# Quasi-inversion of quantum and classical channels
in finite dimensions

Fereshte Shahbeigi  Department of Physics, Sharif University of Technology,
Tehran, Iran  Center for Theoretical Physics, Polish Academy of Sciences,
02-668 Warszawa, Poland  Koorosh Sadri  Department of Physics, Sharif
University of Technology, Tehran, Iran  Morteza Moradi  Department of Physics,
Sharif University of Technology, Tehran, Iran  Karol Życzkowski  Center for
Theoretical Physics, Polish Academy of Sciences, 02-668 Warszawa, Poland
Institute of Theoretical Physics, Uniwersytet Jagielloński, 30-348 Kraków,
Poland  Vahid Karimipour  Department of Physics, Sharif University of
Technology, Tehran, Iran

(June 14, 2021)

###### Abstract

We introduce the concept of quasi-inverse of quantum and classical channels,
prove general properties of these inverses and determine them for a large
class of channels acting in an arbitrary finite dimension. Therefore we extend
the previous results of [1] to arbitrary dimensional channels and to the
classical domain. We demonstrate how application of the proposed scheme can
increase on the average the fidelity between a given random pure state and its
image transformed by the quantum channel followed by its quasi-inversion.

##  1 Introduction

It is generally understood that quantum resources make significant
improvements over the classical ones in most of the information processing
tasks [2]. However, these resources are usually fragile under the noise caused
by inevitable interactions with environment which may drastically neutralize
the quantum advantage mentioned above. Stated in other words, an open quantum
system is usually interacting with an environment, so its dynamics cannot be
described by a unitary evolution, $\rho\longrightarrow U\rho U^{\dagger}$.
This unitary dynamics is an idealization which almost never occurs in reality.
There are always inevitable and unknown couplings with the environment which
destroy the coherence, decrease the purity of the state, and deteriorate
information encoded into a quantum system [3, 4]. One of the central results
in the quantum theory is that a general non-unitary dynamics of an open
quantum system can be characterized by operators acting entirely within the
quantum system [5, 6]. The latter dynamics has long been studied and by now
there is an extensive literature on the subject. The simplest way to describe
a non-unitary dynamics is to use the Kraus form of a channel acting on a
density matrix $\rho$ of order $d$,

$\rho\longrightarrow{\cal E}(\rho)=\sum_{\alpha}K_{\alpha}\rho
K_{\alpha}^{\dagger},$ (1)

which can be interpreted as a generalization of the unitary dynamics. The
standard unitarity condition $U^{\dagger}U=I$ has been replaced above by the
identity resolution, $\sum_{\alpha=1}^{r}K_{\alpha}^{\dagger}K_{\alpha}=I$.
Here $K_{\alpha}$ denotes a Kraus operator of size $d$. The number of these
operators, $r$, may vary, but it is always possible to find representations
with $r\leq d^{2}$. Any map of this class is Completely Positive and Trace-
preserving (CTP) and is called a quantum operation or a quantum channel [5,
6]. It captures the effect of errors (noise, decoherence and dissipation) in a
quantum system caused by interaction with the environment.

It is easy to see that any unitary evolution can be explicitly inverted by
replacing $U$ with $U^{-1}$, so one can get back the original state by turning
the dynamics backward. Even if we set aside practical considerations for
reducing the effect of noise and errors, since after all there exist error
correcting codes and other methods [7, 8, 9, 10, 11, 12, 13, 14, 15, 16] for
dealing with these issues, it is a mathematical curiosity to ask, whether a
general quantum channel can be inverted too [17, 18]. It is however an
established fact that a quantum channel can be exactly inverted only if it is
a unitary transformation.

By inversion of a given quantum channel we mean here using another quantum
channel, which is physically possible. One may therefore ask, whether a
quantum channel can be quasi-inverted, in the sense that another quantum
channel ${\cal E}^{qi}$ exists, such that ${\cal E}^{qi}\circ{\cal E}$ is as
close to the identity map as possible. This was the approach taken in [1],
where the case of qubit channels was studied in detail. As qubit channels are
completely characterized and classified in [19, 20], the authors of [1]
managed to find the quasi-inverse of all qubit channels. It was shown that the
quasi-inverse of every quantum channel, except for a measure zero set, is
nothing but a suitably defined unitary map [1]. An explicit formula for
deriving this unitary map was also derived.

Slightly related issue was earlier considered by Koenig et al. [21] who
analyzed the following problem: given a bipartite quantum state $\rho_{AB}$
one wishes to convert it as closely as possible to a maximally entangled state
by applying a quantum channel only on the system $B$. Due to Jamiołkowski
isomorphism any quantum channel acting on a $d$-dimensional system can be
treated as a bipartite state $\rho_{AB}$ of size $d^{2}$ – see [22]. Thus the
maximal fidelity optimized over all possible channels can be related to the
conditional min-entropy of the state $\rho_{AB}$. A similar approach was later
used by Chiribella and Ebler [23], who demonstrated that an unknown unitary
channel can be optimally inverted with an average fidelity of $1/d^{2}$.

Extension of results obtained in [1] for single-qubit channels for higher
dimensions is not straightforward, as very little is known concerning the
structure of the set of channels acting in dimensions $d\geq 3$. In addition
to the quartic increase of the number of parameters with the dimension $d$,
which defies any kind of geometrical picture for these channels, certain
important and simplifying theorems which hold for qubit channels do not extend
to the higher dimensional case. For instance, for $d=2$ any unital channel,
which leaves the maximally mixed state invariant, ${\cal E}(I)=I$ belongs to
the class of mixed unitary channels, so it can be represented as a mixture of
unitary operations [24]. This property does not hold for general $d$ [25, 26]
and already for $d\geq 3$ there exists the unital channel of Landau and
Streater, which is not mixed unitary [24], see the recent study [27] for
further information on this map. This is also related to another important
difference which concerns the extreme points of a convex set, i.e. those
points which cannot be written as convex combination of other points of the
set. While the extreme points of the convex set of unital qubit channels are
unitary maps, this is no longer the case for higher dimensional channels.

All these properties make the study of quasi-inverse of quantum channels a
non-trivial task. Nevertheless, we obtain here some general results on higher
dimensional channels and their quasi-inverses and substantiate these results
by several examples of general families of $d-$ dimensional channels. We
provide upper and lower bounds for the performance of a channel after it is
compensated by its quasi-inverse, i.e. ${\cal E}^{qi}\circ{\cal E}$. In
particular, we show that the quasi-inverse of a channel of an arbitrary
dimension $d$ need not be a unitary map. We also study a class of self-quasi-
inverse quantum channels including the interesting case of Landau-Streater
[24].

In the second part of this work we study an analogous problem posed for
classical channels, namely for stochastic matrices which map the set of
probability vectors into itself. This question, left open in [1], is of a
special interest in view of the correspondence between quantum channels and
their classical counterparts [28]. We show that several results originally
formulated for the quantum domain find their natural parallels in the
classical scenario.

The structure of this paper is as follows: In Section 2, we collect the
preliminary ingredients, in Section 3 we define the quasi-inverse channel
whose general properties and examples are respectively explored in Sections 4
and 5. These studies are extended to the classical domain in Sections 6 and 7.
We conclude the paper with a discussion. Some of the detailed calculations and
proofs are collected in the Appendices.

##  2 Preliminaries

Let $H_{d}$ be a complex $d-$dimensional Hilbert space for which we choose the
computational basis $\\{|m\rangle,m=1,\cdots,d\\}$. Let $L(H_{d})$ be the
space of linear operators on this Hilbert space. The state of a
$d-$dimensional quantum system (a qudit) is described by a density matrix
$\rho$ which is a positive operator of unit trace acting on this Hilbert
space. The space of all density matrices is denoted by $D(H_{d})$. This is a
convex subset of $L(H_{d})$. Any point of this convex set is described by
$d^{2}-1$ real parameters. Any linear map $A\in L(H_{d})$ can be uniquely
mapped to a vector $|A\rangle\in H_{d}\otimes H_{d}$ in the form

$|A\rangle=d(A\otimes I)|\phi^{+}\rangle,$ (2)

where $|\phi^{+}\rangle=\frac{1}{\sqrt{d}}\sum_{i}|i,i\rangle$ is a maximally
entangled state. This is called vectorization of a matrix
$A=\sum_{i,j}A_{i,j}|i\rangle\langle j|$ into a vector
$|A\rangle=\sum_{i,j}A_{i,j}|i,j\rangle$, with the correspondence between the
inner products:

${\mathrm{Tr}}(A^{\dagger}B)=\langle A|B\rangle.$ (3)

Taking $d^{2}-1$ traceless and Hermitian matrices $\Gamma_{i}$ along with
identity $I$ as a complete basis, one can write any density matrix as

$\rho=\frac{1}{d}\big{(}I+\sum_{i=1}^{d^{2}-1}r_{i}\Gamma_{i}\big{)}=\frac{1}{d}\big{(}I+{\bf
r}\cdot\bm{\Gamma}\big{)}$ (4)

We normalize $\Gamma_{i}$ matrices to satisfy

${\mathrm{Tr}}\left(\Gamma_{i}\Gamma_{j}\right)=d(d-1)\delta_{ij},$ (5)

a concrete choice for this basis is given in Appendix A. We will then have

${\mathrm{Tr}}(\rho^{2})=\frac{1}{d}\big{[}1+(d-1){\bf r}\cdot{\bf r}\big{]}.$
(6)

The $(d^{2}-1)$-dimensional vector
$\bm{r}=(x_{1},x_{2},\dots,x_{d^{2}-1})^{\mathrm{T}}$ is a real vector called
the generalized Bloch vector. In contrast to the qubit case $d=2$, the convex
set of physical states (positive matrices with unit trace) is no longer a unit
ball. In fact the geometry of this convex set can be quite complicated and
only partial facts are known about low dimensions, i.e. for $d=3$ [29, 22].
The set of pure states $\rho=|\psi\rangle\langle\psi|$, where
$|\psi\rangle=\sum_{i=1}^{d}\psi_{i}|i\rangle$ is a sphere of dimension $2d-2$
which we denote by $S_{2d-2}$. On the other hand, any pure state has the
property ${\mathrm{Tr}}(\rho^{2})=1$ which in view of Eqs. (5) and (6) is
equivalent to ${\bf r}\cdot{\bf r}=1$. This is a sphere of dimension
$S_{d^{2}-2}$ which for $d>2$ has a higher dimension than the set of pure
states. Hence the set of pure states is a subset of this larger sphere. This
larger sphere contains other points which are not even states at all, see
Appendix A for concrete examples. The necessary and sufficient condition for
the Bloch vector to describe a pure quantum state can be found in [30]. The
necessary condition for the Bloch vector to produce a general mixed state is
provided in [31].

Consider now a quantum channel ${\cal E}$ represented by its Kraus operators
$K_{\alpha}$ acting on states of dimension $d$ through Eq. (1). The
correspondence (2) leads to the following representation of channels by
matrices acting on vectorized form of density matrices,

${\cal E}(\rho)=\sum_{\alpha}K_{\alpha}\rho K_{\alpha}^{\dagger}\ \ \
\longrightarrow\ \ \ |{\cal E}(\rho)\rangle=\Phi_{\cal E}|\rho\rangle$ (7)

where

$\Phi_{\cal E}=\sum_{\alpha}K_{\alpha}\otimes K_{\alpha}^{*},$ (8)

is called the superoperator of the map ${\cal E}$ [22] in which $*$ denotes
complex conjugation. Although a quantum channel has many different sets of
Kraus operators, connected by
$L_{\beta}=\sum_{\beta}U_{\alpha,\beta}K_{\beta}$, where $U$ is a unitary
matrix, it is straightforward to see that $\Phi$ is unique. It also has the
property that

$\Phi_{\cal E\circ\cal E^{\prime}}=\Phi_{\cal E}\Phi_{\cal E^{\prime}}.$

The quantum channel ${\cal E}$ induces an affine transformation on the
generalized Bloch vector ${\bf r}$

${\bf r}\longrightarrow{\bf r^{\prime}}=M{\bf r}+{\bf t}$ (9)

where

$M_{ij}=\frac{1}{d(d-1)}{\mathrm{Tr}}(\Gamma_{i}{\cal E}(\Gamma_{j}))\ \ \ \
{\rm and}\ \ \ \ t_{i}=\frac{1}{d(d-1)}{\mathrm{Tr}}(\Gamma_{i}{\cal E}(I)).$
(10)

The matrix $M$ of order $d^{2}-1$ is called the distortion matrix. Writing the
basis $\\{I,\Gamma_{i}\\}$ in vectorized form
$\\{|I\rangle,|\Gamma_{i}\rangle\\}$ which are vectors of dimension $d^{2}$,
we have the normalization condition

$\langle I|I\rangle=d,\hskip 28.45274pt\langle
I|\Gamma_{i}\rangle=\langle\Gamma_{i}|I\rangle=0,\hskip
28.45274pt\langle\Gamma_{i}|\Gamma_{j}\rangle=d(d-1)\delta_{i,j}.$ (11)

The $d^{2}$ dimensional vector $|\rho\rangle$ can be written in terms of the
normalized basis vectors $|\tilde{I}\rangle=\frac{1}{\sqrt{d}}|I\rangle$ and
$|\tilde{\Gamma}_{i}\rangle=\frac{1}{\sqrt{d(d-1)}}|\Gamma_{i}\rangle$, in the
symbolic form

$|\rho\rangle=\frac{1}{\sqrt{d}}\left(\begin{array}[]{c}1\\\ \sqrt{d-1}\ {\bf
r}\end{array}\right),$ (12)

where the first component is the coefficient of $|\tilde{I}\rangle$ and the
second component encapsulates the coefficients of $|\tilde{\Gamma}_{i}\rangle$
as a vector. The quantum channel ${\cal E}$ turns this into the vector

$|\rho^{\prime}\rangle=\frac{1}{\sqrt{d}}\left(\begin{array}[]{c}1\\\
\sqrt{d-1}\ M{\bf r}+{\bf t}\end{array}\right).$ (13)

This means that in this basis the superoperator can be written as

$\Phi_{\cal E}=\left(\begin{array}[]{cc}1&0\\\ \sqrt{d-1}\ {\bf
t}&M\end{array}\right).$ (14)

In the case of unital channels the translation vector vanishes, ${\bf t}=0$.
If the entire matrix $M$ also vanishes the corresponding map $\Phi_{*}$
represents the completely depolarizing channel, which sends any state $\rho$
into the maximally mixed state, $\mathcal{E}_{*}(\rho)=I/d$.

The above form of the map $\Phi$, which represents the evolution of the Bloch
vector ${\bf r}$, is also called its Liouville representation [32]. On the
other hand, it can be interpreted as the Fano form [33] of the bi-partite
state representing the map in the Choi-Jamiołkowski isomorphism. Such a state,
proportional to the Choi matrix,

$C_{\cal E}:=d({\cal E}\otimes
I)(|\phi^{+}\rangle\langle\phi^{+}|)=\sum_{i,j}{\cal E}(|i\rangle\langle
j|)\otimes|i\rangle\langle j|,$ (15)

forms a positive operator with ${\mathrm{Tr}}(C_{\cal E})=d$. It is related to
the superoperator of the channel through the simple relation [22]

$C_{\cal E}=\Phi_{\cal E}^{R},$ (16)

where $A^{R}$ denotes the following reshuffling of the entries of a matrix $A$
of order $d^{2}$

$A^{R}_{ij,kl}:=A_{ik,jl}.$ (17)

###  2.1 Average fidelity of a channel

The performance of a quantum channel ${\cal E}$ can be studied in different
ways. Among these we choose the input-output fidelity averaged over a uniform
distribution of pure states, $\overline{F}({\cal E})$, and the entanglement
fidelity $F_{E}({\cal E})$ [34]. These are respectively defined as follows:

$\overline{F}({\cal E})=\int_{S_{2d-2}}d\psi\ \langle\psi|{\cal
E}(|\psi\rangle\langle\psi|)|\psi\rangle,$ (18)

where $\int d\psi=1$ and $d\psi=d(U\psi)$ for any unitary, and

$F_{E}({\cal E})=\langle\phi^{+}|{\cal E}\otimes
I|(|\phi^{+}\rangle\langle\phi^{+}|)|\phi^{+}\rangle.$ (19)

The average fidelity measures how much the input and output states are similar
to each other and the entanglement fidelity measures how much a maximally
entangled state is affected if the channel ${\cal E}$ acts on one part of this
state. As we will see the two quantities are related to each other in a simple
way. Below we calculate these quantities by different methods and each method
sheds light on these quantities from a different angle.

