Available via license: CC BY 4.0
Content may be subject to copyright.
arXiv:2209.09817v1 [quant-ph] 20 Sep 2022
Uncertainty relations
for the support of quantum states
Vincenzo Fiorentino and Stefan Weigert
September 2022
Department of Mathematics, University of York
York YO10 5DD, United Kingdom
vincenzo.fiorentino@york.ac.uk stefan.weigert@york.ac.uk
Abstract
Given a narrow signal over the real line, there is a limit to the localisation of its Fourier
transform. In spaces of prime dimensions, Tao derived a sharp state-independent uncertainty
relation which holds for the support sizes of a pure qudit state in two bases related by a dis-
crete Fourier transform. We generalise Tao’s uncertainty relation to complete sets of mutually
unbiased bases in spaces of prime dimensions. The bound we obtain appears to be sharp for
dimension three only. Analytic and numerical results for prime dimensions up to nineteen sug-
gest that the bound cannot be saturated in general. For prime dimensions two to seven we
construct sharp bounds on the support sizes in (d+ 1) mutually unbiased bases and identify
some of the states achieving them.
Contents
1 Introduction 2
2 Preliminaries 3
2.1 Support inequalities for a Fourier pair of bases . . . . . . . . . . . . . . . . . . . . . 3
2.2 Mutually unbiased bases in prime dimensions . . . . . . . . . . . . . . . . . . . . . 4
3 Support inequalities... 6
3.1 ...for arbitrary pairs of MU bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
3.2 ...for complete sets of (d+ 1) MU bases . . . . . . . . . . . . . . . . . . . . . . . . . 8
4 Saturating support inequalities for MU bases 9
4.1 Constraints on saturating states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
4.2 Dimension d= 3 ...................................... 11
4.3 Dimensions d= 5 and d= 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
4.4 Numerical results for 5≤d≤19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
4.5 Dimensions d > 19 ..................................... 12
5 Sharp lower bounds 13
5.1 Dimension d= 3 ...................................... 13
5.2 Dimension d= 5 ...................................... 13
5.3 Dimension d= 7 ...................................... 14
6 Summary and Conclusions 15
7 Appendix 18
1
1 Introduction
No quantum particle can reside in a state with both its position and momentum distributions
being localised arbitrarily well. For these incompatible observables, Heisenberg’s uncertainty
relation [1,2] establishes a finite lower bound for the product of their variances. This result relies
on a fundamental property of Fourier theory: a real (or complex) function with finite support
on the real line has a Fourier transform which must be non-zero almost everywhere [3]. It is,
however, difficult to quantify the support of functions on unbounded intervals. Using variances
instead of the supports of probability distributions circumvents this difficulty.
The situation is different for quantum systems with finite-dimensional Hilbert spaces since the
support (size) of a pure state—defined as the number of non-zero components in a given orthonor-
mal basis—is always finite. A computational basis state in Cd, say, has support equal to one, and
the support of its (discrete) Fourier transform equals dsince all basis states contribute. Thus, the
product of the support sizes equals dwhich turns out to be its smallest possible value [4].
The underlying product inequality has been generalised in a number of directions [5,6]. Tao
derived an additive inequality [7] which is valid in spaces Cdof prime dimensions d=p: the sum
of the supports of a state and its discrete Fourier transform is bounded from below by the value
(p+ 1). This bound is sharp since a computational basis state and its Fourier transform saturate
it.
Support inequalities and their generalisations have found applications in signal processing
[8,9], for example, and they can be used to identify non-classical quantum states [10] whose
Kirkwood-Dirac quasiprobability distribution [11,12] is not a probability distribution. Such states
provide an advantage in quantum metrology [13] and play a role in weak measurements [14,15,
16] and contextuality [17].
Using the support size of a quantum state as a measure of uncertainty has an unexpected—
and previously unnoticed—operational advantage. Quantum supports take a finite set of integer
values only, in stark contrast to other measures. Variances of observables in a given state or its von
Neumann entropy take real numbers as values which demands many measurements to determine
experimentally. However, a finite number of measurements may already suffice to determine the
exact support size of a quantum state. This situation occurs whenever the state at hand has (i)
“full” support in the basis considered and (ii)each outcome has been registered at least once.
Conformity with a given support inequality may be verified by a finite number of measurements
as small as the bound itself. This property depends, of course, on the assumption that outcomes
with probability zero never occur; limited detection efficiency does not invalidate the argument,
however.
Variance-based uncertainty relations also exist for more than two observables associated with
multiple orthonormal bases [18,19,20,21]: position and momentum may be supplemented by
a third continuous variable which is canonical to each of them. The eigenbases of these three
observables are mutually unbiased and related by fractional Fourier transforms. The product
of their variances satisfies a triple uncertainty relation [18]. Importantly, the lower bound of this
inequality does not follow from the pair uncertainty relations but must be determined indepen-
dently. No quantum state exists which satisfies all three pair uncertainty relations simultaneously.
In a similar vein, entropic uncertainty relations capture the incompatibility of up to (d+1) observ-
ables in finite-dimensional systems, linked to a complete set of mutually unbiased bases known
to exist in prime-power dimension [22,23,24].
The main goal of this paper is to extend Tao’s additive support uncertainty relation to the case
of more than two bases, inspired by the triple uncertainty relation for continuous variables. The
focus will be on prime-dimensional spaces where (p+ 1) mutually unbiased bases exist, known
as complete sets. Being related by discrete analogues of fractional Fourier transforms, the support
sizes of a state in any pair of mutually unbiased bases are expected to satisfy Tao’s bound but it is
unlikely that they will saturate all pair bounds simultaneously.
This paper is structured as follows. Sec. 2sets up notation by briefly describing known prod-
2
uct and sum inequalities for the support of a vector, and the properties of complete sets of MU
bases are summarised. In Sec. 3, Tao’s additive uncertainty relation for the support of quantum
states is shown to hold for any pair of mutually unbiased bases in a complete set, and the gen-
eralised additive support inequality involving all (p+ 1) mutually unbiased bases is established
as a direct consequence. According to Sec. 4the bounds provided by the generalised support
inequality cannot be saturated for prime dimensions 2≤d≤19, except d= 3. Higher achievable
bounds are derived in Sec. 5, for prime numbers up to d= 7. In the last section, we summarise
and discuss the results obtained. The proofs of some lemmata are relegated to an appendix.
2 Preliminaries
2.1 Support inequalities for a Fourier pair of bases
The support (size) of a Hilbert-space vector ψ∈ Hdis given by the number of its non-zero expan-
sion coefficients ψv=hv|ψiin an orthonormal basis B={|vi, v = 0,1, . . . , d −1},
|supp(ψ, B)|= #(ψv6= 0, v = 0 . . . d −1) ∈ {0. . . d}.(1)
The only vector with vanishing support is the zero vector. Due to normalisation, the support of a
quantum state must be at least one, and the maximum is achieved whenever the state ψis a linear
combination of all dbasis states. The support size of a state clearly depends on the chosen basis.
Formally, the support size can be obtained as a limit of the Rényi entropy [25] of the probability
distribution |ψv|2, v = 0 . . . d −1.
Thinking of the support size as the (improper) L0-“norm” of ψ, we will use the notation
|supp(ψ, B)|=kψkB.(2)
The set of expansion coefficients {ψv, v = 0 . . . d −1}has three obvious support-conserving sym-
metries described by (anti-) unitary transformations. The support size is invariant (i)under
rephasing each expansion coefficient separately,
||ψ||B=||Rψ||B, R =diag(eiτ0, eiτ1,...,eiτd−1),(3)
with real numbers τv, v = 0, . . . , d −1;(ii)under permuting the components of any state among
themselves
||ψ||B=||P ψ||B, P ∈Sd,(4)
where Sdis the permutation group acting on sets of delements; and (iii)under the complex
conjugation of some (or all) of its components,
||ψ||B=||Kψ||B, K =Y
some v∈{0... d−1}
Kv,(5)
where each antiunitary operator Kv, v = 0, . . . , d −1, maps one expansion coefficient of the state
ψin the basis Bto its complex conjugate, Kvψv=ψ∗
v, and does not change the others. In the basis
B, the permutations Pare represented by a matrix of order dcontaining exactly one unit entry
in each row and column; hence the unitary invariances are conveniently combined into monomial
matrices M≡RP . The anti-unitary symmetries will play no role.
