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Channel duals and norm closure

Statement

Let $Φ : M_{d} (C) \to M_{e} (C)$ be a finite channel as defined in Finite completely positive trace-preserving maps. There is a unique linear map $Φ^{†} : M_{e} (C) \to M_{d} (C)$ satisfying the trace-pairing identity

$Tr (Φ (X) A) = Tr (X Φ^{†} (A)) (X \in M_{d} (C), A \in M_{e} (C)) .$

If $Φ (X) = \sum_{a} K_{a} X K_{a}^{*}$ , then $Φ^{†} (A) = \sum_{a} K_{a}^{*} A K_{a}$ ; in particular, $Φ^{†}$ is completely positive and unital. Composition and finite convex combinations of channels are channels. A limit in the induced operator norm $∥ Ψ ∥_{\infty \to \infty} : = \sup_{∥ X ∥_{\infty} \leq 1} ∥ Ψ (X) ∥_{\infty}$ , where $∥ X ∥_{\infty} : = \sup_{∥ v ∥ = 1} ∥ X v ∥$ , of channels $M_{d} (C) \to M_{e} (C)$ is also a channel. Moreover, a norm limit of completely positive maps of these fixed finite dimensions is completely positive.

Facts & Assumptions

Given: Integers $d, e \geq 1$ , a finite channel $Φ : M_{d} (C) \to M_{e} (C)$ , and finite matrix traces.

[F1]

Finite completely positive trace-preserving maps: a channel is complex-linear, completely positive under every finite ancilla extension, and trace-preserving.

[F2]

Complete finite Choi-Kraus equivalence: every finite completely positive map has a finite Kraus form $Φ (X) = \sum_{a} K_{a} X K_{a}^{*}$ ; trace preservation is equivalent to $\sum_{a} K_{a}^{*} K_{a} = I_{d}$ .

[F3]

Finite matrix trace cyclicity and nondegeneracy: the finite trace cycles rectangular products, and the matrix-unit pairing shows that ${Tr}_{d} (X D) = 0$ for every $X$ forces $D = 0$ .

Proof

1.1givenF2F3algebra

Using the Kraus form in [F2] and finite trace cyclicity [F3], $Tr (Φ (X) A) = \sum_{a} Tr (K_{a} X K_{a}^{*} A) = Tr (X \sum_{a} K_{a}^{*} A K_{a})$ . Thus $Φ^{†} (A) = \sum_{a} K_{a}^{*} A K_{a}$ satisfies the trace-pairing identity.

1.2givenF2algebra

If $Ψ (Y) = \sum_{b} L_{b} Y L_{b}^{*}$ is another channel, then $(Ψ \circ Φ) (X) = \sum_{a, b} (L_{b} K_{a}) X (L_{b} K_{a})^{*}$ , and $\sum_{a, b} (L_{b} K_{a})^{*} (L_{b} K_{a}) = \sum_{a} K_{a}^{*} (\sum_{b} L_{b}^{*} L_{b}) K_{a} = \sum_{a} K_{a}^{*} K_{a} = I_{d} .$ If $Φ_{r} (X) = \sum_{a} K_{r, a} X K_{r, a}^{*}$ and $p_{r} \geq 0$ , $\sum_{r} p_{r} = 1$ , then $\sum_{r} p_{r} Φ_{r}$ has Kraus matrices $\sqrt{p_{r}} K_{r, a}$ , whose adjoint-square sum is $\sum_{r} p_{r} I_{d} = I_{d}$ . In both cases [F2] gives a channel.

2.1givenF1F2F3step 1.1algebra

If $B, C \in M_{d} (C)$ both satisfy the pairing identity for a fixed $A$ , then $Tr (X (B - C)) = 0$ for every $X$ . By [F3], taking $X = E_{i j}$ gives every entry of $B - C$ equal to zero, so uniqueness follows. The Kraus form in step 1.1 is completely positive by [F2], and $Φ^{†} (I_{e}) = \sum_{a} K_{a}^{*} K_{a} = I_{d}$ by [F1] and [F2].

3.1givenF2algebra∎

Let $Φ_{n} \to Φ$ in the stated induced norm, with each $Φ_{n}$ completely positive. Since $∥ E_{i j} ∥_{\infty} = 1$ , each block $Φ_{n} (E_{i j})$ converges to $Φ (E_{i j})$ ; the finite Choi sums therefore converge in matrix norm. Every $J (Φ_{n}) \geq 0$ by [F2], and for every vector $v$ , $⟨ J (Φ) v, v ⟩ = \lim_{n} ⟨ J (Φ_{n}) v, v ⟩ \geq 0$ , so $J (Φ) \geq 0$ . The equivalence in [F2] gives complete positivity. If each $Φ_{n}$ is trace-preserving, then for every $X$ , continuity of finite matrix trace gives $Tr Φ (X) = \lim_{n} Tr Φ_{n} (X) = Tr X$ ; thus $Φ$ is trace-preserving as well.

Channel duals and norm closure

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources