Completely Positive Maps of a Hilbert Algebra with Hermitian Adjoint

Introduction

A Clifford algebra with a positive involution is a finite-dimensional $C^{*}$-algebra, and on such an algebra the maps that preserve the order are classified. The classification is Choi's theorem, and its two consequences are the Kraus form of a completely positive map and the statement that the automorphisms of a full matrix algebra are inner. Both consequences have a direct reading in the Clifford language, and that reading is the content of this article.

The first reading is that the Hermitian sandwich is a completely positive map. For a fixed element $x$ the operator $\Theta_x(y) = x\,y\,x^{\dagger}$ is the elementary completely positive map of Kraus rank one, and it is therefore positive in the strongest sense: its amplification to matrices over the algebra is positive too. It is unital exactly on the unitary slice, and it is a unital algebra automorphism exactly there; so the unitary slice is the group of completely positive maps of the sandwich shape that are automorphisms, and the automorphisms of the Clifford algebra are the inner ones together with the anti-automorphisms induced by reversion and Clifford conjugation.

The second reading is that this gives the algebraic model of a quantum channel with a Clifford structure: a completely positive, trace-preserving map on the algebra, its Kraus operators, and the unitary slice as the group of unitary channels. The positivity of the involution is Positivity and the Hermitian Cone of a Hilbert Algebra with Hermitian Adjoint; the slice is The Unitary Slice and the Compact Real Form with Hermitian Adjoint; the operator whose complete positivity is at stake is The Hermitian Sandwich on a Hilbert Algebra with Hermitian Adjoint; the antiautomorphisms induced by reversion and Clifford conjugation are Hilbert Algebras; and the general theory of operator algebras is Operator Algebras and K-Theory of Operator Algebras.

Positivity of the involution is assumed in this article. For a Clifford algebra over a definite real form the algebra is a full matrix algebra over $\mathbb{R}$, $\mathbb{C}$ or $\mathbb{H}$ according to the parity of $n$; the completely positive theory is stated in the complexified case $\mathrm{Cl}_{0,n}\otimes_{\mathbb{R}}\mathbb{C}\cong M_{2^{m}}(\mathbb{C})$, where it is the classical theorem, and the real and quaternionic cases are flagged in the summary.

The Completely Positive Maps

Definition. Let $A$ be a finite-dimensional $C^{*}$-algebra with involution ${}^{\dagger}$ and let $\Phi : A \to A$ be linear. For $k \ge 1$ let $A\otimes M_k$ carry the involution $(a_{ij})^{\dagger} = (a_{ji}^{\dagger})$ and let $\Phi_k = \Phi\otimes\mathrm{id}_k$ act entrywise. The map $\Phi$ is positive if $\Phi(a^{\dagger}a) \in P$ for every $a$, and completely positive if $\Phi_k$ is positive for every $k$.

Theorem (the sandwich is completely positive). For every $x \in \mathrm{Cl}(V,q)$ the map

$$ \Theta_x(y) = x\,y\,x^{\dagger} $$

is completely positive, with Kraus form of rank one, $\Theta_x(y) = V y V^{\dagger}$ with $V = x$.

Proof. On the amplification, $\Theta_{x,k}(Y) = (x\otimes 1_k)Y(x\otimes 1_k)^{\dagger}$, so it suffices to prove positivity for each $k$. If $Y = Z^{\dagger}Z$ then

$$ \Theta_{x,k}(Y) = (x\otimes 1)Z^{\dagger}Z(x\otimes 1)^{\dagger} = \bigl(Z\,(x\otimes 1)^{\dagger}\bigr)^{\dagger}\bigl(Z\,(x\otimes 1)^{\dagger}\bigr) \in P , $$

which is positivity in the amplified algebra; the constants $k$ are uniform, so $\Theta_x$ is completely positive.

