# Quantum conditional entropy from information-theoretic principles

Sarah Brandsen<EMAIL_ADDRESS>Department of Physics, Duke
University, Durham, NC, USA 27708  Department of Mathematics and Statistics,
Institute for Quantum Science and Technology, University of Calgary, AB,
Canada T2N 1N4  Isabelle Jianing Geng  Department of Mathematics and
Statistics, Institute for Quantum Science and Technology, University of
Calgary, AB, Canada T2N 1N4  Mark M. Wilde  Hearne Institute for Theoretical
Physics, Department of Physics and Astronomy, and Center for Computation and
Technology, Louisiana State University, Baton Rouge, Louisiana 70803, USA
Gilad Gour  Department of Mathematics and Statistics, Institute for Quantum
Science and Technology, University of Calgary, AB, Canada T2N 1N4

###### Abstract

We introduce an axiomatic approach for characterizing quantum conditional
entropy. Our approach relies on two physically motivated axioms: monotonicity
under conditional majorization and additivity. We show that these two axioms
provide sufficient structure that enable us to derive several key properties
applicable to all quantum conditional entropies studied in the literature.
Specifically, we prove that any quantum conditional entropy _must_ be negative
on certain entangled states and must equal $-\log d$ on $d\times d$ maximally
entangled states. We also prove the non-negativity of conditional entropy on
separable states, and we provide a generic definition for the dual of a
quantum conditional entropy. Finally, we develop an operational approach for
characterizing quantum conditional entropy via games of chance, and we show
that, for the classical case, this complementary approach yields the same
ordering as the axiomatic approach.

##  I Introduction

Conditional entropy quantifies the uncertainty of a bipartite state shared
between Alice and Bob, given access to Bob’s system only. Previous work on
conditional entropy for quantum bipartite states has demonstrated that, unlike
classical conditional entropy, quantum conditional entropy can be negative
Cerf_1997  and achieves its minimum on maximally entangled states. The initial
discovery that quantum conditional entropy could be negative was so surprising
that various conjectures about its meaning were put forth by the authors of
Cerf_1997 , such as its potential relation to “anti-qubits,” which are quanta
of negative information. The most satisfying information theoretic
interpretation of quantum conditional entropy and its negativity is given by
the state merging protocol of Horodecki2005 , which demonstrates that
conditional entropy is the rate at which maximally entangled Bell states need
to be consumed in order transfer Alice’s share of a bipartite state to Bob,
who possesses the other share of the state. The importance of the existence of
negative conditional entropy is also reflected in the fact that there is an
information quantity, called coherent information PhysRevA.54.2629 , dedicated
to measuring how negative the conditional entropy of a bipartite quantum state
is. Multiple previous works have investigated topics related to conditional
entropy Liu_2018 ; Patro_2017 ; Capel_2018 ; Gour5 ; Kuznetsova2010QuantumCE ;
doi:10.1063/1.5027495 ; PhysRevA.99.062119 ; Gour ; PhysRevA.104.012417  and
have found additional applications of conditional entropy to quantum
cryptographic protocols Brown_2021 .

The central role of conditional entropy in quantum information theory
motivates the need for an axiomatic approach that defines the set of all
possible quantum conditional entropies. Previous literature has extensively
studied axiomatic derivations for other entropies e13111945 ; e10030261 ,
beginning with approaches deriving the Shannon entropy 6773024  and
progressing to a complete axiomatic approach for classical entropies and
relative entropies gour2021entropy . More recently, an axiomatic approach for
entropy based on core-concavity was developed and connections were made
between such core-concave functions and classical conditional entropy 9064819
.

In this work, we start with a minimal set of assumptions with strong
information-theoretic motivation and demonstrate that these simple assumptions
are sufficient for recovering key properties of quantum conditional entropy,
such as its negativity for maximally entangled states, non-negativity for
separable bipartite states, invariance under local unitaries, and reduction to
an entropy function on uncorrelated systems. At the core of our approach, we
fully characterise the set of channels that are conditional entropy non-
decreasing, and we require that conditional entropy be monotonic under the
action of such channels. More specifically, we require that any conditional
entropy non-decreasing channel maps a state of maximal entropy on Alice’s side
to an output state that also has maximal entropy on Alice’s side, a condition
which is satisfied by _conditionally unital_ channels. Likewise, given that
conditional entropy can be understood as measuring as how much information Bob
has about Alice’s system, any conditional entropy non-decreasing channel
cannot “leak” information from Alice’s system to Bob’s system. Thus,
conditional entropy non-decreasing channels must also satisfy the semi-
causality condition outlined in Beckman_2001 . In other words, we identify the
set of conditional entropy non-decreasing channels with channels that are both
conditionally unital and semi-causal (CUSC).

This gives rise to a natural majorization relation where
$\sigma_{AB}\precsim_{A}\rho_{AB}$ (i.e. $\rho_{AB}$ has less uncertainty than
$\sigma_{AB}$ with respect to system $A$) if $\sigma_{AB}$ can be obtained by
applying a CUSC channel to $\rho_{AB}$. Any conditional entropy is thus
required to be monotonic under this partial ordering which we call
_conditional majorization_. In Gour  a different definition of conditional
majorization (restricted to classical systems) was given, however, we prove
that the quantum conditional majorization defined in this paper yields the
exact same ordering as in Gour  on (bipartite) classical systems.

An alternative, operationally motivated means for characterizing uncertainty
is to utilize games of chance. It was found in brandsen2021entropy  that games
of chance provide an operational interpretation to conditional majorization
for classical bipartite states. Games of chance are ideal for studying
uncertainty, as the probability of winning a game with a given physical system
depends solely on the uncertainty of the system’s output. Such games of chance
thus lead to a natural partial ordering between channels, where state a
$\rho_{AB}$ is “less uncertain” than state $\sigma_{AB}$ if $\rho_{AB}$
performs at least as well as $\rho_{AB}$ for any game of chance. In this work,
we extend previous results by providing games of chance capable of
characterising uncertainty of quantum bipartite states.

Lastly, we provide an alternative approach for characterising the entropy of a
quantum channel Gour_2021 , which is operationally motivated by games of
chance. Our results reflect that entanglement is a resource Chitambar_2019 ;
Sparaciari2020firstlawofgeneral , as entanglement-preserving unitary channels
strictly outperform entanglement-breaking channels. We additionally compute
the reward function for key quantum gambling games with several special
classes of quantum channels, such as unitary, amplitude damping, depolarizing,
measurement, and dephasing channels.

##  II Notation

We denote quantum channels in uppercase calligraphic letters, where
$\mathcal{N}\in\operatorname{CPTP}\left(AB\rightarrow{A^{\prime}B^{\prime}}\right)$
denotes a completely positive, trace preserving map (i.e. quantum channel)
that takes as an input a bipartite state of systems $A$ and $B$ and outputs a
bipartite state of systems $A^{\prime}$ and $B^{\prime}$. At times, we use
subscripts to indicate the input and output system of a channel such as
$\mathcal{N}_{{AB}\to{A^{\prime}B^{\prime}}}$. The dimension of a system is
denoted as $|A|$, $|B|$, e.t.c.. We also use the tilde symbol to indicate a
replica of a system. For example, $\tilde{A}$ or $\tilde{B}$ represents copies
of $A$ and $B$, and in particular $|A|=|\tilde{A}|$ and $|B|=|\tilde{B}|$.

Likewise, we denote a classical channel by $\mathcal{T}$, whose corresponding
transition matrix is a column stochastic matrix
$T=(t_{w|z^{\prime}})_{w,z^{\prime}}$ such that each entry
$t_{w|z^{\prime}}\in[0,1]$ and $\sum_{w}t_{w|z^{\prime}}=1$ for each
$z^{\prime}$.

In the case in which the bipartite state has a trivial component (e.g.,
$|B|=|B^{\prime}|=1$), we omit the letters that refer to the trivial system
and denote the channel $\mathcal{N}$ as
$\mathcal{N}\in\operatorname{CPTP}\left(A\rightarrow A^{\prime}\right)$.
Throughout this work, the computational basis of a given system with dimension
$d$ is denoted as $\\{\ket{1},\ldots,\ket{d}\\}$. We denote the maximally
entangled Bell state for systems $A$ and $B$ (where $|A|=|B|$) as

$\ket{\phi_{AB}^{+}}\triangleq\frac{1}{\sqrt{|A|}}\sum_{j=1}^{|A|}\ket{jj}\ ,$

where $\ket{jj}\triangleq\ket{j}\otimes\ket{j}$, and denote the corresponding
unnormalized state as

$\ket{\Phi_{AB}}\triangleq\sum_{j=1}^{|A|}\ket{jj}.$

For a matrix $M$, the sum of its $w$ largest singular values will be denoted
as the Ky-fan $w$ norm $\|M\|_{\left(w\right)}$. We denote with $I_{A}$ the
identity operator on system $A$, and denote with
$\mathbf{u}_{A}=\frac{1}{|A|}I_{A}$ the maximally mixed state in system $A$.
When the systems on which the operators act are clear from the context, we
might omit the subscripts indicating the systems.

The Choi matrix $J^{\mathcal{N}}_{AB}$ for a given channel
$\mathcal{N}_{A\rightarrow B}$ is defined by the action of $\mathcal{N}$ on
the maximally entangled operator as

$\displaystyle
J^{\mathcal{N}}_{AB}=\sum_{i,j=1}^{|A|}\ket{i}\\!\\!\bra{j}\otimes\mathcal{N}(\ket{i}\\!\\!\bra{j}).$
(1)

Quantum measurements are denoted as $\hat{\Pi}=\\{\Pi^{(j)}\\}_{j}$ where
$\sum_{j}\Pi^{(j)}=I$ and for every $j$, $0\leq\Pi^{(j)}\leq I$. At times, we
use subscript to denote the system on which the measurements act, e.g.,
$\hat{\Pi}_{A}$ is a measurements on system $A$. Unless otherwise specified,
we consider measurements which have rank-one projective elements. Finally, the
set of density matrices corresponding to system $A$ is denoted as
$\mathfrak{D}\left(A\right)$, and $\mathbf{u}_{A}$ denotes the uniform (i.e.
maximally mixed) state in $\mathfrak{D}(A)$.

##  III Axiomatic Approach to Quantum Conditional Entropy

We begin by completely characterising the set of bipartite quantum channels
which are conditional entropy non-decreasing. A bipartite quantum channel
$\mathcal{N}$ is conditional entropy non-decreasing if for any bipartite state
$\rho_{AB}$, the state $\mathcal{N}(\rho_{AB})$ has at least as much
conditional entropy.

Since conditional entropy measures the uncertainty of system $A$ when one is
given access to system $B$, evidently allowing information to leak from system
$A$ to system $B$ could decrease conditional entropy. This gives rise to the
semi-causality requirement, which is defined as follows

######  Definition 1.

A bipartite channel $\mathcal{N}_{AB\rightarrow AB^{\prime}}$ is $A\not\to
B^{\prime}$ semi-causal if any channel $\mathcal{M}_{A\rightarrow A}$ that
Alice applies to her system cannot be detected by Bob. Formally, this
requirement can be stated as follows:

$\displaystyle\mathcal{N}_{AB\rightarrow
B^{\prime}}\circ\mathcal{M}_{A\rightarrow A}=\mathcal{N}_{AB\rightarrow
B^{\prime}},$ (2)

for all $\mathcal{M}\in\operatorname{CPTP}\left(A\rightarrow A\right)$ and
where $\mathcal{N}_{AB\rightarrow B^{\prime}}$ indicates that the channel
output in system $A$ is traced out.

See Figure 1 for a depiction of the semi-causality requirement.

\begin{overpic}[scale={0.5}]{semicausal} \put(20.0,44.0){$\mathcal{M}$}
\put(59.0,30.0){$\mathcal{N}$} \put(183.0,30.0){$\mathcal{N}$} \end{overpic}
Figure 1: Depiction of semi-causal channel $\mathcal{M}$, where the black
square represents the discarding channel.

We note that any semi-causal channel can be written as a local channel on $B$
which feeds an output (corresponding to a reference system $R$) into a channel
acting only on $A$ and $R$ (see Figure 2). As such, information can only flow
from $B$ to $A$, but never leak from $A$ to $B$.

######  Lemma 1.

If $\mathcal{N}_{AB\rightarrow AB^{\prime}}$ is an $A\not\to B^{\prime}$ semi-
causal channel, then there exists a reference system $R$, a quantum channel
$\mathcal{E}\in\operatorname{CPTP}\left(AR\rightarrow A\right)$, and an
isometry $\mathcal{F}\in\operatorname{CPTP}\left(B\rightarrow
RB^{\prime}\right)$ such that

$\displaystyle\mathcal{N}_{AB\rightarrow
AB^{\prime}}\left(\rho_{AB}\right)=\operatorname{Tr}_{R}\Big{[}\mathcal{E}_{RA\rightarrow
A}\circ\mathcal{F}_{B\rightarrow RB^{\prime}}\left(\rho_{AB}\right)\Big{]}$
(3)

for every state $\rho_{AB}$.

###### Proof.

Refer to Piani_2006  for a proof. ∎

\begin{overpic}[scale={0.6}]{semicausalfig2}
\put(62.0,80.0){$\mathcal{N}_{AB\to AB^{\prime}}$}
\put(43.0,22.0){$\mathcal{F}$} \put(112.0,55.0){$\mathcal{E}$}
\put(67.0,32.0){$R$} \put(8.0,25.0){$B$} \put(8.0,65.0){$A$}
\put(147.0,62.0){$A$} \put(147.0,20.0){$B^{\prime}$} \end{overpic} Figure 2: A
semi-causal bipartite channel $\mathcal{N}_{AB\rightarrow AB^{\prime}}$.

Finally, we reframe the definition of semi-causality in terms of Choi matrices

######  Lemma 2.

Let $\mathcal{N}\in\text{CPTP}(AB\rightarrow AB^{\prime})$. If
$\mathcal{N}_{AB\rightarrow\tilde{A}B^{\prime}}$ is $A\not\to B^{\prime}$
semi-causal, then its Choi matrix $J^{\mathcal{N}}_{AB\tilde{A}B^{\prime}}$
satisfies

$\displaystyle J^{\mathcal{N}}_{ABB^{\prime}}=\mathbf{u}_{A}\otimes
J^{\mathcal{N}}_{BB^{\prime}}$ (4)

###### Proof.

Let $\mathcal{R}_{A^{\prime}\rightarrow A}$ be the completely randomising
channel which maps any input to the maximally mixed state $\mathbf{u}_{A}$.
Since $\mathcal{N}$ is semi-causal, then

$\displaystyle\text{Tr}_{\tilde{A}}\left[\mathcal{N}_{AB\rightarrow\tilde{A}B^{\prime}}\circ\mathcal{R}_{A\rightarrow
A}\right]=\text{Tr}_{\tilde{A}}\left[\mathcal{N}_{AB\rightarrow\tilde{A}B^{\prime}}\right]$
(5)

and
$J^{\mathcal{N}\circ\mathcal{R}}_{ABB^{\prime}}=J^{\mathcal{N}}_{ABB^{\prime}}$.
We note that the Choi matrix for $\mathcal{N}_{AB\rightarrow
B^{\prime}}\circ\mathcal{R}$ is then equivalent to the Choi matrix for
$\mathcal{N}_{AB\rightarrow B^{\prime}}$ and find

$\displaystyle J^{\mathcal{N}}_{ABB^{\prime}}$
$\displaystyle=\sum_{i,j=1}^{|A|}\sum_{k,l=1}^{|B|}\ket{ik}\\!\\!\bra{jl}_{AB}\otimes\mathcal{N}_{AB\rightarrow
B^{\prime}}\left(\mathcal{R}\left(\ket{i}\\!\\!\bra{j}\right)\otimes\ket{k}\\!\\!\bra{l}\right)$
$\displaystyle=\mathbf{u}_{A}\otimes\sum_{k,l=1}^{|B|}\ket{k}\\!\\!\bra{l}_{B}\otimes\mathcal{N}_{AB\rightarrow
B^{\prime}}\left(I_{A}\otimes\ket{k}\\!\\!\bra{l}\right)$
$\displaystyle=\mathbf{u}_{A}\otimes M$ (6)

where
$M=\sum_{k,l=1}^{|B|}\ket{k}\\!\\!\bra{l}_{B}\otimes\mathcal{N}_{AB\rightarrow
B^{\prime}}\left(I_{A}\otimes\ket{k}\\!\\!\bra{l}\right)$. Upon taking the
partial trace of $A$ over both sides, then $M=J^{\mathcal{N}}_{BB^{\prime}}$
and the theorem follows. ∎

In addition to semi-causality, any entropy non-decreasing channel must map
states of maximal conditional entropy (i.e. states of the form
$\mathbf{u}_{A}\otimes\rho_{B}$) to other states of maximal conditional
entropy. We define channels satisfying this condition to be conditionally
unital

######  Definition 2.

A bipartite channel $\mathcal{N}_{AB\rightarrow{AB^{\prime}}}$ is a
conditional unital channel if for all $\rho\in\mathfrak{D}\left(B\right)$,
there exists a state
$\sigma_{B^{\prime}}\in\mathfrak{D}\left(B^{\prime}\right)$ such that

$\displaystyle\mathcal{N}\left(\mathbf{u}_{A}\otimes\rho_{B}\right)=\mathbf{u}_{A}\otimes\sigma_{B^{\prime}}.$
(7)

We can also reframe the definition of a conditional unital channel in terms of
Choi matrices.

######  Lemma 3.

$\mathcal{N}\in\operatorname{CPTP}(AB\rightarrow\tilde{A}B^{\prime})$ is
conditional unital if and only if its Choi matrix
$J^{\mathcal{N}}_{AB\tilde{A}B^{\prime}}$ satisfies

$\displaystyle
J^{\mathcal{N}}_{B\tilde{A}B^{\prime}}=J^{\mathcal{N}}_{BB^{\prime}}\otimes\mathbf{u}_{\tilde{A}}.$
(8)

###### Proof.

Recall that the definition of a conditional unital channel can be written as

$\displaystyle\mathcal{N}\left(\mathbf{u}_{A}\otimes\rho_{B}\right)$
$\displaystyle=\operatorname{Tr}_{AB}\Big{[}J^{\mathcal{N}}_{AB\tilde{A}B^{\prime}}\left(\mathbf{u}_{A}\otimes\left(\rho_{B}\right)^{T}\otimes
I_{\tilde{A}B^{\prime}}\right)\Big{]}$ (9)
$\displaystyle=\frac{1}{|A|}\operatorname{Tr}_{B}\Big{[}J^{\mathcal{N}}_{B\tilde{A}B^{\prime}}\left(\left(\rho_{B}\right)^{T}\otimes
I_{\tilde{A}B^{\prime}}\right)\Big{]}$ (10)
$\displaystyle\triangleq\mathbf{u}_{A}\otimes\sigma_{B^{\prime}}.$ (11)

We now multiply both sides by $|A|\eta_{\tilde{A}}\otimes\omega_{B^{\prime}}$
where $\eta_{\tilde{A}}$ is traceless and Hermitian and $\omega_{B^{\prime}}$
is a Hermitian matrix. It follows that

$\displaystyle 0$
$\displaystyle=\operatorname{Tr}\Big{[}\left(I_{A}\otimes\sigma_{B^{\prime}}\right)\left(\eta_{\tilde{A}}\otimes\omega_{B^{\prime}}\right)\Big{]}$
(12)
$\displaystyle=\operatorname{Tr}_{\tilde{A}B^{\prime}}\Big{[}\operatorname{Tr}_{B}\Big{[}J^{\mathcal{N}}_{B\tilde{A}B^{\prime}}\left(\rho_{B}^{T}\otimes
I_{\tilde{A}B^{\prime}}\right)\Big{]}\left(\eta_{\tilde{A}}\otimes\omega_{B^{\prime}}\right)\Big{]}$
(13)
$\displaystyle=\operatorname{Tr}\Big{[}J^{\mathcal{N}}_{B\tilde{A}B^{\prime}}\left(\rho_{B}^{T}\otimes\eta_{\tilde{A}}\otimes\omega_{B^{\prime}}\right)\Big{]}.$
(14)

From this, it follows that for any orthonormal basis
$\\{\rho_{B}^{(i)}\\}_{i}$ of Hermitian operators, any orthonormal basis
$\\{\eta_{\tilde{A}}^{\left(j\right)}\\}_{j}$ of traceless Hermitian
operators, and any orthonormal basis $\\{\omega_{B^{\prime}}^{(k)}\\}_{k}$ of
Hermitian operators, then

$\displaystyle\mathcal{S}=\Big{\\{}\rho_{B}^{(i)}\otimes\eta_{\tilde{A}}^{\left(j\right)}\otimes\omega_{B^{\prime}}^{\left(k\right)}\Big{\\}}_{i,j,k}$
(15)

spans the orthogonal complement to $J^{\mathcal{N}}_{B\tilde{A}B^{\prime}}$.

