Extreme Points and Factorizability for New Classes of Unital Quantum Channels

Abstract

We introduce and study two new classes of unital quantum channels. The first class describes a 2-parameter family of channels given by completely positive (CP) maps $M_3({\mathbf {C}})\mapsto M_3({\mathbf {C}})$ which are both unital and trace-preserving. Almost every member of this family is factorizable and extreme in the set of CP maps which are both unital and trace-preserving, but is not extreme in either the set of unital CP maps or the set of trace-preserving CP maps. We also study a large class of maps which generalize the Werner-Holevo channel for $d = 3$ in the sense that they are defined in terms of partial isometries of rank $d-1$. Moreover, we extend this to maps whose Kraus operators have the form $t \, | e_j \rangle \langle e_j| \oplus V$ with $V \in M_{d-1} ({\mathbf {C}})$ unitary and $t \in (-1,1)$. We show that almost every map in this class is extreme in both the set of unital CP maps and the set of trace-preserving CP maps. We analyze in detail a particularly interesting family which is extreme unless $t = \tfrac{-1}{d-1}$. For $d = 3$, this includes a pair of channels which have a dual factorization in the sense that they can be obtained by taking the partial trace over different subspaces after using the same unitary conjugation in $M_3({\mathbf {C}}) \otimes M_3({\mathbf {C}})$.

Appendices

Factorizability

1.1 Conditions for Factorizability

Let $\Phi :M_d({\mathbf {C}}) \rightarrow M_d({\mathbf {C}})$ be a UCPT map and ${\mathcal {N}}$ a von Neumann algebra with a faithful trace $\tau$ normalized so that $\tau (I_{{\mathcal {N}}}) = 1$. Following [6], we say (as noted in the introduction) that $\Phi$ has an exact factorization through $M_d({\mathbf {C}}) \otimes {{\mathcal {N}}}$ if there is a unitary ${\mathbf {U}} \in M_d({\mathbf {C}}) \otimes {\mathcal {N}}$ such that for all $\rho \in M_d({\mathbf {C}})$ (2) holds, i.e., $\Phi (\rho ) = ({\mathcal {I}} \otimes \tau ){\mathbf {U}}^*(\rho \otimes I_{{\mathcal {N}}}) {{\mathbf {U}}}$. Furthermore, $\Phi$ is called factorizable if it has an exact factorization through $M_d({\mathbf {C}}) \otimes {\mathcal {N}}$ for some $({\mathcal {N}}, \tau )$.

When ${\mathcal {N}} = M_\nu ({\mathbf {C}})$ is a matrix algebra, (2) becomes $\Phi (\rho ) = ({\mathcal {I}} \otimes \mathrm {Tr}){\mathbf {U}}^*(\rho \otimes \frac{1}{\nu } \mathrm{I}_\nu ) {{\mathbf {U}}}$. Following standard practice used elsewhere in this paper, when ${{{\mathcal {N}}}} = M_d({\mathbf {C}})$, we denote the identity by $\mathrm{I}_d$ rather than the awkward $\mathrm{I}_{M_d({\mathbf {C}})}$. We also follow the standard convention for type I algebras that $\hbox {Tr} \,$ is normalized so that $\hbox {Tr} \, | v \rangle \langle v| = 1$ when v with $\Vert v \Vert ^2 = \langle v, v \rangle = 1$.

The following useful result follows from the factorizability criteria in [5, Theorem 2.2].

Proposition A.1

Necessary and sufficient conditions for a UCPT map $\Phi$ on $M_d({\mathbf {C}})$ of the form $\Phi (\rho ) = \sum _{k=1}^\kappa A_k^* \rho A_k$, with $\{A_1, A_2, \dots , A_\kappa \}$ linearly independent to have an exact factorization through $M_d({\mathbf {C}}) \otimes {\mathcal {N}}$ are that there are $Y_1,Y_2, \dots , Y_\kappa \in {\mathcal {N}}$ such that $\tau ( Y_j Y_k^*) = \delta _{jk}$ and (with $\{|e_s\rangle \}_s$ the standard basis for ${\mathbf {C}}_d$) the following conditions hold for all $1 \le s,t \le d$:

$$\begin{aligned} \sum _{j,k} \langle e_s, A_jA_k^* e_t \rangle Y_jY_k^* = \delta _{st} \, I_{\mathcal {N}}, \quad \sum _{j,k} \langle e_s, A_j^*A_k e_t \rangle Y_j^*Y_k = \delta _{st} \, I_{\mathcal {N}}. \end{aligned}$$

