# Counterexamples in self-testing

Laura Mančinska<EMAIL_ADDRESS>Simon Schmidt<EMAIL_ADDRESS>QMATH,
Department of Mathematical Sciences, University of Copenhagen,
Universitetsparken 5, 2100 Copenhagen Ø, Denmark

###### Abstract

In the recent years self-testing has grown into a rich and active area of
study with applications ranging from practical verification of quantum devices
to deep complexity theoretic results. Self-testing allows a classical verifier
to deduce which quantum measurements and on what state are used, for example,
by provers Alice and Bob in a nonlocal game. Hence, self-testing as well as
its noise-tolerant cousin—robust self-testing—are desirable features for a
nonlocal game to have.

Contrary to what one might expect, we have a rather incomplete understanding
of if and how self-testing could fail to hold. In particular, could it be that
every 2-party nonlocal game or Bell inequality with a quantum advantage
certifies the presence of a specific quantum state? Also, is it the case that
every self-testing result can be turned robust with enough ingeniuty and
effort? We answer these questions in the negative by providing simple and
fully explicit counterexamples. To this end, given two nonlocal games
$\mathcal{G}_{1}$ and $\mathcal{G}_{2}$, we introduce the
$(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game, in which the players get pairs of
questions and choose which game they want to play. The players win if they
choose the same game and win it with the answers they have given. Our
counterexamples are based on this game and we believe this class of games to
be of independent interest.

##  1 Introduction

The notion of self-testing was first introduced by Mayers and Yao [20] and it
has since developed into an active and rich area of study (see [30] for a
review). One major motivation behind self-testing is that it can be used by a
classical verifier to certify that untrusted quantum devices perform according
to their specification. This is accomplished by deriving a quantum mechanical
description of a quantum device merely from classical observations. More
precisely, if Alice and Bob want to know the state and measurements of their
quantum devices, they can play a nonlocal game and check the output
probabilities. If this game is a self-test then after observing the desired
output probabilities, Alice and Bob can conclude that their devices must be
implementing certain measurements on a certain quantum state.

In addition to the early applications of self-testing to certification and
device - independent protocols, in the recent years it has been a key
ingredient for important results in quantum complexity theory [8, 21, 22, 12].
The most notable among these results is the recent breakthrough [12]
establishing that MIP*=RE and resolving Connes’ Embedding Problem which had
resisted all attempts for over 50 years.

It was shown by Bell [2] that the predictions of quantum mechanics are not
compatible with any local hidden-variable theory. He obtained this result by
showing that quantum mechanics predicts violations of what are now known as
_Bell inequalities_. Our definition of self-testing will be given in the
setting of _nonlocal games_ [6] which are another way to explore Bell
inequalities and their violations. A (2-player) nonlocal game is played by two
collaborating players, Alice and Bob, and a referee. The referee gives
questions to Alice and Bob and they each have to respond with an answer. The
players win according to a function on questions and answers, which is known
to the players beforehand. Crucially, Alice and Bob are not allowed to
communicate after they have received their questions. They can, however, agree
on a strategy ahead of time. In a _quantum strategy_ Alice and Bob can share
an entangled state and use local measurements to come up with their answers.
Finding a quantum strategy for a nonlocal game which exceeds the winning
probability of any classical strategy yields a violation of a Bell inequality.
We call quantum strategies with the highest winning probability in a nonlocal
game _optimal_ quantum strategies.

#### Motivation and results.

In a nutshell, _self-testing_ says that any optimal quantum strategy for some
game $\mathcal{G}$ is equal to a reference strategy, up to local isometries.
There are different forms of self-testing. A game can self-test an optimal
quantum strategy or just the shared state of this quantum strategy. Also, a
game can _robustly_ self-test a strategy in which case we in addition require
that any near-optimal quantum strategy must be close to an optimal reference
strategy. For a nonlocal game $\mathcal{G}$, an optimal quantum strategy $S$,
and state $|\psi\rangle$ used in $S$, we have the following relation between
the above forms of self-testing:

$\displaystyle\mathcal{G}\text{ robustly self-tests
}S\Rightarrow\mathcal{G}\text{ self-tests }S\Rightarrow\mathcal{G}\text{ self-
tests }|\psi\rangle.$ (1)

So we see that robust self-tests are the strongest while self-tests of states
are a weaker form of self-testing. It is natural to ask for counterexamples
showing that there are no equivalences above.

Unsurprisingly, it is often easier to prove that a certain game self-tests a
reference strategy $S$ but more care and effort is needed to show that this
self-test is in fact robust. In most known cases, however, the same argument
that is used to establish a self-testing result can also be turned robust by
simply arguing that all the desired relations hold approximately rather than
exactly. In fact, there are no known examples of self-tests that are not
robust! Moreover, if a self-test is proven using a specific technique that
leverages representation theory of finite groups then robustness of such self-
tests follows from the Gowers-Hatami theorem [11, 33]. The idea here is that
all representations of finite groups are stable in a certain precise
mathematical sense. Hence, one might hope that the same holds true for self-
tests and any (exact) self-test is necessarily robust. Regretfully, we show
that this is not the case, _i.e._ , the converse of the first implication from
(1) fails to hold:

######  Theorem A (Theorem 5.2 & Example 5.3).

There exists a nonlocal game $\mathcal{G}$ that exactly self-tests a quantum
strategy $S$, but this self-test is not robust.

The second question we address in this article is whether every 2-party
nonlocal game with quantum advantage self-tests some quantum state. The only
known non-trivial set-ups that do not self-test any quantum state are known in
the many party case [10]. We answer our second question in the negative:

######  Theorem B (Theorem 6.2 & Example 6.13).

There exists a game $\mathcal{G}$ that does not self-test any state.

#### Discussion and outlook.

Another way to get different forms of self-testing is to vary the classical
observations one has access to. In this article we have focused on self-
testing where we assume that a quantum strategy achieves optimal or near-
optimal winning probability in a nonlocal game. Another common form is where
we derive self-testing from optimal or near-optimal Bell inequality
violations. Since every nonlocal game can be cast as a Bell inequality, our
counterexamples also hold in that setting. Finally, rather than self-testing
from a single numerical value (_e.g._ Bell violation or winning probability),
we can prove a self-testing result for quantum strategies that produce a
certain quantum correlation (collection of probabilities). As we explain in
more detail below, our counterexamples do not carry over to the setting of
correlations. We summarize the known counterexamples to self-testing,
including results of this article, in the table below:

|  Non-robust self-test |  Not a measurement self-test |  Not a state self-test
---|---|---|---
Nonlocal games | Theorem A |  [7], [27], [29] | Theorem B
Bell inequalities |  |  [13], [14] |  [1], [9], [10]
Extreme correlations | ? |  [29], [31] | ?
Table 1: Counterexamples for equivalence of different forms of self-testing in
the three settings: nonlocal games, Bell inequalities, and extreme
correlations. Note that every counterexample for nonlocal games also yields a
counterexample for Bell inequalities.

We now comment on Table 1. We present the first example of a game that self-
tests a quantum strategy, but this self-test is not robust (Theorem A) in this
article. Since the correlations of the approximate optimal strategies of our
game do not necessarily converge to the self-tested strategy, we do not get a
non-robust self-test for correlations. We leave this as an open question. For
obtaining an example of a game that does not self-test measurements, one needs
to construct games with inequivalent optimal strategies. In [7] it was shown
that the glued magic square game does not self-test measurements. It turns out
that this game still self-tests a quantum state [16]. This gives an example of
a game that self-tests a state but not the whole strategy. Another, elementary
construction of such a game is due to Storgaard [29, Section 5.6]: Take a game
that is a self-test and add a question to Bob’s question set for which the
players always lose if Bob receives this question. The game is still a state
self-test, but the operators Bob uses for the additional question are unknown
to us. Another counterexample for measurement self-testing was obtained by
Kaniewski [13] in the framework of Bell inequalities. An extreme correlation
obtained by inequivalent strategies was first presented in [31] and studied
further in [29]. In this article, we construct the first examples of two-party
games that do not self-test any state (Theorem B). As shown in the table
above, an example of a three-party Bell inequaltiy that does not self-test any
state has been identified in [10]. Furthermore, Bell inequalities that are
maximally violated by states in an entangled subspace are studied in [1] and
[9], but all their examples use more than two parties. We do not know an
extreme correlation that does not self-test any state.

#### Proof ideas.

We now explain the key ideas behind our two main theorems. For obtaining the
examples of non-robust self-tests and games that do not self-test any state,
we introduce the $(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game for nonlocal
games $\mathcal{G}_{1}$ and $\mathcal{G}_{2}$. In this game the players
receive pairs of questions, one from each game. They choose which question
they answer and win if they answered a question from the same game and
additionally win this game with their answers. We believe that the
$(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game may be of independent interest,
beyond the applications for self-testing results.

For the non-robust self-test (Theorem A), we let $\mathcal{G}_{1}$ be a game
that has no perfect quantum strategy, but a sequence of strategies whose
winning probabilities converge to one. Note that Slofstra constructed such a
game in [28]. We choose $\mathcal{G}_{2}$ to be a pseudo-telepathy game
(_i.e._ a game with perfect quantum strategy, but no perfect classical
strategy) that self-tests some strategy $S_{2}$. Then the
$(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game self-tests a strategy in which
Alice and Bob always choose game $\mathcal{G}_{2}$ and play according to
$S_{2}$. This self-test is not robust, however, as there are near-optimal
strategies for the $(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game coming from the
near-optimal strategies for $\mathcal{G}_{1}$.

To obtain nonlocal games that do not self-test any state (Theorem B), it
suffices to construct a game with two optimal quantum strategies that use
states with coprime Schmidt ranks (see Lemma 6.1). In our case, those will be
perfect quantum strategies. Note that we can get perfect strategies for the
$(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game from perfect quantum strategies
for $\mathcal{G}_{1}$ and $\mathcal{G}_{2}$, respectively, by Alice and Bob
always choosing the same game and playing the perfect quantum strategy of this
game. Thus, to get a game that does not self-test any state, it suffices to
find two pseudo-telepathy games that have perfect strategies using states of
coprime Schmidt ranks. Then the $(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game
will not self-test any state.

The magic square game is a pseudo-telepathy game with a strategy that uses a
state with Schmidt rank $4$. Thus, we are left with finding a pseudo-telepathy
game that has a perfect quantum strategy with a state of odd Schmidt rank. Our
example of such a game is a $(G,t)$-independent set game, where $G$ is the
orthogonality graph of a $3$-dimensional weak Kochen-Specker set and some
$t\in\mathbb{N}$. The perfect quantum strategy uses a state of Schmidt rank
$3$.

#### Structure of the article.

In Section 2 we collect background material on nonlocal games. We then define
all forms of self-tests in Section 3 and provide some known results we will
use later on. Furthermore, we show that the synchronous magic square game
self-tests a quantum strategy. In Section 4 we introduce the
$(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game and establish some of its
properties in case when both $\mathcal{G}_{1}$ and $\mathcal{G}_{2}$ are
pseudo-telepathy games. Using the $(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game,
we construct a non-robust self-test in Section 5. Finally, we obtain games
that do not self-test states in Section 6.

##  2 Preliminaries

We state some basic facts and notions we will use throughout this article. We
write $[m]:=\\{1,\dots,m\\}$. Each Hilbert space $\mathcal{H}$ considered in
this article is finite-dimensional which means that we have
$\mathcal{H}\cong\mathbb{C}^{d}$ for some $d\in\mathbb{N}$. We let
$\||\xi\rangle\|=(\langle\xi|\xi\rangle)^{\frac{1}{2}}$ for
$|\xi\rangle\in\mathcal{H}$. An operator $X\in B(\mathcal{H})$ is _positive_ ,
if $\langle\xi|X|\xi\rangle\geq 0$ for all $|\xi\rangle\in\mathcal{H}$. We
write $X\leq Y$ if $Y-X$ is positive.

