Quasi-inversion of quantum and classical channels in finite dimensions

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, ρ ⟶ UρU^†. 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 ρ of order d,

$$\begin{equation}\rho \rightarrow \mathcal{E}(\rho )=\sum\limits _{\alpha }\;{K}_{\alpha }\rho {K}_{\alpha }^{{\dagger}},\end{equation} \tag{ 1 }$$

which can be interpreted as a generalization of the unitary dynamics. The standard unitarity condition U^† U = I has been replaced above by the identity resolution, ${\sum }_{\alpha =1}^{r}\;{K}_{\alpha }^{{\dagger}}{K}_{\alpha }=I$. Here K_α 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 ⩽ d². Any map of this class is completely positive and trace-preserving 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 [7–16] 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 ${\mathcal{E}}^{qi}$ exists, such that ${\mathcal{E}}^{qi}{\circ}\mathcal{E}$ 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 ρ_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 ρ_AB of size d²—see [22]. Thus the maximal fidelity optimized over all possible channels can be related to the conditional min-entropy of the state ρ_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/d².

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 ⩾ 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, $\mathcal{E}(I)=I$ 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 ⩾ 3 there exists the unital channel of Landau and Streater (LS), 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. ${\mathcal{E}}^{qi}{\circ}\mathcal{E}$. 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 LS [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.

Let H_d be a complex d-dimensional Hilbert space for which we choose the computational basis {|m⟩, m = 1, ..., 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 ρ 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 d² − 1 real parameters. Any linear map A ∈ L(H_d ) can be uniquely mapped to a vector |A⟩ ∈ H_d ⊗ H_d in the form

$$\begin{equation}\vert A\rangle =d(A\otimes I)\vert {\phi }^{+}\rangle ,\end{equation} \tag{ 2 }$$

where $\vert {\phi }^{+}\rangle =\frac{1}{\sqrt{d}}{\sum }_{i}\;\vert i,i\rangle$ is a maximally entangled state. This is called vectorization of a matrix A = ∑_i,j   A_i,j |i⟩⟨j| into a vector |A⟩ = ∑_i,j   A_i,j |i, j⟩, with the correspondence between the inner products:

$$\begin{equation}\mathrm{Tr}({A}^{{\dagger}}B)=\langle A\vert B\rangle .\end{equation} \tag{ 3 }$$

Taking d² − 1 traceless and Hermitian matrices Γ_i along with identity I as a complete basis, one can write any density matrix as

$$\begin{equation}\rho =\frac{1}{d}\left(I+\sum\limits _{i=1}^{{d}^{2}-1}{r}_{i}{{\Gamma}}_{i}\right)=\frac{1}{d}\left(I+\mathbf{r}\cdot \mathbf{\Gamma }\right).\end{equation} \tag{ 4 }$$

We normalize Γ_i matrices to satisfy

$$\begin{equation}\mathrm{Tr}\left({{\Gamma}}_{i}{{\Gamma}}_{j}\right)=d(d-1){\delta }_{ij},\end{equation} \tag{ 5 }$$

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

$$\begin{equation}\mathrm{Tr}({\rho }^{2})=\frac{1}{d}\left[1+(d-1)\mathbf{r}\cdot \mathbf{r}\right].\end{equation} \tag{ 6 }$$

The (d² − 1)-dimensional vector $\boldsymbol{r}={({x}_{1},{x}_{2},\dots ,{x}_{{d}^{2}-1})}^{\mathrm{T}}$ 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 [22, 29]. The set of pure states ρ = |ψ⟩⟨ψ|, where $\vert \psi \rangle ={\sum }_{i=1}^{d}{\psi }_{i}\vert i\rangle$ is a sphere of dimension 2d − 2 which we denote by S_2d−2. On the other hand, any pure state has the property Tr(ρ²) = 1 which in view of equations (5) and (6) is equivalent to r ⋅ 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 $\mathcal{E}$ represented by its Kraus operators K_α acting on states of dimension d through equation (1). The correspondence (2) leads to the following representation of channels by matrices acting on vectorized form of density matrices,

$$\begin{equation}\mathcal{E}(\rho )=\sum\limits _{\alpha }\;{K}_{\alpha }\rho {K}_{\alpha }^{{\dagger}}\enspace \enspace \enspace \rightarrow \enspace \enspace \enspace \vert \mathcal{E}(\rho )\rangle ={{\Phi}}_{\mathcal{E}}\vert \rho \rangle ,\end{equation} \tag{ 7 }$$

where

$$\begin{equation}{{\Phi}}_{\mathcal{E}}=\sum\limits _{\alpha }\;{K}_{\alpha }\otimes {K}_{\alpha }^{{\ast}},\end{equation} \tag{ 8 }$$

is called the superoperator of the map $\mathcal{E}$ [22] in which ∗ denotes complex conjugation. Although a quantum channel has many different sets of Kraus operators, connected by L_β = ∑_β U_α,β K_β , where U is a unitary matrix, it is straightforward to see that Φ is unique. It also has the property that

$$\begin{equation*}{{\Phi}}_{\mathcal{E}{\circ}{\mathcal{E}}^{\prime }}={{\Phi}}_{\mathcal{E}}{{\Phi}}_{{\mathcal{E}}^{\prime }}.\end{equation*}$$

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

$$\begin{equation}\mathbf{r}\rightarrow {\mathbf{r}}^{\prime }=M\mathbf{r}+\mathbf{t},\end{equation} \tag{ 9 }$$

where

$$\begin{equation}{M}_{ij}=\frac{1}{d(d-1)}\enspace \mathrm{Tr}({{\Gamma}}_{i}\mathcal{E}({{\Gamma}}_{j}))\quad \mathrm{a}\mathrm{n}\mathrm{d}\quad {t}_{i}=\frac{1}{d(d-1)}\enspace \mathrm{Tr}({{\Gamma}}_{i}\mathcal{E}(I)).\end{equation} \tag{ 10 }$$

The matrix M of order d² − 1 is called the distortion matrix. Writing the basis {I, Γ_i } in vectorized form {|I⟩, |Γ_i ⟩} which are vectors of dimension d², we have the normalization condition

$$\begin{equation}\langle I\vert I\rangle =d,\quad \langle I\vert {{\Gamma}}_{i}\rangle =\langle {{\Gamma}}_{i}\vert I\rangle =0,\quad \langle {{\Gamma}}_{i}\vert {{\Gamma}}_{j}\rangle =d(d-1){\delta }_{i,j}.\end{equation} \tag{ 11 }$$

The d² dimensional vector |ρ⟩ can be written in terms of the normalized basis vectors $\vert ~{I}\rangle =\frac{1}{\sqrt{d}}\vert I\rangle$ and $\vert {~{{\Gamma}}}_{i}\rangle =\frac{1}{\sqrt{d(d-1)}}\vert {{\Gamma}}_{i}\rangle$, in the symbolic form

$$\begin{equation}\vert \rho \rangle =\frac{1}{\sqrt{d}}\left(\begin{matrix}\hfill 1\hfill \\ \hfill \sqrt{d-1}\enspace \mathbf{r}\hfill \end{matrix}\right),\end{equation} \tag{ 12 }$$

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

$$\begin{equation}\vert {\rho }^{\prime }\rangle =\frac{1}{\sqrt{d}}\left(\begin{matrix}\hfill 1\hfill \\ \hfill \sqrt{d-1}\enspace M\mathbf{r}+\mathbf{t}\hfill \end{matrix}\right).\end{equation} \tag{ 13 }$$

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

$$\begin{equation}{{\Phi}}_{\mathcal{E}}=\left(\begin{matrix}\hfill 1\hfill & \hfill 0\hfill \\ \hfill \sqrt{d-1}\enspace \mathbf{t}\hfill & \hfill M\hfill \end{matrix}\right).\end{equation} \tag{ 14 }$$

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

The above form of the map Φ, which represents the evolution of the Bloch vector 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,

$$\begin{equation}{C}_{\mathcal{E}}{:=}d(\mathcal{E}\otimes I)(\vert {\phi }^{+}\rangle \langle {\phi }^{+}\vert )=\sum\limits _{i,j}\;\mathcal{E}(\vert i\rangle \langle j\vert )\otimes \vert i\rangle \langle j\vert ,\end{equation} \tag{ 15 }$$

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

$$\begin{equation}{C}_{\mathcal{E}}={{\Phi}}_{\mathcal{E}}^{R},\end{equation} \tag{ 16 }$$

where A^R denotes the following reshuffling of the entries of a matrix A of order d²

$$\begin{equation}{A}_{ij,kl}^{R}{:=}{A}_{ik,jl}.\end{equation} \tag{ 17 }$$

2.1. Average fidelity of a channel

The performance of a quantum channel $\mathcal{E}$ can be studied in different ways. Among these we choose the input–output fidelity averaged over a uniform distribution of pure states, $\overline{F}(\mathcal{E})$, and the entanglement fidelity ${F}_{\hspace{-1.5pt}\text{E}}(\mathcal{E})$ [34]. These are respectively defined as follows:

$$\begin{equation}\overline{F}(\mathcal{E})={\int }_{{S}_{2d-2}}\mathrm{d}\psi \langle \psi \vert \mathcal{E}(\vert \psi \rangle \langle \psi \vert )\vert \psi \rangle ,\end{equation} \tag{ 18 }$$

where ∫dψ = 1 and dψ = d(Uψ) for any unitary, and

$$\begin{equation}{F}_{\hspace{-1.5pt}\text{E}}(\mathcal{E})=\langle {\phi }^{+}\vert \mathcal{E}\otimes I\vert (\vert {\phi }^{+}\rangle \langle {\phi }^{+}\vert )\vert {\phi }^{+}\rangle .\end{equation} \tag{ 19 }$$

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 $\mathcal{E}$ 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

$$\begin{align}\overline{F}(\mathcal{E})& =\sum\limits _{\alpha }{\int }_{{S}_{2d-2}}\mathrm{d}\psi \langle \psi \vert {K}_{\alpha }\vert \psi \rangle \langle \psi \vert {K}_{\alpha }^{{\dagger}}\vert \psi \rangle \\ & =\sum\limits _{\alpha }{\int }_{{S}_{2d-2}}\mathrm{d}\psi \langle \psi \vert {K}_{\alpha }\vert \psi \rangle \langle {\psi }^{{\ast}}\vert {K}_{\alpha }^{{\ast}}\vert {\psi }^{{\ast}}\rangle =\mathrm{Tr}(L{{\Phi}}_{\mathcal{E}}),\end{align} \tag{ 20 }$$

where

$$\begin{equation}L{:=}{\int }_{{S}_{2d-2}}\mathrm{d}\psi \enspace \vert \psi \rangle \langle \psi \vert \otimes \vert {\psi }^{{\ast}}\rangle \langle {\psi }^{{\ast}}\vert ,\end{equation} \tag{ 21 }$$

is the isotropic state and ${{\Phi}}_{\mathcal{E}}$ is defined in equation (8). Since L is an isotropic operator in the sense that [L, U ⊗ U*] = 0  ∀U, it can be written in the form [35],

$$\begin{equation}L=\frac{1}{d(d+1)}\left(I+\sum\limits _{i,j}\;\vert i,i\rangle \langle j,j\vert \right).\end{equation} \tag{ 22 }$$

Inserting (22) in (20) and using (16) and (17) one finds the following formula for the average fidelity

$$\begin{equation}\overline{F}(\mathcal{E})=\frac{1}{d(d+1)}\left[d+\mathrm{Tr}{{\Phi}}_{\mathcal{E}}\right].\end{equation} \tag{ 23 }$$

More details on these calculations will be given in appendix B. From (8) and (14) we see that this can also be written as $\overline{F}(\mathcal{E})=\frac{1}{d(d+1)}\left[d+{\sum }_{\alpha }\vert \mathrm{Tr}\enspace {K}_{\alpha }{\vert }^{2}\right]$ and $\overline{F}(\mathcal{E})=\frac{1}{d(d+1)}\left[d+1+\mathrm{Tr}\enspace M\right]$.

