Quantum Relative Lorenz Curves
Francesco Buscemi1 and Gilad Gour2
1

Department of Computer Science and Mathematical Informatics,
Nagoya University, Chikusa-ku, Nagoya, 464-8601, Japan∗
2
Institute for Quantum Science and Technology and Department of Mathematics and Statistics,
University of Calgary, 2500 University Drive NW, Calgary, Alberta, Canada T2N 1N4†

arXiv:1607.05735v2 [quant-ph] 10 Jan 2017

The theory of majorization and its variants, including thermomajorization, have been found to
play a central role in the formulation of many physical resource theories, ranging from entanglement
theory to quantum thermodynamics. Here we formulate the framework of quantum relative Lorenz
curves, and show how it is able to unify majorization, thermomajorization, and their noncommutative analogues. In doing so, we define the family of Hilbert α-divergences and show how it relates
with other divergences used in quantum information theory. We then apply these tools to the problem of deciding the existence of a suitable transformation from an initial pair of quantum states to
a final one, focusing in particular on applications to the resource theory of athermality, a precursor
of quantum thermodynamics.

I.

INTRODUCTION

Lorenz curves, originally introduced to give a quantitative and pictorially clear representation of the inequality
of the wealth distribution in a country [1], have since then
been used also in other contexts in order to effectively
compare different distributions (see for example [2] and
references therein). In their typical formulation, Lorenz
curves fully capture the notion of “nonuniformity” [3] of
a distribution, in the sense that comparing the Lorenz
curves associated to two given distributions (say, p and
q) induces an ordering equivalent to the relation of majorization, which, in turns, is well-known to be equivalent to the existence of a random permutation (i.e., a
bistochastic channel) transforming p into q [2].
More recently, some variants of the original definition
were proposed in order to capture other aspects of a
given distribution, besides its mixedness. In particular, thermomajorization was introduced in [7] to characterize state transitions under thermal operations or
Gibbs preserving operations [8]. Here, the corresponding Lorenz curve characterizes a partial ordering relative
to the Gibbs distribution, rather than the uniform one.
This suggests that Lorenz curves are best understood
not as properties of one given distribution, but rather
of a given pair of distributions, one being the “state”
at hand and the other being the “reference”. For example, the original Lorenz curve contains information about
a given distribution p with respect to the uniform one:
it is in this precise sense, then, that the Lorenz curve
characterizes the degree of nonuniformity of p—exactly
because the reference distribution is chosen to be the uniform one. In the same way, thermomajorization measures
the degree of “athermality” because, in this case, the reference distribution is chosen to be the thermal (Gibbs)
distribution.

∗
†

buscemi@is.nagoya-u.ac.jp
gour@ucalgary.ca

A lot of attention has been devoted recently to the
generalization of the above ideas to the case in which,
rather than comparing distributions, one wants to compare quantum states, namely, density operators defined
on a Hilbert space. This is one of the topics lying at
the core of theories like quantum thermodynamics and,
more generally, quantum resource theories [4–6]. However, a general theory of quantum Lorenz curves would
be interesting in its own right, providing new insights on
the rich analogies existing between quantum theory and
classical probability theory, despite their differences.
In this paper we develop such a theory by introducing the notions of quantum testing region, quantum relative Lorenz curves, and quantum relative majorization
in much analogy with their classical counterparts. We
find equivalent conditions for quantum relative majorization in terms of a new family of divergences that we
call Hilbert α-divergences, with α ∈ (1, ∞), and show
that in the limits α → 1 and α → ∞ the Hilbert
α-divergences are equivalent to the trace-distance and
the max-relative entropy, respectively. As an application to quantum thermodynamics, we show that only the
min- and max-relative entropies are needed to determine
whether it is possible to convert one qubit athermality
resource to another by Gibbs preserving operations. Finally, we show that in higher dimensions, quantum relative Lorenz curves can be used to determine the existence of a test-and-prepare channel converting one pair
of states to another.

II.

QUANTUM RELATIVE LORENZ CURVES

Consider the task of distinguishing which, among two
possible distributions, is the one that originated a set of
observed sample data. This scenario, central in statistics, is usually treated within the framework known as
hypothesis testing [9]: the two distributions are called
the null hypothesis and the alternative hypothesis, respectively, and the task of the statistician is to minimize the
so-called type II error (i.e., the probability of wrongly

2
accepting the null hypothesis, namely, the probability of
false negatives) given that the type I error (i.e., the probability of wrongly rejecting the null hypothesis, namely,
the probability of false positives) falls below a certain
threshold. The whole hypothesis testing problem is hence
“encoded” in the shape of the region of the xy-plane containing all achievable points (x, y) = (type I, type II).
Such a region is, by construction, convex, always contains the points (0, 0) and (1, 1), and is symmetric, in
the sense that (x, y) belongs to the region if and only
if (1 − x, 1 − y) does, as this corresponds to exchanging
the roles of null and alternative hypotheses (see Fig. 1).
Hence, the hypothesis testing is fully characterized by the
upper boundary of the region. In particular, as noticed
by Renes [10], when testing p against the uniform distribution, such boundary coincides with the usual Lorenz
curve; when testing p against the Gibbs distribution, it
coincides with the thermomajorization curve.
The observations in [10] exhibit a fundamental connection between the theory of (thermo)majorization and
hypothesis testing. It is then extremely natural for us
here to introduce the definition of Lorenz curves for pair
of quantum states, leveraging on the fact that hypothesis
testing is well understood in the quantum case too [11–
15]:
Definition 1. Given two density matrices ρ1 and ρ2 on
Cn , the associated testing region T (ρ1 , ρ2 ) ⊂ R2 is defined as the set of achievable points
(x, y) = (Tr[Eρ2 ], Tr[Eρ1 ]) ,
with 0 6 E 6 1n . The quantum Lorenz curve of ρ1 relative to ρ2 is defined as the upper boundary of T (ρ1 , ρ2 ),
see Fig. 2.
Closely related to the testing region, the hypothesis
testing relative entropy (see e.g. [16–18] and references
therein) is defined, for 0 6  6 1, as follows:

DH
(ρ1 kρ2 ) , − log Q (ρ1 kρ2 )

Q (ρ1 kρ2 ) ,

min

06E61n
Tr[ρ1 E]>1−

Tr[ρ2 E].

(1)

As noted in Ref. [17], the computation of Q (ρ1 kρ2 ) can
be solved efficiently by semidefinite linear programming
(SDP). In fact, in what follows (see Eqs. (27) and (28
in Section IV) we show that, using the strong duality
relation of SDP, it is possible to write Q (ρ1 kρ2 ), for any
fixed , as the maximum of a simple function of one real
variable, namely, Q (ρ1 kρ2 ) = maxr>0 f (r), where
f (r) , (1 − )r − Tr(rρ1 − ρ2 )+
i
1h
= 1 + (1 − 2)r − ||rρ1 − ρ2 ||1 .
2

(2)

This observation will play an important role in what follows, by considerably simplifying our analysis.
The above definition of relative Lorenz curve generalizes the classical Lorenz curve to the quantum case. In

FIG. 1. Example of classical testing region in dimension n = 4
with p
~1 = (1/2, 1/4, 1/4, 0)T and p
~2 = (1/6, 1/6, 1/3, 1/3)T .
The Lorenz curve (i.e. the upper boundary) is determined by
the vertices at which the Lorenz curve changes slope.

particular, if ρ1 and ρ2 commute, they can be simultaneously diagonalized, and the testing region in this case
becomes the collection of points
Tcl (p~1 , p~2 ) , {(~t · p~1 , ~t · p~2 ) : ~t ∈ Rn+ , ~t 6 (1, 1, ..., 1)T },
where p~1 and p~2 are the diagonals of ρ1 and ρ2 written
in a vector form. In this case, Blackwell proved a very
strong relation [19, 20]: given two pairs of distributions
(~
p1 , p~2 ) and (~q1 , ~q2 ), the inclusion Tcl (~q1 , ~q2 ) ⊆ Tcl (~
p1 , p~2 )
holds if and only if there exists a column stochastic matrix M such that ~q1 = M p~1 and ~q2 = M p~2 . Known
results about classical (thermo) majorization are therefore special cases of Blackwell’s theorem, even though
Blackwell’s work actually predates some of them (see the
discussion in Refs. [10, 26]).
In the rest of the paper we explore the extent to which
statements similar to Blackwell’s theorem can be proved
in the quantum case. However, our interest here does not
lie as much in the general case, for which we know that
many classical results cease to hold [21–26], but rather in
restricted scenarios of practical relevance, especially for
the growing field of quantum resource theories.
III.