A _positive operator valued measure_ (POVM) consists of a family of positive
operators $\\{M_{i}\in B(\mathcal{H})\,|\,i\in[m]\\}$ such that
$\sum_{i=1}^{m}M_{i}=1_{B(\mathcal{H})}$. If all positive operators are
projections ($M_{i}=M_{i}^{*}=M_{i}^{2}$), then we call $\\{M_{i}\in
B(\mathcal{H})\,|\,i\in[m]\\}$ with $\sum_{i=1}^{m}M_{i}=1_{B(\mathcal{H})}$ a
_projective measurement_ (PVM). A _state_ $|\psi\rangle$ is a unit vector in a
Hilbert space $\mathcal{H}$. Each state
$|\psi\rangle\in\mathcal{H}_{A}\otimes\mathcal{H}_{B}$ admits the so-called
_Schmidt decomposition_ $|\psi\rangle=\sum_{i=1}^{m}\lambda_{i}a_{i}\otimes
b_{i}$, where $\\{a_{i}\,|\,i\in[m]\\}$ and $\\{b_{i}\,|\,i\in[m]\\}$ are
orthonormal sets in $\mathcal{H}_{A}$ and $\mathcal{H}_{B}$, respectively, and
$\lambda_{i}\geq 0$ for all $i$. The strictly positive values $\lambda_{i}>0$
are called _Schmidt coefficients_. The _Schmidt rank_ of a state
$|\psi\rangle$ is the number $n$ of Schmidt coefficients $\lambda_{i}$
(counted with multiplicity) in a Schmidt decomposition. We say that
$|\psi\rangle$ has _full Schmidt rank_ if
$n=\mathrm{dim}(\mathcal{H}_{A})=\mathrm{dim}(\mathcal{H}_{B})$. For
$|\psi\rangle=\sum_{i=1}^{m}\lambda_{i}a_{i}\otimes b_{i}$, we let
$\mathrm{supp}_{A}(|\psi\rangle)=\mathrm{span}\\{a_{i}\\}_{i=1}^{m}$ and
$\mathrm{supp}_{B}(|\psi\rangle)=\mathrm{span}\\{b_{i}\\}_{i=1}^{m}$.
Furthermore, we say that a subspace $\mathcal{K}\subseteq\mathcal{H}$ is an
_invariant subspace_ of $X\in B(\mathcal{H})$ if
$X(\mathcal{K})\subseteq\mathcal{K}$.

In the following, we describe the framework of nonlocal games [6]. A two-
player _nonlocal game_ $\mathcal{G}$ is played between two collaborating
players, Alice and Bob, and a referee. It is specified by finite input sets
$I_{A}$, $I_{B}$ and finite output sets $O_{A}$, $O_{B}$ for Alice and Bob,
respectively as well as a _verification function_ $V:I_{A}\times I_{B}\times
O_{A}\times O_{B}\to\\{0,1\\}$ and a probability distribution $\pi$ on
$I_{A}\times I_{B}$. In the game, the referee samples a pair $(x,y)\in
I_{A}\times I_{B}$ using the distribution $\pi$ and sends $x$ to Alice and $y$
to Bob. Alice and Bob respond with $a\in O_{A}$ and $b\in O_{B}$,
respectively. They win if $V(x,y,a,b)=1$. In our examples, we always use the
uniform distribution $\pi$. The players are not allowed to communicate during
the game, but they can agree on a strategy beforehand. A _classical strategy_
for a nonlocal game is a strategy in which Alice and Bob only have access to
shared randomness. In a _quantum strategy_ , the players are allowed to
perform local measurements on a shared entangled state. Thus, a quantum
strategy may be written as
$S=(|\psi\rangle\in\mathcal{H}_{A}\otimes\mathcal{H}_{B},\\{E_{xa}\\}_{x},\\{F_{yb}\\}_{y})$,
where $|\psi\rangle$ is the shared entangled state, $\\{E_{xa}\\}_{x}$ are the
POVMs for each $x\in I_{A}$ for Alice and $\\{F_{yb}\\}_{y}$ are the POVMs for
each $y\in I_{B}$ for Bob. A nonlocal game $\mathcal{G}$ is called
_synchronous_ [23] if we have $I_{A}=I_{B}$, $O_{A}=O_{B}$ and if Alice and
Bob receive identical inputs, they must answer with identical outputs to win,
i.e. $V(x,x,a,b)=0$ for $a\neq b$.

The _classical value_ $\omega(\mathcal{G})$ of a nonlocal game $\mathcal{G}$
is the greatest probability of winning the game with a classical strategy. The
_quantum value_ $\omega^{*}(\mathcal{G})$ is the supremum over the winning
probabilities of the quantum strategies for the game. An _optimal quantum
strategy_ is a quantum strategy achieving this quantum value. In contrary to
the classical value, the quantum value is not always attained [28]. In
general, the quantum value is bigger than the classical value. We are
especially interested in _pseudo-telepathy_ games, which are games having a
_perfect_ quantum strategy (a strategy that allows the players to win with
probability one), but no perfect classical strategy.

##  3 Self-testing

In this section, we formally introduce the notion of self-testing. The idea of
self-testing is the following. A game self-tests an optimal quantum strategy
$\tilde{S}$ if the strategy is unique, in the sense that every other optimal
strategy $S$ is related the strategy $\tilde{S}$ by a local isometry. As in
[19], we will say that $\tilde{S}$ is a local dilation of $S$.

######  Definition 3.1.

Let
$\tilde{S}=(\tilde{|\psi\rangle}\in\tilde{\mathcal{H}}_{A}\otimes\tilde{\mathcal{H}}_{B},\\{\tilde{E}_{xa}\\}_{x},\\{\tilde{F}_{yb}\\}_{y})$,
$S=(|\psi\rangle\in\mathcal{H}_{A}\otimes\mathcal{H}_{B},\\{E_{xa}\\}_{x},\allowbreak\\{F_{yb}\\}_{y})$
be quantum strategies. We say that $\tilde{S}$ is a _local dilation_ of $S$ if
there exist Hilbert spaces $\mathcal{H}_{A,aux}$ and $\mathcal{H}_{B,aux}$, a
state $|aux\rangle\in\mathcal{H}_{A,aux}\otimes\mathcal{H}_{B,aux}$ and
isometries
$U_{A}:\mathcal{H}_{A}\to\tilde{\mathcal{H}}_{A}\otimes\mathcal{H}_{A,aux}$,
$U_{B}:\mathcal{H}_{B}\to\tilde{\mathcal{H}}_{B}\otimes\mathcal{H}_{B,aux}$
such that with $U:=U_{A}\otimes U_{B}$ it holds

$\displaystyle U|\psi\rangle$
$\displaystyle=|\tilde{\psi}\rangle\otimes|aux\rangle,$ (2) $\displaystyle
U(E_{xa}\otimes 1)|\psi\rangle$ $\displaystyle=[(\tilde{E}_{xa}\otimes
1)|\tilde{\psi}\rangle]\otimes|aux\rangle,$ (3) $\displaystyle U(1\otimes
F_{yb})|\psi\rangle$
$\displaystyle=[(1\otimes\tilde{F}_{yb})|\tilde{\psi}\rangle]\otimes|aux\rangle.$
(4)

Note that if $\tilde{S}$ is a local dilation of $S$, then they induce the same
correlation and thus the same winning probability for a game $\mathcal{G}$.

######  Remark 3.2.

There is a slight abuse of notation in the previous Definition. The lefthand
side of equations (2), (3) and (4) is an element of
$\tilde{\mathcal{H}}_{A}\otimes\mathcal{H}_{A,aux}\otimes\tilde{\mathcal{H}}_{B}\otimes\mathcal{H}_{B,aux}$,
whereas the righthand side is an element of
$\tilde{\mathcal{H}}_{A}\otimes\tilde{\mathcal{H}}_{B}\otimes\mathcal{H}_{A,aux}\otimes\mathcal{H}_{B,aux}$.
We identify those Hilbert spaces via the unitary that flips the second and the
third tensor factors.

######  Remark 3.3.

The equations (3) and (4) are equivalent to

$\displaystyle U(E_{xa}\otimes F_{yb})|\psi\rangle$
$\displaystyle=[(\tilde{E}_{xa}\otimes\tilde{F}_{yb})|\tilde{\psi}\rangle]\otimes|aux\rangle.$

######  Definition 3.4.

Let $\mathcal{G}$ be a nonlocal game and
$\tilde{S}=(\tilde{|\psi\rangle}\in\tilde{\mathcal{H}}_{A}\otimes\tilde{\mathcal{H}}_{B},\\{\tilde{E}_{xa}\\}_{x},\\{\tilde{F}_{yb}\\}_{y})$
be an optimal quantum strategy. We say that $\mathcal{G}$ _self-tests_ the
strategy $\tilde{S}$ if $\tilde{S}$ is a local dilation of any optimal quantum
strategy
$S=(|\psi\rangle\in\mathcal{H}_{A}\otimes\mathcal{H}_{B},\\{E_{xa}\\}_{x},\\{F_{yb}\\}_{y})$.
If only equation (2) holds for all optimal quantum strategies $S$, we say that
$\mathcal{G}$ _self-tests the state_ $|\tilde{\psi}\rangle$.

######  Remark 3.5.

Some authors only impose that $\tilde{S}$ is a local dilation of any
projective strategy $S$. This is a priori a weaker notion of self-testing. We
will see, however, that in some cases (e.g. the magic square game) the PVM
self-test can be used to prove a POVM self-test for the synchronous version of
the game.

We will now state some results of [19] on local dilations. Those will mostly
help to reduce proving self-testing for general quantum strategies to quantum
strategies having states with full Schmidt rank.

######  Lemma 3.6.

[19, Lemma 4.8] Let $X\in B(\mathcal{H}_{A})$, $Y\in B(\mathcal{H}_{B})$ and
$|\psi\rangle\in\mathcal{H}_{A}\otimes\mathcal{H}_{B}$ such that $(X\otimes
1)|\psi\rangle=(1\otimes Y)|\psi\rangle$. Then
$\mathrm{supp}_{A}(|\psi\rangle)$ is invariant under $X$ and
$\mathrm{supp}_{B}(|\psi\rangle)$ is invariant under $Y$.

The next lemma and its corollary split [19, Lemma 4.9] into two parts, making
the lemma slightly more general. We do this since we need to use the lemma in
its more general form later on.

######  Lemma 3.7.

Let $\mathcal{G}$ be a nonlocal game. Let
$S=(|\psi\rangle\in\mathbb{C}^{d_{A}}\otimes\mathbb{C}^{d_{B}},\\{E_{xa}\\}_{x},\\{F_{yb}\\}_{y})$
be a quantum strategy such that $\mathrm{supp}_{A}(|\psi\rangle)$ and
$\mathrm{supp}_{B}(|\psi\rangle)$ are invariant under each $E_{xa}$ and
$F_{yb}$, respectively. Then there exists a quantum strategy
$S^{\prime}=(|\psi^{\prime}\rangle,\\{E_{xa}^{\prime}\\}_{x},\\{F_{yb}^{\prime}\\}_{y})$
such that $|\psi^{\prime}\rangle$ has full Schmidt rank and $S^{\prime}$ is a
local dilation of $S$.

###### Proof.

We follow the proof of [19, Lemma 4.9]. Consider the Schmidt decomposition
$|\psi\rangle=\sum_{i=1}^{r}\alpha_{i}\xi_{i}\otimes\eta_{i}$ and let
$U_{A}:\mathbb{C}^{r}\to\mathbb{C}^{d_{A}}$,
$U_{B}:\mathbb{C}^{r}\to\mathbb{C}^{d_{B}}$ be isometries given by

$\displaystyle U_{A}=\sum_{i=1}^{r}\xi_{i}e_{i}^{*},\quad
U_{B}=\sum_{i=1}^{r}\eta_{i}e_{i}^{*}.$

Then

$\displaystyle E_{xa}^{\prime}=U_{A}^{*}E_{xa}U_{A},\quad
F_{yb}^{\prime}=U_{B}^{*}F_{yb}U_{B}$

are positive operators and $\\{E_{xa}^{\prime}\,|\,x\in[r]\\}$,
$\\{F_{yb}^{\prime}\,|\,y\in[r]\\}$ are POVMs. With
$|\psi^{\prime}\rangle=(U_{A}^{*}\otimes U_{B}^{*})|\psi\rangle$, we get a
quantum strategy $S^{\prime}$, where $|\psi^{\prime}\rangle$ has full Schmidt
rank.