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

$$\begin{equation}{F}_{\hspace{-1.5pt}\text{E}}(\mathcal{E})=\frac{1}{d}\langle {\phi }^{+}\vert \sum\limits _{\alpha }\;{K}_{\alpha }\vert i\rangle \langle j\vert {K}_{\alpha }^{{\dagger}}\otimes \vert i\rangle \langle j\vert {\phi }^{+}\rangle =\frac{1}{{d}^{2}}\sum\limits _{\alpha }\;\vert \mathrm{Tr}\enspace {K}_{\alpha }{\vert }^{2}.\end{equation} \tag{ 24 }$$

Therefore, we find the simple relation [36]

$$\begin{equation}\overline{F}=\frac{1}{d+1}(1+d{F}_{\hspace{-1.5pt}\text{E}}(\mathcal{E})),\end{equation} \tag{ 25 }$$

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

In the quantum communication context, it is usually the case that Alice (the information source), generates a message state ρ_M from a distribution corresponding to her language alphabet which is known to Bob (the receiver). Let us denote by μ 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 ρ_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 ρ_M since Bob does not know it. It only depends on the channel $\mathcal{E}$ and the probability measure μ. 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 ${\mathcal{E}}^{qi}$ fulfilling the following condition [1]

$$\begin{equation}\overline{F}({\mathcal{E}}^{qi}{\circ}\mathcal{E}){\geqslant}\overline{F}({\mathcal{E}}^{\prime }{\circ}\mathcal{E})\quad \forall \enspace {\mathcal{E}}^{\prime },\end{equation} \tag{ 26 }$$

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, $\mathcal{E}(\rho )={\sum }_{i=0}^{3}\;{p}_{i}{\sigma }_{i}\rho {\sigma }_{i}$, 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 μ 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)

$$\begin{equation}F(\rho ,{\rho }^{\prime }){\leqslant}F(\rho ,\rho )\quad \forall \enspace \rho ,{\rho }^{\prime }\in D({H}_{d});\end{equation} \tag{ 27 }$$

the state most similar to a state ρ is the state ρ itself, and (b) concavity:

$$\begin{equation}F(\rho ,\alpha {\rho }_{1}+(1-\alpha ){\rho }_{2}){\geqslant}\alpha F(\rho ,{\rho }_{1})+(1-\alpha )F(\rho ,{\rho }_{2}).\end{equation} \tag{ 28 }$$

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 $\mathcal{E}\equiv (M,\mathbf{t})$, defined by definition 1, is a quantum channel ${\mathcal{E}}^{\prime }$ 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 t). This affine map may not correspond to a legitimate quantum channel and may be outside ${\mathcal{C}}_{d}$. Denote it by ${\mathcal{E}}^{-1}$. However for small enough ɛ, the channel $(1-\varepsilon ){\mathcal{E}}^{\prime }+\varepsilon {\mathcal{E}}^{-1}$ is a CPT $\in {\mathcal{C}}_{d}$. We now note that this channel performs better in quasi-inverting the channel $\mathcal{E}$ since using (27) and (28) we find

$$\begin{align*}& \int \mathrm{d}\mu \enspace F\left(\rho ,\left[(1-\varepsilon ){\mathcal{E}}^{\prime }+\varepsilon {\mathcal{E}}^{-1}\right]{\circ}\mathcal{E}(\rho )\right)\\ & \quad {\geqslant}(1-\varepsilon )\int \mathrm{d}\mu \enspace F\left(\rho ,{\mathcal{E}}^{\prime }{\circ}\mathcal{E}(\rho )\right)+\varepsilon \int \mathrm{d}\mu \enspace F(\rho ,\rho )\\ & \quad =\int \mathrm{d}\mu \enspace F(\rho ,{\mathcal{E}}^{\prime }{\circ}\mathcal{E})\rho \left. \right)+\varepsilon \left(\int \mathrm{d}\mu \left[F(\rho ,\rho )-F\left(\rho ,{\mathcal{E}}^{\prime }{\circ}\mathcal{E}(\rho )\right)\right]\right)\\ & \quad {\geqslant}\int \mathrm{d}\mu \enspace F\left(\rho ,{\mathcal{E}}^{\prime }{\circ}\mathcal{E}(\rho )\right).\end{align*}$$

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 ${\mathcal{C}}_{d}$.

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

$$\begin{equation*}\forall \enspace {\mathcal{E}}^{\prime }\in \mathrm{i}\mathrm{n}\mathrm{t}\enspace {\mathcal{C}}_{d},\exists \enspace \enspace {\mathcal{E}}_{n}^{qi}\in \partial {\mathcal{C}}_{d}\end{equation*}$$

such that

$$\begin{equation*}\langle F\left(\rho ,{\mathcal{E}}_{n}^{qi}{\circ}{\mathcal{E}}_{n}(\rho )\right)\rangle {\geqslant}\langle F\left(\rho ,{\mathcal{E}}^{\prime }{\circ}{\mathcal{E}}_{n}(\rho )\right)\rangle ,\end{equation*}$$

where ⟨X⟩ denotes the average of X over μ. 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 (27) and (28), neither we have assumed the average fidelity to be defined only for pure states. □

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Hereafter we assume the standard fidelity measure, $F(\rho ,{\rho }^{\prime })={\left(\mathrm{Tr}\sqrt{\sqrt{\rho }{\rho }^{\prime }\sqrt{\rho }}\right)}^{2}$, 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 equation (23), the practical method for finding the quasi-inverse is through one of the following maximization problems:

$$\begin{align}\overline{F}({\mathcal{E}}^{qi}{\circ}\mathcal{E})& =\underset{{{\Phi}}^{\prime }}{\mathrm{max}}\enspace \frac{1}{d+1}\left(1+\frac{1}{d}\enspace \mathrm{Tr}\enspace {{\Phi}}^{\prime }{\Phi}\right)\\ & =\underset{\left\{{K}_{\beta }^{\prime }\right\}}{\mathrm{max}}\enspace \frac{1}{d+1}\left(1+\frac{1}{d}\sum\limits _{\alpha ,\beta }\;\vert \mathrm{Tr}\enspace {K}_{\beta }^{\prime }{K}_{\alpha }{\vert }^{2}\right)\\ & =\underset{{M}^{\prime }}{\mathrm{max}}\enspace \frac{1}{d}\left(1+\frac{1}{d+1}\enspace \mathrm{Tr}\enspace {M}^{\prime }M\right).\end{align} \tag{ 29 }$$

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, t ) and (M', t '), respectively, their concatenation, ${\mathcal{E}}^{\prime }{\circ}\mathcal{E}$, is represented by (M'M, M' t + t '). This implies that the translation vector 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 ρ_AB is defined as [40]

$$\begin{equation}{H}_{\mathrm{min}}{(B\vert A)}_{\rho }=-\underset{{\sigma }_{A}}{\mathrm{inf}}\;\underset{\lambda }{\mathrm{inf}}\left\{\lambda \in \mathbb{R}\vert \enspace {\rho }_{AB}{\leqslant}{2}^{\lambda }({\sigma }_{A}\otimes I)\right\},\end{equation} \tag{ 30 }$$

where σ_A is a quantum state. Indeed, it has been proven that [21]

$$\begin{equation}{2}^{-{H}_{\mathrm{min}}{(B\vert A)}_{\rho }}=d\;\underset{{\mathcal{E}}^{\prime }}{\mathrm{max}}\left\langle {\phi }^{+}\right\vert \enspace ({\mathcal{E}}^{\prime }\otimes I)[{\rho }_{AB}]\left\vert {\phi }^{+}\right\rangle .\end{equation} \tag{ 31 }$$

Let us assume that ρ_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

$$\begin{align}{2}^{-{H}_{\mathrm{min}}{(B\vert A)}_{{J}_{\mathcal{E}}}}& =d\;\underset{{\mathcal{E}}^{\prime }}{\mathrm{max}}\left\langle {\phi }^{+}\right\vert ({\mathcal{E}}^{\prime }{\circ}\mathcal{E}\otimes I)\left[\left\vert {\phi }^{+}\right\rangle \left\langle {\phi }^{+}\right\vert \right]\hspace{-2pt}\left\vert {\phi }^{+}\right\rangle \\ & =d\;\underset{{\mathcal{E}}^{\prime }}{\mathrm{max}}\enspace {F}_{\hspace{-1.5pt}\text{E}}({\mathcal{E}}^{\prime }{\circ}\mathcal{E})=d{F}_{\hspace{-1.5pt}\text{E}}({\mathcal{E}}^{qi}{\circ}\mathcal{E}).\end{align} \tag{ 32 }$$

Here ${F}_{\hspace{-1.5pt}\text{E}}({\mathcal{E}}^{\prime }{\circ}\mathcal{E})$ is the entanglement fidelity (24) 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 (25)) are directly related to the conditional min-entropy of the Choi matrix of the 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 Ω is called extreme if it cannot be written as a convex combination of two other points of Ω. 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 $\mathcal{E}$ is in the form

$$\begin{equation}{\mathcal{E}}^{qi}=\sum\limits _{i}\;{\lambda }_{i}{\mathcal{E}}_{i}.\end{equation} \tag{ 33 }$$

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

$$\begin{equation}\overline{F}({\mathcal{E}}_{m}{\circ}\mathcal{E}){\geqslant}\overline{F}\left(\sum\limits _{i}\;{\lambda }_{i}{\mathcal{E}}_{i}{\circ}\mathcal{E}\right)=\overline{F}({\mathcal{E}}^{qi}{\circ}\mathcal{E}),\end{equation} \tag{ 34 }$$

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 ${\mathcal{E}}_{1}^{qi}$ and ${\mathcal{E}}_{2}^{qi}$ are both quasi-inversions of $\mathcal{E}$, then any convex combination of them, $p{\mathcal{E}}_{1}^{qi}+(1-p){\mathcal{E}}_{2}^{qi}$, 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. ${({\mathcal{E}}^{qi})}^{qi}\ne \mathcal{E}$ 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

$$\begin{equation}\underset{{\mathcal{E}}^{\prime \prime }}{\mathrm{max}}\enspace \overline{F}\left({\mathcal{E}}^{\prime \prime }{\circ}({\mathcal{E}}^{qi}{\circ}\mathcal{E})\right){\leqslant}\underset{{\mathcal{E}}^{\prime }}{\mathrm{max}}\enspace \overline{F}\left({\mathcal{E}}^{\prime }{\circ}\mathcal{E}\right)=\overline{F}\left({\mathcal{E}}^{qi}{\circ}\mathcal{E}\right).\end{equation} \tag{ 35 }$$

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 ⩽ p ⩽ 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 afterward.

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 σ₀ = I and other σ_i 's are the Pauli matrices. According to the results of [1], the subset of Pauli channels for which ∀i p₀ ⩾ p_i belongs to ${\mathfrak{C}}_{I}$. This is also a consequence of the result of example 1 in the next section, since σ_i 's satisfy orthogonality. The set of non-correctable Pauli channels is presented in figure 2.

