Distinguishing classically indistinguishable states and channels

A state of a finite-dimensional quantum system is described by a density operator $\rho$ acting on a d-dimensional Hilbert space ${{\mathcal H}}_d$ that is positive, $\rho\geqslant 0$, and normalized by a trace condition, $\newcommand{\re}{{\rm Re}} \newcommand{\tr}[1]{\mathrm{Tr}\left(#1 \right)} \tr{\rho}=1$. A state is pure if $\rho=\rho^2$, so it can be represented by a 1-dimensional (1D) projector, $\newcommand{\ket}[1]{| {#1} \rangle} {\hspace{-2pt}}\! \newcommand{\ketbra}[2]{{\left|#1\right\rangle {\hspace{-2pt}}\! \left\langle#2\right|}} \rho=\ketbra{\psi}{\psi}$; and mixed otherwise. General evolution of quantum states can be described by quantum channels, i.e. completely positive trace preserving (CPTP) maps acting on density matrices of order d. Every quantum channel $\Phi$ admits a Kraus decomposition [5] of the form

$$\begin{align} \renewcommand{\dag}{\dagger} \newcommand{\e}{{\rm e}} \newcommand{\re}{{\rm Re}} \displaystyle \label{eq:kraus_decomposition} \Phi(\cdot)=\sum_k K_k(\cdot)K_k^{\dagger}, \nonumber \end{align} \tag{ 3 }$$

where K_k are called Kraus operators and, due to trace preserving condition, satisfy $\newcommand{\iden}{\mathbb{1}} \newcommand{\re}{{\rm Re}} \renewcommand{\dag}{\dagger}\sum\nolimits_kK_k^{\dagger} K_k=\iden$ with $\newcommand{\iden}{\mathbb{1}} \iden$ denoting the identity matrix of size d. Moreover, with each channel $\Phi$ one can associate a Jamiołkowski state [21], defined by the image of the extended map acting on a maximally entangled state,

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ketbra}[2]{{\left|#1\right\rangle {\hspace{-2pt}}\! \left\langle#2\right|}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \label{eq:jamiol} J_{\Phi} = \frac{1}{d}(\Phi \otimes {{{\mathcal I}}})\ketbra{\Omega}{\Omega}, \nonumber \end{align} \tag{ 4 }$$

with $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\Omega}=\sum\nolimits_k\ket{kk}$ and ${{\mathcal I}}$ denoting the identity channel. Under this isomorphism the CP condition is translated into positivity of $J_\Phi$, and the TP condition is replaced by $\newcommand{\re}{{\rm Re}} \newcommand{\tr}[1]{\mathrm{Tr}\left(#1 \right)} \newcommand{\trr}[2]{\mathrm{Tr}_{#1}\left(#2 \right)} \newcommand{\iden}{\mathbb{1}} \trr{1}{J_\Phi}=\iden/d$.

The subset of classical states is given by quantum states $\rho$ that are incoherent with respect to a given distinguished orthonormal basis $\newcommand{\ket}[1]{| {#1} \rangle} \{\ket{k}\}_{k=1}^d$, i.e. $\newcommand{\matrixel}[3]{{\left\langle #1 \vphantom{#2#3} \right| #2 \left| #3 \vphantom{#1#2} \right\rangle}} \matrixel{k}{\rho}{l}=0$ for $k\neq l$. The choice of the basis is physically motivated by a particular problem under study, e.g. within quantum thermodynamics one is concerned with energy eigenbasis [12, 13]. Classical state can be alternatively represented by a probability distribution $\newcommand{\diag}[1]{\mathrm{diag}\left(#1\right)} \boldsymbol{p}=\diag{\rho}$, where $\newcommand{\diag}[1]{\mathrm{diag}\left(#1\right)} \diag{\rho}$ denotes a mapping of a density matrix $\rho$ into a probability vector $\boldsymbol{p}$ with $p_k=\rho_{kk}$. Moreover, for a general quantum state $\rho$ we call the probability distribution $\newcommand{\diag}[1]{\mathrm{diag}\left(#1\right)} \diag{\rho}$ the classical version of $\rho$. Note that under the completely decohering quantum channel $\newcommand{\D}{{{\mathcal D}}} \D$,

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\matrixel}[3]{{\left\langle #1 \vphantom{#2#3} \right| #2 \left| #3 \vphantom{#1#2} \right\rangle}} \newcommand{\ketbra}[2]{{\left|#1\right\rangle {\hspace{-2pt}}\! \left\langle#2\right|}} \newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\D}{{{\mathcal D}}} \displaystyle \label{eq:decoh_state} \D(\rho)=\sum_k \matrixel{k}{\rho}{k}\ketbra{k}{k}, \nonumber \end{align} \tag{ 5 }$$

every quantum state $\rho$ is mapped to a classical state specified by the classical version of $\rho$.

We also define a subset of classical channels that consists of all channels $\Phi$ whose corresponding Jamiołkowski states are classical, i.e. $\newcommand{\matrixel}[3]{{\left\langle #1 \vphantom{#2#3} \right| #2 \left| #3 \vphantom{#1#2} \right\rangle}} \matrixel{kk'}{J_\Phi}{ll'}=0$ whenever $k\neq l$ or $k'\neq l'$. Classical channel can be alternatively represented by a stochastic transition matrix T given by $\newcommand{\diag}[1]{\mathrm{diag}\left(#1\right)} \frac{1}{d}|T\rangle\rangle=\diag{J_\Phi}$, where $|\cdot\rangle\rangle$ denotes the (row-wise) vectorization of a matrix,

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\iden}{\mathbb{1}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \label{eq:vectorization} |T\rangle\rangle:=(T\otimes\iden)\ket{\Omega}=\left(\mathbb {1} \otimes T^\top \right) \ket{\Omega}, \nonumber \end{align} \tag{ 6 }$$

and T satisfies $T_{kl}\geqslant 0$ and $\sum\nolimits_k T_{kl}=1$. Moreover, for a general quantum channel $\Phi$ we call the corresponding transition matrix T the classical action of $\Phi$. A quantum channel $\Phi$ can be mapped to its classical version via a completely decohering supermap that decoheres the corresponding Jamiołkowski state $J_\Phi$ [16], and is described by the following two-step concatenation

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\D}{{{\mathcal D}}} \displaystyle \Phi \rightarrow \Phi^\D= \D\circ\Phi\circ\D. \nonumber \end{align} \tag{ 7 }$$

Note also that the classical action T of a channel $\Phi$ describes the transition between diagonal states,

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ketbra}[2]{{\left|#1\right\rangle {\hspace{-2pt}}\! \left\langle#2\right|}} \newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\bra}[1]{\langle {#1} |} \displaystyle \label{eq:transition} T_{kl}=\bra{k}\Phi(\ketbra{l}{l})\ket{k}. \nonumber \end{align} \tag{ 8 }$$

Therefore, a classical channel represented by T maps a quantum state with classical version $\boldsymbol{p}$ to a classical state $T\boldsymbol{p}$; and a quantum channel $\Phi$ with classical action T maps a classical state $\boldsymbol{p}$ to a quantum state with classical version given by $T\boldsymbol{p}$. Finally, a stochastic matrix T is called bistochastic if $\sum\nolimits_l T_{kl}=1$; and unistochastic if there exists a unitary matrix U such that $T=U\circ \bar{U}$, with ° representing the entry-wise product (also known as Hadamard or Schur product).

Throughout the paper the dimension of the underlying Hilbert space will be denoted by d, so all operators (matrices) will act on d-dimensional state vectors, while quantum channels will act on $d\times d$ density matrices. The $(d-1)$-dimensional probability simplex that represents the set of d-dimensional classical states will be denoted by $\newcommand{\D}{{{\mathcal D}}} \Delta_d$, while its centre, i.e. the maximally mixed distribution with each entry equal to 1/d, will be denoted by $\newcommand{\e}{{\rm e}} \boldsymbol{\eta}$. Moreover, we introduce a flat probability vector $\boldsymbol{v}^M$ with first M entries equal to 1/M (in particular $\newcommand{\e}{{\rm e}} \boldsymbol{v}^d=\boldsymbol{\eta}$). Beyond the identity matrix and identity channel, $\newcommand{\iden}{\mathbb{1}} \iden$ and ${{\mathcal I}}$, we will make frequent use of the unitary Fourier matrix F and the maximally mixing van der Waerden matrix W defined by

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle F_{kl}=\frac{1}{\sqrt{d}} \exp\left(\frac{2\pi {\rm i}(k-1)(l-1)}{d}\right),\quad W_{kl}=\frac{1}{d}, \nonumber \end{align} \tag{ 9 }$$

so that $|F_{kl}|^2=W_{kl}$. We also define a set of d diagonal unitary matrices D^(k), with the diagonal specified by the columns of F, i.e.

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:diagonal} D^{(k)}_{ll}=\sqrt{d}F_{kl}. \nonumber \end{align} \tag{ 10 }$$

The central problem studied in this work concerns state and channel distinguishability, which are defined as follows. Given a quantum state $\rho$ and a promise that it belongs to a preselected set of M states $\{\rho^{(n)}\}$ (with each one being equally likely), the task is to find the optimal way of deciding n^* satisfying $\rho=\rho^{(n^*)}$. The optimality of the protocol means succeeding with the highest possible probability (and thus the problem is often referred to as the maximum likelihood distinguishability). A similar question can be posed for quantum channels: given a single use of a channel $\Phi$, decide which one from the predefined set of M equally likely channels $\{\Phi^{(n)}\}$ it is. We say that a set of M states (channels) is M-distinguishable if it admits perfect distinguishability, i.e. if there exists a protocol that succeeds with unit probability.

Let us first briefly discuss the simplest case of distinguishability problem for M  =  2. Given two classical states represented by probability distributions $\boldsymbol{p}$ and $\boldsymbol{q}$, one finds that the maximum likelihood probability $P(\boldsymbol{p},\boldsymbol{q})$ of the correct distinction between them is given by

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:classical_state_distinguish} P(\boldsymbol{p},\boldsymbol{q})=\frac{1+\delta(\boldsymbol{p},\boldsymbol{q})}{2},\quad \delta(\boldsymbol{p},\boldsymbol{q}):=\frac{1}{2}\sum_k|p_k-q_k|, \nonumber \end{align} \tag{ 11 }$$

with $\delta$ known as the total variation distance. The optimal protocol simply consists of measuring the system in the distinguished basis, and upon observing outcome k answer $\boldsymbol{p}$ if $p_k\geqslant q_k$, and $\boldsymbol{q}$ otherwise. Similarly, given two quantum states, ${\rho}$ and ${\sigma}$, the optimal measurement leads to probability $P(\rho,\sigma)$ of distinguishing them given by [5]

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\tr}[1]{\mathrm{Tr}\left(#1 \right)} \newcommand{\re}{{\rm Re}} \displaystyle \label{eq:quantum_state_distinguish} P(\rho,\sigma)=\frac{1+D_{{\rm tr}}(\rho,\sigma)}{2},\quad D_{{\rm tr}}(\rho,\sigma)=\frac{1}{2}\tr{|\rho-\sigma|}, \nonumber \end{align} \tag{ 12 }$$

with $D_{{\rm tr}}$ known as the trace distance. We conclude that two states are perfectly distinguishable if and only if they have orthogonal supports. This fact straightforwardly leads to the following statement: a set of M states is M-distinguishable if and only if each pair of states have orthogonal supports.

Let us now proceed to channel distinguishability. To distinguish two classical channels represented by transition matrices T⁽¹⁾ and T⁽²⁾, one has to find a classical state $\boldsymbol{p}$ that optimizes the distinguishability between $T^{(1)}\boldsymbol{p}$ and $T^{(2)}\boldsymbol{p}$. Using convexity one can argue that such an optimal classical state should be sharp, i.e. it has all zero entries except for some k, for which it is equal to 1. Such a state is then transformed by T⁽¹⁾ to the kth column of T⁽¹⁾, denoted by $T^{(1)}_{\star k}$; and by T⁽²⁾ to the kth column of T⁽²⁾, denoted by $T^{(2)}_{\star k}$. Hence, the optimal probability of distinguishing T⁽¹⁾ from T⁽²⁾ is given by

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:classical_channel_distinguish} P(T^{(1)},T^{(2)})=\max_k\frac{1+\delta(T^{(1)}_{\star k},T^{(2)}_{\star k})}{2}. \nonumber \end{align} \tag{ 13 }$$

The problem of distinguishing between general quantum channels $\Phi^{(1)}$ and $\Phi^{(2)}$ is more complicated, due to the possible use of entangled states. Let us thus first consider that one has no access to entanglement. Then, analogously to the classical case, one has to find a quantum state $\rho$ that optimizes the distinguishability between $\Phi^{(1)}(\rho)$ and $\Phi^{(2)}(\rho)$. Again, using convexity argument, one can restrict the optimization to pure states $\psi$ leading to

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \tilde{P}(\Phi^{(1)},\Phi^{(2)})=\max_{\psi}\frac{1+D_{{\rm tr}}(\Phi^{(1)}(\psi),\Phi^{(2)}(\psi))}{2}, \nonumber \end{align} \tag{ 14 }$$

where tilde denotes the constrained optimization with no entanglement. More fundamentally, however, one can make use of entangled states to improve the distinguishability between $\Phi^{(1)}$ and $\Phi^{(2)}$, so that

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle P(\Phi^{(1)},\Phi^{(2)}) =\max_{\Psi}\frac{1+D_{{\rm tr}}[(\Phi^{(1)}\otimes{{\mathcal I}})(\Psi),(\Phi^{(2)}\otimes{{\mathcal I}})(\Psi)]}{2}, \nonumber \end{align} \tag{ 15 }$$

where the optimization is over pure bipartite states $\Psi$.