HILBERT α-DIVERGENCES

In analogy with the notation used for majorization, we
write
(ρ1 , ρ2 )  (ρ01 , ρ02 )
and say that (ρ1 , ρ2 ) relatively majorizes (ρ01 , ρ02 ), whenever the quantum Lorenz curve of ρ1 relative to ρ2 lies
everywhere above the quantum Lorenz curve of ρ01 relative to ρ02 , that is, T (ρ1 , ρ2 ) ⊇ T (ρ01 , ρ02 ).

3
ii) for all α > 1, the data-processing inequality holds:
for any (not necessarily completely) positive tracepreserving map Φ, Hα (Φ(ρ)kΦ(σ)) 6 Hα (ρkσ);
iii) H∞ (ρkσ) , limα→∞ Hα (ρkσ) = Dmax (ρkσ),
namely, the max-relative entropy of Ref. [30];
iv) H1 (ρkσ) , limα→1 Hα (ρkσ) =

1
2 ln(2)

||ρ − σ||1 .

Remark 1. Hα is thus a family of divergences connecting
the trace-distance (when α → 1) with Dmax (when α →
∞). In passing by, we also notice that, while in point (ii)
above the data-processing inequality is stated to hold for
any positive trace-preserving map, Hilbert α-divergences
are in fact monotonically decreasing for an even larger set
of transformations, called 2-statistical morphisms: while
this point is outside the scope of the present work, we
refer the interested reader to Refs. [22–24, 31].
FIG. 2. Numerical example of the quantum testing region for
two random four-dimensional density matrices. Notice that
the curve is only roughly approximated as the sampled measurements are not enough to determine it neatly. Quantum
Lorenz curves are efficiently obtained by semi-definite linear
programming, e.g., using Eq. (2) in the main text.

As a tool to characterize quantum relative majorization, we introduce here a family of divergences as follows:
given two density matrices ρ and σ on Cn , we define, for
all α > 1, the following quantity:
Tr[Eρ]
,
Tr[Eσ]

(3)

α
log2 supα (ρ/σ).
α−1

(4)

supα (ρ/σ) ,

sup
α−1 1n 6E61n

Proof. Properties (ii) and (iii) are direct consequences of
the definition of supα (ρ/σ).
In order to prove property (iv), we start by taking
α > 1 and defining the following two quantities:  , α−1
and δ , sup1+ (ρ/σ) − 1. With these notations, from
the definition of supα (ρ/σ) we obtain
δ >
for all

1
1+ 1

Tr[Eρ]
Tr [E(ρ − σ)]
−1=
,
Tr[Eσ]
Tr [Eσ]

6 E 6 1. Introducing the operator
1
[(1 + )E − 1] ,


∆,

and the corresponding divergence:
Hα (ρkσ) ,

(5)

(6)

we get that Eq. (5) is equivalent to
δ >

The notation used in Eq. (3) is adapted from Refs. [27–
29]: there the quantity
sup(ρ/σ) , inf{λ : λσ − ρ > 0}
= lim supα (ρ/σ)
α→∞

is used to define the Hilbert projective metric

Theorem 1. Let ρ and σ be two density matrices on Cn .
Then:
i) for all α > 1, Hα (ρkσ) > 0, with equality if and
only if ρ = σ;

(7)

for all 0 6 ∆ 6 1. Hence, lim→0 δ = 0. We therefore
have
1
1+
lim
ln(1 + δ )
ln(2) →0 
1
1
=
lim δ
ln(2) →0 
1  Tr [∆(ρ − σ)]
1
lim
>
ln(2) →0  1 +  Tr [∆σ]
1
=
Tr [∆(ρ − σ)] ,
ln(2)

lim H1+ (ρkσ) =

→0

h(ρ, σ) , ln[sup(ρ/σ) sup(σ/ρ)].
We note that, in Ref. [29], the quantity inf(ρ/σ) is also
introduced, as sup{λ : ρ − λσ > 0}: in our notation it coincides with inf 06E61n {Tr[Eρ]/ Tr[Eσ]} = 1/ sup(σ/ρ).
Due to the relation with the Hilbert’s metric, we refer
to the divergences in Eq. (4) as Hilbert α-divergences.
Their main properties are summarized in the following
theorem:

 Tr [∆(ρ − σ)]
,
1 +  Tr [∆σ]

(8)

for all 0 6 ∆ 6 1. We therefore conclude that
H1 (ρkσ) >

1
1
Tr[(ρ − σ)+ ] =
||ρ − σ||1
ln(2)
2 ln(2)

(9)

where we chose ∆ to be the projection to the positive
1
part of ρ − σ. To see that H1 (ρkσ) = 2 ln
2 ||ρ − σ||1 note

4
that, in fact, by definition
δ =

max

1
1+ 16E61

= max

06∆61

Tr [E(ρ − σ)]
Tr [Eσ]

 Tr [∆(ρ − σ)]
1 +  Tr [∆σ]

Proof. The first equivalence holds by definition. Denoting by (p, p̄) and (q, q̄) the generic element of T (ρ1 , ρ2 )
and T (ρ01 , ρ02 ), respectively, the Separation Theorem for
convex sets, applied to T (ρ1 , ρ2 ) and T (ρ01 , ρ02 ), states
that T (ρ1 , ρ2 ) ⊇ T (ρ01 , ρ02 ) if and only if, for any v =
(a, b) ∈ R2 ,
max

=  max Tr [∆(ρ − σ)] + O(2 )

(p,p̄)∈T (ρ1 ,ρ2 )

06∆61

Hence, in the limit  → 0 we get lim→0 1 δ = Tr[(ρ −
σ)+ ] = 21 ||ρ − σ||1 .
Finally, property (i) is proved as follows. Since
supα (ρkσ) > 1 , we always have Hα (ρkσ) > 0. For
α > 1 if Hα (ρkσ) = 0 then supα (ρkσ) = 1. Hence,
Tr[Eρ] 6 Tr[Eσ] for all α−1 1 6 E 6 1. Introducing
α
1
∆,
(E − 1),
α−1
α
we get Tr[∆ρ] 6 Tr[∆σ], namely, Tr[∆ (ρ − σ)] 6 0, for
all 0 6 ∆ 6 1. We therefore must have ρ = σ. The case
α = 1 follows from property (iv).

IV.

RELATIVE MAJORIZATION AS SETS OF
INEQUALITIES

[ap + bp̄] >

Theorem 2. Consider two pairs of density matrices
(ρ1 , ρ2 ) on Cn and (ρ01 , ρ02 ) on Cm . The following are
equivalent:
i) (ρ1 , ρ2 )  (ρ01 , ρ02 );
ii) for all t > 0, ||ρ1 − tρ2 ||1 > ||ρ01 − tρ02 ||1 ;
iii) for all α > 1,
(
Hα (ρ1 kρ2 ) > Hα (ρ01 kρ02 ),
Hα (ρ2 kρ1 ) > Hα (ρ02 kρ01 );


(ρ01 kρ02 ).
(ρ1 kρ2 ) > DH
iv) for all 0 6  6 1, DH

We split the proof into several lemmas.
Lemma 1. Given two pairs of density operators (ρ1 , ρ2 )
and (ρ01 , ρ02 ) on Cn and Cm , respectively, the following
are equivalent:
i) (ρ1 , ρ2 )  (ρ01 , ρ02 );
ii) T (ρ1 , ρ2 ) ⊇ T (ρ01 , ρ02 );
iii) ||t1 ρ1 + t2 ρ2 ||1 > ||t1 ρ01 + t2 ρ02 ||1 for all t1 , t2 ∈ R;
iv) ||ρ1 − tρ2 ||1 > ||ρ01 − tρ02 ||1 , for all t > 0.