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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 Φ₁ and Φ₂ to denote ${\mathcal{E}}_{1}$ and ${\mathcal{E}}_{2}$. Unitary maps, ${\mathcal{E}}_{U}$ and ${\mathcal{E}}_{V}$, are represented by Φ_U and Φ_V . Hence,

$$\begin{align}\mathrm{Tr}\left({{\Phi}}_{2}^{qi}{{\Phi}}_{2}\right)& =\underset{{{\Phi}}^{\prime \prime }}{\mathrm{max}}\enspace \mathrm{Tr}\left({{\Phi}}^{\prime \prime }{{\Phi}}_{2}\right)=\underset{{{\Phi}}^{\prime \prime }}{\mathrm{max}}\enspace \mathrm{Tr}\left({{\Phi}}^{\prime \prime }{{\Phi}}_{U}{{\Phi}}_{1}{{\Phi}}_{V}\right)\\ & =\underset{{{\Phi}}^{\prime \prime }}{\mathrm{max}}\enspace \mathrm{Tr}\left({{\Phi}}_{V}{{\Phi}}^{\prime \prime }{{\Phi}}_{U}{{\Phi}}_{1}\right)=\underset{{{\Phi}}^{\prime }}{\mathrm{max}}\enspace \mathrm{Tr}\left({{\Phi}}^{\prime }{{\Phi}}_{1}\right)=\mathrm{Tr}\left({{\Phi}}_{1}^{qi}{{\Phi}}_{1}\right),\end{align} \tag{ 36 }$$

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, ${\mathcal{E}}_{2}^{qi}={\mathcal{E}}_{{V}^{-1}}{\circ}{\mathcal{E}}_{1}^{qi}{\circ}{\mathcal{E}}_{{U}^{-1}}$, 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 $\mathcal{E}={\mathcal{E}}_{U}{\circ}{\mathcal{E}}_{c}{\circ}{\mathcal{E}}_{V}$, where ${\mathcal{E}}_{c}$ is a canonical map with diagonal distortion matrix M_c = diag(λ₁, λ₂, λ₃). The signed singular values [22] of the matrix M_c confined inside a tetrahedron Δ whose extreme points are unitary operations $\rho \rightarrow {\sigma }_{i}\rho {\sigma }_{i}^{{\dagger}}\;(i=0,1,2,3)$. 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 λ. 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 LS 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:

$$\begin{equation}\frac{f+1}{d+1}{\leqslant}\overline{F}({\mathcal{E}}^{qi}{\circ}\mathcal{E}){\leqslant}\frac{{p}_{m}+1}{d+1},\end{equation} \tag{ 37 }$$

where $f={\mathrm{max}}_{\left\vert \beta \right\rangle }\left\langle \beta \right\vert {C}_{\mathcal{E}}\left\vert \beta \right\rangle$ is the fully entangled fraction of the Choi matrix ${C}_{\mathcal{E}}$ of the channel $\mathcal{E}$ and |β⟩ denotes a maximally entangled state, while p_m is the maximal eigenvalue of ${C}_{\mathcal{E}}$.

Before proceeding with the proof let us mention in view of equation (25) 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

$$\begin{align}\mathrm{Tr}({{\Phi}}_{{\mathcal{E}}^{\prime }}{{\Phi}}_{\mathcal{E}})& =\sum {({{\Phi}}_{{\mathcal{E}}^{\prime }})}_{ij,kl}{({{\Phi}}_{\mathcal{E}})}_{kl,ij}=\sum {({C}_{{\mathcal{E}}^{\prime }})}_{ik,jl}{({C}_{\mathcal{E}})}_{ki,lj}\\ & =\sum {(S{C}_{{\mathcal{E}}^{\prime }}S)}_{ki,lj}{({C}_{\mathcal{E}})}_{ki,lj}=\mathrm{Tr}({~{C}}_{{\mathcal{E}}^{\prime }}{C}_{\mathcal{E}})\\ & {\leqslant}{p}_{m}\mathrm{Tr}({~{C}}_{{\mathcal{E}}^{\prime }})={p}_{m}\enspace \mathrm{Tr}({C}_{{\mathcal{E}}^{\prime }})=d{p}_{m},\end{align} \tag{ 38 }$$

where S = ∑_i,j |i, j⟩⟨j, i| is the swap operator, ${~{C}}_{{\mathcal{E}}^{\prime }}={(S{C}_{{\mathcal{E}}^{\prime }}S)}^{\mathrm{T}}$, and p_m is the largest eigenvalue of the Choi matrix of the channel $\mathcal{E}$. In writing these equations we have used equations (16) and (17), and the fact that ${C}_{\mathcal{E}}$ and ${~{C}}_{{\mathcal{E}}^{\prime }}$ are Hermitian and positive matrices. The above inequality leads to the following upper bound for the improved average fidelity

$$\begin{equation}\overline{F}(\mathcal{E}){\leqslant}\frac{{p}_{m}+1}{d+1}.\end{equation} \tag{ 39 }$$

In order to obtain a lower bound, we note that in view of (26)

$$\begin{equation}\mathrm{Tr}({~{C}}_{{\mathcal{E}}^{\prime }}{C}_{\mathcal{E}})=\mathrm{Tr}({{\Phi}}_{{\mathcal{E}}^{\prime }}{{\Phi}}_{\mathcal{E}}){\leqslant}\mathrm{Tr}({{\Phi}}_{\mathcal{E}}^{qi}{{\Phi}}_{\mathcal{E}})\quad \forall \enspace {C}_{{\mathcal{E}}^{\prime }},\end{equation} \tag{ 40 }$$

and choose for ${~{C}}_{{\mathcal{E}}^{\prime }}$ to be equal to d|β⟩⟨β|, where |β⟩ 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.

There are certain channels whose distortion matrix M in equation (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 figure 3. We will elaborate on this point and its consequences in appendix C.

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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 ρ_AA'BB' = ρ_AB ⊗ ρ_A'B', then one can show the min-entropy (30) is additive [21], i.e. H_min(BB'|AA') = H_min(B|A) + H_min(B'|A'). The proof of the proposition then becomes straightforward using equation (32) 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

$$\begin{equation}\underset{{\mathcal{E}}^{\prime }}{\mathrm{max}}\enspace \overline{F}({\mathcal{E}}^{\prime }{\circ}{\mathcal{E}}^{\otimes N})=\overline{F}\left({({\mathcal{E}}^{qi})}^{\otimes N}{\circ}{\mathcal{E}}^{\otimes N}\right){\leqslant}{\left(\overline{F}({\mathcal{E}}^{qi}{\circ}\mathcal{E})\right)}^{N}.\end{equation} \tag{ 41 }$$

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

$$\begin{equation}\mathrm{Tr}\enspace {{\Phi}}_{{\mathcal{Q}}^{\otimes N}}={(\mathrm{Tr}\enspace {{\Phi}}_{\mathcal{Q}})}^{N}.\end{equation} \tag{ 42 }$$

Let us apply $\mathrm{Tr}({{\Phi}}_{\mathcal{Q}})=x$ for simplicity of notation. In view of equation (23), we prove whenever 1 ⩽ x ⩽ d² the following inequality holds:

$$\begin{equation}\overline{F}({\mathcal{Q}}^{\otimes N})=\frac{{d}^{N}+{x}^{N}}{{d}^{N}({d}^{N}+1)}{\leqslant}{\left(\frac{d+x}{d(d+1)}\right)}^{N}={\left(\overline{F}(\mathcal{Q})\right)}^{N}.\end{equation} \tag{ 43 }$$

To prove this relation we note that both sides are increasing functions of x. When x = 1 or x = d² 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 (43). To prove equation (41), it remains to show that $1{\leqslant}\mathrm{Tr}({{\Phi}}_{{\mathcal{E}}^{qi}{\circ}\mathcal{E}}){\leqslant}{d}^{2}$ for the composition of any quantum channel and its quasi-inverse. The later is, however, a consequence of equation (37) and the fact that fully entangled fraction of the Choi matrix is greater than 1/d.

Taking into account the upper and lower bounds given in (37) and the general theorems of the previous section, in this section we will consider a few classes of examples, as the optimization problem (29) 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

$$\begin{equation}\mathcal{E}(\rho )=\sum\limits _{\alpha =1}^{r}{q}_{\alpha }{V}_{\alpha }\rho {V}_{\alpha }^{{\dagger}},\end{equation} \tag{ 44 }$$

where ${\left\{{V}_{\alpha }\right\}}_{\alpha =1}^{r}$ 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 ${\mathcal{V}}_{m}^{{\dagger}}:\rho \rightarrow {V}_{m}^{{\dagger}}\rho {V}_{m}$, where V_m corresponds to the largest weight q_m in the mixture (44). To see this note that the Choi matrix of this channel reads

$$\begin{equation}{C}_{\mathcal{E}}=\sum\limits _{\alpha }\;{q}_{\alpha }\vert {V}_{\alpha }\rangle \langle {V}_{\alpha }\vert ,\end{equation} \tag{ 45 }$$

where we have used the correspondence (2). In view of the orthogonality of the vectors |V_α ⟩, this is then the spectral decomposition of the Choi matrix ${C}_{\mathcal{E}}$ with eigenvalues equal to p_α = dq_α (note that $\left\vert {V}_{\alpha }\right\rangle$ is not normalized). Let q_m = p_m /d be the largest of coefficients in (44). In view of the diagonal nature of the Choi matrix, p_m is the largest eigenvalue of the Choi matrix. Moreover, by taking |β⟩ = |V_m ⟩, one sees that f = p_m . So the upper and lower bounds of (37) coincide and we find the quasi-inverse is the unitary map induced by ${V}_{m}^{{\dagger}}$, i.e. ${\mathcal{E}}^{qi}={\mathcal{E}}_{{V}_{m}^{{\dagger}}}$, with

$$\begin{equation}\overline{F}({\mathcal{E}}^{qi}{\circ}\mathcal{E})=\frac{{p}_{m}+1}{d+1}.\end{equation} \tag{ 46 }$$

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

$$\begin{equation}\mathcal{E}(\rho )=\sum\limits _{\alpha =1}^{q}{X}_{\alpha }\rho {X}_{\alpha }^{{\dagger}},\end{equation} \tag{ 47 }$$

where

$$\begin{equation}\sum\limits _{\alpha =1}^{q}{X}_{\alpha }^{{\dagger}}{X}_{\alpha }=\sum\limits _{\alpha =1}^{q}{X}_{\alpha }{X}_{\alpha }^{{\dagger}}={I}_{d},\quad \mathrm{a}\mathrm{n}\mathrm{d}\quad \mathrm{Tr}\left({X}_{\alpha }^{{\dagger}}{X}_{\beta }\right)=\frac{d}{q}{\delta }_{\alpha \beta }.\end{equation} \tag{ 48 }$$

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 $\mathcal{E}$ (47) is

$$\begin{equation}{C}_{\mathcal{E}}=\sum\limits _{\alpha }\left\vert {X}_{\alpha }\right\rangle \left\langle {X}_{\alpha }\right\vert ,\end{equation} \tag{ 49 }$$

where $\left\vert {X}_{\alpha }\right\rangle$ is based on the correspondence of equation (2) and it fulfills $\left\langle {X}_{\alpha }\right\vert {X}_{\beta }\rangle =\frac{d}{q}{\delta }_{\alpha \beta }$. 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

$$\begin{equation}{{\Phi}}_{\mathcal{E}}=\sum\limits _{\alpha =1}^{q}{X}_{\alpha }\otimes {X}_{\alpha }^{{\ast}}.\end{equation} \tag{ 50 }$$

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

$$\begin{equation}\enspace \mathrm{Tr}\left({{\Phi}}_{\mathcal{E}}^{{\dagger}}{{\Phi}}_{\mathcal{E}}\right)=\sum\limits _{\alpha ,\beta =1}^{q}\vert \mathrm{Tr}\left({X}_{\alpha }^{{\dagger}}{X}_{\beta }\right){\vert }^{2}=\sum\limits _{\alpha =1}^{q}{\left(\frac{d}{q}\right)}^{2}=\frac{{d}^{2}}{q}=d{p}_{m}.\end{equation} \tag{ 51 }$$

So by such a composition the upper bound of equation (37) is obtained, which completes the proof. □

The fidelity after correction for the channel (47) then reads $\overline{F}({\mathcal{E}}^{{\dagger}}{\circ}\mathcal{E})=\frac{d+q}{q(d+1)}$, while for the case with Tr(X_α ) = 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_α are Hermitian, $\mathcal{E}$ is its own quasi-inverse. Two explicit examples for this case are provided in what follows.

