2022

1]Institute of Theoretical Physics, Jagiellonian University, ul. Łojasiewicza
11, 30–348 Kraków, Poland

2]Center for Theoretical Physics, Polish Academy of Sciences, Al. Lotników
32/46, 02-668 Warszawa, Poland

# Monotonicity of the quantum
2-Wasserstein distance

Rafał Bistroń<EMAIL_ADDRESS>Michał Eckstein
<EMAIL_ADDRESS>Karol Życzkowski<EMAIL_ADDRESS>[  [

###### Abstract

We study a quantum analogue of the 2-Wasserstein distance as a measure of
proximity on the set $\Omega_{N}$ of density matrices of dimension $N$. We
show that such (semi-)distances do not induce Riemannian metrics on the
tangent bundle of $\Omega_{N}$ and are typically not unitary invariant.
Nevertheless, we prove that for $N=2$ dimensional Hilbert space the quantum
2-Wasserstein distance (unique up to rescaling) is monotonous with respect to
any single-qubit quantum operation. Furthermore, for any $N\geq 3$ and the
quantum cost matrix proportional to a projector we demonstrate the
monotonicity under arbitrary mixed unitary channels. Finally, we provide
numerical evidence which allows us to conjecture that the unitary invariant
quantum 2-Wasserstein semi-distance is monotonous with respect to all CPTP
maps in any dimension $N$.

###### keywords:

Quantum transport problem, Monge-Kantorovich distance, monotonicity with
respect to quantum operations. April 14, 2022

##  1 Introduction

The distances on the space of probability vectors are a primary tool to
compare and analyse various statistical distributions. On the mathematical
side, they lie at the heart of information geometry Amari1 ; Amari2 , which
unveils a powerful interplay between Riemannian geometry and statistics. In
particular, with some distances on the space of probability vectors one can
associate a Riemannian metric on the suitable tangent bundle. In the set of
all such distances the one generated by the Fischer–Rao metric is
distinguished as the unique continuous distance monotone under classical
stochastic maps (Cencov theorem cencov ).

In the modern field of quantum information NielsenChuang ; BZ17  the classical
probability vectors are replaced by density matrices. The quantum analogue of
a stochastic map is a completely positive trace preserving (CPTP) linear map,
$\Phi:\Omega_{N}\to\Omega_{M}$, where
$\Omega_{N}\vcentcolon=\\{\rho\in\mathbb{C}^{N\times N}\>|\>\rho=\rho^{*},\
\rho\geq 0,\ \Tr\rho=1\\}$ is the set of density matrices of order $N$. A
semi-distance $D_{\text{mon}}$ on the space of quantum states is called
_monotone_ if the inequality

$\displaystyle D_{\text{mon}}(\Phi(\rho^{A}),\Phi(\rho^{B}))\leq
D_{\text{mon}}(\rho^{A},\rho^{B})$ (1.1)

holds for all states $\rho^{A},\rho^{B}\in\Omega_{N}$, for any dimension $N$,
and all CPTP maps $\Phi$.

The set of all monotone distances on the space of quantum states, inducing a
Riemannian metric on the tangent bundle of $\Omega_{N}$, can be characterised,
via the Morozova–Cencov–Petz theorem Morozova ; Petz1 , through a single-
parameter operator-monotone function. On the other hand, there exist important
monotone distances, which do not induce a Riemannian metric — most notably,
the trace distance $D_{\text{Tr}}(\rho^{A},\rho^{B})=\Tr|\rho^{A}-\rho^{B}|$,
with $|X|=\sqrt{XX^{*}}$.

The property of monotonicity is crucial for applications in quantum
information processing. It implies that the distinguishability, quantified by
a given distance function, cannot increase under an application of any quantum
map.

First attempts to generalise the classical Monge transport distance to the
quantum setting with the use of Husimi function were done in ZS98 ; ZS01 .
More recently, a more general approach of Kantorovich Kan48  and Wasserstein
Vil09  was extended to the quantum case an analysed from the mathematical AF17
; CGGT17 ; BGJ19 ; Fri20 ; Duv20 , physical GMT16 ; GP18 ; CGP20 ; CM20 ;
Ikeda20  and information-theoretical CGNT17 ; PMTL20 ; PT21 ; KdPMLL21
perspective. Among these, a recent proposal transport1 ; transport2  is based
on the optimization over the set of bipartite quantum states with fixed
marginals of the expectation value of a certain ‘cost observable’ (see also
qWGAN ; YZYY18 ; Reira18 ). Concretely, for any classical distance $d$ on the
set of $N$ points one can define transport2  an associated _quantum cost
operator_

$\displaystyle
C^{Q}_{E}=\sum_{j>i=1}^{N}E_{ij}|\psi^{-}_{ij}\rangle\langle\psi^{-}_{ij}|,$
(1.2)

where $E_{ij}=d(x_{i},x_{j})$ is the distance between points $x_{i},x_{j}$,
while $|\psi^{-}_{ij}\rangle\langle\psi^{-}_{ij}|$ is the projector on the
antisymmetric subspace spanned by two base vectors $|i\rangle$, $|j\rangle$,
i.e.
$|\psi^{-}_{ij}\rangle=\tfrac{1}{\sqrt{2}}(|i\rangle|j\rangle-|j\rangle|i\rangle)$.
For any two density matrices $\rho^{A},\rho^{B}\in\Omega_{N}$ one introduces
the set $\Gamma^{Q}(\rho^{A},\rho^{B})$ of bi-partite coupling matrices
$\rho^{AB}$, such that both partial traces are fixed,
$\Tr_{A}\rho^{AB}=\rho^{B}$ and $\Tr_{B}\rho^{AB}=\rho^{A}$. Minimising the
Hilbert–Schmidt scalar product of the cost matrix and $\rho^{AB}$ over all
possible coupling matrices we arrive at the _optimal quantum transport cost_ ,

$\displaystyle
T^{Q}_{E}(\rho^{A},\rho^{B})\vcentcolon=\min_{\rho^{AB}\in\Gamma^{Q}(\rho^{A},\rho^{B})}\Tr\
\big{(}C^{Q}_{E}\rho^{AB}\big{)}.$ (1.3)

It was shown in transport2  that $T^{Q}_{E}$ is a weak metric (i.e. a semi-
metric bounded from below by a genuine metric) on $\Omega_{N}$ for any $N$ and
any quantum cost matrix $C^{Q}_{E}$. Following the classical analogy, one can
thus define transport1  the corresponding _quantum 2-Wasserstein semi-
distance_

$\displaystyle
W_{E}(\rho^{A},\rho^{B})\vcentcolon=\sqrt{\min_{\rho^{AB}\in\Gamma^{Q}(\rho^{A},\rho^{B})}\Tr\
\Big{(}(C^{Q}_{E})^{2}\rho^{AB}\Big{)}}.$ (1.4)

On the 2-point set all classical distance matrices $E$ are equivalent up to
rescaling. In higher dimensions a distinguished role is played by the simplex
geometry, for which $E_{ij}=1-\delta_{ij}$. In this case, the corresponding
quantum cost matrix (1.2) forms a projector onto the antisymmetric subspace of
the Hilbert space $\mathbb{C}^{N\times N}$. For sake of brevity we shall
denote such a quantum cost matrix by

$\displaystyle C^{Q}\vcentcolon=\tfrac{1}{2}(\mathbbm{1}_{N}-S),\quad\text{
where }\quad S(|x\rangle|y\rangle)=|y\rangle|x\rangle$ (1.5)

and, since $(C^{Q})^{2}=C^{Q}$, write

$\displaystyle
W(\rho^{A},\rho^{B})\vcentcolon=\sqrt{T^{Q}(\rho^{A},\rho^{B})},\;\text{ with
}\;T^{Q}(\rho^{A},\rho^{B})\vcentcolon=\min_{\rho^{AB}}\
\Tr\big(C^{Q}\rho^{AB}\big{missing}).$ (1.6)

For single-qubit states, $N=2$, it was shown transport1 ; transport2  that $W$
enjoys the triangle inequality and hence is a genuine distance on the Bloch
ball $\Omega_{2}$. Furthermore, the numerical simulations strongly suggest
that the triangle inequality actually holds for the quantum 2-Wasserstein
semi-distance (1.6) with the cost matrix (1.5) in any dimension $N$. However,
in the problem of monotonicity the triangle inequality does not play any role,
hence we shall — for simplicity — focus on the general semi-distances
$T^{Q}_{E}$ and the specific case of $T^{Q}$ determined by the projection cost
matrix (1.5).