[aq + bq̄] .

(10)

a + b + ||aρ1 − bρ2 ||1
,
2

(11)

The next step is to show that
max

[ap + bp̄] =

(p,p̄)∈T (ρ1 ,ρ2 )

and, analogously, for (ρ01 , ρ02 ). This is done by the following simple passages:
max
(p,p̄)∈T (ρ1 ,ρ2 )

[ap + bp̄] = max {a Tr[ρ1 E] + b Tr[ρ2 E]}
06E61

= max Tr[(aρ1 + bρ2 ) E]
06E61

= Tr (aρ1 + bρ2 )+ ,
(12)
where the last expression denotes the positive part of the
self-adjoint operator aρ1 + bρ2 . Then, since 2 Tr(A)+ =
||A||1 + Tr[A] for any self-adjoint operator, we have that
2

max
(p,p̄)∈T (ρ1 ,ρ2 )

We are now in a position to provide a set of alternative conditions, reformulating the relative majorization
ordering (ρ1 , ρ2 )  (ρ01 , ρ02 ) as sets of inequalities.

max

(q,q̄)∈T (ρ01 ,ρ02 )

[ap + bp̄] = a + b + ||aρ1 + bρ2 ||1 .

(13)

This proves that Eq. (10) is satisfied if and only if
||aρ1 + bρ2 ||1 > ||aρ01 + bρ02 ||1 , for all a, b ∈ R.
We are left to prove that (iii) is equivalent to (iv).
However, since (iv) is a special case of (iii), we only need
to prove that (iv) implies (iii). To this end, we notice
that, whenever t1 , t2 > 0 or t1 , t2 6 0, ||t1 ρ1 + t2 ρ2 ||1 =
||t1 ρ01 − t2 ρ02 ||1 always, simply due to the positivity of
ρ1 , ρ2 , ρ01 , ρ02 . We can hence consider only the cases t1 >
0 > t2 or t2 > 0 > t1 . However, since ||X||1 = ||−X||1 , for
any matrix X, we can further restrict the parameters t1
and t2 to the case t2 < 0 < t1 . The statement is finally
obtained by rescaling both t1 and t2 by the (positive)
factor 1/t1 .
Lemma 1 above shows that statements (i) and (ii) of
Theorem 2 are indeed equivalent. We now move on to
proving the equivalence of the point (iii). We begin with
the following lemma.
Lemma 2. For any choice of density operators ρ and σ,


||λσ − ρ||1
α+1
.
supα (ρ/σ) = inf λ > 1 :
6
λ−1
α−1
Proof. Note first that
supα (ρ/σ)
,

sup

{Tr[Eρ]/Tr[Eσ]}

α−1 16E61

= inf{λ : λ > Tr[Eρ]/Tr[Eσ] for all α−1 1 6 E 6 1}

= inf λ : Tr [E (λσ − ρ)] > 0 for all α−1 1 6 E 6 1





= inf λ ∈ R : α−1 Tr (λσ − ρ)+ > Tr (λσ − ρ)−

,

5
where, in the last equality, we used the decomposition
A = A+ − A− for Hermitian operators and the choice
E = α−1 Π+ + Π− , being Π± the projectors onto the positive and negative parts of (λσ − ρ), respectively. Indeed,
this is choice for the operator E that poses the toughest
constraints compatible with the fixed value of the parameter α. (Equivalently, if Tr[E(λσ − ρ)] > 0 for such
a choice of E, then it is positive for any α−1 1 6 E 6 1.)
Then, using the relations




λ − 1 = Tr (λσ − ρ)+ − Tr (λσ − ρ)−
and
||λσ − ρ||1 = Tr[(λσ − ρ)+ ] + Tr[(λσ − ρ)− ]
gives




 λ−1
supα (ρkσ) = inf λ ∈ R : Tr (λσ − ρ)− 6
.
α−1


λ−1
Then, since Tr (λσ − ρ)− 6 α−1
if and only if
λ−1+||λσ−ρ||1

λ−1
λ−1
||λσ − ρ||1 6 α−1
+ Tr[(λσ − ρ)+ ] = α−1
+
2
after an easy manipulation we obtain


||λσ − ρ||1
α+1
6
.
supα (ρ/σ) = inf λ ∈ R :
λ−1
α−1

,

Lemma 4. Consider two pairs of states (ρ, σ) and
(ρ0 , σ 0 ). Then, the following are equivalent:
i) for all α > 1, supα (ρ/σ) > supα (ρ0 /σ 0 ) and
inf α (ρ/σ) 6 inf α (ρ0 /σ 0 );
ii) ||tσ − ρ||1 > ||tσ 0 − ρ0 ||1 for all t > 0;
Proof. We only need to show that the condition
supα (ρ/σ) > supα (ρ0 /σ 0 ) for all α > 1 is equivalent to ||tσ − ρ||1 > ||tσ 0 − ρ0 ||1 for all t > 1. Then,
if this holds, the remaining statement, namely, that
inf α (ρ/σ) 6 inf α (ρ0 /σ 0 ) for all α > 1 is equivalent to
||tσ − ρ||1 > ||tσ 0 − ρ0 ||1 for all t ∈ [0, 1], simply follows
from the definitions.
Set Mα ≡ supα (ρ/σ), Mα0 ≡ supα (ρ0 /σ 0 ), and recall
the definition of f in Lemma 3. Then, as noticed in
Remark 2 above, it follows that
f (Mα ) =

||Mα0 σ 0 − ρ0 ||1
||Mα σ − ρ||1
α+1
=
=
. (14)
Mα − 1
α−1
Mα0 − 1

Since f is monotonically non-increasing, we get that
Mα > Mα0 implies that
||Mα0 σ 0 − ρ0 ||1
||Mα σ 0 − ρ0 ||1
>
.
0
Mα − 1
Mα − 1

(15)

The statement is finally recovered by noticing that no loss
of generality comes from restricting λ to values greater
than or equal to 1.

Combining the above two equations gives

Lemma 3. For any choice of density operators ρ and σ,
the function

We now make the simple observation that, by definition, the function supα (ρ/σ) = Mα is continuous and
monotonically nondecreasing in α, with supα=1 (ρ/σ) = 1
and supα→∞ (ρ/σ) = sup(ρ/σ). Hence,

f (λ) =

||λσ − ρ||1
λ−1

is monotonically non-increasing in the domain λ > 1 with
f (1) = ∞ and f (∞) = 1.
Remark 2. In particular, Lemma 2 and Lemma 3 above
imply that, for any pair of density operators ρ and σ,
||supα (ρ/σ)σ − ρ||1
α+1
=
.
supα (ρ/σ) − 1
α−1
Proof. Set t = 1/(λ − 1) and define g(t) =
||(1 + t)σ − tρ||1 . Hence, g(t) = f (λ), and it is enough
to show that g(t) is monotonically non-decreasing in its
domain t ∈ [0, ∞). First note that for any 0 < p < 1 and
t1 , t2 ∈ R+ we have
g (pt1 + (1 − p)t2 )
= ||p(1 + t1 )σ − pt1 ρ + (1 − p)(1 + t2 )σ − (1 − p)t2 ρ||1
6 ||p(1 + t1 )σ − pt1 ρ||1 + ||(1 − p)(1 + t2 )σ − (1 − p)t2 ρ||1
= pg(t1 ) + (1 − p)g(t2 ) .
Hence g(t) is a convex function. Moreover, note that
g(0) = 1 6 g(t) for all t > 0. These two properties of g(t)
together imply that it is monotonically non-decreasing in
t.

||Mα σ − ρ||1 > ||Mα σ 0 − ρ0 ||1

||tσ − ρ||1 > ||tσ 0 − ρ0 ||1 ,

∀ α > 1.

