# Characterizing the geometry of the Kirkwood-Dirac positive states

C. Langrenez1, D.R.M. Arvidsson-Shukur2, S. De Bièvre1
$\,{}^{1}$Univ. Lille, CNRS, Inria, UMR 8524, Laboratoire Paul Painlevé,
F-59000 Lille, France
$\,{}^{2}$ Hitachi Cambridge Lab., J.J. Thomson Avenue, Cambridge CB3 0HE, UK
christopher.langrenez@univ-lille.frdrma2@cam.ac.ukstephan.de-bievre@univ-
lille.fr

###### Abstract

The Kirkwood-Dirac (KD) quasiprobability distribution can describe any quantum
state with respect to the eigenbases of two observables $A$ and $B$. KD
distributions behave similarly to classical joint probability distributions
but can assume negative and nonreal values. In recent years, KD distributions
have proven instrumental in mapping out nonclassical phenomena and quantum
advantages. These quantum features have been connected to nonpositive entries
of KD distributions. Consequently, it is important to understand the geometry
of the KD-positive and -nonpositive states. Until now, there has been no
thorough analysis of the KD positivity of mixed states. Here, we characterize
how the full convex set of states with positive KD distributions depends on
the eigenbases of $A$ and $B$. In particular, we identify three regimes where
convex combinations of the eigenprojectors of $A$ and $B$ constitute the only
KD-positive states: $(i)$ any system in dimension $2$; $(ii)$ an open and
dense set of bases in dimension $3$; and $(iii)$ the discrete-Fourier-
transform bases in prime dimension. Finally, we investigate if there can exist
mixed KD-positive states that cannot be written as convex combinations of pure
KD-positive states. We show that for some choices of observables $A$ and $B$
this phenomenon does indeed occur. We explicitly construct such states for a
spin-$1$ system.

##  1 Introduction

In classical mechanics, a joint probability distribution
$\mathcal{P}(\bf{x},\bf{p})$ can describe a system with respect to two
observables, such as position $\bf{x}$ and momentum $\bf{p}$. In quantum
mechanics, however, observables generally do not commute and probabilistic
descriptions of states with respect to more than one observable are often not
available [1, 2, 3, 4, 5, 6]. Nevertheless, one can describe a quantum state
with respect to two joint observables via a quasiprobability distribution.
Quasiprobability distributions obey all but one of Kolmogorov’s axioms for
probability distributions [7]: their entries sum to unity; their marginals
correspond to the probability distributions given by the Born Rule; but
individual quasiprobabilities can take negative or nonreal values.

The quasiprobability formalism provides a useful alternative to other
descriptions of quantum states. The most famous quasiprobability distribution
is the Wigner function. It deals with continuous-variable systems with clear
analogues of position and momentum. Most notably, the Wigner function has
played a pivotal role in the analyses of quantum states of light [8, 9, 10,
11]. The Wigner function, and other quasiprobability distributions [12, 13,
14, 15, 16, 17, 18], allow techniques from statistics and probability theory
to be applied to quantum mechanics.

Most modern quantum-information research is phrased in terms of finite-
dimensional systems—often systems of qubits. Moreover, the observables of
interest are, unlike position and momentum, not necessarily conjugate. The
Wigner function is ill-suited for such systems and observables. Instead,
recent years have seen a different quasiprobability distribution come to the
foreground: the Kirkwood-Dirac (KD) distribution [19, 20, 21, 22, 23, 24].

The KD distribution has proven itself a tremendously versatile tool in
studying and developing quantum-information processing. In its standard form,
the KD distribution describes a quantum state $\rho$ with respect to two
orthonormal bases $\left(|a_{i}\rangle\right)_{i\in\llbracket 1,d\rrbracket}$
and $\left(|b_{j}\rangle\right)_{j\in\llbracket 1,d\rrbracket}$ in a complex
Hilbert space $\mathcal{H}$ of dimension $d$. The KD distribution reads

$\forall(i,j)\in\llbracket 1,d\rrbracket^{2},\;Q_{ij}(\rho)=\langle
b_{j}|a_{i}\rangle\langle a_{i}|\rho|b_{j}\rangle.$ (1.1)

By associating the two bases with the eigenstates of observables of interest,
the KD distribution can be tuned towards a specific problem. So far, the KD
distribution has been used to study, describe or develop: direct state
tomography [17, 25, 26, 27, 28]; quantum metrology [29, 30, 31]; quantum chaos
[32, 21, 33, 34, 35]; weak measurements [36, 37, 38, 39, 40, 21, 41, 42, 43];
quantum thermodynamics [32, 44, 45, 46, 47, 48]; quantum scrambling [21, 33];
Leggett-Garg inequalities [49, 50, 51]; generalised contextuality [39, 45,
46]; consistent-histories interpretations of quantum mechanics [52];
measurement disturbance [53, 40, 54, 23, 24, 55, 56]; and coherence [57]. The
list can be made longer, but the point is clear: the Kirkwood-Dirac
distribution currently experiences great prosperity and growing interest.

Below, a state will be said to be KD positive when its KD distribution only
takes on positive or zero values. Such states have been called KD classical
elsewhere [22, 23, 24]. We prefer to avoid this terminology since the terms
“classical” and “nonclassical” lack unique definitions. The capacity of
quasiprobability distributions to describe quantum phenomena hinges on their
ability to assume negative or nonreal values. An always-positive (probability)
distribution cannot describe all of quantum mechanics. As concerns the KD
distribution, nonpositive quasiprobabilities have been linked to various forms
of quantum advantages in, for example, weak measurements [39, 41], quantum
metrology [29, 30] and quantum thermodynamics [21, 45, 46]. Therefore, it is
important to understand: When does a KD distribution assume only positive or
zero values? While this question has been addressed for pure states [22, 23,
24], a general study of the mixed KD-positive states is lacking. In this work,
we provide such a study.

To analyse how the KD distribution underlies nonclassical phenomena, one must
first understand the geometric structure of the convex set
${\mathcal{E}}_{\mathrm{KD+}}$ of KD-positive states. We know, by the Krein-
Milman theorem [58], that this set is the convex hull of the set
${\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}$ of its extreme points:

${\mathcal{E}}_{\mathrm{KD+}}=\mathrm{conv}\left({\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}\right).$

It is, therefore, desirable to have a full description and a convenient
characterization of ${\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}$. The set
${\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}$ always contains all the basis
states $\left(|a_{i}\rangle\right)_{i\in\llbracket 1,d\rrbracket}$ and
$\left(|b_{j}\rangle\right)_{j\in\llbracket 1,d\rrbracket}$. Additionally,
${\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}$ may contain other pure and also
mixed states. Experience with similar analyses for the Wigner function, where
the mixed-state characterization of Wigner positive states is not fully
solved, indicates that it might be difficult to obtain a full characterization
of ${\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}$ for general KD distributions.

Our results about the convex set ${\mathcal{E}}_{\mathrm{KD+}}$ of KD-positive
states can be summed up as follows. We first identify for what choices of the
bases $\left(|a_{i}\rangle\right)_{i\in\llbracket 1,d\rrbracket}$ and
$\left(|b_{j}\rangle\right)_{j\in\llbracket 1,d\rrbracket}$ the only KD-
positive states are those that are convex mixtures of the basis states. The
following theorem provides a precise statement of these results. We introduce

$\displaystyle{\mathcal{A}}=\\{|a_{i}\rangle\langle a_{i}|\mid i\in\llbracket
1,d\rrbracket\\},\quad{\mathcal{B}}=\\{|b_{j}\rangle\langle b_{j}|\mid
j\in\llbracket 1,d\rrbracket\\},$

which are the families of rank-one projectors associated to the two bases.
Also, we write $U_{ij}=\langle a_{i}|b_{j}\rangle$ for the transition matrix
between the two bases and introduce

$m_{{\mathcal{A}},{\mathcal{B}}}=\min_{i,j}|\langle a_{i}|b_{j}\rangle|.$

######  Theorem 1.1.

The equality

${\mathcal{E}}_{\mathrm{KD+}}=\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right),$
(1.2)

holds under any single one of the following hypotheses:

  1. (i)

If $d=2$ (for qubits) and $m_{{\mathcal{A}},{\mathcal{B}}}>0$;

  2. (ii)

If $d=3$, for all $U$ in a open dense set of probability $1$;

  3. (iii)

If $d$ is prime and $U$ is the discrete Fourier transform (DFT) matrix;

  4. (iv)

If $U$ is sufficiently close to some other $U^{\prime}$ for which Eq. (1.2)
holds.

Note that Eq. (1.2) is equivalent to

${\mathcal{A}}\cup{\mathcal{B}}={\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}={\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}},$
(1.3)

where ${\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}$ denotes the set of pure
KD-positive states. For general ${\mathcal{A}}$ and ${\mathcal{B}}$, one has

${\mathcal{A}}\cup{\mathcal{B}}\subseteq{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}\subseteq{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}.$
(1.4)

In other words, Eq. (1.2) corresponds to the simplest situation, where the set
of extreme states is minimal. In that case, we have an explicit description of
the set of all KD-positive states since the convex set
$\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$ forms a polytope
with a simple geometric structure, detailed in Appendix A. Note that part (iv)
of the theorem guarantees that the property Eq. (1.2) is stable in the sense
that it is verified in an open set of unitary matrices. We conjecture that
part (ii) of the theorem in fact holds in all dimensions. In other words, we
think that the simple structure obtained in Eq. (1.2) is “typically” realized,
meaning that it holds in an open dense set of full measure. We have
numerically checked this conjecture by randomly choosing unitary matrices $U$
for dimensions $d$ up to $10$ (See Section 4 for details). The following
proposition, proven in Section 4, shows a partial result in this direction:

######  Proposition 1.2.

Let $d\geq 2$. There exists an open dense set of unitaries of probability $1$
for which
${\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}={\mathcal{A}}\cup{\mathcal{B}}$.

We stress that we do nevertheless not know if, for the unitaries referred to
in the proposition, the stronger property
${\mathcal{E}}_{\mathrm{KD+}}=\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$
holds.

In general, it is a formidable task to identify
${\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}$ and
${\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}$, given two specific bases
${\mathcal{A}}$ and ${\mathcal{B}}$. Part (iii) of the theorem shows that,
when $U$ is the DFT matrix, and the dimension $d$ is a prime number, one again
satisfies Eq. (1.3). When $d$ is prime and the columns of $U$ form two
mutually unbiased (MUB) bases with the canonical basis, it is still true that
${\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}={\mathcal{A}}\cup{\mathcal{B}}$
(See [59] and Appendix C). But in that case, we have no information about the
possible existence of mixed extreme states. When the dimension is not prime,
and $U$ the DFT matrix, one can identify all pure KD-positive states [23, 59,
24] and one observes that there exist pure KD-positive states that are not
basis states, _i.e._
${\mathcal{A}}\cup{\mathcal{B}}\subsetneq{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}$.
It is again, to the best of our knowledge, not known in that case if there
also exist extreme KD-positive states that are mixed, meaning if
${\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}\subsetneq{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}$.

By analyzing in detail the situation where the transition matrix is real-
valued, we provide below (Section 5) examples for which
${\mathcal{A}}\cup{\mathcal{B}}\subsetneq{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}\subsetneq{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}$
or
${\mathcal{A}}\cup{\mathcal{B}}={\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}\subsetneq{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}.$
In these cases, there therefore exist mixed extreme states, some of which we
explicitly identify. We will highlight such situations with examples where the
bases ${\mathcal{A}}$ and ${\mathcal{B}}$ are the eigenbases of two spin-$1$
components in some particular directions. While this situation is in a sense
exceptional, it has a precise analogue in the analysis of Wigner function.
Indeed, the pure Wigner positive states are known to be the Gaussian states
[2]. But it is also well known that the convex hull of the pure Gaussian
states (which contains all Gaussian states) does not exhaust all Wigner
positive states [60]. As it turns out, even though examples of Wigner positive
states not in this convex hull have been constructed [60, 61, 62], a complete
description of the extreme states of the set of all Wigner positive states is,
to the best of our knowledge, not available. In fact, no mixed extreme states
have been explicitly identified for the Wigner function [63].

The remainder of this paper is structured as follows. In Section 2, we
describe the general framework of our investigation, recall some definitions
and necessary background information, and introduce our notation. In Section
3, we prove several results on the general structure of the geometry of the
set of KD-positive states. These results are essential ingredients for the
proofs of our main results. In Section 4, we prove Theorem 1.1 and Proposition
1.2. In Section 5, we focus on real unitary matrices to construct examples of
mixed states that are KD positive but cannot be written as convex combinations
of pure KD-positve states. Section 6 contains our conclusions and outlook.

##  2 The setting and background

In this section, we introduce some notation, define KD distributions and
recall some of their properties.

Throughout this manuscript, we consider a complex Hilbert space $\mathcal{H}$
of dimension $d$. We consider also two orthonormal bases
$\left(|a_{i}\rangle\right)_{i\in\llbracket 1,d\rrbracket}$ and
$\left(|b_{j}\rangle\right)_{j\in\llbracket 1,d\rrbracket}$ in $\mathcal{H}$.
We denote by $U=\left(\langle a_{i}|b_{j}\rangle\right)_{(i,j)\in\llbracket
1,d\rrbracket^{2}}$ the transition matrix between these two bases. If $\rho$
is a density matrix, we define the KD distribution $Q(\rho)$ to be the
$d\times d$ matrix [19, 20]

$\forall(i,j)\in\llbracket 1,d\rrbracket^{2},Q_{ij}(\rho)=\langle
b_{j}|a_{i}\rangle\langle a_{i}|\rho|b_{j}\rangle.$ (2.1)

Note that, for a given $\rho$, the matrix $Q(\rho)$ depends on the two bases.
Although this will be crucial for our developments below, we do not indicate
this dependence, not to burden the notation. The KD distribution thus
satisfies the following of Kolmogorov’s axioms for joint probability
distributions:

$\sum_{i,j}Q_{ij}(\rho)=\mathrm{Tr}\,\rho=1,\quad\sum_{j}Q_{ij}(\rho)=\langle
a_{i}|\rho|a_{i}\rangle,\quad\sum_{i}Q_{ij}(\rho)=\langle
b_{j}|\rho|b_{j}\rangle.$ (2.2)

However, unlike joint probabilities, $Q(\rho)$ is in general a complex-valued
matrix. We call a state KD positive whenever $Q_{ij}(\rho)\geq 0$ for all
$(i,j)\in\llbracket 1,d\rrbracket^{2}$. The transition matrix $U$ is
determined by the choice of bases and determines whether $\rho$ is KD
positive. For example, if $U=\mathbb{I}_{d}$, the bases are identical. Then,
clearly all states are KD positive.

We will say that two bases, $(|a_{i}\rangle)_{i}$ and
$(|a_{i}^{\prime}\rangle)_{i}$ or $(|b_{j}\rangle)_{j}$ and
$(|b_{j}^{\prime}\rangle)_{j}$, are equivalent if they can be obtained from
each other by permutations of the basis vectors and/or phase rotations. In
that case, the matrices $U$ and $U^{\prime}$ are obtained from one another by
permutations of their columns and rows, and global phase rotations of the rows
and columns. We shall say such matrices are equivalent. The point of these
definitions is that, when the bases (and hence the transition matrices) are
equivalent, then the corresponding sets of KD-positive states are identical.
In particular, we note for later use that, if
$\phi=(\phi_{1},\dots,\phi_{d})\in[0,2\pi)^{d}$,
$\psi=(\psi_{1},\dots,\psi_{d})\in[0,2\pi)^{d}$, and if

$|a_{j}^{\prime}\rangle=\exp(-i\phi_{j})|a_{j}\rangle,\quad|b_{j}^{\prime}\rangle=\exp(-i\psi_{j})|b_{j}\rangle,$

then the transition matrix $U^{\prime}_{ij}=\langle
a_{i}^{\prime}|b_{j}^{\prime}\rangle$ is given by

$U^{\prime}=D(-\phi)UD(\psi),$ (2.3)

where, for any $\phi=(\phi_{1},\dots,\phi_{d})\in[0,2\pi)^{d}$,

$D(\phi)_{jk}=\exp(-i\phi_{j})\delta_{jk}.$

Let us point out that we will often identify a unit vector
$|\psi\rangle\in{\mathcal{H}}$ with its projector $|\psi\rangle\langle\psi|$.

The questions we address in this work are of interest only if the two bases
are in a suitable sense incompatible. For most of this work we will therefore
assume that the unitary matrix $U$ has no zeros:

$m_{{\mathcal{A}},{\mathcal{B}}}=\min_{i,j}|\langle a_{i}|b_{j}\rangle|>0.$
(2.4)

This guarantees that $Q(\rho)$ determines a unique $\rho$ (see Eq. (3.3)). In
addition, it implies that none of the $|a_{i}\rangle\langle a_{i}|$ commutes
with any of the $|b_{j}\rangle\langle b_{j}|$. This is a weak form of
incompatibility between the two bases [24]. Indeed,
$m_{{\mathcal{A}},{\mathcal{B}}}>0$ means that if a measurement in the
${\mathcal{A}}$ basis yields an outcome $i$, then a subsequent measurement in
the ${\mathcal{B}}$ basis may yield any outcome $j$ with a nonvanishing
probability. We recall that a special role is played by mutually unbiased
(MUB) bases, for which $m_{{\mathcal{A}},{\mathcal{B}}}$ takes the maximum
possible value $m_{{\mathcal{A}},{\mathcal{B}}}=\frac{1}{\sqrt{d}}$. All
outcomes $j$ for a ${\mathcal{B}}$-measurement after an initial measurement in
the ${\mathcal{A}}$-basis are then equally probable, and vice versa.

##  3 General structural results

In this section, we prove general results regarding the geometry of KD-
positive states. We work under the assumption that
$m_{{\mathcal{A}},{\mathcal{B}}}>0$.

###  3.1 The KD symbol of an observable

It is well known that the Wigner function can be defined not only for states
$\rho$, but also for arbitrary observables $F$, in which case it is referred
to as the Weyl symbol of $F$. One can proceed similarly with the KD
distribution. Denoting by $\mathcal{S}_{d}$ the set of self-adjoint operators,
we define,

$Q:\left\\{\begin{array}[]{rcl}\mathcal{S}_{d}&\rightarrow&\mathcal{M}_{d}(\mathbb{C})\\\
F&\mapsto&(Q_{ij}(F))_{(i,j)\in\llbracket
1,d\rrbracket^{2}}\end{array}\right.,$ (3.1)

where

$\forall(i,j)\in\llbracket 1,d\rrbracket^{2},\ Q_{ij}(F)=\langle
a_{i}|F|b_{j}\rangle\langle b_{j}|a_{i}\rangle,$

and where $\mathcal{M}_{d}(\mathbb{C})$ is the the space of complex $d$ by $d$
matrices. We shall refer to $Q(F)$ as the KD symbol of $F$. We note that

$\sum_{j=1}^{d}Q_{ij}(F)=\langle
a_{i}|F|a_{i}\rangle\in\mathbb{R},\quad\sum_{i=1}^{d}Q_{ij}(F)=\langle
b_{j}|F|b_{j}\rangle\in\mathbb{R},\quad\sum_{i,j}Q_{ij}(F)=\mathrm{Tr}\,(F).$
(3.2)

Also, for $F,G\in\mathcal{S}_{d}$, we have

$\begin{array}[]{rcl}\mathrm{Tr}(FG)&=&\sum_{(i,j)\in\llbracket
1,d\rrbracket^{2}}\langle a_{i}|F|b_{j}\rangle\langle b_{j}|G|a_{i}\rangle\\\
&=&\displaystyle\sum_{(i,j)\in\llbracket
1,d\rrbracket^{2}}\frac{1}{\left|\langle a_{i}|b_{j}\rangle\right|^{2}}\langle
b_{j}|a_{i}\rangle\langle a_{i}|F|b_{j}\rangle\langle
a_{i}|b_{j}\rangle\langle b_{j}|G|a_{i}\rangle\\\
&=&\displaystyle\sum_{(i,j)\in\llbracket
1,d\rrbracket^{2}}\frac{1}{\left|\langle
a_{i}|b_{j}\rangle\right|^{2}}Q_{ij}(F)\overline{Q_{ij}(G)}.\end{array}$

If ${\mathcal{A}}$ and ${\mathcal{B}}$ are MUB bases, then

$\mathrm{Tr}(FG)=d\sum_{(i,j)\in\llbracket
1,d\rrbracket^{2}}Q_{ij}(F)\overline{Q_{ij}(G)}=d\mathrm{Tr}(Q(F)Q^{\dagger}(G)).$

One may note the analogy between these two identities and the well known
“overlap identity” for the Wigner function/Weyl symbol which expresses
$\mathrm{Tr}\,(FG)$ as a phase space integral of the product of the Wigner
function/Weyl symbol of $F$ and $G$ [10].