We start our analysis by finding necessary conditions for M-distinguishability. Geometrically this problem is equivalent to bounding M-distinguishability regions $\newcommand{\A}{{{\mathcal A}}} \A^M_d$ within the probability simplex $\newcommand{\D}{{{\mathcal D}}} \Delta_d$. First, note that, by definition, we have $\newcommand{\A}{{{\mathcal A}}} \A_{d}^{l} \subset \A_{d}^{k}$ for $1\leqslant k< l\leqslant d$, and $\newcommand{\A}{{{\mathcal A}}} \newcommand{\D}{{{\mathcal D}}} \A_{d}^{1} = \Delta_{d}$. Now, in order to find further non-trivial conditions we introduce the concept of permutohedron [22]:

Definition 3 (Permutohedron). For every $\boldsymbol{x}\in\mathbb{R}^d$ the permutohedron ${{\mathcal P}}_d (\boldsymbol{x})$ is the convex hull of all the permutations of $\boldsymbol{x}$,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \boldsymbol{y}\in {{\mathcal P}}_d(\boldsymbol{x}) \quad \Longleftrightarrow\quad \exists \boldsymbol{\lambda}:~\boldsymbol{y}=\sum_k \lambda_k \Pi_k \boldsymbol{x}, \nonumber \end{align} \tag{ 16 }$$

with $\boldsymbol{\lambda}$ being $d!$-dimensional probability vector and $\{\Pi_k\}$ denoting the set of $d!$ permutation matrices acting on d-dimensional vectors. In particular, we will use a shorthand notation ${{\mathcal P}}_d^M$ while referring to ${{\mathcal P}}_d(\boldsymbol{v}^M)$.

Permutohedra ${{\mathcal P}}_d^M$ form nested convex polytopes in $\mathbb{R}^{d-1}$ with vertices given by $\{\Pi_k \boldsymbol{v}^M\}$ and satisfying ${{\mathcal P}}_d^{M+1}\subset {{\mathcal P}}_d^{M}$, $\newcommand{\D}{{{\mathcal D}}} {{\mathcal P}}_d^1=\Delta_d$ and $\newcommand{\e}{{\rm e}} {{\mathcal P}}_d^d=\{\boldsymbol{\eta}\}$. We illustrate the first few of them in figure 1. Using ${{\mathcal P}}_d^M$ we can now bound $\newcommand{\A}{{{\mathcal A}}} \A_d^M$ via the following result.

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Proposition 4 (Permutohedron bound). The necessary condition for M-distinguishability of a set of quantum states $\{\rho^{(n)}\}^M_{n=1}$ with a fixed classical version $\boldsymbol{p}$ is $\max\nolimits_k p_k \leqslant 1/M$. Equivalently, $\newcommand{\A}{{{\mathcal A}}} \A_d^M \subseteq {{\mathcal P}}_d^M$.

Proof. Since $\{\rho^{(n)}\}^M_{n=1}$ are orthogonal, we have

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\iden}{\mathbb{1}} \displaystyle \label{eq:dist_condition} \sum_{n=1}^M \rho^{(n)} \leqslant \iden. \nonumber \end{align} \tag{ 17 }$$

By taking the matrix element $\newcommand{\bra}[1]{\langle {#1} |} \newcommand{\ket}[1]{| {#1} \rangle} \bra{k}\cdot\ket{k}$ of both sides we get

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle M p_k \leqslant 1, \nonumber \end{align} \tag{ 18 }$$

which holds for all k, so in particular $\max\nolimits_k p_k \leqslant 1/M$. □

Direct application of the above result to time-energy uncertainty scenario, described in the Introduction, leads to the following statement: a state that is able to distinguish M different moments in time satisfies the inequality for min-entropy $H_\infty(\boldsymbol{p})\geqslant \log M$, with $\boldsymbol{p}$ denoting its distribution over energy. We note that this coincides with the particular version of the recent result presented in [23], where the authors studied entropic formulations of energy-time uncertainty relation. Thus, any improvements over the permutohedron bound could also tighten inequalities derived there.

Before proceeding let us also state two useful results concerning M-distinguishability of pure states. First, we can relate it to the existence problem of particular unistochastic matrix.

Lemma 5. M-distinguishability of a set of pure quantum states $\newcommand{\ket}[1]{| {#1} \rangle} \{\ket{\psi}^{(n)}\}^M_{n=1}$ with a fixed classical version $\boldsymbol{p}$ is equivalent to the existence of a unistochastic matrix T with first M columns equal to $\boldsymbol{p}$.

Proof. First, assume that there exists a set of M distinguishable pure states with a fixed classical action $\boldsymbol{p}$, i.e. there exists $\newcommand{\ket}[1]{| {#1} \rangle} \{\ket{\psi^{(n)}}\}_{n=1}^M$ satisfying

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \forall m,n:\quad |\langle k\ket{\psi^{(n)}}|^2=p_k,\quad \langle \psi^{(m)}\ket{\psi^{(n)}}=\delta_{mn}, \nonumber \end{align} \tag{ 19 }$$

with $\delta_{mn}$ denoting Kronecker delta. Now, since $\newcommand{\ket}[1]{| {#1} \rangle} \{\ket{\psi^{(n)}}\}$ form an orthonormal set, one can construct a unitary U with the first M columns given by the components of these states. More precisely, we can define U by

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle U_{kn}=\langle k\ket{\psi^{(n)}} \nonumber \end{align} \tag{ 20 }$$

for $k\in\{1,\dots, d\}$, $n\in\{1,\dots,M\}$, and complete the remaining columns with orthonormal states. Then the stochastic matrix $T=U\circ \bar{U}$ is unistochastic by definition, and $T_{kn}=p_k$ for $n\in\{1,\dots,M\}$.

Conversely, assume that there exists a unistochastic T with $T_{kn}=p_k$ for $n\in\{1,\dots,M\}$. This is equivalent to the existence of a unitary U with the first M columns given by

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \ket{u_n}=\sum_k \sqrt{p_k} {\rm e}^{{\rm i}\phi_{kn}} \ket{k}. \nonumber \end{align} \tag{ 21 }$$

Since the columns of a unitary matrix are orthogonal the set $\newcommand{\ket}[1]{| {#1} \rangle} \{\ket{u_n}\}_{n=1}^M$ forms M perfectly distinguishable quantum states with a fixed diagonal $\boldsymbol{p}$. □

Moreover, we can show that for distributions lying at the boundary of permutohedron ${{\mathcal P}}_d^M$, M-distinguishability of mixed states is equivalent to M-distinguishability of pure states.

Lemma 6. Consider $\boldsymbol{p}$ such that $p_{k^*}=1/M$ for some k^*, i.e. $\boldsymbol{p}$ lies at the boundary of a permutohedron ${{\mathcal P}}_d^M$. Then, $\newcommand{\A}{{{\mathcal A}}} \boldsymbol{p}\in\A_d^M$ implies the existence of M orthogonal pure states with a fixed classical version $\boldsymbol{p}$.

Proof. Assumption $\newcommand{\A}{{{\mathcal A}}} \boldsymbol{p}\in\A_d^M$ means that there exists a set $\{\rho^{(n)}\}_{n=1}^M$ of perfectly distinguishable quantum states. Let us diagonalize each $\rho^{(n)}$,

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\bra}[1]{\langle {#1} |} \displaystyle \rho^{(n)}=\sum_{\alpha=1}^{r_n} \lambda^{(n)}_\alpha \ket{\psi^{(n)}_\alpha} {\hspace{-1pt}} \bra{\psi^{(n)}_\alpha}. \nonumber \end{align} \tag{ 22 }$$

Now, on the one hand we get

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\bra}[1]{\langle {#1} |} \displaystyle \label{eq:lem_cond1} \forall n:~\frac{1}{M}=\bra{k^*}\rho^{(n)}\ket{k^*}=\sum_{\alpha=1}^{r_n}\lambda^{(n)}_\alpha |\bra{k^*}\psi^{(n)}_\alpha\rangle|^2, \nonumber \end{align} \tag{ 23 }$$

resulting in

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\bra}[1]{\langle {#1} |} \displaystyle \sum_{n=1}^M\sum_{\alpha=1}^{r_n}\lambda^{(n)}_\alpha |\bra{k^*}\psi^{(n)}_\alpha\rangle|^2=1. \nonumber \end{align} \tag{ 24 }$$

On the other hand, perfect distinguishability of $\{\rho^{(n)}\}_{n=1}^M$ implies

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\bra}[1]{\langle {#1} |} \displaystyle \bra{\psi^{(m)}_\beta}\psi^{(n)}_\alpha\rangle = \delta_{mn} \delta_{\alpha\beta}, \nonumber \end{align} \tag{ 25 }$$

so that $\newcommand{\ket}[1]{| {#1} \rangle} \{\ket{\psi^{(n)}_\alpha}\}$ can be used to form a unitary matrix U as in the proof of lemma 5. Using analogous argument of the unistochasticity of $T=U\circ \bar{U}$ we then get

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\bra}[1]{\langle {#1} |} \displaystyle \label{eq:lem_cond2} \sum_{n=1}^M \sum_{\alpha=1}^{r_n} |\bra{k^*} \psi^{(n)}_\alpha \rangle|^2\leqslant 1. \nonumber \end{align} \tag{ 26 }$$

Finally, comparing equations (23) and (26), we see that for all n the spectrum $\lambda^{(n)}_\alpha$ is sharp, i.e. each $\rho^{(n)}$ is a pure state. □

We now proceed to analysing how tight the permutohedron bound is. We start with the following tightness result.

Proposition 7. The necessary condition for M-distinguishability, as stated by proposition 4, is also sufficient for M  =  2 and M  =  d. Equivalently, $\newcommand{\A}{{{\mathcal A}}} \A^2_d = {{\mathcal P}}^2_d$ and $\newcommand{\A}{{{\mathcal A}}} \A^d_d = {{\mathcal P}}^d_d$.

Proof. We first show $\newcommand{\A}{{{\mathcal A}}} \A^2_d = {{\mathcal P}}^2_d$. We need to prove that for a given $\boldsymbol{p}$ the condition $\max\nolimits_k p_k\leqslant 1/2$ implies the existence of two perfectly distinguishable quantum states with a fixed classical version $\boldsymbol{p}$. Consider the following two pure states,

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \ket{\psi^{(1)}} = \sum_{k=1}^d \sqrt{p_k} \ket{k},\quad \ket{\psi^{(2)}} = \sum_{k=1}^d \sqrt{p_k} {\rm e}^{{\rm i} \phi_k} \ket{k}, \nonumber \end{align} \tag{ 27 }$$

so that their overlap is given by

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\bra}[1]{\langle {#1} |} \displaystyle | \bra{\psi^{(1)}}\psi^{(2)}\rangle |^2=\sum_{k=1}^d p_k {\rm e}^{{\rm i}\phi_k}. \nonumber \end{align} \tag{ 28 }$$

Now, note that the existence of phases $\{\phi_k\}$ such that the above expression vanishes is equivalent to the possibility of constructing a closed polygon out of d segments of lengths {p _k}. Recall that the generalized triangle inequality states that the longest side of the polygon has to be shorter than the sum of the remaining sides; and its converse ensures that one can build a closed polygon if this condition is satisfied. Therefore, if

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \max_k p_k\leqslant \sum_{k} p_k -\max_k p_k= 1-\max_k p_k, \nonumber \end{align} \tag{ 29 }$$

meaning $\max\nolimits_k p_k\leqslant \frac{1}{2}$, then there exists a choice of phases $\{\phi_k\}$ such that the overlap between $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\psi^{(1)}}$ and $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\psi^{(2)}}$ vanishes.

We now show that $\newcommand{\A}{{{\mathcal A}}} \A^d_d = {{\mathcal P}}^d_d$. For this, we need to prove the existence of d orthogonal states with classical version $\newcommand{\e}{{\rm e}} \boldsymbol{\eta}$. This is simply accomplished by choosing columns of the Fourier matrix, $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\psi^{(k)}}=F\ket{k}$, which are all mutually orthogonal and the corresponding classical states are maximally mixed. □

Although, for the particular cases of M  =  2 and M  =  d, M-distinguishability regions $\newcommand{\A}{{{\mathcal A}}} \A^M_d$ coincide with the corresponding bounding permutohedra ${{\mathcal P}}^M_d$, the following result shows that, in general, the permutohedron bound is not tight.

Proposition 8. The necessary condition for M-distinguishability, as stated by proposition 4, is not sufficient for M  =  d  −  1 and even d  >  2. Equivalently, $\newcommand{\A}{{{\mathcal A}}} \A^{d-1}_d\neq {{\mathcal P}}^{d-1}_d$ for even d  >  2.

The proof of the above result can be found in appendix B. We thus see, that the regions $\newcommand{\A}{{{\mathcal A}}} \A_d^M$ have a more complex structure than permutohedra ${{\mathcal P}}_d^M$. Let us illustrate this using the simplest non-trivial example of d  =  4 and M  =  3. As shown in figure 1, the probability simplex $\newcommand{\D}{{{\mathcal D}}} \Delta_4$ can be represented by a 3D tetrahedron, with maximally mixed distribution $\newcommand{\e}{{\rm e}} \boldsymbol{\eta}$ in the centre and vertices corresponding to sharp probability distributions, i.e. $(1,0,0,0)$ and permutations thereof. Permutohedron ${{\mathcal P}}^3_4$, bounding the region $\newcommand{\A}{{{\mathcal A}}} \A^3_4$, is also a tetrahedron, with vertices $\boldsymbol{f}^i$ located at the centres of the faces of the original tetrahedron, i.e. $\boldsymbol{f}^1=\boldsymbol{v}^3=\frac{1}{3}(1,1,1,0)$ and $\boldsymbol{f}^i$ for i  >  1 are given by permutations of $\boldsymbol{f}^1$. However, as we prove in appendix C, not all points within the tetrahedron ${{\mathcal P}}^3_4$ belong to $\newcommand{\A}{{{\mathcal A}}} \A_4^3$. More precisely, we show that while the points lying on lines connecting edges of ${{\mathcal P}}^3_4$ with its centre belong to $\newcommand{\A}{{{\mathcal A}}} \A_4^3$, the faces of ${{\mathcal P}}^3_4$ and points lying on the lines connecting centres of these faces with the centre of ${{\mathcal P}}^3_4$ do not belong to $\newcommand{\A}{{{\mathcal A}}} \A^3_4$. We illustrate this in figures 2(a) and (b). Moreover, based on numerical evidence, we conjecture that points belonging to $\newcommand{\A}{{{\mathcal A}}} \A^3_4$ have the following product structure:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:3dist_conjecture} a b \boldsymbol{f}^1+a (1-b) \boldsymbol{f}^2+(1-a) b \boldsymbol{f}^3+(1-a) (1-b) \boldsymbol{f}^4, \nonumber \end{align} \tag{ 30 }$$

with $a,b\in[0,1]$. We present this conjectured set in figure 2(c).

