Manuscript En02-C@1 · En02-C-v1.tex · section · 7549c6df

All accessible channels

An exact three-step emission benchmark › All accessible channels · lines 619–657

En02-C@1#eq:logicalreps, En02-C@1#eq:normal, En02-C@1#eq:N3

Direct Clifford conjugation gives

U^{†} X_{2} U = - Z_{1} X_{2} Z_{3} =: L_{X}, U^{†} Z_{2} U = - Y_{1} Z_{2} Y_{3} =: L_{Z}, U^{†} Y_{2} U = - X_{1} Y_{2} X_{3} =: L_{Y} .

The last identity follows also from $Y = i X Z$ . These words all have weight three.

For every message operator $X$ ,

U N_{3} (X) U^{†} = X_{2} \otimes \frac{I _{13}}{4} .

Indeed,

(U \otimes I_{45}) V ∣ ψ ⟩ = ∣ ψ ⟩_{2} \otimes ∣ Φ_{2} ⟩_{14} \otimes ∣ Φ_{2} ⟩_{35} .

For a density operator $ρ = (I + x X + y Y + z Z) /2$ ,

N_{3} (ρ) = \frac{1}{8} (I + x L_{X} + y L_{Y} + z L_{Z}) .

Tracing any one of the three radiation factors removes all three nonidentity terms.

Displayed formulas in this passage

En02-C@1#eq:logicalreps · line 622
$U^{†} X_{2} U = - Z_{1} X_{2} Z_{3} =: L_{X}, U^{†} Z_{2} U = - Y_{1} Z_{2} Y_{3} =: L_{Z}, U^{†} Y_{2} U = - X_{1} Y_{2} X_{3} =: L_{Y} .$
En02-C@1#eq:normal · line 634
$U N_{3} (X) U^{†} = X_{2} \otimes \frac{I _{13}}{4} .$
En02-C@1#eq:N3 · line 649
$N_{3} (ρ) = \frac{1}{8} (I + x L_{X} + y L_{Y} + z L_{Z}) .$
unlabeled · line 641
$(U \otimes I_{45}) V ∣ ψ ⟩ = ∣ ψ ⟩_{2} \otimes ∣ Φ_{2} ⟩_{14} \otimes ∣ Φ_{2} ⟩_{35} .$