# Detection of entanglement in asymmetric quantum networks and multipartite
quantum steering

D. Cavalcanti<EMAIL_ADDRESS>P. Skrzypczyk  ICFO-Institut de
Ciencies Fotoniques, Mediterranean Technology Park, 08860 Castelldefels
(Barcelona), Spain  G. H. Aguilar  R. V. Nery  P. H. Souto Ribeiro  S. P.
Walborn  Instituto de Física, Universidade Federal do Rio de Janeiro, CP
68528, 21941-972, Rio de Janeiro, RJ, Brazil

###### Abstract

The future of quantum communication relies on quantum networks composed by
observers sharing multipartite quantum states. The certification of
multipartite entanglement will be crucial to the usefulness of these networks.
In many real situations it is natural to assume that some observers are more
trusted than others in the sense that they have more knowledge of their
measurement apparatuses. Here we propose a general method to certify all kinds
of multipartite entanglement in this asymmetric scenario and experimentally
demonstrate it in an optical experiment. Our results, which can be seen as a
definition of genuine multipartite quantum steering, gives a method to detect
entanglement in a scenario in-between the standard entanglement and fully
device-independent scenarios, and provides a basis for semi-device-independent
cryptographic applications in quantum networks.

##  I Introduction

The most widely used techniques to detect entanglement rely either on having
knowledge of the quantum state, obtained through quantum state tomography, or
on the use of measurements that constitute an entanglement witness ent
detection . A frequently disregarded assumption behind these methods is that
the measurements and devices used are well characterized. However, a mismatch
between the theoretical description of the measurements and their actual
implementation may lead to erroneous conclusions about the presence of
entanglement Rosset . A way of avoiding this assumption is to use device-
independent techniques Bell review , where the measuring devices are not
trusted to behave as expected, and no specific description of the experimental
observables is assumed. In this approach the measurement devices are
considered as black boxes that the parties can access with classical inputs
(corresponding to the measurement choices) that provide classical outputs
(considered as the measurement results). The presence of entanglement is then
verified analyzing the correlation statistics between the data lists
corresponding to the measurement results. The violation of Bell inequalities
Bell64  certify the presence of entanglement in this scenario, which can be
thought of as a device-independent entanglement witness. The device-
independent approach is especially important in adversarial scenarios, such as
device-independent quantum key distribution DIQKD , where an adversary can use
a mismatch between the real implementation of the protocol and its description
to fake its performance hack1 ; hack2 ; hack3 . However, the violation of a
Bell inequality requires a high degree of correlation between the parties
tolerating then very low levels of noise and demanding highly efficient
detectors and high-quality entangled states Bell review .

An intermediate scenario between the standard and the device-independent cases
is that of quantum steering Schr ; WJD07 . This is the situation where, in the
bipartite case, one of the parties uses a trusted measuring device but the
other does not. As such, we refer to this approach as the semi-device-
independent one. Apart from the fundamental importance of characterizing
separability in different scenarios, quantum steering appears as a practical
situation that is less demanding experimentally than the device-independent
approach. It requires fewer assumptions than the standard case and lower
strength for the quantum correlations to be witnessed or certified. For these
reasons the study of quantum steering, including its applications 1sided ;
SteeringRand  and experimental demonstrations exp steering0 ; exp steering1 ;
exp steering2 ; exp steering3 ; exp steering4 ; exp steering5 ; exp steering6
, have increased rapidly over recent years.

In the multipartite case, much knowledge has been acquired concerning standard
entanglement detection ent detection  and the device-independent case di mult
nl1 ; di mult nl2 ; di mult nl3 ; DIEW ; expDIEW ; McKague ; SelfT1 ; SelfT2 .
However, only few results were found in the semi-device-independent case. For
instance, Ref. CavMulti  provides inequalities to rule out fully separable
states, Ref. Pappa  developed a probabilistic protocol to detect the presence
of a particular multipartite entangled state, and Ref. Reid multi  discussed a
hybrid model where each party is sometimes trusted and sometimes untrusted
(see also Seiji ; Li2015  for recent experimental demonstrations).

Apart from the fundamental problem of understanding multipartite quantum
correlations, extending the semi-device-independent approach to the
multipartite scenario is also relevant for practical purposes. As technology
advances it will be possible to establish large quantum networks. These
networks will be asymmetric in many cases, depending on the experimental
capabilities of each station, the specific architecture of the implemented
protocols, and unavoidable limitations that the setup may impose. Let us give
few examples. Consider for instance prepare-and-measure cryptographic
protocols in which some parties hold the sources of quantum systems and some
others act as the receivers who measure these systems. Since the senders do
not receive any external signals they may consider that no eavesdropper is
manipulating their apparatuses. Thus, any error they observe (as for example
due to detection inefficiencies) can be attributed to the apparatuses’
imperfections. The receivers, on the other hand, are given systems that may
have been intercepted by an eavesdropper, who may use extra degrees of freedom
that are not considered by the receiver (see hack1 ; hack2 ; hack3  for
examples). In this case, the receivers’ apparatuses can not be considered
trusted. Another scenario is that in which no reference frame can be
established by some of the stations RefFrame . In this case, the measurement
directions that some of parties implement are not known, and may as well be
considered as untrusted. Finally, quantum-key-distribution systems and quantum
randomness generators are nowadays at the commercial level. Clearly, the
general consumers of these products are not capable of reverse engineering the
devices, and may not want to trust their providers.

Here we propose a general method to detect all kinds of entanglement that can
be present in a quantum network, where some of the parties use untrusted
measurements and must use data lists. We show how the different types of
entanglement constrain the corresponding observed experimental data and
present an efficient method to obtain semi-device-independent entanglement
witnesses in the form of multipartite steering inequalities. We furthermore
implement this method in a proof-of-principle optical experiment and
demonstrate the violation of multipartite steering inequalities in both
scenarios where either one or two parties perform untrusted measurements.
Finally, we also quantify the advantage that the present approach provides
over the device-independent one in terms of tolerance to noise.

##  II Results

###  II.1 Semi-device-independent test of multipartite entanglement

We start by explaining the scenario considered here, which consists of a
quantum network on $N$ parties sharing an unknown system in state $\rho$ (see
Fig. 1). Some of the parties perform measurements that are uncharacterised, or
untrusted, while others have total control over their measurement apparatuses.
Those parties who do not trust their apparatuses treat them as black boxes in
which they can provide classical inputs (corresponding to the choice of
measurement settings) and receive classical outputs (corresponding to the
measurement outcomes). Notice that not even the Hilbert space dimension of
these systems are assumed. The parties that trust their measurements can
actually implement quantum state tomography, and reconstruct the density
matrix they hold after the untrusted parties announce their measurement
choices and outcomes. Based on this knowledge the goal is to decide if the
original state $\rho$ had some kind of entanglement.

In the general case of $N$ parties there will be several semi-device-
independent cases, depending on which parties are trusted. For simplicity, in
what follows we will explain our method for the case of detecting genuine
multipartite entanglement in a tripartite system. This case contains all the
basic ingredients needed to understand both how to detect other types of
entanglement and how to treat systems composed of more parties. These
procedures are described in detail in the Supplementary Notes I-III.

Let us consider that an unknown tripartite state
$\rho^{\mathrm{A}\mathrm{B}\mathrm{C}}$ is distributed between three parties:
Alice, Bob and Charlie. Two semi-device-independent cases arise: (i) when only
one party’s device is untrusted and (ii) when two parties’ devices are
untrusted. Let us consider the first case, supposing that Alice holds the
untrusted device. In this case, there is no assumptions on Alice’s
measurements and we describe them with some unknown measurement operators
$M_{a|x}$, where the subscript $x$ labels the measurement choices and $a$ the
possible outcomes. Not even the dimension of Alice’s subsystem is assumed.
Since Bob and Charlie trust their apparatuses they can perform tomography and
determine their (unnormalized) conditional states
$\sigma_{a|x}^{\mathrm{B}\mathrm{C}}$ as

$\sigma_{a|x}^{\mathrm{B}\mathrm{C}}=\operatorname{tr}_{\mathrm{A}}(M_{a|x}\otimes\openone_{\mathrm{B}}\otimes\openone_{\mathrm{C}}\rho^{\mathrm{A}\mathrm{B}\mathrm{C}}).$
(1)

The set of unnormalized states
$\\{\sigma_{a|x}^{\mathrm{B}\mathrm{C}}\\}_{a,x}$ is called an assemblage and
contains all the information obtainable in this situation as it encodes both
the probability that Alice obtains the result $a$ given that she made the
measurement $x$, as
$p(a|x)=\operatorname{tr}(\sigma_{a|x}^{\mathrm{B}\mathrm{C}})$, as well as
the corresponding conditional state
$\rho_{a|x}^{\mathrm{B}\mathrm{C}}=\sigma_{a|x}^{\mathrm{B}\mathrm{C}}/p(a|x)$.

The second situation is when two parties, say Alice and Bob, have untrusted
devices. In this situation Bob’s measurement is also treated as a black box
performing measurements associated to unknown measurement operators $M_{b|y}$,
while Charlie can tomographically determine the assemblage

$\sigma_{ab|xy}^{\mathrm{C}}=\operatorname{tr}_{\mathrm{A}\mathrm{B}}(M_{a|x}\otimes
M_{b|y}\otimes\openone_{\mathrm{C}}\rho^{\mathrm{A}\mathrm{B}\mathrm{C}}).$
(2)

The probability distributions of Alice and Bob’s measurements is encoded in
$p(ab|xy)=\operatorname{tr}\sigma_{ab|xy}^{\mathrm{C}}$.

If the initial state $\rho^{\mathrm{A}\mathrm{B}\mathrm{C}}$ contains no
genuine multipartite entanglement, i.e. it is biseparable, then it has the
form

$\displaystyle\rho^{\mathrm{A}\mathrm{B}\mathrm{C}}$ $\displaystyle=$
$\displaystyle\sum_{\lambda}p_{\lambda}^{\mathrm{A}:\mathrm{B}\mathrm{C}}\rho^{\mathrm{A}}_{\lambda}\otimes\rho^{\mathrm{B}\mathrm{C}}_{\lambda}+\sum_{\mu}p_{\mu}^{\mathrm{B}:\mathrm{A}\mathrm{C}}\rho^{\mathrm{B}}_{\mu}\otimes\rho^{\mathrm{A}\mathrm{C}}_{\mu}$
(3) $\displaystyle+$
$\displaystyle\sum_{\nu}p_{\nu}^{\mathrm{A}\mathrm{B}:\mathrm{C}}\rho^{\mathrm{A}\mathrm{B}}_{\nu}\otimes\rho^{C}_{\nu},$

where $p_{\lambda}^{\mathrm{A}:\mathrm{B}\mathrm{C}}$,
$p_{\mu}^{\mathrm{B}:\mathrm{A}\mathrm{C}}$ and
$p_{\nu}^{\mathrm{A}\mathrm{B}:\mathrm{C}}$ are probability distributions.
Then the assemblages (1) and (2) have the form

$\displaystyle\sigma_{a|x}^{\mathrm{B}\mathrm{C}}$ $\displaystyle=$
$\displaystyle\operatorname{tr}(M_{a|x}\otimes\openone_{\mathrm{B}}\otimes\openone_{\mathrm{C}}\rho^{\mathrm{A}\mathrm{B}\mathrm{C}})$
(4) $\displaystyle=$
$\displaystyle\sum_{\lambda}p_{\lambda}^{\mathrm{A}:\mathrm{B}\mathrm{C}}p_{\lambda}(a|x)\rho^{\mathrm{B}\mathrm{C}}_{\lambda}$
$\displaystyle+$
$\displaystyle\sum_{\mu}p_{\mu}^{\mathrm{B}:\mathrm{A}\mathrm{C}}\rho^{\mathrm{B}}_{\mu}\otimes\sigma_{a|x,\mu}^{\mathrm{C}}$
(5) $\displaystyle+$
$\displaystyle\sum_{\nu}p_{\nu}^{\mathrm{A}\mathrm{B}:\mathrm{C}}\sigma_{a|x\nu}^{\mathrm{B}}\otimes\rho^{\mathrm{C}}_{\nu}$
(6)

and

$\displaystyle\sigma_{ab|xy}^{\mathrm{C}}$ $\displaystyle=$
$\displaystyle\operatorname{tr}_{\mathrm{A}\mathrm{B}}(M_{a|x}\otimes
M_{b|y}\otimes\openone_{\mathrm{C}}\rho^{\mathrm{A}\mathrm{B}\mathrm{C}})$ (7)
$\displaystyle=$
$\displaystyle\sum_{\lambda}p_{\lambda}^{\mathrm{A}:\mathrm{B}\mathrm{C}}p_{\lambda}(a|x)\sigma_{b|y,\lambda}^{\mathrm{C}}$
$\displaystyle+$
$\displaystyle\sum_{\lambda}p_{\mu}^{\mathrm{B}:\mathrm{A}\mathrm{C}}p_{\mu}(b|y)\sigma_{a|x,\mu}^{\mathrm{C}}$
(8) $\displaystyle+$
$\displaystyle\sum_{\lambda}p_{\nu}^{\mathrm{A}\mathrm{B}:\mathrm{C}}p_{\nu}(ab|xy)\rho^{\mathrm{C}}_{\nu}$
(9)

respectively.

