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Quantum state discrimination: a geometric approach
Damian Markham∗
Universite´ Paris 7, 175 Rue du Chevaleret, 75013 Paris, France and
Department of Physics, Graduate School of Science, University of Tokyo, Tokyo 113-0033, Japan
Jaros law Adam Miszczak† and Zbigniew Pucha la‡
Institute of Theoretical and Applied Informatics,
Polish Academy of Sciences, Ba ltycka 5, 44-100 Gliwice, Poland
Karol Z˙yczkowski§
Instytut Fizyki im. Smoluchowskiego, Uniwersytet Jagiellon´ski, ul. Reymonta 4, 30-059 Krako´w, Poland and
Centrum Fizyki Teoretycznej, Polska Akademia Nauk, Al. Lotniko´w 32/44, 02-668 Warszawa, Poland
(Dated: November 27, 2007)
We analyse the problem of finding sets of quantum states that can be deterministically discrim-
inated. From a geometric point of view this problem is equivalent to that of embedding a simplex
of points whose distances are maximal with respect to the Bures distance (or trace distance). We
derive upper and lower bounds for the trace distance and for the fidelity between two quantum
states, which imply bounds for the Bures distance between the unitary orbits of both states. We
thus show that when analysing minimal and maximal distances between states of fixed spectra it is
sufficient to consider diagonal states only. Hence considering optimal discrimination, given freedom
up to unitary orbits, it is sufficient to consider diagonal states. This is illustrated geometrically in
terms of Weyl chambers.
PACS numbers: 03.65.Ta
I. INTRODUCTION
The geometry of state space depends on the distance measure chosen. In state discrimination, given a set of
possible states, our task is to find out as ‘best’ as possible which of the states we have in our possession [1, 2, 3].
Finding an optimal procedure of unambiguous discrimination is particularly interesting if the states analyzed are
mixed [4, 5, 6, 7, 8]. The usual approach to the quantum discrimination problem is to begin by considering the
classical case and then extending to the quantum case. Different concepts of ‘best’ induce different measures of
distinguishability in the space of classical probability distributions. In the quantum case, on top of the statistical
uncertainty of states, even pure states cannot be always be perfectly discriminated (if the states are not orthogonal),
meaning that one has to be careful in extending these to the quantum setting. To do this we bring it back to the
classical setting of probability distributions by maximising over all possible discrimination measurements. In this
way the problem of discriminating quantum states has led to several distance measures associated with the ability to
discriminate well (see e.g.[9, 10, 11]). In this work we would like to consider the geometry induced by these measures,
and how the problem of state discrimination can be expressed geometrically.
More precisely, let MN denote the set of mixed quantum states acting on an N dimensional Hilbert space HN . It
is a convex, compact set of dimensionality N2− 1. Its geometric structure depends on the metric used. The following
distances are often used [10, 11]
DHS(ρ1, ρ2) := [Tr(ρ1 − ρ2)2]1/2, (1.1)
Dtr(ρ1, ρ2) :=
1
2Tr|ρ1 − ρ2|, (1.2)
DB(ρ1, ρ2) :=
(
2[1 − Tr|√ρ1
√
ρ2|]
)1/2
, (1.3)
∗Electronic address: markham@phys.s.u-tokyo.ac.jp
†Electronic address: miszczak@iitis.gliwice.pl
‡Electronic address: z.puchala@iitis.gliwice.pl
§Electronic address: karol@cft.edu.pl