The quantum optimal transport problem (1.3) admits a dual formulation qWGAN ;
transport2 . The optimization takes place over the set of pairs $\sigma^{A}$,
$\sigma^{B}\in H_{N}$ of $N\times N$ Hermitian matrices, satisfying a certain
algebraic constraint determined by the cost matrix $C_{E}$. Concretely, let

$\displaystyle\Sigma_{N}\vcentcolon=\\{\sigma^{A},\sigma^{B}\in
H_{N}\;|\;F\vcentcolon=C_{E}-\sigma^{A}\otimes\mathbbm{1}_{N}-\mathbbm{1}_{N}\otimes\sigma^{B}\geq
0\\},$ (1.7)

then

$\displaystyle
T^{Q}_{E}(\rho^{A},\rho^{B})=\sup_{(\sigma^{A},\sigma^{B})\in\Sigma_{N}}\;\Tr(\sigma^{A}\rho^{A}+\sigma^{B}\rho^{B}).$
(1.8)

If neither of the states $\rho^{A},\rho^{B}$ is pure, then the supremum is
achieved transport2 .

The quantum optimal transport cost (1.3) can be efficiently computed using
semidefinite programming transport2 , which makes it a valuable tool for
quantum machine learning. In particular, the quantum 2-Wasserstein semi-
distance (1.6) has been shown to play a key role in the quantum Generative
Adversarial Network scheme qWGAN . From the viewpoint of applications in
machine learning, the monotonicity of a distance is a highly desirable
property, which ensures the noise robustness of the learning algorithms.

The purpose of this work is to study the monotonicity with respect to the
quantum channels of the optimal quantum transport cost $T^{Q}_{E}$ defined
above. We start in Section 2 by considering some general properties of (1.3)
and show that for $N\geq 3$ and a general classical cost matrix $E$ the
corresponding quantum transport cost is not unitarily invariant. Then, we
discuss a quantity associated with the projective transport distance — the
‘SWAP-fidelity’ — and demonstrate, in Section 3, that $T^{Q}_{E}$ does _not_
induce a Riemannian metric, hence the standard Morozova–Cencov–Petz theorem
does not apply. Section 4 includes a proof of monotonicity of $T^{Q}$ under
general mixed unitary channels in any dimension $N$. Then, in Section 5 we
focus on $N=2$ and provide a complete proof of monotonicity of $T^{Q}$ for
arbitrary single-qubits channels. The article is supplemented by two
appendices. Appendix A contains the proofs of the technical results, while in
Appendix B we provide substantial numerical evidence for the monotonicity of
$T^{Q}$ under all qutrit and ququart channels. Basing on these observations,
we are tempted to conjecture that the unitary invariant quantum 2-Wasserstein
semi-distance $W$ is actually monotone for all CPTP maps.

##  2 No unitary invariance for general quantum 2-Wasserstein semi-distances

In the single-qubit case, $N=2$, the only (up to a trivial multiplicative
factor) quantum cost matrix (1.2) is a projector in the antisymmetric subspace
(1.5). But already in the qutrit case, $N=3$, there exists different cost
matrices $C^{Q}_{E}$, for instance the one induced by the classical Euclidean
distance on the line, $E_{12}=E_{23}=1$, $E_{13}=2$ (see the Supplemental
Material in transport1 ).

For the projection matrix $C^{Q}$ the optimal quantum transport cost (1.3) is
invariant under unitary channels:

$\displaystyle
T^{Q}(U\rho^{A}U^{\dagger},U\rho^{B}U^{\dagger})=T^{Q}(\rho^{A},\rho^{B}),\quad\text{
for any }\quad U\in\mathrm{U}(N).$

This stems from the fact that $(U\otimes U)C^{Q}(U^{\dagger}\otimes
U^{\dagger})=C^{Q}$, for any unitary matrix $U\in\mathrm{U}(N)$.

The latter property, however, does not hold for a general quantum cost matrix
(1.2). This implies that one cannot expect the monotonicity to hold in full
generality for $T^{Q}_{E}$, even under unitary channels. Indeed, let us take
two qutrit states

$\displaystyle\rho^{A}=\tfrac{1}{2}\big{(}|1\rangle+|2\rangle\big{)}\big{(}\langle
1|+\langle
2|\big{)},\qquad\rho^{B}=\tfrac{1}{5}\big{(}|1\rangle+2|2\rangle\big{)}\big{(}\langle
1|+2\langle 2|\big{)}$

and the quantum cost matrix induced by the line geometry,

$\displaystyle
C^{Q}_{E}=|\psi^{-}_{12}\rangle\langle\psi^{-}_{12}|+|\psi^{-}_{23}\rangle\langle\psi^{-}_{23}|+2|\psi^{-}_{13}\rangle\langle\psi^{-}_{13}|.$

Consider now a unitary channel, which interchanges the states $|2\rangle$ and
$|3\rangle$,

$U=\begin{pmatrix}1&0&0\\\ 0&0&-1\\\ 0&1&0\\\ \end{pmatrix}$

and yields

$\displaystyle\eta^{A}$
$\displaystyle=U\rho^{A}U^{\dagger}=\tfrac{1}{2}\big{(}|1\rangle+|3\rangle\big{)}\big{(}\langle
1|+\langle 3|\big{)},$ $\displaystyle\eta^{B}$
$\displaystyle=U\rho^{B}U^{\dagger}=\tfrac{1}{5}\big{(}|1\rangle+2|3\rangle\big{)}\big{(}\langle
1|+2\langle 3|\big{)}.$

Since all of the involved states are pure, there is only one coupling matrix
for both pairs (transport2, , Lemma A.3), $\rho^{AB}=\rho^{A}\otimes\rho^{B}$
and $\eta^{AB}=\eta^{A}\otimes\eta^{B}$. We thus have

$\displaystyle T^{Q}_{E}(U\rho^{A}U^{\dagger},U\rho^{B}U^{\dagger})=\Tr
C^{Q}_{E}\eta^{AB}=\tfrac{1}{10}>T^{Q}_{E}(\rho^{A},\rho^{B})=\Tr
C^{Q}_{E}\rho^{AB}=\tfrac{1}{20}.$

Let us note that the lack of unitary invariance of a distance on the space of
quantum states can be a desirable property in certain applications. In
particular, the quantum Wasserstein distance of order 1 proposed in PMTL20 ,
which is not unitarily invariant, offers an improved efficiency in quantum
learning algorithms KdPMLL21 .

