Manuscript En10-A@1 · En10-A.tex · lemma · 0a1d7743

Lemma: Validity of the frozen step

The frozen finite interface › Declared opposite response and relaxation · lines 335–359 · proof included or linked

En10-A@1#lem:step

The map $Φ : L (H_{B}) \to L (H_{B})$ is completely positive and trace preserving.

Proof. The displayed Kraus form proves complete positivity. Moreover,

L_{0}^{†} L_{0} + L_{1}^{†} L_{1} = ∣ 0 ⟩ ⟨ 0 ∣ + h ∣ 1 ⟩ ⟨ 1 ∣ + γ ∣ 1 ⟩ ⟨ 1 ∣ = I_{A} .

Since $U_{ϕ}$ is unitary,

ℓ = 0 \sum^{1} H_{ℓ}^{†} H_{ℓ} = U_{ϕ}^{†} (I_{N} \otimes I_{A}) U_{ϕ} = I_{B} .

This proves trace preservation.

Displayed formulas in this passage

unlabeled · line 344
$L_{0}^{†} L_{0} + L_{1}^{†} L_{1} = ∣ 0 ⟩ ⟨ 0 ∣ + h ∣ 1 ⟩ ⟨ 1 ∣ + γ ∣ 1 ⟩ ⟨ 1 ∣ = I_{A} .$
unlabeled · line 351
$ℓ = 0 \sum^{1} H_{ℓ}^{†} H_{ℓ} = U_{ϕ}^{†} (I_{N} \otimes I_{A}) U_{ϕ} = I_{B} .$