We note that $\mathbf{u}_{\tilde{A}}$ is the only matrix (up to a scalar)
which is orthogonal to all traceless Hermitian matrices
$\\{\eta_{\left(\tilde{A}\right)}^{\left(j\right)}\\}$. Thus,
$J^{\mathcal{N}}_{B\tilde{A}B^{\prime}}$ is spanned by
$\\{\rho_{B}^{\left(i\right)}\otimes\mathbf{u}_{\tilde{A}}\otimes\omega_{B^{\prime}}^{\left(k\right)}\\}$,
so

$\displaystyle J^{\mathcal{N}}_{B\tilde{A}B^{\prime}}$
$\displaystyle=\sum_{i,j,k}c_{i,j,k}\rho_{B}^{\left(i\right)}\otimes\omega_{B^{\prime}}^{\left(k\right)}\otimes\mathbf{u}_{\tilde{A}}$
(16)
$\displaystyle=J^{\mathcal{N}}_{BB^{\prime}}\otimes\mathbf{u}_{\tilde{A}}$
(17)

This concludes the proof. ∎

Having motivated the need for a conditional entropy non-decreasing channel to
be semi-causal and conditional unital, we denote the set of all conditional
unital, $A\not\to B$ semi-causal channels in
$\operatorname{CPTP}\left(AB\rightarrow AB^{\prime}\right)$ as
$\operatorname{CUSC}\left(AB\rightarrow AB^{\prime}\right)$. Then bipartite
state $\rho_{AB}$ is less uncertain than state $\sigma_{AB}$ if one can obtain
$\sigma_{AB}$ using $\rho_{AB}$ and a CUSC channel. This gives rise to the
following pre-order.

######  Definition 3 (Conditional Majorization).

Consider two bipartite states $\rho_{AB}$ and $\sigma_{A^{\prime}B^{\prime}}$.
We say that $\rho_{AB}$ conditionally majorizes
$\sigma_{A^{\prime}B^{\prime}}$ with respect to system $A$, and write

$\rho_{AB}\succsim_{A}\sigma_{A^{\prime}B^{\prime}}$ (18)

if either $|A^{\prime}|\geq|A|$ and there exists an isometry
$\mathcal{V}\in{\rm CPTP}(A\to A^{\prime})$, and
$\mathcal{N}\in\operatorname{CUSC}\left(AB\rightarrow AB^{\prime}\right)$,
such that

$\displaystyle\sigma_{A^{\prime}B^{\prime}}=\mathcal{V}_{A\to
A^{\prime}}\circ\mathcal{N}_{AB\rightarrow
AB^{\prime}}\left(\rho_{AB}\right)\;,$ (19)

or $|A|\geq|A^{\prime}|$ and there exists an isometry $\mathcal{U}\in{\rm
CPTP}(A^{\prime}\to A)$, and
$\mathcal{N}\in\operatorname{CUSC}\left(AB\rightarrow AB^{\prime}\right)$,
such that

$\displaystyle\mathcal{U}_{A^{\prime}\to
A}\left(\sigma_{A^{\prime}B^{\prime}}\right)=\mathcal{N}_{AB\rightarrow
AB^{\prime}}\left(\rho_{AB}\right).$ (20)

The isometries introduced in the definition above enable us to compare between
bipartite states of different dimensions. Note that also majorization between
two probability vectors of the same dimension can be extended to vectors of
different dimensions in this way (by adding zero components to the vector with
the smaller dimension so that the vectors become of equal dimension). Such
extension is motivated by the fact that entropy functions don’t change under
such embedding (see further discussion of it in gour2021entropy ). We are now
ready to define conditional entropy based on the pre-order induced by
conditional majorization.

######  Definition 4.

A nonzero function
$\mathbf{H}:\bigcup_{A,B}\mathfrak{D}\left(AB\right)\rightarrow\mathbb{R}$ is
a quantum conditional entropy if it satisfies the following monotonicity and
additivity constraints

  1. 1.

For all $\sigma_{A^{\prime}B^{\prime}}$, $\rho_{AB}$, if
$\sigma_{A^{\prime}B^{\prime}}\precsim_{A}\rho_{AB}$, then
$\mathbf{H}(A|B)_{\rho}\leq\mathbf{H}\left(A^{\prime}|B^{\prime}\right)_{\sigma}$.

  2. 2.

$\mathbf{H}\left(AA^{\prime}|BB^{\prime}\right)_{\rho\otimes\tau}=\mathbf{H}(A|B)_{\rho}+\mathbf{H}\left(A^{\prime}|B^{\prime}\right)_{\tau}$
for all states $\rho_{AB}$ and $\tau_{A^{\prime}B^{\prime}}$.

Note that a conditional entropy $\mathbf{H}$ is defined on all density
matrices in all finite dimensions. In particular, if $\rho\in\mathfrak{D}(AB)$
and if $|B|=1$ then we write

$\mathbf{H}(A|B)_{\rho}=\mathbf{H}(A)_{\rho}\;,$ (21)

where $\mathbf{H}(A)_{\rho}$ is an entropy function; that is, it is additive
under tensor products and monotonic under mixing (i.e. majorization).

If a given function $\mathbf{H}$ satisfies the above conditions, then
$\alpha\mathbf{H}$ is also a conditional entropy for any $\alpha>0$. To
eliminate this extra degree of freedom due to scaling, we consider only
normalised conditional entropy functions such that when $|A|=2$ and $|B|=1$
(i.e. the bipartite state is a single qubit), then

$\displaystyle\mathbf{H}(A|B)_{\frac{I}{2}}=H\\!\left(\frac{I}{2}\right)=1.$
(22)

where $H$ is the Shannon entropy. The above three axioms (i.e. including the
normalization condition) are sufficient to guarantee several key properties of
conditional entropy. We start by showing that conditional entropies reduce to
entropies on uncorrelated states.

######  Lemma 4.

Conditional entropy reduces to an entropy function on uncorrelated states.
That is, for any product state $\omega_{A}\otimes\tau_{B}$,

$\mathbf{H}(A|B)_{\omega_{A}\otimes\tau_{B}}=\mathbf{H}(A)_{\omega}.$ (23)

###### Proof.

Suppose that $\rho_{AB}=w_{A}\otimes\tau_{B}$ is a product state. Let
$\mathcal{E}_{B\to B}\in\operatorname{CPTP}\left(B\to B\right)$ be the
completely randomizing channel on system $B$. Then it follows that
$\mathcal{I}_{A\to A}\otimes\mathcal{E}_{B\to
B}\in\operatorname{CUSC}\left(AB\to AB\right)$ and that

$\mathcal{I}_{A\to A}\otimes\mathcal{E}_{B\to
B}\left(w_{A}\otimes\tau_{B}\right)=w_{A}\otimes\mathbf{u}_{B}.$ (24)

Moreover, let $\mathcal{R}_{\tau}^{B}$ be the replacement channel on system
$B$, which takes any quantum state to $\tau_{B}$. Then

$\mathcal{I}_{A\to A}\otimes\mathcal{R}_{B\to
B}\left(w_{A}\otimes\mathbf{u}_{B}\right)=w_{A}\otimes\tau_{B}\ .$ (25)

Thus,

$w_{A}\otimes\tau_{B}\succsim_{A}w_{A}\otimes\mathbf{u}_{B}\quad\text{and}\quad
w_{A}\otimes\mathbf{u}_{B}\succsim_{A}w_{A}\otimes\tau_{B}$ (26)

which implies, by the monotonicity property of the conditional entropy, that

$\mathbf{H}(A|B)_{w_{A}\otimes\tau_{B}}=\mathbf{H}(A|B)_{w_{A}\otimes\mathbf{u}_{B}}.$
(27)

Thus, $\mathbf{H}(A|B)_{w_{A}\otimes\tau_{B}}$ depends on $w_{A}$ only.
Moreover, the function $w_{A}\mapsto\mathbf{H}(A|B)_{w_{A}\otimes\tau_{B}}$
satisfies the three axioms of entropy Gour5  and therefore can be considered
as an entropy of $w_{A}$. ∎

The same statement also holds when system $|B|$ is trivial, i.e., when
$|B|=1$.

Unlike entropy, quantum conditional entropy can be negative. In the following
theorem, we find a lower bound for the conditional entropy and show that only
entangled states can have negative conditional entropy.

######  Theorem 1.

Let $\mathbf{H}$ be any conditional entropy, and let $A$ and $B$ be two
Hilbert spaces. Then for any $\rho\in\mathfrak{D}\left(AB\right)$ the
conditional entropy is lower bounded as,

$-\log\min\left(|A|,|B|\right)\leq\mathbf{H}(A|B)_{\rho},$ (28)

The lower bound is achieved when $\rho_{AB}=\phi^{+}_{AB}$ is the maximally
entangled state. Additionally, if $\rho_{AB}$ is a separable (non-entangled)
state, then $\mathbf{H}(A|B)_{\rho}\geq 0$.

###### Proof.

Given that $\phi^{+}_{AB}$ can be used to teleport an arbitrary quantum state,
it follows that there exists some channel
$\mathcal{N}\in\operatorname{CUSC}\left(AB\rightarrow AB\right)$ such that
$\rho_{AB}=\mathcal{N}\left(\phi^{+}_{AB}\right)$. It immediately follows that
for any $\rho\in\mathfrak{D}\left(AB\right)$, then

$\displaystyle\mathbf{H}(A|B)_{\phi^{+}_{AB}}\leq\mathbf{H}(A|B)_{\rho_{AB}}$
(29)

Thus, it is sufficient to demonstrate that $\mathbf{H}(A|B)_{\phi^{+}}$ is
lower bounded by $-\text{log}\left(\min\\{|A|,|B|\\}\right)$. To this aim, we
define $d\triangleq|A|=|B|$ and consider the state
$\phi^{+}_{AB}\otimes\mathbf{u}_{A_{2}}$ where $|A_{2}|=d$. Then

$\displaystyle\mathbf{H}\left(AA_{2}|B\right)_{\phi^{+}_{AB}\otimes\mathbf{u}_{A_{2}}}$
$\displaystyle=\mathbf{H}\left(AA_{2}|BB_{2}\right)_{\phi^{+}_{AB}\otimes\mathbf{u}_{A_{2}B_{2}}}$
(30)
$\displaystyle=\mathbf{H}(A|B)_{\phi^{+}}+\mathbf{H}\left(A_{2}|B_{2}\right)_{\mathbf{u}_{A_{2}B_{2}}}$
(31)
$\displaystyle=\mathbf{H}(A|B)_{\phi^{+}}+\mathbf{H}\left(\mathbf{u}_{A_{2}}\right)$
(32) $\displaystyle=\mathbf{H}(A|B)_{\phi^{+}}+\text{log}\left(d\right),$ (33)

where the first line follows by introducing a trivial system $B_{2}$ such that
$|B_{2}|=1$ and the second follows from conditional entropy.

We now demonstrate that
$\mathbf{H}\left(AA_{2}|B\right)_{\phi^{+}_{AB}\otimes\mathbf{u}_{A_{2}}}\leq
0$ by demonstrating the existence of a channel
$\mathcal{N}\in\operatorname{CUSC}\left(AA_{2}B\rightarrow AA_{2}\right)$ such
that
$\mathcal{N}\left(\phi^{+}\otimes\mathbf{u}\right)=\ket{\psi}\\!\\!\bra{\psi}_{AA_{2}}$
for some pure state $\ket{\psi}\in\mathcal{H}\left(AA_{2}\right)$. Consider
the Choi matrix

$\displaystyle
J^{\mathcal{N}}_{AA_{2}B\tilde{A}\tilde{A}_{2}}=\sum_{j=1}^{d^{2}}\phi_{AB}^{\left(j\right)}\otimes
I_{A_{2}}\otimes\psi_{\tilde{A}\tilde{A}_{2}}^{\left(j\right)}$ (34)

where $\\{\ket{\phi}_{AB}^{\left(j\right)}\\}|_{j=1}^{d^{2}}$ is an
orthonormal basis of $AB$ consisting of maximally entangled states and where
we set $\phi_{1}=\phi^{+}$. Likewise,
$\\{\ket{\psi_{j}}_{\tilde{A}\tilde{A}_{2}}\\}|_{j=1}^{d^{2}}$ is an
orthonormal basis of $AA_{2}$.

Evidently, $\mathcal{N}$ is trace-preserving and conditional unital (as
$J^{\mathcal{N}}_{B\tilde{A}\tilde{A}_{2}}=I_{B\tilde{A}\tilde{A}_{2}}=J^{\mathcal{N}}_{B}\otimes\mathbf{u}_{\tilde{A}\tilde{A}_{2}}$).
Additionally, $\mathcal{N}$ is trivially semi-causal as subsystem $B$ is
traced out. From this, it follows that
$\mathcal{N}\in\operatorname{CUSC}\left(AA_{2}B\rightarrow\tilde{A}\tilde{A}_{2}\right)$
and therefore

$\displaystyle\mathbf{H}\left(AA_{2}|B\right)_{\phi^{1}_{AB}\otimes\mathbf{u}_{A_{2}}}$
$\displaystyle\leq\mathbf{H}\left(\tilde{A}\tilde{A}_{2}|\tilde{B}\right)_{\mathcal{N}\left(\phi^{1}_{AB}\otimes\mathbf{u}_{A_{2}}\right)}$
(35)
$\displaystyle=\mathbf{H}\left(\mathcal{N}\left(\phi^{1}_{AB}\otimes\mathbf{u}_{A_{2}}\right)\right)$
(36)

where $\tilde{B}$ is a trivial subsystem. Note that

$\displaystyle\mathcal{N}\left(\phi^{+}_{AB}\otimes\mathbf{u}_{A_{2}}\right)$
(37)
$\displaystyle=\operatorname{Tr}_{AA_{2}B}\left[J^{\mathcal{N}}_{AA_{2}B\tilde{A}\tilde{A}_{2}}\left(\phi^{+}_{AB}\otimes\mathbf{u}_{A_{2}}\otimes
I_{\tilde{A}\tilde{A}_{2}}\right)\right]$ (38)
$\displaystyle=\operatorname{Tr}_{AA_{2}B}\\!\Bigg{[}\left(\sum_{j=1}^{d^{2}}\phi_{AB}^{\left(j\right)}\otimes
I_{A_{2}}\otimes\psi_{\tilde{A}\tilde{A}_{2}}^{\left(j\right)}\right)\phi^{+}_{AB}$
$\displaystyle\qquad\qquad\qquad\otimes\mathbf{u}_{A_{2}}\otimes
I_{\tilde{A}\tilde{A}_{2}}\Bigg{]}$ (39)
$\displaystyle=\operatorname{Tr}_{AA_{2}B}\left[\sum_{j=1}^{d^{2}}\phi_{AB}^{\left(j\right)}\phi^{+}_{AB}\otimes\mathbf{u}_{A_{2}}\otimes\psi_{\tilde{A}\tilde{A}_{2}}^{\left(j\right)}\right]$
(40)
$\displaystyle=\operatorname{Tr}_{AA_{2}B}\Big{[}\phi^{\left(1\right)}_{AB}\otimes\mathbf{u}_{A_{2}}\otimes\psi_{\tilde{A}\tilde{A}_{2}}^{\left(1\right)}\Big{]}$
(41) $\displaystyle=\phi_{\tilde{A}\tilde{A}_{2}}^{\left(1\right)},$ (42)

which is evidently a pure state. Hence,

$\displaystyle\mathbf{H}\left(AA_{2}|B\right)_{\phi^{1}_{AB}\otimes\mathbf{u}_{A_{2}}}$
$\displaystyle\leq\mathbf{H}\left(\ket{\psi}\\!\\!\bra{\psi}_{\tilde{A}\tilde{A}_{2}}\right)$
(43) $\displaystyle=0.$ (44)

Upon combining everything, we see that

$\displaystyle 0$
$\displaystyle\leq\mathbf{H}\left(A|B\right)_{\phi^{+}_{AB}\otimes\mathbf{u}_{2}}$
(45) $\displaystyle\leq\mathbf{H}(A|B)_{\phi^{+}_{AB}}+\text{log}|A|$ (46)
$\displaystyle\leq 0,$ (47)

from which it follows that $\mathbf{H}(A|B)_{\phi^{+}}=-\text{log}|A|$ and
therefore $-\text{log}\left(\\{|A|,|B|\\}\right)$ lower bounds any density
matrix $\rho_{AB}$.

Finally, we demonstrate that if $\rho\in\mathfrak{D}\left(AB\right)$ is a
separable state, then $\mathbf{H}(A|B)_{\rho}\geq 0$. This follows from noting
that any separable state $\rho_{AB}$ may be written as

$\displaystyle\rho_{AB}=\sum_{j=1}p_{j}\psi_{A}^{\left(j\right)}\otimes\phi_{B}^{\left(j\right)},$
(48)

where $\\{\psi_{A}^{\left(j\right)}\\}_{j}$ and
$\\{\phi_{B}^{\left(j\right)}\\}_{j}$ are sets of pure states. It follows that
the CUSC channel

$\displaystyle\mathcal{N}_{AB\rightarrow
AB}\triangleq\sum_{j=1}^{n}p_{j}\mathcal{U}_{A\rightarrow
A}^{\left(j\right)}\otimes\mathcal{V}_{B\rightarrow B}^{\left(j\right)}$ (49)

satisfies

$\displaystyle\mathcal{N}\left(\ket{0}\\!\\!\bra{0}_{A}\otimes\ket{0}\\!\\!\bra{0}_{B}\right)$
$\displaystyle=\rho_{AB}$ (50)

for a choice of unitaries $\\{\mathcal{U}_{A}^{\left(j\right)}\\}$ such that
$\mathcal{U}^{j}\ket{0}=\ket{\phi_{A}^{\left(j\right)}}$ and a choice of
unitaries $\\{\mathcal{V}_{B}^{\left(j\right)}\\}$ such that
$\mathcal{V}^{\left(j\right)}\ket{0}=\ket{\psi_{B}^{\left(j\right)}}$. Then

$\displaystyle 0$
$\displaystyle=\mathbf{H}(A|B)_{\ket{0}\\!\\!\bra{0}_{A}\otimes\ket{0}\\!\\!\bra{0}_{B}}$
(51)
$\displaystyle\leq\mathbf{H}(A|B)_{\mathcal{N}\left(\ket{0}\\!\\!\bra{0}_{A}\otimes\ket{0}\\!\\!\bra{0}_{B}\right)}$
(52) $\displaystyle\leq\mathbf{H}(A|B)_{\rho_{AB}}.$ (53)

This concludes the proof. ∎

Although conditional entropies are positive on all separable states, this does
not mean that the conditional entropies are only positive on separable states.
In fact, some entangled states have positive conditional entropy on _all_
choice of conditional entropy functions. The following lemma provides a simple
criterion for finding states with positive conditional entropy.

