††thanks: Supported by the NSF grant DMS-1954709.

Self-test, maximally entangled state, device-independent certification, non-
locality, projective quantum measurement

# Constant-sized self-tests for maximally entangled states and single local
projective measurements

Jurij Volčič  Department of Mathematics, Drexel University, Pennsylvania
<EMAIL_ADDRESS>
###### Abstract

Self-testing is a powerful certification of quantum systems relying on
measured, classical statistics. This paper considers self-testing in bipartite
Bell scenarios with small number of inputs and outputs, but with quantum
states and measurements of arbitrarily large dimension. The contributions are
twofold. Firstly, it is shown that every maximally entangled state can be
self-tested with four binary measurements per party. This result extends the
earlier work of Mančinska-Prakash-Schafhauser (2021), which applies to
maximally entangled states of odd dimensions only. Secondly, it is shown that
every single local binary projective measurement can be self-tested with five
binary measurements per party. A similar statement holds for self-testing of
local projective measurements with more than two outputs. These results are
enabled by the representation theory of quadruples of projections that add to
a scalar multiple of the identity. Structure of irreducible representations,
analysis of their spectral features and post-hoc self-testing are the primary
methods for constructing the new self-tests with small number of inputs and
outputs.

##  1 Introduction

Thanks to non-locality of quantum theory, unknown non-communicating quantum
devices measuring an unknown shared entangled state can sometimes be
identified based on classical statistic of their outputs. This phenomenon is
called self-testing, and is the strongest form of device-independent
certification of quantum systems. Self-testing was introduced in [20], and has
been a heavily studied subject ever since; see [26] for a comprehensive review
of major advances on this topic. The immense interest attracted by self-
testing originates from its applications in device-independent quantum
cryptography [1, 12], delegated quantum computation [11], randomness
generation [22, 2], entanglement detection [5], and computational complexity
[13, 17]. For experimental developments, see [16, 25].

This paper focuses on self-testing in bipartite Bell scenarios [6], where two
parties randomly perform measurements on a shared quantum state without
communicating. From these measurements, joint probability distribution of
inputs and outputs of both parties can be constructed as classical data
describing the system. Suppose that each party can perform $N$ measurements,
each of them with $K$ outcomes. Borrowing terminology from quantum games, we
model this setup with bipartite quantum _strategies_. Namely, an $N$-input
$K$-output strategy $\mathcal{S}$ of two parties (subsystems) A and B consists
of a bipartite quantum state $\mathinner{|\psi\rangle}$ in the tensor product
of Hilbert spaces $\mathcal{H}_{A}$ and $\mathcal{H}_{B}$, a measurement
$(\mathcal{M}_{i,a})_{a=1}^{K}$ of positive operators on $\mathcal{H}_{A}$ for
each $i=1,\dots,N$, and a measurement $(\mathcal{N}_{j,b})_{b=1}^{K}$ of
positive operators on $\mathcal{H}_{B}$ for each $j=1,\dots,N$. The
_correlation_ of $\mathcal{S}$ is the array $p$ of probabilities given by the
Born rule
$p(a,b|i,j)=\mathinner{\langle\psi|}\mathcal{M}_{i,a}\otimes\mathcal{N}_{j,b}\mathinner{|\psi\rangle}$,
and is the classically observable data induced by $\mathcal{S}$. There are two
trivial modifications of the strategy $\mathcal{S}$ that do not affect its
correlation: one is a unitary change of local bases, and the other is
extending the state with an ancillary state on which the measurements act
trivially. If any other strategy with correlation $p$ is obtained from
$\mathcal{S}$ using these trivial modifications, then we say that
$\mathcal{S}$ is _self-tested_ by $p$. That is, the state and measurements in
a self-tested strategy are essentially uniquely determined by the correlation.
The most renowned example of a self-tested strategy (with 2 inputs and 2
outputs) consists of maximally entangled qubits and two pairs of Pauli
measurements, which give the maximal quantum violation of the famous CHSH
inequality [8, 28, 20].

The following is a fundamental self-testing problem:

($\star$)_Which states and which measurements can be self-tested, i.e., appear
in a strategy that is self-tested by its correlation? Furthermore, how complex
is such a strategy, e.g., how many inputs and outputs per party are required?_

The breakthrough on ($\star$) for quantum states was achieved in [10], where
the authors showed that every entangled bipartite state can be self-tested.
The number of inputs in the provided self-tests _grows with the local
dimension $n$_ of the quantum state under investigation, which makes these
self-tests rather complicated in large dimensions. The existence result of
[10] was later not only extended to multipartite states in quantum networks
[27] and refined in one-sided device-independent scenarios [23], but also
improved in terms of inputs and outputs needed to self-test certain states. In
[24], the authors show that an $n$-dimensional maximally entangled bipartite
state can be self-tested using $2$ inputs and $n$ outputs. The paper [14] was
the first to provide _constant-sized_ self-tests for some infinite families of
maximally entangled states of _even_ dimension (but not constant-sized self-
tests for all maximally entangled states of even dimension). This result was
complemented by [19], where the authors establish that maximally entangled
state of _any odd_ dimension can be self-tested using 4 inputs and 2 outputs.

In comparison with states, the progress on ($\star$) for measurements has been
more constrained. All two-dimensional projective measurements have been self-
tested [29], and likewise tensor products of Pauli measurements [21, 9].
Recently, it has been established that every projective measurement can be
self-tested [7]. Actually, the self-tests derived in [7] allow for arbitrary
real ensembles of projective measurements to be self-tested simultaneously.
However, self-testing an $n$-dimensional projective measurement in this manner
requires roughly $n^{2}$ inputs.

### Contributions

This paper provides self-tests for _all maximally entangled states_ and _all
single local projective measurements_ , respectively, that are _uniform_ in
number of both inputs and outputs. The first main result concerns maximally
entangled states.

######  Theorem A (Corollary 5.4).

Maximally entangled bipartite state of any local dimension $d$ can be self-
tested using 4 inputs and 2 outputs.

The strategies of Theorem A are given in Definition 5.1. Their construction
and self-testing feature arises from the one-parametric family of universal
C*-algebras $\mathcal{A}_{2-\frac{1}{n}}$ generated by four projections adding
up to $2-\frac{1}{n}$ times the identity. Remarkable results about
representations of these algebras were established by Kruglyak-Rabanovich-
Samoĭlenko using Coxeter functors between representation categories [18].
Their theory is essential in the proof of Theorem A. Representations of
C*-algebras of this type have already been leveraged in [19]. However, their
work uses a different family of parameters ($2-\frac{2}{n}$ for odd $n$,
instead of $2-\frac{1}{n}$ for natural $n$) that leads to simple C*-algebras,
and maximally entangled states of odd dimensions only. On the other hand,
exploiting algebras $\mathcal{A}_{2-\frac{1}{n}}$ for self-testing purposes
requires a more sophisticated analysis of their representations, but applies
to all maximally entangled states.

The second main result of this paper provides constant-sized self-tests for
single local projective measurements with 2 outputs, i.e., _binary_ projective
measurements. Note that a local binary projective measurement $(P,I-P)$ is, up
to unitary change of local basis, given by a real matrix, and determined by
the dimension $n$ and the rank $r$ of the projection $P$.

######  Theorem B (Corollary 5.11).

A single local binary projective measurement of any dimension $n$ and rank $r$
appears in a 5-input 2-output strategy that is self-tested by its correlation.

See Definition 5.9 for the explicit strategies used in Theorem B. A
generalization of Theorem B for local non-binary projective measurements is
given in Corollary 5.13. It is important to stress both the significance and
the limitation of Theorem B. Given a single projective measurement, Theorem B
provides a small self-testing strategy that contains this measurement. Note
that up to a choice of coordinate system, a given projective measurement
always admits a real matrix presentation. However, Theorem B does not address
self-testing of ensembles of projective measurements; from this perspective,
it is weaker than [7], which provides (large) self-tests for all real
ensembles of projective measurements. The strategies of Theorem B are obtained
from the strategies of Theorem A by the principle of _post-hoc self-testing_
[26]. A broad sufficiency criterion for applicability of post-hoc self-testing
was presented in [7]. To apply this criterion in the proof of Theorem B,
certain spectral aspects of representations of $\mathcal{A}_{2-\frac{1}{n}}$
need to be resolved. Namely, we determine the spectrum of the sum of pairs of
projections arising from representations of $\mathcal{A}_{2-\frac{1}{n}}$.

While the derivation of the newly presented self-tests might seem rather
abstract, the resulting correlations admit closed-form expressions, and the
corresponding strategies can recursively constructed using basic tools from
linear algebra (see Appendix A for examples).

### Reader’s guide

Section 2 reviews the standard terminology and notation on quantum strategies
and self-testing. Section 3 presents a construction of four $n\times n$
projections that add to $2-\frac{1}{n}$ times identity, and their basic
properties; these projections are central to this paper, and provide local
projective measurements for the new self-tested strategies. Section 4
establishes certain spectral results about these projections, which are
critical for demonstrating self-testing in this paper. While this section
provides the main new mathematical insight into what is required to establish
the new self-testing results, a reader only interested in main statements may
skip this section. Section 5 presents the new self-tested strategies and their
correlations. Section 6 addresses obstructions to constant-sized self-testing
of arbitrary entangled states and pairs of projective measurements. Lastly,
Appendix A explicitly constructs the distinguished projections appearing in
self-tests for local dimensions up to 6.

### Acknowledgments

The author thanks Ken Dykema for inspiring conversations about self-testing,
and Ricardo Gutierrez-Jauregui for sharing his expertise on experimental
aspects of quantum theory.

##  2 Preliminaries

This section introduces notation and terminology on quantum strategies and
self-testing, following the conventions presented in [19]. For a comprehensive
overview, see [26].

Let $K\in\mathbb{N}$. A $K$-tuple of operators $(P_{a})_{a=1}^{K}$ acting on a
Hilbert space $\mathcal{H}$ is a _positive operator-valued measure (
$K$-POVM)_ if $P_{a}\succeq 0$ and $\sum_{a=1}^{K}P_{a}=I$. If all $P_{a}$ are
projections, then $(P_{a})_{a=1}^{K}$ is a _projection-valued measure (
$K$-PVM)_, or a _projective measurement_. Note that, up to a unitary basis
change, a PVM $(P_{a})_{a=1}^{K}$ is uniquely determined by the ranks
$\operatorname{rk}P_{a}$ for $a=1,\dots,K$. That is, every $K$-PVM with ranks
of projections $r_{1},\dots,r_{K}$ is unitarily equivalent to

$\left(I_{r_{1}}\oplus 0_{r_{2}+\cdots+r_{K}},0_{r_{1}}\oplus I_{r_{2}}\oplus
0_{r_{3}+\cdots+r_{K}},\dots,0_{r_{1}+\cdots+r_{K-1}}\oplus I_{r_{K}}\right).$

A $2$-POVM is also called a _binary_ measurement. Observe that a binary PVM is
simply a pair $(P,I-P)$ where $P$ is a projection, and is determined by the
dimension and the rank of $P$ up to a unitary basis change.

A _(pure bipartite) state_ $\mathinner{|\psi\rangle}$ is a unit vector in
$\mathcal{H}_{A}\otimes\mathcal{H}_{B}$, where
$\mathcal{H}_{A},\mathcal{H}_{B}$ are Hilbert spaces. We say that
$\mathinner{|\psi\rangle}$ has _full Schmidt rank_ if $P\otimes
I\mathinner{|\psi\rangle}=I\otimes Q\mathinner{|\psi\rangle}=0$ for some
projections $P,Q$ implies $P=0$ and $Q=0$. In this case, the Hilbert spaces
$\mathcal{H}_{A}$ and $\mathcal{H}_{B}$ are isomorphic. For $n\in\mathbb{N}$,
the (canonical) _maximally entangled state_ of local dimension $n$ is
$\mathinner{|\phi_{n}\rangle}=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\mathinner{|i\rangle}\\!\mathinner{|i\rangle}\in\mathbb{C}^{n}\otimes\mathbb{C}^{n}$.
For $A,B\in\operatorname{M}_{n}(\mathbb{C})$,

$\mathinner{\langle\phi_{n}|}A\otimes
B\mathinner{|\phi_{n}\rangle}=\tau(AB^{\rm
t})=\frac{1}{n}\operatorname{tr}(AB^{\rm t}),$

where $\tau$ denotes the normalized trace on
$\operatorname{M}_{n}(\mathbb{C})$.

Let $K_{A},K_{B},N_{A},N_{B}\in\mathbb{N}$. An _$(N_{A},N_{B})$ -input
$(K_{A},K_{B})$-output bipartite quantum strategy_ $\mathcal{S}$ is a triple

$\mathcal{S}=\left(\mathinner{|\psi\rangle};\mathcal{M}_{1},\dots,\mathcal{M}_{N_{A}};\mathcal{N}_{1},\dots,\mathcal{N}_{N_{B}}\right)$

where $\mathcal{M}_{i}$ are $K_{A}$-POVMs on a finite-dimensional Hilbert
space $\mathcal{H}_{A}$, $\mathcal{N}_{j}$ are $K_{B}$-POVMs on a finite-
dimensional Hilbert space $\mathcal{H}_{B}$, and
$\mathinner{|\psi\rangle}\in\mathcal{H}_{A}\otimes\mathcal{H}_{B}$ is a state.
When $K=K_{A}=K_{B}$ and $N=N_{A}=N_{B}$, we simply say that $\mathcal{S}$ is
a _$N$ -input $K$-output bipartite strategy_. The _correlation_ of
$\mathcal{S}$ is the $N_{A}\times N_{B}\times K_{A}\times K_{B}$ array $p$
with entries

$\displaystyle
p(a,b|i,j)=\mathinner{\langle\psi|}\mathcal{M}_{i,a}\otimes\mathcal{N}_{j,b}\mathinner{|\psi\rangle}\qquad\qquad$
$\displaystyle 1\leq a\leq K_{A},\ 1\leq b\leq K_{B},$ $\displaystyle 1\leq
i\leq N_{A},\ 1\leq j\leq N_{B}.$

Since $\mathcal{S}$ in particular models non-communication between parties,
the correlation $p$ is _non-signalling_ , meaning that
$p(a|i):=\sum_{b=1}^{K_{B}}p(a,b|i,j)$ and
$p(b|j):=\sum_{a=1}^{K_{A}}p(a,b|i,j)$ are well-defined (the first sum is
independent of $j$ and the second sum is independent of $i$). A correlation
$p$ is called _synchronous_ if $K_{A}=K_{B}$, $N_{A}=N_{B}$ and $p(a,b|i,i)=0$
for all $i$ and $a\neq b$.

Let $\mathcal{S}$ and $\widetilde{\mathcal{S}}$ be $(N_{A},N_{B})$-input
$(K_{A},K_{B})$-output strategies. Then $\widetilde{\mathcal{S}}$ is a _local
dilation_ if $\mathcal{S}$ there exist finite-dimensional Hilbert spaces
$\mathcal{K}_{A},\mathcal{K}_{B}$, a state $\mathinner{|{\rm
aux}\rangle}\in\mathcal{K}_{A}\otimes\mathcal{K}_{B}$ and isometries
$U_{A}:\mathcal{H}_{A}\to\widetilde{\mathcal{H}}_{A}\otimes\mathcal{K}_{A}$
and
$U_{B}:\mathcal{H}_{B}\to\widetilde{\mathcal{H}}_{B}\otimes\mathcal{K}_{B}$
such that

$(U_{A}\otimes
U_{B})(\mathcal{M}_{i,a}\otimes\mathcal{N}_{j,b})\mathinner{|\psi\rangle}=(\widetilde{\mathcal{M}}_{i,a}\otimes\widetilde{\mathcal{N}}_{j,b})\mathinner{|\widetilde{\psi}\rangle}\otimes\mathinner{|{\rm
aux}\rangle}$ (1)

for all $a,b,i,j$. There is a slight abuse of notation in (1); namely, we
identify

$(\widetilde{\mathcal{H}}_{A}\otimes\mathcal{K}_{A})\otimes(\widetilde{\mathcal{H}}_{B}\otimes\mathcal{K}_{B})\equiv(\widetilde{\mathcal{H}}_{A}\otimes\widetilde{\mathcal{H}}_{B})\otimes(\mathcal{K}_{A}\otimes\mathcal{K}_{B}).$

Note that if $\widetilde{\mathcal{S}}$ is a local dilation of $\mathcal{S}$,
then the correlations of $\mathcal{S}$ and $\widetilde{\mathcal{S}}$ coincide.
Finally, we say that a strategy $\widetilde{\mathcal{S}}$ is _self-tested_ by
its correlation if it is a local dilation of any other strategy with the same
correlation.