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Here, we collect the properties of M-distinguishability regions $\newcommand{\A}{{{\mathcal A}}} \A_d^M$ beyond what is stated by the permutohedron bound $\newcommand{\A}{{{\mathcal A}}} \A_d^M\subset {{\mathcal P}}_d^M$. First, we make two obvious observations: the vertices of ${{\mathcal P}}_d^M$ always belong $\newcommand{\A}{{{\mathcal A}}} \A_d^M$ and $\newcommand{\A}{{{\mathcal A}}} \A_d^M$ does not have to be convex. The first one comes from the fact that vertices $\Pi_k\boldsymbol{v}^M$ of ${{\mathcal P}}_d^M$ correspond to maximally mixed states on the M-dimensional subspaces, and we know that then the columns of M-dimensional Fourier matrix form M orthogonal states with a fixed classical version $\boldsymbol{v}^M$. The second observation comes from noting that already $\newcommand{\A}{{{\mathcal A}}} \A_4^3$ is not convex. Despite M-distinguishability regions not being convex, we conjecture that they have a related property of being star-shaped.

Conjecture 9. The M-distinguishability regions $\newcommand{\A}{{{\mathcal A}}} \A_d^M$ form star-shaped domains with the centre point given by $\newcommand{\e}{{\rm e}} \boldsymbol{\eta}$, i.e.

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\A}{{{\mathcal A}}} \displaystyle \boldsymbol{p}\in\A_d^M \quad \Longrightarrow\quad \forall \lambda\in[0,1]:~\lambda\boldsymbol{p}+(1-\lambda)\boldsymbol{\eta}\in\A_d^M. \nonumber \end{align} \tag{ 31 }$$

Remark 10. The above conjecture, via lemma 5, is directly related to the known conjecture about the star-shaped property of the set of unistochastic matrices [24].

The next property allows one to conclude that $\boldsymbol{p}$ belongs to M-distinguishability region, if its coarse-grained version belongs to it. The definition of coarse-graining and the result are as follows.

Definition 11 (Coarse-graining). The set $\newcommand{\G}{{{\mathcal G}}} \G$ of coarse-graining matrices consists of all stochastic matrices with entries in $\{0,1\}$. Moreover, if $\boldsymbol{q}=G\boldsymbol{p}$ for some $\newcommand{\G}{{{\mathcal G}}} G\in\G$, then $\boldsymbol{q}$ is called a coarse-grained version of $\boldsymbol{p}$.

Proposition 12. If there exist M perfectly distinguishable pure states with a classical version $\boldsymbol{q}$ given by coarse-graining of $\boldsymbol{p}$, i.e. $\boldsymbol{q}=G\boldsymbol{p}$, then there exists M perfectly distinguishable pure states with a classical version $\boldsymbol{p}$.

Proof. Assume that there exists a set of M mutually orthogonal states

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \ket{\psi^{(n)}}=\sum_k \sqrt{q_k}\exp(i\phi^{(n)}_{k})\ket{k}, \nonumber \end{align} \tag{ 32 }$$

with $\boldsymbol{q}=G\boldsymbol{p}$. Let us denote by k^* the unique index for which $G_{k^*k}=1$, and construct the set of M states

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \ket{\tilde{\psi}^{(n)}}=\sum_k \sqrt{p_k}\exp(i\phi^{(n)}_{k^*})\ket{k}. \nonumber \end{align} \tag{ 33 }$$

These states all have a fixed classical action $\boldsymbol{p}$ and are mutually orthogonal,

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\bra}[1]{\langle {#1} |} \displaystyle \bra{\tilde{\psi}^{(m)}}\tilde{\psi}^{(n)}\rangle&=\sum_k p_k \exp(i\phi^{(n)}_{k^*}-i\phi^{(m)}_{k^*})\nonumber \\ &=\sum_k q_k \exp(i\phi^{(n)}_{k}-i\phi^{(m)}_{k})\nonumber \\ &=\bra{\psi^{(m)}}\psi^{(n)}\rangle=0, \nonumber \end{align} \tag{ 34 }$$

so that $\newcommand{\A}{{{\mathcal A}}} \boldsymbol{p}\in\A^M_d$. □

In particular, since edges (1-faces) $\boldsymbol{e}^{(kl)}$ of ${{\mathcal P}}^M_d$,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \boldsymbol{e}^{(kl)}=\lambda \Pi_k\boldsymbol{v}^M + (1-\lambda) \Pi_l\boldsymbol{v}^M,\quad \lambda\in[0,1], \nonumber \end{align} \tag{ 35 }$$

can be coarse-grained to a vertex $\boldsymbol{v}^M$, they all belong to the distinguishability region $\newcommand{\A}{{{\mathcal A}}} \A^M_d$. On the other hand, faces (2-faces) of ${{\mathcal P}}^M_d$ do not need to belong to $\newcommand{\A}{{{\mathcal A}}} \A^M_d$, as illustrated in figure 2(b), where no point of the proper 2-face of ${{\mathcal P}}^3_4$ belongs to $\newcommand{\A}{{{\mathcal A}}} \A_4^3$. In fact, the proof of Proposition 8 shows that the centres of 2-faces of ${{\mathcal P}}^{d-1}_d$ do not belong to $\newcommand{\A}{{{\mathcal A}}} \A^{d-1}_d$ for even d  >  2.

Finally, we can say something about the smallest non-trivial distinguishability region $\newcommand{\A}{{{\mathcal A}}} \A_d^{d-1}$. For this, let us first denote by $\newcommand{\B}{{{\mathcal B}}} \newcommand{\e}{{\rm e}} \B(\boldsymbol{p},\epsilon)$ the ball of radius $\newcommand{\e}{{\rm e}} \epsilon$ centred at $\boldsymbol{p}$, so that $\newcommand{\B}{{{\mathcal B}}} \newcommand{\e}{{\rm e}} \boldsymbol{q}\in \B(\boldsymbol{p},\epsilon)$ if and only if $\newcommand{\e}{{\rm e}} \delta(\boldsymbol{p},\boldsymbol{q})\leqslant\epsilon$. We then have the following result.

Proposition 13. For prime d there exists d  −  1 perfectly distinguishable states with classical version $\boldsymbol{p}$ if $\boldsymbol{p}$ is close enough to a maximally mixed distribution, i.e. $\newcommand{\e}{{\rm e}} \delta(\boldsymbol{p},\boldsymbol{\eta})\leqslant\epsilon$ for some $\newcommand{\e}{{\rm e}} \epsilon>0$. Equivalently, for prime d we have

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\B}{{{\mathcal B}}} \newcommand{\A}{{{\mathcal A}}} \displaystyle \exists {\epsilon > 0}:~ \B(\boldsymbol{\eta},\epsilon) \subseteq \A^{d-1}_{d}. \nonumber \end{align} \tag{ 36 }$$

Proof. Due to lemma 5, the existence of d  −  1 orthogonal pure states with classical version $\boldsymbol{p}$, is equivalent to the unistochasticity of the following matrix,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle T(\boldsymbol{p})= \left(\begin{array}{@{}ccccc@{}} p_1 & p_1 & \ldots & p_1 & 1-(d-1)\,p_1 \nonumber \\ p_2 & p_2 & \ldots &p_2 & 1-(d-1)\,p_2 \nonumber \\ \vdots & \vdots& \ddots &\vdots & \vdots \nonumber \\ p_d & p_d & \ldots & p_d & 1-(d-1)\,p_d \end{array}\right). \nonumber \end{align} \tag{ 37 }$$

Now, for $T(\boldsymbol{p})$ lying in the $\newcommand{\e}{{\rm e}} \epsilon'$-ball around $\newcommand{\e}{{\rm e}} T(\boldsymbol{\eta})=W$, $\boldsymbol{p}$ lies in an $\newcommand{\e}{{\rm e}} \epsilon$-ball around $\newcommand{\e}{{\rm e}} \boldsymbol{\eta}$. Hence, if we could show that in the $\newcommand{\e}{{\rm e}} \epsilon'$-ball around W all bistochastic matrices are unistochastic, then we would prove that in the $\newcommand{\e}{{\rm e}} \epsilon$-ball around $\newcommand{\e}{{\rm e}} \boldsymbol{\eta}$ all probability distributions allow for $(d-1)$-distinguishability. Due to result of [25], we know that such a ball exists if there exists an isolated complex Hadamard matrix of dimension d. Finally, since it is known that the Fourier matrix F (which is a complex Hadamard matrix) is isolated for any prime d, it implies the existence of the postulated ball $\newcommand{\A}{{{\mathcal A}}} \newcommand{\B}{{{\mathcal B}}} \newcommand{\e}{{\rm e}} \B(\boldsymbol{\eta},\epsilon) \subset \A^{d-1}_{d}$ for prime d. □

Remark 14. For composite dimensions we know that there exist isolated complex Hadamard matrices for $d \in \{5,\ldots,17 \}$ [26]. Therefore, the above theorem holds in those dimensions. Moreover, there is no isolated complex Hadamard matrix of order 4, so there are probabilities infinitesimally close to $\newcommand{\e}{{\rm e}} \boldsymbol{\eta}$, which do not allow for 3-distinguishability (as can be seen in figure 2(b) with four directions from the centre having no 3-distinguishable states).

We start our study of distinguishability numbers by focusing on channels with a unistochastic classical action T. By definition, there exists at least one unitary channel with a given unistochastic action. In fact, as we now show, for every unistochastic T one can always find d unitary channels that can be perfectly distinguished without using entanglement.

Proposition 15. For every unistochastic T the restricted distinguishability number $\newcommand{\M}{{{\mathcal M}}} \tilde{\M}(T)=d$.

Proof. By definition of a unistochastic matrix T, there exists a unitary matrix U such that $T = U\circ\bar{U}$. Now, consider a set of d unitary channels $V^{(k)}:=D^{(k)}U$, with D^(k) defined in equation (10) and $k\in\{1,\dots,d\}$. Every such channel has the same classical action given by T. Moreover, when acting on a state

$$\begin{align} \renewcommand{\dag}{\dagger} \newcommand{\e}{{\rm e}} \newcommand{\re}{{\rm Re}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \ket{\psi_+}=\frac{1}{\sqrt{d}}U^{\dagger}\sum_l \ket{l} \nonumber \end{align} \tag{ 38 }$$

the set $\{V^{(k)}\}$ produces an orthonormal set of states,

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle D^{(k)}U\ket{\psi_+}=\frac{1}{\sqrt{d}}\sum_l D^{(k)}\ket{l}=\sum_l F_{kl}\ket{l}=F\ket{k}. \nonumber \end{align} \tag{ 39 }$$

□

The obvious next question to ask is whether using entangled input states one can increase this number. As we will show, the answer strongly depends on T, with extreme cases given by $\newcommand{\iden}{\mathbb{1}} T=\iden$ and T  =  W. These correspond to situations where entanglement cannot help at all, and where it raises the number of distinguishable channels all the way to d². Before proving this statement, let us first introduce a family of Schur-product channels defined as follows [27, 28].

Definition 16 (Schur-product channels). The action of a Schur-product channel $\Phi_{X}$ is given by

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Phi_{X} : \rho \mapsto \rho \circ X, \nonumber \end{align} \tag{ 40 }$$

where the entry-wise product is performed in the distinguished basis and X is an arbitrary correlation matrix, i.e. X is positive and has ones on the diagonal.

We can now prove the following result.

Proposition 17. Distinguishability numbers for the identity and van den Waerdan matrix are given by $\newcommand{\M}{{{\mathcal M}}} \newcommand{\iden}{\mathbb{1}} \M(\iden)=d$ and $\newcommand{\M}{{{\mathcal M}}} \M(W)=d^2$.

Proof. To prove the first part we will show that $\newcommand{\M}{{{\mathcal M}}} \newcommand{\iden}{\mathbb{1}} \tilde{\M}(\iden)\geqslant \M(\iden)$ which, together with the condition $\newcommand{\M}{{{\mathcal M}}} \newcommand{\iden}{\mathbb{1}} \tilde{\M}(\iden)\leqslant \M(\iden)$ and proposition 15, leads to $\newcommand{\M}{{{\mathcal M}}} \newcommand{\iden}{\mathbb{1}} \M(\iden)=d$. We start by noting that the most general quantum channel consistent with classical action $\newcommand{\iden}{\mathbb{1}} T=\iden$ is a Schur-product channel (this can be easily seen by comparing their Jamiołkowski states and showing that they are the same). Now, consider n such channels, $\{\Phi^{(n)}\}_{n=1}^M$, each defined via the corresponding correlation matrix X⁽ⁿ⁾. The necessary and sufficient condition for perfect distinguishability between all those channels is the existence of a bipartite state $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\Psi}$ such that that for any two channels, $\Phi^{(m)}$ and $\Phi^{(n)}$, we have [29]

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ketbra}[2]{{\left|#1\right\rangle {\hspace{-2.5pt}}\! \left\langle#2\right|}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \label{eq:dist_schur} [\Phi^{(m)}\otimes{{\mathcal I}} (\ketbra{\Psi}{\Psi})][\Phi^{(n)}\otimes{{\mathcal I}} (\ketbra{\Psi}{\Psi})]=0. \nonumber \end{align} \tag{ 41 }$$

Let us write a general pure bipartite state $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\Psi}$ in the Schmidt basis as

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \ket{\Psi}=\sum_{j=1}^d c_j \ket{a_j b_j}, \nonumber \end{align} \tag{ 42 }$$

with

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \ket{a_j}=\sum_{k=1}^d \alpha_{jk} \ket{k},\quad \ket{b_j}=\sum_{k=1}^d \beta_{jk} \ket{k}. \nonumber \end{align} \tag{ 43 }$$

Now, by straightforward calculation, one can show that equation (41) implies that for all $k,l,p,r$ we have

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \sqrt{c_p c_r} \alpha_{pk} \alpha_{rl}^* \sum_j d_j X^{(m)}_{kj} Y^{(n)}_{jl}=0, \nonumber \end{align} \tag{ 44 }$$

where

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle d_j:=\sum_k |\alpha_{kj}|^2 c_k, \nonumber \end{align} \tag{ 45 }$$

with $d_j\geqslant 0$ and $\sum\nolimits_j d_j=1$. Thus, for every $k,l$ we have that either

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:dist_schur2} \sum_j d_j X^{(m)}_{kj} Y^{(n)}_{jl}=0 \nonumber \end{align} \tag{ 46 }$$

or that $\sqrt{c_p}\alpha_{pk}=0$ for every p , or that $\sqrt{c_r} \alpha_{rl}^*=0$ for every r. The latter two conditions are equivalent to d_k  =  0 and d_l  =  0, which can be seen by squaring and summing the original conditions. We thus conclude that if for some choice of $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\Psi}$ the channels $\{\Phi^{(n)}\}_{n=1}^M$ are all perfectly distinguishable, then for every $k,l$ either d_k  =  0, or d_l  =  0, or equation (46) is true for all $m,n$.