We point out that, when $m_{{\mathcal{A}},{\mathcal{B}}}>0$, the KD symbol
$Q(F)$ determines the observable $F$ uniquely. The reconstruction formula is
[64]

$\forall(i,k)\in\llbracket 1,d\rrbracket^{2},\ \langle
a_{i}|F|a_{k}\rangle=\sum_{j=1}^{d}Q_{ij}(F)\frac{\langle
b_{j}|a_{k}\rangle}{\langle b_{j}|a_{i}\rangle}.$ (3.3)

This property is sometimes referred to as informational completeness. In other
words, the map $Q$ is injective: ${\mathrm{Ker}}Q=\\{0\\}$ if
$m_{{\mathcal{A}},{\mathcal{B}}}>0$, where ${\mathrm{Ker}}Q$ denotes the
kernel of $Q$.

Since the dimension of the real vector space $\mathcal{S}_{d}$ is $d^{2}$, it
follows that $\dim\mathrm{Ran}\left(Q\right)=d^{2},$ where
$\mathrm{Ran}\left(Q\right)$ denotes the image of $Q$. Hence,
$\mathrm{Ran}\left(Q\right)$ is a $d^{2}$-dimensional real vector subspace of
the $2d^{2}$-dimensional real vector space $\mathcal{M}_{d}(\mathbb{C})$. Note
that a matrix
$M\in\mathcal{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}missing}}{\mathcal{M}_{d}(\mathbb{C})}$
belongs to $\mathrm{Ran}\left(Q\right)$ if and only if it satisfies the
$d^{2}$ real linear constraints

$\sum_{j}M_{kj}\frac{\langle b_{j}|a_{i}\rangle}{\langle
b_{j}|a_{k}\rangle}=\sum_{j}\overline{M}_{ij}\frac{\langle
a_{k}|b_{j}\rangle}{\langle a_{i}|b_{j}\rangle}.$

We will further find it useful to consider the imaginary part of $Q$:

$\mathrm{Im}Q:\left\\{\begin{array}[]{rcl}\mathcal{S}_{d}&\rightarrow&\mathcal{M}_{d}(\mathbb{R})\\\
F&\mapsto&(\mathrm{Im}Q_{ij}(F))_{(i,j)\in\llbracket
1,d\rrbracket^{2}}\end{array}\right.,$

which is a real-linear map into the space of real matrices
$\mathcal{M}_{d}(\mathbb{R})$.

To streamline the discussion, we introduce the following terminology. We will
say that a self-adjoint operator $F$ is a _KD-real operator_ whenever its KD
distribution is real-valued. In other words, $F$ is KD real if and only if

$F\in
V_{\mathrm{KDr}}:={\mathrm{Ker}}\,{\mathrm{Im}}Q=Q^{-1}(\mathcal{M}_{d}(\mathbb{R})).$

We will say it is _KD positive_ if its KD distribution takes on real
nonnegative values only. In other words, iff

$F\in V_{\mathrm{KD+}}:=Q^{-1}(\mathcal{M}_{d}(\mathbb{R}^{+}))\subseteq
V_{\mathrm{KDr}}.$

We point out for later use that, since
$V_{\mathrm{KDr}}={\mathrm{Ker}}(\mathrm{Im}Q)\subset\mathcal{S}_{d}$ and
$\dim\mathcal{S}_{d}=d^{2}$,

$\dim V_{\mathrm{KDr}}\leq d^{2}.$ (3.4)

Clearly, if $F_{1},F_{2}\in V_{\mathrm{KD+}}$, then
$\lambda_{1}F_{1}+\lambda_{2}F_{2}\in V_{\mathrm{KD+}}$ for all
$\lambda_{1},\lambda_{2}\geq 0$. In particular, $V_{\mathrm{KD+}}$ is a closed
convex cone. Note that it has no extreme points, except for the origin.

###  3.2 The case
${\mathcal{E}}_{\mathrm{KD+}}=\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$:
a geometric condition

Recall that a density matrix representing a quantum state is a nonnegative
operator $\rho$ satisfying $\mathrm{Tr}\rho=1$. We will write
$\mathcal{S}_{d,+,1}$ for the set of density matrices and $\mathcal{S}_{d,+}$
for the set of positive operators. Hence

${\mathcal{E}}_{\mathrm{KD+}}=V_{\mathrm{KD+}}\cap\mathcal{S}_{d,+,1}.$ (3.5)

Note that ${\mathcal{E}}_{\mathrm{KD+}}$ is compact so that, by the Krein-
Milman theorem,
${\mathcal{E}}_{\mathrm{KD+}}=\mathrm{conv}\left({\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}\right).$

The question we are addressing in this section is under which conditions on
$U$ it is true that

${\mathcal{E}}_{\mathrm{KD+}}=\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right),$
(3.6)

where

$\displaystyle\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$
$\displaystyle=$ $\displaystyle\mathrm{conv}\left(\\{|a_{i}\rangle\langle
a_{i}|,|b_{j}\rangle\langle b_{j}|\mid 1\leq i,j\leq d\\}\right).$ (3.7)

In other words, the question is: Is it true or false that all KD-positive
states are convex mixtures of the basis states? This is equivalent to checking
if the inclusions in Eq. (1.4) are equalities, i.e., if

${\mathcal{A}}\cup{\mathcal{B}}={\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}={\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}.$
(3.8)

One can think of Eq. (3.8) as the situation where the set
${\mathcal{E}}_{\mathrm{KD+}}$ of KD-positive states is the smallest possible.
In a sense then, this corresponds to the choice of two bases ${\mathcal{A}}$
and ${\mathcal{B}}$ that are “most strongly quantum.” Note that, when Eq.
(3.8) holds true, ${\mathcal{E}}_{\mathrm{KD+}}$ is a convex polytope with 2d
known summits $\\{|a_{i}\rangle\langle a_{i}|,|b_{i}\rangle\langle
b_{i}|\\}_{i\in\llbracket 1,d\rrbracket}$. Its geometry is described in
Appendix A.

In Proposition 3.2 we will prove conditions of a geometric nature on the set
of KD-real operators that are equivalent to Eq. (3.6).

We introduce the vector space

$\displaystyle{\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$
$\displaystyle=$
$\displaystyle\mathrm{span}_{\mathbb{R}}\\{|a_{i}\rangle\langle
a_{i}|,|b_{j}\rangle\langle b_{j}|\mid 1\leq i,j\leq d\\}.$ (3.9)

and show the following result:

######  Lemma 3.1.

If $m_{{\mathcal{A}},{\mathcal{B}}}>0$, then

$\mathrm{dim}\left({\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}\right)=2d-1\quad\mathrm{and}\quad{\mathcal{E}}_{\mathrm{KD+}}\cap{\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}=\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right).$

###### Proof.

To prove the first statement, we consider the linear map

$\Gamma:\left\\{\begin{array}[]{rcl}\mathbb{R}^{2d}&\rightarrow&{\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}\\\
((\lambda_{i})_{i\in\llbracket 1,d\rrbracket},(\mu_{j})_{j\in\llbracket
1,d\rrbracket})&\mapsto&\sum_{i=1}^{d}\lambda_{i}|a_{i}\rangle\langle
a_{i}|+\sum_{j=1}^{d}\mu_{j}|b_{j}\rangle\langle b_{j}|\end{array}\right.$

for which the rank theorem gives
$\text{dim}\left({\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}\right)=2d-\text{dim}\left({\mathrm{Ker}}(\Gamma)\right)$.
Now suppose that $F=\sum_{i=1}^{d}\lambda_{i}|a_{i}\rangle\langle
a_{i}|+\sum_{j=1}^{d}\mu_{j}|b_{j}\rangle\langle b_{j}|=0$. Then

$\forall(i,j)\in\llbracket 1,d\rrbracket^{2},\langle
a_{i}|F|b_{j}\rangle=\langle
a_{i}|b_{j}\rangle\left(\lambda_{i}+\mu_{j}\right)=0.$

Now, for any $i\in\llbracket 1,d\rrbracket$, $\langle a_{i}|b_{1}\rangle\neq
0$ because $m_{{\mathcal{A}},{\mathcal{B}}}>0$, hence $\lambda_{i}+\mu_{1}=0$
and finally $\lambda_{i}=-\mu_{1}$ for all $i\in\llbracket 1,d\rrbracket$.
Exchanging the roles $(\lambda_{i})_{i\in\llbracket 1,d\rrbracket}$ and
$(\mu_{j})_{j\in\llbracket 1,d\rrbracket}$, we find that for all
$j\in\llbracket 1,d\rrbracket$, $\mu_{j}=\mu_{1}$. So, the relation stands as

$\mu_{1}\left(-\sum_{i=1}^{d}|a_{i}\rangle\langle
a_{i}|+\sum_{j=1}^{d}|b_{j}\rangle\langle b_{j}|\right)=0,$

which is true for all $\mu_{1}\in\mathbb{R}$. This means that
$\text{dim}\left({\mathrm{Ker}}(\Gamma)\right)=1$ and so

$\text{dim}\left({\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}\right)=2d-1.$

We now turn to the second statement. That
$\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)\subseteq{\mathcal{E}}_{\mathrm{KD+}}\cap{\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$
is immediate. Thus, we only need to prove the other inclusion. Let therefore
$\rho\in{\mathcal{E}}_{\mathrm{KD+}}\cap{\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$.
Hence, there exist $\lambda_{i},\mu_{j}\in\mathbb{R}$ so that

$\rho=\sum_{i=1}^{d}\lambda_{i}|a_{i}\rangle\langle
a_{i}|+\sum_{j=1}^{d}\mu_{j}|b_{j}\rangle\langle b_{j}|.$

Consequently

$\forall(i,j)\in\llbracket 1,d\rrbracket^{2},Q_{ij}(\rho)=\left|\langle
a_{i}|b_{j}\rangle\right|^{2}\left(\lambda_{i}+\mu_{j}\right).$

After a possible reordering of the basis, we can suppose that
$\frac{Q_{11}(\rho)}{|\langle
a_{1}|b_{1}\rangle|^{2}}=\min_{(i,j)\in\llbracket
1,d\rrbracket^{2}}\frac{Q_{ij}(\rho)}{\left|\langle
a_{i}|b_{j}\rangle\right|^{2}}$ so that

$\forall j\in\llbracket
2,d\rrbracket,\mu_{j}-\mu_{1}=\frac{Q_{1j}(\rho)}{\left|\langle
a_{1}|b_{j}\rangle\right|^{2}}-\frac{Q_{11}(\rho)}{\left|\langle
a_{1}|b_{1}\rangle\right|^{2}}\geqslant 0\ \mathrm{and}\
\lambda_{j}-\lambda_{1}=\frac{Q_{j1}(\rho)}{\left|\langle
a_{j}|b_{1}\rangle\right|^{2}}-\frac{Q_{11}(\rho)}{\left|\langle
a_{1}|b_{1}\rangle\right|^{2}}\geqslant 0.$

Moreover, since $\rho$ is KD positive, $Q_{11}(\rho)=\left|\langle
a_{1}|b_{1}\rangle\right|^{2}\left(\lambda_{1}+\mu_{1}\right)\geqslant 0$. So,
either $\mu_{1}$ or $\lambda_{1}$ must be nonnegative. Suppose
$\lambda_{1}\geq 0$, then as $|a_{1}\rangle\langle
a_{1}|=\sum_{j=1}^{d}|b_{j}\rangle\langle
b_{j}|-\sum_{i=2}^{d}|a_{i}\rangle\langle a_{i}|$ we can rewrite $\rho$ as

$\rho=\sum_{i=2}^{d}\left(\lambda_{i}-\lambda_{1}\right)|a_{i}\rangle\langle
a_{i}|+\sum_{j=1}^{d}\left(\mu_{j}+\lambda_{1}\right)|b_{j}\rangle\langle
b_{j}|=\sum_{i=2}^{d}\left(\lambda_{i}-\lambda_{1}\right)|a_{i}\rangle\langle
a_{i}|+\sum_{j=1}^{d}\frac{Q_{1j}(\rho)}{\left|\langle
a_{1}|b_{j}\rangle\right|^{2}}|b_{j}\rangle\langle b_{j}|.$

Hence $\rho\in\mathrm{span}_{\mathbb{R}^{+}}({\mathcal{A}}\cup{\mathcal{B}})$.
Together with the fact that $\mathrm{Tr}\,\rho=1$, this shows that
$\rho\in\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$ and
finalizes our proof. ∎

Recall that $V_{\mathrm{KDr}}={\mathrm{Ker}}(\mathrm{Im}Q)$ and that

${\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}\subset
V_{\mathrm{KDr}}.$

So we conclude that

$2d-1\leq\dim V_{\mathrm{KDr}}=\dim({\mathrm{Ker}}(\mathrm{Im}Q))\leq d^{2}.$
(3.10)

The following proposition shows that the condition $\dim
V_{\mathrm{KDr}}=2d-1$ is equivalent to the requirement that the basis states
are the only extreme KD-positive states which is equivalent to Eq. (3.6).

######  Proposition 3.2.

Suppose $m_{A,B}>0$. Consider the following statements:
(ia)
$V_{\mathrm{KDr}}={\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$;
(ib) $\dim V_{\mathrm{KDr}}=2d-1$;
(iia)
${\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}={\mathcal{A}}\cup{\mathcal{B}}$;
(iib)
${\mathcal{E}}_{\mathrm{KD+}}=\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$.
Then (ia) $\Leftrightarrow$ (ib) $\Leftrightarrow$ (iia) $\Leftrightarrow$
(iib).

###### Proof.

That (ia) $\Leftrightarrow$ (ib) is immediate and so is the equivalence
between (iia) and (iib).

We first show that (ia) implies (iib). Let
$\rho\in{\mathcal{E}}_{\mathrm{KD+}}$. Then it belongs to $V_{\mathrm{KDr}}$
and hence, by (ia),
$\rho\in{\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$. Hence,
by the second statement of Lemma 3.1, as
$\rho\in{\mathcal{E}}_{\mathrm{KD+}}\cap{\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$,
it follows that
$\rho\in\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$. Thus,
${\mathcal{E}}_{\mathrm{KD+}}=\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$.

It remains to show that (iib) implies (ia). We proceed by contraposition.
Suppose that (ia) does not hold so that $\dim V_{\mathrm{KDr}}>2d-1$. Lemma
3.1 then implies that
${\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$ is a proper
subspace of $V_{\mathrm{KDr}}$. So

$V_{\mathrm{KDr}}={\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}\oplus
W,$

with $W$ equal to the orthogonal complement of
${\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$ in
$V_{\mathrm{KDr}}$, which is nontrivial by assumption. Note that $F\in W$
implies that $\langle a_{i}|F|a_{i}\rangle=0=\langle b_{j}|F|b_{j}\rangle$ for
all $(i,j)\in\llbracket 1,d\rrbracket^{2}$. This implies $\mathrm{Tr}\,F=0$.
In addition, $Q(F)$ has only real entries by the definition of
$V_{\mathrm{KDr}}$. Choose $F\in W\backslash\\{0\\}$, and consider, for all
$x\in\mathbb{R}$

$\rho(x)=\rho_{*}+xF\ \mathrm{where}\
\rho_{*}=\frac{1}{d}\mathbb{I}_{d}\in\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right).$

Note that $\mathrm{Tr}\,\rho(x)=1$ for all $x\in\mathbb{R}$ and that, for all
$x\in\mathbb{R}$, one has

$\langle\psi|\rho(x)|\psi\rangle\geq\frac{1}{d}-f_{\mathrm{max}}|x|,$

where $|\psi\rangle$ is any norm-1 vector in ${\mathcal{H}}$. Here,
$f_{\mathrm{max}}=\max\\{|f_{i}||i\in\llbracket 1,d\rrbracket\\}>0$, where the
$f_{i}$ are the eigenvalues of $F$. In particular, if
$|x|\leq\frac{1}{df_{\mathrm{max}}}$, then $\rho(x)$ is a positive operator of
trace $1$. We now show that there exist
$0<x_{+}\leq\frac{1}{df_{\mathrm{max}}}<+\infty$ so that

$\forall x\in[-x_{+},x_{+}],\quad\rho(x)\in{\mathcal{E}}_{\mathrm{KD+}}.$
(3.11)

Since $F\in V_{\mathrm{KDr}}$, we know $\rho(x)\in V_{\mathrm{KDr}}$. One has,
for all $x\in\mathbb{R}$,

$Q_{ij}(\rho(x))=\left|\langle a_{i}|b_{j}\rangle\right|^{2}+xQ_{ij}(F)\geq
m_{{\mathcal{A}},{\mathcal{B}}}^{2}-|x|\max_{i,j}|Q_{ij}(F)|.$

Taking
$x_{+}=\min\\{\frac{1}{df_{\mathrm{max}}},\frac{m_{{\mathcal{A}},{\mathcal{B}}}^{2}}{\max_{i,j}|Q_{ij}(F)|}\\}>0$,
we have Eq. (3.11). This implies that (iib) does not hold since for all
$x\not=0$,
$\rho(x)\not\in{\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$.
∎

###  3.3 Characterizing
$\rho\in\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$

The following proposition is essential to the proof of Theorem 1.1 (iii).

######  Proposition 3.3.

Suppose $m_{{\mathcal{A}},{\mathcal{B}}}>0$. Then,

$\rho\in\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$

if and only if $\rho\in{\mathcal{E}}_{\mathrm{KD+}}$ and

$\quad\forall(i,j,k,l)\in\llbracket
1,d\rrbracket^{4},\frac{Q_{ij}(\rho)}{\left|\langle
a_{i}|b_{j}\rangle\right|^{2}}+\frac{Q_{kl}(\rho)}{\left|\langle
a_{k}|b_{l}\rangle\right|^{2}}=\frac{Q_{il}(\rho)}{\left|\langle
a_{i}|b_{l}\rangle\right|^{2}}+\frac{Q_{kj}(\rho)}{\left|\langle
a_{k}|b_{j}\rangle\right|^{2}}.$ (3.12)

###### Proof.

We first show the reverse implication. Let
$\rho\in{\mathcal{E}}_{\mathrm{KD+}}$ and satisfying Eq. (3.12). We construct
a state $\rho_{2}\in\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$
such that $Q(\rho_{2})=Q(\rho)$. Since $m_{{\mathcal{A}},{\mathcal{B}}}>0$, we
know from Eq. (3.3) that the KD distribution determines the state, such that
$\rho_{2}=\rho$.