##  3 Quadruples of projections adding to a scalar multiple of the identity

In [18], the authors derive several profound results on tuples of projections
that add to a scalar multiple of the identity operator. This is achieved by
studying certain functors between categories of their representations, which
are also the cornerstone of this paper. For our purposes, we focus on
projections $P_{1},P_{2},P_{3},P_{4}$ that add to $(2-\frac{1}{n})I$, where
$n$ is a natural number. First we adopt the language of representations of
C*-algebras, at least to the extent required in this paper. Then we review the
construction of the aforementioned functors from [18, Section 1.2]. Finally,
we refine a part of [18, Proposition 3] to obtain further properties about the
projections $P_{i}$ as above (Proposition 3.1).

For $\alpha\in\mathbb{R}$ define the universal C*-algebra

$\mathcal{A}_{\alpha}={\rm C}^{*}\left\langle x_{1},x_{2},x_{3},x_{4}\colon
x_{i}=x_{i}^{*}=x_{i}^{2},\ x_{1}+x_{2}+x_{3}+x_{4}=\alpha\right\rangle,$

and let $\operatorname{Rep}_{\alpha}$ denote the category of representations
of $\mathcal{A}_{\alpha}$. That is, objects of $\operatorname{Rep}_{\alpha}$
are representations of $\mathcal{A}_{\alpha}$ on Hilbert spaces, and morphisms
of $\operatorname{Rep}_{\alpha}$ are equivariant maps, i.e., bounded linear
operators between Hilbert spaces that intertwine the actions of
representations. For a comprehensive source on C*-algebras and their
representations, see [3]. While the above terminology offers a suitable
mathematical framework for the technical steps in the proofs of this paper,
let us extract the main meaning behind it, sufficient for comprehending the
proofs. Without addressing precisely what a universal C*-algebra is, we can
still say what its representations are. A representation $\pi$ of
$\mathcal{A}_{\alpha}$ is a quadruple of projections $X_{1},X_{2},X_{3},X_{4}$
on a Hilbert space $\mathcal{H}$ that satisfy $X_{1}+X_{2}+X_{3}+X_{4}=\alpha
I$. Thus $\operatorname{Rep}_{\alpha}$ is foremost a collection of such
quadruples; one could think of $\mathcal{A}_{\alpha}$ as their abstract model.
For a $\pi\in\operatorname{Rep}_{\alpha}$ as above we write
$\pi(x_{i})=X_{i}$, and we assign to it a $6$-tuple of numbers
$[\pi]=(\alpha;n;d_{1},d_{2},d_{3},d_{4})$ where $n=\dim\mathcal{H}$ and
$d_{i}=\operatorname{rk}\pi(x_{i})$, the dimension of the range of $X_{i}$ (if
$\mathcal{H}$ is infinite-dimensional, then $n=\infty$; likewise, $d_{i}$ can
be infinite).

Note that representations may be related to each other in several ways. For
example, let $\pi\in\mathcal{A}_{\alpha}$ is given by projections
$X_{1},\dots,X_{4}$ on a Hilbert space $\mathcal{H}$ and
$\rho\in\mathcal{A}_{\alpha}$ is given by projections $Y_{1},\dots,Y_{4}$ on a
Hilbert space $\mathcal{K}$. Then the projections $X_{1}\oplus
Y_{1},\dots,X_{4}\oplus Y_{4}$ act on $\mathcal{H}\oplus\mathcal{K}$ and add
to $\alpha$ times identity, so they determine representation of
$\mathcal{A}_{\alpha}$, called the direct sum of $\pi$ and $\rho$. Next, we
say that $\pi$ and $\rho$ are unitarily equivalent if there is a unitary (that
is, an isometric invertible linear map) $U:\mathcal{H}\to\mathcal{K}$ such
that $Y_{i}=UX_{i}U^{*}$ for $i=1,\dots,4$. Finally, we say that
$\pi\in\operatorname{Rep}_{\alpha}$ is irreducible if it is not unitarily
equivalent to a direct sum of representations. Irreducible representations can
be viewed as the building blocks of $\operatorname{Rep}_{\alpha}$; namely,
every representation is unitarily equivalent to a (possibly infinite) direct
sum of irreducible representations. Without going into technical details,
viewing $\operatorname{Rep}_{\alpha}$ as a category instead of merely a set
encapsulates these relations between representations (e.g., that some of them
are unitarily equivalent, some are direct sums of others, and some are
irreducible).

In this paper, representations of $\mathcal{A}_{\alpha}$ (for certain choices
of $\alpha$) give rise to the projective measurements in self-tested
strategies presented in Section 5. To establish the self-testing property, it
is imperative to have a good handle on $\operatorname{Rep}_{\alpha}$
(concretely, on the irreducible representations within). This is
straightforward for $\alpha=0$ and $\alpha=1$. Indeed, the only quadruples of
projections adding to 0 are tuples of zero operators; these are all direct
sums of the trivial representation $\tau$ given by $\tau(x_{j})=0$ acting on
the one-dimensional Hilbert space. Hence $\operatorname{Rep}_{0}$ contains a
unique irreducible representation. On the other hand, quadruples of
projections adding to 1 are necessarily diagonalizable, and thus unitarily
equivalent to direct sums of $(1,0,0,0)$, $(0,1,0,0)$, $(0,0,1,0)$,
$(0,0,0,1)$ acting on the one-dimensional Hilbert space. Thus
$\operatorname{Rep}_{1}$ contains exactly four unitarily non-equivalent
irreducible representations. For general $\alpha$, representations of
$\mathcal{A}_{\alpha}$ are not yet well-understood; however, the aim of the
next subsection is to leverage the knowledge of the very simple
$\operatorname{Rep}_{1}$ to study $\operatorname{Rep}_{\alpha}$ for certain
values of $\alpha$.

###  3.1 Functors between representation categories

In this subsection we define two functors
$T=T_{\alpha}:\operatorname{Rep}_{\alpha}\to\operatorname{Rep}_{4-\alpha}$
(linear reflection) and
$S=S_{\alpha}:\operatorname{Rep}_{\alpha}\to\operatorname{Rep}_{\frac{\alpha}{\alpha-1}}$
(hyperbolic reflection). The subscripts are omitted when clear from the
context. Before defining $T$ and $S$, let us mention what a reader should
imagine under this terminology. A functor from $\operatorname{Rep}_{\alpha}$
to $\operatorname{Rep}_{\beta}$ is primarily a mapping, that takes each
quadruple of projections adding to $\alpha$ times identity to a quadruple of
projections adding to $\beta$ times identity. However, being a functor means
that this mapping has to respect the additional structure of the categories
$\operatorname{Rep}_{\alpha}$ and $\operatorname{Rep}_{\beta}$; in particular,
it needs to preserve direct sums, and map unitarily equivalent representations
to unitarily equivalent representations. Technically, one encapsulates this by
saying that a functor consists of a map between objects of categories and a
(well-behaved) map between morphisms of categories.

$(T)$: Given a representation $\pi$ of $\mathcal{A}_{\alpha}$ let $T(\pi)$ be
the representation of $\mathcal{A}_{4-\alpha}$ determined by
$T(\pi)(x_{i}):=I-\pi(x_{i})$. Note that $T$ commutes with equivariant maps
between representations, so it extends to a functor
$T:\operatorname{Rep}_{\alpha}\to\operatorname{Rep}_{4-\alpha}$. If
$[\pi]=(\alpha;n;d_{i})$ then $[T(\pi)]=(4-\alpha;n;n-d_{i})$.

($S$): Suppose $\alpha\notin\\{0,1\\}$, and let $\pi$ be a representation of
$\mathcal{A}_{\alpha}$ on $\mathcal{H}$. Denote
$\widehat{\mathcal{H}}=\bigoplus_{i}\operatorname{ran}\pi(x_{i})$. Let
$w_{i}:\operatorname{ran}\pi(x_{i})\to\widehat{\mathcal{H}}$ be the canonical
injections, and let $u_{i}:\operatorname{ran}\pi(x_{i})\to\mathcal{H}$ be
inclusions. Then

$u=\frac{1}{\sqrt{\alpha}}\begin{pmatrix}u_{1}^{*}\\\ \vdots\\\
u_{4}^{*}\end{pmatrix}:\mathcal{H}\to\widehat{\mathcal{H}}$

is an isometry by definition of the algebra $\mathcal{A}_{\alpha}$. Let
$\mathcal{K}=\operatorname{ran}(I-uu^{*})$, with inclusion
$v:\mathcal{K}\to\widehat{\mathcal{H}}$. Note that
$\dim\mathcal{K}=\dim\widehat{\mathcal{H}}-\dim\mathcal{H}$. Define

$S(\pi)(x_{i}):=\frac{\alpha}{\alpha-1}v^{*}w_{i}w_{i}^{*}v.$

Then

$\displaystyle\left(S(\pi)(x_{i})\right)^{2}$
$\displaystyle=\frac{\alpha^{2}}{(\alpha-1)^{2}}v^{*}w_{i}w_{i}^{*}vv^{*}w_{i}w_{i}^{*}v=\frac{\alpha^{2}}{(\alpha-1)^{2}}v^{*}w_{i}w_{i}^{*}(I-uu^{*})w_{i}w_{i}^{*}v$
$\displaystyle=\frac{\alpha^{2}}{(\alpha-1)^{2}}v^{*}w_{i}\left(I-\frac{1}{\alpha}u_{i}^{*}u_{i}\right)w_{i}^{*}v=\frac{\alpha^{2}}{(\alpha-1)^{2}}\left(1-\frac{1}{\alpha}\right)v^{*}w_{i}w_{i}^{*}v$
$\displaystyle=S(\pi)(x_{i})$

and

$\sum_{i=1}^{4}S(\pi)(x_{i})=\sum_{i=1}^{4}\frac{\alpha}{\alpha-1}v^{*}w_{i}w_{i}^{*}v=\frac{\alpha}{\alpha-1}v^{*}\left(\sum_{i=1}^{4}w_{i}w_{i}^{*}\right)v=\frac{\alpha}{\alpha-1}v^{*}v=\frac{\alpha}{\alpha-1}I.$

Therefore $S(\pi)(x_{1}),\dots,S(\pi)(x_{4})$ are projections that give rise
to a representation $S(\pi)$ of $\mathcal{A}_{\frac{\alpha}{\alpha-1}}$ on
$\mathcal{K}$. As described in [18, Section 1.2], one can also extend $S$ to
equivariant maps, resulting in a functor
$S:\operatorname{Rep}_{\alpha}\to\operatorname{Rep}_{\frac{\alpha}{\alpha-1}}$.
If $[\pi]=(\alpha;n;d_{i})$ then
$[S(\pi)]=(\frac{\alpha}{\alpha-1};\sum_{i}d_{i}-n;d_{i})$.

###  3.2 Distinguished quadruples of projections

For $\alpha\in(0,3)$, the (Coxeter) functor

$\Phi^{+}=S\circ T=S_{4-\alpha}\circ
T_{\alpha}:\operatorname{Rep}_{\alpha}\to\operatorname{Rep}_{1+\frac{1}{3-\alpha}}$

define an equivalence of categories (with inverse $T\circ S$) by [18, Theorem
2]. In particular, $\Phi^{+}$ is a bijection between representations of
$\mathcal{A}_{\alpha}$ and $\mathcal{A}_{1+\frac{1}{3-\alpha}}$, which maps
irreducible ones to irreducible ones. If $[\pi]=(\alpha,n,d_{1},\dots,d_{4})$
then $[\Phi^{+}(\pi)]=(1+\frac{1}{3-\alpha};3n-\sum_{i}d_{i};n-d_{i})$. The
functor $\Phi^{+}$ plays an implicit yet crucial role in [18, Proposition 3]
that describes the category $\operatorname{Rep}_{2-\frac{1}{n}}$. For the sake
of completeness, we provide the proof of the part of [18, Proposition 3], and
refine it to extract the additional information needed in this paper. Given a
real number $\beta$ let $\lfloor\beta\rfloor$ denote the largest integer that
is not larger than $\beta$.

The main statement of this section shows that starting with the easily-
understood $\operatorname{Rep}_{1}$ and then repeatedly applying the functor
$\Phi^{+}$, one obtains a good grasp on $\operatorname{Rep}_{2-\frac{1}{n}}$
for every $n\in\mathbb{N}$.

######  Proposition 3.1 ([18, Proposition 3(c)]).

Let $n\in\mathbb{N}$. The C*-algebra $\mathcal{A}_{2-\frac{1}{n}}$ has
precisely four unitarily non-equivalent irreducible representations.
More concretely, there are projections
$\mathfrak{P}^{(n)}_{1},\dots,\mathfrak{P}^{(n)}_{4}\in\operatorname{M}_{n}(\mathbb{R})$
with
$\operatorname{rk}\mathfrak{P}^{(n)}_{1}=\lfloor\frac{n}{2}\rfloor-(-1)^{n}$
and $\operatorname{rk}\mathfrak{P}^{(n)}_{i}=\lfloor\frac{n}{2}\rfloor$ for
$i=2,3,4$, such that given an irreducible representation of
$\mathcal{A}_{2-\frac{1}{n}}$, the quadruple $(\pi(x_{1}),\dots,\pi(x_{4}))$
is unitarily equivalent to one of the

$\displaystyle(\mathfrak{P}^{(n)}_{1},\mathfrak{P}^{(n)}_{2},\mathfrak{P}^{(n)}_{3},\mathfrak{P}^{(n)}_{4}),\quad(\mathfrak{P}^{(n)}_{4},\mathfrak{P}^{(n)}_{1},\mathfrak{P}^{(n)}_{2},\mathfrak{P}^{(n)}_{3}),$
$\displaystyle(\mathfrak{P}^{(n)}_{3},\mathfrak{P}^{(n)}_{4},\mathfrak{P}^{(n)}_{1},\mathfrak{P}^{(n)}_{2}),\quad(\mathfrak{P}^{(n)}_{2},\mathfrak{P}^{(n)}_{3},\mathfrak{P}^{(n)}_{4},\mathfrak{P}^{(n)}_{1}).$

###### Proof.

We prove the statement by induction on $n$. If $n=1$, then
$\mathfrak{P}^{(1)}_{1}=1$ and $\mathfrak{P}^{(1)}_{i}=0$ for $i=2,3,4$ are
the desired $1\times 1$ projections, giving rise to a representation
$\mathcal{A}_{1}\to\mathbb{C}$. Now suppose projections
$\mathfrak{P}^{(n)}_{i}\in\operatorname{M}_{n}(\mathbb{R})$ possess the
desired properties. Then they define an irreducible representation of
$\mathcal{A}_{2-\frac{1}{n}}$ given by $\pi(x_{i})=\mathfrak{P}^{(n)}_{i}$,
and the other three irreducible representations up to unitary equivalence are
obtained by cyclically permuting the generators. Now let
$\mathfrak{P}^{(n+1)}_{i}:=\Phi^{+}(\pi)(x_{i})$. Since
$\Phi^{+}:\operatorname{Rep}_{2-\frac{1}{n}}\to\operatorname{Rep}_{2-\frac{1}{n+1}}$
is an equivalence of categories, $\Phi^{+}(\pi)$ is an irreducible
representation of $\mathcal{A}_{2-\frac{1}{n}}$, and the other three
irreducible representations up unitary equivalence are obtained via cyclic
permutations of generators. The rank values are determined by comparing
$[\pi]$ and $[\Phi^{+}(\pi)]$. ∎

Projections $\mathfrak{P}^{(n)}_{i}$ are central to the self-testing results
in this paper. The intuition behind their applicability to self-tests is the
following: if we momentarily forget irreducibility, they are characterized by
having certain traces and satisfying a linear equation. In a quantum strategy
with a maximally entangled state and projective measurements, traces and
linear relations among the PVMs are encoded by the correlation. This makes
strategies with maximally entangled states and measurements
$(\mathfrak{P}^{(n)}_{i},I-\mathfrak{P}^{(n)}_{i})$ very natural candidates
for the self-testing phenomenon.

