Manuscript VIII-ST@1 · VIII-ST.tex · proposition · fe979f3b

Proposition: What sharp coarse compatibility adds

Named access, residual instruments, and recovery boundaries › Record equality versus residual equality · lines 1172–1204 · proof included or linked

VIII-ST@1#prop:luders

Back to manuscript

Suppose the coarse output instrument is Lüders: $I_{w} (Y) = Q_{w} Y Q_{w}$ , with orthogonal projections $Q_{w}$ . If $T^{*} (Q_{w}) = R_{w}$ , then

T^{*} (Q_{w} B Q_{w}) = R_{w} T^{*} (B) R_{w} .

Thus complete compatibility with the aggregated fine instrument additionally requires

P_{z} T^{*} (B) P_{z^{'}} = 0 whenever z \neq = z^{'}, q (z) = q (z^{'}) = w .

Sharp record compatibility alone does not remove coherences inside an aggregated block.

Proof. Choose Kraus operators for $T$ and the corresponding isometry $V$ , so $T^{*} (B) = V^{†} (B \otimes I) V$ . For projections $Q, R$ with $V^{†} (Q \otimes I) V = R$ ,

∥ (Q \otimes I) V - V R ∥_{on each vector}^{2} = 0

by expansion, using $Q^{2} = Q$ and $R^{2} = R$ . Hence $(Q \otimes I) V = V R$ , proving the first identity. Insert it into Proposition prop:residual-criterion. Multiplication on the left and right by distinct $P_{z}, P_{z^{'}}$ shows necessity of the displayed off-diagonal conditions; their vanishing is also sufficient.

Displayed formulas in this passage

unlabeled · line 1177
$T^{*} (Q_{w} B Q_{w}) = R_{w} T^{*} (B) R_{w} .$
unlabeled · line 1182
$P_{z} T^{*} (B) P_{z^{'}} = 0 whenever z \neq = z^{'}, q (z) = q (z^{'}) = w .$
unlabeled · line 1195
$∥ (Q \otimes I) V - V R ∥_{on each vector}^{2} = 0$

Proposition: What sharp coarse compatibility adds

Displayed formulas in this passage

Nearby in this section