#  Relating incompatibility, noncommutativity, uncertainty and
Kirkwood-Dirac nonclassicality

Stephan De Bièvre
Univ. Lille, CNRS, Inria, UMR 8524
Laboratoire Paul Painlevé, F-59000 Lille, France Stephan.De-Bievre@univ-
lille.fr

###### Abstract

We provide an in-depth study of the recently introduced notion of completely
incompatible observables and its links to the support uncertainty and to the
Kirkwood-Dirac nonclassicality of pure quantum states. The latter notion has
recently been proven central to a number of issues in quantum information
theory and quantum metrology. In this last context, it was shown that a
quantum advantage requires the use of Kirkwood-Dirac nonclassical states. We
establish sharp bounds of very general validity that imply that the support
uncertainty is an efficient Kirkwood-Dirac nonclassicality witness. When
adapted to completely incompatible observables that are close to mutually
unbiased ones, this bound allows us to fully characterize the Kirkwood-Dirac
classical states as the eigenvectors of the two observables. We show
furthermore that complete incompatibility implies several weaker notions of
incompatibility, among which features a strong form of noncommutativity.

##  1 Introduction

Among the salient characteristics of quantum mechanics, distinguishing it from
classical mechanics, feature prominently the incompatibility and
noncommutativity of two observables and the associated uncertainty principles.
They are, together with entanglement, the main ingredients for the study of
the classical-quantum transition within the quantum state space. The subject
continues to attract considerable attention from the viewpoint of foundational
issues [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] as well as in the context of the search
for a quantum advantage in various protocols of quantum information and
metrology [11, 12, 13, 14, 15, 16, 17, 18]. In particular, the use of
Kirkwood-Dirac nonclassical states has recently been proven to be essential in
the latter context [15, 16].

It was pointed out in [19] that when incompatibility is equated with
noncommutativity (as is often the case [3, 6, 17, 9, 18, 10]), only a very
weak notion of incompatibility is obtained and a much stronger notion,
referred to as “complete incompatibility” was proposed. (See Definition 4
below.) The latter provides a mathematical expression to the physical idea
that the measurement of a second observable after the measurement of a first
one always perturbs the result obtained in the first measurement, whatever the
pre-measurement state. Complete incompatibility was then shown to lead to a
strong uncertainty relation for all pure states which in turn was proven to be
linked to the notion of Kirkwood-Dirac nonclassicality. It is the goal of this
paper to further explore the consequences of the complete incompatibility of
two observables and its relation to various weaker notions of incompatibility
(Proposition 7), as well as to deepen the logical links between this notion,
the uncertainty inherent in quantum states, and their Kirkwood-Dirac
nonclassicality. One of the main results of our analysis is that when two
observables are completely incompatible and – in a sense we make precise –
close to mutually unbiased, then all states, except the eigenstates of the two
observables, are Kirkwood-Dirac nonclassical (Theorem 1).

When two observables do not satisfy the stringent complete incompatibility
condition, our results still permit to partially characterize the KD-
nonclassical states (Theorem 12 and Theorem 13).

We shall work in the context of quantum mechanics on a Hilbert space
${\mathcal{H}}$ of finite dimension $d$ and formulate our definitions and
results in terms of two orthonormal bases
${\mathcal{A}}=\\{|a_{i}\rangle|1\leq i\leq d\\}$ and
${\mathcal{B}}=\\{|b_{j}\rangle|1\leq j\leq d\\}$ that one can – but need not
– think of as the eigenbases of two observables $A$ and $B$. We start by
showing (Proposition 7) that the complete incompatibility of two
bases/observables implies a number of weaker forms of incompatibility,
including noncommutativity, that we each interpret physically in terms of
successive repeated measurements. We show in particular that when two
observables are completely incompatible, no spectral projector of $A$ commutes
with any spectral projector of $B$. This is much stronger than saying that the
two observables do not commute, which only implies that at least one spectral
projector of $A$ does not commute with one of $B$. Nevertheless, as we show,
even this very strong form of noncommutativity does not imply complete
incompatibility, which is a stronger notion still.

To study the link of complete incompatibility with uncertainty, we proceed as
follows. Given a state $\psi\in{\mathcal{H}}$, we define
$n_{{\mathcal{A}}}(\psi)$ (respectively $n_{{\mathcal{B}}}(\psi)$) to be the
number of nonvanishing components of $\psi$ on ${\mathcal{A}}$ (respectively
${\mathcal{B}}$):

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

We then introduce the uncertainty diagram of the pair
$({\mathcal{A}},{\mathcal{B}})$ to be the set of points
$(n_{{\mathcal{A}}},n_{{\mathcal{B}}})\in\llbracket 1,d\rrbracket^{2}$ in the
$n_{{\mathcal{A}}}n_{{\mathcal{B}}}$-plane for which there exists
$\psi\in{\mathcal{H}}$ so that $n_{{\mathcal{A}}}(\psi)=n_{{\mathcal{A}}}$,
$n_{{\mathcal{B}}}(\psi)=n_{{\mathcal{B}}}$. Whereas for an arbitrary choice
of bases ${\mathcal{A}}$ and ${\mathcal{B}}$ the uncertainty diagram can have
a complex structure, we show it can be easily characterized in the case the
two bases are completely incompatible. Indeed, in that case, it is composed of
_all_ points $(n_{{\mathcal{A}}},n_{{\mathcal{B}}})\in\llbracket
1,d\rrbracket^{2}$ for which $n_{{\mathcal{A}}}+n_{{\mathcal{B}}}\geq d+1$
(Theorem 1). This last inequality, which is in fact equivalent to complete
incompatibility, can be interpreted as an uncertainty relation: it says
$n_{{\mathcal{A}}}(\psi)$ and $n_{{\mathcal{B}}}(\psi)$ cannot both be small.

To illustrate these first results, we analyse the case where ${\mathcal{A}}$
and ${\mathcal{B}}$ are mutually unbiased bases (MUB), meaning $|\langle
a_{i}|b_{j}\rangle|=\frac{1}{\sqrt{d}}$, for all $i,j$. Those have attracted
considerable attention over the years [20, 21, 22, 23, 24] and many open
questions concerning their classification remain in dimensions $d\geq 6$ [25].
The MUB are sometimes referred to as “complementary” or “maximally
incompatible” [23] because if a measurement in ${\mathcal{A}}$ yields the
outcome $|a_{i}\rangle$, then a subsequent measurement in ${\mathcal{B}}$
yields any of the outcomes $|b_{j}\rangle$ with the same probability. It turns
out, nevertheless, that not all MUB are completely incompatible, as already
pointed out in [19]. Whereas, in dimensions $d=2,3,5$ this is easily seen to
be the case, no MUB are completely incompatible when $d=4$ [19]. In higher
dimension, the situation is unclear: it is in particular not known in which
dimension there exist MUB that are completely incompatible. We show in this
paper (Theorem 10) that, in all dimensions, there exist completely
incompatible bases ${\mathcal{A}},{\mathcal{B}}$ for which
$m_{{\mathcal{A}},{\mathcal{B}}}:=\min_{i,j}|\langle
a_{i}|b_{j}\rangle|\geq\frac{1}{\sqrt{d}}-\delta$, for any $\delta>0$. In
other words, there exist completely incompatible bases that are arbitrary
close to being mutually unbiased.

Having explained that the complete incompatibility of two observables is
equivalent to a strong uncertainty relation for all states
$|\psi\rangle\in{\mathcal{H}}$, we turn to the central question of this paper,
which are the links between complete incompatibility, uncertainty and
“nonclassicality”. For that purpose, we use the notion of nonclassicality
associated to the Kirkwood-Dirac distribution. Recall that, given
${\mathcal{H}}$ and two orthonormal bases ${\mathcal{A}}=\\{|a_{i}\rangle\\}$
and ${\mathcal{B}}=\\{|b_{j}\rangle\\}$, the Kirkwood-Dirac (KD) distribution
of a state $\psi$ [26, 27],

$Q(\psi)_{ij}=\langle a_{i}|\psi\rangle\langle\psi|b_{j}\rangle\langle
b_{j}|a_{i}\rangle,1\leq i,j\leq d,$ (1)

is a quasi-probability distribution somewhat similar in spirit to the Wigner
distribution or Glauber-Sudarshan $P$-function [28, 29] in continuous variable
quantum mechanics, which are intimately linked to the choice of two conjugate
quadratures $X$ and $P$. It is complex-valued and satisfies
$\sum_{ij}Q(\psi)_{ij}=1,$ with marginals $\sum_{j}Q(\psi)_{ij}=|\langle
a_{i}|\psi\rangle|^{2},\sum_{i}Q(\psi)_{ij}=|\langle b_{i}|\psi\rangle|^{2}.$
A state $\psi\in{\mathcal{H}}$ is said to be _KD classical_ if its KD
distribution is real nonnegative everywhere; in other words, if its KD
distribution is a probability distribution. Otherwise it is _KD nonclassical_.
KD nonclassicality is used in quantum tomography [12, 13, 14] as well as in
the theory and applications of weak measurements, (non)contextuality, and
their relation to nonclassical effects in quantum mechanics [2, 5, 4, 8].
Also, KD nonclassicality has been shown to be an essential ingredient
necessary to obtain an operational quantum advantage in postselected metrology
[15, 16].

For these reasons, the question of how to characterize the KD-nonclassical
states poses itself naturally. One way to do this is to identify a
nonclassicality witness. By this, we mean a quantity depending on the state
$|\psi\rangle$ that can be computed or measured without determining the KD-
distribution of the state, and whose value can reveal its KD nonclassicality.
It is crucial to keep in mind that the KD distribution and hence the KD
nonclassicality of $|\psi\rangle$ depend not only on $|\psi\rangle$, but also
on the two bases ${\mathcal{A}},{\mathcal{B}}$ used to define it, the choice
of which is in applications linked to the identification of two observables
$A$ and $B$ of which they are eigenbases. The questions that arise are
therefore: under what conditions on the bases ${\mathcal{A}}$ and
${\mathcal{B}}$ do there exist KD-nonclassical states, how many are there,
where in Hilbert space can they be found and how strongly nonclassical are
they?

A form of incompatibility or of noncommutativity is needed between the
projectors $|a_{i}\rangle\langle a_{i}|$ and $|b_{j}\rangle\langle b_{j}|$ for
KD-nonclassical states to exist. Indeed, if for example $A$ and $B$ have
nondegenerate spectra, and if they are compatible in the usual sense that
$[A,B]=0$, then all those projectors commute, and therefore each
$|a_{i}\rangle$ is up to a phase equal to some $|b_{j}\rangle$. It is then
immediate from (1) that $Q_{ij}(\psi)\geq 0$ for all $\psi$ so that there are
no KD-nonclassical states in ${\mathcal{H}}$. Under the assumption that $A$
and $B$ are incompatible in the sense that they do not commute, a sufficient
but non-optimal condition for a state to be KD nonclassical was given in [16].
(See Theorem 15 below.) In the present paper, we sharpen and optimize that
result in several ways. In particular, our main results under the assumption
that the bases are completely incompatible are summarized in the following
theorem. It is a corollary of Theorem 8 and Theorem 13.

######  Theorem 1.

Two bases ${\mathcal{A}}$ and ${\mathcal{B}}$ are COINC if and only if for all
$\psi\in{\mathcal{H}}$, $n_{{\mathcal{A}}}(\psi)+n_{{\mathcal{B}}}(\psi)\geq
d+1$.
In that case:

  * (i)

For all $1\leq n_{{\mathcal{A}}},n_{{\mathcal{B}}}\leq d$ with
$n_{{\mathcal{A}}}+n_{{\mathcal{B}}}\geq d+1$, $\exists\psi\in{\mathcal{H}}$,
$n_{{\mathcal{A}}}(\psi)=n_{{\mathcal{A}}},n_{{\mathcal{B}}}(\psi)=n_{{\mathcal{B}}}$.

  * (ii)

If $\psi\in{\mathcal{H}}$ satisfies
$n_{{\mathcal{A}},{\mathcal{B}}}(\psi):=n_{{\mathcal{A}}}(\psi)+n_{{\mathcal{B}}}(\psi)>d+1$,
then $\psi$ is KD nonclassical.

  * (iii)

If a state $\psi\in{\mathcal{H}}$ is KD classical, then
$n_{{\mathcal{A}},{\mathcal{B}}}(\psi):=n_{{\mathcal{A}}}(\psi)+n_{{\mathcal{B}}}(\psi)=d+1$.

  * (iv)

If in addition ${\mathcal{A}}$, ${\mathcal{B}}$ are such that

$m_{{\mathcal{A}},{\mathcal{B}}}^{2}:=\min_{ij}|\langle
a_{i}|b_{j}\rangle|^{2}>\frac{1}{d}\left(\frac{\frac{d(d-1)}{2}}{1+\frac{d(d-1)}{2}}\right),$
(2)

and hence in particular when they are mutually unbiased, then the only
classical states are the vectors in ${\mathcal{A}}$ and ${\mathcal{B}}$.

The first part of this result can be suggestively paraphrased as saying that,
when ${\mathcal{A}}$ and ${\mathcal{B}}$ are completely incompatible, all
states, except possible those that have minimal support uncertainty, are KD
nonclassical. Part (ii) in particular shows that the _support uncertainty_
$n_{{\mathcal{A}},{\mathcal{B}}}(\psi):=n_{{\mathcal{A}}}(\psi)+n_{{\mathcal{B}}}(\psi)$
is a KD-nonclassicality witness. According to (iv), if in addition
$m_{{\mathcal{A}},{\mathcal{B}}}$ is close to its maximal possible value
$1/\sqrt{d}$, only the elements of the two bases ${\mathcal{A}}$ and
${\mathcal{B}}$ are classical.

Since complete incompatibility is a stringent condition on the bases
${\mathcal{A}}$ and ${\mathcal{B}}$, not necessarily satisfied in all
situations of interest, it is important to understand what can be said about
the KD-nonclassical states in such situations as well. It was shown in [19]
that, if the bases satisfy the weaker incompatibility condition that

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

then part (ii) and (iii) of Theorem 1 still hold (but not (i)). In other
words, under that condition, the support uncertainty is still a KD-
nonclassicality witness. We show here this remains true in much greater
generality, even if $m_{{\mathcal{A}},{\mathcal{B}}}=0$, but provided not too
many of the overlaps $\langle a_{i}|b_{j}\rangle$ vanish (Theorem 12). We
further establish a sharpening of the bound in (ii) of the above theorem:
provided $m_{{\mathcal{A}},{\mathcal{B}}}$ is sufficiently close to
$1/\sqrt{d}$, any $|\psi\rangle$ that is not an eigenvector of one of the two
observables and satisfies $n_{{\mathcal{A}},{\mathcal{B}}}(\psi)>d$ is
nonclassical (Theorem 13).

The paper is organized as follows. In Section 2 we define the mathematical
framework within which we shall work. Section 3 briefly reviews the standard
notions of incompatibility that appear in the literature and their link with
noncommutativity. Section 4 similarly reviews the various uncertainty
relations available in the literature and introduces the less well known
support uncertainty that we shall consider here, as well as a very general
support uncertainty relation that it satisfies. This sets the stage for our
definition and characterization of complete incompatibility in Section 5 and
for the explanation of its link with a strengthened support uncertainty
relation. A study of the relationships between mutual unbiasedness and
complete incompatibility is proposed in Section 6. In Section 7 we establish
under which conditions the support uncertainty of a state can serve as a KD
nonclassicality witness. This allows us to provide a complete characterization
of all KD-classical states under suitable conditions on ${\mathcal{A}}$ and
${\mathcal{B}}$. Some final conclusions are drawn in Section 8.

##  2 Setup

In this preliminary section, we describe the basic objects entering our study
and fix the notation we shall use. Throughout, we will consider two distinct
orthonormal bases ${\mathcal{A}}=\\{|a_{i}\rangle|1\leq i\leq d\\}$ and
${\mathcal{B}}=\\{|b_{j}\rangle|1\leq j\leq d\\}$ of a $d$-dimensional Hilbert
space ${\mathcal{H}}$. Given two such bases, we define their transition matrix
$U_{ij}=\langle a_{i}|b_{j}\rangle$. Introducing

$m_{{\mathcal{A}},{\mathcal{B}}}:=\min_{i,j}|\langle a_{i}|b_{j}\rangle|,\quad
M_{{\mathcal{A}},{\mathcal{B}}}=\max_{i,j}|\langle a_{i}|b_{j}\rangle|,$

one has

$0\leq m_{{\mathcal{A}},{\mathcal{B}}}\leq d^{-1/2}\leq
M_{{\mathcal{A}},{\mathcal{B}}}\leq 1,$ (3)

which follows immediately from the fact that the transition matrix is unitary
so that

$d=\operatorname{Tr}\mathbb{I}=\sum_{i,j}|\langle b_{j}|a_{i}\rangle|^{2}.$

Clearly

$m_{{\mathcal{A}},{\mathcal{B}}}>0\Rightarrow
M_{{\mathcal{A}},{\mathcal{B}}}<1.$ (4)

The condition $M_{{\mathcal{A}},{\mathcal{B}}}<1$ means ${\mathcal{A}}$ and
${\mathcal{B}}$ have no basis vectors in common. We will see below
(Proposition 7) that both the conditions $M_{{\mathcal{A}},{\mathcal{B}}}<1$
and $m_{{\mathcal{A}},{\mathcal{B}}}>0$ can be viewed as a form of
incompatibility between the two bases, with
$m_{{\mathcal{A}},{\mathcal{B}}}>0$ being the stronger one of the two.

Considering a column of $U$ containing a term for which
$M_{{\mathcal{A}},{\mathcal{B}}}=|\langle a_{i}|b_{j}\rangle|$, one has

$1\geq
M_{{\mathcal{A}},{\mathcal{B}}}^{2}+m_{{\mathcal{A}},{\mathcal{B}}}^{2}(d-1)$
(5)

and similarly, considering a column of $U$ containing a term for which
$m_{{\mathcal{A}},{\mathcal{B}}}=|\langle a_{i}|b_{j}\rangle|$, one has

$1\leq
m_{{\mathcal{A}},{\mathcal{B}}}^{2}+M_{{\mathcal{A}},{\mathcal{B}}}^{2}(d-1).$
(6)

Hence

$m_{{\mathcal{A}},{\mathcal{B}}}=\frac{1}{\sqrt{d}}\Leftrightarrow
M_{{\mathcal{A}},{\mathcal{B}}}=\frac{1}{\sqrt{d}}\Leftrightarrow
m_{{\mathcal{A}},{\mathcal{B}}}=M_{{\mathcal{A}},{\mathcal{B}}};$

any of these conditions characterizes mutually unbiased bases.

When two bases ${\mathcal{A}}$ and ${\mathcal{B}}$ are specified, we can
associate to each $\psi\in{\mathcal{H}}$ its ${\mathcal{A}}$\- and
${\mathcal{B}}$-_representations_ , which are the vectors of their components
$u(\psi)=(\langle a_{1}|\psi\rangle,\dots,\langle
a_{d}|\psi\rangle)\in\mathbb{C}^{d}$ on the ${\mathcal{A}}$ basis and
$v(\psi)=(\langle b_{1}|\psi\rangle,\dots,\langle
b_{d}|\psi\rangle)\in\mathbb{C}^{d}$ on the ${\mathcal{B}}$ basis. One has

$u(\psi)=Uv(\psi)\quad\textrm{and}\quad\sum_{i=1}^{d}|u_{i}(\psi)|^{2}=\langle\psi|\psi\rangle=\sum_{j=1}^{d}|v_{j}(\psi)|^{2}.$
(7)

Given a state $\psi$, so that $\langle\psi|\psi\rangle=1$, $|u_{i}(\psi)|^{2}$
and $|v_{j}(\psi)|^{2}$ form two probabilities representing uncertainties
which can be measured in various ways. We will come back to this extensively
below and in particular study the joint uncertainty with respect to both
bases.

