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Local trace-preserving Kraus maps do not change the opposite reduced state

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H, K$ be complex Hilbert spaces, let $(M_{a})_{a \geq 1}$ be bounded operators on $K$ with $\sum_{a \leq N} M_{a}^{*} M_{a} \to I_{K}$ strongly, and let $R$ be a trace-class operator on $H \hat{\otimes} K$ . In this statement, $R \geq 0$ means that $R$ is self-adjoint and $⟨ R x, x ⟩ \geq 0$ for every $x$ , as in Self-adjoint, positive, unitary and normal operators. Then the series (when either factor is ${0}$ , use the zero partial-trace maps ${Tr}_{K} : S_{1} ({0}) \to S_{1} (H)$ and ${Tr}_{H} : S_{1} ({0}) \to S_{1} (K)$ ; when both factors are nonzero, use the partial traces of Separable partial trace and its basic properties)

$\sum_{a \geq 1} (I_{H} \otimes M_{a}) R (I_{H} \otimes M_{a}^{*})$

converges in trace norm, its sum $Φ (R)$ is trace class with $tr Φ (R) = tr R$ and $Φ (R) \geq 0$ if $R \geq 0$ , and the reduced state on $H$ is unchanged:

${Tr}_{K} Φ (R) = {Tr}_{K} R .$

By symmetry, a local map $R \mapsto \sum_{a} (N_{a} \otimes I_{K}) R (N_{a}^{*} \otimes I_{K})$ on the first factor also leaves ${Tr}_{H} R$ fixed. Selective conditioning (one summand normalized by its nonzero trace) is not covered.

Facts & Assumptions

Given: $H, K$ ; the family $(M_{a})$ ; a trace-class $R$ on $H \hat{\otimes} K$ with nuclear representation $R = \sum_{j} T_{w_{j}, z_{j}}$ ; a bounded $A \in B (H)$ .

[F1]

The amplified family $(I_{H} \otimes M_{a})$ consists of bounded operators on $H \hat{\otimes} K$ with $(I_{H} \otimes M_{a})^{*} (I_{H} \otimes M_{a}) = I_{H} \otimes M_{a}^{*} M_{a}$ , and $\sum_{a \leq N} (I_{H} \otimes M_{a})^{*} (I_{H} \otimes M_{a}) \to I$ strongly: on finite sums of elementary tensors this is the strong convergence of $\sum_{a \leq N} M_{a}^{*} M_{a}$ , and the general case follows by approximation and $∥ I_{H} \otimes S_{N} ∥ \leq 1$ (Bounded tensor maps: extension, norm estimate, adjoint and commutation, Hilbert tensor construction: inner product, completion and product bases).

[F2]

Under the strong completeness hypothesis, $\sum_{a} V_{a} R V_{a}^{*}$ converges in trace norm for every trace-class $R$ , is positive for positive $R$ , and preserves the trace, where $V_{a} : = I_{H} \otimes M_{a}$ ; moreover the identity $tr (\sum_{a} V_{a} R V_{a}^{*}) = tr R$ is the trace-preservation clause (Countable Kraus normalization and trace-norm convergence, Finite-ancilla complete positivity of countable Kraus maps).

[F3]

When $H, K \neq {0}$ , ${Tr}_{K}$ is the unique linear map on $S_{1} (H \hat{\otimes} K)$ with $tr (({Tr}_{K} T) A) = tr (T (A \otimes I_{K}))$ for all bounded $A$ , and it satisfies $∥ {Tr}_{K} T ∥_{1} \leq ∥ T ∥_{1}$ (Separable partial trace and its basic properties); the Statement supplies the zero-map convention when a factor is zero.

[F4]

For a bounded operator $B$ and trace-class $T$ , $B T$ is trace class with $∥ B T ∥_{1} \leq ∥ B ∥ ∥ T ∥_{1}$ , and $tr (B T)$ is linear in $T$ and continuous in trace norm (Trace class is a two sided Banach operator ideal, Trace is absolutely convergent and basis independent). In addition, every trace-class $T$ has a nuclear representation $T = \sum_{j} T_{w_{j}, z_{j}}$ , and for bounded $B$ the trace formulas $tr (T B) = \sum_{j} ⟨ w_{j}, B^{*} z_{j} ⟩ = tr (B T)$ hold with absolute convergence (Nuclear bounded cyclicity for trace-class operators, Facts [F2] and Statement (b)).