Note that the basis states have the following KD distribution:

$\forall(i,j,k)\in\llbracket 1,d\rrbracket^{3},Q_{ij}(|a_{k}\rangle\langle
a_{k}|)=\left|\langle a_{k}|b_{j}\rangle\right|^{2}\delta_{i,k}\text{ and
}Q_{ij}(|b_{k}\rangle\langle b_{k}|)=\left|\langle
a_{i}|b_{k}\rangle\right|^{2}\delta_{j,k}.$

By permuting the order of the vectors in $\mathcal{B}$, we can suppose that
$\frac{Q_{11}(\rho)}{\left|\langle
a_{1}|b_{1}\rangle\right|^{2}}=\min_{j\in\llbracket
1,d\rrbracket}\frac{Q_{1j}(\rho)}{\left|\langle
a_{1}|b_{j}\rangle\right|^{2}}$. We define

$\rho_{2}=\sum_{i=1}^{d}\lambda_{i}|a_{i}\rangle\langle
a_{i}|+\sum_{j=1}^{d}\mu_{j}|b_{j}\rangle\langle b_{j}|$

where $\lambda_{i}=\frac{Q_{i1}(\rho)}{\left|\langle
a_{i}|b_{1}\rangle\right|^{2}}$ for all $i\in\llbracket 1,d\rrbracket$ and
$\mu_{j}=\frac{Q_{1j}(\rho)}{\left|\langle
a_{1}|b_{j}\rangle\right|^{2}}-\frac{Q_{11}(\rho)}{\left|\langle
a_{1}|b_{1}\rangle\right|^{2}}$ for $j\in\llbracket 1,d\rrbracket$ so that
$\mu_{1}=0$. Since $\rho$ is KD positive, $\lambda_{i}\geqslant 0$ and
$\mu_{i}\geqslant 0$ for all $i\in\llbracket 1,d\rrbracket$. Moreover, using
Eq. (3.12), one has that

$\begin{array}[]{rcl}\displaystyle\mathrm{Tr}\,\rho_{2}=\sum_{j=1}^{d}\mu_{j}+\sum_{i=1}^{d}\lambda_{i}&=&\sum_{j=1}^{d}\sum_{i=1}^{d}\left|\langle
a_{i}|b_{j}\rangle\right|^{2}\mu_{j}+\sum_{i=1}^{d}\sum_{j=1}^{d}\left|\langle
a_{i}|b_{j}\rangle\right|^{2}\lambda_{i}\\\
&=&\displaystyle\sum_{j=1}^{d}\sum_{i=1}^{d}\left|\langle
a_{i}|b_{j}\rangle\right|^{2}\left(\frac{Q_{i1}(\rho)}{\left|\langle
a_{i}|b_{1}\rangle\right|^{2}}+\frac{Q_{1j}(\rho)}{\left|\langle
a_{1}|b_{j}\rangle\right|^{2}}-\frac{Q_{11}(\rho)}{\left|\langle
a_{1}|b_{1}\rangle\right|^{2}}\right)\\\
&=&\displaystyle\sum_{j=1}^{d}\sum_{i=1}^{d}Q_{ij}(\rho)=1,\end{array}$

so that $\rho_{2}\in\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$.
Using Eq. (3.12) again, we find $\forall(i,j)\in\llbracket 1,d\rrbracket^{2},$

$\begin{array}[]{rcl}Q_{ij}(\rho_{2})&=&\left|\langle
a_{i}|b_{j}\rangle\right|^{2}\left(\lambda_{i}+\mu_{j}\right)\\\
&=&\left|\langle
a_{i}|b_{j}\rangle\right|^{2}\left(\frac{Q_{i1}(\rho)}{\left|\langle
a_{i}|b_{1}\rangle\right|^{2}}+\frac{Q_{1j}(\rho)}{\left|\langle
a_{1}|b_{j}\rangle\right|^{2}}-\frac{Q_{11}(\rho)}{\left|\langle
a_{1}|b_{1}\rangle\right|^{2}}\right)\\\ &=&Q_{ij}(\rho).\end{array}$

This shows that $\rho_{2}=\rho$ so that
$\rho\in\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$.

For the proof of the direct implication, we note that if
$\rho\in\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$, then
$\rho=\sum_{i=1}^{d}\lambda_{i}|a_{i}\rangle\langle
a_{i}|+\sum_{j=1}^{d}\mu_{j}|b_{j}\rangle\langle b_{j}|$ with
$\sum_{i=1}^{d}\lambda_{i}+\mu_{i}=1$, $\lambda_{i}\geqslant 0$ and
$\mu_{i}\geqslant 0$ for all $i\in\llbracket 1,d\rrbracket$. The KD
distribution of $\rho$ is given by

$\forall(i,j)\in\llbracket 1,d\rrbracket^{2},Q_{ij}(\rho)=\left|\langle
a_{i}|b_{j}\rangle\right|^{2}\left(\lambda_{i}+\mu_{j}\right).$

Hence, $\rho$ is KD-positive and for all $(i,j)\in\llbracket
1,d\rrbracket^{2}$,

$\frac{Q_{11}(\rho)}{\left|\langle
a_{1}|b_{1}\rangle\right|^{2}}+\frac{Q_{ij}(\rho)}{\left|\langle
a_{i}|b_{j}\rangle\right|^{2}}=\left(\lambda_{1}+\mu_{1}\right)+\left(\lambda_{i}+\mu_{j}\right)=\lambda_{1}+\mu_{1}+\lambda_{i}+\mu_{j},$
$\frac{Q_{1j}(\rho)}{\left|\langle
a_{1}|b_{j}\rangle\right|^{2}}+\frac{Q_{i1}(\rho)}{\left|\langle
a_{i}|b_{1}\rangle\right|^{2}}=\left(\lambda_{1}+\mu_{j}\right)+\left(\lambda_{i}+\mu_{1}\right)=\lambda_{1}+\mu_{1}+\lambda_{i}+\mu_{j}.$

This implies Eq. (3.12) with $k=1=l$. For the general case, we write

$\frac{Q_{ij}(\rho)}{\left|\langle
a_{i}|b_{j}\rangle\right|^{2}}+\frac{Q_{kl}(\rho)}{\left|\langle
a_{k}|b_{l}\rangle\right|^{2}}=\frac{Q_{i1}(\rho)}{\left|\langle
a_{i}|b_{1}\rangle\right|^{2}}+\frac{Q_{1j}(\rho)}{\left|\langle
a_{1}|b_{j}\rangle\right|^{2}}-\frac{Q_{11}(\rho)}{\left|\langle
a_{1}|b_{1}\rangle\right|^{2}}+\frac{Q_{k1}(\rho)}{\left|\langle
a_{k}|b_{1}\rangle\right|^{2}}+\frac{Q_{1l}(\rho)}{\left|\langle
a_{1}|b_{l}\rangle\right|^{2}}-\frac{Q_{11}(\rho)}{\left|\langle
a_{1}|b_{1}\rangle\right|^{2}}$

and

$\frac{Q_{il}(\rho)}{\left|\langle
a_{i}|b_{l}\rangle\right|^{2}}+\frac{Q_{kj}(\rho)}{\left|\langle
a_{k}|b_{j}\rangle\right|^{2}}=\frac{Q_{i1}(\rho)}{\left|\langle
a_{i}|b_{1}\rangle\right|^{2}}+\frac{Q_{1l}(\rho)}{\left|\langle
a_{1}|b_{l}\rangle\right|^{2}}-\frac{Q_{11}(\rho)}{\left|\langle
a_{1}|b_{1}\rangle\right|^{2}}+\frac{Q_{k1}(\rho)}{\left|\langle
a_{k}|b_{1}\rangle\right|^{2}}+\frac{Q_{1j}(\rho)}{\left|\langle
a_{1}|b_{j}\rangle\right|^{2}}-\frac{Q_{11}(\rho)}{\left|\langle
a_{1}|b_{1}\rangle\right|^{2}}.$

The right hand sides of these two equations are identical up to a
reorganization of the terms, so

$\frac{Q_{ij}(\rho)}{\left|\langle
a_{i}|b_{j}\rangle\right|^{2}}+\frac{Q_{kl}(\rho)}{\left|\langle
a_{k}|b_{l}\rangle\right|^{2}}=\frac{Q_{il}(\rho)}{\left|\langle
a_{i}|b_{l}\rangle\right|^{2}}+\frac{Q_{kj}(\rho)}{\left|\langle
a_{k}|b_{j}\rangle\right|^{2}}.$

This ends our proof. ∎

The relations (3.12) are simpler for MUB bases and are given in the following
corollary.

######  Corollary 3.4.

Let $\mathcal{A}$ and $\mathcal{B}$ be MUB bases. Then

$\rho\in\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$

if and only if $\rho\in{\mathcal{E}}_{\mathrm{KD+}}$ and

$\quad\forall(i,j,k,l)\in\llbracket
1,d\rrbracket^{4},Q_{ij}(\rho)+Q_{kl}(\rho)=Q_{il}(\rho)+Q_{kj}(\rho).$ (3.13)

###### Proof.

This is a direct consequence of Proposition 3.3. ∎

##  4 Proofs of Theorem 1.1 and of Proposition 1.2.

For convenience, we restate our theorem:

######  Theorem 1.1.

The equality

${\mathcal{E}}_{\mathrm{KD+}}=\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right),$
(1.2)

holds under any single one of the following hypotheses:

  1. (i)

If $d=2$ (for qubits) and $m_{{\mathcal{A}},{\mathcal{B}}}>0$;

  2. (ii)

If $d=3$, for all $U$ in a open dense set of probability $1$;

  3. (iii)

If $d$ is prime and $U$ is the discrete Fourier transform (DFT) matrix;

  4. (iv)

If $U$ is sufficiently close to some other $U^{\prime}$ for which Eq. (1.2)
holds.

_Proof of Theorem 1.1 (i)._ We can, without loss of generality, suppose that
the transition matrix $U$ is a real matrix by executing appropriate phase
changes on the basis vectors. If $U$ has no zeros
($m_{{\mathcal{A}},{\mathcal{B}}}>0$), we can therefore write
$U=\begin{pmatrix}\cos(\theta)&\sin(\theta)\\\
-\sin(\theta)&\cos(\theta)\end{pmatrix}$ for
$\theta\in\mathbb{R}\backslash\frac{\pi}{2}\mathbb{Z}$. To find the dimension
of the space of KD-real operators, we consider $F\in V_{\mathrm{KDr}}$ and
write

$\displaystyle Q_{11}(F)$ $\displaystyle=$ $\displaystyle\langle
b_{1}|a_{1}\rangle\langle a_{1}|F|b_{1}\rangle=\langle
b_{1}|a_{1}\rangle\langle a_{1}|b_{1}\rangle F_{11}+\langle
b_{1}|a_{1}\rangle\langle a_{2}|b_{1}\rangle F_{12}$ $\displaystyle=$
$\displaystyle\cos(\theta)^{2}F_{11}+\cos(\theta)\sin(\theta)F_{12}\in\mathbb{R},$

with $F_{ij}=\langle a_{i}|F|a_{j}\rangle$. Since, by hypothesis, $F$ is self-
adjoint and $\mathrm{Im}Q_{11}(F)=0$, one finds $\mathrm{Im}F_{12}=0$ so
$F_{12}=F_{21}$. Hence $F\in V_{\mathrm{KDr}}$ implies that $F$ is real
symmetric. Conversely, one can check that for any real symmetric $F$, $Q(F)$
is a real matrix. Consequently, $\dim(V_{\mathrm{KDr}})=3$. The result then
follows from Proposition 3.2.

_Proof of Theorem 1.1 (ii)._ This result is restated more explicitly in the
following proposition.

######  Proposition 4.1.

In dimension $d=3$, there exists a set $\mathcal{W}$ of unitary matrices such
that:

  * •

$\forall U\in\mathcal{W}$,
${\mathcal{E}}_{\mathrm{KD+}}=\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$;

  * •

$\mathcal{W}$ is an open and dense subset of the set of unitary matrices;

  * •

$\mathcal{W}$ is a set of probability one for the Haar measure on the unitary
group.

###### Proof.

For any unitary matrix $U$ with $m_{{\mathcal{A}},{\mathcal{B}}}>0$, we write
$U=(A_{kj}e^{i\phi_{kj}})_{(k,j)\in\llbracket 1,3\rrbracket^{2}}$, with
$A_{kj}>0$. We define $\mathcal{W}$ to be the set of unitary matrices in
dimension 3 for which $m_{{\mathcal{A}},{\mathcal{B}}}>0$ and the following
conditions are fulfilled:

$\left\\{\begin{array}[]{rclc}\phi_{21}-\phi_{11}&\neq&0&[\frac{\pi}{2}]\\\
\phi_{22}-\phi_{12}&\neq&0&[\frac{\pi}{2}]\\\
\phi_{31}-\phi_{11}&\neq&0&[\frac{\pi}{2}]\\\
\phi_{32}-\phi_{12}&\neq&0&[\frac{\pi}{2}]\\\
\phi_{21}-\phi_{11}&\neq&\phi_{22}-\phi_{12}&[\pi]\\\
\phi_{31}-\phi_{11}&\neq&\phi_{32}-\phi_{12}&\ [\pi].\end{array}\right.$ (4.1)

Let $U\in\mathcal{W}$. We want to show that
${\mathcal{E}}_{\mathrm{KD+}}=\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$.
According to Proposition 3.2, it is sufficient to show that
$\dim(\mathrm{Ran}\left(\mathrm{Im}Q\right))=4$. (Here and below,
$\mathrm{Ran}\left(T\right)$ stands for the range of a linear map $T$.) For
that purpose, we shall consider the $9\times 9$ matrix $T$ of the linear map
$\mathrm{Im}Q:\mathcal{S}_{d}\rightarrow\mathcal{M}_{d}(\mathbb{R})$ with
respect to the basis

$\left\\{|a_{k}\rangle\langle a_{k}|,(|a_{k}\rangle\langle
a_{j}|+|a_{j}\rangle\langle a_{k}|),i(|a_{k}\rangle\langle
a_{j}|-|a_{j}\rangle\langle a_{k}|)\right\\}_{k\in\llbracket
1,3\rrbracket,k<j}$

of $\mathcal{S}_{d}$ and the canonical basis of $\mathcal{M}_{d}(\mathbb{R})$.
The matrix $T$ can be readily computed but we do not display it here. Note
that, by Eq. (3.10), $\dim({\mathrm{Ker}}{(\mathrm{Im}Q}))\geqslant 5$, so
that
$\dim{(\mathrm{Ran}\left(\mathrm{Im}Q\right))}=9-\dim({\mathrm{Ker}}{(\mathrm{Im}Q}))\leqslant
4$. Equality is obtained, _i.e._
$\dim{(\mathrm{Ran}\left(\mathrm{Im}Q\right))}=4$, if and only if there exists
a $4$ by $4$ submatrix $\Sigma$ of $T$ that has rank $4$. We will show that
the submatrix $\Sigma$, given by

$\Sigma=\begin{pmatrix}A_{11}A_{21}\sin{(\phi_{21}-\phi_{11})}&A_{11}A_{21}\cos{(\phi_{21}-\phi_{11})}&A_{11}A_{31}\sin{(\phi_{31}-\phi_{11})}&A_{11}A_{31}\cos{(\phi_{31}-\phi_{11})}\\\
A_{12}A_{22}\sin{(\phi_{22}-\phi_{12})}&A_{12}A_{22}\cos{(\phi_{22}-\phi_{12})}&A_{12}A_{32}\sin{(\phi_{32}-\phi_{12})}&A_{12}A_{32}\cos{(\phi_{32}-\phi_{12})}\\\
-A_{11}A_{21}\sin{(\phi_{21}-\phi_{11})}&-A_{11}A_{21}\cos{(\phi_{21}-\phi_{11})}&0&0\\\
-A_{12}A_{22}\sin{(\phi_{22}-\phi_{12})}&-A_{12}A_{22}\cos{(\phi_{22}-\phi_{12})}&0&0\end{pmatrix},$

is indeed of rank $4$. To prove this, suppose there exists
$(a_{1},a_{2},a_{3},a_{4})\in\mathbb{R}^{4}$ such that
$(a_{1},a_{2},a_{3},a_{4})\in{\mathrm{Ker}}{\Sigma}$. Then,

$\left\\{\begin{array}[]{rcl}A_{21}(a_{1}\sin{(\phi_{21}-\phi_{11})}+a_{2}\cos{(\phi_{21}-\phi_{11})})&=&-A_{31}(a_{3}\sin{(\phi_{31}-\phi_{11})}+a_{4}\cos{(\phi_{31}-\phi_{11})})\\\
A_{22}(a_{1}\sin{(\phi_{22}-\phi_{12})}+a_{2}\cos{(\phi_{22}-\phi_{12})})&=&-A_{32}(a_{3}\sin{(\phi_{32}-\phi_{12})}+a_{4}\cos{(\phi_{32}-\phi_{12})})\\\
a_{1}\sin{(\phi_{21}-\phi_{11})}+a_{2}\cos{(\phi_{21}-\phi_{11})}&=&0\\\
a_{1}\sin{(\phi_{22}-\phi_{12})}+a_{2}\cos{(\phi_{22}-\phi_{12})}&=&0.\end{array}\right.$

The last two rows simplify to

$\left\\{\begin{array}[]{rcl}a_{1}\tan{(\phi_{21}-\phi_{11})}&=&-a_{2},\\\
a_{1}\tan{(\phi_{22}-\phi_{12})}&=&-a_{2}.\\\ \end{array}\right.$

If $a_{1}\neq 0$, then
$\tan{(\phi_{21}-\phi_{11})}=\tan{(\phi_{22}-\phi_{12})}$, which contradicts
the condition $\phi_{21}-\phi_{11}\neq\phi_{22}-\phi_{12}[\pi]$. So
$a_{1}=a_{2}=0$. Consequently, the first two conditions reduce to

$\left\\{\begin{array}[]{rcl}a_{3}A_{11}A_{31}\sin{(\phi_{31}-\phi_{11})}+a_{4}A_{31}A_{11}\cos{(\phi_{31}-\phi_{11})}=0,\\\
a_{3}A_{12}A_{32}\sin{(\phi_{32}-\phi_{12})}+a_{4}A_{32}A_{12}\cos{(\phi_{32}-\phi_{12})}=0.\\\
\end{array}\right.$

Following the same argument, we find that $a_{3}=a_{4}=0$. Consequently, the
matrix $\Sigma$ has a vanishing kernel and is therefore of rank $4$. In
conclusion, for any unitary matrix in $\mathcal{W}$ it is true that
$\dim{(\mathrm{Ran}\left(\mathrm{Im}Q\right))}=4$, and hence
$\dim{{\mathrm{Ker}}{(\mathrm{Im}Q})}=5$. This concludes the proof of the
first part of the Proposition.

The set $\mathcal{W}$ is clearly open. We now show that it is dense also. For
that purpose, consider an arbitrary unitary matrix $U$. Suppose it does not
belong to $\mathcal{W}$ so that at least one of the six conditions in Eq.
(4.1) is not satisfied for $U$. We write $C_{1},C_{2},C_{3}$ for the columns
of $U$ and remark that $C_{3}=\varepsilon C_{1}\wedge C_{2}$ with
$\varepsilon\in\\{-1,1\\}$; here $\wedge$ denotes the vector product. We then
construct, for $\theta\in\mathbb{R}$, the two columns

$C_{1}(\theta)=\begin{pmatrix}A_{11}e^{i(\phi_{11}+\theta)}\\\
A_{21}e^{i\phi_{21}}\\\ A_{31}e^{i\phi_{31}}\end{pmatrix},\quad
C_{2}(\theta)=\begin{pmatrix}A_{12}e^{i(\phi_{12}-\theta)}\\\
A_{22}e^{i\phi_{22}}\\\ A_{32}e^{i\phi_{32}}\end{pmatrix}.$

They are orthogonal to each other and normalized. Defining
$C_{3}(\theta)=\varepsilon C_{1}(\theta)\wedge C_{2}(\theta)$, we construct
$U(\theta)=(C_{1}(\theta),C_{2}(\theta),C_{3}(\theta))$. This is a family of
unitary matrices for which $U(\theta)\to U$ when $\theta\to 0$. By
construction, for all $\theta\in\mathbb{R}$, the conditions of Eq. (4.1) read:

$\left\\{\begin{array}[]{rclc}\phi_{21}-\phi_{11}&\neq&\theta&[\frac{\pi}{2}]\\\
\phi_{22}-\phi_{12}&\neq&-\theta&[\frac{\pi}{2}]\\\
\phi_{31}-\phi_{11}&\neq&\theta&[\frac{\pi}{2}]\\\
\phi_{32}-\phi_{12}&\neq&-\theta&[\frac{\pi}{2}]\\\ \
\phi_{21}-\phi_{11}-\theta&\neq&\phi_{22}-\phi_{12}+\theta&[\pi]\\\
\phi_{31}-\phi_{11}-\theta&\neq&\phi_{32}-\phi_{12}+\theta&[\pi]\\\
\end{array}\right.$

These conditions are all fulfilled for $\theta\not=0$ small enough. This
implies that the set $\mathcal{W}$ is dense.

To show the set $\mathcal{W}$ is of full Haar measure, we show that its
complement, $\mathcal{W}^{\textrm{c}}$ is of zero Haar measure. The group
$U(3)$ is a $9$-dimensional real manifold. Its Haar measure is absolutely
continuous with respect to the Lebesgue measure in any local coordinate patch
[65]. Now, $\mathcal{W}^{\textrm{c}}$ is the union of the sets where one of
the inequalities in Eq. (4.1) is an equality and of the sets where one of the
matrix elements of $U$ vanishes. Each of these sets is an lower dimensional
submanifold of $U(3)$. Hence it is of zero Lebesgue measure, which concludes
the proof. ∎

_Proof of Theorem 1.1 (iii)._ We write the entries of a DFT transition matrix
$U$ as $U_{kl}=\frac{\omega^{(k-1)(l-1)}}{\sqrt{p}}$ for all
$(k,l)\in\llbracket 1,p\rrbracket^{2}$, where $\omega=e^{-\frac{2i\pi}{p}}$.
In this proof, the indices on the matrix $U$ and on all other matrices
appearing should be thought of as being extended to all integers and as being
periodic with period $p$.