Theorem (unitality and automorphisms). For $x \in \mathrm{Cl}(V,q)$,

$$ \Theta_x(1) = x\,x^{\dagger}, \qquad \Theta_x \text{ unital} \iff x \in U, \qquad \Theta_x \text{ a unital algebra automorphism} \iff x \in U , $$

and on $U$ the map is the inner automorphism $\Theta_x = \mathrm{Ad}_x$.

Proof. $\Theta_x(1) = xx^{\dagger}$, which is $1$ exactly when $x$ is invertible with right inverse $x^{\dagger}$, that is when $x \in U$ (the two conditions $x^{\dagger}x = 1$ and $xx^{\dagger} = 1$ are equivalent in a finite-dimensional algebra). For multiplicativity, $\Theta_x(yz) = xyzx^{\dagger}$ equals $\Theta_x(y)\Theta_x(z) = xyx^{\dagger}xzx^{\dagger}$ for all $y,z$ exactly when $x^{\dagger}x = 1$, taking $y = z = 1$ for the necessity together with unitality; and then $\Theta_x(y) = xyx^{-1}$.

Corollary (the unitary slice as a group of channels). The unitaries of the algebra act on it by unital completely positive maps, and the automorphisms among them form the group $U/Z(U)$ of inner automorphisms; the maps $\Theta_x$ with $x \notin U$ are completely positive, non-unital, and are the Kraus-rank-one maps that are not automorphisms. In particular the automorphism group of $\mathrm{Cl}_{0,n}\otimes\mathbb{C}\cong M_{2^{m}}(\mathbb{C})$ is generated by the inner automorphisms $U/Z(U)$ together with the transposition, and the latter is the anti-automorphism induced by reversion or by Clifford conjugation according to the parity.

Choi's Theorem and the Kraus Form

Theorem (Choi). Let $\Phi : M_d(\mathbb{C}) \to M_d(\mathbb{C})$ be linear, and let

$$ C_{\Phi} = \sum_{i,j=1}^{d} E_{ij}\otimes\Phi(E_{ij}) \in M_d(\mathbb{C})\otimes M_d(\mathbb{C}) $$

be its Choi matrix, with $E_{ij}$ the matrix units. Then $\Phi$ is completely positive if and only if $C_{\Phi} \ge 0$; a completely positive $\Phi$ has a Kraus representation

$$ \Phi(y) = \sum_{r=1}^{R} V_r\,y\,V_r^{\dagger}, $$

of minimal length $R = \operatorname{rank}C_{\Phi}$, and $\Phi$ is unital exactly when $\sum_r V_rV_r^{\dagger} = 1$.

Corollary (the Clifford reading). Under the identification $\mathrm{Cl}_{0,n}\otimes\mathbb{C}\cong M_{2^{m}}(\mathbb{C})$, the sandwich $\Theta_x$ is the completely positive map with Choi matrix of rank one and a single Kraus operator $V = x$, and its complete positivity is the $R=1$ case of the theorem. Every completely positive map of the algebra is a sum of such sandwiches, and every unital completely positive map is a sum $\sum_r\Theta_{x_r}$ with $\sum_r x_rx_r^{\dagger} = 1$, the mixed unitary channels being the case in which each $x_r$ is unitary.

Proof. The Kraus form is the theorem; substituting $R=1$, $V = x$ gives $\Theta_x$, and the unitality condition $\sum_rV_rV_r^{\dagger}=1$ is $\sum_rx_rx_r^{\dagger}=1$.

Remark (Stinespring). Equivalently, a completely positive map on the algebra is the compression of a representation on a larger Hilbert space, $\Phi(y) = P\,\pi(y)\,P$ with $P$ a projection; for the sandwich, $\pi$ is the representation on the tensor product with the Kraus space and $P$ the projection onto the range of $x$. The two descriptions are the same statement in two languages, and the one used in the corpus is the Kraus form.

Channels with a Clifford Structure

Definition. A Clifford channel of the algebra is a completely positive, trace-preserving map $\Phi$ with respect to the trace form $T(x,y) = \operatorname{Tr}(m_{xy})$ of The Blade Form and the Hilbert Structure with Hermitian Adjoint.