First consider the average fidelity where it can be written as

$\displaystyle\overline{F}({\cal E})$ $\displaystyle=$
$\displaystyle\sum_{\alpha}\int_{S_{2d-2}}d\psi\
\langle\psi|K_{\alpha}|\psi\rangle\langle\psi|K_{\alpha}^{\dagger}|\psi\rangle$
(20) $\displaystyle=$ $\displaystyle\sum_{\alpha}\int_{S_{2d-2}}d\psi\
\langle\psi|K_{\alpha}|\psi\rangle\langle\psi^{*}|K_{\alpha}^{*}|\psi^{*}\rangle={\mathrm{Tr}}(L\Phi_{\cal
E}),$ (21)

where

$L:=\int_{S_{2d-2}}d\psi\
|\psi\rangle\langle\psi|\otimes|\psi^{*}\rangle\langle\psi^{*}|,$ (22)

is the isotropic state and $\Phi_{\cal E}$ is defined in Eq. (8). Since $L$ is
an isotropic operator in the sense that $[L,U\otimes U^{*}]=0\ \ \ \forall\
U$, it can be written in the form [35],

$L=\frac{1}{d(d+1)}(I+\sum_{i,j}|i,i\rangle\langle j,j|).$ (23)

Inserting (23) in (20) and using (16) and (17) one finds the following formula
for the average fidelity

$\displaystyle\overline{F}({\cal E})$ $\displaystyle=$
$\displaystyle\frac{1}{d(d+1)}\big{[}d+{\mathrm{Tr}}\Phi_{\cal E}\big{]}.$
(24)

More details on these calculations will be given in Appendix B. From (8) and
(14) we see that this can also be written as $\overline{F}({\cal
E})=\frac{1}{d(d+1)}\left[d+\sum_{\alpha}|{\mathrm{Tr}}K_{\alpha}|^{2}\right]$
and $\overline{F}({\cal E})=\frac{1}{d(d+1)}\left[d+1+{\mathrm{Tr}}M\right]$.

Entanglement fidelity is derived in a simpler way, e.g. by direct expansion of
the state $|\phi^{+}\rangle=\frac{1}{\sqrt{d}}\sum_{i}|i,i\rangle$, and hence
one finds

$F_{E}({\cal
E})=\frac{1}{d}\langle\phi^{+}|\sum_{\alpha}K_{\alpha}|i\rangle\langle
j|K_{\alpha}^{\dagger}\otimes|i\rangle\langle
j|\phi^{+}\rangle=\frac{1}{d^{2}}\sum_{\alpha}|{\mathrm{Tr}}K_{\alpha}|^{2}.$
(25)

Therefore, we find the simple relation [36]

$\overline{F}=\frac{1}{d+1}(1+d\ F_{E}({\cal E})),$ (26)

showing that quantities (18) and (19) are monotone with respect to each other.

##  3 Definition of the quasi-inverse channel

In the quantum communication context, it is usually the case that Alice (the
information source), generates a message state $\rho_{M}$ from a distribution
corresponding to her language alphabet which is known to Bob (the receiver).
Let us denote by $\mu$ this probability measure over $D(H_{d})$, the set of
all quantum states. The state is then fed into a quantum channel $\mathcal{E}$
to reach Bob. The received state is denoted by $\rho_{R}$. Assuming that the
characteristics of the channel are well known to Bob, and before performing
any quantum error correction, he may want to pass the received state through
some other channel in order to process it to a state more $similar$ to the one
Alice has sent. This latter channel clearly cannot depend on $\rho_{M}$ since
Bob does not know it. It only depends on the channel $\mathcal{E}$ and the
probability measure $\mu$. The second channel is applied to somehow invert the
action of $\mathcal{E}$, therefore, it is natural to call it the quasi-inverse
of $\mathcal{E}$ [1] and it is expected that this channel does this inversion
in the best possible way, given the constraint of being a CPT map. Therefore,
we have the formal definition of the quasi-inverse:

######  Definition 1.

The quasi-inverse of a channel $\mathcal{E}$ is defined as a CPT map ${\cal
E}^{qi}$ fulfilling the following condition [1]

$\overline{F}(\mathcal{E}^{qi}\circ\mathcal{E})\geq\overline{F}(\mathcal{E}^{\prime}\circ\mathcal{E})\quad\forall\mathcal{E}^{\prime},$
(27)

where $\overline{F}$ is the average of a proper fidelity function between the
output state and the input pure state.

######  Remark 1.

As we will see in the sequel, the quasi-inverse of a quantum channel is unique
except when the channel falls on a set of measure zero. In the simplest case
of unital qubit channels, in which every channel is unitarily equivalent to a
Pauli channel, ${\cal E}(\rho)=\sum_{i=0}^{3}p_{i}\sigma_{i}\rho\sigma_{i}$,
only the channels with two or more equal weights $p_{i}$ have non-unique
quasi-inverses.

We now prove our first theorem whose validity does not depend on the specific
form of the fidelity function, nor on the input state restricted to be pure,
rather it depends on two very general properties of the fidelity measure, as
described in the proof.

######  Theorem 1.

For any proper similarity function $F$, probability measure $\mu$ and quantum
channel $\mathcal{E}$, a quasi-inverse $\mathcal{E}^{qi}$ can be found on the
boundary of the set of allowed channels.

###### Proof.

A proper similarity measure $F$ should satisfy the following two conditions:
(a)

$F(\rho,\rho^{\prime})\leq F(\rho,\rho)\hskip
11.38109pt\forall\rho,\rho^{\prime}\in D(H_{d});$ (28)

the state most similar to a state $\rho$ is the state $\rho$ itself, and (b)
concavity:

$F(\rho,\alpha\rho_{1}+(1-\alpha)\rho_{2})\geq\alpha
F(\rho,\rho_{1})+(1-\alpha)F(\rho,\rho_{2}).$ (29)

One such measure is the well-known fidelity, but the following argument is
independent of the particular form of this measure. We proceed with a proof by
contradiction.

Let us denote the set of all possible quantum channels on $H_{d}$ by
$\mathcal{C}_{d}$. This is a convex set. Assume the quasi-inverse for the
channel ${\cal E}\equiv(M,{\bf t})$, defined by Definition 1, is a quantum
channel ${\cal E}^{\prime}$ in the interior of $\mathcal{C}_{d}$, figure (1).
Now take the inverse of the above affine map which is given by
$(M^{-1},-M^{-1}{\bf t})$. This affine map may not correspond to a legitimate
quantum channel and may be outside $\mathcal{C}_{d}$. Denote it by ${\cal
E}^{-1}$. However for small enough $\varepsilon$, the channel
$(1-\varepsilon)\mathcal{E}^{\prime}+\varepsilon{\cal E}^{-1}$ is a CPT
$\in\mathcal{C}_{d}$. We now note that this channel performs better in quasi-
inverting the channel ${\cal E}$ since using (28) and (29) we find

$\displaystyle\int{\rm d}\mu\
F\Big{(}\rho,\big{[}(1-\varepsilon)\mathcal{E}^{\prime}+\varepsilon\mathcal{E}^{-1}\big{]}\circ{\cal
E}(\rho)\Big{)}\geq$ $\displaystyle(1-\varepsilon)\int{\rm d}\mu\
F\left(\rho,\mathcal{E}^{\prime}\circ\mathcal{E}(\rho)\right)+\varepsilon\int{\rm
d}\mu\ F(\rho,\rho)=$ $\displaystyle\int{\rm d}\mu\
F(\rho,\mathcal{E}^{\prime}\circ\mathcal{E})\rho)+\varepsilon(\int{\rm d}\mu\
\big{[}F(\rho,\rho)-F\left(\rho,\mathcal{E}^{\prime}\circ\mathcal{E}(\rho)\right)\big{]})\geq$
$\displaystyle\int{\rm d}\mu\
F\left(\rho,\mathcal{E}^{\prime}\circ\mathcal{E}(\rho)\right).$

This means we have found a linear path along which the average fidelity
increases or stays constant as we go toward the boundary of $\mathcal{C}_{d}$.
Clearly, the quasi-inverse has to be in the end of such a path and hence at
the boundary of $\mathcal{C}_{d}$.

Figure 1:  Assume that the requested quasi-inverse of a given channel ${\cal
E}$ is ${\cal E}^{\prime}$, located inside the convex set $\mathcal{C}_{d}$ of
all channels. Consider the true inverse map ${\cal E}^{-1}$ (defined in the
text), which may not belong to $\mathcal{C}_{d}$. Then at the line joing
${\cal E}^{\prime}$ and ${\cal E}^{-1}$ we always find a legitimate quasi-
inverse with a higher average fidelity. Hence the true quasi-inverse ${\cal
E}^{qi}$ lies on the boundary of the set $\mathcal{C}_{d}$ of all channels.

To complete the proof, we need to consider the case of singular channels. This
corresponds to $\det M=0$ and therefore has unit co-dimension. It means for
any singular channel $\mathcal{E}$, we may come up with a sequence of
nonsingular channels $\mathcal{E}_{n}\rightarrow\mathcal{E}$. So far we have
proved that

$\forall\ \ \mathcal{E}^{\prime}\in\textrm{int}\ \mathcal{C}_{d}\hskip
8.53581pt,\hskip 8.53581pt\exists\ \
\mathcal{E}^{qi}_{n}\in\partial\mathcal{C}_{d}\hskip 5.69054pt$

such that

$\hskip 5.69054pt\langle
F\big{(}\rho,\mathcal{E}^{qi}_{n}\circ\mathcal{E}_{n}(\rho)\big{)}\rangle\geq\langle
F\big{(}\rho,\mathcal{E}^{\prime}\circ\mathcal{E}_{n}(\rho)\big{)}\rangle,$

where $\langle X\rangle$ denotes the average of $X$ over $\mu$. Using the
continuity of $F$, implied by its concavity, the proposition is proved for the
singular channels as well. Note that in this proof we have not assumed any
particular form for the fidelity measure, except the two natural properties
(28) and (29), neither we have assumed the average fidelity to be defined only
for pure states. ∎

Hereafter we assume the standard fidelity measure,
$F(\rho,\rho^{\prime})=\Big{(}{\mathrm{Tr}}\sqrt{\sqrt{\rho}\rho^{\prime}\sqrt{\rho}}\Big{)}^{2}$,
of Uhlmann and Jozsa [37, 38], and use the average input-output fidelity (18)
for the performance of the channel. In such a case the average value can be
compared with the mean fidelity between two random quantum states averaged
over the set of all mixed states with an appropriate measure [39].

Under these assumptions, the quasi-inversion is defined as the channel
maximizing $\overline{F}(\mathcal{E}^{\prime}\circ\mathcal{E})$. In view of
Eq. (24), the practical method for finding the quasi-inverse is through one of
the following maximization problems:

$\displaystyle\overline{F}(\mathcal{E}^{qi}\circ\mathcal{E})$ $\displaystyle=$
$\displaystyle\max_{\Phi^{\prime}}\frac{1}{d+1}\left(1+\frac{1}{d}\mathrm{Tr}\Phi^{\prime}\Phi\right)$
$\displaystyle=$
$\displaystyle\max_{\\{K^{\prime}_{\beta}\\}}\frac{1}{d+1}\left(1+\frac{1}{d}\sum_{\alpha,\beta}|{\mathrm{Tr}}K^{\prime}_{\beta}K_{\alpha}|^{2}\right)$
$\displaystyle=$
$\displaystyle\max_{M^{\prime}}\frac{1}{d}\left(1+\frac{1}{d+1}\mathrm{Tr}M^{\prime}M\right).$

These equations immediately imply that for a given channel, the left and right
quasi-inverses are the same, i.e.
$\overline{F}\left(\mathcal{E}^{qi}\circ\mathcal{E}\right)=\overline{F}\left(\mathcal{E}\circ\mathcal{E}^{qi}\right)$.
It is an interesting fact with practical benefit. Either Bob can apply the
quasi-inverse after receiving the state or Alice before sending the state to
Bob. Furthermore, note that for $\mathcal{E}$ and $\mathcal{E}^{\prime}$
denoted by $(M,\bm{t})$ and $(M^{\prime},\bm{t}^{\prime})$, respectively,
their concatenation, $\mathcal{E}^{\prime}\circ\mathcal{E}$, is represented by
$(M^{\prime}M,M^{\prime}\bm{t}+\bm{t}^{\prime})$. This implies that the
translation vector ${\bf t}$, which determines the non-unitality of the
channel, plays no role directly in amount of fidelity and fidelity after
correction. However, it affects the range of the allowed values of the
distortion matrix elements. This is one of the features of quasi-inverse for
qubit channels [1], which survives for higher dimensions.

It is worth mentioning that one may find a relation between the input-output
fidelity after correction and the conditional min-entropy. The latter quantity
for a bipartite state $\rho_{AB}$ is defined as [40]

$H_{\min}(B|A)_{\rho}=-\inf_{\sigma_{A}}\inf_{\lambda}\\{\lambda\in\mathbb{R}|\
\rho_{AB}\leq 2^{\lambda}(\sigma_{A}\otimes I)\\},$ (31)

where $\sigma_{A}$ is a quantum state. Indeed, it has been proven that [21]

$2^{-H_{\min}(B|A)_{\rho}}=d\max_{\mathcal{E}^{\prime}}\langle\phi^{+}|\
(\mathcal{E}^{\prime}\otimes I)[\rho_{AB}]|\phi^{+}\rangle.$ (32)

Let us assume that $\rho_{AB}$ is the Jamiołkowski state (the normalised form
of the Choi matrix (15)) assigned to a quantum channel and denoted by
$J_{\mathcal{E}}=C_{\mathcal{E}}/d$. Then we get

$\displaystyle
2^{-H_{\min}(B|A)_{J_{\mathcal{E}}}}=d\max_{\mathcal{E}^{\prime}}\langle\phi^{+}|(\mathcal{E}^{\prime}\circ\mathcal{E}\otimes
I)[|\phi^{+}\rangle\langle\phi^{+}|]\
|\phi^{+}\rangle=d\max_{\mathcal{E}^{\prime}}F_{E}(\mathcal{E}^{\prime}\circ\mathcal{E})=dF_{E}(\mathcal{E}^{qi}\circ\mathcal{E}).$
(33)

Here $F_{E}(\mathcal{E}^{\prime}\circ\mathcal{E})$ is the entanglement
fidelity (25) of the composed map $\mathcal{E}^{\prime}\circ\mathcal{E}$. So
the above equation shows that entanglement fidelity after correction (thus
input-output fidelity after correction, see (26)) are directly related to the
conditional min-entropy of the Choi matrix of the channel.

##  4 General properties of the quasi-inverse channel

In this section we elaborate on some general properties of the quasi-inverse
of quantum channels and discuss the similarities and the crucial differences
with the qubit case [1]. Thus far, we have seen in Theorem 1 that for any
proper similarity measure the quasi-inverse lies on the boundaries of the set
of quantum channels. In what follows, we show that if this similarity measure
is linear with respect to quantum channels, we can specify the quasi-inverse
in the set of extreme channels, a subset of the boundary points. Recall that a
point of a convex set $\Omega$ is called extreme if it cannot be written as a
convex combination of two other points of $\Omega$. We will use the fact that
the set $\mathcal{C}_{d}$ of quantum channels of dimension $d$ is convex [22]
for any $d$.

######  Theorem 2.

The quasi-inverse of a quantum channel can always be taken to be an extreme
channel.

###### Proof.

Assume that the quasi-inverse of a channel ${\cal E}$ is in the form

${\cal E}^{qi}=\sum_{i}\lambda_{i}{\cal E}_{i}.$ (34)

Let ${\cal E}_{m}$ be the element in the above set for which
$\overline{F}({\cal E}_{m}\circ{\cal E})\geq\overline{F}({\cal
E}_{i}\circ{\cal E})\ \ \forall\ \ i$. Then using the linearity of the quasi-
inverse, we will have

$\overline{F}({\cal E}_{m}\circ{\cal
E})\geq\overline{F}(\sum_{i}\lambda_{i}{\cal E}_{i}\circ{\cal
E})=\overline{F}({\cal E}^{qi}\circ{\cal E}),$ (35)

which means, according to Definition 1, that the quasi-inverse can always be
taken as an extreme channel. ∎

The crucial difference between the qubit case and the higher dimensional case
is that for qubit channels quasi-inversion is unital and the extreme points of
the set of one-qubit unital maps are unitary channels, while this is no longer
the case for $d>2$. Even more than that, not all extreme points of the set of
channels are known for $d>2$, not even for unital channels.