As
$\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)\subseteq{\mathcal{E}}_{\mathrm{KD+}}$,
we only have to prove that
${\mathcal{E}}_{\mathrm{KD+}}\subseteq\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$.
To that end, we use Corollary 3.4. In other words, we need to show that
Eq.(3.13) holds for all $\rho\in{\mathcal{E}}_{\mathrm{KD+}}$; this is
achieved in Eq.(4.8) below.

We need the following lemma, which characterizes $V_{\mathrm{KDr}}$ in the
case where $U$ is the DFT matrix in prime dimension.

######  Lemma 4.2.

Let $U$ be the DFT matrix in prime dimension $d=p$. Then, a self-adjoint
operator $F\in\mathcal{S}_{p}$ belongs to $V_{\mathrm{KDr}}$ if and only if
for all $(i,k)\in\llbracket 1,p\rrbracket^{2}$,

$F_{i(i+k)}=F_{(i-k)i}$ (4.2)

Here, $F_{ik}=\langle a_{i}|F|a_{k}\rangle$ for $(i,k)\in\llbracket
1,p\rrbracket^{2}$.

We remark that Eq.(4.2) means that the matrix $F$ is constant on its $d-1$
off-diagonals.

_Proof of Lemma 4.2._ For all $(i,j)\in\llbracket 1,p\rrbracket^{2}$

$Q_{ij}(F)=\sum_{k=1}^{p}\langle b_{j}|a_{i}\rangle\langle a_{k}|b_{j}\rangle
F_{ik}=\frac{1}{p}\sum_{k=1}^{p}\omega^{(j-1)(1-i)}\omega^{(j-1)(k-1)}F_{ik}=\frac{1}{p}\sum_{k=1}^{p}\omega^{(j-1)(k-i)}F_{ik}.$

In order to compute $\mathrm{Im}(Q_{ij}(F))$, we rewrite $Q_{ij}(F)$ as
follows. Let $i\in\llbracket 2,p\rrbracket$ and $j\in\llbracket
2,p\rrbracket$. Then,

$Q_{ij}(F)=\frac{1}{2}\left(Q_{ij}(F)+Q_{ij}(F)\right)=\frac{1}{2p}\left(\sum_{k=1}^{p}\omega^{(j-1)(k-i)}F_{ik}+\sum_{k=1}^{p}\omega^{(j-1)(k-i)}F_{ik}\right).$

We now rewrite the second sum. We note that
$\overline{\omega}^{(j-1)(k^{\prime}-i)}=\omega^{(j-1)(k-i)}$ if and only if
$(j-1)(k-i)=(j-1)(i-k^{\prime})\ [p]$; as $(j-1)\neq 0\leavevmode\nobreak\
[p]$, it follows that $k+k^{\prime}=2i\leavevmode\nobreak\ [p]$ and thus that
$k^{\prime}=2i-k\leavevmode\nobreak\ [p]$. As the map $k\in\llbracket
1,p\rrbracket\mapsto(2i-k)\leavevmode\nobreak\ [p]\in\llbracket 1,p\rrbracket$
is bijective, one finds that

$Q_{ij}(F)=\frac{1}{2p}\left(\sum_{k=1}^{p}\omega^{(j-1)(k-i)}F_{ik}+\sum_{k=1}^{p}\omega^{(j-1)[(2i-k)-i]}F_{i(2i-k)}\right).$

Note that the indices on $F_{ij}$ are considered modulo $p$. Therefore,

$Q_{ij}(F)=\displaystyle\frac{1}{2p}\left(\sum_{k=1}^{p}\omega^{(j-1)(k-i)}F_{ik}+\overline{\omega}^{(j-1)(k-i)}F_{i(2i-k)}\right).$
(4.3)

By changing the summation index, we have

$Q_{ij}(F)=\frac{1}{2p}\left(\sum_{k^{\prime}=1-i}^{p-i}\omega^{(j-1)k^{\prime}}F_{i(i+k^{\prime})}+\overline{\omega}^{(j-1)k^{\prime}}F_{i(i-k^{\prime})}\right).$

As the indices are considered modulo $p$, the summand is periodic with period
$p$, and we can shift the sum to obtain

$Q_{ij}(F)=\frac{1}{2p}\left(\sum_{k^{\prime}=0}^{p-1}\omega^{(j-1)k^{\prime}}F_{i(i+k^{\prime})}+\overline{\omega}^{(j-1)k^{\prime}}F_{i(i-k^{\prime})}\right).$

If $k^{\prime}\in\llbracket 1,\frac{p-1}{2}\rrbracket$, then
$(p-k^{\prime})\in\llbracket\frac{p+1}{2},p-1\rrbracket$ and

$\omega^{(j-1)(p-k^{\prime})}F_{i[i+(p-k^{\prime})]}+\overline{\omega}^{(j-1)(p-k^{\prime})}F_{i[i-(p-k^{\prime})]}=\omega^{(j-1)k^{\prime}}F_{i(i+k^{\prime})}+\overline{\omega}^{(j-1)k^{\prime}}F_{i(i-k^{\prime})},$

so that we can group these terms together. This leads to

$Q_{ij}(F)=\frac{1}{p}\left(\sum_{k^{\prime}=1}^{\frac{p-1}{2}}\omega^{(j-1)k^{\prime}}F_{i(i+k^{\prime})}+\overline{\omega}^{(j-1)k^{\prime}}F_{i(i-k^{\prime})}\right)+\frac{1}{p}F_{ii}.$
(4.4)

We can then finally compute $\text{Im}(Q_{ij}(F))$ for $(i,j)\in\llbracket
2,p\rrbracket^{2}$:

$\begin{array}[]{rcl}\text{Im}\left(Q_{ij}(F)\right)&=&\displaystyle\frac{1}{p}\sum_{k^{\prime}=1}^{\frac{p-1}{2}}\text{Im}\left(\overline{\omega}^{(j-1)k^{\prime}}F_{i(i-k^{\prime})}+\omega^{(j-1)k^{\prime}}F_{i(i+k^{\prime})}\right)\\\
&=&\displaystyle\frac{1}{2p\sqrt{-1}}\sum_{k^{\prime}=1}^{\frac{p-1}{2}}\overline{\omega}^{(j-1)k^{\prime}}F_{i(i-k^{\prime})}-\omega^{(j-1)k^{\prime}}\overline{F_{i(i-k^{\prime})}}+\omega^{(j-1)k^{\prime}}F_{i(i+k^{\prime})}-\overline{\omega}^{(j-1)k^{\prime}}\overline{F_{i(i+k^{\prime})}}\\\
&=&\displaystyle\frac{1}{2p\sqrt{-1}}\sum_{k^{\prime}=1}^{\frac{p-1}{2}}\omega^{(j-1)k^{\prime}}\left(F_{i(i+k^{\prime})}-\overline{F_{i(i-k^{\prime})}}\right)+\overline{\omega}^{(j-1)k^{\prime}}\left(F_{i(i-k^{\prime})}-\overline{F_{i(i+k^{\prime})}}\right).\end{array}$

Recall that $F\in V_{\mathrm{KDr}}$ if and only if, for any $i\in\llbracket
2,p\rrbracket$, the $p-1$ equations $\text{Im}(Q_{ij}(F))=0$ for
$j\in\llbracket 2,p\rrbracket$ are satisfied. Indeed, as a consequence of Eq.
(3.2), this is equivalent to $\text{Im}(Q_{ij}(F))=0$ for all
$(i,j)\in\llbracket 2,p\rrbracket^{2}$. Hence $F\in V_{\mathrm{KDr}}$ if and
only if

$\sum_{k^{\prime}=1}^{\frac{p-1}{2}}\omega^{(j-1)k^{\prime}}\left(F_{i(i+k^{\prime})}-\overline{F_{i(i-k^{\prime})}}\right)+\overline{\omega}^{(j-1)k^{\prime}}\left(F_{i(i-k^{\prime})}-\overline{F_{i(i+k^{\prime})}}\right)=0.$

This system can be rewritten with
$z_{k}=F_{i(i+k)}-\overline{F_{i(i-k)}}\text{ and
}z_{p-k}=F_{i(i-k)}-\overline{F_{i(i+k)}}$ for $k\in\llbracket
1,\frac{p-1}{2}\rrbracket$:

$A_{\omega}\begin{pmatrix}z_{1}\\\ z_{2}\\\ \vdots\\\ z_{p-1}\end{pmatrix}=0$

where

$A_{\omega}=\begin{pmatrix}\omega&\omega^{2}&\cdots&\omega^{\frac{p-1}{2}}&\overline{\omega}^{\frac{p-1}{2}}&\cdots&\overline{\omega}\\\
\omega^{2}&\omega^{4}&\cdots&\omega^{p-1}&\overline{\omega}^{p-1}&\cdots&\overline{\omega}^{2}\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\\
\omega^{p-1}&\omega^{2(p-1)}&\cdots&\omega^{\frac{(p-1)^{2}}{2}}&\overline{\omega}^{\frac{(p-1)^{2}}{2}}&\cdots&\overline{\omega}^{p-1}\\\
\end{pmatrix}=\begin{pmatrix}\omega&\omega^{2}&\cdots&\omega^{\frac{p-1}{2}}&\omega^{\frac{p+1}{2}}&\cdots&\omega^{p-1}\\\
\omega^{2}&\omega^{4}&\cdots&\omega^{p-1}&\omega^{p+1}&\cdots&\omega^{2(p-1)}\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\\
\omega^{p-1}&\omega^{2(p-1)}&\cdots&\omega^{\frac{(p-1)^{2}}{2}}&\omega^{\frac{(p-1)(p+1)}{2}}&\cdots&\omega^{(p-1)^{2}}\\\
\end{pmatrix}.$

The matrix $A_{\omega}$ is a Vandermonde matrix, written
$V(\omega,\omega^{2},\dots,\omega^{p-1})$ for which all parameters are
different so $A_{\omega}$ is invertible. This means that for all
$k\in\llbracket 1,p-1\rrbracket,z_{k}=0$. Hence, $F\in V_{\mathrm{KDr}}$ if
and only if

$\forall(i,k)\in\llbracket 2,p\rrbracket\times\llbracket
1,\frac{p-1}{2}\rrbracket,F_{i(i+k)}=\overline{F_{i(i-k)}}=F_{(i-k)i}.$ (4.5)

We further rewrite these conditions in a more symmetric form: see Eq. (4.7)
below. Consider $(i,k)\in\llbracket
2,p\rrbracket\times\llbracket\frac{p+1}{2},p-1\rrbracket$. As all indices are
taken modulo $p$,

$F_{i(i+k)}=F_{i(i+k-p)}=F_{i[i-(p-k)]}.$

Since $p-k\in\llbracket 1,\frac{p-1}{2}\rrbracket$, Eq.(4.5) implies that
$F_{i[i-(p-k)]}=\overline{F_{i[i+(p-k)]}}=\overline{F_{i(i-k)}}$. Therefore,
we obtain the following recursion relation:

$\forall(i,k)\in\llbracket 2,p\rrbracket\times\llbracket
1,p-1\rrbracket,F_{i(i+k)}=\overline{F_{i(i-k)}}=F_{(i-k)i}.$ (4.6)

Next, we want to show that the relation also holds for $i=1$ and
$k\in\llbracket 1,p-1\rrbracket$. Suppose $k\in\llbracket 1,p-1\rrbracket$. If
$n\in\llbracket 0,p-2\rrbracket$ and since $F$ is self-adjoint,

$F_{(nk+1)[(n+1)k+1]}=\overline{F_{[(n+1)k+1](nk+1)}}=\overline{F_{[(n+1)k+1][(n+1)k+1-k]}}.$

As $n+1\neq 0\leavevmode\nobreak\ [p]$ and $k\neq 0\leavevmode\nobreak\ [p]$,
it follows that $(n+1)k+1\neq 1\leavevmode\nobreak\ [p]$. We can therefore use
Eq. (4.6) to obtain

$\forall n\in\llbracket 0,p-2\rrbracket,k\in\llbracket 1,p-1\rrbracket,\quad
F_{(nk+1)[(n+1)k+1]}=F_{[(n+1)k+1][(n+1)k+1+k]}=F_{[(n+1)k+1][(n+2)k+1]}.$

It follows from this that for all $n\in\llbracket 1,p-1\rrbracket$,
$F_{1(k+1)}=F_{(nk+1)[(n+1)k+1]}$. And thus, $F_{1(1+k)}=F_{(1-k)1}$ which is
the above relation for $n=p-1$. Thus, this shows that Eq.(4.5) holds for
$i=1$.

Summing up, $F\in{\mathcal{E}}_{\mathrm{KD+}}$ if and only if

$\begin{array}[]{rcl}F_{i(i+k)}&=&F_{(i-k)i}\text{ for all }(i,k)\in\llbracket
1,p\rrbracket\times\llbracket 1,p\rrbracket.\end{array}$ (4.7)

∎

We can now use this result to show that $\rho\in{\mathcal{E}}_{\mathrm{KD+}}$
implies that
$\rho\in\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$, by showing
that Eq. (3.13) holds. Indeed, since $\rho\in{\mathcal{E}}_{\mathrm{KD+}}$
implies that $\rho\in V_{\mathrm{KDr}}$, it follows from Eq. (4.4) and Lemma
4.2 that for all $(i,j)\in\llbracket 1,p\rrbracket^{2}$,

$\begin{array}[]{rcl}Q_{ij}(\rho)+Q_{11}(\rho)&=&\displaystyle\frac{2}{p}\sum_{k=1}^{\frac{p-1}{2}}\text{Re}\left(\omega^{(j-1)k^{\prime}}\rho_{i(i+k^{\prime})}\right)+\frac{1}{p}\rho_{ii}+\frac{2}{p}\sum_{k=1}^{\frac{p-1}{2}}\text{Re}\left(\rho_{1(1+k^{\prime})}\right)+\frac{1}{p}\rho_{11}\\\
&=&\displaystyle\frac{2}{p}\sum_{k=1}^{\frac{p-1}{2}}\text{Re}\left(\omega^{(j-1)k^{\prime}}\rho_{1(1+k^{\prime})}\right)+\frac{1}{p}\rho_{11}+\frac{2}{p}\sum_{k=1}^{\frac{p-1}{2}}\text{Re}\left(\rho_{i(i+k^{\prime})}\right)+\frac{1}{p}\rho_{ii}\\\
&=&Q_{1j}(\rho)+Q_{i1}(\rho).\end{array}$ (4.8)

This establishes the relations (3.13) for $(i,j)\in\llbracket
1,p\rrbracket^{2}$ and $k=l=1$. As in the proof of Proposition 3.3, this
implies that they hold for all $k,l\in\llbracket 1,p\rrbracket^{2}$. Thus, we
have proven that
${\mathcal{E}}_{\mathrm{KD+}}\subseteq\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$.
This ends the proof.

Remark : An alternative proof of Theorem 1.1.(iii) can be obtained as follows.
Lemma 4.2 implies that $F$ is constant on its $(d-1)$ off-diagonals and as $F$
is self-adjoint, only $\frac{d-1}{2}$ of these values are independent. Hence,
the off-diagonals of F are determined by $(d-1)$ real parameters. The diagonal
of $F$ contains $d$ real parameters. Lemma 4.2 implies that $V_{\mathrm{KDr}}$
is a $(2d-1)$ real vector space. Proposition 3.2$(ib)$ then implies that
${\mathcal{E}}_{\mathrm{KD+}}=\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$.

Proof of Theorem 1.1 (iv). Note that this statement means that the set of $U$
for which Eq. (1.2) holds is open. The result follows from the following
Proposition.

######  Proposition 4.3.

Let $U$ be such that $m_{{\mathcal{A}},{\mathcal{B}}}>0$ and
${\mathcal{E}}_{\mathrm{KD+}}=\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$.
Let $U_{\varepsilon}$ be a family of unitary transition matrices between bases
${\mathcal{A}}_{\varepsilon}$ and ${\mathcal{B}}_{\varepsilon}$ satisfying
$\lim_{\varepsilon\to 0}U_{\varepsilon}=U$. Then, for all $\varepsilon$
sufficiently small, one has
${\mathcal{E}}_{\mathrm{KD+}}^{\varepsilon}=\mathrm{conv}\left({\mathcal{A}}_{\varepsilon},{\mathcal{B}}_{\varepsilon}\right)$.

Proposition 4.3 states that the set of $U$ for which
${\mathcal{E}}_{\mathrm{KD+}}=\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$
is an open set, so this proposition proves part (iv) of Theorem 1.1.

###### Proof.

Consider

$\mathcal{L}=\\{M\in\mathcal{M}_{d}(\mathbb{R})\mid\sum_{i}M_{ij}=0,\
\sum_{j}M_{ij}=0\\},$

which is a $(d-1)^{2}$-dimensional real vector space. As a result of Eq.
(3.2), one has

$\mathrm{Im}Q_{\varepsilon}:\mathcal{S}_{d}\to\mathcal{L}.$

Here, $Q_{\varepsilon}(\,\cdot\,)$ is the KD distribution associated to
$U_{\varepsilon}$. Suppose that, for $\varepsilon=0$,
${\mathcal{E}}_{\mathrm{KD+}}=\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$.
Then, according to Proposition 3.2,
$\dim({\mathrm{Ker}}{(\mathrm{Im}Q)})=2d-1$ and hence, since
$\dim\mathcal{S}_{d}=d^{2}$, it follows that $\mathrm{Im}Q$ is surjective. We
now show that, for sufficiently small $\varepsilon$,
$\mathrm{Im}Q_{\varepsilon}$ is also surjective. Writing

$\mathcal{S}_{d}={\mathrm{Ker}}(\mathrm{Im}Q)\oplus[{\mathrm{Ker}}(\mathrm{Im}Q)]^{\perp},$

it follows that
$\widehat{\mathrm{Im}Q}:=\mathrm{Im}Q_{\mid[{\mathrm{Ker}}(\mathrm{Im}Q)]^{\perp}}$
is a linear isomorphism between $[{\mathrm{Ker}}(\mathrm{Im}Q)]^{\perp}$ and
$\mathcal{L}$. It therefore has an inverse $\widehat{\mathrm{Im}Q}^{-1}$. Let
us now write

$\delta U_{\varepsilon}:=U_{\varepsilon}-U,$

with $\delta U_{\varepsilon}\to 0$ as $\varepsilon\to 0$. Then,

$\widehat{\delta\mathrm{Im}Q_{\varepsilon}}:=\widehat{\mathrm{Im}Q_{\varepsilon}}-\widehat{\mathrm{Im}Q},$

with $\delta\mathrm{Im}Q_{\varepsilon}\to 0$. We now consider
$\widehat{\mathrm{Im}Q_{\varepsilon}}:=\mathrm{Im}Q_{\varepsilon\mid[{\mathrm{Ker}}(\mathrm{Im}Q)]^{\perp}}$,
so that

$\widehat{\mathrm{Im}Q_{\varepsilon}}=\widehat{\mathrm{Im}Q}+\delta\widehat{\mathrm{Im}Q_{\varepsilon}}:[{\mathrm{Ker}}(\mathrm{Im}Q)]^{\perp}\to\mathcal{L}.$

To conclude the proof, we show that $\widehat{\mathrm{Im}Q_{\varepsilon}}$ is
a linear isomorphism. One has

$\widehat{\mathrm{Im}Q_{\varepsilon}}=\widehat{\mathrm{Im}Q}(\widehat{\mathbb{I}}+\widehat{\mathrm{Im}Q}^{-1}\delta\widehat{\mathrm{Im}Q_{\varepsilon}}).$

Since
$(\widehat{\mathbb{I}}+\widehat{\mathrm{Im}Q}^{-1}\delta\widehat{\mathrm{Im}Q_{\varepsilon}})$
is a small perturbation of the identity, it is invertible. So
$\widehat{\mathrm{Im}Q_{\varepsilon}}$ is invertible as the composition of two
invertible maps. This implies that $\mathrm{Im}Q_{\varepsilon}$ is surjective
and, hence, that the
$\dim\left({\mathrm{Ker}}({\mathrm{Im}Q_{\varepsilon}})\right)=2d-1$.
Proposition 3.2 then implies that
${\mathcal{E}}_{\mathrm{KD+}}^{\varepsilon}=\mathrm{conv}\left({\mathcal{A}}_{\varepsilon},{\mathcal{B}}_{\varepsilon}\right)$.
∎

Remark. We note that, in dimension $d=2$, if there is a zero in $U$, then $U$
is, up to phase changes, either equal to $\begin{pmatrix}1&0\\\
0&1\end{pmatrix}$ or to $\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}$. In that
case, the two bases are not distinct and all pure states are KD-positive, and
thus all mixed state are also KD-positive. In higher dimension $d\geq 3$, the
presence of zeroes in $U$ considerably complicates the analysis.