2denoting the Hilbert-Schmidt (HS) distance, the trace distance and the Bures distance respectively. The latter
quantity is a function of fidelity [12],
F (ρ1, ρ2) := [Tr|
√
ρ1
√
ρ2|)]2 , (1.4)
and the root fidelity
√
F , (which in some papers is also is called ‘fidelity’). The Bures and the trace distance are
monotone, and do not grow under the action of an arbitrary quantum operation (completely positive, trace preserving
map), while the Hilbert-Schmidt (HS) distance is not monotone. These measures can induce different geometries. For
instance, the set M2 of mixed states of a qubit, is equivalent to the standard Bloch–ball (the Bloch-sphere and its
interior) for the trace or HS metric, and to Uhlmann hemisphere, 12S3, for the Bures distance [13]. For higher N the
geometries induced by the HS and the trace distance also differ.
In the following we consider systems of dimension N greater or equal to two. We begin our discussion of state
discrimination by introducing the diameter of a set of quantum states. The diameter of the set MN is independent of
N , but it does depend on the metric used: the diameter is the maximal distance between any two states, and it reads
DmaxHS =
√
2, Dmaxtr = 1, DmaxB =
√
2, (1.5)
for HS, trace and Bures distances, respectively. Any two states separated by Dmax are supported on orthogonal
subspaces. The reverse implication holds for Bures and trace distances,
supp(ρ1) ⊥ supp(ρ2) ⇔ Dtr(ρ1, ρ2) = 1 ⇔ DB(ρ1, ρ2) =
√
2 , (1.6)
but is not true for the Hilbert-Schmidt distance for N > 2. For instance, the HS distance between two diagonal
density matrices ρ1 = diag(1, 0, 0) and ρ2 = diag(0, 1/2, 1/2) is equal to
√
3/2 < DmaxHS , although they are supported
on orthogonal subspaces. To witness an even more dramatic example consider the Hilbert space of even dimension N
and two diagonal states, ρ1 = diag(N/2, ..., N/2, 0, ..., 0) and ρ2 = diag(0, ..., 0, N/2, ..., N/2). Although they live in
orthogonal subspaces, so their Bures and trace distances are maximal, their HS distance reads 2/
√
N and tends to
zero in the limit of large N . This indicates that when analysing problems of distinguishability, one cannot therefore
rely on the standard Euclidean geometry induced by the Hilbert-Schmidt distance, but rather better use Bures or
trace distances.
The trace distance and the Bures distance are, in several respects, good measures for quantifying the ability to
discriminate states. In [14] Englert introduced the notion of distinguishability between two quantum states and
showed that it is equal to the trace distance between them. Hence two states can be deterministically discriminated
if they can be perfectly distinguished, so their distinguishability is equal to unity. Fuchs and van de Graaf found a
bound between the Bures distance and the trace distance based on the following inequality [10]
1 −
√
F (ρ1, ρ2) ≤ Dtr(ρ1, ρ2) ≤
√
1 − F (ρ1, ρ2) . (1.7)
This implies that if the fidelity between both states is equal to zero (so the states are distinguishable and their Bures
distance is maximal) their trace distance is equal to unity, and is hence maximal. In fact, the trace distance is a
simple function of the probability to successfully discriminate two states in a single shot measurement (optimised
over all allowed quantum measurements) [10]. Similarly, the Bures distance can be seen as the optimised Kullback–
Leibler distance between output statistics over all quantum measurements (again, an optimized cost function for
discrimination) [9].
In the special case where both density matrices are diagonal, and read p and q, the operators commute. Such a case is
often called classical since the distances between quantum states reduce then exactly to their classical analogues: The
trace distance Dtr(p, q) is the equal to the L1 distance (with a normalisation constant 1/2) between both probability
vectors; The Bures distance reads DB(p, q) = [2(1 −B(p, q))]1/2, where
B(p, q) :=
N
∑
i=1
√
piqi (1.8)
denotes the Bhattacharyya coefficient [15], [11]. This quantity is equal to the root fidelity between any two diagonal
states, B(p, q) =
√
F (p, q), so its square B2, is sometimes called classical fidelity between to probability distributions.
In section IV we prove general bounds for the fidelity between arbitrary two quantum states ρ1 and ρ2,
B2(p↑, q↓) ≤ F (ρ1, ρ2) ≤ B2(p↑, q↑) , (1.9)

