{y},\lambda),$
(3)

where $\lambda$ is a hidden variable that follows the probability distribution
$p(\lambda)$. The hidden variable $\lambda$ determines the state
$\rho^{\lambda}_{A}$ received by Alice and also the responses
$p(\mathbf{b}|\mathbf{y},\lambda)$ obtained by $N-1$ Bobs. If the set of
probability distributions $\\{p(a,\mathbf{b}|x,\mathbf{y})\\}$ can not be
described by the above model, it leads to violation of a linear steering
inequality,
$\mathcal{B}\left[\\{p(a,\mathbf{b}|x,\mathbf{y})\\}\right]\leqslant\beta_{L}$,
where $\mathcal{B}$ is a linear combination of
$\\{p(a,\mathbf{b}|x,\mathbf{y})\\}$ and $\beta_{L}$ is the LHS bound , also
called classical bound [41].

In this work we express steering inequalities in terms of generalized
expectation values. These are defined in terms of
$p(a,\mathbf{b}|x,\mathbf{y})$ as

$\displaystyle\langle
A_{k|x}\mathbf{B}_{\mathbf{l}|\mathbf{y}}\rangle=\sum_{a,\mathbf{b}=0}^{d-1}\omega^{ak+b_{1}l_{1}+\ldots+b_{N-1}l_{N-1}}p(a,\mathbf{b}|x,\mathbf{y}),$
(4)

where
$\mathbf{B}_{\mathbf{l}|\mathbf{y}}\equiv(B_{1})_{l_{1}|y_{1}}\ldots(B_{N-1})_{l_{N-1}|y_{N-1}}$,
$\omega=\mathrm{e}^{2\pi\mathbbm{i}/d}$, and
$k,l_{1},\ldots,l_{N-1}\in\\{0,\ldots,d-1\\}$. If the probabilities
$p(a,\mathbf{b}|x,\mathbf{y})$ originate from a quantum measurement, we can
write the expectation values as

$\langle
A_{k|x}\mathbf{B}_{\mathbf{l}|\mathbf{y}}\rangle=\mathrm{Tr}\left[\rho_{A\mathbf{B}}A_{x}^{k}\otimes\bigotimes_{i=1}^{N-1}(B_{i})_{l_{i}|y_{i}}\right],$
(5)

where $A^{k}_{x}$ is the $k$-th power of the unitary observable $A_{x}$,
satisfying $A_{x}^{d}=\mathbb{1}_{d}$, and $(B_{i})_{l_{i}|y_{i}}$ are the
Fourier transforms of the measurement operators $N^{(i)}_{b_{i}|y_{i}}$, i.e.,

$(B_{i})_{l_{i}|y_{i}}=\sum_{b_{i}}\omega^{b_{i}l_{i}}N^{(i)}_{b_{i}|y_{i}}.$
(6)

In fact, since we can write $N^{(i)}_{b_{i}|y_{i}}$ as an inverse Fourier
transform of $(B_{i})_{l_{i}|y_{i}}$, we can also refer to
$\\{(B_{i})_{l_{i}|y_{i}}\\}_{l_{i}=0}^{d-1}$ as the measurements of the
$i$-th Bob. Moreover, if $N_{y_{i}}^{(i)}$ are projective, then
$(B_{i})_{l_{i}|y_{i}}$ are the $l_{i}$-th power of a unitary observable
$B_{i,y_{i}}\coloneqq(B_{i})_{1|y_{i}}$, that is,
$(B_{i})_{l_{i}|y_{i}}=\left(B_{i,y_{i}}\right)^{l_{i}}$ [13].

In what follows, we will present three classes of steering inequalities that
are maximally violated by three classes of GME states, namely the $N$-qudit
graph states, the $N$-qudit Schmidt states, and the generalized $N$-qubit
$W$-states. Recall that a GME state is one that cannot be written as a tensor
product of two other pure states corresponding to two disjoint and nonempty
subsets of the parties. Formally, a GME state is a state that cannot be
written in the form $|\psi\rangle_{A_{1}\ldots
A_{N}}=|\alpha\rangle_{A}\otimes|\beta\rangle_{\overline{A}}$ for any
bipartition $A|\overline{A}$ such that $A$ is a subset of the parties
$A_{1}\ldots A_{N}$, $\overline{A}$ is the complement of $A$ and both $A$ and
$\overline{A}$ are non-empty.

Maximally violating steering inequalities plays a critical role in the
presented below self-testing results. When experimental results $\vec{p}$
exhibit a maximal violation, they allow us to infer both the observer’s state
and the performed measurements up to local isometries. With that in mind, let
us formally define multipartite 1SDI self-testing similar to the one
introduced in [30]. We assume here that the measurements are projective based
on dilation arguments. However, the state $\rho_{N,A\mathbf{B}}$ shared among
all parties is in general mixed, which can be purified to
$|\psi_{N}\rangle_{A\mathbf{B}E}$ by adding an extra system $E$ as
$\rho_{N}=\mathrm{Tr}_{E}(|\psi_{N}\rangle\\!\langle\psi_{N}|)$. Alice’s
measurements are projective and fixed as $A_{x}$, while Bobs measurements are
arbitrary and represented by $B_{i,y_{i}}$. The goal is to certify that the
state shared between Alice and Bob is equivalent to a reference state
$|\psi^{\prime}\rangle_{A\mathbf{B}^{\prime}}$, as well as, that the
measurements performed by Bob are equivalent to the reference measurements
$B_{i,y_{i}}^{\prime}$ that generates the same correlation $\vec{p}$. We say
that $|\psi\rangle_{A\mathbf{B}E}$ and $B_{i,y_{i}}$ are certified by the
statistics $\vec{p}$ if there exists a unitary
$U_{i}:\mathcal{H}_{B_{i}}\to\mathcal{H}_{B_{i}}$ such that

$\displaystyle\left(\mathbbm{1}_{A}\otimes\bigotimes_{i=1}^{N-1}U_{i}\right)|\psi_{N}\rangle_{A\mathbf{B}E}$
$\displaystyle=$
$\displaystyle|\psi^{\prime}\rangle_{A\mathbf{B}^{\prime}}\otimes|\phi\rangle_{\mathbf{B}^{\prime\prime}E},$
(7a) $\displaystyle U_{i}\,B_{i,y_{i}}\,U_{i}^{\dagger}$ $\displaystyle=$
$\displaystyle
B_{i,y_{i}}^{\prime}\otimes\mathbbm{1}_{B_{i}^{\prime\prime}},\qquad\forall i$
(7b)

where Bobs Hilbert space is given by
$\mathcal{H}_{B_{i}}=\mathcal{H}_{B_{i}^{\prime}}\otimes\mathcal{H}_{B_{i}^{\prime\prime}}$,
$|\phi\rangle_{\mathbf{B}^{\prime\prime}E}$ is a junk state in
$\mathcal{H}_{\mathbf{B}^{\prime\prime}}\otimes\mathcal{H}_{E}$. Let us remark
here that the measurements can only be certified on the local supports of the
state, thus, without loss of generality, we assume that the local supports are
full-rank.

##  III Results

###  III.1 Graph states

Figure 2: Examples of graph: (a) ring graph (b) star graph corresponding to
$N-$qudit GHZ state up to local unitaries.

Let us begin by defining graph states of arbitrary local dimension [42, 35,
43]. Consider a multigraph $G=(V,E,\Gamma)$ (cf. Fig. 2), where $V$ is a set
of all the vertices of the graph, whereas $E$ is a set of pairs of vertices
that are connected by edges. Finally, $\Gamma$ is a matrix with elements
$\gamma_{i,j}=0,\ldots,d-1$ denoting the multiplicity of the edges between
vertices $i$ and $j$. For our purpose, we consider graphs that are connected,
that is, those in which there exist of path between any pair of vertices.
Then, the graph state $|G\rangle\in(\mathbbm{C}^{d})^{\otimes N}$
corresponding to the multigraph $G=(V,E,\Gamma)$, where $N$ is the number of
vertices, is constructed as:

  * •

Each vertex of $G$ is associated the state
$|+^{d}_{i}\rangle=\frac{1}{\sqrt{d}}\sum_{i=0}^{d-1}|i\rangle$ where
$\\{|i\rangle\\}$ is the $d-$dimensional computational basis.

  * •

For each edge $e={i,j}$ connecting the vertex $i,j$ of multiplicity
$\gamma_{i,j}$, a controlled unitary

$Z_{i,j}^{\gamma_{i,j}}=\sum_{q=0}^{d-1}|q\rangle\\!\langle
q|_{i}\otimes\left(Z_{j}^{\gamma_{i,j}}\right)^{q}$ (8)

is applied on the state

$|+^{d}_{V}\rangle=\bigotimes_{i=1}^{N}|+_{i}^{d}\rangle.$ (9)

Consequently, a graph state corresponding to the multigraphh $G=(V,E)$ is
defined as

$|G\rangle=\prod_{i,j}Z_{i,j}^{\gamma_{i,j}}|+^{d}_{V}\rangle.$ (10)

An equivalent way to represent graph states is by using the stabilizer
formalism. Given a multigraph $G=(V,E)$, let us consider a vertex $i$ and
associate the operator $X$ to it. All the vertices attached by an edge $e$ to
the $i-$th vertex are associated with the operator $Z$. Thus, the stabilizing
operator corresponding to the $i-$th vertex is given by

$S_{i}(G)=X_{i}\otimes\bigotimes_{j\in
n(i)}Z_{j}^{\gamma_{i,j}}\qquad(i=1,2,\ldots,N),$ (11)

where $n(i)$ is the set of vertices connected to the vertex $i$. The graph
state $|G\rangle$ is the unique eigenstate of all the stabilizing operators
$S_{i}(G)$ corresponding to the eigenvalue $1$. For example, the well-known
Greenberger-Horne-Zeilinger (GHZ) state is a graph state.

Let us now move on to constructing steering inequalities that are maximally
violated by any graph state and then use them later for self-testing. For this
purpose, we construct the steering operator by considering the stabilizing
operators of a graph state of the form (11). Then, we consider the first
vertex, and label it as $0$ vertex, to be trusted and thus the measurement
operators acting on it are known to be $Z_{d}$ and $X_{d}$ (1). Next, we
replace $Z_{d}$ and $X_{d}$ on the $j$-th vertex with $B_{j,0},B_{j,1}$
respectively. The steering operator is then given by

$\displaystyle\mathcal{I}_{d}(G,N)$ $\displaystyle=$ $\displaystyle
X_{A}\otimes\bigotimes_{j\in n(0)}(B_{j,0})^{\gamma_{0,j}}$ (12)
$\displaystyle+\sum_{j\in n(0)}Z_{A}^{\gamma_{j,0}}\otimes
B_{j,1}\otimes\bigotimes_{j^{\prime}\in
n(j)\setminus\\{0\\}}(B_{j^{\prime},0})^{\gamma_{j,j^{\prime}}}$
$\displaystyle+\sum_{j\notin
n(0)\cup{0}}B_{j,1}\otimes\bigotimes_{j^{\prime}\in
n(j)}(B_{j^{\prime},0})^{\gamma_{j,j^{\prime}}}+h.c.,$

where $h.c.$ denotes the Hermitian conjugate of the operator on its left-hand
side with $G,N$ denoting the graph and number of vertices respectively. Notice
that, given a graph $G=(V,E)$, the first term of the above operator
corresponds to the stabilizer when choosing the vertex $0$, the second term
consists of the stabilizers when choosing vertices connected to the
$0^{\text{th}}$ vertex and the third term consists of rest of it. Also notice
that in the above operator $\gamma_{i,j}=\gamma_{j,i}$ for any $i,j$. Thus,
the steering inequality is given by

$\langle\mathcal{I}_{d}(G,N)\rangle\leqslant\beta_{L},$ (13)

where $\beta_{L}$ is the maximum value attainable using the LHS model. In Fact
1 of Appendix A, we provide an upper bound to $\beta_{L}$ in terms of a simple
optimization problem that can be numerically evaluated for any graph $G$ and
any $N,d$. The quantum bound $\beta_{Q}$ of the steering functional
$\langle\mathcal{I}_{d}(G,N)\rangle$ is equal to the number of terms in
$\mathcal{I}_{d}(G,N)$ (12), that is, $\beta_{Q}=2N$ and can be attained by
the graph state $|G\rangle$ and observables $B_{j,0}=Z,B_{j,1}=X$ for any $j$.
Importantly, from Fact 1 of Appendix A, we can conclude that
$\beta_{L}<\beta_{Q}$ and thus the proposed steering inequality is non-
trivial.

Now, if all the parties observe that $\langle\mathcal{I}_{d}(G,N)\rangle=2N$,
then we can establish the following theorem.

######  Theorem 1.

Assume that the steering inequality (13) is maximally violated when the
$0^{\text{th}}$ party Alice chooses the observables $A_{i}$ defined to be
$A_{0}=Z$, $A_{1}=X$ and all the other parties choose their observables as
$B_{j,i}$ for $i\in\\{0,1\\}$ and $j=1,2\ldots,N-1$ acting
$\mathcal{H}_{B_{j}}$. Let us say that the state which attains this violation
is given by
$|\psi_{N}\rangle_{A\mathbf{B}E}\in\mathbbm{C}^{d}\otimes\bigotimes_{j=1}^{N-1}\mathcal{H}_{B_{j}}\otimes\mathcal{H}_{E}$,
then the following statement holds true for any $d$:
$\mathcal{H}_{B_{j}}=\mathcal{H}_{B^{\prime}_{j}}\otimes\mathcal{H}_{B^{\prime\prime}_{j}}$,
where $\mathcal{H}_{B^{\prime}_{j}}\equiv\mathbbm{C}^{d}$,
$\mathcal{H}_{B^{\prime\prime}_{j}}$ is some finite-dimensional Hilbert space,
and there exists a local unitary transformation on Bob’s side
$U_{j}:\mathcal{H}_{B_{j}}\rightarrow\mathcal{H}_{B_{j}}$, such that for any
$j$

$U_{j}\,B_{j,0}\,U_{j}^{\dagger}=Z_{B^{\prime}_{j}}\otimes\mathbbm{1}_{B^{\prime\prime}_{j}},\quad
U_{j}\,B_{j,1}\,U_{j}^{\dagger}=X_{B^{\prime}_{j}}\otimes\mathbbm{1}_{B^{\prime\prime}_{j}}\
\ $ (14)

where ${B^{\prime\prime}}$ denotes Bob’s auxiliary system and the state
$|\psi_{N}\rangle$ is given by,

$\left(\mathbbm{1}_{AE}\otimes\bigotimes_{i=1}^{N-1}U_{j}\right)|\psi_{N}\rangle=|G\rangle_{A\mathbf{B}^{\prime}}\otimes|\xi\rangle_{\mathbf{B}^{\prime\prime}E},$
(15)

where
$|\xi\rangle_{\mathbf{B}^{\prime\prime}E}\in\bigotimes_{j=1}^{N-1}\mathcal{H}_{B_{j}^{\prime\prime}}\otimes\mathcal{H}_{E}$
denotes the auxiliary state.

The proof of the above theorem is given in Appendix A.