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

$$\begin{equation}\mathcal{E}(\rho )=\frac{1}{j(j+1)}\left({J}_{1}\rho {J}_{1}+{J}_{2}\rho {J}_{2}+{J}_{3}\rho {J}_{3}\right),\end{equation} \tag{ 52 }$$

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 equation (47) with q = 3. It is worth mentioning that the LS channel (52) is an extreme point of ${\mathcal{C}}_{d}$ when d ⩾ 3 [24]. However for d = 2, the LS 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 LS channel).  Let G be a Lie group with dimension n, with Lie algebra generators A_α where 1 ⩽ α ⩽ n. Let D_μ be an irreducible unitary representation of the Lie algebra. We define the generalized LS channel as

$$\begin{equation}{\mathcal{E}}_{\mu }(\rho )=\frac{1}{{c}_{\mu }}\sum\limits _{\alpha =1}^{n}{D}_{\mu }({A}_{\alpha })\rho {D}_{\mu }{({A}_{\alpha })}^{{\dagger}},\end{equation} \tag{ 53 }$$

where c_μ is the value of the second Casimir operator in this representation

$$\begin{equation}{c}_{\mu }I=\sum\limits _{\alpha =1}^{n}{D}_{\mu }{({A}_{\alpha })}^{2}.\end{equation} \tag{ 54 }$$

The generators can be made orthogonal so that

$$\begin{equation}\mathrm{Tr}\left[{D}_{\mu }({A}_{\alpha }){D}_{\mu }({A}_{\beta })\right]=\frac{{c}_{\mu }{d}_{\mu }}{n}.\end{equation} \tag{ 55 }$$

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 ${\mathcal{E}}_{w}^{\text{td}}(\rho )=(1-w){\rho }^{\mathrm{T}}+\frac{w}{d}\enspace \mathrm{Tr}(\rho ){I}_{d}$, where $\frac{d}{d+1}{\leqslant}w{\leqslant}\frac{d}{d-1}$ to satisfy complete positivity. The superoperator of this channel is given by:

$$\begin{equation}{{\Phi}}_{w}^{\text{td}}=(1-w)S+w\left\vert {\phi }^{+}\right\rangle \left\langle {\phi }^{+}\right\vert ,\end{equation} \tag{ 56 }$$

and the Choi matrix is equal to

$$\begin{equation}{C}_{w}^{\text{td}}=(1-w)S+\frac{w}{d}I\otimes I.\end{equation} \tag{ 57 }$$

S is the swap operator introduced in the proof of theorem 3. It is a Hermitian unitary so its eigenvalues are ±1. Thus, the largest eigenvalue of ${C}_{w}^{\text{td}}$ is

$$\begin{equation}\begin{cases}_{m}=1-w+\frac{w}{d},\quad \frac{d}{d+1}{\leqslant}w{\leqslant}1,\quad \\ {p}_{m}=w-1+\frac{w}{d},\quad 1{\leqslant}w{\leqslant}\frac{d}{d-1}.\quad \end{cases}\end{equation} \tag{ 58 }$$

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 $\overline{F}({\mathcal{E}}_{+}{\circ}{\mathcal{E}}_{w}^{\text{td}})$ saturates the upper bound of equation (37) when $\frac{d}{d+1}{\leqslant}w{\leqslant}1$, so ${\mathcal{E}}_{+}$ is the quasi-inverse on this domain. However, we find in this region the equality $\overline{F}({\mathcal{E}}_{+}{\circ}{\mathcal{E}}_{w}^{\text{td}})=\overline{F}({\mathcal{E}}_{w}^{\text{td}})$ holds. This implies that for this range of w the transverse-depolarizing channel is non-correctable. On the other hand, for $1{\leqslant}w{\leqslant}\frac{d}{d-1}$, the upper bound of equation (37) is achievable by ${\mathcal{E}}_{-}{\circ}{\mathcal{E}}_{w}^{\text{td}}$ 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 figure 4.

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

$$\begin{equation}{{\Phi}}_{q}^{d}=(1-q)I\otimes I+q\left\vert {\phi }^{+}\right\rangle \left\langle {\phi }^{+}\right\vert ,\qquad {C}_{q}^{d}=d(1-q)\left\vert {\phi }^{+}\right\rangle \left\langle {\phi }^{+}\right\vert +\frac{q}{d}I\otimes I.\end{equation} \tag{ 59 }$$

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.

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Example 4 (Covariant channels).  A quantum channel $\mathcal{E}:L({H}_{d})\rightarrow L({H}_{d})$ is called covariant with respect to a group G, if the following property holds:

$$\begin{equation}\mathcal{E}(U(g)\rho {U}^{{\dagger}}(g))=V(g)\mathcal{E}(\rho ){V}^{{\dagger}}(g),\quad \forall \enspace g\in G.\end{equation} \tag{ 60 }$$

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

$$\begin{equation}{{\Phi}}_{\mathcal{E}}U(g)\otimes {U}^{{\ast}}(g)=V(g)\otimes {V}^{{\ast}}(g){{\Phi}}_{\mathcal{E}}.\end{equation} \tag{ 61 }$$

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

$$\begin{align}{\mathcal{E}}^{qi}& =\underset{{\mathcal{E}}^{\prime }}{\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{m}\mathrm{a}\mathrm{x}}\enspace \mathrm{Tr}({{\Phi}}_{{\mathcal{E}}^{\prime }}{{\Phi}}_{\mathcal{E}})\\ & =\underset{{\mathcal{E}}^{\prime }}{\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{m}\mathrm{a}\mathrm{x}}\enspace \mathrm{Tr}\left({{\Phi}}_{{\mathcal{E}}^{\prime }}({V}^{{\dagger}}(g)\otimes {V}^{\text{T}}(g)){{\Phi}}_{\mathcal{E}}(U(g)\otimes {U}^{{\ast}}(g))\right)\\ & =\underset{{\mathcal{E}}^{\prime }}{\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{m}\mathrm{a}\mathrm{x}}\left((U(g)\otimes {U}^{{\ast}}(g)){{\Phi}}_{{\mathcal{E}}^{\prime }}({V}^{{\dagger}}(g)\otimes {V}^{\text{T}}(g)){{\Phi}}_{\mathcal{E}}\left. \right)\right),\end{align} \tag{ 62 }$$

from which we obtain

$$\begin{equation}{{\Phi}}_{{\mathcal{E}}^{qi}}V(g)\otimes {V}^{{\ast}}(g)=U(g)\otimes {U}^{{\ast}}(g){{\Phi}}_{{\mathcal{E}}^{qi}}.\end{equation} \tag{ 63 }$$

Equivalently this means that

$$\begin{equation}{\mathcal{E}}^{qi}(V(g)\rho {V}^{{\dagger}}(g))=U(g){\mathcal{E}}^{qi}(\rho ){U}^{{\dagger}}(g),\quad \forall \enspace g\in G.\end{equation} \tag{ 64 }$$

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

$$\begin{equation}\mathcal{E}(\rho )=\sum\limits _{k}\;{p}_{k}{U}_{k}\rho {U}_{k}^{{\dagger}}\end{equation} \tag{ 65 }$$

when U_k 's are orthogonal to each other, is the unitary channel ${\mathcal{E}}^{qi}(\rho )={U}_{\mathrm{max}}^{{\dagger}}\rho {U}_{\mathrm{max}}$, where U_max is the unitary corresponding to the maximum probability p_max in (65). 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 (65) when [U_k , U_l ] = 0  ∀k. In the basis in which all the unitaries are diagonal, ${U}_{k}=\mathrm{diag}({\text{e}}^{\text{i}{\theta }_{1}^{(k)}},{\text{e}}^{\text{i}{\theta }_{2}^{(k)}},\dots ,{\text{e}}^{\text{i}{\theta }_{d}^{(k)}})$, the superoperator of the channel reads

$$\begin{equation}{\Phi}=\sum\limits _{k=1}^{M}{p}_{k}{U}_{k}\otimes {U}_{k}^{{\ast}}=\sum\limits _{i,j}^{d}\sum\limits _{k=1}^{M}{p}_{k}\enspace {\text{e}}^{\text{i}({\theta }_{i}^{(k)}-{\theta }_{j}^{(k)})}\vert i,j\rangle \langle i,j\vert =\sum\limits _{i,j}^{d}{w}_{ij}\vert i,j\rangle \langle i,j\vert ,\end{equation} \tag{ 66 }$$

where

$$\begin{equation}{w}_{ij}=\langle {\text{e}}^{\text{i}{\theta }_{ij}}\rangle {:=}\sum\limits _{k}\;{p}_{k}\enspace {\text{e}}^{\text{i}({\theta }_{i}^{(k)}-{\theta }_{j}^{(k)})}.\end{equation} \tag{ 67 }$$

Let G be the unitary group that consists of all diagonal unitary matrices in this basis ⁴ . Since U_k 's commute with any diagonal matrix, the channel $\mathcal{E}=\left\{{U}_{k}\right\}$ is G-covariant.

$$\begin{equation}U\mathcal{E}(\rho ){U}^{{\dagger}}=\mathcal{E}(U\rho {U}^{{\dagger}})\quad \forall \enspace \rho ,\quad \mathrm{a}\mathrm{n}\mathrm{d}\quad \forall \enspace U\in G.\end{equation} \tag{ 68 }$$

According to property (64), this implies that the quasi-inverse of this channel is also G-covariant. Expressed in terms of the superoperators, this means that the superoperator ${{\Phi}}_{\mathcal{E}}$ must commute with the superoperator of all unitary maps ρ ⟶ UρU^†, where U ∈ G. The superoperator of the latter is of the diagonal form

$$\begin{equation}{{\Phi}}_{U}=\sum\limits _{mn}\;{\text{e}}^{\text{i}({\phi }_{m}-{\phi }_{n})}\vert m,n\rangle \langle m,n\vert .\end{equation} \tag{ 69 }$$

Let the superoperator of the quasi-inverse be given by

$$\begin{equation}{{\Phi}}_{{\mathcal{E}}^{qi}}=\sum {{\Phi}}_{ij,kl}\vert ij\rangle \langle kl\vert .\end{equation} \tag{ 70 }$$

Equation (68) now restricts the form of this superoperator to the following simple form

$$\begin{equation}{{\Phi}}_{{\mathcal{E}}^{qi}}=\sum\limits _{i,j}\;{q}_{ij}\vert ij\rangle \langle ij\vert +{r}_{ij}\vert ii\rangle \langle jj\vert .\end{equation} \tag{ 71 }$$

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

$$\begin{equation}{C}_{{\mathcal{E}}^{qi}}=\sum\limits _{i,j}\;{r}_{ij}\vert ij\rangle \langle ij\vert +{q}_{ij}\vert ii\rangle \langle jj\vert .\end{equation} \tag{ 72 }$$

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

$$\begin{equation}F{:=}\mathrm{Tr}({{\Phi}}_{{\mathcal{E}}^{qi}}{{\Phi}}_{\mathcal{E}})=\sum\limits _{i=1}^{d}({q}_{ii}+{r}_{ii})+\sum\limits _{i\ne j=1}^{d}{q}_{ij}{w}_{ij}.\end{equation} \tag{ 73 }$$

Here we have used the equality w_ii = 1  ∀i, subject to the condition that ${C}_{{\mathcal{E}}^{qi}}$ in (72) designates the Choi matrix of a legitimate quantum channel, i.e. it is a positive matrix with partial trace equal to the identity

$$\begin{equation}{C}_{{\mathcal{E}}^{qi}}{\geqslant}0\quad \mathrm{a}\mathrm{n}\mathrm{d}\quad {\mathrm{Tr}}_{1}\enspace {C}_{{\mathcal{E}}^{qi}}={I}_{d}.\end{equation} \tag{ 74 }$$