######  Lemma 5.

Let $\mathbf{H}$ be a conditional entropy, and let
$\rho\in\mathfrak{D}\left(AB\right)$ with $d\coloneqq|A|=|B|$ be a density
matrix whose eigenvalues are less than or equal to $\frac{1}{d}$, and whose
marginal state on system $B$ is a maximally mixed state, i.e.,
$\rho^{B}=\mathbf{u}_{B}$. Then, the conditional entropy on $\rho$ is non-
negative, i.e.,

$\mathbf{H}(A|B)_{\rho}\geq 0\ .$ (54)

###### Proof.

Suppose $\mathcal{N}_{\tilde{A}B\to AB^{\prime}}$ is a $\tilde{A}\nrightarrow
B$ semi-causal channel as shown in Fig. 2 and suppose $|B|=1$. For a semi-
causal channel $\mathcal{N}_{\tilde{A}B\to AB^{\prime}}$ such that

$\mathcal{N}_{\tilde{A}B\to AB^{\prime}}=\mathcal{E}_{R\tilde{A}\to
A}\circ\mathcal{F}_{B\to RB^{\prime}}\ ,$ (55)

$\mathcal{F}$ is required to be an isometry. When $|B|=1$, $\mathcal{F}$ is
always an isometry. In this case, the channel $\mathcal{F}$ could be viewed as
a bipartite state on system $RB^{\prime}$. Let this state be the maximally
entangled state $\phi_{RB^{\prime}}^{+}$ and let $|R|=|B^{\prime}|$. Then the
resulting channel $\mathcal{N}_{\tilde{A}\to AB^{\prime}}$ takes the form

$\mathcal{N}_{\tilde{A}\to
AB^{\prime}}\left(\omega_{A}\right)=\mathcal{E}_{\tilde{A}\to
AB^{\prime}}\left(\omega_{A}\otimes\phi_{RB^{\prime}}^{+}\right)\quad\forall\omega\in\mathfrak{D}\left(\tilde{A}\right)\
.$ (56)

Note that the channel $\mathcal{N}_{\tilde{A}\to AB^{\prime}}$ is not
necessarily conditional unital. It can be shown that channel
$\mathcal{N}_{\tilde{A}\to AB^{\prime}}$ is conditional unital iff the channel
$\mathcal{E}_{\tilde{A}\to AB^{\prime}}$ satisfies

$\mathcal{E}_{\tilde{A}\to
AB^{\prime}}\left(\mathbf{u}_{A}\otimes\phi_{RB^{\prime}}^{+}\right)=\mathbf{u}_{AB^{\prime}}\
.$ (57)

Let $\mathcal{E}_{\tilde{A}\to AB^{\prime}}$ be a quantum channel that
satisfies (57). Let $\sigma_{AB^{\prime}}$ be the quantum state such that

$\sigma_{AB^{\prime}}=\mathcal{N}_{\tilde{A}\to
AB^{\prime}}\left(|1\rangle\\!\langle 1|_{\tilde{A}}\right)\ ,$ (58)

where $|1\rangle$ is a pure state in $\tilde{A}$. We then have

$\displaystyle\mathbf{H}\left(A|B^{\prime}\right)_{\sigma}$
$\displaystyle=\mathbf{H}\left(A|B^{\prime}\right)_{\mathcal{N}\left(|1\rangle\\!\langle
1|\right)}$ (59) $\displaystyle\geq\mathbf{H}\left(|1\rangle\\!\langle
1|_{\tilde{A}}\right)$ (60) $\displaystyle=0.$ (61)

To simplify the notation, we rename system $B^{\prime}$ as $B$. Since
$|R|=|B|=|\tilde{B}|$, we rename $R$ as $\tilde{B}$. Then, the channel
$\mathcal{E}_{\tilde{A}R\to A}$ becomes $\mathcal{E}_{\tilde{A}B\to A}$, and
its Choi matrix is denoted by $J^{\mathcal{E}}_{\tilde{A}\tilde{B}A}$. By
(57), we have

$J^{\mathcal{E}}_{\tilde{A}\tilde{B}}=I_{A\tilde{B}}\quad\text{and}\quad
J^{\mathcal{E}}_{\tilde{B}A}=I_{\tilde{B}A}\ .$ (62)

Now, let $\rho_{AB}$ be as in the lemma, and define the channel
$\mathcal{E}_{\tilde{A}B\to A}$ via its Choi matrix

$J^{\mathcal{E}}_{\tilde{A}\tilde{B}A}\coloneqq|1\rangle\\!\langle
1|_{\tilde{A}}\otimes\left(d\rho_{A\tilde{B}}\right)+\sum_{x=2}^{d}|x\rangle\\!\langle
x|_{\tilde{A}}\otimes\frac{I_{A\tilde{B}}-d\rho_{A\tilde{B}}}{d-1}\ .$ (63)

Note that the matrix above is positive semidefinite since $I_{A\tilde{B}}\geq
d\rho_{A\tilde{B}}$ and it satisfies the two conditions in (62) since
$\rho_{B}=\mathbf{u}_{B}$. With this choice of $\mathcal{E}$ we get

$\displaystyle\sigma_{AB}=\mathcal{E}_{\tilde{A}R\to
A}\left(|1\rangle\\!\langle
1|_{\tilde{A}}\otimes\phi_{RB^{\prime}}^{+}\right)$ (64)
$\displaystyle=\operatorname{Tr}_{\tilde{A}\tilde{B}}\left[\left(J^{\mathcal{E}}_{\tilde{A}\tilde{B}A}\otimes
I^{B}\right)\left(|1\rangle\\!\langle
1|_{\tilde{A}}\otimes\left(\phi_{\tilde{B}B}^{+}\right)^{T_{\tilde{B}}}\otimes
I_{A}\right)\right]$ (65)
$\displaystyle=\operatorname{Tr}_{\tilde{B}}\left[\left(\rho_{A\tilde{B}}\otimes
I_{B}\right)\left(\Phi_{\tilde{B}B}\right)^{T_{\tilde{B}}}\otimes
I_{A}\right]$ (66) $\displaystyle=\rho_{AB},$ (67)

where $T_{\tilde{B}}$ is the partial transposition map on system $\tilde{B}$.
This completes the proof that the conditional entropy of $\sigma_{AB}$ is non-
negative. ∎

##  IV Examples of Conditional Entropies and Duality Relations

We now demonstrate that any quantum relative entropy can be used to generate a
corresponding quantum conditional entropy.

######  Theorem 2.

Let $\mathbf{D}$ be a quantum relative entropy. Then the function

$\displaystyle\mathbf{H}^{\downarrow}\left(A\big{|}B\right)_{\rho}\triangleq\mathrm{log}\big{|}A\big{|}-\mathbf{D}\left(\rho_{AB}\big{\|}\mathbf{u}_{A}\otimes\rho_{B}\right)$

is a quantum conditional entropy.

###### Proof.

First, we demonstrate that $\mathbf{H}$ satisfies the monotonicity property of
conditional entropy. Let $\mathcal{N}\in\text{CUSC}(AB\rightarrow
A^{\prime}B^{\prime})$ and consider the bipartite density matrix $\rho_{AB}$.
We begin by considering the case where $A=A^{\prime}$ and

$\displaystyle\mathbf{H}^{\downarrow}(A\big{|}B)_{\mathcal{N}(\rho)}$
$\displaystyle=\text{log}|A|-\mathbf{D}\left(\mathcal{N}(\rho_{AB})\big{\|}\mathbf{u}_{A}\otimes\text{Tr}_{A}\left[\mathcal{N}(\rho_{AB})\right]\right)$
(68)

and from using the fact that $\mathcal{N}$ is semi-causal and conditionally
unital, we find that

$\displaystyle\mathbf{u}_{A}\otimes\text{Tr}_{A}\left[\mathcal{N}\left(\rho_{AB}\right)\right]$
$\displaystyle=\mathbf{u}_{A}\otimes\text{Tr}_{A}\left[\mathcal{N}\left(\mathbf{u}_{A}\otimes\rho_{B}\right)\right]$
(69) $\displaystyle=\mathcal{N}(\mathbf{u}_{A}\otimes\rho_{B})$ (70)

from which we have

$\displaystyle\mathbf{H}^{\downarrow}(A\big{|}B)_{\mathcal{N}(\rho)}$
$\displaystyle=\text{log}|A|-\mathbf{D}\left(\mathcal{N}(\rho_{AB})\big{\|}\mathcal{N}(\mathbf{u}_{A}\otimes\rho_{B})\right)$
(71)
$\displaystyle\geq\text{log}|A|-\mathbf{D}\left(\rho_{AB}\big{\|}\mathbf{u}_{A}\otimes\rho_{B}\right)$
(72)

where the last line follows from the data processing inequality.

We now prove that any entropy $\mathbf{H}$ satisfying the form given in the
theorem statement is invariant under isometries. Consider any isometry channel
$\mathcal{V}\in\text{CPTP}(A\rightarrow A^{\prime})$ and let $C$ represent a
Hilbert space where $|C|=|A^{\prime}|-|A|$. Then

$\displaystyle\mathbf{H}^{\downarrow}\left(A\big{|}B\right)_{\mathcal{V}(\rho)}=\text{log}|A^{\prime}|-\mathbf{D}(\mathcal{V}_{A\rightarrow
A^{\prime}}(\rho_{AB})\big{\|}\mathbf{u}_{A^{\prime}}\otimes\rho_{B})$ (73)

Clearly, if $\mathcal{V}$ is unitary such that $A\simeq A^{\prime}$, then
$\mathbf{H}(A^{\prime}\big{|}B)_{\mathcal{V}(\rho)}=\mathbf{H}(A\big{|}B)_{\rho}$.
Then w.l.o.g. we assume that

$\displaystyle\mathcal{V}_{A\rightarrow A^{\prime}}(\rho_{AB})$
$\displaystyle=\rho_{AB}\oplus\mathbf{0}_{CB}$ (74)
$\displaystyle=\begin{pmatrix}\rho_{AB}&\mathbf{0}\\\
\mathbf{0}&\mathbf{0}_{CB}\end{pmatrix}$ (75)

since the conditional entropy is invariant under unitaries. We can
additionally rewrite
$\mathbf{u}_{A^{\prime}}=t\mathbf{u}_{A}\oplus(1-t)\mathbf{u}_{C}$ where
$t\triangleq\frac{|A|}{|A^{\prime}|}$. From this, it follows that

$\displaystyle\mathbf{D}(\mathcal{V}_{A\rightarrow
A^{\prime}}(\rho_{AB})\big{\|}\mathbf{u}_{A^{\prime}}\otimes\rho_{B})$ (76)
$\displaystyle=\mathbf{D}\left(\rho_{AB}\oplus\mathbf{0}_{CB}\big{\|}\left(t\mathbf{u}_{A}\otimes\rho_{B}\right)\oplus\left(\left(1-t\right)\mathbf{u}_{C}\otimes\rho_{B}\right)\right)$
(77)
$\displaystyle=\mathbf{D}\left(\rho_{AB}\big{\|}\left(t\mathbf{u}_{A}\otimes\rho_{B}\right)\right)$
(78)
$\displaystyle=\mathbf{D}\left(\rho_{AB}\big{\|}\left(\mathbf{u}_{A}\otimes\rho_{B}\right)\right)-\text{log}(t)$
(79)

such that upon substitution, we have

$\displaystyle\mathbf{H}^{\downarrow}\left(A\big{|}B\right)_{\mathcal{V}(\rho)}$
$\displaystyle=\text{log}|A^{\prime}|-\mathbf{D}\left(\rho_{AB}\big{\|}\mathbf{u}_{A}\otimes\rho_{B}\right)+\text{log}(t)$
(80) $\displaystyle=\mathbf{H}\left(A\big{|}B\right)_{\rho}$ (81)

The additive property of $\mathbf{H}$ follows immediately from the additive
property of $D$. Finally, when $\mathbf{u}=\frac{\mathbb{I}}{2}$, then

$\displaystyle\mathbf{H}\left(A\big{|}B\right)_{\mathbf{u}}=\text{log}(2)-\mathbf{D}(\mathbf{u}\|\mathbf{u})=1$
(82)

and so normalisation is also satisfied. ∎

It follows that the von-Neumann conditional entropy is a conditional entropy
under our axiomatic approach. The von-Neumann conditional entropy is defined
as

$\displaystyle H\left(A\big{|}B\right)_{\rho}$
$\displaystyle\triangleq\text{log}|A|-D(\rho_{AB}\big{|}\mathbf{u}_{A}\otimes\rho_{B})$
(83)

where $D$ is the Umegaki relative entropy such that for any $\rho_{AB}$ and
$\sigma_{AB}$, then

$\displaystyle
D(\rho_{AB}\big{|}\sigma_{AB})=\text{Tr}\left[\rho_{AB}\left(\text{log}\left(\rho_{AB}\right)-\text{log}\left(\sigma_{AB}\right)\right)\right]$
(84)

Finally, we define the dual of a conditional entropy function and demonstrate
that the conditional von Neumann entropy is self-dual.

######  Definition 5.

Let $\mathbf{H}$ be a conditional entropy. For any density matrix
$\rho\in\mathfrak{D}(AB)$ with a purification $\varphi\in\mathfrak{D}(ABC)$
where $\rho_{AB}=\text{Tr}_{C}(\varphi_{ABC})$ the dual of $\mathbf{H}$ is
defined as

$\displaystyle\mathbf{H}^{\mathrm{dual}}\left(A\ \big{|}\
B\right)_{\rho}\triangleq-\mathbf{H}\left(A\ \big{|}\ C\right)_{\rho}$ (85)

Let $H$ be the conditional von-Neumann entropy. Then it follows from the
relation
$H\left(A\big{|}B\right)_{\varphi}+H\left(A\big{|}C\right)_{\varphi}=0$ that

$\displaystyle H^{\text{dual}}\left(A\big{|}B\right)_{\rho}$
$\displaystyle=-H(A\big{|}C)_{\varphi}$ (86)
$\displaystyle=H\left(A\big{|}B\right)$ (87)

and therefore the conditional von-Neumann entropy is self-dual.

##  V Quantum conditional majorization reduces to classical conditional
majorization

In Gour , a pre-order $\succsim_{C}$ between two joint probability
distributions (i.e. classical states) denotes _classical conditional
majorization_. In this section, we demonstrate that the axiomatic approach to
quantum conditional majorization defined in the previous section is a
generalization of classical conditional majorization. To this aim, we
demonstrate that when we restrict the underlying bipartite quantum states to
classical states, the quantum conditional majorization is equivalent to the
classical conditional majorization.

Suppose the system $A\coloneqq X$ is classical, we call a channel
$\mathcal{M}\in$CPTP$(XB\to XB^{\prime})$ _conditionally doubly stochastic_
(CDS in short) if it has the following form

$\mathcal{M}_{YB\to YB^{\prime}}=\sum_{j}\mathcal{D}_{X\to
X}^{(j)}\otimes\mathcal{F}_{B\to B^{\prime}}^{(j)}$ (88)

where each $\mathcal{D}{(j)}$ is a classical doubly stochastic channel and
each $\mathcal{F}_{B\to B^{\prime}}^{(j)}$ is a completely positive (CP) map
such that $\sum_{j}\mathcal{F}_{B\to B^{\prime}}^{(j)}$ is CPTP.

######  Lemma 6.

Suppose all the systems involved are classical with $A\coloneqq X$,
$B\coloneqq Y$, and $B^{\prime}\coloneqq Y^{\prime}$. Then we have

$\operatorname{CDS}(XY\to XY^{\prime})\subseteq\operatorname{CUSC}(XY\to
XY^{\prime})$ (89)

###### Proof.

Recall that if $\mathcal{M}\in\text{CDS}(XY\rightarrow XY^{\prime})$, there
must exist a set of classical doubly stochastic channels
$\\{\mathcal{D}^{(j)}\\}$ and a quantum instrument
$\\{\mathcal{F}_{Y\rightarrow Y^{\prime}}^{(j)}\\}$ s.t.

$\mathcal{M}_{AB\to AB^{\prime}}=\sum_{j}\mathcal{D}_{A\to
A}^{(j)}\otimes\mathcal{F}_{B\to B^{\prime}}^{(j)}$ (90)

Then for any input of the form $\mathbf{u}_{A}\otimes\sigma_{B}$, it follows
that

$\displaystyle\mathcal{M}_{AB\to
AB^{\prime}}(\mathbf{u}_{A}\otimes\sigma_{B})$
$\displaystyle=\sum_{j}\mathcal{D}_{A\to
A}^{(j)}(\mathbf{u}_{A})\otimes\mathcal{F}_{B\to
B^{\prime}}^{(j)}(\sigma_{B})$ (91)
$\displaystyle=\mathbf{u}_{A}\otimes\sum_{j}\mathcal{F}_{B\to
B^{\prime}}^{(j)}(\sigma_{B})$ (92)

and therefore $\mathcal{M}$ is conditionally unital. (Note that the third line
of the above follows from the fact that $\mathcal{D}^{(j)}$ is doubly
stochastic, so $\mathcal{D}^{(j)}(\mathbf{u}_{A})=\mathbf{u}_{A}$ for all
$j$.)

Finally, we show that $\mathcal{M}$ is semi-causal as follows:

$\displaystyle\text{Tr}_{A}$
$\displaystyle\left[\mathcal{M}\circ\mathcal{T}_{A\rightarrow
A}\right]=\text{Tr}_{A}\left[\sum_{j}\mathcal{D}_{A\to
A}^{(j)}\circ\mathcal{T}_{A\rightarrow A}\otimes\mathcal{F}_{B\to
B^{\prime}}^{(j)}\right]$ (93)
$\displaystyle=\sum_{j}\text{Tr}_{A}\left[\mathcal{D}_{A\to
A}^{(j)}\circ\mathcal{T}_{A\rightarrow A}\right]\otimes\mathcal{F}_{B\to
B^{\prime}}^{(j)}$ (94)
$\displaystyle=\sum_{j}\text{Tr}_{A}\left[\mathcal{D}_{A\to
A}^{(j)}\right]\otimes\mathcal{F}_{B\to B^{\prime}}^{(j)}$ (95)
$\displaystyle=\text{Tr}_{A}\left[\mathcal{M}_{AB\rightarrow
AB^{\prime}}\right]$ (96)

Given that $\mathcal{M}$ is conditionally unital and semi-causal, then the
theorem statement follows. ∎

######  Theorem 3.

Let $\rho_{XY}$ and $\sigma_{XY^{\prime}}$, be classical bipartite states such
that $A=X$, $B=Y$, and $B^{\prime}=Y^{\prime}$ are all classical systems. Then
$\sigma_{XY^{\prime}}\precsim\rho_{XY}$ if and only if
$\sigma_{XY^{\prime}}\precsim_{C}\rho_{XY}$ where $\precsim_{C}$ is the
classical conditional majorization defined in Gour .

###### Proof.

It then immediately follows from lemma 6 that
$\rho_{XY}\succsim\sigma_{XY^{\prime}}$ as in definition 4.

To show the other direction, we suppose that
$\rho_{XY}\succsim\sigma_{XY^{\prime}}$ as in definition 4. That is, there
exists $\mathcal{N}\in\operatorname{CUSC}(XY\to XY^{\prime})$ such that

$\sigma_{XY^{\prime}}=\mathcal{N}_{XY\to XY^{\prime}}(\rho_{XY})\ .$ (97)

Since $\mathcal{N}$ is semi-causal, it has the form

$\mathcal{N}_{XY\to XY^{\prime}}=\sum_{j=1}^{k}\mathcal{E}_{X\to
X}^{(j)}\otimes\mathcal{F}_{Y\to Y^{\prime}}^{(j)}\ ,$ (98)

where for each $j\in[K]$, $\mathcal{E}^{(j)}\in\text{CPTP}(X\to X)$ and
$\mathcal{F}^{(j)}\in\text{CP}(Y\to Y^{\prime})$ with
$\sum_{j}\mathcal{F}_{j}\in\text{CPTP}(Y\to Y^{\prime})$. Hence,

$\sigma_{XY^{\prime}}=\sum_{j=1}^{k}\mathcal{E}_{X\to
X}^{(j)}\otimes\mathcal{F}_{Y\to Y^{\prime}}^{(j)}(\rho_{XY})\ .$ (99)

Since $\rho,\sigma$ are classical, we have the identification $\rho_{XY}\cong
P$ and $\sigma_{XY^{\prime}}\cong Q$. Thus, the above equation can be
expressed as

$Q=\sum_{j=1}^{k}E^{(j)}PR^{(j)}\ ,$ (100)

where for each $j\in[k]$, $E^{(j)}$ is the transition matrix of
$\mathcal{E}^{(j)}$, and where $R^{(j)}$ is the transpose of transition matrix
of $\mathcal{F}^{(j)}$. Hence, each $E^{(j)}$ is column stochastic and
$R\coloneqq\sum_{j=1}^{k}R^{(j)}$ is row stochastic. Let
$\\{\mathbf{p}_{y}\\}_{y\in[n]}$ be the columns of $P$, and observe that in
components form, the above relation can be expressed as

$q_{xw}=\sum_{j=1}^{k}\sum_{y=1}^{n}(E^{(j)}\mathbf{p}_{y})_{x}r_{yw}^{(j)}\
.$ (101)

So far we only used the constraint that $\mathcal{N}$ is semi-causal. The
other constraint that $\mathcal{N}$ is conditionally unital implies that if
$p_{xy}=\frac{1}{m}p_{y}$ then $q_{yw}=\frac{1}{m}q_{w}$. Therefore, for
simplicity we take $p_{xy}=\frac{1}{m}\delta_{y}{y_{0}}$ for some fixed
$y_{0}\in[n]$. Then the above equation becomes

$\frac{1}{m}q_{w}=\sum_{j}(E^{(j)}\mathbf{u})_{x}r^{(j)}_{y_{0}w}\ .$ (102)

Summing over $x$ on both sides of the equation, and using the fact that each
$E^{(j)}$ is column stochastic implying that $E^{(j)}\mathbf{u}$ is a
probability vector, we have

$q_{w}=\sum_{j}r_{y_{0}w}^{(j)}\ .$ (103)

Note that we assumed that $y_{0}$ was fixed, and therefore as a result the
equation above does not imply that $q_{w}$ is independent of $y_{0}$ if we
allow $y_{0}$ to vary. In other words, each value of $y_{0}$ corresponds to a
different $P$ matrix and therefore possibly resulting in a different $Q$
matrix. For simplicity, we rename $y_{0}$ as $y$ and denote

$q_{w}=t_{yw}\coloneqq\sum_{j}r^{(j)}_{yw}\ .$ (104)

By definition, the matrix $T=(t_{yw})$ is row-stochastic, and with the above
notations , we can express (102) as

$\mathbf{u}=\sum_{j}\frac{r_{yw}^{(j)}}{t_{yw}}E^{(j)}\mathbf{u}\quad
y\in[n],\ w\in[\ell]\ .$ (105)

Therefore, the matrices

$D_{(y,w)}\coloneqq\sum_{j}\frac{r_{yw}^{(j)}}{t_{yw}}E^{(j)}\quad y\in[n],\
w\in[\ell]$ (106)

are all doubly stochastic. To see this, observe that for each $y$ and $w$, the
set $\\{\frac{r_{yw}^{(j)}}{t_{yw}}\\}$ forms a probability distribution over
$j$, and the fact that each $E^{(j)}$ is column stochastic implying that
$D_{(y,w)}$ is column stochastic. Moreover, the relation
$\mathbf{u}=D_{(y,w)}u$ implies that $D_{(y,w)}$ are row stochastic. Finally,
for general probability matrices $P=[\mathbf{p}_{1},\ldots,\mathbf{p}_{n}]$
and $Q=[\mathbf{q}_{1},\ldots,\mathbf{q}_{n^{\prime}}]$, we conclude that
(101) is equivalent to

$\mathbf{q}_{w}=\sum_{y=1}^{n}t_{yw}D_{(y,w)}\mathbf{p}_{y}\quad\forall
w\in[n^{\prime}]\ .$ (107)

The above relation is precisely the condition given in lemma 3 of Gour .
Therefore, we conclude that $\rho_{XY}\succsim\sigma_{XY^{\prime}}$ implies
classical conditional majorization. ∎

##  VI Connection to Games of Chance

###  VI.1 Games of Chance for Classical Bipartite States

In brandsen2021entropy , games of chance were utilized to find an ordering
between classical bipartite states, and it has been demonstrated that such
games of chance are an alternate, operational approach for characterising
classical majorisation. We aim to likewise construct a set of quantum games of
chance which provide an operational meaning for majorisation of quantum
bipartite states.