We conjecture that Theorem 1.1 (ii) is true in all dimensions $d\geq 2$. We
numerically checked this conjecture in dimensions $d$ up to $10$. For that
purpose, we sampled random unitary matrices according to the Haar measure on
the unitary group and computed numerically the rank of $\mathrm{Im}Q$ for each
such matrix. When it equals $(d-1)^{2}$, Proposition 3.2 (ib) guarantees that
${\mathcal{E}}_{\mathrm{KD+}}=\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$.
We did not find any instance where this condition was not satisfied.

We now prove Proposition 1.2, which can be seen as a first step in the proof
of this conjecture. For convenience, we repeat it here:

######  Proposition 1.2.

Let $d\geq 2$. There exists an open dense set of unitaries of probability $1$
for which
${\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}={\mathcal{A}}\cup{\mathcal{B}}$.

###### Proof.

We define $\mathcal{V}$ to be the set of $d$ by $d$ unitary matrices $U$
satisfying $m_{{\mathcal{A}},{\mathcal{B}}}>0$ and

$\forall(k,k^{\prime},j,j^{\prime})\in\llbracket 1,d\rrbracket^{4},k\neq
k^{\prime},j\neq j^{\prime},\
\phi_{k,j}-\phi_{k^{\prime},j}\neq\phi_{k,j^{\prime}}-\phi_{k^{\prime},j^{\prime}}[2\pi].$
(4.9)

Here we wrote, as before, $U_{kj}=\langle
a_{k}|b_{j}\rangle=A_{kj}\exp(i\phi_{kj})$, with $A_{kj}>0$. We will first
show that, if $U\in\mathcal{V}$, then the only pure KD-positive states are the
basis states. We proceed by contradiction. Suppose that there exists a pure
KD-classical state $|\psi\rangle\in\mathcal{H}$ that is not a basis state. By
reordering the two bases, we can suppose that
$S_{{\mathcal{A}}}(\psi)=\llbracket 1,n_{{\mathcal{A}}}(\psi)\rrbracket$ and
$S_{{\mathcal{B}}}(\psi)=\llbracket 1,n_{{\mathcal{B}}}(\psi)\rrbracket$ with
$n_{{\mathcal{A}}}(\psi)\geqslant 2$ and $n_{{\mathcal{B}}}(\psi)\geqslant 2$.
Then, we change the phases of the basis states as follows: $|a_{j}\rangle$ is
changed to $e^{-i\alpha_{j}}|a_{j}\rangle$ for $j\in\llbracket
1,n_{{\mathcal{A}}}(\psi)\rrbracket$ where $\alpha_{j}=\arg(\langle
a_{j}|\psi\rangle)$ and $|b_{j}\rangle$ is changed to
$e^{i\beta_{j}}|b_{j}\rangle$ for $j\in\llbracket
1,n_{{\mathcal{B}}}(\psi)\rrbracket$ where $\beta_{j}=\arg(\langle
b_{j}|\psi\rangle)$. Thus, for all $(i,j)\in\llbracket
1,n_{{\mathcal{A}}}(\psi)\rrbracket\times\llbracket
1,n_{{\mathcal{B}}}(\psi)\rrbracket$,

$\langle
a_{i}|\psi\rangle\in\mathbb{R}^{+}_{*},\langle\psi|b_{j}\rangle\in\mathbb{R}^{+}_{*}\
\mathrm{and}\ Q_{ij}(\psi)=A_{ij}e^{i(\phi_{ij}+\alpha_{i}+\beta_{j})}\langle
a_{i}|\psi\rangle\langle\psi|b_{j}\rangle\in\mathbb{R}^{+}_{*}.$

Consequently,

$\forall(i,j)\in\llbracket 1,n_{{\mathcal{A}}}(\psi)\rrbracket\times\llbracket
1,n_{{\mathcal{B}}}(\psi)\rrbracket,\phi_{ij}+\alpha_{i}+\beta_{j}=0\leavevmode\nobreak\
[2\pi].$

As $n_{{\mathcal{A}}}(\psi)\geqslant 2$ and $n_{{\mathcal{B}}}(\psi)\geqslant
2$, it follows that

$\phi_{11}-\phi_{12}=\beta_{2}-\beta_{1}\leavevmode\nobreak\ [2\pi]\
\mathrm{and}\ \phi_{21}-\phi_{22}=\beta_{2}-\beta_{1}\leavevmode\nobreak\
[2\pi].$

Thus,

$\phi_{11}-\phi_{12}=\phi_{21}-\phi_{22}\leavevmode\nobreak\ [2\pi],$

which is a contradiction because $U\in\mathcal{V}$. Therefore, the only KD-
classical pure states associated to $U\in\mathcal{V}$ are the basis states.

We now show that the set $\mathcal{V}$ is an open and dense set. That it is
open follows directly from its definition. It remains to show that it is
dense. For that purpose, we will show below that the set $\mathcal{Z}$ defined
by

$\phi_{11}-\phi_{12}\neq\phi_{21}-\phi_{22}\leavevmode\nobreak\ [2\pi]\
\mathrm{and}\ m_{{\mathcal{A}},{\mathcal{B}}}>0$

is dense. Reordering the basis elements, it then follows that all sets defined
by

$\phi_{k,j}-\phi_{k^{\prime},j}\neq\phi_{k,j^{\prime}}-\phi_{k^{\prime},j^{\prime}}\leavevmode\nobreak\
[2\pi]$

are dense. One concludes that $\mathcal{V}$ is dense as a finite intersection
of dense sets. It remains to prove that $\mathcal{Z}$ is dense. Let
$U\in\mathcal{Z}$; hence $m_{{\mathcal{A}},{\mathcal{B}}}>0$ and
$\phi_{11}-\phi_{12}=\phi_{21}-\phi_{22}\leavevmode\nobreak\ [2\pi]$. We
denote by $(C_{i})_{i\in\llbracket 1,d\rrbracket}$ the columns of $U$. We
define, for $\theta\in\mathbb{R}$

$C_{1}(\theta)=\begin{pmatrix}A_{11}e^{i(\phi_{11}+\theta)}\\\
A_{21}e^{i\phi_{21}}\\\ \vdots\\\ A_{d1}e^{i\phi_{d1}}\end{pmatrix}\
\mathrm{and}\ C_{2}(\theta)=\begin{pmatrix}A_{12}e^{i(\phi_{12}-\theta)}\\\
A_{22}e^{i\phi_{22}}\\\ \vdots\\\ A_{d2}e^{i\phi_{d2}}\end{pmatrix}.$ (4.10)

Note that $C_{1}(\theta)$ and $C_{2}(\theta)$ are normalized and orthogonal.
By applying the Gram-Schmidt algorithm to
$\begin{pmatrix}C_{1}(\theta)&C_{2}(\theta)&C_{3}&\cdots&C_{d}\end{pmatrix}$,
we obtain a unitary matrix

$U_{\theta}=\begin{pmatrix}C_{1}(\theta)&C_{2}(\theta)&C_{3}(\theta)&\cdots&C_{d}(\theta)\end{pmatrix}$

such that $U(\theta)\rightarrow U$ when $\theta\rightarrow 0$. Therefore, for
$\theta\not=0$ small enough, $m_{{\mathcal{A}},{\mathcal{B}}}(\theta)>0$ and

$\phi_{21}-\phi_{11}-\theta\neq\phi_{22}-\phi_{12}+\theta\ [2\pi].$

This proves $\mathcal{V}$ is dense.

We finally show that $\mathcal{V}$ is a set of probability one for the unique
normalized Haar measure on the unitary group. Note that $\mathcal{V}$ is
contained in the complement of the union of the subsets

$\phi_{k,j}-\phi_{k^{\prime},j}=\phi_{k,j^{\prime}}-\phi_{k^{\prime},j^{\prime}}\
[2\pi]$

and of the subsets where one of the elements of $U$ vanishes. Each of those
subsets is a lower dimensional submanifold of the unitary group. Since the
Haar measure is known to be absolutely continuous with respect to Lebesgue
measure in any local coordinate system on the unitary group [65], this implies
that the Haar measure of these manifolds vanishes. The same property therefore
holds for their union, so that indeed $\mathcal{V}$ has measure $1$.

∎

Proving Eq. (1.2) for a particular $U$ can be hard, as the result on the DFT
in prime dimensions (Theorem 1.1 (iii)) shows. It is certainly not always
true. To see this, one may first note that for the DFT in non-prime
dimensions, it is well known (see for example [24, 59]) that
${\mathcal{A}}\cup{\mathcal{B}}\subsetneq{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}.$
We do not know, however, if in this case
${\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}\subsetneq{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}$
. In the next section we construct examples where

${\mathcal{A}}\cup{\mathcal{B}}\subsetneq{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}\subsetneq{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}.$

We further point out again that Eq. (1.2) is notably different from what
happens for the continuous-variable Wigner positivity. Indeed, there exist
Wigner positive states outside the convex hull of the pure Wigner positive
states [2, 66]. We note also that the discrete-variable Wigner function in
$d=3$ has positive states outside the convex hull of its pure positive states
[63]. This is not the case for the KD distribution associated to the DFT, as a
result of Theorem 1.1 (iii).

##  5 Extreme KD-positive states are not necessarily pure

Below, we construct examples where

${\mathcal{A}}\cup{\mathcal{B}}\subsetneq{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}\subsetneq{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}\quad\textrm{or}\quad{\mathcal{A}}\cup{\mathcal{B}}={\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}\subsetneq{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}.$

In other words, in these cases there exist mixed extreme KD-positive states.
In particular, when such states are viewed as a convex combination of pure
states, at least one of those pure states must be KD-negative. In fact, the
convex combination of any state in

${\mathcal{E}}_{\mathrm{KD+}}\setminus\mathrm{conv}\left({\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}\right),$

must include pure KD-negative states. Proposition 3.2 states that
${\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}={\mathcal{A}}\cup{\mathcal{B}}$
is true if and only if the space $V_{\mathrm{KDr}}$ of KD-real operators is of
its smallest possible dimension: $\dim V_{\mathrm{KDr}}=2d-1$ (see Eq.
(3.10)). In Section 5.1, we show that this is never the case if $U$ is a real
(hence orthogonal) matrix and $d\geq 3$ (Lemma 5.1).

Let us point out that this statement does not contradict Theorem 1.1 (ii),
which states that in dimension $d=3$, the equality
${\mathcal{E}}_{\mathrm{KD+}}=\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$
holds with probability one; nor does it contradict our conjecture that it
holds for all $d\geq 3$. Indeed, the space of unitary matrices is of dimension
$d^{2}$; the space of orthogonal matrices is only of dimension
$\frac{d(d+1)}{2}$, i.e., a “thin” subset. More formally, the space of
orthogonal matrices is a hypersurface of lower dimension with empty interior
and is therefore of zero probability among all unitary matrices.

The result of Lemma 5.1 allows us to construct examples of bases
${\mathcal{A}}$ and ${\mathcal{B}}$ for which
${\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}\subsetneq{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}$.
In Section 5.2, we provide an explicit example of an orthogonal matrix
$U^{\star}$ in $d=3$ for which

${\mathcal{A}}\cup{\mathcal{B}}\subsetneq{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}\subsetneq{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}.$

In Section 5.3, we first show that the transition matrix between the bases of
two different spin components of a spin-$s$ system is always (equivalent to) a
real transition matrix. We then show that the a matrix $U^{\star}$ constructed
in Secion 5.2 arises for a spin-$1$ system, in which the two bases
${\mathcal{A}}$ and ${\mathcal{B}}$ correspond to the eigenvectors of the spin
component in the $z$ direction and another, specific direction, respectively.

Finally, in Section 5.4, we show that there exist examples in all dimensions
$d=2^{n}$ (for integer $n$) and in all dimensions $d=4m$ (for integer $m\leq
166$) for which

${\mathcal{A}}\cup{\mathcal{B}}={\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}\subsetneq{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}.$

In these cases, the two bases are perturbations of real MUB bases.

For completeness, we mention that we did not find any example where

${\mathcal{A}}\cup{\mathcal{B}}\subsetneq{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}={\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}.$

###  5.1 Real transition matrices: structural results

######  Lemma 5.1.

If $U$ is real and $m_{{\mathcal{A}},{\mathcal{B}}}>0$, then
$V_{\mathrm{KDr}}={\mathcal{S}_{d,\mathrm{r}}}$, where
${\mathcal{S}_{d,\mathrm{r}}}$ is the set of self-adjoint operators that have
a real and symmetric matrix on the ${\mathcal{A}}$ (and hence also on the
${\mathcal{B}}$) basis. Hence, $\dim V_{\mathrm{KDr}}=\frac{d(d+1)}{2}$. If in
addition $d\geq 3$, then
${\mathcal{A}}\cup{\mathcal{B}}\subsetneq{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}$.

Note that $\frac{d(d+1)}{2}$ is strictly larger than $2d-1$ for all $d>2$. We
can further interpret this lemma as follows. If $U$ is real, then the
observables $F$ that have a real KD symbol are precisely those described by a
real symmetric matrix on the ${\mathcal{A}}$ and ${\mathcal{B}}$ bases. This
constitutes a concrete identification of $V_{\mathrm{KDr}}$, which is not
available in general. In this situation, the kernel of $\textrm{Im}Q$, which
is $V_{\mathrm{KDr}}$, is large and in particular, larger than
${\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$.

###### Proof.

Suppose that $F\in V_{\mathrm{KDr}}$, so

$\forall(i,j)\in\llbracket 1,d\rrbracket^{2},\;\mathrm{Im}(Q_{ij}(F))=0.$

We know also that $\forall(i,j)\in\llbracket
1,d\rrbracket^{2},\;Q_{ij}(F)=\langle b_{j}|a_{i}\rangle\langle
a_{i}|F|b_{j}\rangle=\langle b_{j}|a_{i}\rangle\sum_{k=1}^{d}\langle
a_{k}|b_{j}\rangle\langle a_{i}|F|a_{k}\rangle$. As $\langle
b_{j}|a_{i}\rangle\langle a_{k}|b_{j}\rangle\in\mathbb{R}$, we finally find
that

$\mathrm{Im}(Q_{ij}(F))=\sum_{k=1,k\neq i}^{d}\langle
b_{j}|a_{i}\rangle\langle a_{k}|b_{j}\rangle\mathrm{Im}\left(\langle
a_{i}|F|a_{k}\rangle\right)=\langle b_{j}|a_{i}\rangle\left(\sum_{k=1,k\neq
i}^{d}\mathrm{Im}\left(\langle a_{i}|F|a_{k}\rangle\right)\langle
a_{k}|b_{j}\rangle\right)=0.$

Since this equation holds for all $j$, by fixing $i\in\llbracket
1,d\rrbracket$ we can write

$\forall j\in\llbracket 1,d\rrbracket,\sum_{k=1,k\neq
i}^{d}\mathrm{Im}\left(\langle a_{i}|F|a_{k}\rangle\right)\langle
a_{k}|b_{j}\rangle=0$

and thus

$\sum_{k=1,k\neq i}^{d}\mathrm{Im}\left(\langle
a_{i}|F|a_{k}\rangle\right)|a_{k}\rangle=0.$

So, for all $k\in\llbracket 1,d\rrbracket,\mathrm{Im}\left(\langle
a_{i}|F|a_{k}\rangle\right)=0$ and thus, $F\in\mathcal{S}_{d,\mathrm{r}}$.

For the last statement, note that, if
${\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}={\mathcal{A}}\cup{\mathcal{B}}$,
then
${\mathcal{E}}_{\mathrm{KD+}}=\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$
and hence, by Proposition 3.2, $\dim V_{\mathrm{KDr}}=2d-1$, which is a
contradiction with the fact that dim$V_{\mathrm{KDr}}=\frac{d(d+1)}{2}$. ∎

As announced above, it is the goal of this section to exhibit mixed KD-
positive states that cannot be written as convex combinations of the pure KD-
positive states. In order to find and analyse such states, we will concentrate
on dimension $d=3$ where the analysis is tractable.

Note that, when $d=3$, then $\dim V_{\mathrm{KDr}}=6$ and
$\dim({\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})})=5$, so
that ${\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$ is a co-
dimension $1$ subspace of $V_{\mathrm{KDr}}$. We characterize the one-
dimensional subspace of $V_{\mathrm{KDr}}$ perpendicular to
${\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$, as follows. We
use the Hilbert-Schmidt inner product on $\mathcal{S}_{d,\mathrm{r}}$. Then,
$F_{\perp}\in\mathcal{S}_{d,\mathrm{r}}$ is a unit vector orthogonal to
${\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$ if and only if

$\forall i\in\llbracket 1,3\rrbracket,\
\mathrm{Tr}\,(F_{\perp}|a_{i}\rangle\langle a_{i}|)=0,\
\mathrm{Tr}\,(F_{\perp}|b_{i}\rangle\langle b_{i}|)=0,\ \mathrm{and}\
\mathrm{Tr}\,F^{2}_{\perp}=1.$

It follows that the matrix of $F_{\perp}$ in the ${\mathcal{A}}$-basis is of
the form

$F_{\perp}=\frac{1}{\sqrt{2}}\begin{pmatrix}0&f_{1}&f_{2}\\\ f_{1}&0&f_{3}\\\
f_{2}&f_{3}&0\end{pmatrix}$ (5.1)

with $f=(f_{1},f_{2},f_{3})\in\mathbb{R}^{3}$ and

$f_{1}^{2}+f_{2}^{2}+f_{3}^{2}=1.$

The vector $f$ is, up to a sign, uniquely determined by the conditions
$\mathrm{Tr}\,(F_{\perp}|b_{j}\rangle\langle b_{j}|)=0$ for all
$j\in\llbracket 1,3\rrbracket$. Since $\mathrm{Tr}\,F_{\perp}=0$, $F_{\perp}$
can be neither positive nor negative. We can assume that $f_{-}<0\leq
f_{0}\leq f_{+}$ are its three eigenvalues. Consequently,

$V_{\mathrm{KDr}}={\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}\oplus\mathbb{R}F_{\perp}.$
(5.2)

Any $F\in V_{\mathrm{KDr}}$ can then be decomposed uniquely as
$F=F_{\mathrm{b}}+xF_{\perp}$, with
$F_{\mathrm{b}}\in{\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$
and $x\in\mathbb{R}$. Note that

$\mathrm{Tr}\,F=\mathrm{Tr}\,F_{\mathrm{b}},\quad\mathrm{Tr}\,F^{2}=\mathrm{Tr}\,F_{\mathrm{b}}^{2}+x^{2}.$
(5.3)

We finally note that it follows from results in [24] that, when $d=3$ and
$m_{{\mathcal{A}},{\mathcal{B}}}>0$, there is a finite set of pure KD-positive
states. Further structural information on the set
${\mathcal{E}}_{\mathrm{KD+}}$ and its extreme points, for real orthogonal
$U$, is given in Lemma B.1 and Proposition B.2.