Proof

1.1givenF1F2

[Convergence of the local map] If $H = {0}$ or $K = {0}$ , then $R = 0$ and every summand $(I_{H} \otimes M_{a}) R (I_{H} \otimes M_{a})^{*}$ is zero; hence $Φ (R) = 0$ and both sides of ${Tr}_{K} Φ (R) = {Tr}_{K} R$ are zero by the convention in the Statement. Assume henceforth $H, K \neq {0}$ . By [F1] the amplified family satisfies the strong completeness hypothesis of [F2] on the Hilbert space $H \hat{\otimes} K$ , so the series $\sum_{a} V_{a} R V_{a}^{*}$ with $V_{a} = I_{H} \otimes M_{a}$ converges in trace norm, its sum $Φ (R)$ is trace class with $tr Φ (R) = tr R$ , and $Φ (R) \geq 0$ for $R \geq 0$ .

2.1step 1.1F4algebra

[Pairing computation] For bounded $A$ and $V_{a} = I_{H} \otimes M_{a}$ , apply the bounded cyclicity of [F4] to the trace-class operator $T = R V_{a}^{*} (A \otimes I_{K})$ and bounded $V_{a}$ . It gives $tr (V_{a} R V_{a}^{*} (A \otimes I_{K})) = tr (R V_{a}^{*} (A \otimes I_{K}) V_{a})$ with $V_{a}^{*} (A \otimes I_{K}) V_{a} = (I_{H} \otimes M_{a}^{*}) (A \otimes I_{K}) (I_{H} \otimes M_{a}) = A \otimes M_{a}^{*} M_{a}$ . For each $N$ , sum this identity over $a \leq N$ ; trace-norm convergence of the series in step 1.1 and trace-norm continuity of the bounded trace pairing [F4] give $tr (Φ (R) (A \otimes I_{K})) = \lim_{N \to \infty} tr (R (A \otimes S_{N}))$ , where $S_{N} = \sum_{a \leq N} M_{a}^{*} M_{a}$ .

3.1step 2.1F1F4algebra

[Passing to the limit in the completeness sum] Let $R = \sum_{j} T_{w_{j}, z_{j}}$ be nuclear. For each $N$ , $tr (R (A \otimes \sum_{a \leq N} M_{a}^{*} M_{a})) = \sum_{j} ⟨ w_{j}, (A \otimes S_{N})^{*} z_{j} ⟩$ with $S_{N} : = \sum_{a \leq N} M_{a}^{*} M_{a}$ , an absolutely convergent series with $∣ ⟨ w_{j}, (A \otimes S_{N})^{*} z_{j} ⟩ ∣ \leq ∥ A ∥ ∥ w_{j} ∥ ∥ z_{j} ∥$ ; since $(A \otimes S_{N})^{*} \to A^{*} \otimes I$ strongly, each term converges to $⟨ w_{j}, (A^{*} \otimes I) z_{j} ⟩$ , and the summable majorant gives $\sum_{j} ⟨ w_{j}, (A \otimes S_{N})^{*} z_{j} ⟩ \to \sum_{j} ⟨ w_{j}, (A^{*} \otimes I) z_{j} ⟩ = tr (R (A \otimes I_{K}))$ . Hence $tr (Φ (R) (A \otimes I_{K})) = tr (R (A \otimes I_{K}))$ .

4.1step 3.1F3∎

[Conclusion] By [F3], $tr (({Tr}_{K} Φ (R)) A) = tr (Φ (R) (A \otimes I_{K})) = tr (R (A \otimes I_{K})) = tr (({Tr}_{K} R) A)$ for every bounded $A$ ; since the pairings with all $A$ determine a trace-class operator by the uniqueness in [F3], ${Tr}_{K} Φ (R) = {Tr}_{K} R$ . The statement for the first factor is obtained by exchanging the roles of $H$ and $K$ .

Local trace-preserving Kraus maps do not change the opposite reduced state

Statement

Facts & Assumptions

Proof

Depends on

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