1 6 ∀t 6 sup(ρ/σ) ,

(16)

(17)

and the above is enough to conclude that the same ordering holds in fact for all t > 1.
Conversely, suppose the inequality above holds for all
t > 1. This implies that, if
||λσ − ρ||1
α+1
6
,
λ−1
α−1

(18)

||λσ 0 − ρ0 ||1
α+1
6
.
λ−1
α−1

(19)

then also

But from Lemma 2 this implies that supα (ρ0 /σ 0 ) 6
supα (ρ/σ).
Lemma 4 above hence proves the equivalence of
point (ii) and point (iii) of Theorem 2, because
supα (ρ1 /ρ2 ) > supα (ρ01 /ρ02 ) if and only if Hα (ρ1 kρ2 ) >
Hα (ρ01 kρ02 ), and inf α (ρ1 /ρ2 ) 6 inf α (ρ01 /ρ02 ) if and only if
Hα (ρ2 kρ1 ) > Hα (ρ02 kρ01 ).
The proof of Theorem 2 is complete if we prove the
equivalence of the remaining point (iv). Also in this case,

6

rather than proving the statement for DH
, we will prove


it for the corresponding Q , related with DH
as given in
Eq. (1) of the main text. We recall the definition: given
two density operators ρ1 and ρ2 on Cn , for any  ∈ [0, 1],

Q (ρ1 kρ2 ) =

min

06A61n
Tr[ρ1 A]>1−

Tr[ρ2 A].

=

Lemma 5. Given two pairs of density operators (ρ1 , ρ2 )
and (ρ01 , ρ02 ) on Cn and Cm , respectively, the following
are equivalent:


i) Q (ρ1 kρ2 ) 6 Q
ii) ||ρ2 − rρ1 ||1 >

(ρ01 kρ02 ),

||ρ02

−

for all  ∈ [0, 1];

rρ01 ||1 ,

max

hv1 , xi1 ,

max

x∈K1
v2 −Γ(x)∈K2

(21)

hv2 , yi2 ,

(22)

where Γ∗ is defined by the relation hy, Γ(x)i = hΓ∗ (y), xi,
for all x ∈ K1 and all y ∈ K2 .
In our case, denote
n
o
V1 , R ⊕ Hn = (r, A) r ∈ R ; A ∈ Hn
with inner product
h(r, A), (t, B)i1 , rt + Tr[AB] .
Further, define K1 = R+ ⊕ Mn,+ to be the positive cone
in V1 . Similarly, set V2 = Hn and K2 = Mn,+ . The linear
map Γ : V1 → V2 is given by:
(23)

∗

hence, the corresponding dual map Γ : V2 → V1 is given
by
Γ∗ (B) = (Tr[ρ1 B], −B) .

(24)

Finally, set v1 = (1 − , −1n ) and v2 = ρ2 .
Since, for these choices, y ∈ K2 if and only if y > 0 and
Γ∗ (y) − v1 ∈ K1 if and only if Tr[ρy] > 1 −  and y 6 1n ,
the dual form (22) becomes exactly the right-hand side
of Eq. (20), namely,
min

y∈K2
Γ (y)−v1 ∈K1
∗

hv2 , yi2 = Q (ρ1 kρ2 ) .

(26)

hv1 , xi1 = max {(1 − )r − Tr(rρ1 − ρ2 )+ } ,
r>0

Q (ρ1 kρ2 ) = max f (r),

(27)

f (r) , (1 − )r − Tr(rρ1 − ρ2 )+ .

(28)

where

Note that
Tr(rρ1 − ρ2 )+ =

(25)

||rρ1 − ρ2 ||1 + r − 1
,
2

that is,
f (r) =

Γ(r, A) = rρ1 − A ;

{(1 − )r − Tr[A]} .

r>0

the dual form involves the adjoint map Γ : V2 → V1 :
min

A>rρ1 −ρ2
r,A>0

The right-hand side of the above equation can be further
simplified as follows. We first fix r and optimize over
A. Since the A > 0 with minimum trace such that A >
rρ1 − ρ2 is exactly A = (rρ1 − ρ2 )+ , we conclude that

∗

y∈K2
Γ∗ (y)−v1 ∈K1

max

hv1 , xi1

that is,

for all r > 0.

Proof. Consider the following setting of linear programming. Let V1 and V2 be two (inner product) vector spaces
with two cones K1 ⊂ V1 and K2 ⊂ V2 . Consider two vectors v1 ∈ V1 and v2 ∈ V2 , and a linear map Γ : V1 → V2 .
Given a problem in its primal form:
x∈K1
v2 −Γ(x)∈K2

max

x∈K1
v2 −Γ(x)∈K2

(20)

For later convenience, we introduce the following notation: Hn to denote the set of n-by-n Hermitian matrices
on Cn , and Mn,+ to denote the set of n-by-n complex
positive semi-definite matrices.



For the primal form, since x = (r, A) ∈ K1 if and only
if r > 0 and A > 0, and v2 − Γ(x) ∈ K1 if and only if
ρ2 − rρ1 + A > 0, we obtain

1 + (1 − 2)r − ||rρ1 − ρ2 ||1
.
2

Therefore, denoting f0 (r) = 2−1 {1 + (1 − 2)r −
||rρ01 − ρ02 ||1 }, we have that
||rρ1 − ρ2 ||1 > ||rρ01 − ρ02 ||1 =⇒ f (r) 6 f0 (r),
independently of r and . We thus have proved that (ii)
implies (i).
To show that (i) implies (ii), suppose Q (ρ1 kρ2 ) 6
 0
Q (ρ1 kρ02 ) for all  ∈ [0, 1]. Let r > 0 be the minimum
value of r achieving Q (ρ01 kρ02 ), in formula,
r , min{r > 0 : f0 (r) = Q (ρ01 kρ02 )}.

(29)

In all such points r , definition (27) together with the assumption (i) guarantee that ||r ρ1 − ρ2 ||1 > ||r ρ01 − ρ02 ||1 .
This fact can be simply shown by the following chain of
inequalities:
1 + (1 − 2)r − ||r ρ01 − ρ02 ||1
2
= Q (ρ01 kρ02 )
> Q (ρ1 kρ2 )
1 + (1 − 2)r − ||rρ1 − ρ2 ||1
= max
r
2
1 + (1 − 2)r − ||r ρ1 − ρ2 ||1
>
.
2

7
The crucial observation now is that the points r , representing the solutions of (29) for varying  ∈ [0, 1], coincide
with the points where the quantity ||rρ01 − ρ02 ||1 , thought
as a function of r, changes its slope (see Fig. 3 below).
For example, for  = 0, we have to consider the function
f00 (r) =

Due to the linearity in r of the expression above we conclude that in the classical case
Q (ρkσ) =

f (rk )



k
k


X
X
1 −  −
= max
pj  rk +
qj ,

k∈{1,...,n} 

1 + r − ||rρ01 − ρ02 ||1
,
2

and this achieves its maximum value 1
r0 = sup(ρ02 /ρ01 ). But then, if we know
||rρ1 − ρ2 ||1 is not below ||rρ01 − ρ02 ||1 in
where the latter changes its slope, this
conclude that
||rρ01 − ρ02 ||1 6 ||rρ1 − ρ2 ||1 ,

j=1

for r > r∗ ≡
that the curve
all the points
is sufficient to

V.

FIG. 3. Typical behavior of ||rρ1 − ρ2 ||1 , for two random density matrices ρ1 and ρ2 on C3 , as a function of r ∈ R (continuous line). For r 6 r∗ ≡ inf(ρ2 /ρ1 ) = sup{λ : λρ1 − ρ2 6 0},
the curve becomes equal to 1 − r (dashed line). For r > r∗ ≡
sup(ρ2 /ρ1 ) = inf{λ : λρ1 − ρ2 > 0}, the curve becomes equal
to r − 1 (dotted line).

The classical case

As a “consistency check” we separately consider the
classical case here. Suppose ρ and σ are both diagonal
with elements p1 , ..., pn and q1 , ..., qn , respectively. Denote rj ≡ qj /pj if pj > 0 and otherwise rj = 0. W.l.o.g.
suppose r1 6 r2 6 · · · 6 rn . We therefore get
⇐⇒

APPLICATIONS

In this section, we study how the conditions in Theorem 2 are logically related to the existence of a suitable
transformation mapping (ρ1 , ρ2 ) into (ρ01 , ρ02 ).
A.

rpj − qj > 0

j=1

thus reconstructing the Blackwell criterion for pairs
of probability distributions (including majorization and
thermomajorization).