Given two distinct orthonormal bases Band B′of Hd, one may ask to which extent a state
can be “localised” in both of them. Clearly, the product of its support sizes in Band B′may take
values between one and d2. If the bases are related by B′=FB, where Fis the discrete Fourier
transform with matrix elements (in the B-basis)
Fvv′=1
√de−2πivv′/d v , v′∈ {0. . . d −1},(6)
3
then the product of the support sizes of a state ψand its Fourier transform e
ψ=F†ψis bounded
from below [4],
kψkB||ψ||B′≥d , (7)
where we use the fact that the support size of the Fourier transformed state e
ψin the basis Bis
equal to the support size of the state ψin the basis B′, i.e.
||e
ψ||B=||F†ψ||B=||ψ||B′.(8)
The inequality (7) represents a finite-dimensional equivalent of Heisenberg’s uncertainty relation
for position and momentum observables of a quantum particle: quantum states localised in posi-
tion, say, necessarily come with a broad variance in momentum, the Fourier-transformed position
observable.
For spaces Hdwith prime dimensions d, an additive inequality for the supports of a quantum
state in a pair of Fourier-related bases is known [7],
kψkB+||ψ||B′≥d+ 1 ,(9)
which is stronger than the multiplicative relation (7), as follows from the inequality d+ 1 −x≥
d/x, for x∈[1, d]. In the terminology of [10], any two bases Band B′are said to be completely
incompatible if and only if the support sizes of the expansion coefficients of any (non-zero) vector
ψ∈ Hdsatisfy this bound.
The inequality (9) is a special case of a theorem valid for finite additive Abelian groups Gwith
|G|elements and trivial subgroups only which necessitates the restriction to prime dimensions
[7]. Consider a complex-valued function f:G→Cand its transform e
f:G→C, defined by
e
f(v′) = 1
p|G|X
v∈G
f(v)e(v, v′),(10)
where e(v, v′)is a “bi-character” of Gsatisfying e(v1+v2, v′) = e(v1, v ′)e(v2, v′)and an analogous
relation for its second argument. In the particular case of e(v, v′) = e−2πivv′/d, one obtains an
inequality for the supports of fand e
f≡F†f.
Theorem 1 (Tao’s theorem ).If f:G→Cis a non-zero function and the cardinality |G|of the group G
is prime, then
|supp(f)|+|supp(e
f)| ≥ |G|+ 1 .(11)
Upon identifying f(v)with hv|ψiand e
f(v′)with hv′|ψi, respectively, we obtain the inequality
(9) relative to the bases Band B′introduced via Fin Eq. (6).
The main ingredient of Tao’s proof is a fundamental property of the Fourier matrix in prime
dimensions [26,27,7] which dates back to the 1920s: all its square submatrices are invertible.
Theorem 2 (Chebotarëv’s theorem).If dis prime, then all minors of the Fourier matrix Fin Eq. (6) are
non-zero.
The inequalities (7) and (9) involve a pair of mutually unbiased bases of Hd, namely the com-
putational basis Band its Fourier transform. We will now introduce larger sets of mutually un-
biased bases to formulate more general support inequalities. Not surprisingly, Chebotarëv’s the-
orem must be generalised to other matrices which emerge when establishing bounds on support
sizes of quantum states in multiple bases (cf. Sec. 3.1).
2.2 Mutually unbiased bases in prime dimensions
Two orthonormal bases of the space Hd=Cdare said to be mutually unbiased (MU) if the inner
products between any two states (not of the same basis) have modulus 1/√d. Then, to know the
outcome of a projective measurement performed in one basis implies complete uncertainty about
the outcome of a subsequent projective measurement performed in the other.
4
When dis a power of a prime number p, i.e. d=pn, sets of (d+ 1) mutually unbiased
bases have been constructed [28,29,30,31]. Such complete sets are both maximal—in the sense
that no additional MU basis can be added to it—and tomographically complete: the probability
distributions of outcomes in the (d+ 1) bases uniquely encode an unknown quantum state. It is
an open problem whether complete sets of MU bases exist in composite dimensions, d6=pn.
For d= 2, the eigenstates of the Pauli operators Z2,X2and X2Z2=−iY2form a complete
set which has a simple structure. The two Hadamard matrices encoding the bases MU to the
computational basis B0are
H1=F=1
√21 1
1−1, H2=DF where D=1 0
0i.(12)
If dis an odd prime, the eigenstates of the (d+ 1) generalised Pauli operators Zd,Xd,XdZd,
XdZ2
d, ..., XdZd−1
d, represent a maximal set of MU bases. The k-th state of the j-th basis is given
by
|φj
ki=1
√d
d−1
X
x=0
ω−kxω(j−1)x2|φ0
xi, j ∈ {1... d}, k ∈ {0... d −1},(13)
where the states |φ0
xi ≡ |xi, x = 0 . . . d −1form the computational basis B0and ω≡e2iπ/d is a
d-th root of the number 1[31]. For each value of j, the equimodular expansion coefficients
[Hj]xk =hx|φj
ki=1
√dω−kxω(j−1)x2, j ∈ {1... d}, x, k ∈ {0... d −1},(14)
define a complex-valued Hadamard matrix. These are unitary matrices since their columns are given
by the components (in the computational basis B0) of dorthogonal vectors. The relations (13)
suggest to think of the matrices Hjas rotations about finite angles, hence as discrete analogs of
fractional Fourier transforms. Other inequivalent complete sets of MU bases may exist in some
dimensions but throughout this paper we will always assume them to be given by Eq. (13),
forming the complete standard set of MU bases.
The Hadamard matrix H1in (14) coincides with the discrete Fourier matrix Fgiven in Eq. (6).
The remaining Hadamard matrices Hjmap the computational basis B0of Hdto other orthonor-
mal bases denoted by Bj. From an active view of these transformations, the state ψis mapped to
the state H†
jψ. The relation between the supports of the state ψin B0and the j-th MU basis Bj
reads,
||H†
jψ||0=||ψ||j,(15)
abbreviating the notation introduced in (8), i.e. ||ψ||Bj≡ ||ψ||j,j∈ {0. . . d}.
The columns of the dHadamard matrices Hjin (14) are related in a simple way to each other,
namely by
|φj
ki=Dj−1Bk|φ1
0i, j ∈ {1... d}, k ∈ {0... d −1},(16)
with two diagonal (d×d)matrices Band D; in other words, all states of a complete set of MU
bases can be generated easily from any given state such as |φ1
0i—except for the states of the com-
putational basis B0. Within each Hadamard matrix, the matrix Bcyclically shifts a given column
to the right,
B|φj
ki=(|φj
k+1i, k = 0, . . . , d −2,
|φj
0i, k =d−1 ; (17)
its entries are given by the components of the first column of the Fourier matrix F=H1,
B=diag 1, ω−1,...,ω−(d−1),(18)
except for the factor √d.
5
The matrix Dis given by the components of the first column of the second Hadamard matrix
H2, i.e.
D=diag 1, ω1,...,ω(d−1)2,(19)
cyclically mapping a state of the j-th MU basis to the corresponding one of the MU basis with
label (j+ 1),
D|φj
ki=(|φj+1
ki, j = 1, . . . , d −1,
|φ1
ki, j =d . (20)
In terms of Hadamard matrices, this property reads
DHj=(Hj+1 , j = 1, . . . , d −1,
H1, j =d . (21)
Writing Hj=Dj−1H1≡Dj−1F, Chebotarëv’s theorem is seen to imply that the minors of
all Hadamard matrices Hj,j= 1 . . . d, are non-zero: the ranks of the minors of Fdo not change
upon multiplying their rows with non-zero scalars. In view of Eq. (12), this generalisation is also
valid for the case of d= 2.
3 Support inequalities...
How well can one localise quantum states simultaneously in (d+ 1) MU bases? To answer this
question, we need to minimise the support sizes of a state relative to the complete set. In a first
step, we now show that the support sizes of a quantum state relative to any pair of MU bases
also satisfy Tao’s bound (9). Second, by combining the resulting pair inequalities, we establish a
state-independent lower bound.
3.1 ...for arbitrary pairs of MU bases
Tao’s result establishes—for a space of prime dimensions dsupporting a cyclic abelian group—a
sharp inequality for the support sizes of a quantum state and its Fourier transform. Most pairs
of the MU bases introduced in Eq. (13) are, however, not related by a Fourier transform. Never-
theless, Tao’s bound also holds for the supports of the images of any quantum state generated by
two Hadamard matrices as we will show now.
Theorem 3. Given any pair of distinct MU bases associated with matrices Hjand Hk,j, k ∈ {0. . . d},
the support sizes of a quantum state |ψi ∈ Hdsatisfy the state-independent sharp bound,
kψkj+kψkk≥d+ 1 ,(22)
where dis any prime number.
It is important to realise that Theorem 3does not cover arbitrary pairs of MU bases in prime
dimensions but only those defined in Eq. (14). Nevertheless, all pairs of MU bases in dimensions
d= 2,3and 5are found to be completely incompatible since in these dimensions all Hadamard
matrices are equivalent to the Fourier matrix. Already for the next prime, d= 7, other types of
Hadamard matrices exist [32].
