Manuscript Oai280-B@1 · Oai280-B.tex · lemma · 6d92ab75

Lemma: Two-qubit privacy

Privacy, perfectness, and exact access-dependent recovery › A complete syndrome check · lines 408–442 · proof included or linked

Oai280-B@1#lem:privacy, Oai280-B@1#eq:privacy

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For every subset $B$ with $∣ B ∣ \leq 2$ and every matrix $M \in B (C)$ ,

Tr_{B ˉ} (Π_{B} V M V^{†} Π_{B}^{†}) = Tr (M) \frac{I _{B}}{2 ^{∣ B ∣}} .

The identity holds on every matrix unit and after amplification by every independent reference.

Proof. The printed syndrome table follows by counting local Pauli anticommutations in (eq:stabilizers). Its $15$ entries are exactly the $15$ nonzero four-bit vectors.

A nonidentity Pauli of weight one has nonzero syndrome. A weight-two Pauli is a product of single-qubit Paulis on distinct legs. Its syndrome is the sum of two distinct entries and is therefore nonzero.

If $P$ anticommutes with some $g_{j}$ , then

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

so $V^{†} P V = 0$ . Expanding an arbitrary observable on $B$ in its Pauli basis shows that the reduced channel has only its identity coefficient. Trace preservation fixes that coefficient as the right-hand side of (eq:privacy).

This is an operator identity for all $M$ , not just a statement about diagonal input states. Tensoring the channel identity with a reference proves the last assertion.

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Oai280-B@1#eq:privacy · line 411
$Tr_{B ˉ} (Π_{B} V M V^{†} Π_{B}^{†}) = Tr (M) \frac{I _{B}}{2 ^{∣ B ∣}} .$
unlabeled · line 431
$V^{†} P V = V^{†} g_{j} P g_{j} V = - V^{†} P V,$

Lemma: Two-qubit privacy

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