##  3 The lack of Riemannian structure

Let us now focus on the case of the projective cost matrix $C^{Q}$. The
corresponding transport cost (1.6) can be associated with _SWAP-fidelity_
transport1 ,

$F_{S}(\rho^{A},\rho^{B})\vcentcolon=\max_{\rho^{AB}\in\Gamma^{Q}(\rho^{A},\rho^{B})}\big{(}\Tr\,S\rho^{AB}\big{)}=1-2T^{Q}(\rho^{A},\rho^{B}).$
(3.1)

Clearly, the monotonicity of $T^{Q}$ under CPTP maps is equivalent to the
reverse monotonicity of the SWAP-fidelity,
$F_{S}(\Phi(\rho^{A}),\Phi(\rho^{B}))\geq F_{S}(\rho^{A},\rho^{B})$. The SWAP-
fidelity shares many properties with the standard Uhlmann–Jozsa quantum
fidelity Uh76 ; Jo94  and equals to the latter if either of the states
$\rho^{A},\rho^{B}$ is pure transport1 .

The quantum fidelity $F$ is indeed reverse monotone under all CPTP maps Jo94 .
Consequently, any distance on the space of quantum state determined by a
strictly decreasing function $h:[0,1]\to\mathbb{R}^{+}$,
$D_{h,F}(\rho^{A},\rho^{B})=h(F(\rho^{A},\rho^{B}))$, will automatically be
monotone. The prominent examples include the root infidelity,
$I\vcentcolon=\sqrt{1-F}$, the Bures distance,
$B\vcentcolon=\sqrt{2(1-\sqrt{F})}$ and the Bures angle,
$A\vcentcolon=\tfrac{2}{\pi}\arccos\sqrt{F}$. As the SWAP-fidelity (3.1)
shares many important properties with the standard Uhlmann–Jozsa fidelity, one
might expect it to be reverse monotone as well.

All of the above-mentioned fidelity-based distances generate Riemannian
metrics on the tangent bundle of $\Omega_{N}$, for any $N$. More precisely
(see e.g. SIOR17  and Amari1 ), a semi-distance
$d:\Omega_{N}\times\Omega_{N}\to\mathbb{R}^{+}$ generates a Riemannian metric
$g$ on $T_{\rho}\Omega_{N}$ if the expansion

$\displaystyle d(\rho,\rho+tv)^{2}=g_{\rho}(v,v)t^{2}+o(t^{2}),\text{ as }t\to
0^{+},$ (3.2)

holds for any $v\in\Omega_{N}$ and any $v\in T_{\rho}\Omega_{N}=\\{v\in
H_{N},\text{ with }\Tr v=0\\}$.

All monotone distances on the space of quantum states which generate a
Riemannian metric, are characterised via the Morozova–Cencov–Petz theorem
Morozova ; Petz1 . The latter gives an explicit formula for the metric $g$ in
terms of a single operator monotone function. However, there are important
distances, which do _not_ generate a Riemannian metric, but are nevertheless
monotone. A classical example is the $l_{1}$-distance (the ‘taxicab
distance’), which corresponds to the trace distance in the quantum case. It
turns out that the quantum 2-Wasserstein distance (1.6) shares this feature
with the trace distance.

Let us focus on the case $N=2$ and show the failure of (3.2) for the quantum
2-Wasserstein distance (1.6). Because of the unitary invariance of $W$ we can
restrict ourselves to the states in the real slice of the Bloch ball,
$\Omega_{2}^{\mathbb{R}}\subset\Omega_{2}$,

$\displaystyle\rho(r,\theta)=\tfrac{1}{2}\big{[}\mathbbm{1}+(2r-1)(\sigma_{1}\sin\theta+\sigma_{3}\cos\theta)\big{]},\quad\text{
with }r\in[0,1],\theta\in[0,\pi),$ (3.3)

where $\sigma_{i}$ denote the Pauli matrices. Consequently,
$T_{\rho(r,\theta)}\Omega_{2}^{\mathbb{R}}$ is a real vector space
diffeomorphic to $\mathbb{R}^{2}$. Furthermore, by unitary invariance, it is
sufficient to consider the tangent space at the point $\rho(r,0)$. Basing on
the results of transport2  we can compute the square of the metric derivative
(see e.g. AGS05  for a precise definition) of the quantum 2-Wasserstein
distance $W$ on $\Omega_{2}$ in the direction of a vector $v\in
T_{\rho(r,0)}\Omega_{2}^{\mathbb{R}}$.

######  Proposition 3.1.

For any $r\in(0,1)$ and any tangent vector $v=(v_{1},v_{2})\in
T_{\rho(r,0)}\Omega_{2}^{\mathbb{R}}$ we have

$\displaystyle G(r,v)$ $\displaystyle\vcentcolon=\lim_{t\to
0^{+}}\frac{T^{Q}\big{(}\rho(r,0),\rho(r+tv_{1},tv_{2})\big{)}}{t^{2}}$
$\displaystyle=\max_{\phi\in[0,2\pi)}\frac{\big{(}2v_{1}\cos(\phi)-(2r-1)v_{2}\sin(\phi)\big{)}^{2}}{16\big{(}1+(2r-1)\cos(\phi)\big{)}}$
(3.4)

Proof: The proof, which bases on a semi-analytical formula derived in
transport2 , can be found in Appendix A.

If $W$ would generate a Riemannian metric $g$ on $T\Omega_{2}$, then formula
(3.2) would imply that $g$ can be recovered from the equality

$\displaystyle
G(r,v)=g_{\rho(r,0)}(v,v)=g_{11}(r)v_{1}^{2}+2g_{12}(r)v_{1}v_{2}+g_{22}(r)v_{2}^{2},$
(3.5)

which should be valid for all $r\in(0,1)$ and $v\in\mathbb{R}^{2}$. The
functions $g_{11}$ and $g_{22}$ can be computed explicitly from the analytic
formulae for the 2-Wasserstein distance between commuting and isospectral
qubits, (A.2) and (A.3), respectively. On the other hand, one can convince
oneself with the help of formula (3.4) that $g_{12}$ depends not only on $r$,
but also on the tangent vector $v$ – see Fig. 1. Hence, formula (3.5) fails
and the square of the metric derivative (3.4) of the 2-Wasserstein distance
does _not_ induce a Riemannian metric on $T\Omega_{2}$.

Figure 1: The plot of
$h(r,v_{2})\vcentcolon=G\big{(}r,(1,v_{2})\big{)}-g_{11}(r)-g_{22}(r)v_{2}^{2}$
as a function of $v_{2}$ for $r=1/3$, $r=1/4$ and $r=1/5$ — colours orange,
red and purple, respectively. If Formula (3.5) would be true then, for any
$r$, $h(r,v_{2})$ would be a linear function of $v_{2}$, which clearly is not
the case.

Finally, let us note that the failure of formula (3.5) extends to the general
case of $\Omega_{N}$ and any quantum cost matrix $C^{Q}_{E}$. This is because
$\Omega_{2}$ is embedded isometrically in $\Omega_{N}$ for any $N\geq 3$ and
the restriction of any $C^{Q}_{E}$ to a suitable four-dimensional subspace of
$\mathbb{C}^{N\times N}$ is proportional to $C^{Q}$ — see (transport2, , Prop.
2.4).

##  4 Monotonicity under general mixed unitary channels

Recall that the CPTP maps can be conveniently characterised with the help of
Kraus operators. For any channel $\Phi:\Omega_{N}\to\Omega_{M}$ there exists a
set of $N\times M$ matrices $\\{K_{i}\\}_{i=1}^{R}$, with $R\leq NM$, called
Kraus operators, such that

$\sum_{i}K_{i}^{\dagger}K_{i}=\mathbbm{1}_{N}$ (4.1)

and for any $\rho\in\Omega_{N}$

$\Phi(\rho)=\sum_{i}K_{i}\rho K_{i}^{\dagger}.$ (4.2)

Conversely, quantum channels can be constructed through Kraus operators, since
every set of $N\times M$ matrices satisfying (4.1) defines a CPTP map via
(4.2).