Proof. The case of dimension d= 2 is straightforward. If a state |ψi ∈ H2has support one in one
MU basis, it must have support two in both other bases, due to being MU to their members. Thus,
the sum of the supports of any state in two bases must be at least three.
For odd primes d,we will consider two cases separately: either (i)one of the bases in Eq. (22)
is the computational basis, so that j= 0, say, or (ii)neither of them.
6
(i)Defining the vector φ=D1−kψ, we obtain
kψk0+kψkk=kψk0+kH†
kψk0=kψk0+kF†D1−kψk0=kDk−1φk0+kF†φk0,(23)
recalling that Hk=Dk−1Fholds according to Eq. (21). Since Dis a diagonal unitary hence a
support-conserving unitary matrix (cf. (3)), we obtain
kψk0+kψkk=kφk0+kF†φk0≥d+ 1 ,(24)
where Tao’s theorem was used in the last step.
(ii)Now consider the case where the non-zero labels jand kdiffer from each other. Defining
the vector φ=H†
jψ, the sum of the support sizes can be written as
kψkj+kψkk=kH†
jψk0+kH†
kψk0=kφk0+kH†
kHjφk0.(25)
The product H†
kHjof two distinct Hadamard matrices is, in fact, always equal to another Hadamard
matrix H†
t,t6= 0, up to a monomial matrix M(j, k); this result is the content of Lemma 1stated
directly after the proof. As the matrix M(j, k)is support-conserving (cf. Eqs. (3) and (4)) for all
states of the space Hd, we find
kφk0+kM(j, k)H†
tφk0=kφk0+kH†
tφk0≥d+ 1 ,(26)
where (24) of Part (i)has been used in the final step.
The proof just relies on dissolving products of the Hadamard matrices Hjwhich encode a
complete set of MU bases. Clearly, products of the form H†
kHj,j6=k, are Hadamard matrices
since their matrix elements, being overlaps of MU vectors, have modulus 1/√d,
hH†
kHjiℓℓ′=hφk
ℓ|φj
ℓ′i.(27)
When d= 2, one finds explicitly that H†
1H2=MH†
2and H†
2H1=M′H†
2, with monomial matrices
Mand M′. In other words, the phases of the matrix elements (27) coincide with those of the ad-
joint of another transition matrix after permuting and rephasing its rows. This property actually
holds for any prime dimension.
Lemma 1. Let dbe an odd prime and j, k ∈ {1. . . d}with j6=k. Then
H†
kHj=M(j, k)H†
t(28)
for a monomial matrix M(j, k)if and only if t= 1 + χ∈ {1,...,d}where the integer χsatisfies
4 (j−k)χ= 1 mod d.
Proof. See Appendix.
Furthermore, Lemma 1allows us to generalise Chebotarëv’s theorem to the product matrices
H†
kHj, for distinct indices jand k.
Corollary 1. If dis prime, then all minors of the Hadamard matrices H†
kHj,j, k ∈ {0. . . d}and j6=k,
are non-zero.
Proof. Let dbe an odd prime. Any Ht,t∈ {1. . . d}, has only non-zero minors, as was mentioned
after Eq. (21), as do the adjoints H†
t. Therefore, the claim holds if one of the labels j, k, is zero.
For both j, k 6= 0, Lemma 1applies. Since rephasing and permuting the rows of a matrix do not
change the rank of any submatrix, we can conclude that the matrices H†
kHj,j, k 6= 0 and j6=k
also have non-zero minors only.
In dimension d= 2, the result follows from inspecting the products H†
1H2=MH†
2and
H†
2H1=M′H†
2.
7
According to Corollary 1the vectors formed by the columns (or rows) of all squaresubmatrices
of the Hadamard matrices H†
kHj,j6=k, are linearly independent. What is more, up to dvectors
taken from any two MU bases are linearly independent.
Corollary 2. Given a complete set of MU bases in the space Hdof prime dimension d, up to dvectors
taken from any two MU bases are linearly independent.
Proof. See Appendix.
Corollary 1also demonstrates that all pairs of MU bases taken from the complete standard set
in prime dimension are completely incompatible, in the sense of Sec. 2.1.This statement is stronger
than the results of [7,10] since the bases we consider are not necessarily related by the Fourier
matrix F.
3.2 ...for complete sets of (d+ 1) MU bases
Let us denote the sum of the numbers of non-zero expansion coefficients of a state ψ∈ Hdin a
complete standard set of MU bases by
S(d) = ||ψ||0+||ψ||1+···+||ψ||d.(29)
Then, the inequalities (22) imply that the overall support size S(d)cannot fall below a certain
threshold. This is a central result of our paper.
Theorem 4. For any prime d, the overall support S(d)of a quantum state |ψi ∈ Hdin a complete standard
set of MU bases bases satisfies the additive state-independent bound
S(d)≥(d+ 1)2
2≡T(d).(30)
Proof. Write down (d+1) copies of the support inequality (22) with indices (j, j +1),j= 0,... d−1,
and (d, 0), respectively. Adding them up, the right-hand-sides of (22) give (d+ 1)2, and since each
term ||ψ||j≡ kH†
jψk0,j= 0, . . . , d, occurs twice on the left, one obtains the inequality (30).
An alternative approach treats all pair supports equally: write down (22) for all d(d+1) distinct
pairs of indices (j, k)and consider the sum of the supports. After removing common factors, the
bound (30) on S(d)follows.
Support inequalities other than Eqs. (22) and (30) exist. They may involve any number be-
tween two and (d+ 1) MU bases. For example, picking the first three MU bases and combining
the associated pair inequalities from (22) leads to the additive “triple support inequality”,
S(d; 3) ≡ kψk0+kF†ψk0+kH†
2ψk0≥3
2(d+ 1) .(31)
Clearly, this inequality cannot be saturated for dimension d= 2 because the overall support S(2)
of a state is always an integer number. Taking only two possible values, the smallest achiev-
able value of the triple support size S(2; 3) ≡ S(2) equals Ts(2) = 5; here and in the following,
achievable—or sharp—bounds of S(d)are denoted by Ts(d).
The lower bound on the triple uncertainty relation for continuous variables [18], derived sim-
ilarly by combining pair uncertainty relations, can also not be reached. Theorem 4is not con-
structive, hence it is not obvious whether the case of d= 2 represents an exception or whether
the inequalities (30) are never sharp. In the next section we will first derive some general results
about multiple-support inequalities, followed by a closer look at dimensions 3≤d≤19.
8
4 Saturating support inequalities for MU bases
To saturate the bound of the inequality (30) means to identify states that minimise all support
pair relations simultaneously. We present a number of rigorous results for prime dimensions
d≤7. Numerical methods are then used to determine whether the generalised inequality can be
saturated for dimensions up to d= 19.
4.1 Constraints on saturating states
Our first general result is a necessary and sufficient condition that the support inequality involv-
ing a complete set of (d+ 1) MU bases Eq. (30) be saturated.
Theorem 5 (Equal support sizes).The additive support inequality for a complete set of MU bases (30)
is saturated by a state ψ∈ Hdif and only if it has the same support in all (d+ 1) MU bases, i.e.
||ψ||j=d+ 1
2, j ∈ {0... d},(32)
where dis an odd prime.
Proof. Substituting the values (32) into (30) directly produces the lower bound.
For the converse, we show that the supports must have the values given in (32) if equality
holds in Eq. (30). Noting that the support of any state ψ∈ Hdranges from 1to d, i.e.
kψk0=d+ 1
2±∆,∆∈0,1,..., 1
2(d−1),(33)
we will proceed by exhausting all its values in the computational basis B0. It turns out that the the
minimum in (30) cannot be reached if the support is either (i)smaller or (ii)larger than (d+ 1)/2,
leaving (iii)the values in (32) as the only option.
(i)If kψk0= (d+ 1) /2−∆,∆>0, then (22) implies that kψkj≥(d+ 1) /2 + ∆,j={1. . . d}.
Hence, the sum of the supports in all (d+ 1) MU bases equals
S(d) =
d
X
j=0 kψkj≥d+ 1
2−∆ + dd+ 1
2+ ∆
≥(d+ 1)2
2+ (d−1)∆ >(d+ 1)2
2.
(34)
Therefore, the inequality cannot be saturated by a state which has support smaller than (d+ 1) /2
in the basis B0.
(ii)Assume that kψk0= (d+ 1 ) /2 + ∆,∆>0. Clearly, the lower bound of the sum in Eq. (30)
can only be reached if the support of the state ψis smaller than (d+ 1)/2in at least one of the MU
bases, kψkj∗<(d+ 1) /2,j∗∈ {1. . . d}, say. Repeating the argument from (i)relative to the MU
basis Bj∗instead of B0implies that the inequality (30) cannot be saturated.