In view of the block-diagonal form of the Choi matrix, it turns out that its eigenvalues are of the form {r_i≠j } ∪ Eigenvalues  of  Y, where Y is the d-dimensional matrix

$$\begin{equation}Y=\sum\limits _{i,j=1}{q}_{ij}\vert i\rangle \langle j\vert +\sum\limits _{i}\;{r}_{ii}\vert i\rangle \langle i\vert .\end{equation} \tag{ 75 }$$

The second condition in (74) leads to the following set of equalities

$$\begin{equation}{q}_{ii}+\sum\limits _{j}\;{r}_{ij}=1\quad \forall \enspace i,\end{equation} \tag{ 76 }$$

which can be rewritten as

$$\begin{equation}{q}_{ii}+{r}_{ii}+\sum\limits _{j\ne i}\;{r}_{ij}=1\quad \forall \enspace i.\end{equation} \tag{ 77 }$$

Note that r_i≠j , being the eigenvalues of the Choi matrix are non-negative. Therefore in order to maximize the right-hand side of (73), we can take all of them to be zero, reducing (77) to

$$\begin{equation}{q}_{ii}+{r}_{ii}=1\quad \forall \enspace i,\end{equation} \tag{ 78 }$$

which further simplifies the expression (73) and reduces our problem to maximization of the expression

$$\begin{equation}F=d+\sum\limits _{i\ne j=1}^{d}{q}_{ij}{w}_{ij}\equiv \mathrm{Tr}({X}^{\mathrm{T}}W),\end{equation} \tag{ 79 }$$

subject to the positivity of the following matrix

$$\begin{equation}X=I+\sum\limits _{i\ne j}{q}_{ij}\vert i,j\rangle \langle i,j\vert .\end{equation} \tag{ 80 }$$

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

In general, 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 two dimensional unital channels are unitary maps. Here we show that for channels defined by the Choi matrix (72), the extreme points are unitary maps if d ⩽ 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 Ω is depicted in figure 5. In this figure X₀ and ${X}_{0}^{\prime }$ are extreme points while Y₀ is not. We note that any line (no matter how short) passing through an extreme point like X₀ or ${X}_{0}^{\prime }$ contains points which do not belong to Ω, while for the non-extreme point Y₀, there exists a sufficiently short line (namely the one lying on the edge) which lies entirely in Ω. We present this more formally in the following statement.

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Definition 2. Let Y₀ be a non-extreme point of Ω. Then there is a $Z\in \mathcal{S}$, and > 0, such that Y₀ + t(Z − Y₀) ∈ Ω for all t ∈ (−, ). We call Z a witness of non-extremality of Y₀.

This definition implies its equivalent form.

Lemma 1. Let X₀ be an extreme point of Ω and let $Z\in \mathcal{S}$ be an arbitrary element in $\mathcal{S}$ such that for all t ∈ (−, ), ${X}_{0}+t(Z-{X}_{0})\in \mathcal{S}$. Then Z = X₀.

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 Ω is bounded as $r{\leqslant}\sqrt{{d}^{2}-m}$.

Proof. Let Y₀ be any element Ω with rank r. We show that if $r{ >}\sqrt{{d}^{2}-m}$, then we can always find an element Z ∈ Ω different from Y₀ such that the sufficiently short line segment Y₀ + t(Z − Y₀) belongs entirely to Ω. This shows that such points cannot be extreme points of Ω. To this end, let us expand Y₀ in its eigenbasis as

$$\begin{equation}{Y}_{0}=\sum\limits _{\alpha =1}^{r}{\lambda }_{\alpha }\vert {u}_{\alpha }\rangle \langle {u}_{\alpha }\vert ,\end{equation} \tag{ 81 }$$

where λ_α > 0  ∀α. We choose $Z-{Y}_{0}\in {\mathbb{M}}_{d}$ in the form

$$\begin{equation}Z-{Y}_{0}=\sum\limits _{\alpha ,\beta }{z}_{\alpha \beta }\vert {u}_{\alpha }\rangle \langle {u}_{\beta }\vert ,\end{equation} \tag{ 82 }$$

and let t be so small that Y₀ + t(Z − Y₀) ⩾ 0. The only other requirement that is needed for this matrix to belong to Ω is to satisfy d² − m linear homogeneous equations which define the subspace Ω. This is a system of d² − m linear homogeneous equations on r² variables and if d² − m < r², it has always a non-zero solution. This means that the point Y₀ is not an extreme point of the set Ω. 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 (80) which is a subset of a d² − d dimensional space, have rank $r{< }\sqrt{d}$. This means that for d = 3, the extreme points of the set (80) 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 (78), the superoperator is a diagonal matrix given by ${{\Phi}}_{{\mathcal{E}}^{qi}}=\mathrm{diag}[1,{q}_{12},{q}_{21},1,{q}_{23},{q}_{31},{q}_{32},1]$ and its Choi matrix is as follows:

$$\begin{equation}{C}_{{\mathcal{E}}^{qi}}=\left(\begin{matrix}\hfill 1\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill {q}_{12}\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill {q}_{13}\hfill \\ \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill \\ \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill \\ \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill \\ \hfill {q}_{21}\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill 1\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill {q}_{23}\hfill \\ \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill \\ \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill \\ \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill \\ \hfill {q}_{31}\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill {q}_{32}\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill 1\hfill \end{matrix}\right).\end{equation} \tag{ 83 }$$

The Hermitian matrices W and X, which appear in equation (79), are now three dimensional and take the form

$$\begin{equation}X=\left(\begin{matrix}\hfill 1\hfill & \hfill {q}_{12}\hfill & \hfill {q}_{13}\hfill \\ \hfill {q}_{21}\hfill & \hfill 1\hfill & \hfill {q}_{23}\hfill \\ \hfill {q}_{31}\hfill & \hfill {q}_{32}\hfill & \hfill 1\hfill \end{matrix}\right),\quad \mathrm{a}\mathrm{n}\mathrm{d}\quad W=\left(\begin{matrix}\hfill 1\hfill & \hfill {w}_{12}\hfill & \hfill {w}_{13}\hfill \\ \hfill {w}_{21}\hfill & \hfill 1\hfill & \hfill {w}_{23}\hfill \\ \hfill {w}_{31}\hfill & \hfill {w}_{32}\hfill & \hfill 1\hfill \end{matrix}\right).\end{equation} \tag{ 84 }$$

Having proved that the matrix X₀ which maximizes the expression (79) is of unit rank, we can write it as X = |ϕ⟩⟨ϕ|, where $\vert \phi \rangle ={\left(\begin{matrix}\hfill {q}_{1}\hfill & \hfill {q}_{2}\hfill & \hfill {q}_{3}\hfill \end{matrix}\right)}^{\mathrm{T}}$ maximizes Tr(W|ϕ⟩⟨ϕ|), 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 ${\mathcal{E}}^{qi}(\rho )=U\rho {U}^{{\dagger}}$, where the unitary operator U is given by $U=\mathrm{diag}\left({q}_{1},{q}_{2},{q}_{3}\right)$. In view of (29) and (79), the final average fidelity becomes

$$\begin{equation}\overline{F}=\frac{1}{4}\left(1+\frac{1}{3}\enspace \mathrm{Tr}({{\Phi}}_{{\mathcal{E}}^{qi}}{{\Phi}}_{\mathcal{E}})\right)=\frac{1}{4}\left(1+\frac{1}{3}\enspace {\text{max}}_{\vert \phi \rangle }\langle \phi \vert W\vert \phi \rangle \right).\end{equation} \tag{ 85 }$$

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

$$\begin{equation}\mathcal{E}(\rho )=\int \mathrm{d}\tau \enspace f(\tau )U(\tau )\rho {U}^{{\dagger}}(\tau ),\quad U(\tau )=\left(\begin{matrix}\hfill {\text{e}}^{\text{i}\tau }\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill 1\hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill {\text{e}}^{-\text{i}\tau }\hfill \end{matrix}\right),\end{equation} \tag{ 86 }$$

for some distribution f(τ). For this channel we have

$$\begin{equation}{w}_{12}={w}_{23}=\langle {\text{e}}^{\text{i}\tau }\rangle ,\qquad {w}_{13}=\langle {\text{e}}^{2i\tau }\rangle ,\end{equation} \tag{ 87 }$$

where the matrix W is Hermitian with unit diagonal entries and the averages are taken with respect to the distribution function f(τ). The vector |ϕ⟩ is then of the form $\vert \phi \rangle ={\left(\begin{matrix}\hfill {\text{e}}^{\text{i}{\tau }_{m}}\hfill & \hfill 1\hfill & \hfill {\text{e}}^{-\text{i}{\tau }_{m}}\hfill \end{matrix}\right)}^{\mathrm{T}}$, where τ_m is chosen to maximize ⟨ϕ|W|ϕ⟩ which is

$$\begin{equation}\langle \phi \vert W\vert \phi \rangle =\mathrm{R}\mathrm{e}\left[2\langle {\text{e}}^{\text{i}\tau }\rangle {\text{e}}^{-\text{i}{\tau }_{m}}+\langle {\text{e}}^{2\text{i}\tau }\rangle {\text{e}}^{-2\text{i}{\tau }_{m}}\right].\end{equation} \tag{ 88 }$$

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

$$\begin{equation}V=\left(\begin{matrix}\hfill {\text{e}}^{\text{i}{\tau }_{m}}\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill 1\hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill {\text{e}}^{-\text{i}{\tau }_{m}}\hfill \end{matrix}\right),\end{equation} \tag{ 89 }$$

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

$$\begin{equation}\overline{F}=\frac{3+\left\langle {\left(1+2\enspace \mathrm{cos}(\tau -{\tau }_{m})\right)}^{2}\right\rangle }{12}.\end{equation} \tag{ 90 }$$

Figure 6 shows the average fidelity and the increase in average fidelity for a simple discrete distribution P(τ = 0) = 1 − p and P(τ = τ₀) = p.

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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 p whose entries are non-negative and add up to one. The set of all probability vectors of length d, forms the simplex Δ_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 $\left\vert i\right\rangle$ for i ∈ {1, ..., d} denotes the pure probability vector whose all components are zero but the ith 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 $\left\{\left\vert i\right\rangle \right\}$, i.e. $\left\vert \mathbf{p}\right\rangle ={\sum }_{i}\;{p}_{i}\left\vert i\right\rangle$.

A classical channel is represented by a stochastic transition matrix T of order d with non-negative elements where 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 (d² − 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 ρ_diag = diag(p). Let us order the generators of SU(d) matrices used in (5) in such a way that Γ_i are diagonal for i = 1, ..., d − 1. Then any diagonal matrix ρ_diag, representing the classical state p, is represented in the Bloch form (4), where the Bloch vector t_cl has now only d − 1 components.

Using such a Bloch representation for point p of the probability simplex Δ_d one can represent the action of an arbitrary stochastic matrix in the form [44],

$$\begin{equation}T=\left(\begin{matrix}\hfill 1\hfill & \hfill 0\hfill \\ \hfill \sqrt{d-1}\enspace {\mathbf{t}}_{\text{cl}}\hfill & \hfill {M}_{\text{cl}}\hfill \end{matrix}\right).\end{equation} \tag{ 91 }$$

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 d² − 1, while the classical translation vector t_cl consists of d − 1 components of the original translation vector t of size d² − 1. Hence in the Bloch basis the classical transition matrix T forms a block of a matrix representing a quantum operation ${{\Phi}}_{\mathcal{E}}$, which decoheres to it [28].