Note that the linearity of
$\overline{F}(\mathcal{E}^{\prime}\circ\mathcal{E})$ over
$\mathcal{E}^{\prime}$ implies that if $\mathcal{E}^{qi}_{1}$ and
$\mathcal{E}^{qi}_{2}$ are both quasi-inversions of $\mathcal{E}$, then any
convex combination of them, $p\mathcal{E}^{qi}_{1}+(1-p)\mathcal{E}^{qi}_{2}$,
is the inverse as well. In accordance with the above theorem we arrive at the
following result.

######  Corollary 1.

The quasi-inverse is either unique or an infinite number of them exist where
at least two of them are extreme channels.

An immediate result of Theorem 2 is that the quasi-inversion is not an
involution, i.e. $(\mathcal{E}^{qi})^{qi}\neq\mathcal{E}$ for a general
$\mathcal{E}$. It is because a quasi-inverse map is an extreme point. So even
if one takes into account the non-uniqueness of quasi-inverse, see Remark 1
and Corollary 1, there are always non-extremal maps which are not quasi-
inverse of any other maps.

######  Proposition 1.

Let $\mathcal{E}^{qi}$ denote the quasi-inverse of $\mathcal{E}$. Then
$(\mathcal{E}^{qi}\circ\mathcal{E})^{qi}$ can be taken to be the identity.

###### Proof.

The quasi-inverse $(\mathcal{E}^{qi}\circ\mathcal{E})^{qi}$ is the map which
maximizes the input-output fidelity

$\max_{\mathcal{E}^{\prime\prime}}\overline{F}\left(\mathcal{E}^{\prime\prime}\circ(\mathcal{E}^{qi}\circ\mathcal{E})\right)\leq\max_{\mathcal{E}^{\prime}}\overline{F}\left(\mathcal{E}^{\prime}\circ\mathcal{E}\right)=\overline{F}\left(\mathcal{E}^{qi}\circ\mathcal{E}\right).$
(36)

This inequality is saturated if we take $\mathcal{E}^{\prime\prime}$ equal to
the identity map. ∎

On the other hand, let $\mathfrak{C}_{\mathcal{E}}$ denote the set of quantum
channels for which $\mathcal{E}$ defines the quasi-inverse. We argue that such
a set is convex because for any
$\mathcal{E}_{1},\mathcal{E}_{2}\in\mathfrak{C}_{\mathcal{E}}$ we have
$p\mathcal{E}_{1}+(1-p)\mathcal{E}_{2}\in\mathfrak{C}_{\mathcal{E}}$, where
$0\leq p\leq 1$. In this sense, $\mathfrak{C}_{I}$ is a special convex subset
of quantum channels which are not correctable, i.e. the identity map is its
quasi-inverse. According to the above proposition, applying the quasi-
inversion we actually send a given quantum channel to this special subset
since we cannot correct the fidelity afterwards.

As an example consider the tetrahedron of Pauli channels,
$\Phi=\sum_{i}p_{i}\Phi_{i}=\sum_{i=0}^{3}p_{i}\sigma_{i}\otimes\overline{\sigma}_{i}$
where $\sigma_{0}=I$ and other $\sigma_{i}$’s are the Pauli matrices.
According to the results of [1], the subset of Pauli channels for which
$\forall i\ \ p_{0}\geq p_{i}$ belongs to $\mathfrak{C}_{I}$. This is also a
consequence of the result of Example 1 in the next section, since
$\sigma_{i}$’s satisfy orthogonality. The set of non-correctable Pauli
channels is presented in Fig 2.

Figure 2: The set of Pauli channels. The blue region shows $\mathfrak{C}_{I}$,
the convex subset of the channels whose quasi-inversion is the identity map.
$\Phi_{*}$ is the center of the tetrahedron defined by
$\frac{1}{4}\sum_{i=0}^{3}\Phi_{i}$.

Two unitarily equivalent channels are defined as $\mathcal{E}_{1}$ and
$\mathcal{E}_{2}$ such that
$\mathcal{E}_{2}=\mathcal{E}_{U}\circ\mathcal{E}_{1}\circ\mathcal{E}_{V}$
where $\mathcal{E}_{U}$ and $\mathcal{E}_{V}$ are two unitary maps. We use
$\Phi_{1}$ and $\Phi_{2}$ to denote $\mathcal{E}_{1}$ and $\mathcal{E}_{2}$.
Unitary maps, $\mathcal{E}_{U}$ and $\mathcal{E}_{V}$, are represented by
$\Phi_{U}$ and $\Phi_{V}$. Hence,

$\displaystyle{\mathrm{Tr}}\left(\Phi^{qi}_{2}\Phi_{2}\right)$
$\displaystyle=$
$\displaystyle\max_{\Phi^{\prime\prime}}{\mathrm{Tr}}\left(\Phi^{\prime\prime}\Phi_{2}\right)=\max_{\Phi^{\prime\prime}}{\mathrm{Tr}}\left(\Phi^{\prime\prime}\Phi_{U}\Phi_{1}\Phi_{V}\right)$
(37) $\displaystyle=$
$\displaystyle\max_{\Phi^{\prime\prime}}{\mathrm{Tr}}\left(\Phi_{V}\Phi^{\prime\prime}\Phi_{U}\Phi_{1}\right)=\max_{\Phi^{\prime}}{\mathrm{Tr}}\left(\Phi^{\prime}\Phi_{1}\right)={\mathrm{Tr}}\left(\Phi^{qi}_{1}\Phi_{1}\right),$

where we have used the fact that the set of all quantum channels,
$\mathcal{C}_{d}$, is invariant under unitary transformations. This relation
proves the quasi-inverse channels of $\mathcal{E}_{1}$ and $\mathcal{E}_{2}$
are also unitarily equivalents,
$\mathcal{E}^{qi}_{2}=\mathcal{E}_{V^{-1}}\circ\mathcal{E}^{qi}_{1}\circ\mathcal{E}_{U^{-1}}$,
and they can reach the same amount of fidelity after correction. This fact
expands the result of [1] related to unitarily equivalent channels of the form
$\mathcal{E}_{2}=\mathcal{E}_{U}\circ\mathcal{E}_{1}\circ\mathcal{E}_{U^{-1}}$.

Here a crucial difference between the qubit channels and higher dimensional
channels shows up. In the case of qubits, a complete characterization of qubit
channels exists and it is known that any qubit channel has the decomposition
${\cal E}={\cal E}_{U}\circ{\cal E}_{c}\circ{\cal E}_{V}$, where ${\cal
E}_{c}$ is a canonical map with diagonal distortion matrix
$M_{c}=\operatorname{diag}(\lambda_{1},\lambda_{2},\lambda_{3})$. The signed
singular values [22] of the matrix $M_{c}$ confined inside a tetrahedron
$\Delta$ whose extreme points are unitary operations
$\rho\longrightarrow\sigma_{i}\rho\sigma_{i}^{\dagger}\ \ \ (i=0,1,2,3)$.
Therefore the task of finding quasi-inverse of any qubit channel is
considerably easy compared with higher dimensional channels where such a
canonical decomposition does not exist and even if there was, with a
presumably diagonal matrix $M$, we were faced with a highly complex
characterization of the vector $\lambda$. It is known that the structure of
the convex set of higher dimensional channels is far more complex than that of
a simple tetrahedron, and in particular it is known that the extreme points of
this set are not necessarily unitary channels. A well-known counter example is
the Landau-Streater channel [24] which will be discussed in Section 5. Let us
now put general bounds on the improved average fidelity.

######  Theorem 3.

The average input-output fidelity of a channel after correction has the
following upper and lower bounds:

$\frac{f+1}{d+1}\leq\overline{F}(\mathcal{E}^{qi}\circ\mathcal{E})\leq\frac{p_{m}+1}{d+1},$
(38)

where $f=\max_{|\beta\rangle}\langle\beta|C_{\mathcal{E}}|\beta\rangle$ is the
fully entangled fraction of the Choi matrix ${C}_{\cal E}$ of the channel
${\mathcal{E}}$ and $|\beta\rangle$ denotes a maximally entangled state, while
$p_{m}$ is the maximal eigenvalue of ${C}_{\cal E}$.

Before proceeding with the proof let us mention in view of Eq. (26) and the
definition of fully entangled fraction, the lower bound in above equation is
actually an upper bound for the input-output fidelity before we correct it
with quasi-inverse map, $\overline{F}(\mathcal{E})$.

###### Proof.

To prove the upper bound, we note that

$\displaystyle{\mathrm{Tr}}(\Phi_{{\cal E}^{\prime}}\Phi_{\cal E})$
$\displaystyle=$ $\displaystyle\sum(\Phi_{{\cal
E}^{\prime}})_{ij,kl}(\Phi_{\cal E})_{kl,ij}=\sum(C_{{\cal
E}^{\prime}})_{ik,jl}(C_{\cal E})_{ki,lj}$ (39) $\displaystyle=$
$\displaystyle\sum(SC_{{\cal E}^{\prime}}S)_{ki,lj}(C_{\cal
E})_{ki,lj}={\mathrm{Tr}}(\tilde{C}_{{\cal E}^{\prime}}C_{\cal E})$
$\displaystyle\leq$ $\displaystyle p_{m}{\mathrm{Tr}}(\tilde{C}_{{\cal
E}^{\prime}})=p_{m}{\mathrm{Tr}}(C_{{\cal E}^{\prime}})=dp_{m},$

where $S=\sum_{i,j}|i,j\rangle\langle j,i|$ is the swap operator,
$\tilde{C}_{\mathcal{E}^{\prime}}=(SC_{\mathcal{E}^{\prime}}S)^{\mathrm{T}}$,
and $p_{m}$ is the largest eigenvalue of the Choi matrix of the channel ${\cal
E}$. In writing these equations we have used Eqs. (16) and (17), and the fact
that $C_{\mathcal{E}}$ and $\tilde{C}_{\mathcal{E}^{\prime}}$ are Hermitian
and positive matrices. The above inequality leads to the following upper bound
for the improved average fidelity

$\overline{F}({\cal E})\leq\frac{p_{m}+1}{d+1}.$ (40)

In order to obtain a lower bound, we note that in view of (27)

${\mathrm{Tr}}(\tilde{C}_{{\cal E}^{\prime}}C_{\cal
E})={\mathrm{Tr}}(\Phi_{{\cal E}^{\prime}}\Phi_{\cal
E})\leq{\mathrm{Tr}}(\Phi_{\cal E}^{qi}\Phi_{\cal E})\ \ \ \forall\ {C}_{{\cal
E}^{\prime}},$ (41)

and choose for $\tilde{C}_{{\cal E}^{\prime}}$ to be equal to
$d|\beta\rangle\langle\beta|$, where $|\beta\rangle$ is a maximally entangled
state. This gives the lower bound. ∎

One of the main differences between the qubit case and the higher dimensional
channels is that the singular values of $M$ in the qubit case are always less
than or equal to one. As a result, the Bloch vector cannot be stretched by
applying the distortion matrix $M$ in the qubit case, while as we will show in
an explicit example, this is not necessarily the case for higher dimensional
channels.

Figure 3: (color online) Left: In $d=2$ every Bloch vector ${\bf r}$ is
rotated or shrunk when it is acted on by the distortion matrix of a qubit
channel. Right: for $d>2$, the distortion matrix of a quantum channel can also
stretch the Bloch vector.

There are certain channels whose distortion matrix $M$ in Eq. (14) can stretch
the generalized Bloch vector r. This is due to the non-spherical shape of the
space of quantum states in higher dimensions, see Fig. 3. We will elaborate on
this point and its consequences in Appendix C.

######  Proposition 2.

The quasi-inverse of the tensor product of quantum channels is the tensor
product of the quasi-inverses, i.e. if
$\mathcal{E}=\mathcal{E}_{1}\otimes\mathcal{E}_{2}$ then
$\mathcal{E}^{qi}=\mathcal{E}_{1}^{qi}\otimes\mathcal{E}_{2}^{qi}$.

###### Proof.

Let
$\rho_{AA^{\prime}BB^{\prime}}=\rho_{AB}\otimes\rho_{A^{\prime}B^{\prime}}$,
then one can show the min-entropy (31) is additive [21], i.e.
$H_{\min}(BB^{\prime}|AA^{\prime})=H_{\min}(B|A)+H_{\min}(B^{\prime}|A^{\prime})$.
The proof of the proposition then becomes straightforward using Eq. (33) and
the fact that for the tensor product of quantum channels, the Jamiołkowski
state is in the tensor product shape. ∎

Applying this proposition we show that for $N$ copies of a given quantum
channel in a multipartite setting

$\max_{\mathcal{E}^{\prime}}\overline{F}(\mathcal{E}^{\prime}\circ\mathcal{E}^{\otimes
N})=\overline{F}\left((\mathcal{E}^{qi})^{\otimes N}\circ\mathcal{E}^{\otimes
N}\right)\leq\left(\overline{F}(\mathcal{E}^{qi}\circ\mathcal{E})\right)^{N}.$
(42)

To prove this equation we note that for any quantum channel $\mathcal{Q}$

${\mathrm{Tr}}\Phi_{{\cal Q}^{\otimes N}}=({\mathrm{Tr}}\Phi_{\cal Q})^{N}.$
(43)

Let us apply ${\mathrm{Tr}}(\Phi_{\cal Q})=x$ for simplicity of notation. In
view of Eq. (24), we prove whenever $1\leq x\leq d^{2}$ the following
inequality holds:

$\overline{F}(\mathcal{Q}^{\otimes
N})=\frac{d^{N}+x^{N}}{d^{N}(d^{N}+1)}\leq(\frac{d+x}{d(d+1)})^{N}=\left(\overline{F}(\mathcal{Q})\right)^{N}.$
(44)

To prove this relation we note that both sides are increasing functions of
$x$. When $x=1$ or $x=d^{2}$ both sides are equal. Comparing their derivatives
with respect to $x$ at $x=1$, we see the right hand side function grows faster
at $x=1$ which proves the inequality (44). To prove Eq. (42), it remains to
show that $1\leq{\mathrm{Tr}}(\Phi_{\mathcal{E}^{qi}\circ\mathcal{E}})\leq
d^{2}$ for the composition of any quantum channel and its quasi-inverse. The
later is, however, a consequence of Eq. (38) and the fact that fully entangled
fraction of the Choi matrix is greater than $1/d$.

##  5 Examples

Taking into account the upper and lower bounds given in (38) and the general
theorems of the previous section, in this section we will consider a few
classes of examples, as the optimization problem (3) cannot be solved
analytically in the general case. In the case of single-qubit systems a
complete classification of quantum channels [19, 20] leads to an explicit
description of their quasi-inverse which can be unitary [1]. For higher
dimensional channels, the quasi-inverse may not necessarily be a unitary map.
Thus identifying the quasi-inverse is related to finding the extreme points of
the set $\mathcal{C}_{d}$ of quantum channels, which remains an open problem.

######  Example 1 (Mixed unitary channels with orthogonal unitaries).

A mixed unitary channel is defined [41, 42] by

$\mathcal{E}(\rho)=\sum_{\alpha=1}^{r}q_{\alpha}V_{\alpha}\rho
V_{\alpha}^{\dagger},$ (45)

where $\\{V_{\alpha}\\}_{\alpha=1}^{r}$ is an arbitrary set of $r$ unitary
transformations. We restrict ourselves to the case where the unitaries are
orthogonal with respect to the Hilbert-Schmidt scalar product,
${\mathrm{Tr}}(V_{\alpha}^{\dagger}V_{\beta})=d\delta_{\alpha,\beta}$. In
analogy to the construction of an approximate time reversal proposed in [17],
the quasi-inverse of $\mathcal{E}$ is then the unitary channel ${\cal
V}_{m}^{\dagger}:\rho\longrightarrow V_{m}^{\dagger}\rho V_{m}$, where $V_{m}$
corresponds to the largest weight $q_{m}$ in the mixture (45). To see this
note that the Choi matrix of this channel reads

$C_{\cal E}=\sum_{\alpha}q_{\alpha}|V_{\alpha}\rangle\langle V_{\alpha}|,$
(46)

where we have used the correspondence (2). In view of the orthogonality of the
vectors $|V_{\alpha}\rangle$, this is then the spectral decomposition of the
Choi matrix $C_{\cal E}$ with eigenvalues equal to $p_{\alpha}=dq_{\alpha}$
(note that $|V_{\alpha}\rangle$ is not normalized). Let $q_{m}=p_{m}/d$ be the
largest of coefficients in (45). In view of the diagonal nature of the Choi
matrix, $p_{m}$ is the largest eigenvalue of the Choi matrix. Moreover, by
taking $|\beta\rangle=|V_{m}\rangle$, one sees that $f=p_{m}$. So the upper
and lower bounds of (38) coincide and we find the quasi-inverse is the unitary
map induced by $V_{m}^{\dagger}$, i.e. ${\cal
E}^{qi}=\mathcal{E}_{V_{m}^{\dagger}}$, with

$\overline{F}({\cal E}^{qi}\circ{\cal E})=\frac{p_{m}+1}{d+1}.$ (47)

Let us emphasize again that this result is valid only for mixture of unitary
maps corresponding to unitary matrices mutually orthogonal in sense of the
Hilbert-Schmidt scalar product. If this assumption is not satisfied the quasi-
inverse is not the inverse of one of the unitaries. As the Example 5 shows.