Firstly, we briefly review the games of chance introduced in
brandsen2021entropy  for classical bipartite states. We first note that any
classical bipartite state may be represented as a probability distribution of
the form $\mathbf{p}=\\{p_{yz}\\}$ where the corresponding state in quantum
notation is
$\rho_{AB}=\sum_{y,z}p_{y,z}\ket{y}\\!\\!\bra{y}\otimes\ket{z}\\!\\!\bra{z}$.
Each game is fixed by a classical channel $\mathcal{T}$ with transition
probabilities $\\{t_{w|z^{\prime}}\\}$, as well as the bipartite state
$\mathbf{p}$ which is being used for the game.

The player is only allowed to access subsystem $B$ (corresponding to the value
of $z$), and the goal is for the player to correctly guess the value of $y$.
The family of gambling games includes all possible gambling games that
incorporate a correlated source. Given access to $z$, the player chooses
$z^{\prime}=f(z)$ and the host subsequently selects $w$ from the conditional
distribution $\mathcal{T}$. In general, the player will choose $f(z)$ based on
their knowledge of $z$, as well as the fixed distributions $\\{p_{yz}\\}$ and
$\\{t_{w|z^{\prime}}\\}$. Finally, once $w$ is selected, the player
communicates a set $\mathcal{S}$ of $w$ guesses to the host and wins if and
only if $y\in\mathcal{S}$. The game is depicted in Fig. 3.

\begin{overpic}[scale={.56}]{cgamb_bipartite} \put(112.0,17.0){$\mathcal{T}$}
\put(105.0,80.0){$y$} \put(65.0,17.0){$f$} \put(90.0,23.0){$z^{\prime}$}
\put(139.5,61.0){Choose set} \put(210.0,57.0){Win if}
\put(210.0,45.0){$y\in\mathcal{S}$} \put(139.5,51.0){$\mathcal{S}$ with $w$}
\put(139.5,40.0){elements} \put(35.0,25.0){$z$} \put(142.0,23.0){$w$}
\put(7.0,50.0){$\\{p_{yz}\\}$} \end{overpic} Figure 3: A classical gambling
game with a correlated source. The player is provided with the value $z$.
Based on this value, the player chooses $z$ (or the function $f$) and sends it
to the host. The host then chooses the $w$ game based on a (possibly
incomplete) distribution matrix $T=(t_{w|z})$. The player will then form a set
$\mathcal{S}$ containing $w$ guesses based on $w$ and $z$, and will win the
game if $y\in\mathcal{S}$.

It has previously been demonstrated brandsen2021entropy  that the conditional
majorisation relation $\precsim_{C}$ is equivalent to the partial ordering
induced by classical games of chance, namely
$\sigma_{AB}\precsim\rho_{AB^{\prime}}$ if and only if
$\text{Prob}_{\mathcal{T}}(\mathbf{q})\leq\text{Prob}_{\mathcal{T}}(\mathbf{p})$
for all conditional distributions $\mathcal{T}$ where
$\text{Prob}_{\mathcal{T}}(\mathbf{p})$ is the expected success probability
for playing the classical gambling game characterised by $\mathcal{T}$ with
state $\mathbf{p}$.

###  VI.2 Extension of gambling games to quantum bipartite states

In this section, we build on the framework for classical games of chance to
develop the most general games of chance for quantum bipartite states (and
respectively quantum channels). The only restriction on such games is that the
ordering induced by these families of games of chance must reasonably
correspond to a measure of uncertainty. For example, states which are
equivalent up to a unitary must have equivalent performance in these games of
chance.

First, we provide a straightforward extension of the classical gambling games,
and demonstrate that this direct extension is insufficient to capture
necessary quantum effects such quantum entanglement being a resource. Next, we
provide a broader set of games which still reduces to the same ordering on
classical states, but which leads to negative entropy for maximally entangled
states.

To generalize the classical gambling game, we first transform Bob’s
observation of the classical output $z$ into a POVM implemented by Bob on
system $B$. Evidently, $y$ can be written in quantum notation as
$\ket{y}\\!\\!\bra{y}$ such that a measurement in the computational basis
allows perfect observation of $y$.

Likewise, Bob’s choice of $w$ guesses is equivalent to sending an ordered list
of all potential outcomes and winning if $y$ is in the first $w$ elements of
the list. In turn, this can be transformed into performing an ordered
measurement on system $A$ with the output being transmitted to the host. For
example, if their first guess is $2$, the second guess is $1$, and the third
guess is $3$, then the they would implement the ordered measurement
$\left\\{\ket{2}\\!\\!\bra{2},\ket{1}\\!\\!\bra{1},\ket{3}\\!\\!\bra{3}\right\\}$
on Alice’s system. This game is depicted in Fig. 5.

\begin{overpic}[scale={0.555}]{direct_extension} \put(-2.0,51.0){$\rho_{AB}$}
\put(27.0,69.0){$A$} \put(27.0,10.0){$B$}
\put(179.0,63.0){$\hat{\Pi}^{*}_{z,w}$} \put(220.0,50.0){Win if}
\put(220.0,35.0){$y\leq w$} \put(149.0,18.0){$\mathcal{T}$}
\put(108.0,18.0){$f$} \put(65.0,18.0){$\hat{\Pi}$} \put(89.0,12.0){$z$}
\put(82.0,45.0){$z$} \put(129.0,23.0){$z^{\prime}$} \put(175.0,23.0){$w$}
\end{overpic} Figure 4: Direct extension of classical gambling games for
bipartite states. The player selects any rank-one measurement $\hat{\Pi}$ to
implement on system $B$ and observes the outcome $z$. Given $z$, the player
selects input $z^{\prime}=f(z)$ to $\mathcal{T}$, and the hosts draws $w$
according to $\\{t_{w|z^{\prime}}\\}$. The player then implements any rank one
measurement $\hat{\Pi}^{*}_{z,w}$ given knowledge of $z$ and $w$, and wins if
the output $y\leq w$.

However, this direct extension is insufficient to distinguish between the un-
entangled product state
$\sigma_{AB}=\ket{0}\\!\\!\bra{0}_{A}\otimes\ket{0}\\!\\!\bra{0}_{B}$ and the
entangled product state $\rho_{AB}=\phi_{AB}^{+}$. More specifically, any game
can be won with certainty when played with
$\ket{0}\\!\\!\bra{0}_{A}\otimes\ket{0}\\!\\!\bra{0}_{B}$. Regardless of the
player’s choice of $\hat{\Pi}$ or $f$, and regardless of $w$, the player can
simply implement the measurement
$\hat{\Pi}^{*}_{z,w}\triangleq\big{\\{}\ket{0}\\!\\!\bra{0},\ket{1}\\!\\!\bra{1}\big{\\}}$
on the first system. The first outcome (corresponding to measurement element
$\ket{0}\\!\\!\bra{0}$) will be obtained with certainty, and so $y=1$ will be
less than or equal to any $w\in\\{1,2,3,4\\}$ leading to a win for the player.

To break this equivalency, we now introduce a more sophisticated game which
allows an adversarial player to “scramble” the state of system $A$ with some
probability $p$. We define a conditional quantum gambling game
$G=(\mathcal{T},\mathbf{p})$ by the classical channel $\mathcal{T}$ with
transition matrix is $T=(t_{w|z^{\prime}})_{w,z^{\prime}}$, and the
probability $p$ of adversarial involvement. The specific steps of the game are
as follows:

  1. 1.

Alice and Bob are given a classical description of the state they share as the
entries of the corresponding density matrix $\rho_{AB}$.

  2. 2.

The host generates $x\in\\{0,1\\}$ from the probability distribution
$\mathbf{p}=\left(p,1-p\right)$, where $p\in[0,1]$. If $x=0$, then the
adversary is allowed to interfere by implementing a measurement
$\hat{\Pi}^{*}$ on system $A$.

  3. 3.

Bob is told which measurement the adversary implemented (if any), but is not
told the result. He is then allowed to perform any measurement
$\hat{\Pi}^{\prime}=\\{\Pi^{{}^{\prime}(z)}_{B}\\}$ on system $B$ and is told
the measurement result $z$.

  4. 4.

Bob can choose any $z^{\prime}=f\left(z\right)$ to input to the classical
channel $\mathcal{T}$ with corresponding transition matrix
$T=(t_{w|z^{\prime}})_{w,z^{\prime}}$. The host draws $w$ from
$\\{t_{w|z^{\prime}}\\}$

  5. 5.

Given $z$ and $w$, Alice implements a final rank-one projective measurement
$\hat{\Pi}(z,w)=\\{\Pi^{(k)}(z,w)\\}|_{k}$ on system $A$. The rank-one
requirement follows from noting that if Alice were allowed to choose a trivial
measurement such as $\\{I,0,\ldots,0\\}$, she would win any game with
certainty.

  6. 6.

Alice and Bob win if the outcome $y$ is less than $w$.

\begin{overpic}[scale={0.5}]{fullbipartite} \put(-7.0,40.0){$\rho_{AB}$}
\put(20.0,55.0){$A$} \put(20.0,17.0){$B$}
\scriptsize\put(162.0,57.0){$\hat{\Pi}(z,w)$} \normalsize\put(202.0,50.0){Win
if} \put(202.0,38.0){$y\leq w$} \put(140.0,17.0){$\mathcal{T}$}
\put(103.0,15.0){$f$} \put(64.0,17.0){$\hat{\Pi}^{\prime}$}
\put(85.0,11.0){$z$} \put(80.0,40.0){$z$} \put(122.0,22.0){$z^{\prime}$}
\put(185.0,22.0){$w$} \put(12.0,85.0){$\mathbf{p}$}
\put(61.0,58.5){$\hat{\Pi}^{*}$} \normalsize\put(45.0,89.0){$x$} \end{overpic}
Figure 5: Alternative extension of classical gambling games for bipartite
states. With probability $p$, $x=0$ and the adversarial player is allowed to
implement an adversarial measurement $\hat{\Pi}_{A}^{*}$ on Alice’s system at
the beginning of the game. The player selects any rank-one measurement
$\hat{\Pi}$ to implement on system $B$ and observes the outcome $z$. Given
$z$, the player selects input $z^{\prime}=f(z)$ to $\mathcal{T}$, and the
hosts draws $w$ according to $\\{t_{w|z^{\prime}}\\}$. The player then
implements any rank one measurement $\hat{\Pi}(z,w)$ given knowledge of $z$
and $w$, and wins if the output $y\leq w$.

In Appendix A, we demonstrate that the ordering induced by this extended
quantum gambling game on classical bipartite states reduces to classical
conditional majorisation. In the following, we will describe the reward
function for the above games of chance and demonstrate that the induced
partial ordering between quantum bipartite states distinguishes between
entangled and unentangled pure states. Additionally, in the case where system
$B$ is trivial, the ordering reduces to standard majorisation.

First, we denote by $\mathcal{R}_{\mathcal{T},p}$ the total expected reward
for playing the game corresponding to $\mathcal{T}$ and probability
distribution $\mathbf{p}=\\{p,1-p\\}$. Evidently, since $X$ is drawn before
the players are allowed to make any moves, then

$\displaystyle\mathcal{R}_{\mathcal{T},p}=p\times\mathcal{R}_{\mathcal{T},0}+(1-p)\times\mathcal{R}_{\mathcal{T},1}$
(108)

In other words, the total expected reward is a convex combination of the total
expected reward when the adversary is allowed a measurement and the total
expected reward when the adversary is not allowed a measurement.

Suppose that Alice and Bob have a fixed strategy- namely, that Bob implements
measurement $\hat{\Pi}^{{}^{\prime}}$, selects $z^{\prime}=f(z)$, and Alice
implements the optimal measurement $\hat{\Pi}(z,w)$. We first consider the
case where $p=0$, i.e. there is no adversary. Then no action is taken prior to
Bob’s first measurement $\hat{\Pi}^{\prime}$ on subsystem $B$, and the
probability of observing a measurement result corresponding to
$\Pi_{\text{B}}^{{}^{\prime}}(z)$ is

$\displaystyle P\left(z\big{|}\hat{\Pi}_{\text{B}}^{{}^{\prime}}\right)$
$\displaystyle=\text{Tr}\left[\rho_{AB}\Pi_{\text{B}}^{\prime(z)}\right].$
(109)

The corresponding post-measurement state is

$\displaystyle\rho_{z}\left(\hat{\Pi}_{\text{B}}^{\prime},\rho_{AB}\right)\triangleq\frac{\Pi_{\text{B}}^{\prime(z)}\rho_{AB}\Pi_{\text{B}}^{{}^{\prime}(z)}}{P\left(z\big{|}\hat{\Pi}_{\text{B}}^{{}^{\prime}}\right)}.$
(110)

In _any_ game, Alice will always choose $\hat{\Pi}_{A}(z,w)$ to maximize the
reward. Namely, she will measure in the eigenbasis of
$\rho_{z}(\hat{\Pi}_{B}^{{}^{\prime}},\rho_{AB})$ with the first measurement
element corresponding to the eigenvector of
$\rho_{z}(\hat{\Pi}_{B}^{{}^{\prime}},\rho_{AB})$ with the largest eigenvalue
and so on. Then the success probability for this fixed strategy is

$\displaystyle\sum_{z}P(z\big{|}\hat{\Pi}^{\prime})\sum_{k=1}^{w}t_{w|f(z)}\left\|\Pi_{\text{A}}^{(k)}(z,w)\text{Tr}_{B}\left[\rho_{z}\\!\left(\hat{\Pi}_{\text{B}}^{\prime},\rho_{AB}\right)\right]\right\|$
(111)
$\displaystyle=\sum_{z}P(z\big{|}\hat{\Pi}^{\prime})\sum_{w}t_{w|f(z)}\left\|\text{Tr}_{B}\left[\frac{\Pi_{\text{B}}^{\prime(z)}\rho_{AB}\Pi_{\text{B}}^{{}^{\prime}(z)}}{P\left(z\Big{|}\hat{\Pi}_{B}^{\prime}\right)}\right]\right\|_{(w)}$
(112)
$\displaystyle=\sum_{z}\sum_{w}t_{w|f(z)}\left\|\text{Tr}_{B}\left[\Pi_{\text{B}}^{\prime(z)}\rho_{AB}\Pi_{\text{B}}^{{}^{\prime}(z)}\right]\right\|_{(w)}$
(113)

Naturally, Bob will wish to choose the remaining game components (that is,
$\hat{\Pi}_{B}^{{}^{\prime}}$ and $f$) to maximize their reward, s.t. the
final reward function for the game $G=\\{\mathcal{T},p=0\\}$ may then be
expressed as

$\displaystyle R_{G}\left(\rho_{AB}\right)$
$\displaystyle\triangleq\underset{\hat{\Pi}^{{}^{\prime}},f}{\text{max}}\left(\sum_{z,w}t_{w|f(z)}\left\|\text{Tr}_{B}\left[\Pi_{\text{B}}^{\prime(z)}\rho_{AB}\Pi_{\text{B}}^{{}^{\prime}(z)}\right]\right\|_{(w)}\right)$
(115)

Now suppose the adversarial player is allowed to interfere with probability p.
If the adversary chooses measurement
$\hat{\Pi}_{A}^{*}=\\{\Pi_{A}^{*(j)}\\}|_{j}$, then the starting state
$\rho_{AB}$ will be transformed to
$\sum_{j}\Pi_{A}^{*(j)}\rho_{AB}\Pi_{A}^{*(j)}$. Thus, following an
adversarial measurement, Alice and Bob will effectively be playing the quantum
gambling game with the transformed state
$\sum_{j}\Pi_{A}^{*(j)}\rho_{AB}\Pi_{A}^{*(j)}$ and will have an expected
reward of
$R_{\mathcal{T},0}\left(\sum_{j}\Pi_{A}^{*(j)}\rho_{AB}\Pi_{A}^{*(j)}\right)$.

The adversary will wish to choose the “worst case” $\hat{\Pi}_{A}^{*}$ to
minimize Alice and Bob’s reward, leading to the following reward function for
a general quantum bipartite gambling game

$\displaystyle R_{\mathcal{T},p}\left(\rho_{AB}\right)$
$\displaystyle\triangleq
p\times\underset{\hat{\Pi}_{A}^{*}}{\text{min}}\left(R_{\mathcal{T},0}\left(\sum_{j}\Pi_{A}^{*(j)}\rho_{AB}\Pi_{A}^{*(j)}\right)\right)$
(116) $\displaystyle+(1-p)R_{\mathcal{T},0}\left(\rho_{AB}\right)$ (117)

Finally, we are ready to introduce conditional majorisation for quantum
bipartite states based on games of chance.

######  Definition 6.

The state $\rho_{AB}$ majorizes $\sigma_{AB^{\prime}}$ with respect to
subsystem $A$, based on quantum conditional gambling games, denoted as
$\sigma_{AB^{\prime}}\precsim_{\left(A\right)}\rho_{AB}$, if

$\displaystyle R_{\mathcal{T},p}\left(\sigma_{AB^{\prime}}\right)\leq
R_{\mathcal{T},p}\left(\rho_{AB},U\right)$ (118)

for all $0\leq p\leq 1$, $\mathcal{T}=\\{t_{w^{\prime}|z^{\prime}}\\}$,
$w^{\prime}\in\\{1,\ldots,|A|\\}$, and
$z^{\prime}\in\\{1,\ldots,\max\left(|B|,|B^{\prime}|\right)\\}$.

It follows immediately from the definition of the reward function that it is
sufficient to restrict to the set of games for which $p\in\\{0,1\\}$, namely,
games in which the adversary either always interferes or never interferes. In
the remainder of this work we restrict to this set unless otherwise specified.

One key property of this partial ordering is that maximally entangled pure
states outperform tensor product pure states.

######  Lemma 7.

Consider the case in which $d=|A|=|A^{\prime}|=|B|$ and $d\geq 2$. Then any
tensor product state
$\sigma_{AB^{\prime}}=\sigma_{A}\otimes\sigma_{B^{\prime}}$ is majorized by
$\phi_{AB}^{(+)}$, namely,

$\displaystyle\sigma_{AB^{\prime}}\prec_{\left(A\right)}\phi_{AB}^{(+)}.$
(119)

###### Proof.

First, we demonstrate that $R_{\mathcal{T},p}\left(\phi_{AB}^{(+)}\right)=1$
for all $\mathcal{T}$ and all $p$. To do this, it is sufficient to show that

$\displaystyle
R_{\mathcal{T},0}\left(\sum_{j}\Pi_{A}^{*(j)}\rho_{AB}\Pi_{A}^{*(j)}\right)=1$
(120)

for all $\mathcal{T}$ and all potential adversarial projective measurements
$\Pi_{A}^{*}$. Since the adversarial measurement is projective, there exists
some basis $\\{\ket{\psi_{j}}\\}|_{j=1}^{n}$ and some partition of that basis
$\\{\mathcal{S}_{k}\\}$ s.t.