Given bases ${\mathcal{A}}$ and ${\mathcal{B}}$, we introduce a family of
orthogonal projectors as follows. For every $S,T\subset\llbracket
1,d\rrbracket:=\\{1,2,\dots,d\\}$,

${\Pi_{{\mathcal{A}}}}(S)=\sum_{i\in S}|a_{i}\rangle\langle
a_{i}|,\quad{\Pi_{{\mathcal{B}}}}(T)=\sum_{j\in T}|b_{j}\rangle\langle
b_{j}|.$

In some cases the two bases ${\mathcal{A}},{\mathcal{B}}$ arise as the
eigenbases of two self-adjoint operators $A$ and $B$ on ${\mathcal{H}}$. If
their eigenvalues are non-degenerate, then the corresponding bases are unique
(up to a global phase and possible reordering of the basis vectors), but not
otherwise. In what follows, given two observables $A$ and $B$, we shall always
assume a particular eigenbasis has been chosen for each. In many examples of
interest, however, the two bases ${\mathcal{A}}$ and ${\mathcal{B}}$ are not
associated to observables, as we shall see. If a basis is given, we will say a
self-adjoint operator $A$ with eigenvalues
$\alpha_{1}<\alpha_{2}<\dots<\alpha_{L}$ is adapted to the basis
${\mathcal{A}}$ if the latter is a basis of eigenvectors for $A$. We can
associate a projective partition of unity to such an adapted observable by
considering the partition of $\llbracket 1,d\rrbracket$ defined by

$S_{\ell}=\\{i\in\llbracket 1,d\rrbracket\mid
A|a_{i}\rangle=\alpha_{\ell}|a_{i}\rangle\\},\quad 1\leq\ell\leq L.$ (8)

The ${\Pi_{{\mathcal{A}}}}(S_{\ell})$ are then the eigenprojectors of $A$ and
one has

$\cup_{\ell}S_{\ell}=\llbracket
1,d\rrbracket,\quad\sum_{\ell}{\Pi_{{\mathcal{A}}}}(S_{\ell})=\mathbb{I}.$

Finally, there is a unique projection-valued measure associated to $A$ in the
usual way. Let $I\subset\mathbb{R}$. Then we introduce the spectral projector
of $A$ for $I$ to be the operator $\Pi(A\in I)$ defined as follows:

$\Pi(A\in I)=\sum_{\alpha_{\ell}\in I}{\Pi_{{\mathcal{A}}}}(S_{\ell}).$

This is the sum of the projectors onto the basis vectors whose corresponding
eigenvalue is in $I$. Note that these spectral projectors are all of the type
${\Pi_{{\mathcal{A}}}}(S)$ for some $S\subset\llbracket 1,d\rrbracket$.

These considerations lead to the following description of quantum measurements
associated to a basis ${\mathcal{A}}$, that we need for our discussion of both
uncertainty and incompatibility. We only need some basic elements of the
quantum theory of projective measurements as discussed in in [1, 30, 31], that
we briefly recall.

We in particular only consider selective measurements: the outcome of an
individual measurement is always recorded. Given a basis ${\mathcal{A}}$, one
can associate to every partition $S_{1},\dots,S_{L}$ of the index set
$\llbracket 1,d\rrbracket$ a family of projectors
${\Pi_{{\mathcal{A}}}}(S_{1}),\dots\Pi_{{\mathcal{A}}}(S_{L})$ forming a
projective partition of unity in the sense that

$\sum_{\ell=1}^{L}{\Pi_{{\mathcal{A}}}}(S_{\ell})=\mathbb{I}.$

To such a family corresponds a selective projective measurement the outcomes
of which are the $S_{\ell}$. In the case where the $S_{\ell}$ are as defined
in (8), this is commonly referred to as a measurement of $A$. The outcomes
$S_{\ell}$ are then identified with the eigenvalues $\alpha_{\ell}$. If
initially the system is in the state $\psi\in{\mathcal{H}}$ and the outcome
$S_{\ell}$ is realized then the post-measurement state is the pure state
${\Pi_{{\mathcal{A}}}}(S_{\ell})\psi/\|{\Pi_{{\mathcal{A}}}}(S_{\ell})\psi\|$.
The probability of this outcome is
$\|{\Pi_{{\mathcal{A}}}}(S_{\ell})\psi\|^{2}$. This is what we shall refer to
as a measurement in the basis ${\mathcal{A}}$. When $L=d$ and
$S_{\ell}=\\{\ell\\}$, for $\ell=1,\dots,d$, we will say the measurement is
finegrained. Otherwise it is coarsegrained. When $L=2$, a case we will
frequently consider, the corresponding measurement has only two possible
outcomes and one has $S_{1}=S,S_{2}=S^{\textrm{c}}$, for some
$S\subset\llbracket 1,d\rrbracket$. In line with the preceding terminology, we
can also refer to this as a measurement of ${\Pi_{{\mathcal{A}}}}(S)$. Indeed,
the projector ${\Pi_{{\mathcal{A}}}}(S)$ has two eigenvalues, $1$ and $0$,
that correspond to the outcomes $S$ and $S^{\textrm{c}}$.

Note that the transition matrix $U$ depends on the order in which the basis
vectors are listed. None of the definitions or results in this work will
depend on that order and we will freely reorder the bases when it is
convenient. Similarly, the results only depend on the basis vectors up to a
global phase. In particular, if $k$ of the basis vectors of ${\mathcal{A}}$
coincide with $k$ of those of ${\mathcal{B}}$ then one can consider
$|a_{i}\rangle=|b_{i}\rangle$ for $1\leq i\leq k$, after a possible reordering
and adjusting of the phases. From the point of view of incompatibility,
uncertainty and KD nonclassicality, nothing interesting then happens on the
subspace of ${\mathcal{H}}$ spanned by the first $k$ basis vectors and the
entire analysis in this paper can then be done on the $d-k$-dimensional
complementary subspace. In particular, one then has

$Q_{ij}(\psi)=|\langle\psi|a_{i}\rangle|^{2}\delta_{ij},\quad\forall 1\leq
i,j\leq k.$

More generally, when $U$ is block-diagonal, then so is $Q(\psi)$ and the
analysis should be performed for each block separately.

For later use, we introduce the following notation. For $S,T\subset\llbracket
1,d\rrbracket$ we introduce the $|S|\times|T|$ matrix $U(S,T)$ with entries

$U(S,T)_{k\ell}=\langle a_{i_{k}}|b_{j_{\ell}}\rangle,$ (9)

where $1\leq k\leq|S|,1\leq\ell\leq|T|$, and $i_{k}\in S,j_{\ell}\in T$. Note
that Rank $U(S,T)\leq\min\\{|S|,|T|\\}$. Also, in what follows

${\mathcal{H}}(S,T)={\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}\cap{\Pi_{{\mathcal{B}}}}(T){\mathcal{H}}.$
(10)

##  3 (In)compatibility and (non)commutativity

To study (in)compatibility, we need to discuss repeated successive
measurements in the bases ${\mathcal{A}}$ and ${\mathcal{B}}$, a subject we
turn to in this section. We start with a brief but critical analysis of the
standard relationship between compatibility and commutativity of observables
[32, 31], which is helpful in motivating the notion of complete
incompatibility introduced in Section 5.

Suppose on a state $\psi$ successive measurements in ${\mathcal{A}}$, then
${\mathcal{B}}$, yield the outcomes $S$, then $T$. We will say these outcomes
are compatible if, when measuring again in ${\mathcal{A}}$ after having
obtained $T$, the outcome $S$ occurs with probability one. This means that the
measurement of ${\mathcal{B}}$ after the first measurement of ${\mathcal{A}}$
has not disturbed the first outcome. Repeating measurements in ${\mathcal{A}}$
and/or ${\mathcal{B}}$ will then systematically yield the same outcomes,
always with probability one. As a result, we adopt the following definition:

######  Definition 2.

$S,T$ are compatible outcomes for a measurement in ${\mathcal{A}}$ followed by
one in ${\mathcal{B}}$ if

${\Pi_{{\mathcal{B}}}}(T){\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}\subset{\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}.$
(11)

Indeed, if the pre-measurement state is $\psi$, the (nonnormalized) state
after the first two measurements is
${\Pi_{{\mathcal{B}}}}(T){\Pi_{{\mathcal{A}}}}(S)\psi$. For the second
measurement in ${\mathcal{A}}$ to yield $S$ with probability one for all such
pre-measurement states $\psi$, one needs that
${\Pi_{{\mathcal{B}}}}(T){\Pi_{{\mathcal{A}}}}(S)\psi\in{\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}$
for all such $\psi$. This justifies the definition. For brevity, we will
simply say $S$ and $T$ are compatible.

Eq. (11) is equivalent to
$[{\Pi_{{\mathcal{A}}}}(S),{\Pi_{{\mathcal{B}}}}(T)]=0$:

${\Pi_{{\mathcal{B}}}}(T){\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}\subset{\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}\Leftrightarrow[{\Pi_{{\mathcal{A}}}}(S),{\Pi_{{\mathcal{B}}}}(T)]=0.$
(12)

Indeed, Eq. (11) means ${\Pi_{{\mathcal{B}}}}(T)$ leaves
${\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}$ invariant, which implies it also
leaves its orthogonal complement
${\Pi_{{\mathcal{A}}}}(S^{\textrm{c}}){\mathcal{H}}$ invariant so that
${\Pi_{{\mathcal{A}}}}(S){\Pi_{{\mathcal{B}}}}(T){\Pi_{{\mathcal{A}}}}(S^{\textrm{c}})=0$.
Hence

$\displaystyle{\Pi_{{\mathcal{B}}}}(T){\Pi_{{\mathcal{A}}}}(S)\psi$
$\displaystyle=$
$\displaystyle{\Pi_{{\mathcal{A}}}}(S){\Pi_{{\mathcal{B}}}}(T){\Pi_{{\mathcal{A}}}}(S)\psi={\Pi_{{\mathcal{A}}}}(S){\Pi_{{\mathcal{B}}}}(T)\psi-{\Pi_{{\mathcal{A}}}}(S){\Pi_{{\mathcal{B}}}}(T){\Pi_{{\mathcal{A}}}}(S^{\textrm{c}})\psi$
$\displaystyle=$
$\displaystyle{\Pi_{{\mathcal{A}}}}(S){\Pi_{{\mathcal{B}}}}(T)\psi.$

The converse implication is obvious. One then also has

${\Pi_{{\mathcal{B}}}}(T){\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}\subset{\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}\Leftrightarrow[{\Pi_{{\mathcal{A}}}}(S),{\Pi_{{\mathcal{B}}}}(T)]=0\Leftrightarrow{\Pi_{{\mathcal{A}}}}(S){\Pi_{{\mathcal{B}}}}(T){\mathcal{H}}\subset{\Pi_{{\mathcal{B}}}}(T){\mathcal{H}}.$
(13)

As a result, if first a measurement in ${\mathcal{B}}$ is made, and the
outcome $T$ obtained, then a subsequent measurement of ${\mathcal{A}}$ with
outcome $S$ does not perturb the first measurement. In other words, if $S,T$
are compatible outcomes for a measurement in ${\mathcal{A}}$ followed by one
in ${\mathcal{B}}$, then $T,S$ are compatible outcomes for a measurement in
${\mathcal{B}}$ followed by one in ${\mathcal{A}}$.

We will say two observables $A$ and $B$ are compatible when all outcomes
associated to their eigenvalues $\alpha_{\ell},\beta_{k}$ (see Eq. (8)) are
compatible in the sense of Eq. (11). In view of what precedes, this means $A$
and $B$ are compatible if and only if they commute. Mathematically, this is a
very strong, restrictive property. It entails a number of further physical
properties such as joint measurability and measurement non-disturbance [3, 6,
17].

Note that $S$ and $T$ are mutually exclusive when, after a measurement in
${\mathcal{A}}$ yielded the outcome $S$, a measurement in ${\mathcal{B}}$
cannot yield the outcome $T$. This is equivalent to
${\Pi_{{\mathcal{B}}}}(T){\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}=0$. This, in
turn, implies Eq. (11), so that, in this sense, mutually exclusive events are
compatible. While this sounds odd, considering the standard everyday meaning
of the words “compatible” and “mutually exclusive”, it is implicit in all
discussions on quantum mechanics and we shall also adopt this practice.
Indeed, it often occurs that two observables commute, while many outcomes
$\alpha_{\ell},\beta_{k}$ are mutually exclusive.

What happens if $[{\Pi_{{\mathcal{A}}}}(S),{\Pi_{{\mathcal{B}}}}(T)]\not=0$?
Suppose a measurement in ${\mathcal{A}}$ on the pre-measurement state $\psi$
yields the outcome $S$ so that the post-measurement state is, up to
normalization, ${\Pi_{{\mathcal{A}}}}(S)\psi$. Suppose that a subsequent
measurement in $B$ yields the outcome $T$, and the post-measurement state is
${\Pi_{{\mathcal{B}}}}(T){\Pi_{{\mathcal{A}}}}(S)\psi$. As we saw, this second
measurement has not perturbed the first if and only if

${\Pi_{{\mathcal{A}}}}(S){\Pi_{{\mathcal{B}}}}(T){\Pi_{{\mathcal{A}}}}(S)\psi={\Pi_{{\mathcal{B}}}}(T){\Pi_{{\mathcal{A}}}}(S)\psi,$
(14)

which is equivalent to
${\Pi_{{\mathcal{B}}}}(T){\Pi_{{\mathcal{A}}}}(S)\psi\in{\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}$.
Do such states exist? The answer is yes, if and only if

${\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}\cap{\Pi_{{\mathcal{B}}}}(T){\mathcal{H}}\not=\\{0\\}.$
(15)

In that case, taking
$0\not=\phi\in{\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}\cap{\Pi_{{\mathcal{B}}}}(T){\mathcal{H}}$,
$\phi^{\prime}\in{\Pi_{{\mathcal{A}}}}(S^{\textrm{c}}){\mathcal{H}}$, and
setting $\psi=\phi+\phi^{\prime}$, one has indeed Eq. (14). Note that Eq. (14)
is equivalent to the statement

${\Pi_{{\mathcal{A}}}}(S)\psi\in{\mathrm{Ker}}[{\Pi_{{\mathcal{A}}}}(S),{\Pi_{{\mathcal{B}}}}(T)].$

Of course, even though
$[{\Pi_{{\mathcal{A}}}}(S),{\Pi_{{\mathcal{B}}}}(T)]\not=0$, generally the
kernel of $[{\Pi_{{\mathcal{A}}}}(S),{\Pi_{{\mathcal{B}}}}(T)]$ is not
trivial. In fact, it is always true that

${\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}\cap{\Pi_{{\mathcal{B}}}}(T){\mathcal{H}}\subset{\mathrm{Ker}}[{\Pi_{{\mathcal{A}}}}(S),{\Pi_{{\mathcal{B}}}}(T)].$

In conclusion, if
${\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}\cap{\Pi_{{\mathcal{B}}}}(T){\mathcal{H}}\not=\\{0\\}$,
then, for any state
$\psi\in{\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}\cap{\Pi_{{\mathcal{B}}}}(T){\mathcal{H}}$,
repeated successive measurements in ${\mathcal{A}}$ and ${\mathcal{B}}$ will
give with probability one the outcomes $S$ and $T$. In other words, even if
the commutator does not vanish, it is still possible for the measurement in
${\mathcal{B}}$ to yield the outcome $T$, without perturbing the previous
outcome $S$. However, if
$[{\Pi_{{\mathcal{A}}}}(S),{\Pi_{{\mathcal{B}}}}(T)]\not=0$, there also exist
pre-measurement states $\psi$ for which this is not true. To see this, note
that, since the projectors don’t commute, we know
${\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}$ is not invariant under
${\Pi_{{\mathcal{B}}}}(T)$. So there exists
$\psi\in{\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}$ so that
${\Pi_{{\mathcal{B}}}}(T)\psi\not\in{\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}$. If
initially the system is in such a state $\psi$, the outcome of an
${\mathcal{A}}$ measurement is $S$. Since
${\Pi_{{\mathcal{B}}}}(T)\psi\not\in{\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}$, we
know ${\Pi_{{\mathcal{B}}}}(T)\psi\not=0$, so that the probability that a
measurement in ${\mathcal{B}}$ yields $T$ does not vanish. When this occurs,
the post-measurement state is, up to normalization,
${\Pi_{{\mathcal{B}}}}(T)\psi\not\in{\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}$.
Hence the probability that a subsequent measurement of $A$ yields $S$ is
strictly less than one.

In conclusion, when two projectors ${\Pi_{{\mathcal{A}}}}(S)$ and
${\Pi_{{\mathcal{B}}}}(T)$ do not commute, it may still happen for some pre-
measurement states that repeated successive measurements in ${\mathcal{A}}$
and ${\mathcal{B}}$ systematically yield the same outcomes $S$ and $T$.
However, this does no longer happen for every pre-measurement state that
yields the outcomes $S$ and $T$ in the first two measurements.

A simple but instructive example illustrating this situation is this:

$\displaystyle A$ $\displaystyle=$ $\displaystyle a_{1}(|a_{1}\rangle\langle
a_{1}|+|a_{2}\rangle\langle a_{2}|)+a_{3}|a_{3}\rangle\langle a_{3}|,\quad
a_{1}<a_{3},$ $\displaystyle B$ $\displaystyle=$ $\displaystyle
b_{1}(|b_{1}\rangle\langle b_{1}|+|b_{2}\rangle\langle
b_{2}|)+b_{3}|b_{3}\rangle\langle b_{3}|,\quad b_{1}<b_{3},$

with

$|b_{1}\rangle=\frac{1}{\sqrt{2}}(|a_{1}\rangle+|a_{2}\rangle),\quad|b_{2}\rangle=\frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}(|a_{1}\rangle-|a_{2}\rangle)+|a_{3}\rangle\right),\quad|b_{3}\rangle=\frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}(|a_{1}\rangle-|a_{2}\rangle)-|a_{3}\rangle\right).$

Let $S=\\{1,2\\},T=\\{1,2\\}$, so that ${\Pi_{{\mathcal{A}}}}(S)$ and
${\Pi_{{\mathcal{B}}}}(T)$ are the eigenprojectors of $A$, $B$ with eigenvalue
$a_{1},b_{1}$. They do not commute but
${\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}\cap{\Pi_{{\mathcal{B}}}}(T){\mathcal{H}}=\mathbb{C}|b_{1}\rangle$.
Now, let $|\psi\rangle=\frac{1}{\sqrt{2}}(|b_{1}\rangle+|a_{3}\rangle)$. Then
a measurement of $A$ yields the outcome $a_{1}$ with probability $1/2$; the
post-measurement state is then $|b_{1}\rangle$. A subsequent measurement of
$B$ yields the outcome $b_{1}$ without altering the state and hence without
altering the outcomes of any further measurements of $A$, which will always
yield $a_{1}$. In this situation, the measurement of $B$ has not disturbed the
one of $A$. If, on the other hand, the pre-measurement state is
$|\psi\rangle=\frac{1}{\sqrt{2}}(|a_{2}\rangle+|a_{3}\rangle)$, then a first
measurement of $A$ still yields $a_{1}$ with probability $1/2$, but now the
resulting state is $|a_{2}\rangle$ and

$|a_{2}\rangle=\frac{1}{\sqrt{2}}\left(|b_{1}\rangle-\frac{1}{\sqrt{2}}(|b_{2}\rangle+|b_{3}\rangle)\right).$

A subsequent measurement of $B$ yields $b_{1}$ with probability $3/4$ and the
resulting state is then

$\sqrt{\frac{2}{3}}(|b_{1}\rangle-\frac{1}{\sqrt{2}}|b_{2}\rangle).$

This is not an eigenvector of $A$ and a second measurement of $A$ only yields
$a_{1}$ again with probability $5/6$ (and $a_{3}$ with probability $1/6$). In
other words, this time the measurement of $B$ has perturbed the initial value
obtained for $A$. This example thus illustrates that indeed, when two
operators do not commute, successive measurements may or may not perturb
outcomes previously obtained.