We will now show that $S^{\prime}$ is a local dilation of $S$. Set
$H_{A,aux}=\mathbb{C}^{d_{A}}$, $H_{B,aux}=\mathbb{C}^{d_{B}}$ and define
isometries
$V_{A}:\mathbb{C}^{d_{A}}\to\mathbb{C}^{r}\otimes\mathbb{C}^{d_{A}}$,
$V_{B}:\mathbb{C}^{d_{B}}\to\mathbb{C}^{r}\otimes\mathbb{C}^{d_{B}}$ by

$\displaystyle V_{A}(v)$
$\displaystyle=U_{A}^{*}(v)\otimes\xi_{1}+e_{1}\otimes(1_{d_{A}}-U_{A}U_{A}^{*})(v),$
$\displaystyle V_{B}(w)$
$\displaystyle=U_{B}^{*}(w)\otimes\eta_{1}+e_{1}\otimes(1_{d_{B}}-U_{B}U_{B}^{*})(w),$

where $v\in\mathbb{C}^{d_{A}}$, $w\in\mathbb{C}^{d_{B}}$. Since
$\mathrm{supp}_{A}(|\psi\rangle)$ and $\mathrm{supp}_{B}(|\psi\rangle)$ are
invariant under each $E_{xa}$ and $F_{yb}$, respectively, and $U_{A}U_{A}^{*}$
and $U_{B}U_{B}^{*}$ are the projections onto
$\mathrm{supp}_{A}(|\psi\rangle)$ and $\mathrm{supp}_{B}(|\psi\rangle)$,
respectively, we have

$\displaystyle(V_{A}\otimes V_{B})(E_{xa}\otimes
F_{yb})|\psi\rangle=(U_{A}^{*}\otimes U_{B}^{*})((E_{xa}\otimes
F_{yb})|\psi\rangle)\otimes(\xi_{1}\otimes\eta_{1}).$

Furthermore, it holds

$\displaystyle(U_{A}^{*}\otimes U_{B}^{*})(E_{xa}\otimes F_{yb})|\psi\rangle$
$\displaystyle=(U_{A}^{*}\otimes U_{B}^{*})(E_{xa}\otimes
F_{yb})(U_{A}U_{A}^{*}\otimes U_{B}U_{B}^{*})|\psi\rangle$
$\displaystyle=(E_{xa}^{\prime}\otimes F_{yb}^{\prime})|\psi^{\prime}\rangle,$

since $U_{A}U_{A}^{*}$ and $U_{B}U_{B}^{*}$ are the projections onto
$\mathrm{supp}_{A}(|\psi\rangle)$ and $\mathrm{supp}_{B}(|\psi\rangle)$,
respectively. This concludes the proof. ∎

######  Corollary 3.8.

Let $\mathcal{G}$ be a synchronous game and let
$S=(|\psi\rangle,\\{E_{xa}\\}_{x},\\{F_{yb}\\}_{y})$ a perfect quantum
strategy of $\mathcal{G}$. Then there exists a perfect quantum strategy
$S^{\prime}=(|\psi^{\prime}\rangle,\\{E_{xa}^{\prime}\\}_{x},\allowbreak\\{F_{yb}^{\prime}\\}_{y})$
of $\mathcal{G}$ such that $|\psi^{\prime}\rangle$ has full Schmidt rank and
$S^{\prime}$ is a local dilation of $S$.

###### Proof.

By [19, Corollary 3.6 (a)], we know $(E_{xa}\otimes 1)|\psi\rangle=(1\otimes
F_{xa})|\psi\rangle$. Lemma 3.6 yields that $\mathrm{supp}_{A}(|\psi\rangle)$
and $\mathrm{supp}_{B}(|\psi\rangle)$ are invariant under each $E_{xa}$ and
$F_{yb}$, respectively. We obtain the result by Lemma 3.7. ∎

We also have a transitivity result for local dilations. One can use it for
proving self-testing results in the follwing way. First show that a general
optimal quantum strategy dilates to a quantum strategy having a state of full
Schmidt rank. Then check if the reference strategy is a local dilation of any
optimal strategy with a state of full Schmidt rank.

######  Lemma 3.9.

[19, Lemma 4.7] Let $S_{1}$, $S_{2}$ and $S_{3}$ be quantum strategies. If
$S_{1}$ is a local dilation of $S_{2}$ and $S_{2}$ is a local dilation of
$S_{3}$, then $S_{1}$ is a local dilation of $S_{3}$.

Finally, we give the definition of robust self-testing. Roughly speaking, a
game robustly self-tests a quantum strategy $\tilde{S}$ if it self-tests the
strategy and additionally, for every almost optimal strategy $S$, a local
dilation of $S$ is close to $\tilde{S}$. For $\delta>0$, we say that a quantum
strategy is _$\delta$ -optimal_ for a game $\mathcal{G}$ if the winning
probability is greater or equal to $\omega^{*}(\mathcal{G})-\delta$, where
$\omega^{*}(\mathcal{G})$ is the quantum value of $\mathcal{G}$.

######  Definition 3.10.

Let $\mathcal{G}$ be a nonlocal game,
$\tilde{S}=(\tilde{|\psi\rangle}\in\tilde{\mathcal{H}}_{A}\otimes\tilde{\mathcal{H}}_{B},\\{\tilde{E}_{xa}\\}_{x},\\{\tilde{F}_{yb}\\}_{y})$
an optimal quantum strategy. Then $\mathcal{G}$ is a _robust self-test_ for
$\tilde{S}$ if $\mathcal{G}$ is a self-test for $\tilde{S}$ and for any
$\varepsilon>0$, there exists a $\delta>0$ such that for any $\delta$-optimal
strategy
$S=(|\psi\rangle\in\mathcal{H}_{A}\otimes\mathcal{H}_{B},\\{E_{xa}\\}_{x},\\{B_{yb}\\}_{y})$,
there exists Hilbert spaces $\mathcal{H}_{A,aux}$ and $\mathcal{H}_{B,aux}$, a
state $|aux\rangle\in\mathcal{H}_{A,aux}\otimes\mathcal{H}_{B,aux}$ and
isometries
$U_{A}:\mathcal{H}_{A}\to\tilde{\mathcal{H}}_{A}\otimes\mathcal{H}_{A,aux}$,
$U_{B}:\mathcal{H}_{B}\to\tilde{\mathcal{H}}_{B}\otimes\mathcal{H}_{B,aux}$
such that with $U:=U_{A}\otimes U_{B}$ it holds

$\displaystyle\|U|\psi\rangle-\tilde{|\psi\rangle}\otimes|aux\rangle\|$
$\displaystyle\leq\varepsilon,$ $\displaystyle\|U(E_{xa}\otimes
1)|\psi\rangle-[(\tilde{E}_{xa}\otimes
1)\tilde{|\psi\rangle}]\otimes|aux\rangle\|$ $\displaystyle\leq\varepsilon,$
$\displaystyle\|U(1\otimes\tilde{F}_{yb})|\psi\rangle-[(1\otimes\tilde{F}_{yb})\tilde{|\psi\rangle}]\otimes|aux\rangle\|$
$\displaystyle\leq\varepsilon.$

###  3.1 The synchronous magic square game self-tests a quantum strategy

In this subsection, we show that the synchronous magic square game self-tests
a quantum strategy $\tilde{S}$ with maximally entangled state
$|\psi_{4}\rangle$. In [34] it was shown that the magic square game self-tests
a strategy $S$. Note that their general strategies consist of projective
measurements, the case of strategies with POVMs is not considered. We will
deduce a POVM self-testing result for the synchronous magic square game from
the PVM self-test of the magic square.

We will first describe the magic square game of [5, Section 5] and then its
synchronous version. Note that there are also other versions of the magic
square game. We choose the one with minimal input and output sets. Consider
the following set of equations

$\displaystyle x_{1}x_{2}x_{3}=1\quad(r_{1}),$ $\displaystyle
x_{1}x_{4}x_{7}=-1\quad(c_{1}),$ $\displaystyle
x_{4}x_{5}x_{6}=1\quad(r_{2}),$ $\displaystyle
x_{2}x_{5}x_{8}=-1\quad(c_{2}),$ $\displaystyle
x_{7}x_{8}x_{9}=1\quad(r_{3}),$ $\displaystyle
x_{3}x_{6}x_{9}=-1\quad(c_{3}).$

In the magic square game, the referee sends one of the three equations on the
left hand side to Alice and one of the three equations on the right hand side
to Bob. Alice answers with a $\\{-1,1\\}$-assignment of the variables such
that their product is $1$ and Bob answers with a $\\{-1,1\\}$-assignment of
the variables such that their product is $-1$. The players win the game, if
their assignments coincide in the common variable of the equations they
received. More formally, we have

$\displaystyle I_{A}=R:=\\{r_{1},r_{2},r_{3}\\},$ $\displaystyle
I_{B}=C:=\\{c_{1},c_{2},c_{3}\\},$ $\displaystyle
O_{A}=O_{1}:=\\{(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)\\},$ $\displaystyle
O_{B}=O_{-1}:=\\{(-1,-1,-1),(-1,1,1),(1,-1,1),(1,1,-1)\\},$ $\displaystyle
V_{MS}(r_{i},c_{j},a,b)=\begin{cases}1\text{ if }a_{j}=b_{i},\\\ 0\text{
otherwise.}\end{cases}$

Here $a=(a_{1},a_{2},a_{3}),b=(b_{1},b_{2},b_{3})$. Note that it is shown in
[5, Section 5] that this game is a pseudo-telepathy game.

The synchronous version of the magic square game is played as follows. The
referee sends one of the six equations above to Alice and one of them to Bob.
They answer with $\\{-1,1\\}$-assignments such that the equations are
fulfilled. Alice and Bob win the game if their assignments coincide on the
common variables of the equations they received. Here, we have

$\displaystyle I_{A}=I_{B}=R\cup C,\quad O_{A}=O_{B}=O_{1}\cup O_{-1},$
$\displaystyle V(x,y,a,b)=\begin{cases}V_{MS}(x,y,a,b)\text{ if }x\in R,y\in
C,a\in O_{1},b\in O_{-1},\\\ V_{MS}(y,x,b,a)\text{ if }x\in C,y\in R,a\in
O_{-1},b\in O_{1},\\\ 1\text{ if }(x,y\in R,x\neq y,a,b\in O_{1})\text{ or
}(x,y\in C,x\neq y,a,b\in O_{-1}),\\\ \delta_{ab}\text{ if }x=y\in R,a,b\in
O_{1}\text{ or }x=y\in C,a,b\in O_{-1},\\\ 0\text{ otherwise.}\end{cases}$

The following self-testing statement for the magic square game was shown in
[34]. We will use this theorem to deduce a self-testing statement for the
synchronous magic square game. Recall the Pauli matrices

$\displaystyle\sigma_{X}=\begin{pmatrix}0&1\\\
1&0\end{pmatrix},\,\sigma_{Y}=\begin{pmatrix}0&-i\\\ i&0\end{pmatrix}\text{
and }\sigma_{Z}=\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix}.$

######  Theorem 3.11 ([34]).

Restricting to projective measurements, the magic square game self-tests a
perfect quantum strategy
$S=(|\psi_{4}\rangle,\\{E_{xa}\\}_{x},\\{F_{yb}\\}_{y})$. Here
$|\psi_{4}\rangle=\frac{1}{2}\sum_{i=1}^{4}e_{i}\otimes e_{i}$ and
$E_{xa}=\frac{1}{8}(1+a_{1}X_{1})(1+a_{2}X_{2})(1+a_{3}X_{3})$,
$F_{yb}=\frac{1}{8}(1+b_{1}Y_{1}^{{}^{\mathsf{T}}})(1+b_{2}Y_{2}^{{}^{\mathsf{T}}})(1+b_{3}Y_{3}^{{}^{\mathsf{T}}})$,
where $X_{i}$ and $Y_{i}$ are the i-th entries in row $x$ and column $y$ of

$I\otimes\sigma_{Z}$$\sigma_{Z}\otimes
I$$\sigma_{Z}\otimes\sigma_{Z}$$\sigma_{X}\otimes
I$$I\otimes\sigma_{X}$$\sigma_{X}\otimes\sigma_{X}$$-\sigma_{X}\otimes\sigma_{Z}$$-\sigma_{Z}\otimes\sigma_{X}$$\sigma_{Y}\otimes\sigma_{Y}$

######  Corollary 3.12.

Let $S$ be as in Theorem 3.11. The synchronous magic square game self-tests
the perfect quantum strategy
$\tilde{S}=(|\psi_{4}\rangle,\\{\tilde{E}_{xa}\\}_{x},\\{\tilde{F}_{yb}\\}_{y})$,
where

$\displaystyle\tilde{E}_{xa}=\begin{cases}E_{xa}\text{ if }x\in R,a\in
O_{1}\\\ F_{xa}^{{}^{\mathsf{T}}}\text{ if }x\in C,a\in O_{-1},\\\ 0\text{
otherwise,}\end{cases}$ $\displaystyle\tilde{F}_{yb}=\begin{cases}F_{yb}\text{
if }y\in C,b\in O_{-1}\\\ E_{yb}^{{}^{\mathsf{T}}}\text{ if }y\in R,b\in
O_{1},\\\ 0\text{ otherwise.}\end{cases}$

###### Proof.

Let
$S^{\prime}=(|\psi^{\prime}\rangle,\\{E_{xa}^{\prime}\\},\\{F_{yb}^{\prime}\\})$
be a perfect quantum strategy for the synchronous magic square game. By
Corollary 3.8, there exists a perfect quantum strategy
$\hat{S}=(\hat{|\psi\rangle},\\{\hat{E}_{xa}\\},\\{\hat{F}_{yb}\\})$ that is a
local dilation of $S^{\prime}$, where $\hat{|\psi\rangle}$ has full Schmidt
rank, $\hat{|\psi\rangle}=\sum_{i=1}^{d}\hat{\lambda}_{i}e_{i}\otimes e_{i}$
with $\hat{\lambda}_{i}>0$ for all $i\in[d]$. Furthermore, by [19, Corollary
3.6], the operators $\hat{E}_{xa}$ and $\hat{F}_{yb}$ are projections.