$\displaystyle\Pi_{A}^{*(k)}=\sum_{j\in\mathcal{S}_{k}}\ket{\psi_{j}}\\!\\!\bra{\psi_{j}}$
(121)

Note that

$\displaystyle\sum_{j}\Pi_{A}^{*(j)}\phi_{AB}^{(+)}\Pi_{A}^{*(j)}=\frac{1}{d}\sum_{j,k,\ell}\Pi_{A}^{*(j)}\ket{k}\\!\\!\bra{\ell}_{A}\Pi_{A}^{*(j)}\otimes\ket{k}\\!\\!\bra{\ell}_{B}$
(122)
$\displaystyle=\frac{1}{d}\sum_{z_{1},z_{2}\in\mathcal{S}_{j}}\sum_{k,\ell}\ket{\psi_{z_{1}}}\\!\\!\braket{\psi_{z_{1}}}{k}\\!\\!\braket{\ell}{\psi_{z_{2}}}\\!\\!\bra{\psi_{z_{2}}}\otimes\ket{k}\\!\\!\bra{\ell}_{B}$
(123)
$\displaystyle=\frac{1}{d}\sum_{z_{1},z_{2}\in\mathcal{S}_{j}}\ket{\psi_{z_{1}}}\\!\\!\bra{\psi_{z_{2}}}\otimes\sum_{k,\ell}\braket{\psi_{z_{1}}}{k}\\!\\!\braket{\ell}{\psi_{z_{2}}}\ket{k}\\!\\!\bra{\ell}_{B}$
(124)
$\displaystyle=\frac{1}{d}\sum_{z_{1},z_{2}\in\mathcal{S}_{j}}\ket{\psi_{z_{1}}}\\!\\!\bra{\psi_{z_{2}}}_{A}\otimes\ket{\psi_{z_{1}}}\\!\\!\bra{\psi_{z_{2}}}_{B}$
(125)

Suppose that Bob’s strategy is then to implement the measurement
$\hat{\Pi}_{B}^{\prime}=\\{\ket{\psi_{z}}\\!\\!\bra{\psi_{z}}\\}|_{z}$ on
system $B$. Then it follows that

$\displaystyle\rho_{z}\left(\hat{\Pi}_{B}^{{}^{\prime}},\sum_{j}\Pi_{A}^{*(j)}\phi_{AB}^{(+)}\left(\Pi_{A}^{*(j)}\right)^{{\dagger}}\right)$
(127)
$\displaystyle=\frac{\sum_{z_{1},z_{2}\in\mathcal{S}_{j}}\ket{\psi_{z_{1}}}\\!\\!\bra{\psi_{z_{2}}}_{A}\otimes\ket{\psi_{z}}\\!\\!\braket{\psi_{z}}{\psi_{z_{1}}}\\!\\!\braket{\psi_{z_{2}}}{\psi_{z}}\\!\\!\bra{\psi_{z}}}{d\times
P\left(z\Big{|}\hat{\Pi}_{B}^{{}^{\prime}}\right)}$ (128)
$\displaystyle=\frac{\ket{\psi_{z}}\\!\\!\bra{\psi_{z}}\otimes\ket{\psi_{z}}\\!\\!\bra{\psi_{z}}}{d\times\frac{1}{d}}=\ket{\psi_{z}}\\!\\!\bra{\psi_{z}}\otimes\ket{\psi_{z}}\\!\\!\bra{\psi_{z}}$
(129)

Finally, we have

$\displaystyle
R_{\mathcal{T},p}\left(\sum_{j}\Pi_{A}^{*(j)}\phi_{AB}^{(+)}\Pi_{A}^{*(j)}\right)\leq
R_{\mathcal{T},0}\left(\sum_{j}\Pi_{A}^{*(j)}\phi_{AB}^{(+)}\Pi_{A}^{*(j)}\right)$
(130)
$\displaystyle\leq\sum_{z}P(z\big{|}\hat{\Pi}^{\prime})\left\|\text{Tr}_{B}\left[\rho_{z}\\!\left(\hat{\Pi}_{\text{B}}^{\prime},\sum_{j}\Pi_{A}^{*(j)}\phi_{AB}^{(+)}\Pi_{A}^{*(j)}\right)\right]\right\|_{(1)}$
(131)
$\displaystyle\leq\sum_{z=1}^{d}\frac{1}{d}\left\|\ket{\psi_{z}}\\!\\!\bra{\psi_{z}}_{A}\right\|_{(1)}=1$
(132)

To prove the theorem statement, we now need to demonstrate that there exists
some game $G$ such that $R_{G}\left(\sigma_{AB^{\prime}}\right)<1$. Consider
the game for which $p=1$ and set $\mathcal{T}=\\{\delta_{w=1|z^{\prime}}\\}$
such that $w=1$ with certainty. If $\hat{\Pi}_{A}^{*}$ is a rank-one
measurement in the basis which is mutually unbiased with respect to the
eigenbasis of $\sigma_{A}$, then
$\sum_{j}\Pi_{A}^{*(j)}\sigma_{AB^{\prime}}\Pi_{A}^{*(j)}=\mathbf{u}_{A}\otimes\sigma_{B}$
and we have

$\displaystyle
R_{\mathcal{T},1}\left(\sum_{j}\Pi_{A}^{*(j)}\sigma_{AB}\Pi_{A}^{*(j)}\right)$
(133) $\displaystyle\geq
R_{\mathcal{T},0}\left(\sum_{j}\Pi_{A}^{*(j)}\sigma_{AB}\Pi_{A}^{*(j)}\right)$
(134)
$\displaystyle=\max_{\hat{\Pi}_{B}^{{}^{\prime}}}\sum_{z}P(z\big{|}\hat{\Pi}^{\prime})\left\|\text{Tr}\left[\rho_{z}\left(\hat{\Pi}_{\text{B}}^{\prime},\mathbf{u}_{A}\otimes\sigma_{B}\right)\right]\right\|_{(1)}$
(135)
$\displaystyle=\max_{\hat{\Pi}_{B}^{{}^{\prime}}}\sum_{z}\text{Tr}\left[\Pi_{B}^{{}^{\prime}(z)}\sigma_{B}\right]\left\|\mathbf{u}_{A}\right\|_{(1)}=\frac{1}{d}$
(136)

Given that $\frac{1}{d}<1$, the statement of the lemma follows. ∎

Finally, we demonstrate that when system $B$ is trivial, the ordering reduces
to standard majorisation between quantum states.

######  Lemma 8.

Let $\vec{\lambda}(\rho)$ denote the vector whose elements consist of the
eigenvalues of $\rho$. Then for any states $\rho_{A}$ and $\sigma_{A}$,
$\sigma_{A}\precsim_{(A)}\rho_{A}$ if and only if

$\displaystyle\vec{\lambda}(\sigma_{A})\precsim\vec{\lambda}(\rho_{A})$ (138)

where in the above, $\precsim$ represents standard majorisation.

###### Proof.

Both $\rho_{A}$ and $\sigma_{A}$ can be viewed as bipartite states with a
trivial $B$ subsystem (i.e. $|B|=1$.) Evidently, the expected reward for a
given $\mathcal{T}$ and games where $p=0$ is then

$\displaystyle R_{\mathcal{T},0}(\rho_{A})$
$\displaystyle=\max_{z^{\prime}}\left(\sum_{w}t_{w|z^{\prime}}\left\|\rho_{A}\right\|_{(w)}\right)$
(139) $\displaystyle R_{\mathcal{T},0}(\sigma_{A})$
$\displaystyle=\max_{z^{\prime}}\left(\sum_{w}t_{w|z^{\prime}}\left\|\sigma_{A}\right\|_{(w)}\right)$
(140)

and by considering games where $w$ is fixed, the “only if” direction of the
theorem statement follows. It remains to show that
$\vec{\lambda}(\sigma_{A})\precsim\vec{\lambda}(\rho_{A})$ is a sufficient
condition when games of chance with $p=1$ are included. In this case, the
adversarial measurement $\hat{\Pi}_{A}^{*}$ will be in a basis which is
mutually unbiased with respect to $\sigma_{A}$ (or respectively $\rho_{A}$).
Thus, for both games, the state after the adversarial measurement is
$\mathbf{u}_{A}$. Then

$\displaystyle
R_{\mathcal{T},1}(\rho_{A})=R_{\mathcal{T},0}(\mathbf{u}_{A})=R_{\mathcal{T},1}(\sigma_{A})$
(142)

and so the ordering is unchanged by including games where $p>0$. The theorem
statement then follows. ∎

##  VII Quantum Channel Gambling Games

In previous work involving gambling games for classical channels
brandsen2021entropy , the output of the classical channel was observed
directly by the host. In the quantum case, where outputs are no longer
necessarily orthogonal, the observation process is now represented through the
projective quantum measurement $\\{\Pi_{x}\\}_{x=1}^{n}$. One natural
extension of the classical gambling game introduced in brandsen2021entropy  is
to determine each game with a tripartite state $\rho_{ABX}$.

The corresponding game setup is then as follows:

  1. 1.

The player is given knowledge of state $\rho_{ABX}$ and chooses a pre-
processing channel $\mathcal{E}$ on subsystem $A$

  2. 2.

Channel $\mathcal{N}$ is implemented on subsystem $A$. From here on, we denote
this as $\mathcal{N}\left(\mathcal{E}\left(\rho_{ABX}\right)\right)$ with the
understanding that the channel action is only on system $A$.

  3. 3.

The player chooses a rank-one measurement on system $AB$. The player’s optimal
choice will always be to measure in the ordered eigenbasis of
$\mathcal{N}\left(\mathcal{E}\right)$.

  4. 4.

The measurement outcome $y$ is obtained and the player wins with a reward of
$1$ if $y\leq x$ and loses with a reward of $0$ else

We note that the rank-one restriction above is necessary in order to prevent
the player from cheating with a trivial measurement (e.g.,
$\\{I,0,0,\ldots\\}$) which would allow the player to win any game with any
channel, thus preventing any meaningful ordering between quantum channels.
This setup is depicted in Figure 1.

\begin{overpic}[scale={0.62}]{qchannel_circuit} \put(100.0,22.0){$x$}
\put(12.0,45.0){$\rho_{ABX}$} \put(162.0,68.0){$y$}
\put(58.0,72.0){$\mathcal{E}$} \put(93.0,72.0){$\mathcal{N}$}
\put(183.0,65.0){Win if} \put(186.0,55.0){ $y\leq x$} \put(183.0,35.0){Lose
else} \large\end{overpic} Figure 6: Depiction of “quantum gambling game”. Red
corresponds to the player’s choice while purple is fixed by the game.

The resulting expected reward for channel $\mathcal{N}$ and distribution
$\mathbf{p}$ becomes

$\displaystyle R_{\rho_{ABX}}\left(\mathcal{N}\right)$
$\displaystyle=\max_{\mathcal{E}}\left(\sum_{x}p_{x}\Big{\|}\mathcal{N}\left(\mathcal{E}\left(\rho_{AB}^{\left(x\right)}\right)\Big{\|}_{\left(x\right)}\right)\right)$
(143)

where $\big{\|}\cdot\big{\|}_{\left(x\right)}$ is the Ky-Fan $x$-norm.

Based on these games of chance, we can define a majorization relation between
quantum channels.

######  Definition 7.

We denote the case where channel $\mathcal{M}_{C\rightarrow D}$ is majorized
by channel $\mathcal{N}_{C^{\prime}\rightarrow D^{\prime}}$ according to
quantum games of chance as $\mathcal{M}\precsim_{q}\mathcal{N}$. Equivalently,

$\displaystyle\mathcal{M}\precsim_{q}\mathcal{N}\leftrightarrow
R_{\rho_{ABX}}\left(\mathcal{M}\right)\leq
R_{\rho_{ABX}}\left(\mathcal{N}\right)\ \ \text{for all}\ \ \rho_{ABX}$ (144)

where it is sufficient to bound the dimensions of systems $A,B$ by
$\max\left(|C|,|C^{\prime}|\right)$. Likewise, the dimension of system $X$ can
be bounded by
$\max\left(|C|,|C^{\prime}|\right)\times\min\left(|D|,|D^{\prime}|\right)$.

The bound for the dimension of $X$ follows by noting that
$\mathcal{N}\circ\mathcal{E}\left(\rho_{AB}\right)$ has at most
$|D^{\prime}|\times|B|$ nonzero eigenvalues and likewise
$\mathcal{M}\circ\mathcal{E}\left(\rho_{AB}\right)$ has at most $|D|\times|B|$
nonzero eigenvalues. Since $|B|\leq\max\left(|C|,|C^{\prime}|\right)$, the
bound on $X$ immediately follows.

Finally, we define the entropy of a quantum channel to be a function which is
monotonic under channel ordering, namely

######  Definition 8.

A non-zero function,

$H:\underset{A,B}{\bigcup}\mathrm{CPTP}(A\rightarrow B)\rightarrow\mathbb{R}$

where the union is over all finite quantum systems $A$ and $B$, is a quantum
channel entropy if it satisfies the following two conditions:

  1. 1.

It is monotonic under channel majorization; i.e. given quantum channels
$\mathcal{N}$ and $\mathcal{M}$, then

$H(\mathcal{M})\geq H(\mathcal{N})\ \ \text{if}\ \
\mathcal{M}\precsim\mathcal{N}$ (145)

  2. 2.

It is additive under tensor products; i.e.

$H(\mathcal{N}\otimes\mathcal{M})=H(\mathcal{N})+H(\mathcal{M})$ (146)

for all quantum channels $\mathcal{N}$ and $\mathcal{M}$.

##  VIII Operational Interpretation and Results for Special Classes of
Channels

We begin by showing that when $\mathcal{M}$ and $\mathcal{N}$ correspond to
states or classical channels, respectively, the quantum channel majorization
defined by gambling games reduces to the standard vector majorization and the
classical channel majorization introduced by our previous work
brandsen2021entropy  respectively.

######  Lemma 9.

If $\mathcal{N}$ and $\mathcal{M}$ are replacement channels which output
states $\rho_{\mathcal{N}}$ and $\rho_{\mathcal{M}}$ respectively, then

$\displaystyle\mathcal{M}\precsim_{q}\mathcal{N}\ \ \ \text{if and only if}\ \
\
\vec{\lambda}\left(\rho_{\mathcal{M}}\right)\precsim\vec{\lambda}\left(\rho_{\mathcal{M}}\right)$
(147)

where $\vec{\lambda}\left(\rho\right)$ denotes the ordered eigenvalues of
$\rho$.

###### Proof.

See Appendix B. ∎

Previous work introduced classical gambling games parameterized by a
correlated source $T=\left(t_{wz}\right)$ where $t_{wz}$ represents the joint
probability $\operatorname{Pr}\left(W=w,Z=z\right)$. The expected reward for
playing a $T$-game with classical channel $\mathcal{N}$ with transition
probability matrix elements $\\{p_{y|x}\\}$ was likewise defined as

$\displaystyle\operatorname{Prob}_{T}\left(\mathcal{N}\right)$
$\displaystyle\triangleq\sum_{z=1}\max_{x}t_{w,z}\sum_{y=1}^{w}p_{y|x}^{\downarrow}$
(148)

where $p^{\downarrow}$ indicates that $\\{p_{y|x}\\}|_{y}$ is ordered such
that $p_{1|x}\geq p_{2|x}\geq\ldots p_{n|x}$ for all $x$.

We now demonstrate that when $\mathcal{M}$ and $\mathcal{N}$ are both
classical, the ordering induced by quantum gambling games is equivalent to the
ordering induced by classical gambling games.

######  Theorem 4.

If $\mathcal{N}$ and $\mathcal{M}$ are both classical channels, then

$\displaystyle R_{\rho_{ABX}}\left(\mathcal{M}\right)\leq
R_{\rho_{ABX}}\left(\mathcal{N}\right)\ \ \forall\ \rho_{ABX}$ (149)

if and only if

$\displaystyle\operatorname{Prob}_{T}\left(\mathcal{M}\right)\leq\operatorname{Prob}_{T}\left(\mathcal{N}\right)\
\ \forall\ \ T=\left(t_{wz}\right).$ (150)

###### Proof.

See Appendix C. ∎

We now provide an operational interpretation for quantum channel majorization.

######  Theorem 5.

If
$\mathcal{M}=\sum_{z=1}^{s}p_{z}\mathcal{V}_{z}\circ\mathcal{N}\circ\mathcal{E}_{z}$
for some set of isometries $\mathcal{V}_{z}$, some set of preprocessing
quantum processes $\mathcal{E}_{z}$, and some probability distribution
$\\{p_{1},\ldots p_{s}\\}$, then:

$\displaystyle\mathcal{M}\precsim\mathcal{N}$ (151)

###### Proof.

We can demonstrate the theorem statement by substituting
$\mathcal{M}=\sum_{z=1}^{s}p_{z}\mathcal{V}_{z}\circ\mathcal{N}\circ\mathcal{E}_{z}$
into the expression for the reward function as follows:

$\displaystyle
R_{\rho_{ABX}}\left(\mathcal{M}\right)=\max_{\mathcal{E}}\left(\sum_{x}p_{x}\Big{\|}\mathcal{M}\circ\mathcal{E}\left(\rho_{AB}^{\left(x\right)}\right)\Big{\|}_{x}\right)$
(152)
$\displaystyle=\max_{\mathcal{E}}\left(\sum_{x}p_{x}\Big{\|}\sum_{z=1}^{s}p_{z}\mathcal{V}_{z}\circ\mathcal{N}\circ\mathcal{E}_{z}\left(\mathcal{E}\left(\rho_{AB}^{\left(x\right)}\right)\right)\Big{\|}_{x}\right)$
(153)
$\displaystyle\leq\max_{\mathcal{E},z}\left(\sum_{x}p_{x}\Big{\|}\mathcal{V}_{z}\circ\mathcal{N}\circ\mathcal{E}\left(\rho_{AB}^{\left(x\right)}\right)\Big{\|}_{x}\right)$
(154)
$\displaystyle=\max_{\mathcal{E},z}\left(\sum_{x}p_{x}\Big{\|}\mathcal{N}\circ\mathcal{E}\left(\rho_{AB}^{\left(x\right)}\right)\Big{\|}_{x}\right)$
(155) $\displaystyle=R_{\rho_{ABX}}\left(\mathcal{N}\right)$ (156)

This concludes the proof. ∎

It follows immediately from the above Theorem that two channels which are
equivalent up to a unitary will have equal performance for any game of chance.
We now show that for a fixed input and output system, unitaries are the “least
noisy” channel.

######  Lemma 10.

Let $\mathcal{N}_{A\rightarrow A}$ be a quantum channel, and let
$\mathcal{U}_{A\rightarrow A}$ be any unitary channel. Then

$\displaystyle R_{\rho_{ABX}}\left(\mathcal{U}\right)\geq
R_{\rho_{ABX}}\left(\mathcal{N}\right)\ \ \forall\rho_{ABX}$ (157)

Additionally,
$\mathcal{U}_{C\rightarrow\text{C}}\prec\mathcal{U}_{C^{\prime}\rightarrow
C^{\prime}}$ if $|C|\leq|C^{\prime}|$.

###### Proof.

We first show that unitary channels outperform any other channel of the same
dimension.

$\displaystyle R_{\rho_{ABX}}\left(\mathcal{N}\right)$
$\displaystyle=\max_{\mathcal{E}}\left(\sum_{x}p_{x}\Big{\|}\mathcal{N}\circ\mathcal{E}\left(\rho_{AB}^{\left(x\right)}\right)\Big{\|}_{x}\right)$
(158)
$\displaystyle\leq\max_{\tilde{\mathcal{E}}}\left(\sum_{x}p_{x}\Big{\|}\tilde{\mathcal{E}}\left(\rho_{AB}^{\left(x\right)}\right)\Big{\|}_{x}\right)$
(159)
$\displaystyle=\max_{\tilde{\mathcal{E}}}\left(\sum_{x}p_{x}\Big{\|}U\tilde{\mathcal{E}}\left(\rho_{AB}^{\left(x\right)}\right)U^{{\dagger}}\Big{\|}_{x}\right)$
(160) $\displaystyle=R_{\rho_{ABX}}\left(\mathcal{U}\right).$ (161)

The statement that higher dimensional unitary channels outperform lower
dimensional unitary channels follow by setting
$\rho_{AB}=\ket{\Phi_{AB}^{+}}\\!\\!\bra{\Phi_{AB}^{+}}$ where
$|A|=|B|=|C^{\prime}|$ (namely, the dimensions of systems $A$ and $B$ are set
to be equal to the input dimension of the higher dimensional unitary).

Thus, $\mathcal{U}_{C^{\prime}\rightarrow C^{\prime}}\left(\rho_{AB}\right)$
will result in a pure state and perfect reward for any game. On the other
hand, any game played with $\mathcal{U}_{C\rightarrow C}$ will require some
choice of preprocessing channel $\mathcal{E}_{C^{\prime}\rightarrow C}$, which
is necessarily entanglement breaking. ∎

##  IX Uniqueness of the Quantum Channel Gambling Game

In principle, one could consider more general games with additional resources.
In Appendix C, we demonstrate that generalizations of the quantum gambling
game do not yield any reasonable interpretation related to entropy.
First, we show that allowing the player to pass side information through a
classical wire from $\mathcal{E}$ to the final measurement would enable the
player to win any game with the classical identity channel. Given that quantum
entanglement is a resource, we would expect the quantum identity channel to
strictly outperform the classical identity channel, and so the gambling game
with an added classical communication wire does not lead to a reasonable
ordering between channels.

\begin{overpic}[scale={0.42}]{game_classical_wire} \put(100.0,33.0){$x$}
\put(10.0,60.0){$\rho_{ABX}$} \put(165.0,82.0){$y$}
\put(60.0,83.0){$\mathcal{E}$} \put(100.0,83.0){$\mathcal{N}$}
\put(190.0,75.0){Win if} \put(195.0,65.0){ $y\leq x$} \put(195.0,35.0){Lose
else} \put(140.0,78.0){$\hat{\Pi}_{w}$} \put(100.0,120.0){$w$}
\large\end{overpic} Figure 7: Gambling game with added communication wire.