It is common practice in quantum mechanics to identify the incompatibility of
observables with their noncommutativity [31]. In view of the previous
discussion, this yields a rather weak notion of incompatible since it means
that in some – but not in all – circumstances the measurement of the second
observable may perturb the outcome of a previous measurement of the first
observable. This is of course the result of the fact that incompatibility is
defined here as the negation of compatibility, which is a very strong notion:
it requires that a measurement of $B$ _never_ disturbs a previous measurement
of $A$.

In what follows, we will say two bases ${\mathcal{A}}$, ${\mathcal{B}}$ are
compatible if $[{\Pi_{{\mathcal{A}}}}(S),{\Pi_{{\mathcal{B}}}}(T)]=0$ for all
$S,T\in\llbracket 1,d\rrbracket$. And we will say two such bases are
incompatible if this is not true; in other words, if _there exist_
$S,T\in\llbracket 1,d\rrbracket$ so that
$[{\Pi_{{\mathcal{A}}}}(S),{\Pi_{{\mathcal{B}}}}(T)]\not=0$. We will see below
that, to connect incompatibility to uncertainty and nonclassicality, a
stronger notion of incompatibility is needed.

##  4 Uncertainties

It is well known that one way to understand the Heisenberg uncertainty
relation is to view it as a property of the Fourier transform saying, loosely
speaking, that a function and its Fourier transform cannot both be sharply
localized. When made precise, this statement can take many forms, thoroughly
reviewed in [33] (See also [34]). The Heisenberg uncertainty relation (in the
form proven by Weyl [35]) expresses the localization of the wave function in
terms of the variances of its position and momentum distributions. Another
form was proven much more recently and states that a wave function and its
Fourier transform cannot both have their support in sets of finite volume. In
other words, they cannot both vanish outside a region of finite volume. This
formulation is very similar to the general support uncertainty principles that
we will consider below (Section 4.3) in the finite dimensional setting. They
have attracted attention more recently, in the context of signal analysis [36,
37], in the theory of the Fourier analysis on finite groups [38, 34], and in
the study of nonclassicality in quantum systems with a finite dimensional
Hilbert space [19], which is our focus here.

In order to give perspective on the notion of support uncertainty, we first
briefly review some aspects of the Heisenberg, Robertson, and entropic
uncertainty relations. In Section 4.1 we recall the well-known link between
noncommutativity and uncertainty as expressed in the Robertson uncertainty
relation and show that noncommutativity alone does not lead to very strong
information about the uncertainty inherent in a state. We briefly review
entropic uncertainty relations in Section 4.2 and turn to the support
uncertainty relation in Section 4.3.

###  4.1 The Heisenberg and Robertson uncertainty relations

Rather than viewing the original Heisenberg uncertainty relation as a property
of the Fourier transform, one can alternatively (and equivalently) see it as a
bound on the uncertainties of an arbitrary state $|\psi\rangle$ with respect
to two conjugate variables $Q$ and $P$ satisfying the canonical commutation
relation $[Q,P]=i$. This alternative viewpoint leads to the generalization
shown by Robertson [39], and which concerns arbitrary noncommuting observables
$A$ and $B$:

$\Delta A\Delta
B\geq\frac{1}{2}|\langle\psi|C|\psi\rangle|,\quad\textrm{where}\ C=-i[A,B].$
(16)

It relates the precision with which we can know the value of two noncommuting
observables $A$ and $B$ when the system under consideration is in a quantum
state $|\psi\rangle$. More precisely, when the right hand side of this
uncertainty relation does not vanish, the inequality can be interpreted as
providing a limit on the statistical variation of the outcomes of measurements
of $A$ and $B$ on an ensemble of systems in state $|\psi\rangle$ [30, 40]. In
other words, it quantifies to what extent the two observables are incompatible
in this state: the sharper the knowledge on one, the less sharp the knowledge
on the other. However, in finite dimensional Hilbert spaces, this can never
give a useful lower bound on the product of the variances _for all states
$|\psi\rangle\in{\mathcal{H}}$ at once_. Indeed, when $|\psi\rangle$ is an
eigenstate of $A$ or of $B$, the right hand side vanishes and the inequality
does not give a useful lower bound on the variances at all. In other words, in
a finite dimensional Hilbert space, one always has that

$\min_{\langle\psi|\psi\rangle=1}|\langle\psi|C|\psi\rangle|=0.$ (17)

In fact, since $\operatorname{Tr}C=0$, the self-adjoint operator $C$ cannot
have purely positive or purely negative spectrum: if it has a positive
eigenvalue, it must also have a negative one, and vice-versa. As a result, the
quadratic form $\langle\psi|C|\psi\rangle$ cannot be positive or negative
definite, which again implies (17). As a result, to get any information on the
uncertainty on the knowledge of the values of the observables $A$ and $B$ in
the state $|\psi\rangle$, one needs to know something about the state
$|\psi\rangle$, namely the value of $\langle\psi|C|\psi\rangle$. When it
vanishes, no information is obtained. In this sense, the noncommutativity of
$A$ and $B$ alone does not provide automatically an information on the degree
of incompatibility of the two observables for any state $|\psi\rangle$. Again
we see here that the common practice of equating the “incompatibility” of two
quantum observables with the noncommutativity of the corresponding
observables, recalled in the previous section, yields only limited information
and suggests stronger notions of incompatibility are needed. We will
extensively come back to this point below.

One may think the above observations are particular to the finite dimensional
situation, but as we now briefly indicate, they are not. Let us therefore
consider two observables represented by bounded self-adjoint operators $A$ and
$B$ on an infinite dimensional Hilbert space. They may not have any
normalizable eigenvectors, and only continuous spectrum. The above arguments
leading to (18) do then not apply but one still has the following simple
result, closely related to Putnam’s theorem [41]. The proof is given in
Appendix A.

######  Proposition 3.

Let $A$ and $B$ be self-adjoint bounded operators on a finite or infinite
dimensional Hilbert space and let $C=-i[A,B]$. Then

$\inf_{\langle\psi|\psi\rangle=1}|\langle\psi|C|\psi\rangle|=0.$ (18)

Note that (18) again implies that the Heisenberg-Robertson uncertainty
relation in Eq. (16) does not give a uniform bound on the variances for all
$|\psi\rangle$ at once so that, here as well, noncommutativity alone does not
imply an information on the degree of incompatibility of the observables $A$
and $B$.

Note that, if $A=Q$ and $B=P$, then $C=\mathbb{I}$ and of course (18) does not
hold in that case. A slight variation on this example is $A=h(Q)$, $B=P$, so
that $C=h^{\prime}(Q)$. Assuming $h^{\prime}(q)>\delta>0$ for all $q$, again
(18) does not hold. As a concrete example, one may consider $h(q)=q+a\sin q$
with $a<1$. Then $h^{\prime}(q)\geq(1-a)>0$ and
$\inf_{\langle\psi|\psi\rangle=1}|\langle\psi|C|\psi\rangle|=1-a$. Of course,
in none of these cases are $A$ and $B$ bounded. These examples show that, in
infinite dimension, the commutator $C$ can be a positive definite operator
which has positive spectrum bounded away from zero. This cannot happen
however, if both $A$ and $B$ are bounded, as a result of (18). In that case,
it is still possible for $C$ to be non-negative, meaning that its spectrum is
non-negative, and includes $0$ so that (18) still holds. Examples of such
operators are given in [42, 43, 44].

The conclusion we draw is therefore that, generally, the fact that two
observables do not commute does not necessarily provide a useful bound on the
uncertainties inherent in their measurement on an arbitrary state, as measured
by the variances. As in the discussion in the previous section, this implies
again that the standard notion of incompatibility of two observables (namely
their noncommutativity) does not imply much about such uncertainty. These
observations provide a first motivation to look for other means than the
variance to express the uncertainty inherent in a state with regards to the
measurement of two observables.

Before doing so, let us point out that, in finite dimension, there is a second
drawback to the use of variances to measure uncertainties. Indeed, when
measurements are made in two bases ${\mathcal{A}}$ and ${\mathcal{B}}$ that
are not naturally associated with observables $A$ and $B$, then mean values
and variances cannot be meaningfully defined [45, 46]. Different approaches to
the quantification of uncertainty are then needed. The entropic and support
uncertainty relations provide such alternatives, as discussed next.

###  4.2 Entropic uncertainty relations

The two shortcomings of (16) in finite dimensional Hilbert space pointed out
above are one of the reasons that entropic uncertainty relations have been
developed in this context [45, 46, 40]. The best known of those is the
Maassen-Uffink uncertainty relation which reads

$H_{{\mathcal{A}},{\mathcal{B}}}(\psi)=H_{\mathcal{A}}(\psi)+H_{\mathcal{B}}(\psi)\geq\ln
M_{{\mathcal{A}},{\mathcal{B}}}^{-2},$ (19)

where ${\mathcal{A}},{\mathcal{B}}$ are two orthonormal bases of
${\mathcal{H}}$, as above, and

$H_{\mathcal{A}}(\psi)=-\sum_{i}|\langle a_{i}|\psi\rangle|^{2}\ln|\langle
a_{i}\psi\rangle|^{2},\quad H_{\mathcal{B}}(\psi)=-\sum_{i}|\langle
a_{i}|\psi\rangle|^{2}\ln|\langle a_{i}\psi\rangle|^{2}$

are the Shannon entropies associated to the ${\mathcal{A}}$\- and
${\mathcal{B}}$-representations of $|\psi\rangle$. Note that, when
${\mathcal{A}},{\mathcal{B}}$ are the eigenbases of two observables $A$ and
$B$, the entropic uncertainty $H_{{\mathcal{A}},{\mathcal{B}}}(\psi)$ does not
depend on the spectra of $A$ and $B$, only on the eigenbases, contrary to the
variances $\Delta A,\Delta B$. In addition, $H_{\mathcal{A}}$ and
$H_{\mathcal{B}}$ are insensitive to permutations of the basis vectors and to
global phase changes.

The entropy $H_{\mathcal{A}}(\psi)$ vanishes if and only if $|\psi\rangle$
equals one of the basis vectors; in that case the uncertainty inherent in
$|\psi\rangle$ with respect to an ${\mathcal{A}}$-measurement vanishes. An
analogous statement holds for $H_{\mathcal{B}}$. The entropy
$H_{{\mathcal{A}},{\mathcal{B}}}(\psi)$ yields a measure of the joint
uncertainty inherent in $|\psi\rangle$ with respect to ${\mathcal{A}}$ and
${\mathcal{B}}$ measurements. It vanishes iff $|\psi\rangle$ is a common basis
vector of ${\mathcal{A}}$ and ${\mathcal{B}}$. In fact, such a vector exists
if and only if $M_{{\mathcal{A}},{\mathcal{B}}}=1$. In that case, the Maassen-
Uffink uncertainty relation provides no information, since the Shannon
entropies are always nonnegative. When $M_{{\mathcal{A}},{\mathcal{B}}}<1$, on
the other hand, it provides a uniform lower bound valid for all
$|\psi\rangle$, similar to the Heisenberg uncertainty relation for conjugate
observables. We will see below that the condition
$M_{{\mathcal{A}},{\mathcal{B}}}<1$ can be interpreted as a – very weak –
incompatibility condition between the bases: see Proposition 7. The smaller
$M_{{\mathcal{A}},{\mathcal{B}}}$, the stronger the bound obtained. From (3)
one concludes therefore that the strongest entropic uncertainty relation is
obtained when $M_{{\mathcal{A}},{\mathcal{B}}}=\frac{1}{\sqrt{d}}$, in which
case all overlaps have the same amplitude and the bases are mutually unbiased.
The lower bound is then reached (at least) by the basis vectors. We note
however that, even when $M_{{\mathcal{A}},{\mathcal{B}}}<1$, the bound is not
always optimal and may not be reached, as extensively discussed in [47].

###  4.3 The support uncertainty relation

Another uncertainty relation that has been used in various contexts other than
quantum mechanics is the support uncertainty relation (see Eq. (21) below),
first introduced in [36] in the context of the Fourier transform on finite
groups. It has been shown to have much wider validity, however: see [37, 34].

To describe it, we introduce first the ${\mathcal{A}}$-support and
${\mathcal{B}}$-support of $|\psi\rangle$:

$S_{\psi}=\\{i\in\llbracket 1,d\rrbracket\mid\langle
a_{i}|\psi\rangle\not=0\\},\quad\mathrm{and}\quad T_{\psi}=\\{j\in\llbracket
1,d\rrbracket\mid\langle b_{j}|\psi\rangle\not=0\\}.$

They are the supports of the state $|\psi\rangle$ in the ${\mathcal{A}}$\- and
${\mathcal{B}}$-representations introduced in Eq. (7). One can also think of
$S_{\psi}$ as the set of possible outcomes of a fine-grained measurement in
the ${\mathcal{A}}$-basis with the pre-measurement state $|\psi\rangle$. We
then define

$n_{\mathcal{A}}(\psi)=|S_{\psi}|\not=0,\
n_{\mathcal{B}}(\psi)=|T_{\psi}|\not=0,$ (20)

where $|S|$ denotes the number of elements in $S$. Then
$n_{\mathcal{A}}(\psi)$ (respectively $n_{\mathcal{B}}(\psi)$) is the number
of nonvanishing entries of $u(\psi)$ (respectively $v(\psi)$);
$n_{\mathcal{A}}(\psi)$ is therefore the size of the support or the “spread”
of the probability distribution $|u_{i}(\psi)|^{2}$ (respectively
$|v_{j}(\psi)|^{2}$) of the state $|\psi\rangle$ in the ${\mathcal{A}}$\-
(respectively ${\mathcal{B}}$)-representations. As such
$n_{{\mathcal{A}}}(\psi)$ and $n_{{\mathcal{B}}}(\psi)$ are measures of the
localization of $|\psi\rangle$ in the respective representations. They can be
thought of as providing a measure of the uncertainty on the outcomes of a
measurement in the ${\mathcal{A}}$-basis, respectively ${\mathcal{B}}$-basis
when the system is in the state $|\psi\rangle$.

In our notation the support uncertainty relation then reads

$n_{\mathcal{A}}(\psi)n_{\mathcal{B}}(\psi)\geq
M_{{\mathcal{A}},{\mathcal{B}}}^{-2}.$ (21)

The analogy with the Heisenberg-Robinson uncertainty relation is clear: it
says that the supports of $|\psi\rangle$ in the ${\mathcal{A}}$\- and in the
${\mathcal{B}}$-representation cannot both be small since their product is
bounded below. Like the Robertson uncertainty principle, it has a very simple
proof, that we provide for completeness in Appendix B. (See also [19] for an
alternatiive argument.) It can also be seen as an immediate consequence of the
more sophisticated Maassen-Uffink bound since the Shannon entropy is well-
known to have the property that

$H_{\mathcal{A}}(\psi)\leq\ln(n_{{\mathcal{A}}}(\psi)),\quad
H_{\mathcal{B}}(\psi)\leq\ln(n_{{\mathcal{B}}}(\psi)).$

Unlike what happens in the Heisenberg-Robinson uncertainty relation, and
similarly to what happens for the Maassen-Uffink entropic inequality, the
lower bound in Eq. (21) has the advantage that it does not depend on the state
$|\psi\rangle$, and only depends on ${\mathcal{A}}$ and ${\mathcal{B}}$
through the maximum value $M_{{\mathcal{A}},{\mathcal{B}}}$ of the matrix
elements of $U$. It is also not necessarily saturated, as we will see below.
Like the Maassen-Uffink bound, it is without interest if
$M_{{\mathcal{A}},{\mathcal{B}}}=1$, since both $n_{{\mathcal{A}}}(\psi)$ and
$n_{{\mathcal{B}}}(\psi)$ are positive integers.

For the purpose of relating uncertainty to incompatibility and KD
nonclassicality, it turns out to be more instructive to consider lower bounds
on the sum

$n_{{\mathcal{A}},{\mathcal{B}}}(\psi):=n_{{\mathcal{A}}}(\psi)+n_{{\mathcal{B}}}(\psi),$
(22)

that we shall refer to as the _support uncertainty_ of $|\psi\rangle$, rather
than on the product $n_{{\mathcal{A}}}(\psi)n_{{\mathcal{B}}}(\psi)$. Both
such bounds yield information on the tradeoff between uncertainty in the
${\mathcal{A}}$\- and ${\mathcal{B}}$-representations of the state, but, as we
shall establish, the sum provides a more natural and direct link with a strong
form of incompatibility and KD nonclassicality. The situation is not unlike
the one in quantum optics, where the sum of the variances $(\Delta Q)^{2}$ and
$(\Delta P)^{2}$ of two conjugate quadratures $Q,P$ of the field yields a
measure of the uncertainty and of the optical nonclassicality of the state
referred to as its _total noise_ [48]. It satisfies the lower bound $(\Delta
Q)^{2}+(\Delta P)^{2}\geq 1$ for all states.

For later purposes, we also introduce the _minimal support uncertainty_
$n_{{\mathcal{A}},{\mathcal{B}}}^{\min}$ of the two bases
${\mathcal{A}},{\mathcal{B}}$ as

$n_{{\mathcal{A}},{\mathcal{B}}}^{\min}=\min_{|\psi\rangle\not=0}n_{{\mathcal{A}},{\mathcal{B}}}(\psi).$
(23)

Clearly

$2\leq n_{{\mathcal{A}},{\mathcal{B}}}^{\min}\leq d+1.$

The lower bound is obvious since for any nonzero
$|\psi\rangle\in{\mathcal{H}}$, $n_{{\mathcal{A}}}(\psi)\geq 1$ and
$n_{{\mathcal{B}}}(\psi)\geq 1$. For the upper bound, consider any basis
vector $|a_{i}\rangle$. Then $n_{{\mathcal{A}}}(|a_{i}\rangle)=1$,
$n_{{\mathcal{B}}}(|a_{i}\rangle)\leq d$. So
$n_{{\mathcal{A}},{\mathcal{B}}}(|a_{i}\rangle)\leq d+1$, which yields the
upper bound. Note that

$n_{{\mathcal{A}},{\mathcal{B}}}^{\min}\geq 3\Leftrightarrow
M_{{\mathcal{A}},{\mathcal{B}}}<1.$ (24)

Indeed, $n_{{\mathcal{A}},{\mathcal{B}}}^{\min}\geq 3$ is equivalent to the
fact that each ${\mathcal{A}}$ basis vector is the linear combination of at
least two ${\mathcal{B}}$-basis vectors and vice versa. Hence
$M_{{\mathcal{A}},{\mathcal{B}}}<1$. The converse is obvious. More generally,
Eq. (21) implies that

$n_{{\mathcal{A}},{\mathcal{B}}}^{\min}\geq\frac{2}{M_{{\mathcal{A}},{\mathcal{B}}}},$
(25)

which is again not sharp in general, as can be observed in Fig. 1. See also
the next subsection.