The above results can be summarized as follows. There exists a convex subset
$\mathcal{D}$ of
${\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$, real numbers
$x_{\min}\leq 0\leq x_{\max}$, a concave function

$x_{+}:{\mathcal{D}}\to[x_{\min},x_{\max}]\subset\mathbb{R},$

as well as a convex function

$x_{-}:\mathcal{D}\to[x_{\min},x_{\max}]\subset\mathbb{R},$

so that

${\mathcal{E}}_{\mathrm{KD+}}=\\{\sigma+[x_{-}(\sigma),x_{+}(\sigma)]F_{\perp}\mid\sigma\in{\mathcal{D}}\\}.$

In other words, ${\mathcal{E}}_{\mathrm{KD+}}$ can be seen as the intersection
of the subgraph of $x_{+}$ and of the supergraph of $x_{-}$. As a result, the
extreme points of ${\mathcal{E}}_{\mathrm{KD+}}$ lie on the graphs of $x_{-}$
and of $x_{+}$:

${\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}\subseteq x_{+}(\mathcal{D})\cup
x_{-}(\mathcal{D}).$

Since the precise form of the functions $x_{\pm}$ as well as of the set
$\mathcal{D}$ depend in a nontrivial manner on the matrix $U$, determining
explicitly the nature of the set of extreme KD-positive states is far from
straigthforward for general real $U$, even in dimension $3$. In the next
subsection, we study a special example where the nature of the extreme KD-
positive states can be mapped out in more detail. In particular, we show that
${\mathcal{E}}_{\mathrm{KD+}}$ is not necessarily a polytope.

###  5.2 Extreme KD-positive states can be mixed: an example in $d=3$.

We now provide an example of an orthogonal $U$ in $d=3$ for which

${\mathcal{A}}\cup{\mathcal{B}}\subsetneq{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}\subsetneq{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}.$

The first strict inclusion follows from Proposition 3.2 and says that there
exist additional pure KD-positive states distinct from the basis states. In
our example, there will, in addition, be mixed extreme KD-positive states. In
other words, the polytope
$\mathrm{conv}\left({\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}\right)$ does
not exhaust all of ${\mathcal{E}}_{\mathrm{KD+}}$.

To motivate our choice of $U$, we first recall that it was shown in [24]
(Theorem 13) that, when $d=3$, any $U$ for which
$\frac{m_{{\mathcal{A}},{\mathcal{B}}}}{M_{{\mathcal{A}},{\mathcal{B}}}}>\frac{1}{\sqrt{2}}$
has the property that
${\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}={\mathcal{A}}\cup{\mathcal{B}}$.
If there existed such $U$ among the real orthogonal matrices, this would imply
by Lemma 5.1 that
${\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}\subsetneq{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}$.
However, such real orthogonal matrices do not exist. The largest value of
$\frac{m_{{\mathcal{A}},{\mathcal{B}}}}{M_{{\mathcal{A}},{\mathcal{B}}}}$ that
can be obtained for real orthogonal matrices is $\frac{1}{2}$, attained for
the matrix

$U^{\star}=\frac{1}{3}\begin{pmatrix}-1&2&2\\\ 2&-1&2\\\ 2&2&-1\end{pmatrix}.$
(5.4)

We further know from [23] that in $d=3$, a pure KD-positive state
$|\psi\rangle$ of any transition matrix with
$m_{{\mathcal{A}},{\mathcal{B}}}>0$ satisfies
$n_{{\mathcal{A}},{\mathcal{B}}}(\psi)=n_{\mathcal{A}}(\psi)+n_{\mathcal{B}}(\psi)=4$,
where

$n_{{\mathcal{A}}}(\psi)=\sharp\\{i\in\llbracket 1,d\rrbracket\mid\langle
a_{i}|\psi\rangle\not=0\\},\ \mathrm{and}\
n_{{\mathcal{B}}}(\psi)=\sharp\\{j\in\llbracket 1,d\rrbracket\mid\langle
b_{j}|\psi\rangle\not=0\\}.$ (5.5)

Here, $\sharp\\{\cdot\\}$ denotes the cardinality of $\\{\cdot\\}$. One can
construct all such states and check if they are KD-positive. Doing so, we
find, in addition to the basis states, the following three KD-positive states:

$|\varphi_{3}\rangle=\frac{|a_{1}\rangle-|a_{2}\rangle}{\sqrt{2}},\
|\varphi_{2}\rangle=\frac{|a_{3}\rangle-|a_{1}\rangle}{\sqrt{2}},\
\mathrm{and}\
|\varphi_{1}\rangle=\frac{|a_{2}\rangle-|a_{3}\rangle}{\sqrt{2}}.$

Their associated KD distribution s are

$Q(\varphi_{3})=\frac{1}{6}\begin{pmatrix}1&2&0\\\ 2&1&0\\\
0&0&0\end{pmatrix},\ Q(\varphi_{2})=\frac{1}{6}\begin{pmatrix}1&0&2\\\
0&0&0\\\ 2&0&1\end{pmatrix},\ \mathrm{and}\
Q(\varphi_{1})=\frac{1}{6}\begin{pmatrix}0&0&0\\\ 0&1&2\\\
0&2&1\end{pmatrix},$

respectively. Here, to lighten the notation, we write
$Q(\varphi_{i}):=Q(|\varphi_{i}\rangle\langle\varphi_{i}|)$. The operator
$F_{\perp}$ in Eq. (5.1) is readily computed to be

$F_{\perp}=\frac{1}{\sqrt{6}}\begin{pmatrix}0&1&1\\\ 1&0&1\\\
1&1&0\end{pmatrix}.$

A simple computation shows that

$|\varphi_{i}\rangle\langle\varphi_{i}|=F_{b}^{i}+x_{i}F_{\perp},$

with
$x_{i}=\langle\varphi_{i}|F_{\perp}|\varphi_{i}\rangle=-\frac{1}{\sqrt{6}}$
and

$F_{b}^{i}=\frac{5}{6}\sum_{k\neq i}|b_{k}\rangle\langle
b_{k}|+\frac{1}{12}|b_{i}\rangle\langle b_{i}|-\frac{3}{4}|a_{i}\rangle\langle
a_{i}|.$

Hence, that the triangle with vertices
$|\varphi_{i}\rangle\langle\varphi_{i}|$ lies in a plane parallel to
${\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$ at a distance
$|x_{i}|$ from it. It follows that
$\mathrm{conv}\left({\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}\right)$ lies
between $x=-\frac{1}{\sqrt{6}}$ and $x=0$. Indeed, if

$\rho=\sum_{i=1}^{3}\lambda_{i}|a_{i}\rangle\langle
a_{i}|+\mu_{i}|b_{i}\rangle\langle
b_{i}|+\delta_{i}|\varphi_{i}\rangle\langle\varphi_{i}|$

with $\sum_{i=1}^{3}\lambda_{i}+\mu_{i}+\delta_{i}=1$ where
$\lambda_{i}\geqslant 0,\mu_{i}\geqslant 0,\ \mathrm{and}\ \delta_{i}\geqslant
0$ for all $i\llbracket 1,3\rrbracket$, then

$x_{\rho}=\mathrm{Tr}(\rho
F_{\perp})=\sum_{i=1}^{3}\delta_{i}\langle\varphi_{i}|F_{\perp}|\varphi_{i}\rangle=-\frac{1}{\sqrt{6}}\sum_{i=1}^{3}\delta_{i}.$

Thus, $-\frac{1}{\sqrt{6}}\leqslant x_{\rho}\leqslant 0$. It follows that any
KD-positive states for which $x>0$ cannot belong to
$\mathrm{conv}\left({\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}\right)$. We
now construct such states.

We consider $\rho({x})=\frac{1}{3}I_{d}+xF_{\perp}$. Its eigenvalues are
$\left\\{\frac{2(1+x\sqrt{6})}{6},\frac{2-x\sqrt{6}}{6},\frac{2-x\sqrt{6}}{6}\right\\}$
which are positive if and only if
$x\in[-\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}}]$. Only in these cases is
$\rho(x)$ a quantum state. Moreover,

$Q(F_{\perp})=\frac{\sqrt{6}}{27}\begin{pmatrix}-2&1&1\\\ 1&-2&1\\\
1&1&-2\end{pmatrix},$

and consequently

$Q(\rho(x))=\frac{1}{27}\begin{pmatrix}1-2x\sqrt{6}&4+x\sqrt{6}&4+x\sqrt{6}\\\
4+x\sqrt{6}&1-2x\sqrt{6}&4+x\sqrt{6}\\\
4+x\sqrt{6}&4+x\sqrt{6}&1-2x\sqrt{6}\end{pmatrix}.$

So $\rho(x)$ is KD-positive if and only if
$x\in[-\frac{4}{\sqrt{6}},\frac{1}{2\sqrt{6}}]$. Finally, $\rho(x)$ is a KD-
positive state provided that $x\in[-\frac{1}{\sqrt{6}},\frac{1}{2\sqrt{6}}]$.

Thus, for all $x\in]0,\frac{2}{3}]$, $\rho(x)$ is a KD-positive mixed state
and from what precedes, it follows that $\rho(x)$ is not in the convex hull of
the KD-positive pure states. This construction therefore exhibits explicit KD-
positive mixed states that are not in the convex hull of
${\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}$. In addition, Lemma B.1 allows
the generalisation of this construction, which shows that, for every
$\rho\in\mathrm{Int}(\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right))$,
there exists a continuous family of states $\rho(x)=\rho+xF_{\perp}$ which are
not in the convex hull of KD-positive pure states. Here and elsewhere,
$\mathrm{Int}(A)$ stands for the interior of $A$.

Since ${\mathcal{E}}_{\mathrm{KD+}}$ is a convex and compact set it follows
from the Krein-Milman Theorem [58] that it is the convex hull of its extreme
points. We thus reach the conclusion that ${\mathcal{E}}_{\mathrm{KD+}}$ has
extreme points that are not pure and not in the convex hull of the pure KD-
positive states. We identify some of them explicitly in Appendix B.2. It is
known that mixed extreme states exist also for the discrete variable Wigner
function (at least in $d=3$ [63]), as well as in the continuous variable
Wigner function [60, 61, 62]. However, to the best of our knowledge, such
states have never been explicitly identified.

Comment. We point out that the set ${\mathcal{E}}_{\mathrm{KD+}}$ is in the
example above not a polytope, since it has an infinite number of extreme
points. To see this, note that, if ${\mathcal{E}}_{\mathrm{KD+}}$ was a
polytope, any 2-dimensional section of it would be a polygon. However, if we
consider the 2-dimensional plane containing $F_{\perp}$,
$\frac{1}{2}(|a_{1}\rangle\langle a_{1}|+|a_{2}\rangle\langle a_{2}|)$ and
$|a_{3}\rangle\langle a_{3}|$ and we intersect it with
${\mathcal{E}}_{\mathrm{KD+}}$, then simple computations show that this
intersection is not a polygon, as illustrated in Fig.1. Hence,
${\mathcal{E}}_{\mathrm{KD+}}$ is not a polytope.

Figure 1: A point $(k,x)$ on this graph represents the state
$\frac{k}{2}(|a_{1}\rangle\langle a_{1}|+|a_{2}\rangle\langle
a_{2}|)+(1-k)|a_{3}\rangle\langle a_{3}|+xF_{\perp}$. A state is KD positive
if and only if it lies inside the region drawn. Thus, this figure displays a
2-dimensional section of ${\mathcal{E}}_{\mathrm{KD+}}$ that is not a polygon.
This shows that ${\mathcal{E}}_{\mathrm{KD+}}$ is not a polytope.

###  5.3 An application: the case of spin-$s$

In this subsection, we show how the transition matrix $U^{\star}$ of the
previous subsection arises naturally in a spin-$1$ system.

First, we show that the transition matrices between spin component bases for
spin-$s$ systems are (equivalent to) a real matrix. Let $|z,m\rangle$ be the
standard basis vectors of $J_{z}$, with eigenvalues $m=-s,\dots,s$. Let

$R(\alpha,\beta,\gamma)=R_{z}(\alpha)R_{y}(\beta)R_{z}(\gamma),$

be the rotation matrix with Euler angles $\alpha,\beta,\gamma$. Further, let

$\hat{R}(\alpha,\beta,\gamma)=\exp(-i\alpha J_{z})\exp(-i\beta
J_{y})\exp(-i\gamma J_{z}),$

be the irreducible unitary action of the rotation group on the spin-$s$ space
$\mathcal{H}^{(s)}$. Then the states

$\hat{R}(\alpha,\beta,\gamma)|z,m\rangle$

are the eigenstates of

$\hat{R}(\alpha,\beta,\gamma)e_{z}\cdot\vec{J},$

where $e_{z}$ is the unit vector along the $z$-axis and
$\vec{J}=(J_{x},J_{y},J_{z})$ [67]. We define

$|a_{m}\rangle=\exp(-i\alpha m)|z,m\rangle,\quad|b_{m}\rangle=\exp(i\gamma
m)\hat{R}(\alpha,\beta,\gamma)|z,m\rangle.$ (5.6)

Let us write ${\mathcal{E}}_{\mathrm{KD+}}(\alpha,\beta,\gamma)$ for the
corresponding space of KD-positive states. Then the transition matrix
$U(\alpha,\beta,\gamma)$ between these two bases has matrix elements

$U_{m^{\prime}m}(\alpha,\beta,\gamma)=d^{(s)}_{m^{\prime}m}(\beta),$

where $d^{(s)}(\beta)$ is Wigner’s small $d$-matrix for spin-$s$ [67]. Note
that the transition matrix depends only on the Euler angle $\beta$, not on the
two others. Consequently, the same is true for
${\mathcal{E}}_{\mathrm{KD+}}(\alpha,\beta,\gamma)={\mathcal{E}}_{\mathrm{KD+}}(\beta)$.
Wigner’s small $d$-matrix is real-valued so that the theory of the previous
subsections applies. In particular, one never has
${\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}(\alpha,\beta,\gamma)={\mathcal{A}}\cup{\mathcal{B}}$
in this situation.

Let us now concentrate on the case $s=1$. One can then check that, with
$\alpha_{0}=-\gamma_{0}=\frac{\pi}{4}$ and $\beta_{0}=\arccos(-1/3)$,

$R(\alpha_{0},\beta_{0},\gamma_{0})=\frac{1}{3}\begin{pmatrix}-1&2&2\\\
2&-1&2\\\ 2&2&-1\end{pmatrix}.$

This unitary can be interpreted as a rotation by the angle $\pi$ about the
axis $n=\frac{1}{\sqrt{3}}\begin{pmatrix}1\\\ 1\\\ 1\end{pmatrix}$. The
$\\{|b_{m}\rangle\\}$ therefore forms, for $m=-1,0,1$, the eigenbasis of the
observable $e_{z^{\prime}}\cdot J$ where

$e_{z^{\prime}}=R(\alpha_{0},\beta_{0},\gamma_{0})e_{z},$

so that $e_{z^{\prime}}\cdot J=\frac{1}{3}(2J_{x}+2J_{y}-J_{z})$. Hence,

$U(\beta_{0})=d^{(1)}(\beta_{0})=\begin{pmatrix}\frac{1+\cos(\beta_{0})}{2}&-\frac{\sin(\beta_{0})}{\sqrt{2}}&\frac{1-\cos(\beta_{0})}{2}\\\
\frac{\sin(\beta_{0})}{\sqrt{2}}&\cos(\beta_{0})&-\frac{\sin(\beta_{0})}{\sqrt{2}}\\\
\frac{1-\cos(\beta_{0})}{2}&\frac{\sin(\beta_{0})}{\sqrt{2}}&\frac{1+\cos(\beta_{0})}{2}\end{pmatrix}=\frac{1}{3}\begin{pmatrix}1&-2&2\\\
2&-1&-2\\\ 2&2&1\end{pmatrix}.$

Furthermore,

$U(\beta_{0})=d^{(1)}(\beta_{0})=\begin{pmatrix}1&0&0\\\ 0&-1&0\\\
0&0&-1\end{pmatrix}U^{\star}\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\
0&0&1\end{pmatrix},$

where $U^{\star}$ is as in Eq. (5.4). In conclusion, it then follows from Eq.
(2.3) that the matrix $U(\beta_{0})$ is equivalent to the matrix $U^{\star}$.
As a result, if, for spin-$1$, the bases ${\mathcal{A}}$ and ${\mathcal{B}}$
are as in Eq. (5.6), with $\beta=\beta_{0}$, then the set of KD-positive
states ${\mathcal{E}}_{\mathrm{KD+}}(\beta_{0})$ is as described in the
previous subsection. In particular, there then exist mixed KD-positive states
that are not mixtures of pure KD-positive states.

###  5.4 Examples where
$\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)=\mathrm{conv}\left({\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}\right)\subsetneq{\mathcal{E}}_{\mathrm{KD+}}$

In this subsection, we show that there exist bases ${\mathcal{A}}$ and
${\mathcal{B}}$ for which

${\mathcal{A}}\cup{\mathcal{B}}={\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}\subsetneq{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}},$
(5.7)

so that

$\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)=\mathrm{conv}\left({\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}\right)\subsetneq{\mathcal{E}}_{\mathrm{KD+}}.$
(5.8)

In words, in this situation, the only pure KD-positive states are the basis
states, but those do not exhaust all extreme KD-positive states: there exist
also mixed extreme KD-positive states. With the following proposition we show
the occurrence of such situations in a large class of examples in dimensions
$d=4n$ (with integer $n\leq 166$) or $2^{m}$ with $m\in\mathbb{N}^{*}$ [68,
69]. These examples are less explicit than the three-dimensional example of
the previous subsections.

######  Proposition 5.2.

Suppose that $U$ is a real-valued transition matrix for MUB bases in dimension
$d\geqslant 4$. Then, there exists a real-valued and unitary matrix $W$ close
to $U$, such that $W$ satisfies Eq. (5.7).

###### Proof.

We write $U=(\langle a_{i}^{U}|b_{j}^{U}\rangle)_{(i,j)\in\llbracket
1,d\rrbracket^{2}}$. We have that

$m_{{\mathcal{A}},{\mathcal{B}}}^{U}=\min_{(i,j)\in\llbracket
1,d\rrbracket^{2}}\left|\langle
a_{i}^{U}|b_{j}^{U}\rangle\right|=\frac{1}{\sqrt{d}}$

and

$M_{{\mathcal{A}},{\mathcal{B}}}^{U}=\max_{(i,j)\in\llbracket
1,d\rrbracket^{2}}\left|\langle
a_{i}^{U}|b_{j}^{U}\rangle\right|=\frac{1}{\sqrt{d}}$

because $U$ is the transition matrix for MUB bases.

It follows from Theorem 13 and (the proof of) Theorem 5 in [24] that for all
$\varepsilon>0$ small enough, there exists a real unitary matrix $W=(\langle
a_{i}^{W}|b_{j}^{W}\rangle)_{(i,j)\in\llbracket 1,d\rrbracket^{2}}$ such that
$\left|\left|U-W\right|\right|_{\infty}\leqslant\varepsilon$ and with the
property that the only KD-positive pure states for $W$ are the basis states:
${\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}=\mathcal{A}\cup\mathcal{B}$.

Thus,
$\mathrm{span}_{\mathbb{R}}({\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}})={\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$.
On the other hand, since $W$ is real, we know from Lemma 5.1 that
$V_{\mathrm{KDr}}=\mathcal{S}_{d,\mathrm{r}}$. Hence
$\mathrm{dim}(V_{\mathrm{KDr}})=\frac{d(d+1)}{2}>2d-1=\mathrm{dim}({\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})})$,
so that it follows from Proposition 3.2 that
$\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)\subsetneq{\mathcal{E}}_{\mathrm{KD+}}$.
Thus, Eq. (5.7) holds. ∎

##  6 Conclusions and discussion

In recent years, the Kirkwood-Dirac quasiprobability distribution has risen as
a versatile and powerful tool to study and develop protocols in discrete-
variable quantum-information processing. Given two bases $\mathcal{A}$ and
$\mathcal{B}$ and a state $\rho$, the associated KD distribution can be either
a probability distribution or not. As reviewed in our Introduction, the
existence of negative or nonreal entries in a KD distribution has been linked
to quantum phenomena in several areas of quantum mechanics. This motivated us
to investigate the divide between positive and nonpositive KD distributions.
Previous studies [22, 23, 24] have mapped out sufficient and necessary
conditions for a pure state to assume a nonpositive KD distribution. But, to
the best of our knowledge, no previous work has provided such an analysis for
mixed states.

In this work, we have presented the first thorough analysis of the set of
mixed states that assume positive KD distributions. Our results can be grouped
in two categories.