3where the vectors p and q represent the spectra of ρ1 and ρ2, while the arrows up (down) indicate that the eigenvalues
are put in the nondecreasing (nonincreasing) order. These results imply equivalent bounds for the Bures distance
√
2 − 2
√
p↑.
√
q↑ ≤ DB(ρ1, ρ2) ≤
√
2 − 2
√
p↑.
√
q↓ . (1.10)
Analogous bounds for the trace distance proved in the same section read
Dtr(p↑, q↑) ≤ Dtr(ρ1, ρ2) ≤ Dtr(p↑, q↓) , (1.11)
where the symbols p↑ and q↓ denote here diagonal density matrices with all eigenvalues in the increasing (decreasing)
order.
In this paper we set out to give a geometric interpretation to the problem of state discrimination in terms of the
geometries induced by the trace and Bures distance. We begin in section II by giving a set of conditions on states
such that they may be perfectly discriminated. In section III we present some geometrical consequences of these
conditions and phrase the problem of state discrimination in terms of the embedding of simplices with respect to
different distance functions. In section IV we investigate the distance between states under unitary orbits and its
geometric interpretation, and prove the above bounds. We finish in section V with conclusions.
II. PERFECT DISCRIMINATION OF STATES
We begin by looking at some conditions on the set of states that can be perfectly discriminated. Our condition
will follow from simple analysis of the measurements (in terms of the associated positive operator valued measure
(POVM)), and give general conditions which, in the next section, will be used to give some geometrical consequences
of the problem.
Theorem 1 Two states ρ1 and ρ2 can be deterministically discriminated iff their supports do not overlap.
Proof. Any perfect state discrimination strategy for two states ρ1, ρ2 can be written as a three element POVM
{A1, A2, A?}, where the outcomes correspond to concluding it is the state ρ1, ρ2 and allowing for inconclusive outcome
respectively.
Note that although in general we can have far more possible outcomes than three, this formalism does include all
possible discrimination strategies - this is because we can always group the outcomes corresponding to state ρ1 to
give A1, and those to state ρ2 to give A2, and the remaining elements we group to give A?. The probability of success
of the strategy can always be written in terms of such POVMs, thus we can restrict ourselves to only these three
element POVMs for perfect discrimination.
The conditions on the POVM for deterministic state discrimination are
Tr(A1ρ1) = 1 (2.1)
Tr(A2ρ2) = 1 (2.2)
A1 +A2 +A? = I1 (2.3)
I1 ≥ Ai ≥ 0 (2.4)
(this is the same logic as in [16], only without the separability condition). The first two are necessary for perfect state
discrimination, and the last two are just the conditions for {Ai} to be a POVM.
Conditions (2.1) and (2.2) imply that the elements A1 and A2 include projections onto the support of ρ1 and ρ2
respectively. To see this, rewrite (2.1) in the eigenbasis of ρ1 =
∑
i λi|i〉〈i| (we extend this basis to the full space for
writing A1 in (2.6))
Tr(A1ρ1) =
∑
i
λi〈i|A1|i〉
=
∑
i
λiqi = 1, (2.5)
where qi := 〈i|A1|i〉 is a probability, hence
∑
i λiqi ≤ 1 and equality is obtained only when qi = 1 for all i such that
λi 6= 0. If we also demand conditions (2.3),(2.4) the most general Ak can be written
Ak = Pk +
∑
i,j /∈Supp(ρ1),Supp(ρ2)
αi,j |i〉〈j| (2.6)

4where Pk =
∑
i∈Supp(ρk) |i〉〈i| is the projector onto the support of state ρk. The support of a state ρ, with eigen-
decomposition ρ = ∑j αj |j〉〈j| is given by P =
∑
j |j〉〈j| . From here, condition (2.3) clearly says
P1 + P2 ≤ I1
⇒ Tr(P1P2) = 0
⇒ Tr(ρ1ρ2) = 0
⇒ Tr|ρ1 − ρ2|/2 = 1. (2.7)
Hence the supports have zero overlap.

The theorem is easily extended to sets of states {ρi}Mi=1.
Theorem 2 The states {ρi}Mi=1 can be deterministically discriminated iff their supports do not overlap.
This directly leads to
Proposition 1 Consider K states acting on the N dimensional Hilbert space, which can be discriminated determin-
istically. Then
K
∑
i=1
rank(ρi) ≤ N. (2.8)
This proposition is clear from the theorem, but also can be derived from the result in [16]. This is done by taking
the zero entanglement case of the main result presented there. Specifically, the left hand inequality in equation (8)
for zero entanglement, along with equation (1) in [16] give exactly (2.8).
III. SOME GEOMETRICAL CONSEQUENCES
We now look at what the above results have to say in terms of the geometric interpretation of the problem of state
discrimination. Due to property (1.6) the above theorem can also be formulated as the condition that the trace (or
Bures) distance between states are maximal. This fact has an immediate geometric implication. Let us start to work
with the trace distance and denote by ∆k ∈ Rk a maximal regular k − simplex defined by k + 1 points with mutual
trace distance between points equal to Dmaxtr = 1.
Proposition 2 Let R be an arbitrary convex subset of MN . Assume that there exists a simplex ∆k ⊂ R and assume
that R does not contain ∆k+1. Then the maximal number of states of R which can be discriminated deterministically
is equal to k + 1.
An analogous of the Proposition 2 may be formulated for the geometry induced by the Bures distance.
FIG. 1: Set of positive operators ρ1, . . . ρk with a) k = 2 and b) k = 3 distinguishable states which form a maximal simplex of
size k with side length Dmax, with respect to the Bures (or the trace) metric.