We have already pointed out that the Maassen-Uffink and support uncertainty
relations provide no information when $M_{{\mathcal{A}},{\mathcal{B}}}=1$.
This is not surprising. Indeed, when $M_{{\mathcal{A}},{\mathcal{B}}}=1$,
${\mathcal{A}}$ has $k\geq 1$ basis vectors $|a_{i}\rangle$ that are equal (up
to an irrelevant phase) to $k$ basis vectors $|b_{j}\rangle$. It is then not
possible to get a nontrivial bound constraining the uncertainties
$n_{{\mathcal{A}}}(\psi),n_{{\mathcal{B}}}(\psi)$ since they can both be
simultaneously equal to $1$. As anticipated at the end of Section 2, after
reordering the basis vectors and adjusting their phase, we can then assume
$|a_{1}\rangle=|b_{1}\rangle,\dots,|a_{k}\rangle=|b_{k}\rangle$ and split the
Hilbert space ${\mathcal{H}}$ into the subspace ${\mathcal{H}}_{k}$ generated
by the first $k$ basis vectors and its orthogonal complement
${\mathcal{H}}_{k}^{\perp}$, of dimension $d^{\prime}=d-k$. The analysis of
the uncertainty inherent in a state $|\psi\rangle$ can then be done separately
for its component in each of these subspaces.

###  4.4 Uncertainty diagrams

Given two bases ${\mathcal{A}}$ and ${\mathcal{B}}$, their _uncertainty
diagram_ UNCD$({\mathcal{A}},{\mathcal{B}})$ was defined in [19] to be the
subset of all points $(n_{{\mathcal{A}}},n_{{\mathcal{B}}})$ in the first
quadrant of the $n_{{\mathcal{A}}}n_{{\mathcal{B}}}$-plane for which there
exists a state $|\psi\rangle\in{\mathcal{H}}$ so that

$n_{{\mathcal{A}}}=n_{{\mathcal{A}}}(\psi),\quad
n_{{\mathcal{B}}}=n_{{\mathcal{B}}}(\psi),$ (26)

In other words,

${\mathrm{UNCD}}({\mathcal{A}},{\mathcal{B}})=\\{(n_{{\mathcal{A}}},n_{{\mathcal{B}}})\in\mathbb{N}^{2}|\exists\psi\in{\mathcal{H}},n_{{\mathcal{A}}}(\psi)=n_{{\mathcal{A}}},n_{{\mathcal{B}}}(\psi)=n_{{\mathcal{B}}}\\}.$
(27)

One can also view the uncertainty diagram is the image of the map
$|\psi\rangle\to(n_{{\mathcal{A}}}(\psi),n_{{\mathcal{B}}}(\psi))\in\llbracket
1,d\rrbracket^{2}$. As we will see, the uncertainty diagram is a useful
practical and visual tool to study the mutual uncertainty of a state with
respect to the ${\mathcal{A}}$\- and ${\mathcal{B}}$\- representations.
Analogous uncertainty diagrams for entropic uncertainty measures were
introduced in [47]. In view of Eq. (21), the uncertainty diagram always lies
above the hyperbola

$n_{{\mathcal{A}}}n_{{\mathcal{B}}}=M_{{\mathcal{A}},{\mathcal{B}}}^{-2}.$

This lower bound is not necessarily reached, of course, since in general
$M_{{\mathcal{A}},{\mathcal{B}}}^{-2}$ is not an integer.

Three examples of uncertainty diagrams are presented in Fig. 1. In the left
panel a spin $1$ system is considered, for which
${\mathcal{H}}=\mathbb{C}^{3}$. We choose for ${\mathcal{A}}$ and
${\mathcal{B}}$ the eigenbases of $J_{z}$ and $J_{x}$, where

$J_{z}=\begin{pmatrix}1&0&0\\\ 0&0&0\\\ 0&0&-1\end{pmatrix},\quad
J_{x}=\frac{1}{\sqrt{2}}\begin{pmatrix}0&1&0\\\ 1&0&1\\\ 0&1&0\end{pmatrix}.$

We write $|z,\epsilon\rangle$ and $|x,\epsilon\rangle$, $\epsilon=1,0,-1$ for
the corresponding eigenvectors. The transition matrix from the $J_{z}$ basis
to the $J_{x}$ basis is then

$U=\frac{1}{2}\begin{pmatrix}1&\sqrt{2}&1\\\ \sqrt{2}&0&-\sqrt{2}\\\
1&-\sqrt{2}&1\end{pmatrix}.$

In this case, $m_{{\mathcal{A}},{\mathcal{B}}}=0$,
$M_{{\mathcal{A}},{\mathcal{B}}}=1/\sqrt{2}$ and the support uncertainty
relation reads $n_{z}n_{x}(\psi)\geq 2$. The corresponding lower bound on the
uncertainty diagram is therefore $n_{z}n_{x}=2$. Note that $J_{x}$ and $J_{z}$
do not commute so that the bases are incompatible. Details of the computations
underlying the left panel of Fig. 1 are given in Appendix C.

Figure 1: Uncertainty diagrams. Dot-dashed line:
$n_{\mathcal{A}}(\psi)+n_{\mathcal{B}}(\psi)=d+1$. Dashed curve:
$n_{\mathcal{A}}(\psi)n_{\mathcal{B}}(\psi)=M_{{\mathcal{A}},{\mathcal{B}}}^{-2}$.
Diamonds (blue): KD-nonclassical states. Squares (red): KD-classical states.
Left panel: Case of the spin $1$ transition matrix ($d=3$) for which
$m_{{\mathcal{A}},{\mathcal{B}}}=0,M_{{\mathcal{A}},{\mathcal{B}}}=1/\sqrt{2}$.
Middle panel. Case of the Tao matrix $(d=6)$ which is a MUB for which
therefore
$m_{{\mathcal{A}},{\mathcal{B}}}=\frac{1}{\sqrt{d}}=M_{{\mathcal{A}},{\mathcal{B}}}$.
Note the absence of states for which
$(n_{{\mathcal{A}}},n_{{\mathcal{B}}})=(2,5),(3,3)$ or $(5,2)$. Right panel.
Case of the DFT matrix for $d=7$ which is both a MUB and COINC.

The middle panel of Fig. 1 shows the uncertainty diagram for two basis
${\mathcal{A}},{\mathcal{B}}$ for which the transition matrix is given by the
Tao matrix [49], which is the transition matrix for two MUB in dimension $d=6$
given by

$U=\frac{1}{\sqrt{d}}\\\ \begin{pmatrix}1&1&1&1&1&1\\\
1&1&\omega&\omega&\omega^{2}&\omega^{2}\\\
1&\omega&1&\omega^{2}&\omega^{2}&\omega\\\
1&\omega&\omega^{2}&1&\omega&\omega^{2}\\\
1&\omega^{2}&\omega^{2}&\omega&1&\omega\\\
1&\omega^{2}&\omega&\omega^{2}&\omega&1\end{pmatrix},\quad\omega=\exp(i\frac{2\pi}{3}).$
(28)

The details of the computations underlying the middle panel of Fig. 1 are
given in Appendix E.

The right hand panel of Fig. 1 shows the uncertainty diagram of two bases
related by the discrete Fourier transform (DFT) in dimension $d=7$. Its
structure is fully explained by Theorem 8 below, as we shall see.

In general, given two bases ${\mathcal{A}}$ and ${\mathcal{B}}$, it is not
clear what the precise shape of its uncertainty diagram is. For example, it is
not clear what its lower edge $n_{\mathcal{A}}\in\llbracket 1,d\rrbracket\to
L(n_{{\mathcal{A}}})$ is, defined as

$L(n_{{\mathcal{A}}})=\min\\{n_{{\mathcal{B}}}(\psi)|\psi\in{\mathcal{H}},n_{{\mathcal{A}}}(\psi)=n_{{\mathcal{A}}},\|\psi\|=1\\}.$
(29)

We know from the support uncertainty relation that $L(n_{{\mathcal{A}}})\geq
M_{{\mathcal{A}},{\mathcal{B}}}^{-2}n_{{\mathcal{A}}}^{-1}$, but this bound
may not be reached for all values of $n_{{\mathcal{A}}}$. Also, there may be
“holes” in the uncertainty diagram, meaning values
$n_{{\mathcal{B}}}>L(n_{{\mathcal{A}}})$ for which there exists no
$|\psi\rangle$ so that
$n_{{\mathcal{A}}}(\psi)=n_{{\mathcal{A}}},n_{{\mathcal{B}}}(\psi)=n_{{\mathcal{B}}}$.
Both these phenomena are illustrated in Fig. 1. For example, for the Tao
matrix, one has

$L(1)=6,\ L(2)=4=L(3),\ L(4)=2,\ L(5)=3,\ L(6)=1.$

Further examples of uncertainty diagrams are given in [19]. Similar
difficulties arise with the uncertainty diagrams for entropic uncertainty, as
studied in [47].

It turns out that for completely incompatible bases, to which we turn our
attention next (Definition 4), the lower edge of the uncertainty diagram is
easily determined to be

$L(n_{{\mathcal{A}}})=d+1-n_{{\mathcal{A}}}.$

In addition, it turns out the uncertainty diagram then has no “holes”. This is
shown in Theorem 8 below.

##  5 Complete incompatibility

###  5.1 Complete incompatibility: definition, interpretation.

The previous sections make it clear that incompatiblity, understood as
noncommutativity, is a rather weak notion. On the one hand, it only implies
that a measurement in ${\mathcal{B}}$ _may_ perturb a previously obtained
outcome in a measurement in ${\mathcal{A}}$. On the other hand, it provides
only limited information on the (minimal) uncertainty inherent in an arbitrary
given state. To avoid these shortcomings, the following very stringent notion
of incompatibility was introduced in [19].

######  Definition 4.

We say that two bases ${\mathcal{A}}$ and ${\mathcal{B}}$ are completely
incompatible (COINC) if and only if all index sets $S,T$ in $\llbracket
1,d\rrbracket$ for which $|S|+|T|\leq d$ have the property that
${\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}\cap{\Pi_{{\mathcal{B}}}}(T){\mathcal{H}}=\\{0\\}$.

It follows from the discussion in Section 3, and particularly the one
concerning Eq. (15), that this mathematical property has a direct
interpretation in terms of the incompatibility of repeated successive
measurements in the ${\mathcal{A}}$ and ${\mathcal{B}}$ bases. In short,
complete incompatibility of ${\mathcal{A}}$ and ${\mathcal{B}}$ means the
following. Suppose that the system is prepared in _any_ state $|\psi\rangle$
and a measurement in the ${\mathcal{A}}$ basis yielding the outcome $S$ is
followed by one in the ${\mathcal{B}}$ basis yielding the outcome $T$. In that
case, this second measurement has _always_ a nonvanishing probability to
perturb the result of the first, provided $|S|+|T|\leq d$. Since Eq. (15)
allows for the possibility of the second measurement not to perturb the first
for certain pre-measurement states, it is by negating this condition that a
very strong notion of incompatibility is obtained. This is all the more true
since the condition is imposed for all outcomes $S,T$, with $|S|+|T|\leq d$.

Note that the restriction $|S|+|T|\leq d$ is needed since, when $|S|+|T|>d$,
the intersection
${\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}\cap{\Pi_{{\mathcal{B}}}}(T){\mathcal{H}}$
is necessarily nontrivial because the dimension of
${\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}$ is $|S|$ and that of
${\Pi_{{\mathcal{B}}}}(T){\mathcal{H}}$ is $|T|$. What this expresses is the
intuitively expected fact that if the measurements are too coarsegrained, then
it may happen that one does not perturb the other. Note also that it would be
sufficient in the definition to require
${\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}\cap{\Pi_{{\mathcal{B}}}}(T){\mathcal{H}}=\\{0\\}$
for all $S,T$ so that $|S|+|T|=d$, since the more general requirement follows
from this immediately.

The condition
$\Pi_{\mathcal{A}}(S){\mathcal{H}}\cap\Pi_{\mathcal{B}}(T){\mathcal{H}}=\\{0\\}$
can also be understood geometrically, a viewpoint that sometimes helps the
intuition. Given $S$, respectively $T$, one can think of
${\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}$, respectively
${\Pi_{{\mathcal{B}}}}(T){\mathcal{H}}$ as “coordinate vector spaces” for the
${\mathcal{A}}$-, respectively ${\mathcal{B}}$-representations. Indeed,
analogously with the “$xy$-plane”, which corresponds to points in
$\mathbb{R}^{3}$ that have vanishing $z$-coordinate, these spaces contain all
states that have vanishing coordinates corresponding to indices in $S^{c}$,
respectively $T^{c}$. The condition that
$\Pi_{\mathcal{A}}(S){\mathcal{H}}\cap\Pi_{\mathcal{B}}(T){\mathcal{H}}=\\{0\\}$
for all $|S|+|T|\leq d$ can then be understood as saying that these coordinate
spaces for the ${\mathcal{A}}$\- and ${\mathcal{B}}$-representation have a
trivial intersection. To put it differently, the coordinate spaces for
${\mathcal{A}}$ sit askew in the coordinate spaces for ${\mathcal{B}}$, and
vice versa.

The definition of COINC bases depends on the two bases ${\mathcal{A}}$ and
${\mathcal{B}}$ only through the unitary transition matrix between them,
defined as $U|a_{j}\rangle=|b_{j}\rangle$, with matrix elements
$U_{ij}=\langle a_{i}|b_{j}\rangle$ in the ${\mathcal{A}}$-basis. This can be
seen as follows. If ${\mathcal{A}},{\mathcal{B}}$ have transition matrix $U$
and ${\mathcal{A}}^{\prime},{\mathcal{B}}^{\prime}$ have transition matrix
$U^{\prime}$, then $U=U^{\prime}$ if and only if there exists a unitary map
$V$ so that $|a^{\prime}_{i}\rangle=V|a_{i}\rangle$ and
$|b^{\prime}_{i}\rangle=V|b_{i}\rangle$. But in that case, clearly,
${\mathcal{A}}$ and ${\mathcal{B}}$ are COINC if and only if
${\mathcal{A}}^{\prime}$ and ${\mathcal{B}}^{\prime}$ are. Conversely, given
an arbitrary unitary $d\times d$ matrix $U$, one can always view it as the
transition matrix between the canonical basis of
${\mathcal{H}}=\mathbb{C}^{d}$ (taken to be ${\mathcal{A}}$) and the basis
composed of the columns of $U$ (taken to be ${\mathcal{B}}$). In view of this,
we will say a unitary matrix $U$ is COINC whenever these two bases
${\mathcal{A}}$ and ${\mathcal{B}}$ are. Depending on the situation at hand,
it is sometimes more convenient to think in terms of two bases
${\mathcal{A}},{\mathcal{B}}$ and sometimes in terms of a unitary matrix $U$,
without reference to the bases. Note finally that the complete incompatibility
of two bases is also not affected by a renumbering, nor by global phase
changes of the basis vectors.

###  5.2 Complete incompatibility: characterization and examples

A number of examples of COINC bases were given in [19]. It is first of all
easy to see that in dimension $d=2,3$, two bases are COINC if and only if
$m_{{\mathcal{A}},{\mathcal{B}}}>0$, in other words when the transition matrix
has no zeros. This is no longer true in higher dimension as we shall show in
Proposition 7 below. For example, it is shown in [19] using result of Tao [49]
that, when the transition matrix $U$ is the discrete Fourier transform (DFT),
the bases are COINC if and only if $d$ is prime. Other explicit examples of
bases that are COINC don’t seem to be very easy to come by. It was
nevertheless illustrated numerically in [19] that small perturbations of MUB
for $d=4$ tend to be COINC. Theorem 5 below explains this numerical
observation in much more generality.

######  Theorem 5.

The set of completely incompatible unitary $d\times d$ matrices is open and
dense in the set of all unitary matrices.

The proof, given in Appendix F, relies on the following algebraic
characterization of complete incompatibility proven in [19].

######  Lemma 6.

Two bases ${\mathcal{A}}$ and ${\mathcal{B}}$ are COINC if and only if none of
the minors of their transition matrix $U$ vanishes.

To make this paper self-contained we also provide a proof of the lemma in
Appendix F.

The theorem may seem somewhat surprising at first sight, since we insisted on
the fact that complete incompatibility is a strong notion of incompatibility.
To understand why it is true, one may consider Lemma 6. A minor is a multi-
variable polynomial in the matrix elements of $U$ and its zeros form a very
“thin” set: essentially it is a hypersurface of co-dimension $1$. Hence, since
small perturbations of $U$ will typically change the values of all its minors,
those that initially vanish will typically become nonvanishing. On the other
hand, if the perturbation is small enough, those that do not vanish initially
will remain nonvanishing. This suggest that the set of COINC transition
matrices is dense. This is precisely what we prove in Appendix F. The proof is
somewhat technical because of the need to ensure that the perturbations
considered preserve the unitarity. By the same reasoning, the set of COINC
transition matrices is dense. Indeed, if no minor of $U$ vanishes, then this
will remain true for all small perturbations of $U$. The situation is similar
to what happens when one considers commuting self-adjoint $d\times d$
matrices, which are considered compatible. Those also form a very thin subset
of all pairs of self-adjoint matrices and again, a slight perturbation will
typically render them noncommuting, hence incompatible. And in this context
also, when two observables do not commute, small perturbations of them will
still not commute. This being said, the set of completely incompatible
observables is considerably smaller than the set of merely incompatible ones,
as will become clear in the following section.