###  III.2 Schmidt states

Every pure bipartite state $|\psi\rangle_{AB}$ admits a Schmidt decomposition,
that is, there exist orthonormal bases $|i\rangle_{A}$ and $|i\rangle_{B}$ in
$\mathcal{H}_{A}$ and $\mathcal{H}_{B}$, respectively, such that
$|\psi\rangle_{AB}=\sum_{i}\alpha_{i}|i\rangle_{A}\otimes|i\rangle_{B}$, where
$\alpha_{i}>0$ and $\sum_{i}\alpha_{i}^{2}=1$. In the multipartite setting, an
analogous decomposition is in general impossible. The multipartite pure states
which can be represented via Schmidt decomposition are referred to as Schmidt
states [17]. Under the action of local unitary operators, a multipartite
Schmidt state $|\psi(\boldsymbol{\alpha})\rangle_{A\mathbf{B}}$ belonging to
$\mathbb{C}^{d_{A}}\otimes\mathbb{C}^{d_{B_{1}}}\ldots\otimes\mathbb{C}^{d_{B_{N-1}}}$
can be reduced to the following form

$\displaystyle|\psi(\boldsymbol{\alpha})\rangle_{A\mathbf{B}}=\sum_{i=0}^{d-1}\alpha_{i}|i\rangle^{\otimes
N},$ (16)

where $d=\min\\{d_{A},d_{B_{1}},\ldots,d_{B_{N-1}}\\}$ and
$\boldsymbol{\alpha}\equiv\alpha_{1},\ldots,\alpha_{N}$, such that
$0<\alpha_{i}<1$ for all $i$ and $\sum_{i=0}^{d-1}\alpha_{i}^{2}=1$. Because
$\alpha_{i}>0$ for all $i$, the Schmidt rank of the state is maximal, which
ensures that it is a $N$-qudit state.

Inspired from [31], let us now introduce a general class of steering
inequalities that are maximally violated by states of the form (16) for a
given $\boldsymbol{\alpha}$. For this purpose, we consider the following
steering operator

$\displaystyle\mathcal{S}_{d}(\boldsymbol{\alpha},N)$ $\displaystyle=$
$\displaystyle\sum_{k=1}^{d-1}\left[\left(\sum_{j=1}^{N-1}A_{0}^{k}\otimes
B_{j,0}^{k}\right)\right.$ (17)
$\displaystyle+\left.\gamma(\boldsymbol{\alpha})A_{1}^{k}\otimes\bigotimes_{j=1}^{N-1}B_{j,1}^{k}+\delta_{k}(\boldsymbol{\alpha})A_{0}^{k}\right],$

where the coefficients $\gamma(\boldsymbol{\alpha})$ and
$\delta_{k}(\boldsymbol{\alpha})$ are given by

$\displaystyle\gamma(\boldsymbol{\alpha})$ $\displaystyle=$ $\displaystyle
d\left[\sum_{\begin{subarray}{c}i,j=0\\\ i\neq
j\end{subarray}}^{d-1}\frac{\alpha_{i}}{\alpha_{j}}\right]^{-1},$ (18)
$\displaystyle\delta_{k}(\boldsymbol{\alpha})$ $\displaystyle=$
$\displaystyle-\frac{\gamma(\boldsymbol{\alpha})}{d}\sum_{\begin{subarray}{c}i,j=0\\\
i\neq j\end{subarray}}^{d-1}\frac{\alpha_{i}}{\alpha_{j}}\omega^{k(d-j)}.$
(19)

In our scenario, we consider that all the $N-1$ Bobs perform two arbitrary
measurements denoted by $B_{i,0}$ and $B_{i,1}$. Also, we assume that the
local dimensions of the state shared by them are unknown. At the same time,
because Alice’s device is trusted, her measurements are completely
characterized and the local dimension of the shared state on Alice’s side is
known. Alice’s observables are the generalized Pauli operators (1), that is,
$A_{0}=Z_{d}$ and $A_{1}=X_{d}$.

The corresponding steering inequality is given by

$\displaystyle\langle\mathcal{S}_{d}(\boldsymbol{\alpha},N)\rangle\leqslant\beta_{L}.$
(20)

Again $\beta_{L}$ is the maximum value of
$\langle\mathcal{S}_{d}(\boldsymbol{\alpha},N)\rangle$ attainable with LHS
models. In Fact 3 of Appendix B, we provide a form of $\beta_{L}$ in terms of
an optimization problem that can be numerically evaluated for any
$d,N,\boldsymbol{\alpha}$. Then, the quantum bound $\beta_{Q}$ of
$\langle\mathcal{S}_{d}(\boldsymbol{\alpha},N)\rangle$ is given by
$\beta_{Q}=(d-1)(N-1)+1$ and can be attained by
$|\psi(\boldsymbol{\alpha})\rangle$ and observables $B_{j,0}=Z_{d}$ and
$B_{j,1}=X_{d}$ for any $j$. Importantly, from Fact 3 we can conclude that
$\beta_{L}<\beta_{Q}$ and thus the proposed steering inequality is non-
trivial.

Now, if all the parties observe that
$\langle\mathcal{S}_{d}(\boldsymbol{\alpha},N)\rangle=(d-1)(N-1)+1$, then we
can establish the following theorem.

######  Theorem 2.

Assume that the steering inequality (20) is maximally violated when the
$0^{\text{th}}$ party Alice chooses the observables $A_{i}$ defined to be
$A_{0}=Z_{d}$, $A_{1}=X_{d}$ and all the other parties choose their
observables as $B_{j,i}$ for $i=0,1$ and $j=1,2\ldots,N-1$. Let us say that
the state which attains this violation is given by
$|\psi_{N}\rangle_{A\mathbf{B}E}\in\mathbbm{C}^{d}\otimes\bigotimes_{j=1}^{N-1}\mathcal{H}_{B_{j}}\otimes\mathcal{H}_{E}$,
then the following statement holds true for any $d$:
$\mathcal{H}_{B_{i}}=\mathcal{H}_{B^{\prime}_{i}}\otimes\mathcal{H}_{B^{\prime\prime}_{i}}$,
where $\mathcal{H}_{B^{\prime}_{i}}\equiv\mathbbm{C}^{d}$,
$\mathcal{H}_{B^{\prime\prime}_{i}}$ is some finite-dimensional Hilbert space,
and there exists a local unitary transformation on Bob’s side
$U_{i}:\mathcal{H}_{B_{i}}\rightarrow\mathbbm{C}^{d}\otimes\mathcal{H}_{B^{\prime\prime}_{i}}$
such that for any $i$

$U_{i}\,B_{i,0}\,U_{i}^{\dagger}=Z_{d}\otimes\mathbbm{1}_{B^{\prime\prime}_{i}},\quad
U_{i}\,B_{i,1}\,U_{i}^{\dagger}=X_{d}\otimes\mathbbm{1}_{B^{\prime\prime}_{i}},$
(21)

where ${B^{\prime\prime}}$ denotes Bob’s auxiliary system and the state
$|\psi_{N}\rangle$ is given by,

$\left(\mathbbm{1}_{AE}\otimes\bigotimes_{i=1}^{N-1}U_{j}\right)|\psi_{N}\rangle=|\psi(\boldsymbol{\alpha})\rangle_{A\mathbf{B}^{\prime}}\otimes|\xi\rangle_{\mathbf{B}^{\prime\prime}E},$
(22)

where
$|\xi\rangle_{\mathbf{B}^{\prime\prime}E}\in\bigotimes_{j=1}^{N-1}\mathcal{H}_{B_{j}^{\prime\prime}}\otimes\mathcal{H}_{E}$
denotes the auxiliary state.

The proof of the above theorem can be found in Appendix B.

###  III.3 $N$-qubit generalized $W$ states

The third and last class of states that we are going to consider comprises the
$N$-qubit generalized $W$ states. As the name suggests, they represent a
generalization of the well-known $W$ state
$|\psi_{W}\rangle=(|100\rangle+|010\rangle+|001\rangle)/\sqrt{3}$ and have the
following form

$\displaystyle|\psi_{W_{N}}\rangle=\sum_{l=0}^{N-1}\alpha_{l+1}|0\ldots,0,1_{l},0,\ldots,0\rangle,$
(23)

such that $\alpha_{i}>0$ for every $i$ and $\sum_{i=1}^{N}\alpha_{i}^{2}=1$.
We introduce a steering inequality that is maximally violated by the above
state through the following steering operator

$\displaystyle\mathcal{W}_{N}(\boldsymbol{\alpha})$ $\displaystyle=$
$\displaystyle-2Z_{A}\otimes\bigotimes_{k=1}^{N-1}B_{k,0}+\sum_{l=1}^{N-1}\left[Z_{A}\otimes\mathbbm{1}_{l}\otimes\left(\mathbbm{1}-P_{l}\right)\right.$
$\displaystyle\left.+\left(\gamma_{l}X_{A}\otimes
B_{l,1}+\delta_{l}Z_{A}\otimes\mathbbm{1}_{l}\right)\otimes P_{l}\right]$
$\displaystyle+\sum_{l=1}^{N-1}\left[\mathbbm{1}_{A}\otimes
B_{l,0}\otimes\left(\mathbbm{1}-P_{l}\right)+\mathbbm{1}_{A}\otimes\mathbbm{1}_{l}\otimes
P_{l}\right],$

where

$\displaystyle\gamma_{l}=\frac{2\alpha_{l+1}\alpha_{1}}{\alpha_{l+1}^{2}+\alpha_{1}^{2}},\quad\delta_{l}=\frac{\alpha_{l+1}^{2}-\alpha_{1}^{2}}{\alpha_{l+1}^{2}+\alpha_{1}^{2}},$
(25)

for every $l\in\\{1,\ldots,N-1\\}$,

$\displaystyle P_{l}=\frac{1}{2^{N-2}}\bigotimes_{k=1,k\neq
l}^{N-1}\left(\mathbbm{1}_{k}+B_{k,0}\right),$ (26)

and $B_{k,0}$ and $B_{k,1}$ are the measurements acting on the $k$-th qubit.
In the same way, $\mathbbm{1}_{l}$ is the identity operation acting on the
$l$-th qubit. The corresponding steering inequality is given by

$\displaystyle\langle\mathcal{W}_{N}(\boldsymbol{\alpha})\rangle\leqslant\beta_{L}.$
(27)

where $\beta_{L}$ is the LHS bound. Unfortunately, we were not able to
calculate the LHS bound for this case. The quantum bound, nonetheless, is
given by $\beta_{Q}=2N$ and is achieved by the state $|\psi_{W_{N}}\rangle$
and measurements $B_{j,0}=Z$ and $B_{j,1}=X$, as we prove in Fact 5 of
Appendix C. By observing this value, the parties can certify the state and
measurements through the following statement:

######  Theorem 3.

Assume that the steering inequality (27) is maximally violated when the
$0^{\text{th}}$ party Alice chooses the observables $A_{i}$ defined to be
$A_{0}=Z$, $A_{1}=X$ and all the other parties choose their observables as
$B_{j,i}$ for $i\in\\{0,1\\}$ and $j=1,2\ldots,N-1$ acting
$\mathcal{H}_{B_{j}}$. Let us say that the state which attains this violation
is given by
$|\psi_{N}\rangle_{A\mathbf{B}E}\in\mathbbm{C}^{2}\otimes\bigotimes_{j=1}^{N-1}\mathcal{H}_{B_{j}}\otimes\mathcal{H}_{E}$,
then the following statement holds true:
$\mathcal{H}_{B_{j}}=\mathcal{H}_{B^{\prime}_{j}}\otimes\mathcal{H}_{B^{\prime\prime}_{j}}$,
where $\mathcal{H}_{B^{\prime}_{j}}\equiv\mathbbm{C}^{2}$,
$\mathcal{H}_{B^{\prime\prime}_{j}}$ is some finite-dimensional Hilbert space,
and there exists a local unitary transformation on Bob’s side
$U_{j}:\mathcal{H}_{B_{j}}\rightarrow\mathcal{H}_{B_{j}}$, such that

$U_{j}\,B_{j,0}\,U_{j}^{\dagger}=Z_{B^{\prime}_{j}}\otimes\mathbbm{1}_{B^{\prime\prime}_{j}},\quad
U_{j}\,B_{j,1}\,U_{j}^{\dagger}=X_{B^{\prime}_{j}}\otimes\mathbbm{1}_{B^{\prime\prime}_{j}}$
(28)

for every $j$, where ${B^{\prime\prime}}$ denotes Bob’s auxiliary system and
the state $|\psi_{N}\rangle$ is given by,

$\left(\mathbbm{1}_{AE}\otimes\bigotimes_{i=1}^{N-1}U_{j}\right)|\psi_{N}\rangle_{A\mathbf{B}E}=|\psi_{W_{N}}\rangle_{A\mathbf{B}^{\prime}}\otimes|\xi\rangle_{\mathbf{B}^{\prime\prime}E},$
(29)

where
$|\xi\rangle_{\mathbf{B}^{\prime\prime}E}\in\mathcal{H}_{B_{1}^{\prime\prime}}\otimes\ldots\otimes\mathcal{H}_{B_{N-1}^{\prime\prime}}\otimes\mathcal{H}_{E}$
denotes the auxiliary state.

The reader will find the proof for the above theorem in Appendix C.

##  IV Discussion

In our work, we demonstrated that although it’s very difficult to create self-
testing methods with the fewest measurements possible, having trust in one of
the parties makes it possible to create 1SDI schemes for many types of GME
states. We focused on graph states, Schmidt states, and generalized $W$
states. Another important finding is that we can certify a pair of operators
that are mutually unbiased with minimal entanglement across any number of
parties in one process.

Our work raises several follow-up questions. An obvious one is how to expand
our approach to certify more types of genuinely entangled multipartite states.
In fact, it would be interesting to see whether the assumption considered here
that one of the measurement device is trusted allows one to design
certification scheme for any GME state based on the minimal number of
measurements per observes. A related question that arises here is whether the
ideas presented above can be combined with the results of Ref [44, 45] to
design self-testing schemes for multipartite states which are based on finite
statistics. What is more, it would have been interesting to explore whether
the 1SDI certification scheme for graph states of any local dimension provided
here could be generalized to stabilizer subspaces (cf. Ref. [46]). At the same
time, as far as the experimental testing of our schemes is concerned, it is
crucial to understand how resilient they are to experimental errors.
Additionally, we can use our self-testing results for secure multipartite
randomness generation.

##  V Acknowledgments

This project was funded within the QuantERA II Programme (VERIqTAS project)
that has received funding from the European Union’s Horizon 2020 research and
innovation programme under Grant Agreement No 101017733 and from the Polish
National Science Center (project No 2021/03/Y/ST2/00175). Funded by the
European Union. Views and opinions expressed are however those of the
author(s) only and do not necessarily reflect those of the European Union or
the European Commission. Neither the European Union nor the granting authority
can be held responsible for them. This project has received funding from the
European Union’s Horizon Europe research and innovation programme under grant
agreement No 101080086 NeQST.

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Supplementary Material: Almost device-independent certification of GME states
with minimal measurements

##  Appendix A Appendix A: Self-testing any Graph state

Consider a connected graph $G$ with $N$ vertices and edges denoted as $e$ with
their multiplicity denoted by $\gamma_{i,j}$. Then the graph state $|G\rangle$
corresponding to graph $G$ is the unique state that stabilizes the following
$N$ operators

$\displaystyle S(G)_{d,i}=X_{i}\otimes\bigotimes_{j\in
n(i)}Z_{j}^{\gamma_{i,j}}\qquad i=1,2,\ldots,N$ (30)

Let us now recall the steering operator introduced in the main text.