∀r > 0.

This completes the proof of the lemma.

1.

max

k∈{1,...,n}

r > rj

(30)

Coherent energy transitions with
Gibbs-preserving operations

We consider here the resource theory of athermality [7, 8]. In this theory, quantum systems that are not
in thermal equilibrium with their environment are considered resources (e.g., work can be extracted from such
systems). Hence, free systems are those prepared in the
Pd
Gibbs state, i.e., γ = Z −1 x=1 e−βEx |xihx|, and permitted operations must preserve γ. Consider now two
possibly non-commuting quantum states ρ and σ both
with same 2-dimensional support spanned by the same
two energy eigenstates, say |xi and |yi. Such states form
the building blocks of quantum thermodynamics as they
contain the smallest units of athermality [8]. Here we
find both necessary and sufficient conditions under which
a Gibbs-preserving transition between ρ and σ is possible. We call such transitions coherent energy transitions,
since not only the transitions |xi → |yi and |yi → |xi
are considered, but also transitions between any linear
superpositions of such energy eigenstates.
The main result about coherent energy transitions is
the following:
Theorem 3. With γ, ρ, and σ as above, assuming that
γ > 0 (i.e., non-zero temperature), the following are
equivalent:
i) ρ can be transformed into σ by a γ-preserving
CPTP operation (i.e., Gibbs-preserving operation);
ii) (ρ, γ)  (σ, γ);

Hence, for r ∈ (rk , rk+1 ]
iii)
f (r) = (1 − )r − Tr(rρ − σ)+ = (1 − )r − r

k
X
j=1

pj +

k
X

qj

j=1

(31)

(

Dmax (ρkγ) > Dmax (σkγ),
Dmax (γkρ) > Dmax (γkσ).

(32)

8
In other words, in this case, we do not need to check
the validity of point (iii) of Theorem 2 for all values of
α, but only in the limit α → ∞. Moreover, in this case,
we know that a CPTP map between the two pairs of
states exists. In the case of zero temperature, i.e., if
γ > 0, a third condition has to be added to the above list,
namely, Dmin (γkρ) > Dmin (γkσ), where Dmin (γkρ) =
− log Tr[Πγ ρ] denotes the min-relative entropy [30] and
Πγ is the projector onto the support of γ. Finally, we
did not include the condition Dmin (ρkγ) > Dmin (σkγ)
since it is trivial, unless σ is rank-one (i.e., a pure state).
However, as shown below in the proof, it turns out that
in this case the other conditions implies this one.
Theorem 3 generalizes an earlier work given in [32] to
the generic case in which the Gibbs state is not pure. It
demonstrates that three athermality monotones (given
in terms of the min/max relative entropies) provide both
necessary and sufficient conditions for the existence of a
Gibbs preserving map connecting two non-thermal states
with the same two-dimensional support. Since the set
of Gibbs-preserving operations is strictly larger than the
set of thermal-operations [8], these three monotones, in
general, will not be sufficient to determine convertibility
under thermal operations [33].
The Gibbs state is given by
d
1 X
γ=
exp(−βEx )|xihx|,
Z x=1

(33)

where β = 1/kT is the inverse temperature, d is the
dimension of the quantum system, {|xi}dx=1 are the complete set of eigenstates of the Hamiltonian, Ex the eigenvalues of the Hamiltonian, and Z is the partition function
Pd
x=1 exp(−βEx ). Consider now two quantum states ρ
and σ both with the same 2-dimensional support given
by, e.g.,
supp(ρ) = supp(σ) = span {|xi}x=1,2 ,

(34)

but it does not matter which two energy eigenstates are
chosen (with the condition E2 > E1 ). Denote further by
γ (2) the Gibbs state projected onto this two dimensional
subspace:
γ (2) ≡ p|1ih1| + (1 − p)|2ih2|

(35)

where
exp(−βE1 )
1
=
,
exp(−βE1 ) + exp(−βE2 )
1 + exp(−β∆E)
(36)
with ∆E = E2 − E1 > 0 so that p > 1/2. We start with
the following lemma:
p≡

Lemma 6. Let ρ and σ be as in (34), and let γ be as
in (33) with γ (2) as in (35). Then, there exists a CPTP
map Φ such that Φ(ρ) = σ and Φ(γ) = γ if and only
if there exists a CPTP map E such that E(ρ) = σ and
E(γ (2) ) = γ (2) .

Proof. Suppose there exists Φ such that Φ(ρ) = σ and
Φ(γ) = γ. Then, for all t > 0 we have
||σ − tγ||1 = ||Φ(ρ) − Φ(γ)||1 6 ||ρ − tγ||1 ,

(37)

since the trace norm is contractive. Next, denoting by P
the projection onto span{|1i, |2i} and r ≡ exp(−βE1 ) +
exp(−βE2 ), we have
||ρ − tγ||1 = ||P ρP − tP γP ||1 + t ||(I − P )γ(I − P )||1
r
r
= ρ − t γ (2) + t γ − γ (2) ,
Z
Z
1
1
and, analogously,
||σ − tγ||1 = σ − t

r (2)
γ
Z

+t γ−
1

r (2)
γ
Z

.
1

Therefore, since r/Z > 0, we can introduce the new parameter t0 , t · Zr so that the inequality in Eq. (37) can
be rewritten as
σ − t0 γ (2)
1

6 ρ − t0 γ (2)

∀ t0 > 0 .

(38)

1

From the Alberti–Uhlmann result on qubits [34] there
exists E as in the lemma.
Conversely, suppose there exists a CPTP map E such
that E(ρ) = σ and E(γ (2) ) = γ (2) . Then, define Φ as
follows. Let P = |1ih1| + |2ih2| be the projector onto the
support of ρ and σ, and define
Φ(·) := E (P (·)P ) + (Id − P )(·)(Id − P ) .

(39)

By construction, Φ is CPTP since E is CPTP, and it
is easy to verify that Φ(ρ) = σ and Φ(γ) = γ. This
completes the proof.
Lemma 7. Let ρ and σ be two qubit density matrices
and let γ (2) be the Gibbs state given in (35). Then, ρ
can be converted to σ by Gibbs preserving operations if
and only if the following three inequalities simultaneously
hold:

(2)
(2)

Dmax (ρkγ ) > Dmax (σkγ ),
(2)
(2)
(40)
Dmax (γ kρ) > Dmax (γ kσ),

D (γ (2) kρ) > D (γ (2) kσ).
min
min
Before proceeding, we notice that, while in Eq. (40)
above the projected Gibbs state γ (2) appears, in Eq. (32)
of Theorem 3 we use the original γ. However, since
P ρP = ρ, P σP = σ, and P γP = cγ (2) (for some c > 0),
and since both Dmax and Dmin in this case only depend
on what there is on the support of P , the two set of
conditions are clearly equivalent.
Denote by
m(ρ, γ (2) ) , inf(ρ/γ (2) ) = sup{t ∈ R : tγ (2) − ρ 6 0}
M (ρ, γ (2) ) , sup(ρ/γ (2) ) = inf{t ∈ R : tγ (2) − ρ > 0} .
(41)

9
Note that m(ρ, γ (2) ) 6 1 6 M (ρ, γ (2) ). Since we consider here the qubit case, it follows that m(ρ, γ (2) ) and
M (ρ, γ (2) ) are the roots to the quadratic polynomial
Det(ρ − tγ (2) ). A straightforward calculation gives (assuming det(γ (2) ) > 0)
h
ih
i
det(ρ−tγ (2) ) = det(γ (2) ) t − m(ρ, γ (2) ) t − M (ρ, γ (2) )
(42)
with m and M given explicitly below after we introduce
a few notations.
Without loss of generality we can assume that the offdiagonal terms of ρ are non-negative real numbers since
γ (2) is invariant under conjugation by any 2 × 2 unitary
matrix which is diagonal on the energy eigenbasis (i.e.,
commutes with γ (2) ). Hence, we can write
p


a
ε a(1 − a)
ρ= p
(43)
ε a(1 − a)
1−a
with ε, a ∈ [0, 1]. Taking γ (2) as in (35) we get the following explicit expressions for m(ρ, γ (2) ) and M (ρ, γ (2) )
assuming det(γ (2) ) > 0 (i.e. 0 < p < 1)
i
p
1h
r0 + r1 − (r0 − r1 )2 + 4r0 r1 2
m(ρ, γ (2) ) =
2
i
p
1h
M (ρ, γ (2) ) =
r0 + r1 + (r0 − r1 )2 + 4r0 r1 2 (44)
2
where
1−a
a
and r1 ≡
.
(45)
r0 ≡
p
1−p
By definition, both m and M are monotonic in the
sense that