###  5.3 A hierarchy of incompatibilities

As a first illustration of the strength of the notion of complete
incompatibility, let us consider the case where $A$ and $B$ are observables
with non-degenerate spectrum and eigenbases ${\mathcal{A}}$ and
${\mathcal{B}}$. Saying that $A$ and $B$ are incompatible in the standard
sense that they do not commute then means that there exist at least one $i,j$
so that $[|a_{i}\rangle\langle a_{i}|,\langle b_{j}\rangle\langle
b_{j}|]\not=0$. This is clearly much weaker than requiring that this needs to
be true for all $i,j$ which in turn is weaker than requiring that
$[{\Pi_{{\mathcal{A}}}}(S),{\Pi_{{\mathcal{B}}}}(T)]\not=0$ for all
$S,T\subset\llbracket 1,d\rrbracket$ with $|S|+|T|\leq d$. This is readily
illustrated on a spin $1$ system, so that ${\mathcal{H}}=\mathbb{C}^{3}$,
taking ${\mathcal{A}}$ to be the eigenbasis of $J_{z}$ and ${\mathcal{B}}$ of
$J_{x}$. In spite of the fact that $J_{x}$ and $J_{z}$ do not commute, the
corresponding eigenbases are not COINC. To see this, one may note that
$|x,0\rangle\in{\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}$, with $S=\\{-1,1\\}$.
So, if first a coarse-grained measurement is made of $J_{z}$ on an arbitrary
state $|\psi\rangle$, with two possible outcomes “$J_{z}=0$” and
“$J_{z}\not=0$”, and the outcome is “$J_{z}\not=0$”, then a subsequent fine-
grained measurement of $J_{x}$ yielding the outcome “$J_{x}=0$” will not
perturb the first measurement. This is reflected in the fact that the outcomes
$T=\\{0\\}$ and $S=\\{-1,1\\}$ are compatible in the sense of Definition 2:
$[{\Pi_{{\mathcal{A}}}}(S),{\Pi_{{\mathcal{B}}}}(T)]=0$.

To shed further light on the definition of complete incompatibility and to
justify the use of this terminology, we will now show that Definition 4
implies three properties that can be seen as weaker manifestations of
incompatibility.

######  Proposition 7.

Let ${\mathcal{A}},{\mathcal{B}}$ be two orthonormal bases on a $d\geq 2$
dimensional Hilbert space. Consider the following statements:

  1. (i)

${\mathcal{A}}$ and ${\mathcal{B}}$ are COINC.

  2. (ii)

For all $S,T\subset\llbracket 1,d\rrbracket$, with $1\leq|S|,|T|<d$,
$[{\Pi_{{\mathcal{A}}}}(S),{\Pi_{{\mathcal{B}}}}(T)]\not=0$.

  3. (iii)

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

  4. (iv)

$M_{{\mathcal{A}},{\mathcal{B}}}<1$.

Then (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii) $\Rightarrow$ (iv).
In addition

  * •

When $d=2$, one has (i) $\Leftrightarrow$ (ii) $\Leftrightarrow$ (iii)
$\Leftrightarrow$ (iv);

  * •

When $d=3$, one has (i) $\Leftrightarrow$ (ii) $\Leftrightarrow$ (iii)
$\Rightarrow$ (iv), but (iv) does not imply (iii);

  * •

When $d\geq 4$, one has (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii)
$\Rightarrow$ (iv), but (iv) does not imply (iii) and (iii) does not imply
(ii);

  * •

When $d=4$ or $d=6$, (ii) does not imply (i).

The proof of the proposition is given below, but we first discuss its meaning
by interpreting each of the four statements (i)-(iv) as a manifestation of
incompatibility. The interpretation of (i) was given above. We need to
understand the meaning of (ii), which is a strong statement of
“noncommutativity”. Indeed, when two operators $A$ and $B$ commute, then all
their spectral projectors commute. Saying that they don’t commute means
therefore that there exist some spectral projectors that don’t commute; (ii)
is much stronger, since it requires no spectral projectors to commute. For
example, the spin 1 bases discussed above clearly do not satisfy (ii), as
explained above.

To see that (ii) implies also a strong notion of incompatibility, first recall
that we showed two outcomes $S$ and $T$ are compatible (Definition 2) if and
only if $[{\Pi_{{\mathcal{A}}}}(S),{\Pi_{{\mathcal{B}}}}(T)]=0$ (see Eq.
(12)). So (ii) can be paraphrased by saying that no two outcomes $S$ and $T$
are compatible for a measurement in ${\mathcal{A}}$ followed by one in
${\mathcal{B}}$. This is indeed a strong notion of incompatibility. It can in
fact be viewed as the negation of a weak form of compatibility given by the
negation of (ii): there exist nontrivial $S,T$ so that
$[{\Pi_{{\mathcal{A}}}}(S),{\Pi_{{\mathcal{B}}}}(T)]=0$. However, note that
(ii) only means that for any two such outcomes, it _may_ happen – depending on
the pre-measurement state – that the occurrence of the second one perturbs the
first one. In other words, the second statement guarantees the possible
perturbation of the first measurement by the second for all outcomes, but not
for all states. One then understands why the second assertion is a weaker form
of incompatibility than the first. Indeed, when (i) holds, the second outcome
_always_ perturbs the first one, as explained above. In Appendix
LABEL:s:counterexamples we give examples of bases ${\mathcal{A}}$ and
${\mathcal{B}}$ satisfying (ii) but not (i) in dimensions $4$ and $6$. We
conjecture it is true that (ii) does not imply (i) in all dimensions $d\geq
4$, but we have not produced such examples in other dimensions than $4$ and
$6$. Note that checking condition (ii) in higher dimension is necessarily more
involved since there are then many more choices of $S$ and $T$.

Statement (iv), namely $M_{{\mathcal{A}},{\mathcal{B}}}<1$ guarantees that
none of the basis vectors of ${\mathcal{A}}$ is a basis vector of
${\mathcal{B}}$, and vice versa. This means that, when a first fine-grained
measurement in ${\mathcal{A}}$ leaves the system in a basis state of
${\mathcal{A}}$, a subsequent measurement of ${\mathcal{B}}$ has at least two
possible outcomes with nonzero probability and therefore necessarily disturbs
the first measurement, which is indeed a manifestation of incompatibility. So
a fine-grained measurement in ${\mathcal{B}}$ then necessarily perturbs a
previous fine-grained measurement in ${\mathcal{A}}$. In particular, if
${\mathcal{A}}$ and ${\mathcal{B}}$ are eigenbases of two observables $A$ and
$B$ with nondegenerate spectra, then (iv) implies the noncommutativity of all
$|a_{i}\rangle\langle a_{i}|$ and $|b_{j}\rangle\langle b_{j}|$ which implies
the noncommutativity of $A$ and $B$ [50]. The converse is not true: $A$ and
$B$ may not commute and nevertheless have some eigenvectors in common, in
which case $M_{{\mathcal{A}},{\mathcal{B}}}=1$.

Statement (iii) is a strengthening of (iv). It is sometimes referred to as a
condition of “complementarity” for the two bases. This is because the
condition $m_{{\mathcal{A}},{\mathcal{B}}}>0$ implies that, when the system is
in one of the basis states of ${\mathcal{A}}$ and a fine-grained measurement
is made in ${\mathcal{B}}$, then any outcome $|b_{j}\rangle$ may occur with
nonvanishing probability. And vice versa. In that sense, there is uncertainty
in the outcome of this second measurement, and this uncertainty is larger when
$m_{{\mathcal{A}},{\mathcal{B}}}$ is larger. Some authors [23] reserve the
term “complementary” or “maximally incompatible” for mutually unbiased bases,
for which these probabilities are the same for all $j$ so that
$m_{{\mathcal{A}},{\mathcal{B}}}$ takes on its maximal possible value
$1/\sqrt{d}$. When (iii) holds, repeated successive fine-grained measurements
of ${\mathcal{A}}$ and ${\mathcal{B}}$ do not yield the same result with
probability one. In fact, they can give any result and in that sense there is
incompatibility between ${\mathcal{A}}$ and ${\mathcal{B}}$. The proposition
asserts this form of incompatibility does not imply the stronger condition on
the commutators in (ii), in any dimension $d\geq 4$.

We now turn to the proof of Proposition 7 which contains instructive examples
and counterexamples.
_Proof._ (i) $\Rightarrow$ (ii). We prove the contrapositive. Suppose there
exist $S,T$, with $1\leq|S|,|T|<d$,
$[{\Pi_{{\mathcal{A}}}}(S),{\Pi_{{\mathcal{B}}}}(T)]=0$. Then also
$[{\Pi_{{\mathcal{A}}}}(S),{\Pi_{{\mathcal{B}}}}(T^{c})]=0=[{\Pi_{{\mathcal{A}}}}(S^{c}),{\Pi_{{\mathcal{B}}}}(T)]=[{\Pi_{{\mathcal{A}}}}(S^{c}),{\Pi_{{\mathcal{B}}}}(T^{c})]$.
So we can assume without loss of generality that $1\leq|S|\leq|S^{c}|<d$,
$1\leq|T|\leq|T^{c}|<d$, and $1\leq|S|\leq|T|<d$. It follows that
$|S|+|T|\leq|S|+|T^{c}|\leq d$. Suppose there exists $\phi\in{\mathcal{H}}$ so
that
$|\psi\rangle={\Pi_{{\mathcal{A}}}}(S){\Pi_{{\mathcal{B}}}}(T)\phi\not=0$.
Then
$0\not=\psi\in{\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}\cap{\Pi_{{\mathcal{B}}}}(T){\mathcal{H}}$,
so that ${\mathcal{A}},{\mathcal{B}}$ are not COINC. If such a $\phi$ does not
exist, then ${\Pi_{{\mathcal{A}}}}(S)\Pi_{\mathcal{B}}(T)=0$. Now let
$0\not=\psi\in{\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}$. Then
$|\psi\rangle={\Pi_{{\mathcal{A}}}}(S)\psi={\Pi_{{\mathcal{A}}}}(S){\Pi_{{\mathcal{B}}}}(T^{c})\psi={\Pi_{{\mathcal{B}}}}(T^{c}){\Pi_{{\mathcal{A}}}}(S)\psi$.
Hence
$0\not=\psi\in{\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}\cap{\Pi_{{\mathcal{B}}}}(T^{c}){\mathcal{H}}$,
implying ${\mathcal{A}},{\mathcal{B}}$ are not COINC.
(ii) $\Rightarrow$ (iii). We prove the contrapositive. Suppose
$m_{{\mathcal{A}},{\mathcal{B}}}=0$. Then there exist $i,j$ so that $\langle
a_{i}|b_{j}\rangle=0$. Possibly renumbering the bases, we can assume $i=1=j$.
Hence $|b_{1}\rangle\in{\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}$, with
$S=\\{2,3,\dots,d\\}$. Setting $T=\\{1\\}$, this means
${\Pi_{{\mathcal{B}}}}(T){\mathcal{H}}\subset{\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}$.
Hence $[{\Pi_{{\mathcal{A}}}}(S),{\Pi_{{\mathcal{B}}}}(T)]=0$ (Lemma 16 (iii))
and so (ii) does not hold.
(iii) $\Rightarrow$ (iv). See Eq. (4).

We now consider the reverse implications. We first show that, if $d=2$ or $3$,
then (iii) implies (i) (and hence (ii) by the above). We will use Lemma 6 for
that purpose. Note that, if (iii) holds, then $U^{\dagger}$, which is the
inverse of $U$, also has no zero elements. In addition, the matrix elements of
$U^{\dagger}$ are non-zero multiples of $(d-1)$-minors of $U$. This implies
that, when $d=2,3$, no minors of $U$ vanish. Hence the bases are COINC.

That (iv) does not imply (iii) is obvious for all $d\geq 3$ and is illustrated
by the spin $1$ example given above for $d=3$.

We now show that (iii) does not imply (ii) if $d\geq 4$. Treating first the
case $d=4$, we consider

$U(s)=\frac{1}{2}\begin{pmatrix}1&1&1&1\\\ 1&1&-1&-1\\\ 1&-1&s&-s\\\
1&-1&-s&s\end{pmatrix},\quad|s|=1;U^{\dagger}(s)=U(\overline{s}).$

These are the most general transition matrices for MUB in dimension $d=4$, up
to permutations of rows and columns, and global phase [25]. Taking
$S=\\{1,2\\}=T$, we have

${\Pi_{{\mathcal{A}}}}(S)=\begin{pmatrix}1&0&0&0\\\ 0&1&0&0\\\ 0&0&0&0\\\
0&0&0&0\end{pmatrix}\ \textrm{and}\
{\Pi_{{\mathcal{B}}}}(T)=\frac{1}{2}\begin{pmatrix}1&1&0&0\\\ 1&1&0&0\\\
0&0&1&1\\\ 0&0&1&1\end{pmatrix}$

These clearly commute so (ii) does not hold. An analogous construction yields
the result in all dimensions $d\geq 4$ as we now show. Let $n=d-2$ and set,
for some $a,\delta\in\mathbb{C}$,

$U=\begin{pmatrix}a&a&\delta&\dots&\delta\\\ a&a&-\delta&\dots&-\delta\\\
\delta&-\delta&u_{11}&\dots&u_{1n}\\\ \vdots&\vdots&\vdots&\vdots&\vdots\\\
\delta&-\delta&u_{n1}&\dots&u_{nn}\end{pmatrix}$

The normalization and orthogonality of the first two columns and rows implies

$|a|=\frac{1}{2},\quad|\delta|=\frac{1}{\sqrt{2n}}.$

Now write $u_{i}=(u_{i1}\dots u_{in})^{T}\in\mathbb{R}^{n}$, then the matrix
$U$ will be unitary provided one has, for all $1\leq i<j\leq n$,

$\langle 1|u_{i}\rangle=0,\quad\langle
u_{i}|u_{i}\rangle=1-2\delta^{2},\quad\langle
u_{i}|u_{j}\rangle=-2\delta^{2},$

where $|1\rangle=(1\dots 1)^{T}\in\mathbb{R}^{n}$. Normalizing the $u_{i}$,
one has $\tilde{u}_{i}=u_{i}/\|u_{i}\|$ and, for all $1\leq i<j\leq n$,

$\langle\tilde{u}_{i}|\tilde{u}_{j}\rangle=-\frac{1}{n-1}.$

This means the $\tilde{u}_{i}$ form a regular simplex of $n$ equidistant
points on the unit sphere in the $(n-1)$-dimensional Euclidean plane
perpendicular to $|1\rangle\in\mathbb{R}^{n}$. When $d=4$, $n=2$ and
$\tilde{u}_{2}=-\tilde{u}_{1}$. When $d=5$, $n=3$ and
$\tilde{u}_{1},\tilde{u}_{2},\tilde{u}_{3}$ form an equilateral triangle. When
$d=6$, $n=4$, and $\tilde{u}_{1},\dots\tilde{u}_{4}$ form a regular
tetrahedron. The matrix $U$ so constructed clearly satisfies (iii) but, for
the same reason as in the case $d=4$, it does not satisfy (ii).

It remains to show that in dimension $d=4$ and $d=6$ it is not true that (ii)
implies (i). We start with $d=4$. Let

$|c\rangle=\frac{1}{\sqrt{2}}(|1-2\rangle+|3+4\rangle),\quad|c^{\prime}\rangle=\frac{1}{\sqrt{2}}(|1-2\rangle-|3+4\rangle),$

where

$|1\pm 2\rangle=\frac{1}{\sqrt{2}}(|a_{1}\rangle\pm|a_{2}\rangle),$

and similarly for $|3\pm 4\rangle$. Then define, for
$\theta,\theta^{\prime}\in[0,\pi]$,

$\displaystyle|b_{1}\rangle$ $\displaystyle=$
$\displaystyle\cos\theta|1+2\rangle+\sin\theta|c\rangle$
$\displaystyle|b_{2}\rangle$ $\displaystyle=$
$\displaystyle-\sin\theta|1+2\rangle+\cos\theta|c\rangle$
$\displaystyle|b_{3}\rangle$ $\displaystyle=$
$\displaystyle\cos\theta^{\prime}|3-4\rangle+\sin\theta^{\prime}|c^{\prime}\rangle$
$\displaystyle|b_{4}\rangle$ $\displaystyle=$
$\displaystyle-\sin\theta^{\prime}|3-4\rangle+\cos\theta^{\prime}|c^{\prime}\rangle$

The $|b_{i}\rangle$ clearly form an orthonormal basis for all
$\theta,\theta^{\prime}$. Also,
$|1+2\rangle\in{\Pi_{{\mathcal{B}}}}(\\{1,2\\})$ so that
${\Pi_{{\mathcal{A}}}}(\\{1,2\\})\cap{\Pi_{{\mathcal{B}}}}(\\{1,2\\})\not=\\{0\\}$.
Hence the bases ${\mathcal{A}}$ and ${\mathcal{B}}$ are not COINC. The
transition matrix between the two bases is

$\displaystyle
U=\begin{pmatrix}\frac{1}{\sqrt{2}}\cos\theta+\frac{1}{2}\sin\theta&-\frac{1}{\sqrt{2}}\sin\theta+\frac{1}{2}\cos\theta&\frac{1}{2}\sin\theta^{\prime}&\frac{1}{2}\cos\theta^{\prime}\\\
\frac{1}{\sqrt{2}}\cos\theta-\frac{1}{2}\sin\theta&-\frac{1}{\sqrt{2}}\sin\theta-\frac{1}{2}\cos\theta&-\frac{1}{2}\sin\theta^{\prime}&-\frac{1}{2}\cos\theta^{\prime}\\\
\frac{1}{2}\sin\theta&\frac{1}{2}\cos\theta&\frac{1}{\sqrt{2}}\cos\theta^{\prime}-\frac{1}{2}\sin\theta^{\prime}&-\frac{1}{\sqrt{2}}\sin\theta^{\prime}-\frac{1}{2}\cos\theta^{\prime}\\\
\frac{1}{2}\sin\theta&\frac{1}{2}\cos\theta&-\frac{1}{\sqrt{2}}\cos\theta^{\prime}-\frac{1}{2}\sin\theta^{\prime}&\frac{1}{\sqrt{2}}\sin\theta^{\prime}-\frac{1}{2}\cos\theta^{\prime}\end{pmatrix}$

To show (ii) holds, we need to show none of the commutators
$[{\Pi_{{\mathcal{A}}}}(S),{\Pi_{{\mathcal{B}}}}(T)]$ vanish, if
$1<|S|,|T|<d$. Choosing $\theta,\theta^{\prime}$ so that
$m_{{\mathcal{A}},{\mathcal{B}}}>0$, it is immediate from Lemma 16 (v) that,
for $|S|=1$ or $|T|=1$,
$[{\Pi_{{\mathcal{A}}}}(S),{\Pi_{{\mathcal{B}}}}(T)]\not=0$. Indeed, since the
transition matrix has no zeros, no eigenvector $|a_{i}\rangle$ is contained in
any of the spaces ${\Pi_{{\mathcal{B}}}}(T){\mathcal{H}}$, nor is it
perpendicular to it. Since
$[{\Pi_{{\mathcal{A}}}}(S),{\Pi_{{\mathcal{B}}}}(T)]=[{\Pi_{{\mathcal{A}}}}(S^{\textrm{c}}),{\Pi_{{\mathcal{B}}}}(T)]=[{\Pi_{{\mathcal{A}}}}(S),{\Pi_{{\mathcal{B}}}}(T^{\textrm{c}})]=[{\Pi_{{\mathcal{A}}}}(S^{\textrm{c}}),{\Pi_{{\mathcal{B}}}}(T^{\textrm{c}})]$,
the same holds when $|S|=3$ or $|T|=3$. It remains therefore only to check the
cases where $|S|=2=|T|$, of which there are 36. However, since when the
statement is true for some $S,T$, it also is for $S,T^{\textrm{c}}$, _etc._ ,
we need to only check nine cases, which we take to be
$S=\\{1,2\\},\\{1,3\\},\\{1,4\\}$ and $T=\\{1,2\\},\\{1,3\\},\\{1,4\\}$. We
will use Lemma 17 and therefore need to check, for each of those choices, that
${\Pi_{{\mathcal{A}}}}(S^{\textrm{c}}){\Pi_{{\mathcal{B}}}}(T){\Pi_{{\mathcal{A}}}}(S)\not=0$.
If $S=\\{i,j\\}$, $S^{\textrm{c}}=\\{m,n\\}$ and $T=\\{k,\ell\\}$, we have

$\displaystyle{\Pi_{{\mathcal{A}}}}(S^{\textrm{c}}){\Pi_{{\mathcal{B}}}}(T){\Pi_{{\mathcal{A}}}}(S)$
$\displaystyle=$ $\displaystyle\hskip-56.9055pt|a_{m}\rangle\left[\langle
a_{m}|b_{k}\rangle\langle b_{k}|a_{i}\rangle+\langle
a_{m}|b_{\ell}\rangle\langle b_{\ell}|a_{i}\rangle\right]\langle
a_{i}|+|a_{m}\rangle\left[\langle a_{m}|b_{k}\rangle\langle
b_{k}|a_{j}\rangle+\langle a_{m}|b_{\ell}\rangle\langle
b_{\ell}|a_{j}\rangle\right]\langle a_{j}|$
$\displaystyle\hskip-56.9055pt+|a_{n}\rangle[\langle a_{n}|b_{k}\rangle\langle
b_{k}|a_{i}\rangle+\langle a_{n}|b_{\ell}\rangle\langle
b_{\ell}|a_{i}\rangle]\langle a_{i}|+|a_{n}\rangle[\langle
a_{n}|b_{k}\rangle\langle b_{k}|a_{j}\rangle+\langle
a_{n}|b_{\ell}\rangle\langle b_{\ell}|a_{j}\rangle]\langle a_{j}|$

Since all four terms appearing here are linearly independent, it is enough to
check for each choice of $S$ and $T$ that one of the four coefficients in
square brackets does not vanish. The computations are straightforward but
tedious and we don’t reproduce the details here. Setting
$\theta=\frac{\pi}{3}=\theta^{\prime}$, one does indeed find
${\Pi_{{\mathcal{A}}}}(S^{\textrm{c}}){\Pi_{{\mathcal{B}}}}(T){\Pi_{{\mathcal{A}}}}(S)\not=0$
for the above choices of $T$ and $S$.