Let us mention explicitly two distinguished classical maps. The first is the permutation map P_α where P_α |i⟩ = |α_i ⟩, where 1 ⩽ i ⩽ d, while α_i denote permutations of {1, 2, ..., d}. The next one is the flat, van der Wearden matrix, denoted by T_*, which sends all input states to the uniform state. It implies ${({T}_{{\ast}})}_{ij}=1/d$ for any 1 ⩽ i, j ⩽ 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 |p⟩ and |q⟩ which is defined as

$$\begin{equation}\langle \mathbf{p}\vert \mathbf{q}\rangle {:=}{\left(\sum\limits _{i=1}^{d}\sqrt{{p}_{i}{q}_{i}}\right)}^{2}.\end{equation} \tag{ 92 }$$

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

$$\begin{equation}\overline{\mathcal{F}}(T){:=}\frac{1}{d}\sum\limits _{i=1}^{d}F(\vert i\rangle ,T\vert i\rangle )=\frac{1}{d}\enspace \mathrm{Tr}(T).\end{equation} \tag{ 93 }$$

Under these assumptions, the quasi-inversion is the channel (the stochastic matrix) increasing equation (93) as much as possible:

$$\begin{equation}\mathrm{Tr}({T}^{qi}T){\geqslant}\mathrm{Tr}({T}^{\prime }T),\quad \forall \enspace {T}^{\prime }\in {\mathcal{S}}_{d}.\end{equation} \tag{ 94 }$$

To emphasize even further similarity to the quantum case, consider the Bloch representation (91) of the classical map T involving its distortion matrix M_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,

$$\begin{equation}\overline{\mathcal{F}}({T}^{qi}T)=\underset{{M}_{\text{cl}}^{\prime }}{\mathrm{max}}\enspace \frac{1}{d}\left(1+\mathrm{Tr}{M}_{\text{cl}}^{\prime }{M}_{\text{cl}}\right),\end{equation} \tag{ 95 }$$

which is in analogy to equation (29). However, the difference in prefactor with respect to equation (29) 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 (27) and (28) 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 $T={\left({\mathbf{t}}_{1},{\mathbf{t}}_{2},\dots ,{\mathbf{t}}_{n}\right)}^{\mathrm{T}}$ where ${\mathbf{t}}_{i}^{\mathrm{T}}$ denotes the ith row as a vector. Denote its quasi-inverse as a matrix T^qi = (x₁, x₂, ..., x_n ), where x_i is the ith column as a vector. The aim is to maximize Tr(TT^qi ) = ∑_i t_i ⋅ x_i . We can maximize this sum if we maximize each inner product t_i ⋅ x_i independently. Since none of the components of the vectors x_i can be larger than one, the maximization is achieved if we choose each x_i = (0, 0, ..., 1, 0, 0) such that 1 stands in the position of maximum component of t_i . This proves the theorem. □

For example assume a general two dimensional stochastic matrix parameterized as

$$\begin{equation}T=\left(\begin{matrix}\hfill 1-x\hfill & \hfill y\hfill \\ \hfill x\hfill & \hfill 1-y\hfill \end{matrix}\right),\quad 0{\leqslant}x,y{\leqslant}1.\end{equation} \tag{ 96 }$$

In this case, the quasi-inverse is either I, σ_x , or a mixture of these two for x + y < 1, x + y > 1 and x + y = 1, respectively. This is depicted in figure 7. In higher dimensions, however, more options are possible.

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Example 6. As examples, the quasi-inverse of the stochastic matrices

$$\begin{equation}{T}_{1}=\frac{1}{8}\left(\begin{matrix}\hfill 2\hfill & \hfill 6\hfill & \hfill 2\hfill \\ \hfill 4\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 5\hfill \end{matrix}\right),\quad \mathrm{a}\mathrm{n}\mathrm{d}\quad {T}_{2}=\frac{1}{48}\left(\begin{matrix}\hfill 12\hfill & \hfill 6\hfill & \hfill 3\hfill \\ \hfill 4\hfill & \hfill 36\hfill & \hfill 42\hfill \\ \hfill 32\hfill & \hfill 6\hfill & \hfill 3\hfill \end{matrix}\right)\end{equation} \tag{ 97 }$$

are given by

$$\begin{equation}{T}_{1}^{qi}=\left(\begin{matrix}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{matrix}\right),\quad \mathrm{a}\mathrm{n}\mathrm{d}\quad {T}_{2}^{qi}=\left(\begin{matrix}\hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \end{matrix}\right).\end{equation} \tag{ 98 }$$

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 T₁ and T₂ increase from 1/3 and 5/18 to 35/72 and 43/72, respectively.

Corollary 3. If T denotes a stochastic matrix for which ∀i, j: T_ii ⩾ 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

$$\begin{equation}T=\frac{1}{24}\left(\begin{matrix}\hfill 8\hfill & \hfill 3\hfill & \hfill 8\hfill \\ \hfill 12\hfill & \hfill 6\hfill & \hfill 6\hfill \\ \hfill 4\hfill & \hfill 15\hfill & \hfill 10\hfill \end{matrix}\right)\end{equation} \tag{ 99 }$$

has as its quasi-inverse

$$\begin{equation}{T}^{qi}=\lambda \left(\begin{matrix}\hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{matrix}\right)+(1-\lambda )\left(\begin{matrix}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{matrix}\right)=\left(\begin{matrix}\hfill \lambda \hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 1-\lambda \hfill & \hfill 0\hfill & \hfill 0\hfill \end{matrix}\right).\end{equation} \tag{ 100 }$$

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

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^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 ρ into its diagonal $\mathcal{D}(\rho )={\rho }_{d}=\sum {\rho }_{ii}\left\vert i\right\rangle \left\langle i\right\vert$, 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 Δ, 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

$$\begin{equation}{({T}_{\mathcal{E}})}_{ij}=\left\langle i\right\vert \mathcal{E}(\left\vert j\right\rangle \left\langle j\right\vert )\left\vert i\right\rangle ={({{\Phi}}_{\mathcal{E}})}_{ii,jj}={({C}_{\mathcal{E}})}_{ij,ij}.\end{equation} \tag{ 101 }$$

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 (91) forms a block of the quantum distortion matrix M of size d² − 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_α }, then ${T}_{\mathcal{E}}=\sum {K}_{\alpha }\odot {K}_{\alpha }^{{\ast}}$ where ⊙ 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^{iθ n⋅ σ} , where θ and n depend on the channel. On the other hand, as shown in figure 7, the quasi-inverse of any two-dimensional classical map is either identity I or the permutation σ_x , which arise due to super-decoherence of a small subset of the unitary channels [45].

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 T₁ and T₂ which are related by two arbitrary permutations P_α and P_β in the following way T₂ = P_α ◦T₁◦P_β . Let us call them permutationaly equivalent stochastic matrices. Pursuing the same approach adopted in obtaining equation (36), and noting that the set of stochastic matrices, ${\mathcal{S}}_{d}$, is invariant under permutations, we get

$$\begin{equation}{T}_{2}^{qi}={P}_{\beta }^{-1}{\circ}{T}_{1}^{qi}{\circ}{P}_{\alpha }^{-1}.\end{equation} \tag{ 102 }$$

Moreover, T₁ and T₂ 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

$$\begin{equation}\mathrm{Tr}({P}_{m}^{-1}{P}_{n})=d{\delta }_{n,m}.\end{equation} \tag{ 103 }$$

Equation (103) indicates that

$$\begin{equation}\forall \enspace \enspace k\;{P}_{m}\vert k\rangle \ne {P}_{n}\vert k\rangle ,\quad \text{if}\enspace m\ne n.\end{equation} \tag{ 104 }$$

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

$$\begin{equation}\langle j\vert \sum\limits _{m}{P}_{m}\vert k\rangle =0,1\quad \forall \enspace j,\enspace k.\end{equation} \tag{ 105 }$$

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, ..., d − 1}, where $X={\sum }_{n=0}^{d-1}\vert n+1\rangle \langle n\vert$ is the full-cycle permutation, or any subset thereof. By definition we have P_m |k⟩ = |k + m⟩ which clearly satisfies (105) as ∑_m ⟨j|P_m |k⟩ = ∑_m δ_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:

$$\begin{equation}{P}_{1}=\left(\begin{matrix}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 3\hfill \end{matrix}\right),\quad {P}_{2}=\left(\begin{matrix}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill 2\hfill & \hfill 1\hfill \end{matrix}\right),\quad {P}_{3}=\left(\begin{matrix}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill 1\hfill & \hfill 2\hfill \end{matrix}\right),\end{equation} \tag{ 106 }$$

with matrix representations

$$\begin{equation}{P}_{1}=\left(\begin{matrix}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{matrix}\right),\quad {P}_{2}=\left(\begin{matrix}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{matrix}\right),\quad {P}_{3}=\left(\begin{matrix}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{matrix}\right).\end{equation} \tag{ 107 }$$

One can see that P₁ and P₂ are orthogonal while P₁ and P₃ are not. Also one can see that

$$\begin{equation}{P}_{1}+{P}_{2}=\left(\begin{matrix}\hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \end{matrix}\right),\end{equation} \tag{ 108 }$$

satisfying the condition (105) while

$$\begin{equation}{P}_{1}+{P}_{3}=\left(\begin{matrix}\hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \end{matrix}\right),\end{equation} \tag{ 109 }$$

violates (105). Furthermore one can check that in the group S₃ with generators ${\sigma }_{1}=\left(\begin{matrix}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 3\hfill \end{matrix}\right)$ and ${\sigma }_{2}=\left(\begin{matrix}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill 2\hfill \end{matrix}\right)$, each of the two sets of even and odd permutations, respectively given by (I, σ₁ σ₂, σ₂ σ₁) and (σ₁, σ₂, σ₁ σ₂ σ₁) comprise orthogonal permutations.

Consider now a bi-stochastic matrix of the form

$$\begin{equation}B=\sum {\lambda }_{m}{P}_{m},\end{equation} \tag{ 110 }$$

where {P_m } are orthogonal. In what follows we will show the quasi-inverse of any such bi-stochastic matrix is ${P}_{{m}_{0}}^{-1}$ where m₀ refers to the index of the greatest coefficient λ_m . To see this we note that

$$\begin{align}\overline{\mathcal{F}}(T{\circ}B)& =\frac{\mathrm{Tr}(TB)}{d}=\frac{{\sum }_{m}\;{\lambda }_{m}\enspace \mathrm{Tr}(T{P}_{m})}{d}\\ & {\leqslant}\frac{{\lambda }_{{m}_{0}}\enspace \mathrm{Tr}\left(T{\sum }_{m}\;{P}_{m}\right)}{d}=\frac{{\lambda }_{{m}_{0}}{\sum }_{k,j}\langle k\vert T\vert j\rangle \langle j\vert {\sum }_{m}\;{P}_{m}\vert k\rangle }{d}.\end{align} \tag{ 111 }$$

Using equation (105) for orthogonal permutations, we find

$$\begin{equation}\overline{\mathcal{F}}(T{\circ}B){\leqslant}\frac{{\lambda }_{{m}_{0}}{\sum }_{k,j}\langle k\vert T\vert j\rangle }{d}={\lambda }_{{m}_{0}},\end{equation} \tag{ 112 }$$

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 ${T}^{qi}={P}_{{m}_{0}}^{-1}$ 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

$$\begin{equation}{P}_{1}=I\oplus I\oplus I,\quad {P}_{2}={\sigma }_{x}\oplus {\sigma }_{x}\oplus I,\quad {P}_{3}={\sigma }_{x}\oplus I\oplus {\sigma }_{x},\end{equation} \tag{ 113 }$$

where $I=\left(\begin{matrix}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{matrix}\right)$ and ${\sigma }_{x}=\left(\begin{matrix}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{matrix}\right)$. These permutations are obviously non-orthogonal in the sense that $\mathrm{Tr}({P}_{i}{P}_{j}^{\mathrm{T}})\ne 0$. Consider now the following convex combination

$$\begin{equation}P={\lambda }_{1}{P}_{1}+{\lambda }_{2}{P}_{2}+{\lambda }_{3}{P}_{3},\end{equation} \tag{ 114 }$$

where λ₁ + λ₂ + λ₃ = 1 and all ${\lambda }_{i}{< }\frac{1}{2}$. In explicit form this permutation matrix is given by

$$\begin{equation}P=\left(\begin{matrix}\hfill {\lambda }_{1}\hfill & \hfill {\lambda }_{2}+{\lambda }_{3}\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill \\ \hfill {\lambda }_{2}+{\lambda }_{3}\hfill & \hfill {\lambda }_{1}\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill \\ \hfill .\hfill & \hfill .\hfill & \hfill {\lambda }_{1}+{\lambda }_{3}\hfill & \hfill {\lambda }_{2}\hfill & \hfill .\hfill & \hfill .\hfill \\ \hfill .\hfill & \hfill .\hfill & \hfill {\lambda }_{2}\hfill & \hfill {\lambda }_{1}+{\lambda }_{3}\hfill & \hfill .\hfill & \hfill .\hfill \\ \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill {\lambda }_{1}+{\lambda }_{2}\hfill & \hfill {\lambda }_{3}\hfill \\ \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill .\hfill & \hfill {\lambda }_{3}\hfill & \hfill {\lambda }_{1}+{\lambda }_{2}\hfill \end{matrix}\right).\end{equation} \tag{ 115 }$$

Since ${\lambda }_{i}{< }\frac{1}{2}$ for all i, we find ${\lambda }_{i}+{\lambda }_{j}{ >}\frac{1}{2}$ for all (i ≠ j) (due to the requirement that λ₁ + λ₂ + λ₃ = 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 = σ_x ⊕ I ⊕ I which is not equal to inverse of any of the permutations P₁, P₂ or P₃.