######  Remark 3.2.

Proposition 3.1 does not provide a closed-form expression for projections
$\mathfrak{P}^{(n)}_{1},\dots,\mathfrak{P}^{(n)}_{4}\in\operatorname{M}_{n}(\mathbb{R})$
as functions of $n$. Nevertheless, definitions of functors $T$ and $S$ give
rise to a recursive procedure for constructing
$\mathfrak{P}^{(n)}_{i}\in\operatorname{M}_{n}(\mathbb{R})$ from
$\mathfrak{P}^{(n-1)}_{i}\in\operatorname{M}_{n-1}(\mathbb{R})$. This
procedure requires only matrix arithmetic and Gram-Schmidt orthogonalization.
Basis of recursion $n=1$: set $\mathfrak{P}^{(1)}_{1}:=1$ and
$\mathfrak{P}^{(1)}_{i}:=0$ for $i=2,3,4$.
Recursive step $n\to n+1$: given
$\mathfrak{P}^{(n)}_{1},\dots,\mathfrak{P}^{(n)}_{4}$ let

  * •

$U_{i}$ be an $n\times\operatorname{rk}(n-\mathfrak{P}^{(n)}_{i})$ matrix
whose columns form an orthonormal basis of the column space of
$I-\mathfrak{P}^{(n)}_{i}$;

  * •

$V_{i}$ be an $(\operatorname{rk}\mathfrak{P}^{(n)}_{i})\times(n+1)$ matrix
such that the columns of

$\begin{pmatrix}V_{1}\\\ \vdots\\\ V_{4}\end{pmatrix}$

form an orthonormal basis of the column space of

$I-\frac{1}{2+\frac{1}{n}}\begin{pmatrix}U_{1}^{*}\\\ \vdots\\\
U_{4}^{*}\end{pmatrix}\begin{pmatrix}U_{1}&\cdots&U_{4}\end{pmatrix}.$

Then set $\mathfrak{P}^{(n+1)}_{i}:=(2-\frac{1}{n+1})V_{i}^{*}V_{i}$.

Using the above procedure, we obtain the following projections for $n=1,2,3$:

$\displaystyle\mathfrak{P}^{(1)}_{1}=(1),\ \mathfrak{P}^{(1)}_{2}=(0),\
\mathfrak{P}^{(1)}_{3}=(0),\ \mathfrak{P}^{(1)}_{4}=(0)$
$\displaystyle\mathfrak{P}^{(2)}_{1}=\begin{pmatrix}0&0\\\ 0&0\end{pmatrix},\
\mathfrak{P}^{(2)}_{2}=\begin{pmatrix}1&0\\\ 0&0\end{pmatrix},\
\mathfrak{P}^{(2)}_{3}=\begin{pmatrix}\frac{1}{4}&\frac{-\sqrt{3}}{4}\\\
\frac{-\sqrt{3}}{4}&\frac{3}{4}\\\ \end{pmatrix},\
\mathfrak{P}^{(2)}_{4}=\begin{pmatrix}\frac{1}{4}&\frac{\sqrt{3}}{4}\\\
\frac{\sqrt{3}}{4}&\frac{3}{4}\\\ \end{pmatrix}$
$\displaystyle\mathfrak{P}^{(3)}_{1}=\begin{pmatrix}1&0&0\\\ 0&1&0\\\
0&0&0\end{pmatrix},\ \mathfrak{P}^{(3)}_{2}=\begin{pmatrix}0&0&0\\\
0&\frac{4}{9}&\frac{-2\sqrt{5}}{9}\\\
0&\frac{-2\sqrt{5}}{9}&\frac{5}{9}\end{pmatrix},\ $
$\displaystyle\qquad\mathfrak{P}^{(3)}_{3}=\begin{pmatrix}\frac{1}{3}&\frac{1}{3\sqrt{3}}&\frac{\sqrt{5}}{3\sqrt{3}}\\\
\frac{1}{3\sqrt{3}}&\frac{1}{9}&\frac{\sqrt{5}}{9}\\\
\frac{\sqrt{5}}{3\sqrt{3}}&\frac{\sqrt{5}}{9}&\frac{5}{9}\end{pmatrix},\
\mathfrak{P}^{(3)}_{4}=\begin{pmatrix}\frac{1}{3}&\frac{-1}{3\sqrt{3}}&\frac{-\sqrt{5}}{3\sqrt{3}}\\\
\frac{-1}{3\sqrt{3}}&\frac{1}{9}&\frac{\sqrt{5}}{9}\\\
\frac{-\sqrt{5}}{3\sqrt{3}}&\frac{\sqrt{5}}{9}&\frac{5}{9}\end{pmatrix}$

The linear-algebraic nature of this procedure allows for a feasible
implementation using exact arithmetic. For concrete matrices in cases
$n=4,5,6$, see Appendix A.

For later use we record a technical fact.

######  Lemma 3.3.

The $4\times 4$ matrix

$\begin{pmatrix}\operatorname{rk}\mathfrak{P}^{(n)}_{1}&\operatorname{rk}\mathfrak{P}^{(n)}_{2}&\operatorname{rk}\mathfrak{P}^{(n)}_{3}&\operatorname{rk}\mathfrak{P}^{(n)}_{4}\\\
\operatorname{rk}\mathfrak{P}^{(n)}_{4}&\operatorname{rk}\mathfrak{P}^{(n)}_{1}&\operatorname{rk}\mathfrak{P}^{(n)}_{2}&\operatorname{rk}\mathfrak{P}^{(n)}_{3}\\\
\operatorname{rk}\mathfrak{P}^{(n)}_{3}&\operatorname{rk}\mathfrak{P}^{(n)}_{4}&\operatorname{rk}\mathfrak{P}^{(n)}_{1}&\operatorname{rk}\mathfrak{P}^{(n)}_{2}\\\
\operatorname{rk}\mathfrak{P}^{(n)}_{2}&\operatorname{rk}\mathfrak{P}^{(n)}_{3}&\operatorname{rk}\mathfrak{P}^{(n)}_{4}&\operatorname{rk}\mathfrak{P}^{(n)}_{1}\end{pmatrix}=-(-1)^{n}I_{4}+\left\lfloor\frac{n}{2}\right\rfloor\begin{pmatrix}1&1&1&1\\\
1&1&1&1\\\ 1&1&1&1\\\ 1&1&1&1\end{pmatrix}$

is invertible for every $n\in\mathbb{N}$.

######  Remark 3.4.

Let us determine the normalized traces of $\mathfrak{P}^{(n)}_{i}$ and their
products; these values will appear in the self-testing correlations of this
paper. Clearly,

$\tau\left(\mathfrak{P}^{(n)}_{1}\right)=\frac{1}{2}-\frac{1+3(-1)^{n}}{4n},\qquad\tau\left(\mathfrak{P}^{(n)}_{i}\right)=\frac{1}{2}-\frac{1-(-1)^{n}}{4n},\quad\text{for
}i=2,3,4.$

Next, by Proposition 3.1, for every permutation $\sigma$ of $\\{2,3,4\\}$
there exists a unitary $U\in\operatorname{M}_{n}(\mathbb{C})$ such that

$U\mathfrak{P}^{(n)}_{1}U^{*}=\mathfrak{P}^{(n)}_{1},\qquad
U\mathfrak{P}^{(n)}_{i}U^{*}=\mathfrak{P}^{(n)}_{\sigma(i)},\quad\text{for
}i=2,3,4.$

Therefore $\tau(\mathfrak{P}^{(n)}_{1}\mathfrak{P}^{(n)}_{i})$ is independent
of $i\in\\{2,3,4\\}$, and $\tau(\mathfrak{P}^{(n)}_{i}\mathfrak{P}^{(n)}_{j})$
is independent of $i,j\in\\{2,3,4\\}$ with $i\neq j$. From the equation
$\sum_{j=1}^{4}\mathfrak{P}^{(n)}_{i}\mathfrak{P}^{(n)}_{j}=(2-\frac{1}{n})\mathfrak{P}^{(n)}_{i}$
for $i=1,\dots,4$ we then obtain

$\displaystyle\tau\left(\mathfrak{P}^{(n)}_{1}\mathfrak{P}^{(n)}_{i}\right)$
$\displaystyle=\frac{1}{3}\left(1-\frac{1}{n}\right)\tau\left(\mathfrak{P}^{(n)}_{1}\right)\quad\text{for
}i=2,3,4,$
$\displaystyle\tau\left(\mathfrak{P}^{(n)}_{i}\mathfrak{P}^{(n)}_{j}\right)$
$\displaystyle=\frac{1}{2}\left(1-\frac{1}{n}\right)\left(\tau\left(\mathfrak{P}^{(n)}_{i}\right)-\frac{1}{3}\tau\left(\mathfrak{P}^{(n)}_{1}\right)\right)\quad\text{for
}i,j=2,3,4\text{ and }i\neq j.$

##  4 Spectral results

Let $n\in\mathbb{N}$. The projections
$\mathfrak{P}^{(n)}_{1},\dots,\mathfrak{P}^{(n)}_{4}$ of Proposition 3.1 play
a central role in self-tests of Section 5 below. Namely, they appear as
projective measurements in a self-tested strategy in Subsection 5.1; the fact
that they are determined by a linear relation
$\mathfrak{P}^{(n)}_{1}+\cdots+\mathfrak{P}^{(n)}_{4}=(2-\frac{1}{n})I$ is
beneficial for deducing the measurements from the correlation. Nevertheless,
to obtain a self-test, one still needs to be able to deduce the quantum state
from the correlation. Furthermore, in Subsection 5.2, the presented strategies
contain an additional projective measurement, which, while related to the
$\mathfrak{P}^{(n)}_{i}$, is itself not a part of quadruple adding to a scalar
multiple of identity. To help with the identification of the quantum state and
the additional measurements from the correlation, we first require some
information on eigenvalues and eigenvectors of certain tensor combinations and
sums of pairs of the matrices $\mathfrak{P}^{(n)}_{i}$. Concretely,
Proposition 4.2 shows how the maximally entangled state is related to
$\mathfrak{P}^{(n)}_{1},\dots,\mathfrak{P}^{(n)}_{4}$, and Proposition 4.4
shows that $\mathfrak{P}^{(n)}_{3}+\mathfrak{P}^{(n)}_{4}$ has pairwise
distinct eigenvalues, which enables post-hoc self-testing techniques [26, 7].

###  4.1 Role of the maximally entangled state

First, we identify the largest eigenvalue of
$\sum_{i}\mathfrak{P}^{(n)}_{i}\otimes\mathfrak{P}^{(n)}_{i}$ and the
corresponding eigenvector (cf. [19, Lemma 5.7]), and bound the spectrum of
$\sum_{i}\mathfrak{P}^{(n)}_{i}\otimes\mathfrak{P}^{(n)}_{\sigma(i)}$ for a
nontrivial cyclic permutation $\sigma$ of $(1,2,3,4)$. Given
$\mathinner{|\psi\rangle}=\sum_{i,j}\alpha_{ij}\mathinner{|i\rangle}\\!\mathinner{|j\rangle}\in\mathbb{C}^{n}\otimes\mathbb{C}^{n}$
let
$\operatorname{mat}(\mathinner{|\psi\rangle})=\sum_{i,j}\alpha_{ij}\mathinner{|i\rangle}\\!\mathinner{\langle
j|}\in\operatorname{M}_{n}(\mathbb{C})$ denote its matricization; note that
$\operatorname{mat}(\mathinner{|\phi_{n}\rangle})=\frac{1}{\sqrt{n}}I$, and

$\operatorname{mat}\big{(}A\otimes
B\mathinner{|\psi\rangle}\big{)}=A\operatorname{mat}(\mathinner{|\psi\rangle})B^{\rm
t}$

for $A,B\in\operatorname{M}_{n}(\mathbb{C})$.

######  Lemma 4.1.

Let $n\in\mathbb{N}$ and let $\sigma$ be a cyclic permutation $\sigma$ of
$(1,2,3,4)$. Denote
$M=\frac{n}{2n-1}\sum_{i=1}^{4}\mathfrak{P}^{(n)}_{i}\otimes\mathfrak{P}^{(n)}_{\sigma(i)}$.

  1. (i)

If $\sigma=\operatorname{id}$, then the largest eigenvalue of $M$ is 1, with
the eigenspace $\mathbb{C}\mathinner{|\phi_{n}\rangle}$.

  2. (ii)

If $\sigma\neq\operatorname{id}$, then all eigenvalues of $M$ are strictly
smaller than 1.

###### Proof.

Let $\mathinner{|\psi\rangle}\in\mathbb{C}^{n}\otimes\mathbb{C}^{n}$ be an
arbitrary state. Then

$\begin{split}\mathinner{\langle\psi|}I\otimes
I-M\mathinner{|\psi\rangle}&\geq\mathinner{\langle\psi|}I\otimes
I-\frac{n}{2n-1}\sum_{i=1}^{4}\mathfrak{P}^{(n)}_{i}\otimes
I\mathinner{|\psi\rangle}\\\
&=\mathinner{\langle\psi|}\left(I-\frac{n}{2n-1}\sum_{i=1}^{4}\mathfrak{P}^{(n)}_{i}\right)\otimes
I\mathinner{|\psi\rangle}=0.\end{split}$ (2)

Therefore the largest eigenvalue of $M$ is at most 1. Since

$\mathinner{\langle\phi_{n}|}I\otimes
I-\frac{n}{2n-1}\sum_{i=1}^{4}\mathfrak{P}^{(n)}_{i}\otimes\mathfrak{P}^{(n)}_{i}\mathinner{|\phi_{n}\rangle}=\tau\left(I-\frac{n}{2n-1}\sum_{i=1}^{4}\mathfrak{P}^{(n)}_{i}\right)=0,$

$\mathinner{|\phi_{n}\rangle}$ is an eigenvector of $M$ for eigenvalue 1 if
$\sigma=\operatorname{id}$. Suppose
$\mathinner{|\psi\rangle}\in\mathbb{C}^{n}\otimes\mathbb{C}^{n}$ satisfies
$M\mathinner{|\psi\rangle}=\mathinner{|\psi\rangle}$. Then (2) gives

$\mathinner{\langle\psi|}M\mathinner{|\psi\rangle}=\mathinner{\langle\psi|}\frac{n}{2n-1}\sum_{i=1}^{4}\mathfrak{P}^{(n)}_{i}\otimes
I\mathinner{|\psi\rangle}$

and therefore

$\mathinner{\langle\psi|}\sum_{i=1}^{4}\mathfrak{P}^{(n)}_{i}\otimes(I-\mathfrak{P}^{(n)}_{\sigma(i)})\mathinner{|\psi\rangle}=0.$

Positive semidefinitness then implies
$\mathfrak{P}^{(n)}_{i}\otimes(I-\mathfrak{P}^{(n)}_{\sigma(i)})\mathinner{|\psi\rangle}=0$,
and analogously
$(I-\mathfrak{P}^{(n)}_{i})\otimes\mathfrak{P}^{(n)}_{\sigma(i)}\mathinner{|\psi\rangle}=0$.
In particular, $\mathfrak{P}^{(n)}_{i}\otimes
I\mathinner{|\psi\rangle}=I\otimes\mathfrak{P}^{(n)}_{\sigma(i)}\mathinner{|\psi\rangle}$
for $i=1,\dots,4$. Therefore

$\mathfrak{P}^{(n)}_{i}\operatorname{mat}(\mathinner{|\psi\rangle})=\operatorname{mat}(\mathinner{|\psi\rangle})\mathfrak{P}^{(n)}_{\sigma(i)}\qquad\text{for
}i=1,\dots,4.$ (3)

Note that $\mathfrak{P}^{(n)}_{1},\dots,\mathfrak{P}^{(n)}_{4}$ and
$\mathfrak{P}^{(n)}_{\sigma(1)},\dots,\mathfrak{P}^{(n)}_{\sigma(4)}$ give
rise to two irreducible representations of $\mathcal{A}_{2-\frac{1}{n}}$ by
Proposition 3.1, which are unitarily equivalent if and only if
$\sigma=\operatorname{id}$. Since
$\operatorname{mat}(\mathinner{|\psi\rangle})$ intertwines these two
irreducible representations, Schur’s lemma implies that
$\operatorname{mat}{\mathinner{|\psi\rangle}}=\gamma I$ for some
$\gamma\in\mathbb{C}$ if $\sigma=\operatorname{id}$, and
$\operatorname{mat}{\mathinner{|\psi\rangle}}=0$ if if
$\sigma\neq\operatorname{id}$. Therefore $\mathinner{|\psi\rangle}$ is a
scalar multiple of $\mathinner{|\phi_{n}\rangle}$ if
$\sigma=\operatorname{id}$, and $1$ is not an eigenvalue of $M$ if
$\sigma\neq\operatorname{id}$. ∎

The following proposition shows how the maximally entangled state
$\mathinner{|\phi_{n}\rangle}$ is intrinsically connected to representations
of $\mathcal{A}_{2-\frac{1}{n}}$.