Thus, the fact that the original state is biseparable imposes constraints on
the observed assemblages. For instance, in (4) the dependence on the variables
$a$ and $x$ is only through the distribution $p_{\lambda}(a|x)$ and not
through the quantum states. This is a typical instance of an unsteerable
bipartite assemblage WJD07 . The assemblage in (5) satisfies two constraints:
each conditional state is a separable state, and the dependence in $a$ and $x$
is due only to Charlie’s system, and not Bob’s. The assemblage in (6) is
similar to the one in (5), only with Bob’s and Charlie’s roles exchanged.
Thus, in order to test if a given assemblage $\sigma_{a|x}^{obs}$ has the form
(4) consistent with having been produced by a biseparable state one could run
the following program:

$\displaystyle\text{find}~{}\Gamma_{a|x}^{\mathrm{A}:\mathrm{B}\mathrm{C}},\Gamma_{a|x}^{\mathrm{B}:\mathrm{A}\mathrm{C}},\Gamma_{a|x}^{\mathrm{C}:\mathrm{A}\mathrm{B}},$
(10) such that
$\displaystyle\sigma_{a|x}^{obs}=\Gamma_{a|x}^{\mathrm{A}:\mathrm{B}\mathrm{C}}+\Gamma_{a|x}^{\mathrm{B}:\mathrm{A}\mathrm{C}}+\Gamma_{a|x}^{\mathrm{C}:\mathrm{A}\mathrm{B}},$
$\displaystyle\Gamma_{a|x}^{\mathrm{A}:\mathrm{B}\mathrm{C}}\geq
0,~{}\Gamma_{a|x}^{\mathrm{B}:\mathrm{A}\mathrm{C}}\geq
0,\Gamma_{a|x}^{\mathrm{C}:\mathrm{A}\mathrm{B}}\geq 0,$
$\displaystyle\Gamma_{a|x}^{\mathrm{A}:\mathrm{B}\mathrm{C}}\text{is
unsteerable},$
$\displaystyle\Gamma_{a|x}^{\mathrm{B}:\mathrm{A}\mathrm{C}}\text{is separable
and unsteerable from $A$ to $B$},$
$\displaystyle\Gamma_{a|x}^{\mathrm{C}:\mathrm{A}\mathrm{B}}\text{is separable
and unsteerable from $A$ to $C$}.$

If no such triple of assemblages exists, then the underlying state was
definitely not biseparable, and therefore genuine multipartite entangled. The
main problem with this method is that, apart from systems with dimension lower
than $6$, testing separability is computationally demanding gurvitz . As we
show in the Supplementary Notes I and II, we can overcome this problem by
considering approximations of the set of separable states which relax the
above program into a semidefinite program (SDP) DPS ; DPSmulti , for which
efficient numerical methods exist.

A similar analysis can be made for the decomposition (7) (see Supplementary
Note II) and other types of entanglement (see Supplementary Table I). For
instance, (7) refers to an assemblage that is unsteerable from $A$ to $C$ and
(8) to one that is unsteerable from $B$ to $C$. The assemblage (9) has two
properties: it is unsteerable, and the probability distributions
$p_{\nu}(ab|xy)$ must have quantum realisations, i.e. must come from
measurements on quantum states. Again, this last requirement is in general
difficult to test. However, we can once again make use of relaxations of the
set of quantum probability distributions NPA  to transform the program into a
semidefinite program.

All in all, for each semi-device-independent scenario the type of entanglement
in the distributed state will impose constraints on the assemblages one could
observe. These constraints allow the parties holding the trusted devices to
determine if this state must have been genuine multipartite entangled (e.g.if
the observed data admits no decomposition of the form (4) or (7) then there
exists no biseparable state that could explain it). Therefore, even not
knowing the initial state or what type of measurements the untrusted parties
performed, it is possible to discriminate the assemblages that were produced
by states containing some type of entanglement.

Finally, in each case the program can be seen as a membership test for the
observed assemblage to be contained inside a convex set. It is always possible
to certify that a point lies outside a convex set by finding a separating
hyperplane between the set and the point. As we show in the Supplementary Note
II, in each case we can find the lagrange dual program to the set membership
test, which always amounts to finding such a separating hyperplane. Such
separating hyperplanes are precisely multipartite steering inequalities, which
can alternatively be thought of as semi-device independent entanglement
witnesses. Thus our method naturally generates steering inequalities which can
then be used as witnesses for multipartite entanglement.

Figure 1: Asymmetric tripartite networks where untrusted devices are treated
as black boxes with classical inputs and outputs a) One untrusted party
scenario: Alice, who holds an untrusted device treats it as a black box in
which she inputs $x$ (the measurement choice) and receives an output $a$ (the
measurement outcome). This procedure corresponds mathematically to applying
some unknown measurement operator Ma|x to the shared tripartite quantum state,
which produces a post-measured state $\sigma^{\mathrm{B}\mathrm{C}}_{a|x}$ at
Bob and Charlie’s locations. b) A similar situation occurs in two untrusted
parties scenario, when both Alice and Bob perform untrusted measurements
(corresponding to unkown measurement operators Ma|x and Mb|y respectively)
preparing quantum states $\sigma^{\mathrm{C}}_{ab|xy}$ on Charlie’s system.

###  II.2 Practical considerations

Due to experimental errors and finite statistics the experimentally observed
data is not strictly compatible with any physical state and local
measurements. In particular, all assemblages which exactly reproduce the
experimental data in general do not satisfy the no-signalling constraint that
$\sum_{a}\sigma_{a|x}^{\mathrm{B}\mathrm{C}}=\sum_{a}\sigma_{a|x^{\prime}}^{\mathrm{B}\mathrm{C}}$
for $x\neq x^{\prime}$. Since the present methods are tailored to detect
entanglement of physical states we can not use the observed data directly to
test for the presence of entanglement.

We thus propose to proceed with the following steps: First, given the
experimental data, generate a physical assemblage that best approximates it
through, for instance, a maximum likelihood reconstruction method. Second,
having obtained the best physical approximation to the actual data, use the
SDP method discussed in the Supplementary Note II to check for any type of
entanglement. This method also generates an inequality that is satisfied by
all assemblages coming from states which do not have the type of entanglement
tested. Finally check that the observed data violates this inequality.

###  II.3 Examples: GHZ and W states

As examples, we used our method to produce the following inequality that is
satisfied by all assemblages of the form (1) (see also Supplementary IV):

$1+0.1547\langle
Z_{\mathrm{B}}Z_{\mathrm{C}}\rangle-\tfrac{1}{3}\big{(}\langle
A_{3}Z_{\mathrm{B}}\rangle+\langle A_{3}Z_{\mathrm{C}}\rangle+\langle
A_{1}X_{\mathrm{B}}X_{\mathrm{C}}\rangle\\\ -\langle
A_{1}Y_{\mathrm{B}}Y_{\mathrm{C}}\rangle-\langle
A_{2}X_{\mathrm{B}}Y_{\mathrm{C}}\rangle-\langle
A_{2}Y_{\mathrm{B}}X_{\mathrm{C}}\rangle\big{)}\geq 0,$ (11)

with $A_{i}$ for $i=1,2,3$, being observables in Alice’s system with outcomes
labelled $\pm 1$ and $X,Y$ and $Z$ representing the Pauli operators. The GHZ
state $(|000\rangle+|111\rangle)/\sqrt{2}$ violates this inequality by
$-0.8453\ngeq 0$ when Alice’s measurements are also $X$, $Y$ and $Z$, which
numerical optimization suggests are the optimal choices for Alice.

In the case Alice and Bob perform untrusted measurements we have derived the
following inequality which is satisfied by assemblages of the form (2):

$1-\alpha\langle A_{3}B_{3}\rangle-\alpha\langle A_{3}Z\rangle-\alpha\langle
B_{3}Z\rangle-\beta\langle A_{1}B_{1}X\rangle\\\ +\beta\langle
A_{1}B_{2}Y\rangle+\beta\langle A_{2}B_{1}Y\rangle+\beta\langle
A_{2}B_{2}X\rangle\geq 0$ (12)

where $\alpha=0.1831$ and $\beta=0.2582$, and similarly $B_{i}$ for $i=1,2,3$
represent Bob’s measurement which we assume to have $\pm 1$ outcomes. The GHZ
state achieves a violation $-0.5820\ngeq 0$ now when both Alice and Bob
perform $X$, $Y$ and $Z$ measurements.

Similar inequalities for two untrusted parties and for the W state, given by
$|W\rangle=(|001\rangle+|010\rangle+|100\rangle)/\sqrt{3}$, are presented in
the Supplementary Note IV.

We have also considered noisy versions of the GHZ and W states given by

$\displaystyle\rho_{\psi}$
$\displaystyle=w|\psi\rangle\langle\psi|+(1-w)\openone/8,$ (13)

where $|\psi\rangle$ can be either the GHZ or the W state. We computed how
much white noise can be added to these states until we are unable to detect
genuine multipartite entanglement. Specifically, we quantify the minimum $w$
for which our method guarantees that following states are genuinely
multipartite entangled. The results are summarized in Table 3, together with
the known bounds for standard entanglement tests GuhSee10 ; JunMorGuh12  and
the device-independent case DIEW . One can see that trusting some of the
parties offers a significant advantage in terms of noise tolerance.

###  II.4 Experimental violation of genuine tripartite steering witnesses

Figure 2: Experimental setup and results a) An ultraviolet laser oscillating
at 325nm pumps two cross axis BBO crystals, each 1mm long. Probabilistically,
two photons are produced in the state (14) via parametric down conversion
kwiat99 . The polarization entangled state is obtained by the superposition of
signal and idler beams produced in the first crystal with vertical
polarization and the ones produced in the second crystal with horizontal
polarization. Signal photons in B are sent to beam displacer BD1 which
transmits the vertical polarization component and deviates the horizontally
polarized component. This results in the production of a GHZ state right after
BD1. Two qubits in this state are represented by the polarization degrees of
freedom of photons A and B, and one qubit is encoded in the spatial (or path)
degree of freedom of photon B. Photons in mode A are sent directly to
detection after polarization projection, which is done using the quarter
waveplate $\text{Q}_{\text{A}_{\text{p}}}$, half waveplate
H${}_{\text{A}_{\text{p}}}$, and polarizing beam splitter PBS${}_{\text{A}}$.
We perform a joint analysis of the polarization and path degree of freedom of
photon B using the sequence of devices given by quarter waveplate
Q${}_{\text{B}_{\text{p}}}$, half waveplate H${}_{\text{B}_{\text{p}}}$, beam
displacer BD2, quarter waveplate Q${}_{\text{B}_{\text{s}}}$, half waveplate
H${}_{\text{B}_{\text{s}}}$, and polarizing beam spliter PBS${}_{\text{B}}$.
For a given set of adjustments of the quarter and half waveplates, we perform
one specific joint projection in both polarization and path degrees of freedom
basis. Since there is a coherent combination of spatial modes 0 and 1 in BD2,
the measure of the spatial degrees of freedom of photon B is done by mapping
the spatial qubit state before BD2 in the polarization state at the output of
BD2. Even though the projection is made simultaneously for both degrees of
freedom in this case, they are independent, or in other words, all
combinations of projections are possible farias12b ; aguilar2014 . b)
Histograms obtained by computing the semi-device-independent entanglement
witness from the experimental data (see main text and the Supplementary Note V
for more details about the witness). We measured the value of each witness 215
independent times and plot the number of times $N$ the value $S$ of the
witness felt within a given interval. From this statistical procedure, an
average value and a variance is obtained for the witness. The upper histogram
is for the case of one untrusted party, resulting in the average value of
-0.82 and standard deviation of 0.05. The lower histogram is for the case of
two untrusted parties, resulting in the average of -0.56 and an standard
deviation of 0.04.