We now turn again to the Tao matrix in $d=6$. We know it is not COINC and will
now show it satisfies nevertheless condition (ii). Using Lemma 17, one easily
sees that the latter is equivalent to

$\forall S,T\in\llbracket
1,d\rrbracket,1\leq|S|,|T|<d,U(S^{\textrm{c}},T)U(S,T)^{\dagger}\not=0.$

Because of the very larg number of index sets $S,T$ to check, we resorted to a
numerical computation to show that, when $U$ is the Tao matrix, this is indeed
true. ∎

###  5.4 The uncertainty diagram of COINC bases

In this section we analyze what the complete incompatibility of two bases
implies for their uncertainty diagram: we will see it can be completely
determined and that it has a very simple form.

######  Theorem 8.

Let ${\mathcal{A}},{\mathcal{B}}$ be two orthonormal bases of ${\mathcal{H}}$.
Then the following statements are equivalent:
(i) ${\mathcal{A}}$ and ${\mathcal{B}}$ are COINC; (ii)
${\mathrm{UNCD}}({\mathcal{A}},{\mathcal{B}})\subset\\{n_{{\mathcal{A}}}+n_{{\mathcal{B}}}\geq
d+1\\}\cap\llbracket 1,d\rrbracket^{2}$; (iii)
${\mathrm{UNCD}}({\mathcal{A}},{\mathcal{B}})=\\{n_{{\mathcal{A}}}+n_{{\mathcal{B}}}\geq
d+1\\}\cap\llbracket 1,d\rrbracket^{2}$
In particular, $n_{{\mathcal{A}},{\mathcal{B}}}^{\min}=d+1$ and the lower edge
$L(n_{{\mathcal{A}}})$, defined in (29), satisfies
$L(n_{{\mathcal{A}}})=d+1-n_{{\mathcal{A}}}$.

Part (ii) of the theorem is essentially a rephrasing of the definition of the
total incompatibility of two bases in terms of the localization properties of
all states in the ${\mathcal{A}}$\- and ${\mathcal{B}}$-representations.
Precisely, as already explained in [19], it ensures that ${\mathcal{A}}$ and
${\mathcal{B}}$ are COINC if and only if all states $|\psi\rangle$ have a
support uncertainty that is at least equal to $d+1$. This means, for example,
that if $|\psi\rangle$ is the superposition of two (respectively three, …)
basis vectors $|a_{i}\rangle$, then it has nonvanishing coefficients on at
least $d-1$ (respectively $d-2$, $d-3$,…) basis vectors $|b_{j}\rangle$. The
uncertainty diagram of COINC bases therefore forms a triangle whose lower edge
has equation $n_{{\mathcal{A}}}+n_{{\mathcal{B}}}=d+1$. The relation
$n_{{\mathcal{A}}}(\psi)+n_{{\mathcal{B}}}(\psi)\geq d+1$ is clearly an
uncertainty relation. It is much stronger than the multiplicative uncertainty
relation Eq. 21. But of course, it only holds for a restricted class of bases,
namely the COINC ones.

Part (iii) of the theorem gives a stronger assertion. It says that, if
${\mathcal{A}}$ and ${\mathcal{B}}$ are COINC, given any two index sets
$S,T\subset\llbracket 1,d\rrbracket$ such that $|S|+|T|\geq d+1$, there exist
a state $|\psi\rangle$ with $S_{\psi}=S,T_{\psi}=T$. In fact, the proof shows
that the set of such states is an open and dense subset of
$\Pi_{\mathcal{A}}(S){\mathcal{H}}\cap\Pi_{\mathcal{B}}(T){\mathcal{H}}$,
which is a $(|S|+|T|-d)$-dimensional subspace of ${\mathcal{H}}$. (See Lemma 9
below.) In particular, a randomly chosen $|\psi\rangle$ in
$\Pi_{\mathcal{A}}(S){\mathcal{H}}\cap\Pi_{\mathcal{B}}(T){\mathcal{H}}$ has
these support properties. The theorem is illustrated graphically in the third
panel of Fig. 1 for the example of the DFT with $d=7$, which is COINC as
already pointed out. Note the absence of states with
$(n_{\mathcal{A}}(\psi),n_{\mathcal{B}}(\psi))$ in the region of the
$(n_{\mathcal{A}},n_{\mathcal{B}})$-plane above or on the curve
$n_{\mathcal{A}}n_{\mathcal{B}}=d$ and strictly below the line
$n_{\mathcal{A}}+n_{\mathcal{B}}=d+1$.

The theorem also provides a link between total incompatibility and minimal
uncertainty. Suppose $|\psi\rangle$ is a minimal support uncertainty state,
for which therefore
$n_{{\mathcal{A}},{\mathcal{B}}}(\psi)=n_{{\mathcal{A}}}(\psi)+n_{{\mathcal{B}}}(\psi)=d+1$.
Suppose now we wish to reduce the uncertainty in an
${\mathcal{A}}$-measurement by considering a state $\phi$ for which
$n_{{\mathcal{A}}}(\phi)=n_{{\mathcal{A}}}(\psi)-1$. Then necessarily, for
this state, $n_{{\mathcal{B}}}(\phi)\geq n_{{\mathcal{B}}}(\psi)+1$. So the
gain in precision (or the decrease of uncertainty) of the
${\mathcal{A}}$-measurement is compensated by at least an equal loss in
precision for the ${\mathcal{B}}$-measurement. In this sense, when the bases
are COINC, the increase of precision on the variable associated to
${\mathcal{A}}$ is constrained optimally by the loss of precision on the one
associated to ${\mathcal{B}}$.

_Proof._ (i) $\Rightarrow$ (ii). This is an immediate consequence of the
definition of COINC. Indeed, clearly
$|\psi\rangle\in{\Pi_{{\mathcal{A}}}}(S_{\psi}){\mathcal{H}}\cap{\Pi_{{\mathcal{B}}}}(T_{\psi}){\mathcal{H}}$.
Hence $|S_{\psi}|+|T_{\psi}|>d$.

(i) $\Leftarrow$ (ii). Let us prove the contrapositive. Suppose
${\mathcal{A}}$ and ${\mathcal{B}}$ are not COINC. Then there exist $S,T$,
with $|S|+|T|\leq d$ and
$\Pi_{\mathcal{A}}(S){\mathcal{H}}\cap\Pi_{\mathcal{B}}(T){\mathcal{H}}\not=\\{0\\}$.
Hence there exist a state
$|\psi\rangle\in\Pi_{\mathcal{A}}(S){\mathcal{H}}\cap\Pi_{\mathcal{B}}(T){\mathcal{H}}$.
For this state
$n_{{\mathcal{A}}}(\psi)\leq|S|,n_{{\mathcal{B}}}(\psi)\leq|T|$. Hence
$n_{{\mathcal{A}},{\mathcal{B}}}(\psi)\leq d$.

Since clearly (iii) implies (ii), it remains to prove (i) implies (iii). For
that purpose, let $1\leq R\leq d$, and consider $S,T\in\llbracket
1,d\rrbracket$ so that $|S|+|T|=d+R$. Reorganizing the labels on the basis
vectors we can assume, without loss of generality, that

$T=\\{1,2,\dots,K\\},\quad S=\\{1,2,\dots,d-(K-R)\\},$

with $0\leq K-R\leq d-1$. Note that $R\leq K$.

We need to construct
$|\psi\rangle\in\Pi_{\mathcal{A}}(S){\mathcal{H}}\cap\Pi_{\mathcal{B}}(T){\mathcal{H}}$
with the condition that $S_{\psi}=S,T_{\psi}=T.$ Lemma 9 provides the result.∎

######  Lemma 9.

If ${\mathcal{A}}$ and ${\mathcal{B}}$ are COINC, and $S,T\in\llbracket
1,d\rrbracket$ so that $|S|+|T|=d+R$, $R\geq 1$, then
$\Pi_{\mathcal{A}}(S){\mathcal{H}}\cap\Pi_{\mathcal{B}}(T){\mathcal{H}}$ is
$R$-dimensional and
$\Pi_{\mathcal{A}}(S){\mathcal{H}}+\Pi_{\mathcal{B}}(T){\mathcal{H}}={\mathcal{H}}$.
In addition, the subset of
$|\psi\rangle\in\Pi_{\mathcal{A}}(S){\mathcal{H}}\cap\Pi_{\mathcal{B}}(T){\mathcal{H}}$
for which $T_{\psi}=T,S_{\psi}=S$ is dense and open in
$\Pi_{\mathcal{A}}(S){\mathcal{H}}\cap\Pi_{\mathcal{B}}(T){\mathcal{H}}$.

###### Proof.

Note that if $K=R$, then $S=\llbracket 1,d\rrbracket$ and the first statement
is immediate. We therefore consider the case where $R<K$. Let
$|\psi\rangle=\sum_{j=1}^{K}d_{j}|b_{j}\rangle\in\Pi_{\mathcal{B}}(T){\mathcal{H}}$,
with $d\in\mathbb{C}^{K}$. Then
$|\psi\rangle\in\Pi_{\mathcal{A}}(S){\mathcal{H}}$ if and only if

$\langle a_{i}|\psi\rangle=0=\sum_{j=1}^{K}\langle a_{i}|b_{j}\rangle
d_{j},\quad\forall 1\leq d-(K-R)+1\leq i\leq d.$ (30)

These are $K-R$ homogeneous equations in the $K$ unknows $d_{j}$. Then, by
Lemma 6, the $(K-R)$ by $K$ matrix $(\langle a_{i}|a_{j}\rangle)_{ij}$, with
$1\leq d-(K-R)+1\leq i\leq d,1\leq j\leq K$ is of maximal rank $K-R$. As a
result, the solutions of the homogeneous equation (30) form an $R$-dimensional
subspace of $\Pi_{\mathcal{B}}(T){\mathcal{H}}$.

By the dimension theorem for the sum of two vector spaces we have

$\dim({\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}+{\Pi_{{\mathcal{B}}}}(T){\mathcal{H}})=\dim{\Pi_{{\mathcal{A}}}}(S){\mathcal{H}}+\dim{\Pi_{{\mathcal{B}}}}(T){\mathcal{H}}-\dim({\Pi_{{\mathcal{A}}}}{\mathcal{H}}(S)\cap{\Pi_{{\mathcal{B}}}}(T){\mathcal{H}}),$

which proves the first statement.

Now fix $1\leq j\leq K$ and let $T^{\prime}_{j}=T\setminus\\{j\\}$. Then, by
the same reasoning $\Pi_{\mathcal{A}}(S)\cap\Pi_{\mathcal{B}}(T^{\prime}_{j})$
is an $(R-1)$-dimensional subspace of
$\Pi_{\mathcal{A}}(S)\cap\Pi_{\mathcal{B}}(T)$. It is the subspace of its
states for which $\langle b_{j}|\psi\rangle=0$. Then consider

$\psi\in{\mathcal{O}}_{T}:=\left(\Pi_{\mathcal{A}}(S)\right)\cap\Pi_{\mathcal{B}}(T)\setminus\left(\cup_{j=1}^{K}\Pi_{\mathcal{A}}(S)\cap\Pi_{\mathcal{B}}(T^{\prime}_{j})\right).$

Then $\langle b_{j}|\psi\rangle\not=0$, for all $1\leq j\leq K$. So
$T_{\psi}=T$. Note that ${\mathcal{O}}_{T}$ is an open dense set because it is
obtained by removing from the $R$-dimensional vector space
$\Pi_{\mathcal{A}}(S)\cap\Pi_{\mathcal{B}}(T)$ a finite number of
$(R-1)$-dimensional vector spaces. We can similarly define

${\mathcal{O}}_{S}:=\left(\Pi_{\mathcal{A}}(S)\right)\cap\Pi_{\mathcal{B}}(T)\setminus\left(\cup_{i=1}^{d-(K-R)}\Pi_{\mathcal{A}}(S^{\prime}_{i})\cap\Pi_{\mathcal{B}}(T)\right).$

Then $\langle a_{i}|\psi\rangle\not=0$, for all $1\leq i\leq d-(K-R)$. So
$S_{\psi}=S$. Taking $|\psi\rangle\in{\mathcal{O}}_{T}\cap{\mathcal{O}}_{S}$,
which is still open and dense, we obtain the desired result. ∎

##  6 Relating complete incompatibility and mutual unbiasedness

We have justified our definition of complete incompatibility of two bases
${\mathcal{A}}$ and ${\mathcal{B}}$ (Definition 4) through an analysis of the
effect of a ${\mathcal{B}}$-measurement on previously obtained information in
an ${\mathcal{A}}$-measurement. We have seen that, essentially, two bases are
COINC if and only if the measurement in ${\mathcal{B}}$ always perturbs this
previous information, whatever the outcome of the ${\mathcal{A}}$-measurement,
which can be coarse grained or fine grained.

One can, alternatively, consider only the situations where the
${\mathcal{A}}$-measurement is fine grained, so that the state after the
${\mathcal{A}}$-measurement is $|a_{i}\rangle$, and require that the
subsequent fine-grained ${\mathcal{B}}$ measurement has a maximally uncertain
outcome, which means that $|\langle a_{i}|b_{j}\rangle|=1/\sqrt{d}$, so that
all outcomes $b_{j}$ are equally likely. This idea has lead to the definition
of MUB which are, as already mentioned, bases ${\mathcal{A}}$ and
${\mathcal{B}}$ for which the transition matrix has the property that
$M_{{\mathcal{A}},{\mathcal{B}}}=1/\sqrt{d}=m_{{\mathcal{A}},{\mathcal{B}}}$.
MUB have attracted attention because of their potential use in various quantum
information protocols [20, 21, 22, 23]. One may note that, when dealing with
two conjugate (and hence continuous) variables $X$ and $P$, an analogous
property holds; indeed $|\langle x|p\rangle|=\frac{1}{\sqrt{2\pi}}$ in that
case, expressing the idea that when the localization in $X$ is perfect, the
uncertainty in $P$ is maximal and vice versa. The definition of MUB captures
this same phenomenon: perfect localization in the
${\mathcal{A}}$-representation leads to maximal uncertainty in the
${\mathcal{B}}$-representation. However, conjugate variables have an
additional property: their support in the $X$\- and $P$-representations cannot
be both finite. Indeed, it is well known that there do not exist states
$|\psi\rangle$ for which the $X$-representation $\langle x|\psi\rangle$ is
localized inside some bounded set $S\subset\mathbb{R}$ and the
$P$-representation $\langle p|\hat{\psi}\rangle$ is localized in some bounded
set $T\subset\mathbb{R}$ [33]. In fact, if the $X$-support of $|\psi\rangle$
is bounded, its $P$-support must be unbounded. The definition of complete
incompatibility naturally transcribes this second property to a somewhat
analogous statement in the finite-dimensional case. However, for the DFT in
finite dimension, it follows from [38] and Theorem 8 that it is COINC if and
only if the dimension is prime. This shows not all MUB are COINC. We point out
that, on the contrary, the noncommutativity/incompatibility measures
introduced in [18] are maximal on MUB and therefore do not distinguish between
them.

More generally, one may therefore ask the question under what circumstances
MUB are COINC. It is easy to see that in dimenion $d=2,3,5$ all MUB are COINC,
whereas it was shown in [19] that in dimension $d=4$, none of them are. In
fact, we showed above that, in dimension $d=4$, no MUB satisfy the weaker
condition (ii) of Proposition 7. In higher dimension, the situation is not
clear, in particular because a complete characterization of all MUB is not
known for $d\geq 6$.

Two questions arise therefore naturally:

“In which dimensions do their exist MUB that are COINC?”

and

“In which dimensions are all MUB COINC?”

The answer to the first question certainly includes all prime dimensions
since, as already mentioned, the DFT is COINC in prime dimension. Answering
this question is probably not easy since a complete characterization of all
MUB does not exist. Similarly, the only dimensions eligible for a positive
answer to the second question are the prime dimensions. Again, in dimensions
$2,3,5$ the assertion is valid, but in higher prime dimensions the answer is
not known, to the best of our knowledge.

While these questions seem difficult, as an immediate consequence of Theorem
5, one nevertheless has that, arbitrarily close to any MUB, there are always
bases that are COINC, in any dimension. This is the content of the following
theorem, proven in Appendix F.

######  Theorem 10.

For all $d\geq 2$ and for all $0<m<\frac{1}{\sqrt{d}}$ there exist bases
${\mathcal{A}},{\mathcal{B}}$ that are COINC and whose transition matrix $U$
satisfies $m\leq m_{{\mathcal{A}},{\mathcal{B}}}\leq 1/{\sqrt{d}}$.

##  7 Support uncertainty as a KD-nonclassicality witness

We now turn to the connection between the support uncertainty of states
$|\psi\rangle$ and their KD-nonclassical nature, as defined in the
Introduction. Our most general result, under the weakest conditions on the
transition matrix $U$ (Theorem 12), can be paraphrased as stating that -
provided $U$ does not have too many zeros - states $|\psi\rangle$ with a large
support uncertainty are KD nonclassical. In other words, the support
uncertainty is a KD-nonclassicality witness. The natural lower bound on the
support uncertainty to obtain this conclusion is, as we shall see, the line
$n_{{\mathcal{A}},{\mathcal{B}}}=d+1$, that we shall refer to as the _KD-
nonclassicality edge_. Under such general circumstances, the support
uncertainty is not a faithful witness of nonclassicality: there may be states
with a small support uncertainty that are nevertheless KD nonclassical. An
example of this situation can be observed in the central panel of Fig. 1,
where the uncertainty diagram of the Tao matrix (which is not COINC, as shown
above) is displayed.

