Theorem: Exact flagged one-particle optima

Suppose $\mathrm{Ran}W\subset {\bigwedge }^1{\mathbb{C}}^n$, with code dimension $d\ge 2$. Then, for every $t$ and $M$,

Proof. Put $s=s_M(t)$. On the active collected and residual mode subspaces, the complete isometry has the form

where $A$ and $B$ are isometries from the code to orthogonal one-particle label spaces. Terms with zero weight are omitted.

Its two reduced channels are

Decode the occupied collected block with $A^{†}$, and prepare $I_C/d$ on vacuum or any unused output block. The decoded channel is

Lemma lem:replacement gives error $s(1-d^{-2})$.

For any decoder, the Bell-input output from the vacuum branch has overlap $d^{-2}$ with the target Bell state. The occupied branch has overlap at most one. Therefore total Bell overlap is at most $1-s+s/d^2$. The Bell-projector witness gives the matching lower bound on recovery.

For leakage, choose

The difference from its replacement channel is $sB({\mathrm{id}}_C-{\mathcal{T}}_{I_C/d})(\cdot )B^{†}$, so its half-diamond norm is $s(1-d^{-2})$.

It remains to show that a different replacement state cannot do better. First apply a CPTP projection onto the active one-particle code block and vacuum, sending all unused residual blocks to vacuum. This fixes $\mathcal{C}$ and cannot increase distance. Next average over all code unitaries, represented by $BU{B}^{†}$ on the active block and by the identity on vacuum. Covariance and convexity cannot increase the distance. The averaged replacement state has the form

The trace distance for a Bell input is

If $r\le s$, this is $s-r/d^2\ge s(1-d^{-2})$. If $s\le r\le d^{2}s$, it is $r(1-d^{-2})$. If $r\ge d^{2}s$, it is $r-s$, which is also at least $s(1-d^{-2})$. These cases cover the allowed interval. Thus every replacement channel has distance at least the claimed value.