In order to illustrate the utility and efficiency of our approach, we use this
technique to violate genuine multipartite steering witnesses in a real
laboratory setting where one or two parties perform untrusted measurements.
The experimental set-up is shown in Fig 2 and is set to produce a GHZ state
encoded in the polarization and path degree of freedom of two photons
farias12b ; aguilar2014  with high fidelity. The experimental procedure starts
preparing photons in a state close to

$|\Psi\rangle=\left(\frac{|00\rangle+|11\rangle}{\sqrt{2}}\right)_{{\text{A}_{p}}{\text{B}_{p}}}\otimes|0\rangle_{\text{B}_{s}},$
(14)

where $\text{A}_{p}$ and $\text{B}_{p}$ represent the polarization qubit of
photons A and B respectively, where 0 and 1 stand for horizontal and vertical
polarization states, and $\text{B}_{s}$ represents the spatial degree of
freedom of photon B. To obtain a GHZ state, we couple the spatial degree of
freedom with the polarization using Beam Displacer (BD1) which transforms
$|0\rangle_{\text{B}_{p}}|0\rangle_{\text{B}_{s}}\rightarrow|0\rangle_{\text{B}_{p}}|0\rangle_{\text{B}_{s}}$
and
$|1\rangle_{\text{B}_{p}}|0\rangle_{\text{B}_{s}}\rightarrow|1\rangle_{\text{B}_{p}}|1\rangle_{\text{B}_{s}}$.
Once we obtain the desired state, every qubit is measured in the eigenstates
of the three Pauli operators. For the polarization degrees of freedom this is
carried out using a quarter-wave-plate (QWP), a half-wave plate (HWP) and a
polarizing beam splitter (PBS) or BD2, depending on the photon. For the
spatial degrees of freedom this is carried out using the interferometer
described in Fig. 2 farias12b ; aguilar2014 .

Although this experiment is tailored to produce a GHZ state and perform
measurements corresponding to the Pauli operators, the analysis we perform on
the experimental data makes no assumption about the state nor the untrusted
measurements. We consider two cases, one where part $\text{A}_{p}$ is
untrusted and parts $\text{B}_{p}$ and $\text{B}_{s}$ hold the trusted
devices, and another when parts $\text{B}_{p}$ and $\text{B}_{s}$ hold the
untrusted devices and part $\text{A}_{p}$ the trusted one. For the two cases
we follow the procedure described in Sec. II.2 (using a least-squares
optimisation to provide physical assemblages), which provides inequalities of
the form $S\geq 0$ (whose exact form can be found in the Supplementary Note V)
whose violation certify that the corresponding assemblages cannot be written
in the biseparable form (4) or (7), respectively. We finally observe a
violation of these inequalities by the experimental data (see Fig. 2 b)). We
have performed each experiment (i.e. measuring all correlators) 215
independent times, from which we calculate an average value of $S=-0.82\pm
0.05$ for one untrusted party and $S=-0.56\pm 0.04$ for two untrusted parties.
This proves that there exists no biseparable tripartite state and measurements
performed by the untrusted parties that could have generated the observed
assemblages.

##  III Discussion

We have derived a method to detect multipartite entanglement when some of the
apparatuses used in a quantum network are untrusted or uncharacterized. This
method allows the detection of of all kinds of entanglement in quantum
networks where some of the observers use their measurement apparatuses simply
as data lists. This scenario is experimentally less demanding than the
nonlocality scenario, as it tolerates more noise, for instance. We have
performed a proof-of-principle experiment demonstrating the existence of
genuine tripartite entanglement, without any assumption on the source or the
measurements being performed in some of the subsystems.

Our results provide a feasible test for multipartite entanglement in quantum
networks and bridges the two well known cases of multipartite entanglement and
multipartite Bell nonlocality. Moreover the scenario considered is a natural
generalization of bipartite quantum steering WJD07  (see CavMulti ; Reid multi
for alternative definitions) . Since steering has found applications in
cryptographic protocols 1sided ; SteeringRand , we believe that our results
can be used as a starting point to define semi-device-independent
cryptographic applications in future quantum networks.

##  IV Acknowledgements

We thank A. Acín, M. Navascués and L. Aolita for discussions. DC was supported
by the Beatriu de Pinós fellowship (BP-DGR 2013) and PS by the Marie Curie
COFUND action through the ICFOnest program. Financial support was provided by
Brazilian agencies FAPERJ, CNPq, CAPES, and the National Institute of Science
and Technology for Quantum Information.

no. untr. meas. | 0 | 1 | 2 | 3
---|---|---|---|---
GHZ | $\frac{3}{7}\approx 0.429$ | $\approx 0.54$ | $\approx 0.63$ | $2/3\approx 0.67$
W | $\approx 0.479$ | $\approx 0.57$ | $\approx 0.67$ | $\approx 0.72$
Table 1: Critical robustness to white noise $w$. We provide a comparison between the known bounds on critical robustness to white noise of the GHZ and W states above which genuine multipartite entanglement can be detected in 4 different scenarios: when no party is untrusted (i.e. the standard entanglement scenario GuhSee10 ; JunMorGuh12  ), when 1 and 2 parties hold untrusted devices, for which we used the semi-device-independent method developed here and when all devices are untrusted, i.e. the device-independent case developed in DIEW . In the Supplementary Table II we also display the bounds concerning the detection of (not necesarily genuine multipartite) entanglement in these states. Form of state |  Unstrusted parties |  Known objects |  SDP
---|---|---|---
∑_λp_λρ_λ^A⊗ρ_λ^B⊗ρ_λ^C |  A |  σ_a—x^BC |  max $\displaystyle\quad p$ s.t $\displaystyle\quad\textstyle{\sum}_{\mu}D_{\mu}(a|x)\sigma_{\mu}^{\mathrm{B}\mathrm{C}}=\sigma_{a|x}^{\mathrm{B}\mathrm{C}}-p\,\mathrm{id}_{a|x}^{\mathrm{B}\mathrm{C}},$ (15) $\displaystyle\quad\left(\sigma_{\mu}^{\mathrm{B}\mathrm{C}}\right)^{T_{\mathrm{B}}}\geq 0,\quad\sigma_{\mu}^{\mathrm{B}\mathrm{C}}\geq 0.$
|  A and B |  σ_ab—xy^C |  max $\displaystyle\quad p$ s.t. $\displaystyle\quad\textstyle{\sum}_{\mu,\lambda}D_{\mu}(a|x)D_{\lambda}(b|y)\sigma_{\mu\lambda}^{\mathrm{C}}=\sigma_{ab|xy}^{\mathrm{C}}-p\,\mathrm{id}_{ab|xy}^{\mathrm{C}},$ (16) $\displaystyle\quad\sigma_{\mu\lambda}^{\mathrm{C}}\geq 0.$
∑_λp_λρ^A_λ⊗ρ^BC_λ |  A |  σ_a—x^BC |  max $\displaystyle\quad p$ s.t $\displaystyle\quad\textstyle{\sum}_{\mu}D_{\mu}(a|x)\sigma_{\mu}^{\mathrm{B}\mathrm{C}}=\sigma_{a|x}^{\mathrm{B}\mathrm{C}}-p\,\mathrm{id}_{a|x}^{\mathrm{B}\mathrm{C}},$ (17) $\displaystyle\quad\sigma_{\mu}^{\mathrm{B}\mathrm{C}}\geq 0.$
|  B |  σ_b—y^AC |  max $\displaystyle\quad p$ s.t $\displaystyle\quad\Gamma_{b|y}^{\mathrm{A}\mathrm{C}}=\sigma_{b|y}^{\mathrm{A}\mathrm{C}}-p\,\mathrm{id}_{b|y}^{\mathrm{A}\mathrm{C}},$ (18) $\displaystyle\quad\operatorname{tr}_{\mathrm{C}}\Gamma_{b|y}^{\mathrm{A}\mathrm{C}}=\textstyle{\sum}_{\mu}D_{\mu}(b|y)\sigma_{\mu}^{\mathrm{A}},$ $\displaystyle\quad(\Gamma_{b|y}^{\mathrm{A}\mathrm{C}})^{T_{\mathrm{A}}}\geq 0,\quad\Gamma_{b|y}^{\mathrm{A}\mathrm{C}}\geq 0,\quad\sigma_{\mu}^{\mathrm{A}}\geq 0.$
|  A and  B |  σ_ab—xy^C |  max $\displaystyle\quad p$ s.t $\displaystyle\quad\textstyle{\sum}_{\mu}D_{\mu}(a|x)\sigma_{b|y,\mu}^{\mathrm{C}}=\sigma_{ab|xy}^{\mathrm{C}}-p\,\mathrm{id}_{ab|xy}^{\mathrm{C}},$ (19) $\displaystyle\quad\sigma_{b|y,\mu}^{\mathrm{C}}\geq 0.$
|  B and C |  σ_bc—yz^A |  max $\displaystyle\quad p$ s.t. $\displaystyle\quad\textstyle{\sum}_{\mu}D^{\mathrm{NS}}(bc|yz,\mu)\sigma_{\mu}^{\mathrm{A}}=\sigma_{bc|yz}^{\mathrm{A}}-p\,\mathrm{id}_{bc|yz}^{\mathrm{A}}$ (20) $\displaystyle\quad\textstyle{\sum}_{\mu}D^{\mathrm{NS}}(bc|yz,\mu)\sigma_{\mu}^{\mathrm{A}}\in\mathcal{Q}^{(k)}_{\mathrm{A}},\quad\sigma_{\mu}^{\mathrm{A}}\geq 0$
$\displaystyle\sum_{\lambda}p_{\lambda}^{\mathrm{A}:\mathrm{B}\mathrm{C}}\rho^{\mathrm{A}}_{\lambda}\otimes\rho^{\mathrm{B}\mathrm{C}}_{\lambda}$ $\displaystyle+\sum_{\lambda}p_{\lambda}^{\mathrm{B}:\mathrm{A}\mathrm{C}}\rho^{\mathrm{B}}_{\lambda}\otimes\rho^{\mathrm{A}\mathrm{C}}_{\lambda}$ $\displaystyle+\sum_{\lambda}p_{\lambda}^{\mathrm{A}\mathrm{B}:\mathrm{C}}\rho^{\mathrm{A}\mathrm{B}}_{\lambda}\otimes\rho^{C}_{\lambda}$ |  A |  σ_a—x^BC |  max $\displaystyle\quad p$ s.t. $\displaystyle\quad\Gamma_{a|x}^{\mathrm{A}:\mathrm{B}\mathrm{C}}+\Gamma_{a|x}^{\mathrm{B}:\mathrm{A}\mathrm{C}}+\Gamma_{a|x}^{\mathrm{C}:\mathrm{A}\mathrm{B}}=\sigma_{a|x}^{\mathrm{B}\mathrm{C}}-p\,\mathrm{id}_{a|x}^{\mathrm{B}\mathrm{C}}$ $\displaystyle\quad\Gamma_{a|x}^{\mathrm{A}:\mathrm{B}\mathrm{C}}=\textstyle{\sum}_{\mu}D_{\mu}(a|x)\sigma_{\mu}^{\mathrm{B}\mathrm{C}},\quad\sigma_{\mu}^{\mathrm{B}\mathrm{C}}\geq 0$ $\displaystyle\quad\operatorname{tr}_{\mathrm{C}}\Gamma_{a|x}^{\mathrm{B}:\mathrm{A}\mathrm{C}}=\textstyle{\sum}_{\mu}D_{\mu}(a|x)\sigma_{\mu}^{\mathrm{B}},\quad\sigma_{\mu}^{\mathrm{B}}\geq 0,$ (21) $\displaystyle\quad\operatorname{tr}_{\mathrm{B}}\Gamma_{a|x}^{\mathrm{C}:\mathrm{A}\mathrm{B}}=\textstyle{\sum}_{\mu}D_{\mu}(a|x)\sigma_{\mu}^{\mathrm{C}},\quad\sigma_{\mu}^{\mathrm{C}}\geq 0,$ $\displaystyle\quad\left(\Gamma_{a|x}^{\mathrm{B}:\mathrm{A}\mathrm{C}}\right)^{T_{\mathrm{B}}}\geq 0,\quad\left(\Gamma_{a|x}^{\mathrm{C}:\mathrm{A}\mathrm{B}}\right)^{T_{\mathrm{B}}}\geq 0.$ $\displaystyle\quad\Gamma_{a|x}^{\mathrm{B}:\mathrm{A}\mathrm{C}}\geq 0,\quad\Gamma_{a|x}^{\mathrm{C}:\mathrm{A}\mathrm{B}}\geq 0,\quad\sum_{a}\Gamma_{a|x}^{\mathrm{B}:\mathrm{A}\mathrm{C}}=\sum_{a}\Gamma_{a|x^{\prime}}^{\mathrm{B}:\mathrm{A}\mathrm{C}}.$
|  A and B |  σ_ab—xy^C |  max $\displaystyle\quad p$ s.t. $\displaystyle\quad\Pi_{ab|xy}^{\mathrm{A}:\mathrm{B}\mathrm{C}}+\Pi_{ab|xy}^{\mathrm{B}:\mathrm{A}\mathrm{C}}+\Pi_{ab|xy}^{\mathrm{C}:\mathrm{A}\mathrm{B}}=\sigma_{ab|xy}^{\mathrm{C}}-p\,\mathrm{id}_{ab|xy}^{\mathrm{C}}$ $\displaystyle\quad\Pi_{ab|xy}^{\mathrm{A}:\mathrm{B}\mathrm{C}}=\textstyle{\sum}_{\mu}D_{\mu}(a|x)\sigma_{b|y,\mu}^{\mathrm{C}},\quad\sigma_{b|y,\mu}^{\mathrm{C}}\geq 0$ $\displaystyle\quad\Pi_{ab|xy}^{\mathrm{B}:\mathrm{A}\mathrm{C}}=\textstyle{\sum}_{\mu}D_{\mu}(b|y)\sigma_{a|x,\mu}^{\mathrm{C}},\quad\sigma_{a|x,\mu}^{\mathrm{C}}\geq 0$ (22) $\displaystyle\quad\Pi_{ab|xy}^{\mathrm{C}:\mathrm{A}\mathrm{B}}=\textstyle{\sum}_{\mu}D_{\nu}^{\mathrm{NS}}(ab|xy)\sigma_{\nu}^{\mathrm{C}},\quad\sigma_{\nu}^{\mathrm{C}}\geq 0$ $\displaystyle\quad\Pi_{ab|xy}^{\mathrm{C}:\mathrm{A}\mathrm{B}}\in\mathcal{Q}_{\mathrm{C}}^{(k)},\quad\sum_{b}\sigma_{b|y,\mu}^{\mathrm{C}}=\sum_{b}\sigma_{b|y^{\prime},\mu}^{\mathrm{C}}$
Table 2: Collection of all SDP tests for the tripartite case. All expressions
with indices should be understood to hold for each value of the index. Various
SDPs depend upon a parameter $k$, such that for larger values of $k$ we obtain
a better approximate characterisation of the set, and therefore a more
stringent test. All programs are strictly feasible and are such that a
negative optimal value $p^{*}<0$ certifies that the assemblage has the
corresponding type of entanglement. In this case the dual provides a semi-
device-indepenent entanglement witness in the form of a steering inequality.
An optimal solution $p^{*}\geq 0$ indicates that the assemblage is inside the
corresponding set, i.e. that one cannot conclude that the state contains the
desired type of entanglement.