When $U$ is COINC and $m_{{\mathcal{A}},{\mathcal{B}}}$ close to $1/\sqrt{d}$,
we obtain a complete characterization of the KD-classical states: we show that
the only KD-classical states are the basis states, all others being
nonclassical (Theorem 13).

To state these results, we need some further terminology. Let $Z_{\textrm{c}}$
be the maximum total number of zeros that can be found in any two distinct
columns of the transition matrix $U$. Let $Z_{\textrm{r}}$ be the maximum
total number of zeros in two distinct rows of $U$. If $Z$ is the total number
of zeros, then of course $Z_{\textrm{r}},Z_{\textrm{c}}\leq Z$. If $Z=0,1$ or
$2$, then so are $Z_{\textrm{c}}$ and $Z_{\textrm{r}}$. But $Z$ can be
considerably higher than $Z_{\textrm{r}}$ and $Z_{\textrm{c}}$, as we will see
in examples below. Theorem 12 is an immediate consequence of the following
more technical statement, which is of interest in its own right.

######  Proposition 11.

Let ${\mathcal{A}},{\mathcal{B}}$ be two orthonormal bases on a $d\geq 2$
dimensional Hilbert space ${\mathcal{H}}$, with transition matrix $U$. Let
$|\psi\rangle\in{\mathcal{H}}$ and suppose

$\displaystyle\max\\{n_{{\mathcal{A}}}(\psi),n_{{\mathcal{B}}}(\psi\\}>\max\\{Z_{\textrm{r}},Z_{\textrm{c}}\\},$
(31) $\displaystyle
n_{{\mathcal{A}},{\mathcal{B}}}(\psi)=n_{\mathcal{A}}(\psi)+n_{\mathcal{B}}(\psi)>d+1,$
(32)

then $|\psi\rangle\in{\mathcal{H}}$ is KD nonclassical.

The proof of this result is given below.

The first condition on $|\psi\rangle$, Eq. (31), is implied by the second in
those cases where there are not “too many” zeros in $U$. This idea is made
precise in the following theorem.

######  Theorem 12.

Let ${\mathcal{A}},{\mathcal{B}}$ be two orthonormal bases on a Hilbert space
${\mathcal{H}}$ of dimension $d\geq 2$, with transition matrix $U$. Suppose
$\max\\{Z_{\textrm{r}},Z_{\textrm{c}}\\}\leq\frac{d+1}{2}$. Then, if
$|\psi\rangle\in{\mathcal{H}}$ satisfies Eq. (32), it is KD nonclassical.

This result was proven in [19] for all $d\geq 2$, but only under the
restrictive hypothesis that $U$ has no zeros, in which case Eq. (31) is
automatically satisfied for any $|\psi\rangle$. The case $d=2$ is particular
and fully covered by the following remarks. Note that when $d=2$, then
$Z=\max\\{Z_{\textrm{r}},Z_{\textrm{c}}\\}$. Since a unitary two by two matrix
cannot have exactly one zero, it has either one or two zeros. The hypothesis
$\max\\{Z_{\textrm{r}},Z_{\textrm{c}}\\}\leq\frac{d+1}{2}=3/2$ therefore
corresponds to the case where $U$ has no zeros at all. In that case the
theorem, together with the fact that the basis states are classical, implies
that the only nonclassical states are those with
$n_{{\mathcal{A}}}(\psi)=2=n_{{\mathcal{B}}}(\psi)$. If on the contrary $U$
does have two zeros, then it is (equivalent to) the identity matrix, and then
there are no nonclassical states at all.

Proof of Theorem 12. We use Proposition 11. It is enough to show that when
$\max\\{Z_{\textrm{r}},Z_{\textrm{c}}\\}\leq\frac{d+1}{2}$, Eq. (31) is
implied by Eq. (32). Suppose therefore that Eq. (32) holds. Then
$\max\\{n_{{\mathcal{A}}}(\psi),n_{{\mathcal{B}}}(\psi)\\}>\frac{d+1}{2}\geq\max\\{Z_{\textrm{r}},Z_{\textrm{c}}\\}$,
which proves the result. ∎

Here are two examples with $d=6$, that satisfy the hypotheses of the theorem:

$U_{6}=\frac{1}{\sqrt{5}}\begin{pmatrix}0&1&1&1&1&1\\\ 1&0&1&-1&1&-1\\\
1&1&0&-1&-1&1\\\ 1&-1&-1&0&1&1\\\ 1&1&-1&1&0&-1\\\
1&-1&1&1&-1&0\end{pmatrix},\quad
U^{\prime}_{6}=\frac{1}{\sqrt{5}}\begin{pmatrix}0&1&1&1&1&1\\\
1&0&1&-1&1&-1\\\ 1&1&0&1&-1&-1\\\ 1&-1&1&0&-1&1\\\ 1&1&-1&-1&0&1\\\
1&-1&-1&1&1&0\end{pmatrix}.$ (33)

Indeed, $Z=6,Z_{\textrm{r}}=2=Z_{\textrm{c}}$. These examples can be extended
to arbitrarily high dimension, as follows. Let us write
$|a_{i}\rangle,|b_{j}\rangle$ for two bases of $\mathbb{C}^{6}$ having $U_{6}$
as transition matrix. Let ${\mathcal{H}}=\mathbb{C}^{6}\otimes\mathbb{C}^{k}$
and let $V$ be the unitary $k$ by $k$ transition matrix of two bases
$|a^{\prime}_{i^{\prime}}\rangle,|b^{\prime}_{j^{\prime}}\rangle$ of
$\mathbb{C}^{k}$, that we assume to not have any zeros. Hence $U_{6}\otimes V$
is the transition matrix between the bases
$|a_{i},a^{\prime}_{i^{\prime}}\rangle,|b_{i},b^{\prime}_{j^{\prime}}\rangle$.
Then each basis vector $|b_{j},b^{\prime}_{j^{\prime}}\rangle$ has exactly $k$
zero matrix elements on the basis $|a_{i},a^{\prime}_{i^{\prime}}\rangle$. So
the total number of zeros in the transition matrix is $Z=6k^{2}$ and any $2$
such basis vectors have $2k$ vanishing components implying
$Z_{\textrm{c}}=2k$. Similarly $Z_{\textrm{r}}=2k$. Since the dimension of the
Hilbert space is $d=6k$, the condition in Theorem 12 holds for any
$k\in\mathbb{N}_{*}$. These examples show that the number of zeros in any
given column can in fact be very large, and proportional to the length of the
column (which is $6k$ in these examples).

When the two bases ${\mathcal{A}}$ and ${\mathcal{B}}$ are mutually unbiased,
or in a suitable sense close to mutually unbiased, a stronger result can be
proven, that we now turn to.

######  Theorem 13.

Let ${\mathcal{A}},{\mathcal{B}}$ be two orthonormal bases on a Hilbert space
${\mathcal{H}}$ of dimension $d\geq 3$, with transition matrix $U$. Suppose

$\frac{d-1}{d+1}<\left(\frac{m_{{\mathcal{A}},{\mathcal{B}}}}{M_{{\mathcal{A}},{\mathcal{B}}}}\right)^{2}\leq
1.$ (34)

Suppose $|\psi\rangle$ is KD classical. Then either
$|\psi\rangle\in{\mathcal{A}}\cup{\mathcal{B}}$ or
$n_{{\mathcal{A}}}(\psi)+n_{{\mathcal{B}}}(\psi)\leq d$. Consequently, if
${\mathcal{A}}$ and ${\mathcal{B}}$ are in addition COINC, then the only
classical states are the basis states.

We excluded the case $d=2$ from the statement because in that case, if
$m_{{\mathcal{A}},{\mathcal{B}}}>0$, we know from Theorem 12 that, if
$|\psi\rangle$ is classical, then
$n_{{\mathcal{A}}}(\psi)+n_{{\mathcal{B}}}(\psi)\leq 3$. This, in turn,
implies that $|\psi\rangle\in{\mathcal{A}}\cup{\mathcal{B}}$, since $d=2$,
proving the result. So the additional constraint on
$m_{{\mathcal{A}},{\mathcal{B}}}/M_{{\mathcal{A}},{\mathcal{B}}}$ is not
relevant when $d=2$.

The theorem implies that for the DFT in prime dimension, which is both
mutually unbiased and COINC, only the basis vectors are classical. This can be
observed in the rightmost panel of Fig. 1 for $d=7$.

Note that $m_{{\mathcal{A}},{\mathcal{B}}}=1/\sqrt{d}$ is equivalent to
$m_{{\mathcal{A}},{\mathcal{B}}}=M_{{\mathcal{A}},{\mathcal{B}}}$ and hence to
the bases being mutually unbiased. The theorem completely characterizes the
classical states of COINC bases that have a large
$m_{{\mathcal{A}},{\mathcal{B}}}$, meaning that
$m_{{\mathcal{A}},{\mathcal{B}}}$ is close to its maximal possible value
$1/\sqrt{d}$, attained only for MUB. More precisely, suppose that, for some
$0\leq\delta<1$,

$m_{{\mathcal{A}},{\mathcal{B}}}^{2}>\frac{1}{d}(1-\delta).$

Then the normalization of the columns of $U$ implies that

$m_{{\mathcal{A}},{\mathcal{B}}}^{2}(d-1)+M_{{\mathcal{A}},{\mathcal{B}}}^{2}\leq
1,$

so that

$\left(\frac{m_{{\mathcal{A}},{\mathcal{B}}}}{M_{{\mathcal{A}},{\mathcal{B}}}}\right)^{2}>\frac{1}{1+\frac{\delta
d}{1-\delta}}.$

Hence, provided

$\delta<\frac{2}{d(d-1)}\frac{1}{1+\frac{2}{d(d-1)}},$

hypothesis Eq. (34) is satisfied. This proves Theorem 1 (iv). Note that, while
this is an increasingly restrictive condition as the dimension $d$ grows, we
know such $U$ exist by Theorem 10.

The proof of Theorem 13 relies on the arguments in the proof of Proposition 11
that we therefore prove first.

_Proof (of Proposition 11)_ The proof follows the strategy used in [16] and
[19]. We proceed by contradiction and suppose $|\psi\rangle$ is KD classical
and that hypothesis (31) holds. We need to prove that

$n_{{\mathcal{A}}}(\psi)+n_{{\mathcal{B}}}(\psi)\leq d+1.$

When $n_{{\mathcal{A}}}(\psi)=1$, or $n_{{\mathcal{B}}}(\psi)=1$,
$|\psi\rangle$ is one of the basis states and the result is then immediate. We
can therefore assume that $n_{{\mathcal{A}}}(\psi),n_{{\mathcal{B}}}(\psi)\geq
2$. Since the KD distribution (see Eq. (1)) is insensitive to global phase
rotations
$|a_{i}\rangle\to\exp(i\phi_{i})|a_{i}\rangle,|b_{j}\rangle\to\exp(i\phi_{j}^{\prime})|b_{j}\rangle$,
we can suppose that all $\langle a_{i}|\psi\rangle$ and
$\langle\psi|b_{j}\rangle$ are nonnegative (hence real) for $1\leq i,j\leq d$.
Possibly reordering the basis vectors, we can suppose that $\langle
a_{i}|\psi\rangle\not=0\not=\langle b_{j}|\psi\rangle$ for $1\leq i\leq
n_{\mathcal{A}}(\psi),1\leq j\leq n_{\mathcal{B}}(\psi)$ whereas all other
$\langle a_{i}|\psi\rangle$, $\langle b_{j}|\psi\rangle$ vanish. By
hypothesis, the KD distribution of $|\psi\rangle$ is real and nonnegative.
Hence, for the same range of $i$ and $j$, we can conclude $\langle
a_{i}|b_{j}\rangle$ is real and nonnegative.

Now, assume $2\leq n_{{\mathcal{B}}}(\psi)\leq n_{{\mathcal{A}}}(\psi)=d$.
Then the matrix $U$ contains two columns with real nonnegative entries. Since
by hypothesis (31), there are at most $Z_{c}<d$ zeros in these two columns of
$U$ this is in contradiction with the fact that those columns are orthogonal.

Let us therefore assume that $2\leq n_{\mathcal{B}}(\psi)\leq
n_{\mathcal{A}}(\psi)<d$. Then, for $1\leq j<j^{\prime}\leq
n_{\mathcal{B}}(\psi)$, we have

$0=\langle
b_{j}|b_{j^{\prime}}\rangle=\sum_{i=1}^{n_{\mathcal{A}}(\psi)}\langle
b_{j}|a_{i}\rangle\langle
a_{i}|b_{j^{\prime}}\rangle+\sum_{i=n_{\mathcal{A}}(\psi)+1}^{d}\langle
b_{j}|a_{i}\rangle\langle a_{i}|b_{j^{\prime}}\rangle.$ (35)

We first show that for all $1\leq j<j^{\prime}\leq n_{\mathcal{B}}(\psi)$, one
has

$\sum_{i=1}^{n_{\mathcal{A}}(\psi)}\langle b_{j}|a_{i}\rangle\langle
a_{i}|b_{j^{\prime}}\rangle>0.$ (36)

To see this, note first that, for any fixed pair $j\not=j^{\prime}$, the sum
in (36) contains $n_{\mathcal{A}}(\psi)$ nonnegative terms. If at least one of
those is positive, the sum is positive. But this is the case, since at most
$Z_{c}$ of these terms can vanish, and $Z_{c}<n_{\mathcal{A}}(\psi)$ by
hypothesis (31). This proves Eq. (36).

It then follows from Eq. (35) that, for all $1\leq j<j^{\prime}\leq
n_{\mathcal{B}}(\psi)$, one has

$\sum_{i=n_{\mathcal{A}}(\psi)+1}^{d}\langle b_{j}|a_{i}\rangle\langle
a_{i}|b_{j^{\prime}}\rangle<0.$

Defining, for each $1\leq j\leq n_{\mathcal{B}}(\psi)$ the vector
$v_{j}=(\langle a_{n_{\mathcal{A}}(\psi)+1}|b_{j}\rangle,\dots,\langle
a_{d}|b_{j}\rangle)\in\mathbb{C}^{d-n_{\mathcal{A}}(\psi)}$ we see from the
above that $\langle v_{j}|v_{j^{\prime}}\rangle<0$. It then follows from Lemma
14 below that

$n_{\mathcal{B}}(\psi)\leq d-n_{\mathcal{A}}(\psi)+1.$

This proves the result.

The case where $n_{\mathcal{A}}(\psi)\leq n_{\mathcal{B}}(\psi)<d$ is treated
similarly, inverting the roles of the columns and the rows. ∎

_Proof (of Theorem 13)_ We proceed as in the proof of Proposition 11. We can
again suppose $2\leq n_{{\mathcal{B}}}(\psi)\leq n_{{\mathcal{A}}}(\psi)<d$.
Note that this implies that $|\psi\rangle$ is not one of the basis vectors
since $m_{{\mathcal{A}},{\mathcal{B}}}>0$. Eq. (35) now yields

$m_{{\mathcal{A}},{\mathcal{B}}}^{2}n_{{\mathcal{A}}}(\psi)\leq|\sum_{i=n_{\mathcal{A}}(\psi)+1}^{d}\langle
b_{j}|a_{i}\rangle\langle a_{i}|b_{j^{\prime}}\rangle|\leq
M_{{\mathcal{A}},{\mathcal{B}}}^{2}(d-n_{{\mathcal{A}}}(\psi)).$

Using that $n_{{\mathcal{A}}}(\psi)\leq n_{{\mathcal{B}}}(\psi)$, this implies

$n_{{\mathcal{A}}}(\psi)+n_{{\mathcal{B}}}(\psi)\leq\frac{2d}{1+\left(\frac{m_{{\mathcal{A}},{\mathcal{B}}}}{M_{{\mathcal{A}},{\mathcal{B}}}}\right)^{2}}<\frac{2d}{1+\frac{d-1}{d+1}}=d+1.$

It follows that $n_{{\mathcal{A}}}(\psi)+n_{{\mathcal{B}}}(\psi)\leq d$, which
is the desired result. The second statement of the Theorem now follows from
Theorem 8. ∎

The following lemma can be understood geometrically as putting an upper bound
on the number of vectors in $\mathbb{C}^{n}$ that can have an obtuse angle
between them, two by two. It is a refinement of a result in [16], where only
part (i) of the Lemma was proven. It appeared in the Supplementary Material of
[19], we repeat the proof here for completeness.

######  Lemma 14.

Let $n,k\in\mathbb{N}_{*}$ and
$v_{1},v_{2},\dots,v_{k}\in\mathbb{C}^{n}\setminus\\{0\\}$. Then the following
holds:
(i) If $\langle v_{i}|v_{j}\rangle\leq 0$ for all $1\leq i<j\leq k$, then
$k\leq 2n$.
(ii) If $\langle v_{i}|v_{j}\rangle<0$ for all $1\leq i<j\leq k$, then $k\leq
n+1$.