5Thus the problem of finding the maximal number of distinguishable states on a certain set is equivalent to the
problem of embedding inside it a regular simplex of maximal dimensionality with the diameter given by Dmax (see
Fig. 1).
At this point it is worth mentioning a different quantum problem of finding ‘symmetric, informationally complete
positive operational valued measures’ (SIC POVM) [17]. This has a similar geometric interpretation of inscribing
inside the set MN of mixed states an N2 − 1 dimensional simplex spanned by N2 pure states |φj〉, the overlap of
which is constant, F = |〈φi|φj〉|2 = 1/(N + 1) for any i 6= j. Therefore, in this case, the side of the simplex with
respect to the Bures distance reads DSICB =
√
2(1 −
√
F ) =
√
2 − 2/
√
N + 1, and for a finite dimension N , this is
smaller than DmaxB =
√
2.
So in the distinguishability problem we wish to embed into the set MN of mixed states a simplex of the maximal
side length DmaxB with dimensionality not larger than N , while in the SIC POVM problem we try to inscribe inside
the same set a higher dimensional simplex of a smaller side length DSICB .
IV. DISTANCES BETWEEN UNITARY ORBITS
In this section we shall be concerned with the distances between orbits generated from quantum states by unitaries.
That is, given two states ρ1 and ρ2 with fixed spectra, we wish to know how “far” or how “close” we can make these
states by unitary action. We will find that for the Bures and trace distance, the closest and the farthest that can be
achieved is given when both states are diagonal in the same basis. This has a geometric interpretation in terms of the
Weyl chambers as will be discussed.
This problem can be interesting in many areas of quantum information. Operationally the problem of finding the
best unitary separation of two density matrices may be interesting if we are restricted to certain spectra or mixedness.
For example in coding for noisy channels. If we know that the output of some channel will imply a certain mixedness
(or even specific spectra), we naturally want to choose to encode on states that are least affected by this. If we are
encoding classical information, this would be those states which remain most distinguishable afterwards. A simple
example of such a channel would be one which probabilistically adds white noise. Freedom of the input state would
correspond to unitary freedom of the outputs states which we wish to optimise over, hence considering the optimum
over unitary orbits of the output mixed states is equal to finding the optimum encoding. We will see that in such
cases, when only the spectra are restricted, the worst and best cases are given by taking them diagonal in the same
basis.
Consider first two classical, N–point, normalised probability distributions, p = (p1, . . . , pN ) and q = (q1, . . . , qN )
such that pi, qi ≥ 0 and
∑
i pi =
∑
i qi = 1. As earlier, let p
↓
i denote the vector ordered decreasingly, p
↓
i ≥ p
↓
i+1, while
let p↑i represent components of the probability vector in the increasing order: p
↑
i ≤ p
↑
i+1.
Any quantum state ρ1 generates an orbit of unitarily equivalent states, Uρ1U †. Two states ρ1 and Uρ1U † are
sometimes called geometrically uniform and they have been recently considered in the context of unambiguous dis-
crimination [4, 5, 8].
We are going to discuss another problem of distinguishing states from two orbits. Consider two diagonal quantum
states, ρ1 = diag(p) and ρ2 = diag(q), from which we two orbits of unitarily equivalent states. We shall analyze the
minimal and maximal distance Dx between the orbits,
M(ρ1, ρ2) := max
U,V
Dx(Uρ1U †, V ρ2V †) = max
W
Dx(ρ1,Wρ2W †), (4.1)
m(ρ1, ρ2) := min
U,V
Dx(Uρ1U †, V ρ2V †) = min
W
Dx(ρ1,Wρ2W †), (4.2)
since performing maximization over two unitary matrices U and V is equivalent to find a single unitary matrix
W = U †V . Here Dx stands for one of the monotone distances DB or Dtr. A similar statement for the non monotone
Hilbert-Schmidt distance (1.1) was already proved in [18].
We conjecture that extrema for these distances are obtained for diagonal matrices. Then the extremization has to
be performed only over the group P of permutation matrices, which change the order of the spectra,
M(ρ1, ρ2) = max
P
Dx(p, q) = Dx(p↓, q↑) = Dx(p↑, q↓), (4.3)
m(ρ1, ρ2) = min
P
Dx(p, q) = Dx(p↓, q↓) = Dx(p↑, q↑). (4.4)
The minimum is then achieved for the same order of components in both vectors, while the maximum occurs for
opposite ordering so using the above formula one can evaluated analytically the extremal distances for both distances
in consideration.