######  Proposition 4.2.

Let $n\in\mathbb{N}$, let $a_{1},\dots,a_{4},b_{1},\dots,b_{4}$ be nonnegative
integers with $a_{1}+\cdots+a_{4}=b_{1}+\cdots+b_{4}$, and let
$\sigma_{1},\dots,\sigma_{4}$ be the distinct cyclic permutations of
$(1,2,3,4)$. Consider the identification

$\mathbb{C}^{(a_{1}+\cdots+a_{4})n}\otimes\mathbb{C}^{(b_{1}+\cdots+b_{4})n}\equiv\left(\bigoplus_{j,k=1}^{4}\mathbb{C}^{a_{j}}\otimes\mathbb{C}^{b_{k}}\right)\otimes(\mathbb{C}^{n}\otimes\mathbb{C}^{n}).$

Then the largest eigenvalue of

$\frac{n}{2n-1}\sum_{i=1}^{4}\left(\bigoplus_{j=1}^{4}I_{a_{j}}\otimes\mathfrak{P}^{(n)}_{\sigma_{j}(i)}\right)\otimes\left(\bigoplus_{j=1}^{4}I_{b_{j}}\otimes\mathfrak{P}^{(n)}_{\sigma_{j}(i)}\right)$

is 1, with the eigenspace

$\Big{\\{}\big{(}\mathinner{|{\rm aux}_{1}\rangle}\oplus\mathinner{|{\rm
aux}_{2}\rangle}\oplus\mathinner{|{\rm aux}_{3}\rangle}\oplus\mathinner{|{\rm
aux}_{4}\rangle}\big{)}\otimes\mathinner{|\phi_{n}\rangle}\colon\mathinner{|{\rm
aux}_{j}\rangle}\in\mathbb{C}^{a_{j}}\otimes\mathbb{C}^{b_{j}}\Big{\\}}.$

###### Proof.

Follows from the distributivity of tensor product over direct sum, and Lemma
4.1. ∎

###  4.2 Spectrum of the sum of two distinguished projections

Next, we analyze the spectrum of the matrix
$\mathfrak{P}^{(n)}_{3}+\mathfrak{P}^{(n)}_{4}$ for every $n$. To do this, we
return to the functors between categories $\operatorname{Rep}_{\alpha}$. Given
a finite-dimensional representation $\pi$ of $\mathcal{A}_{\alpha}$, let
$\Lambda_{\pi}\subset[0,2]$ denote the set of eigenvalues of
$\pi(x_{3}+x_{4})$.

######  Lemma 4.3.

Let $\pi$ be an $n$-dimensional representation of $\mathcal{A}_{\alpha}$.

  1. (i)

$\Lambda_{T(\pi)}=2-\Lambda_{\pi}$.

  2. (ii)

Let $\alpha\notin\\{0,1\\}$.

    1. (ii.a)

If
$\operatorname{rk}\pi(x_{1})+\operatorname{rk}\pi(x_{2})>n=\operatorname{rk}\pi(x_{3})+\operatorname{rk}\pi(x_{4})$
then

$\Lambda_{S(\pi)}=\\{0\\}\cup\left(\tfrac{\alpha}{\alpha-1}-\tfrac{1}{\alpha-1}\Lambda_{\pi}\right).$

    2. (ii.b)

If
$\operatorname{rk}\pi(x_{3})+\operatorname{rk}\pi(x_{4})>n=\operatorname{rk}\pi(x_{1})+\operatorname{rk}\pi(x_{2})$
then

$\Lambda_{S(\pi)}=\left\\{\tfrac{\alpha}{\alpha-1}\right\\}\cup\left(\tfrac{\alpha}{\alpha-1}-\tfrac{1}{\alpha-1}\Lambda_{\pi}\right).$

  3. (iii)

Let $\alpha\in(0,3)$.

    1. (iii.a)

If
$\operatorname{rk}\pi(x_{1})+\operatorname{rk}\pi(x_{2})<n=\operatorname{rk}\pi(x_{3})+\operatorname{rk}\pi(x_{4})$
then

$\Lambda_{\Phi^{+}(\pi)}=\\{0\\}\cup\left(1-\tfrac{1}{3-\alpha}+\tfrac{1}{3-\alpha}\Lambda_{\pi}\right).$

    2. (iii.b)

If
$\operatorname{rk}\pi(x_{3})+\operatorname{rk}\pi(x_{4})<n=\operatorname{rk}\pi(x_{1})+\operatorname{rk}\pi(x_{2})$
then

$\Lambda_{\Phi^{+}(\pi)}=\left\\{1+\tfrac{1}{3-\alpha}\right\\}\cup\left(1-\tfrac{1}{3-\alpha}+\tfrac{1}{3-\alpha}\Lambda_{\pi}\right).$

###### Proof.

Equation (i) follows immediately from $T(\pi)(x_{i})=I-\pi(x_{i})$. Equations
(iii) are consequences of (i) and (ii) because $\Phi^{+}=S\circ T$.

Equations (ii): Suppose $\pi$ act on $\mathcal{H}$ with $\dim\mathcal{H}=n$,
and let

$\displaystyle u_{i}$
$\displaystyle:\operatorname{ran}\pi(x_{i})\to\mathcal{H},$ $\displaystyle
w_{i}$
$\displaystyle:\operatorname{ran}\pi(x_{i})\to\operatorname{ran}\pi(x_{1})\oplus\cdots\oplus\operatorname{ran}\pi(x_{4}),$
$\displaystyle v=\left(\begin{smallmatrix}v_{1}\\\ \vdots\\\
v_{4}\end{smallmatrix}\right)$
$\displaystyle:\operatorname{ran}\left(I-\frac{1}{\alpha}\left(\begin{smallmatrix}u_{1}^{*}\\\
\vdots\\\
u_{4}^{*}\end{smallmatrix}\right)\left(\begin{smallmatrix}u_{1}&\cdots&u_{4}\end{smallmatrix}\right)\right)\to\operatorname{ran}\pi(x_{1})\oplus\cdots\oplus\operatorname{ran}\pi(x_{4})$

be inclusions as in the construction of $S$. Then
$S(\pi)(x_{i})=\frac{\alpha}{\alpha-1}v^{*}w_{i}w_{i}^{*}v$, and the
characteristic polynomial of $S(\pi)(x_{3}+x_{4})$ equals

$\displaystyle\det\big{(}\lambda I-S(\pi)(x_{3}+x_{4})\big{)}$
$\displaystyle=\ $ $\displaystyle\det\left(\lambda
I-\tfrac{\alpha}{\alpha-1}v^{*}(w_{3}w_{3}^{*}+w_{4}w_{4}^{*})v\right)$
$\displaystyle=\ $ $\displaystyle\det\left(\lambda
I-\tfrac{\alpha}{\alpha-1}\left(\begin{smallmatrix}v_{3}^{*}&v_{4}^{*}\end{smallmatrix}\right)\left(\begin{smallmatrix}v_{3}\\\
v_{4}\end{smallmatrix}\right)\right)$ $\displaystyle=\ $
$\displaystyle\lambda^{\operatorname{rk}\pi(x_{1})+\operatorname{rk}\pi(x_{2})-n}\det\left(\lambda
I-\tfrac{\alpha}{\alpha-1}\left(\begin{smallmatrix}v_{3}\\\
v_{4}\end{smallmatrix}\right)\left(\begin{smallmatrix}v_{3}^{*}&v_{4}^{*}\end{smallmatrix}\right)\right)$
$\displaystyle=\ $
$\displaystyle\lambda^{\operatorname{rk}\pi(x_{1})+\operatorname{rk}\pi(x_{2})-n}\det\left(\lambda
I-\tfrac{\alpha}{\alpha-1}\left(I-\tfrac{1}{\alpha}\left(\begin{smallmatrix}u_{3}^{*}\\\
u_{4}^{*}\end{smallmatrix}\right)\left(\begin{smallmatrix}u_{3}&u_{4}\end{smallmatrix}\right)\right)\right)$
$\displaystyle=\ $
$\displaystyle\lambda^{\operatorname{rk}\pi(x_{1})+\operatorname{rk}\pi(x_{2})-n}\det\left(\big{(}\lambda-\tfrac{\alpha}{\alpha-1}\big{)}I+\tfrac{1}{\alpha-1}\left(\begin{smallmatrix}u_{3}^{*}\\\
u_{4}^{*}\end{smallmatrix}\right)\left(\begin{smallmatrix}u_{3}&u_{4}\end{smallmatrix}\right)\right)$
$\displaystyle=\ $
$\displaystyle\lambda^{\operatorname{rk}\pi(x_{1})+\operatorname{rk}\pi(x_{2})-n}\left(\lambda-\tfrac{\alpha}{\alpha-1}\right)^{\operatorname{rk}\pi(x_{3})+\operatorname{rk}\pi(x_{4})-n}\det\left(\big{(}\lambda-\tfrac{\alpha}{\alpha-1}\big{)}I+\tfrac{1}{\alpha-1}\left(\begin{smallmatrix}u_{3}&u_{4}\end{smallmatrix}\right)\left(\begin{smallmatrix}u_{3}^{*}\\\
u_{4}^{*}\end{smallmatrix}\right)\right)$ $\displaystyle=\ $
$\displaystyle\lambda^{\operatorname{rk}\pi(x_{1})+\operatorname{rk}\pi(x_{2})-n}\left(\lambda-\tfrac{\alpha}{\alpha-1}\right)^{\operatorname{rk}\pi(x_{3})+\operatorname{rk}\pi(x_{4})-n}\det\left(\big{(}\lambda-\tfrac{\alpha}{\alpha-1}\big{)}I+\tfrac{1}{\alpha-1}\pi(x_{3}+x_{4})\right).$

Therefore

$\Lambda_{S(\pi)}=\\{0\\}\cup\big{(}\tfrac{\alpha}{\alpha-1}-\tfrac{1}{\alpha-1}\Lambda_{\pi}\big{)}$

if
$\operatorname{rk}\pi(x_{1})+\operatorname{rk}\pi(x_{2})>n=\operatorname{rk}\pi(x_{3})+\operatorname{rk}\pi(x_{4})$,
and

$\Lambda_{S(\pi)}=\left\\{\tfrac{\alpha}{\alpha-1}\right\\}\cup\big{(}\tfrac{\alpha}{\alpha-1}-\tfrac{1}{\alpha-1}\Lambda_{\pi}\big{)}$

if
$\operatorname{rk}\pi(x_{3})+\operatorname{rk}\pi(x_{4})>n=\operatorname{rk}\pi(x_{1})+\operatorname{rk}\pi(x_{2})$.
∎

The following proposition identifies all eigenvalues of the matrix
$\mathfrak{P}_{3}^{(n)}+\mathfrak{P}_{4}^{(n)}$; in particular, they are all
simple (pairwise distinct).

######  Proposition 4.4.

Eigenvalues of $n(\mathfrak{P}_{3}^{(n)}+\mathfrak{P}_{4}^{(n)})$ are
$\\{0,2,\dots,2n-2\\}$ if $n$ is odd, and $\\{1,3,\dots,2n-1\\}$ if $n$ is
even.

###### Proof.

Let $\pi_{1}:\mathcal{A}_{1}\to\mathbb{C}$ be given as $\pi_{1}(x_{1})=1$ and
$\pi_{1}(x_{2})=\pi(x_{3})=\pi(x_{4})=0$. For $n\geq 2$ denote
$\pi_{n}=\Phi^{+}(\pi_{1})$. By Proposition 3.1 we have
$\operatorname{rk}\pi_{n}(x_{1})+\operatorname{rk}\pi_{n}(x_{2})<n=\operatorname{rk}\pi_{n}(x_{3})+\operatorname{rk}\pi_{n}(x_{4})$
if $n$ is even, and
$\operatorname{rk}\pi_{n}(x_{3})+\operatorname{rk}\pi_{n}(x_{4})<n=\operatorname{rk}\pi_{n}(x_{1})+\operatorname{rk}\pi_{n}(x_{2})$
if $n$ is odd. By Lemma 4.3,

$\displaystyle\Lambda_{\pi_{n+1}}$
$\displaystyle=\\{0\\}\cup\left(\tfrac{1}{n+1}+\tfrac{n}{n+1}\Lambda_{\pi_{n}}\right)\qquad$
$\displaystyle\text{if }n\text{ is even,}$ $\displaystyle\Lambda_{\pi_{n+1}}$
$\displaystyle=\\{2-\tfrac{1}{n+1}\\}\cup\left(\tfrac{1}{n+1}+\tfrac{n}{n+1}\Lambda_{\pi_{n}}\right)\qquad$
$\displaystyle\text{if }n\text{ is odd.}$

Therefore

$\displaystyle(n+1)\Lambda_{\pi_{n+1}}$
$\displaystyle=\\{0\\}\cup\left(1+n\Lambda_{\pi_{n}}\right)\qquad$
$\displaystyle\text{if }n\text{ is even,}$
$\displaystyle(n+1)\Lambda_{\pi_{n+1}}$
$\displaystyle=\\{2n+1\\}\cup\left(1+n\Lambda_{\pi_{n}}\right)\qquad$
$\displaystyle\text{if }n\text{ is odd.}$

Since $\Lambda_{\pi_{1}}=\\{0\\}$, induction on $n$ shows that

$\displaystyle n\Lambda_{\pi_{n}}$ $\displaystyle=\\{0,2,\dots,2n-2\\}\qquad$
$\displaystyle\text{if }n\text{ is odd,}$ $\displaystyle n\Lambda_{\pi_{n}}$
$\displaystyle=\\{1,3,\dots,2n-1\\}\qquad$ $\displaystyle\text{if }n\text{ is
even.}$

Finally, $\mathfrak{P}_{3}^{(n)},\mathfrak{P}_{4}^{(n)}$ are simultaneously
unitarily equivalent to $\pi_{n}(x_{3}),\pi_{n}(x_{4})$. ∎

Lastly, we determine how eigenvectors of
$\mathfrak{P}_{3}^{(n)}+\mathfrak{P}_{4}^{(n)}$ interact with
$\mathfrak{P}_{1}^{(n)}$ and $\mathfrak{P}_{2}^{(n)}$.