######  Example 2 (Uniform mixture of orthogonal conjugations).

In the following two examples, we will bring some quantum channels which are
self-inverse.

######  Theorem 4.

Let $\mathcal{E}$ be a unital channel obtained by the uniform mixture of
conjugation (not necessarily unitary ones) by matrices which are orthogonal to
each other. Such a channel is specified by

$\mathcal{E}(\rho)=\sum_{\alpha=1}^{q}X_{\alpha}\rho X^{\dagger}_{\alpha},$
(48)

where

$\sum_{\alpha=1}^{q}X_{\alpha}^{\dagger}X_{\alpha}=\sum_{\alpha=1}^{q}X_{\alpha}X_{\alpha}^{\dagger}=I_{d},\qquad{\rm
and}\qquad{\mathrm{Tr}}\left(X_{\alpha}^{\dagger}X_{\beta}\right)=\frac{d}{q}\delta_{\alpha\beta}.$
(49)

We now show that the quasi-inverse of this map is given by its dual, i.e.
$\mathcal{E}^{qi}(\rho)=\sum X_{\alpha}^{\dagger}\rho X_{\alpha}$.

###### Proof.

The Choi matrix of the channel $\mathcal{E}$ (48) is

$C_{\mathcal{E}}=\sum_{\alpha}|X_{\alpha}\rangle\langle X_{\alpha}|,$ (50)

where $|X_{\alpha}\rangle$ is based on the correspondence of Eq. (2) and it
fulfills $\langle
X_{\alpha}|X_{\beta}\rangle=\frac{d}{q}\delta_{\alpha\beta}$. Thus we find
that the above equation is indeed the spectral decomposition of the
degenerated Choi matrix with the $q$-fold degenerated largest eigenvalue equal
to $p_{m}=\frac{d}{q}$. The superoperator of this channel is given by

$\Phi_{\mathcal{E}}=\sum_{\alpha=1}^{q}X_{\alpha}\otimes X_{\alpha}^{\ast}.$
(51)

Note that since $\mathcal{E}$ is assumed to be unital,
$\Phi^{\dagger}_{\mathcal{E}}$ is also a valid quantum channel corresponding
to the dual of $\mathcal{E}$. Composing $\Phi_{\mathcal{E}}$ and
$\Phi_{\mathcal{E}}^{\dagger}$, we get

${\mathrm{Tr}}\left(\Phi^{\dagger}_{\mathcal{E}}\Phi_{\mathcal{E}}\right)=\sum_{\alpha,\beta=1}^{q}|{\mathrm{Tr}}\left(X_{\alpha}^{\dagger}X_{\beta}\right)|^{2}=\sum_{\alpha=1}^{q}(\frac{d}{q})^{2}=\frac{d^{2}}{q}=dp_{m}.$
(52)

So by such a composition the upper bound of Eq. (38) is obtained, which
completes the proof. ∎

The fidelity after correction for the channel (48) then reads
$\overline{F}(\mathcal{E}^{\dagger}\circ\mathcal{E})=\frac{d+q}{q(d+1)}$,
while for the case with ${\mathrm{Tr}}(X_{\alpha})=0$ we have
$\overline{F}(\mathcal{E})=\frac{1}{d+1}$ before applying quasi-inversion
which admits significant improvement specially in higher dimensions and when
$q$ is not large. Note that for $q=1$ the average fidelity after correction is
$1$, as it is expected. Moreover, it is obvious that if the operators
$X_{\alpha}$ are Hermitian, $\mathcal{E}$ is its own quasi-inverse. Two
explicit examples for this case are provided in what follows.

Example 2.1 (Landau-Streater (LS) channel). Consider the Landau-Streater
channel [24],

$\mathcal{E}(\rho)=\frac{1}{j(j+1)}\left(J_{1}\rho J_{1}+J_{2}\rho
J_{2}+J_{3}\rho J_{3}\right),$ (53)

where $J_{i}$ are the Hermitian generators of $SU(2)$ in its irreducible
representation in dimension $d=2j+1$ and they satisfy
${\mathrm{Tr}}(J_{m}^{\dagger}J_{n})=\frac{1}{3}j(j+1)(2j+1)\delta_{mn}$. It
is clear then that LS channel is a special case of Eq. (48) with $q=3$. It is
worth mentioning that the Landau-Streater channel (53) is an extreme point of
$\mathcal{C}_{d}$ when $d\geq 3$ [24]. However for $d=2$, the Landau-Streater
channel is not an extreme point of $\mathcal{C}_{2}$, so in view of Theorem 2
there exist several quasi-inverse channels.

Example 2.2 (Generalized Landau-Streater channel). Let $G$ be a Lie group with
dimension $n$, with Lie algebra generators $A_{\alpha}$ where $1\leq\alpha\leq
n$. Let $D_{\mu}$ be an irreducible unitary representation of the Lie algebra.
We define the generalized Landau-Streater channel as

$\mathcal{E}_{\mu}(\rho)=\frac{1}{c_{\mu}}\sum_{\alpha=1}^{n}D_{\mu}(A_{\alpha})\rho
D_{\mu}(A_{\alpha})^{\dagger},$ (54)

where $c_{\mu}$ is the value of the second Casimir operator in this
representation

$c_{\mu}I=\sum_{\alpha=1}^{n}D_{\mu}(A_{\alpha})^{2}.$ (55)

The generators can be made orthogonal so that

${\mathrm{Tr}}\left[D_{\mu}(A_{\alpha})D_{\mu}(A_{\beta})\right]=\frac{c_{\mu}d_{\mu}}{n}.$
(56)

Such a channel satisfies the assumption of Theorem 4 and is hence its own
inverse.

######  Example 3 (The transverse-depolarizing and depolarizing channel).

The transverse-depolarizing channel is defined as
$\mathcal{E}_{w}^{td}(\rho)=(1-w)\rho^{\mathrm{T}}+\frac{w}{d}{\mathrm{Tr}}(\rho)I_{d}$,
where $\frac{d}{d+1}\leq w\leq\frac{d}{d-1}$ to satisfy complete positivity.
The superoperator of this channel is given by:

$\Phi_{w}^{td}=(1-w)S+w|\phi^{+}\rangle\langle\phi^{+}|,$ (57)

and the Choi matrix is equal to

$C_{w}^{td}=(1-w)S+\frac{w}{d}I\otimes I.$ (58)

$S$ is the swap operator introduced in the proof of Theorem 3. It is a
Hermitian unitary so its eigenvalues are $\pm 1$. Thus, the largest eigenvalue
of $C_{w}^{td}$ is

$\begin{cases}p_{m}=1-w+\frac{w}{d},\quad\quad\frac{d}{d+1}\leq w\leq 1,\\\
\\\ p_{m}=w-1+\frac{w}{d},\quad\quad 1\leq w\leq\frac{d}{d-1}.\end{cases}$
(59)

Let $\mathcal{E}_{\pm}(\rho)=\frac{1}{d\pm
1}\left({\mathrm{Tr}}(\rho)I_{d}\pm\rho^{\mathrm{T}}\right)$. The channel
$\mathcal{E}_{+}$ is the transverse-depolarizing channel with
$w=\frac{d}{d+1}$ and $\mathcal{E}_{-}$, also called Werner-Holevo channel
[43], is equal to $\mathcal{E}_{w}^{td}$ for $w=\frac{d}{d-1}$. Indeed, any
transverse-depolarizing channel is a convex combination of $\mathcal{E}_{+}$
and $\mathcal{E}_{-}$. Now it is straightforward to see
$\overline{F}(\mathcal{E}_{+}\circ\mathcal{E}_{w}^{td})$ saturates the upper
bound of Eq. (38) when $\frac{d}{d+1}\leq w\leq 1$, so $\mathcal{E}_{+}$ is
the quasi-inverse on this domain. However, we find in this region the equality
$\overline{F}(\mathcal{E}_{+}\circ\mathcal{E}_{w}^{td})=\overline{F}(\mathcal{E}_{w}^{td})$
holds. This implies that for this range of $w$ the transverse-depolarizing
channel is non-correctable. On the other hand, for $1\leq w\leq\frac{d}{d-1}$,
the upper bound of Eq. (38) is achievable by
$\mathcal{E}_{-}\circ\mathcal{E}_{w}^{td}$ confirming that $\mathcal{E}_{-}$
is the quasi-inverse in this interval and it can indeed improve the fidelity
by $\Delta\overline{F}=\frac{2(w-1)}{d+1}$. The parameter space of the
transverse-depolarizing channel is shown in Fig. 4.

Figure 4: (Color Online)A rough sketch of the parameter space of the
transverse-depolarizing channel (57). The colored region defines the valid
range of the parameters for this map to be complete positive. The orange part
belongs to the set of non-correctable maps, also can be quasi-inverted by
$\mathcal{E}_{+}$, the boundary of the orange part. The blue part is the set
of correctable channels, invertible by $\mathcal{E}_{-}$, the boundary of the
blue part. Fixing $d$, any transverse-depolarizing channel is a convex
combination of $\mathcal{E}_{+}$ and $\mathcal{E}_{-}$.

We can also consider the depolarizing channel defined as
$\mathcal{E}_{q}^{d}=(1-q)\rho+\frac{q}{d}{\mathrm{Tr}}(\rho)I_{d}$, where $q$
is a probability. The superoperator and the Choi matrix are given by

$\Phi_{q}^{d}=(1-q)I\otimes I+q|\phi^{+}\rangle\langle\phi^{+}|,\quad
C_{q}^{d}=d(1-q)|\phi^{+}\rangle\langle\phi^{+}|+\frac{q}{d}I\otimes I.$ (60)

It is now obvious that the largest eigenvalue of the Choi matrix is given by
$p_{m}=d(1-q)+\frac{q}{d}$ which is actually equal to
${\mathrm{Tr}}(\Phi_{q}^{d})/d$. This implies that quasi-inverse is the
identity map. So the depolarizing channel is not correctable.

######  Example 4 (Covariant Channels).

A quantum channel ${\cal E}:L(H_{d})\longrightarrow L(H_{d})$ is called
covariant with respect to a group $G$, if the following property holds:

${\cal E}(U(g)\rho U^{\dagger}(g))=V(g){\cal E}(\rho)V^{\dagger}(g),\hskip
28.45274pt\forall\ g\in G.$ (61)

in which $U(g)$ and $V(g)$ are two not necessarily equivalent representations
of $g$. From the vectorization (2), we find

$\Phi_{{\cal E}}U(g)\otimes U^{*}(g)=V(g)\otimes V^{*}(g)\Phi_{\cal E}.$ (62)

This property has implications for the quasi-inverse. To see this we note that

$\displaystyle{\cal E}^{qi}$ $\displaystyle=$
$\displaystyle\underset{\mathcal{E}^{\prime}}{\mathrm{argmax}}\
{\mathrm{Tr}}(\Phi_{\cal E^{\prime}}\Phi_{\cal
E})=\underset{\mathcal{E}^{\prime}}{\mathrm{argmax}}\
{\mathrm{Tr}}\Big{(}\Phi_{{\cal E}^{\prime}}(V^{\dagger}(g)\otimes
V^{T}(g))\Phi_{\cal E}(U(g)\otimes U^{*}(g))\Big{)}$ (63) $\displaystyle=$
$\displaystyle\underset{\mathcal{E}^{\prime}}{\mathrm{argmax}}\
\Big{(}(U(g)\otimes U^{*}(g))\Phi_{{\cal E}^{\prime}}(V^{\dagger}(g)\otimes
V^{T}(g))\Phi_{\cal E})\Big{)},$ (64)

from which we obtain

$\Phi_{{\cal E}^{qi}}V(g)\otimes V^{*}(g)=U(g)\otimes U^{*}(g)\Phi_{{\cal
E}^{qi}}.$ (65)

Equivalently this means that

${\cal E}^{qi}(V(g)\rho V^{\dagger}(g))=U(g){\cal
E}^{qi}(\rho)U^{\dagger}(g),\hskip 28.45274pt\forall\ g\in G.$ (66)

Therefore the quasi-inverse of a covariant channel is also covariant except
that the order of the two representations of the group is reversed.

######  Example 5 (mixed unitary channels with commuting unitaries).

In contrast to the qubit case where a classification of quantum channels
facilitates the study of various aspects of them including their quasi-
inverses, for higher dimensional channels, many aspects do not easily yield an
analytical treatment. In this section we pose a simple problem whose solution,
as we will see, is quite nontrivial and yet instructive. We have seen in
Example 1, that the quasi-inverse of a mixed unitary channel of the form

${\cal E}(\rho)=\sum_{k}p_{k}U_{k}\rho U_{k}^{\dagger}$ (67)

when $U_{k}^{\prime}s$ are orthogonal to each other, is the unitary channel
${\cal E}^{qi}(\rho)=U_{max}^{\dagger}\rho U_{max}$, where $U_{max}$ is the
unitary corresponding to the maximum probability $p_{max}$ in (67). When the
unitaries are not orthogonal to each other, then we know the complete answer
only for qubit channels [1]. For higher dimensional channels we can tackle the
simplified version of the problem, if all unitary matrices commute, so that
they are diagonal in a certain basis. Under such assumptions we will prove
that for qutrit channels the quasi-inverse is a unitary map. Our analysis also
reveals certain facts about higher dimensional channels which may be of
interest in their own right. At the end of this section we will consider a
concrete case for $d=3$ and the reader can follow the general arguments here
by looking at that special case.

We aim to find the quasi-inverse of the channel in (67) when $[U_{k},U_{l}]=0\
\ \ \forall\ k$. In the basis in which all the unitaries are diagonal,
$U_{k}={\rm diag}(e^{i\theta^{(k)}_{1}},e^{i\theta^{(k)}_{2}},\cdots
e^{i\theta^{(k)}_{d}})$, the superoperator of the channel reads

$\Phi=\sum_{k=1}^{M}p_{k}U_{k}\otimes
U^{*}_{k}=\sum_{i,j}^{d}\sum_{k=1}^{M}p_{k}e^{i(\theta^{(k)}_{i}-\theta^{(k)}_{j})}|i,j\rangle\langle
i,j|=\sum_{i,j}^{d}w_{ij}|i,j\rangle\langle i,j|,$ (68)

where

$w_{ij}=\langle
e^{i\theta_{ij}}\rangle:=\sum_{k}p_{k}e^{i(\theta_{i}^{(k)}-\theta_{j}^{(k)})}.$
(69)

Let $G$ be the unitary group that consists of all diagonal unitary matrices in
this basis.111This is also known as the phase group. Since $U_{k}$’s commute
with any diagonal matrix, the channel $\mathcal{E}=\\{U_{k}\\}$ is
$G$-covariant.