M Φ(ρ), Φ(γ (2) ) 6 M (ρ, γ (2) )


m Φ(ρ), Φ(γ (2) ) > m(ρ, γ (2) ) .
(46)
Note that m(ρ, γ (2) ) = 1/M (γ (2) , ρ) and M is related to
the max relative entropy:
Dmax (ρkγ (2) ) = log M (ρ, γ (2) )
Dmax (γ

(2)

kρ) = log M (γ

(2)

, ρ) = − log m(ρ, γ

Lemma 8. (Alberti–Uhlmann) Let ρ, σ, η, and τ be
qubit density matrices. Then, there exists a CPTP map
Φ such that σ = Φ(ρ) and η = Φ(τ ) if and only if
M (ρ, τ ) > M (σ, η) > m(σ, η) > m(ρ, τ ) ,

(49)

and
det(σ − tη) > det(ρ − tτ ) ∀ m(σ, η) 6 t 6 M (σ, η) .
(50)
We now apply the above lemma above to the case τ =
η = γ (2) .

1.

First case: non-zero temperature (γ (2) > 0).

We first assume det(γ (2) ) > 0. The necessity of (40)
follows from the fact that the min/max relative entropies
both satisfies the data processing inequality. We therefore need to show that they are sufficient. With the
choice τ = η = γ (2) the conditions (49) are equivalent
to the conditions (40) (recall the last condition of (40)
is trivial since we assume for now that γ (2) is full rank).
It is therefore left to show that the conditions (50) hold
automatically if Eqs. (40) hold. Indeed, recall that for
det(γ (2) ) > 0 we have
h
ih
i
det(ρ−tγ (2) ) = det(γ (2) ) t − m(ρ, γ (2) ) t − M (ρ, γ (2) )
(51)
and
h
ih
i
det(σ−tγ (2) ) = det(γ (2) ) t − m(σ, γ (2) ) t − M (σ, γ (2) )
(52)
Hence, the inequality det(σ − tγ (2) ) > det(ρ − tγ (2) ) is
equivalent to
h
i
t M (ρ, γ (2) ) + m(ρ, γ (2) ) − M (σ, γ (2) ) − m(σ, γ (2) )
+ m(σ, γ (2) )M (σ, γ (2) ) − m(ρ, γ (2) )M (ρ, γ (2) ) > 0 (53)

(2)

) (47)

A dual definition is the min-relative entropy defined by:


Dmin (γ (2) kρ) = − log Tr ρΠγ (2) ,
(48)
where Πγ (2) is the projection to the support of γ (2) .
Clearly, the if det(γ (2) ) > 0 then Dmin (γ (2) , ρ) = 0. Summarizing, we showed that the conditions in Eq. (40) are
equivalent to (remember the assumption here γ (2) > 0;
the case of rank-one γ (2) will be considered separately
below)
(
M (ρ, γ (2) ) > M (σ, γ (2) )
m(ρ, γ (2) ) 6 m(σ, γ (2) ).
To see how the above conditions can be used to prove
Theorem 3, we need the following lemma from Ref. [34]:

We therefore need to show that the above inequality
holds for all m(σ, γ (2) ) 6 t 6 M (σ, γ (2) ). It is therefore sufficient to show that it holds at the two extreme
points of the interval. Indeed, for t = m(σ, γ (2) ) after
some algebra the expression in (53) becomes
h

ih
i
M (ρ, γ (2) ) − m(σ, γ (2) ) m(σ, γ (2) ) − m(ρ, γ (2) )

which is non-negative due to (40). Similarly, substituting
t = M (σ, γ (2) ) in (53) gives
h

ih
i
M (σ, γ (2) ) − m(ρ, γ (2) ) M (ρ, γ (2) ) − M (σ, γ (2) )

which is again non-negative due to (40). This completes
the proof of the theorem for the case det(γ (2) ) > 0.

10
2.

Second case: zero temperature (det(γ (2) ) = 0).

In this case direct calculation gives

det(ρ − tγ (2) ) = det(ρ) − t 1 − Tr[ρΠγ (2) ]

(54)


det(σ − tγ (2) ) = det(σ) − t 1 − Tr[σΠγ (2) ] .

(55)

and

Note that Πγ (2) is either the projection |1ih1| or |2ih2|.
Now, assuming ρ 6= σ 6= γ (2) (otherwise the problem
becomes trivial), since γ (2) is rank 1 we have M (ρ, γ (2) ) =
M (σ, γ (2) ) = ∞. On the other hand, in this case a simple
calculation gives
m(ρ, γ (2) ) =

det(ρ)
1 − Tr[ρΠγ (2) ]

m(σ, γ (2) ) =

det(σ)
1 − Tr[σΠγ (2) ]

where we used ρ 6= γ (2) and σ 6= γ (2) , so that together
with γ (2) being rank 1 gives Tr[ρΠγ (2) ] < 1 and similarly Tr[σΠγ (2) ] < 1. Hence, in this case exploring the
behaviour of det(σ − tγ (2) ) > det(ρ − tγ (2) ) in the limit
t → M (σ, γ (2) ) = ∞ we must have

for some effect 0 6 E 6 1 and some density matrices ξ1 and ξ2 . Test-and-prepare channels are, in other
words, measure-and-prepare channels for which the measurement has only two possible outcomes. Although restricted, this class seems quite natural in the framework
of quantum relative Lorenz curves, which are defined only
in terms of binary measurements (i.e., hypothesis tests).
Indeed, a necessary and sufficient condition for the existence of a test-and-prepare channel between two pairs of
density matrices can be expressed in terms of quantum
relative Lorenz curves as follows:
Theorem 4. Given two pairs of density matrices (ρ1 , ρ2 )
and (ρ01 , ρ02 ) on Cn and Cm , respectively, there exists a
test-and-prepare channel E such that E(ρ1 ) = ρ01 and
E(ρ2 ) = ρ02 , if and only if the quantum Lorenz curve of
ρ1 relative to ρ2 is nowhere below the segments joining
the points (0, 0), (1, 1) and passing through either


1 − m0 M 0 (1 − m0 )
(x, y) =
,
M 0 − m0 M 0 − m0
or

(x, y) =

M 0 − 1 m0 (M 0 − 1)
,
M 0 − m0 M 0 − m0

0

which is satisfied since m(ρ, γ (2) ) 6 m(σ, γ (2) ). Hence,
det(σ − tγ (2) ) > det(ρ − tγ (2) ) for all t with m(σ, γ (2) ) 6
t < ∞. This completes the proof.
Test-and-prepare channels

The proof of Theorem 3 above relies on a lemma proved
by Alberti and Uhlmann [34], which, together with Theorem 2, implies that, if n = m = 2 (i.e., for qubits)
then (ρ1 , ρ2 )  (ρ01 , ρ02 ) if and only if there exists a completely positive trace-preserving (CPTP) map Φ such
that Φ(ρi ) = ρ0i (i = 1, 2). However, explicit counterexamples exist, showing that as soon as one leaves the qubit
case, already when n = 3 and m = 2, this is not true
anymore [25]. Hence, leaving aside the general case, we
focus instead on a special class of CPTP maps, namely,
test-and-prepare channels of the form:
E(ρ) , Tr[Eρ]ξ1 + Tr[(1 − E)ρ]ξ2 ,

0

M 0 , 2Dmax (ρ1 kρ2 ) = sup(ρ01 /ρ02 )

(2)

which is equivalent to Dmin (γ kρ) > Dmin (γ kσ). At
the point t = m(σ, γ (2) ), the inequality det(σ − tγ (2) ) >
det(ρ − tγ (2) ) becomes

0 > det(ρ) − m(σ, γ (2) ) 1 − Tr[ρΠγ (2) ]



det(ρ)
= 1 − Tr[ρΠγ (2) ]
− m(σ, γ (2) )
1 − Tr[ρΠγ (2) ]


= 1 − Tr[ρΠγ (2) ] m(ρ, γ (2) ) − m(σ, γ (2) )

B.