6Let us first show that this conjecture holds for the Bures distance.
Theorem 3 The maximum and minimum Bures distance between the unitary orbits of two states are given by diagonal
states with
M(ρ1, ρ2) = max
P
DB(p, q) = DB(p↓, q↑) = DB(p↑, q↓), (4.5)
and
m(ρ1, ρ2) = min
P
DB(p, q) = DB(p↓, q↓) = DB(p↑, q↑). (4.6)
Proof. a) We start by providing an upper bound for the Bures distance (1.3):
Let us start with the inequality
√
p↑.
√
q↑ ≥ Tr√ρ1
√
ρ2 ≥
√
p↑.
√
q↓ (4.7)
which is a particular case of (A1) from lemma 3 proved in appendix A. Since Tr|√ρ1
√
ρ2| ≥ Tr
√
ρ1
√
ρ2, we immediately
infer that the root fidelity is bounded from below by the Bhattacharayya coefficient between the spectra put in an
opposite order,
√
F (ρ1, ρ2) = Tr|
√
ρ1
√
ρ2| ≥ Tr
√
ρ1
√
ρ2 ≥
√
p↑ ·
√
q↓ = B(p↑ · q↓) . (4.8)
This implies an upper bound for the Bures distance which is clearly achievable, M(ρ1, ρ2) = DB(p↑, q↓).

In this way we obtain a general upper bound (4.5) for the Bures distance between any two density operators with
spectra p and q,
DB(ρ1, ρ2) ≤ DB(p↑, q↓) =
[
2(1 −
√
p↑.
√
q↓)
]1/2
. (4.9)
b) Next we provide a lower bound for the Bures distance (1.3):
To prove the case for minimisation our task is to show
√
p↑.
√
q↑ ≥ Tr|√ρ1
√
ρ2|, (4.10)
or equivalently, to get an upper bound for the root fidelity
√
F (ρ1, ρ2).
First we note that for any operator A we have [10, 19, 24]
max
U
|TrUA| = Tr
√
AA† ≡ Tr|A| ≡ ||A||1, (4.11)
where the maximum is taken over all unitaries U . We will also use the inequality of von Neumann inequality [20],
which concerns absolute value of the trace of a product of two matrices and their singular values.
Lemma 1 (von Neumann inequality) Let σ1(A), . . . , σn(A) and σ1(B), . . . , σn(B) denote singular values of the
matrices A and B arranged in nonincreasing order. For any matrices A and B the following inequality holds
|TrAB| ≤
n
∑
i=1
σi(A)σi(B) . (4.12)
For a recent exposition see [21] and [22].
Without loosing the generality we can assume that ρ1 is diagonal, ρ1 = diag(p) and ρ2 = V diag(q)V †. Then
max
V
√
F (ρ1, ρ2) = max
V
Tr|√ρ1
√
ρ2| = max
V
Tr|√pV√qV †|. (4.13)
Using (4.11) and the cyclic property of trace we get
max
V
√
F (ρ1, ρ2) = max
V,U
|TrU√pV√qV †| = max
V,U
|Tr√pV√qV †U | (4.14)
= max
V,W
|Tr√pV√qW | (4.15)

7where W = V †U is unitary. Since the vectors √p and √q contain singular values of matrices √pV and √qW ,
respectively, it follows from (4.12) that
|Tr√pV√qW | ≤
n
∑
i=1
σ↑i (
√
pV )σ↑i (
√
qW ) . (4.16)
Thus we get the bound for the maximal root fidelity at the unitary orbit,
max
V
√
F (ρ1, ρ2) ≤
n
∑
i=1
σ↑i (
√
pV )σ↑i (
√
qW ) (4.17)
=
√
p↑ ·
√
q↑ . (4.18)
This result implies the desired upper bound for the root fidelity,
√
F (ρ1, ρ2) ≤
√
p↑ ·
√
q↑ , (4.19)
which finishes the proof of the lower bound (4.6). Squaring the relations (4.8) and (4.19) we establish the inequalities
(1.9) and (1.10).

Now we are going to formulate and prove an analogous conjecture for the trace distance.
Theorem 4 The maximum and minimum trace distance between the unitary orbits of two states are given by diagonal
states with
M(ρ1, ρ2) = max
P
Dtr(p, q) = Dtr(p↓, q↑) = Dtr(p↑, q↓), (4.20)
and
m(ρ1, ρ2) = min
P
Dtr(p, q) = Dtr(p↓, q↓) = Dtr(p↑, q↑). (4.21)
Proof. The above theorem can be expressed in term of singular values as
n
∑
i=1
|σi(ρ1) − σi(ρ2)| ≤
n
∑
i=1
σi(ρ1 − ρ2) ≤
n
∑
i=1
|σi(ρ1) − σn+1−i(ρ2)|. (4.22)
Here σi(ρ1) and σi(ρ2) denote decreasingly ordered singular values of both operators.
The lower bound follows from the special case (k = n) of the following lemma from [24].
Lemma 2 Let A,B ∈ Mn, and suppose A,B,A − B have decreasingly ordered singular values σ1(A) ≥ . . . ≥
σn(A), σ1(B) ≥ . . . ≥ σn(B), σ1(A − B) ≥ . . . ≥ σn(A − B). Define si(A,B) ≡ |σi(A) − σi(B)| and let
s[1](A,B) ≥ . . . ≥ s[n](A,B) denote a decreasingly ordered rearrangement of the values si(A,B). Then
k
∑
i=1
s[i](A,B) ≤
k
∑
i=1
σi(A−B) for k = 1, 2, . . . , n. (4.23)
The upper bound in (4.22) follows from lemma 5 in appendix B if A and B are positive semidefinite. Since any two
density matrices, ρ1 and ρ2, are hermitian and positive, their eigenvalues and singular values are equal. Making use
of the definition (1.2) we obtain therefore required bounds for the trace distance
2Dtr(p↓, q↓) ≤ Tr|ρ1 − ρ2| ≤ 2Dtr(p↓, q↑) (4.24)
equivalent to eq. (1.11).