$U\mathcal{E}(\rho)U^{\dagger}=\mathcal{E}(U\rho U^{\dagger})\hskip
14.22636pt\forall\rho,\ \ {\rm and}\ \ \forall\ U\in G.$ (70)

According to property (66), this implies that the quasi-inverse of this
channel is also $G$-covariant. Expressed in terms of the superoperators, this
means that the superoperator ${\Phi}_{\cal E}$ must commute with the
superoperator of all unitary maps $\rho\longrightarrow U\rho U^{\dagger}$,
where $U\in G,$. The superoperator of the latter is of the diagonal form

$\Phi_{U}=\sum_{mn}e^{i(\phi_{m}-\phi_{n})}|m,n\rangle\langle m,n|.$ (71)

Let the superoperator of the quasi-inverse be given by

$\Phi_{{\cal E}^{qi}}=\sum\Phi_{ij,kl}|ij\rangle\langle kl|.$ (72)

Equation (70) now restricts the form of this superoperator to the following
simple form

$\Phi_{{\cal E}^{qi}}=\sum_{i,j}q_{ij}|ij\rangle\langle
ij|+r_{ij}|ii\rangle\langle jj|.$ (73)

The Choi-matrix of the quasi-inverse is obtained by reshuffling the entries of
the superoperator, see (16), which amounts to

$C_{{\cal E}^{qi}}=\sum_{i,j}r_{ij}|ij\rangle\langle
ij|+q_{ij}|ii\rangle\langle jj|.$ (74)

According to
$\overline{F}(\mathcal{E}^{qi}\circ\mathcal{E})=\frac{1}{d+1}\left(1+\frac{1}{d}\mathrm{{\mathrm{Tr}}}\Phi_{{\cal
E}^{qi}}\Phi_{\cal E}\right)$, the quasi-inverse is the channel which
maximizes the following quantity

$F:={\mathrm{Tr}}(\Phi_{{\cal E}^{qi}}\Phi_{\cal
E})=\sum_{i=1}^{d}(q_{ii}+r_{ii})+\sum_{i\neq j=1}^{d}{q_{ij}w_{ij}}.$ (75)

Here we have used the equality $w_{ii}=1\ \forall\ i$, subject to the
condition that $C_{\mathcal{E}^{qi}}$ in (74) designates the Choi matrix of a
legitimate quantum channel, i.e. it is a positive matrix with partial trace
equal to the identity

$C_{\mathcal{E}^{qi}}\geq 0\ \ \ \ {\rm and}\ \ \
{\mathrm{Tr}}_{1}C_{\mathcal{E}^{qi}}=I_{d}.$ (76)

In view of the block-diagonal form of the Choi matrix, it turns out that its
eigenvalues are of the form $\\{r_{i\neq j}\\}\cup{\rm Eigenvalues\ of\ Y}$,
where $Y$ is the $d$-dimensional matrix

$Y=\sum_{i,j=1}q_{ij}|i\rangle\langle j|+\sum_{i}r_{ii}|i\rangle\langle i|.$
(77)

The second condition in (76) leads to the following set of equalities

$q_{ii}+\sum_{j}r_{ij}=1\hskip 28.45274pt\forall\ i,$ (78)

which can be rewritten as

$q_{ii}+r_{ii}+\sum_{j\neq i}r_{ij}=1\hskip 28.45274pt\forall\ i.$ (79)

Note that $r_{i\neq j}$, being the eigenvalues of the Choi matrix are non-
negative. Therefore in order to maximize the right hand side of (75), we can
take all of them to be zero, reducing (79) to

$q_{ii}+r_{ii}=1\hskip 28.45274pt\forall\ i,$ (80)

which further simplifies the expression (75) and reduces our problem to
maximization of the expression

$F=d+\sum_{i\neq j=1}^{d}{q_{ij}w_{ij}}\equiv{\mathrm{Tr}}(X^{\mathrm{T}}W),$
(81)

subject to the positivity of the following matrix

$X=I+\sum_{i\neq j}q_{ij}|i,j\rangle\langle i,j|.$ (82)

The set of all matrices $X$, denoted by $\Omega$ is a convex set. Therefore
the linear function $F$ takes its maximum at the extreme points of the set
$\Omega$.

In genera, the problem of finding the extreme points of $d-$dimensional
channels is a difficult and rather non-trivial. It is only known that the
extreme points of the set of $2$ dimensional unital channels are unitary maps.
Here we show that for channels defined by the Choi matrix (74), the extreme
points are unitary maps if $d\leq 3$. We show also a stronger result: for any
dimension $d$ the rank of any extreme point of this set is less than
$\sqrt{d}$. To this end, we first need to clarify a few definitions and a
lemma. In what follows, $\mathcal{S}$ is a vector space and
$\Omega\subset\mathcal{S}$ is a convex subset of $\mathcal{S}$.

A basic property of extreme points of a convex set $\Omega$ is depicted in
figure (5). In this figure $X_{0}$ and $X^{\prime}_{0}$ are extreme points
while $Y_{0}$ is not. We note that any line (no matter how short) passing
through an extreme point like $X_{0}$ or $X^{\prime}_{0}$ contains points
which do not belong to $\Omega$, while for the non-extreme point $Y_{0}$,
there exists a sufficiently short line (namely the one lying on the edge)
which lies entirely in $\Omega$. We present this more formally in the
following statement.

######  Definition 2.

Let $Y_{0}$ be a non-extreme point of $\Omega$. Then there is a
$Z\in\mathcal{S}$, and $\epsilon>0$, such that $Y_{0}+t(Z-Y_{0})\in\Omega$ for
all $t\in(-\epsilon,\epsilon)$. We call $Z$ a witness of non-extremality of
$Y_{0}$.

This definition implies its equivalent form.

######  Lemma 1.

Let $X_{0}$ be an extreme point of $\Omega$ and let $Z\in\mathcal{S}$ be an
arbitrary element in $\mathcal{S}$ such that for all
$t\in(-\epsilon,\epsilon)$, $X_{0}+t(Z-X_{0})\in\mathcal{S}$. Then $Z=X_{0}$.

Figure 5: $X_{0}$ and $X^{\prime}_{0}$ are extreme points of the convex set
$\Omega$. Any short segment passing through these points contains points
outside $\Omega$. On the other hand $Y_{0}$ is not an extreme point and there
always exists a short segment passing through it which lies entirely inside
$\Omega$.

Now we can state and prove the following result.

######  Theorem 5.

Let $\mathbb{V}_{m}\subset\mathbb{M}_{d}$ be an $m-$dimensional subspace of
the set of $d-$dimensional complex matrices and let
$\Omega\subset\mathbb{V}_{m}$ be the cone of semi-definite positive matrices.
Then the rank $r$ of any extreme point of $\Omega$ is bounded as
$r\leq\sqrt{d^{2}-m}$.

###### Proof.

Let $Y_{0}$ be any element $\Omega$ with rank $r$. We show that if
$r>\sqrt{d^{2}-m}$, then we can always find an element $Z\in\Omega$ different
from $Y_{0}$ such that the sufficiently short line segment $Y_{0}+t(Z-Y_{0})$
belongs entirely to $\Omega$. This shows that such points cannot be extreme
points of $\Omega$. To this end, let us expand $Y_{0}$ in its eigenbasis as

$Y_{0}=\sum_{\alpha=1}^{r}\lambda_{\alpha}|u_{\alpha}\rangle\langle
u_{\alpha}|,$ (83)

where $\lambda_{\alpha}>0\ \ \forall\alpha.$ We choose
$Z-Y_{0}\in\mathbb{M}_{d}$ in the form

$Z-Y_{0}=\sum_{\alpha,\beta}z_{\alpha\beta}|u_{\alpha}\rangle\langle
u_{\beta}|,$ (84)

and let $t$ be so small that $Y_{0}+t(Z-Y_{0})\geq 0$. The only other
requirement that is needed for this matrix to belong to $\Omega$ is to satisfy
$d^{2}-m$ linear homogeneous equations which define the subspace $\Omega$.
This is a system of $d^{2}-m$ linear homogeneous equations on $r^{2}$
variables and if $d^{2}-m<r^{2}$, it has always a non-zero solution. This
means that the point $Y_{0}$ is not an extreme point of the set $\Omega$.
Hence the rank of any extreme point of this set should be less than or equal
to $\sqrt{d^{2}-m}$. ∎

######  Corollary 2.

As a corollary we find that the extreme points of the set of matrices (82)
which is a subset of a $d^{2}-d$ dimensional space, have rank $r<\sqrt{d}$.
This means that for $d=3$, the extreme points of the set (82) have unit rank
and hence the quasi-inverse of mixed unitary channels with commuting unitaries
is a unitary channel. This theorem by itself does not preclude the existence
of quasi-inverses which are unitary maps in higher dimensions.

Consider now the case of $d=3$. Here after setting $r_{ij}=0$ and satisfying
the constraint (80), the superoperator is a diagonal matrix given by
$\Phi_{{\cal
E}^{qi}}=\operatorname{diag}[1,q_{12},q_{21},1,q_{23},q_{31},q_{32},1]$ and
its Choi matrix is as follows:

$C_{{\cal
E}^{qi}}=\left(\begin{array}[]{ccccccccc}1&.&.&.&q_{12}&.&.&.&q_{13}\\\
.&.&.&.&.&.&.&.&.\\\ .&.&.&.&.&.&.&.&.\\\ .&.&.&.&.&.&.&.&.\\\
q_{21}&.&.&.&1&.&.&.&q_{23}\\\ .&.&.&.&.&.&.&.&.\\\ .&.&.&.&.&.&.&.&.\\\
.&.&.&.&.&.&.&.&.\\\ q_{31}&.&.&.&q_{32}&.&.&.&1\end{array}\right).$ (85)

The Hermitian matrices $W$ and $X$, which appear in Eq. (81), are now $3$
dimensional and take the form

$X=\left(\begin{array}[]{ccc}1&q_{12}&q_{13}\\\ q_{21}&1&q_{23}\\\
q_{31}&q_{32}&1\end{array}\right),\qquad{\rm and}\qquad
W=\left(\begin{array}[]{ccc}1&w_{12}&w_{13}\\\ w_{21}&1&w_{23}\\\
w_{31}&w_{32}&1\end{array}\right).$ (86)

Having proved that the matrix $X_{0}$ which maximizes the expression (81) is
of unit rank, we can write it as $X=|\phi\rangle\langle\phi|$, where
$|\phi\rangle=\left(\begin{array}[]{ccc}q_{1}&q_{2}&q_{3}\end{array}\right)^{\mathrm{T}}$
maximizes ${\mathrm{Tr}}(W|\phi\rangle\langle\phi|)$, with $q_{i}$ being
unimodular complex numbers. Note that the maximum will be smaller than the
largest eigenvalue of $W$, if the corresponding eigenvectors is not built of
unimodular entries. Therefore the quasi-inverse is the unitary map ${\cal
E}^{qi}(\rho)=U\rho U^{\dagger}$, where the unitary operator $U$ is given by
$U=\operatorname{diag}\left(q_{1},q_{2},q_{3}\right)$. In view of (3) and
(81), the final average fidelity becomes

$\overline{F}=\frac{1}{4}\bigl{(}1+\frac{1}{3}{\mathrm{Tr}}(\Phi_{{\cal
E}^{qi}}\Phi_{\cal
E})\bigr{)}=\frac{1}{4}\bigl{(}1+\frac{1}{3}max_{|\phi\rangle}\langle\phi|W|\phi\rangle\bigr{)}.$
(87)

As a concrete example, consider a spin-1 particle subject to a magnetic field
in the $z$ direction, where the strength of the magnetic field or the exposure
time is random. The evolution of the state is given by the following channel

${\cal E}(\rho)=\int d\tau f(\tau)U(\tau)\rho U^{\dagger}(\tau),\hskip
28.45274ptU(\tau)=\left(\begin{array}[]{ccc}e^{i\tau}&&\\\ &1&\\\
&&e^{-i\tau}\end{array}\right),$ (88)

for some distribution $f(\tau)$. For this channel we have

$w_{12}=w_{23}=\langle e^{i\tau}\rangle,\hskip 28.45274ptw_{13}=\langle
e^{2i\tau}\rangle,$ (89)

where the matrix $W$ is Hermitian with unit diagonal entries and the averages
are taken with respect to the distribution function $f(\tau)$. The vector
$|\phi\rangle$ is then of the form
$|\phi\rangle=\left(\begin{array}[]{ccc}e^{i\tau_{m}}&1&e^{-i\tau_{m}}\end{array}\right)^{\mathrm{T}}$,
where $\tau_{m}$ is chosen to maximize $\langle\phi|W|\phi\rangle$ which is

$\langle\phi|W|\phi\rangle=Re\left[2\langle e^{i\tau}\rangle
e^{-i\tau_{m}}+\langle e^{2i\tau}\rangle e^{-2i\tau_{m}}\right].$ (90)

In view of the form of $X$, the quasi-inverse will be given by

$V=\left(\begin{array}[]{ccc}e^{i\tau_{m}}&&\\\ &1&\\\
&&e^{-i\tau_{m}}\end{array}\right),$ (91)

and the average fidelity after application of the quasi-inverse is given by

$\overline{F}=\frac{3+\big{\langle}\left(1+2\cos(\tau-\tau_{m})\right)^{2}\big{\rangle}}{12}.$
(92)

Figure 6 shows the average fidelity and the increase in average fidelity for a
simple discrete distribution $P(\tau=0)=1-p$ and $P(\tau=\tau_{0})=p$.

Figure 6: (Left) The average fidelity, $\overline{F}$, and (Right) the
increase of the average fidelity, $\Delta\overline{F}$, of the channel (88)
after correction with quasi-inverse (91).

##  6 Quasi-inversion of classical channels

There are several known parallels between probability distributions and
stochastic matrices on the one hand and their quantum counterparts, namely
density matrices and quantum channels, on the other hand. For instance,
stochastic and bi-stochastic matrices acting on probability vectors form
classical analogues of quantum channels and unital quantum channels.
Furthermore, the discrete group of permutations is the analog of the
continuous group of unitary channels. As the concept of convexity is critical
in both domains, it is instructive to analyze, how the notion of quasi-inverse
works in the classical setup.

In more concrete terms, the state of classical stochastic system of dimension
$d$ is a real vector $\mathbf{p}$ whose entries are non-negative and add up to
one. The set of all probability vectors of length $d$, forms the simplex
$\Delta_{d}$ which is a $(d-1)$-dimensional compact and convex set. For the
sake of simplicity, let us use the bra-ket notation here to mention a
probability vector. In this sense, let $|i\rangle$ for $i\in\\{1,\dots,d\\}$
denotes the pure probability vector whose all components are zero but the
$i$-th element which is equal to $1$. Therefore, a general mixed probability
vector $\mathbf{p}=(p_{1},\dots,p_{d})^{\mathrm{T}}$ can be stated as a convex
combination of $\\{|i\rangle\\}$, i.e.
$|\mathbf{p}\rangle=\sum_{i}p_{i}|i\rangle$.

A classical channel is represented by a stochastic transition matrix $T$ of
order $d$ with non-negative elements where the the sum of all elements in each
column is equal to $1$. This is the analog of trace-preserving property. The
space of stochastic matrices of order $d$ is a $(d^{2}-d)$-dimensional convex
and compact set which will be denoted by $\mathcal{S}_{d}$.

Making use of the analogy to the quantum case consider the generalized Bloch
representation (4) of a diagonal density matrix $\rho_{\rm diag}={\rm
diag}(\mathbf{p})$. Let us order the generators of $SU(d)$ matrices used in
(5) in such a way that $\Gamma_{i}$ are diagonal for $i=1,\dots,d-1$. Then any
diagonal matrix $\rho_{\rm diag}$, representing the classical state
$\mathbf{p}$, is represented in the Bloch form (4), where the Bloch vector
${\bf t}_{cl}$ has now only $d-1$ components.

Using such a Bloch representation for point $\mathbf{p}$ of the probability
simplex $\Delta_{d}$ one can represent the action of an arbitrary stochastic
matrix in the form [44],

$T=\left(\begin{array}[]{cc}1&0\\\ \sqrt{d-1}\ {\bf t}_{\rm
cl}&M_{cl}\end{array}\right).$ (93)

Note that this representation mimics the Liouville form (14) of a quantum
operation, with the only difference that the classical distortion matrix
$M_{cl}$ forms a $(d-1)$ dimensional truncation of the quantum distortion
matrix $M$ or order $d^{2}-1$, while the classical translation vector ${\bf
t}_{\rm cl}$ consists of $d-1$ components of the original translation vector
${\bf t}$ of size $d^{2}-1$. Hence in the Bloch basis the classical transition
matrix $T$ forms a block of a matrix representing a quantum operation
$\Phi_{\cal E}$, which decoheres to it [28].

Let us mention explicitly two distinguished classical maps. The first is the
permutation map ${P}_{\alpha}$ where
${P}_{\alpha}|i\rangle=|\alpha_{i}\rangle$, where $1\leq i\leq d$, while
$\alpha_{i}$ denote permutations of $\\{1,2,\cdots,d\\}$. The next one is the
flat, van der Wearden matrix, denoted by $T_{*}$, which sends all input states
to the the uniform state. It implies $(T_{*})_{ij}={1/d}$ for any $1\leq
i,j\leq d$. As a result, the summation of elements of the columns of an
assumed matrix $A$ is equal to $1$ if and only if $T_{*}A=T_{*}$. Accordingly,
if $A$ is an invertible matrix, the summation of elements on the columns of
${A}^{-1}$ is also equal to $1$.