,

whichever is higher, where

Tr[σΠγ (2) ] > Tr[ρΠγ (2) ]
(2)



and
0

0

m0 , 2−Dmax (ρ2 kρ1 ) = inf(ρ01 /ρ02 ).
Proof. Consider a test-and-prepare channel of the form:
E(ρ) = Tr[Eρ]σ1 + Tr [(1 − E)ρ] σ2

(56)

where σ1 , σ2 are density matrices (i.e. positivie semidefinite matrices with trace 1) and 0 6 E 6 1. If E(ρj ) =
ρ0j for j = 1, 2, then
ρ01 = e1 σ1 + (1 − e1 )σ2
ρ02 = e2 σ1 + (1 − e2 )σ2
where ej ≡ Tr[Eρj ] for j = 1, 2. Assuming e1 6= e2
(otherwise, ρ01 = ρ02 ), the above equations are equivalent
to
1
[(1 − e2 )ρ01 − (1 − e1 )ρ02 ]
e1 − e2
1
σ2 =
[−e2 ρ01 + e1 ρ02 ]
e1 − e2

σ1 =

Note that σ1 and σ2 have trace 1 since ρ01 and ρ02 have
trace 1. W.l.o.g. we can assume e1 > e2 . With this
choice, σ1 and σ2 are positive semi-definite if and only if
ρ01 −

1 − e1 0
ρ >0
1 − e2 2

and ρ02 −

e2 0
ρ >0.
e1 1

(57)

11
Note that since we assume e1 > e2 we have 1 − e2 > 0
and e1 > 0. Denote by m0 ≡ inf(ρ01 /ρ02 ) and by M 0 ≡
sup(ρ01 /ρ02 ), and note that inf(ρ02 /ρ01 ) = 1/M 0 . We therefore get that σ1 and σ2 are positive semi-definite if and
only if
1 − e1
6 m0
1 − e2

and

e2
1
6 0 .
e1
M

Finally, set v1 = ρ1 − m0 ρ2 and v2 = (0, 1n ). With these
choices, v2 − Γ(x) = (Tr [x (ρ1 − M 0 ρ2 )] , 1n − x), so that
the primal problem becomes
max

x∈K1
v2 −Γ(x)∈K2

=

(58)

The above inequalities are equivalent to
(
Tr [E (ρ1 − m0 ρ2 )] > 1 − m0 ,
Tr [E (ρ1 − M 0 ρ2 )] > 0 .

hv1 , xi1

max

06E61n
Tr[E (ρ1 −M 0 ρ2 )]>0

Tr [E (ρ1 − m0 ρ2 )]

(66)

where we renamed x with E. The dual problem is given
by
min

y∈K2
Γ (y)−v1 ∈K1

hv2 , yi2

∗

We therefore arrive at the following lemma:

=

Lemma 9. There exists a channel E of the form (56)
such that E(ρ1 ) = ρ01 and E(ρ2 ) = ρ02 , if and only if
W (ρ1 , ρ2 , ρ01 , ρ02 ) > 0

(59)

where the witness W is defined as
W (ρ1 , ρ2 , ρ01 , ρ02 ) ,
max
m0 − 1 +

06E61
Tr[E (ρ1 −M 0 ρ2 )]>0

Tr [E (ρ1 − m0 ρ2 )]

max

(61)

The dual form involves the adjoint map Γ∗ : V2 → V1 :
min

hv2 , yi2

(62)

For our purposes, we take V1 = Hn the space of n × n
Hermitian matrices, and we take K1 = Hn,+ the cone of
positive semi-definite matrices in Hn . We further define
the vector space
n
o
V2 ≡ R ⊕ Hn = (r, A) r ∈ R ; A ∈ Hn ,
(63)
with inner product h(r, A), (t, B)i1 := rt + Tr[AB]. Further, define K2 = R+ ⊕ Hn,+ to be the positive cone in
V2 . The linear map Γ : V1 → V2 is defined as follows:
0

Γ(A) = (− Tr [A (ρ1 − M ρ2 )] , A) .

F > r(ρ1 − M 0 ρ2 ) + (ρ1 − m0 ρ2 )
= (1 + r)ρ1 − (rM 0 + m0 )ρ2

0

Γ (r, A) = A − r (ρ1 − M ρ2 ) .

F = [(1 + r)ρ1 − (rM 0 + m0 )ρ2 ]+ .

min Tr [(1 + r)ρ1 − (rM 0 + m0 )ρ2 ]+
r>0

= min
r>0

1 − m0 − r(M 0 − 1) + ||(1 + r)ρ1 − (rM 0 + m0 )ρ2 ||1
.
2
(69)

By plugging the above equation into (60), we therefore
conclude that
1
W (ρ1 , ρ2 , ρ01 , ρ02 ) = m0 − 1 + (1 − m0 )
2


1
0
+ min ||(1 + r)ρ1 − (rM + m0 )ρ2 ||1 − r(M 0 − 1)
2 r>0
1
= (m0 − 1)
2


1
+ min ||(1 + r)ρ1 − (rM 0 + m0 )ρ2 ||1 − r(M 0 − 1) ,
2 r>0
namely,
2W (ρ1 , ρ2 , ρ01 , ρ02 ) = −(1 − m0 )


+ min ||(1 + r)ρ1 − (rM 0 + m0 )ρ2 ||1 − r(M 0 − 1) .
r>0

(70)
Introducing

(65)

(68)

We therefore conclude that the dual problem is equivalent
to

(64)

Note that the dual map Γ∗ : V2 → V1 is given by
∗

(67)

is of course the positive part of the left-hand side:

hv1 , xi1

y∈K2
Γ∗ (y)−v1 ∈K1

Tr[F ]

where we took y = (r, F ). We can further simplify the
above expression. First note that, for any given r, the
positive semi-definite matrix F with the smallest trace
that satisfies

(60)

The calculation of W can be simplified using the following dual formulation of linear programming, analogously
to what we did in the proof of Lemma 5. Let V1 and
V2 be two (inner product) vector spaces with two cones
K1 ⊂ V1 and K2 ⊂ V2 . Consider two vectors v1 ∈ V1 and
v2 ∈ V2 , and a linear map Γ : V1 → V2 . Then, the primal
form is:
x∈K1
v2 −Γ(x)∈K2

min

r,F >0
F >r(ρ1 −M 0 ρ2 )+(ρ1 −m0 ρ2 )

t,

rM 0 + m0
,
1+r

12
and noting that t ∈ [m0 , M 0 ) we obtain that there exists
a channel E of the form (56) such that E(ρ1 ) = ρ01 and
E(ρ2 ) = ρ02 , if and only if
||ρ1 − tρ2 ||1
(M 0 − t)(1 − m0 ) + (t − m0 )(M 0 − 1)
>
M 0 − m0
0
0
m + M − 2m0 M 0 + t (m0 + M 0 − 2))
,
=
M 0 − m0

E(ρ1 ) = p1 ρ01 ,
(71)

∀t>0,

(72)

where
 0

1
M (1 − m0 )
0
σ1 ≡ 0
0
m0 (M 0 − 1)
M − m0


1
1 − m0
0
σ2 ≡ 0
0
M0 − 1
M − m0

(73)

are two diagonal 2 × 2 density matrices with the property
that sup(σ1 /σ2 ) = M 0 ≡ sup(ρ01 /ρ02 ) and inf(σ1 /σ2 ) =
m0 ≡ inf(ρ01 /ρ02 ).
Finally, the statement of Theorem 4 is obtained noticing that condition (72) is equivalent, due to Theorem 2,
to saying that the Lorenz curve of ρ1 relative to ρ2 is
never below that of σ1 relative to σ2 . However, since the
latter is the quantum Lorenz curve of two classical probability distributions, it is just made of two segments joining
0) with
either
 the points (0, 
 (1, 1), passing through