Let $\hat{\varphi}:=\mathrm{diag}(\hat{\lambda}_{i})\in\mathbb{C}^{d\times
d}$, i.e. $\hat{\varphi}$ is a diagonal matrix with entries
$\hat{\lambda}_{i}$. Let $x\in R$ and $a\in O_{-1}$. Then

$\displaystyle\mathrm{Tr}(\hat{E}_{xa}\hat{\varphi}^{2})=\langle\hat{\psi}|\hat{E}_{xa}\otimes
1|\hat{\psi}\rangle=\langle\hat{\psi}|\hat{E}_{xa}\otimes\sum_{b}\hat{F}_{yb}|\hat{\psi}\rangle=0,$

since $\langle\hat{\psi}|\hat{E}_{xa}\otimes\hat{F}_{yb}|\hat{\psi}\rangle=0$
for all $y,b$ as $a\in O_{-1}$. Since $\hat{\varphi}$ is invertible, we
conclude $\hat{E}_{xa}=0$. We similarly get $\hat{E}_{xa}=0$ for $x\in C,a\in
O_{1}$ and $\hat{F}_{yb}=0$ for $y\in C,b\in O_{1}$ or $y\in R$, $b\in
O_{-1}$.

By the previous argument, we get that $\\{\hat{E}_{xa}\,|\,a\in O_{1}\\}$ and
$\\{\hat{F}_{yb}\,|\,b\in O_{-1}\\}$ are PVMs for $x\in R$, $y\in C$. Together
with $\hat{|\psi\rangle}$, they form a perfect quantum strategy for the magic
square game. By Theorem 3.11, there exists Hilbert spaces
$\mathcal{H}_{A,aux}$ and $\mathcal{H}_{B,aux}$, a state
$|aux\rangle\in\mathcal{H}_{A,aux}\otimes\mathcal{H}_{B,aux}$ and isometries
$U_{A}:\hat{\mathcal{H}}_{A}\to\mathcal{H}_{A}\otimes\mathcal{H}_{A,aux}$,
$U_{B}:\hat{\mathcal{H}}_{B}\to\mathcal{H}_{B}\otimes\mathcal{H}_{B,aux}$ such
that with $U:=U_{A}\otimes U_{B}$ it holds

$\displaystyle U|\hat{\psi}\rangle$
$\displaystyle=|\psi_{4}\rangle\otimes|aux\rangle,$ $\displaystyle
U(\hat{E}_{xa}\otimes 1)|\hat{\psi}\rangle$ $\displaystyle=[(E_{xa}\otimes
1)|\psi_{4}\rangle]\otimes|aux\rangle,$ $\displaystyle
U(1\otimes\hat{F}_{yb})|\hat{\psi}\rangle$ $\displaystyle=[(1\otimes
F_{xa})|\psi_{4}\rangle]\otimes|aux\rangle$ (5)

for $x\in R$, $y\in C$ and $a\in O_{1}$, $b\in O_{-1}$.

Now, let $x\in C$ and $a\in O_{-1}$. By [19, Corollary 3.6 (a)], we have

$\displaystyle U(\hat{E}_{xa}\otimes
1)|\hat{\psi}\rangle=U(1\otimes\hat{F}_{xa})|\hat{\psi}\rangle.$

Furthermore, it holds

$\displaystyle U(1\otimes\hat{F}_{xa})|\hat{\psi}\rangle=[(1\otimes
F_{yb})|\psi_{4}\rangle]\otimes|aux\rangle$

by (5). We conclude

$\displaystyle U(\hat{E}_{xa}\otimes 1)|\hat{\psi}\rangle=[(1\otimes
F_{yb})|\psi_{4}\rangle]\otimes|aux\rangle=[(F_{xa}^{{}^{\mathsf{T}}}\otimes
1)|\psi_{4}\rangle]\otimes|aux\rangle$ (6)

for all $x\in C$ and $a\in O_{-1}$. One similarly obtains

$\displaystyle U(1\otimes\hat{F}_{yb})|\hat{\psi}\rangle=[(1\otimes
E_{yb}^{{}^{\mathsf{T}}})|\psi_{4}\rangle]\otimes|aux\rangle$ (7)

for all $y\in R$ and $a\in O_{1}$. Since we know $\hat{E}_{xa}=0=E_{xa}$ and
$\hat{F}_{yb}=0=F_{yb}$ for the remaining $x,y,a,b$, we deduce from (5), (6)
and (7) that the synchronous magic square game self-tests the perfect quantum
strategy $\tilde{S}$. ∎

##  4 The $(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game

Let $\mathcal{G}_{1}$ and $\mathcal{G}_{2}$ be nonlocal games. The
$(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game is played as follows: The referee
sends Alice and Bob a pair of questions $(x_{1},x_{2})$ and $(y_{1},y_{2})$,
respectively, where $x_{1},y_{1}$ are questions in $\mathcal{G}_{1}$ and
$x_{2},y_{2}$ in $\mathcal{G}_{2}$. Each of them chooses one of the questions
they received and responds with an answer from the corresponding game. To win
the game, two conditions have to be fulfilled:

  * (1)

Alice and Bob have to choose questions from the same game,

  * (2)

their answers have to win the corresponding game.

More formally, suppose the nonlocal games $\mathcal{G}_{1}$ and
$\mathcal{G}_{2}$ have input sets $I_{A,i}$, $I_{B,i}$, output sets $O_{A,i}$,
$O_{B,i}$, verification functions $V_{\mathcal{G}_{i}}$ and probability
distributions $\pi_{i}$ on $I_{A,i}\times I_{B,i}$ for $i=1,2$. Then the
$(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game has input sets $I_{A,1}\times
I_{A,2}$, $I_{B,1}\times I_{B,2}$, output sets $O_{A,1}\dot{\cup}O_{A,2}$,
$O_{B,1}\dot{\cup}O_{B,2}$ and verification function

$\displaystyle
V((x_{1},x_{2}),(y_{1},y_{2}),a,b)=\begin{cases}V_{\mathcal{G}_{i}}(x_{i},y_{i},a,b)\,\text{
if }a\in O_{A,i}\text{ and }b\in O_{B,i}\text{ for some $i=1,2$,}\\\ 0\,\text{
otherwise.}\end{cases}$

For the probability distribution, we take $\pi=\pi_{1}\times\pi_{2}$ on
$(I_{A,1}\times I_{A,2})\times(I_{B,1}\times I_{B,2})$, _i.e._
$\pi((x_{1},x_{2}),(y_{1},y_{2}))=\pi_{1}(x_{1},y_{1})\pi_{2}(x_{2},y_{2})$.
In this article, we assume that the probability distributions of
$\mathcal{G}_{1}$ and $\mathcal{G}_{2}$ are uniform, so this will also be the
case for the $(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game. The
$(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game will be used for our
counterexamples in this article, we believe that it can also be of independent
interest.

We first check that the $(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game keeps its
quantum advantage if one of the games is a pseudo-telepathy game and the other
one does not have a perfect quantum strategy.

######  Lemma 4.1.

Let $\mathcal{G}_{1}$ be a pseudo-telepathy game and let $\mathcal{G}_{2}$ be
a game with $\omega(G)<1$. Then the
$(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game is a pseudo-telepathy game.

###### Proof.

It is easy to see that the $(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game has a
perfect strategy. Indeed, Alice and Bob can always choose to answer the
question from $\mathcal{G}_{1}$ and then use a perfect quantum strategy for
$\mathcal{G}_{1}$ to come up with their answers.

It remains to show that there is no perfect classical strategy for the
$(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game. First note that it suffices to
show that there is no perfect deterministic strategy. For contradiction assume
that there is a perfect deterministic strategy $S$. If in this strategy Alice
chooses a question $x_{1}$ for some pair $(x_{1},x_{2})$, then we know that
Bob has to choose $y_{1}$ for all pairs $(y_{1},y_{2})$, as otherwise there
are questions for which the players lose the game. A similar argument shows
that Alice also always has to choose the questions from $\mathcal{G}_{1}$.
This shows that in strategy $S$ the players always choose to answer questions
from $\mathcal{G}_{1}$. Hence, $S$ can be used to construct a perfect
classical strategy for game $\mathcal{G}_{1}$ which is a contradiction. Since
$\mathcal{G}_{2}$ also does not admit a perfect classical strategy, a similar
argument works if in strategy $S$ Alice chooses a question $x_{2}$ for some
tuple $(x_{1},x_{2})$. ∎

The following lemma will be used to show that restricting a POVM measurement
with a projection yields a POVM measurement on a smaller Hilbert space.

######  Lemma 4.2.

[4, Proposition II, 3.3.2] Let $A$ be a $C^{*}$-algebra, let $a\in A$ be a
positive element and $p\in A$ a projection. If $a\leq p$, then $ap=pa=a$.

The next lemma shows that if we have a perfect quantum strategy for the
$(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game, where Alice and Bob have non-zero
probability of answering with some outputs of $\mathcal{G}_{1}$, then there is
a perfect quantum strategy for $\mathcal{G}_{1}$. In particular, this shows
that if $\mathcal{G}_{2}$ has a perfect quantum strategy and $\mathcal{G}_{1}$
does not, the players will always choose to play $\mathcal{G}_{2}$ and never
$\mathcal{G}_{1}$.

######  Lemma 4.3.

Let $\mathcal{G}_{1}$ and $\mathcal{G}_{2}$ be nonlocal games and consider the
$(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game as above. If there is a perfect
quantum strategy
$S=(|\psi\rangle,\\{E_{(x_{1},x_{2})a}\\}_{(x_{1},x_{2})},\\{F_{(y_{1},y_{2})b}\\}_{(y_{1},y_{2})})$
of the $(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game, where additionally

$\displaystyle\langle\psi|E_{(x_{1},x_{2})a}\otimes
F_{(y_{1},y_{2})b}|\psi\rangle>0$ (8)

for some $a\in O_{A,1}$ and $b\in O_{B,1}$, then $\mathcal{G}_{1}$ has a
perfect quantum strategy.

###### Proof.

Let
$S^{\prime}=(|\psi^{\prime}\rangle\in\mathcal{H}_{A}^{\prime}\otimes\mathcal{H}_{B}^{\prime},\\{E_{(x_{1},x_{2})a}^{\prime}\\}_{(x_{1},x_{2})},\\{F_{(y_{1},y_{2})b}^{\prime}\\}_{(y_{1},y_{2})})$
be a perfect quantum strategy of the
$(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game. By restricting the state and the
operators from $S^{\prime}$ to
$\mathrm{supp}_{A}(|\psi^{\prime}\rangle)\otimes\mathrm{supp}_{B}(|\psi^{\prime}\rangle)$,
we get a perfect quantum strategy
$S=(|\psi\rangle\in\mathcal{H}_{A}\otimes\mathcal{H}_{B},\\{E_{(x_{1},x_{2})a}\\}_{(x_{1},x_{2})},$$\\{F_{(y_{1},y_{2})b}\\}_{(y_{1},y_{2})})$
such that

$\displaystyle\langle\psi|E_{(x_{1},x_{2})a}\otimes
F_{(y_{1},y_{2})b}|\psi\rangle=\langle\psi^{\prime}|E_{(x_{1},x_{2})a}^{\prime}\otimes
F_{(y_{1},y_{2})b}^{\prime}|\psi^{\prime}\rangle$ (9)

and $|\psi\rangle$ has full Schmidt rank, _i.e._
$|\psi\rangle=\sum_{i=1}^{d}\lambda_{i}e_{i}\otimes e_{i}$ with
$\lambda_{i}>0$ for all $i\in[d]$. We will first show

$\displaystyle\sum_{a\in O_{A,i}}E_{(x_{1},x_{2})a}=\sum_{a\in
O_{A,i}}E_{(s_{1},s_{2})a},\qquad\sum_{b\in
O_{B,j}}F_{(y_{1},y_{2})b}=\sum_{b\in O_{B,j}}F_{(t_{1},t_{2})b}$

for all $(x_{1},x_{2}),(s_{1},s_{2})\in I_{A,1}\times I_{A,2}$, $i=1,2$ and
$(y_{1},y_{2}),(t_{1},t_{2})\in I_{B,1}\times I_{B,2}$, $j=1,2$. Note that we
have

