Manuscript En02-B@1 · En02-B-v1.tex · lemma · ca28dee4

Lemma: Two-qubit privacy and erasure certificate

The exact five-qubit perfect-code building block › Access classification · lines 636–670 · proof included or linked

En02-B@1#lem:privacy, En02-B@1#eq:privacy, En02-B@1#eq:KL

For $C \subseteq {1, \dots, 5}$ with $∣ C ∣ \leq 2$ ,

Tr_{C ˉ} (V ρ V^{†}) = tr (ρ) \frac{I _{C}}{2 ^{∣ C ∣}} .

If $A = C ˉ$ , and $E_{e} = (I_{A} \otimes ⟨ e ∣_{C}) V$ , then

E_{e}^{†} E_{f} = \frac{δ _{e f}}{2 ^{∣ C ∣}} I_{2} .

Proof. A nonzero syndrome supplies a stabilizer $g_{j}$ anticommuting with the Pauli $Q$ . Since $g_{j} V = V$ ,

V^{†} Q V = V^{†} g_{j} Q g_{j} V = - V^{†} Q V,

so it vanishes. Expanding operators on $C$ in the Pauli basis gives (eq:privacy), including off-diagonal logical matrix units. Equation (eq:KL) follows by applying the same identity to $I_{A} \otimes ∣ e ⟩ ⟨ f ∣_{C}$ .

Displayed formulas in this passage

En02-B@1#eq:privacy · line 639
$Tr_{C ˉ} (V ρ V^{†}) = tr (ρ) \frac{I _{C}}{2 ^{∣ C ∣}} .$
En02-B@1#eq:KL · line 647
$E_{e}^{†} E_{f} = \frac{δ _{e f}}{2 ^{∣ C ∣}} I_{2} .$
unlabeled · line 658
$V^{†} Q V = V^{†} g_{j} Q g_{j} V = - V^{†} Q V,$

Lemma: Two-qubit privacy and erasure certificate

Displayed formulas in this passage

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