Example: Complete preservation without either local inverse

Let $0\le c\le 1$, $z=\sqrt{1-c^2}$, and choose $0\le \theta \le \pi /4$ with $\cos (2\theta )=c$. Define

and the isometry

Then

The complete output is exactly reversible. For $0<c<1$, neither local output exactly recovers the code.

Proof. Orthogonal $B$-labels give (eq:controlN). Put ${\sigma }_i=|r_i⟩⟨r_i|$. Their trace distance is $z$.

For any qubit decoder, its Bell overlap is

The two diagonal output tests pull back to complementary effects. Optimal binary discrimination therefore gives

The Bell-projector test yields error at least $(3-z)/4$.

The two states are interchanged by $Z$. Symmetrizing an optimal discrimination measurement makes its two error probabilities equal to $(1-z)/2$. Measuring and preparing the inferred logical basis state then realizes

Its Pauli probabilities are

Lemma lem:unitary-mixture gives error $1-p_I=(3-z)/4$, proving (eq:controlR).

Tracing $R$ gives

The identity decoder has error $(1-c)/2$. The outputs of $|+⟩$ and $|-⟩$ have trace distance $c$. If a decoder has error $\epsilon$, contraction and the triangle inequality give $1\le 2\epsilon +c$. This proves (eq:controlB).

The channel ${\Delta }_c$ is Pauli covariant. Averaging a replacement state over Pauli conjugations cannot increase its distance from this channel and gives $I_2/2$. That replacement is therefore optimal. Relative to the uniform Pauli representation of ${\mathcal{T}}_{I_2/2}$, the coefficient differences in the order $I,Z,X,Y$ are

Lemma lem:unitary-mixture gives

which is (eq:controlleak). Complete restoration is Theorem thm:global.