Finally we note the fidelity of two probability vectors $|\mathbf{p}\rangle$
and $|\mathbf{q}\rangle$ which is defined as

$\langle\mathbf{p}|\mathbf{q}\rangle:=(\sum_{i=1}^{d}\sqrt{p_{i}q_{i}})^{2}.$
(94)

Proceeding in the same way that we did for the quantum case, here we should
define the average fidelity of a classical channel to be the fidelity of an
output state with the input pure state averaged over all input states. Thus we
define

$\mathcal{\overline{F}}(T):=\frac{1}{d}\sum_{i=1}^{d}F(|i\rangle,T|i\rangle)=\frac{1}{d}{\mathrm{Tr}}(T).$
(95)

Under these assumptions, the quasi-inversion is the channel (the stochastic
matrix) increasing Eq. (95) as much as possible:

${\mathrm{Tr}}(T^{qi}T)\;\geq\;{\mathrm{Tr}}(T^{\prime}T),\hskip
11.38109pt\forall T^{\prime}\in\mathcal{S}_{d}.$ (96)

To emphasize even further similarity to the quantum case, consider the Bloch
representation (93) of the classical map $T$ involving its distortion matrix
$M_{\rm cl}$. Then the average fidelity of the corrected classical
transformation $T^{qi}T$ can be expressed as the maximum over the set of
allowed classical distortion matrices,

$\overline{\mathcal{F}}(T^{qi}T)=\max_{M^{\prime}_{\rm
cl}}\frac{1}{d}\left(1+\mathrm{Tr}M^{\prime}_{\rm cl}M_{\rm cl}\right),$ (97)

which is in analogy to Eq. (3). However, the difference in prefactor with
respect to Eq. (3) is due to the averaging over the set of classical
probabilities of a dimension smaller than the set of quantum states.

All the arguments of Theorem 1, based on the linearity of the fidelity
function and its two properties (28) and (29) are also valid here and
therefore Theorems 1 and 2 and the corollary 1 hold true also if we replace
the quantum channels with classical ones. In particular, the fact that the
quasi-inverse of a quantum channel can be taken to correspond to an extreme
point is very important, since compared with the quantum case, we have a much
better knowledge of the convex set of classical channels and its extreme
points. In the general case, our basic theorem is the following:

######  Theorem 6.

Let the stochastic matrix $T$ be such that in each row $i$, the element in the
$a_{i}$-th column be the maximum. Then its quasi-inverse is found by replacing
that single element by $1$ and setting all the other elements in that row
equal to zero and then transposing the matrix.

###### Proof.

Let the matrix $T$ be written as $T=\left({\bf t}_{1},{\bf t}_{2},\cdots,{\bf
t}_{n}\right)^{\mathrm{T}}$ where ${\bf t}_{i}^{\mathrm{T}}$ denotes the
$i-$th row as a vector. Denote its quasi-inverse as a matrix $T^{qi}=({\bf
x}_{1},{\bf x}_{2},\cdots,{\bf x}_{n})$, where ${\bf x}_{i}$ is the $i-th$
column as a vector. The aim is to maximize
${\mathrm{Tr}}(TT^{qi})=\sum_{i}{\bf t}_{i}\cdot{\bf x}_{i}$. We can maximize
this sum if we maximize each inner product ${\bf t}_{i}\cdot{\bf x}_{i}$
independently. Since none of the components of the vectors ${\bf x_{i}}$ can
be larger than one, the maximization is achieved if we choose each ${\bf
x}_{i}=(0,0,\cdots 1,0,0)$ such that $1$ stands in the position of maximum
component of ${\bf t}_{i}$. This proves the theorem. ∎

For example assume a general two dimensional stochastic matrix parameterized
as

$T=\left(\begin{array}[]{cc}1-x&y\\\ x&1-y\end{array}\right),\hskip
28.45274pt0\leq x,y\leq 1.$ (98)

In this case, the quasi-inverse is either $I$, $\sigma_{x}$, or a mixture of
these two for $x+y<1$, $x+y>1$ and $x+y=1$, respectively. This is depicted in
Fig. 7.

Figure 7: The set of all stochastic matrices of dimension $2$. Transitions
located in the gray region cannot be corrected in sense of input-output
fidelity. For other stochastic maps one can apply $\sigma_{x}$ to modify the
fidelity.

In higher dimensions, however, more options are possible.

######  Example 6.

As examples, the quasi-inverse of the stochastic matrices

$T_{1}=\frac{1}{8}\left(\begin{array}[]{ccc}2&6&2\\\ 4&1&1\\\
2&1&5\end{array}\right),\hskip 28.45274pt{\rm and}\hskip
28.45274ptT_{2}=\frac{1}{48}\left(\begin{array}[]{ccc}12&6&3\\\ 4&36&42\\\
32&6&3\end{array}\right)$ (99)

are given by

$T_{1}^{qi}=\left(\begin{array}[]{ccc}0&1&0\\\ 1&0&0\\\
0&0&1\end{array}\right),\hskip 28.45274pt{\rm and}\hskip
28.45274ptT_{2}^{qi}=\left(\begin{array}[]{ccc}1&0&1\\\ 0&0&0\\\
0&1&0\end{array}\right).$ (100)

The second example shows that the quasi-inverse of a classical map is not
necessarily a permutation. In the examples above, the average fidelities of
the stochastic maps $T_{1}$ and $T_{2}$ increase from ${1}/{3}$ and ${5}/{18}$
to ${35}/{72}$ and ${43}/{72}$, respectively.

######  Corollary 3.

If $T$ denotes a stochastic matrix for which $\hskip 5.69054pt\forall i,j:\
T_{ii}\geq T_{ij}$, then its quasi-inverse is $I_{d}$ which means it is not
possible to increase the average fidelity.

As it is seen, these quasi-inverses are at extreme points of the space of
stochastic matrices. To have quasi-inverses which are not necessarily at the
extreme points, we should consider the case where there are more than one
maximum entry in each row. In this case if we follow the argument leading to
Theorem 6, we see the quasi-inverse can be the convex combination of all
quasi-inverses which we construct when we consider only one of these elements.
The next example illustrates this point.

######  Example 7.

The stochastic matrix

$T=\frac{1}{24}\left(\begin{array}[]{ccc}8&3&8\\\ 12&6&6\\\
4&15&10\end{array}\right)$ (101)

has as its quasi-inverse

$T^{qi}=\lambda\left(\begin{array}[]{ccc}1&1&0\\\ 0&0&1\\\
0&0&0\end{array}\right)+(1-\lambda)\left(\begin{array}[]{ccc}0&1&0\\\ 0&0&1\\\
1&0&0\end{array}\right)=\left(\begin{array}[]{ccc}\lambda&1&0\\\ 0&0&1\\\
1-\lambda&0&0\end{array}\right).$ (102)

Here the average fidelity increases from the value $\frac{1}{3}$ to
$\frac{35}{72}$ which is expectedly independent from $\lambda$.

######  Theorem 7.

Among the stochastic matrices with a unique quasi-inverse, only symmetric
permutations are their own quasi-inverse.

###### Proof.

Uniqueness of the quasi-inverse suggests that in each row of $T$ there exists
a single entry which is strictly larger than other elements in the row. Hence,
we have exactly one non-zero array (which is equal to $1$) in each column of
$T^{qi}$. Suppose that the classical channel $T$ is a self-quasi-inverse
stochastic matrix, i.e. $T^{qi}=T$. This equality implies the existence of
exactly one non-zero element equal to unity in each row of $T^{qi}$. Because
we would have had more than one leading value in a row of $T$ which violates
the uniqueness of $T^{qi}$, otherwise. So $T^{qi}$ and consequently $T$ are
the same permutation matrix. However, a permutation matrix has a real inverse
which is equal to its transpose. So we have $T^{qi}=T=T^{\mathrm{T}}$
suggesting that $T$ is a symmetric permutation matrix. ∎

###  6.1 On the commutativity of super-decoherence and quasi-inverse.

Now that we have discussed the quasi-inverses of both quantum and classical
channels, a natural question is whether or not through super-decoherence [28]
of quasi-inverse of a quantum channel $\mathcal{E}$, the quasi-inverse of a
classical channel $T_{\mathcal{E}}$ can be obtained. In other words, we want
to see if the action of taking quasi-inverse commutes with super-decoherence.
As we will see below, in general the answer is negative. For convenience we
first remind the concept of super-decoherence [28].

The decoherence channel removes off-diagonal elements of density matrices and
sends any quantum state $\rho$ into its diagonal
$\mathcal{D}(\rho)=\rho_{d}=\sum\rho_{ii}|i\rangle\langle i|$, i.e. a
classical state. Defining an analogous process in the space of quantum
channels, one may extract a classical map, a stochastic matrix, from any
quantum channel. This process is called super-decoherence, noted by $\Delta$,
to emphasize that it acts on quantum channels and not states. For a quantum
channel $\mathcal{E}$, the assigned classical transition matrix obtained by
super-decoherence is defined by $T_{\mathcal{E}}=\Delta(\mathcal{E})$ where

$(T_{\mathcal{E}})_{ij}=\langle i|\mathcal{E}(|j\rangle\langle
j|)|i\rangle=(\Phi_{\mathcal{E}})_{ii,jj}=(C_{\mathcal{E}})_{ij,ij}.$ (103)

The last equality above shows that this stochastic matrix is actually gained
by decohering the Choi matrix and clarifies why it is called super-
decoherence. It is straightforward to see stochasticity of $T_{\mathcal{E}}$
is guaranteed by the fact that $\mathcal{E}$ is a positive and trace
preserving map. This is related to the fact that the classical distortion
matrix $M_{cl}$ of size $(d-1)$ used in (93) forms a block of the quantum
distortion matrix $M$ of size $d^{2}-1$ present in the Liouville form (14) of
any corresponding quantum operation [44].

Moreover, one can show that if the channel $\mathcal{E}$ is described by the
set of Kraus operators $\\{K_{\alpha}\\}$, then $T_{\mathcal{E}}=\sum
K_{\alpha}\odot K_{\alpha}^{*}$ where $\odot$ defines Hadamard (entry-wise)
product. Through this relation it is easy to see by super-decohering a unital
channel we get a bistochastic matrix, while a unistochastic matrix is obtained
if the input channel is unitary.

However, a simple counter example shows that the answer to the question posed
at the beginning of this subsection is negative. Consider the case of a
single-qubit channel, where the quasi-inverse of a channel is in general a
unitary map, $U=e^{i{\theta\bf n}\cdot{\bm{\sigma}}}$, where $\theta$ and
${\bf n}$ depend on the channel. On the other hand, as shown in Fig. 7, the
quasi-inverse of any 2-dimensional classical map is either identity $I$ or the
permutation $\sigma_{x}$, which arise due to super-decoherence of a small
subset of the unitary channels [45].

##  7 Further results and examples

Although Theorem 6 gives a complete prescription for finding the quasi-inverse
of any stochastic matrix, it is instructive to consider a few special classes.
These examples illustrate further the parallels between classical and quantum
notions of maps and their quasi-inverses. The first result is the analog of
unitarily equivalent quantum maps.

Consider two transition matrices $T_{1}$ and $T_{2}$ which are related by two
arbitrary permutations $P_{\alpha}$ and $P_{\beta}$ in the following way
$T_{2}=P_{\alpha}\circ T_{1}\circ P_{\beta}$. Let us call them permutationaly
equivalent stochastic matrices. Pursuing the same approach adopted in
obtaining Eq. (37), and noting that the set of stochastic matrices,
$\mathcal{S}_{d}$, is invariant under permutations, we get

$T_{2}^{qi}=P^{-1}_{\beta}\circ T^{qi}_{1}\circ P^{-1}_{\alpha}.$ (104)

Moreover, $T_{1}$ and $T_{2}$ have the same amount of fidelity after
correction. The next example illustrates the connection with its quantum
analog, Example 1.

######  Example 8 (Convex combination of orthogonal permutations).

A bi-stochastic matrix $T$ is a stochastic matrix with the extra property that
the sum of entries of each row equals unity. It is a well-known result due to
Birkhoff’s theorem that any such matrix can be written as a convex combination
of permutations. For dimension $d$, there are $d!$ such permutations which
form the extreme points of the convex set of these matrices, conventionally
called the Birkhoff polytope. Note however that a bi-stochastic matrix has
$(d-1)^{2}$ independent parameters and hence the convex decomposition of an
arbitrary bi-stochastic matrix in terms of $d!$ permutations is not unique.
Consider now a special class of bi-stochastic matrices which are convex
combination of orthogonal permutations. These permutations are defined by the
property that

${\mathrm{Tr}}(P^{-1}_{m}P_{n})=d\delta_{n,m}.$ (105)

Equation (105) indicates that

$\forall\ \ k\hskip 28.45274ptP_{m}|k\rangle\neq P_{n}|k\rangle,\ \ \ if\ \ \
m\neq n.$ (106)

This means that there are at most $d$ orthogonal permutations in the group
$S_{d}$ of all permutations of $d$ objects. Furthermore, it implies that

$\langle j|\sum_{m}P_{m}|k\rangle=0,1\hskip 28.45274pt\forall\ \ j,\ k.$ (107)

Here the sum runs over all the permutations in the convex combination, which
may be a subset of all the $d$ orthogonal permutations. A simple example
consists of the set of permutations of the form $\\{P_{m}=X^{m},\ \ m=0,\cdots
d-1\\}$, where $X=\sum_{n=0}^{d-1}|n+1\rangle\langle n|$ is the full-cycle
permutation, or any subset thereof. By definition we have
$P_{m}|k\rangle=|k+m\rangle$ which clearly satisfies (107) as $\sum_{m}\langle
j|P_{m}|k\rangle=\sum_{m}\delta_{j,k+m}=0,1.$ In this example all the
permutations commute with each other. As an example consisting of non-
commuting but orthogonal permutations consider the following:

$P_{1}=\left(\begin{array}[]{ccc}1&2&3\\\ 2&1&3\end{array}\right),\hskip
28.45274ptP_{2}=\left(\begin{array}[]{ccc}1&2&3\\\
3&2&1\end{array}\right),\hskip
28.45274ptP_{3}=\left(\begin{array}[]{ccc}1&2&3\\\ 3&1&2\end{array}\right),$
(108)

with matrix representations

$P_{1}=\left(\begin{array}[]{ccc}0&1&0\\\ 1&0&0\\\
0&0&1\end{array}\right),\hskip
28.45274ptP_{2}=\left(\begin{array}[]{ccc}0&0&1\\\ 0&1&0\\\
1&0&0\end{array}\right),\hskip
28.45274ptP_{3}=\left(\begin{array}[]{ccc}0&1&0\\\ 0&0&1\\\
1&0&0\end{array}\right).$ (109)

One can see that $P_{1}$ and $P_{2}$ are orthogonal while $P_{1}$ and $P_{3}$
are not. Also one can see that

$P_{1}+P_{2}=\left(\begin{array}[]{ccc}0&1&1\\\ 1&1&0\\\
1&0&1\end{array}\right),$ (110)

satisfying the condition (107) while

$P_{1}+P_{3}=\left(\begin{array}[]{ccc}0&2&0\\\ 1&0&1\\\
1&0&1\end{array}\right),$ (111)

violates (107). Furthermore one can check that in the group $S_{3}$ with
generators $\sigma_{1}=\left(\begin{array}[]{ccc}1&2&3\\\
2&1&3\end{array}\right)$ and $\sigma_{2}=\left(\begin{array}[]{ccc}1&2&3\\\
1&3&2\end{array}\right)$, each of the two sets of even and odd permutations,
respectively given by $(I,\ \sigma_{1}\sigma_{2},\ \sigma_{2}\sigma_{1})$ and
$(\sigma_{1},\ \sigma_{2},\ \sigma_{1}\sigma_{2}\sigma_{1})$ comprise
orthogonal permutations.

Consider now a bi-stochastic matrix of the form

$B=\sum\lambda_{m}P_{m},$ (112)

where $\\{P_{m}\\}$ are orthogonal. In what follows we will show the quasi-
inverse of any such bi-stochastic matrix is $P^{-1}_{m_{0}}$ where $m_{0}$
refers to the index of the greatest coefficient $\lambda_{m}$. To see this we
note that

$\displaystyle\mathcal{\overline{F}}(T\circ B)$ $\displaystyle=$
$\displaystyle\frac{{\mathrm{Tr}}(TB)}{d}=\frac{\sum_{m}\lambda_{m}{\mathrm{Tr}}(TP_{m})}{d}$
(113) $\displaystyle\leq$
$\displaystyle\frac{\lambda_{m_{0}}{\mathrm{Tr}}(T\sum_{m}P_{m})}{d}=\frac{\lambda_{m_{0}}\sum_{k,j}\langle
k|T|j\rangle\langle j|\sum_{m}P_{m}|k\rangle}{d}.$ (114)

Using Eq. (107) for orthogonal permutations, we find

$\mathcal{\overline{F}}(T\circ B)\leq\frac{\lambda_{m_{0}}\sum_{k,j}\langle
k|T|j\rangle}{d}=\lambda_{m_{0}}.$ (115)

where we used the fact that for every stochastic matrix, the sum of its
elements is equal to the dimension $d$. Now, it is straightforward to see
$T^{qi}=P^{-1}_{m_{0}}$ recovers the average fidelity of $T$ to this upper
bound.