We now consider what this means geometrically, and we will do this in terms of the so called Weyl chamber. A
Weyl chamber is a simplex of ordered eigenvalues (see, e.g. [11]). Any unitary orbit is generated from an ordered
spectrum of the density matrix, which corresponds to a point inside a Weyl chamber, i.e. the asymmetric 1/N ! part

8FIG. 2: The minimal distance m between the orbits of unitarily similar states stemming from two quantum states are equal
to the distances between the corresponding spectra a and b belonging to the same Weyl chamber shown for a) N = 2 and b)
N = 3. The maximal distance M is achieved for points a and b′ belonging to the opposite Weyl chambers.
of the simplex of eigenvalues. Thus the minimal distance between a diagonal state ρ1 and a unitary orbit stemming
from ρ2 is obtained if the orbit intersects the Weyl chamber distinguished by ρ1. On the other hand the maximum
is achieved also for a diagonal ρ2 with permuted spectrum, which belongs to another Weyl chamber (see Fig. 2 for
N = 2 and N = 3).
Let us analyze the simplest case N = 2, for which the simplex of eigenvalues is equivalent to an interval [0, 1],
while the intervals [0, 1/2) and (1/2, 1] form two Weyl chambers. A unitary orbit generated by each point of a Weyl
chamber has the structure of the sphere, S2. The above statement has an intuitive interpretation: the minimal
distance between two concentric spheres is equal to the distance between two of their points belonging to the same
radius of the ball. The maximal distance between these spheres equals to the distance between their points placed at
the diameter of the ball on the other sides of its center. For example consider two states in the Bloch ball. The radius
is given by the entropy, in this case completely defining the spectrum also. So two orbits are given by two concentric
spheres of different radius. Common eigenbases corresponds to a common axis, hence the closest and furthest states
both lie on the same axis, either both on the same side or opposite sides of the center respectively.
The above property shows that looking for a set of perfectly distinguishable states in a certain set S of mixed states
which is invariant with respect to the unitary rotations, it is enough to analyze the subset of diagonal matrices.
Proposition 3 Let R∆ be an arbitrary convex subset of the (N − 1) dimensional simplex of the eigenvalues. Let
R denote the set of quantum states obtained from this set by any unitary rotation, R := {ρ ∈ MN : ρ =
U [diag(p)]U †, and p ∈ R∆}. Let k ≤ N be the number such that ∆k−1 ∈ R∆ and there exists no ∆k ∈ R∆.
Then the maximal number of perfectly distinguishable states in R is equal to k, so it is equal to the maximal number
of diagonal distinguishable states.
As before, the symbol ∆k represents a regular k dimensional simplex containing k + 1 points separated by the
maximal distance Dmax with respect to the trace (or Bures) distance. Let us emphasise again that the geometry
induced by the Bures metric differs considerably with respect to the flat Euclidean geometry induced by the HS
metric. For instance, the simplex of eigenvalues for N = 3 forms a flat equilateral triangle (of side
√
2) in the HS
case, while it is equivalent to the octant of a sphere S2 for the Bures distance.
V. CONCLUSIONS
In this work we have commenced with the analysis of the geometry of the problem of quantum distinguishability.
We have shown that the problem of finding the maximal number of perfectly distinguishable states in a certain set R
containing quantum states is equivalent to finding the dimension of the largest simplex of a fixed side size which can
be embedded inside the set R. For this purpose one cannot use Euclidean simplices defined by the HS distance, but
use simplices with respect to Bures or trace distances.
Fidelity between any two quantum states is shown to be bounded by the classical fidelities between both spectra
put in the same order (upper bound) or in the opposite order (lower bound). This observation implies bounds for the
Bures distance between two quantum states are achieved for diagonal states. Thus looking for distinguishable states