$\mathcal{I}_{d}(G,N)=X_{A}\otimes\bigotimes_{j\in
n(0)}(B_{j,0})^{\gamma_{0,j}}+\sum_{j\in n(0)}Z_{A}^{\gamma_{j,0}}\otimes
B_{j,1}\otimes\bigotimes_{j^{\prime}\in
n(j)\setminus\\{0\\}}(B_{j^{\prime},0})^{\gamma_{j,j^{\prime}}}+\sum_{j\notin
n(0)\cup{0}}B_{j,1}\otimes\bigotimes_{j^{\prime}\in
n(j)}(B_{j^{\prime},0})^{\gamma_{j,j^{\prime}}}+h.c.$ (31)

where $h.c.$ denotes the Hermitian conjugate of the operator on its left-hand
side. Also notice that in the above operator $\gamma_{i,j}=\gamma_{j,i}$ for
any $i,j$ where $\gamma_{i,j}$ is the number of edges connecting vector $i,j$
and $n(j)$ is the set of vertices connected to $j$ vertex. The cardinality of
$n(j)$ is denoted as $\overline{n}_{j}$. Now, the steering inequality is given
by

$\langle\mathcal{I}_{d}(G,N)\rangle\leqslant\beta_{L}.$ (32)

The classical bound $\beta_{L}$ is given below.

######  Fact 1.

The local hidden state (LHS) bound $\beta_{L}$ of the steering functional
$\langle\mathcal{I}_{d}(G,N)\rangle$ where $\mathcal{I}_{d}(G,N)$ is given in
(31) is upper bounded by

$\displaystyle\beta_{L}\leqslant\max_{\rho}\left(2|\langle
X_{A}\rangle_{\rho}|+2\sum_{j\in n(0)}|\langle
Z_{A}^{\gamma_{j,0}}\rangle_{\rho}|\right)+2(N-\overline{n}_{0}-1)$ (33)

where $Z$ and $X$ are given in eq. (1). Furthermore, $\beta_{L}<2N$.

###### Proof.

As shown in [30], for any LHS model with trusted Alice, the expectation values
of the joint observables take the form

$\displaystyle\langle A^{k}_{x}\otimes
B^{l}_{y}\rangle_{LHS}=\sum_{\lambda}p(\lambda)\langle
A^{k}_{x}\rangle_{\rho_{\lambda}}\langle B^{l}_{y}\rangle_{\lambda}.$ (34)

Consequently, we can express the steering inequality
$\langle\mathcal{I}_{d}(G,N)\rangle$, where $\mathcal{I}_{d}(G,N)$ is given in
(31), for any LHS model as

$\displaystyle\langle\mathcal{I}_{d}(G,N)\rangle_{LHS}=\sum_{\lambda}p(\lambda)\langle\mathcal{I}_{d}(G,N)\rangle_{\lambda,LHS},$
(35)

where $\langle\mathcal{I}_{d}(G,N)\rangle_{\lambda,LHS}$ is

$\displaystyle\langle\mathcal{I}_{d}(G,N)\rangle_{\lambda,LHS}=\langle
X_{A}\rangle_{\rho_{\lambda}}\langle\bigotimes_{j\in
n(0)}(B_{j,0})^{\gamma_{0,j}}\rangle_{{\lambda}}+\sum_{j\in n(0)}\langle
Z_{A}^{\gamma_{j,0}}\rangle_{\rho_{\lambda}}\langle
B_{j,1}\otimes\bigotimes_{j^{\prime}\in
n(j)\setminus\\{0\\}}(B_{j^{\prime},0})^{\gamma_{j,j^{\prime}}}\rangle_{{\lambda}}$
$\displaystyle+\sum_{j\notin n(0)\cup{0}}\langle
B_{j,1}\rangle_{\rho_{\lambda}}\langle\bigotimes_{j^{\prime}\in
n(j)}(B_{j^{\prime},0})^{\gamma_{j,j^{\prime}}}\rangle_{{\lambda}}+\langle
h.c\rangle_{LHS}.\qquad$ (36)

As $\langle B^{l}_{y}\rangle_{\lambda}\leq 1$, we obtain that

$\displaystyle\langle\mathcal{I}_{d}(G,N)\rangle_{LHS}\leqslant\sum_{\lambda}p(\lambda)\left(2|\langle
X_{A}\rangle_{\rho_{\lambda}}|+2\sum_{j\in n(0)}|\langle
Z_{A}^{\gamma_{j,0}}\rangle_{\rho_{\lambda}}|\right)+2(N-\overline{n}_{0}-1),$
(37)

which can be upper bounded as

$\displaystyle\langle\mathcal{I}_{d}(G,N)\rangle_{LHS}\leqslant\sum_{\lambda}p(\lambda)\max_{\rho}\left(2|\langle
X_{A}\rangle_{\rho}|+2\sum_{j\in n(0)}|\langle
Z_{A}^{\gamma_{j,0}}\rangle_{\rho}|\right)+2(N-\overline{n}_{0}-1).$ (38)

By using the fact that $\sum_{\lambda}p(\lambda)=1$, the above finally gives
us

$\displaystyle\langle\mathcal{I}_{d}(G,N)\rangle_{LHS}\leqslant\max_{\rho}\left(2|\langle
X_{A}\rangle_{\rho}|+2\sum_{j\in n(0)}|\langle
Z_{A}^{\gamma_{j,0}}\rangle_{\rho}|\right)+2(N-\overline{n}_{0}-1).$ (39)

One can numerically evaluate the above quantity for any $d$ and any graph. Let
us finally observe from the above expression that the LHS bound of the
steering functional $\mathcal{I}_{d}(G,N)$ satisfies $\beta_{L}<2N$ because
$Z$ and $X$ are mutually incompatible and there is no quantum state that can
simultaneously stabilize $Z$ and $X$. ∎

Let us now compute the quantum bound of the steering functional
$\langle\mathcal{I}_{d}(G,N)\rangle$.

######  Fact 2.

The quantum bound of the steering functional
$\langle\mathcal{I}_{d}(G,N)\rangle$ where $\mathcal{I}_{d}(G,N)$ is given in
(31) is given by $\beta_{Q}=2N$ and is achieved by the graph state $|G\rangle$
with the observables

$B_{j,0}=Z,\qquad B_{j,1}=X.$ (40)

###### Proof.

Let us first observe that the number of terms in the steering operator (31) is
equal to twice the number of stabilizers $S(G)_{d,i}$ (11), that is, $2N$.
Since the expectation value of each term is upper-bounded by $1$, thus, we
have that $\langle\mathcal{I}_{d}(G,N)\rangle\leq 2N$.

Consider now that the unknown observables $B_{i,j}$ are given by

$\displaystyle B_{j,0}=Z,\qquad B_{j,1}=X$ (41)

along with the graph state $|G\rangle$, the state that is uniquely stabilized
by the $S(G)_{i}$ operators. It is straightforward to observe that using these
states and observables one can attain the quantum bound. ∎

Now notice that the steering operator consists $2N$ terms and consequently,
the algebraic bound of $\langle\mathcal{I}_{d}(G,N)\rangle$ is also $2N$ as
all the operators $B^{\dagger}_{i,j}B_{i,j}=\mathbbm{1}$. Thus, it is
necessary that any state $|\psi_{N}\rangle$ and observables $B_{i,j}$ that
achieves the quantum bound must satisfy the following conditions

$\langle\psi_{N}|\overline{S}(G)_{i}|\psi_{N}\rangle=1,\qquad
i={1,2,\ldots,N},$ (42)

where $\overline{S}(G)_{i}$ represents each term of the steering operator
(31). Now, using Cauchy-Schwarz inequality, one can conclude from here that

$\forall i\qquad\overline{S}(G)_{i}|\psi_{N}\rangle=|\psi_{N}\rangle.$ (43)

The above condition would be extremely useful for certifying any graph state.
Let us now establish the following theorem.

######  Theorem 1.

Assume that the steering inequality (32) is maximally violated when the
$0^{\text{th}}$ party Alice chooses the observables $A_{i}$ defined to be
$A_{0}=Z$, $A_{1}=X$ and all the other parties choose their observables as
$B_{j,i}$ for $i\in\\{0,1\\}$ and $j=1,2\ldots,N-1$ acting
$\mathcal{H}_{B_{j}}$. Let us say that the state which attains this violation
is given by
$|\psi_{N}\rangle_{A\mathbf{B}}\in\mathbbm{C}^{d}\otimes\bigotimes_{j=1}^{N-1}\mathcal{H}_{B_{j}}\otimes\mathcal{H}_{E}$,
then the following statement holds true for any $d$:
$\mathcal{H}_{B_{j}}=\mathcal{H}_{B^{\prime}_{j}}\otimes\mathcal{H}_{B^{\prime\prime}_{j}}$,
where $\mathcal{H}_{B^{\prime}_{j}}\equiv\mathbbm{C}^{d}$,
$\mathcal{H}_{B^{\prime\prime}_{j}}$ is some finite-dimensional Hilbert space,
and there exists a local unitary transformation on Bob’s side
$U_{j}:\mathcal{H}_{B_{j}}\rightarrow\mathcal{H}_{B_{j}}$, such that

$\displaystyle\forall j,\quad
U_{j}\,B_{j,0}\,U_{j}^{\dagger}=Z_{B^{\prime}_{j}}\otimes\mathbbm{1}_{B^{\prime\prime}_{j}},\qquad
U_{j}\,B_{j,1}\,U_{j}^{\dagger}=X_{B^{\prime}_{j}}\otimes\mathbbm{1}_{B^{\prime\prime}_{j}}$
(44)

where ${B^{\prime\prime}}$ denotes Bob’s auxiliary system and the state
$|\psi_{N}\rangle$ is given by,

$\displaystyle\left(\mathbbm{1}_{AE}\otimes\bigotimes_{i=1}^{N-1}U_{j}\right)|\psi_{N}\rangle_{A\mathbf{B}E}=|G\rangle_{A\mathbf{B}^{\prime}}\otimes|\xi\rangle_{\mathbf{B}^{\prime\prime}E},$
(45)

where
$|\xi\rangle_{\mathbf{B}^{\prime\prime}E}\in\bigotimes_{j=1}^{N-1}\mathcal{H}_{B_{j}^{\prime\prime}}\otimes\mathcal{H}_{E}$
denotes the auxiliary state.

###### Proof.

Let us begin by considering the first term and second term of the steering
operator $\eqref{graphsteeop}$ corresponding to the stabilizers when choosing
the $0^{\text{th}}$ vertex and vertices connected to it respectively. As
concluded in Eq. (43), we have that

$\displaystyle X_{A}\otimes\bigotimes_{j\in
n(0)}(B_{j,0})^{\gamma_{0,j}}|\psi_{N}\rangle=|\psi_{N}\rangle,$ (46)

and

$\displaystyle Z_{A}^{\gamma_{j,0}}\otimes
B_{j,1}\otimes\bigotimes_{j^{\prime}\in
n(j)\setminus\\{0\\}}(B_{j^{\prime},0})^{\gamma_{j,j^{\prime}}}|\psi_{N}\rangle=|\psi_{N}\rangle\qquad\forall
j\in n(0).$ (47)

Let us now choose a particular $k\in n(0)$ and rewrite the above conditions
as,

$\displaystyle X_{A}\otimes(B_{k,0})^{\gamma_{0,k}}\otimes\bigotimes_{j\in
n(0)\setminus\\{k\\}}(B_{j,0})^{\gamma_{0,j}}|\psi_{N}\rangle=|\psi_{N}\rangle,$
(48)

and

$\displaystyle Z_{A}^{\gamma_{k,0}}\otimes
B_{k,1}\otimes\bigotimes_{j^{\prime}\in
n(k)\setminus\\{0\\}}(B_{j^{\prime},0})^{\gamma_{k,j^{\prime}}}|\psi_{N}\rangle=|\psi_{N}\rangle.$
(49)

Since the observables $X$ and $Z$ are unitary and $\gamma_{0,k}=\gamma_{k,0}$,
we have that

$\displaystyle(B_{k,0})^{\gamma_{0,k}}\otimes\bigotimes_{j\in
n(0)\setminus\\{k\\}}(B_{j,0})^{\gamma_{0,j}}|\psi_{N}\rangle=X_{A}^{-1}|\psi_{N}\rangle$
(50)

and

$\displaystyle B_{k,1}\otimes\bigotimes_{j^{\prime}\in
n(k)\setminus\\{0\\}}(B_{j^{\prime},0})^{\gamma_{k,j^{\prime}}}|\psi_{N}\rangle=Z_{A}^{-\gamma_{0,k}}|\psi_{N}\rangle.$
(51)

Now, multiplying $Z_{A}^{-\gamma_{0,k}}$ to the condition (50) and then using
Eq. (51), we get that

$\displaystyle(B_{k,0})^{\gamma_{0,k}}B_{k,1}\otimes\left(\bigotimes_{j\in
n(0)\setminus\\{k\\}}(B_{j,0})^{\gamma_{0,j}}\right)\left(\bigotimes_{j^{\prime}\in
n(k)\setminus\\{0\\}}(B_{j^{\prime},0})^{\gamma_{k,j^{\prime}}}\right)|\psi_{N}\rangle=Z_{A}^{-\gamma_{0,k}}X_{A}^{-1}|\psi_{N}\rangle.$
(52)

Then, multiplying $X_{A}^{-1}$ to the condition (51) and then using Eq. (50),
we get that

$\displaystyle
B_{k,1}(B_{k,0})^{\gamma_{0,k}}\otimes\left(\bigotimes_{j^{\prime}\in
n(k)\setminus\\{0\\}}(B_{j^{\prime},0})^{\gamma_{k,j^{\prime}}}\right)\left(\bigotimes_{j\in
n(0)\setminus\\{k\\}}(B_{j,0})^{\gamma_{0,j}}\right)|\psi_{N}\rangle=X_{A}^{-1}Z_{A}^{-\gamma_{0,k}}|\psi_{N}\rangle.$
(53)

Notice, that terms inside the large brackets in both the above formulas
commute. Now, by using the fact that
$\omega^{\gamma_{0,k}}Z_{A}^{-\gamma_{0,k}}X_{A}^{-1}=X_{A}^{-1}Z_{A}^{-m_{e,k0}}$,
we can multiply (52) by $\omega^{\gamma_{0,k}}$ and subtract (53) to obtain
that

$\displaystyle\left(\omega^{\gamma_{0,k}}(B_{k,0})^{\gamma_{0,k}}B_{k,1}-B_{k,1}(B_{k,0})^{\gamma_{0,k}}\right)\otimes\left(\bigotimes_{j\in
n(0)\setminus\\{k\\}}(B_{j,0})^{\gamma_{0,j}}\right)\left(\bigotimes_{j^{\prime}\in
n(k)\setminus\\{0\\}}(B_{j^{\prime},0})^{\gamma_{k,j^{\prime}}}\right)|\psi_{N}\rangle=0.$
(54)