######  Proposition 4.5.

Let $\lambda$ be an eigenvalue of
$\mathfrak{P}_{3}^{(n)}+\mathfrak{P}_{4}^{(n)}$, with a corresponding unit
eigenvector $\mathinner{|e\rangle}\in\mathbb{R}^{n}$.

  1. (i)

If $\lambda\neq 1-\frac{1}{n}$ then

$\mathinner{\langle
e|}\mathfrak{P}^{(n)}_{1}\mathinner{|e\rangle}=\mathinner{\langle
e|}\mathfrak{P}^{(n)}_{2}\mathinner{|e\rangle}=1-\frac{1}{2n}-\frac{\lambda}{2}.$

  2. (ii)

If $\lambda=1-\frac{1}{n}$ then

$\mathinner{\langle
e|}\mathfrak{P}^{(n)}_{1}\mathinner{|e\rangle}=\left\\{\begin{array}[]{ll}0&\text{if
}n\text{ even,}\\\ 1&\text{if }n\text{
odd,}\end{array}\right.\qquad\mathinner{\langle
e|}\mathfrak{P}^{(n)}_{2}\mathinner{|e\rangle}=\left\\{\begin{array}[]{ll}1&\text{if
}n\text{ even,}\\\ 0&\text{if }n\text{ odd.}\end{array}\right.$

###### Proof.

(i) By the defining relation of $\mathfrak{P}^{(n)}_{i}$,

$\mathfrak{P}^{(n)}_{1}\mathinner{|e\rangle}+\mathfrak{P}^{(n)}_{2}\mathinner{|e\rangle}+\lambda\mathinner{|e\rangle}=\left(2-\frac{1}{n}\right)\mathinner{|e\rangle}.$
(4)

Multiplying (4) on the left with $\mathinner{\langle
e|}\mathfrak{P}^{(n)}_{i}$ for $i=1,2$ results in

$\displaystyle\mathinner{\langle
e|}\mathfrak{P}^{(n)}_{1}\mathinner{|e\rangle}+\mathinner{\langle
e|}\mathfrak{P}^{(n)}_{1}\mathfrak{P}^{(n)}_{2}\mathinner{|e\rangle}$
$\displaystyle=\left(2-\frac{1}{n}-\lambda\right)\mathinner{\langle
e|}\mathfrak{P}^{(n)}_{1}\mathinner{|e\rangle},$
$\displaystyle\mathinner{\langle
e|}\mathfrak{P}^{(n)}_{2}\mathfrak{P}^{(n)}_{1}\mathinner{|e\rangle}+\mathinner{|e\rangle}\mathfrak{P}^{(n)}_{2}\mathinner{|e\rangle}$
$\displaystyle=\left(2-\frac{1}{n}-\lambda\right)\mathinner{\langle
e|}\mathfrak{P}^{(n)}_{2}\mathinner{|e\rangle}.$

Therefore $\mathinner{\langle
e|}\mathfrak{P}^{(n)}_{1}\mathinner{|e\rangle}=\mathinner{\langle
e|}\mathfrak{P}^{(n)}_{2}\mathinner{|e\rangle}$ if $\lambda\neq
1-\frac{1}{n}$. Multiplying (4) on the left with $\mathinner{\langle e|}$ then
gives $\mathinner{\langle
e|}\mathfrak{P}^{(n)}_{1}\mathinner{|e\rangle}=\mathinner{\langle
e|}\mathfrak{P}^{(n)}_{2}\mathinner{|e\rangle}=1-\frac{1}{2n}-\frac{\lambda}{2}$.

(ii) Note that $\mathfrak{P}_{3}^{(n)}+\mathfrak{P}_{4}^{(n)}$ admits $n$
orthonormal eigenvectors
$\mathinner{|e_{1}\rangle},\dots,\mathinner{|e_{n}\rangle}\in\mathbb{R}^{n}$
by Proposition 4.4. Hence

$\operatorname{tr}\mathfrak{P}_{i}^{(n)}=\sum_{k=1}^{n}\mathinner{\langle
e_{k}|}\mathfrak{P}^{(n)}_{i}\mathinner{|e_{k}\rangle}$

for $i=1,2$. By (ii) and Proposition 3.1 we therefore have

$\displaystyle\mathinner{\langle
e|}\mathfrak{P}^{(n)}_{i}\mathinner{|e\rangle}$
$\displaystyle=\operatorname{tr}\mathfrak{P}_{i}^{(n)}-(n-1)\left(1-\frac{1}{2n}\right)+\frac{1}{2}\left(\operatorname{tr}\Big{(}\mathfrak{P}_{3}^{(n)}+\mathfrak{P}_{4}^{(n)}\Big{)}-1+\frac{1}{n}\right)$
$\displaystyle=2\left\lfloor\frac{n}{2}\right\rfloor-n+1-\left\\{\begin{array}[]{ll}(-1)^{n}&\text{if
}i=1\\\ 0&\text{if }i=2\end{array}\right.$

since
$\operatorname{tr}\mathfrak{P}^{(n)}_{i}=\operatorname{rk}\mathfrak{P}^{(n)}_{i}$.
∎

##  5 Constant-sized self-tests

In this section we derive the main results of the paper: every maximally
entangled state is self-tested by a 4-input 2-output strategy (Subsection
5.1), and every single binary PVM is self-tested by a 5-input 2-output
strategy (Subsection 5.2).

###  5.1 Self-testing maximally entangled states

First we introduce a family of 4-input 2-output strategies that self-test
maximally entangled states of all dimensions (Theorem 5.2).

######  Definition 5.1.

For $n\in\mathbb{N}$ let $\mathfrak{P}^{(n)}_{i}$ be the $n\times n$
projections as in Proposition 3.1. Let $\mathcal{S}_{n}$ be the 4-input
2-output bipartite strategy

$\mathcal{S}_{n}=\left(\mathinner{|\phi_{n}\rangle};\left(\mathfrak{P}^{(n)}_{i},I-\mathfrak{P}^{(n)}_{i}\right)_{i=1}^{4};\left(\mathfrak{P}^{(n)}_{i},I-\mathfrak{P}^{(n)}_{i}\right)_{i=1}^{4}\right).$

Note that the correlation of $\mathcal{S}_{n}$ is synchronous,
$p(a,b|i,i)=\tau\left(\mathfrak{P}^{(n)}_{i}(I-\mathfrak{P}^{(n)}_{i})\right)=0$
for $a\neq b$. Furthermore,

$\displaystyle p(1,1|i,j)$
$\displaystyle=\mathinner{\langle\phi_{n}|}\mathfrak{P}^{(n)}_{i}\otimes\mathfrak{P}^{(n)}_{j}\mathinner{|\phi_{n}\rangle}=\tau\left(\mathfrak{P}^{(n)}_{i}\mathfrak{P}^{(n)}_{j}\right),$
$\displaystyle p(1|i)$
$\displaystyle=\mathinner{\langle\phi_{n}|}\mathfrak{P}^{(n)}_{i}\otimes
I\mathinner{|\phi_{n}\rangle}=\mathinner{\langle\phi_{n}|}I\otimes\mathfrak{P}^{(n)}_{i}\mathinner{|\phi_{n}\rangle}=\tau\left(\mathfrak{P}^{(n)}_{i}\right)$

for $i,j=1,\dots,4$, and these values are computed in Remark 3.4. Comprising
everything together, the correlation of $\mathcal{S}_{n}$ is determined by the
vector

$\big{(}p(1|i)\big{)}_{i=1}^{4}=\begin{pmatrix}\frac{\lfloor\frac{n}{2}\rfloor-(-1)^{n}}{n}&\frac{\lfloor\frac{n}{2}\rfloor}{n}&\frac{\lfloor\frac{n}{2}\rfloor}{n}&\frac{\lfloor\frac{n}{2}\rfloor}{n}\end{pmatrix}$

and the symmetric matrix $\big{(}p(1,1|i,j)\big{)}_{i,j=1}^{4}$

$\left(\begin{array}[]{cccc}\frac{\lfloor\frac{n}{2}\rfloor-(-1)^{n}}{n}&\frac{(n-1)(\lfloor\frac{n}{2}\rfloor-(-1)^{n})}{3n^{2}}&\frac{(n-1)(\lfloor\frac{n}{2}\rfloor-(-1)^{n})}{3n^{2}}&\frac{(n-1)(\lfloor\frac{n}{2}\rfloor-(-1)^{n})}{3n^{2}}\\\\[5.0pt]
\cdot&\frac{\lfloor\frac{n}{2}\rfloor}{n}&\frac{(n-1)(2n-1+3(-1)^{n})}{12n^{2}}&\frac{(n-1)(2n-1+3(-1)^{n})}{12n^{2}}\\\\[5.0pt]
\cdot&\cdot&\frac{\lfloor\frac{n}{2}\rfloor}{n}&\frac{(n-1)(2n-1+3(-1)^{n})}{12n^{2}}\\\\[5.0pt]
\cdot&\cdot&\cdot&\frac{\lfloor\frac{n}{2}\rfloor}{n}\\\ \end{array}\right).$

Notice that while a closed-form expression for the strategy $\mathcal{S}_{n}$
has not been given (instead, the projections in $\mathcal{S}_{n}$ can be
recursively constructed as in Remark 3.2), its correlation admits a closed-
form expression (as a function of $n$).

The next theorem establishes that $\mathcal{S}_{n}$ is a local dilation of any
strategy $\mathcal{S}$ that produces the same correlation as
$\mathcal{S}_{n}$. The blueprint for the proof is threefold. Firstly, the
correlation manages to encode the defining linear relation of measurements in
$\mathcal{S}_{n}$, which leads to measurements of $\mathcal{S}$ essentially
forming a representation of $\mathcal{A}_{2-\frac{1}{n}}$. Secondly, the
established relationship between the maximally entangled state and
representations of $\mathcal{A}_{2-\frac{1}{n}}$ (Proposition 4.2) allows one
to identify the state in $\mathcal{S}$. Thirdly, the finer look at the
correlation shows that the representation of $\mathcal{A}_{2-\frac{1}{n}}$
arising from measurements of $\mathcal{S}$ cannot be an direct sum of the
different irreducible representations, but is actually a direct copy of the
irreducible representation coming from $\mathcal{S}_{n}$.

######  Theorem 5.2.

The strategy $\mathcal{S}_{n}$ is self-tested by its correlation for every
$n\in\mathbb{N}$.

###### Proof.

Let $p$ be the correlation of $\mathcal{S}_{n}$. Suppose

$\mathcal{S}=\left(\mathinner{|\psi\rangle};(P_{i},I-P_{i})_{i=1}^{4};(Q_{i},I-Q_{i})_{i=1}^{4}\right)$

is another strategy with the correlation $p$. Since $p$ is synchronous and
local dilations are transitive, by [19, Lemma 4.9 and Corollary 3.6] it
suffices to assume that the state
$\mathinner{|\psi\rangle}\in\mathcal{H}\otimes\mathcal{H}$ has full Schmidt
rank, $P_{i},Q_{i}$ are projections on $\mathcal{H}$, and

$P_{i}\otimes I\mathinner{|\psi\rangle}=I\otimes
Q_{i}\mathinner{|\psi\rangle}$ (5)

for $i=1,\dots,4$. By equality of correlations and (5),

$\displaystyle\mathinner{\langle\psi|}\left(\frac{2n-1}{n}I-\sum_{i=1}^{4}P_{i}\right)^{2}\otimes
I\mathinner{|\psi\rangle}$ $\displaystyle=\,$
$\displaystyle\mathinner{\langle\psi|}\left(\frac{2n-1}{n}I-\sum_{i=1}^{4}P_{i}\right)\otimes\left(\frac{2n-1}{n}I-\sum_{i=1}^{4}Q_{i}\right)\mathinner{|\psi\rangle}$
$\displaystyle=\,$
$\displaystyle\mathinner{\langle\phi_{n}|}\left(\frac{2n-1}{n}I-\sum_{i=1}^{4}\mathfrak{P}^{(n)}_{i}\right)\otimes\left(\frac{2n-1}{n}I-\sum_{i=1}^{4}\mathfrak{P}^{(n)}_{i}\right)\mathinner{|\phi_{n}\rangle}=0,$

and analogously for $Q_{i}$. Since $\mathinner{|\psi\rangle}$ has full rank,
we obtain

$\frac{2n-1}{n}I-\sum_{i=1}^{4}P_{i}=0=\frac{2n-1}{n}I-\sum_{i=1}^{4}Q_{i}.$
(6)

Furthermore,

$\mathinner{\langle\psi|}\frac{n}{2n-1}\sum_{i=1}^{4}P_{i}\otimes
Q_{i}\mathinner{|\psi\rangle}=\mathinner{\langle\phi_{n}|}\frac{n}{2n-1}\sum_{i=1}^{4}\mathfrak{P}^{(n)}_{i}\otimes\mathfrak{P}^{(n)}_{i}\mathinner{|\phi_{n}\rangle}=1.$
(7)

Let $\sigma_{1},\dots,\sigma_{4}$ be the distinct cyclic permutations of
$(1,2,3,4)$, with $\sigma_{1}=\operatorname{id}$. By (6) and Proposition 3.1
there exist nonnegative integers $a_{1},\dots,a_{4},b_{1},\dots,b_{4}$ with
$a_{1}+\cdots+a_{4}=b_{1}+\cdots+b_{4}$, and unitaries $U$ and $V$ on
$\mathcal{H}$, such that

$UP_{i}U^{*}=\bigoplus_{j=1}^{4}I_{a_{j}}\otimes\mathfrak{P}^{(n)}_{\sigma_{j}(i)},\qquad
VQ_{i}V^{*}=\bigoplus_{j=1}^{4}I_{b_{j}}\otimes\mathfrak{P}^{(n)}_{\sigma_{j}(i)}$

for $i=1,\dots,4$. By (7) and Proposition 4.2,

$U\otimes V\mathinner{|\psi\rangle}=\big{(}\mathinner{|{\rm
aux}_{1}\rangle}\oplus\mathinner{|{\rm aux}_{2}\rangle}\oplus\mathinner{|{\rm
aux}_{3}\rangle}\oplus\mathinner{|{\rm
aux}_{4}\rangle}\big{)}\otimes\mathinner{|\phi_{n}\rangle}$

for some $\mathinner{|{\rm
aux}_{j}\rangle}\in\mathbb{C}^{a_{j}}\otimes\mathbb{C}^{b_{j}}$, where we
identified

$\mathcal{H}\otimes\mathcal{H}\equiv\left(\bigoplus_{j,k=1}^{4}\mathbb{C}^{a_{j}}\otimes\mathbb{C}^{b_{k}}\right)\otimes(\mathbb{C}^{n}\otimes\mathbb{C}^{n}).$

Then

$\mathinner{\langle\phi_{n}|}\mathfrak{P}^{(n)}_{i}\otimes
I\mathinner{|\phi_{n}\rangle}=\mathinner{\langle\psi|}P_{i}\otimes
I\mathinner{|\psi\rangle}=\sum_{j=1}^{4}\mathinner{\langle{\rm aux}_{j}|{\rm
aux}_{j}\rangle}\mathinner{\langle\phi_{n}|}\mathfrak{P}^{(n)}_{\sigma_{j}(i)}\otimes
I\mathinner{|\phi_{n}\rangle}$

gives rise to a linear system of equations in $\mathinner{\langle{\rm
aux}_{j}|{\rm aux}_{j}\rangle}$,

$\operatorname{rk}\mathfrak{P}^{(n)}_{i}=\sum_{j=1}^{4}\operatorname{rk}\mathfrak{P}^{(n)}_{\sigma_{j}(i)}\cdot\mathinner{\langle{\rm
aux}_{j}|{\rm aux}_{j}\rangle}\qquad\text{for }i=1,2,3,4.$ (8)

By Lemma 3.3, the system (8) has a unique solution; since
$\sigma_{1}=\operatorname{id}$, we obtain $\mathinner{\langle{\rm
aux}_{1}|{\rm aux}_{1}\rangle}=1$ and $\mathinner{\langle{\rm aux}_{j}|{\rm
aux}_{j}\rangle}=0$ for $j=2,3,4$. Since $\mathinner{|\psi\rangle}$ is a
faithful state, it follows that $a_{j}=b_{j}=0$ for $j=2,3,4$, and
$a_{1}=b_{1}$. Therefore

$UP_{i}U^{*}=I_{a_{1}}\otimes\mathfrak{P}^{(n)}_{i},\qquad
VQ_{i}V^{*}=I_{a_{1}}\otimes\mathfrak{P}^{(n)}_{i},\qquad U\otimes
V\mathinner{|\psi\rangle}=\mathinner{|{\rm
aux}_{1}\rangle}\otimes\mathinner{|\phi_{n}\rangle},$

so $\mathcal{S}_{n}$ is a local dilation of $\mathcal{S}$. ∎

######  Remark 5.3.

The proof of Theorem 5.2 follows the core ideas of the proof of [19, Corollary
7.1], which treats maximally entangled states of odd dimension. The main
difference arises from applying the representation theory of C*-algebras
$\mathcal{A}_{\alpha}$ for different values of $\alpha$. Namely, in [19] the
authors focus on $\mathcal{A}_{2-\frac{2}{n}}$ for odd $n$ (and their analogs
on more than four generators), since $\mathcal{A}_{2-\frac{2}{n}}$ for odd $n$
is simple and isomorphic to $\operatorname{M}_{n}(\mathbb{C})$ (i.e., it has a
unique irreducible representation, which is $n$-dimensional). On the other
hand, algebras $\mathcal{A}_{2-\frac{1}{n}}$ for $n\in\mathbb{N}$ are not
simple, as they are isomorphic to
$\mathbb{C}^{4}\otimes\operatorname{M}_{n}(\mathbb{C})$. Non-simplicity is the
origin of intricacies in the proof of Theorem 5.2 and auxiliary results.