$\displaystyle\sum_{a\in
O_{A,1}\dot{\cup}O_{A,2}}E_{(x_{1},x_{2})a}=1=\sum_{b\in
O_{B,1}\dot{\cup}O_{B,2}}F_{(y_{1},y_{2})b}$ (10)

and $\langle\psi|E_{(x_{1},x_{2})a}\otimes F_{(y_{1},y_{2})b}|\psi\rangle=0$
for all $a\in O_{A,i}$, $b\in O_{B,j}$ with $i\neq j$, since $S$ is a perfect
quantum strategy. Let
$\varphi:=\mathrm{diag}(\lambda_{i})\in\mathbb{C}^{d\times d}$, _i.e._
$\varphi$ is the diagonal matrix with entries $\lambda_{i}$. Let
$p_{(x_{1},x_{2})i}:=\sum_{a\in O_{A,i}}E_{(x_{1},x_{2})a}$, $i=1,2$ and
$q_{(y_{1},y_{2})j}:=\sum_{b\in O_{B,j}}F_{(y_{1},y_{2})b}$, $j=1,2$. It holds

$\displaystyle\mathrm{Tr}(p_{(x_{1},x_{2})i}\varphi(q_{(y_{1},y_{2})j})^{{}^{\mathsf{T}}}\varphi)$
$\displaystyle=\mathrm{Tr}(\varphi^{*}p_{(x_{1},x_{2})i}\varphi(q_{(y_{1},y_{2})j})^{{}^{\mathsf{T}}})$
$\displaystyle=\langle\psi|\sum_{a\in
O_{A,i}}E_{(x_{1},x_{2})a}\otimes\sum_{b\in
O_{B,j}}F_{(y_{1},y_{2})b}|\psi\rangle=0$

for $i\neq j$, since $S$ is a perfect quantum strategy for the
$(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game. Thus, we have

$\displaystyle
p_{(x_{1},x_{2})i}\varphi(q_{(y_{1},y_{2})j})^{{}^{\mathsf{T}}}\varphi=0$ (11)

for all $i\neq j$. Using equations (10) and (11) several times, we get

$\displaystyle p_{(x_{1},x_{2})i}\varphi^{2}$
$\displaystyle=p_{(x_{1},x_{2})i}\varphi((q_{(y_{1},y_{2})i})^{{}^{\mathsf{T}}}+(q_{(y_{1},y_{2})j})^{{}^{\mathsf{T}}})\varphi$
$\displaystyle=p_{(x_{1},x_{2})i}\varphi(q_{(y_{1},y_{2})i})^{{}^{\mathsf{T}}}\varphi$
$\displaystyle=(p_{(x_{1},x_{2})i}+p_{(x_{1},x_{2})j})\varphi(q_{(y_{1},y_{2})i})^{{}^{\mathsf{T}}}\varphi$
$\displaystyle=\varphi(q_{(y_{1},y_{2})i})^{{}^{\mathsf{T}}}\varphi$
$\displaystyle=p_{(s_{1},s_{2})i}\varphi(q_{(y_{1},y_{2})i})^{{}^{\mathsf{T}}}\varphi$
$\displaystyle=p_{(s_{1},s_{2})i}\varphi^{2}$

for all $(x_{1},x_{2}),(s_{1},s_{2})\in I_{A,1}\times I_{A,2}$ and
$i,j\in\\{1,2\\},i\neq j$. Since $|\psi\rangle$ has full Schmidt rank,
$\varphi$ is invertible and we obtain $p_{(x_{1},x_{2})i}=p_{(s_{1},s_{2})i}$
for all $(x_{1},x_{2}),(s_{1},s_{2})\in I_{A,1}\times I_{A,2}$. One similarly
shows $q_{(y_{1},y_{2})j}=q_{(t_{1},t_{2})j}$. From now on, we let
$p_{i}:=p_{(x_{1},x_{2})i}$, $q_{j}:=q_{(y_{1},y_{2})j}$ for $i=1,2$, $j=1,2$.

We will now show that $p_{i}$ and $q_{j}$ are projections. Using equations
(10) and (11), we obtain

$\displaystyle p_{1}p_{2}\varphi^{2}$
$\displaystyle=p_{1}p_{2}\varphi(q_{1}^{{}^{\mathsf{T}}}+q_{2}^{{}^{\mathsf{T}}})\varphi$
$\displaystyle=p_{1}p_{2}\varphi(q_{2}^{{}^{\mathsf{T}}})\varphi$
$\displaystyle=p_{1}(p_{1}+p_{2})\varphi(q_{2}^{{}^{\mathsf{T}}})\varphi$
$\displaystyle=p_{1}\varphi(q_{2}^{{}^{\mathsf{T}}})\varphi$
$\displaystyle=0.$

Since $\varphi$ is invertible, we obtain $p_{1}p_{2}=0$. In particular, we
have $p_{1}=p_{1}(p_{1}+p_{2})=p_{1}^{2}$ and since we already know
$p_{1}=p_{1}^{*}$, we get that $p_{1}$ is a projection. It immediately follows
that $p_{2}=1-p_{1}$ is a projection. One can similarly show that $q_{1}$ and
$q_{2}$ are projections.

In the next step, we show $p_{1}\neq 0\neq q_{1}$. By (8) and (9), we have

$\displaystyle\langle\psi|E_{(x_{1},x_{2})a}\otimes
F_{(y_{1},y_{2})b}|\psi\rangle=\langle\psi^{\prime}|E_{(x_{1},x_{2})a}^{\prime}\otimes
F_{(y_{1},y_{2})b}^{\prime}|\psi^{\prime}\rangle>0$

for some $a\in O_{A,1}$ and $b\in O_{B,1}$. Since all $E_{(x_{1},x_{2})a}$,
$F_{(y_{1},y_{2})b}$ are positive, it holds

$\displaystyle\langle\psi|p_{1}\otimes
q_{1}|\psi\rangle\geq\langle\psi|E_{(x_{1},x_{2})a}\otimes
F_{(y_{1},y_{2})b}|\psi\rangle>0$

which implies $p_{1}\neq 0\neq q_{1}$.

In the last step, we construct a perfect quantum strategy for
$\mathcal{G}_{1}$. Fix some $x_{2}\in O_{A,2}$ and $y_{2}\in O_{B,2}$ and
define the quantum strategy $\tilde{S}$ for $\mathcal{G}_{1}$ as follows.

$\displaystyle|\tilde{\psi}\rangle$ $\displaystyle:=(p_{1}\otimes
q_{1})|\psi\rangle\in p_{1}\mathcal{H}_{A}\otimes q_{1}\mathcal{H}_{B},$
$\displaystyle\tilde{E}_{x_{1}a}$ $\displaystyle:=E_{(x_{1},x_{2})a},$
$\displaystyle\tilde{F}_{y_{1}b}$ $\displaystyle:=F_{(y_{1},y_{2})b}$

for $x_{1}\in I_{A,1}$, $y_{1}\in I_{B,1}$ and $a\in O_{A,1}$, $b\in O_{B,1}$.
Note that $p_{1}$ is the identity in $B(p_{1}\mathcal{H}_{A})$ and similarly
$q_{1}$ is the identity in $B(q_{1}\mathcal{H}_{B})$. Using Lemma 4.2, we see
that $\tilde{E}_{x_{1}a}\in B(p_{1}\mathcal{H}_{A})$ and
$\tilde{F}_{y_{1}b}\in B(q_{1}\mathcal{H}_{B})$. Thus
$\\{\tilde{E}_{x_{1}a}\,|\,a\in O_{A,1}\\}$ and
$\\{\tilde{F}_{y_{1}b}\,|\,b\in O_{B,1}\\}$ are POVMs on the space
$p_{1}\mathcal{H}_{A}\otimes q_{1}\mathcal{H}_{B}$ for all $x_{1}$, $y_{1}$.
Furthermore, we have

$\displaystyle\langle\tilde{\psi}|\tilde{E}_{x_{1}a}\otimes\tilde{F}_{y_{1}b}|\tilde{\psi}\rangle$
$\displaystyle=\langle\psi|p_{1}E_{(x_{1},x_{2})a}p_{1}\otimes
q_{1}F_{(y_{1},y_{2})b}q_{1}|\psi\rangle$
$\displaystyle=\langle\psi|E_{(x_{1},x_{2})a}\otimes
F_{(y_{1},y_{2})b}|\psi\rangle$

by Lemma 4.2. Since
$S=(|\psi\rangle,\\{E_{(x_{1},x_{2})a}\\}_{(x_{1},x_{2})},\\{F_{(y_{1},y_{2})b}\\}_{(y_{1},y_{2})})$
is a perfect quantum strategy for the
$(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game, we know that

$\displaystyle\langle\psi|E_{(x_{1},x_{2})a}\otimes
F_{(y_{1},y_{2})b}|\psi\rangle=0$

whenever $V((x_{1},x_{2}),(y_{1},y_{2}),a,b)=0$. But since we have $a\in
O_{A,1}$ and $b\in O_{B,1}$, this is by definition equivalent to
$V_{\mathcal{G}_{1}}(x_{1},y_{1},a,b)=0$. We conclude that $\tilde{S}$ is a
perfect quantum strategy for $\mathcal{G}_{1}$. ∎

We believe that the $(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game is of
independent interest. We now list a few natural directions of further inquiry.
The first one that comes to mind is what happens if we consider games with
quantum advantage, but not necessarily perfect quantum strategies in Lemma 4.1
and Lemma 4.3. Furthermore, one might ask about the non-signalling strategies
of the $(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game. We leave those as open
questions here. In this article, we will use the
$(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game in the next sections to construct
non-robust self-tests and games that do not self-test any states.

##  5 A non-robust self-test

In this section, we construct a game that non-robustly self-tests a perfect
quantum strategy. The idea is to consider the
$(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game with the following games. We let
$\mathcal{G}_{1}$ be a game that has no perfect quantum strategy, but a
sequence of quantum strategies whose winning probabilities converge to $1$.
Note that such a game was constructed by Slofstra in [28]. For
$\mathcal{G}_{2}$ we take a pseudo-telepathy game that self-tests a quantum
strategy. Then the $(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game still self-
tests this strategy, because $\mathcal{G}_{1}$ has no perfect quantum
strategy. This self-test is not robust, however, since we can construct a
near-optimal strategies of $(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game from
the ones of $\mathcal{G}_{1}$ and these strategies are not close to the self-
tested strategy.

For our proof technique to go through, we need to choose a game
$\mathcal{G}_{2}$ that is synchronous. In this case we can ensure that if one
of the players gets a pair of questions $(x_{1},x_{2})$ and chooses to play
game $\mathcal{G}_{1}$, the question $x_{2}$ does not matter for the output
(see part $(ii)$ below).

######  Lemma 5.1.

Let $\mathcal{G}_{1}$ be a nonlocal game and let $\mathcal{G}_{2}$ be a
synchronous nonlocal game. Furthermore, let
$S=(|\psi\rangle,\\{E_{(x_{1},x_{2})a}\\}_{(x_{1},x_{2})},\\{F_{(y_{1},y_{2})b}\\}_{(y_{1},y_{2})})$
be a perfect quantum strategy for the
$(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game.

  * (i)

It holds $(E_{(x_{1},x_{2})a}\otimes 1)|\psi\rangle=(1\otimes
F_{(y_{1},x_{2})a})|\psi\rangle$ for all $x_{1}\in I_{A,1},y_{1}\in I_{B,1}$,
$x_{2}\in I_{2}$ and $a\in O_{2}$.

  * (ii)

For all $x_{1},x_{3}\in I_{A,1}$, $x_{2}\in I_{2}$, $a\in O_{2}$ and
$y_{1},y_{3}\in I_{B,1}$, $y_{2}\in I_{2}$, $b\in O_{2}$, we have

$\displaystyle(E_{(x_{1},x_{2})a}\otimes
1)|\psi\rangle=(E_{(x_{3},x_{2})a}\otimes 1)|\psi\rangle\,\text{ and
}\,(1\otimes F_{(y_{1},y_{2})b})|\psi\rangle=(1\otimes
F_{(y_{3},y_{2})b})|\psi\rangle.$

  * (iii)

If $|\psi\rangle$ has full Schmidt rank, then
$E_{(x_{1},x_{2})a}=E_{(x_{3},x_{2})a}$ for $x_{1},x_{3}\in I_{A,1}$,
$x_{2}\in I_{2}$, $a\in O_{2}$ and $F_{(y_{1},y_{2})b}=F_{(y_{3},y_{2})b}$ for
$y_{1},y_{3}\in I_{B,1}$, $y_{2}\in I_{2}$, $b\in O_{2}$.