A particular class of quantum channels arises from the statistical mixtures of
unitary maps. In such a case, we have $K_{i}=\sqrt{p_{i}}U_{i}$, where
$U_{i}\in U(N)$ and $p_{i}\geq 0$, $\sum_{i}p_{i}=1$. The general properties
of the projective quantum transport cost imply the following result:

######  Proposition 4.1.

Let $\Psi$ be a mixed unitary channel on $\Omega_{N}$, then for any
$\rho^{A},\rho^{B}\in\Omega_{N}$ we have

$\displaystyle T^{Q}\big{(}\Psi(\rho^{A}),\Psi(\rho^{B})\big{)}\leq
T^{Q}(\rho^{A},\rho^{B}).$

Proof: This follows directly from the joint convexity of the transport cost
(transport2, , Proposition 2.2) and the unitary invariance of $T^{Q}$.

The convexity of the optimal quantum transport cost $T^{Q}_{E}$ implies a more
general result, which holds for any quantum cost matrix (1.2).

######  Proposition 4.2.

Let $\Phi_{1},\ldots,\Phi_{R}$ be quantum channels under which the optimal
quantum transport cost is monotonous:

$T_{E}^{Q}\big{(}\Phi_{i}(\rho^{A}),\Phi_{i}(\rho^{B})\big{)}\leq
T_{E}^{Q}(\rho^{A},\rho^{B})\quad\text{ for all
}\;\rho^{A},\rho^{B}\in\Omega_{N}\text{ and }i\in\\{1,\ldots,R\\}.$

Then, $T^{Q}_{E}$ is monotonous under any convex combination of the channels
$\Phi_{1},\ldots,\Phi_{R}$,

$T_{E}^{Q}\big{(}\Psi(\rho^{A}),\Psi(\rho^{B})\big{)}\leq
T_{E}^{Q}(\rho^{A},\rho^{B})\quad\text{ for all
}\;\Psi=\sum_{i}p_{i}\Phi_{i},\text{ with }p_{i}\geq 0,\sum_{i}p_{i}=1.$

The last result implies that it is sufficient to study the monotonicity of the
optimal quantum transport under the _extremal channels_ , i.e. the ones which
cannot be decomposed as a convex combination of CPTP maps.

We can also invoke the dual formulation (1.8) to derive the following useful
result:

######  Proposition 4.3.

Let $\Phi:\Omega_{N}\to\Omega_{M}$ be a CPTP map characterised by Kraus
operators $\\{K_{i}\\}$ and denote by $\Phi^{*}:H_{M}\to H_{N}$ the map dual
to $\Phi$, ie. $\Phi^{*}(\sigma)=\sum_{i}K_{i}^{\dagger}\sigma K_{i}$. If for
any pair of matrices $\big{(}\sigma^{A},\sigma^{B}\big{)}\in\Sigma_{M}$, with
$\Sigma_{M}$ defined in (1.7), we have
$\big{(}\Phi^{*}(\sigma^{A}),\Phi^{*}(\sigma^{B})\big{)}\in\Sigma_{N}$, then
$T^{Q}_{E}$ is monotone with respect to the map $\Phi$,
$T^{Q}_{E}\big{(}\Phi(\rho^{A}),\Phi(\rho^{B})\big{)}\leq
T^{Q}_{E}(\rho^{A},\rho^{B})$.

Proof: Clearly, if $\sigma$ is a Hermitian matrix, then so is
$\Phi^{*}(\sigma)$. From the dual formulation of the optimal quantum transport
problem (1.8), we deduce

$\displaystyle T^{Q}_{E}\big{(}\Phi(\rho^{A}),\Phi(\rho^{B})\big{)}$
$\displaystyle=\sup_{(\sigma^{A},\sigma^{B})\in\Sigma_{M}}\Tr(\sigma^{A}\Phi(\rho^{A})+\sigma^{B}\Phi(\rho^{B}))$
$\displaystyle=\sup_{(\sigma^{A},\sigma^{B})\in\Sigma_{M}}\Tr(\Phi^{*}(\sigma^{A})\rho^{A}+\Phi^{*}(\sigma^{B})\rho^{B})$
$\displaystyle\leq\sup_{(\eta^{A},\eta^{B})\in\Sigma_{N}}\Tr\big(\eta^{A}\rho^{A}+\eta^{B}\rho^{B}\big{missing})=T^{Q}_{E}(\rho^{A},\rho^{B}).$

##  5 Monotonicity under arbitrary single-qubit channels

Let us now focus on the single qubit case, $N=2$. If we write a state
$\rho\in\Omega_{2}$ using its Bloch vector,
$\rho(\vec{r})=\frac{1}{2}\big{(}\mathbbm{1}+\vec{r}\cdot\vec{\sigma}\big{)}$,
then the action of any quantum channel $\Phi:\Omega_{2}\to\Omega_{2}$ can be
expressed KR01  in terms of a real 3 $\times$ 3 matrix $M$ and a translation
vector $\vec{t}\in\mathbb{R}^{3}$,

$\displaystyle\Phi(\rho(\vec{r}))=\rho\big{(}M\vec{r}+\vec{t}\,\big{)},$

such that $\|M\vec{r}+\vec{t}\,\|\leq 1$. Furthermore, the matrix $M$ can be
diagonalised with the help of two unitary rotations $U,V\in\mathrm{U}(2)$ as
follows,

$\displaystyle\Phi(\rho)=U\big{[}\Phi_{d}\big{(}V\rho
V^{\dagger}\big{)}\big{]}U^{\dagger},$ (5.1)

where $\Phi_{d}$ is a quantum channel with a diagonal matrix $M$.

In RSW01  it was shown that any such single-qubit map $\Phi_{d}$, which
belongs to the closure of the set of extremal quantum channels, can be
realised with two Kraus operators

$\displaystyle K_{1}$
$\displaystyle=\left[\cos\left(\tfrac{1}{2}v\right)\cos\left(\tfrac{1}{2}u\right)\right]\mathbbm{1}+\left[\sin\left(\tfrac{1}{2}v\right)\sin\left(\tfrac{1}{2}u\right)\right]\sigma_{3}\
,$ (5.2) $\displaystyle K_{2}$
$\displaystyle=\left[\sin\left(\tfrac{1}{2}v\right)\cos\left(\tfrac{1}{2}u\right)\right]\sigma_{1}+i\left[\cos\left(\tfrac{1}{2}v\right)\sin\left(\tfrac{1}{2}u\right)\right]\sigma_{2}\
,$

with $u\in[0,2\pi)$ and $v\in[0,\pi)$. For such quantum channels we have the
following technical result proven in Appendix A.

######  Lemma 5.1.

Let $\Phi:\Omega_{2}\to\Omega_{2}$ be a CPTP map determined by Kraus operators
(5.2) and assume that $\sigma^{A},\sigma^{B}\in H_{2}$ are diagonal. If
$F=C^{Q}-\sigma^{A}\otimes\mathbbm{1}_{N}-\mathbbm{1}_{N}\otimes\sigma^{B}\geq
0$ then also
$F^{\Phi}=C^{Q}-\Phi^{*}(\sigma^{A})\otimes\mathbbm{1}_{N}-\mathbbm{1}_{N}\otimes\Phi^{*}(\sigma^{B})\geq
0$.

It turns out that the matrices $\sigma^{A},\sigma^{B}$ yielding the maximum in
(1.8) actually commute and thus can be simultaneously diagonalised.

