J Asymptotic Structure of Quantum Channels: Supporting Results

This appendix supports Chapter 25. It proves the positivity, trace-preservation, unitality, and trace-adjoint properties of the mean-ergodic projections used there, together with the weighted-trace and idempotent-retraction results used for fixed-point algebras and Choi–Effros absorption. It also records the Gram-matrix unitary-extension criterion used in the Chapter 16 Stinespring construction and its zero-extension consequence for the fixed-point decomposition.

J.1 Preservation under the mean-ergodic projection

This section proves the positivity, trace-preservation, and unitality assertions used in Theorem 25.1.7. For a bounded-orbit endomorphism \(T\), let \(P_T\) denote the mean-ergodic projection of Definition 25.1.4.

Theorem J.1.1 Positive mean-ergodic projection

✓

Let \(T:M_{D}(\mathbb {C})\to M_{D}(\mathbb {C})\) be positive and have bounded orbits. Then its mean-ergodic projection \(P_T\) is positive.

Proof ▶

Every Cesàro average is positive. Therefore, for \(X\ge 0\), closedness of the positive cone and convergence in (??) give \(P_T(X)\ge 0\).

Theorem J.1.2 Trace-preserving mean-ergodic projection

✓

Let \(T:M_{D}(\mathbb {C})\to M_{D}(\mathbb {C})\) be trace-preserving and have bounded orbits. Then its mean-ergodic projection \(P_T\) is trace-preserving.

Proof ▶

Every nonempty Cesàro average preserves the trace. Convergence of the averages and continuity of the trace therefore give \(\operatorname{tr}(P_T(X))=\operatorname{tr}(X)\).

Theorem J.1.3 Identity under the mean-ergodic projection

✓

Let \(T:M_{D}(\mathbb {C})\to M_{D}(\mathbb {C})\) have bounded orbits. If \(T(\mathbb {1})=\mathbb {1}\), then \(P_T(\mathbb {1})=\mathbb {1}\).

Proof ▶

The identity is a fixed point of \(T\), so the fixed-point characterization of the mean-ergodic projection gives the conclusion.

These three preservation statements give the corresponding properties of the Cesàro projection in Theorem 25.1.7.

J.2 Trace adjoints of ergodic projections

This section proves the two adjoint facts used in Theorem 25.1.8. For a linear map \(S\), write \(S^*\) for its adjoint with respect to the trace pairing; for a bounded-orbit map \(T\), write \(P_T\) for its mean-ergodic projection.

Theorem J.2.1 Trace preservation and the trace adjoint

✓

On any finite full matrix algebra, a linear endomorphism \(S\) is trace-preserving if and only if its trace-pairing adjoint fixes the identity:

\begin{align} (\forall X,\ \operatorname{tr}(S(X))=\operatorname{tr}(X)) & \quad \Longleftrightarrow \quad S^*(\mathbb {1})=\mathbb {1}. \label{eq:channel_tp_adjoint_unital} \end{align}

Proof ▶

The trace-pairing identity gives \(\operatorname{tr}(S^*(\mathbb {1})X)=\operatorname{tr}(S(X))\). The equivalence (??) follows from nondegeneracy of the trace pairing.

Theorem J.2.2 Adjoint mean-ergodic projection

✓

Let \(T:M_{D}(\mathbb {C})\to M_{D}(\mathbb {C})\) have bounded orbits, let \(P_T\) be its mean-ergodic projection, and let \(P_T^*\) be the trace-pairing adjoint of \(P_T\). Then

\begin{align} (P_T^*)^2 & =P_T^*, \notag \\ \operatorname{range}(P_T^*) & =\ker (T^*-\mathbb {1}). \label{eq:channel_adjoint_mean_ergodic} \end{align}

and \(P_T^*(Y)=Y\) if and only if \(T^*(Y)=Y\).

Proof ▶

Since \(P_T^2=P_T\), the trace-pairing identity gives \((P_T^*)^2=P_T^*\). If \(T^*(Y)=Y\), decompose \(X-P_T(X)=(T-\mathbb {1})(Z)\). Then

\begin{align} \operatorname{tr}(Y(X-P_T(X))) & =\operatorname{tr}((T^*(Y)-Y)Z)=0. \notag \end{align}

Nondegeneracy of the trace pairing gives \(P_T^*(Y)=Y\). Conversely, \(P_TT=P_T\) implies \(T^*P_T^*=P_T^*\), so every fixed point of \(P_T^*\) is a fixed point of \(T^*\):

\begin{align} P_T^*(Y)=Y & \quad \Longrightarrow \quad T^*(Y)=T^*(P_T^*(Y))=P_T^*(Y)=Y. \notag \end{align}

Hence

\begin{align} Y\in \operatorname{range}(P_T^*) & \quad \Longleftrightarrow \quad P_T^*(Y)=Y \quad \Longleftrightarrow \quad T^*(Y)=Y \quad \Longleftrightarrow \quad Y\in \ker (T^*-\mathbb {1}). \notag \end{align}

The first equivalence uses idempotence, and the second is the fixed-point equivalence proved above. This proves (??).

The first theorem gives unitality of \(P_T^*\) from trace preservation of \(P_T\); the second identifies its range and fixed points, completing the adjoint retraction argument in Theorem 25.1.8.

J.3 Weighted traces and idempotent retractions

This section proves the weighted-trace lemma used in Theorem 25.2.4 and the three absorption identities complementing Theorem 25.2.8. In the absorption statements, \(E\) is a unital idempotent Kraus map whose adjoint has a positive definite fixed point, as in Theorem 25.2.7.