Finally, with a considerable effort, the authors of [19] also establish that
their self-tests are _robust_. Such robustness analysis is omitted in this
paper; nevertheless, there is no obstruction for the techniques of [19,
Section 6] to imply robust versions of the newly presented self-tests.

######  Corollary 5.4.

The following states and binary projective measurements can be self-tested by
4-input 2-output bipartite strategies for every $n\in\mathbb{N}$:

  1. (a)

maximally entangled state of local dimension $n$;

  2. (b)

binary projective measurement determined by an $n\times n$ projection with
rank in

$\left\\{\left\lceil\frac{n}{2}\right\rceil,\
\left\lfloor\frac{n}{2}\right\rfloor-(-1)^{n},\
\left\lceil\frac{n}{2}\right\rceil+(-1)^{n}\right\\}.$

###  5.2 Self-testing local projective measurements

Next we introduce a two-parametric family of 5-input 2-output strategies that
self-test binary PVMs of all dimensions and ranks (Theorem 5.10). These
strategies are obtained from the 4-input 2-output strategies of Subsection 5.1
by adding an additional binary PVM. The phenomenon, where a self-tested
strategy is extended to a new one while preserving the self-testing feature,
is called post-hoc self-testing [26]. The key sufficiency condition for post-
hoc self-testing was derived in [7], and is presented next.

Given an invertible hermitian matrix $X\in\operatorname{M}_{n}(\mathbb{C})$
let $\operatorname{sgn}(X)\in\operatorname{M}_{n}(\mathbb{C})$ be the unique
hermitian unitary matrix that commutes with $X$, and
$\operatorname{sgn}(X)X\succ 0$. Equivalently, $\operatorname{sgn}(X)$ is the
unitary part of the polar decomposition of $X$. In other words,
$\operatorname{sgn}$ is the matrix extension of the usual sign function via
functional calculus. This map plays a role in the following post-hoc self-
testing criterion established in [7].

######  Proposition 5.5.

[7, Proposition 3.7] Suppose
$P,P_{i},Q_{j}\in\operatorname{M}_{n}(\mathbb{R})$ for $i=1,\dots,N_{A}$ and
$j=1,\dots,N_{B}$ are projections, and the $(N_{A},N_{B})$-input
$(2,2)$-output strategy

$\left(\mathinner{|\phi_{n}\rangle};\left(P_{i},I-P_{i}\right)_{i=1}^{N_{A}};\left(Q_{i},I-Q_{i}\right)_{i=1}^{N_{B}}\right)$

is self-tested by its correlation. If

$2P-I\in\operatorname{sgn}\Big{(}\operatorname{GL}_{n}(\mathbb{R})\cap\operatorname{span}_{\mathbb{R}}\\{I,Q_{1},\dots,Q_{N_{B}}\\}\Big{)},$

then the $(N_{A}+1,N_{B})$-input $(2,2)$-output strategy

$\left(\mathinner{|\phi_{n}\rangle};\left(P_{i},I-P_{i}\right)_{i=1}^{N_{A}},(P,I-P);\left(Q_{i},I-Q_{i}\right)_{i=1}^{N_{B}}\right)$

is self-tested by its correlation.

As mentioned at the beginning of the subsection, Proposition 5.5 will be used
to obtain a self-tested strategy by extending $\mathcal{S}_{n}$ from
Subsection 5.1. Recall that $\mathfrak{P}^{(n)}_{3}+\mathfrak{P}^{(n)}_{4}$
has pairwise distinct eigenvalues by Proposition 4.4. This gives rise to a
family of projections that satisfy the sufficiency condition in Proposition
5.5.

######  Proposition 5.6.

Let $n,r\in\mathbb{N}$ with $r\leq n$. The matrix

$\mathfrak{Q}^{(n,r)}:=\frac{1}{2}\bigg{(}I+\operatorname{sgn}\Big{(}(2r-\tfrac{1}{2})I-n\big{(}\mathfrak{P}^{(n)}_{3}+\mathfrak{P}^{(n)}_{4}\big{)}\Big{)}\bigg{)}\in\operatorname{M}_{n}(\mathbb{R})$

is a projection of rank $r$, and satisfies

$2\mathfrak{Q}^{(n,r)}-I\in\operatorname{sgn}\Big{(}\operatorname{GL}_{n}(\mathbb{R})\cap\operatorname{span}_{\mathbb{R}}\\{I,\mathfrak{P}^{(n)}_{3},\mathfrak{P}^{(n)}_{4}\\}\Big{)}.$

###### Proof.

The matrix $\mathfrak{Q}^{(n,r)}$ is a projection by definition of the map
$\operatorname{sgn}$. By Proposition 4.4, the matrix
$n(\mathfrak{P}^{(n)}_{3}+\mathfrak{P}^{(n)}_{4})$ has eigenvalues
$\\{0,2,\dots,2n-2\\}$ if $n$ is odd and $\\{1,3,\dots,2n-1\\}$ if $n$ is
even. Therefore
$(2r-\tfrac{1}{2})I-n(\mathfrak{P}^{(n)}_{3}+\mathfrak{P}^{(n)}_{4})$ has $r$
positive eigenvalues and $n-r$ negative eigenvalues. Consequently, the
multiplicities of eigenvalues $1$ and $-1$ of
$\operatorname{sgn}((2r-\tfrac{1}{2})I-n(\mathfrak{P}^{(n)}_{3}+\mathfrak{P}^{(n)}_{4}))$
are $r$ and $n-r$, respectively. Hence the rank of $\mathfrak{Q}^{(n,r)}$ is
$r$. ∎

######  Remark 5.7.

For $r\leq n$ let
$\mathinner{|e_{1}\rangle},\dots,\mathinner{|e_{r}\rangle}\in\mathbb{R}^{n}$
be unit eigenvectors of $\mathfrak{P}^{(n)}_{3}+\mathfrak{P}^{(n)}_{4}$
corresponding to the smallest $r$ eigenvalues in increasing order (note that
$\mathinner{|e_{i}\rangle}$ are uniquely determined up to a sign because
$\mathfrak{P}^{(n)}_{3}+\mathfrak{P}^{(n)}_{4}$ has $n$ distinct eigenvalues).
Then

$\mathfrak{Q}^{(n,r)}=\mathinner{|e_{1}\rangle}\\!\mathinner{\langle
e_{1}|}+\cdots+\mathinner{|e_{r}\rangle}\\!\mathinner{\langle e_{r}|}.$

For concrete matrix representations of $\mathfrak{Q}^{(n,r)}$ when $1\leq
r<n\leq 6$, see Appendix A. While this is arguably a simpler and
computationally more available definition of $\mathfrak{Q}^{(n,r)}$ than the
original in Proposition 5.6, the presentation in terms of the
$\operatorname{sgn}$ map is critical in establishing the self-test of Theorem
5.10 below.

######  Remark 5.8.

Let us determine the normalized traces of
$\mathfrak{P}^{(n)}_{i}\mathfrak{Q}^{(n,r)}$ for $r<\frac{n}{2}$. Clearly,
$\tau\left(\mathfrak{Q}^{(n,r)}\right)=\frac{r}{n}$. By Proposition 3.1 there
exists a unitary $U\in\operatorname{M}_{n}(\mathbb{C})$ such that
$U\mathfrak{P}^{(n)}_{3}U^{*}=\mathfrak{P}^{(n)}_{4}$ and
$U\mathfrak{P}^{(n)}_{4}U^{*}=\mathfrak{P}^{(n)}_{3}$, and therefore
$\operatorname{tr}\left(\mathfrak{P}^{(n)}_{3}\mathfrak{Q}^{(n,r)}\right)=\operatorname{tr}\left(\mathfrak{P}^{(n)}_{4}\mathfrak{Q}^{(n,r)}\right)$.
Thus

$\tau\left(\mathfrak{P}^{(n)}_{i}\mathfrak{Q}^{(n,r)}\right)=\frac{1}{2}\tau\left(\mathfrak{Q}^{(n,r)}\big{(}\mathfrak{P}^{(n)}_{3}+\mathfrak{P}^{(n)}_{4}\big{)}\mathfrak{Q}^{(n,r)}\right)=\frac{r}{2n^{2}}\left(r-\frac{1-(-1)^{n}}{2}\right)$

for $i=3,4$ by Proposition 4.4, since
$\operatorname{tr}(\mathfrak{Q}^{(n,r)}(\mathfrak{P}^{(n)}_{3}+\mathfrak{P}^{(n)}_{4})\mathfrak{Q}^{(n,r)})$
is the sum of smallest $r$ eigenvalues of
$\mathfrak{P}^{(n)}_{3}+\mathfrak{P}^{(n)}_{4}$ by Remark 5.7. Since
$r<\frac{n}{2}$, Proposition 4.5 and Remark 5.7 imply
$\operatorname{tr}(\mathfrak{Q}^{(n,r)}\mathfrak{P}^{(n)}_{1}\mathfrak{Q}^{(n,r)})=\operatorname{tr}(\mathfrak{Q}^{(n,r)}\mathfrak{P}^{(n)}_{2}\mathfrak{Q}^{(n,r)})$.
By the defining relation of $\mathfrak{P}^{(n)}_{i}$ we then obtain

$\tau\left(\mathfrak{P}^{(n)}_{i}\mathfrak{Q}^{(n,r)}\right)=\frac{1}{2}\left(\left(2-\frac{1}{n}\right)\tau\left(\mathfrak{Q}^{(n,r)}\right)-\tau\left(\mathfrak{P}^{(n)}_{3}\mathfrak{Q}^{(n,r)}\right)-\tau\left(\mathfrak{P}^{(n)}_{4}\mathfrak{Q}^{(n,r)}\right)\right)$

for $i=1,2$.

######  Definition 5.9.

Given $n,r\in\mathbb{N}$ with $r<n$, let $\mathfrak{P}^{(n)}_{i}$ be as in
Proposition 3.1, and let $\mathfrak{Q}^{(n,r)}$ be as in Proposition 5.6. Let
$\mathcal{S}_{n,r}$ be the $(5,4)$-input $(2,2)$-output bipartite strategy

$\left(\mathinner{|\phi_{n}\rangle};\left(\mathfrak{P}^{(n)}_{i},I-\mathfrak{P}^{(n)}_{i}\right)_{i=1}^{4},(\mathfrak{Q}^{(n,r)},I-\mathfrak{Q}^{(n,r)});\left(\mathfrak{P}^{(n)}_{i},I-\mathfrak{P}^{(n)}_{i}\right)_{i=1}^{4}\right).$

Since $\mathcal{S}_{n,r}$ is an extension of $\mathcal{S}_{n}$, its
correlation is determined by that of $\mathcal{S}_{n}$ and

$\displaystyle p(1|5)$
$\displaystyle=\mathinner{\langle\phi_{n}|}\mathfrak{Q}^{(n,r)}\otimes
I\mathinner{|\phi_{n}\rangle}=\tau\left(\mathfrak{Q}^{(n,r)}\right),$
$\displaystyle p(1,1|i,5)$
$\displaystyle=\mathinner{\langle\phi_{n}|}\mathfrak{Q}^{(n,r)}\otimes\mathfrak{P}^{(n)}_{j}\mathinner{|\phi_{n}\rangle}=\tau\left(\mathfrak{P}^{(n)}_{i}\mathfrak{Q}^{(n,r)}\right)$

for $i=1,\dots,4$, which are computed in Remark 5.8.

Let $n,r\in\mathbb{N}$ with $r<n$. If $r=\frac{n}{2}$, then a binary
projective measurement of dimension $n$ and rank $r$ is up to a unitary basis
change contained in the self-tested strategy $\mathcal{S}_{n}$. Otherwise, a
binary projective measurement of dimension $n$ and rank $r$ is contained, up
to a unitary basis change and a reordering of outputs, in $\mathcal{S}_{n,r}$
or $\mathcal{S}_{n,n-r}$. For this reason, let us explicitly determine the
correlation of $\mathcal{S}_{n,r}$ only for $r<\frac{n}{2}$. Since
$\mathcal{S}_{n,r}$ is an extension of $\mathcal{S}_{n}$ (whose correlation is
given in Subsection 5.1) and Remark 5.8 computes the additional inner products
(for $r<\frac{n}{2}$), the correlation of $\mathcal{S}_{n,r}$ is determined by
the vector

$\big{(}p(1|j)\big{)}_{j=1}^{5}=\begin{pmatrix}\frac{\lfloor\frac{n}{2}\rfloor-(-1)^{n}}{n}&\frac{\lfloor\frac{n}{2}\rfloor}{n}&\frac{\lfloor\frac{n}{2}\rfloor}{n}&\frac{\lfloor\frac{n}{2}\rfloor}{n}&\frac{r}{n}\end{pmatrix}$

and the $5\times 4$ matrix $\big{(}p(1,1|i,j)\big{)}_{i,j}$

$\left(\begin{array}[]{cccc}\frac{\lfloor\frac{n}{2}\rfloor-(-1)^{n}}{n}&\frac{(n-1)(\lfloor\frac{n}{2}\rfloor-(-1)^{n})}{3n^{2}}&\frac{(n-1)(\lfloor\frac{n}{2}\rfloor-(-1)^{n})}{3n^{2}}&\frac{(n-1)(\lfloor\frac{n}{2}\rfloor-(-1)^{n})}{3n^{2}}\\\\[5.0pt]
\cdot&\frac{\lfloor\frac{n}{2}\rfloor}{n}&\frac{(n-1)(2n-1+3(-1)^{n})}{12n^{2}}&\frac{(n-1)(2n-1+3(-1)^{n})}{12n^{2}}\\\\[5.0pt]
\cdot&\cdot&\frac{\lfloor\frac{n}{2}\rfloor}{n}&\frac{(n-1)(2n-1+3(-1)^{n})}{12n^{2}}\\\\[5.0pt]
\cdot&\cdot&\cdot&\frac{\lfloor\frac{n}{2}\rfloor}{n}\\\\[5.0pt]
\frac{r(4n-2r-1-(-1)^{n})}{4n^{2}}&\frac{r(4n-2r-1-(-1)^{n})}{4n^{2}}&\frac{r(2r-1+(-1)^{n})}{4n^{2}}&\frac{r(2r-1+(-1)^{n})}{4n^{2}}\\\
\end{array}\right)$

where the missing entries are determined by $p(1,1|i,j)=p(1,1|j,i)$ for
$i,j\leq 4$.