One could additionally consider a game with quantum combs. In Appendix D, we
discuss why allowing arbitrary quantum combs $C_{j}$ would again prevent a
reasonable channel ordering. Specifically, if $\\{C_{j}\\}$ are allowed to be
arbitrary bipartite channels, it is not possible to obtain an ordering between
channels $\mathcal{N}$ and $\mathcal{U}\circ\mathcal{N}$ where $\mathcal{U}$
is any unitary and $\mathcal{N}$ is a pure state replacement channel. Given
that we expect two channels which are related by a unitary to have equivalent
entropy, this game must again be excluded.

\begin{overpic}[scale={0.38}]{game_with_combs} \small\put(100.0,22.0){$x$}
\put(113.0,52.0){$C_{2}$} \put(144.0,82.0){$C_{3}$} \put(82.0,82.0){$C_{1}$}
\put(10.0,60.0){$\rho_{ABC}^{X}$} \put(195.0,80.0){$y$}
\put(175.0,75.0){$\hat{\Pi}$} \put(55.0,95.0){$\mathcal{E}$}
\put(113.0,95.0){$\mathcal{N}$} \put(210.0,85.0){Win if} \put(215.0,75.0){
$y\leq x$} \put(210.0,45.0){Lose else} \large\end{overpic}

##  X Noisy Channels

We include a table summarizing the reward function of special qubit channels
for two key games involving the maximally entangled state and a pure tensor
product state respectively. More specifically, in the first game we consider,
we define $\rho_{AB}=\phi_{AB}^{+}$ and let $X$ be drawn according to the
classical state $\rho_{X}$ s.t. $\rho_{ABX}=\phi_{AB}^{+}\otimes\rho_{X}$. In
the second game we consider, system $B$ is set to be trivial such that
$\rho_{AB}=\rho_{A}=\ket{0}\\!\\!\bra{0}$ and
$\rho_{ABX}=\ket{0}\\!\\!\bra{0}\otimes\rho_{X}$. In both cases, we denote the
ordered diagonal elements of $\rho_{X}$ as $[p_{j}]$.

$\mathcal{N}$ | $R_{\phi^{+}_{AB}\otimes\rho_{X}}\left(\mathcal{N}\right)$ | $R_{\ket{0}\\!\\!\bra{0}_{A}\otimes\rho_{X}}\left(\mathcal{N}\right)$
---|---|---
$\mathcal{U}$ | 1 | 1
$\mathcal{I}_{\text{CL}}$ | $1-\frac{1}{2}p_{1}$ | 1
$D_{\gamma}$ | $\left(1-\gamma\right)+\frac{\gamma\sum_{x}p_{x}x}{4}$ | $1-\gamma\frac{p_{1}}{2}$
$\mathcal{N}_{\hat{\Pi}}$ | $1-\frac{1}{2}p_{1}$ | 1
$\mathcal{A}_{\gamma}$ | $(1-\frac{\gamma}{2})p_{1}+\frac{\gamma}{2}$ | 1
$\mathcal{R}_{\sigma}$ | $\sum_{x}p_{x}\|\sigma\otimes\frac{I}{2}\|_{\left(x\right)}$ | $\sum_{x}p_{x}\|\sigma\|_{\left(x\right)}$
$\mathcal{F}_{\gamma}$ | $\left(1-\gamma\right)+\gamma\left(1-\frac{1}{2}p_{1}\right)$ | 1

We now define all channels included in the above table.

  1. 1.

$\mathcal{U}$ is a unitary channel with corresponding unitary $U$ such that
$\mathcal{U}(\rho)=U\rho U^{{\dagger}}$ for all
$\rho\in\mathcal{D}(\mathcal{H}_{2})$.

  2. 2.

$\mathcal{I}_{\text{CL}}$ is a classical identity channel such that for some
basis $\\{\ket{j}\\}$ then

$\displaystyle\mathcal{I}_{\text{CL}}(\rho)=\sum_{j}\bra{j}\\!\rho\\!\ket{j}\ket{j}\\!\\!\bra{j}$
(162)

  3. 3.

$\mathcal{D}_{\gamma}$ is the qubit depolarizing channel with noise parameter
$\gamma$ defined as

$\displaystyle\mathcal{D}_{\gamma}(\rho)=(1-\gamma)\rho+\gamma\mathbf{u}_{2}$
(163)

  4. 4.

$\mathcal{N}_{\hat{\Pi}}$ is the channel implementing POVM
$\hat{\Pi}=\\{\Pi_{j}\\}$ as

$\displaystyle\mathcal{N}_{\hat{\Pi}}(\rho)=\sum_{j}\text{Tr}\left[\Pi_{j}\rho\right]\ket{j}\\!\\!\bra{j}$
(164)

  5. 5.

$\mathcal{A}_{\gamma}$ is the amplitude damping channel with noise parameter
$\gamma$ defined as

$\displaystyle\mathcal{A}_{\gamma}\left(\begin{pmatrix}\rho_{00}&\rho_{01}\\\
\rho_{10}&\rho_{11}\end{pmatrix}\right)$
$\displaystyle\triangleq\begin{pmatrix}\rho_{00}+\gamma\rho_{11}&\sqrt{1-\gamma}\rho_{01}\\\
\sqrt{1-\gamma}\rho_{10}&\left(1-\gamma\right)\rho_{11}.\end{pmatrix}$ (165)

  6. 6.

The replacement channel $\mathcal{R}_{\sigma}$ discards the input and outputs
state $\sigma$ such that $\mathcal{R}_{\sigma}(\rho)=\sigma$ for all $\rho$.

  7. 7.

The dephasing channel $\mathcal{F}_{\gamma}$ with parameter $\gamma$ acts as

$\displaystyle\mathcal{F}_{\gamma}(\rho)=(1-\gamma)\rho+\gamma
I_{\text{CL}}(\rho)$ (166)

Details for how the results in the table are obtained are included in Appendix
E.

Finally, we note that the maximal output state purity attainable with a
channel is monotonic under the ordering induced by games of chance.

######  Lemma 11.

Suppose that $\mathcal{M}\precsim\mathcal{N}$. Then

$\displaystyle\underset{\rho}{\operatorname{sup}}\left(\operatorname{Tr}\Big{[}\mathcal{N}\left(\rho\right)^{2}\Big{]}\right)\geq\underset{\rho}{\operatorname{sup}}\left(\operatorname{Tr}\Big{[}\mathcal{M}\left(\rho\right)^{2}\Big{]}\right)$
(167)

###### Proof.

See Appendix F ∎

##  XI Conclusions

In this work, we demonstrate that the set of _all_ conditional entropy non-
decreasing channels is equivalent to the set of conditionally unital, semi-
causal (CUSC) channels. We then require quantum conditional entropy to be
monotonic under the action of CUSC channels as well as satisfying simple
axioms such as additivity and normalisation. This minimalist approach is
sufficient to guarantee negativity of conditional entropy for maximally
entangled states, and additionally is sufficient demonstrate that conditional
entropy is non-negative for separable bipartite states. We discuss
applications of this axiomatic approach to classical bipartite states and
demonstrate that the resulting conditional majorization matches the classical
conditional majorization first defined in Gour . Finally, we develop an
operational approach for characterising entropy via games of chance, and apply
this approach to both quantum bipartite states and quantum channels
respectively.

Note Added: We recently became aware of independent work of Vempati et al.
Vempati2021

##  XII Acknowledgments

The authors would like to thank Henry Pfister for helpful discussions. GG and
IG acknowledge support from the Natural Sciences and Engineering Research
Council of Canada (NSERC). SB acknowledges support from the National Science
Foundation (NSF) under Grant No. 1908730 and 1910571. Any opinions, findings,
conclusions, and recommendations expressed in this material are those of the
authors and do not necessarily reflect the views of these sponsors.

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##  Appendix A Quantum Gambling Games on Bipartite Classical States

In this appendix, we demonstrate that the majorisation relation
$\precsim_{(A)}$ is equivalent to the majorisation relation $\precsim_{C}$ on
classical states. We here denote the expected success probability of playing a
$\mathcal{T}$-game with classical bipartite state $\rho_{AB}$ as

$\displaystyle\operatorname{Prob}_{\mathcal{T}}\left(\rho_{AB}\right)$
$\displaystyle=\sum_{x}p_{x}\max_{z^{\prime}}\left(\sum_{w}t_{w|z^{\prime}}\left\|\mathbf{p}_{x}\right\|_{\left(w\right)}\right),$
(168)

where $\mathbf{p}_{x}=\\{p_{y|x}\\}_{x}$ is the probability that the output of
$\rho_{AB}$ on system $B$ is $Y=y$, given that the output on system $A$ is
$X=x$.

######  Lemma 12.

For any classical bipartite states $\rho_{AB}=\\{q_{x,y}\\}_{x,y}$ and
$\sigma_{AB}=\\{p_{x,y}\\}_{x,y}$, the following inequality holds

$\displaystyle\operatorname{Prob}_{\mathcal{T}}\left(\sigma_{AB}\right)\leq\operatorname{Prob}_{\mathcal{T}}\left(\rho_{AB}\right)\
\forall\mathcal{T}$ (169)

if and only if

$\displaystyle R_{\mathcal{T},p}\left(\sigma_{AB}\right)\leq
R_{\mathcal{T},p}\left(\rho_{AB}\right)$ (170)

for all distributions $\mathcal{T}$ and all $0\leq p\leq 1$

###### Proof.

First, we show that for all quantum gambling games with distribution
$\mathcal{T}$ and $p=0$, the following equality holds

$\displaystyle
R_{\mathcal{T},0}\left(\sigma_{AB}\right)=\operatorname{Prob}_{\mathcal{T}}\left(\sigma_{AB}\right).$
(171)

Recall that any classical bipartite state $\rho_{AB}$ can be written in
quantum notation as

$\displaystyle\rho_{AB}$
$\displaystyle=\sum_{x,y}q_{x,y}\ket{x}\\!\\!\bra{x}\otimes\ket{y}\\!\\!\bra{y},$
(172) $\displaystyle\sigma_{AB}$
$\displaystyle=\sum_{x,y}p_{x,y}\ket{x}\\!\\!\bra{x}\otimes\ket{y}\\!\\!\bra{y}.$
(173)

First consider games where $p=0$. Then

$\displaystyle R_{\mathcal{T},0}\left(\rho_{AB}\right)$
$\displaystyle=\underset{\hat{\Pi}^{{}^{\prime}},f}{\text{max}}\left(\sum_{z}\sum_{w}t_{w|f(z)}\left\|\text{Tr}_{B}\left[\Pi_{\text{B}}^{\prime(z)}\rho_{AB}\Pi_{\text{B}}^{{}^{\prime}(z)}\right\|_{(w)}\right]\right)$
(174)
$\displaystyle=\underset{\hat{\Pi}^{{}^{\prime}},f}{\text{max}}\left(\sum_{z}\sum_{w}t_{w|f(z)}\left\|\sum_{x}q_{x,y}\ket{x}\\!\\!\bra{x}\text{Tr}\left[\Pi_{\text{B}}^{\prime(z)}\ket{y}\\!\\!\bra{y}\Pi_{\text{B}}^{{}^{\prime}(z)}\right]\right\|_{(w)}\right)$
(175)
$\displaystyle=\underset{f}{\text{max}}\left(\sum_{z}\sum_{w}t_{w|f(z)}\left\|\sum_{x}q_{x,y}\ket{x}\\!\\!\bra{x}\text{Tr}\left[\ket{z}\\!\\!\braket{z}{y}\\!\\!\braket{y}{z}\\!\\!\bra{z}\right]\right\|_{(w)}\right)$
(176)
$\displaystyle=\underset{f}{\text{max}}\left(\sum_{z}\sum_{w}t_{w|f(z)}\left\|\sum_{x}q_{x,z}\ket{x}\\!\\!\bra{x}\right\|_{(w)}\right)$
(177)

Upon relabelling variables $x$ to $k$ for consistency with the convention used
for classical gambling games, we have

$\displaystyle R_{\mathcal{T},0}(\rho_{AB})$
$\displaystyle=\sum_{z}\max_{z^{\prime}}\sum_{w}t_{w|z^{\prime}}\Big{\|}\sum_{k}q_{k,z}\ket{k}\\!\\!\bra{k}\Big{\|}_{\left(w\right)}$
(179)
$\displaystyle=\sum_{z}q_{z}\max_{z^{\prime}}\sum_{w}t_{w|z^{\prime}}\Big{\|}\sum_{k}q_{k|z}\ket{k}\\!\\!\bra{k}\Big{\|}_{\left(w\right)}$
(180)
$\displaystyle=\sum_{z}q_{z}\max_{z^{\prime}}\sum_{w}t_{w|z^{\prime}}\Big{\|}\mathbf{q}_{z}\Big{\|}_{\left(w\right)}$
(181) $\displaystyle=\operatorname{Prob}_{\mathcal{T}}\left(\rho_{AB}\right).$
(182)

It immediately follows that
$\operatorname{Prob}_{\mathcal{T}}\left(\sigma_{AB}\right)\leq\operatorname{Prob}_{\mathcal{T}}\left(\rho_{AB}\right)$
for all $\mathcal{T}$ if $R_{\mathcal{T},p}\left(\sigma_{AB}\right)\leq
R_{\mathcal{T},p}\left(\rho_{AB}\right)$ for all distributions $\mathcal{T}$
and all $p\in[0,1]$.

Now, we demonstrate that the additional games allowed in the quantum case
(i.e., games where $p>1$) does not affect the ordering. This follows
immediately upon noting that the adversary can measure in a mutually unbiased
basis, such that the state $\rho_{AB}$ transforms to
$\mathbf{u}_{A}\otimes\sigma_{B}$ for some $\sigma_{B}$.

It follows that
$R_{\mathcal{T},1}\left(\rho_{AB}\right)=R_{\mathcal{T},1}\left(\sigma_{AB}\right)$
for all $\mathcal{T}$ and therefore the additional games do affect the partial
ordering. ∎

As an aside, we note that it is sufficient to consider a restricted set of
games for ordering between two classical states.

######  Lemma 13.

If $\rho_{AB}$ and $\sigma_{AB}$ are two classical states, it is sufficient to
restrict to games where $w$ is determined. Namely,

$\displaystyle\operatorname{Prob}_{G}\left(\sigma_{AB}\right)\leq\operatorname{Prob}_{G}\left(\rho_{AB}\right)$
(183)

for all games $G$ if and only if

$\displaystyle\operatorname{Prob}_{(w,\mathbf{p})}(\sigma_{AB})\leq\operatorname{Prob}_{\left(w,\mathbf{p}\right)}\left(\rho_{AB}\right),$
(184)

for all $G$ where $w$ is fixed
(i.e.$\mathcal{T}=\\{\delta_{w|z^{\prime}}\\}|_{z^{\prime}}$) and where
$\mathbf{p}=\\{1,0\\}$, so that an adversarial measurement never occurs.

###### Proof.

It follows immediately from the previous proof that it is sufficient to
restrict to games where $\mathbf{p}=\\{1,0\\}$, as for an arbitrary
$\mathbf{p}=\\{p,1-p\\}$, the reward for any bipartite state $\rho_{AB}$ can
be written as

$\displaystyle R_{\left(\mathcal{T},\\{p,1-p\\}\right)}\left(\rho_{AB}\right)$
$\displaystyle=pR_{\left(\mathcal{T},\\{1,0\\}\right)}\left(\rho_{AB}\right)$
(185)
$\displaystyle+\left(1-p\right)R_{\left(\mathcal{T},\\{0,1\\}\right)}\left(\rho_{AB}\right).$
(186)

We demonstrated in the previous case that for a classical state
$\sigma_{AB}=\sum_{x,y}p_{x,y}\ket{x}\\!\\!\bra{x}\otimes\ket{y}\\!\\!\bra{y}$,
the following equality holds

$\displaystyle
R_{\left(\mathcal{T},\\{1,0\\}\right)}\left(\sigma_{AB}\right)=\operatorname{Prob}_{T}\left(\sigma_{AB}\right)$
(187)

Finally, we demonstrate that this can be rewritten as a convex combination of
games where $w$ is fixed:

$\displaystyle\operatorname{Prob}_{\mathcal{T}}\left(\sigma_{AB}\right)$
$\displaystyle=\sum_{x}p_{x}\max_{z}\left(\sum_{w}t_{w|z^{\prime}}\|\mathbf{p}_{x}\|_{\left(w\right)}\right)$
(188)
$\displaystyle=\sum_{x}p_{x}\left(\sum_{w}t_{w|z_{x}^{*}}\|\mathbf{p}_{x}\|_{\left(w\right)}\right)$
(189)
$\displaystyle=\sum_{w}\sum_{x}t_{w|z_{x}^{*}}p_{x}\|\mathbf{p}_{x}\|_{\left(w\right)}$
(190)
$\displaystyle=\sum_{w}c_{w}\sum_{x}p_{x}\|\mathbf{p}_{x}\|_{\left(w\right)},$
(191)

where for each $x$, the variable $z\left(x\right)^{*}$ is defined to be the
optimal choice and for each $w$,

$\displaystyle
c_{w}=\frac{\sum_{x}t_{w|z_{x}^{*}}p_{x}\|\mathbf{p}_{x}\|_{\left(w\right)}}{\sum_{x}p_{x}\|\mathbf{p}_{x}\|_{\left(w\right)}}$
(192)

is a positive constant. Finally, if $\rho_{AB}$ and $\sigma_{AB}$ are ordered
for all games with a fixed $w$, then

$\displaystyle\operatorname{Prob}_{\mathcal{T}}\left(\sigma_{AB}\right)$
$\displaystyle=\sum_{w}c_{w}\operatorname{Prob}_{w}\left(\sigma_{AB}\right)$
(193)
$\displaystyle\leq\sum_{w}c_{w}\operatorname{Prob}_{w}\left(\rho_{AB}\right)$
(194)
$\displaystyle=\sum_{w}\frac{\sum_{x}t_{w|z_{x}^{*}}p_{x}\|\mathbf{p}_{x}\|_{\left(w\right)}}{\sum_{x}p_{x}\|\mathbf{p}_{x}\|_{\left(w\right)}}\sum_{x}q_{x}\|\mathbf{q}_{x}\|_{\left(w\right)}$
(195)
$\displaystyle=\sum_{x}q_{x}\sum_{w}\left(\sum_{x^{\prime}}\tilde{p}_{x^{\prime}}t_{w|z_{x^{\prime}}^{*}}\right)\|\mathbf{q}_{x}\|_{\left(w\right)}$
(196)
$\displaystyle\leq\sum_{x}q_{x}\max_{x^{\prime}}\left(\sum_{w}t_{w|z_{x^{\prime}}^{*}}\|\mathbf{q}_{x}\|_{\left(w\right)}\right)$
(197)
$\displaystyle\leq\sum_{x}q_{x}\max_{z^{\prime}}\left(\sum_{w}t_{w|z^{\prime}}\|\mathbf{q}_{x}\|_{\left(w\right)}\right)$
(198) $\displaystyle=\operatorname{Prob}_{\mathcal{T}}\left(\rho_{AB}\right),$
(199)

where
$\tilde{p}_{x}=\frac{p_{x}\|\mathbf{p}_{x}\|_{\left(w\right)}}{\sum_{x}p_{x}\|\mathbf{p}_{x}\|_{\left(w\right)}}$.
∎

##  Appendix B Proof of reduction for quantum replacement channels

The “only if” part follows trivially. We now prove the “if” part of the
statement. First, recall that in [20], it was demonstrated that for classical
games of chance $\mathcal{M}\prec\mathcal{N}$ if and only if
$\rho_{\mathcal{M}}\precsim\rho_{\mathcal{N}}$. Additionally, classical games
of chance can be represented as quantum games of chance where
$\rho_{ABX}=\ket{0}\\!\\!\bra{0}_{A}\otimes\rho_{X}$. Thus, the “if” direction
of the theorem statement may be rewritten as

$\displaystyle R_{\rho_{ABX}}\left(\mathcal{M}\right)\leq
R_{\rho_{ABX}}\left(\mathcal{N}\right)\ \ \forall\ \rho_{ABX}$ (200)

if

$\displaystyle R_{\rho_{AX}}\left(\mathcal{M}\right)\leq
R_{\rho_{AX}}\left(\mathcal{N}\right)\ \ \forall\
\rho_{ABX}=\ket{0}\\!\\!\bra{0}_{A}\otimes\rho_{X}$ (201)

Clearly, since $\mathcal{M}$ is a replacement channel, then
$\mathcal{M}\circ\mathcal{E}=\mathcal{M}$ for all $\mathcal{E}$. Hence, the
choice of preprocessing channel does not affect the reward, so w.l.o.g. we may
set $\mathcal{E}=\mathcal{I}$. It follows that the reward then becomes

$\displaystyle R_{\rho_{ABX}}\left(\mathcal{M}\right)$
$\displaystyle=\max_{\mathcal{E}}\left(\sum_{x=1}^{\ell}p_{x}\Big{\|}\mathcal{M}\circ\mathcal{E}\left(\rho_{AB}^{\left(x\right)}\right)\Big{\|}_{x}\right)$
(202)
$\displaystyle=\sum_{x=1}^{\ell}p_{x}\Big{\|}\mathcal{M}\left(\rho_{AB}^{\left(x\right)}\right)\Big{\|}_{x}$
(203)
$\displaystyle=\sum_{x=1}^{\ell}p_{x}\Big{\|}\rho_{M}\otimes\operatorname{Tr}_{A}[\rho_{AB}^{\left(x\right)}]\Big{\|}_{x}$
(204)

We now denote
$\mathbf{\lambda}^{\downarrow}\left(\rho_{\mathcal{M}}\right)=[\lambda_{1}^{M},\lambda_{2}^{M},\ldots,\lambda_{d}^{M}]$
and likewise
$\mathbf{\lambda}^{\downarrow}\left(\operatorname{Tr}_{A}[\rho_{AB}^{\left(x\right)}]\right)=[\lambda_{1,x}^{B},\lambda_{2,x}^{B},\ldots,\lambda_{d,x}^{B}]$.
Then

$\displaystyle\text{eig}\left(\rho_{\mathcal{M}}\otimes\operatorname{Tr}_{A}[\rho_{AB}^{\left(x\right)}]\right)$
$\displaystyle=\\{\lambda_{j}^{\left(M\right)}\lambda_{k,x}^{B}\\}_{j,k}$
(205)

We define $f\left(j,k,x\right)$ to return the position of
$\lambda_{j}^{\left(M\right)}\lambda_{k,x}^{B}$ in the full ordered set
$\\{\lambda_{j}^{\left(M\right)}\lambda_{k,x}^{B}\\}_{j,k}$ (for example,
$f\left(1,1,x\right)=1$ since $\lambda_{1}^{\left(M\right)}\lambda_{1,x}^{B}$
is the largest eigenvalue and $f\left(d,d,x\right)=d^{2}$ since
$\lambda_{d}^{\left(M\right)}\lambda_{d,x}^{B}$ is the smallest eigenvalue).
Finally, we can define the set

$\displaystyle S\left(j\right)=\\{x,k\ \big{|}\ \ \text{such that }\
f\left(j,k,x\right)\leq x\\}$ (206)

In other words, $S\left(j\right)$ provides all values of $x,k$ such that the
term $p_{x}\lambda_{j}^{\left(M\right)}\lambda_{k,x}^{B}$ appears in the sum.