Consider now the action of channels $\{\Phi^{(n)}\}_{n=1}^M$ on a separable state $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\psi}$ defined by

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \ket{\psi}=\sum_j \sqrt{d_j} \ket{j}. \nonumber \end{align} \tag{ 47 }$$

For the output states to be perfectly distinguishable we need that for every pair $m,n$ the following condition is satisfied [29]

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ketbra}[2]{{\left|#1\right\rangle {\hspace{-2pt}}\! \left\langle#2\right|}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \Phi^{(m)}(\ketbra{\psi}{\psi})\Phi^{(n)}(\ketbra{\psi}{\psi})=0. \nonumber \end{align} \tag{ 48 }$$

This means that for every $k,l$ we require

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \sqrt{d_k d_l} \sum_j d_j X^{(m)}_{kj} X^{(n)}_{jl}=0. \nonumber \end{align} \tag{ 49 }$$

The above conditions are precisely the same as for distinguishability with arbitrary entangled state, i.e. if there exists an entangled state $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\Psi}$ allowing for perfect distinguishability of all channels $\{\Phi^{(n)}\}_{n=1}^M$, then there also exists a separable state $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\psi}$ allowing for perfect distinguishability. Therefore, as entanglement does not increase the maximal number of distinguishable channels with classical action $\newcommand{\iden}{\mathbb{1}} T=\iden$, and without entanglement this number is equal to d, we conclude that $\newcommand{\M}{{{\mathcal M}}} \newcommand{\iden}{\mathbb{1}} \M(\iden)=d$.

We now turn to T  =  W case. Note that all d² unitary matrices U^(kl) defined by

$$\begin{align} \renewcommand{\dag}{\dagger} \newcommand{\e}{{\rm e}} \newcommand{\re}{{\rm Re}} \displaystyle U^{(kl)}:=D^{(k)} F D^{(l)\dagger}, \nonumber \end{align} \tag{ 50 }$$

have the same classical action given by W. Moreover, the action of each of these channels on one half of the maximally entangled state $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\Omega}$ produces an orthonormal set of states $\newcommand{\ket}[1]{| {#1} \rangle} \{\ket{\Psi_{kl}}\}$ with

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\iden}{\mathbb{1}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \ket{\Psi_{kl}} = (U^{(kl)}\otimes\iden)\ket{\Omega}. \nonumber \end{align} \tag{ 51 }$$

To see that are $\newcommand{\ket}[1]{| {#1} \rangle} \{\ket{\Psi_{kl}}\}$ are indeed orthogonal, note that

$$\begin{align} \renewcommand{\dag}{\dagger} \newcommand{\e}{{\rm e}} \newcommand{\re}{{\rm Re}} \newcommand{\ketbra}[2]{{\left|#1\right\rangle \left\langle#2\right|}} \newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\bra}[1]{\langle {#1} |} \displaystyle \langle \Psi_{k'l'} \ket{\Psi_{kl}}&=\frac{1}{d} \sum_{j} \bra{j} D^{(l')} F^{\dagger} D^{(k')\dagger} D^{(k)} F D^{(l)\dagger}\ket{j} =\sum_{j,j'} \bra{j} F_{l'j} F^{\dagger} D^{(k')\dagger}\ketbra{j'}{j'} D^{(k)} F F_{lj}^*\ket{j}\nonumber \\ &=\sum_{j,j'} F_{l'j} F_{k'j'}^* F_{kj'} F_{lj}^*=[FF^{\dagger}]_{l'l} [FF^{\dagger}]_{kk'} =\delta_{ll'}\delta_{kk'}. \nonumber \end{align} \tag{ 52 }$$

Thus, we conclude that all d² channels U^(kl) with classical action T  =  W are perfectly distinguishable, and so $\newcommand{\M}{{{\mathcal M}}} \M(W)=d^2$. □

Corollary 18. The maximal set of perfectly distinguishable Schur-product channels is d.

For a general matrix T one expects that $\newcommand{\M}{{{\mathcal M}}} \M(T)$ can take all values between the above extremes given by d and d². We will now show how the construction used while proving $\newcommand{\M}{{{\mathcal M}}} \M(W)=d^2$ can be generalized, opening a way to construct N  >  d perfectly distinguishable channels, and thus finding lower bounds on $\newcommand{\M}{{{\mathcal M}}} \M(T)$ for general unistochastic T. First, we restrict our search for the maximal set of perfectly distinguishable channels with a fixed unistochastic classical action T to unitary channels. Since the moduli of every entry for all these unitaries are equal, we may further restrict our considerations to unitaries of the following form

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle U^{(mn)}=L^{(m)} U R^{(n)}, \nonumber \end{align} \tag{ 53 }$$

where L^(m) and R⁽ⁿ⁾ are general diagonal unitaries with

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle L^{(m)}_{kk}=\exp\left({\rm i}\phi^{(m)}_k\right),\quad R^{(n)}_{kk}=\exp\left({\rm i}\theta^{(n)}_k\right), \nonumber \end{align} \tag{ 54 }$$

and U is any unitary matrix satisfying $U\circ U=T$. Finally, we assume that the input state used in the distinguishability protocol is the maximally entangled state $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\Omega}$. The set of unitaries {U^(mn)} is then perfectly distinguishable if for all pairs $m,n$ and $m',n'$ we have that the following expression vanishes,

$$\begin{align} \renewcommand{\dag}{\dagger} \newcommand{\e}{{\rm e}} \newcommand{\re}{{\rm Re}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \label{eq:ort_unitaries} \langle \langle U^{(mn)}\ket{U^{(m'n')}} \rangle=\sum_{k,k'} [L^{(n')\dagger} L^{(n)} T R^{(m)}R^{(m')\dagger}]_{kk'}. \nonumber \end{align} \tag{ 55 }$$

We note the close resemblance of the above problem to pure state distinguishability. There one needed to find phases $\{\phi^{(n)}_{k}\}$, so that $\newcommand{\e}{{\rm e}} \sum\nolimits_k p_k \exp(i\phi^{(n)}_k-i\phi^{(m)}_{k})$ vanishes for all $m,n$; here, one is looking for phases $\{\phi^{(m)}_k\}$ and $\{\theta^{(n)}_k\}$, so that equation (55) is satisfied. In appendix D we show how the above method can be used to find d  +  1 perfectly distinguishable channels with a particular classical action.

Although for unistochastic action one could always find d perfectly distinguishable channels, it is no longer the case when one considers general stochastic action T. In fact, there exist T for which one cannot construct even 2 distinguishable channels. As a particular example consider the completely contractive classical action T defined by T_1k  =  1 for all k, and T_jk  =  0 for all k and $j\neq 1$. This classical action uniquely defines a quantum channel $\newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\ketbra}[2]{{\left|#1\right\rangle {\hspace{-3pt}}\! \left\langle#2\right|}} \Phi(\cdot)=\ketbra{1}{1}$, and thus $\newcommand{\M}{{{\mathcal M}}} \M(T)=1$.

As channel distinguishability ultimately depends on state distinguishability, we can employ the results from section 3.

Proposition 19. Consider a classical action T and denote the probability distribution formed from the entries of its lth column by $T_{\star l}$. If, for any l, we have $\newcommand{\A}{{{\mathcal A}}} T_{\star l}\in\A^M_d$, then $\newcommand{\M}{{{\mathcal M}}} \tilde{\M}(T)\geqslant M$.

Proof. Define a set of channels $\{\Phi^{(n)}\}$ by

$$\begin{align} \renewcommand{\dag}{\dagger} \newcommand{\e}{{\rm e}} \newcommand{\re}{{\rm Re}} \displaystyle \Phi^{(n)}(\rho):=\sum_{l=1}^d K^{(n)}_{l}\rho K^{(n)\dagger}_{l} \nonumber \end{align} \tag{ 56 }$$

with

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\bra}[1]{\langle {#1} |} \displaystyle K^{(n)}_{l}:=\ket{\psi^{(n)}_l}{\hspace{-1pt}}\bra{l} \nonumber \end{align} \tag{ 57 }$$

and

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \ket{\psi^{(n)}_l}=\sum_{k=1}^d \sqrt{T_{kl}}{\rm e}^{{\rm i}\phi^{(n)}_{k}}\ket{k}. \nonumber \end{align} \tag{ 58 }$$

The classical action of each of $\Phi^{(n)}$ is given by T for every choice of phases $\{\phi^{(n)}_{k}\}$,

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ketbra}[2]{{\left|#1\right\rangle{\hspace{-3pt}}\! \left\langle#2\right|}} \newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\bra}[1]{\langle {#1} |} \displaystyle \bra{k}{\Phi^{(n)}(\ketbra{l}{l})}\ket{k}=|\langle{k}\ket{\psi^{(n)}_l}|^2=T_{kl}. \nonumber \end{align} \tag{ 59 }$$

At the same time the state $\newcommand{\ket}[1]{| {#1} \rangle} \ket{l}$ is mapped by $\Phi^{(n)}$ to $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\psi^{(n)}_l}$, whose classical version is $T_{\star l}$ independently of $\{\phi^{(n)}_{k}\}$. Therefore, if there exists M perfectly distinguishable states with classical version $T_{\star l}$, then it is possible to choose phases $\{\phi^{(n)}_{k}\}$ so that each $\Phi^{(n)}$ maps $\newcommand{\ket}[1]{| {#1} \rangle} \ket{l}$ to an orthogonal state $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\psi_l^{(n)}}$. □

The above result, together with proposition 7, imply the following corollary.

Corollary 20. If the entries of at least one column of the stochastic matrix T satisfy the triangle inequality, i.e. the largest entry is smaller than the sum of the remaining entries, then $\newcommand{\M}{{{\mathcal M}}} \tilde{\M}(T)\geqslant 2$.

Even if no column of T satisfies the triangle inequality, there can still exist two perfectly distinguishable channels. This time the distinguishability protocol will require the use of entanglement, but before we state the result, we first need to introduce a particular swap procedure $S_{kl}^{\,j\alpha}$. Given a stochastic matrix T the matrix $T'=S_{kl}^{\,j\alpha}(T)$ is obtained by multiplying column k of T by a real number $\alpha$, which is then followed by a transposition of two elements in row j , one belonging to column k and the other to column l.

Proposition 21. Assume that the classical action T can be transformed by some swap procedure $S_{kl}^{\,j\alpha}$ into a matrix $T'$, such that both columns k and l of $T'$ satisfy the triangle inequality. Then, $\newcommand{\M}{{{\mathcal M}}} \M(T)\geqslant 2$.

The proof of the above result can be found in appendix E.

Finally, we proceed to the results concerning distinguishability of quantum channels with a fixed classical action T that is bistochastic. Our main result states that one can always find at least two perfectly distinguishable channels with a given bistochastic classical action.

Proposition 22. For every bistochastic matrix T we have $\newcommand{\M}{{{\mathcal M}}} \M(T)\geqslant 2$.

Proof. First, if there exists a column of T that satisfies the triangle inequality, i.e. the largest entry is smaller than the sum of the remaining entries, then $\newcommand{\M}{{{\mathcal M}}} \M(T)\geqslant 2$ due to lemma 20. Otherwise, we deal with T such that each column contains an element larger than $\frac{1}{2}$. Without loss of generality we can assume those elements are placed on the diagonal of the matrix T. Similarly, without loss of generality we may assume that the largest of the non-diagonal elements of T is T₂₁. The plan now is to show that for a proper choice of $\alpha$, the swap procedure $S_{21}^{2\alpha}$ transforms T into $T'$ such that the triangle inequality is satisfied by columns 1 and 2 of $T'$. This will allow us to use proposition 21 and conclude that $\newcommand{\M}{{{\mathcal M}}} \M(T)\geqslant 2$. To achieve this we will separately consider two situations: $T_{11}>T_{22}$ and $T_{11} \leqslant T_{22}$.

First assume $T_{11}<T_{22}$ and choose $S_{21}^{2\alpha}$ with $\alpha =1$, so that

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle T = \left(\begin{array}{@{}ccc@{}} T_{11} & T_{12} & \ldots \nonumber \\ T_{21} & T_{22} &\ldots \nonumber \\ T_{31} &T_{32}& \ldots \nonumber \\ \vdots & \vdots & \ddots \end{array}\right) \xrightarrow{S_{21}^{2\alpha}} T'= \left(\begin{array}{@{}ccc@{}} T_{11} & T_{12} & \ldots \nonumber \\ T_{22} & T_{21} &\ldots \nonumber \\ T_{31} &T_{32}& \ldots \nonumber \\ \vdots & \vdots & \ddots\end{array}\right) . \nonumber \end{align} \tag{ 60 }$$

As T is bistochastic, we have

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\im}{{\rm Im}} \displaystyle T_{21} + \sum_{k \neq 1} T_{2k} = 1 \implies T_{21} \leqslant 1 - T_{22}=\sum_{k \neq 2} T_{k2}, \nonumber \end{align} \tag{ 61 }$$

therefore the second column of $T'$ satisfies the triangle inequality (because the largest element in column 2 of $T'$ is T₂₁). Similarly, we have

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle T_{22}\leqslant 1-T_{21}=\sum_{k \neq 2} T_{k1}, \nonumber \end{align} \tag{ 62 }$$

so the first column of $T'$ also satisfies the triangle inequality (because the largest element in column 1 of $T'$ is T₂₂). We conclude that, due to proposition 21, there exist two perfectly distinguishable channels with classical action T.