M 0 (1−m0 )
1−m0
M 0 −m0 , M 0 −m0

or

0
0
M 0 −1 m (M −1)
M 0 −m0 , M 0 −m0

, whichever

where pi = Tr[(E + F )ρi ], and m, M , m0 , and M 0 are as
in Theorem 4.
When the protocol fails, we just prepare the state ρ02
independently of the input. In this way, we realized a
channel that deterministically transforms ρ2 into ρ02 , but
is also able to transform ρ1 into ρ01 , whenever the successful event is recorded.
In Ref. [29], it is shown that a probabilistic transformation from (ρ1 , ρ2 ) to (ρ01 , ρ02 ) exists if and only if
M 0 /m0 6 M/m. The above theorem hence slightly extends that. For example, Theorem 5 implies that the
success probability p1 be bounded as
p1 6 M/M 0 ≡ e−∆Fmax ,
where we define the max-free energy difference as
∆Fmax , Dmax (ρ01 kρ02 ) − Dmax (ρ1 kρ2 ).
In order to prove Theorem 5, let us consider a probabilistic test-and-prepare channel of the form:
Φ(ρ) = Tr[Eρ]σ1 + Tr [F ρ] σ2

Probabilistic transformations

By mixing ρ01 with a sufficient fraction of ρ02 , while
keeping ρ02 unchanged, it is always possible to decrease
the gap between m0 and M 0 , until the conditions of Theorem 4 are met. In this way, with sufficient mixing, any
output pair can be obtained, but the noise due to mixing
cannot be undone afterwards.
One way to overcome this problem is to relax the assumptions made on the channel, in particular, the condition of trace-preservation. We hence consider probabilistic channels of the following form:
E(ρ) , Tr[Eρ]ξ1 + Tr[F ρ]ξ2 ,

(75)

exists if and only if

determines the steepest curve.

C.

E(ρ2 ) = p2 ρ02 ,

p1
m
M
6
6 0,
0
m
p2
M

for all t ∈ [m0 , M 0 ], or equivalently,
||ρ1 − tρ2 ||1 > ||σ1 − tσ2 ||1 ,

Theorem 5. Consider two pairs of density matrices
(ρ1 , ρ2 ) and (ρ01 , ρ02 ) on Cn and Cm , respectively. Then,
a channel of the form (74), such that

(74)

where E, F > 0, E + F 6 1, and ξ1 , ξ2 are two (normalized) density matrices. The above transformation constitutes a heralded probabilistic transformation, in the sense
that we know if the protocol succeeded or not, with success probability given by Psucc = Tr[(E +F )ρ]. The main
result of this subsection is given by the following

(76)

where σ1 , σ2 are normalized density matrices and E, F >
0 with E + F 6 1. If E(ρj ) = pj ρ0j for j = 1, 2 with
0 < p1 , p2 6 1, then
p1 ρ01 = e1 σ1 + f1 σ2
p2 ρ02 = e2 σ1 + f2 σ2
where ej ≡ Tr[Eρj ] and fj ≡ Tr[F ρj ] for j = 1, 2. Assuming e1 f2 6= e2 f1 the above equations are equivalent
to
1
[f2 p1 ρ01 − f1 p2 ρ02 ]
e 1 f2 − e 2 f1
1
σ2 =
[−e2 p1 ρ01 + e1 p2 ρ02 ]
e 1 f2 − e 2 f1

σ1 =

Note that σ1 and σ2 have trace 1 since ρ01 and ρ02 have
trace 1. Without loss of generality, we can assume
e1 /e2 > f1 /f2 . With this choice, σ1 and σ2 are positive semi-definite if and only if
ρ01 − q −1

f1 0
ρ >0
f2 2

and q −1

e1 0
ρ − ρ01 > 0 ,
e2 2

(77)

13
where q , p1 /p2 . Again, denote by m0 = inf(ρ01 /ρ02 ) and
by M 0 = sup(ρ01 /ρ02 ). We therefore get that σ1 and σ2
are positive semi-definite if and only if
f1
6 qm0
f2

and

e1
> qM 0 .
e2

Tr [(E + F )(qρ2 − ρ1 )] = 0
⇐⇒ Tr [E(qρ2 − ρ1 )] = 0
⇐⇒ Tr [Eρ1 ] = qTr[Eρ2 ] .

(78)

Recalling the definitions of e1 , e2 , f1 , f2 , the above inequalities are equivalent to
Tr [F (qm0 ρ2 − ρ1 )] > 0
Tr [E (ρ1 − qM 0 ρ2 )] > 0 .

unless Tr[F ρ1 ] = Tr[F ρ2 ] = 0. We therefore must have
Tr[F ρ1 ] = Tr[F ρ2 ] = 0. This later condition gives

(79)

We therefore arrive at the following lemma:
Lemma 10. There exists a CP map Φ of the form (76)
such that Φ(ρ1 ) = p1 ρ01 and Φ(ρ2 ) = p2 ρ02 , if and only if
Tr [F (qm0 ρ2 − ρ1 )] > 0

(80)

Tr [E (ρ1 − qM 0 ρ2 )] > 0 .

(81)

But this last equality gives
Tr [E (ρ1 − qM 0 ρ2 )]
= (1 − M 0 )Tr[Eρ1 ]
(83)
60,
since M 0 > 1. We therefore must have Tr[Eρ1 ] =
Tr[Eρ2 ] = 0. Together with Tr[F ρ1 ] = Tr[F ρ2 ] = 0,
it gives Pmax (q) = 0. Following similar lines we get
Pmax (q) = 0 for q > M/M 0 .
As a consequence of the above discussion, we obtain
the following corollary, which is consistent with a result
in Ref. [29], but slightly more general:

and

Corollary 1.
Our goal is to maximize p1 under these constraints
along with the constraint Tr [(E + F )(qρ2 − ρ1 )] = 0 that
defines q. Therefore, the maximum value of p1 , with a
fixed value of q ∈ R+ , is given by
Pmax (q) =

max

Tr[(E+F )(qρ2 −ρ1 )]=0
Tr[E (ρ1 −qM 0 ρ2 )]>0

Tr [(E + F )ρ1 ]

(82)

Tr[F (qm0 ρ2 −ρ1 )]>0
E,F >0 , E+F 61n

This is an optimization problem that can be solved efficiently and algorithmically using SDP. Moreover, in the
lemma below we show that if q is not in the right interval
then Pmax (q) = 0.
Lemma 11. Pmax (q) > 0 implies that
M
m
6q6 0 .
m0
M
Proof. Suppose q <
ρ1 6 0. Moreover,

m
m0 .

Then, qm0 < m so that qm0 ρ2 −

max Pmax (q) > 0

q∈R+

VI.

⇐⇒

h(ρ1 , ρ2 ) > h(ρ01 , ρ02 ) .

DISCUSSION.

In the present work we introduced quantum relative
Lorenz curves and Hilbert α-divergences, studied their
properties, and applied them to the problem of characterizing necessary and sufficient conditions for the existence
of a suitable transformation from an initial pair of states
(ρ1 , ρ2 ) to a final one (ρ01 , ρ02 ). In particular, a strong
equivalence has been proved in the case of coherent energy transitions with Gibbs-preserving maps, a paradigm
that has immediate applications in quantum thermodynamics and the resource theory of athermality. Finally,
we also considered the cases of test-and-prepare channels
and probabilistic transformations, giving necessary and
sufficient conditions for both.

Tr [F (qm0 ρ2 − ρ1 )]
= Tr [F (mρ2 − ρ1 )] − (m − m0 q)Tr[F ρ2 ]
<0,

Acknowledgments.—The authors are grateful to Mark
Girard for his help with the Figures 1 and 2. F.B. acknowledges financial support from the JSPS KAKENHI,
No. 26247016. G.G. acknowledges financial support from
NSERC.

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