(iii)If kψk0= (d+ 1) /2then (22) implies that kψkj≥(d+ 1) /2,j={1. . . d}. However,
given these bounds, the minimum of S(d)in (30) can be achieved only if the support of the state
ψ∈ Hdtakes the value (d+ 1) /2in all other MU bases as well.
The second general result states that a specific d-th root of unity can appear at most twice in the
columns of the Hadamard matrices Hj,j= 2 . . . d, given in (14). The proof of another necessary–
but not sufficient–condition for saturating the generalised inequality (30) will rely on this limit of
the occurrences of roots. The property clearly does not apply to H0≡Inor to H1≡Fas is seen
by inspecting (6).
Lemma 2 (Frequency of roots).Let dbe prime and consider the states |φj
ki,j= 2, . . . , d, in Eq. (13)
forming the bases Bjwhich are MU to both the identity and the Fourier matrix. Any d-th root ωn,n∈
{0. . . d}, figures at most twice among the numbers √dhx|φj
ki,x∈ {0. . . d −1}.
9
Proof. We need to determine the number of solutions of the equation ω−kx+(j−1)x2=ωnwhich
becomes (j−1) x2−kx −nmod d= 0 upon taking the logarithm and rearranging. Since j6=
1, the equation is quadratic for each nand there can be at most two integer solutions for the
unknown x. The extension to the special case of d= 2 is trivial.
According to Theorem 5, a state saturating (30) must have (d−1)/2vanishing expansion
coefficients in each MU basis of a complete set, in any odd prime dimension. A third general
result is that there are constraints on the distributions of these zeroes when expanded in the MU
bases of a complete set.
To spell out these constraints, let us introduce the zero distributions Zjof a state ψ∈ Hdwhich
list the indices of the vanishing expansion coefficients in the (d+ 1) bases of the complete set,
Zj=κ∈ {0. . . d −1}:hφj
κ|ψi= 0, j = 0, . . . , d . (35)
Using the relation hφj
κ|ψi=hφ0
κ|H†
j|ψi, one can also think of Zjas the set of vanishing coefficients
of the state H†
j|ψiin the computational basis.
Two zero distributions of vectors in the same Hilbert space are said to be compatible,Z ∼ Z′,
if they are equal up to a cyclic shift. In other words, two distributions Z={κ1, κ2, ..., κδ}and
Z′={κ′
1, κ′
2, ..., κ′
δ}must have the same number δof elements and satisfy κ′
i=κi+µ(mod d), for
all i= 1, . . . , δ, with some fixed integer µ. Compatibility of zero distributions is an equivalence
relation between classes of delements.
The extension of Chebotarëv’s Theorem shown in Sec. 3.1 and Lemma 2imply a constraint on
zero distributions for all prime dimensions d > 3. This property will be used in Sec. 4.3 to prove
that the support inequality (30)cannot be saturated in dimensions d= 5 and d= 7.
Theorem 6. Let d > 3be prime and ψ∈ Hdbe a state with (d−1)/2expansion coefficients vanishing in
the computational basis and in two more MU bases, i.e.
kψ0k=kψkj1=kψkj2=d+ 1
2, j16=j2, j1, j26= 0 .(36)
Then the zero distributions associated with the vectors H†
j1|ψiand H†
j2|ψi, respectively, are incompatible.
Proof. Since the state ψhas d−≡(d−1)/2vanishing components in three bases with labels
j= 0, j1, j2, it satisfies 3d−conditions,
hφ0
κ0
1|ψi=hφ0
κ0
2|ψi=...= 0 ,Z0=nκ0
1, κ0
2,...,κ0
d−o,
hφj1
κ1
1|ψi=hφj1
κ1
2|ψi=...= 0 ,Zj1=nκ1
1, κ1
2,...,κ1
d−o,(37)
hφj2
κ2
1|ψi=hφj2
κ2
2|ψi=...= 0 ,Zj2=nκ2
1, κ2
2,...,κ2
d−o.
We proceed by contradiction. To assume that the zero distributions Zj1and Zj2are compatible
means that they are related by a cyclic shift by some integer µ∈ {0. . . d −1},
κ2
i=κ1
i+µmod dfor all i∈ {1. . . d−}.(38)
Then, according to Eq. (16), the corresponding states must be related by powers of the matrices
Dand B,
|φj2
κ2
ii=Dj2−1Bκ2
i|φ1
0i=Dj2−1Bκ2
iD−j1+1B−κ1
i|φj1
κ1
ii=Dj2−j1Bµ|φj1
κ1
ii,(39)
where we have used the fact that Dand Bcommute. Defining V†
µ=Dj2−j1Bµ, the third set of
conditions in (37) turns into
hφj2
κ2
i|ψi=hφj1
κ1
i|Vµψi= 0 for all i∈ {1...d−}.(40)
10
Since Vµis diagonal in the computational basis, we have
hφ0
κ0
i|ψi=hφ0
κ0
i|Vµψi= 0 for all i∈ {1. . . d−},(41)
which means that Z0and Zj1are zero distributions for the pair of vectors ψand Vµψ. In other
words, these two states are both orthogonal to the same set of 2d−= (d−1) vectors
φ0
κ0
1,...,φ0
κ0
d−
, φj1
κ1
1
,...,φj1
κ1
d−,(42)
stemming from the computational basis B0and the basis Bj1. According to Corollary 2, this is
a set of (d−1) linearly independent vectors so that only one unique ray in Hdcan exist that is
orthogonal to all of them. Therefore, the vectors ψand Vµψmust be collinear, i.e. Vµψ=λψ for
some non-zero scalar λ∈C.
Since Vµ=Dj1−j2B−µis diagonal in B0, its eigenvectors are computational basis states. By
assumption, the state ψhas d+≡(d+ 1) /2non-zero coefficients in this basis. Thus, the state
ψwill be an eigenvector of the unitary Vµonly if λis an eigenvalue with multiplicity of d+(at
least). However, this is impossible for prime dimensions d > 3: the non-zero matrix elements
on the diagonal of Vµcoincide with the components of the state |φj1−j2
−µiin the computational
basis but for j16=j2no more than two of the components may coincide according to Lemma 2(if
j1−j2= 1 (mod d)then all components differ). Thus, at most two of the eigenvalues of Vµcan
coincide. No contradiction arises for dimension d= 3 where ψhas exactly two non-vanishing
coefficients in the computational basis.
4.2 Dimension d= 3
To prove that the bound (30) can be achieved in the space H3, we exhibit the states which minimise
the support inequality.
Theorem 7. The state ψsaturates the generalised support inequality (30) in dimension d= 3 if and only
if it is one of the following nine (non-normalised) qutrit states,
1
−ωm
0
,
1
0
−ωm
,
0
1
−ωm
, m ∈ {0,1,2},(43)
with ω≡e2iπ/3a third root of unity.
Proof. Theorem 5implies that a state ψsaturates Eq. (30) w.r.t. a complete set of MU bases if
and only if it has support two in each of them, i.e. ||ψ||j≡ kH†
jψk0= 2,j= 0 ...3. First, we
assume that the third component of a candidate state vanishes in the computational basis, i.e.
ψ=a, b, 0T, with non-zero complex numbers aand b. Applying the matrices H†
j,j= 1,2,3,
to it, we find four vectors,
a
b
0
,
a+b
a+ωb
a+ω2b
,
a+ω2b
a+b
a+ωb
,
a+ωb
a+ω2b
a+b
.(44)
The components of the last three vectors agree, except for permutations. Hence, support size two
can occur in three different ways: one component of each vector vanishes if
b=−ωma , m ∈ {0,1,2},(45)
holds for some value of m. After removing an irrelevant phase, we obtain the first three vectors
given in Eq. (43). Second, repeating this argument for initial vectors of the form ψ=a, 0, bT
and ψ=0, a, bT, respectively, leads to the remaining six vectors in (43).
Having exhausted all three-component vectors in the computational basis with support two,
we have shown that the nine vectors in (43) are the only states saturating the support inequality
(30) for d= 3.
11
4.3 Dimensions d= 5 and d= 7
We will show that it is impossible to reach the lower bound of the support inequality (30) in
dimensions d= 5 and d= 7. The proof relies on a property of the zero distributions of the vectors
H†
jψ,j= 0 . . . d, which were introduced in Sec. 4.1.
Theorem 8. The additive support uncertainty relation (30) cannot be saturated in dimensions d= 5 and
d= 7.
Proof. Let Zd
nbe the set of the zero distributions with nzeroes among the computational-basis
coefficients of qudit states in the Hilbert space Hd. These distributions are determined by choos-
ing nout of dindices; hence, there are |Zd
n|=d
nsuch sets. Recalling that compatible sets
of zero distributions form equivalence classes, obtained from rigidly shifting a given one, only
|Zd
n/∼ | =d
n/d inequivalent zero distributions exist.