######  Theorem 5.10.

The strategy $\mathcal{S}_{n,r}$ is self-tested by its correlation for all
$n,r\in\mathbb{N}$ with $r<n$.

###### Proof.

By Theorem 5.2, the strategy $\mathcal{S}_{n}$ is self-tested by its
correlation. Note that the projection $\mathfrak{Q}^{(n,r)}$ lies in the image
of the span of $\\{\mathfrak{P}_{i}^{(n)}\\}_{i=1}^{4}$ under the map
$\operatorname{sgn}$. Therefore $\mathcal{S}_{n,r}$ is self-tested by its
correlation by Proposition 5.5. ∎

######  Corollary 5.11.

Every local binary projective measurement appears in a 5-input 2-output
strategy that is self-tested by its correlation.

###### Proof.

Every binary PVM is, up to unitary basis change, determined by its dimension
and ranks of its projections. Therefore it suffices to consider measurements
$(\mathfrak{Q}^{(n,r)},I-\mathfrak{Q}^{(n,r)})$, and these appear in the
5-input 2-output strategies $\mathcal{S}_{n,r}$, self-tested by Theorem 5.10.
∎

Finally, we generalize Theorem 5.10 to arbitrary $K$-PVMs. Given
$r_{1},\dots,r_{K},n\in\mathbb{N}$ with $n=r_{1}+\cdots+r_{K}$, Remark 5.7
shows that

$\mathfrak{Q}^{(r_{1},\dots,r_{K})}_{a}:=\mathfrak{Q}^{(n,r_{1}+\cdots+r_{a})}-\mathfrak{Q}^{(n,r_{1}+\cdots+r_{a-1})}$

is a projection of rank $r_{a}$ for every $a=1,\dots,K$, and

$\left(\mathfrak{Q}^{(r_{1},\dots,r_{K})}_{a}\right)_{a=1}^{K}$

is a $K$-PVM. To it we assign a certain bipartite strategy with a mixed number
of inputs and outputs.

######  Definition 5.12.

Let $r_{1},\dots,r_{K},n\in\mathbb{N}$ with $n=r_{1}+\cdots+r_{K}$. We define
a bipartite strategy $\mathcal{S}_{r_{1},\dots,r_{K}}$ that has 4 inputs with
2 outputs and 1 input with $K$ outputs for the first party, and 4 inputs with
2 outputs for the second party:

$\mathcal{S}_{r_{1},\dots,r_{K}}=\left(\mathinner{|\phi_{n}\rangle};\left(\mathfrak{P}^{(n)}_{i},I-\mathfrak{P}^{(n)}_{i}\right)_{i=1}^{4},\left(\mathfrak{Q}^{(r_{1},\dots,r_{K})}_{a}\right)_{a=1}^{K};\left(\mathfrak{P}^{(n)}_{i},I-\mathfrak{P}^{(n)}_{i}\right)_{i=1}^{4}\right).$

As for the correlation of $\mathcal{S}_{n,r}$ from Definition 5.9, one can
derive similar (yet more involved) formulae for the correlation of
$\mathcal{S}_{r_{1},\dots,r_{K}}$ using Remark 5.7, and Propositions 4.4 and
4.5.

######  Corollary 5.13.

Let $r_{1},\dots,r_{K},n\in\mathbb{N}$ with $n=r_{1}+\cdots+r_{K}$ be
arbitrary. Then the strategy $\mathcal{S}_{r_{1},\dots,r_{K}}$ is self-tested
by its correlation.

In particular, every single local $K$-PVM appears in a self-tested strategy
that has 8 inputs with 2 outputs and 1 input with $K$ outputs.

###### Proof.

Let

$\mathcal{S}=\left(\mathinner{|\psi\rangle};\left(P_{i},I-P_{i}\right)_{i=1}^{4},\left(R_{a}\right)_{a=1}^{K};\left(Q_{i},I-Q_{i}\right)_{i=1}^{4}\right)$

be a bipartite strategy with the same correlation as
$\mathcal{S}_{r_{1},\dots,r_{K}}$. Define bipartite strategies that have $3+K$
inputs with 2 outputs for the first party, and 4 inputs with 2 outputs for the
second party:

$\displaystyle\widetilde{\mathcal{S}}$
$\displaystyle=\left(\mathinner{|\phi_{n}\rangle};\left(\mathfrak{P}^{(n)}_{i},I-\mathfrak{P}^{(n)}_{i}\right)_{i=1}^{4},\left(\mathfrak{Q}^{(n,r_{1}+\cdots+r_{a})}_{a},I-\mathfrak{Q}^{(n,r_{1}+\cdots+r_{a})}_{a}\right)_{a=1}^{K-1};\right.$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\quad\left.\left(\mathfrak{P}^{(n)}_{i},I-\mathfrak{P}^{(n)}_{i}\right)_{i=1}^{4}\right),$
$\displaystyle\mathcal{S}^{\prime}$
$\displaystyle=\left(\mathinner{|\psi\rangle};\left(P_{i},I-P_{i}\right)_{i=1}^{4},\left(R_{1}+\cdots+R_{a},I-(R_{1}+\cdots+R_{a})\right)_{a=1}^{K-1};\right.$
$\displaystyle\qquad\qquad\qquad\qquad\quad\
\quad\left.\left(Q_{i},I-Q_{i}\right)_{i=1}^{4}\right).$

Since the projections $\mathfrak{Q}^{(n,r_{1}+\cdots+r_{a})}$ lie in the image
of the span of $\\{\mathfrak{P}_{i}^{(n)}\\}_{i=1}^{4}$ under the map
$\operatorname{sgn}$ by Proposition 5.6, and the strategy $\mathcal{S}_{n}$ is
self-tested by Theorem 5.2, the strategy $\widetilde{\mathcal{S}}$ is self-
tested by a repeated application of Proposition 5.5. Therefore
$\widetilde{\mathcal{S}}$ is a local dilation of $\mathcal{S}^{\prime}$. The
same local isometries and the ancillary state show that
$\mathcal{S}_{r_{1},\dots,r_{K}}$ is a local dilation of $\mathcal{S}$. ∎

##  6 Obstructions to constant-sized self-tests

In a sense, maximally entangled states of all dimensions and single binary
projective measurements of all dimensions and ranks can be self-tested with a
constant number of inputs and outputs because they form discrete families of
objects (i.e., they are parameterized by one and two natural parameters,
respectively). On the other hand, there are no constant-sized self-tests for
all entangled states, nor for all pairs of binary projective measurements, as
implied by the results of this section (for self-tests with varying numbers of
inputs, see [11] and [7]). The local dimension of subsystems in a quantum
strategy is not directly responsible for the absence of constant-sized self-
tests; rather, dimensions of parameter spaces describing states and pairs of
binary projective measurements are the obstructions to existence of uniform
self-tests. The proofs of statements in this section rely on notions from real
algebraic geometry [4].

By the singular value decomposition, every bipartite
$\mathinner{|\psi\rangle}\in\mathbb{C}^{n}\otimes\mathbb{C}^{n}$ is, up to a
left-right unitary basis change, equal to

$\sum_{i=1}^{n}c_{i}\mathinner{|i\rangle}\\!\mathinner{|i\rangle}$

for $c_{i}\geq 0$ and $\sum_{i=1}^{n}c_{i}^{2}=1$. The numbers $c_{i}$ are the
_Schmidt coefficients_ of $\mathinner{|\psi\rangle}$. For example, all the
Schmidt coefficients of $\mathinner{|\phi_{n}\rangle}$ are
$\frac{1}{\sqrt{n}}$. Note that $\mathinner{|\psi\rangle}$ has full Schmidt
rank if and only if $c_{i}>0$ for all $i$.

######  Proposition 6.1.

Let $L,K,N\in\mathbb{N}$ satisfy

$L>(N(K-1)+1)^{2}.$

Then for all $d_{1},\dots,d_{L}\in\mathbb{N}$ there exists a bipartite state
with $L$ distinct Schmidt coefficients of multiplicities $d_{1},\dots,d_{L}$
that cannot be self-tested by $N$-inputs and $K$-outputs.

###### Proof.

Let $\mathbf{A}$ denote the set of all $N$-input $K$-output bipartite quantum
strategies whose states are of the form

$\mathinner{|\psi\rangle}=\sum_{\ell=1}^{L}\lambda_{\ell}\sum_{i=d_{\ell-1}+1}^{d_{\ell}}\mathinner{|i\rangle}\\!\mathinner{|i\rangle},\qquad\lambda_{1}<\cdots<\lambda_{L}$
(9)

where $d_{0}:=0$. In particular, the states in strategies from $\mathbf{A}$
have full Schmidt rank and $L$ distinct Schmidt coefficients of multiplicities
$d_{1},\dots,d_{L}$. Consider the action of
$G:=\operatorname{U}_{d_{1}}(\mathbb{C})\times\cdots\times\operatorname{U}_{d_{L}}(\mathbb{C})$
on $\mathbf{A}$, given by

$U\cdot\big{(}\mathinner{|\psi\rangle};(\mathcal{M}_{i})_{i};(\mathcal{N}_{j})_{j}\big{)}=\Big{(}U\otimes
U\mathinner{|\psi\rangle};(U\mathcal{M}_{i}U^{*})_{i};(U\mathcal{N}_{j}U^{*})_{j}\Big{)}$

for $U=\oplus_{\ell=1}^{L}U_{\ell}\in G$. Note that $G$ encodes precisely all
actions of local unitaries that preserve the form (9) of states in strategies
from $\mathcal{S}$. Let $\mathbf{B}$ be the quotient of $\mathbf{A}$ with
respect to the action of $G$, and let $\pi:\mathbf{A}\to\mathbf{B}$ be the
canonical projection. Given $\mathcal{S}\in\mathbf{A}$ let
$f(\mathcal{S})\in\mathbb{R}^{d_{1}+\cdots+d_{L}}\otimes\mathbb{R}^{d_{1}+\cdots+d_{L}}$
be its state (i.e., $f$ is the projection onto the first component of the
strategy). To
$\mathcal{S}=(\mathinner{|\psi\rangle};(\mathcal{M}_{i})_{i};(\mathcal{N}_{j})_{j})$
we also assign a tuple $g(\mathcal{S})\in\mathbb{R}^{(N(K-1)+1)^{2}-1}$
consisting of

$\displaystyle\mathinner{\langle\psi|}\mathcal{M}_{i,a}\otimes\mathcal{N}_{j,b}\mathinner{|\psi\rangle},\qquad
i,j=1,\dots,N,\ a,b=1,\dots,K-1,$
$\displaystyle\mathinner{\langle\psi|}\mathcal{M}_{i,a}\otimes
I\mathinner{|\psi\rangle},\qquad i=1,\dots,N,\ a=1,\dots,K-1,$
$\displaystyle\mathinner{\langle\psi|}I\otimes\mathcal{N}_{j,b}\mathinner{|\psi\rangle},\qquad
j=1,\dots,N,\ b=1,\dots,K-1.$

Note that $g(\mathcal{S})$ determines the correlation of $\mathcal{S}$. The
set $\mathbf{A}$ is semialgebraic and the maps $f,g$ are semialgebraic [4,
Section 2]. Furthermore, $\mathbf{B}$ is semialgebraic by [4, Proposition
2.2.4] since $G$ is a semialgebraic group. The maps $f,g$ factor through
$\pi$, in the sense that there are semialgebraic maps $f^{\prime},g^{\prime}$
on $\mathbf{B}$ satisfying $f^{\prime}\circ\pi=f$ and $g^{\prime}\circ\pi=g$.
Let $\mathbf{C}\subseteq\mathbf{B}$ be the set of equivalence classes
$[\mathcal{S}]$ such that
$g^{\prime-1}(\\{g^{\prime}([\mathcal{S}])\\})=\\{[\mathcal{S}]\\}$. Then
$\mathbf{C}$ is also semialgebraic by [4, Proposition 2.2.4]. Note that if
$\mathcal{S}\in\mathbf{A}$ is self-tested by its correlation then
$\pi(\mathcal{S})\in\mathbf{C}$. Observe that $\dim
f^{\prime}(\mathbf{B})=L-1$, and $\dim\mathbf{C}=\dim
g^{\prime}(\mathbf{C})\leq(N(K-1)+1)^{2}-1$ by [4, Theorem 2.8.8] since
$g^{\prime}|_{\mathbf{C}}$ is injective. Surjectivity of
$f^{\prime}|_{\mathbf{C}}$ would imply $\dim\mathbf{C}\geq L-1$, contradicting
$L-1>(N(K-1)+1)^{2}-1$. Therefore $f^{\prime}|_{\mathbf{C}}$ is not
surjective. In particular, there exists a state $\mathinner{|\psi\rangle}$ of
the form (9) such that $\pi(\mathcal{S})\notin\mathbf{C}$ for every
$\mathcal{S}\in f^{-1}(\\{\mathinner{|\psi\rangle}\\})$. In particular, no
$N$-input $K$-output strategy containing $\mathinner{|\psi\rangle}$ is self-
tested by its correlation. ∎

By the renowned theorem of Halmos [15], a pair of projections
$P_{1},P_{2}\in\operatorname{M}_{n}(\mathbb{C})$ is, up ot a unitary basis
change, equal to

$\begin{split}P_{1}&=\varepsilon_{1}\oplus\cdots\oplus\varepsilon_{o}\oplus\begin{pmatrix}1&0\\\
0&0\end{pmatrix}\oplus\cdots\oplus\begin{pmatrix}1&0\\\ 0&0\end{pmatrix},\\\
P_{2}&=\varepsilon^{\prime}_{1}\oplus\cdots\oplus\varepsilon^{\prime}_{o}\oplus\begin{pmatrix}\tfrac{1+\cos\alpha_{1}}{2}&\tfrac{\sin\alpha_{1}}{2}\\\
\tfrac{\sin\alpha_{1}}{2}&\tfrac{1-\cos\alpha_{1}}{2}\end{pmatrix}\oplus\cdots\oplus\begin{pmatrix}\tfrac{1+\cos\alpha_{L}}{2}&\tfrac{\sin\alpha_{L}}{2}\\\
\tfrac{\sin\alpha_{L}}{2}&\tfrac{1-\cos\alpha_{L}}{2}\end{pmatrix},\end{split}$
(10)

where $\varepsilon_{i},\varepsilon^{\prime}_{i}\in\\{0,1\\}$ and
$\alpha_{\ell}\in(0,\frac{\pi}{2})$. The number of distinct $2\times 2$ blocks
in (10) equals the number of distinct positive eigenvalues of
$i(P_{1}P_{2}-P_{2}P_{1})$.

######  Proposition 6.2.

Let $L,N\in\mathbb{N}$ satisfy $L+1>(N+1)^{2}$. Then for all
$d_{0},d_{1},\dots,d_{L}\in\mathbb{N}$ there exists a pair of binary
projective measurements $(P_{1},I-P_{1}),(P_{2},I-P_{2})$ with $L$ distinct
$2\times 2$ blocks in (10) with multiplicities $d_{1},\dots,d_{L}$ and $d_{0}$
$1\times 1$ blocks, that cannot be self-tested by $N$-inputs and $2$-outputs.