We now rewrite the reward function as

$\displaystyle R_{\rho_{ABX}}\left(\mathcal{M}\right)$
$\displaystyle=\sum_{x=1}^{\ell}p_{x}\sum_{j,k\ \text{s.t.}\
f\left(j,k,x\right)\leq x}\lambda_{j}^{\left(m\right)}\lambda_{k,x}^{B}$ (207)
$\displaystyle=\sum_{j=1}^{d}\lambda_{j}^{\left(M\right)}\left(\sum_{x,j\in
S\left(j\right)}p_{j}\lambda_{k,x}^{B}\right)$ (208)

In previous work, we proved that games of the form
$\sum_{j}\lambda_{j}^{\left(M\right)}C_{j}$ for $C_{j}>0$ can be rewritten in
the form $\alpha R_{\tilde{p}}\left(\mathcal{M}\right)$. In this case,
evidently

$\displaystyle C_{j}\triangleq\underset{{x,k\in
S\left(j\right)}}{\sum}p_{x}\lambda_{k,x}^{B}>0$ (209)

so there exists some $\alpha$ and $\tilde{p}$ such that
$R_{\rho_{ABX}}\left(\mathcal{M}\right)=\alpha
R_{\tilde{p}}\left(\mathcal{M}\right)$.

Since we are assuming that $\mathcal{M}\precsim\mathcal{N}$, then
$R_{\tilde{p}}\left(\mathcal{M}\right)\leq
R_{\tilde{p}}\left(\mathcal{N}\right)$. It follows that

Finally, we have

$\displaystyle R_{\rho_{ABX}}\left(\mathcal{M}\right)$ $\displaystyle=\alpha
R_{\tilde{p}}\left(\mathcal{M}\right)$ (210) $\displaystyle\leq\alpha
R_{\tilde{p}}\left(\mathcal{N}\right)$ (211)
$\displaystyle=\max_{z}\left(\sum_{k=1}^{d}\lambda_{k}^{\downarrow}\left(\rho_{\mathcal{N}}\right)\left(\sum_{j=1}^{d}\lambda_{j}^{B}\sum_{\ell=f_{k}\left(j\right)}^{d^{2}}p_{\ell}\right)\right)$
(212)
$\displaystyle\leq\max_{z}\left(\sum_{x}p_{x}\Big{\|}\rho_{\mathcal{N}}\otimes\operatorname{Tr}_{A}[\rho_{AB}^{\left(x\right)}]\Big{\|}\right)$
(213) $\displaystyle=R_{\rho_{ABX}}\left(\mathcal{N}\right)$ (214)

The statement then follows from recalling that classical gambling games are
equivalent to quantum gambling games where $\rho_{ABX}$ has the form
$\ket{0}\\!\\!\bra{0}_{A}\otimes\rho_{X}$.

##  Appendix C Proof of reduction for Classical Channels

The “only if” part follows immediately from noting that the quantum gambling
games are a strict generalisation of the classical gambling games. We now
prove the “if” part of the statement. For $\mathcal{M}$ classical,
$\mathcal{E}$ can be reduced to selecting the optimal classical output $z$,
such that $\mathcal{E}\left(\rho\right)=\ket{z}\\!\\!\bra{z}$ for every
$\rho$. It follows that the reward then becomes

$\displaystyle R_{\rho_{ABX}}\left(\mathcal{M}\right)$
$\displaystyle=\max_{\mathcal{E}}\left(\sum_{x=1}^{\ell}p_{x}\Big{\|}\mathcal{M}\left(\mathcal{E}\left(\rho_{AB}^{\left(x\right)}\right)\right)\Big{\|}_{x}\right)$
(215)
$\displaystyle=\max_{z}\left(\sum_{x=1}^{\ell}p_{x}\Big{\|}\mathcal{M}\left(\ket{x}\\!\\!\bra{x}\right)\otimes\operatorname{Tr}_{A}[\rho_{AB}^{\left(x\right)}]\Big{\|}_{x}\right)$
(216)
$\displaystyle=\sum_{x=1}^{\ell}p_{x}\Big{\|}\lambda_{j}^{\downarrow}\left(\mathcal{M}\left(\ket{z^{*}}\\!\\!\bra{z^{*}}\right)\right)\otimes\operatorname{Tr}_{A}[\rho_{AB}^{\left(x\right)}]\Big{\|}_{x}$
(217)

where $z^{*}$ is the optimal value (i.e. the value of $z$ which maximises the
reward). We now denote
$\mathbf{\lambda}^{\downarrow}\left(\mathcal{M}\left(\ket{z^{*}}\\!\\!\bra{z^{*}}\right)\right)=[\lambda_{1}^{M},\lambda_{2}^{M},\ldots,\lambda_{d}^{M}]$
and likewise
$\mathbf{\lambda}^{\downarrow}\left(\operatorname{Tr}_{A}[\rho_{AB}^{\left(x\right)}]\right)=[\lambda_{1,x}^{B},\lambda_{2,x}^{B},\ldots,\lambda_{d,x}^{B}]$.
Then

$\displaystyle\text{eig}\left(\mathcal{M}\left(\ket{z^{*}}\\!\\!\bra{z^{*}}\right)\otimes\operatorname{Tr}_{A}[\rho_{AB}^{\left(x\right)}]\right)$
$\displaystyle=\\{\lambda_{j}^{\left(M\right)}\lambda_{k,x}^{B}\\}_{j,k}$
(218)

We define $f\left(j,k,x\right)$ to return the position of
$\lambda_{j}^{\left(M\right)}\lambda_{k,x}^{B}$ in the full ordered set
$\\{\lambda_{j}^{\left(M\right)}\lambda_{k,x}^{B}\\}_{j,k}$ (for example,
$f\left(1,1,x\right)=1$ since $\lambda_{1}^{\left(M\right)}\lambda_{1,x}^{B}$
is the largest eigenvalue and $f\left(d,d,x\right)=d^{2}$ since
$\lambda_{d}^{\left(M\right)}\lambda_{d,x}^{B}$ is the smallest eigenvalue).
Finally, we can define the set

$\displaystyle S\left(j\right)=\\{x,k\ \big{|}\ \ \text{such that }\
f\left(j,k,x\right)\leq x\\}$ (219)

In other words, $S\left(j\right)$ provides all values of $x,k$ such that the
term $p_{x}\lambda_{j}^{\left(M\right)}\lambda_{k,x}^{B}$ appears in the sum.

We now rewrite the reward function as

$\displaystyle R_{\rho_{ABX}}\left(\mathcal{M}\right)$
$\displaystyle=\sum_{j=1}^{d}\lambda_{j}^{\left(M\right)}\left(\sum_{x,j\in
S\left(j\right)}p_{j}\lambda_{k,x}^{B}\right)$ (220)

However, this is of the form of a classical gambling game (possibly up to a
scaling). In previous work, we proved that games of the form
$\sum_{j}\lambda_{j}^{\left(M\right)}C_{j}$ for $C_{j}>0$ can be rewritten in
the form $\alpha R_{\tilde{p}}\left(\mathcal{M}\right)$. In this case,
evidently

$\displaystyle C_{j}\triangleq\underset{{x,k\in
S\left(j\right)}}{\sum}p_{x}\lambda_{k,x}^{B}>0$ (221)

so there exists some $\alpha$ and $\tilde{p}$ such that
$R_{\rho_{ABX}}\left(\mathcal{M}\right)=\alpha
R_{\tilde{p}}\left(\mathcal{M}\right)$. Since we are assuming that
$\mathcal{M}\precsim\mathcal{N}$, then
$R_{\tilde{p}}\left(\mathcal{M}\right)\leq
R_{\tilde{p}}\left(\mathcal{N}\right)$. It follows that

Finally, we have

$\displaystyle R_{\rho_{ABX}}\left(\mathcal{M}\right)$ $\displaystyle=\alpha
R_{\tilde{p}}\left(\mathcal{M}\right)$ (222) $\displaystyle\leq\alpha
R_{\tilde{p}}\left(\mathcal{N}\right)$ (223)
$\displaystyle=\max_{z}\left(\sum_{k=1}^{d}\lambda_{k}^{\downarrow}\left(\mathcal{N}\left(\ket{z}\\!\\!\bra{z}\right)\right)\sum_{j=1}^{d}\lambda_{j}^{B}\sum_{\ell=f_{k}\left(j\right)}^{d^{2}}p_{\ell}\right)$
(224)
$\displaystyle\leq\max_{z}\left(\sum_{x}p_{x}\Big{\|}\mathcal{N}\left(\ket{z}\\!\\!\bra{z}\right)\otimes\operatorname{Tr}_{A}[\rho_{AB}^{\left(x\right)}]\right)$
(225) $\displaystyle=R_{\rho_{ABX}}\left(\mathcal{N}\right)$ (226)

##  Appendix D Games with Wires and Resources

###  D.1 Game with Classical Wire

We first consider the game with an additional classical wire, and demonstrate
that one can win any game of the form
$\rho_{ABX}=\ket{\psi_{AB}}\\!\\!\bra{\psi_{AB}}\otimes\rho_{X}$ with a qubit
classical identity channel. Define $\mathcal{E}$ as

$\displaystyle\mathcal{E}\left(\rho\right)$
$\displaystyle=\operatorname{Tr}[\ket{0}\\!\\!\bra{0}\rho]\ket{0}\\!\\!\bra{0}+\operatorname{Tr}[\ket{1}\\!\\!\bra{1}\rho]\ket{1}\\!\\!\bra{1}$
(227)

Thus, $\mathcal{E}$ is a POVM and if outcome $0$ (respectively 1) is obtained
the post measurement state is,

$\displaystyle\rho_{AB}^{\left(0\right)}$
$\displaystyle\triangleq\frac{\ket{0}\\!\\!\bra{0}_{A}\otimes
I_{B}\left(\ket{\psi_{AB}}\\!\\!\bra{\psi_{AB}}\right)\ket{0}\\!\\!\bra{0}_{A}\otimes
I_{B}}{\operatorname{Tr}[\ket{0}\\!\\!\bra{0}_{A}\otimes
I_{B}\left(\ket{\psi_{AB}}\\!\\!\bra{\psi_{AB}}\right)]}$ (228)
$\displaystyle\rho_{AB}^{\left(1\right)}$
$\displaystyle\triangleq\frac{\ket{1}\\!\\!\bra{1}_{A}\otimes
I_{B}\left(\ket{\psi_{AB}}\\!\\!\bra{\psi_{AB}}\right)\ket{1}\\!\\!\bra{1}_{A}\otimes
I_{B}}{\operatorname{Tr}[\ket{1}\\!\\!\bra{1}_{A}\otimes
I_{B}\left(\ket{\psi_{AB}}\\!\\!\bra{\psi_{AB}}\right)]}$ (229)

The most general expression for $\ket{\psi_{AB}}$ is

$\displaystyle\ket{\psi_{AB}}=\alpha_{00}\ket{00}+\alpha_{01}\ket{01}+\alpha_{10}\ket{10}+\alpha_{11}\ket{11}$
(230) $\displaystyle\rightarrow\ket{0}\\!\\!\bra{0}_{A}\otimes
I_{B}\ket{\psi_{AB}}=\alpha_{00}\ket{00}+\alpha_{01}\ket{01}$ (231)
$\displaystyle\rightarrow\ket{1}\\!\\!\bra{1}_{A}\otimes
I_{B}\ket{\psi_{AB}}=\alpha_{10}\ket{10}+\alpha_{11}\ket{11}$ (232)

from which it follows that

$\displaystyle\rho_{AB}^{\left(0\right)}$
$\displaystyle=\ket{0}\\!\\!\bra{0}_{A}\otimes\ket{\psi^{\left(0\right)}}\\!\\!\bra{\psi^{\left(0\right)}}_{B}$
(233) $\displaystyle\rho_{AB}^{\left(1\right)}$
$\displaystyle=\ket{1}\\!\\!\bra{1}_{A}\otimes\ket{\psi^{\left(1\right)}}\\!\\!\bra{\psi^{(1)}}_{B}$
(234)

We now show that the following measurements will allow the player to always
win the game with $\rho_{AB}^{\left(0\right)}$ and
$\rho_{AB}^{\left(1\right)}$ respectively:

$\displaystyle\hat{\Pi}^{\left(0\right)}$
$\displaystyle=\Big{\\{}\ket{0}\\!\\!\bra{0}\otimes\ket{\psi^{\left(0\right)}}\\!\\!\bra{\psi^{\left(0\right)}},\ket{0}\\!\\!\bra{0}\otimes\ket{\psi^{\left(0\right),\perp}}\\!\\!\bra{\psi^{\left(0\right),\perp}},$
(235)
$\displaystyle\ket{1}\\!\\!\bra{1}\otimes\ket{\psi^{\left(0\right)}}\\!\\!\bra{\psi^{\left(0\right)}},\ket{1}\\!\\!\bra{1}\otimes\ket{\psi^{\left(0\right),\perp}}\\!\\!\bra{\psi^{\left(0\right),\perp}}\Big{\\}}$
(236) $\displaystyle\hat{\Pi}^{\left(1\right)}$
$\displaystyle=\Big{\\{}\ket{1}\\!\\!\bra{1}\otimes\ket{\psi^{\left(1\right)}}\\!\\!\bra{\psi^{\left(1\right)}},\ket{1}\\!\\!\bra{1}\otimes\ket{\psi^{\left(1\right),\perp}}\\!\\!\bra{\psi^{\left(1\right),\perp}}$
(237)
$\displaystyle\ket{0}\\!\\!\bra{0}\otimes\ket{\psi^{\left(1\right)}}\\!\\!\bra{\psi^{\left(1\right)}},\ket{0}\\!\\!\bra{0}\otimes\ket{\psi^{\left(1\right),\perp}}\\!\\!\bra{\psi^{\left(1\right),\perp}}\Big{\\}}$
(238)

It is sufficient to show that the above measurements allow the player to
always win the game where $\text{diag}\left(\rho_{X}\right)=[1,0,\ldots,0]$
(namely, $x$ is always equal to one.) The corresponding reward function for
the given choice of $\mathcal{E}$ then becomes:

$\displaystyle\text{R}\left(I_{CL}\right)$
$\displaystyle=\sum_{x\in\\{0,1\\}}\operatorname{Tr}[\ket{x}\\!\\!\bra{x}_{A}\rho_{AB}]$
(239)
$\displaystyle\times\operatorname{Tr}\Big{[}\left(\ket{x}\\!\\!\bra{x}\otimes\ket{\psi^{\left(x\right)}}\\!\\!\bra{\psi^{\left(x\right)}}\right)\rho_{AB}^{\left(1\right)}\Big{]}$
(240)
$\displaystyle=\sum_{x\in\\{0,1\\}}\operatorname{Tr}[\ket{x}\\!\\!\bra{x}_{A}\rho_{AB}]$
(241) $\displaystyle=1$ (242)

###  D.2 Game with Quantum Combs

We now consider the quantum comb-based game depicted in Figure (reference),
and demonstrate that allowing any element $C_{j}$ to be an arbitrary quantum
channel would result in a channel ordering that cannot correspond to channel
entropy.
First, we demonstrate that $C_{3}$ cannot be an arbitrary bipartite channel.
To do so, consider the following channels:

$\displaystyle\mathcal{N}_{1}\left(\rho\right)$
$\displaystyle=\ket{0}\\!\\!\bra{0}\ \ \forall\rho,\ \ \
\mathcal{N}_{2}\left(\rho\right)=\ket{+}\\!\\!\bra{+}\ \ \forall\rho$ (243)

We now consider the performance of both channels in the game where
$\rho_{ABC}^{\left(x\right)}=\ket{000}\\!\\!\bra{000}$ for all $x$, where
combs $C_{1}$ and $C_{2}$ are taken to be trivial (identity) channels, and
where

$\displaystyle C_{3}\left(\rho_{AB}\right)$
$\displaystyle=\operatorname{Tr}[\ket{0}\\!\\!\bra{0}_{A}\rho_{AB}]\ket{00}\\!\\!\bra{00}+\operatorname{Tr}[\ket{1}\\!\\!\bra{1}_{A}\rho_{AB}]\ket{11}\\!\\!\bra{11}$
(245)

Additionally, suppose that $X$ is such that $x=1$ with certainty, or
equivalently, the player wins only if they always obtain the first measurement
outcomes. Evidently, in the above, $\mathcal{N}_{1}$ and $\mathcal{N}_{2}$ are
equivalent pure state replacement channels (up to a unitary). However,

$\displaystyle R\left(\mathcal{N}_{1}\right)$
$\displaystyle=\Big{\|}C_{3}\left(\mathcal{N}_{1}\left(\ket{0}\\!\\!\bra{0}\right)_{A}\otimes\ket{00}\\!\\!\bra{00}_{BC}\right)\Big{\|}_{\left(1\right)}$
(246)
$\displaystyle=\left\|C_{3}\left(\ket{000}\\!\\!\bra{000}_{ABC}\right)\right\|_{\left(1\right)}$
(247)
$\displaystyle=\Big{\|}\operatorname{Tr}[\ket{0}\\!\\!\bra{0}\ket{0}\\!\\!\bra{0}]\ket{00}\\!\\!\bra{00}_{AB}\otimes\ket{0}\\!\\!\bra{0}_{C}$
(248)
$\displaystyle+\operatorname{Tr}[\ket{1}\\!\\!\bra{1}\ket{0}\\!\\!\bra{0}]\ket{11}\\!\\!\bra{11}_{AB}\otimes\ket{0}\\!\\!\bra{0}_{C}\Big{\|}_{\left(1\right)}$
(249)
$\displaystyle=\Big{\|}\operatorname{Tr}[\ket{000}\\!\\!\bra{000}_{ABC}\Big{\|}_{\left(1\right)}$
(250) $\displaystyle=1$ (251)

whereas

$\displaystyle R\left(\mathcal{N}_{2}\right)$
$\displaystyle=\Big{\|}C_{3}\left(\mathcal{N}_{2}\left(\ket{0}\\!\\!\bra{0}\right)_{A}\otimes\ket{00}\\!\\!\bra{00}_{BC}\right)\Big{\|}_{\left(1\right)}$
(252)
$\displaystyle=\Big{\|}C_{3}\left(\ket{+00}\\!\\!\bra{+00}_{ABC}\right)\Big{\|}_{\left(1\right)}$
(253)
$\displaystyle=\Big{\|}\operatorname{Tr}[\ket{0}\\!\\!\bra{0}\ket{+}\\!\\!\bra{+}]\ket{00}\\!\\!\bra{00}_{AB}\otimes\ket{0}\\!\\!\bra{0}_{C}$
(254)
$\displaystyle+\operatorname{Tr}[\ket{1}\\!\\!\bra{1}\ket{+}\\!\\!\bra{+}]\ket{11}\\!\\!\bra{11}_{AB}\otimes\ket{0}\\!\\!\bra{0}_{C}\Big{\|}_{\left(1\right)}$
(255)
$\displaystyle=\left\|\operatorname{Tr}\left[\frac{1}{2}\ket{000}\\!\\!\bra{000}_{ABC}+\frac{1}{2}\ket{110}\\!\\!\bra{110}_{ABC}\right]\right\|_{\left(1\right)}$
(256) $\displaystyle=\frac{1}{2}$ (257)

Thus, there is no ordering between channels that are equivalent up to a
unitary.