According to Theorem 5, a state |ψi ∈ Hdsaturating (30) for d > 3, must have n= (d−1)/2
zeroes in each basis. In addition, a saturating state requires the existence of at least dincompatible
zero distributions as Theorem 6does not allow compatible zero distributions for more than two
bases. In other words, the inequality |Zd
(d−1)/2/∼ | ≥ dmust hold. Clearly, this does not happen
for d= 5 and d= 7 since |Z5
2/∼ | = 2 <5and |Z7
3/∼ | = 5 <7, respectively. When d≥11,
however, the inequality is satisfied, with |Z11
5/∼ | = 42 >11, for example.
4.4 Numerical results for 5≤d≤19
For prime numbers dgreater than seven, more than dincompatible zero distributions exist which
removes the bottleneck we exploited to prove Theorem 8. In the absence of an analytic handle on
the problem, we will use numerical means to check whether the bound imposed by (30) can be
reached for dimensions larger than d= 7.
A saturating state necessarily has (d−1)/2zeroes in each MU basis. Thus, if one picks two
distinct MU bases with labels j1, j2∈ {0. . . d}, say, with corresponding zero distributions Zj1
and Zj2, the state will have vanishing scalar products with a total of (d−1) states which—in
view of Corollary 2—are known to be linearly independent. Consequently, there is a unique ray
ψ⊥∈ Hdassociated with any two zero distributions of the type considered. If the support size
of the states ψ⊥generated in this way (i.e. for all possible choices of initial zero distributions Zj1
and Zj2) is always larger than (d+ 1)/2in some third MU basis, then the support inequality (30)
cannot be saturated: if no state with support size (d+ 1)/2in three MU bases exists, then no state
with support size (d+ 1)/2in (d+ 1) MU bases will exist. Since only a finite number of zero
distributions needs to be checked for a given dimension d, this approach actually represents an
algorithm to check whether the lower bound can be reached.
Running the program for prime numbers with 5≤d≤19 means to check an exponentially
increasing number of cases. On a standard PC, the program ran about a second for d= 5 and
d= 7 while it took about a week for d= 17. No state has been found which would display
(d−1)/2zeroes in three MU bases. For dimensions d= 5 and d= 7, this result is stronger
than that of Sec. 4.3 since the non-existence of a state with two and three zeroes, respectively, is
sufficient to derive Theorem 8, but not vice versa. Due to the exponential increase in the number
of zero distributions, dimensions larger than d= 19 were out of of reach.
4.5 Dimensions d > 19
To satisfy the additive support inequality (30) relative to (d+ 1) MU bases, a state needs to satisfy
more than one pair relation (22) simultaneously which seems unlikely. It is all the more surprising
that for dimension d= 3 the bound T(3) = 8 is actually sharp, i.e. Ts(3) ≡T(3). Our results for
prime dimensions up to d= 19 suggest that this case is exceptional.
We conjecture that the generalised uncertainty relation (30) in prime dimensions can only be saturated
when d= 3.Here is a plausibility argument to support this view. Assume that a saturating state
12
ψ∈ Hdexists for some prime dimension d≥3. According to Theorem 5, the state must be
orthogonal to exactly (d−1) /2elements from each of the (d+ 1) MU bases. Corollary 2implies
that orthogonality with respect to just two such sets—i.e. (d−1) vectors—already determines a
unique state. Therefore, the remaining (d−1)2/2vectors (one set of (d−1)/2vectors is associated
with each of the (d−1) MU bases not yet considered) must all lie in the same (d−1)-dimensional
subspace orthogonal to the state ψ. This is known to to happen for d= 3 but seems hard to satisfy
for larger dimensions.
5 Sharp lower bounds
According to the results presented in Sec. 4, no states exist which would saturate the lower
bounds (30) for the support sizes in dimensions up to d= 19, with the exception of d= 3. The
focus of this section will be on identifying achievable bounds.
5.1 Dimension d= 3
Theorem 7in Sec. 4.2 displays the states which achieve the lower bound (30) in dimension d= 3.
In other words, the bound for the overall support of qutrit states ψis sharp, S(3) ≥8, where
S(d)≡Pd
j=0 kψkjfor ψ∈ Hd.
5.2 Dimension d= 5
Theorem 8shows that, for any state ψ∈ H5, the overall support of the states H†
jψ,j={0. . . d},
must satisfy S(5) >18. In this section, we will prove a sharp lower bound, namely S(5) ≥22 ≡
Ts(5).
To begin, we generalise Lemma 2which will be necessary for the proof of Lemma 4.
Lemma 3. Let dbe an odd prime and ω≡e2iπ
d. Consider two states |φj1
k1i,|φj2
k2i ∈ Hdtaken from
different MU bases, j1, j26= 0, and let {|xi} be the computational basis. Then there can be at most two
values of x∈ {0. . . d −1}such that
hx|φj1
k1i=ωnhx|φj2
k2i(46)
for the same value of n∈ {0...d −1}. If two different states are taken from the same basis, j1=j2, then
the equation has exactly one solution for each value of n.
Proof. See Appendix.
Now consider a state with two vanishing expansion coefficients in both the computational
basis and a second basis of the complete set. It turns out that such a state can have only non-zero
coefficients in the remaining four bases, resulting in a total support size of 26.
Lemma 4. If the support of a state ψ∈ H5equals three in both the computational basis and another MU
basis with label j6= 0, i.e. kψk0=kψkj= 3, then its support size in each of the remaining four bases
equals five, kψkj′= 5, with j′6= 0, j.
Proof. The proof, given in Appendix, uses Corollary 1and Lemma 3.
This result allows us to determine a sharp bound Ts(5) for the support size S(5).
Theorem 9. Given a state ψ∈ H5, the sharp bound on its overall support size S(5) in a complete set of
MU bases is given by Ts(5) = 22.
Proof. To construct the bound, we go through all possible values of the support size of the state ψ
in the computational basis, i.e. kψk0∈ {1... 5}.
13
For kψk0= 1, the pair inequalities (22) imply that the state ψmust have full support in all
other five MU bases, i.e. kψkj6=0 = 5. Hence, the overall support of a computational basis state is
given by S(5) = 26.
For kψk0= 2, the pair inequalities (22) imply that the state ψcan have at most one zero in each
of the other five MU bases, i.e. kψkj6=0 = 4. Hence, the overall support of ψis given by S(5) = 22.
All 300 states of the form
|ψi=1
√2|φj
k1i − ωn|φj
k2i, j ∈ {0...5}, k1, k2, n ∈ {0...4}, k16=k2.(47)
achieve this bound. More generally, for primes d > 3, there are
(d+ 1) d2
d=1
2d2−1d2(48)
such states as jtakes (d+ 1) values, ntakes dvalues and there are 2
ddifferent pairs of k1and k2.
The three-dimensional case is an exception, as demonstrated by Theorem 7.
For kψk0= 3, we apply Lemma 4: the support size of ψcan equal three in only one other
MU basis while the state must have full support in the others, leading to S(5) = 26. It is also
possible to have kψkj= 4 in all bases but the first one, i.e. for j6= 0. In this case seven expansion
coefficients would vanish over the complete set, resulting in an overall support size of S(5) = 23.
This bound is larger than the one already obtained for the case of kψk0= 2.
If kψk0= 4 and all other support sizes are also equal to four, the resulting overall support of
S(5) = 24 is again larger that the previous bound of S(5) = 22 obtained for kψk0= 2. To improve
on the value of S(5) = 24, at least one of the other norms must fall below four, i.e. 1≤ ||ψ||j∗≤3
for some j∗6= 0. This assumption, however, sends us back to one of the cases already discussed:
we formally map j∗7→ 0and repeat the arguments given for 1≤ ||ψ||0≤3.
Similarly, full support in all six MU bases cannot beat any of the bounds given so far. Improv-
ing on the value of S(5) = 30 is only possible by decreasing some of the support sizes, so that
we will end up in one of the previously discussed cases. Having considered all support sizes of
a state in a basis, we have exhausted all possibilities and conclude that the bound on the overall
support of a state ψ∈ H5in six MU bases is indeed given by Ts(5) = 22.
5.3 Dimension d= 7
Our aim is to identify states which minimise the overall support S(7) = P7
j=0 kψkj. To deter-
mine the sharp bound for d= 7, we will proceed as in the previous section. However, since no
equivalent to Lemma 4is known, we will partly rely on numerical results.
For kψk0= 1, the pair inequalities (22) imply that the state ψmust have full support in all
other seven MU bases,kψkj6=0 = 7. Hence, the overall support of ψis given by S(7) = 50.