|  | EW | SW | SW | DIEW
---|---|---|---|---|---
no. untr. | 0 | 1 | 2 | 3
no. meas. | n/a | 2 | 3 | 2 | 3 | 2 | 3
GHZ | ent. |  | $\tfrac{1}{5}$
---
= 0.2
DurCir00
| $\approx$
---
$0.2613$
| $\approx$
---
$0.2500$
| $\tfrac{1}{2}=$
---
$0.5$
| $\tfrac{3}{7}\approx$
---
$0.4286$
| $\tfrac{1}{2}=$
---
$0.5$
GruLasZuk10
| $\tfrac{1}{2}=$
---
$0.5$
GruLasZuk10
GME |  | $\tfrac{3}{7}\approx$
---
0.4286
GuhSee10
| $\approx$
---
$0.6307$
| $\approx$
---
$0.5420$
| $\tfrac{1}{\sqrt{2}}\approx$
---
$0.7071$
| $\approx$
---
$0.6322$
| $\tfrac{1}{\sqrt{2}}\approx$
---
$0.7071$
DIEW
| $\tfrac{2}{3}\approx$
---
$0.6667$
DIEW
W | ent. |  | $\approx$
---
0.2096
Sza11
| $\tfrac{3}{11}\approx$
---
$0.2727$
| $\approx$
---
$0.2698$
| $\approx$
---
$0.5765$
| $\approx$
---
$0.4434$
| $\approx$
---
$0.6442$
GruLasZuk10
| $\approx$
---
$0.6048$
GruLasZuk10
GME |  | $\approx$
---
0.479
JunMorGuh12
| $\approx$
---
$0.6440$
| $\approx$
---
$0.5684$
| $\approx$
---
$0.7218$
| $\approx$
---
$0.6757$
| 3/4
---
= 0.75
DIEW
| $\approx$
---
$0.7158$
DIEW
Table 3: Critical robustness to white noise $w^{*}$ for different scenarios.
We provide a comparison between the critical robustness to white noise of the
GHZ and W state, in the 4 different scenarios: All 3 parties with trusted
devices, i.e. using an entanglement witness (EW), 2 or 1 parties with trusted
devices, i.e. using a steering witness (SW) and no parties with trusted
devices i.e. nonlocality (DIEW). We give critical values for both detecting
entanglement and for detecting genuine multipartite entanglement ($\approx$
refers to numerical results).

Supplementary note 1

Characterising multipartite assemblages

Here we determine the constraints that different types of entanglement in the
initial state $\rho^{\mathrm{A}\mathrm{B}\mathrm{C}}$ impose on the
assemblages produced by untrusted measurements and define the corresponding
sets that they characterise. We will first consider the question of whether or
not there is any entanglement in the state, before moving on to entanglement
in a given bipartition, for which there are a number of different cases, given
the asymmetry of the scenario, and finish with the question of detecting
genuine multipartite entanglement.

### Case 1. Multipartite entanglement

Let us then start by considering a state
$\rho^{\mathrm{A}:\mathrm{B}:\mathrm{C}}$ that is fully separable i.e.

$\rho^{\mathrm{A}:\mathrm{B}:\mathrm{C}}=\sum_{\lambda}p_{\lambda}\rho_{\lambda}^{\mathrm{A}}\otimes\rho_{\lambda}^{\mathrm{B}}\otimes\rho_{\lambda}^{\mathrm{C}},$
(23)

where $p_{\lambda}$ defines a probability distribution.

#### Case 1A. Multipartite entanglement with one untrusted party

We treat first the case where a single party, taken to be $\mathrm{A}$,
performs a set of untrusted measurements $\\{M_{a|x}\\}_{a,x}$ on her share of
the state, providing the parties $\mathrm{B}$ and $\mathrm{C}$ with the
assemblage

$\displaystyle\sigma_{a|x}^{\mathrm{B}\mathrm{C}}$ $\displaystyle=$
$\displaystyle\operatorname{tr}_{\mathrm{A}}(M_{a|x}\otimes\openone_{\mathrm{B}}\otimes\openone_{\mathrm{C}}\rho^{\mathrm{A}:\mathrm{B}:\mathrm{C}})$
(24) $\displaystyle=$
$\displaystyle\sum_{\lambda}p(a|x,\lambda)\rho_{\lambda}^{\mathrm{B}}\otimes\rho_{\lambda}^{\mathrm{C}}.$

where
$p(a|x,\lambda)=p_{\lambda}\operatorname{tr}(M_{a|x}\rho^{\mathrm{A}}_{\lambda})$.
Notice first that the dependence of Bob and Charlie’s assemblage on $a$ and
$x$ comes only from the common pre-shared variable $\lambda$. This is a
typical instance of an unsteerable assemblage, or, in other words, this is a
local hidden state (LHS) model for the assemblage
$\sigma_{a|x}^{\mathrm{B}\mathrm{C}}$ WJD07 . Notice further that it is
composed only by separable (unnormalised) states for $\mathrm{B}$ and
$\mathrm{C}$. Thus, in this case, testing for the presence of multipartite
entanglement reduces to testing if the assemblage
$\sigma_{a|x}^{\mathrm{B}\mathrm{C}}$ is steerable and separable at the same
time. This type of assemblages forms a set
$\Sigma^{\mathrm{A}:\mathrm{B}:\mathrm{C}}_{\mathrm{B}\mathrm{C}}$, given by

$\Sigma^{\mathrm{A}:\mathrm{B}:\mathrm{C}}_{\mathrm{B}\mathrm{C}}=\Big{\\{}\sigma_{a|x}^{\mathrm{B}\mathrm{C}}\Big{|}\sigma_{a|x}^{\mathrm{B}\mathrm{C}}=\textstyle{\sum_{\mu}}D(a|x,\mu)\rho_{\mu}^{\mathrm{B}\mathrm{C}},\rho_{\mu}^{\mathrm{B}\mathrm{C}}\in\mathrm{SEP}\Big{\\}}$
(25)

where we have used the fact that any probability distribution $p(a|x,\lambda)$
can be written as a convex combination of deterministic ones $D(a|x,\mu)$,
i.e. $p_{\lambda}(a|x)=\sum_{\mu}q(\mu|\lambda)D(a|x,\mu)$, and denote the set
of (unnormalised) separable quantum states by $\mathrm{SEP}$.

#### Case 1B. Multipartite entanglement with two untrusted parties

Now we consider that two of the parties, $\mathrm{A}$ and $\mathrm{B}$, have
untrusted measuring devices. In this case they prepare an assemblage for
Charlie, which is given by

$\displaystyle\sigma_{ab|xy}^{\mathrm{C}}$ $\displaystyle=$
$\displaystyle\operatorname{tr}_{\mathrm{A}\mathrm{B}}(M_{a|x}\otimes
M_{b|y}\otimes\openone_{\mathrm{C}}\rho^{\mathrm{A}:\mathrm{B}:\mathrm{C}})$
(26) $\displaystyle=$
$\displaystyle\sum_{\lambda}p_{\lambda}(ab|xy)\rho_{\lambda}^{\mathrm{C}}.$

Once again, since the only dependence of the assemblage on $a$, $b$, $x$ and
$y$ is through $\lambda$ this assemblage is also unsteerable. Moreover,
because the set of probability distributions (also called a behaviour)
$p_{\lambda}(ab|xy)$ arises from local measurements on a separable state it
must be local, i.e. it can be written as
$p_{\lambda}(ab|xy)=\sum_{\mu\nu}q(\mu\nu|\lambda)D(a|x,\mu)D(b|y,\nu)$ Bell
review . Therefore, the relevant set of assemblages
$\Sigma^{\mathrm{A}:\mathrm{B}:\mathrm{C}}_{\mathrm{C}}$ is now given by

$\Sigma^{\mathrm{A}:\mathrm{B}:\mathrm{C}}_{\mathrm{C}}=\Big{\\{}\sigma_{ab|xy}^{\mathrm{C}}\Big{|}\sigma_{ab|xy}^{\mathrm{C}}=\textstyle{\sum_{\mu\nu}}D(a|x,\mu)D(b|y,\nu)\rho_{\mu\nu}^{\mathrm{C}},\\\
\rho_{\mu\nu}^{\mathrm{C}}\geq 0\Big{\\}}.$ (27)

### Case 2. Entanglement in a bipartition

Let us now consider the case where the state $\rho^{ABC}$ is separable with
respect to a single given bipartition. Choosing this partition to be
$\mathrm{A}:\mathrm{B}\mathrm{C}$ we now consider states of the form

$\rho^{\mathrm{A}:\mathrm{B}\mathrm{C}}=\sum_{\lambda}p_{\lambda}\rho^{\mathrm{A}}_{\lambda}\otimes\rho^{\mathrm{B}\mathrm{C}}_{\lambda}.$
(28)

Crucially, given the asymmetry of picking a bipartition as well as the
asymmetry of picking the trusted party (or parties), we will see below that we
have two inequivalent situations to consider, for both the cases of one or of
two untrusted parties.