(65)

Proof

By [5, Theorem 2.2], if $\Phi$ has an exact factorization through $M_d({\mathbf {C}}) \otimes {{\mathcal {N}}}$, then one can find $Y_1,Y_2, \dots , Y_\kappa$ in ${\mathcal {N}}$ such that ${\mathbf {U}} := \sum _{k=1}^\kappa A_k \otimes Y_k \in M_d({\mathbf {C}}) \otimes {\mathcal {N}}$ is unitary. Hence

$$\begin{aligned} {\mathbf {U}}{\mathbf {U}}^* = I_d \otimes I_{{\mathcal {N}}} = \sum _{j,k} A_jA_k^* \otimes Y_jY_k^*, \quad {\mathbf {U}}^*{\mathbf {U}} = I_d \otimes I_{{\mathcal {N}}} = \sum _{j,k} A_j^*A_k \otimes Y_j^*Y_k. \end{aligned}$$

(66)

Now observe that for any $A \in M_d({\mathbf {C}})$ and $Y \in {\mathcal {N}}$ one has

$$\begin{aligned} A \otimes Y = \sum _{s,t} \langle e_s, A e_t \rangle \, | e_s \rangle \langle e_t | \otimes Y = \sum _{s,t} | e_s \rangle \langle e_t | \otimes \langle e_s, A e_t \rangle \, Y. \end{aligned}$$

Applying this to both sides of the equations in (66) and using the fact that $\{ |e_s \rangle \langle e_t |\}_{s,t}$ is a basis for $M_d({\mathbf {C}})$, gives the equations in (65). $\square$

1.2 Pairs of Factorizable Maps

Definition A.2

Let $d = p\cdot q$ be a product of primes and ${\mathbf {U}}\in M_d({\mathbf {C}}) \simeq M_p({\mathbf {C}}) \otimes M_q({\mathbf {C}})$ unitary. Then $\Phi (\rho ) = ( {{\mathcal {I}}}\otimes \mathrm{Tr} ){\mathbf {U}}^* \big (\rho \otimes \tfrac{1}{q} \mathrm{I}_q \big ) {\mathbf {U}}$ and $\Psi (\gamma ) = ( \mathrm{Tr} \otimes {{\mathcal {I}}}){\mathbf {U}}^* \big ( \tfrac{1}{p } \mathrm{I}_p \otimes \gamma \big ) {\mathbf {U}}$ are UCPT maps on $M_p ({\mathbf {C}})$ and $M_q ({\mathbf {C}})$, respectively. air The channels $\Phi , \Psi$ are said to be dual channels associated with the unitary ${\mathbf {U}}$.

Theorem 4.8 essentially says that the channels $\Phi , \Psi$ in Sect. 4.5 are dual channels associated with the unitary ${\mathbf {U}}$ given by (47) and $p = q = 3$.

Let $\{ U_j \}$ be a set of q unitary matrices in $M_p({\mathbf {C}})$ and ${{\mathbf {X}}} = \bigoplus _{j=1}^q U_j = \sum _{j=1} U_j \otimes | e_k \rangle \langle e_k|$.

Then ${{\mathbf {X}}} \in M_p({\mathbf {C}}) \otimes M_q ({\mathbf {C}})$ is unitary and, as essentially observed in [6, Proposition 2.8], the associated pair of dual channels are $\Phi (\rho ) = \tfrac{1}{q} \sum _{j=1}^q U_j^* \rho \, U_j$ and $\Upsilon (\gamma ) = C \circledast \gamma$, where $\circledast$ denotes the Schur or Hadamard product and $C \in M_q ({\mathbf {C}})$ is the matrix with elements $c_{jk} = \tfrac{1}{p} \hbox {Tr} \, U_j^* U_k$. When the $U_j$ are also orthogonal so that $\hbox {Tr} \, U_j^* U_k = p \, \delta _{jk}$, then $\Upsilon (\gamma ) = \mathrm{I}_q \circledast \gamma = \sum _{j=1}^q \gamma _{jj} | e_j \rangle \langle e_j|$ is diagonal.