For kψk0= 2, the pair inequalities (22) imply that the state ψcan have at most one zero in
each of the other seven MU bases, i.e. kψkj6=0 = 6. Hence, the overall support of ψis given by
S(7) = 44, achieved by states of the form
|ψi=1
√2|φj
k1i − ωn|φj
k2i, j ∈ {0...7}, k1, k2, n ∈ {0...6}, k16=k2.(49)
According to Eq. (48), there are 1176 such states.
For kψk0= 3, the smallest possible value of S(7) compatible with the pair inequalities is
S(7) = 38, as the states in the other bases must have support size at least five each, i.e. kψkj6=0 = 5.
However, no state achieving this bound has been found (numerically). The computations show
that a state with support sizes three and five in two MU bases must have full support in the
remaining six MU bases so that S(7) = 50. Assuming support size six in all but the first MU basis,
the overall support would be S(7) = 45 which is higher than the bound of S(7) = 44 achievable
for kψk0= 2 .
14
Given a support size of four in the first MU basis, kψk0= 4, not all other support sizes can
be equal to four according to Theorem 8. One case corresponds a state having support size four
in the first and one other MU basis. It is possible to (numerically) construct states for which the
remaining six supports sizes must be equal to six, leading to S(7) = 44. We neither know analytic
expressions for these states nor their total number. The other scenario compatible with kψk0= 4
corresponds to the remaining seven support sizes each equalling five, i.e.kψkj6=0 = 5, leading to
S(7) = 39 but we have obtained no evidence for a state achieving this value.
Assume now that kψk0= 5 and that the support of ψin the other MU bases is also at least five
(we exclude all cases with ||ψ||j∗<5for some j∗6= 0 since—upon relabeling the MU bases—they
have effectively already been considered). An overall support of S(7) = 40 results, below the
previously obtained value of S(7) = 44 for kψk0= 2. Numerically searching for states achieving
this bound, we find that pairs of states with support size five in two MU bases exist but no triples,
ruling out the value S(7) = 40. Assuming support size five in two bases and at least six in the
remaining six MU bases leads to a higher support size, S(7) = 46.
Starting out with a support size of kψk0>5, no smaller lower bound will exist if all other
support sizes take a value of at least six as S(7) ≥48 follows immediately. If not all support sizes
take a value of at least six we are being sent back to a previously discussed case. Thus, we have
established the sharp bound of Ts(7) = 44 on the overall support of seven-component vectors in
eight MU bases, partly relying on numerics.
6 Summary and Conclusions
Tao’s uncertainty relation provides a lower bound on the sum of the support sizes of a state
ψ∈ Hdin the standard basis and its Fourier transform, for prime dimensions d. By generalising
the bound to arbitrary pairs of mutually unbiased bases (cf. Theorem 3), we show in Theorem
4that the sum of the support sizes of a state ψin a complete set of (d+ 1) MU bases cannot fall
below T(d)≡(d+ 1)2/2. The bound is found to be sharp for d= 3, and proofs were given that it
cannot be saturated for dimensions d= 2,5and 7. Numerical results indicate that no states exist
which achieve the bound for prime numbers up to d≤19. Table 1summarises these results. We
conjecture that the inequality is saturated in dimension d= 3 only.
d2 3 5 7 11 13 17 19
T(d)9/2 8 18 32 72 98 162 200
T(d)achievable? ×X× × (×) (×) (×) (×)
Ts(d)5 8 22 (44) ? ? ? ?
Table 1: Lower and sharp bounds T(d)and Ts(d), respectively, on the support sizes of states ψ∈
Hdwhen expanded in complete sets of (d+ 1) MU bases, for small prime dimensions (numerical
results in parentheses).
Tao’s pair support inequality has been used to identify KD-nonclassical states, i.e. states whose
Kirkwood-Dirac quasiprobability distribution has negative or complex contributions [10]. Given
two orthonormal bases of a finite-dimensional space Hd, d ∈N,with no common elements, a
state ψis found to be KD-nonclassical if the sum of its support sizes in these bases is greater than
(d+ 1). KD-classicality is readily generalised to complete sets of MU bases instead of pairs only.
In this context, the results of Sec. 4mean that no states exist which are KD-classical with respect
to the full set of (d+ 1) MU bases in small prime dimensions. When d= 3, the claim follows by
directly computing the complex KD distributions of the nine minimal uncertainty states of Eq.
(43).
The uncertainty of quantum states involving more than two MU bases has been studied be-
fore. Building on a result for a pair of mutually unbiased observables [33], entropic uncertainty
15
relations have been found which involve (d+ 1) MU bases [22,23]. Similarly, Heisenberg’s un-
certainty relation for continuous variables has a counterpart based on three observables satisfying
the canonical commutation relation pairwise [18]. Often, the generalisations are straightforward
but the resulting inequalities tend not to be achievable. Sharp bounds and the minimising states
are usually difficult to find (see e.g. [34,19] and the review [35]). In this respect, the additive
inequality proposed here is no exception.
Support uncertainty relations for multiple MU bases have many interesting features. As for
the pair inequalities, a finite number of measurements can be sufficient to confirm that a quan-
tum state satisfies a specific bound. The minimum number of required measurements is simply
given by the value of the relevant bound, be it sharp or not: it is sufficient that T(d)different
outcomes be registered when measurements in the MU bases are performed on the state ψ. This
property also ensures that KD-nonclassicality may sometimes be detected with a finite number of
measurements.
Furthermore, the lower bounds of support inequalities neither depend on the state considered
nor the value of Planck’s constant. The absence of ~as a parameter suggests that no support in-
equalities for continuous variables will emerge in the limit of systems with ever larger dimensions
d. The maximal support size of a quantum state grows without bound and, therefore, does not
approach a well-defined quantitative measure for uncertainty. Finally, we would like to point out
that determining bounds on support sizes is technically difficult since they are basis-dependent
quantities.
Establishing sharp bounds for dimensions d≥11 remains an open question which will require
new insights since numerical approaches become unfeasible with increasing dimensions. Other
directions of future work will be to study support uncertainty relations for smaller sets of MU
bases such as triples, for example. The simplification stems from the considerably smaller number
of parameters in comparison to complete MU sets. Preliminary analytical and numerical results
for small prime dimensions 3≤d≤19 suggest that no state can saturate the bound T(d; 3) on the
triple uncertainty relation (31) for d6= 3.
References
[1] W. Heisenberg. Über den anschaulichen Inhalt der quantentheoretischen Kinematik und
Mechanik. Zeitschrift für Physik, 43(3): 172-198, 1927.
[2] E.H. Kennard. Zur Quantenmechanik einfacher Bewegungstypen. Zeitschrift für Physik,
44(4):326-352, 1927.
[3] G.B. Folland and A. Sitaram. The Uncertainty Principle: A Mathematical Survey. Journal of
Fourier Analysis and Applications, 3(3):207-238, 1997.
[4] D.L. Donoho and P.B. Stark. Uncertainty Principles and Signal Recovery. SIAM Journal on
Applied Mathematics, 49(3):906-931, 1989.
[5] R. Meshulam. An Uncertainty Inequality for Groups of Order pq.European Journal of Combi-
natorics, 13(5):401-407, 1992.
[6] A. Wigderson and Y. Wigderson. The Uncertainty Principle: Variations on a Theme. Bulletin
of the American Mathematical Society, 58(2):225-261, 2021.
[7] T. Tao. An Uncertainty Principle for Cyclic Groups of Prime Order. Mathematical Research
Letters, 12(1):121-127, 2005.
[8] E.J. Candes, J. Romberg, and T. Tao. Robust Uncertainty Principles: Exact Signal Reconstruc-
tion from Highly Incomplete Frequency Information. IEEE Transactions on Information Theory,
52(2):489-509, 2006.
16
[9] E.J. Candes and J. Romberg. Quantitative Robust Uncertainty Principles and Optimally
Sparse Decompositions. Foundations of Computational Mathematics, 6(2):227-254, 2006.
[10] S. De Bièvre. Complete Incompatibility, Support Uncertainty, and Kirkwood-Dirac Nonclas-
sicality. Physical Review Letters, 127(19):190404, 2021.
[11] J.G. Kirkwood. Quantum Statistics of Almost Classical Assemblies. Physical Review, 44(1):31-
37, 1933.
[12] P.A.M. Dirac. On the Analogy Between Classical and Quantum Mechanics. Reviews of Modern
Physics, 17(2):195-199, 1945.
[13] D.R.M. Arvidsson-Shukur, N.Y. Halpern, H.V. Lepage, A.A. Lasek, C.H.W. Barnes, and S.
Lloyd. Quantum Advantage in Postselected Metrology. Nature Communications, 11(1):3775,
2020.