Lemma J.3.1 Faithfulness of a positive-definite weighted trace

✓

If \(M\succeq 0\), \(\rho {\gt}0\), and \(\operatorname{tr}(\rho M)=0\), then \(M=0\).

Proof ▶

Write \(\rho =S^\dagger S\) with \(S\) invertible. The matrix \(SMS^\dagger \) is positive semidefinite and \(\operatorname{tr}(SMS^\dagger )=\operatorname{tr}(S^\dagger SM)=\operatorname{tr}(\rho M)=0\). A positive semidefinite matrix of trace zero vanishes, so \(SMS^\dagger =0\). Since \(S\) and \(S^\dagger \) are invertible, \(SMS^\dagger =0\) implies \(MS^\dagger =0\), and hence \(M=0\).

Theorem J.3.2 Choi–Effros left absorption

✓

Under the same hypotheses, for all \(X,Y\in M_{D}(\mathbb {C})\), \(E(E(X)Y)=E(E(X)E(Y))\).

Proof ▶

Theorem 25.2.7 places \(E(X)\) in the multiplicative domain, so \(E(E(X)Y)=E(E(X))E(Y)=E(X)E(Y)\). Applying the same multiplicativity to \(E(X)\) and \(E(Y)\) gives \(E(E(X)E(Y))=E(E(X))E(E(Y))=E(X)E(Y)\) by idempotence.

Theorem J.3.3 Choi–Effros right absorption

✓

Under the same hypotheses, for all \(X,Y\in M_{D}(\mathbb {C})\), \(E(XE(Y))=E(E(X)E(Y))\).

Proof ▶

Theorem 25.2.7 places \(E(Y)\) in the multiplicative domain, so \(E(XE(Y))=E(X)E(E(Y))=E(X)E(Y)\). Applying the same multiplicativity to \(E(X)\) and \(E(Y)\) gives \(E(E(X)E(Y))=E(E(X))E(E(Y))=E(X)E(Y)\) by idempotence.

Theorem J.3.4 Choi–Effros symmetric absorption

✓

Under the same hypotheses, for all \(X,Y\in M_{D}(\mathbb {C})\), \(E(E(X)Y)=E(XE(Y))\).

Proof ▶

By Theorems J.3.2 and J.3.3, both sides equal \(E(E(X)E(Y))\).

The left and right identities identify both absorbed products with the projected product, and the symmetric identity equates them.

J.4 Unitary extensions for fixed-point decompositions

The fixed-point decomposition uses an ambient unitary to identify an isometric support inclusion with a standard direct summand. The Gram-matrix criterion below supplies this unitary and also serves the Stinespring-unitary construction in Chapter 18, Theorem 18.6.15. Its zero-extension consequence completes the fixed-point decomposition in Theorem 25.8.5.

Lemma J.4.1 Rectangular Gram equality gives a unitary

✓

Let \(A,B:\mathbb {C}^n\to \mathbb {C}^m\) be linear maps with \(B^\dagger B=A^\dagger A\). Then there is a unitary \(U\) on \(\mathbb {C}^m\) such that \(B=UA\).

Proof ▶

Let \(\mathcal R_A=\operatorname {ran}A\) and \(\mathcal R_B=\operatorname {ran}B\). Define \(V:\mathcal R_A\to \mathcal R_B\) by \(V(Ax)=Bx\). This is well defined: if \(Ax=Ay\), then

\begin{align} \lVert B(x-y)\rVert ^2 & =\langle x-y,B^\dagger B(x-y)\rangle =\langle x-y,A^\dagger A(x-y)\rangle =\lVert A(x-y)\rVert ^2=0. \notag \end{align}

The same calculation, with two vectors and polarization, gives \(\langle V(Ax),V(Ay)\rangle =\langle Ax,Ay\rangle \). Thus \(V\) is an isometry onto \(\mathcal R_B\). In particular, \(\dim \mathcal R_A=\dim \mathcal R_B\), so their orthogonal complements in \(\mathbb {C}^m\) have equal dimension. Choose a unitary \(V^\perp :\mathcal R_A^\perp \to \mathcal R_B^\perp \) and set \(U=V\oplus V^\perp \) with respect to the two orthogonal decompositions of \(\mathbb {C}^m\). Then \(U\) is unitary and \(UAx=Bx\) for every \(x\in \mathbb {C}^n\); hence \(B=UA\).

Theorem J.4.2 Unitary completion of an isometric inclusion

✓

Let \(W:\mathbb {C}^n\to \mathbb {C}^D\) satisfy \(W^\dagger W=\mathbb {1}_n\). There are an identification \(\mathbb {C}^D\simeq \mathbb {C}^{D-n}\oplus \mathbb {C}^n\) and a unitary \(U\) on \(\mathbb {C}^D\) such that, for every \(A\in M_{n}(\mathbb {C})\),

\begin{align} WAW^\dagger & =U(0_{D-n}\oplus A)U^\dagger . \label{eq:representations_unitary_completion} \end{align}

Proof ▶

The isometry \(W\) implies \(n\leq D\). Let \(J:\mathbb {C}^n\to \mathbb {C}^{D-n}\oplus \mathbb {C}^n\) be the canonical inclusion into the second summand. Since \(W^\dagger W=J^\dagger J=\mathbb {1}_n\), Lemma J.4.1 gives a unitary \(U\) with \(W=UJ\). Substituting this identity gives (??) because \(JAJ^\dagger =0_{D-n}\oplus A\).

Equality of the two inclusion Gram matrices therefore supplies exactly the unitary conjugation used to adjoin the complementary zero block in Theorem 25.8.5.