Thus the quasi-inverse of convex combination of orthogonal permutations is the
inverse of the single permutation which has the largest share in the convex
combination. This is however not the case if the permutations are not
orthogonal. This is shown in the next example.

######  Example 9 (Convex combination of non-orthogonal permutations).

Consider the following permutations

$P_{1}=I\oplus I\oplus I,\ \ \ \ P_{2}=\sigma_{x}\oplus\sigma_{x}\oplus I,\ \
\ \ P_{3}=\sigma_{x}\oplus I\oplus\sigma_{x},$ (116)

where $I=\left(\begin{array}[]{cc}1&0\\\ 0&1\end{array}\right)$ and
$\sigma_{x}=\left(\begin{array}[]{cc}0&1\\\ 1&0\end{array}\right)$. These
permutations are obviously non-orthogonal in the sense that
${\mathrm{Tr}}(P_{i}P_{j}^{\mathrm{T}})\neq 0.$ Consider now the following
convex combination

$P=\lambda_{1}P_{1}+\lambda_{2}P_{2}+\lambda_{3}P_{3},$ (117)

where $\lambda_{1}+\lambda_{2}+\lambda_{3}=1$ and all
$\lambda_{i}<\frac{1}{2}.$ In explicit form this permutation matrix is given
by

$P=\left(\begin{array}[]{cccccc}\lambda_{1}&\lambda_{2}+\lambda_{3}&.&.&.&.\\\
\lambda_{2}+\lambda_{3}&\lambda_{1}&.&.&.&.\\\
.&.&\lambda_{1}+\lambda_{3}&\lambda_{2}&.&.\\\
.&.&\lambda_{2}&\lambda_{1}+\lambda_{3}&.&.\\\
.&.&.&.&\lambda_{1}+\lambda_{2}&\lambda_{3}\\\
.&.&.&.&\lambda_{3}&\lambda_{1}+\lambda_{2}\end{array}\right).$ (118)

Since $\lambda_{i}<\frac{1}{2}$ for all $i$, we find
$\lambda_{i}+\lambda_{j}>\frac{1}{2}$ for all $(i\neq j)$ (due to the
requirement that $\lambda_{1}+\lambda_{2}+\lambda_{3}=1$). Thus according to
Theorem 6, the quasi-inverse of $P$ is given by replacing the largest entry of
each row with unity and then replacing the resulting matrix, hence
$P^{qi}=\sigma_{x}\oplus I\oplus I$ which is not equal to inverse of any of
the permutations $P_{1}$, $P_{2}$ or $P_{3}$.

##  8 Concluding remarks

We have extended the concept of quasi-inversion of qubit channels [1] to
quantum channels in arbitrary dimensions and to the classical domain, i.e.
Markov processes in discrete time. In both cases, a quasi-inverse is a map
which when combined with the original channel increases the average input-
output fidelity in an optimal way. While the complete classification of qubit
channels [19, 20], makes a complete characterization of their quasi-inverse
possible, in higher dimensions the lack of such classification makes the
problem a highly non-trivial one. The most notable difference is that in the
qubit case, the extreme points of all unital channels are the unitary maps
while in higher dimensions this is not the case any more and no general
theorem is known on extreme points. Therefore in this paper, we have
established certain general theorems on the nature of the quasi-inverse, and
have provided certain bounds on the average fidelity after quasi-inversion.
Applying these general results to some concrete cases, we have found in
examples 1 through 5 the quasi-inverse for a large class of quantum channels.
Moreover, we have done a parallel analysis for the classical channels,
represented by stochastic matrices.

As shown in the Appendix D, we have also obtained exact expressions for the
average input-output fidelity for classical channels in any dimension $d$ and
have shown that the quasi-inversion increases this quantity from $d^{-1}$ to,
approximately, $d^{-1/2}$. In the quantum case, we analyzed numerically in
Appendix E the improvement of the average fidelity of a random channel after
correction with quasi-inversion and after applications of the best possible
unitary evolution – see Fig. 10. If the dimension of the systems increases,
the non-unitarity of quasi-inversion becomes larger. In other words, for a
generic channel acting in higher dimensions, its quasi-inverse is usually non-
unitary.

It is noteworthy to mention that by applying quasi-inversion, we aim to get as
close as possible to the identity map in the sense of the maximizing average
fidelity and so by the reduction of the average Bures distance. We do not
expect the notion of quasi-inversion to improve some other properties of
channels, for which the identity map is not the most distinguished channel.
There are other figures of merit, like the average output purity of a channel

$\overline{P}(\mathcal{E}):=\int{\rm d}\psi\
{\mathrm{Tr}}\left[(\mathcal{E}(|\psi\rangle\langle\psi|)\right)^{2}],$ (119)

or the unitarity of a channel [46]

$\overline{u}(\mathcal{E}):=\frac{d}{d-1}\int{\rm d}\psi\
{\mathrm{Tr}}\left[\mathcal{E}(|\psi\rangle\langle\psi|)-\mathcal{E}(\frac{I}{d})\right]^{2},$
(120)

which can also be studied in the same way, leading to a different version of
the quasi-inverse, both in the quantum and the classical domain. We are aware
of examples which show that these notions of quasi-inverse are different. That
is, a quasi-inverse which increases the average fidelity can increase,
decrease or keep constant the unitarity of a quantum channel. One can also
study the effect of quasi-inversion on the cohering power of quantum channels
as defined in [47, 48, 49].

Finally, an interesting by-product of our study is Appendix C, where we have
constructed special types of channels with affine maps $(M,\mathbf{t})$, where
the distortion matrix $M$ can stretch the Bloch vectors. Such channels are
possible only in dimensions higher than two. These are of course different
from channels of the form ${\cal E}(\rho)=|\psi\rangle\langle\psi|$, where the
distortion matrix $M$ vanishes and the stretching is due only to the
translation vector $\mathbf{t}$.

We hope that this study can be pursued in different directions, i.e. in
obtaining more information about the extreme points of the space of quantum
channels, and hence the quasi-inverse of larger classes of channels, those
channels which do not have unique quasi-inverses, and those which are their
own quasi-inverse. And also more importantly in finding connections with the
recovery maps [50, 51, 52, 53].

It is a pleasure to thank Seyed Javad Akhtarshenas, Erik Aurell, Giulio
Chiribella, Sergey Filippov, Kamil Korzekwa, and Łukasz Pawela for several
discussions and helpful remarks. This research was partially supported by the
grant number G98024071 from Iran National Science Foundation. Financial
support by Narodowe Centrum Nauki under the grant number
DEC-2015/18/A/ST2/00274 and by the Foundation for Polish Science under the
Team-Net NTQC project is gratefully acknowledged.

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##  Appendix A An explicit basis for $L(H_{d})$

In this appendix we choose an explicit representation for the matrices
$\Gamma_{i}$ belonging to $L(H_{d})$. Let $E_{ij}:=|i\rangle\langle j|$ be the
standard basis of matrices, then the set

$H_{k}=\sqrt{\frac{d(d-1)}{k(k+1)}}(\sum_{i=1}^{k}E_{ii}-kE_{k+1,k+1})\hskip
28.45274ptk=1,\cdots d-1$ (121)

and

$X_{ij}=\sqrt{\frac{d(d-1)}{2}}(E_{ij}+E_{ji})\hskip
28.45274ptY_{ij}=\sqrt{\frac{d(d-1)}{2}}(-iE_{ij}+iE_{ji})$ (122)

form a basis of Hermitian matrices with the proper normalization in (5). They
are nothing but the standard Gell-Mann matrices, properly normalized to
satisfy (5). To work in parallel with the Bloch representation of any
classical probability vector $\bf p$ it is convenient to order the matrix
basis in such a way that the $d-1$ diagonal matrices appear first,
$\Gamma_{k}=H_{k}$, for $k=1,\dots,d-1$ – see Section 6.

A density matrix can be written as $\rho=\frac{1}{d}(I+{\bf
r}\cdot{\bm{\Gamma}})$, where ${\mathbf{r}}$ is the generalized Bloch vector
and ${\bm{\Gamma}}$ is a vector constructed by the elements $\Gamma_{i}$. We
note, as stated after Eq. (6), that any pure state corresponds to unit vector
${\bf r}$. However the converse is not true. In fact one can easily verify
that a matrix like $\frac{1}{d}(I-H_{d-1})=|d\rangle\langle d|$ is a pure
state while a state like $\frac{1}{d}(I+H_{d-1})$ is not a state at all, since
it has negative eigenvalues. Both correspond to antipodal points on the sphere
$S_{d^{2}-2}$.

##  Appendix B Some details on the calculations of average input-output
fidelity

As shown in Eq. (20), the input-output fidelity is given by
$\overline{F}({\cal E})={\mathrm{Tr}}(L\Phi_{\cal E})$, where $L$ is given by

$\displaystyle L$ $\displaystyle=$ $\displaystyle\int{\rm d}\psi\
|\psi\rangle\langle\psi|\otimes|\psi^{\ast}\rangle\langle\psi^{\ast}|$ (123)
$\displaystyle=$ $\displaystyle\int{\rm d}U\ (U\otimes
U^{\ast})|00\rangle\langle 00|(U\otimes U^{\ast})^{\dagger}.$

The last line shows $L=T\left(|00\rangle\langle 00|\right)$ is an isotropic
state obtained by twirling the state $|00\rangle$. In general for any state
$\rho$, its twirling gives [35]:

$\displaystyle T(\rho)$ $\displaystyle=$ $\displaystyle\int{\rm d}U\ (U\otimes
U^{\ast})\rho(U\otimes U^{\ast})^{\dagger}$ (124) $\displaystyle=$
$\displaystyle\frac{d^{2}}{d^{2}-1}\left((1-f_{\rho})\frac{I}{d^{2}}+(f_{\rho}-\frac{1}{d^{2}})P_{+}\right),$

where $f_{\rho}={\mathrm{Tr}}\left(P_{+}\rho\right)$ and
$P_{+}=|\phi_{+}\rangle\langle\phi_{+}|$ is the maximally entangled state. So
for the state $|00\rangle\langle 00|$ of a $d$-dimensional system, one has:

$f_{|00\rangle}={\mathrm{Tr}}\left(P_{+}|00\rangle\langle
00|\right)=\frac{1}{d},$ (125)

and

$L=T\left(|00\rangle\langle
00|\right)=\frac{1}{d(d+1)}I+\frac{1}{(d+1)}P_{+},$ (126)

which coincides with Eq. (23) in the text.

One can also write the average fidelity (20) in the form

$\overline{F}({\cal E})={\mathrm{Tr}}(\Lambda\Psi_{\cal E}),$ (127)

where $\Psi_{\cal E}=\sum_{\alpha}K_{\alpha}\otimes K_{\alpha}^{\dagger}$ and

$\displaystyle\Lambda=\int{\rm d}\psi\
|\psi\rangle\langle\psi|\otimes|\psi\rangle\langle\psi|$

is a symmetric Werner state

$\Lambda=\frac{1}{d(d+1)}(I+P_{sym})=\frac{1}{d(d+1)}(I+\sum_{i,j}|i,j\rangle\langle
j,i|).$ (128)

This will also lead to the same result as in (24).

Finally let us calculate the average fidelity in yet another way, by
expressing the pure states as

$|\psi\rangle\langle\psi|=\frac{1}{d}(I+{\bf n}\cdot\bm{\Gamma}).$ (129)

From

${\cal E}(|\psi\rangle\langle\psi|)=\frac{1}{d}\bigl{(}I+(M{\bf n}+{\bf
t})\cdot\bm{\Gamma}\bigr{)}.$ (130)

and Eq. (129), one finds

$\langle\psi|{\cal
E}(|\psi\rangle\langle\psi|)|\psi\rangle=\frac{1}{d}\bigl{(}1+(d-1){\bf
n}\cdot(M{\bf n}+{\bf t})\bigr{)}.$ (131)

If we now assume a uniform distribution of pure states on the sphere
$S_{d^{2}-2}$ (which is of course not dense) and hence use the relations

$\int d{\bf n}\ n_{i}=0,\hskip 28.45274pt\int d{\bf n}\
n_{i}n_{j}=\frac{1}{(d^{2}-1)}\delta_{ij}$ (132)

we arrive at

$\overline{F}({\cal
E})=\frac{1}{d}\Bigl{(}1+\frac{1}{d+1}{\mathrm{Tr}}M\Bigr{)},$ (133)

which coincides with the result (24) which we obtained by integrating over the
invariant volume $d\psi$ of all pure states. Here we have not used any
specific measure of volume over the sphere $S_{d^{2}-2}$ and only have assumed
that pure states are distributed symmetrically (albeit in a parse way) on this
sphere.

##  Appendix C Quantum channels which stretch the Bloch vector

In two dimension, to satisfy the positivity condition, the affine matrix $M$
of any positive map, and thus any quantum channel, fulfills
$M^{{\mathrm{T}}}M\leq I$. As a result, the Bloch vector corresponding to a
qubit state is always shrunk when $M$ acts on it. Here we show that this is no
longer the case in higher dimensions. This is one of the strange or un-
expected properties of higher dimensional channels which makes the study of
their quasi-inverses, among other things, difficult. As describing this class
of quantum maps is not straightforward we construct an example of such a
channel in this Appendix.

Let $H_{d}$ be a $d-$dimensional Hilbert space and let $P_{1},P_{2},Q_{1}$ and
$Q_{2}$ be orthogonal projectors of rank $d_{1},d_{2},m_{1}$ and $m_{2}$
respectively, with ($d=d_{1}+d_{2}=m_{1}+m_{2}$), i.e.

$P_{1}+P_{2}=Q_{1}+Q_{2}=I,\quad P_{1}P_{2}=Q_{1}Q_{2}=0.$

The following operators are Hermitian and traceless

$A:=\frac{m_{2}Q_{1}-m_{1}Q_{2}}{\sqrt{m_{1}m_{2}}},\hskip
28.45274ptB=\frac{d_{2}P_{1}-d_{1}P_{2}}{\sqrt{d_{1}d_{2}}},$ (134)

and we have

${\mathrm{Tr}}(AB)=0,\hskip 28.45274pt{\mathrm{Tr}}(A^{2})=d,\hskip
28.45274pt{\mathrm{Tr}}(B^{2})=d.$ (135)

Consider now the following measure-and-prepare CPT map

${\cal
E}(\rho)=\frac{1}{d_{1}}{\mathrm{Tr}}(Q_{1}\rho)P_{1}+\frac{1}{d_{2}}{\mathrm{Tr}}(Q_{2}\rho)P_{2}.$
(136)

By expanding the projectors as

$\displaystyle P_{1}=\sum_{i=1}^{d_{1}}|i\rangle\langle i|,\hskip
48.93877ptP_{2}=\sum_{i=d_{1}+1}^{d}|i\rangle\langle i|,$ (137) $\displaystyle
Q_{1}=\sum_{\alpha=1}^{m_{1}}|\varphi_{\alpha}\rangle\langle\varphi_{\alpha}|,\hskip
28.45274ptQ_{2}=\sum_{\alpha=m_{1}+1}^{d}|\varphi_{\alpha}\rangle\langle\varphi_{\alpha}|.$
(138)

a set of Kraus operators for this map is given by

$K_{i,\alpha}:=\frac{1}{\sqrt{d_{1}}}|i\rangle\langle\varphi_{\alpha}|\hskip
28.45274pti=1,\dots,d_{1},\ \ \ \alpha=1,\dots,m_{1}$ (139)

and

$\hskip
28.45274ptL_{i,\alpha}:=\frac{1}{\sqrt{d_{2}}}|i\rangle\langle\varphi_{\alpha}|\hskip
28.45274pti=d_{1}+1,\dots,d,\ \ \ \alpha=\mu_{1}+1,\dots,d.$ (140)

Note that

$\sum_{i,\alpha}K_{i,\alpha}^{\dagger}K_{i,\alpha}+\sum_{i,\alpha}L_{i,\alpha}^{\dagger}L_{i,\alpha}=I_{d},$
(141)

but

$\sum_{i,\alpha}K_{i,\alpha}K_{i,\alpha}^{\dagger}+\sum_{i,\alpha}L_{i,\alpha}L_{i,\alpha}^{\dagger}=\frac{m_{1}}{d_{1}}P_{1}+\frac{m_{2}}{d_{2}}P_{2},$
(142)

hence the channel is non-unital, unless $m_{1}=d_{1}$ and $m_{2}=d_{2}$.