#### Case 2A. Entanglement in a bipartition with one untrusted party

When only one of the parties performs uncharacterised measurements the
asymmetry of (28) leads to two different situations: (i) when the lone party
$\mathrm{A}$ has the untrusted devices and (ii) when $\mathrm{B}$ (or
equivalently $\mathrm{C}$) does. In the first case Bob and Charlie’s
assemblage is given by

$\displaystyle\sigma_{a|x}^{\mathrm{B}\mathrm{C}}$ $\displaystyle=$
$\displaystyle\operatorname{tr}_{\mathrm{A}}(M_{a|x}\otimes\openone_{\mathrm{B}}\otimes\openone_{\mathrm{C}}\rho^{\mathrm{A}:\mathrm{B}\mathrm{C}})$
(29) $\displaystyle=$
$\displaystyle\sum_{\lambda}p(a|x,\lambda)\rho^{\mathrm{B}\mathrm{C}}_{\lambda},$

Once again the dependence of Bob and Charlie’s assemblage on $a$ and $x$ comes
only from the common variable $\lambda$, and so this assemblage is
unsteerable. In comparison to previously, the states distributed to Bob and
Charlie are now arbitrary entangled states, and hence there is no additional
structure that the decomposition imposes. The set
$\Sigma^{\mathrm{A}:\mathrm{B}\mathrm{C}}_{\mathrm{B}\mathrm{C}}$ defined by
assemblages of the form (29) is therefore given by

$\Sigma^{\mathrm{A}:\mathrm{B}\mathrm{C}}_{\mathrm{B}\mathrm{C}}=\Big{\\{}\sigma_{a|x}^{\mathrm{B}\mathrm{C}}\Big{|}\sigma_{a|x}^{\mathrm{B}\mathrm{C}}=\textstyle{\sum_{\mu}}D(a|x,\mu)\rho_{\mu}^{\mathrm{B}\mathrm{C}},\rho_{\mu}^{\mathrm{B}\mathrm{C}}\geq
0\Big{\\}}.$ (30)

In the second case, where Bob is the one not trusting his measurements, Alice
and Charlie are left with the following assemblage:

$\displaystyle\sigma_{b|y}^{\mathrm{A}\mathrm{C}}$ $\displaystyle=$
$\displaystyle\operatorname{tr}_{\mathrm{B}}(\openone_{\mathrm{A}}\otimes
M_{b|y}\otimes\openone_{\mathrm{C}}\rho^{\mathrm{A}:\mathrm{B}\mathrm{C}})$
(31) $\displaystyle=$
$\displaystyle\sum_{\lambda}p_{\lambda}\rho^{\mathrm{A}}_{\lambda}\otimes\sigma_{b|y,\lambda}^{\mathrm{C}},$

which is a fundamentally different situation. This assemblage now has two main
features: (i) the only dependence on the variables $b$ and $y$ are due to
Charlie’s states. In other words, $\sigma_{b|y}^{\mathrm{A}\mathrm{C}}$ is
unsteerable from Bob to Alice but not necessarily from Bob to Charlie (note
however that the no-signalling condition still holds from Bob to Charlie).
This implies that if we trace out system C (or apply any quantum-to-classical
map to it) the resulting assemblage for Alice alone will be unsteerable; (ii)
it is composed by separable (unnormalised) states. The relevant set,
$\Sigma^{\mathrm{A}:\mathrm{B}\mathrm{C}}_{\mathrm{A}\mathrm{C}}$, is now
given by

$\Sigma^{\mathrm{A}:\mathrm{B}\mathrm{C}}_{\mathrm{A}\mathrm{C}}=\Big{\\{}\sigma_{b|y}^{\mathrm{A}\mathrm{C}}\Big{|}\operatorname{tr}_{\mathrm{C}}\sigma_{b|y}^{\mathrm{A}\mathrm{C}}=\textstyle{\sum_{\mu}}D(b|y,\mu)\rho_{\mu}^{\mathrm{A}},\\\
\rho_{\mu}^{\mathrm{A}}\geq
0,\sigma_{b|y}^{\mathrm{A}\mathrm{C}}\in\mathrm{SEP},\sum_{b}\sigma_{b|y}^{\mathrm{A}\mathrm{C}}=\sum_{b}\sigma_{b|y^{\prime}}^{\mathrm{A}\mathrm{C}}\Big{\\}}$
(32)

#### Case 2B. Entanglement in a bipartition with two untrusted parties

Again the asymmetry of the decomposition (28) leads to two different
situations: In one the untrusted measurements are at $\mathrm{A}$ and
$\mathrm{B}$ (or similarly $\mathrm{A}$ and $\mathrm{C}$), whilst in the other
they are at $\mathrm{B}$ and $\mathrm{C}$. In the first case the assemblage
obtained is given by

$\displaystyle\sigma_{ab|xy}^{\mathrm{C}}$ $\displaystyle=$
$\displaystyle\operatorname{tr}_{\mathrm{A}\mathrm{B}}(M_{a|x}\otimes
M_{b|y}\otimes\openone_{\mathrm{C}}\rho^{\mathrm{A}:\mathrm{B}\mathrm{C}})$
(33) $\displaystyle=$
$\displaystyle\sum_{\lambda}p(a|x,\lambda)\sigma_{b|y,\lambda}^{\mathrm{C}}.$

This assemblage has only one main feature, that it may contain only steering
from Bob to Charlie, and not from Alice to Charlie. It then defines the set
$\Sigma^{\mathrm{A}:\mathrm{B}\mathrm{C}}_{\mathrm{C}}$ as

$\Sigma^{\mathrm{A}:\mathrm{B}\mathrm{C}}_{\mathrm{C}}=\Big{\\{}\sigma_{ab|xy}^{\mathrm{C}}\Big{|}\sigma_{ab|xy}^{\mathrm{C}}=\textstyle{\sum_{\mu}}D(a|x,\mu)\sigma_{b|y,\mu}^{\mathrm{C}},\\\
\sigma_{b|y,\mu}^{\mathrm{C}}\geq
0,\sum_{b}\sigma_{b|y,\mu}^{\mathrm{C}}=\sum_{b}\sigma_{b|y^{\prime},\mu}^{\mathrm{C}}\Big{\\}}$
(34)

In the second case the resulting assemblage is given by

$\displaystyle\sigma_{bc|yz}^{\mathrm{A}}$ $\displaystyle=$
$\displaystyle\operatorname{tr}_{\mathrm{B}\mathrm{C}}(\openone_{\mathrm{A}}\otimes
M_{b|y}\otimes M_{c|z}\rho^{\mathrm{A}:\mathrm{B}\mathrm{C}})$ (35)
$\displaystyle=$
$\displaystyle\sum_{\lambda}p(bc|yz,\lambda)\rho^{\mathrm{A}}_{\lambda}.$

Here, there are two main features: (i) this assemblage is unsteerable; (ii)
The behaviour $p(bc|yz,\lambda)$ arises from local measurements on a possibly
entangled state $\rho_{\lambda}^{\mathrm{B}\mathrm{C}}$, it may contain
nonlocal quantum correlations Bell review . The final set we define is
therefore $\Sigma^{\mathrm{A}:\mathrm{B}\mathrm{C}}_{\mathrm{A}}$, given by

$\Sigma^{\mathrm{A}:\mathrm{B}\mathrm{C}}_{\mathrm{A}}=\Big{\\{}\sigma_{bc|yz}^{\mathrm{A}}\Big{|}\sigma_{bc|yz}^{\mathrm{A}}=\textstyle{\sum_{\lambda}}p(bc|yz,\lambda)\sigma_{\lambda}^{\mathrm{A}},\\\
\sigma_{\lambda}^{\mathrm{A}}\geq 0,p(bc|yz,\lambda)\in\mathcal{Q}\Big{\\}}$
(36)

where we have denoted by $\mathcal{Q}$ the set of probability distributions
which can arise from local measurements on quantum states.

### Case 3. Genuine multipartite entanglement

Let us now turn to the question of genuine multipartite entanglement (GME)
detection. Genuine tripartite entangled states are the ones that can not be
written as

$\displaystyle\rho^{\mathrm{bisep}}$ $\displaystyle=$
$\displaystyle\sum_{\lambda}p_{\lambda}^{\mathrm{A}:\mathrm{B}\mathrm{C}}\rho^{\mathrm{A}}_{\lambda}\otimes\rho^{\mathrm{B}\mathrm{C}}_{\lambda}+\sum_{\lambda}p_{\lambda}^{\mathrm{B}:\mathrm{A}\mathrm{C}}\rho^{\mathrm{B}}_{\lambda}\otimes\rho^{\mathrm{A}\mathrm{C}}_{\lambda}$
(37) $\displaystyle+$
$\displaystyle\sum_{\lambda}p_{\lambda}^{\mathrm{A}\mathrm{B}:\mathrm{C}}\rho^{\mathrm{A}\mathrm{B}}_{\lambda}\otimes\rho^{C}_{\lambda},$

where $p_{\lambda}^{\mathrm{A}:\mathrm{B}\mathrm{C}}$,
$p_{\lambda}^{\mathrm{B}:\mathrm{A}\mathrm{C}}$ and
$p_{\lambda}^{\mathrm{A}\mathrm{B}:\mathrm{C}}$ are probability distributions.
Our goal once again is to determine what constraints the form (37) imposes on
the obtained assemblages, which will now follow straightforwardly given the
analysis made before.

#### Case 3A. Genuine multipartite entanglement with one untrusted party

When Alice is the one holding the untrusted devices, Bob and Charlie’s
assemblage is given by

$\displaystyle\sigma_{a|x}^{\mathrm{B}\mathrm{C}}$ $\displaystyle=$
$\displaystyle\operatorname{tr}(M_{a|x}\otimes\openone_{\mathrm{B}}\otimes\openone_{\mathrm{C}}\rho^{\mathrm{bisep}})$
(38) $\displaystyle=$
$\displaystyle\underbrace{\sum_{\lambda}p_{\lambda}^{\mathrm{A}:\mathrm{B}\mathrm{C}}p(a|x,\lambda)\rho^{\mathrm{B}\mathrm{C}}_{\lambda}}_{\Gamma_{a|x}^{\mathrm{A}:\mathrm{B}\mathrm{C}}\in\Sigma^{\mathrm{A}:\mathrm{B}\mathrm{C}}_{\mathrm{B}\mathrm{C}}}+\underbrace{\sum_{\lambda}p_{\lambda}^{\mathrm{B}:\mathrm{A}\mathrm{C}}\rho^{\mathrm{B}}_{\lambda}\otimes\sigma_{a|x,\lambda}^{\mathrm{C}}}_{\Gamma_{a|x}^{\mathrm{B}:\mathrm{A}\mathrm{C}}\in\Sigma^{\mathrm{B}:\mathrm{A}\mathrm{C}}_{\mathrm{B}\mathrm{C}}}+\underbrace{\sum_{\lambda}p_{\lambda}^{\mathrm{A}\mathrm{B}:\mathrm{C}}\sigma_{a|x\lambda}^{\mathrm{B}}\otimes\rho^{\mathrm{C}}_{\lambda}}_{\Gamma_{a|x}^{\mathrm{C}:\mathrm{A}\mathrm{B}}\in\Sigma^{\mathrm{C}:\mathrm{A}\mathrm{B}}_{\mathrm{B}\mathrm{C}}}.$

The terms $\Gamma_{a|x}^{\mathrm{A}:\mathrm{B}\mathrm{C}}$ and
$\Gamma_{a|x}^{\mathrm{B}:\mathrm{A}\mathrm{C}}$ can be seen as assemblages
having the same structure as the assemblages (29) and (31) respectively, while
the assemblage $\Gamma_{a|x}^{\mathrm{C}:\mathrm{A}\mathrm{B}}$ is identical
to $\Gamma_{a|x}^{\mathrm{B}:\mathrm{A}\mathrm{C}}$, except that the role of
Bob and Charlie is interchanged.