######  Lemma 5.2.

Let $\rho^{A}$, $\rho^{B}\in\Omega_{2}$ be mixed quantum states of full rank
and with different spectra, and let $\underline{\sigma}^{A}$,
$\underline{\sigma}^{B}$ be two Hermitian matrices saturating the dual problem
(1.8), i.e.

$T^{Q}(\rho^{A},\rho^{B})=\Tr\big(\underline{\sigma}^{A}\rho^{A}+\underline{\sigma}^{B}\rho^{B}\big{missing}).$

Then, $\underline{\sigma}^{A}$ and $\underline{\sigma}^{B}$ commute.

Proof: This fact is shown in the course of the proof of Theorem 5.1 in
transport2 .

We are now in a position to present the main result of this paper.

######  Theorem 5.3.

The optimal quantum transport cost is monotonous under all CPTP maps
$\Phi:\Omega_{2}\to\Omega_{2}$.

Proof: Thanks to the unitary invariance of the transport cost $T^{Q}$ we can
restrict ourselves to channels $\Phi_{d}$ defined by Eq. (5.1). Furthermore,
by Proposition 4.2, it is sufficient to check the monotonicity for extremal
single-qubit channels. Such channels are completely characterised by two Kraus
operators (5.2), with $u\in[0,2\pi)$, $v\in[0,\pi)$ such that either $\sin
u\cdot\sin v\neq 0$ or $|\cos u\cdot\cos v|<1$ (see Theorem 1.4 in RSW01 ). We
can thus use the characterisation (5.2) and apply Lemma 5.2.

Now, assume that the states $\rho^{A},\rho^{B}\in\Omega_{2}$ are non-
isospectral and neither of them is pure. By Lemma 5.2 we know that the
Hermitian matrices saturating the dual problem,
$\underline{\sigma}^{A},\underline{\sigma}^{B}$, commute. Let then $V\in U(2)$
be the matrix diagonalizing simultaneously $\underline{\sigma}^{A}$ and
$\underline{\sigma}^{B}$. Using the unitary invariance of $T^{Q}$ we obtain

$\displaystyle T^{Q}\big{(}\Phi(\rho^{A}),\Phi(\rho^{B})\big{)}$
$\displaystyle=T^{Q}\Big{(}V^{\dagger}\Phi(\rho^{A})V,V^{\dagger}\Phi(\rho^{B})V\Big{)}$
$\displaystyle=\max_{(\sigma^{A},\sigma^{B})\in\Sigma_{2}}\Tr(\sigma^{A}V^{\dagger}\Phi(\rho^{A})V+\sigma^{B}V^{\dagger}\Phi(\rho^{B})V)$
$\displaystyle=\Tr(\underline{\sigma}^{A}V^{\dagger}\Phi(\rho^{A})V+\underline{\sigma}^{B}V^{\dagger}\Phi(\rho^{B})V)$
$\displaystyle=\Tr(\Phi^{*}(V\underline{\sigma}^{A}V^{\dagger})\rho^{A}+\Phi^{*}(V\underline{\sigma}^{B}V^{\dagger})\rho^{B}).$

Now, Lemma 5.1 implies that
$\big{(}\Phi^{*}(V\underline{\sigma}^{A}V^{\dagger}),\Phi^{*}(V\underline{\sigma}^{B}V^{\dagger})\big{)}\in\Sigma_{2}$,
hence we conclude that

$\displaystyle T^{Q}\big{(}\Phi(\rho^{A}),\Phi(\rho^{B})\big{)}$
$\displaystyle=\Tr(\Phi^{*}(V\underline{\sigma}^{A}V^{\dagger})\rho^{A}+\Phi^{*}(V\underline{\sigma}^{B}V^{\dagger})\rho^{B})$
$\displaystyle\leq\sup_{(\eta^{A},\eta^{B})\in\Sigma_{2}}\Tr\big(\sigma^{A}\rho^{A}+\sigma^{B}\rho^{B}\big{missing})=T^{Q}(\rho^{A},\rho^{B}).$

This inequality extends by the continuity of $T^{Q}$ and $\Phi$ to the
limiting cases of isospectral and pure states $\rho^{A},\rho^{B}$.

Acknowledgements: It is a pleasure to thank Tomasz Miller for his helpful
remarks. Financial support by Narodowe Centrum Nauki under the Maestro grant
number DEC- 2015/18/A/ST2/00274 and by Foundation for Polish Science under the
Team-Net project no. POIR.04.04.00-00-17C1/18-00 is gratefully acknowledged.

##  Appendix A Proofs of technical results

###  A.1 $N=2$: Semi-analytic formulae for distance $W$

In this Appendix we present the proof of Proposition 3.1, along with the semi-
analytic and analytic formulae for the optimal quantum transport cost for
qubits, which we invoked in Section 3.

We first recall a semi-analytic formula for $T^{Q}$ between two single-qubit
states in the Bloch parametrisation (3.3) derived in transport2 :

$T^{Q}\big{(}\rho(s,0),\rho(r,\theta)\big{)}\\\
=\max_{\phi\in[0,2\pi)}\tfrac{1}{4}\left(\sqrt{1+(2s-1)\cos\phi}-\sqrt{1+(2r-1)\cos(\theta+\phi)}\right)^{2}.$
(A.1)

For the special case of commuting or isospectral states one can derive
transport1  explicit analytic formulae:

$\displaystyle T^{Q}\big{(}\rho(s,0),\rho(r,0)\big{)}$
$\displaystyle=\tfrac{1}{2}\max\left\\{\left(\sqrt{r}-\sqrt{s}\right)^{2},\left(\sqrt{1-r}-\sqrt{1-s}\right)^{2}\right\\},$
(A.2) $\displaystyle T^{Q}\big{(}\rho(r,0),\rho(r,\theta)\big{)}$
$\displaystyle=\left(\tfrac{1}{2}-\sqrt{r(1-r)}\right)\sin^{2}(\theta/2).$
(A.3)

We shall now prove Proposition (3.1) via the following lemma:

######  Lemma A.1.

For any $r\in(0,1)$ and any tangent vector $v=(v_{1},v_{2})\in
T_{\rho(r,0)}\Omega_{2}^{\mathbb{R}}$ we have

$\displaystyle
T^{Q}\big{(}\rho(r,0),\rho(r+v_{1}t,v_{2}t)\big{)}=G(r,v)\,t^{2}+O(t^{3}),\quad\text{
as }t\downarrow 0,$ (A.4)

where the function $G$ is defined by formula (3.4),

$\displaystyle
G(r,v)=\max_{\phi\in[0,2\pi)}\frac{\big{(}2v_{1}\cos(\phi)-(2r-1)v_{2}\sin(\phi)\big{)}^{2}}{16\big{(}1+(2r-1)\cos(\phi)\big{)}}.$

Proof: In order to slightly simplify the notation we set
$\xi\vcentcolon=2r-1\in(-1,1)$ and assume that $\xi\neq 0$, i.e. $r\neq 1/2$.
Note that the maximally mixed state $\rho(1/2,0)$ commutes with any other
state $\rho^{B}$ in the Bloch ball, hence one can use Formula (A.2) to compute
the value of $T^{Q}\big{(}\rho(1/2,0),\rho^{B}\big{)}$ (cf. Eq. (12) in
transport1 ).

