Proposition: Exact pair distances and complementary sandwich

For a unit Bloch vector $n=(n_x,n_y,n_z)$, let ${\rho }_{\pm }=(I\pm n\cdot (X,Y,Z))/2$. Then

Consequently,

and

The equatorial input pair gives the additional elementary bound ${\epsilon }_{\mathrm{rec}}({\mathcal{N}}_{\eta })\ge t/2$.

Proof. The input difference is $n\cdot (X,Y,Z)$. Its complementary image has odd-sector diagonal entries $-\eta t{n}_z$, twice, and an even-sector block

The two eigenvalues of this block are $t(\eta n_z+1)$ and $t(\eta n_z-1)$. Its trace norm is $2t$; the odd entries add $2\eta t|n_z|$. This proves (eq:environmentPair). Interchanging $\eta ,t$ gives (eq:accessiblePair).

Any difference of two qubit states is $rn\cdot (X,Y,Z)$ with $0\le r\le 1$. Thus the largest pair distances occur at $r=1,|n_z|=1$, which proves (eq:pairDiameters).

The computational-basis accessible outputs have distance $1-t^2$. If a decoder has error $\epsilon$, contraction and the triangle inequality give

so $\epsilon \ge t^2/2$. The complementary pair in the same basis attains $L_E$; Lemma lem:leakageBell gives its lower bound. For $n_z=0$, the accessible distance is $\eta$, yielding $1\le 2\epsilon +\eta$, hence $\epsilon \ge t/2$.

The upper bounds follow from the specified decoder of Section sec:specified and from ignoring the input and preparing $I_{M^{\prime }}/2$. The latter is the uniform Pauli channel and has error $3/4$ by Lemma lem:witnesses.