Let us now turn to the second case, $T_{11}\geqslant T_{22}$. Again, we obtain $T'$ by a swap procedure $S_{21}^{2\alpha}$,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle T= \left(\begin{array}{@{}ccc@{}} T_{11} & T_{12}& \ldots \nonumber \\ T_{21} & T_{22} &\ldots \nonumber \\ T_{31} &T_{32}& \ldots \nonumber \\ \vdots & \vdots & \ddots \end{array}\right) \rightarrow T'= \left(\begin{array}{@{}ccc@{}} T_{11} & \alpha T_{12}& \ldots \nonumber \\ \alpha T_{22} & T_{21} &\ldots \nonumber \\ T_{31} & \alpha T_{32}& \ldots \nonumber \\ \vdots & \vdots & \ddots \end{array}\right) \nonumber \end{align} \tag{ 63 }$$

with

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \alpha = \frac{2 T_{11} + T_{21} -1}{T_{22}}. \nonumber \end{align} \tag{ 64 }$$

Due to bistochasticity of T we have $T_{21}\leqslant 1-T_{11}$, so

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \alpha T_{22} = 2T_{11} + T_{21}-1 \leqslant T_{11}, \nonumber \end{align} \tag{ 65 }$$

meaning that T₁₁ is the largest element in the first column of $T'$. This column satisfies the triangle inequality, because

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \sum_{k \neq 1} T'_{k1} = \alpha T_{22} + (1 - T_{11} -T_{21})=T_{11}. \nonumber \end{align} \tag{ 66 }$$

It remains to show that the second column of $T'$ satisfies the triangle inequality. We denote the second largest element in the second column of T by x. If $T_{21} \geqslant \alpha x$ then T₂₁ is the greatest element in the second column of $T'$. Then, the triangle inequality has the following form

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle T_{21} \leqslant \sum_{k \neq 2} \alpha T_{k2}= \alpha (1- T_{22}), \nonumber \end{align} \tag{ 67 }$$

which is equivalent to

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle (2T_{11} - 1)(1-T_{22}) - (2T_{22} - 1)T_{21}\geqslant 0. \nonumber \end{align} \tag{ 68 }$$

As $2T_{11} -1 \geqslant 2T_{22} -1$ and $1-T_{22} \geqslant T_{21}$, the above inequality holds.

If $T_{21} < \alpha x$ then $\alpha x$ is the greatest element in the second column of $T'$. Then, the triangle inequality has the following form

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \alpha x \leqslant \sum_{k \neq 2} \alpha T_{k2} - \alpha x + T_{21} =\alpha(1-T_{22}-x)+T_{21}, \nonumber \end{align} \tag{ 69 }$$

which is equivalent to

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \alpha(1-T_{22})-2\alpha x + T_{21}\geqslant 0. \nonumber \end{align} \tag{ 70 }$$

Since $T_{21} \geqslant x$ it is sufficient to check that the function f  defined by

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle f(T_{21}) := \alpha(1-T_{22}) - T_{21}(2\alpha -1) \nonumber \end{align} \tag{ 71 }$$

is greater or equal 0 for $T_{21} \in [0,1-T_{11}]$. The function f  is concave, i.e.

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle f(\,ps + (1-p)t) \geqslant p f(s) + (1-p)\,f(t), \nonumber \end{align} \tag{ 72 }$$

with $p \in [0,1]$. It is thus sufficient to check that $f(0) \geqslant 0$ and $f(1-T_{11}) \geqslant 0$. By straightforward calculation, one obtains

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle f(0) = \frac{1}{T_{22}}(2T_{11}-1)(1-T_{22})\geqslant 0, \nonumber \end{align} \tag{ 73a }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle f(1-T_{11}) = \frac{1}{T_{22}}(2T_{11}-1)(T_{11}-T_{22})\geqslant 0. \nonumber \end{align} \tag{ 73b }$$

Therefore $f(T_{21})\geqslant 0$ for all $T_{21} \in [0,1-T_{11}]$. This means that the triangle inequality is satisfied for the second column of $T'$ and, due to proposition 21, ends the proof. □

The above result can be further refined for a particular subset of bistochastic matrices defined in the following way.

Definition 23 (Circulant matrix). A stochastic matrix T is called circulant if it is of the form:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:circulant} T = \sum_{k=1}^{d} \lambda_k X^k, \nonumber \end{align} \tag{ 74 }$$

with $\newcommand{\D}{{{\mathcal D}}} \boldsymbol{\lambda}\in\Delta_d$ and

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ketbra}[2]{{\left|#1\right\rangle{\hspace{-2.5pt}}\! \left\langle#2\right|}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle X=\sum_{k=1}^{d} \ketbra{k\oplus 1}{k}, \nonumber \end{align} \tag{ 75 }$$

where $\oplus$ denotes addition modulo d.

For this particular family of bistochastic matrices we can prove a result analogous to proposition 15 that concerns unistochastic matrices.

Proposition 24. For every circulant T the restricted distinguishability number $\newcommand{\M}{{{\mathcal M}}} \tilde{\M}(T)=d$.

Proof. For a given circulant matrix T, define a set of d quantum channels $\{\Phi^{(n)}\}_{n=1}^d$ through their Jamiołkowski states,

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\bra}[1]{\langle {#1} |} \displaystyle J_{\Phi^{(n)}} = \sum_{\alpha=1}^{d} \lambda_\alpha \ket{\psi_\alpha^{(n)}}{\hspace{-1pt}}\bra{\psi_\alpha^{(n)}}, \nonumber \end{align} \tag{ 76 }$$

with $\boldsymbol{\lambda}$ defining T through equation (74) and $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\psi_\alpha^{(n)}}$ given by

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \ket{\psi_\alpha^{(n)}}=\frac{1}{\sqrt{d}} \sum_{k=1}^{d} F_{nk}\ket{k,k \oplus \alpha}. \nonumber \end{align} \tag{ 77 }$$

By direct inspection one can check that

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\bra}[1]{\langle {#1} |} \displaystyle \bra{kl}J_{\Phi^{(n)}}\ket{kl}=\frac{1}{d}T_{kl}, \nonumber \end{align} \tag{ 78 }$$

so that for all n the classical action of $\Phi^{(n)}$ is given by T. Moreover,

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\bra}[1]{\langle {#1} |} \displaystyle \bra{\psi_\beta^{(m)}}{\psi_{\alpha}^{(n)}}\rangle =\delta_{\alpha\beta} \delta_{mn} , \nonumber \end{align} \tag{ 79 }$$

with the first Kronecker delta coming from orthogonality of supports and the second one from orthogonality of columns of the Fourier matrix F. This implies orthogonality of the Jamiołkowski states $J_{\Phi^{(n)}}$, and thus quantum channels $\Phi^{(n)}$ sharing the same classical action T are perfectly distinguishable. □

In this final section we provide a solution for the problem of distinguishing classically indistinguishable qubit channels. We start by noting that a classical action of a general qubit channel is given by

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:qubit_classical} T=\left(\begin{array}{@{}cc@{}} a & 1-b \nonumber \\ 1-a & b \end{array}\right), \nonumber \end{align} \tag{ 80 }$$

with $0\leqslant a,b\leqslant 1$. For a  =  b we deal with bistochastic matrices that for two-dimensional systems coincide with unistochastic matrices. Without loss of generality, we may assume that $a \geqslant b$ and introduce $\newcommand{\D}{{{\mathcal D}}} \Delta:=a - b\geqslant 0$.

The full characterization of restricted distinguishability numbers for qubit channels is given by the following Proposition and is illustrated in figure 3(a).

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Proposition 25. Restricted distinguishability number $\newcommand{\M}{{{\mathcal M}}} \tilde{\M}(T)$ for a qubit classical action T parametrized as in equation (80) is given by

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\M}{{{\mathcal M}}} \displaystyle \tilde{\M}(T)=\left\{\begin{array}{@{}ll@{}} 1&:~\mathrm{for}~\frac{1}{2}<|a-b|\leqslant 1, \nonumber \\ 2&:~\mathrm{for}~0\leqslant |a-b|\leqslant \frac{1}{2}. \nonumber \\ \end{array}\right. \nonumber \end{align} \tag{ 81 }$$

Proof. We first consider the case $\newcommand{\D}{{{\mathcal D}}} \Delta \in \left[0 , \frac{1}{2} \right]$. We define two quantum channels, $\Phi^{(+)}$ and $\Phi^{(-)}$, in the following way,

$$\begin{align} \renewcommand{\dag}{\dagger} \newcommand{\e}{{\rm e}} \newcommand{\re}{{\rm Re}} \displaystyle \Phi^{(\pm)}(\cdot)= \Psi^{(\pm)}(U(\cdot)U^{\dagger}), \nonumber \end{align} \tag{ 82 }$$

with a unitary U given by

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\D}{{{\mathcal D}}} \displaystyle U = \frac{1}{\sqrt{1-\Delta}} \left(\begin{array}{@{}cc@{}} \sqrt{1-a} & -\sqrt{b} \nonumber \\ \sqrt{b} & \sqrt{1-a} \end{array}\right) \nonumber \end{align} \tag{ 83 }$$

and quantum channels $\Psi^{(\pm)}$ defined by their Jamiołkowski states,

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\D}{{{\mathcal D}}} \displaystyle J_{\Psi^{(\pm)}} = \frac{1}{2} \left(\begin{array}{@{}cccc@{}} \Delta & 0 & \pm \Delta & 0 \nonumber \\ 0 & 1 &\pm\sqrt{1 - 2\Delta} &0 \nonumber \\ \pm \Delta & \pm\sqrt{1 - 2\Delta} & 1 - \Delta &0 \nonumber \\ 0&0&0&0 \end{array}\right). \nonumber \end{align} \tag{ 84 }$$

Using the fact that the Jamiołkowski states of $\Phi^{(\pm)}$ are related to those of $\Psi^{(\pm)}$ by

$$\begin{align} \renewcommand{\dag}{\dagger} \newcommand{\e}{{\rm e}} \newcommand{\re}{{\rm Re}} \displaystyle J_{\Phi^{(\pm)}}=\left(\mathbb {1} \otimes U \right) J_{\Psi^{(\pm)}} \left(\mathbb {1} \otimes U^{\dagger} \right), \nonumber \end{align} \tag{ 85 }$$

it is straightforward to verify that the classical action of $\Phi^{(\pm)}$ (encoded on the diagonal of $J_{\Phi^{(\pm)}}$) is given by T parametrized as in equation (80). Moreover, a state $\newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\ketbra}[2]{{\left|#1\right\rangle \left\langle#2\right|}} \newcommand{\re}{{\rm Re}} \renewcommand{\dag}{\dagger}\rho= U^{\dagger} \ketbra{\psi}{\psi} U$ with

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\D}{{{\mathcal D}}} \displaystyle \ket{\psi} = \frac{\ket{0}+\sqrt{1 - 2 \Delta}\ket{1}}{\sqrt{2(1-\Delta)}} \nonumber \end{align} \tag{ 86 }$$

is mapped by $\Phi^{(\pm)}$ to orthogonal states,

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ketbra}[2]{{\left|#1\right\rangle {\hspace{-2.5pt}}\! \left\langle#2\right|}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \Phi^{(\pm)}(\rho)=\Psi^{(\pm)}(\ketbra{\psi}{\psi})=\ketbra{\pm}{\pm}, \nonumber \end{align} \tag{ 87 }$$

so that $\Phi^{(+)}$ and $\Phi^{(-)}$ are perfectly distinguishable.

We will now show that for $\newcommand{\D}{{{\mathcal D}}} \Delta >\frac{1}{2}$ we have $\newcommand{\M}{{{\mathcal M}}} \M(T)=1$. The image of a qubit channel $\Phi$ is an ellipsoid inside a Bloch ball. Antipodal points $\newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\ketbra}[2]{{\left|#1\right\rangle {\hspace{-2.5pt}}\! \left\langle#2\right|}} \ketbra{0}{0}$ and $\newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\ketbra}[2]{{\left|#1\right\rangle {\hspace{-2.5pt}}\! \left\langle#2\right|}} \ketbra{1}{1}$ are mapped onto antipodal points,

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ketbra}[2]{{\left|#1\right\rangle {\hspace{-2.5pt}}\! \left\langle#2\right|}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \Phi\left(\ketbra{0}{0} \right) = \frac{1}{2} \left(\begin{array}{@{}cc@{}} 1 + z_{0} & x_{0} - {\rm i} y_{0} \nonumber \\ x_{0} + {\rm i} y_{0} & 1 - z_{0} \end{array}\right), \nonumber \end{align} \tag{ 88a }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ketbra}[2]{{\left|#1\right\rangle {\hspace{-2.5pt}}\! \left\langle#2\right|}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \Phi\left(\ketbra{1}{1} \right) = \frac{1}{2} \left(\begin{array}{@{}cc@{}} 1 + z_{1} & x_{1} - {\rm i} y_{1} \nonumber \\ x_{1} + {\rm i} y_{1} &1 - z_{1} \end{array}\right). \nonumber \end{align} \tag{ 88b }$$

Fixing T corresponds to fixing z₀ and z₁,

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ketbra}[2]{{\left|#1\right\rangle {\hspace{-2.5pt}}\! \left\langle#2\right|}} \newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\bra}[1]{\langle {#1} |} \displaystyle \bra{0} \Phi\left(\ketbra{0}{0} \right) \ket{0} = \frac{1}{2} \left(1 + z_{0} \right)=a, \nonumber \end{align} \tag{ 89a }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ketbra}[2]{{\left|#1\right\rangle {\hspace{-2pt}}\! \left\langle#2\right|}} \newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\bra}[1]{\langle {#1} |} \displaystyle \bra{1} \Phi\left(\ketbra{1}{1} \right) \ket{1} = \frac{1}{2} \left(1 - z_{1} \right)=b, \nonumber \end{align} \tag{ 89b }$$

so that z₀  =  2a  −  1 and z₁  =  1  −  2b. Now, the centre of the ellipsoid lies in the middle between the antipodal points and its z coordinate is equal to $\newcommand{\D}{{{\mathcal D}}} z_{c} = \Delta$. The z coordinate of any point belonging to ellipsoid must lie between $z_{c} + \zeta$ and $z_{c} - \zeta$, for some $\zeta\geqslant 0$. If $\newcommand{\D}{{{\mathcal D}}} z_c=\Delta > \frac{1}{2}$ then $\zeta < \frac{1}{2}$, because the ellipsoid has to lie inside the Bloch ball. Thus, $z_{c} - \zeta > 0$ and the entire ellipsoid lies inside the northern hemisphere of the Bloch ball. Analogously, if $\newcommand{\D}{{{\mathcal D}}} z_c=\Delta < - \frac{1}{2}$ then $\zeta < \frac{1}{2}$ and the entire ellipsoid lies inside the southern hemisphere. In either case, regardless of the choice of $\Phi$, the image of $\Phi$ lies entirely inside one of the hemispheres and does not contain two orthogonal states. This implies that one cannot construct two channels with the same classical action T that will be perfectly distinguishable without using entangled states. □