From the above relations and the fact that $P_{1}+P_{2}=Q_{1}+Q_{2}=I$, we
find

$P_{1}=\frac{1}{d}(d_{1}I+\sqrt{d_{1}d_{2}}B)\hskip
28.45274ptP_{2}=\frac{1}{d}(d_{2}I-\sqrt{d_{1}d_{2}}B)$ (143)

and

$Q_{1}=\frac{1}{d}(m_{1}I+\sqrt{m_{1}m_{2}}A)\hskip
28.45274ptQ_{2}=\frac{1}{d}(m_{2}I-\sqrt{m_{1}m_{2}}A)$ (144)

Consider now a state

$\rho=\frac{1}{d}(1+xA)$ (145)

where $x$ represents the Bloch vector of this state, since
$x={\mathrm{Tr}}(\rho A).$ We will then find

${\cal E}(\rho)=\frac{1}{d}(I+x^{\prime}B),$ (146)

where

$x^{\prime}=\sqrt{\frac{m_{1}m_{2}}{d_{1}d_{2}}}x+\frac{m_{1}d_{2}-m_{2}d_{1}}{d\sqrt{d_{1}d_{2}}}.$
(147)

Straightforward calculation completes the reasoning. Hence we find that if
$m_{1}m_{2}>d_{1}d_{2}$, then the Bloch vector can be stretched. It is
interesting to note that the inhomogeneous translation does not compensate the
stretching of $x$, rather it enhances it. Obviously, this is not possible in
dimensions $d=2,3$ but it is possible in dimensions $d\geq 4$, where for
example we can take $(m_{1},m_{2})=(2,2)$ and $(d_{1},d_{2})=(1,3)$. Moreover
we see that if the channel is unital, then $x^{\prime}=x$.

To see how the existence of these channels may affect quasi-inversion, let us
denote by $\mathcal{M}_{s}$ a subset of quantum channels whose distortion
matrices $M_{s}$ can only shrink the Bloch vector, i.e.
$\mathcal{M}_{s}=\\{\mathcal{E}_{s}|\ M_{s}^{\mathrm{T}}M_{s}\leq I\\}$. Let
us highlight two remarks related to the set $\mathcal{M}_{s}$. First,
according to Russo-Dye theorem any linear positive map obtains its norm at the
identity [54]. Thus, singular values of any unital map are less than or equal
to one which implies all unital maps belong to $\mathcal{M}_{s}$. Moreover,
for any $M_{s}$ included in the set $\mathcal{M}_{s}$ it is always possible to
find unitary operators $U_{1}$ and $U_{2}$ such that $M_{s}=(U_{1}+U_{2})/2$.
For now, let us assume that quasi-inverse for a quantum channel $\mathcal{E}$
belongs to $\mathcal{M}_{s}$. In that case, one has (3)

$\displaystyle\overline{F}(\mathcal{E}^{qi}\circ\mathcal{E})$ $\displaystyle=$
$\displaystyle\max_{\mathcal{E}^{\prime}}\overline{F}(\mathcal{E}^{\prime}\circ\mathcal{E})=\max_{\mathcal{E}_{s}\in\mathcal{M}_{s}}\overline{F}(\mathcal{E}_{s}\circ\mathcal{E})$
(148) $\displaystyle=$
$\displaystyle\max_{M_{s}}\frac{1}{d}\left(1+\frac{1}{d+1}{\mathrm{Tr}}M_{s}M\right)$
$\displaystyle\leq$
$\displaystyle\max_{U_{1},U_{2}}\frac{1}{d}\left(1+\frac{1}{2(d+1)}{\mathrm{Tr}}[(U_{1}+U_{2})M]\right)$
$\displaystyle=$
$\displaystyle\frac{1}{d}\left(1+\frac{1}{d+1}{\mathrm{Tr}}|M|\right).$

This inequality imposes an upper bound on the corrected fidelity whenever
quasi-inversion lies in the set $\mathcal{M}_{s}$. An example of which is a
qubit channel where not only quasi-inversion is a unital and unitary map in
$\mathcal{M}_{s}$, but also the entire set of qubit channels are contained in
$\mathcal{M}_{s}$. However, in higher dimensions $\mathcal{M}_{s}$ is a
nontrivial subset of quantum channels as our above example shows. It is a
question then whether quasi-inversion of any quantum channel belongs to the
set $\mathcal{M}_{s}$. In what follows, we show with an example it is possible
to exceed the bound in (148), which implies a negative answer to the question.
Moreover, as the bound (148) is valid when quasi-inversion is a unital map,
one infers in higher dimensions, against qubit channel, quasi-inverse is not
necessarily unital.

Before presenting the example, we mention explicitly taking
$\Gamma_{1}=\sqrt{d-1}A$ and $\Gamma_{2}=\sqrt{d-1}B$, see Eq. (134), along
with the identity and $d^{2}-3$ other Hermitian and traceless operators
orthogonal to $A$ and $B$ as the set of basis, the distortion matrix $M$ and
the translation vector $\mathbf{t}$ of the channel $\mathcal{E}$ in the
beginning of this section are given by their entries as

$M_{ij}=\sqrt{\frac{m_{1}m_{2}}{d_{1}d_{2}}}\delta_{i2}\delta_{j1},\quad\quad\quad
t_{i}=\frac{m_{1}d_{2}-m_{2}d_{1}}{d\sqrt{d-1}\sqrt{d_{1}d_{2}}}\delta_{i2}.$
(149)

Now assume $\lambda>0$ is sufficiently small so the $d$-dimensional map
$\mathcal{E}^{\prime}$ described by the following affine parameters is a
quantum channel

$M^{\prime}_{ij}=\lambda\delta_{i1}\delta_{j2},\quad\quad\quad
t^{\prime}_{i}=0.$ (150)

Note that
${\mathrm{Tr}}(MM^{\prime})=\sqrt{\frac{m_{1}m_{2}}{d_{1}d_{2}}}\lambda$. This
amount is certainly larger than ${\mathrm{Tr}}|M^{\prime}|=\lambda$ if
$m_{1}m_{2}>d_{1}d_{2}$, i.e. if the affine matrix $M$ can stretch the Bloch
vector and $\mathcal{E}$ is not contained in $\mathcal{M}_{s}$. So for the
channel $\mathcal{E}^{\prime}$ specified in Eq. (150) we have through Eq. (3)

$\overline{F}\left(\mathcal{E}^{\prime
qi}\circ\mathcal{E}^{\prime}\right)\geq\frac{1}{d}\left[1+\frac{\lambda}{d+1}\max\sqrt{\frac{m_{1}m_{2}}{d_{1}d_{2}}}\right].$
(151)

For an even $d$ one has
$\max\sqrt{\frac{m_{1}m_{2}}{d_{1}d_{2}}}=\frac{d}{2\sqrt{d-1}}$, while
$\max\sqrt{\frac{m_{1}m_{2}}{d_{1}d_{2}}}=\frac{\sqrt{d+1}}{2}$ for an odd
$d$. In any case, assuming $d>3$ we get

$\overline{F}\left(\mathcal{E}^{\prime
qi}\circ\mathcal{E}^{\prime}\right)\geq\frac{1}{d}\left[1+\frac{\lambda}{2\sqrt{d+1}}\right]>\frac{1}{d}\left[1+\frac{\lambda}{d+1}\right]=\frac{1}{d}\left[1+\frac{{\mathrm{Tr}}|M^{\prime}|}{d+1}\right].$
(152)

##  Appendix D Statistical Properties of Classical Channels and their quasi-
inverses

In this section we discuss some of the properties of classical channels and
their inverses. A random classical channel is a stochastic matrix $T$ in which
the columns are independently picked from the uniform ensemble over the
probabilistic simplex

$\Delta_{d}\equiv\big{\\{}\mathbf{p}\in\mathbb{R}^{d}\hskip
2.84526pt\big{|}\hskip 2.84526ptp_{i}\geq 0;\hskip
2.84526pt\sum_{i}p_{i}=1\big{\\}}$ (153)

Let the Probability Distribution Function (PDF) for any component $p_{i}$ be
denoted by $f_{d}(x)$, that is $\mathbb{P}[x\leq p_{i}\leq x+dx]=f_{d}(x)dx\ \
\ \forall\ i$. Noting as an example figure (8) for the $d=3$ case, we see that
this PDF is proportional to the area of the narrow slab on the triangle which
defines the probability simplex.

Figure 8: A probability simplex in 3 dimensions. Each point on the large
triangle represents a valid probability vector.

For the general case, this function is given by

$\phi_{d}(x)\equiv(d-1)(1-x)^{d-2};\hskip 11.38109ptx\in[0,1].$ (154)

It is also useful to have an expression for the corresponding CDF

$\Phi_{d}(x)=\mathbb{P}[p_{i}\leq x]=\int_{0}^{x}\phi_{d}(t)dt=1-(1-x)^{d-1}.$
(155)

The average fidelity for any classical channel $T$ is given by
$\overline{\mathcal{F}}(T)\equiv\frac{1}{d}{\mathrm{Tr}}T$. If we now take the
average over the uniform ensemble of all stochastic matrices, we find

$\langle\overline{\mathcal{F}}\rangle=\frac{1}{d}\sum_{i}\langle
T_{ii}\rangle=\int_{0}^{1}(d-1)(1-x)^{d-2}xdx=\frac{1}{d},$ (156)

where we have used the statistical independence of the columns of $T$ in the
uniform ensemble. Moreover we can also find the variance of this average
fidelity which turns out to be

$\operatorname{Var}\overline{\mathcal{F}}=\frac{1}{d^{2}}\sum_{i}\operatorname{Var}T_{ii}=\frac{d-1}{d^{3}(d+1)}.$
(157)

The interesting question is how much this average fidelity increases for
classical channels when we apply the quasi-inversion. To find this we note
that the average fidelity after quasi inversion is

$\overline{\mathcal{F}}(T^{qi}\circ T)=\frac{1}{d}\sum_{i}\max_{j}T_{ij},$
(158)

that is the average fidelity after quasi-inversion is the average of the
maximum element in each row of $T$. Assuming that the largest element of two
different rows do not occur on the same column (which can happen only for a
subset of measure zero in the ensemble), the ensemble average of the improved
average fidelity becomes

$\langle\overline{\mathcal{F}}(T^{qi}\circ T)\rangle=\langle p_{m}\rangle,$
(159)

where $p_{m}$ is the largest element in a single probability vector chosen
uniformly. To find the average of this quantity for the uniform ensemble, we
invoke the proposition that only on a set of measure zero, the maximum values
of two different rows may occur on the same column. Therefore we first find
the following Cumulative Distribution Function (CDF), i.e. the probability
that the maximum values in all columns are less than $x$

$\Psi_{d}(x):=\mathbb{P}[p_{m}\leq x,\hskip 2.84526pt\forall
j]=\Phi_{d}(x)^{d}=\big{[}1-(1-x)^{d-1}\big{]}^{d}$

this leads to the following PDF, i.e. the probability that the maximum value
is between $x$ and $x+dx$,

$\psi_{d}(x)=\frac{d\Psi_{d}(x)}{dx}=d(d-1)\big{[}1-(1-x)^{d-1}\big{]}^{d-1}(1-x)^{d-2}$

Now it is easy to see that

$\displaystyle\langle\overline{\mathcal{F}}(T^{qi}\circ T)\rangle=\langle
p_{m}\rangle$ $\displaystyle=$ $\displaystyle 1-\langle
1-p_{m}\rangle=1-d(d-1)\int_{0}^{1}dx\big{[}x(1-x^{d-1})\big{]}^{d-1}$ (160)
$\displaystyle=$ $\displaystyle
1+\sum_{n=0}^{d}\frac{(-1)^{n}(d-1)n}{(d-1)n+1}\begin{pmatrix}d\\\
n\end{pmatrix}.$ (161)

Figure 9 shows the average fidelity after quasi-inversion versus dimension.
The curve fits the equation

$\langle\overline{\mathcal{F}}(T^{qi}\circ T)\rangle=cd^{x}\sim d^{-1/2},$
(162)

with $c\approx 1.055$ and the exponent $x\approx-0.5042$. Therefore, the
average fidelity of a classical channel appears to be increasing due to quasi-
inversion from $1/d$ to $1/\sqrt{d}$.

Figure 9: The average fidelity of a classical channel before
$\langle\overline{\mathcal{F}}(T)\rangle$ (blue cross) and after
$\langle\overline{\mathcal{F}}(T^{qi}\circ T)\rangle$ (red triangle) it is
compensated by the quasi-inversion.

##  Appendix E  A note on statistical Properties of Quantum Channels and their
quasi-inverses

Investigating the same problem in the set of quantum channels seems a highly
difficult task since in quantum case the quasi-inversion is not generally
known. Nonetheless, numerical analysis is possible in lower dimensions and it
shows improvement in amount of fidelity for a typical channel. To see that,
one may notice for random quantum channels distributed uniformly in the set of
quantum channels [44]

$\langle\Phi_{\mathcal{E}}\rangle=\Phi_{*},$ (163)

where $\Phi_{*}$ is the maximally depolarizing channel, see Section 2. This
implies $\langle{\mathrm{Tr}}\Phi_{\mathcal{E}}\rangle=1$. Thus, the average
amount of input-output fidelity over random channels distributed uniformly in
$d$ dimension is given by (24)

$\langle\overline{F}(\mathcal{E})\rangle=\frac{1}{d}.$ (164)

To get the fidelity after correction averaged over the set of uniformly
distributed quantum channels, we applied numerically searching for the quasi-
inverse. The sketch of our method is based on the observation that any quantum
channel ${\cal E}:D(H_{d})\longrightarrow D(H_{d})$ should satisfy a
continuous family of inequalities

$\langle\phi|({\cal E}\otimes I)(|\psi\rangle\langle\psi|)|\phi\rangle\geq
0\hskip 28.45274pt|\phi\rangle,\ |\psi\rangle\ \in H_{d}\otimes H_{d}.$ (165)

To approximate the quasi-inverse of a channel numerically, we first
approximate the set of CP operators by only including a finite subset of such
constraints obtained by random entangled states in $H_{d}\otimes H_{d}$. Then
the quasi-inverse will be calculated as the answer to a finite linear
programming problem using the simplex method or other efficient algorithms.
The results for dimensions $d=2,3,4,5$ are presented in Fig. 10 and they
confirm improvement in the average fidelity after correction for random
channels.

Figure 10: The amount of input-output fidelity
$\langle\overline{F}(\mathcal{E})\rangle$ (blue cross), fidelity after
correction with the best possible unitary
$\langle\overline{F}(\mathcal{U}_{\mathcal{E}}\circ\mathcal{E})\rangle$ (black
circle), and fidelity after correction with quasi-inversion
$\langle\overline{F}(\mathcal{E}^{qi}\circ\mathcal{E})\rangle$ (red triangle)
averaged over the set of random channels distributed uniformly for
$d=2,\dots,5$.

One may also think about applying a unitary evolution to correct the input-
output fidelity. However, our numerical results show that the best unitary
quantum channel, which will be denoted by $\mathcal{U}_{\mathcal{E}}$, cannot
modify the fidelity in the best possible way, see Fig. 10 for a comparison.
Indeed, we can take the purity of Jamiołkowski state as a signature of
unitarity of a given channel $\mathcal{E}$, i.e. this quantity is equal to $1$
if and only if the channel is unitary and any deviation from $1$ shows non-
unitarity. Adopting such a function, one finds unitarity for the quasi-inverse
of random channels averaged over the set of uniformly distributed CPT maps is
equal to $0.99\pm 0.01$ for $d=3$, $0.68\pm 0.05$ for $d=4$, and $0.52\pm
0.05$ for $d=5$. This interesting fact counter-intuitively shows in higher
dimensions for a typical channel quasi inversion is almost a non-unitary
channel. However, if we measure unitality by $1-|\mathbf{t}|^{2}$
($\mathbf{t}$ is the translation vector, see (9)) averaged over the set of
random channels, we see for $d=3,4,5$ it is respectively given by: $0.98\pm
0.01$, $0.97\pm 0.01$, and $0.97\pm 0.01$, which confirms that the quasi-
inverse of a typical channel is close to be unital.