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 example 1 through example 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 figure 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

$$\begin{equation}\overline{P}(\mathcal{E}){:=}\int \mathrm{d}\psi \enspace \mathrm{Tr}\left[{(\mathcal{E}(\left\vert \psi \right\rangle \left\langle \psi \right\vert ))}^{2}\right],\end{equation} \tag{ 116 }$$

or the unitarity of a channel [46]

$$\begin{equation}\overline{u}(\mathcal{E}){:=}\frac{d}{d-1}\int \mathrm{d}\psi \enspace \mathrm{Tr}{\left[\mathcal{E}(\left\vert \psi \right\rangle \left\langle \psi \right\vert )-\mathcal{E}\left(\frac{I}{d}\right)\right]}^{2},\end{equation} \tag{ 117 }$$

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 [47–49].

Finally, an interesting by-product of our study is appendix C, where we have constructed special types of channels with affine maps (M, 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 $\mathcal{E}(\rho )=\vert \psi \rangle \langle \psi \vert$, where the distortion matrix M vanishes and the stretching is due only to the translation vector 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 [50–53].

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.

The data that support the findings of this study are available upon reasonable request from the authors.

In this appendix we choose an explicit representation for the matrices Γ_i belonging to L(H_d ). Let E_ij := |i⟩⟨j| be the standard basis of matrices, then the set

$$\begin{equation}{H}_{k}=\sqrt{\frac{d(d-1)}{k(k+1)}}\left(\sum\limits _{i=1}^{k}{E}_{ii}-k{E}_{k+1,k+1}\right)\quad k=1,\dots ,d-1\end{equation} \tag{ 118 }$$

and

$$\begin{equation}{X}_{ij}=\sqrt{\frac{d(d-1)}{2}}({E}_{ij}+{E}_{ji})\qquad {Y}_{ij}=\sqrt{\frac{d(d-1)}{2}}(-i{E}_{ij}+i{E}_{ji})\end{equation} \tag{ 119 }$$

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 p it is convenient to order the matrix basis in such a way that the d − 1 diagonal matrices appear first, Γ_k = H_k , for k = 1, ..., d − 1—see section 6.

A density matrix can be written as $\rho =\frac{1}{d}(I+\mathbf{r}\cdot \mathbf{\Gamma })$, where r is the generalized Bloch vector and Γ is a vector constructed by the elements Γ_i . We note, as stated after equation (6), that any pure state corresponds to unit vector r. However the converse is not true. In fact one can easily verify that a matrix like $\frac{1}{d}(I-{H}_{d-1})=\vert \mathrm{d}\rangle \langle \mathrm{d}\vert$ 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 ${S}_{{d}^{2}-2}$.

As shown in equation (20), the input–output fidelity is given by $\overline{F}(\mathcal{E})=\mathrm{Tr}(L{{\Phi}}_{\mathcal{E}})$, where L is given by

$$\begin{align}L& =\int \mathrm{d}\psi \enspace \left\vert \psi \right\rangle \left\langle \psi \right\vert \otimes \left\vert {\psi }^{{\ast}}\right\rangle \left\langle {\psi }^{{\ast}}\right\vert \\ & =\int \mathrm{d}U(U\otimes {U}^{{\ast}})\left\vert 00\right\rangle \left\langle 00\right\vert {(U\otimes {U}^{{\ast}})}^{{\dagger}}.\end{align} \tag{ 120 }$$

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

$$\begin{align}T(\rho )& =\int \mathrm{d}U(U\otimes {U}^{{\ast}})\rho {(U\otimes {U}^{{\ast}})}^{{\dagger}}\\ & =\frac{{d}^{2}}{{d}^{2}-1}\left((1-{f}_{\hspace{-1.5pt}\rho })\frac{I}{{d}^{2}}+\left({f}_{\hspace{-1.5pt}\rho }-\frac{1}{{d}^{2}}\right){P}_{+}\right),\end{align} \tag{ 121 }$$

where ${f}_{\hspace{-1.5pt}\rho }=\mathrm{Tr}\left({P}_{+}\rho \right)$ and ${P}_{+}=\left\vert {\phi }_{+}\right\rangle \left\langle {\phi }_{+}\right\vert$ is the maximally entangled state. So for the state $\left\vert 00\right\rangle \left\langle 00\right\vert$ of a d-dimensional system, one has:

$$\begin{equation}{f}_{\hspace{-1.5pt}\left\vert 00\right\rangle }=\mathrm{Tr}\left({P}_{+}\left\vert 00\right\rangle \left\langle 00\right\vert \right)=\frac{1}{d},\end{equation} \tag{ 122 }$$

and

$$\begin{equation}L=T\left(\left\vert 00\right\rangle \left\langle 00\right\vert \right)=\frac{1}{d(d+1)}I+\frac{1}{(d+1)}{P}_{+},\end{equation} \tag{ 123 }$$

which coincides with equation (22) in the text.

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

$$\begin{equation}\overline{F}(\mathcal{E})=\mathrm{Tr}({\Lambda}{{\Psi}}_{\mathcal{E}}),\end{equation} \tag{ 124 }$$

where ${{\Psi}}_{\mathcal{E}}={\sum }_{\alpha }{K}_{\alpha }\otimes {K}_{\alpha }^{{\dagger}}$ and

$$\begin{equation*}{\Lambda}=\int \mathrm{d}\psi \enspace \left\vert \psi \right\rangle \left\langle \psi \right\vert \otimes \left\vert \psi \right\rangle \left\langle \psi \right\vert \end{equation*}$$

is a symmetric Werner state

$$\begin{equation}{\Lambda}=\frac{1}{d(d+1)}(I+{P}_{\text{sym}})=\frac{1}{d(d+1)}\left(I+\sum\limits _{i,j}\vert i,j\rangle \langle j,i\vert \right).\end{equation} \tag{ 125 }$$

This will also lead to the same result as in (23).

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

$$\begin{equation}\vert \psi \rangle \langle \psi \vert =\frac{1}{d}(I+\mathbf{n}\cdot \mathbf{\Gamma }).\end{equation} \tag{ 126 }$$

From

$$\begin{equation}\mathcal{E}(\vert \psi \rangle \langle \psi \vert )=\frac{1}{d}\left(I+(M\mathbf{n}+\mathbf{t})\cdot \mathbf{\Gamma }\right).\end{equation} \tag{ 127 }$$

and equation (126), one finds

$$\begin{equation}\langle \psi \vert \mathcal{E}(\vert \psi \rangle \langle \psi \vert )\vert \psi \rangle =\frac{1}{d}\left(1+(d-1)\mathbf{n}\cdot (M\mathbf{n}+\mathbf{t})\right).\end{equation} \tag{ 128 }$$

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

$$\begin{equation}\int \mathrm{d}\mathbf{n}\enspace \enspace {n}_{i}=0,\quad \int \mathrm{d}\mathbf{n}\enspace \enspace {n}_{i}{n}_{j}=\frac{1}{({d}^{2}-1)}{\delta }_{ij}\end{equation} \tag{ 129 }$$

we arrive at

$$\begin{equation}\overline{F}(\mathcal{E})=\frac{1}{d}\left(1+\frac{1}{d+1}\enspace \mathrm{Tr}\enspace M\right),\end{equation} \tag{ 130 }$$

which coincides with the result (23) which we obtained by integrating over the invariant volume dψ 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.

In two dimension, to satisfy the positivity condition, the affine matrix M of any positive map, and thus any quantum channel, fulfills M^T M ⩽ 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 P₁, P₂, Q₁ and Q₂ be orthogonal projectors of rank d₁, d₂, m₁ and m₂ respectively, with (d = d₁ + d₂ = m₁ + m₂), i.e.

$$\begin{equation*}{P}_{1}+{P}_{2}={Q}_{1}+{Q}_{2}=I,\quad {P}_{1}{P}_{2}={Q}_{1}{Q}_{2}=0.\end{equation*}$$

The following operators are Hermitian and traceless

$$\begin{equation}A{:=}\frac{{m}_{2}{Q}_{1}-{m}_{1}{Q}_{2}}{\sqrt{{m}_{1}{m}_{2}}},\quad B=\frac{{d}_{2}{P}_{1}-{d}_{1}{P}_{2}}{\sqrt{{d}_{1}{d}_{2}}},\end{equation} \tag{ 131 }$$

and we have

$$\begin{equation}\mathrm{Tr}(AB)=0,\quad \mathrm{Tr}({A}^{2})=d,\quad \mathrm{Tr}({B}^{2})=d.\end{equation} \tag{ 132 }$$

Consider now the following measure-and-prepare CPT map

$$\begin{equation}\mathcal{E}(\rho )=\frac{1}{{d}_{1}}\enspace \mathrm{Tr}({Q}_{1}\rho ){P}_{1}+\frac{1}{{d}_{2}}\enspace \mathrm{Tr}({Q}_{2}\rho ){P}_{2}.\end{equation} \tag{ 133 }$$

By expanding the projectors as

$$\begin{align}& {P}_{1}=\sum\limits _{i=1}^{{d}_{1}}\vert i\rangle \langle i\vert ,\qquad {P}_{2}=\sum\limits _{i={d}_{1}+1}^{d}\vert i\rangle \langle i\vert ,\\ & {Q}_{1}=\sum\limits _{\alpha =1}^{{m}_{1}}\vert {\varphi }_{\alpha }\rangle \langle {\varphi }_{\alpha }\vert ,\qquad {Q}_{2}=\sum\limits _{\alpha ={m}_{1}+1}^{d}\vert {\varphi }_{\alpha }\rangle \langle {\varphi }_{\alpha }\vert .\end{align} \tag{ 134 }$$

a set of Kraus operators for this map is given by

$$\begin{equation}{K}_{i,\alpha }{:=}\frac{1}{\sqrt{{d}_{1}}}\vert i\rangle \langle {\varphi }_{\alpha }\vert \quad i=1,\dots ,{d}_{1},\enspace \enspace \enspace \alpha =1,\dots ,{m}_{1}\end{equation} \tag{ 135 }$$

and

$$\begin{equation}{L}_{i,\alpha }{:=}\frac{1}{\sqrt{{d}_{2}}}\vert i\rangle \langle {\varphi }_{\alpha }\vert \quad i={d}_{1}+1,\dots ,d,\quad \alpha ={\mu }_{1}+1,\dots ,d.\end{equation} \tag{ 136 }$$

Note that

$$\begin{equation}\sum\limits _{i,\alpha }\;{K}_{i,\alpha }^{{\dagger}}{K}_{i,\alpha }+\sum\limits _{i,\alpha }\;{L}_{i,\alpha }^{{\dagger}}{L}_{i,\alpha }={I}_{d},\end{equation} \tag{ 137 }$$

but

$$\begin{equation}\sum\limits _{i,\alpha }\;{K}_{i,\alpha }{K}_{i,\alpha }^{{\dagger}}+\sum\limits _{i,\alpha }\;{L}_{i,\alpha }{L}_{i,\alpha }^{{\dagger}}=\frac{{m}_{1}}{{d}_{1}}{P}_{1}+\frac{{m}_{2}}{{d}_{2}}{P}_{2},\end{equation} \tag{ 138 }$$

hence the channel is non-unital, unless m₁ = d₁ and m₂ = d₂.