We now proceed to entangled-assisted distinguishability protocols. Our results on distinguishability numbers for qubit channels are captured by the following Proposition and are illustrated in figure 3(b)

Proposition 26. Distinguishability number $\newcommand{\M}{{{\mathcal M}}} \M(T)$ for a qubit classical action T parametrized as in equation (80) satisfies

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\M}{{{\mathcal M}}} \displaystyle \M(T)=\left\{\begin{array}{@{}ll@{}} 2&:~{\rm for}~0<|a-b|\leqslant \frac{1}{2}, \nonumber \\ 3&:~{\rm for}~a=b~ {\rm and}~ a\in\left[\frac{1}{3},\frac{2}{3}\right]\setminus\left\{\frac{1}{2}\right\}, \nonumber \\ 4&:~{\rm for}~ a=b=\frac{1}{2}. \end{array}\right. \nonumber \end{align} \tag{ 90 }$$

Proof. First, we will focus on classical action T that is not bistochastic, i.e. $a\neq b$ implying $\newcommand{\D}{{{\mathcal D}}} \Delta>0$. A general entangled two-qubit state is given by

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \ket{\psi}=\sum_{j,k=0}^1 c_{jk} \ket{jk}. \nonumber \end{align} \tag{ 91 }$$

Now, the output $\rho^{(n)}$ of a channel $\Phi^{(n)}$ acting on one part of this state can be written as

$$\begin{align} \renewcommand{\dag}{\dagger} \newcommand{\e}{{\rm e}} \newcommand{\re}{{\rm Re}} \newcommand{\iden}{\mathbb{1}} \newcommand{\ketbra}[2]{{\left|#1\right\rangle {\hspace{-2.25pt}}\! \left\langle#2\right|}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \label{eq:rank_argument1} \rho^{(n)}:=\Phi^{(n)}\otimes{{\mathcal I}} (\ketbra{\psi}{\psi})=(\iden\otimes[\psi]^\top)J_{\Phi^{(n)}} (\iden\otimes[\psi]^\top)^{\dagger}, \nonumber \end{align} \tag{ 92 }$$

where $[\psi]$ is a matrix obtained from $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\psi}$ via mapping $\newcommand{\ket}[1]{| {\1} \rangle} |\hspace{1pt} {jk}\rangle \rightarrow |\hspace{1pt} {j}\rangle \hspace{-1pt} \langle {k}|$. Since we analyse entanglement-assisted discrimination, we may assume that the Schmidt number of $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\psi}$ is 2, and thus $[\psi]$ is invertible. Moreover, as T is not unistochastic, $\Phi^{(n)}$ cannot be a unitary [16] and thus the rank of $J_{\Phi^{(n)}}$ has to be at least 2. Therefore, for any channel $\Phi^{(n)}$ whose classical action T is not bistochastic, the rank of the output state $\rho^{(n)}$ must be at least 2. However, the necessary condition for the set $\{\rho^{(n)}\}_{n=1}^M$ to be mutually orthogonal is

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:rank_argument2} \sum_{n=1}^M \mathrm{rank}(\rho^{(n)})\leqslant 4, \nonumber \end{align} \tag{ 93 }$$

and so $M\leqslant 2$. This proves that for $0<|a-b|\leqslant 1/2$ we have $\newcommand{\M}{{{\mathcal M}}} \M(T)=2$, due to

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\M}{{{\mathcal M}}} \displaystyle 2=\tilde{\M}(T)\leqslant \M(T) \leqslant 2, \nonumber \end{align} \tag{ 94 }$$

where the first equality comes from proposition 25.

We now proceed to bistochastic classical actions, a  =  b. We will first show that if two qubit unitary channels, U and $V$, are perfectly distinguishable with the use of some entangled state $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\Psi}$, they are mutually orthogonal, $\newcommand{\tr}[1]{\mathrm{Tr}\left(#1 \right)} \newcommand{\re}{{\rm Re}} \renewcommand{\dag}{\dagger}\tr{UV^{\dagger}}=0$, meaning also that they are perfectly distinguishable with the use of maximally entangled state $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\Omega}$. As a result, looking for M perfectly distinguishable unitary channels, we may only focus on a single input state $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\Omega}$. To see this, note that perfect distinguishability of unitaries U and $V$ with the use of state $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\Psi}$ means

$$\begin{align} \renewcommand{\dag}{\dagger} \newcommand{\e}{{\rm e}} \newcommand{\re}{{\rm Re}} \newcommand{\tr}[1]{\mathrm{Tr}\left(#1 \right)} \newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\bra}[1]{\langle {#1} |} \displaystyle 0 = \bra{\Psi} \left(V^{\dagger} \otimes \mathbb {1} \right) \left(U \otimes \mathbb {1} \right) \ket{\Psi} = \tr {V^{\dagger} U \rho}, \nonumber \end{align} \tag{ 95 }$$

where $\newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\ketbra}[2]{{\left|#1\right\rangle {\hspace{-2.5pt}}\! \left\langle#2\right|}} \newcommand{\re}{{\rm Re}} \newcommand{\tr}[1]{\mathrm{Tr}\left(#1 \right)} \newcommand{\trr}[2]{\mathrm{Tr}_{#1}\left(#2 \right)} \rho = \trr{2}{\ketbra{\Psi}{\Psi}}$. Since the matrix $\newcommand{\re}{{\rm Re}} \renewcommand{\dag}{\dagger}V^{\dagger} U$ has a spectral decomposition

$$\begin{align} \renewcommand{\dag}{\dagger} \newcommand{\e}{{\rm e}} \newcommand{\re}{{\rm Re}} \displaystyle V^{\dagger} U = U_{0} \left(\begin{array}{@{}cc@{}} {\rm e}^{{\rm i} \phi_1} & 0 \nonumber \\ 0 & {\rm e}^{{\rm i} \phi_2} \end{array}\right) U_{0}^{\dagger} = U \Lambda U_0^{\dagger}, \nonumber \end{align} \tag{ 96 }$$

we can rewrite the distinguishability condition as

$$\begin{align} \renewcommand{\dag}{\dagger} \newcommand{\e}{{\rm e}} \newcommand{\re}{{\rm Re}} \newcommand{\tr}[1]{\mathrm{Tr}\left(#1 \right)} \displaystyle \tr {V^{\dagger} U \rho} =\tr{\Lambda U_{0}^{\dagger} \rho U_0} = 0. \nonumber \end{align} \tag{ 97 }$$

Defining $\newcommand{\re}{{\rm Re}} \renewcommand{\dag}{\dagger}\rho^{\prime} = U_{0}^{\dagger} \rho U_0$, we obtain

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle {\rm e}^{{\rm i} \phi_1} \rho^{\prime}_{11} + {\rm e}^{{\rm i} \phi_2} \rho^{\prime}_{22} = 0. \nonumber \end{align} \tag{ 98 }$$

This implies $\rho^{\prime}_{11} =\rho^{\prime}_{22}$ and ${\rm e}^{{\rm i} \phi_1} = - {\rm e}^{{\rm i} \phi_2}$. Thus,

$$\begin{align} \renewcommand{\dag}{\dagger} \newcommand{\e}{{\rm e}} \newcommand{\re}{{\rm Re}} \newcommand{\tr}[1]{\mathrm{Tr}\left(#1 \right)} \displaystyle \tr {V^{\dagger} U} = \tr{\Lambda} = 0. \nonumber \end{align} \tag{ 99 }$$

But, this implies perfect distinguishability between U and $V$ with the use of maximally entangled state $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\Omega}$, because

$$\begin{align*} \renewcommand{\dag}{\dagger} \newcommand{\e}{{\rm e}} \newcommand{\re}{{\rm Re}} \newcommand{\tr}[1]{\mathrm{Tr}\left(#1 \right)} \newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\bra}[1]{\langle {#1} |} \displaystyle \bra{\Omega} \left(V^{\dagger} \otimes \mathbb {1} \right) \left(U \otimes \mathbb {1} \right) \ket{\Omega} = \tr{V^{\dagger}{U}}=0. \nonumber \end{align*}$$

We will now find necessary conditions for $\newcommand{\M}{{{\mathcal M}}} \M(T)=4$. Using an analogous rank argument as before (captured by equations (92) and (93)), we see that all four channels must be rank 1, i.e. be unitary. As explained above, these unitaries U_i must be orthogonal, meaning that there must exist four mutually orthogonal states $\frac{1}{\sqrt{2}}|U_i\rangle\rangle$ with classical version $\frac{1}{2}(a,1-a,1-a,a)$. From the permutohedron bound, proposition 4, we know that a necessary condition for this is a  =  1/2. Moreover, this condition is sufficient, as the following four unitaries with classical action T are all mutually orthogonal:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:van_der_coher1} U_1=\frac{1}{\sqrt{2}}\left(\begin{array}{@{}cc@{}} -1&1 \nonumber \\ 1&1 \end{array}\right),\quad U_2=\frac{1}{\sqrt{2}}\left(\begin{array}{@{}cc@{}} 1&-1 \nonumber \\ 1&1 \end{array}\right), \nonumber \end{align} \tag{ 100a }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle U_3=\frac{1}{\sqrt{2}}\left(\begin{array}{@{}cc@{}} 1&1 \nonumber \\ -1&1 \end{array}\right),\quad U_4=\frac{1}{\sqrt{2}}\left(\begin{array}{@{}cc@{}} 1&1 \nonumber \\ 1&-1 \end{array}\right). \label{eq:van_der_coher2} \nonumber \end{align} \tag{ 100b }$$

We proceed to finding necessary conditions for $\newcommand{\M}{{{\mathcal M}}} \M(T)=3$. Again, from the rank argument, the considered three channels are either all unitary, or two of them are unitary and one has rank 2. In the first case, we can use the orthogonality condition, so that the existence of three perfectly distinguishable unitary channels is equivalent to the existence of three mutually orthogonal states $\frac{1}{\sqrt{2}}|U_i\rangle\rangle$ with classical version $\frac{1}{2}(a,1-a,1-a,a)$. From the permutohedron bound, we clearly see that it is possible only if $a\in[1/3,2/3]$. In the second case, we have two unitary channels U and $V$, and the third channel is a mixed unitary channel

$$\begin{align} \renewcommand{\dag}{\dagger} \newcommand{\e}{{\rm e}} \newcommand{\re}{{\rm Re}} \displaystyle \label{eq:mixed_unitary} \Phi(\cdot)=\lambda S(\cdot)S^{\dagger} + (1-\lambda) T(\cdot)T^{\dagger}, \nonumber \end{align} \tag{ 101 }$$

with $S,T$ unitary and $\lambda\in(0,1)$, because all unital (bistochastic) qubit channels are mixed-unitary channels [30]. Perfect distinguishability between U and $\Phi$ implies then that one can perfectly distinguish between U and S, and between U and T. Analogous implication holds for $V$. Therefore, perfect distinguishability between U, $V$ and $\Phi$ is equivalent to the existence of two sets of mutually orthogonal vectors: $\frac{1}{\sqrt{2}}\{|\mathbf{U}\rangle\rangle,|\mathbf{V}\rangle\rangle,|\mathbf{S}\rangle\rangle\}$ and $\frac{1}{\sqrt{2}}\{|\mathbf{U}\rangle\rangle,|\mathbf{V}\rangle\rangle,|\mathbf{T}\rangle\rangle\}$. Applying the permutohedron bound to these two sets yields:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle (a,1-a,1-a,a)+\frac{1}{2}(s_1,s_2,s_3,s_4)\leqslant (1,1,1,1) \label{eq:perm_add_1}, \nonumber \end{align} \tag{ 102a }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle (a,1-a,1-a,a)+\frac{1}{2}(t_1,t_2,t_3,t_4)\leqslant (1,1,1,1), \label{eq:perm_add_2} \nonumber \end{align} \tag{ 102b }$$

where the inequalities are elementwise and vectors $\boldsymbol{s}$ and $\boldsymbol{t}$, according to equation (101), satisfy

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \lambda \boldsymbol{s} + (1-\lambda) \boldsymbol{t} =(a,1-a,1-a,a). \nonumber \end{align} \tag{ 103 }$$

We clearly see that if a  >  2/3 then either equation (102a) or (102a) does not hold, because either s₁ or t₁ must be larger than a. Similarly, one of these equations does not hold for a  <  1/3, because either s₂ or t₂ must be larger than 1  −  a.

