TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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Finite-dimensional channels compose and tensor

Statement

Let $Φ : L (H) \to L (K)$ and $Ψ : L (K) \to L (L)$ be finite-dimensional quantum channels. Then $Ψ \circ Φ$ is a channel. If $Θ : L (H^{'}) \to L (K^{'})$ is another finite-dimensional channel, the tensor map $Φ \otimes Θ : L (H \otimes H^{'}) \to L (K \otimes K^{'})$ is the linear map specified on product matrix units by $(Φ \otimes Θ) (E_{u v} \otimes F_{s t}) = Φ (E_{u v}) \otimes Θ (F_{s t})$ . The basis result Tensor products of operator matrix units form a basis makes this map well-defined and unique. It is also a channel, with the displayed factor order.

Facts & Assumptions

Given: Finite-dimensional complex inner-product spaces and channels $Φ, Ψ, Θ$ with the domains and codomains in the Statement.

[F1]

Finite-dimensional quantum channel: each channel is completely positive and trace-preserving.

[F2]

Kraus and Choi characterizations of complete positivity: every completely positive map has a finite Kraus representation.

[F3]

Trace preservation is Kraus normalization: a Kraus representation is trace-preserving exactly when its adjoint-square sum is the identity.

[F4]

A finite Kraus sum is completely positive: every finite Kraus sum is completely positive.

[F5]

Finite Kraus representation: the family $(A_{i})$ represents $Φ$ when $Φ (X) = \sum_{i} A_{i} X A_{i}^{†}$ , and likewise for the other maps.

[F6]

Tensor products of finitely many linear maps: tensor maps act on elementary tensors by applying the factor maps in order.

[F7]

Tensor maps preserve composition, identities, and adjoints: tensor maps preserve composition and identities, and the tensor adjoint is the tensor of the adjoints.

[F8]

Tensor products of operator matrix units form a basis: the product matrix units form a basis of each operator space on a tensor product.

[F9]

Adjoints satisfy $(S + T)^{*} = S^{*} + T^{*}$ , $(λ T)^{*} = \overline{λ} T^{*}$ , $(S T)^{*} = T^{*} S^{*}$ , and $T^{* *} = T$ : for compatible maps, the adjoint reverses composition, so $(B_{j} A_{i})^{†} = A_{i}^{†} B_{j}^{†}$ .

Proof

1.1givenF1F2F3F4F5F9algebrachoose

By [F1] and [F2], choose finite Kraus families $(A_{i})$ for $Φ$ and $(B_{j})$ for $Ψ$ . By [F9], $(B_{j} A_{i})^{†} = A_{i}^{†} B_{j}^{†}$ for each $i, j$ , so for every $X \in L (H)$ , $(Ψ \circ Φ) (X) = \sum_{j, i} (B_{j} A_{i}) X (B_{j} A_{i})^{†}$ . Thus the composite has a finite Kraus representation and is completely positive by [F4]. By [F1] and [F3], $\sum_{i} A_{i}^{†} A_{i} = I_{H}$ and $\sum_{j} B_{j}^{†} B_{j} = I_{K}$ . Therefore $\sum_{j, i} (B_{j} A_{i})^{†} (B_{j} A_{i}) = \sum_{i} A_{i}^{†} (\sum_{j} B_{j}^{†} B_{j}) A_{i} = \sum_{i} A_{i}^{†} A_{i} = I_{H}$ . By [F3] the composite is trace-preserving, hence a channel by [F1].

1.2givenF1F2F4F5F6F7F8algebrachoose

By [F1] and [F2], choose finite Kraus families $(A_{i})$ and $(C_{j})$ for $Φ$ and $Θ$ . For any product matrix unit $E_{u v} \otimes F_{s t}$ , [F6] and [F7] give $\sum_{i, j} (A_{i} \otimes C_{j}) (E_{u v} \otimes F_{s t}) (A_{i} \otimes C_{j})^{†} = (\sum_{i} A_{i} E_{u v} A_{i}^{†}) \otimes (\sum_{j} C_{j} F_{s t} C_{j}^{†}) = Φ (E_{u v}) \otimes Θ (F_{s t}) = (Φ \otimes Θ) (E_{u v} \otimes F_{s t})$ by the definition in the Statement. By [F8] these product matrix units form a basis of $L (H \otimes H^{'})$ ; linearity therefore gives a finite Kraus representation of $Φ \otimes Θ$ on every input operator. Hence this tensor map is completely positive by [F4].

2.1givenF1F3F4F7step 1.2algebra∎

By [F1] and [F3], $\sum_{i} A_{i}^{†} A_{i} = I_{H}$ and $\sum_{j} C_{j}^{†} C_{j} = I_{H^{'}}$ . Using [F7], $\sum_{i, j} (A_{i} \otimes C_{j})^{†} (A_{i} \otimes C_{j}) = (\sum_{i} A_{i}^{†} A_{i}) \otimes (\sum_{j} C_{j}^{†} C_{j}) = I_{H} \otimes I_{H^{'}} = I_{H \otimes H^{'}}$ . Thus [F3] makes the tensor map trace-preserving, and [F4] gives complete positivity; it is a channel by [F1]. Empty Kraus families are covered: their normalization equals the input identity only when that input space is zero, where all displayed sums and identities are empty or zero.

Depends on

Used by

Dependency tree · two levels

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Sources

John Watrous, The Theory of Quantum Information (2018), §§2.2.1–2.2.3 (standard reference, not scraped)
John Preskill, Quantum Computation, Chapter 3, §§3.2–3.3 (standard reference, not scraped)