Let us start with writing, for sufficiently small $t>0$,

$\displaystyle W\big{(}\rho(r,0),\rho(r+v_{1}t,v_{2}t)\big{)}$
$\displaystyle=\tfrac{1}{2}\max_{\phi\in[0,2\pi)}\left|\sqrt{1+\xi\cos\phi}-\sqrt{1+(\xi+2v_{1}t)\cos(\phi+v_{2}t)}\right|$
$\displaystyle\leq\sqrt{G(r,v)}\,t+\tfrac{1}{2}\,\max_{\phi\in[0,2\pi)}R(\xi,v,\phi,t),$
(A.5)

where

$\displaystyle R(\xi,v,\phi,t)$ (A.6)
$\displaystyle\qquad\vcentcolon=\left|\sqrt{1+(\xi+2v_{1}t)\cos(\phi+v_{2}t)}-\sqrt{1+\xi\cos\phi}-\frac{2v_{1}t\cos\phi-\xi
v_{2}t\sin\phi}{2\sqrt{1+\xi\cos\phi}}\right|.$

Our goal is to establish a uniform bound on $R(\xi,v,\phi,t)$, which is at
least quadratic in $t$ and does not depend on $\phi$.

Let us identify the three terms in Formula (A.6),

$\displaystyle A$ $\displaystyle\vcentcolon=1+(\xi+2v_{1}t)\cos(\phi+v_{2}t),$
$\displaystyle B\vcentcolon=1+\xi\cos\phi,$ $\displaystyle C$
$\displaystyle\vcentcolon=2v_{1}t\cos\phi-\xi v_{2}t\sin\phi.$

Then, we have

$\displaystyle R$
$\displaystyle=\left|\sqrt{A}-\sqrt{B}-\frac{C}{2\sqrt{B}}\right|=\left|\frac{A-B-C}{\sqrt{A}+\sqrt{B}}+\frac{C(B-A)}{2\sqrt{B}(\sqrt{A}+\sqrt{B})^{2}}\right|$
$\displaystyle\leq\frac{|A-B-C|}{\sqrt{B}}+\frac{|C|\cdot|A-B|}{2B^{3/2}}.$

The most obvious estimates are

$\displaystyle B\geq 1-|\xi|,$ $\displaystyle|C|\leq\big{(}2|v_{1}|+|\xi
v_{2}|\big{)}t.$

In order to establish suitable bounds on the remaining terms we will use the
sum-to-product trigonometric identities AS72 . We have

$\displaystyle|A-B-C|$
$\displaystyle=\Big{|}(\xi+2v_{1}t)\big{[}\cos(\phi+v_{2}t)-\cos\phi\big{]}+\xi
v_{2}t\sin\phi\Big{|}$
$\displaystyle=\Big{|}2(\xi+2v_{1}t)\sin\left(\phi+\tfrac{v_{2}t}{2}\right)\sin\left(\tfrac{v_{2}t}{2}\right)-\xi
v_{2}t\sin\phi\Big{|}$ $\displaystyle\leq
2|\xi|\cdot\Big{|}\sin\left(\phi+\tfrac{v_{2}t}{2}\right)\sin\left(\tfrac{v_{2}t}{2}\right)-\tfrac{1}{2}\xi
v_{2}t\sin\phi\Big{|}+4t|v_{1}|\cdot\big{|}\sin\left(\tfrac{v_{2}t}{2}\right)\big{|}$
$\displaystyle\leq
2|\xi|\cdot\Big{[}\big{|}\sin\left(\phi+\tfrac{v_{2}t}{2}\right)-\sin\phi\big{|}\cdot\big{|}\sin\left(\tfrac{v_{2}t}{2}\right)\big{|}+$
$\displaystyle\qquad\qquad\qquad\qquad\qquad+|\sin\phi|\cdot\big{|}\sin\left(\tfrac{v_{2}t}{2}\right)-\tfrac{v_{2}t}{2}\big{|}\Big{]}+2|v_{1}v_{2}|t^{2}$
$\displaystyle\leq
2|\xi|\cdot\Big{[}\big{|}\sin\left(\tfrac{v_{2}t}{4}\right)\big{|}\cdot\big{|}\cos\left(\phi+\tfrac{v_{2}t}{4}\right)\big{|}\cdot|v_{2}|t+\tfrac{|v_{2}|^{3}t^{3}}{48}\Big{]}+2|v_{1}v_{2}|t^{2}$
$\displaystyle\leq\tfrac{1}{2}\big{(}|\xi
v_{2}^{2}|+4|v_{1}v_{2}|\big{)}t^{2}+\tfrac{1}{24}|v_{2}|^{3}t^{3}.$

Similarly, we obtain

$\displaystyle|A-B|$
$\displaystyle=\Big{|}\xi\big{[}\cos(\phi+v_{2}t)-\cos\phi\big{]}+2v_{1}t\,\cos(\phi+v_{2}t)\Big{|}$
$\displaystyle\leq
2|\xi|\cdot\big{|}\sin\left(\phi+\tfrac{v_{2}t}{2}\right)\big{|}\cdot\big{|}\sin\left(\tfrac{v_{2}t}{2}\right)\big{|}+2t|v_{1}|$
$\displaystyle\leq\big{(}|\xi v_{2}|+2|v_{1}|\big{)}t.$

Summa summarum, we arrive at the following estimate

$\displaystyle R(\xi,v,\phi,t)\leq\left(\frac{|\xi
v_{2}^{2}|+4|v_{1}v_{2}|}{2\sqrt{1-|\xi|}}+\frac{\big{(}|\xi
v_{2}|+2|v_{1}|\big{)}^{2}}{(1-|\xi|)^{3/2}}\right)t^{2}+\frac{|v_{2}|^{3}}{24\sqrt{1-|\xi|}}\,t^{3}.$

This provides the desired uniform bound, which we can plug into Formula (A.5)
and conclude that

$\displaystyle W\big{(}\rho(r,0),\rho(r+v_{1}t,v_{2}t)\big{)}$
$\displaystyle\leq\sqrt{G(r,v)}\,t+c_{1}t^{2}+c_{2}t^{3}.$

The square of the above equality implies Eq. (A.4).

One can show (cf. (transport2, , Appendix B)) that the angle $\phi$ yielding
the maximum in Eq. (3.4) defining the function $G$ can be obtained from the
solution to a quartic equation. For the specific cases of two commuting or
isospectral qubits we have

$\displaystyle
G\big{(}(r,(1,0)\big{)}=\tfrac{1}{8}\max\big{\\{}\tfrac{1}{1-r},\tfrac{1}{r}\big{\\}},$
$\displaystyle
G\big{(}(r,(0,1)\big{)}=\tfrac{1}{8}\left(1-2\sqrt{r(1-r)}\right).$ (A.7)

Equipped with these formulae one can reproduce Fig. 1 and convince oneself
that Eq. (3.5) fails.

###  A.2 Proof of Lemma 5.1

In this Appendix we present the proof of Lemma 5.1, which is the key to
Theorem 5.3. Let us fix a single-qubit channel $\Phi$ determined by two Kraus
operators (5.2). The channel, as well as its dual,
$\Phi^{*}(\sigma)=\sum_{i=1,2}K_{i}^{\dagger}\sigma K_{i}$, is parametrised by
two angles $u\in[0,2\pi)$ and $v\in[0,\pi)$.

It claims that if $\sigma^{A}$ and $\sigma^{B}$ are two diagonal Hermitian 2
$\times$ 2 matrices, then the positivity of the operator
$F=C^{Q}-\sigma^{A}\otimes\mathbbm{1}-\mathbbm{1}\otimes\sigma^{B}$, appearing
in the dual optimisation problem (1.8), implies the positivity of the operator
$F^{\Phi}=C^{Q}-\Phi^{*}(\sigma^{A})\otimes\mathbbm{1}-\mathbbm{1}\otimes\Phi^{*}(\sigma^{B})$,
for the extremal single-qubit channel $\Phi$ determined by Eq. (5.2).

Since both matrices $\sigma^{A}$ and $\sigma^{B}$ are diagonal, we have four
independent real parameters. Note, however, that a gauge transformation
$\sigma^{A}\mapsto\sigma^{A}+a~{}\mathbbm{1}$,
$\sigma^{B}\mapsto\sigma^{B}-a~{}\mathbbm{1}$, with any $a\in\mathbb{R}$, does
not affect neither the operator $F$, nor the value of the optimal quantum
transport cost (1.8). Consequently, we can eliminate one of the parameters are
left with three of them. It is convenient to use the following
parametrisation:

$\displaystyle c\vcentcolon=\Tr\big(\sigma^{A}\sigma_{3}\big{missing})\,,$
$\displaystyle d\vcentcolon=\Tr\big(\sigma^{B}\sigma_{3}\big{missing})\,,$
$\displaystyle x\vcentcolon=\Tr\big(\sigma^{A}+\sigma^{B}\big{missing}),$

where $\sigma_{3}$ denotes the third Pauli matrix. Then, the matrices $F$ and
$F^{\Phi}$ take the following form,

$\displaystyle F=\frac{1}{2}\,\begin{pmatrix}F_{11}&0&0&0\\\
0&F_{22}&F_{23}&0\\\ 0&F_{23}&F_{33}&0\\\ 0&0&0&F_{44}\end{pmatrix},$
$\displaystyle F^{\Phi}=\frac{1}{2}\begin{pmatrix}F^{\Phi}_{11}&0&0&0\\\
0&F^{\Phi}_{22}&F^{\Phi}_{23}&0\\\ 0&F^{\Phi}_{23}&F^{\Phi}_{33}&0\\\
0&0&0&F^{\Phi}_{44}\end{pmatrix},$ (A.8)

with entries:

$\displaystyle F_{11}$ $\displaystyle=-\left(c+d+x\right),$ $\displaystyle
F_{22}$ $\displaystyle=1-c+d-x,$ $\displaystyle F_{23}=F_{32}=-1,$ (A.9)
$\displaystyle F_{33}$ $\displaystyle=1+c-d-x,$ $\displaystyle F_{44}$
$\displaystyle=c+d-x,$ $\displaystyle F_{11}^{\Phi}$
$\displaystyle=-\left(c+d\right)\cos(u-v)-x,$ (A.10) $\displaystyle
F_{22}^{\Phi}$ $\displaystyle=1-c\cos(u-v)+d\cos(u+v)-x,$ $\displaystyle
F_{23}^{\Phi}=F_{32}^{\Phi}=-1,$ $\displaystyle F_{33}^{\Phi}$
$\displaystyle=1+c\cos(u+v)-d\cos(u-v)-x,$ $\displaystyle F_{44}^{\Phi}$
$\displaystyle=\left(c+d\right)\cos(u+v)-x.$

The demand $F\geq 0$ implies, in particular, the non-negativity of the its
diagonal elements, which yields the following constraints on the parameters,

$\displaystyle x\leq 0,$ $\displaystyle x\leq c+d\leq-x,$ $\displaystyle
x-1\leq c-d\leq 1-x.$ (A.11)

The non-negativity of the central 2 $\times$ 2 minor of $F$, which is the only
non-trivial one, gives an additional constraint

$\displaystyle(c-d)^{2}\leq x(x-2).$ (A.12)

To simplify notation let us set

$\displaystyle\alpha\vcentcolon=\cos(u-v)\quad\text{ and
}\quad\beta\vcentcolon=\cos(u+v).$

We start with showing that the diagonal elements of $F^{\Phi}$ are non-
negative. If $\alpha\in[0,1]$ then (A.11) yields

$F^{\Phi}_{11}=-\left(c+d\right)\alpha-x\geq x\big{(}\alpha-1\big{)}\geq 0,$

while if $\alpha\in[-1,0]$ then we obtain

$F^{\Phi}_{11}\geq-x(\alpha+1)\geq 0.$

Analogously, we deduce that $F^{\Phi}_{44}\geq 0$.

Let us now rewrite

$\displaystyle F_{22}^{\Phi}$
$\displaystyle=1-x-\tfrac{1}{2}(c+d)(\alpha-\beta)+\tfrac{1}{2}(d-c)(\alpha+\beta)$

and assume that $(\alpha-\beta)\geq 0$ and $(\alpha+\beta)\geq 0$. Then,
inequalities (A.11) yield,

$\displaystyle F_{22}^{\Phi}$ $\displaystyle\geq
1-x+\tfrac{1}{2}x(\alpha-\beta)+\tfrac{1}{2}(x-1)(\alpha+\beta)=\big{[}1-\tfrac{1}{2}(\alpha+\beta)\big{]}+\tfrac{1}{2}x(\alpha-1)\geq
0,$

because $\alpha\in[0,1]$ and $(\alpha+\beta)\in[0,2]$. Analogously, one shows
that $F_{22}^{\Phi}\geq 0$ under three other possible assumptions about the
signs of $\alpha-\beta$ and $\alpha+\beta$. Along the same lines, one can
prove that $F_{33}^{\Phi}\geq 0$.

Finally, let us consider the central 2 $\times$ 2 minor of $F^{\Phi}$. With
$\pm$ denoting the sign of $\alpha+\beta$ and using the previous estimates on
$F_{22}^{\Phi}$ and $F_{33}^{\Phi}$, along with constraint (A.12), we have

$\displaystyle F_{22}^{\Phi}F_{33}^{\Phi}-1$
$\displaystyle=\Big{[}1-x-\tfrac{1}{2}(c+d)(\alpha-\beta)-\tfrac{1}{2}(c-d)(\alpha+\beta)\Big{]}\times$
$\displaystyle\qquad\times\Big{[}1-x-\tfrac{1}{2}(c+d)(\alpha-\beta)+\tfrac{1}{2}(c-d)(\alpha+\beta)\Big{]}-1$
$\displaystyle\geq\Big{[}1-x\pm\tfrac{1}{2}x(\alpha-\beta)-\tfrac{1}{2}(c-d)(\alpha+\beta)\Big{]}\times$
$\displaystyle\qquad\times\Big{[}1-x\pm\tfrac{1}{2}x(\alpha-\beta)+\tfrac{1}{2}(c-d)(\alpha+\beta)\Big{]}-1$
$\displaystyle=\Big{[}1-x\Big{(}1\mp\tfrac{1}{2}(\alpha-\beta)\Big{)}\Big{]}^{2}-\tfrac{1}{4}(c-d)^{2}(\alpha+\beta)^{2}-1$
$\displaystyle\geq-2x\Big{(}1\mp\tfrac{1}{2}(\alpha-\beta)\Big{)}+x^{2}\Big{(}1\mp\tfrac{1}{2}(\alpha-\beta)\Big{)}^{2}-\tfrac{1}{4}x(x-2)(\alpha+\beta)^{2}$
$\displaystyle=x^{2}(1\mp\alpha)(1\pm\beta)-x\Big{[}\tfrac{1}{2}(1\mp\alpha)^{2}+\tfrac{1}{2}(1\pm\beta)^{2}+1+\alpha\beta\Big{]}\geq
0.$

The last inequality holds because $\alpha,\beta\in[-1,1]$, while $x\leq 0$
from constraints (A.11).

Since the central minor of $F^{\Phi}$ is the only non-trivial one we conclude
that, indeed, if $F$ is positive semi-definite, then $F^{\Phi}$ is so, for any
quantum channel determined by Kraus operators (5.2).

##  Appendix B Monotonicity: numerical results

In this section the results of numerical calculations are presented. We
performed Monte Carlo simulations in which we drew pairs of quantum states and
a quantum channel and checked how the optimal quantum transport cost between
the selected states changes after the application the channel.

We examined the monotonicity of the optimal quantum transport for qutrits
($N=3$) and ququarts ($N=4$) using two different types of tests. In the first
one we picked random quantum channels corresponding to dynamical Choi matrices
of a fixed rank. In the second one we employed random extremal quantum
channels, defined up to unitary pre- and post-processing of the quantum
states. Because of the unitary invariance of $T^{Q}$ and its convexity (recall
Prop. 4.2), the monotonicity of $T^{Q}$ under such channels implies the
monotonicity with respect to all CPTP maps.

To generate a random quantum state of order $N$ we first drew a rectangular
complex random matrix $X$ of size $N\times k$, with $k\leq N$. Setting
$\rho=XX^{\dagger}/\Tr[XX^{\dagger}]$ we ensure that a random matrix $\rho$
size $N$ and rank $k$ is non-negative and $\Tr\rho=1$. Our calculations show
that the rank of states does not affect significantly the results. Therefore
we focus on full rank states to maximize the explored space.

Random quantum channels $\Phi:\Omega_{N}\to\Omega_{N}$ of a given rank $k$
were generate via the Choi–Jamiołkowski isomorphism. We generated a random
$N^{2}\times k$ complex matrix $X$ and created an auxiliary $N^{2}\times
N^{2}$ positive matrix $\widetilde{Y}=XX^{\dagger}\geq 0$. By the
Choi–Jamiołkowski isomorphism a quantum state corresponds to a quantum channel
if the second of its partial traces is a totally mixed state. To assure this
property we defined $Y=\Tr_{1}\widetilde{Y}$, where $\Tr_{1}$ is the partial
trace on the first subsystem. Then, we set
$\rho_{\Phi}=\left(\mathbbm{1}_{N}\otimes
Y^{-1/2}\right)\widetilde{Y}\left(\mathbbm{1}_{N}\otimes Y^{-1/2}\right)$. For
further details on random quantum channel generation we encourage the Reader
to consult rand_quant ; ZPNC11 .

Having normalized $\rho_{\Phi}$, to calculate the action of channel $\Phi$ on
any state $\rho$ we used the definition of Choi–Jamiołkowski isomorphism:

$\displaystyle
N\Tr_{2}[\rho_{\Phi}~{}(\mathbbm{1}_{N}\otimes\rho)]=N\Tr_{2}\left[(\Phi\otimes\mathbbm{1})(|\psi^{+}\rangle\langle\psi^{+}|)~{}(\mathbbm{1}_{N}\otimes\rho)\right]=$
(B.1) $\displaystyle=\Tr_{2}\left[(\Phi\otimes\mathbbm{1}_{N\times
N})\left(\left(\sum_{i=1}^{N}|ii\rangle\right)\left(\sum_{j=1}^{N}\langle
jj|\right)\right)~{}(\mathbbm{1}_{N}\otimes\rho)\right]=$
$\displaystyle=\sum_{ij}\Phi(|i\rangle\langle j|)\Tr[|i\rangle\rho\langle
j|]=\Phi(\rho),$

where $|\psi^{+}\rangle=\frac{1}{\sqrt{N}}\sum_{i=1}^{N}|ii\rangle$ is a
maximally entangled state.

In the test of the second type we focused on extremal quantum channels. Using
the methods presented in rand_extr_chann  we described all such channels by
their Kraus operators, $K_{i}=U_{i}D_{i}$ with $i=1,\ldots,N$, where $U_{i}$
and $D_{i}$ denote, respectively, the unitary matrices and diagonal matrices
of order $N$. An explicit form of $U_{i}$ and $D_{i}$ for a qutrit reads,

$\displaystyle
D_{1}=\operatorname{diag}(a,b,c)~{},~{}D_{2}=\operatorname{diag}(d,e,f)~{},~{}D_{3}=\sqrt{1-D_{1}^{2}-D_{2}^{2}}~{},$
$\displaystyle U_{1}=\begin{pmatrix}1&0&0\\\ 0&0&1\\\
0&1&0\end{pmatrix},~{}~{}U_{2}=\begin{pmatrix}0&0&1\\\ 0&1&0\\\
1&0&0\end{pmatrix},~{}~{}U_{3}=\begin{pmatrix}0&1&0\\\ 1&0&0\\\
0&0&1\end{pmatrix}.$

An analogous parametrisation works for $N=4$ (see rand_extr_chann ). All of
the parameters in the above formulae are non-negative and chosen so that the
matrices $D_{3}$ (for the qutrit case) and $D_{4}$ (for the ququart case) are
positive semi-definite.

In each Monte Carlo simulation we generated a random pair of initial states
$\rho^{A},\rho^{B}$ of order $N$ and a quantum channel $\Phi$ using the method
presented above. Then, we calculated the quantum transport cost before and
after the application of the channel, $T^{Q}(\rho^{A},\rho^{B})$ and
$T^{Q}\big{(}\Phi(\rho^{A}),\Phi(\rho^{B})\big{)}$, respectively and compared
the results.

The simulations were programmed in the Python language, using Numpy library
for algebraic calculations, Cvxpy library for the transport costs calculations
and a solver supplied by the MOSEK Optimizer for solving semidefinite
programming problems. The plots were generated with help of the matplotlib
library. The numerical precision of calculations was set to $10^{-12}$. We
checked the accuracy of the optimisation algorithm by generating random
unitary channels, which correspond to $k=1$ in procedure described above. The
deviation between the obtained value an the actual value of $T^{Q}$ was not
greater than $10^{-9}$, hence we can take this value as the numerical accuracy
of the algorithm.

The results of the simulations are presented in Table 1 and Figure 2. In all
of the considered cases the difference
$T^{Q}(\rho^{A},\rho^{B})-T^{Q}(\Phi(\rho^{A}),\Phi(\rho^{B}))$ was positive
and a few orders of magnitude larger than the numerical precision. These
results allow us to conjecture that optimal quantum transport cost $T^{Q}$ for
dimensions $3$ and $4$ is monotonous under all CPTP maps.

|  $N=3$, qutrit |  $N=4$, ququart
---|---|---
Random channels | rank $k$ | samples | rank $k$ | samples
| 2 | 300 000 | 2 | 200 000
| 3 | 300 000 | 3 | 200 000
| 9 | 600 000 | 4 | 200 000
|  |  | 16 | 400 000
$\min\,(T^{Q}-T^{Q}\circ\Phi)$ |  | 0.00146 |  | 0.00644
Extremal channels | samples | 7 500 000 | samples | 7 500 000
$\min\,(T^{Q}-T^{Q}\circ\Phi)$ |  | 0.00023485 |  | 0.006197901
Table 1: The results of the Monte Carlo simulations discussed in the text.
‘Samples’ refers to number of drawn triples $(\rho^{A},\rho^{B},\Phi)$,
whereas ‘$\min(T^{Q}-T^{Q}\circ\Phi)$’ refers to the smallest difference
$T^{Q}(\rho^{A},\rho^{B})-T^{Q}(\Phi(\rho^{A}),\Phi(\rho^{B}))$ found in the
samples.

Figure 2: The results of the Monte Carlo simulations random density matrices
of order $N=3$ (top) and $N=4$ (bottom) discussed in the text. The optimal
quantum transport cost, $T^{Q}\circ\Phi$, after the action of a quantum
channel $\Phi$ is plotted against the initial value of $T^{Q}$. Results
obtained for channels of a fixed rank $k$ are presented on the left and the
extremal channels on the right. The dashed line marks the equality
$T^{Q}(\rho^{A},\rho^{B})=T^{Q}\big{(}\Phi(\rho^{A}),\Phi(\rho^{B})\big{)}$,
which is saturated for unitary channels ($k=1$).

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