We can thus conclude that the necessary condition for $\newcommand{\M}{{{\mathcal M}}} \M(T)=3$ is $a\in[1/3,2/3]$. Moreover, this condition is sufficient, since one can find three unitary channels with a fixed classical action T that, when acting on one part of a maximally entangled state $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\Omega}$, map it to three orthogonal states. More precisely, consider the following unitaries

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:unitary_endpoint1} U_{1} = \left(\begin{array}{@{}cc@{}} \sqrt{a} & \sqrt{1-a} \nonumber \\ \sqrt{1-a} & - \sqrt{a} \end{array}\right), \nonumber \end{align} \tag{ 104a }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:unitary_endpoint2} U_{2} = \left(\begin{array}{@{}cc@{}} \sqrt{a} & \sqrt{1-a} {\rm e}^{{\rm i} \theta} \nonumber \\ \sqrt{1-a} {\rm e}^{{\rm i} \phi} & -\sqrt{a} {\rm e}^{{\rm i} (\phi+\theta)} \end{array}\right), \nonumber \end{align} \tag{ 104b }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:unitary_endpoint3} U_{3} = \left(\begin{array}{@{}cc@{}} \sqrt{a} & \sqrt{1-a} {\rm e}^{{\rm i} 2\theta} \nonumber \\ \sqrt{1-a} {\rm e}^{2 {\rm i} \phi} & -\sqrt{a} {\rm e}^{2 {\rm i} (\phi+\theta)} \end{array}\right) \nonumber \end{align} \tag{ 104c }$$

with $\phi$ and $\theta$ specified by:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle 2a-1 = \left(\cot \frac{\phi}{2} \right) \sqrt{\frac{1 - \cot^2 \frac{\phi}{2}}{1 + 3\cot^2 \frac{\phi}{2}}}, \nonumber \end{align} \tag{ 105 }$$

and

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \cot^2 \frac{\theta}{2} = \frac{1 - \cot^2 \frac{\phi}{2}}{1 + 3\cot^2 \frac{\phi}{2}}. \nonumber \end{align} \tag{ 106 }$$

One can check by direct calculation that when $a\in[1/3,2/3]$ the above unitaries are indeed orthogonal. □

As a final remark, let us comment on the most well-studied qubit channels, the phase-damping channel and the (generalised) amplitude-damping channel, from the perspective of our work. It is straightforward to notice that the classical action of a phase-damping channel, specified by Kraus operators

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:phase_damp} K^{\rm PD}_1=\left(\begin{array}{@{}cc@{}} 1 & 0 \nonumber \\ 0 & \sqrt{1-\lambda} \end{array}\right),\quad K^{\rm PD}_2=\left(\begin{array}{@{}cc@{}} 0 & 0 \nonumber \\ 0 & \sqrt{\lambda} \end{array}\right), \nonumber \end{align} \tag{ 107 }$$

is given by the identity matrix for any value of the damping parameter $\lambda\in[0,1]$. Thus, through proposition 17, there exist two perfectly distinguishable channels with the same classical action as the phase-damping channel, e.g. the identity and the phase flip channels. On the other hand, the classical action of the amplitude-damping channel, specified by Kraus operators

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:amplitude_damp} K^{\rm AD}_1=\left(\begin{array}{@{}cc@{}} 1 & 0 \nonumber \\ 0 & \sqrt{1-\gamma} \end{array}\right),\quad K^{\rm AD}_2=\left(\begin{array}{@{}cc@{}} 0 & \sqrt{\gamma} \nonumber \\ 0 & 0 \end{array}\right), \nonumber \end{align} \tag{ 108 }$$

is given by the matrix T from equation (80), with a  =  1 and $b=1-\gamma$. From figures 3(a) and (b), we see that for small damping parameters $\gamma\leqslant 1/2$ there are two perfectly distinguishable channels with the same classical action as the amplitude damping channel, but for $\gamma>1/2$ there exists only one such channel. Finally, the classical action of the generalised amplitude damping channel, specified by Kraus operators

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:gen_amplitude_damp} K^{\rm GAD}_1=\sqrt{p}K_1^{\rm AD},\quad K^{\rm GAD}_2=\sqrt{p}K_2^{\rm AD}, \nonumber \end{align} \tag{ 109a }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle K^{\rm GAD}_3=\sqrt{1-p}\left(\begin{array}{@{}cc@{}} \sqrt{1-\gamma} & 0 \nonumber \\ 0 & 1 \end{array}\right), \nonumber \end{align} \tag{ 109b }$$

$$\begin{align} \renewcommand{\dag}{\dagger} \newcommand{\e}{{\rm e}} \newcommand{\re}{{\rm Re}} \displaystyle K^{\rm GAD}_4=\sqrt{1-p}K_2^{{\rm AD}\dagger}, \nonumber \end{align} \tag{ 109c }$$

is given by T with $a=p+(1-p)(1-\gamma)$ and $b=1-p+p(1-\gamma)$. This means that, depending on $p\in[0,1]$ and $\gamma\in[0,1]$, the parameters of the classical action can take the values $a\in[0,1]$ and $b\in[1-a,1]$ (the upper-right half of figures 3(a) and (b)). Therefore, the number of perfectly distinguishable channels with the same classical action as the generalised amplitude-damping channel can vary between 1 (e.g. for p   =  1 and $\gamma=1$) and 4 (only for p   =  1/2 and $\gamma=1$).

Looking for a coherification of a classical state $\boldsymbol{p}$ we aim at finding its preimage with respect to the completely decohering channel $\newcommand{\D}{{{\mathcal D}}} \D$, i.e. a quantum state $\rho$ such that $\newcommand{\diag}[1]{\mathrm{diag}\left(#1\right)} \diag{\rho}=\boldsymbol{p}$ [16]. In other words, different coherifications of $\boldsymbol{p}$ correspond to different quantum states with a fixed diagonal (representing populations in the distinguished basis $\newcommand{\ket}[1]{| {#1} \rangle} \{\ket{k}\}$) but different off-diagonal terms (representing coherences with respect to $\newcommand{\ket}[1]{| {#1} \rangle} \{\ket{k}\}$). The strength of coherification can be measured, for example, by the purity or l₁-norm of coherence of the coherified state. However, positivity of $\rho$ constrains the off-diagonal terms and the extreme case, which we refer to as complete coherification, corresponds to a pure state $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\psi}$. Now, a set of all complete coherifications of $\boldsymbol{p}$ is a set of all pure states $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\psi}$ satisfying $\newcommand{\diag}[1]{\mathrm{diag}\left(#1\right)} \newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\ketbra}[2]{{\left|#1\right\rangle {\hspace{-2.5pt}}\! \left\langle#2\right|}} \boldsymbol{p}=\diag{\ketbra{\psi}{\psi}}$. Therefore, the complete coherification procedure can be seen as quantization of the simplex of classical probability vectors—the set of all classical input states is mapped to the set of all pure quantum states while preserving the measurement statistics in the distinguished basis. Similarly, if we constrain the strength of coherification, we will map the classical simplex to a set of mixed quantum states corresponding to partially decohered pure states.

Let us visualize this concept using the simplest example of a qubit system. A distribution over a classical bit can be represented by a unit segment with extremal points corresponding to sharp distributions $(1,0)$ and $(0,1)$, see figure A1. We can now embed this classical state space into a quantum one, i.e. the interval $[0,1]$ representing classical probabilistic states becomes embedded inside the 3-dimensional Bloch ball containing the density matrices of size d  =  2. This is visualized in figure A1 as inserting the unit segment into a balloon. Now, the coherification procedure can be understood as inflating the balloon, effectively expanding the state space. Observe that the classical pure states, $(1,0)$ and $(0,1)$, do not change their positions and become quantum basis states, $|0\rangle$ and $|1\rangle$. From the presented picture it is clear that coherification can be seen as inverse of the decoherence process, which leads to the diminishing of the off-diagonal entries of the density matrix, see figure A1.

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Looking for a coherification of a classical stochastic matrix T we aim at finding the preimage of its Jamiołkowski state with respect to the completely decohering channel $\newcommand{\D}{{{\mathcal D}}} \D$, i.e. we look for a quantum channel $\Phi$ such that its Jamiołkowski state $J_\Phi$ satisfies $\newcommand{\diag}[1]{\mathrm{diag}\left(#1\right)} \diag{J_\Phi}=|\mathbf{T}\rangle\rangle\$ [16]. In other words, different coherifications of T correspond to different quantum channels with a fixed classical action (representing population transitions in the distinguished basis $\newcommand{\ket}[1]{| {#1} \rangle} \{\ket{k}\}$) but varying otherwise, e.g. with different action on the off-diagonal terms. Note that, although coherification of quantum channels is defined via the coherification of corresponding quantum states, due to an additional trace-preserving constraint the process is more involved, and complete coherification is generally impossible [16]. Nevertheless, the coherification procedure can again be seen as quantization of the classical space of stochastic matrices—first, one embeds this space in the space of quantum channels, and then maps every stochastic matrix T into a channel with classical action T.

Unlike the set of one-qubit quantum states, which has only three dimensions and can thus be conveniently visualized, the set of all one-qubit quantum channels has 12 dimensions, which makes it hard to analyze. Fortunately, every unital channel acting on a qubit system is unitarily equivalent to a Pauli channel,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Psi_{\boldsymbol{p}}(\cdot)=\sum_{j=0}^3 \lambda_{j} \sigma_j (\cdot) \sigma_j, \nonumber \end{align} \tag{ A.1 }$$

with $\sigma_j$ denoting three Pauli matrices appended by the identity matrix, $\newcommand{\iden}{\mathbb{1}} \sigma_0=\iden$, and $\boldsymbol \lambda$ being a classical probability vector of length four. Thus the set all Pauli channels (and, hence, the set of all unital channels) can be represented by a regular 3D tetrahedron, which can be easily visualized, see figure A2.

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Classical unital channels correspond to bistochastic matrices which, in the case of a 2-dimensional system, can be parametrized by a single number $a\in[0,1]$,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle B_a= \left(\begin{array}{@{}cc@{}} a & 1-a \nonumber \\ 1-a & a \end{array}\right). \nonumber \end{align} \tag{ A.2 }$$

These classical channels, after embedding in the space of unital quantum channels, form an interval within the tetrahedron of unital quantum channels, see figure A2. The endpoints of the interval, B₁ and B₀, correspond to a completely decohering channel $\newcommand{\D}{{{\mathcal D}}} \D$ and $\newcommand{\D}{{{\mathcal D}}} \D$ followed by the Pauli x channel. To see this, note that the Jamiołkowski state of a classical channel B₁ is given by $\newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\ketbra}[2]{{\left|#1\right\rangle {\hspace{-2.15pt}}\! \left\langle#2\right|}} J_{B_1}\propto \ketbra{00}{00}+\ketbra{11}{11}$, while $\newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\ketbra}[2]{{\left|#1\right\rangle {\hspace{-2.25pt}}\! \left\langle#2\right|}} J_{{\mathcal I}}=\ketbra{\Omega}{\Omega}$ and $\newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\ketbra}[2]{{\left|#1\right\rangle {\hspace{-2.25pt}}\! \left\langle#2\right|}} J_{\sigma_z}=\ketbra{\Omega'}{\Omega'}$ with $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\Omega'}\propto\ket{00}-\ket{11}$. It is thus clear that $J_{B_1}=\frac{1}{2}(J_{{\mathcal I}}+J_{\sigma_z})$, so that the classical channel B₁ is given by the equal mixture of identity and Pauli z channels, which in turn is equal to $\newcommand{\D}{{{\mathcal D}}} \D$. Similarly, the Jamiołkowski state of a classical channel B₀ is given by $\newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\ketbra}[2]{{\left|#1\right\rangle{\hspace{-2.25pt}}\! \left\langle#2\right|}} J_{B_0}\propto \ketbra{01}{01}+\ketbra{10}{10}$, while $\newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\ketbra}[2]{{\left|#1\right\rangle {\hspace{-2.5pt}}\! \left\langle#2\right|}} J_{\sigma_x}=\ketbra{\omega}{\omega}$ and $\newcommand{\ket}[1]{| {#1} \rangle} \newcommand{\ketbra}[2]{{\left|#1\right\rangle {\hspace{-2.5pt}}\! \left\langle#2\right|}} J_{\sigma_y}=\ketbra{\omega'}{\omega'}$ with $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\omega}\propto\ket{01}+\ket{10}$ and $\newcommand{\ket}[1]{| {#1} \rangle} \ket{\omega'}\propto\ket{01}-\ket{10}$. Analogously, we have that $J_{B_0}=\frac{1}{2}(J_{\sigma_x}+J_{\sigma_y})$, so that the classical channel B₀ is given by the equal mixture of Pauli x and y  channels, which in turn is equal to $\newcommand{\D}{{{\mathcal D}}} \sigma_x\D(\cdot)\sigma_x$.

To visualize the coherification procedure of the set of classical bistochastic maps we may again imagine inserting the unit interval (representing classical channels) inside a balloon and inflating it. This time, however, the balloon is confined inside the regular tetrahedron of unital channels, see figure A3. In practice, it is hardly possible to inflate the balloon so that it reaches the corners of the tetrahedron, which corresponds to complete coherification of classical bistochastic channels to unitary channels. Note also that, similarly to quantum states, coherification of quantum channels can be seen as the inverse process to the decohering supermap (decohering the Jamiołkowski state of a channel), which in the current case sends all elements of the tetrahedron back to the unit interval of classical bistochastic matrices, see figure A3.

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The permutohedron ${{\mathcal P}}_4^3$ is a tetrahedron with vertices given by $\boldsymbol{f}^1=\frac{1}{3}(1,1,1,0)$ and $\boldsymbol{f}^i$ for $i\in\{2,3,4\}$ are given by permutations of $\boldsymbol{f}^1$. Without loss of generality, a point on the edge of this tetrahedron has the form $\boldsymbol{p}^{(s)} = \frac{1}{3} \left(s,1-s,1,1\right)$ with $s \in \left[0,\frac{1}{2}\right]$. We will consider a point $\boldsymbol{p}^{(s,t)}$ on a line connecting $\boldsymbol{p}^{(s)}$ and the centre $\newcommand{\e}{{\rm e}} \boldsymbol{\eta}$ of ${{\mathcal P}}_4^3$,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \boldsymbol{p}^{(s,t)} = 3\left(\frac{1}{3} - t \right)\boldsymbol{\eta} + 3 t \boldsymbol{p}^{(s)}, \nonumber \end{align} \tag{ C.1 }$$

with $t \in \left[0,\frac{1}{3}\right]$. We will now show that all such points belong to the distinguishability region $\newcommand{\A}{{{\mathcal A}}} \A_4^3$. In order to achieve this we will consider the following three pure states

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \ket{\psi_1} = x_1\ket{1}+x_2\ket{2}+ x_3 \ket{3}+x_4\ket{4}, \nonumber \end{align} \tag{ C.2 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \ket{\psi_2} = x_1\ket{1}+x_2{\rm e}^{{\rm i} \alpha_2}\ket{2}+ x_3{\rm e}^{{\rm i} \alpha_3} \ket{3}+x_4{\rm e}^{{\rm i} \alpha_4}\ket{4}, \nonumber \end{align} \tag{ C.3 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \ket{\psi_3} = x_1\ket{1}+x_2{\rm e}^{{\rm i} \alpha_2}\ket{2}+ x_3{\rm e}^{{\rm i} \alpha_4} \ket{3}+x_4{\rm e}^{{\rm i} \alpha_3}\ket{4}, \nonumber \end{align} \tag{ C.4 }$$

with $x_i = \sqrt{p^{(s,t)}_{i}}$ and prove that for all $s\in \left[0,\frac{1}{2}\right]$ and $t \in \left[0,\frac{1}{3}\right]$ there exists a choice of phases $\{\alpha_2,\alpha_3,\alpha_4\}$, such that the above states are mutually orthogonal.