As the terms inside the large brackets in the above expression are unitary and
thus invertible, we arrive at a simple condition

$\displaystyle\left(\omega^{\gamma_{0,k}}(B_{k,0})^{\gamma_{0,k}}B_{k,1}-B_{k,1}(B_{k,0})^{\gamma_{0,k}}\right)\otimes\mathbbm{1}_{A\mathbf{B}E\setminus\\{B_{k}\\}}|\psi_{N}\rangle=0.$
(55)

By taking a partial trace over all the subsystems except the $k^{\text{th}}$
one, we get that

$\displaystyle\left(\omega^{\gamma_{0,k}}(B_{k,0})^{\gamma_{0,k}}B_{k,1}-B_{k,1}(B_{k,0})^{\gamma_{0,k}}\right)\rho_{k,N}=0,$
(56)

where
$\rho_{k,N}=\mathrm{Tr}_{A\mathbf{B}E\setminus\\{B_{k}\\}}(|\psi_{N}\rangle\\!\langle\psi_{N}|)$.
As discussed before, the local states of each party are full-rank and thus
invertible. Consequently, we finally have that

$\displaystyle\omega^{\gamma_{0,k}}(B_{k,0})^{\gamma_{0,k}}B_{k,1}=B_{k,1}(B_{k,0})^{\gamma_{0,k}}.$
(57)

As proven in [13], if two unitary observables $B_{k,0}$ and $B_{k,1}$, for
which $B_{k,0}^{d}=B_{k,1}^{d}=\mathbbm{1}$, satisfy relation (57), then there
exists a unitary $U_{k}:\mathcal{H}_{B_{k}}\rightarrow\mathcal{H}_{B_{k}}$
such that

$\displaystyle\quad
U_{k}\,B_{k,0}\,U_{k}^{\dagger}=Z_{B^{\prime}_{k}}\otimes\mathbbm{1}_{B^{\prime\prime}_{k}},\qquad
U_{k}\,B_{k,1}\,U_{k}^{\dagger}=X_{B^{\prime}_{k}}\otimes\mathbbm{1}_{B^{\prime\prime}_{k}}.$
(58)

Similarly, we can reach the above conclusion for any vertex $j\in n(0)$. Using
these derived observables, we then find the observables for vertices connected
with any vertex $j\in n(0)$ in exactly the same manner as done above. Since
any vertex is connected to at least one other vertex, we continue this until
all the observables are found and we get that

$\displaystyle
U_{j}\,B_{j,0}\,U_{j}^{\dagger}=Z_{B^{\prime}_{j}}\otimes\mathbbm{1}_{B^{\prime\prime}_{j}},\qquad
U_{j}\,B_{j,1}\,U_{j}^{\dagger}=X_{B^{\prime}_{j}}\otimes\mathbbm{1}_{B^{\prime\prime}_{j}}\qquad\forall
j.$ (59)

Let us now characterize the state $|\psi_{N}\rangle$ that achieves the quantum
bound of the steering functional $\langle\mathcal{I}_{d}(G,N)\rangle$ (31).
Notice now that by putting into the condition (43) the derived observables
from (59), we get that

$\displaystyle\forall
i\qquad\overline{S}(G)_{i}|\psi_{N}\rangle=\left(\bigotimes_{i=1}^{N-1}U_{j}^{\dagger}\right)S(G)_{d,i,AB_{1}^{\prime\prime}\ldots
B_{N}^{\prime\prime}}\otimes\mathbbm{1}_{B_{1}^{\prime\prime}\ldots
B_{N}^{\prime\prime}E}\left(\bigotimes_{i=1}^{N-1}U_{j}\right)|\psi_{N}\rangle=|\psi_{N}\rangle,$
(60)

where $S(G)_{d,i}$ are the stabilizing operators of a graph state $|G\rangle$
of the form (11). As discussed above, the graph state $|G\rangle$ is the
unique eigenstate of such operators with eigenvalue $1$. Consequently, we have
that

$\displaystyle\left(\bigotimes_{i=1}^{N-1}U_{j}\right)|\psi_{N}\rangle_{A\mathbf{B}E}=|G\rangle_{A\mathbf{B}^{\prime}}\otimes|\xi\rangle_{\mathbf{B}^{\prime\prime}E},$
(61)

where
$|\xi\rangle_{\mathbf{B}^{\prime\prime}E}\in\bigotimes_{j=1}^{N-1}\mathcal{H}_{B_{j}^{\prime\prime}}\otimes\mathcal{H}_{E}$
denotes the auxiliary state. ∎

##  Appendix B Appendix B: Self-testing Schmidt states

The Schmidt states we consider have the form

$\displaystyle|\psi(\boldsymbol{\alpha})\rangle_{A\mathbf{B}}=\sum_{i=0}^{d-1}\alpha_{i}|i\rangle^{\otimes
N},$ (62)

where $d=\min\\{d_{A},d_{B_{i}}\\}$ and $\boldsymbol{\alpha}$ is a vector
composed of the Schmidt coefficients $\alpha_{i}$, such that $0<\alpha_{i}<1$
for all $i$ and $\sum_{i=0}^{d-1}\alpha_{i}^{2}=1$. Next, we are going to
introduce a steering operator that provides an inequality maximally violated
by states of the form (62).

Let us now recall the steering operator for Schmidt states introduced in the
main text:

$\displaystyle\mathcal{S}_{d}(\boldsymbol{\alpha},N)=\sum_{k=1}^{d-1}\left[\left(\sum_{j=1}^{N-1}A_{0}^{k}\otimes
B_{j,0}^{k}\right)+\gamma(\boldsymbol{\alpha})A_{1}^{k}\otimes\bigotimes_{j=1}^{N-1}B_{j,1}^{k}+\delta_{k}(\boldsymbol{\alpha})A_{0}^{k}\right],$
(63)

where

$\displaystyle\gamma(\boldsymbol{\alpha})=\frac{d}{\sum_{\begin{subarray}{c}i,j=0\\\
i\neq
j\end{subarray}}^{d-1}\frac{\alpha_{i}}{\alpha_{j}}},\quad\delta_{k}(\boldsymbol{\alpha})=-\frac{\gamma(\boldsymbol{\alpha})}{d}\sum_{\begin{subarray}{c}i,j=0\\\
i\neq j\end{subarray}}^{d-1}\frac{\alpha_{i}}{\alpha_{j}}\omega^{k(d-j)}.$
(64)

In our scenario, we consider that all the $N-1$ Bobs perform two arbitrary
measurements denoted by $B_{i_{0}}$ and $B_{i_{1}}$. Also, we assume that the
local dimensions of the state they share are unknown. At the same time,
because Alice’s device is trusted, her measurements are completely
characterized and the local dimension of the shared state on Alice’s side is
known. Alice’s observables are the generalized Pauli $Z$ and $X$ operators

$\displaystyle Z_{d}=\sum_{i=0}^{d-1}\omega^{i}|i\rangle\langle i|,\qquad
X_{d}=\sum_{i=0}^{d-1}|i+1\rangle\langle i|.$ (65)

The corresponding steering inequality is given by

$\displaystyle\langle\mathcal{S}_{d}(\boldsymbol{\alpha},N)\rangle\leqslant\beta_{L}$
(66)

where $\beta_{L}$ is the LHS bound given below.

######  Fact 3.

The local hidden state (LHS) bound $\beta_{L}$ of the steering functional
$\langle S_{d}(\boldsymbol{\alpha})\rangle$ where $S_{d}(\boldsymbol{\alpha})$
is given in (63) is upper bounded by

$\displaystyle\beta_{L}\leqslant\max_{\begin{subarray}{c}|\eta_{0}|,\ldots,|\eta_{d-1}|\\\
|\eta_{0}|^{2}+\ldots+|\eta_{d-1}|^{2}=1\end{subarray}}\left(d(N-1)\max_{a}{|\eta_{a}|^{2}}+\gamma(\boldsymbol{\alpha})\left[\left(\sum_{i=0}^{d-1}|\eta_{i}|\right)^{2}-\sum_{i=0}^{d-1}\alpha_{i}\sum_{a=0}^{d-1}\frac{|\eta_{a}|^{2}}{\alpha_{a}}\right]\right)-(N-2).$
(67)

Furthermore, $\beta_{L}<(d-1)(N-1)+1$.

###### Proof.

In order to determine the LHS bound $\beta_{L}$ of the inequality in Eq. (63),
we first express the functional $\mathcal{S}_{d}(\boldsymbol{\alpha},N)$ in
the probability picture as following

$\displaystyle S_{d}(\boldsymbol{\alpha})=\ $ $\displaystyle
d\sum_{j=1}^{N-1}\sum_{a,b_{j}=0}^{d-1}c_{a,b_{j}}p(a,b_{j}|0,0)+\gamma(\boldsymbol{\alpha})\left(d\sum_{a,\mathbf{b}=0}^{d-1}c_{a,\mathbf{b}}p(a,\mathbf{b}|1,1,\ldots,1)-\sum_{\begin{subarray}{c}i,a=0\\\
i\neq a\end{subarray}}^{d-1}\alpha_{i}\frac{p(a|0)}{\alpha_{a}}\right)$
$\displaystyle-(N-1)-\gamma(\boldsymbol{\alpha})-\delta_{0}(\boldsymbol{\alpha}),$
(68)

where $c_{a,b}=1$ if $a\oplus_{d}b=0$ and 0 otherwise, where $\oplus_{d}$ is
the addition modulo $d$. Also, $c_{a,\mathbf{b}}=1$ if
$a\oplus_{d}b_{1}\oplus_{d}\ldots\oplus_{d}b_{N-1}=0$ and 0 otherwise. We
implement this notation because
$\sum_{k=0}^{d-1}\omega^{kn(a-b)}=d\delta_{a,b}$ for any $n$. Likewise,
$\sum_{k=0}^{d-1}\omega^{k(d+a-b)}=d\delta_{a,b}$. We note, from Eq. (64) that
$\delta_{0}(\boldsymbol{\alpha})=-1$, which results in

$\displaystyle S_{d}(\boldsymbol{\alpha})=\ $ $\displaystyle
d\sum_{j=1}^{N-1}\sum_{a,b_{j}=0}^{d-1}c_{a,b_{j}}p(a,b_{j}|0,0)+\gamma(\boldsymbol{\alpha})\left(d\sum_{a,\mathbf{b}=0}^{d-1}c_{a,\mathbf{b}}p(a,\mathbf{b}|1,1,\ldots,1)-\sum_{i=0}^{d-1}\alpha_{i}\sum_{a=0}^{d-1}\frac{p(a|0)}{\alpha_{a}}\right)-(N-2),$
(69)

We consider the LHS model in which the correlations are given as

$\displaystyle
p(a,\mathbf{b}|x,\mathbf{y})=\sum_{\lambda\in\Lambda}p(\lambda)p(a|x,\rho^{\lambda}_{A})p(\mathbf{b}|\mathbf{y},\lambda),$
(70)

where $\Lambda$ is some set of hidden variables $\lambda$ shared between Alice
and all the Bobs with probability distribution $p(\lambda)$, whereas
$p(a|x,\rho^{\lambda}_{A})$ and $p(\mathbf{b}|\mathbf{y},\lambda)$ are local
probability distributions. Recall that Alice’s distributions are quantum and
depend on the hidden state $\rho^{\lambda}_{A}$. The LHS model transforms our
steering functional as

$\displaystyle S_{d}(\boldsymbol{\alpha})=$ $\displaystyle
d\sum_{j=1}^{N-1}\sum_{a=0}^{d-1}\sum_{\lambda\in\Lambda}p(\lambda)p(a|0,\rho^{\lambda}_{A})p(d-a|0,\lambda)+\gamma(\boldsymbol{\alpha})\Biggl{(}d\sum_{a=0}^{d-1}\sum_{\lambda\in\Lambda}p(\lambda)p(a|1,\rho^{\lambda}_{A})p(\mathbf{b}|1,\ldots,1,\lambda)$
$\displaystyle-\sum_{i=0}^{d-1}\alpha_{i}\sum_{a=0}^{d-1}\frac{p(a|0,\rho_{A})}{\alpha_{a}}\Biggr{)}-(N-2),$
(71)

where $b_{j}=d-a$ for every $j$ in the first term,
$a\oplus_{d}b_{1}\oplus_{d}\ldots\oplus_{d}b_{N-1}=0$ in the second term, and
in the last term we have used the fact that Alice’s distributions are quantum
and substituted $\rho_{A}=\sum_{\lambda}\rho^{\lambda}_{A}$. Now, we note that
the first two terms in the above expression are bounded from above in the
following way

$\displaystyle\sum_{a=0}^{d-1}\sum_{\lambda\in\Lambda}p(\lambda)p(a|0,\rho^{\lambda}_{A})p(d-a|0,\lambda)\leqslant\sum_{\lambda\in\Lambda}p(\lambda)\max_{a}{p(a|0,\rho^{\lambda}_{A})}\leqslant\max_{|\psi\rangle\in\mathcal{C}^{d}}\max_{a}{p(a|0,|\psi\rangle)},$
$\displaystyle\sum_{a=0}^{d-1}\sum_{\lambda\in\Lambda}p(\lambda)p(a|1,\rho^{\lambda}_{A})p(\mathbf{b}|1,\ldots,1,\lambda)\leqslant\sum_{\lambda\in\Lambda}p(\lambda)\max_{a}{p(a|1,\rho^{\lambda}_{A})}\leqslant\max_{|\psi\rangle\in\mathcal{C}^{d}}\max_{a}{p(a|1,|\psi\rangle)},$
(72)

where to get the first inequalities in both cases we used the fact that
probabilities are normalized. To get the second inequality we exploit the fact
that Alice’s probability distributions are quantum and hence linear functions
of the state. Substituting this in our function, we get

$\displaystyle
S_{d}(\boldsymbol{\alpha})\leqslant\max_{\psi\in\mathcal{C}^{d}}\left[d(N-1)\max_{a}{p(a|0,|\psi\rangle)}+\gamma(\boldsymbol{\alpha})\left(d\max_{a}p(a|1,|\psi\rangle)-\sum_{i=0}^{d-1}\alpha_{i}\sum_{a=0}^{d-1}\frac{p(a|0,|\psi\rangle)}{\alpha_{a}}\right)\right]-(N-2),$
(73)

where we have used the fact that the maximization over the state
$|\psi\rangle$ is happening over the whole expression. We can further write
the state $|\psi\rangle$ in the computational basis as
$|\psi\rangle=\sum_{i}\eta_{i}|i\rangle$ and then substitute the explicit
forms of Alice’s observables $X$ and $Z$, which results in

$\displaystyle
S_{d}(\boldsymbol{\alpha})\leqslant\max_{\begin{subarray}{c}|\eta_{0}|,\ldots,|\eta_{d-1}|\\\
|\eta_{0}|^{2}+\ldots+|\eta_{d-1}|^{2}=1\end{subarray}}\left(d(N-1)\max_{a}{|\eta_{a}|^{2}}+\gamma(\boldsymbol{\alpha})\left[\left(\sum_{i=0}^{d-1}|\eta_{i}|\right)^{2}-\sum_{i=0}^{d-1}\alpha_{i}\sum_{a=0}^{d-1}\frac{|\eta_{a}|^{2}}{\alpha_{a}}\right]\right)-(N-2).$
(74)

If we can now show that the terms inside the square bracket are less than
zero, we can conclude that $S_{d}(\boldsymbol{\alpha})\leqslant(d-1)(N-1)+1$.
To do this, we notice the following

$\displaystyle\left(\sum_{i=0}^{d-1}|\eta_{i}|\right)^{2}=\left(\sum_{i=0}^{d-1}\sqrt{\alpha_{i}}\frac{|\eta_{i}|}{\sqrt{\alpha_{i}}}\right)^{2}\leqslant\sum_{i=0}^{d-1}\alpha_{i}\sum_{j=0}^{d-1}\frac{|\eta_{j}|^{2}}{\alpha_{j}},$
(75)

where we have used the Cauchy-Schwarz inequality to get the last expression.
Consequently, we have $S_{d}(\boldsymbol{\alpha})\leqslant(d-1)(N-1)+1$.
However, from Eq. (74) we see that $S_{d}(\boldsymbol{\alpha})=(d-1)(N-1)+1$
is possible iff $\max_{a}|\eta_{a}|^{2}=1$ as well as the terms inside the
square brackets should also vanish. However, this is possible only if
$\alpha_{i}=\lambda\eta_{i}$. Because
$\sum_{i}\alpha_{i}^{2}=\sum_{i}|\eta_{i}|^{2}=1$, we must have $\lambda=1$.
However, $\alpha_{i}<1$ for al $i$, which does not allow for
$\max_{a}|\eta_{a}|^{2}=1$ to hold. Therefore,
$S_{d}(\boldsymbol{\alpha})=(d-1)(N-1)+1$ is never possible, which implies
$S_{d}(\boldsymbol{\alpha})<(d-1)(N-1)+1$. ∎

Let us now find the quantum bound of the steering functional
$\langle\mathcal{S}_{d}(\boldsymbol{\alpha},N)\rangle$.