###### Proof.

Let $S$ be as above and $x_{1}\in I_{A,1}$, $x_{2}\in I_{2}$, $a\in O_{2}$.
Since $\mathcal{G}_{2}$ is synchronous, it holds

$\displaystyle\langle\psi|E_{(x_{1},x_{2})a}\otimes
1|\psi\rangle=\langle\psi|E_{(x_{1},x_{2})a}\otimes
F_{(y_{1},x_{2})a}|\psi\rangle=\langle\psi|1\otimes
F_{(y_{1},x_{2})a}|\psi\rangle$

for all $y_{1}\in I_{B,1}$. Using the equation above and the fact that
$E_{(x_{1},x_{2})a}^{2}\leq E_{(x_{1},x_{2})a}$ and
$F_{(y_{1},x_{2})a}^{2}\leq F_{(y_{1},x_{2})a}$, we obtain

$\displaystyle\|(E_{(x_{1},x_{2})a}\otimes 1)|\psi\rangle-(1\otimes
F_{(y_{1},x_{2})a})|\psi\rangle\|^{2}$
$\displaystyle=\langle\psi|E_{(x_{1},x_{2})a}^{2}\otimes
1|\psi\rangle+\langle\psi|1\otimes F_{(y_{1},x_{2})a}^{2}|\psi\rangle$
$\displaystyle\quad-2\langle\psi|E_{(x_{1},x_{2})a}\otimes
F_{(y_{1},x_{2})a}|\psi\rangle$
$\displaystyle\leq\langle\psi|E_{(x_{1},x_{2})a}\otimes
1|\psi\rangle+\langle\psi|1\otimes F_{(y_{1},x_{2})a}|\psi\rangle$
$\displaystyle\quad-2\langle\psi|E_{(x_{1},x_{2})a}\otimes
F_{(y_{1},x_{2})a}|\psi\rangle$ $\displaystyle=0.$

This yields $(E_{(x_{1},x_{2})a}\otimes 1)|\psi\rangle=(1\otimes
F_{(y_{1},x_{2})a})|\psi\rangle$ and thus proves part $(i)$. Since this
equation holds for every $x_{1}\in I_{A,1}$, we get
$(E_{(x_{1},x_{2})a}\otimes 1)|\psi\rangle=(E_{(x_{3},x_{2})a}\otimes
1)|\psi\rangle$, $x_{1},x_{3}\in I_{A,1}$ and similarly $(1\otimes
F_{(y_{1},x_{2})a})|\psi\rangle=(1\otimes F_{(y_{3},x_{2})a})|\psi\rangle$ for
$y_{1},y_{3}\in I_{B,1}$ which yields $(ii)$. Part $(iii)$ follows from $(ii)$
since $|\psi\rangle\langle\psi|$ is invertible if $|\psi\rangle$ has full
Schmidt rank. ∎

We can now present the non-robust self-test.

######  Theorem 5.2.

Let $\mathcal{G}_{1}$ be a nonlocal game that has no perfect quantum strategy,
but there exists a sequence of quantum strategies whose winning probabilities
converge to $1$. Let $\mathcal{G}_{2}$ be a synchronous pseudo-telepathy game
that self-tests a strategy
$S_{2}=(\tilde{|\psi\rangle}\in\tilde{\mathcal{H}}_{A}\otimes\tilde{\mathcal{H}}_{B},\\{\hat{E}_{x_{2}a}\\}_{x_{2}},\\{\hat{F}_{y_{2}b}\\}_{y_{2}})$.
In this case the $(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game is a non-robust
self-test for the strategy
$\tilde{S}_{2}=(\tilde{|\psi\rangle},\\{\tilde{E}_{(x_{1},x_{2})a}\\}_{(x_{1},x_{2})},$
$\\{\tilde{F}_{(y_{1},y_{2})b}\\}_{(y_{1},y_{2})})$, where

$\displaystyle\tilde{E}_{(x_{1},x_{2})a}=\begin{cases}\hat{E}_{x_{2}a}\text{
for }a\in O_{A,2},\\\ 0\text{
otherwise,}\end{cases}\tilde{F}_{(y_{1},y_{2})b}=\begin{cases}\hat{F}_{y_{2}b}\text{
for }b\in O_{B,2},\\\ 0\text{ otherwise.}\end{cases}$

###### Proof.

Let
$S^{\prime}=(|\psi^{\prime}\rangle,\\{E^{\prime}_{(x_{1},x_{2})a}\\}_{(x_{1},x_{2})},\\{F^{\prime}_{(y_{1},y_{2})b}\\}_{(y_{1},y_{2})})$
be a perfect quantum strategy for the
$(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game. To show that the
$(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game is a self-test for the strategy
$\tilde{S}_{2}$, we have to prove that $\tilde{S}_{2}$ is a local dilation of
$S^{\prime}$.

By assumption, the game $\mathcal{G}_{1}$ has no perfect quantum strategy,
thus we get $\langle\psi^{\prime}|E_{(x_{1},x_{2})a}^{\prime}\otimes
F_{(y_{1},y_{2})b}^{\prime}|\psi^{\prime}\rangle=0$ for all $a\in O_{A,1}$,
$b\in O_{B,1}$ by Lemma 4.3. Summing over all $b\in O_{B,1}\dot{\cup}O_{B,2}$
and $a\in O_{A,1}\dot{\cup}O_{A,2}$, respectively, yields
$\langle\psi^{\prime}|E_{(x_{1},x_{2})a}^{\prime}\otimes
1|\psi^{\prime}\rangle=0$, $\langle\psi^{\prime}|1\otimes
F_{(y_{1},y_{2})b}^{\prime}|\psi^{\prime}\rangle=0$ for $a\in O_{A,1}$, $b\in
O_{B,1}$. Therefore, using $(E^{\prime}_{(x_{1},x_{2})a})^{2}\leq
E^{\prime}_{(x_{1},x_{2})a}$, we have

$\displaystyle\|(E^{\prime}_{(x_{1},x_{2})a}\otimes
1)|\psi^{\prime}\rangle\|^{2}$
$\displaystyle=\langle\psi^{\prime}|(E^{\prime}_{(x_{1},x_{2})a})^{2}\otimes
1)|\psi^{\prime}\rangle$
$\displaystyle\leq\langle\psi^{\prime}|E^{\prime}_{(x_{1},x_{2})a}\otimes
1)|\psi^{\prime}\rangle$ $\displaystyle=0.$

We conclude $(E^{\prime}_{(x_{1},x_{2})a}\otimes 1)|\psi^{\prime}\rangle=0$
for all $a\in O_{A,1}$. We similarly get $(1\otimes
F^{\prime}_{(y_{1},y_{2})b})|\psi^{\prime}\rangle=0$ for all $b\in O_{B,1}$.
This especially yields that $\mathrm{supp}_{A}(|\psi^{\prime}\rangle)$ and
$\mathrm{supp}_{B}(|\psi^{\prime}\rangle)$ are invariant under those
$E_{(x_{1},x_{2})a}^{\prime}$ and $F_{(y_{1},y_{2})b}^{\prime}$, respectively.
By Lemma 5.1 $(i)$, we know $(E^{\prime}_{(x_{1},x_{2})a}\otimes
1)|\psi^{\prime}\rangle=(1\otimes
F^{\prime}_{(y_{1},x_{2})a})|\psi^{\prime}\rangle$ for all $x_{1}\in
I_{A,1},y_{1}\in I_{B,1}$, $x_{2}\in I_{2}$ and $a\in O_{2}$. Therefore, Lemma
3.6 yields that $\mathrm{supp}_{A}(\psi^{\prime})$ is invariant under
$E_{(x_{1},x_{2})a}^{\prime}$ and $\mathrm{supp}_{B}(\psi^{\prime})$ is
invariant under $F_{(y_{1},y_{2})b}^{\prime}$ in those cases. We conclude that
$\mathrm{supp}_{A}(|\psi^{\prime}\rangle)$ and
$\mathrm{supp}_{B}(|\psi^{\prime}\rangle)$ are invariant under all
$E_{(x_{1},x_{2})a}^{\prime}$ and $F_{(y_{1},y_{2})b}^{\prime}$, respectively.
Lemma 3.7 now yields that there exists a perfect quantum strategy
$S=(|\psi\rangle,\\{E_{(x_{1},x_{2})a}\\}_{(x_{1},x_{2})},\\{F_{(y_{1},y_{2})b}\\}_{(y_{1},y_{2})})$
such that $S$ is a local dilation of $S^{\prime}$ and $|\psi\rangle$ has full
Schmidt rank, _i.e._ $|\psi\rangle=\sum_{i=1}^{d}\lambda_{i}e_{i}\otimes
e_{i}$ with $\lambda_{i}>0$ for all $i\in[d]$.

Let $\varphi=\mathrm{diag}(\lambda_{i})$, where $\lambda_{i}$ are the Schmidt
coefficients of $|\psi\rangle$. Once again, by Lemma 4.3, we have
$\langle\psi|E_{(x_{1},x_{2})a}\otimes F_{(y_{1},y_{2})b}|\psi\rangle=0$ for
all $a\in O_{A,1}$, $b\in O_{B,1}$. Summing over all $b\in
O_{B,1}\dot{\cup}O_{B,2}$ and $a\in O_{A,1}\dot{\cup}O_{A,2}$, respectively,
yields

$\displaystyle\mathrm{Tr}(E_{(x_{1},x_{2})a}\varphi^{2})$
$\displaystyle=\langle\psi|E_{(x_{1},x_{2})a}\otimes 1|\psi\rangle=0,$
$\displaystyle\mathrm{Tr}((\varphi^{*})^{2}(F_{(y_{1},y_{2})b})^{{}^{\mathsf{T}}})$
$\displaystyle=\langle\psi|1\otimes F_{(y_{1},y_{2})b}|\psi\rangle=0$

for $a\in O_{A,1}$, $b\in O_{B,1}$. Since $|\psi\rangle$ has full Schmidt
rank, we obtain $E_{(x_{1},x_{2})a}=0$ and $F_{(y_{1},y_{2})b}=0$ for all
$a\in O_{A,1}$, $b\in O_{B,1}$. Therefore, $\\{E_{(x_{1},x_{2})a}\,|\,a\in
O_{A,2}\\}$ and $\\{F_{(y_{1},y_{2})b}\,|\,b\in O_{B,2}\\}$ are POVMs. For
fixed $x_{1}\in I_{A,1}$, $y_{1}\in I_{B,1}$, we get that
$(|\psi\rangle,\\{E_{(x_{1},x_{2})a}\\}_{(x_{1},x_{2})},\allowbreak\\{F_{(y_{1},y_{2})b}\\}_{(y_{1},y_{2})})$
is a perfect quantum strategy for $\mathcal{G}_{2}$. Because $\mathcal{G}_{2}$
is a self-test for $S_{2}$, we know that for fixed $x_{1},y_{1}$ there exists
Hilbert spaces $\mathcal{H}_{A,aux}$ and $\mathcal{H}_{B,aux}$, a state
$|aux\rangle\in\mathcal{H}_{A,aux}\otimes\mathcal{H}_{B,aux}$ and isometries
$U_{A}:\mathcal{H}_{A}\to\tilde{\mathcal{H}}_{A}\otimes\mathcal{H}_{A,aux}$,
$U_{B}:\mathcal{H}_{B}\to\tilde{\mathcal{H}}_{B}\otimes\mathcal{H}_{B,aux}$
such that with $U:=U_{A}\otimes U_{B}$ it holds

$\displaystyle U|\psi\rangle$
$\displaystyle=|\tilde{\psi}\rangle\otimes|aux\rangle,$ $\displaystyle
U(E_{(x_{1},x_{2})a}\otimes 1)|\psi\rangle$
$\displaystyle=[(\hat{E}_{x_{2}a}\otimes
1)\tilde{|\psi\rangle}]\otimes|aux\rangle,$ $\displaystyle U(1\otimes
F_{(y_{1},y_{2})b})|\psi\rangle$
$\displaystyle=[(1\otimes\hat{F}_{(y_{1},y_{2})b})\tilde{|\psi\rangle}]\otimes|aux\rangle.$

Note that from Lemma 5.1 $(iii)$, we have
$E_{(x_{1},x_{2})a}=E_{(x_{3},x_{2})a}$ for $x_{1},x_{3}\in I_{A,1}$,
$x_{2}\in I_{2}$, $a\in O_{2}$ and $F_{(y_{1},y_{2})b}=F_{(y_{3},y_{2})b}$ for
$y_{1},y_{3}\in I_{B,1}$, $y_{2}\in I_{2}$, $b\in O_{2}$. Thus, we get

$\displaystyle U(E_{(x_{3},x_{2})a}\otimes 1)|\psi\rangle$
$\displaystyle=U(E_{(x_{1},x_{2})a}\otimes 1)|\psi\rangle$
$\displaystyle=[(\hat{E}_{x_{2}a}\otimes
1)\tilde{|\psi\rangle}]\otimes|aux\rangle$
$\displaystyle=[(\tilde{E}_{(x_{3},x_{2})a}\otimes
1)\tilde{|\psi\rangle}]\otimes|aux\rangle$

for all $(x_{3},x_{2})\in I_{A,1}\times I_{A,2}$, $a\in O_{2}$. We similarly
get $U(1\otimes
F_{(y_{3},y_{2})b})|\psi\rangle=[(1\otimes\tilde{F}_{(y_{3},y_{2})b})\tilde{|\psi\rangle}]\otimes|aux\rangle$
for all $(y_{3},y_{2})\in I_{B,1}\times I_{B,2}$, $a\in O_{2}$. Since we know
$E_{(x_{3},x_{2})a}=0$ and $F_{(y_{3},y_{2})b}=0$ for all $a\in O_{1}$, $b\in
O_{1}$, we deduce that $\tilde{S_{2}}$ is a local dilation of $S$.
Summarizing, we have that $\tilde{S_{2}}$ is a local dilation of $S$ and $S$
is a local dilation of $S^{\prime}$. By Lemma 3.9, we get that $\tilde{S_{2}}$
is a local dilation of $S^{\prime}$, thus the
$(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game is a self-test for
$\tilde{S}_{2}$.