The overlap $\newcommand{\bra}[1]{\langle {#1} |} \newcommand{\braket}[2]{{\left\langle#1\right| \left.#2\right\rangle}} \braket{\psi_2}{\psi_3}$ reads

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\braket}[2]{{\left\langle#1\right| \left.#2\right\rangle}} \newcommand{\bra}[1]{\langle {#1} |} \displaystyle \braket{\psi_2}{\psi_3} = \frac{1}{2}\left(1 - t + (1+t) \cos(\alpha_3 - \alpha_4) \right), \nonumber \end{align} \tag{ C.5 }$$

so that orthogonality condition, $\newcommand{\bra}[1]{\langle {#1} |} \newcommand{\braket}[2]{{\left\langle#1\right| \left.#2\right\rangle}} \braket{\psi_2}{\psi_3}=0$, gives

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{alpha3_condition} \alpha_3 = \arccos\frac{t-1}{t+1} + \alpha_4. \nonumber \end{align} \tag{ C.6 }$$

The remaining overlaps are equal, $\newcommand{\bra}[1]{\langle {#1} |} \newcommand{\braket}[2]{{\left\langle#1\right| \left.#2\right\rangle}} \braket{\psi_1}{\psi_2} = \braket{\psi_1}{\psi_3} =: F$, and given by

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle F=&~\frac{1}{4}[1 + \left(4s - 3 \right)t +(1+t-4st){\rm e}^{{\rm i} \alpha_2} +\left. \left(1 + t\right)({\rm e}^{{\rm i} \alpha_3} +{\rm e}^{{\rm i} \alpha_4})\right]. \nonumber \end{align} \tag{ C.7 }$$

Using equation (C.6) we can simplify the above expression to arrive at

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:annulus} F = &~\frac{1}{4}[1 + \left(4s - 3 \right)t + (1 + t -4st){\rm e}^{{\rm i} \alpha_2} + 2\left({\rm i}\sqrt{t} +t \right){\rm e}^{{\rm i} \alpha_4}]. \nonumber \end{align} \tag{ C.8 }$$

We now note that equation (C.8) for all $\alpha_2,\alpha_4 \in [0,2\pi)$ describes an annulus $\mathrm{ann}(x;R,r)$ with the centre x, larger radius R and smaller radius r equal to

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle x = \frac{1}{4} \left(1 + \left(4s - 3 \right)t \right),\label{eq:ann1} \nonumber \end{align} \tag{ C.9 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle R = \frac{1}{4}\left| \left| 1 + t -4st \right|+ 2 \left| i\sqrt{t} +t \right| \right|,\label{eq:ann2} \nonumber \end{align} \tag{ C.10 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle r = \frac{1}{4}\left|\left|1 + t -4st\right| - 2\left|i\sqrt{t} +t \right| \right| .\label{eq:ann3} \nonumber \end{align} \tag{ C.11 }$$

The existence of phases such that $\newcommand{\ket}[1]{| {#1} \rangle} \{\ket{\psi_1},\ket{\psi_2},\ket{\psi_3}\}$ are mutually orthogonal is thus equivalent to $0 \in \mathrm{ann}(x;R,r)$, which can be verified by the following elementary calculations.

First, we need to prove that the distance between the centre of the annulus x and 0 is smaller than the larger radius R (see figure C1). Substituting the expressions for x and R into $x \leqslant R$ yields

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle 4t(1-2s) + 2\sqrt{t(t+1)}\geqslant 0, \nonumber \end{align} \tag{ C.12 }$$

which is always satisfied since $s\leqslant 1/2$ and $t\geqslant 0$. Next, we need to prove that the distance between the centre of the annulus x and 0 is larger than the smaller radius r. Here we have two cases: one when $\left|1 + t -4st\right| \geqslant 2\left|i\sqrt{t} +t \right|$ holds, and one when the opposite holds. In both cases the condition $x\geqslant r$ simplifies to $t \leqslant 1/3$ which is always true, because $t\in[0,1/3]$.

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Without loss of generality, a point $\boldsymbol{p}$ lying in the interior of the face of permutohedron ${{\mathcal P}}_4^3$ can be expressed by

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \boldsymbol{p} = \frac{1}{3}(1,r+s,q+s,q+r), \nonumber \end{align} \tag{ C.13 }$$

with q  +  r  +  s  =  1 and $q,r,s>0$. Changing variables according to

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle a_1 =\frac{r+s}{2},\quad a_2 =\frac{q+s}{2},\quad a_3 =\frac{q+r}{2}, \nonumber \end{align} \tag{ C.14 }$$

we have

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \boldsymbol{p}=\frac{1}{3}(1,2a_1,2a_2,2a_3), \nonumber \end{align} \tag{ C.15 }$$

with $a_1+a_2+a_3=1$, $0<a_j<\frac{1}{2}$.

We will now prove that one cannot find three mutually orthogonal states $\newcommand{\ket}[1]{| {#1} \rangle} \{\ket{\psi_1},\ket{\psi_2},\ket{\psi_3}\}$ with the classical version given by $\boldsymbol{p}$. The general form of such states is given by

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \ket{\psi_1} = \frac{1}{\sqrt{3}} \left[\ket{0}+\sqrt{2}\sum_{j=1}^3 \sqrt{a_j}\ket{j} \right] \nonumber \end{align} \tag{ C.16 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \ket{\psi_2} = \frac{1}{\sqrt{3}} \left[\ket{0}+\sqrt{2}\sum_{j=1}^3 \sqrt{a_j}{\rm e}^{{\rm i}\phi_j}\ket{j} \right] \nonumber \end{align} \tag{ C.17 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \ket{\psi_3} = \frac{1}{\sqrt{3}} \left[\ket{0}+\sqrt{2}\sum_{j=1}^3 \sqrt{a_j}{\rm e}^{{\rm i}\theta_j}\ket{j} \right]. \nonumber \end{align} \tag{ C.18 }$$

Orthogonality conditions can be now rewritten in terms of truncated vectors,

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \ket{\tilde{\psi_1}} = \sum_{j=1}^{3}\sqrt{a_j} \ket{j},. \nonumber \end{align} \tag{ C.19 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \ket{\tilde{\psi_2}} = \sum_{j=1}^{3}\sqrt{a_j} {\rm e}^{{\rm i} \phi_j}\ket{j},\label{psi2} \nonumber \end{align} \tag{ C.20 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \ket{\tilde{\psi_3}} = \sum_{j=1}^{3}\sqrt{a_j} {\rm e}^{{\rm i} \theta_j}\ket{j} , \nonumber \end{align} \tag{ C.21 }$$

as:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:ortho} \langle\tilde{\psi_1}|\tilde{\psi_2}\rangle = \langle\tilde{\psi_1}|\tilde{\psi_3}\rangle = \langle\tilde{\psi_2}|\tilde{\psi_3}\rangle = -\frac{1}{2}. \nonumber \end{align} \tag{ C.22 }$$

We will now prove, by contradiction, that for the orthogonality condition to hold the set of vectors $\newcommand{\ket}[1]{| {#1} \rangle} \{\ket{\tilde{\psi_1}},\ket{\tilde{\psi_2}},\ket{\tilde{\psi_3}}\}$ must be linearly independent. Assume that these vectors are linearly dependent, i.e. there exists complex numbers $\alpha,\beta,\gamma$ such that $|\alpha| + |\beta|+ |\gamma| > 0$ and

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \alpha \ket{\tilde{\psi_1}} + \beta \ket{\tilde{\psi_2}} + \gamma \ket{\tilde{\psi_3}} = 0. \nonumber \end{align} \tag{ C.23 }$$

By the orthogonality condition, equation (C.22), we have

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \alpha - \frac{\beta}{2} - \frac{\gamma}{2} = 0, \nonumber \end{align} \tag{ C.24 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle -\frac{\alpha}{2} + \beta -\frac{\gamma}{2} = 0, \nonumber \end{align} \tag{ C.25 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle -\frac{\alpha}{2} - \frac{\beta}{2} + \gamma =0, \nonumber \end{align} \tag{ C.26 }$$

which leads to conclusion that $\alpha = \beta = \gamma$. Since at least one of them is nonzero, all are nonzero. Thus, we have

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \ket{\tilde{\psi_3}} = - \ket{\tilde{\psi_1}} - \ket{\tilde{\psi_2}}. \nonumber \end{align} \tag{ C.27 }$$

But this means:

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \sum_{j=1}^{3} \sqrt{a_j} {\rm e}^{{\rm i} \theta_j} \ket{j} = \ket{\tilde{\psi_3}} = - \sum_{j=1}^{3} \sqrt{a_j} \left(1 + {\rm e}^{{\rm i} \phi_j} \right)\ket{j}. \nonumber \end{align} \tag{ C.28 }$$

Hence,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \forall j \in \{1,2,3\}: \quad {\rm e}^{{\rm i} \theta_j}= - \left(1 + {\rm e}^{{\rm i} \phi_j} \right). \nonumber \end{align} \tag{ C.29 }$$

In particular, focusing on the absolute value,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{cos} \forall j \in \{1,2,3\}: \quad \cos \phi_j = -\frac{1}{2}, \nonumber \end{align} \tag{ C.30 }$$

which implies

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{sin} \forall j \in \{1,2,3\}: \quad \sin \phi_j = \epsilon_j \frac{\sqrt{3}}{2}, \nonumber \end{align} \tag{ C.31 }$$

where $\newcommand{\e}{{\rm e}} \epsilon_j \in \{-1,1\}$. Inserting equations (C.30) and (C.31) into (C.20), we obtain

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\ket}[1]{| {#1} \rangle} \displaystyle \ket{\tilde{\psi_2}} = \sum_{j=1}^{3} \sqrt{a_j} \left(-\frac{1}{2} + \epsilon_j i \frac{\sqrt{3}}{2} \right) \ket{j}. \nonumber \end{align} \tag{ C.32 }$$

The overlap $\langle\tilde{\psi_1}|\tilde{\psi_2}\rangle$ is therefore equal to

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \langle\tilde{\psi_1}|\tilde{\psi_2}\rangle & = \sum_{j=1}^{3} {a_j} \left(-\frac{1}{2} + \epsilon_j i \frac{\sqrt{3}}{2} \right) = - \frac{1}{2} + {\rm i} \frac{\sqrt{3}}{2} \sum_{j=1}^{3} a_j \epsilon_j. \nonumber \end{align} \tag{ C.33 }$$

Since we know that $\langle\tilde{\psi_1}|\tilde{\psi_2}\rangle=-\frac{1}{2}$ the imaginary part must vanish, meaning that

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \sum_{j=1}^{3} a_j \epsilon_j = 0. \nonumber \end{align} \tag{ C.34 }$$

But this is only possible if a_i are a permutation of $\left(\frac{1}{2},\frac{1}{4},\frac{1}{4}\right)$. As we know that $a_j < \frac{1}{2}$ we arrive at a contradiction, which disproves the assumption of linear dependence of $\newcommand{\ket}[1]{| {#1} \rangle} \{\ket{\tilde{\psi_1}},\ket{\tilde{\psi_2}},\ket{\tilde{\psi_3}}\}$.

We now know that the set $\newcommand{\ket}[1]{| {#1} \rangle} \{\ket{\tilde{\psi_1}},\ket{\tilde{\psi_2}},\ket{\tilde{\psi_3}}\}$ must be linearly independent but, on the other hand, the Gram matrix for these vectors reads

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \left(\begin{array}{@{}ccc@{}} 1 & -\frac{1}{2} & -\frac{1}{2} \nonumber \\ -\frac{1}{2}& 1 & -\frac{1}{2} \nonumber \\ -\frac{1}{2}& -\frac{1}{2}&1 \end{array}\right). \nonumber \end{align} \tag{ C.35 }$$

It is straightforward to check that it is of rank 2, which means that vectors $\newcommand{\ket}[1]{| {#1} \rangle} \{\ket{\tilde{\psi_1}},\ket{\tilde{\psi_2}},\ket{\tilde{\psi_3}}\}$ are linearly dependent. This finishes the proof that orthogonal $\newcommand{\ket}[1]{| {#1} \rangle} \{\ket{\tilde{\psi_1}},\ket{\tilde{\psi_2}},\ket{\tilde{\psi_3}}\}$ cannot exist.

We now proceed to states with classical version $\boldsymbol{q}$ lying on the lines connecting the centre of the permutohedron ${{\mathcal P}}_4^3$ with the centres of its faces. Without loss of generality such a point can be expressed by

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \boldsymbol{q} = \frac{1}{4}(1-3t,1+t,1+t,1+t), \nonumber \end{align} \tag{ C.36 }$$

with $t\in[-1/9,0]$. Now, via lemma 5, we know that the existence of mutually orthogonal $\newcommand{\ket}[1]{| {#1} \rangle} \{\ket{{\psi_1}},\ket{{\psi_2}},\ket{{\psi_3}}\}$ with classical version $\boldsymbol{q}$ is equivalent to the existence of a unistochastic matrix

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle T(\boldsymbol{q})= \frac{1}{4}\left(\begin{array}{@{}cccc@{}} 1-3t & 1-3t& 1-3t & 1+9t \nonumber \\ 1+t & 1+t & 1+t &1-3t \nonumber \\ 1+t & 1+t & 1+t &1-3t \nonumber \\ 1+t & 1+t & 1+t & 1-3t \end{array}\right). \nonumber \end{align} \tag{ C.37 }$$

The above family of bistochastic matrices has been studied in [24], where the authors showed that for no values of $t\in[-1/9,0]$ is $T(\boldsymbol{q})$ unistochastic (see equation (33) of [24]). Therefore, there cannot exist three orthogonal pure states with classical action given by $\boldsymbol{q}$.

Due to lemma 5, the problem of finding d  −  1 orthogonal vectors with the same classical version $\boldsymbol{p}$ is equivalent to verifying whether a particular bistochastic matrix B, with the first d  −  1 columns given by $\boldsymbol{p}$, is unistochastic. By employing the algorithm proposed by Uffe Haagerup and described in [36] we numerically verified whether matrices B corresponding to probability vectors within the permutohedron ${{\mathcal P}}_4^3$ are unistochastic, and this way obtained a numerical approximation of the distinguishability region $\newcommand{\A}{{{\mathcal A}}} \A_4^3$. Its form suggested the conjecture described by equation (30) and the supporting evidence is presented in figure C2.

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