######  Fact 4.

The quantum bound of the steering functional
$\langle\mathcal{S}_{d}(\boldsymbol{\alpha},N)\rangle$ where
$\mathcal{S}_{d}(\boldsymbol{\alpha},N)$ is given in (63) is given by
$\beta_{Q}=(d-1)(N-1)+1$ and is achieved by the Schmidt state
$|\psi(\alpha)\rangle$ (62) with the observables

$B_{j,0}=Z,\qquad B_{j,1}=X.$ (76)

###### Proof.

Let us rewrite our functional as

$\displaystyle\mathcal{S}_{d}(\boldsymbol{\alpha},N)=\sum_{k=1}^{d-1}\left(\sum_{j=1}^{N-1}A_{0}^{k}\otimes
B_{j,0}^{k}\right)+T_{d}(\boldsymbol{\alpha},N)$ (77)

with

$\displaystyle
T_{d}(\boldsymbol{\alpha},N)=\sum_{k=1}^{d-1}\left(\gamma(\boldsymbol{\alpha})A_{1}^{k}\otimes\bigotimes_{j=1}^{N-1}B_{j,1}^{k}+\delta_{k}(\boldsymbol{\alpha})A_{0}^{k}\right).$
(78)

Since, $A_{0}$ is unitary and
$(B^{k}_{j|0})^{\dagger}B^{k}_{j|0}\leqslant\mathbbm{1}$ for any $k,j$, we
have $|\langle\psi_{N}|A_{0}^{k}\otimes B_{j,0}^{k}|\psi_{N}\rangle|\leqslant
1$ for all $k\in\\{1,\ldots,d-1\\}$, $j\in\\{1,\ldots,N-1\\}$ and
$|\psi_{N}\rangle$. This implies that

$\displaystyle\max_{|\psi_{N}\rangle}\langle\psi_{N}|\mathcal{S}_{d}(\boldsymbol{\alpha},N)|\psi_{N}\rangle\leqslant(d-1)(N-1)+\max_{|\psi_{N}\rangle}\langle\psi_{N}|T_{d}(\boldsymbol{\alpha},N)|\psi_{N}\rangle$
(79)

Therefore, we only need to prove that
$\langle\psi_{N}|T_{d}(\boldsymbol{\alpha},N)|\psi_{N}\rangle\leqslant 1$. We
can represent any pure state in the Hilbert space
$\mathcal{H}_{A}\bigotimes_{i=1}^{N-1}\mathcal{H}_{B_{i}}$ as

$\displaystyle|\psi_{N}\rangle_{A\mathbf{B}}=\sum_{i=0}^{d-1}\lambda_{i}|i\rangle_{A}|e_{i}\rangle_{\mathbf{B}},$
(80)

where $\lambda_{i}\geqslant 0$ and
$\lambda_{0}^{2}+\ldots+\lambda_{d-1}^{2}=1$, and $|e_{i}\rangle$ are some
vectors in the Hilbert space $\bigotimes_{i=1}^{N-1}\mathcal{H}_{B_{j}}$ such
that $|e_{i}\rangle$ are not necessarily orthogonal. With this representation
we can rewrite $\langle\psi_{N}|T_{d}(\boldsymbol{\alpha},N)|\psi_{N}\rangle$
as

$\displaystyle\langle\psi_{N}|T_{d}(\boldsymbol{\alpha},N)|\psi_{N}\rangle$
$\displaystyle=$
$\displaystyle\sum_{k=1}^{d-1}\sum_{i,j=0}^{d-1}\bigg{[}\gamma(\boldsymbol{\alpha})\lambda_{i}\lambda_{j}\langle
i|A_{1}^{k}|j\rangle\langle
e_{i}|\bigotimes_{j=1}^{N-1}B_{j,1}^{k}|e_{j}\rangle+\delta_{k}(\boldsymbol{\alpha})\lambda_{i}\lambda_{j}\langle
i|A_{0}^{k}|j\rangle\langle{e_{i}}|{e_{j}}\rangle\bigg{]},$ (81)
$\displaystyle=$
$\displaystyle\sum_{k=1}^{d-1}\sum_{i=0}^{d-1}\bigg{[}\gamma(\boldsymbol{\alpha})\lambda_{i}\lambda_{i-k}\langle
e_{i}|\bigotimes_{j=1}^{N-1}B_{j,1}^{k}|e_{j}\rangle+\delta_{k}(\boldsymbol{\alpha})\lambda_{i}^{2}\omega^{ik}\bigg{]},$

where for the second equality we have substituted $A_{i}$ from (1). Next, we
note that

$\displaystyle\sum_{k=1}^{d-1}\sum_{i=0}^{d-1}\delta_{k}(\boldsymbol{\alpha})\lambda_{i}^{2}\omega^{ik}=1+\sum_{k=0}^{d-1}\sum_{i=0}^{d-1}\delta_{k}(\boldsymbol{\alpha})\lambda_{i}^{2}\omega^{ik}=1+\gamma(\boldsymbol{\alpha})-\gamma(\boldsymbol{\alpha})\sum_{i,j=0}^{d-1}\frac{\alpha_{i}}{\alpha_{j}}\lambda_{j}^{2}.$
(82)

where we get the first equality by using $\delta_{0}(\boldsymbol{\alpha})=-1$
and the second equality by using the explicit form of
$\delta_{k}(\boldsymbol{\alpha})$.By substituting the above expression in Eq.
(81), we get

$\displaystyle\langle\psi_{N}|T_{d}(\boldsymbol{\alpha},N)|\psi_{N}\rangle=1+\gamma(\boldsymbol{\alpha})\sum_{k=0}^{d-1}\sum_{i=0}^{d-1}\lambda_{i}\lambda_{i-k}\text{Re}\left(\langle
e_{i}|\bigotimes_{j=1}^{N-1}B_{j,1}^{k}|e_{j}\rangle\right)-\gamma(\boldsymbol{\alpha})\sum_{i,j=0}^{d-1}\frac{\alpha_{i}}{\alpha_{j}}\lambda_{j}^{2},$
(83)

since $\mathcal{S}_{d}$ is an Hermitian operator and every $\lambda_{i}$ is
positive. In this expression we use the fact that Re($z$)$\leqslant|z|$ and
$(B^{k}_{j|0})^{\dagger}B^{k}_{j|0}=\mathbbm{1}$ for any $k,j$, to obtain

$\displaystyle\langle\psi_{N}|T_{d}(\boldsymbol{\alpha},N)|\psi_{N}\rangle\leqslant
1+\gamma(\boldsymbol{\alpha})\left[\left(\sum_{i=0}^{d-1}\lambda_{i}^{2}\right)^{2}-\sum_{i=0}^{d-1}\alpha_{i}\sum_{j=0}^{d-1}\frac{\lambda_{j}^{2}}{\alpha_{j}}\right].$
(84)

We see that the term in square brackets has the same form as the terms in
square brackets of Eq. (74), which is upper bounded by 0. This implies that

$\displaystyle\langle\psi_{N}|T_{d}(\boldsymbol{\alpha},N)|\psi_{N}\rangle\leqslant
1\implies\mathcal{S}_{d}(\boldsymbol{\alpha},N)\leqslant(d-1)(N-1)+1.$ (85)

It is now simple to observe that with the Schmidt state
$|\psi(\boldsymbol{\alpha})\rangle$ from (62) and observables

$B_{j,0}=Z,\qquad B_{j,1}=X,$ (86)

one can attain the value
$\langle\mathcal{S}_{d}(\boldsymbol{\alpha},N)\rangle=(d-1)(N-1)+1$. This
completes the proof. ∎

Consequently, if all the parties attain the quantum bound $(d-1)(N-1)+1$ of
$\mathcal{S}_{d}(\boldsymbol{\alpha},N)$, the following conditions hold true

$\displaystyle\langle\psi_{N}|A_{0}^{k}\otimes
B_{j,0}^{k}|\psi_{N}\rangle=1,\qquad\langle\psi_{N}|T_{d}(\boldsymbol{\alpha},N)|\psi_{N}\rangle=1,$
(87)

for every $j\in\\{1,\ldots,N-1\\}$ and every $k\in\\{1,\ldots,d-1\\}$, where
$T_{d}(\boldsymbol{\alpha},N)$ is given in (78). These conditions hold because
of Eqs. (77), (79), and (85). As a result, we have the following conditions
for the state $|\psi_{N}\rangle$ and observables $A_{i}$ and $B_{j_{i}}$

$\displaystyle A_{0}^{k}\otimes B_{j,0}^{k}|\psi_{N}\rangle=|\psi_{N}\rangle,$
(88)
$\displaystyle\sum_{k=1}^{d-1}\left(\gamma(\boldsymbol{\alpha})A_{1}^{k}\begin{subarray}{c}{N-1}\\\
\bigotimes\\\
j=1\end{subarray}B_{j,1}^{k}+\delta_{k}(\boldsymbol{\alpha})A_{0}^{k}\right)|\psi_{N}\rangle=|\psi_{N}\rangle.$
(89)

for every $j\in\\{1,\ldots,N-1\\}$ and every $k\in\\{1,\ldots,d-1\\}$. Since
$A_{i}^{\prime}s$ and $B_{i}^{\prime}s$ are unitary observables, we can
rewrite Eq. (88) as

$\displaystyle\left(A_{0}^{(-k)}\otimes\mathbbm{1}\right)|\psi_{N}\rangle=\left(\mathbbm{1}\otimes
B_{j,0}^{(k)}\right)|\psi_{N}\rangle$ (90)

The measurements chosen by Alice are $A_{0}=Z$ and $A_{1}=X$. Thus, we have
the following relation among Alice’s observables

$\displaystyle A_{1}A_{0}=\omega A_{0}A_{1}$ (91)

where $\omega=e^{\frac{2\pi i}{d}}$. As we are about to present, the above
relations will be particularly useful for self-testing the Schmidt states.

######  Theorem 2.

Assume that the steering inequality (66) is maximally violated when the
$0^{\text{th}}$ party Alice chooses the observables $A_{i}$ defined to be
$A_{0}=Z$, $A_{1}=X$ and all the other parties choose their observables as
$B_{j,i}$ for $i\in\\{0,1\\}$ and $j=1,2\ldots,N-1$ acting
$\mathcal{H}_{B_{j}}$. Let us say that the state which attains this violation
is given by
$|\psi_{N}\rangle_{A\mathbf{B}E}\in\mathbbm{C}^{d}\otimes\bigotimes_{j=1}^{N-1}\mathcal{H}_{B_{j}}\otimes\mathcal{H}_{E}$,
then the following statement holds true for any $d$:
$\mathcal{H}_{B_{j}}=\mathcal{H}_{B^{\prime}_{j}}\otimes\mathcal{H}_{B^{\prime\prime}_{j}}$,
where $\mathcal{H}_{B^{\prime}_{j}}\equiv\mathbbm{C}^{d}$,
$\mathcal{H}_{B^{\prime\prime}_{j}}$ is some finite-dimensional Hilbert space,
and there exists a local unitary transformation on Bob’s side
$U_{j}:\mathcal{H}_{B_{j}}\rightarrow\mathcal{H}_{B_{j}}$, such that

$\displaystyle\forall j,\quad
U_{j}\,B_{j,0}\,U_{j}^{\dagger}=Z_{B^{\prime}_{j}}\otimes\mathbbm{1}_{B^{\prime\prime}_{j}},\qquad
U_{j}\,B_{j,1}\,U_{j}^{\dagger}=X_{B^{\prime}_{j}}\otimes\mathbbm{1}_{B^{\prime\prime}_{j}}$
(92)

where ${B^{\prime\prime}}$ denotes Bob’s auxiliary system and the state
$|\psi_{N}\rangle$ is given by,

$\displaystyle\left(\mathbbm{1}_{AE}\otimes\bigotimes_{i=1}^{N-1}U_{j}\right)|\psi_{N}\rangle_{A\mathbf{B}E}=|\psi(\boldsymbol{\alpha})\rangle_{A\mathbf{B}^{\prime}}\otimes|\xi\rangle_{\mathbf{B}^{\prime\prime}E},$
(93)

where
$|\xi\rangle_{\mathbf{B}^{\prime\prime}E}\in\bigotimes_{j=1}^{N-1}\mathcal{H}_{B_{j}^{\prime\prime}}\otimes\mathcal{H}_{E}$
denotes the auxiliary state.