Proposition. The sandwich $\Theta_x$ preserves the trace functional $\mathrm{Sc}$ exactly when $x \in U$:

$$ \mathrm{Sc}\bigl(\Theta_x(y)\bigr) = \mathrm{Sc}(y) \ \text{ for all } y \iff x \in U . $$

Proof. By cyclicity of the scalar part, $\mathrm{Sc}(xyx^{\dagger}) = \mathrm{Sc}(y\,x^{\dagger}x)$, so the condition is $\mathrm{Sc}\bigl(y\,(x^{\dagger}x)\bigr) = \mathrm{Sc}(y)$ for all $y$. Since $x^{\dagger}x = 1$ for $x \in U$ this is satisfied, and conversely the trace form $T(y,z) = 2^{n}\mathrm{Sc}(yz)$ of The Blade Form and the Hilbert Structure with Hermitian Adjoint is non-degenerate, so the vanishing of $\mathrm{Sc}\bigl(y(x^{\dagger}x-1)\bigr)$ for all $y$ forces $x^{\dagger}x = 1$.

Corollary. A sandwich is trace-preserving exactly when it is unital, and then it is the unitary channel $\mathrm{Ad}_x$; the trace form is invariant under the automorphisms $\mathrm{Ad}_x$, $x \in U$, because $T(xyx^{-1}, xzx^{-1}) = \operatorname{Tr}(m_ym_z) = T(y,z)$ by cyclicity of the trace.

Remark. The statement is the trace-preservation condition, and it is the same as the unitarity condition because the trace and the involution are compatible when the involution is positive. The channels of the algebra are thus the completely positive maps whose Kraus operators have $\sum_rx_rx_r^{\dagger} = \sum_rx_r^{\dagger}x_r = 1$; the unitary slice is the group of invertible channels, and the non-invertible ones have Kraus rank larger than one.

Summary

When the dagger is a positive involution, the Clifford algebra is a finite-dimensional $C^{*}$-algebra and the Hermitian sandwich $\Theta_x(y) = xyx^{\dagger}$ is a completely positive map of Kraus rank one, with the single Kraus operator $V=x$. It is positive in the strongest sense: the amplified map $\Theta_x\otimes\mathrm{id}_k$ sends $(z^{\dagger}z)$ to $\bigl(z(x\otimes1)^{\dagger}\bigr)^{\dagger}\bigl(z(x\otimes1)^{\dagger}\bigr)$, a square in the amplified algebra. It is unital if and only if $x$ lies in the unitary slice, and it is a unital algebra automorphism exactly there, where it becomes the inner automorphism $\mathrm{Ad}_x$; so the unitary slice is the group of complete-positivity maps of the sandwich shape that are automorphisms, and the automorphism group of the complexified algebra is generated by the inner ones together with the transposition induced by reversion or Clifford conjugation. Choi's theorem with its Choi matrix and Kraus form gives the general completely positive map as a sum $\sum_r\Theta_{x_r}$, unital exactly when $\sum_rx_rx_r^{\dagger}=1$, and the trace-preserving case is the Clifford channel; the unitary channels are exactly the sandwiches of the unitary slice.

Summary of Notation

Symbol	Meaning
$\Phi_k = \Phi\otimes\mathrm{id}_k$	Amplification of a map to $A\otimes M_k$
$\Theta_x(y) = xyx^{\dagger}$	Sandwich, completely positive of Kraus rank one
$V_r$	Kraus operators, $\Phi(y)=\sum_rV_ryV_r^{\dagger}$
$C_{\Phi}$	Choi matrix, $\sum_{ij}E_{ij}\otimes\Phi(E_{ij})$
$U$	Unitary slice, group of unitary channels
$U/Z(U)$	Group of inner automorphisms
$T(x,y) = \operatorname{Tr}(m_{xy})$	Trace form, defining trace-preserving channels