Whenever a UCPT map $\Phi$ has an exact factorization through $M_d({\mathbf {C}}) \otimes M_\nu ({\mathbf {C}})$, there is a unitary ${\mathbf {U}}\in M_d({\mathbf {C}}) \otimes M_\nu ({\mathbf {C}})$ for which $\Phi (\rho ) = ({\mathcal {I}} \otimes \mathrm {Tr}){\mathbf {U}}^*(\rho \otimes \frac{1}{\nu } \mathrm{I}_\nu ) {{\mathbf {U}}}$ so that there is another UCPT map $\Psi$ such that $\Phi , \Psi$ are the dual pair associated with that unitary. However, a UCPT map can have an exact factorization in more than one way. An example is given by the channel $\Psi$ in Sect. 4.5 which corresponds to $t = 1$ in (16). This channel has an exact factorization through $M_3({\mathbf {C}}) \otimes M_3({\mathbf {C}})$ with the unitary ${\mathbf {W}}$ given by (48) and another with ${\mathbf{X}} = \sum _{k = 1}^3 B_j \otimes | e_j \rangle \langle e_j|$ as in Remark 3.1. The dual pair associated with ${\mathbf {W}}$ is $\Psi , \Phi$ as in Section 4.5; the dual pair associated with ${\mathbf{X}}$ is $\Psi , \Upsilon$ where $\Upsilon (\gamma ) =\mathrm{I}_3 \circledast \gamma$ as above.

One can extend this to situations in which d is a product of more than two primes. However, each way of writing d as a product of two integers will give a different pair of dual channels. Finally, we note that when $p = q$, it is possible to have a self-dual channel. For example, when ${\mathbf {U}}=\tfrac{1}{\sqrt{2}} \big ( \sigma _x \otimes \sigma _z + i \sigma _z \otimes \sigma _x )$, one finds $\Phi (\rho ) = \Psi (\rho ) = \tfrac{1}{2}\big ( \sigma _x \rho \sigma _x + \sigma _z \rho \sigma _z )$. with $\sigma _{x,y.z}$ denoting the usual Pauli matrices.

It is worth noting that ${\mathbf {U}}= \sum _k A_k \otimes B_k$ with $A_k \in M_p({\mathbf {C}})$ and $B_k \in M_q({\mathbf {C}})$ does not imply that $\Phi (\rho ) = \sum _k A_k^* \rho A_k$ and $\Psi (\gamma ) = \sum _k B_k^* \gamma B_k$. This only holds if $\hbox {Tr} \, B_j^* B_k = q \, \delta _{jk}$ in the first case and $\hbox {Tr} \, A_j^*A_k =p \, \delta _{jk}$ in the latter. Thus, for example, when ${\mathbf {U}}$ is given by (5), $\Psi$ in the dual pair $\Phi _{\alpha ,\beta } , \Psi _{\alpha ,\beta }$ is not given by $\Psi (\gamma ) = \tfrac{1}{2}\sum _{j,k =1}^2 |e_j \rangle \langle e_k | \gamma |e_k \rangle \langle e_j | = \tfrac{1}{2}( \hbox {Tr} \, \gamma ) \mathrm{I}_2$, but by

$$\begin{aligned} \Psi _{\alpha ,\beta }&= \tfrac{|\alpha |^2 + 1}{3} \big ( | e_1 \rangle \langle e_1| \gamma | e_1 \rangle \langle e_1| + | e_2 \rangle \langle e_2| \gamma | e_2 \rangle \langle e_2| \big ) \\&\quad + \tfrac{|\beta |^2 + 1}{3} \big ( |e_1 \rangle \langle e_2 | \gamma |e_2 \rangle \langle e_1 | + |e_2 \rangle \langle e_1 | \gamma |e_1 \rangle \langle e_2 | \big ) \\&= \tfrac{1}{3} (\hbox {Tr} \, \gamma ) \mathrm{I}_2 +\tfrac{1}{3} \begin{pmatrix} a^2 \gamma _{11} + b^2 \gamma _{22} &{} 0 \\ 0 &{} b^2 \gamma _{11} + a^2 \gamma _{22} \end{pmatrix} . \end{aligned}$$

It should be emphasized that this duality is not equivalent to the notion of a pair of “complementary channels” used in the quantum information literature [5, 8, 14], which is defined in terms of the Stinespring representation, and goes back to Arveson [2, Section 1.3] who used the term “lifting”. In that case, the auxiliary space is interpreted as the environment and the complementary channel maps the input state $\rho$ to a state for the environment. For a channel $\Phi : {\mathbf {C}}_{d_A} \mapsto {\mathbf {C}}_{d_B}$ of the form $\Phi (\rho _A) = \sum _{k = 1}^{d_E} A_k^* \rho A_k$, one can regard the Stinespring representation as mapping $\rho _A \mapsto \rho _{BE} = \sum _{jk} A_j ^*\rho _A A_k \otimes |e_j \rangle \langle e_k |$ with $|e_j \rangle$ the standard basis for ${\mathbf {C}}_{d_E}$. Then $\Phi (\rho _A) = \hbox {Tr} _E \, \rho _{BE}$ and the complementary channel $\Phi ^C : {\mathbf {C}}_{d_A} \mapsto {\mathbf {C}}_{d_E}$ is