[14] H.F. Hofmann. On the Role of Complex Phases in the Quantum Statistics of Weak Measure-
ments. New Journal of Physics, 13(10):103009, 2011.
[15] J. Dressel and A.N. Jordan. Significance of the Imaginary Part of the Weak Value. Physical
Review A, 85(1):012107, 2012.
[16] D.R.M. Arvidsson-Shukur, J.C. Drori, and N.Y. Halpern. Conditions Tighter than Non-
commutation Needed for Nonclassicality. Journal of Physics A: Mathematical and Theoretical,
54(28):284001, 2021.
[17] R. Kunjwal, M. Lostaglio, and M.F. Pusey. Anomalous Weak Values and Contextuality: Ro-
bustness, Tightness, and Imaginary Parts. Physical Review A, 100(4):042116, 2019.
[18] S. Kechrimparis and S. Weigert. Heisenberg Uncertainty Relation for Three Canonical Ob-
servables. Physical Review A, 90(6):062118, 2014.
[19] S. Kechrimparis and S. Weigert. Geometry of Uncertainty Relations for Linear Combinations
of Position and Momentum. Journal of Physics A: Mathematical and Theoretical, 51(2):025303,
2017.
[20] V.V. Dodonov, E.V. Kurmyshev, and V.I. Man’ko. Generalized Uncertainty Relation and Cor-
related Coherent States. Physics Letters A, 79(2):150-152, 1980.
[21] V.V. Dodonov. Variance Uncertainty Relations without Covariances for Three and Four Ob-
servables. Physical Review A, 97(2):022105, 2018.
[22] I.D. Ivanovi´c. An Inequality for the Sum of Entropies of Unbiased Quantum Measurements.
Journal of Physics A: Mathematical and General, 25(7):L363-L364, 1992.
[23] J. Sánchez. Entropic Uncertainty and Certainty relations for Complementary Observables.
Physics Letters A, 173(3):233-239, 1993.
[24] M.A. Ballester and S. Wehner. Entropic Uncertainty Relations and Locking: Tight bounds for
Mutually Unbiased Bases. Physical Review A, 75(2):022319, 2007.
[25] A. Rényi. On measures of entropy and information. Proceedings of the Fourth Berkeley Sympo-
sium on Mathematical Statistics and Probability, 1:547-561, 1961.
[26] P. Stevenhagen and H.W. Lenstra. Chebotarëv and his Density Theorem. The Mathematical
Intelligencer, 18(2):26-37, 1996.
[27] P.E. Frenkel. Simple Proof of Chebotarëv’s theorem on Roots of Unity. arXiv:math/0312398,
2004.
17
[28] I.D. Ivanovi´c. Geometrical Description of Quantal State Determination. Journal of Physics A.
Mathematical and General, 14(12):3241-3245, 1981.
[29] W.K. Wootters and B.D. Fields. Optimal State-Determination by Mutually Unbiased Mea-
surements. Annals of Physics, 191(2):363-381, 1989.
[30] S. Bandyopadhyay, P.O. Boykin, V. Roychowdhury, and F. Vatan. A New Proof for the Exis-
tence of Mutually Unbiased Bases. Algorithmica, 34(4):512-528, 2002.
[31] T. Durt. About Mutually Unbiased Bases in Even and Odd Prime Power Dimensions. Journal
of Physics A. Mathematical and General, 38(23):5267-5283, 2005.
[32] W. Bruzda, W. Tadej, and K. ˙
Zyczkowski. Catalogue of Complex Hadamard Matrices. Avail-
able online at https://chaos.if.uj.edu.pl/karol/hadamard, 2006.
[33] H. Maassen and J.B.M. Uffink. Generalized Entropic Uncertainty Relations. Physical Review
Letters, 60(12):1103-1106, 1988.
[34] S. Wu, S. Yu, and K. Mølmer. Entropic Uncertainty Relation for Mutually Unbiased Bases.
Physical Review A, 79(2):022104, 2009.
[35] S. Wehner and A. Winter. Entropic Uncertainty Relations — A Survey. New Journal of Physics,
12(2):025009, 2010.
[36] B.C. Berndt, K.S. Williams, and R.J. Evans. Gauss and Jacobi Sums. Wiley, 1998.
7 Appendix
We present proofs of Lemma 1, Corollary 2and Lemmata 3and 4in this order.
Lemma 1. Let dbe an odd prime and j, k ∈ {1. . . d}with j6=k. Then
H†
kHj=M(j, k)H†
t(28)
for a monomial matrix M(j, k)if and only if t= 1 + χ∈ {1,...,d}where the integer χsatisfies
4 (j−k)χ= 1 mod d.
Proof. Using Eqs. (14), we calculate the matrix elements of the product H†
kHj, with j, k 6= 0 and
j6=k,hH†
kHjiℓℓ′=hφk
ℓ|φj
ℓ′i=1
√dGd(j−k, ℓ −ℓ′),(50)
with the generalised Gauss sum [36]
Gd(j, ℓ) = 1
√d
d−1
X
x=0
ωjx2+ℓx .(51)
Using 1 = ω(ℓ−ℓ′)2χω−(ℓ−ℓ′)2χin (50) and letting χbe an integer satisfying 4 (j−k)χ≡1 (mod d),
we obtain a standard Gauss sum Gd(j−k , 0) with known closed form. Explicitly, for j6=k, we
obtain
Gd(j−k, ℓ −ℓ′) = ω−(ℓ−ℓ′)2χ1
√dX
x
ω(j−k)x2+(ℓ−ℓ′)xω(ℓ−ℓ′)2χ
=ω−(ℓ−ℓ′)2χ1
√dX
x
ω(j−k)[x+2(j−k)(ℓ−ℓ′)]2
=ω−(ℓ−ℓ′)2χ"1
√dX
x
ω(j−k)x2#=ω−(ℓ−ℓ′)2χGd(j−k, 0)
=ω−(ℓ−ℓ′)2χj−k
dεd,(52)
18
where ¯adenotes the multiplicative inverse of a,¯aa ≡1mod d, while a
bdenotes the Jacobi
symbol of the integers aand b, and
εd=(1if d≡1 (mod 4) ,
iif d≡3 (mod 4) .(53)
The sum Gd(j−k, ℓ −ℓ′)in (52) reduces to a phase factor as it should since the components of
the matrix H†
kHjare given by the overlap of states stemming from different MU bases.
Combining (50) and (14), we now determine the elements of the matrix V≡H†
kHjHtfor
arbitrary t6= 0:
Vℓℓ′=
d−1
X
ℓ′′=0hφk
ℓ|φj
ℓ′′ ihℓ′′|φt
ℓ′i=1
d
d−1
X
ℓ′′=0
Gd(j−k, ℓ −ℓ′′)ω−ℓ′ℓ′′ +(t−1)ℓ′′ 2.(54)
We can simplify this expression by substituting (52) into it, to find
Vℓℓ′=1
dj−k
dεd
d−1
X
ℓ′′=0
ω−(ℓ−ℓ′′)2χω−ℓ′ℓ′′ +(t−1)ℓ′′ 2
=1
dj−k
dεdω−ℓ2χ
d−1
X
ℓ′′=0
ω(t−1−χ)ℓ′′2+(2ℓχ−ℓ′)ℓ′′
.
(55)
Letting t= 1 + χ, we obtain sums over all d-th roots of one which vanish unless the exponents of
ωvanish,
d−1
X
ℓ′′=0
ω(2ℓχ−ℓ′)ℓ′′
=(dif ℓ′= 2ℓχ mod d ,
0otherwise .(56)
Thus, for this value of t, the matrix elements of Vtake the form
Vℓℓ′=
j−k
dεdω−ℓ2χif ℓ′= 2ℓχ mod d ,
0otherwise ,(57)
so that the only non-zero elements of the matrix Vare those with indices (ℓ, 2ℓχ mod d). Each
row ℓhas exactly one non-zero entry and the map ℓ7→ 2ℓχ mod dconstitutes a permutation of the
elements of {0. . . d −1}since dis a prime number and 2χ6= 0 mod d. (Assume this was not the
case, i.e. 2χx mod d= 2χy mod dfor some x, y ∈ {0. . . d −1},x6=y. Then 2χ(x−y) = nd for
some integer nwhich is never the case whenever dis prime and χ6= 0 mod d.) As a consequence,
each column will also display exactly one non-zero entry.
Therefore, the product Vof three Hadamard matrices is equal to a monomial matrix M(j, k)
if t= 1 + χ, i.e.