  * •

Firstly, we have established that in several scenarios the set of KD positive
states equals the convex combinations of the bases’ states $\mathcal{A}$ and
$\mathcal{B}$. In particular, we have proven this to be the case for: $(i)$
any qubit ($d=2$) system provided that $m_{{\mathcal{A}},{\mathcal{B}}}>0$;
$(ii)$ an open dense set of probability $1$ of possible choices of bases
${\mathcal{A}}$ and ${\mathcal{B}}$ in dimension $3$; $(iii)$ prime
dimensions, when the unitary transition matrix between the two bases
${\mathcal{A}}$ and ${\mathcal{B}}$ is the discrete Fourier transform; and
$(iv)$ any two bases that are sufficiently close to some other pair of bases
for which the property holds. In addition to having shown that
${\mathcal{E}}_{\mathrm{KD+}}=\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$
for randomly chosen bases in dimension $d=3$, we conjecture that this is true
also in higher dimensions $d\geq 4$. We have given analytical and numerical
evidence to that effect.

  * •

Secondly, we have proven that there exist scenarios where the set of KD-
positive states includes mixed states that cannot be written as convex
combinations of pure KD-positive states: in other words, we have shown that in
such cases
${\mathcal{E}}_{\mathrm{KD+}}\neq\mathrm{conv}\left({\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}\right)$.
This mirrors what happens for mixed Wigner positive states which are known not
to all be mixtures of pure Wigner positive states [60, 61, 62]. However, we go
further by explicitly constructing, for a specific spin-$1$ system, extreme
KD-positive states that cannot be written as convex mixtures of pure KD-
positive states. For the Wigner distribution, such extreme positive states
have not yet been constructed.

Having a good understanding of the KD-positive states is a prerequisite for an
efficient study of states for which the KD distribution takes negative or
nonreal values. The latter are known to be related to nonclassical phenomena
in various applications. To analyze the connection between mixed nonpositive
states and nonclassicality, one should investigate measures and monotones of
KD nonpositivity. In an upcoming paper, currently under construction, we use
the findings of this work to analyze the so-called KD negativity [34] of mixed
states. Furthermore, our follow-up paper extends to mixed states the
characterization of KD-positive states via their support uncertainty, as was
done for pure states in [22, 23, 24].

Acknowledgments: This work was supported in part by the Agence Nationale de la
Recherche under grant ANR-11-LABX-0007-01 (Labex CEMPI), by the Nord-Pas de
Calais Regional Council and the European Regional Development Fund through the
Contrat de Projets État-Région (CPER), and by the CNRS through the MITI
interdisciplinary programs. We thank Girton College, Cambridge, for support of
this work. D.R.M. Arvidsson-Shukur thanks Nicole Yunger Halpern for useful
discussions.

##  Appendix A The geometric structure of
$\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$

The polytope $\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$ has
$2d$ vertices and lies in the $2(d-1)$-dimensional affine subspace of
${\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$ determined by
the constraint $\mathrm{Tr}\,F=1$. In this Appendix, we identify its interior
and the facets making up its boundary. We will see the interior is not empty,
so that the polytope is $2(d-1)$-dimensional. Its facets are therefore
$(2d-3)$-dimensional. They are polytopes with $2(d-1)$ vertices. In
particular, when $d=3$,
$\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$ is a four-
dimensional object. Its boundary has $9$ facets, which are three-dimensional
tetrahedra.

######  Lemma A.1.

Let $m_{{\mathcal{A}},{\mathcal{B}}}>0$. Let
$\rho=\sum_{i}\lambda_{i}|a_{i}\rangle\langle
a_{i}|+\sum_{j}\mu_{j}|b_{j}\rangle\langle
b_{j}|\in\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$, with
$\lambda_{i}\geq 0,\mu_{j}\geq 0$. Then

$\rho\in\mathrm{Int}\left(\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)\right)\Leftrightarrow\lambda_{1}\lambda_{2}\dots\lambda_{d}\not=0\
\textrm{or}\ \mu_{1}\mu_{2}\dots\mu_{d}\not=0.$ (A.1)

Note that the expression of $\rho$ as a convex combination of the basis
vectors is not unique. What we are saying is that, if $\rho$ can be expressed
in the manner stated, then it belongs to the interior of the polytope, and
vice versa.

###### Proof.

$\Rightarrow$ We will show the contrapositive. Suppose therefore that
$\lambda_{1}\lambda_{2}\dots\lambda_{d}=0=\mu_{1}\mu_{2}\dots\mu_{d}$. We need
to show
$\rho\not\in\mathrm{Int}(\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right))$.
We can assume, without loss of generality, that $\lambda_{d}=0=\mu_{d}$. Then
$Q(\rho)_{dd}=0.$ Now consider, for $\varepsilon>0$,
$\delta\rho=\varepsilon(|a_{1}\rangle\langle a_{1}|-|a_{d}\rangle\langle
a_{d}|)$. Then, $\mathrm{Tr}(\rho+\delta\rho)=1$,
$\rho+\delta\rho\in{\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$
and

$Q(\rho+\delta\rho)_{dd}=-\varepsilon|\langle a_{d}|b_{d}\rangle|^{2}<0.$

Hence
$\rho+\delta\rho\not\in\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$
so that $\rho$ belongs to the boundary of
$\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$.
$\Leftarrow$ We consider the case where
$\lambda_{1}\lambda_{2}\dots\lambda_{d}\not=0$, the other case being
analogous. We can suppose without loss of generality that
$\lambda_{1}=\min\lambda_{i}>0$. Then

$\sum_{i}\lambda_{i}|a_{i}\rangle\langle
a_{i}|=\frac{\lambda_{1}}{2}|a_{1}\rangle\langle
a_{1}|+\sum_{i\not=1}(\lambda_{i}-\frac{\lambda_{1}}{2})|a_{i}\rangle\langle
a_{i}|+\frac{\lambda_{1}}{2}\sum_{j}|b_{j}\rangle\langle b_{j}|.$

Hence

$\rho=\frac{\lambda_{1}}{2}|a_{1}\rangle\langle
a_{1}|+\sum_{i\not=1}(\lambda_{i}-\frac{\lambda_{1}}{2})|a_{i}\rangle\langle
a_{i}|+\sum_{j}(\mu_{j}+\frac{\lambda_{1}}{2})|b_{j}\rangle\langle b_{j}|.$

Note that all coefficients are strictly positive. Now consider a perturbation

$\delta\rho=\sum_{i}\varepsilon_{i}|a_{i}\rangle\langle
a_{i}|+\sum_{j}\delta_{j}|b_{j}\rangle\langle b_{j}|$

with $\varepsilon_{i},\delta_{j}\in\mathbb{R}$ and
$\mathrm{Tr}\,\delta\rho=0$, then

$\rho+\delta\rho\in\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$

provided the $\varepsilon_{i},\delta_{j}$ are small enough. In other words,
there is a small ball centered on $\rho$ that belongs to
$\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$. ∎

As a consequence of the proof, we also have the following result:

######  Corollary A.2.

Let $m_{{\mathcal{A}},{\mathcal{B}}}>0$ and let $\rho$ be a density matrix.
Then
$\rho\in\mathrm{Int}(\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right))$
if and only if there exist $\lambda_{i}>0,\mu_{j}>0$ so that
$\rho=\sum_{i}\lambda_{i}|a_{i}\rangle\langle
a_{i}|+\sum_{j}\mu_{j}|b_{j}\rangle\langle b_{j}|.$

Let us introduce the notation

$[i_{1},i_{2},\dots,i_{k};j_{1},j_{2},\dots,j_{\ell}]=\mathrm{conv}\left(|a_{i_{1}}\rangle\langle
a_{i_{1}}|,\dots,|a_{i_{k}}\rangle\langle a_{i_{k}}|,|b_{j_{1}}\rangle\langle
b_{j_{1}}|,\dots,|b_{j_{\ell}}\rangle\langle b_{j_{\ell}}|\right),$

for any choice $1\leq i_{1}<i_{2}<\dots i_{k}\leq d$, $1\leq
j_{1}<j_{2}<\dots<j_{\ell}\leq d$. Also, for $i\in\llbracket 1,d\rrbracket$,
we write $\bar{i}=\llbracket 1,d\rrbracket\setminus\\{i\\}$. Then the Lemma
implies that

$\partial(\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right))=\cup_{i,j}[\bar{i};\bar{j}].$

When $d=3$, this becomes

$\partial(\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right))=\cup_{i_{1}<i_{2};j_{1}<j_{2}}[i_{1},i_{2};j_{1},j_{2}].$

The boundary is then the union of $9$ tetrahedra. They are glued together
along 18 triangles of one of the following forms: $[i_{1},i_{2};j]$ with
$i_{1}<i_{2}$ or $[i;j_{1},j_{2}]$ with $j_{1}<j_{2}$.

##  Appendix B The geometry of ${\mathcal{E}}_{\mathrm{KD+}}$: the case of
real orthogonal $U$ in $d=3$

In this section, we give some more details about the geometry of the convex
set ${\mathcal{E}}_{\mathrm{KD+}}$ of all KD-positive states in the particular
case where $U$ is a real orthogonal matrix in dimension $d=3$ (Section B.1).
We then analyze in detail the example of Section 5.2 for which both
$\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)\subsetneq{\mathcal{E}}_{\mathrm{KD+}}$
and
$\mathrm{conv}({\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}})\subsetneq{\mathcal{E}}_{\mathrm{KD+}}$
(Section B.2).

###  B.1 Identifying ${\mathcal{E}}_{\mathrm{KD+}}$.

We recall that, as in Section 5, in dimension 3, if $U$ is real, we can write

$V_{\mathrm{KDr}}={\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}\oplus\mathbb{R}F_{\perp}.$

Here, $F_{\perp}$ is orthogonal to the
${\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$. We denote by
$P_{{\mathcal{A}},{\mathcal{B}}}$ the orthogonal projection on
${\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$ associated to
this decomposition, and by $T_{\perp}:F\in
V_{\mathrm{KDr}}\mapsto\mathrm{Tr}(FF_{\perp})$. Hence, the orthogonal
projection of $F$ on $\mathbb{R}F_{\perp}$ is given by
$T_{\perp}(F)F_{\perp}$. The following technical lemma and proposition collect
the main properties of the set ${\mathcal{E}}_{\mathrm{KD+}}$ in this
particular situation.

######  Lemma B.1.

Let $\sigma\in{\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$.
Then we have either
$\left(\sigma+\mathbb{R}F_{\perp}\right)\cap{\mathcal{E}}_{\mathrm{KD+}}=\emptyset$
or there exists $-\infty<x_{-}(\sigma)\leqslant x_{+}(\sigma)<+\infty$ such
that

$\left(\sigma+\mathbb{R}F_{\perp}\right)\cap{\mathcal{E}}_{\mathrm{KD+}}=\sigma+[x_{-}(\sigma),x_{+}(\sigma)]F_{\perp}.$

###### Proof.

If $\sigma\in{\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$,
then $\left(\sigma+\mathbb{R}F_{\perp}\right)\cap{\mathcal{E}}_{\mathrm{KD+}}$
is a compact convex set. Suppose the set is not empty. Therefore, as
$T_{\perp}$ is continuous,
$T_{\perp}(\left(\sigma+\mathbb{R}F_{\perp}\right)\cap{\mathcal{E}}_{\mathrm{KD+}})$
is a non-empty compact interval of $\mathbb{R}$. This interval can be written
as
$\left(\sigma+\mathbb{R}F_{\perp}\right)\cap{\mathcal{E}}_{\mathrm{KD+}}=\sigma+[x_{-}(\sigma),x_{+}(\sigma)]F_{\perp}$
with $-\infty<x_{-}(\sigma)\leqslant x_{+}(\sigma)<+\infty$. ∎

Let
$\mathcal{D}=P_{{\mathcal{A}},{\mathcal{B}}}({\mathcal{E}}_{\mathrm{KD+}})$
which is a subset of
${\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$. Note that the
second alternative of Lemma B.1 happens if and only if $\sigma\in\mathcal{D}$.
In other words, $\mathcal{D}$ is the domain of definition of $x_{-}$ and
$x_{+}$. We will designate by $\mathrm{Int}(\mathcal{D})$ the interior of
$\mathcal{D}$ as a subset of
${\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$.

######  Proposition B.2.

We have the following properties:

  1. (i)

If
$\sigma\in\mathrm{Int}(\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right))$,
$-\infty<x_{-}(\sigma)<0<x_{+}(\sigma)<+\infty$;

  2. (ii)

If $\sigma\in\mathcal{D}$ and
$\sigma\notin\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$, then
either $0<x_{-}(\sigma)\leq x_{+}(\sigma)<+\infty$ or
$-\infty<x_{-}(\sigma)\leq x_{+}(\sigma)<0$;

  3. (iii)

If $\sigma\in{\mathcal{A}}\cup{\mathcal{B}}$ then
$x_{-}(\sigma)=x_{+}(\sigma)=0$;

  4. (iv)

The function $x_{+}$ has a maximum $x_{\max}$ on $\mathcal{D}$. Moreover, the
extreme points of

$\mathcal{Y}_{+}=\\{\sigma+x_{\max}F_{\perp}\mid\sigma\in\mathcal{D},x_{+}(\sigma)=x_{\max}\\}$

are extreme points of ${\mathcal{E}}_{\mathrm{KD+}}$;

  5. (v)

The function $x_{-}$ has a minimum $x_{\min}$ on $\mathcal{D}$ . Moreover, the
extreme points of

$\mathcal{Y}_{-}=\\{\sigma+x_{\min}F_{\perp}\mid\sigma\in\mathcal{D},x_{-}(\sigma)=x_{\min}\\}$

are extreme points of ${\mathcal{E}}_{\mathrm{KD+}}$;

  6. (vi)

The function $x_{+}$ (resp. $x_{-}$) is concave (resp. convex) on
$\mathcal{D}$. Thus, it is continuous on $\mathrm{Int}(\mathcal{D}).$

In particular, the proposition implies that ${\mathcal{E}}_{\mathrm{KD+}}$
lies between the “bounding planes”:

$\\{F\in
V_{\mathrm{KDr}}\mid\mathrm{Tr}\,FF_{\perp}=x_{\max}\\}\quad\textrm{and}\quad\\{F\in
V_{\mathrm{KDr}}\mid\mathrm{Tr}\,FF_{\perp}=x_{\min}\\}.$

###### Proof.

  1. (i)

Suppose
$\sigma\in\mathrm{Int}(\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right))$,
then, by Corollary A.2, $\sigma=\sum_{i=1}^{d}\lambda_{i}|a_{i}\rangle\langle
a_{i}|+\mu_{i}|b_{i}\rangle\langle b_{i}|$ with $\lambda_{i}>0$ and
$\mu_{i}>0$ for all $i\in\llbracket 1,d\rrbracket$. Thus,
$\min_{(i,j)\in\llbracket 1,d\rrbracket^{2}}Q_{i,j}(\sigma)>0$. Then, for
$\varepsilon>0$,

$\forall(i,j)\in\llbracket 1,d\rrbracket^{2},\ Q_{i,j}(\sigma+\varepsilon
F_{\perp})=Q_{i,j}(\sigma)+\varepsilon
Q_{i,j}(F_{\perp})>\min_{(i,j)\in\llbracket
1,d\rrbracket^{2}}Q_{i,j}(\sigma)-\varepsilon\max_{(i,j)\in\llbracket
1,d\rrbracket^{2}}\left|Q_{i,j}(F_{\perp})\right|.$

Thus, there exists an $\varepsilon_{1}>0$ such that for
$\varepsilon\in[0,\varepsilon_{1}]$, $\sigma+\varepsilon F_{\perp}\in
V_{\mathrm{KD+}}$. Here, we recall that $V_{\mathrm{KD+}}$ is the set of self-
adjoint operators with positive KD distributions.

Moreover, for $|\psi\rangle\in\mathcal{H}_{1}$, for all
$\varepsilon\in\mathbb{R}^{+}$,

$\langle\psi|\sigma+\varepsilon
F_{\perp}|\psi\rangle\geqslant\langle\psi|\sigma|\psi\rangle-\varepsilon\max_{|\phi\rangle\in\mathcal{H}_{1}}\left|\langle\phi|F_{\perp}|\phi\rangle\right|$

and

$\begin{array}[]{rcl}\langle\psi|\sigma|\psi\rangle=\sum_{i=1}^{d}\lambda_{i}\left|\langle\psi|a_{i}\rangle\right|^{2}+\mu_{i}\left|\langle\psi|b_{i}\rangle\right|^{2}&\geqslant&\min_{i\in\llbracket
1,d\rrbracket}\\{\lambda_{i},\mu_{i}\\}\sum_{i=1}^{d}\left|\langle\psi|a_{i}\rangle\right|^{2}+\left|\langle\psi|b_{i}\rangle\right|^{2}\\\
&\geqslant&2\min_{i\in\llbracket
1,d\rrbracket}\\{\lambda_{i},\mu_{i}\\}>0.\end{array}$

Consequently, there exists an $\varepsilon_{2}>0$ such that for
$\varepsilon\in[0,\varepsilon_{2}]$, for all $|\psi\rangle\in\mathcal{H}_{1}$,
$\langle\psi|\sigma+\varepsilon F_{\perp}|\psi\rangle\geqslant 0$. Therefore,
for $\varepsilon\in[0,\min(\varepsilon_{1},\varepsilon_{2})]$,
$\sigma+\varepsilon F_{\perp}$ is a density matrix with a positive KD
distribution so that $0<x_{+}(\sigma)$.

By changing $\varepsilon$ to $-\varepsilon$ in the previous lines, it follows
that $x_{-}(\sigma)<0$.

  2. (ii)

If $\sigma\notin\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$,
then by Lemma 3.1, $\sigma\notin{\mathcal{E}}_{\mathrm{KD+}}$ and thus, either
$0<x_{-}(\sigma)$ or $0>x_{+}(\sigma)$.

  3. (iii)

Suppose $\sigma=|a_{1}\rangle\langle a_{1}|$, then
$\mathrm{Tr}((\sigma+xF_{\perp})^{2})=1+x^{2}$. For $x\neq 0$,
$\mathrm{Tr}((\sigma+xF_{\perp})^{2})>1$ implying that $\sigma+xF_{\perp}$ is
not a state. Hence, $\sigma+xF_{\perp}\notin{\mathcal{E}}_{\mathrm{KD+}}$ for
all $x\neq 0$. Thus, $x_{+}(\sigma)=0=x_{-}(\sigma)$.

  4. (iv)

As ${\mathcal{E}}_{\mathrm{KD+}}$ is a compact set, $\mathrm{T}_{\perp}$ is
bounded and reaches its bounds on ${\mathcal{E}}_{\mathrm{KD+}}$. Especially,
it reaches its maximum $x_{\max}$ which is strictly positive. Note that

$\mathcal{Y}_{+}=\left\\{\rho\in\mathcal{S}_{d,+,1}\mid\mathrm{T}_{\perp}(\rho)=\mathrm{Tr}(F_{\perp}\rho)=x_{\max}\right\\}\cap{\mathcal{E}}_{\mathrm{KD+}}.$
(B.1)

Thus, $\mathcal{Y}_{+}$ is compact, convex and not empty so it has an extreme
point. Let $\rho_{e}$ be such an extreme point. We show, by contradiction,
that $\rho_{e}$ is also an extreme point of ${\mathcal{E}}_{\mathrm{KD+}}$.
Suppose that $\rho_{e}$ is not, and write
$\rho_{e}=\lambda\rho_{1}+(1-\lambda)\rho_{2}$ with
$\rho_{1},\rho_{2}\in{\mathcal{E}}_{\mathrm{KD+}}$ and $\lambda\in(0,1)$. So,
$\mathrm{T}_{\perp}(\rho_{1})\leqslant x_{\max}$ and
$\mathrm{T}_{\perp}(\rho_{2})\leqslant x_{\max}$. Now, suppose
$\mathrm{T}_{\perp}(\rho_{1})<x_{\max}$. Then,
$\mathrm{T}_{\perp}(\rho_{e})<x_{\max}$, which is a contradiction. Thus,
$\mathrm{T}_{\perp}(\rho_{1})=\mathrm{T}_{\perp}(\rho_{2})=x_{\max}$, which
show that $\rho_{1},\rho_{2}\in\mathcal{Y}_{+}.$ As $\rho_{e}$ is an extreme
point of $\mathcal{Y}_{+}$, $\rho_{1}=\rho_{2}=\rho_{e}$ and so $\rho_{e}$ is
an extreme point of ${\mathcal{E}}_{\mathrm{KD+}}$.

  5. (v)

The proof is analogous to the one of (iv).