###### Proof.

We proceed analogously as in the proof of Proposition 6.1. The set
$\mathbf{A}$ consists of $N$-input $2$-output strategies whose first two
measurements are given by projections of the form (10) with $L$ angles
$\alpha_{\ell}$ of multiplicities $d_{1},\dots,d_{L}$. Let
$f:\mathbf{A}\to\operatorname{M}_{d_{0}+2(d_{1}+\cdots+d_{L})}(\mathbb{R})^{2}$
be the projection onto the pair of projections defining the first two
measurements in a strategy. The group $G$ consists of all unitaries preserving
the structure of (10). Then $g,\mathbf{B},\mathbf{C}$ are defined similarly as
in the proof of Proposition 6.1, and the same dimension arguments apply. ∎

##  Appendix A Distinguished projections in low dimensions

As a demonstration of Remark 3.2, we construct
$\mathfrak{P}^{(n)}_{1},\dots\mathfrak{P}^{(n)}_{4}$ for $n\leq 6$.
$n=1$: $(1),(0),(0),(0)$
$n=2$:

$\begin{pmatrix}0&0\\\ 0&0\end{pmatrix},\begin{pmatrix}1&0\\\
0&0\end{pmatrix},\begin{pmatrix}\frac{1}{4}&\frac{-\sqrt{3}}{4}\\\
\frac{-\sqrt{3}}{4}&\frac{3}{4}\\\
\end{pmatrix},\begin{pmatrix}\frac{1}{4}&\frac{\sqrt{3}}{4}\\\
\frac{\sqrt{3}}{4}&\frac{3}{4}\\\ \end{pmatrix}$

$n=3$:

$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 0&0&0\end{pmatrix},\begin{pmatrix}0&0&0\\\
0&\frac{4}{9}&\frac{-2\sqrt{5}}{9}\\\
0&\frac{-2\sqrt{5}}{9}&\frac{5}{9}\end{pmatrix},\begin{pmatrix}\frac{1}{3}&\frac{1}{3\sqrt{3}}&\frac{\sqrt{5}}{3\sqrt{3}}\\\
\frac{1}{3\sqrt{3}}&\frac{1}{9}&\frac{\sqrt{5}}{9}\\\
\frac{\sqrt{5}}{3\sqrt{3}}&\frac{\sqrt{5}}{9}&\frac{5}{9}\end{pmatrix},\begin{pmatrix}\frac{1}{3}&\frac{-1}{3\sqrt{3}}&\frac{-\sqrt{5}}{3\sqrt{3}}\\\
\frac{-1}{3\sqrt{3}}&\frac{1}{9}&\frac{\sqrt{5}}{9}\\\
\frac{-\sqrt{5}}{3\sqrt{3}}&\frac{\sqrt{5}}{9}&\frac{5}{9}\end{pmatrix}$

$n=4$:

$\displaystyle\begin{pmatrix}1&0&0&0\\\ 0&0&0&0\\\ 0&0&0&0\\\
0&0&0&0\end{pmatrix},\begin{pmatrix}\frac{1}{4}&0&\frac{-\sqrt{3}}{4}&0\\\
0&1&0&0\\\ \frac{-\sqrt{3}}{4}&0&\frac{3}{4}&0\\\ 0&0&0&0\end{pmatrix},$
$\displaystyle\begin{pmatrix}\frac{1}{4}&\frac{-\sqrt{15}}{16}&\frac{\sqrt{3}}{8}&\frac{\sqrt{21}}{16}\\\
\frac{-\sqrt{15}}{16}&\frac{3}{8}&\frac{-3\sqrt{5}}{16}&0\\\
\frac{\sqrt{3}}{8}&\frac{-3\sqrt{5}}{16}&\frac{1}{2}&\frac{-\sqrt{7}}{16}\\\
\frac{\sqrt{21}}{16}&0&\frac{-\sqrt{7}}{16}&\frac{7}{8}\end{pmatrix},\begin{pmatrix}\frac{1}{4}&\frac{\sqrt{15}}{16}&\frac{\sqrt{3}}{8}&\frac{-\sqrt{21}}{16}\\\
\frac{\sqrt{15}}{16}&\frac{3}{8}&\frac{3\sqrt{5}}{16}&0\\\
\frac{\sqrt{3}}{8}&\frac{3\sqrt{5}}{16}&\frac{1}{2}&\frac{\sqrt{7}}{16}\\\
\frac{-\sqrt{21}}{16}&0&\frac{\sqrt{7}}{16}&\frac{7}{8}\end{pmatrix}$

$n=5$:

$\displaystyle\begin{pmatrix}1&0&0&0&0\\\ 0&1&0&0&0\\\ 0&0&1&0&0\\\
0&0&0&0&0\\\ 0&0&0&0&0\end{pmatrix},\begin{pmatrix}0&0&0&0&0\\\
0&\frac{4}{25}&0&\frac{-2\sqrt{21}}{25}&0\\\
0&0&\frac{16}{25}&0&\frac{-12}{25}\\\
0&\frac{-2\sqrt{21}}{25}&0&\frac{21}{25}&0\\\
0&0&\frac{-12}{25}&0&\frac{9}{25}\end{pmatrix},$
$\displaystyle\begin{pmatrix}\frac{2}{5}&\frac{3}{5\sqrt{5}}&0&\frac{\sqrt{21}}{5\sqrt{5}}&0\\\
\frac{3}{5\sqrt{5}}&\frac{8}{25}&\frac{\sqrt{7}}{25}&\frac{\sqrt{21}}{25}&\frac{3\sqrt{7}}{25}\\\
0&\frac{\sqrt{7}}{25}&\frac{2}{25}&\frac{-\sqrt{3}}{25}&\frac{6}{25}\\\
\frac{\sqrt{21}}{5\sqrt{5}}&\frac{\sqrt{21}}{25}&\frac{-\sqrt{3}}{25}&\frac{12}{25}&\frac{-3\sqrt{3}}{25}\\\
0&\frac{3\sqrt{7}}{25}&\frac{6}{25}&\frac{-3\sqrt{3}}{25}&\frac{18}{25}\end{pmatrix},\begin{pmatrix}\frac{2}{5}&\frac{-3}{5\sqrt{5}}&0&\frac{-\sqrt{21}}{5\sqrt{5}}&0\\\
\frac{-3}{5\sqrt{5}}&\frac{8}{25}&\frac{-\sqrt{7}}{25}&\frac{\sqrt{21}}{25}&\frac{-3\sqrt{7}}{25}\\\
0&\frac{-\sqrt{7}}{25}&\frac{2}{25}&\frac{\sqrt{3}}{25}&\frac{6}{25}\\\
\frac{-\sqrt{21}}{5\sqrt{5}}&\frac{\sqrt{21}}{25}&\frac{\sqrt{3}}{25}&\frac{12}{25}&\frac{3\sqrt{3}}{25}\\\
0&\frac{-3\sqrt{7}}{25}&\frac{6}{25}&\frac{3\sqrt{3}}{25}&\frac{18}{25}\end{pmatrix}$

$n=6$:

$\displaystyle\begin{pmatrix}1&0&0&0&0&0\\\ 0&1&0&0&0&0\\\ 0&0&0&0&0&0\\\
0&0&0&0&0&0\\\ 0&0&0&0&0&0\\\
0&0&0&0&0&0\end{pmatrix},\begin{pmatrix}\frac{1}{9}&0&0&\frac{-2\sqrt{2}}{9}&0&0\\\
0&\frac{4}{9}&0&0&\frac{-2\sqrt{5}}{9}&0\\\ 0&0&1&0&0&0\\\
\frac{-2\sqrt{2}}{9}&0&0&\frac{8}{9}&0&0\\\
0&\frac{-2\sqrt{5}}{9}&0&0&\frac{5}{9}&0\\\ 0&0&0&0&0&0\end{pmatrix},$
$\displaystyle\begin{pmatrix}\frac{13}{36}&\frac{1}{4\sqrt{3}}&\frac{-\sqrt{\frac{35}{3}}}{12}&\frac{\sqrt{2}}{9}&\frac{\sqrt{\frac{5}{3}}}{4}&0\\\
\frac{1}{4\sqrt{3}}&\frac{7}{36}&0&\frac{-1}{4\sqrt{6}}&\frac{\sqrt{5}}{9}&\frac{\sqrt{\frac{55}{6}}}{12}\\\
\frac{-\sqrt{\frac{35}{3}}}{12}&0&\frac{5}{12}&\frac{-\sqrt{\frac{35}{6}}}{6}&0&0\\\
\frac{\sqrt{2}}{9}&\frac{-1}{4\sqrt{6}}&\frac{-\sqrt{\frac{35}{6}}}{6}&\frac{17}{36}&\frac{-\sqrt{\frac{5}{6}}}{4}&0\\\
\frac{\sqrt{\frac{5}{3}}}{4}&\frac{\sqrt{5}}{9}&0&\frac{-\sqrt{\frac{5}{6}}}{4}&\frac{23}{36}&\frac{-\sqrt{\frac{11}{6}}}{12}\\\
0&\frac{\sqrt{\frac{55}{6}}}{12}&0&0&\frac{-\sqrt{\frac{11}{6}}}{12}&\frac{11}{12}\end{pmatrix},\begin{pmatrix}\frac{13}{36}&\frac{-1}{4\sqrt{3}}&\frac{\sqrt{\frac{35}{3}}}{12}&\frac{\sqrt{2}}{9}&\frac{-\sqrt{\frac{5}{3}}}{4}&0\\\
\frac{-1}{4\sqrt{3}}&\frac{7}{36}&0&\frac{1}{4\sqrt{6}}&\frac{\sqrt{5}}{9}&\frac{-\sqrt{\frac{55}{6}}}{12}\\\
\frac{\sqrt{\frac{35}{3}}}{12}&0&\frac{5}{12}&\frac{\sqrt{\frac{35}{6}}}{6}&0&0\\\
\frac{\sqrt{2}}{9}&\frac{1}{4\sqrt{6}}&\frac{\sqrt{\frac{35}{6}}}{6}&\frac{17}{36}&\frac{\sqrt{\frac{5}{6}}}{4}&0\\\
\frac{-\sqrt{\frac{5}{3}}}{4}&\frac{\sqrt{5}}{9}&0&\frac{\sqrt{\frac{5}{6}}}{4}&\frac{23}{36}&\frac{\sqrt{\frac{11}{6}}}{12}\\\
0&\frac{-\sqrt{\frac{55}{6}}}{12}&0&0&\frac{\sqrt{\frac{11}{6}}}{12}&\frac{11}{12}\end{pmatrix}$

To obtain $\mathfrak{Q}_{n,r}$, one computes
$\mathfrak{Q}_{n,r}=\sum_{i=1}^{r}\mathinner{|e_{i}\rangle}\\!\mathinner{\langle
e_{i}|}$ where $\mathinner{|e_{i}\rangle}$ are unit eigenvectors of
$\mathfrak{P}^{(n)}_{3}+\mathfrak{P}^{(n)}_{4}$ corresponding to the $r$
smallest eigenvalues in increasing order. Examples for $r<n\leq 5$ are given
below.
$n=2$, $r=1$:

$\begin{pmatrix}1&0\\\ 0&0\end{pmatrix}$

$n=3$, $r=1,2$:

$\begin{pmatrix}0&0&0\\\ 0&\frac{5}{6}&\frac{-\sqrt{5}}{6}\\\
0&\frac{-\sqrt{5}}{6}&\frac{1}{6}\end{pmatrix},\begin{pmatrix}1&0&0\\\
0&\frac{5}{6}&\frac{-\sqrt{5}}{6}\\\
0&\frac{-\sqrt{5}}{6}&\frac{1}{6}\end{pmatrix}$

$n=4$, $r=1,2,3$:

$\begin{pmatrix}\frac{3}{4}&0&\frac{-\sqrt{3}}{4}&0\\\ 0&0&0&0\\\
\frac{-\sqrt{3}}{4}&0&\frac{1}{4}&0\\\
0&0&0&0\end{pmatrix},\begin{pmatrix}\frac{3}{4}&0&\frac{-\sqrt{3}}{4}&0\\\
0&1&0&0\\\ \frac{-\sqrt{3}}{4}&0&\frac{1}{4}&0\\\
0&0&0&0\end{pmatrix},\begin{pmatrix}1&0&0&0\\\ 0&1&0&0\\\ 0&0&1&0\\\
0&0&0&0\end{pmatrix}$

$n=5$, $r=1,2,3,4$:

$\displaystyle\begin{pmatrix}0&0&0&0&0\\\ 0&0&0&0&0\\\
0&0&\frac{9}{10}&0&\frac{-3}{10}\\\ 0&0&0&0&0\\\
0&0&\frac{-3}{10}&0&\frac{1}{10}\end{pmatrix},\begin{pmatrix}0&0&0&0&0\\\
0&\frac{7}{10}&0&\frac{-\sqrt{21}}{10}&0\\\
0&0&\frac{9}{10}&0&\frac{-3}{10}\\\
0&\frac{-\sqrt{21}}{10}&0&\frac{3}{10}&0\\\
0&0&\frac{-3}{10}&0&\frac{1}{10}\end{pmatrix},$
$\displaystyle\begin{pmatrix}1&0&0&0&0\\\
0&\frac{7}{10}&0&\frac{-\sqrt{21}}{10}&0\\\
0&0&\frac{9}{10}&0&\frac{-3}{10}\\\
0&\frac{-\sqrt{21}}{10}&0&\frac{3}{10}&0\\\
0&0&\frac{-3}{10}&0&\frac{1}{10}\end{pmatrix},\begin{pmatrix}1&0&0&0&0\\\
0&1&0&0&0\\\ 0&0&\frac{9}{10}&0&\frac{-3}{10}\\\ 0&0&0&1&0\\\
0&0&\frac{-3}{10}&0&\frac{1}{10}\end{pmatrix}$

$n=6$, $r=1,2,3,4,5$:

$\displaystyle\begin{pmatrix}0&0&0&0&0&0\\\
0&\frac{5}{6}&0&0&\frac{-\sqrt{5}}{6}&0\\\ 0&0&0&0&0&0\\\ 0&0&0&0&0&0\\\
0&\frac{-\sqrt{5}}{6}&0&0&\frac{1}{6}&0\\\
0&0&0&0&0&0\end{pmatrix},\begin{pmatrix}\frac{2}{3}&0&0&\frac{-\sqrt{2}}{3}&0&0\\\
0&\frac{5}{6}&0&0&\frac{-\sqrt{5}}{6}&0\\\ 0&0&0&0&0&0\\\
\frac{-\sqrt{2}}{3}&0&0&\frac{1}{3}&0&0\\\
0&\frac{-\sqrt{5}}{6}&0&0&\frac{1}{6}&0\\\ 0&0&0&0&0&0\end{pmatrix},$
$\displaystyle\begin{pmatrix}\frac{2}{3}&0&0&\frac{-\sqrt{2}}{3}&0&0\\\
0&\frac{5}{6}&0&0&\frac{-\sqrt{5}}{6}&0\\\ 0&0&1&0&0&0\\\
\frac{-\sqrt{2}}{3}&0&0&\frac{1}{3}&0&0\\\
0&\frac{-\sqrt{5}}{6}&0&0&\frac{1}{6}&0\\\
0&0&0&0&0&0\end{pmatrix},\begin{pmatrix}1&0&0&0&0&0\\\
0&\frac{5}{6}&0&0&\frac{-\sqrt{5}}{6}&0\\\ 0&0&1&0&0&0\\\ 0&0&0&1&0&0\\\
0&\frac{-\sqrt{5}}{6}&0&0&\frac{1}{6}&0\\\
0&0&0&0&0&0\end{pmatrix},\begin{pmatrix}1&0&0&0&0&0\\\ 0&1&0&0&0&0\\\
0&0&1&0&0&0\\\ 0&0&0&1&0&0\\\ 0&0&0&0&1&0\\\ 0&0&0&0&0&0\end{pmatrix}$

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