From the above relations and the fact that P₁ + P₂ = Q₁ + Q₂ = I, we find

$$\begin{equation}{P}_{1}=\frac{1}{d}({d}_{1}I+\sqrt{{d}_{1}{d}_{2}}B)\qquad {P}_{2}=\frac{1}{d}({d}_{2}I-\sqrt{{d}_{1}{d}_{2}}B)\end{equation} \tag{ 139 }$$

and

$$\begin{equation}{Q}_{1}=\frac{1}{d}({m}_{1}I+\sqrt{{m}_{1}{m}_{2}}A)\qquad {Q}_{2}=\frac{1}{d}({m}_{2}I-\sqrt{{m}_{1}{m}_{2}}A)\end{equation} \tag{ 140 }$$

Consider now a state

$$\begin{equation}\rho =\frac{1}{d}(1+xA),\end{equation} \tag{ 141 }$$

where x represents the Bloch vector of this state, since x = Tr(ρA). We will then find

$$\begin{equation}\mathcal{E}(\rho )=\frac{1}{d}(I+{x}^{\prime }B),\end{equation} \tag{ 142 }$$

where

$$\begin{equation}{x}^{\prime }=\sqrt{\frac{{m}_{1}{m}_{2}}{{d}_{1}{d}_{2}}}x+\frac{{m}_{1}{d}_{2}-{m}_{2}{d}_{1}}{d\sqrt{{d}_{1}{d}_{2}}}.\end{equation} \tag{ 143 }$$

Straightforward calculation completes the reasoning. Hence we find that if m₁ m₂ > d₁ d₂, 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 ⩾ 4, where for example we can take (m₁, m₂) = (2, 2) and (d₁, d₂) = (1, 3). Moreover we see that if the channel is unital, then x' = 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. ${\mathcal{M}}_{s}=\left\{{\mathcal{E}}_{s}\vert \enspace {M}_{s}^{\mathrm{T}}{M}_{s}{\leqslant}I\right\}$. 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 U₁ and U₂ such that M_s = (U₁ + U₂)/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 (29)

$$\begin{align}\overline{F}({\mathcal{E}}^{qi}{\circ}\mathcal{E})& =\underset{{\mathcal{E}}^{\prime }}{\mathrm{max}}\enspace \overline{F}({\mathcal{E}}^{\prime }{\circ}\mathcal{E})=\underset{{\mathcal{E}}_{s}\in {\mathcal{M}}_{s}}{\mathrm{max}}\enspace \overline{F}({\mathcal{E}}_{s}{\circ}\mathcal{E})\\ & =\underset{{M}_{s}}{\mathrm{max}}\enspace \frac{1}{d}\left(1+\frac{1}{d+1}\enspace \mathrm{Tr}\enspace {M}_{s}M\right)\\ & {\leqslant}\underset{{U}_{1},{U}_{2}}{\mathrm{max}}\enspace \frac{1}{d}\left(1+\frac{1}{2(d+1)}\enspace \mathrm{Tr}[({U}_{1}+{U}_{2})M]\right)\\ & =\frac{1}{d}\left(1+\frac{1}{d+1}\enspace \mathrm{Tr}\vert M\vert \right).\end{align} \tag{ 144 }$$

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 (144), which implies a negative answer to the question. Moreover, as the bound (144) 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 equation (131), along with the identity and d² − 3 other Hermitian and traceless operators orthogonal to A and B as the set of basis, the distortion matrix M and the translation vector t of the channel $\mathcal{E}$ in the beginning of this section are given by their entries as

$$\begin{equation}{M}_{ij}=\sqrt{\frac{{m}_{1}{m}_{2}}{{d}_{1}{d}_{2}}}{\delta }_{i2}{\delta }_{j1},\quad {t}_{i}=\frac{{m}_{1}{d}_{2}-{m}_{2}{d}_{1}}{d\sqrt{d-1}\sqrt{{d}_{1}{d}_{2}}}{\delta }_{i2}.\end{equation} \tag{ 145 }$$

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

$$\begin{equation}{M}_{ij}^{\prime }=\lambda {\delta }_{i1}{\delta }_{j2},\quad {t}_{i}^{\prime }=0.\end{equation} \tag{ 146 }$$

Note that $\mathrm{Tr}(M{M}^{\prime })=\sqrt{\frac{{m}_{1}{m}_{2}}{{d}_{1}{d}_{2}}}\lambda$. This amount is certainly larger than Tr|M'| = λ if m₁ m₂ > d₁ d₂, 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 equation (146) we have through equation (29)

$$\begin{equation}\overline{F}\left({\mathcal{E}}^{\prime qi}{\circ}{\mathcal{E}}^{\prime }\right){\geqslant}\frac{1}{d}\left[1+\frac{\lambda }{d+1}\enspace \mathrm{max}\sqrt{\frac{{m}_{1}{m}_{2}}{{d}_{1}{d}_{2}}}\right].\end{equation} \tag{ 147 }$$

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

$$\begin{equation}\overline{F}\left({\mathcal{E}}^{\prime qi}{\circ}{\mathcal{E}}^{\prime }\right){\geqslant}\frac{1}{d}\left[1+\frac{\lambda }{2\sqrt{d+1}}\right]{ >}\frac{1}{d}\left[1+\frac{\lambda }{d+1}\right]=\frac{1}{d}\left[1+\frac{\mathrm{Tr}\vert {M}^{\prime }\vert }{d+1}\right].\end{equation} \tag{ 148 }$$

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

$$\begin{equation}{{\Delta}}_{d}\equiv \left\{\mathbf{p}\in {\mathbb{R}}^{d}\vert {p}_{i}{\geqslant}0;\sum\limits _{i}\;{p}_{i}=1\right\}\end{equation} \tag{ 149 }$$

Let the probability distribution function (PDF) for any component p_i be denoted by f_d (x), that is $\mathbb{P}[x{\leqslant}{p}_{i}{\leqslant}x+\mathrm{d}x]={f}_{\hspace{-1.5pt}d}(x)\mathrm{d}x\;\forall \enspace i$. 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.

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For the general case, this function is given by

$$\begin{equation}{\phi }_{d}(x)\equiv (d-1){(1-x)}^{d-2};\quad x\in [0,1].\end{equation} \tag{ 150 }$$

It is also useful to have an expression for the corresponding cumulative distribution function (CDF)

$$\begin{equation}{{\Phi}}_{d}(x)=\mathbb{P}[{p}_{i}{\leqslant}x]={\int }_{0}^{x}{\phi }_{d}(t)\mathrm{d}t=1-{(1-x)}^{d-1}.\end{equation} \tag{ 151 }$$

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

$$\begin{equation}\langle \overline{\mathcal{F}}\rangle =\frac{1}{d}\sum\limits _{i}\langle {T}_{ii}\rangle ={\int }_{0}^{1}(d-1){(1-x)}^{d-2}x\enspace \mathrm{d}x=\frac{1}{d},\end{equation} \tag{ 152 }$$

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

$$\begin{equation}\mathrm{Var}\enspace \overline{\mathcal{F}}=\frac{1}{{d}^{2}}\sum\limits _{i}\mathrm{Var}\enspace {T}_{ii}=\frac{d-1}{{d}^{3}(d+1)}.\end{equation} \tag{ 153 }$$

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

$$\begin{equation}\overline{\mathcal{F}}({T}^{qi}{\circ}T)=\frac{1}{d}\sum\limits _{i}\;{\mathrm{max}}_{j}\enspace {T}_{ij},\end{equation} \tag{ 154 }$$

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

$$\begin{equation}\langle \overline{\mathcal{F}}({T}^{qi}{\circ}T)\rangle =\langle {p}_{m}\rangle ,\end{equation} \tag{ 155 }$$

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 CDF, i.e. the probability that the maximum values in all columns are less than x

$$\begin{equation*}{{\Psi}}_{d}(x){:=}\mathbb{P}[{p}_{m}{\leqslant}x,\forall \enspace j]={{\Phi}}_{d}{(x)}^{d}={\left[1-{(1-x)}^{d-1}\right]}^{d}\end{equation*}$$

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

$$\begin{equation*}{\psi }_{\mathit{\text{d}}}(x)=\frac{\mathrm{d}{{\Psi}}_{d}(x)}{\mathrm{d}x}=d(d-1){\left[1-{(1-x)}^{d-1}\right]}^{d-1}{(1-x)}^{d-2}\end{equation*}$$

Now it is easy to see that

$$\begin{align}\langle \overline{\mathcal{F}}({T}^{qi}{\circ}T)\rangle =\langle {p}_{m}\rangle & =1-\langle 1-{p}_{m}\rangle =1-d(d-1){\int }_{0}^{1}\mathrm{d}x{\left[x(1-{x}^{d-1})\right]}^{d-1}\\ & =1+\sum\limits _{n=0}^{d}\frac{{(-1)}^{n}(d-1)n}{(d-1)n+1}\left(\begin{matrix}\hfill d\hfill \\ \hfill n\hfill \end{matrix}\right).\end{align} \tag{ 156 }$$

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

$$\begin{equation}\langle \overline{\mathcal{F}}({T}^{qi}{\circ}T)\rangle =c{d}^{x}\sim {d}^{-1/2},\end{equation} \tag{ 157 }$$

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

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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]

$$\begin{equation}\langle {{\Phi}}_{\mathcal{E}}\rangle ={{\Phi}}_{{\ast}},\end{equation} \tag{ 158 }$$

where Φ_* is the maximally depolarizing channel, see section 2. This implies $\langle \mathrm{Tr}\enspace {{\Phi}}_{\mathcal{E}}\rangle =1$. Thus, the average amount of input–output fidelity over random channels distributed uniformly in d dimension is given by (23)

$$\begin{equation}\langle \overline{F}(\mathcal{E})\rangle =\frac{1}{d}.\end{equation} \tag{ 159 }$$

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 $\mathcal{E}:D({H}_{d})\rightarrow D({H}_{d})$ should satisfy a continuous family of inequalities

$$\begin{equation}\langle \phi \vert (\mathcal{E}\otimes I)(\vert \psi \rangle \langle \psi \vert )\vert \phi \rangle {\geqslant}0\qquad \vert \phi \rangle ,\hspace{2pt}\vert \psi \rangle \in {H}_{d}\otimes {H}_{d}.\end{equation} \tag{ 160 }$$

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 ⊗ 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 figure 10 and they confirm improvement in the average fidelity after correction for random channels.

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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 figure 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 ± 0.01 for d = 3, 0.68 ± 0.05 for d = 4, and 0.52 ± 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 − |t|² (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 ± 0.01, 0.97 ± 0.01, and 0.97 ± 0.01, which confirms that the quasi-inverse of a typical channel is close to be unital.