#### Case 3B. Genuine multipartite entanglement with two untrusted parties

Consider now that Alice and Bob perform untrusted measurements, leading to:

$\displaystyle\sigma_{ab|xy}^{\mathrm{C}}$ $\displaystyle=$
$\displaystyle\operatorname{tr}_{\mathrm{A}\mathrm{B}}(M_{a|x}\otimes
M_{b|y}\otimes\openone_{\mathrm{C}}\rho^{\mathrm{bisep}})$ (39)
$\displaystyle=$
$\displaystyle\underbrace{\sum_{\lambda}p_{\lambda}^{\mathrm{A}:\mathrm{B}\mathrm{C}}p(a|x,\lambda)\sigma_{b|y,\lambda}^{\mathrm{C}}}_{\Pi_{ab|xy}^{\mathrm{A}:\mathrm{B}\mathrm{C}}\in\Sigma^{\mathrm{A}:\mathrm{B}\mathrm{C}}_{\mathrm{C}}}+\underbrace{\sum_{\lambda}p_{\lambda}^{\mathrm{B}:\mathrm{A}\mathrm{C}}p(b|y,\lambda)\sigma_{a|x,\lambda}^{\mathrm{C}}}_{\Pi_{ab|xy}^{\mathrm{B}:\mathrm{A}\mathrm{C}}\in\Sigma^{\mathrm{B}:\mathrm{A}\mathrm{C}}_{\mathrm{C}}}+\underbrace{\sum_{\lambda}p_{\lambda}^{\mathrm{A}\mathrm{B}:\mathrm{C}}p(ab|xy,\lambda)\rho^{\mathrm{C}}_{\lambda}}_{\Pi_{ab|xy}^{\mathrm{C}:\mathrm{A}\mathrm{B}}\in\Sigma^{\mathrm{C}:\mathrm{A}\mathrm{B}}_{\mathrm{C}}}.$

Again, the assemblages $\Pi_{ab|xy}^{\mathrm{A}:\mathrm{B}\mathrm{C}}$ and
$\Pi_{ab|xy}^{\mathrm{B}:\mathrm{A}\mathrm{C}}$ are seen to have the same
structure as (33), while the assemblage
$\Pi_{ab|xy}^{\mathrm{C}:\mathrm{A}\mathrm{B}}$ has the same structure as
(35).

Supplementary note 2

SDP tests and semi-device-independent entanglement witnesses

We have previously determined the constraints that each kind of entanglement
imposes, and defined the corresponding sets of assemblages these constraints
define. We now turn to the following practical question: given that we have
observed a specific assemblage, can we test for a certain type of entanglement
by checking whether or not the assemblage belongs to one of the previously
defined sets?

Crucially, it turns out that all of the sets defined above are either
specified solely in terms of positive semi-definite (PSD) constraints and
linear matrix inequalities (LMIs), or can be approximated from the outside by
a set with such a specification. Testing for membership inside such a set is
an optimisation problem known as a semi-definite program (SDP), for which
efficient numerical methods exist for the case of small systems, allowing for
an answer to this question SDP . Moreover, due to the theory of duality, the
dual SDP provides us with a semi-device-independent witness that allow us to
certify the presence of the different types of entanglement solely from the
knowledge of the assemblage. This is similar to the ideas of entanglement
witnesses and Bell inequalities in the standard and fully device-independent
scenarios respectively.

###  .1 Deriving the SDP tests

In those cases where the the sets defined above are not specified solely in
terms of PSD constraints and LMIs our strategy is to show that there exist
suitable relaxations which are, that is to define bigger sets which are
specified solely in terms of such constraints.

Working through in the order that they appeared, we shall consider each set in
turn. The first set is
$\Sigma^{\mathrm{A}:\mathrm{B}:\mathrm{C}}_{\mathrm{B}\mathrm{C}}$, given in
equation (25). It is the final constraint,
$\rho_{\mu}^{\mathrm{B}\mathrm{C}}\in\mathrm{SEP}$, that does not have the
desired form, as the set of (unnormalised) separable states has in general a
complicated structure. The one case where the set in fact has a simple
characterisation is if the dimensions satisfy
$d_{\mathrm{B}}d_{\mathrm{C}}\leq 6$, in which case the set of separable
states is exactly the set of states positive under partial transposition (PPT)
HorPLA . In this simple case we can rewrite
$\rho_{\mu}^{\mathrm{B}\mathrm{C}}\in\mathrm{SEP}$ as
$\left(\rho_{\mu}^{\mathrm{B}\mathrm{C}}\right)^{\mathrm{T}_{\mathrm{B}}}\geq
0$, where $\mathrm{T}_{\mathrm{B}}$ denotes the partial transposition with
respect to system B. Since this operation is a linear map (on the state), this
is now a PSD constraint, and the set is in fact in the desired form.

In all other dimensions, we can use the relaxation of the separable states to
those that have a $k$-symmetric PPT extension DPS . That is, we define the set
$\mathrm{SYM}^{(k)}_{\mathrm{B}\mathrm{C}}$

$\mathrm{SYM}^{(k)}_{\mathrm{B}\mathrm{C}}=\Big{\\{}\rho^{\mathrm{B}\mathrm{C}}\Big{|}\rho^{\mathrm{B}\mathrm{C}}=\operatorname{tr}_{\mathrm{B}_{2}\cdots\mathrm{B}_{k}}\rho^{\mathrm{B}_{1}\cdots\mathrm{B}_{k}\mathrm{C}},\\\
S_{ij}\rho^{\mathrm{B}_{1}\cdots\mathrm{B}_{k}\mathrm{C}}S_{ij}^{\dagger}=\rho^{\mathrm{B}_{1}\cdots\mathrm{B}_{k}\mathrm{C}}\quad\forall
i\neq j,\\\
\big{(}\rho^{\mathrm{B}_{1}\cdots\mathrm{B}_{k}\mathrm{C}}\big{)}^{T_{\mathrm{C}}}\geq
0,\rho^{\mathrm{B}_{1}\cdots\mathrm{B}_{k}\mathrm{C}}\geq 0\Big{\\}}$ (40)

where $S_{ij}$ is the swap operator between $\mathrm{B}_{i}$ and
$\mathrm{B}_{j}$. This demands that $\rho^{\mathrm{B}\mathrm{C}}$ can be
extended to a state with $k$ Bobs, which is symmetric under interchange and
PPT, such that the reduced state of a single Bob and Charlie is the original
state. Such sets are all specified in terms of PSD constraints and LMIs, and
converge to the set of separable states as $k\to\infty$ DPS ; DPSmulti . For
the case $k=1$ it also reduces to the set of PPT state. We thus define the
sequence of relaxations

$\Sigma^{\mathrm{A}:\mathrm{B}:\mathrm{C}(k)}_{\mathrm{B}\mathrm{C}}=\Big{\\{}\sigma_{a|x}^{\mathrm{B}\mathrm{C}}\Big{|}\sigma_{a|x}^{\mathrm{B}\mathrm{C}}=\textstyle{\sum_{\mu}}D(a|x,\mu)\rho_{\mu}^{\mathrm{B}\mathrm{C}},\\\
\rho_{\mu}^{\mathrm{B}\mathrm{C}}\in\mathrm{SYM}^{(k)}_{\mathrm{B}\mathrm{C}}\Big{\\}},$
(41)

which now have the desired structure for each $k$.

Moving on, the set $\Sigma^{\mathrm{A}:\mathrm{B}:\mathrm{C}}_{\mathrm{C}}$
given in (27) already has the desired structure. This is also true for the set
$\Sigma^{\mathrm{A}:\mathrm{B}\mathrm{C}}_{\mathrm{B}\mathrm{C}}$ defined in
(30). The set
$\Sigma^{\mathrm{A}:\mathrm{B}\mathrm{C}}_{\mathrm{A}\mathrm{C}}$ (32)
contains the requirement that
$\sigma_{b|y}^{\mathrm{A}\mathrm{C}}\in\mathrm{SEP}$, which is dealt with in
exactly the same way as above. Thus we define the relaxed set

$\Sigma^{\mathrm{A}:\mathrm{B}\mathrm{C}(k)}_{\mathrm{A}\mathrm{C}}=\Big{\\{}\sigma_{b|y}^{\mathrm{A}\mathrm{C}}\Big{|}\operatorname{tr}_{\mathrm{C}}\sigma_{b|y}^{\mathrm{A}\mathrm{C}}=\textstyle{\sum_{\mu}}D(b|y,\mu)\rho_{\mu}^{\mathrm{A}},\\\
\rho_{\mu}^{\mathrm{A}}\geq
0,\sigma_{b|y}^{\mathrm{A}\mathrm{C}}\in\mathrm{SYM}^{(k)}_{\mathrm{A}\mathrm{C}},\sum_{b}\sigma_{b|y}^{\mathrm{A}\mathrm{C}}=\sum_{b}\sigma_{b|y^{\prime}}^{\mathrm{A}\mathrm{C}}\Big{\\}}$
(42)

The set $\Sigma^{\mathrm{A}:\mathrm{B}\mathrm{C}}_{\mathrm{C}}$ given in (34)
has the desired structure.

Finally, the set $\Sigma^{\mathrm{C}:\mathrm{A}\mathrm{B}}_{\mathrm{C}}$
(which is has the same structure as (36)) is less straightforward because of
the constraint $p(ab|xy,\lambda)\in\mathcal{Q}$, i.e. that the behaviours
$p(ab|xy,\lambda)$ should have a quantum realisation. As in the nonlocality
scenario of deciding if a behaviour has a quantum realisation, the exact
answer to this problem is in general intractable. However, we can use the idea
introduced in Pus13  (see also NavTorVer14 ) to obtain a semi-definite
relaxation. The basic idea is to apply the method of the NPA hierarchy NPA
only to the untrusted devices, whilst leaving Charlie quantum, using also the
fact that the state is separable on the $\mathrm{A}\mathrm{B}:\mathrm{C}$
partition. We thus relax to
$\Pi_{ab|xy}^{\mathrm{C}:\mathrm{A}\mathrm{B}}\in\mathcal{Q}_{\mathrm{C}}^{(k)}$,
where $\mathcal{Q}_{\mathrm{C}}^{(k)}$ is defined by

$\mathcal{Q}_{\mathrm{C}}^{(k)}=\Big{\\{}\sigma_{ab|xy}^{\mathrm{C}}\Big{|}\Gamma^{(k)}_{\mathrm{C}}\geq
0,\left(\Gamma^{(k)}_{\mathrm{C}}\right)^{\mathrm{T}_{\mathrm{C}}}\geq 0\\\
\operatorname{tr}\big{(}G_{j}\Gamma^{(k)}_{\mathrm{C}}\big{)}=\operatorname{tr}\big{(}h_{j}\sigma_{ab|xy}^{\mathrm{C}}\big{)}\forall
j\Big{\\}}$ (43)

for some sets of operators $\\{G_{j}^{(k)}\\}$ and $\\{h_{j}^{(k)}\\}$, which
encode the constraints that arise in the original NPA hierarchy NPA , coming
from (i) orthogonality of measurement outcomes, and (ii) commutivity of Alices
and Bob measurements. Also, we can apply the idea from Pus13  and impose that
the matrices $\Gamma_{\mathrm{C}}^{(k)}$ are positive under partial
transposition of Charlie (the trusted party), as a relaxation of the
separability criteria across the bipartition. The main difference with the NPA
approach is that whereas previously the elements of the matrix $\Gamma^{(k)}$
were complex numbers with certain ones equal to the nonlocal behaviour, now
one should think of the elements as matrices (of the dimension of Charlie),
with certain ones equal to members of the assemblage.