It remains to show that this self-test is not robust. Since there exists a
sequence of quantum strategies for $\mathcal{G}_{1}$ whose winning probability
converges to $1$, for every $\delta>0$ there is a quantum strategy
$\hat{S}_{\delta}=(|\psi^{(\delta)}\rangle,\\{\hat{E}^{(\delta)}_{x_{1}a}\\}_{x_{1}},\\{\hat{F}^{(\delta)}_{y_{1}b}\\}_{y_{1}}))$
with winning probability at least $1-\delta$. By defining

$\displaystyle
E^{(\delta)}_{(x_{1},x_{2})a}:=\begin{cases}\hat{E}^{(\delta)}_{x_{1}a}\text{
for }a\in O_{A,1},\\\ 0\text{
otherwise,}\end{cases}F^{(\delta)}_{(y_{1},y_{2})b}:=\begin{cases}\hat{F}^{(\delta)}_{y_{1}b}\text{
for }b\in O_{B,1},\\\ 0\text{ otherwise,}\end{cases}$

we obtain a strategy
$S_{\delta}=(|\psi^{(\delta)}\rangle,\\{E^{(\delta)}_{(x_{1},x_{2})a}\\}_{(x_{1},x_{2})},\\{F^{(\delta)}_{(y_{1},y_{2})b}\\}_{(y_{1},y_{2})}))$
for the $(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game with the same winning
probability as $\hat{S}^{(\delta)}$ (_i.e._ at least $1-\delta$). Since we
have $E^{(\delta)}_{(x_{1},x_{2})a}=0$ for all $a\in O_{A,2}$ and all
$\delta>0$, we see that

$\displaystyle\|U(E^{(\delta)}_{(x_{1},x_{2})a}\otimes
1)|\psi^{(\delta)}\rangle-[(\tilde{E}_{(x_{1},x_{2})a}\otimes
1)\tilde{|\psi\rangle}]\otimes|aux^{\prime}\rangle\|=\|[(\tilde{E}_{(x_{1},x_{2})a}\otimes
1)\tilde{|\psi\rangle}]\otimes|aux^{\prime}\rangle\|$

for all $a\in O_{A,2}$, all $\delta>0$, all suitable isometries $U$ and all
auxiliary states $|aux^{\prime}\rangle$. Since we have
$\tilde{E}_{(x_{1},x_{2})a}=0$ for all $a\in O_{A,1}$, we know that there is
$a_{0}\in O_{A,2}$ such that $\tilde{E}_{(x_{1},x_{2})a_{0}}\neq 0$ and thus

$\displaystyle\|[(\tilde{E}_{(x_{1},x_{2})a_{0}}\otimes
1)\tilde{|\psi\rangle}]\otimes|aux^{\prime}\rangle\|>\varepsilon^{\prime}$

for some $\varepsilon^{\prime}>0$. Summarizing, we found an
$\varepsilon^{\prime}>0$ such that for all $\delta>0$, we have a
$\delta$-optimal strategy $S^{(\delta)}$ such that

$\displaystyle\|U(E^{(\delta)}_{(x_{1},x_{2})a_{0}}\otimes
1)|\psi^{(\delta)}\rangle-[(\tilde{E}_{(x_{1},x_{2})a_{0}}\otimes
1)\tilde{|\psi\rangle}]\otimes|aux^{\prime}\rangle\|>\varepsilon^{\prime}$

for all isometries $U$. This shows that the
$(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game is a non-robust self-test for
$\tilde{S}_{2}$. ∎

######  Example 5.3.

By [28], there exists a linear constraint system game that has no perfect
quantum strategy, but a sequence of strategies whose winning probabilities
converge to $1$. The proof is constructive, the linear system has $184$
equations and $235$ variables. Let $\mathcal{G}_{1}$ be this linear constraint
system game. We have $|I_{A,1}|=184$, $|I_{B,1}|=235$ and $|O_{A,1}|=8$,
$|O_{B,1}|=2$. We let $\mathcal{G}_{2}$ be the synchronous version of the
magic square game (see Subsection 3.1), for which we know that it is
synchronous pseudo-telepathy game. We have $|I_{A,2}|=|I_{B,2}|=6$ and
$|O_{A,2}|=|O_{B,2}|=8$. By Corollary 3.12, we know that it self-tests the
perfect quantum strategy $S_{2}$. Theorem 5.2 shows that the
$(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game is a non-robust self-test for the
strategy $\tilde{S}_{2}$. For this game, we have $|I_{A}|=1104$,
$|I_{B}|=1410$ and $|O_{A}|=16$, $|O_{B}|=10$.

##  6 Games that do not self-test states

We will now construct games that do not self-test any state. We will once more
use the $(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game. We first show that a game
does not self-test any state if it has two optimal strategies using states of
coprime Schmidt rank.

######  Lemma 6.1.

Let $\mathcal{G}$ be a nonlocal game such that
$\omega^{*}(\mathcal{G})>\omega(\mathcal{G})$. Let

$S_{1}=\left(|\psi_{1}\rangle\in\mathcal{H}_{A}^{(1)}\otimes\mathcal{H}_{B}^{(1)},\\{E_{xa}^{(1)}\\}_{x},\\{F_{yb}^{(1)}\\}_{y}\right),\quad
S_{2}=\left(|\psi_{2}\rangle\in\mathcal{H}_{A}^{(2)}\otimes\mathcal{H}_{B}^{(2)},\\{E_{xa}^{(2)}\\}_{x},\\{F_{yb}^{(2)}\\}_{y}\right)$

be two optimal quantum strategies. If the Schmidt ranks of $|\psi_{1}\rangle$
and $|\psi_{2}\rangle$ are coprime (i.e. $\mathrm{gcd}(n_{1},n_{2})=1$ for
Schmidt ranks $n_{1},n_{2}$), then $\mathcal{G}$ does not self-test any state
$|\tilde{\psi}\rangle$ of an optimal quantum strategy.

###### Proof.

We prove this lemma by contradiction. Assume that $\mathcal{G}$ self-tests the
state
$|\tilde{\psi}\rangle\in\tilde{\mathcal{H}}_{A}\otimes\tilde{\mathcal{H}}_{B}$.
The state $|\tilde{\psi}\rangle$ has Schmidt rank $d>1$ as otherwise the
classical value and the quantum value of $\mathcal{G}$ coincide. Let
$n_{1},n_{2}$ be the Schmidt ranks of the states
$|\psi_{1}\rangle\in\mathcal{H}_{A}^{(1)}\otimes\mathcal{H}_{B}^{(1)}$ and
$|\psi_{2}\rangle\in\mathcal{H}_{A}^{(2)}\otimes\mathcal{H}_{B}^{(2)}$. Since
$\mathcal{G}$ self-tests $|\tilde{\psi}\rangle$, we get isometries
$U_{A}^{(i)}:\mathcal{H}_{A}^{(i)}\to\tilde{\mathcal{H}}_{A}\otimes\mathcal{H}_{A,aux}^{(i)}$,
$U_{B}:\mathcal{H}_{B}^{(i)}\to\tilde{\mathcal{H}}_{B}\otimes\mathcal{H}_{B,aux}^{(i)}$
and states
$|aux_{i}\rangle\in\mathcal{H}_{A,aux}^{(i)}\otimes\mathcal{H}_{B,aux}^{(i)}$,
$i=1,2$, such that with $U_{i}=U_{A}^{(i)}\otimes U_{B}^{(i)}$ it holds that

$\displaystyle
U_{1}|\psi_{1}\rangle=|\tilde{\psi}\rangle\otimes|aux_{1}\rangle$ and
$\displaystyle
U_{2}|\psi_{2}\rangle=|\tilde{\psi}\rangle\otimes|aux_{2}\rangle.$

Note that since we have $U_{i}=U_{A}^{(i)}\otimes U_{B}^{(i)}$, the states
$U_{i}|\psi_{i}\rangle$ have the same the Schmidt rank with respect to the
bipartition $\tilde{\mathcal{H}}_{A}\otimes\mathcal{H}_{A,aux}^{(i)}$,
$\tilde{\mathcal{H}}_{B}\otimes\mathcal{H}_{B,aux}^{(i)}$ as
$|\psi_{i}\rangle$ with respect to $\mathcal{H}_{A}^{(i)}$,
$\mathcal{H}_{B}^{(i)}$. Furthermore, the Schmidt rank of
$|\tilde{\psi}\rangle\otimes|aux_{i}\rangle$ with respect to the bipartition
$\tilde{\mathcal{H}}_{A}\otimes\mathcal{H}_{A,aux}^{(i)}$,
$\tilde{\mathcal{H}}_{B}\otimes\mathcal{H}_{B,aux}^{(i)}$ is equal to $dc_{i}$
for $c_{i}$ being the Schmidt rank of $|aux_{i}\rangle$. Thus, comparing the
Schmidt ranks with respect to
$\tilde{\mathcal{H}}_{A}\otimes\mathcal{H}_{A,aux}^{(i)}$,
$\tilde{\mathcal{H}}_{B}\otimes\mathcal{H}_{B,aux}^{(i)}$ in the equations
above yields $n_{1}=dc_{1}$ and $n_{2}=dc_{2}$. This shows that $d>1$ is a
common divisor of $n_{1}$ and $n_{2}$ which contradicts our assumption that
$n_{1}$ and $n_{2}$ are coprime. ∎

The next theorem shows that the $(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game
has two perfect quantum strategies using states with coprime Schmidt rank if
the games $\mathcal{G}_{1}$ and $\mathcal{G}_{2}$ have perfect quantum
strategies with states of coprime Schmidt rank. Thus we are just left with
finding two nonlocal games whose perfect quantum strategies have states with
coprime Schmidt rank.

######  Theorem 6.2.

Let $\mathcal{G}_{1}$ and $\mathcal{G}_{2}$ be pseudo-telepathy games with
perfect quantum strategies

$S_{1}=\left(|\psi_{1}\rangle\in\mathcal{H}_{A}^{(1)}\otimes\mathcal{H}_{B}^{(1)},\\{E_{xa}^{(1)}\\}_{x},\\{F_{yb}^{(1)}\\}_{y}\right),S_{2}=\left(|\psi_{2}\rangle\in\mathcal{H}_{A}^{(2)}\otimes\mathcal{H}_{B}^{(2)},\\{E_{xa}^{(2)}\\}_{x},\\{F_{yb}^{(2)}\\}_{y}\right),$

respectively. Assume that $|\psi_{1}\rangle$ and $|\psi_{2}\rangle$ have
coprime Schmidt ranks. Then the $(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game
does not self-test any state.