$$\begin{aligned} \Phi ^C(\rho _A) = \hbox {Tr} _B \, \rho _{BE} = \sum _{jk} \hbox {Tr} \, (A_j ^*\rho _A A_k ) |e_j \rangle \langle e_k | . \end{aligned}$$

Even when $d_A = d_B$ these concepts are quite different. Complementary channels are defined with the implicit assumption of a pure ancilla rather than a maxiamlly mixed ancilla. In the notion of dual pairs introduced above, the roles of the input and environment are interchanged in terms of both the subspace over which the trace is taken and the space in which the (maximally mixed) ancilla resides.

1.3 Factorizability of $\Phi \circ \Phi ^*$ with Choi Rank $\le 4$

When a UCPT map $\Phi$ is factorizable, the adjoint $\Phi ^*$ is also factorizable. Moreover, the maps $\Phi \circ \Phi$, $\Phi ^* \circ \Phi$, and $\Phi \circ \Phi ^*$ are also factorizable. However, there are some special circumstances in which $\Phi$ is not factorizable, but $\Phi ^* \circ \Phi$ and $\Phi \circ \Phi ^*$ are factorizable. This includes the Arveson–Ohno channel (3) and the channels in Sect. 3.3.1 for $d = 4$ and all $t \in (-1, 1)$.

The next result is a straightforward generalization of [6, Remark 5.6]. It follows from [6, Lemma 5.5] that the condition Choi rank $\le 4$ is critical.

Proposition A.3

Let $\Phi : M_d({\mathbf {C}}) \mapsto M_d({\mathbf {C}})$ be a UCPT map with Choi rank $\le 4$. Then the maps $\Phi \circ \Phi ^*$ and $\Phi ^* \circ \Phi$ each have exact factorizations through $M_d({\mathbf {C}}) \otimes M_4({\mathbf {C}})$.

Proof

Let $\{A_k \in M_d({\mathbf {C}} ) : k =1, 2, 3, 4 \}$ satisfy $\sum _{k = 1}^4 A_k^*A_k = \sum _{k = 1}^4 A_k A_k^* = \mathrm{I}_d$ and define the UCPT map $\Phi (\rho ) = \sum _{k = 1}^4 A_k^* \rho A_k$. As observed in [6], one can always choose some $A_k = 0$ so that the result follows if there is a unitary map ${\mathbf {U}}\in M_{4d}({\mathbf {C}})$ such that

$$\begin{aligned} ({{\mathcal {I}}}\otimes \hbox {Tr} \, ) {\mathbf {U}}^* ( \rho \otimes \tfrac{1}{4} \mathrm{I}_4 ) {\mathbf {U}}= (\Phi \circ \Phi )^*(\rho ) = \sum _{j = 1}^4 \sum _{k = 1}^4 A_j^* A_k \rho A_k^* A_j . \end{aligned}$$

(67)

Following the strategy in [6, Remark 5.6], define ${\mathbf {U}}= \sum _{j,k = 1}^4 A_j^* A_k \otimes (2 | e_j \rangle \langle e_k | - \delta _{jk} \mathrm{I}_4 )$. Then, by repeatedly using $\sum _{k = 1}^4 A_k^*A_k = \sum _{k = 1}^4 A_k A_k^* = \mathrm{I}_d$ one finds

$$\begin{aligned} {\mathbf {U}}{\mathbf {U}}^*&= \sum _{j,k = 1}^4 \sum _{m,n = 1}^4 A_j^* A_k A_n^* A_m \otimes ( 2 | e_j \rangle \langle e_k | -\delta _{jk} \mathrm{I}_4 ) (2 | e_n \rangle \langle e_m | - \delta _{mn} \mathrm{I}_4 ) \\&= 4 \sum _{jkm} A_j^* A_k A_k^* A_m \otimes |e_j \rangle \langle e_m | - 2 \sum _{j,m,n} A_j^* A_j A_n^* A_m \otimes | e_n \rangle \langle e_m | \\&\quad - 2\sum _{jkm} A_j^* A_k A_m^* A_m \otimes | e_j \rangle \langle e_k | + \sum _{k,n} A_k^* A_k A_n^* A_n \otimes \mathrm{I}_4 \\&= 4 \sum _{jm} A_j^* A_m \otimes |e_j \rangle \langle e_m | - 2 \sum _{m,n} A_n^* A_m \otimes | e_n \rangle \langle e_m |\\&\quad - 2 \sum _{jk} A_j^* A_k \otimes | e_j \rangle \langle e_k | + \mathrm{I}_d \otimes \mathrm{I}_4 \\&= \mathrm{I}_d \otimes \mathrm{I}_4 \end{aligned}$$