9in a rotationally invariant subset of the set of quantum states it is sufficient to restrict analysis to a smaller set of
classical states, which correspond to diagonal density matrices.
Acknowledgments
It is a pleasure to thank I. Bengtsson and P. Horodecki for inspiring discussions and to C.R. Johnson for helpful
correspondence.
We acknowledge financial support by the Polish Ministry of Science and Information Technology and by the Euro-
pean Research Project SCALA. DM acknowledges support from QICS.
This project was also partially funded by Polish Ministry of Science and Higher Education grant number N519 012
31/1957.
APPENDIX A: BOUND FOR THE TRACE OF A PRODUCT OF STATES
Let ρ = ρ† and σ = σ† denote two Hermitian operators acting on an N–dimensional Hilbert space. As throughout
the paper, their spectra will be denoted by p = eig(ρ) and q = eig(σ) respectively. Let p↓, q↓ denote the N -element
vector of eigenvalues ordered in decreasing order, while the same spectra ordered increasingly will be written as p↑
and q↑. The symbol (p↑)s denotes the vector consisting of ordered elements of p↑, each component raised to power s.
Lemma 3 Let ρ ≥ 0 and σ ≥ 0 and let s, t denote positive real numbers. Then
(ps)↑ · (qt)↓ ≤ Tr ρsσt ≤ (ps)↑ · (qt)↑ . (A1)
Proof. Let |µi〉 and |νj〉 denote the eigenvectors of the states ρ and σ. We will start by finding a form of Trρsσt,
in terms of overlaps with a doubly stochastic matrix.
Trρsσt = Tr


∑
i,j
psi q
t
j |µi〉〈µi|νj〉〈νj |

 (A2)
=
∑
i,j
psi q
t
j |〈µi|U |µj〉|2 (A3)
=
∑
i,j
psi q
t
jBi,j , (A4)
where U is the unitary relating the two eigenbases U |µi〉 = |νi〉, ∀i and B :=
∑
i,j |Ui,j |2|µi〉〈µj | so that Bij = |Uij |2.
Hence matrix B is by construction unistochastic [25] and thus bistochastic.
It is convenient to introduce at this place two non-normalised vectors, |ψ〉 := ∑ p′i|µi〉, and |φ〉 :=
∑
q′j |µj〉, where
p′i = psi and q′j = qtj are non–negative. Then the trace can be rewritten in the form
Trρsσt = 〈ψ|B|φ〉. (A5)
Birkhoff’s theorem [23] states that any doubly stochastic matrix can be written as a finite convex combination of
permutation matrices Oi, hence we write B =
∑
i riOi,
∑
i ri = 1. Thus the extremum of a linear function of the
bistochastic matrix B will be realized at one of its extremal points. There are exactly N ! of them, and among all
possible permutations Oi the maximum is obtained if the orders of elements of both vectors are the same, while the
minimum is achieved if both spectra are in opposite order,
〈ψ|B|φ〉 =
∑
i
ri〈ψ|Oi|φ〉 ≥ 〈ψ|Omin|φ〉 = (p↑)s · (q↓)t (A6)
≤ 〈ψ|Omax|φ〉 = (p↑)s · (q↑)t . (A7)
Since all components of the vector p (and q) are non–negative raising each element to a positive exponent s (or t)
will not change the order of a vector, (p↑)s = (ps)↑. Putting it all together we arrive at (A1) and complete the proof.


10
For concreteness let us write down explicitly some special cases. In the simplest case s = t = 1 one obtains
p↑ · q↓ ≤ Trρσ ≤ p↑ · q↑ , (A8)
while setting s = t = 1/2 one becomes inequality (4.7) used in the proof of inequality (4.9).
An analogue of lemma 2 may be obtained in the case one of the two operators is not positive.
Lemma 4 Consider a positive number s > 0 a state ρ ≥ 0 and an Hermitian operator σ = σ† not necessarily positive.
Then
(ps)↑ · q↓ ≤ Trρsσ ≤ (ps)↑ · q↑ . (A9)
Proof of this lemma is similar to the proof of lemma 2. In this case the vector q of eigenvalues of operator σ
contains in general also negative entries, so the vector |φ〉 := ∑j qj |µj〉, is given by a pseudomixture with some
weights negative. Constructing unitary bases U and bistochastic matrix M one may write the analyzed trace in the
form (A5) and make use of the Birkhoff theorem. Since operator ρ with spectrum p is positive, raising its components
to a positive power will not change the order, (p↑)s = (ps)↑. Therefore we may perform the last step analogous to
(A7) obtaining the desired result.