###### Proof.

We begin by taking (89) and multiplying it by $B_{l,1}\otimes\mathbbm{1}$ on
both sides,

$\displaystyle\left(\sum_{k=1}^{d-1}\gamma(\boldsymbol{\alpha})A_{1}^{k}\otimes
B_{l,1}^{k+1}\begin{subarray}{c}{N-1}\\\ \bigotimes\\\ j=1\\\ j\neq
l\end{subarray}B_{j,1}^{k}\right)|\psi_{N}\rangle=\left(\mathbbm{1}-\sum_{k=1}^{d-1}\delta_{k}(\boldsymbol{\alpha})A_{0}^{k}\right)\otimes
B_{l,1}|\psi_{N}\rangle$ (94)

Now, we multiply the above by $B_{l,0}\otimes\mathbbm{1}$ on both the sides,

$\displaystyle\left(\sum_{k=1}^{d-1}\delta_{k}(\boldsymbol{\alpha})A_{1}^{k}\otimes
B_{l,0}B_{l,1}^{k+1}\begin{subarray}{c}{N-1}\\\ \bigotimes\\\ j=1\\\ j\neq
l\end{subarray}B_{j,1}^{k}\right)|\psi_{N}\rangle=\left(\mathbbm{1}-\sum_{k=1}^{d-1}\delta_{k}(\boldsymbol{\alpha})A_{0}^{k}\right)\otimes
B_{l,0}B_{l,1}|\psi_{N}\rangle$ (95)

Similarly, we multiply both the sides of (94) by
$A_{0}^{-1}\otimes\mathbbm{1}$ to have

$\displaystyle\left(\sum_{k=1}^{d-1}\gamma(\boldsymbol{\alpha})A_{0}^{-1}A_{1}^{k}\otimes
B_{l,1}^{k+1}\begin{subarray}{c}{N-1}\\\ \bigotimes\\\ j=1\\\ j\neq
l\end{subarray}B_{j,1}^{k}\right)|\psi_{N}\rangle=\left(\mathbbm{1}-\sum_{k=1}^{d-1}\delta_{k}(\boldsymbol{\alpha})A_{0}^{k}\right)A_{0}^{-1}\otimes
B_{l,1}|\psi_{N}\rangle.$ (96)

By using the commutation relation (91), and since $B_{j,1}\otimes\mathbbm{1}$
and $A_{0}\otimes\mathbbm{1}$ commute, we get

$\displaystyle\left(\sum_{k=1}^{d-1}\gamma(\boldsymbol{\alpha})\omega^{k}A_{1}^{k}\otimes
B_{l,1}^{k+1}\begin{subarray}{c}{N-1}\\\ \bigotimes\\\ j=1\\\ j\neq
l\end{subarray}B_{j,1}^{k}\right)A_{0}^{-1}|\psi_{N}\rangle=\left(\mathbbm{1}-\sum_{k=1}^{d-1}\delta_{k}(\boldsymbol{\alpha})A_{0}^{k}\right)A_{0}^{-1}\otimes
B_{l,1}|\psi_{N}\rangle$ (97)

In the above equation we use (90) for $j=l$, which gives

$\displaystyle\left(\sum_{k=1}^{d-1}\gamma(\boldsymbol{\alpha})\omega^{k}A_{1}^{k}\otimes
B_{l,1}^{k+1}B_{l,0}\begin{subarray}{c}{N-1}\\\ \bigotimes\\\ j=1\\\ j\neq
l\end{subarray}B_{j,1}^{k}\right)|\psi_{N}\rangle=\left(\mathbbm{1}-\sum_{k=1}^{d-1}\delta_{k}(\boldsymbol{\alpha})A_{0}^{k}\right)\otimes
B_{l,1}B_{l,0}|\psi_{N}\rangle.$ (98)

We can proceed now by multiplying (98) by $\omega$ and subtracting it from
(95) as follows

$\displaystyle\left(\sum_{k=1}^{d-1}\gamma(\boldsymbol{\alpha})A_{1}^{k}\otimes(B_{l,0}B_{l,1}^{k+1}-\omega^{k+1}B_{l,1}^{k+1}B_{l,0})\begin{subarray}{c}{N-1}\\\
\bigotimes\\\ j=1\\\ j\neq
l\end{subarray}B_{j,1}^{k}\right)|\psi_{N}\rangle=\left(\mathbbm{1}-\sum_{k=1}^{d-1}\delta_{k}(\boldsymbol{\alpha})A_{0}^{k}\right)\otimes(B_{l,0}B_{l,1}-\omega
B_{l,1}B_{l,0})|\psi_{N}\rangle$ (99)

Next, we again take (89) and multiply it by $A_{1}^{-1}\otimes\mathbbm{1}$ on
both sides,

$\displaystyle\left(\sum_{k=1}^{d-1}\gamma(\boldsymbol{\alpha})A_{1}^{k-1}\otimes
B_{l,1}^{k}\begin{subarray}{c}{N-1}\\\ \bigotimes\\\ j=1\\\ j\neq
l\end{subarray}B_{j,1}^{k}-\delta_{k}(\boldsymbol{\alpha})A_{1}^{-1}A_{0}^{k}\right)|\psi_{N}\rangle=A_{1}^{-1}|\psi_{N}\rangle.$
(100)

We multiply the above by $B_{l,0}\otimes\mathbbm{1}$, use the fact that
$B_{l,0}\otimes\mathbbm{1}$ commutes with $A_{1}\otimes\mathbbm{1}$, and then
use (90) for $k=1$ to obtain

$\displaystyle\left(\sum_{k=1}^{d-1}\gamma(\boldsymbol{\alpha})A_{1}^{k-1}\otimes
B_{l,0}B_{l,1}^{k}\begin{subarray}{c}{N-1}\\\ \bigotimes\\\ j=1\\\ j\neq
l\end{subarray}B_{j,1}^{k}-\delta_{k}(\boldsymbol{\alpha})A_{1}^{-1}A_{0}^{k-1}\right)|\psi_{N}\rangle=A_{1}^{-1}A_{0}^{-1}|\psi_{N}\rangle.$
(101)

Now, we multiply by $A_{0}^{-1}\otimes\mathbbm{1}$ on both the sides of (100)

$\displaystyle\left(\sum_{k=1}^{d-1}\gamma(\boldsymbol{\alpha})A_{0}^{-1}A_{1}^{k-1}\otimes
B_{l,1}^{k}\begin{subarray}{c}{N-1}\\\ \bigotimes\\\ j=1\\\ j\neq
l\end{subarray}B_{j,1}^{k}-\delta_{k}(\boldsymbol{\alpha})A_{0}^{-1}A_{1}^{-1}A_{0}^{k}\right)|\psi_{N}\rangle=A_{0}^{-1}A_{1}^{-1}|\psi_{N}\rangle$
(102)

Using the commutation relation (91), the condition (90), and the fact that
$B_{j,1}\otimes\mathbbm{1}$ and $A_{0}\otimes\mathbbm{1}$ commute, we obtain
that

$\displaystyle\left(\sum_{k=1}^{d-1}\gamma(\boldsymbol{\alpha})\omega^{k-1}A_{1}^{k-1}\otimes
B_{l,1}^{k}B_{l,0}\begin{subarray}{c}{N-1}\\\ \bigotimes\\\ j=1\\\ j\neq
l\end{subarray}B_{j,1}^{k}-\omega^{-1}\delta_{k}(\boldsymbol{\alpha})A_{1}^{-1}A_{0}^{k-1}\right)|\psi_{N}\rangle=\omega^{-1}A_{1}^{-1}A_{0}^{-1}|\psi_{N}\rangle.$
(103)

We proceed now by multiplying (101) by $\omega$ and subtracting it from (103)
to obtain

$\displaystyle\left(\sum_{k=1}^{d-1}\gamma(\boldsymbol{\alpha})A_{1}^{k-1}\otimes(B_{l,0}B_{l,1}^{k}-\omega^{k}B_{l,1}^{k}B_{l,0})\begin{subarray}{c}{N-1}\\\
\bigotimes\\\ j=1\\\ j\neq
l\end{subarray}B_{j,1}^{k}\right)|\psi_{N}\rangle=0,$ (104)

which can be broken up as

$\displaystyle\left(\sum_{k=2}^{d-1}\gamma(\boldsymbol{\alpha})A_{1}^{k-1}\otimes(B_{l,0}B_{l,1}^{k}-\omega^{k}B_{l,1}^{k}B_{l,0})\begin{subarray}{c}{N-1}\\\
\bigotimes\\\ j=1\\\ j\neq
l\end{subarray}B_{j,1}^{k}\right)|\psi_{N}\rangle=-\gamma(\boldsymbol{\alpha})\left((B_{l,0}B_{l,1}-\omega
B_{l,1}B_{l,0})\begin{subarray}{c}{N-1}\\\ \bigotimes\\\ j=1\\\ j\neq
l\end{subarray}B_{j,1}\right)|\psi_{N}\rangle.$ (105)

The above can be further simplified as

$\displaystyle\left(\sum_{k=2}^{d-1}\gamma(\boldsymbol{\alpha})A_{1}^{k-1}\otimes(B_{l,0}B_{l,1}^{k}-\omega^{k}B_{l,1}^{k}B_{l,0})\begin{subarray}{c}{N-1}\\\
\bigotimes\\\ j=1\\\ j\neq
l\end{subarray}B_{j,1}^{k-1}\right)|\psi_{N}\rangle=-\gamma(\boldsymbol{\alpha})\left(B_{l,0}B_{l,1}-\omega
B_{l,1}B_{l,0}\right)|\psi_{N}\rangle.$ (106)

Now, we can see that the l.h.s. of (99) and (106) are the same. Thus, we can
conclude that

$\displaystyle\left(\mathbbm{1}-\sum_{k=1}^{d-1}\delta_{k}(\boldsymbol{\alpha})A_{0}^{k}\right)\otimes(B_{l,0}B_{l,1}-\omega
B_{l,1}B_{l,0})|\psi_{N}\rangle=-\gamma(\boldsymbol{\alpha})\left(B_{l,0}B_{l,1}-\omega
B_{l,1}B_{l,0}\right)|\psi_{N}\rangle$ (107)

We finally get

$\displaystyle\left(\mathbbm{1}(1+\gamma(\boldsymbol{\alpha}))-\sum_{k=1}^{d-1}\delta_{k}(\boldsymbol{\alpha})A_{0}^{k}\right)\otimes(B_{l,0}B_{l,1}-\omega
B_{l,1}B_{l,0})|\psi_{N}\rangle=0.$ (108)

It can be shown that if $(P\otimes Q).v=0$ for any $v\neq 0$ and $P$ is
invertible, then we can conclude that $Q=0$. We can use this fact to show that
$\left(\mathbbm{1}(1+\gamma(\boldsymbol{\alpha}))-\sum_{k=1}^{d-1}\delta_{k}(\boldsymbol{\alpha})A_{0}^{k}\right)$
is invertible. Indeed, by expanding it term by term, we get

$\displaystyle\left(\mathbbm{1}(1+\gamma(\boldsymbol{\alpha}))-\sum_{k=1}^{d-1}\delta_{k}(\boldsymbol{\alpha})A_{0}^{k}\right)=\sum_{l=0}^{d-1}\left((1+\gamma(\boldsymbol{\alpha}))-\sum_{k=1}^{d-1}\delta_{k}(\boldsymbol{\alpha})\omega^{kl}\right)|l\rangle\langle
l|.$ (109)

Notice that the above matrix is not invertible if any of the elements is 0.
Thus, we solve the r.h.s. of the above equation for the $l^{\text{th}}$
element

$\displaystyle(1+\gamma(\boldsymbol{\alpha}))+\frac{\gamma(\boldsymbol{\alpha})}{d}\left(\sum_{k=1}^{d-1}\sum_{\begin{subarray}{c}i,j=0\\\
i\neq
j\end{subarray}}^{d-1}\frac{\alpha_{i}}{\alpha_{j}}\omega^{k(l-j)}\right)$
$\displaystyle=(1+\gamma(\boldsymbol{\alpha}))+\frac{\gamma(\boldsymbol{\alpha})}{d}\left(\sum_{\begin{subarray}{c}i=0\\\
i\neq
l\end{subarray}}^{d-1}\frac{\alpha_{i}}{\alpha_{l}}(d-1)-\sum_{\begin{subarray}{c}i,j=0\\\
i\neq j,j\neq l\end{subarray}}^{d-1}\frac{\alpha_{i}}{\alpha_{j}}\right)$
$\displaystyle=(1+\gamma(\boldsymbol{\alpha}))+\gamma(\boldsymbol{\alpha})\left(\sum_{\begin{subarray}{c}i=0\end{subarray}}^{d-1}\frac{\alpha_{i}}{\alpha_{l}}-\frac{1}{d}\sum_{\begin{subarray}{c}i,j=0\end{subarray}}^{d-1}\frac{\alpha_{i}}{\alpha_{j}}\right)$
(110)

where we have used the fact that
$\sum_{k=0}^{d-1}\omega^{k(l-j)}=d\delta_{l,j}$. Substituting
$\gamma(\boldsymbol{\alpha})$ from (64), we get

$\displaystyle(1+\gamma(\boldsymbol{\alpha}))+\gamma(\boldsymbol{\alpha})\left(\sum_{\begin{subarray}{c}i=0\end{subarray}}^{d-1}\frac{\alpha_{i}}{\alpha_{l}}-\frac{1}{d}\sum_{\begin{subarray}{c}i,j=0\end{subarray}}^{d-1}\frac{\alpha_{i}}{\alpha_{j}}\right)$
$\displaystyle=\gamma(\boldsymbol{\alpha})\left(1+\frac{1}{d}\sum_{\begin{subarray}{c}i,j=0\\\
i\neq
j\end{subarray}}^{d-1}\frac{\alpha_{i}}{\alpha_{j}}-\frac{1}{d}\sum_{\begin{subarray}{c}i,j=0\end{subarray}}^{d-1}\frac{\alpha_{i}}{\alpha_{j}}+\sum_{\begin{subarray}{c}i=0\end{subarray}}^{d-1}\frac{\alpha_{i}}{\alpha_{l}}\right),$
$\displaystyle=\gamma(\boldsymbol{\alpha})\left(\frac{1}{d}\sum_{\begin{subarray}{c}i,j=0\end{subarray}}^{d-1}\frac{\alpha_{i}}{\alpha_{j}}-\frac{1}{d}\sum_{\begin{subarray}{c}i,j=0\end{subarray}}^{d-1}\frac{\alpha_{i}}{\alpha_{j}}+\sum_{\begin{subarray}{c}i=0\end{subarray}}^{d-1}\frac{\alpha_{i}}{\alpha_{l}}\right)=\gamma(\boldsymbol{\alpha})\left(\sum_{\begin{subarray}{c}i=0\end{subarray}}^{d-1}\frac{\alpha_{i}}{\alpha_{l}}\right)>0.$
(111)

Thus, we can conclude that the matrix
$\left(\mathbbm{1}(1+\gamma(\boldsymbol{\alpha}))-\sum_{k=1}^{d-1}\delta_{k}(\boldsymbol{\alpha})A_{0}^{k}\right)$
is invertible. Therefore, we have

$\displaystyle\mathbbm{1}\otimes(B_{l,0}B_{l,1}-\omega
B_{l,1}B_{l,0})|\psi_{N}\rangle=0$ (112)

Finally, we can conclude that $B_{l,0}B_{l,1}=\omega B_{l,1}B_{l,0}$. As was
shown in [13], if two observables $B_{0}$ and $B_{1}$ follow the above
relation, then there exists a unitary matrix $U$ such that

$\displaystyle
UB_{l,0}U^{\dagger}=Z^{\dagger}_{B_{l}^{\prime}}\otimes\mathbbm{1}_{B_{l}^{\prime\prime}},\quad
UB_{l,1}U^{\dagger}=X_{B_{l}^{\prime}}\otimes\mathbbm{1}_{B_{l}^{\prime\prime}}.$
(113)

From the above relations, we can use (88) and (89) to show that the state
which saturates the quantum bound is the Schmidt state of local dimension $d$.