so that ${\mathbf {U}}$ is unitary. Similarly, one finds (using $\hbox {Tr} \, |e_j \rangle \langle e_k | = \delta _{jk}$ and $\hbox {Tr} \, \mathrm{I}_4 = 4$)

$$\begin{aligned} ({{\mathcal {I}}}\otimes \hbox {Tr} \, ) {\mathbf {U}}^* ( \rho \otimes \tfrac{1}{4} \mathrm{I}_4 ) {\mathbf {U}}&= \tfrac{1}{4} \sum _{jkmn} A_j^* A_k \rho A_n^* A_m \big ( 4 \delta _{jm} \delta _{kn} -2 \delta _{jk} \delta _{mn}\\&\quad -2 \delta _{jk} \delta _{mn} + 4 \delta _{jk} \delta _{mn} \big ) \\&= \sum _{jk} A_j^* A_k \rho A_k^* A_j = (\Phi \circ \Phi )^*(\rho ) \end{aligned}$$

$\square$

Linear Dependence of $\{ A_m^* A_n \}$ versus $\{ A_m A_n^* \}$

We give an explicit example to show that $\{ A_m^* A_n \}$ can be linearly independent, but $\{ A_m A_n^* \}$ linearly dependent. Let $d = 4$ and $W = \tfrac{1}{21} \begin{pmatrix} 8 &{} -11 &{} 16 \\ -19 &{} -8 &{} 4 \\ -4 &{} 16 &{} 13 \end{pmatrix}$. When $A_m$ is constructed as in (11) with all $V_m = W$, we found that

• $\{ A_m^* A_n \}$ is linearly independent unless $t = \pm 1, t = \tfrac{-13}{3} , t = \tfrac{-59}{84} , t = \tfrac{19}{21}$ , or $t = \tfrac{107}{21}$.

• $\{ A_m A_n^* \}$ is linearly independent unless $t = \pm 1 , t = \tfrac{-59}{84} ,~ t = \tfrac{-1}{7} , t= \tfrac{19}{21}$ , or $t = \tfrac{107}{21}$.

Thus, when $t = -\tfrac{1}{7}$, $\{ A_m^* A_n^* \}$ is linearly independent but $\{ A_m A_n^* \}$ is linearly dependent. When $t = - \tfrac{13}{3}$, $\{ A_m A_n^* \}$ is linearly independent but $\{ A_m^* A_n \}$ is linearly dependent. At $t = \pm 1, ~t = \tfrac{-59}{84} ,~ t = \tfrac{19}{21}, ~ t = \tfrac{107}{21}$ both sets are linearly dependent. For all other values of $t \in {{\mathbf {R}}}$ both sets are linearly independent.

This result might seem counter-intuitive because the cyclicity of the trace implies

$$\begin{aligned} \hbox {Tr} \, A_j^* A_k (A_m^* A_n )^* = \hbox {Tr} \, A_m A_j^* A_k A_n^* = \hbox {Tr} \, A_m A_j^* (A_n A_k^*)^* \end{aligned}$$

so that the Gram matrices for the two sets have the same elements, albeit arranged differently. Let G and H denote these Gram matrices and consider the elements $g_{11,kk} = \hbox {Tr} \, A_1^* A_1 (A_k^* A_k )^* = \hbox {Tr} \, A_k A_1^* (A_k A_1^* )^* = h_{k1,k1}$. Since $g_{11,kk}$ all lie in the first row of G, $\det G$ will not contain any terms with $g_{11,kk} \cdot g_{11,jj}$ when $j \ne k$. However, since $h_{k1,k1}$ lies on the diagonal of H, $\det H$ will contain a term which includes $\prod _{k = 2}^d h_{k1,k1} = \prod _{k = 2}^d g_{11,kk}$.

We conjecture that if $\{ A_m A_n^* \}$ is linearly dependent for all t, then $\{ A_m A_n^* \}$ should also be linearly dependent for all t.