APPENDIX B: BOUND FOR THE TRACE OF A DIFFERENCE OF TWO STATES
In this appendix we prove the following lemma.
Lemma 5 Let A and B denote hermitian matrices of size n. Let us order decreasingly their eigenvalues, λ1(A) ≥
. . . ≥ λn(A) and λ1(B) ≥ . . . ≥ λn(B). Then the following upper bound for the trace of the absolute value of the
difference holds
Tr|A−B| =
n
∑
i=1
σi(A−B) ≤
n
∑
i=1
|λi(A) − λn+1−i(B)|. (B1)
Proof. Let us express both operators in their eigen representation, A = ∑ni pi|µi〉〈µi| and B =
∑n
i qi|νi〉〈νi|,
where for convenience we have introduced the notation pi = λi(A) and qi = λi(B). Making use of Eq. (4.11) and
basic properties of the trace we get
Tr|A−B| = max
U
|TrAU − TrBU | (B2)
= max
U
|
n
∑
i=1
pi〈µi|U |µi〉 − qi〈νi|U |νi〉|. (B3)
Since |〈µi|U |µi〉| ≤ 1, |〈νi|U |νi〉| ≤ 1 and TrU =
∑n
i=1〈µi|U |µi〉 =
∑n
i=1〈νi|U |νi〉 we have
Tr|A−B| ≤ max
{
|
n
∑
i=1
ξipi − ζiqi| : |ξi| ≤ 1, |ζi| ≤ 1 for i = 1, . . . , n ,
n
∑
i=1
ξi =
n
∑
i=1
ζi
}
. (B4)
For fixed values of ξi and ζi we denote s =
∑n
i=1 ξipi − ζiqi. Let s = ceiϕ, we have
c = |s| = | s
eiϕ
| = |
n
∑
i=1
ξi
eiϕ
pi −
ζi
eiϕ
qi|.
Because | ξieiϕ | ≤ 1 and |
ζi
eiϕ | ≤ 1 we can without loss of generality assume that s ∈ IR. Note now that under this
assumption we have
max
{
|
n
∑
i=1
ξipi − ζiqi| : |ξi| ≤ 1, |ζi| ≤ 1 for i = 1, . . . , n ,
n
∑
i=1
ξi =
n
∑
i=1
ζi,
}
(B5)
= max
{
|
n
∑
i=1
Re(ξi)pi −Re(ζi)qi| : |ξi| ≤ 1, |ζi| ≤ 1 for i = 1, . . . , n ,
n
∑
i=1
ξi =
n
∑
i=1
ζi,
}
(B6)
= max
{
|
n
∑
i=1
ξipi − ζiqi| : −1 ≤ ξi ≤ 1,−1 ≤ ζi ≤ 1 for i = 1, . . . , n ,
n
∑
i=1
ξi =
n
∑
i=1
ζi,
}
(B7)

11
The term ∑ni=1 ξipi − ζiqi is a linear function of 2n variables ξ1, . . . ξn, ζ1, . . . , ζn, so it reaches its extreme value at
the edges of the polygon defined by
{
−1 ≤ ξi ≤ 1,−1 ≤ ζi ≤ 1 for i = 1, . . . , n ,
n
∑
i=1
ξi =
n
∑
i=1
ζi
}
. (B8)
Thus we can focus on the edges of the polygon
{
ξi ∈ {−1, 1}, ζi ∈ {−1, 1} for i = 1, . . . , n ,
n
∑
i=1
ξi =
n
∑
i=1
ζi
}
. (B9)
Note that we obtain the maximum if in the sum ∑ni=1 ξipi+(−ζi)qi the n maximum values of {p1, . . . , pn, q1, . . . , qn}
will be equipped with +1 coefficient and n minimum values with −1. Because p1 ≥ p2 ≥ · · · ≥ pn and q1 ≥ q2 ≥ · · · ≥
qn, we can thus write the n maximum values as
max{p1, qn},max{p2, qn−1}, . . . ,max{pn, q1}, (B10)
and the n minimum values as
min{p1, qn},min{p2, qn−1}, . . . ,min{pn, q1}. (B11)
So the maximum value of
max
{
|
n
∑
i=1
ξipi − ζiqi| : ξi ∈ {−1, 1}, ζi ∈ {−1, 1} for i = 1, . . . , n ,
n
∑
i=1
ξi =
n
∑
i=1
ζi
}
(B12)
is equal
|
n
∑
i=1
max{pi, qn−i+1} − min{pi, qn−i+1}| =
n
∑
i=1
|pi − qn−i+1|. (B13)
This gives us required upper bound (B1).

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