Now, let us prove that the state maximally violating the steering inequality
must be equivalent to the Schmidt state, up to local unitary transformations.
By taking a general state of the form
$|\psi_{N}\rangle=\sum_{i,j_{1},j_{2},\ldots,j_{N-1}=0}^{d-1}|ij_{1}j_{2}\ldots
j_{N-1}\rangle|\psi_{ij_{1}j_{2}\ldots j_{N-1}}\rangle$, we can plug them in
(88) with $k=1$ for the $l^{\text{th}}$ Bob to obtain

$\displaystyle(Z\otimes
Z^{\dagger}_{l}\otimes\mathbbm{1})\left(\sum_{i,j_{1},j_{2},\ldots,j_{N-1}=0}^{d-1}|ij_{1}j_{2}\ldots
j_{N-1}\rangle|\psi_{ij_{1}j_{2}\ldots
j_{N-1}}\rangle\right)=\sum_{i,j_{1},j_{2},\ldots,j_{N-1}=0}^{d-1}|ij_{1}j_{2}\ldots
j_{N-1}\rangle|\psi_{ij_{1}j_{2}\ldots j_{N-1}}\rangle,$ (114)

which is equivalent to,

$\displaystyle\sum_{i,j_{1},j_{2},\ldots,j_{N-1}=0}^{d-1}\omega^{i-j_{l}}|ij_{1}j_{2}\ldots
j_{N-1}\rangle|\psi_{ij_{1}j_{2}\ldots
j_{N-1}}\rangle=\sum_{i,j_{1},j_{2},\ldots,j_{N-1}=0}^{d-1}|ij_{1}j_{2}\ldots
j_{N-1}\rangle|\psi_{ij_{1}j_{2}\ldots j_{N-1}}\rangle.$ (115)

For $i\neq j_{l}$, the above identity implies
$\omega^{j_{l}-i}|\psi_{ij_{1}j_{2}\ldots
j_{N-1}}\rangle=|\psi_{ij_{1}j_{2}\ldots j_{N-1}}\rangle$. This indicates that

$\forall i,j_{l},\text{ such that }i\neq j_{l},\quad|\psi_{ij_{1}j_{2}\ldots
j_{N-1}}\rangle=0.$ (116)

Consequently, we must have $i=j_{l}$ for all $l$. Next, we can proceed
similarly with Eq. (89) to obtain

$\displaystyle\left(\sum_{k=1}^{d-1}\gamma(\boldsymbol{\alpha})X^{k}\begin{subarray}{c}{N-1}\\\
\bigotimes\\\
j=1\end{subarray}X^{k}\otimes\mathbbm{1}+\delta_{k}(\boldsymbol{\alpha})Z^{k}\right)\left(\sum_{i=0}^{d-1}|ii\ldots
i\rangle|\psi_{ii\ldots i}\rangle\right)=\sum_{i=0}^{d-1}|ii\ldots
i\rangle|\psi_{ii\ldots i}\rangle.$ (117)

After performing the operations, we get

$\displaystyle\sum_{i=0}^{d-1}\left(\sum_{k=1}^{d-1}\gamma(\boldsymbol{\alpha})|i+k\rangle|i+k\rangle\ldots|i+k\rangle+\omega^{ki}\delta_{k}(\boldsymbol{\alpha})|ii\ldots
i\rangle\right)|\psi_{ii\ldots i}\rangle=\sum_{i=0}^{d-1}|ii\ldots
i\rangle|\psi_{ii\ldots i}\rangle.$ (118)

If we take the inner product of the above relation with $\langle ss\ldots s|$,
we obtain

$\displaystyle\sum_{k=1}^{d-1}\left(\gamma(\boldsymbol{\alpha})|\psi_{s-k,s-k\ldots,s-k}\rangle+\omega^{ks}\delta_{k}(\boldsymbol{\alpha})\right)|\psi_{ss\ldots
s}\rangle=|\psi_{ss\ldots s}\rangle.$ (119)

From (64) we have that

$\displaystyle\sum_{k=1}^{d-1}\left(|\psi_{s-k,s-k\ldots,s-k}\rangle-\frac{\omega^{ks}}{d}\sum_{\begin{subarray}{c}i,j=0\\\
i\neq
j\end{subarray}}^{d-1}\frac{\alpha_{i}}{\alpha_{j}}\omega^{k(d-j)}\right)|\psi_{ss\ldots
s}\rangle=\frac{1}{d}\sum_{\begin{subarray}{c}i,j=0\\\ i\neq
j\end{subarray}}^{d-1}\frac{\alpha_{i}}{\alpha_{j}}|\psi_{ss\ldots s}\rangle,$
(120)

which can be simplified in the following way

$\displaystyle\sum_{k=1}^{d-1}|\psi_{s-k,s-k\ldots,s-k}\rangle=\frac{\sum_{i=0}^{d-1}\alpha_{i}}{\alpha_{s}}|\psi_{ss\ldots
s}\rangle.$ (121)

Thus, for any $s$ we have

$\displaystyle|\psi_{ss\ldots
s}\rangle=\frac{\sum_{k=1}^{d-1}|\psi_{s-k,s-k\ldots,s-k}\rangle}{\sum_{i=0}^{d-1}\alpha_{i}}\alpha_{s},\quad\forall
s\in\\{0,1,\ldots,d-1\\}.$ (122)

If we define

$|\xi\rangle_{\mathbf{B}^{\prime\prime}E}\coloneqq\frac{\sum_{k=1}^{d-1}|\psi_{s-k,s-k\ldots,s-k}\rangle}{\sum_{i=0}^{d-1}\alpha_{i}},$
(123)

then we can write

$\displaystyle\left(\mathbbm{1}_{AE}\otimes\bigotimes_{i=1}^{N-1}U_{j}\right)|\psi_{N}\rangle_{A\mathbf{B}E}=\left(\sum_{i=0}^{d-1}\alpha_{i}|i\rangle^{\otimes
N}\right)\otimes|\xi\rangle_{\mathbf{B}^{\prime\prime}E},$ (124)

which completes the proof. ∎

##  Appendix C Appendix C: Self-test $N$-qubit generalized $W$-States

The $N$-qubit generalized $W$ states have the following form

$\displaystyle|\psi_{W_{N}}\rangle=\sum_{l=0}^{N-1}\alpha_{l+1}|0\ldots,0,1_{l},0,\ldots,0\rangle=\alpha_{1}|10\ldots
00\rangle+\alpha_{2}|01\ldots 00\rangle+\ldots+\alpha_{N}|00\ldots 01\rangle,$
(125)

such that $\alpha_{i}>0$ for every $i$ and $\sum_{i=1}^{N}\alpha_{i}^{2}=1$.
We introduce a steering inequality that is maximally violated by the above
state through the following steering operator

$\displaystyle\mathcal{W}_{N}(\boldsymbol{\alpha})=-2Z_{A}\otimes\bigotimes_{k=1}^{N-1}B_{k,0}$
$\displaystyle+\sum_{l=1}^{N-1}\left[Z_{A}\otimes\mathbbm{1}_{l}\otimes\left(\mathbbm{1}-P_{l}\right)+\left(\gamma_{l}X_{A}\otimes
B_{l,1}+\delta_{l}Z_{A}\otimes\mathbbm{1}_{l}\right)\otimes P_{l}\right]$
$\displaystyle+\sum_{l=1}^{N-1}\left[\mathbbm{1}_{A}\otimes
B_{l,0}\otimes\left(\mathbbm{1}-P_{l}\right)+\mathbbm{1}_{A}\otimes\mathbbm{1}_{l}\otimes
P_{l}\right],$ (126)

where

$\displaystyle\gamma_{l}=\frac{2\alpha_{l+1}\alpha_{1}}{\alpha_{l+1}^{2}+\alpha_{1}^{2}},\quad\delta_{l}=\frac{\alpha_{l+1}^{2}-\alpha_{1}^{2}}{\alpha_{l+1}^{2}+\alpha_{1}^{2}},\qquad\forall
l\in\\{1,\ldots,N-1\\},$ (127)

with

$\displaystyle P_{l}=\frac{1}{2^{N-2}}\bigotimes_{k=1,k\neq
l}^{N-1}\left(\mathbbm{1}_{k}+B_{k,0}\right)$ (128)

and $B_{k,0}$ and $B_{k,1}$ are the measurements acting on the $k$-th qubit.
In the same way, $\mathbbm{1}_{l}$ is the identity operation acting on the
$l$-th qubit. The corresponding steering inequality is given by

$\displaystyle\langle\mathcal{W}_{N}(\boldsymbol{\alpha})\rangle\leqslant\beta_{L}.$
(129)

where $\beta_{L}$ is the LHS bound.

Let us now find the quantum bound of the steering functional
$\langle\mathcal{W}_{N}(\boldsymbol{\alpha})\rangle$.

######  Fact 5.

The quantum bound of the steering functional
$\langle\mathcal{W}_{N}(\boldsymbol{\alpha})\rangle$ where
$\mathcal{W}_{N}(\boldsymbol{\alpha})$ is given in (C) is given by
$\beta_{Q}=2N$ and is achieved by the $N$-qubit generalized $W$ state
$|\psi_{W_{N}}\rangle$ (125) with the observables

$B_{j,0}=Z,\qquad B_{j,1}=X.$ (130)

###### Proof.

Before proceeding, let us recall here that $B_{j,0}$ are unitary observables
as they correspond to projective measurements. Now, let us observe that the
first term in the steering operator $\mathcal{W}_{N}$ (C) is

$\displaystyle Z_{A}\otimes\bigotimes_{k=1}^{N-1}B_{k,0}\leqslant\mathbbm{1}$
(131)

where we used the fact that $|B_{k,0}|\leqslant\mathbbm{1}$. Similarly, one
can straightforwardly conclude that for any $l$

$\displaystyle\mathbbm{1}_{A}\otimes
B_{l,0}\otimes\left(\mathbbm{1}-P_{l}\right)+\mathbbm{1}_{A}\otimes\mathbbm{1}_{l}\otimes
P_{l}\leqslant\mathbbm{1}$ (132)

Let us now observe that for any $l$, $B_{l,1}=\Pi_{l}^{+}-\Pi_{l}^{-}$ where
$\Pi_{l}^{+},\Pi_{l}^{-}$ denote the projector onto the positive and negative
eigenspace respectively. Now, we observe that

$\displaystyle|\gamma_{l}X_{A}\otimes
B_{l,1}+\delta_{l}Z_{A}\otimes\mathbbm{1}_{l}|\leqslant|(\gamma_{l}X_{A}+\delta_{l}Z_{A})\otimes\Pi_{l}^{+}|+|(-\gamma_{l}X_{A}+\delta_{l}Z_{A})\otimes\Pi_{l}^{-}|.$
(133)

As $\Pi_{l}^{+},\Pi_{l}^{-}$ are positive matrices, we can conclude from the
above formula that

$\displaystyle|\gamma_{l}X_{A}\otimes
B_{l,1}+\delta_{l}Z_{A}\otimes\mathbbm{1}_{l}|\leqslant\lambda^{(1)}_{\mathrm{max}}\mathbbm{1}_{A}\otimes\Pi_{l}^{+}+\lambda^{(2)}_{\mathrm{max}}\mathbbm{1}_{A}\otimes\Pi_{l}^{-}$
(134)

where $\lambda^{(1)}_{\mathrm{max}},\lambda^{(2)}_{\mathrm{max}}$ are the
maximum eigenvalues of
$|\gamma_{l}X_{A}+\delta_{l}Z_{A}|,|-\gamma_{l}X_{A}+\delta_{l}Z_{A}|$
respectively. One can straightaway evaluate that
$\lambda^{(1)}_{\mathrm{max}}=\lambda^{(2)}_{\mathrm{max}}=1$ and thus we
obtain that

$\displaystyle|\gamma_{l}X_{A}\otimes
B_{l,1}+\delta_{l}Z_{A}\otimes\mathbbm{1}_{l}|\leqslant\mathbbm{1}_{A,l}$
(135)

Finally considering the second term in (C) for any $l$, we have that

$\displaystyle
Z_{A}\otimes\mathbbm{1}_{l}\otimes\left(\mathbbm{1}-P_{l}\right)+\left(\gamma_{l}X_{A}\otimes
B_{l,1}+\delta_{l}Z_{A}\otimes\mathbbm{1}_{l}\right)\otimes
P_{l}\leqslant\mathbbm{1}_{A,l}\otimes\left(\mathbbm{1}-P_{l}\right)+|\gamma_{l}X_{A}\otimes
B_{l,1}+\delta_{l}Z_{A}\otimes\mathbbm{1}_{l}|\otimes
P_{l}\leqslant\mathbbm{1}.$ (136)

Thus, the maximum quantum value of the steering inequality (C) is $2N$ and can
be achieved when $B_{j,0}=Z$ and $B_{j,1}=X$ with the generalised $W$ state
$|\psi_{W_{N}}\rangle$. ∎

To achieve the maximal violation of a steering inequality through operator
(C), the following conditions need to be satisfied:

$\displaystyle
Z_{A}\otimes\bigotimes_{k=1}^{N-1}B_{k,0}|\psi\rangle_{A\mathbf{B}}=-|\psi\rangle_{A\mathbf{B}}$
(137)

and for all $l=1,\ldots,N-1$,

$\displaystyle\left[Z_{A}\otimes\mathbbm{1}_{l}\left(\mathbbm{1}-P_{l}\right)+\left(\gamma_{l}X_{A}\otimes
B_{l,1}+\delta_{l}Z_{A}\otimes\mathbbm{1}_{l}\right)\otimes
P_{l}\right]|\psi\rangle_{A\mathbf{B}}=|\psi\rangle_{A\mathbf{B}}$ (138)

and,

$\displaystyle\left[\mathbbm{1}_{A}\otimes
B_{l,0}\otimes\left(\mathbbm{1}-P_{l}\right)+\mathbbm{1}_{A}\otimes\mathbbm{1}_{l}\otimes
P_{l}\right]|\psi\rangle_{A\mathbf{B}}=|\psi\rangle_{A\mathbf{B}}.$ (139)

Now, let us proceed to the self-testing statement.

######  Theorem 3.

Assume that the steering inequality (129) is maximally violated when the
$0^{\text{th}}$ party Alice chooses the observables $A_{i}$ defined to be
$A_{0}=Z$, $A_{1}=X$ and all the other parties choose their observables as
$B_{j,i}$ for $i\in\\{0,1\\}$ and $j=1,2\ldots,N-1$ acting
$\mathcal{H}_{B_{j}}$. Let us say that the state which attains this violation
is given by
$|\psi_{N}\rangle_{A\mathbf{B}E}\in\mathbbm{C}^{2}\otimes\bigotimes_{j=1}^{N-1}\mathcal{H}_{B_{j}}\otimes\mathcal{H}_{E}$,
then the following statement holds true:
$\mathcal{H}_{B_{j}}=\mathcal{H}_{B^{\prime}_{j}}\otimes\mathcal{H}_{B^{\prime\prime}_{j}}$,
where $\mathcal{H}_{B^{\prime}_{j}}\equiv\mathbbm{C}^{2}$,
$\mathcal{H}_{B^{\prime\prime}_{j}}$ is some finite-dimensional Hilbert space,
and there exists a local unitary transformation on Bob’s side
$U_{j}:\mathcal{H}_{B_{j}}\rightarrow\mathcal{H}_{B_{j}}$, such that

$\displaystyle\forall j,\quad
U_{j}\,B_{j,0}\,U_{j}^{\dagger}=Z_{B^{\prime}_{j}}\otimes\mathbbm{1}_{B^{\prime\prime}_{j}},\qquad
U_{j}\,B_{j,1}\,U_{j}^{\dagger}=X_{B^{\prime}_{j}}\otimes\mathbbm{1}_{B^{\prime\prime}_{j}}$
(140)

where ${B^{\prime\prime}}$ denotes Bob’s auxiliary system and the state
$|\psi_{N}\rangle$ is given by,

$\displaystyle\left(\mathbbm{1}_{AE}\otimes\bigotimes_{i=1}^{N-1}U_{j}\right)|\psi_{N}\rangle_{A\mathbf{B}E}=|\psi_{W_{N}}\rangle_{A\mathbf{B}^{\prime}}\otimes|\xi\rangle_{\mathbf{B}^{\prime\prime}E},$
(141)

where
$|\xi\rangle_{\mathbf{B}^{\prime\prime}E}\in\bigotimes_{j=1}^{N-1}\mathcal{H}_{B_{j}^{\prime\prime}}\otimes\mathcal{H}_{E}$
denotes the auxiliary state.