  6. (vi)

We show that $x_{+}$ is concave on its domain of definition. As $\mathcal{D}$
is the projection of a compact convex set, it is a compact convex set. Take
$\sigma,\sigma^{\prime}\in\mathcal{D}$ and $\lambda\in[0,1]$. We will show
that $x_{+}(\lambda\sigma+(1-\lambda)\sigma^{\prime})\geqslant\lambda
x_{+}(\sigma)+(1-\lambda)x_{+}(\sigma^{\prime})$. We have that

$\lambda\sigma+(1-\lambda)\sigma^{\prime}+\left[\lambda
x_{+}(\sigma)+(1-\lambda)x_{+}(\sigma^{\prime})\right]F_{\perp}=\lambda(\sigma+x_{+}(\sigma)F_{\perp})+(1-\lambda)(\sigma^{\prime}+x_{+}(\sigma^{\prime})F_{\perp}).$

As a convex combination of KD-positive states, it is a KD-positive state such
that $\lambda
x_{+}(\sigma)+(1-\lambda)x_{+}(\sigma^{\prime})\in[x_{-}(\lambda\sigma+(1-\lambda)\sigma^{\prime}),x_{+}(\lambda\sigma+(1-\lambda)\sigma^{\prime})]$.
Therefore, $\lambda x_{+}(\sigma)+(1-\lambda)x_{+}(\sigma^{\prime})\leqslant
x_{+}(\lambda\sigma+(1-\lambda)\sigma^{\prime})$. Thus, $x_{+}$ is a concave
function on $\mathcal{D}$. It is then continuous on
$\mathrm{Int}(\mathcal{D})$ [58]. The same argument shows that $x_{-}$ is
convex and thus also continuous on $\mathrm{Int}{\mathcal{D}}$.

∎

###  B.2 Identifying ${\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}$ : an
example of mixed extreme states.

We now identify some extreme mixed states of ${\mathcal{E}}_{\mathrm{KD+}}$
for the unitary matrix

$U=\frac{1}{3}\begin{pmatrix}-1&2&2\\\ 2&-1&2\\\ 2&2&-1\end{pmatrix},$ (B.2)

introduced in Section 5.2, using Proposition B.2 (iv). Note that we identified
all pure KD-positive states for $U$ in Section 5.2 and that they all lie below
${\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$: if
$\rho\in{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{pure}}$, then
$\mathrm{Tr}\,(\rho F_{\perp})\leq 0$. So any
$\rho\in{\mathcal{E}}_{\mathrm{KD+}}^{\mathrm{ext}}$ for which
$\mathrm{Tr}\,(\rho F_{\perp})>0$ is a mixed extreme KD-positive state. We
explicitly find some of those states as follows. We first determine in Lemma
B.3 the maximum $x_{\max}$ of the function $x_{+}$, which is strictly
positive. This allows us to give a precise description of the set
$\mathcal{Y}_{+}$ in Proposition B.2, and in particular of its extreme points.

######  Lemma B.3.

Let $U$ be as in Eq. (B.2). Then, the maximum $x_{\max}$ of $x_{+}$ on
$\mathcal{D}$ is

$x_{\max}=\max_{\sigma\in\mathcal{D}}x_{+}(\sigma)=\frac{1}{2\sqrt{6}}.$

This value is reached for
$\sigma=\frac{1}{3}\mathbb{I}_{3}\in\mathrm{conv}\left({\mathcal{A}}\cup{\mathcal{B}}\right)$.

###### Proof.

We set $\sigma\in{\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$
with $\mathrm{Tr}\,(\sigma)=1$ so that

$\sigma=\sum_{i=1}^{3}\lambda_{i}|a_{i}\rangle\langle
a_{i}|+\mu_{i}|b_{i}\rangle\langle b_{i}|\ \mathrm{and}\
\sum_{i=1}^{3}\lambda_{i}+\mu_{i}=1.$

For all $x\in\mathbb{R}$, we compute

$Q(\sigma+xF_{\perp})=\frac{1}{27}\begin{pmatrix}3(\mu_{1}+\lambda_{1})-2x\sqrt{6}&12(\mu_{2}+\lambda_{1})+x\sqrt{6}&12(\mu_{3}+\lambda_{1})+x\sqrt{6}\\\
12(\mu_{1}+\lambda_{2})+x\sqrt{6}&3(\mu_{2}+\lambda_{2})-2x\sqrt{6}&12(\mu_{3}+\lambda_{2})+x\sqrt{6}\\\
12(\mu_{1}+\lambda_{3})+x\sqrt{6}&12(\mu_{2}+\lambda_{3})+x\sqrt{6}&3(\mu_{3}+\lambda_{3})-2x\sqrt{6}\end{pmatrix}.$

Thus, if

$3(\mu_{1}+\lambda_{1})-2x\sqrt{6}<0\ \mathrm{or}\
3(\mu_{2}+\lambda_{2})-2x\sqrt{6}<0\ \mathrm{or}\
3(\mu_{3}+\lambda_{3})-2x\sqrt{6}<0,$

or equivalently, if

$x>\frac{3}{2\sqrt{6}}\min_{i\in\llbracket
1,3\rrbracket}(\mu_{i}+\lambda_{i}),$

then $\sigma+xF_{\perp}$ is not KD-positive. Hence, since
$\sigma+x_{+}(\sigma)F_{\perp}$ is KD-positive,

$x_{+}(\sigma)\leqslant\frac{3}{2\sqrt{6}}\min_{i\in\llbracket
1,3\rrbracket}(\mu_{i}+\lambda_{i}).$ (B.3)

Thus,

$x_{+}(\sigma)\leqslant\frac{3}{2\sqrt{6}}\inf_{\\{\lambda_{i},\mu_{i}\\}}\min_{i\in\llbracket
1,3\rrbracket}(\mu_{i}+\lambda_{i}),$

where the infimum is taken over all the $(\lambda_{i},\mu_{i})_{i\in\llbracket
1,3\rrbracket}$ such that
$\sigma=\sum_{i=1}^{3}\lambda_{i}|a_{i}\rangle\langle
a_{i}|+\mu_{i}|b_{i}\rangle\langle b_{i}|.$ Moreover, $\min_{i\in\llbracket
1,3\rrbracket}(\mu_{i}+\lambda_{i})\leqslant\frac{1}{3}$ for any
$\sigma=\sum_{i=1}^{3}\lambda_{i}|a_{i}\rangle\langle
a_{i}|+\mu_{i}|b_{i}\rangle\langle
b_{i}|\in{\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$ with
$\mathrm{Tr}\,(\sigma)=1$. Indeed, suppose there exists a $\sigma$ such that
$\min_{i\in\llbracket 1,3\rrbracket}(\mu_{i}+\lambda_{i})>\frac{1}{3}$, then
$\mathrm{Tr}(\sigma)\geqslant 3\min_{i\in\llbracket
1,3\rrbracket}(\mu_{i}+\lambda_{i})>1$, which is a contradiction. Thus,

$\inf_{\\{\lambda_{i},\mu_{i}\\}}\min_{i\in\llbracket
1,3\rrbracket}(\mu_{i}+\lambda_{i})\leqslant\frac{1}{3}.$

Therefore,

$x_{+}(\sigma)\leqslant\frac{1}{2\sqrt{6}}.$

Consequently, as the bound does not depend on $\sigma$, we obtain that

$x_{\max}\leqslant\frac{1}{2\sqrt{6}}.$

It is proven, in Section 5.2, that
$x_{+}(\frac{1}{3}\mathbb{I}_{3})=\frac{1}{2\sqrt{6}}$. Consequently,
$x_{\max}=\frac{1}{2\sqrt{6}}$.

∎

######  Proposition B.4.

The set
$\mathcal{Y}_{+}={\mathcal{E}}_{\mathrm{KD+}}\cap\\{\rho\in\mathcal{S}_{d,+,1}\mid\mathrm{Tr}(\rho
F_{\perp})=\frac{1}{2\sqrt{6}}\\}$ is of the form

$\left\\{\frac{1}{3}\mathbb{I}_{d}+\frac{1}{2\sqrt{6}}F_{\perp}+\lambda_{1}(|a_{1}\rangle\langle
a_{1}|-|b_{1}\rangle\langle b_{1}|)+\lambda_{2}(|a_{2}\rangle\langle
a_{2}|-|b_{2}\rangle\langle b_{2}|)\right\\},$

with $\left|\lambda_{2}-\lambda_{1}\right|\leqslant\frac{3}{8}$,
$\left|\lambda_{1}\right|\leqslant\frac{3}{8}$ and
$\left|\lambda_{2}\right|\leqslant\frac{3}{8}$. Its extreme points are
obtained for the following values of $(\lambda_{1},\lambda_{2})$:

$\left\\{\left(0,\frac{3}{8}\right),\left(0,-\frac{3}{8}\right),\left(\frac{3}{8},0\right),\left(-\frac{3}{8},0\right),\left(\frac{3}{8},\frac{3}{8}\right),\left(-\frac{3}{8},-\frac{3}{8}\right)\right\\}.$
(B.4)

###### Proof.

Let $\rho\in\mathcal{Y}_{+}$. Then there exists $\sigma\in\mathcal{D}$ so that
$\rho=\sigma+\frac{1}{2\sqrt{6}}F_{\perp}$. In addition,
$x_{+}(\sigma)=\frac{1}{2\sqrt{6}}$. Then, Eq. (B.3) implies that there exist
$\lambda_{i},\mu_{i}\in\mathbb{R}$ so that

$\sigma=\sum_{i}(\lambda_{i}|a_{i}\rangle\langle
a_{i}|+\mu_{i}|b_{i}\rangle\langle b_{i}|)$

and so that $\min_{i\in\llbracket
1,3\rrbracket}\mu_{i}+\lambda_{i}=\frac{1}{3}$. Indeed, if such a
decomposition does not exist, as
$\sigma\in{\mathrm{span}_{\mathbb{R}}({\mathcal{A}}\cup{\mathcal{B}})}$, all
decompositions

$\sigma=\sum_{i}(\alpha_{i}|a_{i}\rangle\langle
a_{i}|+\beta_{i}|b_{i}\rangle\langle b_{i}|)$

satisfy $\min_{i\in\llbracket
1,3\rrbracket}(\alpha_{i}+\beta_{i})<\frac{1}{3}$. Thus, as shown in Eq (B.3),

$x_{+}(\sigma)\leqslant\frac{3}{2\sqrt{6}}\min_{i\in\llbracket
1,3\rrbracket}(\alpha_{i}+\beta_{i})<\frac{1}{2\sqrt{6}},$

which is a contradiction.

Since $\sum_{i}(\lambda_{i}+\mu_{i})=1$, this implies
$\lambda_{i}+\mu_{i}=\frac{1}{3}$ for $i=1,2,3$. It follows that there exist
$(\lambda_{i})_{i\in\llbracket 1,3\rrbracket}$ so that

$\sigma=\frac{1}{3}\mathbb{I}_{3}+\lambda_{1}(|a_{1}\rangle\langle
a_{1}|-|b_{1}\rangle\langle b_{1}|)+\lambda_{2}(|a_{2}\rangle\langle
a_{2}|-|b_{2}\rangle\langle b_{2}|)+\lambda_{3}(|a_{3}\rangle\langle
a_{3}|-|b_{3}\rangle\langle b_{3}|).$

As $\sum_{i=1}^{3}|a_{i}\rangle\langle
a_{i}|=\sum_{i=1}^{3}|b_{i}\rangle\langle b_{i}|$, we can simplify the
expression to obtain

$\sigma=\frac{1}{3}\mathbb{I}_{3}+\lambda_{1}(|a_{1}\rangle\langle
a_{1}|-|b_{1}\rangle\langle b_{1}|)+\lambda_{2}(|a_{2}\rangle\langle
a_{2}|-|b_{2}\rangle\langle b_{2}|).$ (B.5)

The KD distribution of $\rho=\sigma+\frac{1}{2\sqrt{6}}F_{\perp}$ is

$Q(\sigma+\frac{1}{2\sqrt{6}}F_{\perp})=\frac{1}{27}\begin{pmatrix}0&4+12(\lambda_{1}-\lambda_{2})+\frac{1}{2}&4+12\lambda_{1}+\frac{1}{2}\\\
4-12(\lambda_{1}-\lambda_{2})+\frac{1}{2}&0&4+12\lambda_{2}+\frac{1}{2}\\\
4-12\lambda_{1}+\frac{1}{2}&4-12\lambda_{2}+\frac{1}{2}&0\end{pmatrix}.$

Since $\rho\in\mathcal{Y}_{+}$, it is KD positive, which is equivalent to

$-4+12|\lambda_{1}|\leqslant\frac{1}{2},\
-4+12|\lambda_{2}|\leqslant\frac{1}{2},\ \mathrm{and}\
-4+12|\lambda_{1}-\lambda_{2}|\leqslant\frac{1}{2},$

or

$|\lambda_{1}|\leqslant\frac{3}{8},|\lambda_{2}|\leqslant\frac{3}{8},\
\mathrm{and}\ |\lambda_{1}-\lambda_{2}|\leqslant\frac{3}{8}.$ (B.6)

The eigenvalues of $\rho=\sigma+\frac{1}{2\sqrt{6}}F_{\perp}$ are :

$\left\\{\frac{1}{4},\frac{3}{8}+\frac{1}{24}\sqrt{9+8^{3}(\lambda_{1}^{2}+\lambda_{2}^{2}-\lambda_{1}\lambda_{2})},\frac{3}{8}-\frac{1}{24}\sqrt{9+8^{3}(\lambda_{1}^{2}+\lambda_{2}^{2}-\lambda_{1}\lambda_{2}})\right\\}.$

Consequently, $\rho=\sigma+\frac{1}{2\sqrt{6}}F_{\perp}$ is a positive
operator if and only if

$\frac{3}{8}-\frac{1}{24}\sqrt{9+8^{3}(\lambda_{1}^{2}+\lambda_{2}^{2}-\lambda_{1}\lambda_{2})}\geqslant
0.$

This equation is equivalent to

$\lambda_{1}^{2}+\lambda_{2}^{2}+(\lambda_{1}-\lambda_{2})^{2}\leqslant
2\left(\frac{3}{8}\right)^{2}.$ (B.7)

We have therefore established that $\rho\in\mathcal{Y}_{+}$ if and only if it
can be written as $\rho=\sigma+\frac{1}{2\sqrt{6}}F_{\perp}$ with $\sigma$ as
in Eq. (B.5) and with $\lambda_{1},\lambda_{2}$ satisfying Eq. (B.6)-(B.7).
The set
$\mathcal{C}^{\prime}=\\{(\lambda_{1},\lambda_{2})\in\mathbb{R}^{2}\mid\lambda_{1}^{2}+\lambda_{2}^{2}+(\lambda_{1}-\lambda_{2})^{2}\leqslant
2\left(\frac{3}{8}\right)^{2}\\}$ is bounded by an ellipse and is as such
convex. The set
$\mathcal{C}=\\{(\lambda_{1},\lambda_{2})\in\mathbb{R}^{2}\mid|\lambda_{1}|\leqslant\frac{3}{8},|\lambda_{2}|\leqslant\frac{3}{8}\
\mathrm{and}\ |\lambda_{1}-\lambda_{2}|\leqslant\frac{3}{8}\\}$ is a convex
hexagon. Its extreme points are identified to be those given in Eq. (B.4).

Noting that these extreme points lie on the ellipse bounding
$\mathcal{C}^{\prime}$, we conclude that in fact
$\mathcal{C}\subsetneq\mathcal{C^{\prime}}$; see Fig. 2. This proves our
Proposition.

∎

Figure 2: The convex regions $\mathcal{C}$ and $\mathcal{C}^{\prime}$ in the
$(\lambda_{1},\lambda_{2})$-plane defined by Eq. (B.6)-(B.7). Dots represent
the extreme points of $\mathcal{Y}_{+}$

##  Appendix C Pure KD-positive states for MUBs

The pure KD-positive states of MUB bases can be characterized as follows.

######  Theorem C.1.

Suppose ${\mathcal{A}}$ and ${\mathcal{B}}$ are MUB bases. Then:
(i) A pure state $|\psi\rangle$ is KD positive iff
$n_{{\mathcal{A}}}(\psi)n_{{\mathcal{B}}}(\psi)=d$;
(ii) If $d$ is a prime number, then the only pure KD-positive states are the
basis states.

This result is implicit in [59]. We provide a simple proof below. Note that
this result implies that, when $d$ is a prime number, then the only pure KD-
positive states of MUB bases are their basis states. This last result was
proven for the DFT in [24], where the same result is also obtained for
perturbations of MUB bases that are completely incompatible, a notion
introduced in [23]. It is not known, to the best of our knowledge, if under
the hypotheses of the theorem, there do also exist mixed KD-positive states.

###### Proof.

Suppose $|\psi\rangle$ is a KD-positive state. By permuting the order and
changing the phases of basis states, we can suppose that

$S_{{\mathcal{A}}}(\psi)=\llbracket
1,n_{{\mathcal{A}}}(\psi)\rrbracket,S_{{\mathcal{B}}}(\psi)=\llbracket
1,n_{{\mathcal{B}}}(\psi)\rrbracket,\left(\langle
a_{i}|\psi\rangle\right)_{i\in\llbracket
1,d\rrbracket}\in\left(\mathbb{R}^{+}\right)^{d}\text{ and }\left(\langle
b_{j}|\psi\rangle\right)_{j\in\llbracket
1,d\rrbracket}\in\left(\mathbb{R}^{+}\right)^{d}.$

Here $S_{{\mathcal{A}}}(\psi)=\\{i\in\llbracket 1,d\rrbracket,\langle
a_{i}|\psi\rangle\neq 0\\}$ and $n_{{\mathcal{A}}}(\psi)=\sharp
S_{{\mathcal{A}}}(\psi)$. The same definitions hold for ${\mathcal{B}}$.
Hence, since $U$ is the transition matrix for MUB bases and since
$Q(\psi)\in\left(\mathbb{R}^{+}\right)^{d^{2}}$, one concludes that
$\forall(i,j)\in\llbracket 1,n_{{\mathcal{A}}}(\psi)\rrbracket\times\llbracket
1,n_{{\mathcal{B}}}(\psi)\rrbracket,\langle
a_{i}|b_{j}\rangle=\frac{1}{\sqrt{d}}$. By construction, one has

$\forall i\in\llbracket 1,n_{{\mathcal{A}}}(\psi)\rrbracket,\langle
a_{i}|\psi\rangle=\sum_{j=1}^{n_{{\mathcal{B}}}(\psi)}\langle
a_{i}|b_{j}\rangle\langle
b_{j}|\psi\rangle=\frac{1}{\sqrt{d}}\sum_{i=1}^{n_{{\mathcal{B}}}(\psi)}\langle
b_{j}|\psi\rangle$

which is independent of $i$. Similarly,

$\forall j\in\llbracket 1,n_{{\mathcal{B}}}(\psi)\rrbracket,\langle
b_{j}|\psi\rangle=\sum_{i=1}^{n_{{\mathcal{A}}}(\psi)}\langle
b_{j}|a_{i}\rangle\langle
a_{i}|\psi\rangle=\frac{1}{\sqrt{d}}\sum_{i=1}^{n_{{\mathcal{A}}}(\psi)}\langle
a_{i}|\psi\rangle$

which is independent of $j$ so that

$\langle
b_{1}|\psi\rangle=\frac{1}{\sqrt{d}}\sum_{i=1}^{n_{{\mathcal{A}}}(\psi)}\langle
a_{i}|\psi\rangle=\frac{1}{d}\sum_{i=1}^{n_{{\mathcal{A}}}(\psi)}\sum_{i=1}^{n_{{\mathcal{B}}}(\psi)}\langle
b_{j}|\psi\rangle=\frac{\langle
b_{1}|\psi\rangle}{d}\sum_{i=1}^{n_{{\mathcal{A}}}(\psi)}\sum_{i=1}^{n_{{\mathcal{B}}}(\psi)}1=\frac{\langle
b_{1}|\psi\rangle}{d}n_{{\mathcal{A}}}(\psi)n_{{\mathcal{B}}}(\psi).$

As $\langle b_{1}|\psi\rangle\neq 0$, one finds that

$n_{{\mathcal{A}}}(\psi)n_{{\mathcal{B}}}(\psi)=d.$

In particular, when $d$ is prime, this implies that
$n_{{\mathcal{A}}}(\psi)=1$ or $n_{{\mathcal{B}}}(\psi)=1$. In either case,
$|\psi\rangle$ is a basis state. ∎

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