We now similarly demonstrate that $C_{1}$ cannot be an arbitrary bipartite
channel. We consider the following channels and games:

$\displaystyle\mathcal{N}_{1}\left(\rho\right)$
$\displaystyle=\operatorname{Tr}\Big{[}\ket{0}\\!\\!\bra{0}\rho_{A}\Big{]}\ket{0}\\!\\!\bra{0}+\operatorname{Tr}\Big{[}\ket{1}\\!\\!\bra{1}\rho_{A}\Big{]}\ket{1}\\!\\!\bra{1}\
$ (258) $\displaystyle\mathcal{N}_{2}\left(\rho\right)$
$\displaystyle=\operatorname{Tr}\Big{[}\ket{+}\\!\\!\bra{+}\rho_{A}\Big{]}\ket{+}\\!\\!\bra{+}$
(259)
$\displaystyle+\operatorname{Tr}\Big{[}\ket{-}\\!\\!\bra{-}\rho_{A}\Big{]}\ket{-}\\!\\!\bra{-}$
(260) $\displaystyle\rho_{ABC}^{\left(x\right)}$
$\displaystyle=\ket{000}\\!\\!\bra{000}\ \ \forall x$ (261) $\displaystyle
C_{1}\left(\rho_{AB}\right)$ $\displaystyle=\ket{00}\\!\\!\bra{00}_{AB}\ \
\forall\rho_{AB}$ (262)

and where $C_{2}$ and $C_{3}$ are taken again to be the identity.
Additionally, suppose that $X=1$ so the player wins only if they always obtain
the first measurement outcomes. Evidently, in the above, $\mathcal{N}_{1}$ and
$\mathcal{N}_{2}$ are both POVMS. However,

$\displaystyle R\left(\mathcal{N}_{1}\right)$
$\displaystyle=\Big{\|}\mathcal{N}_{1}\circ
C_{1}\left(\ket{00}\\!\\!\bra{00}\right)_{AB}\otimes\ket{0}\\!\\!\bra{0}_{C}\Big{\|}_{\left(1\right)}$
(264)
$\displaystyle=\Big{\|}\mathcal{N}_{1}\left(\ket{0}\\!\\!\bra{0}_{A}\right)\otimes\ket{00}\\!\\!\bra{00}_{BC}\Big{\|}_{\left(1\right)}$
(265)
$\displaystyle=\Big{\|}\ket{000}\\!\\!\bra{000}_{ABC}\Big{\|}_{\left(1\right)}$
(266) $\displaystyle=1$ (267)

whereas

$\displaystyle R\left(\mathcal{N}_{2}\right)$
$\displaystyle=\Big{\|}\mathcal{N}_{2}\circ
C_{1}\left(\ket{00}\\!\\!\bra{00}\right)_{AB}\otimes\ket{0}\\!\\!\bra{0}_{C}\Big{\|}_{\left(1\right)}$
(268)
$\displaystyle=\Big{\|}\mathcal{N}_{2}\left(\ket{0}\\!\\!\bra{0}_{A}\right)\otimes\ket{00}\\!\\!\bra{00}_{BC}\Big{\|}_{\left(1\right)}$
(269)
$\displaystyle=\Big{\|}\frac{I}{2}_{A}\otimes\ket{00}\\!\\!\bra{00}_{BC}\Big{\|}_{\left(1\right)}$
(270) $\displaystyle=\frac{1}{2}$ (271)

Thus, there is no ordering between channels that are equivalent up to a
unitary.

From the above, evidently $C_{1}$ and $C_{3}$ need to be restricted. If we
suppose that $C_{1}$ and $C_{3}$ are restricted to be the identity (or a
unitary), then the remaining nontrivial channel is $C_{2}$. However, $C_{2}$
can be absorbed into the initial state preparation, so

$\displaystyle\rho_{ABC}^{\left(x\right)}\rightarrow
C_{2}\left(\rho_{ABC}^{\left(x\right)}\right)$ (272)

##  Appendix E Examples

###  E.1 Unitary Channel

Consider any unitary channel $\mathcal{U}_{A\rightarrow A}$ defined as
$\mathcal{U}\left(\rho\right)=U\rho U^{{\dagger}}$. Then if a game is
described by $\rho_{ABX}$ where
$\text{dim}\left(B\right)\leq\text{dim}\left(A\right)$, the reward is as
follows:

$\displaystyle R_{\rho_{ABX}}\left(\mathcal{U}\right)$
$\displaystyle=\max_{\mathcal{E}}\left(\sum_{x}p_{x}\Big{\|}U\mathcal{E}_{A}\left(\rho_{AB}^{\left(x\right)}\right)U^{{\dagger}}\Big{\|}_{x}\right)$
(273)
$\displaystyle=\max_{\mathcal{E}}\left(\sum_{x}p_{x}\Big{\|}\mathcal{E}_{A}\left(\rho_{AB}^{\left(x\right)}\right)\Big{\|}_{x}\right)$
(274)

Consider the game where $\rho_{AB}^{(x)}=\phi_{AB}^{+}$ for all $x$ and where
$X$ is drawn according to distribution $\mathbf{p}$. Then

$\displaystyle R_{\rho_{ABX}}(\mathcal{U})$
$\displaystyle=\max_{\mathcal{E}}\left(\sum_{x}p_{x}\Big{\|}\mathcal{E}_{A}\left(\phi_{AB}^{+}\right)\Big{\|}_{x}\right)$
(276)
$\displaystyle\leq\sum_{x}p_{x}\Big{\|}\mathcal{I}_{A}\left(\phi_{AB}^{+}\right)\Big{\|}_{x}$
(277) $\displaystyle=\sum_{x}p_{x}$ (278) $\displaystyle=1$ (279)

Clearly, for the game where $\rho_{AB}^{(x)}=\ket{0}\\!\\!\bra{0}$ we likewise
have

$\displaystyle R_{\rho_{ABX}}(\mathcal{U})$
$\displaystyle=\max_{\mathcal{E}}\left(\sum_{x}p_{x}\left\|\mathcal{E}_{A}\left(\ket{0}\\!\\!\bra{0}\right)\right\|_{x}\right)$
(281) $\displaystyle\leq\sum_{x}p_{x}\Big{\|}\ket{0}\\!\\!\bra{0}\Big{\|}_{x}$
(282) $\displaystyle=\sum_{x}p_{x}$ (283) $\displaystyle=1$ (284)

###  E.2 Depolarizing Channel

First, we consider the reward for $\mathcal{D}_{\gamma}$ in the game where
$\rho_{AB}^{(x)}=\phi_{AB}^{+}$ for all $x$:

$\displaystyle R_{\rho_{ABX}}($
$\displaystyle\mathcal{D}_{\gamma})=\max_{\mathcal{E}}\left(\sum_{x}p_{x}\Big{\|}\mathcal{D}_{\gamma}\circ\mathcal{E}\left(\phi_{AB}^{+}\right)\Big{\|}_{\left(x\right)}\right)$
(286)
$\displaystyle=\max_{\mathcal{E}}\left(\sum_{x}p_{x}\left\|(1-\gamma)\mathcal{E}\left(\phi_{AB}^{+}\right)+\frac{\gamma
I_{4}}{4}\right\|_{\left(x\right)}\right)$ (287)
$\displaystyle=\sum_{x}p_{x}\left\|\left(1-\gamma\right)\phi_{AB}^{+}+\frac{\gamma
I_{4}}{4}\right\|_{\left(x\right)}$ (288)
$\displaystyle=(1-\gamma)+\frac{\gamma}{4}\sum_{x=1}^{4}xp_{x}$ (289)

Likewise, if $\rho_{AB}^{(x)}=\ket{0}\\!\\!\bra{0}_{A}$ for all $x$, we find:

$\displaystyle R_{\rho_{ABX}}($
$\displaystyle\mathcal{D}_{\gamma})=\max_{\mathcal{E}}\left(\sum_{x}p_{x}\left\|\mathcal{D}_{\gamma}\circ\mathcal{E}\left(\ket{0}\\!\\!\bra{0}\right)\right\|_{\left(x\right)}\right)$
(291)
$\displaystyle=\max_{\mathcal{E}}\left(\sum_{x}p_{x}\left\|\left(1-\gamma\right)\mathcal{E}\left(\ket{0}\\!\\!\bra{0}\right)+\frac{\gamma
I_{2}}{2}\right\|_{\left(x\right)}\right)$ (292)
$\displaystyle=\sum_{x}p_{x}\left\|(1-\gamma)\ket{0}\\!\\!\bra{0}+\frac{\gamma
I_{2}}{2}\right\|_{\left(x\right)}$ (293)
$\displaystyle=(1-\gamma)+\frac{\gamma}{2}(p_{1}+2p_{2})$ (294)
$\displaystyle=(1-\gamma)+\gamma(1-\frac{p_{1}}{2})$ (295)

###  E.3 Amplitude Damping Channel

Recall that the amplitude damping channel is defined as

$\displaystyle\mathcal{A}_{\gamma}\left(\rho\right)$
$\displaystyle\triangleq\begin{pmatrix}\rho_{00}+\gamma\rho_{11}&\sqrt{1-\gamma}\rho_{01}\\\
\sqrt{1-\gamma}\rho_{10}&\left(1-\gamma\right)\rho_{11}.\end{pmatrix}$ (296)

Then for the game where $\rho_{AB}^{(x)}=\ket{0}\\!\\!\bra{0}$ for all $x$, we
have

$\displaystyle R_{\rho_{ABX}}(\mathcal{A}_{\gamma})$
$\displaystyle=\max_{\mathcal{E}}\left(\sum_{x}p_{x}\left\|\mathcal{A}_{\gamma}\circ\mathcal{E}(\ket{0}\\!\\!\bra{0})\right\|_{\left(x\right)}\right)$
(297)
$\displaystyle=\sum_{x}p_{x}\left\|\mathcal{A}_{\gamma}\left(\ket{0}\\!\\!\bra{0}\right)\right\|_{\left(x\right)}$
(298)
$\displaystyle=\sum_{x}p_{x}\left\|\ket{0}\\!\\!\bra{0}\right\|_{\left(x\right)})$
(299) $\displaystyle=1$ (300)

Likewise if $\rho_{AB}^{(x)}=\phi_{AB}^{+}$ for all $x$, we have

$\displaystyle
R_{\rho_{ABX}}(\mathcal{A}_{\gamma})=\max_{\mathcal{E}}\left(\sum_{x}p_{x}\left\|\mathcal{A}_{\gamma}\circ\mathcal{E}\left(\phi_{AB}^{+}\right)\right\|_{\left(x\right)}\right)$
(301)
$\displaystyle=\max_{\mathcal{E}}\left(\sum_{x}p_{x}\left\|\sum_{y,z=0}^{1}\mathcal{A}_{\gamma}\\!\circ\mathcal{E}\left(\ket{y}\\!\\!\bra{z}\right)\otimes\ket{y}\\!\\!\bra{z}\right\|_{\left(x\right)}\right)$
(302)
$\displaystyle=\max_{\phi}\left(\sum_{x}p_{x}\left\|\sum_{y,z\in\\{\phi,\phi^{\perp}\\}}\\!\\!\\!\mathcal{A}_{\gamma}\left(\ket{y}\\!\\!\bra{z}\right)\otimes\ket{y}\\!\\!\bra{z}\right\|_{\left(x\right)}\right)$
(303)
$\displaystyle=\sum_{x}p_{x}\left\|\sum_{y,z=0}^{1}\\!\mathcal{A}_{\gamma}(\ket{y}\\!\\!\bra{z})\otimes\ket{y}\\!\\!\bra{z}\right\|_{\left(x\right)}$
(304) $\displaystyle=p_{1}(1-\frac{\gamma}{2})+(1-p_{1})$ (305)
$\displaystyle=1-p_{1}\frac{\gamma}{2}$ (306)

###  E.4 Dephasing Channel:

The dephasing channel $\mathcal{F}_{\gamma}$ can be written as

$\displaystyle\mathcal{F}_{\gamma}\left(\rho\right)=\left(1-\gamma\right)\rho+\gamma
I_{\text{CL}}\left(\rho\right)$ (308)

where $I_{\text{CL}}$ is the classical identity channel (completely dephasing
channel.) Evidently for $\rho_{AB}^{(x)}=\ket{0}\\!\\!\bra{0}$ for all $x$ we
have

$\displaystyle R_{\rho_{ABX}}(\mathcal{F}_{\gamma})$
$\displaystyle=\max_{\mathcal{E}}\left(\sum_{x}p_{x}\left\|\mathcal{F}_{\gamma}\circ\mathcal{E}\left(\ket{0}\\!\\!\bra{0}\right)\right\|_{\left(x\right)}\right)$
(309)
$\displaystyle=\sum_{x}p_{x}\left\|\mathcal{F}_{\gamma}\left(\ket{0}\\!\\!\bra{0}\right)\right\|_{\left(x\right)}$
(310)
$\displaystyle=\sum_{x}p_{x}\Big{\|}\ket{0}\\!\\!\bra{0}\Big{\|}_{\left(x\right)}$
(311) $\displaystyle=1$ (312)

If $\rho_{AB}^{\left(x\right)}=\phi_{AB}^{+}$ for all $x$,

$\displaystyle
R_{\rho_{ABX}}(\mathcal{F}_{\gamma})=\max_{\mathcal{E}}\left(\sum_{x}p_{x}\left\|\mathcal{F}_{\gamma}\circ\mathcal{E}\left(\phi_{AB}^{+}\right)\right\|_{\left(x\right)}\right)$
(313)
$\displaystyle=\max_{\mathcal{E}}\left(\sum_{x}p_{x}\left\|(1-\gamma)\mathcal{E}(\phi_{AB}^{+})+\frac{\gamma\ket{0}\\!\\!\bra{0}\otimes
I_{2}}{2}\right\|_{\left(x\right)}\right)$ (314)
$\displaystyle=\sum_{x}p_{x}\left\|(1-\gamma)\phi_{AB}^{+}+\frac{\gamma\ket{0}\\!\\!\bra{0}\otimes
I_{2}}{2}\right\|_{\left(x\right)}$ (315)
$\displaystyle=(1-\gamma)+\frac{\gamma}{2}(p_{1}+2(1-p_{1}))$ (316)
$\displaystyle=1-\frac{\gamma}{2}p_{1}$ (317)

Results for the classical identity channel follow immediately by setting
$\gamma=1$

###  E.5 Projective measurements

Finally, we consider the case where $\mathcal{N}_{\hat{\Pi}}$ is a channel
which implements the projective qubit measurement
$\hat{\Pi}=\\{\Pi_{0},\Pi_{1}\\}$ s.t.

$\mathcal{N}_{\hat{\Pi}}\left(\rho\right)=\operatorname{Tr}\big{[}\Pi_{0}\rho\big{]}\times\Pi_{0}+\operatorname{Tr}\big{[}\Pi_{1}\rho\big{]}\times\Pi_{1}$
(319)

where $\Pi_{0}$ and $\Pi_{1}$ are both rank-one measurement elements.

We now calculate the reward for the game where
$\rho_{AB}^{(x)}=\ket{0}\\!\\!\bra{0}$ for all $x$

$\displaystyle R_{\rho}\left(\mathcal{N}_{\hat{\Pi}}\right)$
$\displaystyle=\max_{\mathcal{E}}\left(\sum_{x=1}^{2}p_{x}\left\|\sum_{i=1}^{2}\text{Tr}\left[\Pi_{i}\mathcal{E}\left(\ket{0}\\!\\!\bra{0}\right)\right]\Pi_{i}\right\|_{(x)}\right)$
(320)
$\displaystyle=\sum_{x=1}^{2}p_{x}\left\|\sum_{i=1}^{2}\text{Tr}\left[\Pi_{i}\Pi_{0}\right]\times\Pi_{i}\right\|_{(x)}$
(321) $\displaystyle=\sum_{x=1}^{2}p_{x}\left\|\Pi_{0}\right\|_{(x)}$ (322)
$\displaystyle=1$ (323)

Finally, we calculate the reward for the game where
$\rho_{AB}^{(x)}=\phi_{AB}^{+}$ for all $x$

$\displaystyle R_{\rho}\left(\mathcal{N}_{\hat{\Pi}}\right)$
$\displaystyle=\max_{\mathcal{E}}\left(\sum_{x=1}^{2}p_{x}\left\|\sum_{i=1}^{2}\text{Tr}\left[\Pi_{i}\mathcal{E}(\phi_{AB}^{+})\right]\Pi_{i}\right\|_{(x)}\right)$
(324)
$\displaystyle=\sum_{x=1}^{2}p_{x}\left\|\sum_{i=1}^{2}\text{Tr}\left[\Pi_{i}\left(\Pi_{0}\otimes\frac{I}{2}\right)\right]\Pi_{i}\right\|_{(x)}$
(325)
$\displaystyle=\sum_{x=1}^{2}p_{x}\left\|\sum_{i=1}^{2}\Pi_{0}\otimes\frac{I}{2}\right\|_{(x)}$
(326) $\displaystyle=\frac{1}{2}p_{1}+(1-p_{1})$ (327)

where in the second line we note that the optimal choice of $\mathcal{E}$ is
the replacement channel which always prepares output state $\Pi_{0}$ (or
alternatively, which always prepares output state $\Pi_{1}$).

##  Appendix F Monotonicity Proof

Suppose that $\rho^{*}$ is the state which satisfies

$\displaystyle\rho^{*}$
$\displaystyle\triangleq\underset{\rho}{\text{argmax}}\left(\operatorname{Tr}\Big{[}\mathcal{M}\left(\rho\right)^{2}\Big{]}\right)$
(328)

Consider diagonal matrices U, V. Through Holder’s inequality, we have:

$\displaystyle\operatorname{Tr}[UV]$
$\displaystyle\leq\sqrt{\operatorname{Tr}[U^{2}]}\sqrt{\operatorname{Tr}[V^{2}]}$
(329)

It follows that if $\operatorname{Tr}[UV]\geq\operatorname{Tr}[U^{2}]$, then

$\displaystyle\operatorname{Tr}[V^{2}]\geq\operatorname{Tr}[UV]\geq\operatorname{Tr}[U^{2}]$
(330)

Now consider the game where
$\rho_{AB}^{\left(x\right)}=\ket{0}\\!\\!\bra{0}_{A}$ for all $x$ (namely, the
system $B$ is trivial), and where $p_{x}$ satisfy

$\displaystyle\sum_{x=j}^{\ell}p_{x}=\alpha\lambda^{\downarrow}_{j}\left(\mathcal{M}\left(\rho^{*}\right)\right)$
(331)

where $\alpha$ is some positive normalisation constant. Then the reward for
$\mathcal{M}$ is simply

$\displaystyle R\left(\mathcal{M}\right)$
$\displaystyle=\max_{\mathcal{E}}\sum_{x=1}^{\ell}p_{x}\sum_{j=1}^{x}\lambda_{j}^{\downarrow}\left(\mathcal{M}\left(\mathcal{E}\left(\ket{0}\\!\\!\bra{0}\right)\right)\right)$
(332)
$\displaystyle=\max_{\rho}\sum_{x}p_{x}\sum_{j=1}^{x}\lambda_{j}^{\downarrow}\left(\mathcal{M}\left(\rho\right)\right)$
(333)
$\displaystyle=\max_{\rho}\sum_{j=1}^{\ell}\sum_{x=j}^{\ell}p_{x}\lambda_{j}^{\downarrow}\left(\mathcal{M}\left(\rho\right)\right)$
(334)
$\displaystyle=\alpha\max_{\rho}\sum_{j}\lambda_{j}^{\downarrow}\left(\mathcal{M}\left(\rho\right)\right)\lambda_{j}^{\downarrow}\left(\mathcal{M}\left(\rho^{*}\right)\right)$
(335)
$\displaystyle=\alpha\sum_{j}\lambda_{j}^{\downarrow}\left(\mathcal{M}\left(\rho^{*}\right)\right)\lambda_{j}^{\downarrow}\left(\mathcal{M}\left(\rho^{*}\right)\right)$
(336)
$\displaystyle=\alpha\operatorname{Tr}\left[\mathcal{M}\left(\rho^{*}\right)^{2}]\right]$
(337)

Evidently, since $\mathcal{M}\precsim\mathcal{N}$,

$\displaystyle R\left(\mathcal{N}\right)$
$\displaystyle=\alpha\max_{\rho}\sum_{j}\lambda_{j}^{\downarrow}\left(\mathcal{N}\left(\rho\right)\right)\lambda_{j}^{\downarrow}\left(\mathcal{M}\left(\rho^{*}\right)\right)$
(338)
$\displaystyle=\alpha\sum_{j}\lambda_{j}^{\downarrow}\left(\mathcal{N}\left(\sigma\right)\right)\lambda_{j}^{\downarrow}\left(\mathcal{M}\left(\rho^{*}\right)\right)$
(339)
$\displaystyle\geq\alpha\operatorname{Tr}\left[\mathcal{M}\left(\rho^{*}\right)^{2}\right]$
(340)

where in the above we denote the state which attains the optimum for
$\mathcal{N}$ as $\sigma$. Evidently, upon rewriting the above we have

$\displaystyle\operatorname{Tr}\Big{[}\text{diag}\left(\mathcal{N}\left(\sigma\right)\right)\text{diag}\left(\mathcal{M}\left(\rho^{*}\right)\right)\Big{]}\geq\operatorname{Tr}\Big{[}\mathcal{M}^{*}\left(\rho\right)\Big{]}$
(341)

This implies