H†
kHj=M(j, k)H†
1+χ.(58)
We complete the proof by showing that the matrix V=H†
kHjHtis not monomial for any other
value of t. For t6= 1 + χ, the sum on the right-hand side of (55) represents another generalised
Gauss sum so that
Vℓℓ′=1
√dj−k
dεdω−ℓ2χGd(t−1−χ, 2ℓχ −ℓ′)(59)
with Gd(t−1−χ, 2ℓχ −ℓ′) = √dhφ1+χ
2ℓχ |φt
ℓ′i. We can now substitute the expression (52) and
obtain
Vℓℓ′=1
√dε2
dj−k
dt−1−χ
dω−ℓ2χω−(2ℓχ−ℓ′)2eχ(60)
where eχ∈ {1. . . d}is an integer satisfying 4 (t−1−χ)eχ= 1 mod d. Hence, the matrix elements
Vℓℓ′are all non-zero confirming that the matrix Vis not monomial unless t= 1 + χ.
19
Corollary 2. Given a complete set of MU bases in the space Hdof prime dimension d, up to dvectors
taken from any two MU bases are linearly independent.
Proof. Construct a matrix Mof order d×(d1+d2)from any (d1+d2)≤dcolumn vectors—
expressed in the computational basis—from the two MU bases Bj1and Bj2. Then left-multiply M
by H†
j1. Since hx|H†
j1|φj1
ki=hx|ki, the first d1columns will be elements of the computational basis,
while the remaining d2columns will be taken from H†
j1Hj2since hx|H†
j1|φj2
ki=hx|H†
j1Hj2|ki.
By swapping rows appropriately via a permutation operator Pwhich does not change linear
independence of column vectors, the top left square can be mapped to the d1-dimensional identity.
For example, if we consider d= 5 and d1=d2= 2, we obtain a 5×4matrix,
P H †
j1M=
1 0 ∗ ∗
0 1 ∗ ∗
0 0 ∗ ∗
0 0 ∗ ∗
0 0 ∗ ∗
,(61)
where the asterisks refer to the elements of H†
j1Hj2.
The (d1+d2)≤dvectors are linearly dependent only if Mdoes not have full rank, i.e.
rank (M)< d1+d2. Since P H †
j1is unitary, it follows that rank P H †
j1M< d1+d2. Given
the form of the matrix (61), the bottom-right part of P H †
j1Mmust contain a (d2×d2)submatrix
with vanishing determinant. However, this is prohibited by Corollary 1which ensures for all
prime numbers dthat H†
j1Hj2has non-vanishing minors if j16=j2. Thus, all (d1+d2)column
vectors of Mmust be linearly independent.
Lemma 3. Let dbe an odd prime and ω≡e2iπ
d. Consider two states |φj1
k1i,|φj2
k2i ∈ Hdtaken from
different MU bases, j1, j26= 0, and let {|xi} be the computational basis. Then there can be at most two
values of x∈ {0. . . d −1}such that
hx|φj1
k1i=ωnhx|φj2
k2i(46)
for the same value of n∈ {0...d −1}. If two different states are taken from the same basis, j1=j2, then
the equation has exactly one solution for each value of n.
Proof. By substituting (13) into (46) and taking the logarithm, one obtains ax2+bx −n= 0 mod d
where a=j1−j2and b=k2−k1. This quadratic equation can have no more than two integer
solutions. Thus, at most two components of the states |φj1
k1iand |φj2
k2ican be identical in the
computational basis, up to multiplication by ωn. If j1=j2, then a= 0 and the equation is linear
with a single solution for each value of n.
Lemma 4. If the support of a state ψ∈ H5equals three in both the computational basis and another MU
basis with label j6= 0, i.e. kψk0=kψkj= 3, then its support size in each of the remaining four bases
equals five, kψkj′= 5, with j′6= 0, j.
Proof. Four scalar products with the state ψvanish,
hx1|ψi=hy1|ψi=hφj
x2|ψi=hφj
y2|ψi= 0 ,(62)
two for each of the bases. Hence, the zero distributions of the states ψand H†
jψ, are given by
Z0={x1, y1}and Zj={x2, y2}respectively, with four integer numbers x1,...,y2∈ {0...4}.
Now suppose that there is a third MU basis Bj′, different from both B0and Bj, in which the
state ψdoes not have full support. In other words, there is at least one vanishing scalar product,
hφj′
x3|ψi= 0, say, where x3∈ {0...4}. Expressing the components of these five vectors with
respect to the computational basis and arranging them into a 5×5matrix, we find after permuting
20
the rows and rephasing the last three vectors
M=1
√5
√5 0 ∗ ∗ ∗
0√5∗ ∗ ∗
0 0 1 1 1
0 0 ωaωcωe
0 0 ωbωdωf
, a, . . . , f ∈ {0...4}.(63)
Being elements of the Hadamard matrices Hjand Hj′, the entries of the last three columns are
powers of ω, a fifth root of one. Corollary 2—upon setting d= 5,d1=d2= 2 and j1= 0, j2=j—
ensures the linear independence of the first four vectors.
If the determinant of Mdoes not vanish, det M6= 0,then the five column vectors forming it
are linearly independent, thus spanning H5. However, the only state being orthogonal to all of H5
is ψ= 0 which does not represent a quantum state. Thus, for an acceptable state ψproducing the
given five vanishing expansion coefficients, the five vectors involved must be linearly dependent,
i.e. det M= 0. Consequently, the determinant of the bottom right 3×3matrix of Mmust vanish,
∆≡det
1 1 1
ωaωcωe
ωbωdωf
=ωa+d+ωe+b+ωc+f−ωc+b−ωe+d−ωa+f= 0 .(64)
Each of the six terms in this expression is a power of a fifth root ωof unity, hence non-zero. It
is well known that the set {ωn|n= 0, ..., 3}is linear independent over the rational numbers Q. As
a consequence, every non-zero complex number that is expressible as a linear combination (over
Q) of these roots of unity has a unique expression. Since ω4=−1−ω−ω2−ω3, it must follow that
the only decomposition of zero over Qin terms of fifth roots of 1is 0 = c1 + ω+ω2+ω3+ω4,
where c∈Q. In other words, for the sum to vanish all five roots must be multiplied to the same
rational coefficient.
We distinguish two cases: either c6= 0 or c= 0. Since (64) involves six terms with coefficients
±1, we conclude that the case of c6= 0 cannot be realised: it is impossible to get all five roots to
appear with the same non-zero coefficient. For example, let (a+d) = (e+b) mod 5, then Eq. (64)
reduces to
∆ = 2ωa+d+ωc+f−ωc+b−ωe+d−ωa+f(65)
Since all roots must appear, the exponents in (65) are all different. However, the coefficients are
not equal throughout and the sum cannot vanish. A similar argument holds for any other equality
between exponents.
The case of c= 0 must therefore apply: the determinant ∆vanishes if and only if the six terms
in Eq. (64) cancel each other in pairs, i.e. the the powers of ωmust occur an even number of times,
and with an equal number of positive and negative coefficients. Hence, the first term in (64) is
necessarily paired up with one of the powers with a negative coefficient leading. Three cases arise
which we will consider separately.
(i)For the first and the fourth term to cancel, we must have (a+d) = (c+b) mod 5 ,or
(a−b) = (c−d) mod 5 ,(66)
relating the expansion coefficients of two vectors of the same basis, namely |φj
x2iand |φj
y2i. Con-
sequently, the third and fourth column vectors in the matrix Min (63) have (at least) two equal
entries in identical positions, up to an irrelevant common phase factor. This would result in a
vanishing 2×2submatrix of Hjcontradicting Corollary 1(and Lemma 3). Thus the determinant
∆cannot vanish in this case.
(ii)For the first and the fifth term to cancel, we must have (a+d) = (e+d) mod 5 ,or
a=emod 5 ,(67)
21
relating the expansion coefficients of two vectors of different bases, namely |φj
x2iand |φj′
x3i. Corol-
lary 1does not apply to this case. We do know, however, that the fourth term in the sum (64)
must pair up with either the second or the third term of the sum in (64). In the first case, we find
(c+b) = (e+b) mod 5 ,or
c=emod 5 .(68)
Given the constraint (67), we we obtain the identity
a=cmod 5 ,(69)
again relating the expansion coefficients of two vectors of the same basis, namely |φj
x2iand |φj
y2i.
As in the Case (i), a contradiction to Corollary 1arises.
In the second case, we pair up terms three and four of the sum (64), leading to the identity
(c+b) = (c+f) mod 5, or
b=fmod 5 .(70)
Together with Eq. (67), it follows that the last three elements of the third and and fifth columns of
Mare identical. However, according to Lemma 3, two vectors stemming from two different bases
MU to the computational basis can have at most two identical components.
(iii): Assuming that the first and the sixth term of the sum (64) cancel again leads to a contra-
diction along the lines of the argument considered in Case (ii).
Thus, we are forced to conclude that the determinant ∆cannot not vanish for any j′6= 0, j
and any x3, which implies that the state ψmust have full support.
22