###### Proof.

We get perfect quantum strategies
$\tilde{S_{1}}=(|\psi_{1}\rangle,\\{\tilde{E}_{(x_{1},x_{2})a}^{(1)}\\}_{(x_{1},x_{2})},\allowbreak\\{\tilde{F}_{(y_{1},y_{2})b}^{(1)}\\}_{(y_{1},y_{2})})$
and
$\tilde{S_{2}}=(|\psi_{2}\rangle,\\{\tilde{E}_{(x_{1},x_{2})a}^{(2)}\\}_{(x_{1},x_{2})},\\{\tilde{F}_{(y_{1},y_{2})b}^{(2)}\\}_{(y_{1},y_{2})})$
for the $(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game by defining

$\displaystyle\tilde{E}_{(x_{1},x_{2})a}^{(i)}:=\begin{cases}E_{x_{i}a}\text{
for }a\in O_{A,i},\\\ 0\text{
otherwise,}\end{cases}\tilde{F}_{(y_{1},y_{2})b}^{(i)}:=\begin{cases}F_{y_{i}a}\text{
for }a\in O_{B,i},\\\ 0\text{ otherwise,}\end{cases}$

for $i=1,2$. Since $|\psi_{1}\rangle$ and $|\psi_{2}\rangle$ have coprime
Schmidt rank, Lemma 6.1 shows that the
$(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game does not self-test any state. ∎

In the following, we will construct an explicit game fulfilling the conditions
of Theorem 6.2. Note that if we look at strategies involving maximally
entangled states on $\mathbb{C}^{d_{i}\times d_{i}}$, then it suffices to find
two such strategies with coprime $d_{1}$ and $d_{2}$. In our case, we will
have $d_{1}=3$, $d_{2}=4$. We can get a perfect quantum strategy with
$d_{2}=4$ from the magic square game (see Subsection 3.1). For $d_{1}=3$, we
will use an independent set game with a graph coming from a $3$-dimensional
weak Kochen-Specker set. This will be explained in the next subsections.

###  6.1 Independent set game

In this subsection we discuss the independent set game, which was introduced
in [17]. We will see the connection between quantum independent sets and
perfect quantum strategies for the game. For us, a graph $G$ is always finite,
simple and undirected. Thus, it consists of a finite vertex set $V(G)$ and an
edge set $E(G)$ which is a set of unordered pairs of vertices.

######  Definition 6.3.

Let $G$ be a graph. An independent set of size $t$ in the graph $G$ is a set
of vertices $\\{v_{1},\dots,v_{t}\\}\in V(G)$ such that $(v_{i},v_{j})\notin
E(G)$ for all $i\neq j$. The _independence number_ $\alpha(G)$ denotes the
size of the largest independent set in $G$.

For a natural number $t\in\mathbb{N}$ and a graph $G$, the
$(G,t)$-_independent set game_ is played with two players Alice and Bob, and a
referee. Alice and Bob try to convince the referee that they know an
independent set of size $t$ of the graph $G$. The game is played as follows.
The referee sends the players natural numbers $x_{A},x_{B}\in[t]$ and the
players answer with vertices $v_{A},v_{B}\in V(G)$. In order to win the
$(G,t)$-independent set game, the following conditions must be met:

  * (1)

If $x_{A}=x_{B}$, then $v_{A}=v_{B}$,

  * (2)

If $x_{A}\neq x_{B}$, then $v_{A}\neq v_{B}$ and $(v_{A},v_{B})\notin E(G)$.

The players can agree on a strategy beforehand, but cannot communicate during
the game.

Note that the $(G,t)$-independent set game can be thought of as the
$(K_{t},\bar{G})$ \- homomorphism game, see [17].

######  Definition 6.4.

Let $G$ be a graph. A _quantum independent set_ of size $t$ in $G$ is a
collection $P=\\{P_{xu}\\}_{x\in[t],u\in V(G)}$ of projections
$P_{xu}\in\mathbb{C}^{d\times d}$ such that

  * (i)

$\sum_{u\in V(G)}P_{xu}=1_{\mathbb{C}^{d\times d}}$ for all $x\in[t]$,

  * (ii)

$P_{xu}P_{yv}=0$ for $(u,v)\in E(G)$ and all $x,y\in[t]$,

  * (iii)

$P_{xu}P_{yu}=0$ for all $x\neq y$, $u\in V(G)$.

The quantum independence number $\alpha_{q}(G)$ denotes the maximum number $t$
such that there exists a quantum independent set of size $t$ in $G$.

The following lemma follows from [17, Section 2.1& 2.2].

######  Lemma 6.5.

Let $G$ be a graph and let $P=\\{P_{xu}\\}_{x\in[t],u\in V(G)}$ be a quantum
independent set of size $t$ in $G$ of projections
$P_{xu}\in\mathbb{C}^{d\times d}$.

  * (i)

The strategy $(\frac{1}{\sqrt{d}}\sum_{i=1}^{d}e_{i}\otimes
e_{i},\\{P_{xu}\\}_{x},\\{(P_{yv})^{{}^{\mathsf{T}}}\\}_{y})$ is a perfect
quantum strategy of the $(G,t)$-independent set game.

  * (ii)

If $t>\alpha(G)$, then the $(G,t)$-independent set game is a pseudo-telepathy
game.

By the previous lemma, we see that quantum independent sets yield perfect
quantum strategies with maximally entangled state for the independent set
game. Thus, our goal is to construct a graph that has quantum independent sets
with projections in odd dimension. This will be done in the next subsection
using odd-dimensional Kochen-Specker sets.

###  6.2 Kochen-Specker sets

To get a counterexample for state self-testing, we use Kochen-Specker sets to
construct an explicit independent set game having a perfect quantum strategy
with a state of Schmidt rank $3$. Kochen-Specker sets are sets of vectors that
provide proofs of the (Bell-)Kochen-Specker theorem [3, 15]. Let
$S\subseteq\mathbb{C}^{n}$ be a set of vectors. A function $f:S\to\\{0,1\\}$
is a _marking function_ for $S$ if for all orthonormal bases $B\subseteq S$,
we have $\sum_{v\in B}f(v)=1$.

######  Definition 6.6.

Let $S\subseteq\mathbb{C}^{n}$ be a set of unit vectors.

  * (i)

The set $S$ is a _Kochen-Specker set_ if there is no marking function for $S$.

  * (ii)

The set $S$ is a _weak Kochen-Specker set_ [25] if for all marking functions
$f$ for $S$ there exist orthogonal vectors $u,v\in S$ such that $f(u)=f(v)=1$.

The above notions were generalized in [18] to sets of projections. Let
$\mathcal{Q}_{n}\subseteq\mathbb{C}^{n\times n}$ be the set of all $n\times n$
projections. A marking function $f$ for $S\subsetneq\mathcal{Q}_{n}$ is a
function $f:S\to\\{0,1\\}$ such that for all $M\subseteq S$ with $\sum_{p\in
M}p=1_{\mathbb{C}^{n\times n}}$, we have $\sum_{p\in M}f(p)=1$.

######  Definition 6.7.

[18] A set $S\subsetneq\mathcal{Q}_{n}$ is a _projective Kochen-Specker set_
if for all marking functions $f$ for $S$, there exists $p,p^{\prime}\in S$ for
which $pp^{\prime}=0$ and $f(p)=f(p^{\prime})=1$.

Let $\\{v_{1},\dots,v_{k}\\}\subseteq\mathbb{C}^{n}$ be a weak Kochen-Specker
set. Then we get a projective Kochen-Specker set by considering the rank one
projections
$\\{v_{1}v_{1}^{\dagger},\dots,v_{k}v_{k}^{\dagger}\\}\subseteq\mathbb{C}^{n\times
n}$.

######  Definition 6.8.

Let $S$ be a projective Kochen-Specker set. Let
$S_{1}=\\{p_{11},\dots,p_{1i_{1}}\\}$, …,
$S_{k}=\\{p_{k1},\dots,p_{ki_{k}}\\}$ be all subsets of $S$ such that
$\sum_{b\in[i_{a}]}p_{ab}=1$. We define the graph $G_{S}$ as follows: Let
$V(G_{S})=\\{(a,b)\,|\,a\in[k],b\in[i_{a}]\\}$, where $((a,b),(c,d))\in
E(G_{S})$ if and only $p_{ab}p_{cd}=0$. Note that $G_{S}$ is the orthogonality
graph of the multiset $S_{1}\dot{\cup}\dots\dot{\cup}S_{k}$.

######  Remark 6.9.

Note that we may have $p_{ab}=p_{cd}$ for some $a,b,c,d$ in the above
definition.

The next lemma shows that given a projective Kochen-Specker set $S$ we can
construct a quantum independent set in the orthogonality graph $G_{S}$.

######  Lemma 6.10.

Let $S\subsetneq\mathcal{Q}_{n}$ be a projective Kochen-Specker set. Let
$S_{1}=\\{p_{11},\dots,p_{1i_{1}}\\}$, …,
$S_{k}=\\{p_{k1},\dots,p_{ki_{k}}\\}$ be all subsets of $S$ such that
$\sum_{b\in[i_{a}]}p_{ab}=1$ and $G_{S}$ as in Definition 6.8. Then the
collection $Q=\\{Q_{j(a,b)}\\}_{j\in[k],(a,b)\in V(G_{S})}$ with
$Q_{j(a,b)}:=\delta_{aj}p_{ab}$ is a quantum independent set of size $k$ of
$G_{S}$.

###### Proof.

We check conditions (i)-(iii) of Definition 6.4. For (i), we compute

$\displaystyle\sum_{(a,b)\in V(G_{S})}Q_{j(a,b)}=\sum_{b\in[i_{j}]}p_{jb}=1$

by definition of $Q$ and choice of $S_{j}$. Condition (ii) is fulfilled since
for $((a,b),(c,d))\in E(G_{S})$, we know
$Q_{j(a,b)}Q_{l(c,d)}=\delta_{aj}\delta_{cl}p_{ab}p_{cd}=0$, since
$p_{ab}p_{cd}=0$ by definition of $G_{S}$. For (iii), we have
$Q_{j(a,b)}Q_{l(a,b)}=\delta_{aj}\delta_{al}p_{ab}=0$ for $j\neq l$, since
$\delta_{aj}\delta_{al}=0$ for $j\neq l$. ∎

From previous work it is already known that the size of the quantum
independent set $Q$ from Lemma 6.10 is larger than the independence number of
the orthogonality graph.

######  Theorem 6.11.

[26, Theorem 3.4.4] Let $S$ be a projective Kochen-Specker set and let
$S_{1}$, …, $S_{k}$ and $G_{S}$ be as in Definition 6.8. Then
$k>\alpha(G_{S})$.

Combining Lemma 6.5, Lemma 6.10 and Theorem 6.11, we have the following
corollary.

######  Corollary 6.12.

Let $S\subsetneq\mathcal{Q}_{n}$ be a projective Kochen-Specker set and
$G_{S}$ as in Definition 6.8. Then the $(G_{S},k)$-independent set game is a
pseudo-telepathy game with a perfect quantum strategy using the maximally
entangled state $|\psi_{n}\rangle=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}e_{i}\otimes
e_{i}$.

We will now give an example of a $(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game
that does not self-test any state. Note that a $3$-dimensional weak Kochen-
Specker set yields a projective Kochen-Specker set
$S\subsetneq\mathcal{Q}_{3}$ and thus the $(G_{S},k)$-independent set game has
a quantum strategy using the state $|\psi_{3}\rangle$.

######  Example 6.13.

Consider Peres’ $3$-dimensional weak Kochen-Specker set $S_{1}$ with $33$
vectors forming $16$ bases [24]. We get a projective Kochen-Specker set by
considering the associated rank-$1$ projections. The orthogonality graph
$G_{S_{1}}$ has $48$ vertices. Furthermore, we get
$\alpha(G_{S_{1}})=15,\alpha_{q}(G_{S_{1}})\geq 16$ by using Sage [32] to
compute $\alpha(G_{S_{i}})$ and Lemma 6.10 for a lower bound on
$\alpha_{q}(G_{S_{i}})$. By Corollary 6.12, we know that the
$(G_{S_{1}},16)$-independent set game has a perfect quantum strategy with a
state of Schmidt rank $3$. We let $\mathcal{G}_{1}$ be the
$(G_{S_{1}},16)$-independent set game and let $\mathcal{G}_{2}$ be the magic
square game considered in Subsection 3.1. We know that $\mathcal{G}_{2}$ has a
perfect quantum strategy with a state of Schmidt rank $4$ from Theorem 3.11.
Therefore, the $(\mathcal{G}_{1}\lor\mathcal{G}_{2})$-game does not self-test
any state by Theorem 6.2. In this case, we have $|I_{A}|=|I_{B}|=48$,
$|O_{A}|=|O_{B}|=52$.

#### Acknowledgments.

a
The authors thank the referees for useful comments and suggestions. S.S. has
received funding from the European Union’s Horizon 2020 research and
innovation programme under the Marie Sklodowska-Curie grant agreement No.
101030346. L.M. is funded by the European Union under the Grant Agreement No
101078107, QInteract and VILLUM FONDEN via the QMATH Centre of Excellence
(Grant No 10059) and Villum Young Investigator grant (No 37532). This project
was funded within the QuantERA II Programme that has received funding from the
European Union’s Horizon 2020 research and innovation programme under Grant
Agreement No 101017733.

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