###### Proof.

For simplicity, we drop the lower indices from the state
$|\psi\rangle_{A\mathbf{B}E}$. First, let us consider (137) and rewrite it as

$\displaystyle Z_{A}\otimes B_{l,0}\otimes\bigotimes_{k=1,k\neq
l}^{N-1}B_{k,0}|\psi\rangle=-|\psi\rangle$ (142)

and multiply it by $B_{l^{\prime},0}$ for $l\neq l^{\prime}$ and use the fact
that $B_{l^{\prime},0}^{2}=\mathbbm{1}$. By adding the result to (137) we
obtain

$Z_{A}\otimes
B_{l,0}\otimes(\mathbbm{1}+B_{l^{\prime},0})\otimes\bigotimes_{k=1,k\neq
l,l^{\prime}}^{N-1}B_{k,0}|\psi\rangle=-\mathbbm{1}\otimes(\mathbbm{1}+B_{l^{\prime},0})|\psi\rangle.$
(143)

Then, multiplying (143) with $B_{l^{\prime\prime},0}$ and then adding the
resulting formula with (143), we obtain

$\displaystyle Z_{A}\otimes
B_{l,0}\otimes(\mathbbm{1}+B_{l^{\prime},0})\otimes(\mathbbm{1}+B_{l^{\prime\prime},0})\otimes\bigotimes_{k=1,k\neq
l,l^{\prime},l^{\prime\prime}}^{N-1}B_{k,0}|\psi\rangle=-\mathbbm{1}\otimes(\mathbbm{1}+B_{l^{\prime},0})\otimes(\mathbbm{1}+B_{l^{\prime\prime},0})|\psi\rangle.$
(144)

Continuing in a similar manner, we get that

$\displaystyle Z_{A}\otimes B_{l,0}\otimes
P_{l}|\psi\rangle=-\mathbbm{1}\otimes P_{l}|\psi\rangle$ (145)

where $P_{l}$ is given in Eq. (128). Denoting $|\tilde{\psi}\rangle\coloneqq
P_{l}|\psi\rangle$, we have

$\displaystyle Z_{A}\otimes
B_{l,0}|\tilde{\psi}\rangle=-|\tilde{\psi}\rangle.$ (146)

Next, we consider (138) for any $l$ and multiply it by $P_{l}$, which using
the fact that $P_{l}^{2}=P_{l}$ gives us

$\displaystyle\left(\gamma_{l}X_{A}\otimes
B_{l,1}+\delta_{l}Z_{A}\otimes\mathbbm{1}_{l}\right)|\tilde{\psi}\rangle=|\tilde{\psi}\rangle$
(147)

which, using the fact that $B_{l,1}$ is unitary, can be rearranged to

$\displaystyle\gamma_{l}X_{A}|\tilde{\psi}\rangle=\left(\mathbbm{1}-\delta_{l}Z_{A}\otimes\mathbbm{1}_{l}\right)\otimes
B_{l,1}|\tilde{\psi}\rangle$ (148)

Now, let us multiply the above by $Z_{A}\otimes\mathbbm{1}_{l}$ and use the
fact that this term commutes with
$\left(\mathbbm{1}_{A}-\delta_{l}Z_{A}\right)\otimes B_{l,1}$. That gives us

$\gamma_{l}Z_{A}X_{A}\otimes\mathbbm{1}_{l}|\tilde{\psi}\rangle=\left[\left(\mathbbm{1}_{A}-\delta_{l}Z_{A}\right)\otimes
B_{l,1}\right](Z_{A}\otimes\mathbbm{1}_{l})|\tilde{\psi}\rangle$ (149)

which utilising (146) gives us

$\gamma_{l}Z_{A}X_{A}\otimes\mathbbm{1}_{l}|\tilde{\psi}\rangle=-\left(\mathbbm{1}_{A}-\delta_{l}Z_{A}\right)\otimes
B_{l,1}B_{l,0}|\tilde{\psi}\rangle.$ (150)

Now, let us consider (146) and multiply it by
$\gamma_{l}X_{A}\otimes\mathbbm{1}_{l}$. Similarly, we obtain

$\gamma_{l}X_{A}Z_{A}\otimes\mathbbm{1}_{l}|\tilde{\psi}\rangle=-(\mathbbm{1}_{A}\otimes
B_{l,0})(\gamma_{l}X_{A}\otimes\mathbbm{1}_{l})|\tilde{\psi}\rangle.$ (151)

Using(148), the above formula can be expressed as

$\gamma_{l}X_{A}Z_{A}\otimes\mathbbm{1}_{l}|\tilde{\psi}\rangle=-\left(\mathbbm{1}_{A}-\delta_{l}Z_{A}\right)\otimes
B_{l,0}B_{l,1}|\tilde{\psi}\rangle.$ (152)

By adding (152) to (150) and taking into account that $ZX+XZ=0$, we obtain

$(\mathbbm{1}_{A}-\delta_{l}Z_{A})\otimes(B_{l,1}B_{l,0}+B_{l,0}B_{l,1})|\tilde{\psi}\rangle=0.$
(153)

Because $\alpha_{i}\neq 0$ for any $i$, we have that $\delta_{l}\neq 1$ for
any $l$. Therefore, $\mathbbm{1}_{A}-\delta_{l}Z_{A}$ is invertible, which
then using the fact that the local states are full-rank allows us to conclude
from the above formula that

$\\{B_{l,0},B_{l,1}\\}=0.$ (154)

Since the operators $B_{l,0}$ and $B_{l,1}$ anti-commute, we can use the
result from [13] to guarantee that there exist unitaries $U_{B_{l}}$ for every
$l=1,\ldots,N-1$ such that

$\displaystyle U_{B_{l}}B_{l,0}U_{B_{l}}^{\dagger}=Z\otimes\mathbbm{1},\qquad
U_{B_{l}}B_{l,1}U_{B_{l}}^{\dagger}=X\otimes\mathbbm{1}.$ (155)

Let us now find the state that maximally violates the steering inequality
provided by the operator (C). As the Hilbert spaces of all parties decompose
as $\mathbb{C}^{2}\otimes H_{B_{j}^{\prime\prime}}$, we consider a state in a
general form as

$\displaystyle\bigotimes_{l=1}^{N-1}U_{B_{l}}|\psi\rangle=\sum_{i_{0},i_{1},\ldots,i_{N-1}=0,1}|i_{0}i_{1}\ldots
i_{N-1}\rangle|\psi_{i_{0}i_{1}\ldots i_{N-1}}\rangle,$ (156)

where $U_{B_{l}}$ are the unitaries in (155) and $|\psi_{i_{0}i_{1}\ldots
i_{N-1}}\rangle$ are some unnormalized states. We can multiply (137) by
$\bigotimes_{l=1}^{N-1}U_{B_{l}}$ to obtain

$Z_{A}\otimes\bigotimes_{l=1}^{N-1}\left(U_{B_{l}}B_{l,0}U_{B_{l}}^{\dagger}\right)U_{B_{l}}|\psi\rangle=-\bigotimes_{l=1}^{N-1}U_{B_{l}}|\psi\rangle.$
(157)

Now using Eq. (155) and putting in the state from (156), the above formula can
be expressed as

$\sum_{i_{0},i_{1},\ldots,i_{N-1}=0,1}(-1)^{1+i_{0}+i_{1}+\ldots+i_{N-1}}|i_{0}i_{1}\ldots
i_{N-1}\rangle|\psi_{i_{0}i_{1}\ldots
i_{N-1}}\rangle=\sum_{i_{0},i_{1},\ldots,i_{N-1}=0,1}|i_{0}i_{1}\ldots
i_{N-1}\rangle|\psi_{i_{0}i_{1}\ldots i_{N-1}}\rangle.$ (158)

The above relation implies that the state (156) is of the form

$\bigotimes_{l=1}^{N-1}U_{B_{l}}|\psi\rangle=\sum_{i_{0},i_{1},\ldots,i_{N-1}=0,1}|i_{0}i_{1}\ldots
i_{N-1}\rangle|\psi_{i_{0}i_{1}\ldots i_{N-1}}\rangle\quad\text{with}\quad
i_{0}+i_{1}+\ldots+i_{N-1}=2n+1,$ (159)

for some non-negative integer $n$.

Now, we consider relation (138) for any $l$ and multiply it with
$(\mathbbm{1}-P_{l})$ to obtain for any $l$

$\displaystyle
Z_{A}\otimes\mathbbm{1}_{l}\otimes\left(\mathbbm{1}-P_{l}\right)|\psi\rangle=\left(\mathbbm{1}-P_{l}\right)|\psi\rangle.$
(160)

As $B_{j,0}$ are certified in (155), we obtain that $P_{l}=|0\ldots
0\rangle\\!\langle 0\ldots 0|\otimes\mathbbm{1}_{B_{1}^{\prime\prime}\ldots
B_{N-1}^{\prime\prime}}$. Now plugging in the state (159) in the above
formula, we get that

$\displaystyle\sum_{\begin{subarray}{c}i_{0},i_{1},\ldots,i_{N-1}=0,1\\\
i_{1}+\ldots i_{N-1}-i_{l}\neq 0\end{subarray}}(-1)^{i_{0}}|i_{0}i_{1}\ldots
i_{N-1}\rangle|\psi_{i_{0}i_{1}\ldots
i_{N-1}}\rangle=\sum_{\begin{subarray}{c}i_{0},i_{1},\ldots,i_{N-1}=0,1\\\
i_{1}+\ldots i_{N-1}-i_{l}\neq 0\end{subarray}}|i_{0}i_{1}\ldots
i_{N-1}\rangle|\psi_{i_{0}i_{1}\ldots i_{N-1}}\rangle.$ (161)

This implies from (159) that

$\displaystyle|\psi_{1i_{1}\ldots i_{N-1}}\rangle=0\quad\mathrm{s.t.}\quad
i_{1}+\ldots+i_{N-1}=2n(n\neq 0).$ (162)

Now, considering (139) for any $l$ and multiplying it with
$\mathbbm{1}-P_{l}$, we obtain

$\displaystyle\mathbbm{1}_{A}\otimes
B_{l,0}\otimes\left(\mathbbm{1}-P_{l}\right)|\psi\rangle=\left(\mathbbm{1}-P_{l}\right)|\psi\rangle.$
(163)

Now plugging in the state (159) with the certified observables (155), we
obtain

$\displaystyle\sum_{\begin{subarray}{c}i_{0},i_{1},\ldots,i_{N-1}=0,1\\\
i_{1}+\ldots i_{N-1}-i_{l}\neq 0\end{subarray}}(-1)^{i_{l}}|i_{0}i_{1}\ldots
i_{N-1}\rangle|\psi_{i_{0}i_{1}\ldots
i_{N-1}}\rangle=\sum_{\begin{subarray}{c}i_{0},i_{1},\ldots,i_{N-1}=0,1\\\
i_{1}+\ldots i_{N-1}-i_{l}\neq 0\end{subarray}}|i_{0}i_{1}\ldots
i_{N-1}\rangle|\psi_{i_{0}i_{1}\ldots i_{N-1}}\rangle.$ (164)

This implies from (159) that

$\displaystyle|\psi_{i_{0}i_{1}\ldots,1_{l},\ldots,i_{N-1}}\rangle=0\quad\mathrm{s.t.}\quad
i_{0}+i_{1}+\ldots+i_{N-1}-i_{l}=2n(n\neq 0).$ (165)

It is now straightforward to observe from Eqs. (159),(162), and (165) that the
state $|\psi\rangle$ that attains the quantum bound of
$\langle\mathcal{W}_{N}\rangle$ is of the form

$\displaystyle\bigotimes_{l=1}^{N-1}U_{B_{l}}|\psi\rangle=\sum_{i_{0},i_{1},\ldots,i_{N-1}=0,1}|i_{0}i_{1}\ldots
i_{N-1}\rangle|\psi_{i_{0}i_{1}\ldots i_{N-1}}\rangle\quad\text{with}\quad
i_{0}+i_{1}+\ldots+i_{N-1}=1.$ (166)

Let us now consider (138) and multiply it with $P_{l}$ to obtain for any $l$

$\displaystyle\left(\gamma_{l}X_{A}\otimes
B_{l,1}+\delta_{l}Z_{A}\otimes\mathbbm{1}_{l}\right)\otimes
P_{l}|\psi\rangle=P_{l}|\psi\rangle.$ (167)

Plugging in the state from (166) and observables from (155), we obtain the
following two conditions

$\displaystyle\gamma_{l}|\psi_{0\ldots 1_{l}\ldots
0_{N-1}}\rangle-\delta_{l}|\psi_{1\ldots 0_{l}\ldots
0_{N-1}}\rangle=|\psi_{1\ldots 0_{l}\ldots 0_{N-1}}\rangle$
$\displaystyle\gamma_{l}|\psi_{1\ldots 0_{l}\ldots
0_{N-1}}\rangle+\delta_{l}|\psi_{0\ldots 1_{l}\ldots
0_{N-1}}\rangle=|\psi_{0\ldots 1_{l}\ldots 0_{N-1}}\rangle.$

Using the values of $\gamma_{l},\delta_{l}$ from (127), the solution of these
two relations is $|\psi_{0\ldots 1_{l}\ldots
0_{N-1}}\rangle=\frac{\alpha_{l+1}}{\alpha_{1}}|\psi_{1\ldots 0_{l}\ldots
0_{N-1}}\rangle$ for all $l=1,\ldots,N-1$. Therefore, the state which
maximally violated our steering inequalities has the following form

$\displaystyle\left(\mathbbm{1}_{AE}\otimes\bigotimes_{l=1}^{N-1}U_{B_{l}}\right)|\psi\rangle_{A\mathbf{B}E}=\left(\sum_{l=0}^{N-1}\alpha_{l+1}|0\ldots,0,1_{l},0,\ldots,0\rangle\right)\otimes\left(\frac{1}{\alpha_{1}}|\psi_{1,0,\ldots,0}\rangle\right),$
(168)

where
$|\xi\rangle_{\mathbf{B}^{\prime\prime}E}=\frac{1}{\alpha_{1}}|\psi_{1,0,\ldots,0}\rangle$.
∎