Last, in order to be able to impose semidefinite constraints to the set
$\Sigma^{\mathrm{C}:\mathrm{A}\mathrm{B}}_{\mathrm{C}}$ we need to constrain
the number of terms in the summation in $\lambda$. We do this by noticing that
any quantum behaviour can be written as a convex combination of extremal non-
signalling behaviours $D^{\mathrm{NS}}(bc|yz,\nu)$ Bell review .

Given the above, set $\Sigma^{\mathrm{C}:\mathrm{A}\mathrm{B}}_{\mathrm{A}}$
is relaxed to

$\Sigma^{\mathrm{C}:\mathrm{A}\mathrm{B}(k)}_{\mathrm{C}}=\Big{\\{}\sigma_{ab|xy}^{\mathrm{C}}\Big{|}\sigma_{ab|xy}^{\mathrm{C}}=\textstyle{\sum_{\nu}}D^{\mathrm{NS}}(ab|xy,\nu)\sigma_{\nu}^{\mathrm{C}},\\\
\sigma_{ab|xy}^{\mathrm{C}}\in\mathcal{Q}^{(k)}_{\mathrm{C}}\Big{\\}}.$ (44)

Having found appropriate relaxations of all of the sets which we wish to
consider, we can now straightforwardly write down an approximate optimisation
problem in the form of an SDP that needs to be solved to check for the desired
type of entanglement in each given scenario. Let us describe explicitly the
approximate test that checks for the existence of a decomposition of the form
(39), i.e. that checks for genuine multipartite entanglement with two
untrusted parties. We provide all the other SDP tests for the other
decompositions described above in Supplementary Table 2.

In (39) we see that we have to find 3 assemblages, each contained in a
different set, with each set either in a form directly usable, or for which we
just gave an outer approximation above. Thus, by introducing the maximally
mixed assemblage
$\mathrm{id}_{ab|xy}^{\mathrm{C}}=\frac{1}{m_{\mathrm{A}}m_{\mathrm{B}}}\openone_{\mathrm{C}}/d_{\mathrm{C}}$
we arrive at the following SDP test for genuine multipartite entanglement with
2 untrusted parties

$\displaystyle\mathrm{max}\quad$ $\displaystyle p$
$\displaystyle\mathrm{s.t.}\quad$
$\displaystyle\Pi_{ab|xy}^{\mathrm{A}:\mathrm{B}\mathrm{C}}+\Pi_{ab|xy}^{\mathrm{B}:\mathrm{A}\mathrm{C}}+\Pi_{ab|xy}^{\mathrm{C}:\mathrm{A}\mathrm{B}}=\sigma_{ab|xy}^{\mathrm{obs}}-p\,\mathrm{id}_{ab|xy}^{\mathrm{C}}$
$\displaystyle\Pi_{ab|xy}^{\mathrm{A}:\mathrm{B}\mathrm{C}}\in\Sigma^{\mathrm{A}:\mathrm{B}\mathrm{C}}_{\mathrm{C}},\Pi_{ab|xy}^{\mathrm{B}:\mathrm{A}\mathrm{C}}\in\Sigma^{\mathrm{B}:\mathrm{A}\mathrm{C}}_{\mathrm{C}},$
(45)
$\displaystyle\quad\quad\quad\Pi_{ab|xy}^{\mathrm{C}:\mathrm{A}\mathrm{B}}\in\Sigma^{\mathrm{C}:\mathrm{A}\mathrm{B}(k)}_{\mathrm{C}}$

where $\sigma_{ab|xy}^{\mathrm{obs}}$ is the observed assemblage of Charlie.
Since $\mathrm{id}_{ab|xy}^{\mathrm{C}}$ is clearly contained in all 3 sets,
being producible from the maximally mixed state, a sufficiently large negative
$p$ will always be a solution, hence the SDP is strictly feasible. A strictly
negative optimal solution $p^{*}<0$ certifies that
$\sigma_{ab|xy}^{\mathrm{obs}}$ being measured does not have the desired
decomposition, i.e. that the state is genuinely multipartite entangled. On the
other hand an optimal value $p^{*}=0$ indicates that a decomposition can be
found. Note however that in this case, given the relaxation of the problem,
one is not able to conclude anything regarding the separability of the state.
One can take the parameter $k$ larger to obtain a better approximation to the
original problem.

###  .2 Semi device-independent entanglement witnesses

The dual of the SDP (.1) is also readily written down SDP , and is given by

$\displaystyle\min$
$\displaystyle\quad\operatorname{tr}\sum_{abxy}F_{ab|xy}\sigma_{ab|xy}^{\mathrm{obs}}$
s.t.
$\displaystyle\quad\operatorname{tr}\sum_{abxy}F_{ab|xy}\sigma_{ab|xy}^{\mathrm{C}}\geq
0$
$\displaystyle\quad\quad\quad\quad\forall\,\,\sigma_{ab|xy}^{\mathrm{C}}\in\Sigma^{\mathrm{A}:\mathrm{B}\mathrm{C}}_{\mathrm{C}}\cup\Sigma^{\mathrm{B}:\mathrm{A}\mathrm{C}}_{\mathrm{C}}\cup\Sigma^{\mathrm{C}:\mathrm{A}\mathrm{B}(k)}_{\mathrm{C}}$
(46)
$\displaystyle\quad\operatorname{tr}\sum_{abxy}F_{ab|xy}\mathrm{id}_{ab|xy}^{\mathrm{C}}=1$

which is seen to constitute a witness for genuine multipartite entanglement.
That is, the dual provides a set of operators $\\{F_{ab|xy}\\}_{abxy}$ such
that the linear functional $\beta=\sum_{abxy}F_{ab|xy}\sigma_{ab|xy}$ is
greater than zero for all assemblages which arise from measurements on a bi-
separable state. An observed value $\beta^{\mathrm{obs}}<0$ thus provides a
witness which certifies the genuine multipartite entanglement of the state in
a semi-device-independent manner. The final condition,
$\operatorname{tr}\sum_{abxy}F_{ab|xy}\mathrm{id}_{ab|xy}^{\mathrm{C}}=1$ is a
convention, which simply defines an overall scale for the witness.

More generally, the dual of each SDP in Supplementary Table 2 provides witness
operators $\\{F_{a|x}\\}_{ax}$, for the case of Alice untrusted, or
$\\{F_{ab|xy}\\}_{abxy}$, for the case of Alice and Bob untrusted (or a
permutation of the parties) which constitute a semi-device-independent
entanglement witnesses of the form

$\displaystyle\operatorname{tr}\sum_{ax}F_{a|x}\sigma_{a|x}\geq
0\quad\forall\sigma_{a|x}\in\Sigma$
$\displaystyle\operatorname{tr}\sum_{abxy}F_{ab|xy}\sigma_{ab|xy}\geq
0\quad\forall\sigma_{ab|xy}\in\Sigma^{\prime}$ (47)

with corresponding violations
$\beta^{\mathrm{obs}}=\operatorname{tr}\sum_{ax}F_{a|x}\sigma_{a|x}^{\mathrm{obs}}<0$
or
$\beta^{\mathrm{obs}}=\operatorname{tr}\sum_{ax}F_{ab|xy}\sigma_{ab|xy}^{\mathrm{obs}}<0$
respectively, where $\Sigma$ and $\Sigma^{\prime}$ are sets, or union of sets
(depending upon the type of entanglement one is checking for), as defined
above.

Finally, we note that it is possible to put these witnesses into two more
friendly forms in the case of binary measurement outcomes, so-called
_observable_ and _coefficient_ forms. Starting with the observable form, we
use the definition of the observed assemblage, and introduce the observables
$A_{x}=M_{0|x}-M_{1|x}$ to arrive at

$\displaystyle\operatorname{tr}\sum_{ax}F_{a|x}\sigma_{a|x}^{\mathrm{obs}}=\operatorname{tr}\sum_{ax}M_{a|x}\otimes
F_{a|x}\rho$
$\displaystyle=\tfrac{1}{2}\operatorname{tr}\sum_{ax}(\openone_{\mathrm{A}}+(-1)^{a}A_{x})\otimes
F_{a|x}\rho$
$\displaystyle=\operatorname{tr}\big{(}\openone_{\mathrm{A}}\otimes\tfrac{1}{2}\sum_{ax}F_{a|x}+\sum_{x}A_{x}\otimes\tfrac{1}{2}\sum_{a}(-1)^{a}F_{a|x}\big{)}\rho$
$\displaystyle=\operatorname{tr}\big{(}\openone_{\mathrm{A}}\otimes
J_{\emptyset}+\sum_{x}A_{x}\otimes J_{x}\big{)}\rho$ (48)

for the case of one untrusted party, where we have defined the observables
$J_{\emptyset}=\tfrac{1}{2}\sum_{ax}F_{a|x}$ and
$J_{x}=\tfrac{1}{2}\sum_{a}(-1)^{a}F_{ab|xy}$ for Bob and Charlie. For the
coefficient form we further expand these matrices in the (complete) basis of
Pauli operators, namely,

$\displaystyle J_{\emptyset}$
$\displaystyle=\sum_{yz}f_{0yz}\sigma^{y}\otimes\sigma^{z},$ $\displaystyle
J_{x}=\sum_{yz}f_{xyz}\sigma^{y}\otimes\sigma^{z}$ (49)

where $\sigma^{0}=\openone$, $\sigma^{1}=X$, $\sigma^{2}=Y$ and $\sigma^{3}=Z$
and denoting $A_{0}=\openone$ then we arrive at the compact form

$\displaystyle\operatorname{tr}\sum_{ax}F_{a|x}\sigma_{a|x}^{\mathrm{obs}}=\operatorname{tr}\left(\sum_{xyz}(f_{xyz}A_{x}\otimes\sigma^{y}\otimes\sigma^{z})\rho\right)$
(50)

For the case of two untrusted parties, an analogous but longer calculation
gives for the observable form

$\operatorname{tr}\sum_{abxy}F_{ab|xy}\sigma_{ab|xy}^{\mathrm{obs}}\\\
=\operatorname{tr}\big{(}\openone_{\mathrm{A}}\otimes\openone_{\mathrm{B}}\otimes
K_{\emptyset}+\sum_{x}A_{x}\otimes\openone_{\mathrm{B}}\otimes K_{x}\\\
\quad+\sum_{y}\openone_{\mathrm{A}}\otimes B_{y}\otimes
K^{\prime}_{y}+\sum_{xy}A_{x}\otimes B_{y}\otimes K_{xy}\big{)}\rho$ (51)

where we have introduced the observables $B_{y}$ for Bob, as well as the
observables $K_{\emptyset}=\tfrac{1}{4}\sum_{abxy}F_{ab|xy}$,
$K_{x}=\tfrac{1}{4}\sum_{aby}(-1)^{a}F_{ab|xy}$,
$K^{\prime}_{y}=\tfrac{1}{4}\sum_{abx}(-1)^{b}F_{ab|xy}$ and
$K_{xy}=\tfrac{1}{4}\sum_{ab}(-1)^{a+b}F_{ab|xy}$ for the trusted party
Charlie, that need to be measured in the corresponding configurations, given
above. For the coefficient form, again by expanding these matrices in the
complete basis of Pauli matrices,

$\displaystyle K_{\emptyset}$ $\displaystyle=\sum_{z}f_{00z}\sigma^{z},$
$\displaystyle K_{x}=\sum_{z}f_{x0z}\sigma^{z}$ $\displaystyle K^{\prime}_{y}$
$\displaystyle=\sum_{z}f_{0yz}\sigma^{z},$ $\displaystyle
K_{xy}=\sum_{z}f_{xyz}\sigma^{z}$ (52)

and by denoting $B_{0}=\openone$, we obtain the analogous compact form as
previously,

$\displaystyle\operatorname{tr}\sum_{abxy}F_{ab|xy}\sigma_{ab|xy}^{\mathrm{obs}}=\operatorname{tr}\left(\sum_{xyz}(f_{xyz}A_{x}\otimes
B_{y}\otimes\sigma^{z})\rho\right)$ (53)

Supplementary note 3

Generalisation to more parties

We have presented our main results in the tripartite case. Notice however that
the same procedure can readily be followed to derive SDPs to test the presence
of different kinds of entanglement for general $N$-partite systems.