The proof follows from an induction argument, given below. For $n=1$, there is
no difference between (i) and (ii). Indeed, one may note that one can always
take $v_{1}>0$, by applying a common phase rotation to all $v_{i}$, which does
not change the inner products $\langle v_{i}|v_{j}\rangle$ between them. Hence
$v_{j}<0$ for all $j\not=1$. But if $k>2$, then this contradicts the
requirement that $v_{2}v_{3}\leq 0$. So $k\leq 2$ when $n=1$. For arbitrary
$n$, it is clear the upper bound in (i) is reached by taking for example
$v_{1}=e_{1}=-v_{2},v_{3}=e_{2}=-v_{4},\dots v_{2n}=-e_{n}$. Some of the
$v_{i}$ are then orthogonal, so that this set does not satisfy the hypothesis
of (ii). When $n=2$, one can understand geometrically why in (ii) the upper
bound is only $k=3$ and not $k=4$, as in (i). For that purpose, let us reason
as if we were working in $\mathbb{R}^{2}$, not $\mathbb{C}^{2}$. By applying a
rotation to all $v_{i}\in\mathbb{R}^{2}$, we can consider $v_{1}=e_{1}$. Let
$v_{j}=(\cos\theta_{j},\sin\theta_{j})$. Since we are in the plane, the
hypothesis then implies that the angles between $v_{j}$ and $v_{1}$ must be
larger than $\pi/2$ for all $2\leq j\leq k$, so that
$\theta_{j}\in]-\pi/2,\pi/2[$. But the hypothesis further implies
$|\theta_{j}-\theta_{j^{\prime}}|>\frac{\pi}{2}$, which can only be true if
$j$ takes at most $2$ values so that $k\leq 3$.
_Proof (of Lemma 14)_ (i) The proof goes by induction. We have seen the result
holds for $n=1$. Suppose the result holds for some $n\in\mathbb{N}_{*}$. We
show it holds for $n+1$. Let $v_{1},\dots,v_{k}\in\mathbb{C}^{n+1}$. As above,
we can suppose $v_{1}=a_{1}e_{1}$, $a_{1}>0$. Write $v_{j}=a_{j}e_{1}+w_{j}$,
with $\langle w_{j},e_{1}\rangle=0$, for all $j=2,\dots,k$. By hypothesis,
$a_{j}\leq 0$ for all $2\leq j\leq k$. As a result, for all $2\leq i<j\leq k$,

$0\geq\langle v_{i}|v_{j}\rangle=a_{i}a_{j}+\langle w_{i}|w_{j}\rangle.$ (37)

Hence, for all $2\leq i<j\leq k$, $\langle w_{i}|w_{j}\rangle\leq 0$. Note
that at most one of the $w_{j}$ can vanish. Indeed, if two of them vanish, say
$w_{2}=0=w_{3}$, then $a_{2}\not=0\not=a_{3}$ and hence $\langle
v_{2}|v_{3}\rangle=a_{2}a_{3}>0$, which contradicts the hypothesis. We
conclude that among the $(k-1)$ vectors $w_{j}\in\mathbb{C}^{n}$, there are at
least $(k-2)$ that don’t vanish and since their mutual inner products are all
non-positive, the induction hypothesis allows to conclude that $k-2\leq 2n$ so
that $k\leq 2(n+1)$.
(ii) The argument is similar and proceeds again by induction. This time, by
hypothesis, all $a_{j}<0$, for $j=2,\dots,k$: none of the $a_{j}$ can vanish.
Now suppose one of the $w_{j}$ vanishes: $w_{2}=0$. Then, for $j=3,\dots,k$
since $\langle v_{1}|v_{j}\rangle=a_{1}a_{j}<0,\langle
v_{2}|v_{j}\rangle=a_{2}a_{j}<0$, which is impossible since $a_{1}>0,a_{2}<0$.
Consequently, none of the $w_{j},j=2,\dots k$ vanishes. We have found
therefore $(k-1)$ non vanishing vectors in $\mathbb{C}^{n}$ with a negative
inner product. So $k-1\leq n+1$ or $k\leq(n+1)+1$. ∎

Theorem 12 provides an improvement on existing results that we now further
analyze. First, as already mentioned, the result was proven in [19] under the
much stricter hypothesis that $Z=0$, _i.e._
$m_{{\mathcal{A}},{\mathcal{B}}}>0$. Recall from Proposition 7 that this
constitutes a weak incompatibility condition on ${\mathcal{A}}$ and
${\mathcal{B}}$.

In [16], on the other hand, the following closely related result was obtained,
which imposes no condition on the zeros of $U$:

######  Theorem 15.

Let ${\mathcal{A}}$ and ${\mathcal{B}}$ be orthonormal bases in a
$d$-dimensional Hilbert space ${\mathcal{H}}$. Suppose
$M_{{\mathcal{A}},{\mathcal{B}}}<1$. Then, if $|\psi\rangle\in{\mathcal{H}}$
satisfies

$n_{{\mathcal{A}},{\mathcal{B}}}(\psi)=n_{{\mathcal{A}}}(\psi)+n_{{\mathcal{B}}}(\psi)>\lfloor
3d/2\rfloor,$ (38)

then $|\psi\rangle$ is KD nonclassical.

To compare Theorem 12 and Theorem 15, we first remark that, in the latter
result, the hypothesis on $U$ is essentially empty, so that it has a very
general applicability. Indeed, if $M_{{\mathcal{A}},{\mathcal{B}}}=1$, some of
the $|a_{i}\rangle$ are equal, up to a phase, to some of the $|b_{j}\rangle$,
and then one can split the Hilbert space into a direct sum and study the
nonclassicality on the subspace orthogonal to the common basis vectors, as
already pointed out above. On the other hand, the conclusion of KD
nonclassicality in Theorem 15 is obtained only on states $|\psi\rangle$ for
which $n_{{\mathcal{A}},{\mathcal{B}}}(\psi)>\lfloor 3d/2\rfloor$; this is a
more restrictive family of states than those satisfying the bound
$n_{{\mathcal{A}},{\mathcal{B}}}(\psi)>d+1$ required in Theorem 12 as soon as
$d\geq 4$ and it becomes increasingly restrictive as $d$ grows. In fact is
easy to show that, when $d=3$, Theorem 12 and Theorem 15 are equivalent. To
see this, one needs to remark that in that case, the condition on the zeros of
the matrix $U$ in Theorem 12 does not actually constitute a restriction. We
refer to Appendix G for the details of the argument.

In the high-dimensional examples above, where $d=6k$, $k\in\mathbb{N}_{*}$,
and $U=U_{6}\otimes V$, Theorem 12 guarantees the KD nonclassicality of all
states $|\psi\rangle$ for which $n_{{\mathcal{A}},{\mathcal{B}}}(\psi)>6k+1$,
whereas Theorem 15 only concludes this when
$n_{{\mathcal{A}},{\mathcal{B}}}(\psi)>9k$.

##  8 Conclusion

We have analyzed the Kirkwood-Dirac quasi-probability distributions associated
with two observables $A$ and $B$, with eigenbases ${\mathcal{A}}$ and
${\mathcal{B}}$ on a finite dimensional Hilbert space of states
${\mathcal{H}}$. We have characterized the Kirkwood-Dirac (non)classical
states in terms of their support uncertainty and shown the special role played
by the complete incompatibility of the observables in this analysis. In
particular, when the observables are both completely incompatible and (close
to) mutually unbiased, we have shown that the only Kirkwood-Dirac classical
states are the basis vectors of ${\mathcal{A}}$ and ${\mathcal{B}}$.

A number of questions remain open. Some, relating to the precise link between
mutual unbiasedness and complete incompatibility, have been suggested in
Section 6. We have also not attempted to establish a hierarchy among the
Kirkwood-Dirac nonclassical states, a question of obvious interest. Finally,
the extension of our results to mixed states remains a subject for further
research.

Acknowledgements This work was supported in part by the Agence Nationale de la
Recherche under grant ANR-11-LABX-0007-01 (Labex CEMPI) and by the Nord-Pas de
Calais Regional Council and the European Regional Development Fund through the
Contrat de Projets État-Région (CPER).

APPENDICES

##  Appendix A Proof of Proposition 3

To see this, we write
${\mathcal{H}}={\mathcal{H}}_{-}\oplus{\mathrm{Ker}}C\oplus{\mathcal{H}}_{+},$
where ${\mathcal{H}}_{-},{\mathcal{H}}_{+}$ are the negative and positive
spectral subspaces for $C$ and $C_{-}<0,C_{+}>0$ are the restrictions of $C$
to ${\mathcal{H}}_{-},{\mathcal{H}}_{+}$. So

$C=C_{-}\oplus 0\oplus C_{+}.$

Note that, in finite dimension, one always has $\operatorname{Tr}C=0$, as can
be seen by using the cyclicity of the trace. So, in finite dimension, either
$C=0$, or both $C_{-}\not=0$ and $C_{+}\not=0$.

In infinite dimension, this argument does not work, since $A$ and/or $B$ may
not be trace class. In fact, it turns out that in infinite dimension it is
possible that $C_{-}=0,C_{+}\not=0$ and Ker$C=\\{0\\}$. Examples can be found
in [42, 43, 44] where $C$ is trace class and positive. It always has zero in
its spectrum then, but not necessarily as an eigenvalue. In some examples $C$
is finite rank and nonnegative in which case Ker$C\not=\\{0\\}$.

We now turn to the general proof of (18). If ${\mathrm{Ker}}C\not=\\{0\\}$ it
trivially holds. We therefore suppose ${\mathrm{Ker}}C=\\{0\\}$. Let us first
consider the case where the commutator $C$ has both positive and negative
spectrum, so that $C_{-}\not=0\not=C_{+}$. The proof of (18) then goes as
follows. Let $0\not=\psi_{\pm}\in{\mathcal{H}}_{\pm}$ and let
$|\psi\rangle=\alpha\psi_{-}+\beta\psi_{+}$, with
$\alpha,\beta\in\mathbb{R}_{*}^{+}$. Then

$\parallel\psi\parallel^{2}=\alpha^{2}\parallel\psi_{-}\parallel^{2}+\beta^{2}\parallel\psi_{+}\parallel^{2},\quad\textrm{and}\quad\langle\psi|C|\psi\rangle=-\alpha^{2}\langle\psi_{-}||C_{-}||\psi_{-}\rangle+\beta^{2}\langle\psi_{+}|C_{+}\psi_{+}\rangle.$

Imposing $\langle\psi|C|\psi\rangle=0,\parallel\psi\parallel=1$, one finds a
unique nontrivial solution for $(\alpha^{2},\beta^{2})$. This proves (18).

Suppose now that ${\mathrm{Ker}}C=\\{0\\},C_{-}=0,C_{+}\not=0$. We prove (18)
by contradiction. Suppose there exists $\delta>0$ so that
$\langle\psi|C|\psi\rangle=\langle\psi|C_{+}|\psi\rangle\geq\delta$ for all
$|\psi\rangle\in{\mathcal{H}}$. Then, for all $t\in\mathbb{R}^{+}$,

$\langle\psi,\exp(iAt)B\exp(-iAt)\psi\rangle=\langle\psi,B\psi\rangle+i\int_{0}^{t}{\mathrm{d}}s\langle\psi_{t},iC\psi_{t}\rangle\leq\langle\psi,B\psi\rangle-\delta
t.$

But this is impossible if $B$ is bounded. ∎

It follows from this that, in infinite dimension, if $A$ and $B$ are bounded
and have a positive commutator $C\geq 0$, then $0\in\sigma(C)$ and (18) holds.
Indeed, if not $\inf\sigma(C)=\delta>0$. And that leads to the above
contradiction.

When both $A$ and $B$ are unbounded, the commutator can of course be positive.
The standard case is when they are canonically conjugate so that
$C=i\mathbb{I}$ and (16) reduces to the Heisenberg uncertainty principle,
which provides a uniform lower bound for all $|\psi\rangle$.

##  Appendix B Proof of the support uncertainty principle (21)

Let $|\psi\rangle$ be an arbitrary state in ${\mathcal{H}}$. Then

$\displaystyle|\langle a_{i}|\psi\rangle|$ $\displaystyle=$
$\displaystyle\left|\sum_{j}\langle a_{i}|b_{j}\rangle\langle
b_{j}|\psi\rangle\right|\leq M_{{\mathcal{A}},{\mathcal{B}}}\sum_{j}|\langle
b_{j}|\psi\rangle|$ $\displaystyle\leq$ $\displaystyle
M_{{\mathcal{A}},{\mathcal{B}}}\sqrt{n_{\mathcal{B}}(\psi)}\left(\sum_{j}|\langle
b_{j}|\psi\rangle|^{2}\right)^{1/2}$ $\displaystyle=$ $\displaystyle
M_{{\mathcal{A}},{\mathcal{B}}}\sqrt{n_{\mathcal{B}}(\psi)}\left(\sum_{k}|\langle
a_{k}|\psi\rangle|^{2}\right)^{1/2}$ $\displaystyle\leq$ $\displaystyle
M_{{\mathcal{A}},{\mathcal{B}}}\sqrt{n_{\mathcal{B}}(\psi)}\sqrt{n_{\mathcal{A}}(\psi)}\max_{k}|\langle
a_{k}|\psi\rangle|$

Taking the maximum over $i$, the result follows.

$T$ $S$ | $\\{1,0\\}$ | $\\{1,-1\\}$ | $\\{0,-1\\}$
---|---|---|---
$\\{1,0\\}$ |  $\psi_{-1,-1}$ ✓ | $\psi_{-1,0}$ |  $\psi_{-1,1}$✓
$\\{1,-1\\}$ | $\psi_{0,-1}$ |  $\psi_{0,0}$✓ | $\psi_{0,1}$
$\\{0,-1\\}$ |  $\psi_{1,-1}$ ✓ | $\psi_{1,0}$ |  $\psi_{1,1}$✓

Figure 2: Uncertainty and KD nonclassicality for spin $1$ system. (a) Left
panel. The curve $n_{z}n_{x}=2$ (dashed (black) curve) and KD-nonclassicality
edge $n_{z}+n_{x}=4$ (dot-dashed (black) line). The (blue) diamonds indicate
the $(n_{z}(\psi),n_{x}(\psi))$ values corresponding to KD-nonclassical states
$\psi$. The (red) squares correspond to KD-classical states. (b) In the top
row and in the first column of the table, one has the possible sets
$S,T\subset\\{1,0,-1\\}$ with exactly two elements. And in each cell of the
table, the state generating the intersection
$\Pi_{z}(S){\mathcal{H}}\cap\Pi_{x}(T){\mathcal{H}}$, which is one-
dimensional. The state $\psi_{\epsilon,\epsilon^{\prime}}$ is perpendicular to
both $|x,\epsilon\rangle$ and $|z,\epsilon^{\prime}\rangle$. The check marks
(✓) indicate that the corrsponding state has
$n_{x,z}({\psi_{\epsilon,\epsilon^{\prime}}})=4$, which corresponds to the KD-
nonclassicality edge. Explicit computation of their KD distribution shows they
are KD classical. The four other states are equal to either $|x,0\rangle$ or
$|z,0\rangle$ and as such are KD classical as well. For them
$n_{z,x}({\psi_{\epsilon,\epsilon^{\prime}}})=3$.

##  Appendix C Spin $1$

We analyze here the KD nonclassicality of all spin $1$ states. The results are
represented graphically in Fig. 2. Since $n_{z}(\psi)$ and $n_{x}(\psi)$ are
integers, the support uncertainty relation rules out only the case
$n_{z}(\psi)=1=n_{x}(\psi)$, but that is obvious from the matrix $U$ at any
rate: the two observables don’t have common eigenvectors. There are exactly
two states on the curve $n_{z}n_{x}=2$: $|z,0\rangle,|x,0\rangle$. They are KD
classical, since they are basis vectors. Also, for them $n_{z,x}(\psi)=3$. To
study the KD nonclassicality of the other states, remark that, since $Z=1$, it
follows from Theorem 12 that all states with
$n_{z,x}(\psi)=n_{z}(\psi)+n_{x}(\psi)>4$ are KD nonclassical. This leaves
only the states with $n_{z,x}(\psi)=4$ to analyze. The four eigenvectors
$|z,\pm\rangle,|x,\pm\rangle$ are in this case. They are KD classical, since
they are basis vectors. The list of states $|\psi\rangle$ so that
$n_{z}(\psi)=2=n_{x}(\psi)$ is given in Fig. 2. Explicit computations not
reproduced here show they are all classical.

In conclusion, in a spin $1$ system all states $|\psi\rangle$ above the
nonclassicality edge $n_{z,x}(\psi)=4$ are KD nonclassical. All states on or
below the nonclassicality edge are KD classical. Note that the overwhelming
majority of states has $n_{\psi}=6$ and is therefore KD nonclassical.

##  Appendix D Commuting and noncommuting projectors

We collect here some elementary results on commuting and noncommuting
projectors needed in the main part of the article. In what follows, $\Pi$ and
$\Pi^{\prime}$ are two orthogonal projectors on a Hilbert space
${\mathcal{H}}$. They project onto $\Pi{\mathcal{H}}$ and
$\Pi^{\prime}{\mathcal{H}}$ respectively. The following lemma concerns
commuting projectors.

######  Lemma 16.

Let $\Pi$ and $\Pi^{\prime}$ be two orthogonal projectors on a Hilbert space
${\mathcal{H}}$. Then

  * (i)

$[\Pi,\Pi^{\prime}]=0$ if and only if $\Pi\Pi^{\prime}$ is the projector onto
$\Pi{\mathcal{H}}\cap\Pi^{\prime}{\mathcal{H}}$.

  * (ii)

$[\Pi,\Pi^{\prime}]=0$ if and only if $(\mathbb{I}-\Pi)\Pi^{\prime}\Pi=0$ if
and only if $\Pi^{\prime}$ leaves $\Pi{\mathcal{H}}$ invariant.

  * (iii)

$\Pi{\mathcal{H}}\subset\Pi^{\prime}{\mathcal{H}}$ if and only if
$\Pi\Pi^{\prime}=\Pi=\Pi^{\prime}\Pi$.

  * (iv)

$\Pi\Pi^{\prime}=0$ if and only if $\Pi{\mathcal{H}}$ is orthogonal to
$\Pi^{\prime}{\mathcal{H}}$.

  * (v)

If $\textrm{dim}\,\Pi{\mathcal{H}}=1$, then $[\Pi,\Pi^{\prime}]=0$ if and only
if either $\Pi{\mathcal{H}}\subset\Pi^{\prime}{\mathcal{H}}$ or
$\Pi{\mathcal{H}}$ is orthogonal to $\Pi^{\prime}{\mathcal{H}}$.

_Proof._ (i) If the commutator vanishes, it is clear that $\Pi\Pi^{\prime}$ is
a projector. In that case, one easily checks that
$\Pi\Pi^{\prime}{\mathcal{H}}=\Pi{\mathcal{H}}\cap\Pi^{\prime}{\mathcal{H}}$.
Conversely, if $\Pi\Pi^{\prime}$ is a projector, then it is self-adjoint and
therefore $[\Pi,\Pi^{\prime}]=0$.
(ii) This is obvious.
(iii) Suppose $\Pi{\mathcal{H}}\subset\Pi^{\prime}{\mathcal{H}}$. Let
$|\psi\rangle\in{\mathcal{H}}$ and write
$\psi=\Pi^{\prime}\psi+(\mathbb{I}-\Pi^{\prime})\psi$. Then
$\Pi\psi=\Pi\Pi^{\prime}\psi+\Pi(\mathbb{I}-\Pi^{\prime})\psi$. The second
term vanishes since $(\mathbb{I}-\Pi^{\prime})\psi$ is orthogonal to
$\Pi^{\prime}{\mathcal{H}}$ and therefore to $\Pi{\mathcal{H}}$. So
$\Pi=\Pi\Pi^{\prime}$. Taking adjoints, we also have $\Pi=\Pi^{\prime}\Pi$.
The converse is obvious.
(iv) This is obvious.
(v) We know from (i) that the commutator vanishes if and only if
$\Pi\Pi^{\prime}$ projects onto
$\Pi{\mathcal{H}}\cap\Pi^{\prime}{\mathcal{H}}$. Also, one either has
$\Pi{\mathcal{H}}\cap\Pi^{\prime}{\mathcal{H}}=\Pi{\mathcal{H}}$ or
$\Pi{\mathcal{H}}\cap\Pi^{\prime}{\mathcal{H}}=\\{0\\}$. In the first case
$\Pi{\mathcal{H}}\subset\Pi^{\prime}{\mathcal{H}}$. In the second case,
according to (i), $\Pi\Pi^{\prime}$ is the projector onto $\\{0\\}$, which
means $\Pi\Pi^{\prime}=0$. But this equivalent to $\Pi{\mathcal{H}}$ being
orthogonal to $\Pi^{\prime}{\mathcal{H}}$ by (iv). ∎

We now turn to noncommuting projectors.

######  Lemma 17.

Let $\Pi,\Pi^{\prime}$ be two orthogonal projectors Then
$[\Pi,\Pi^{\prime}]\not=0$ if and only if there exists
$|\psi\rangle\in{\mathcal{H}}$ so that
$\Pi^{\prime}\Pi\psi\not\in\Pi{\mathcal{H}}$, if and only if
$(\mathbb{I}-\Pi)\Pi^{\prime}\Pi\not=0$.