First of all one specifies the scenarios by fixing (i) a particular type of
entanglement and (ii) the pattern of trusted and untrusted parties. The
entanglement can be chosen arbitrarily, for example one may ask that the state
is not fully separable, be separable across a given number of fixed
bipartitions, or be a convex combination of states separable over a given
number of partitions (but not necessarily fixed). The patten may also be
chosen arbitrarily, ranging from all but one party trusted, to all but one
untrusted.

Given the specification, one then enumerates the list of properties which the
corresponding assemblages have. These properties will fall into two classes -
those which impose constraints which are directly applicable, i.e. are in the
form of PSD constraints and LMIs, and those which are not. As in the
tripartite case, the objective is then to relax the non-directly applicable
constraints to find an approximate SDP test.

The main difficultly in our approach is that as the number of parties
increases, and the local dimension of the Hilbert space, we expect that the
difficulty of the problem will grow to the point where current numerical
techniques are unable to solve efficiently the tests. For example, one class
of constraints that will arise is that multipartite assemblages will need to
have quantum realisations. In principle such a constraint can still be imposed
by applying the NPA hierarchy NPA  to the untrusted devices, however in the
multipartite setting this soon becomes intractable. Alternatively, one may
have constraints that a multipartite quantum state is separable. One can again
relax this using the generalisation of the $k$-shareability condition DPSmulti
.

In summary, the approach presented here is most suitable to scenarios
involving relatively small numbers of parties, where it provides powerful
tests for multipartite entanglement (and explicitly provides witnesses in each
case). This is however expected as this is also the case in standard
entanglement detection techniques ent detection  (due to the increase of the
Hilbert space dimension) and in the fully device-independent approach Bell
review  (due to the number of the space of local probability distributions).

Supplementary note 4

Examples: GHZ and W states

In order to demonstrate the usefulness of our previous characterisation we
apply the above SDP to two exemplary genuine multipartite states, namely the
GHZ and W states. More specifically we are interested in how much white noise
can be added to these states until our method fails to detect either
entanglement or GME, i.e. we want find the minimum $w$, denoted by $w^{*}$,
allowing us to detect either entanglement or GME in the states

$\displaystyle\rho_{\mathrm{GHZ}}$
$\displaystyle=w|\mathrm{GHZ}\rangle\langle\mathrm{GHZ}|+(1-w)\openone/8;$
(54) $\displaystyle\rho_{W}$ $\displaystyle=w|W\rangle\langle
W|+(1-w)\openone/8,$

where $|\mathrm{GHZ}\rangle=(|000\rangle+|111\rangle)/{\sqrt{2}}$ and
$|W\rangle=(|001\rangle+|010\rangle+|100\rangle)/{\sqrt{3}}$. Supplementary
Table 3 gives a summary of the results, in terms of the numbers provided by
our methods and a comparison to what was known regarding entanglement
witnesses and Bell inequalities. All results were obtained using cvx SDP  for
Matlab to solve the SDP, and the optimisation toolbox to numerically search
for the best choices of measurements for Alice (and Bob). Since such a search
over measurements choices provides no guarantee that the global optimum is
obtained, all results constitute upper bounds. However, all of our numerical
evidence suggests that the values obtained cannot be improved.

As we can see the values of $w^{*}$ lies in between the bound for
entanglement, where the largest number of assumptions are made, and the bound
from nonlocality, where no assumptions are made. Furthermore, as one would
expect, stronger bounds are possible with 2 parties trust their devices
compared to the case of only 1.

We end by presenting the steering witnesses we obtain in the above for the GHZ
and W states that certify genuine tripartite entanglement in a semi-device-
independent fashion.

Starting with the GHZ state and the case of two untrusted parties (and three
measurements), the optimal witness is

$1-\alpha\langle A_{2}B_{2}\rangle-\alpha\langle A_{2}Z\rangle-\alpha\langle
B_{2}Z\rangle-\beta\langle A_{0}B_{0}X\rangle\\\ +\beta\langle
A_{0}B_{1}Y\rangle+\beta\langle A_{1}B_{0}Y\rangle+\beta\langle
A_{1}B_{1}X\rangle\geq 0$ (55)

where $\alpha=0.1831$ and $\beta=0.2582$, and the pure GHZ state achieves a
violation $-0.5821\ngeq 0$. For the case of the GHZ state and only a single
untrusted party, the witness is

$1+0.1547\langle
Z_{\mathrm{B}}Z_{\mathrm{C}}\rangle-\tfrac{1}{3}\big{(}\langle
A_{2}Z_{\mathrm{B}}\rangle+\langle A_{2}Z_{\mathrm{C}}\rangle+\langle
A_{0}X_{\mathrm{B}}X_{\mathrm{C}}\rangle\\\ -\langle
A_{0}Y_{\mathrm{B}}Y_{\mathrm{C}}\rangle-\langle
A_{1}X_{\mathrm{B}}Y_{\mathrm{C}}\rangle-\langle
A_{1}Y_{\mathrm{B}}X_{\mathrm{C}}\rangle\big{)}\geq 0$ (56)

with the pure GHZ state now achieving a violation of $-0.8453\ngeq 0$.
Interestingly, we note first that the structure of both witnesses is the same,
the only difference being in the coefficients. Furthermore the only terms
which appear are those which arise from the stabiliser relations of the GHZ
state.

Moving on to the W state, for two untrusted parties the optimal witness is

$1+0.2517\big{(}\langle A_{2}\rangle+\langle B_{2}\rangle\big{)}+0.3520\langle
Z\rangle-0.1112\big{(}\langle A_{0}X\rangle\\\ +\langle A_{1}Y\rangle+\langle
B_{0}X\rangle+\langle B_{1}Y\rangle\big{)}+0.1296\big{(}\langle
A_{2}Z\rangle+\langle B_{2}Z\rangle\big{)}\\\ -0.1943\big{(}\langle
A_{0}B_{0}\rangle+\langle A_{1}B_{1}\rangle\big{)}+0.2277\langle
A_{2}B_{2}\rangle\\\ -0.1590\big{(}\langle A_{0}B_{0}Z\rangle+\langle
A_{1}B_{1}Z\rangle\big{)}+0.2228\langle A_{2}B_{2}Z\rangle\\\
-0.2298\big{(}\langle A_{0}B_{2}X\rangle+\langle A_{1}B_{2}Y\rangle+\langle
A_{2}B_{0}X\rangle+\langle A_{2}B_{1}Y\rangle\big{)}\geq 0$ (57)

and the pure W state obtains the violation $-0.4803\ngeq 0$. For one untrusted
party the witness is

$1+0.4405\big{(}\langle Z_{\mathrm{B}}\rangle+\langle
Z_{\mathrm{C}}\rangle\big{)}-0.0037\langle
Z_{\mathrm{B}}Z_{\mathrm{C}}\rangle-0.1570\big{(}\langle
X_{\mathrm{B}}X_{\mathrm{C}}\rangle\\\ +\langle
Y_{\mathrm{B}}Y_{\mathrm{C}}\rangle+\langle
A_{2}X_{\mathrm{B}}X_{\mathrm{C}}\rangle+\langle
A_{2}Y_{\mathrm{B}}Y_{\mathrm{C}}\rangle\big{)}+0.2424\big{(}\langle
A_{2}\rangle\\\ +\langle
A_{2}Z_{\mathrm{B}}Z_{\mathrm{C}}\rangle\big{)}+0.1848\big{(}\langle
A_{2}Z_{\mathrm{B}}\rangle+\langle
A_{2}Z_{\mathrm{C}}\rangle\big{)}-0.2533\big{(}\langle A_{0}X_{B}\rangle\\\
+\langle A_{0}X_{\mathrm{C}}\rangle+\langle A_{1}Y_{\mathrm{B}}\rangle+\langle
A_{1}Y_{\mathrm{C}}\rangle+\langle
A_{0}X_{\mathrm{B}}Z_{\mathrm{C}}\rangle+\langle
A_{0}Z_{\mathrm{B}}X_{\mathrm{C}}\rangle\\\ +\langle
A_{1}Y_{\mathrm{B}}Z_{\mathrm{C}}\rangle+\langle
A_{1}Z_{\mathrm{B}}Y_{\mathrm{C}}\rangle\big{)}\geq 0$ (58)

with he pure W state achieving the violation $-0.7594\ngeq 0$. Again, we note
that structurally the witnesses are the same in the case of one and two
untrusted parties.

Supplementary note 5

Experimental inequalities

In this section we give the semi-device independent entanglement witnesses
that were used to optimally certify the presence of genuine multipartite
entanglement of the GHZ state. Starting with the case of one untrusted party,
in coefficient form the inequality is given by

$\displaystyle
f_{xy0}=\left(\begin{array}[]{rrrr}1.0000&-0.0183&0.1079&0.0130\\\
0.1522&0.1518&-0.0734&0.0251\\\ -0.2870&-0.1125&-0.1229&0.0189\\\
-0.0658&0.0095&-0.0151&-0.2142\end{array}\right),$ (63) $\displaystyle
f_{xy1}=\left(\begin{array}[]{rrrr}0.2096&0.0941&0.1565&0.0077\\\
0.0633&0.3006&-0.0195&0.0132\\\ -0.0388&-0.0133&-0.3040&-0.0033\\\
0.0614&0.0040&-0.0389&-0.0387\\\ \end{array}\right),$ (68) $\displaystyle
f_{xy2}=\left(\begin{array}[]{rrrr}-0.1487&-0.1841&0.0862&-0.0061\\\
-0.1839&0.0323&0.2801&0.0028\\\ 0.1036&0.2920&-0.0423&0.0071\\\
0.0066&0.0683&0.0277&0.0242\\\ \end{array}\right),$ (73) $\displaystyle
f_{xy3}=\left(\begin{array}[]{rrrr}0.0189&0.0114&0.0363&-0.2187\\\
-0.0259&-0.0144&0.0255&-0.0204\\\ -0.0714&-0.0027&0.0099&0.0505\\\
-0.0189&0.1263&-0.2037&0.0145\\\ \end{array}\right),$ (78)

where $x$ labels the rows and $y$ the columns. For two untrusted parties the
inequality is given by

$\displaystyle
f_{xy0}=\left(\begin{array}[]{rrrr}1.0000&0.2375&-0.2613&0.0023\\\
0.0026&-0.0041&-0.0055&0.0241\\\ -0.0039&-0.0055&-0.0114&0.0291\\\
-0.0012&0.0014&0.0032&-0.1515\\\ \end{array}\right),$ (83) $\displaystyle
f_{xy1}=\left(\begin{array}[]{rrrr}0.0041&-0.0033&-0.0022&0.0275\\\
0.1673&0.2676&-0.0317&0.0044\\\ 0.1710&-0.0015&-0.2737&-0.0072\\\
0.0009&0.0059&0.0004&0.0001\\\ \end{array}\right),$ (88) $\displaystyle
f_{xy2}=\left(\begin{array}[]{rrrr}-0.0063&-0.0091&0.0053&0.0106\\\
-0.1669&-0.0004&0.2651&0.0073\\\ 0.1675&0.2569&-0.0307&0.0036\\\
0.0019&0.0161&0.0127&0.0004\\\ \end{array}\right),$ (93) $\displaystyle
f_{xy3}=\left(\begin{array}[]{rrrr}-0.0013&0.0054&0.0011&-0.1523\\\
0.0024&0.0089&0.0053&0.0002\\\ 0.0023&0.0053&0.0047&0.0001\\\
-0.1527&-0.0050&-0.0085&-0.0014\\\ \end{array}\right).$ (98)

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