Manuscript X-A-conv-ST@1 · X-A-conv-ST.tex · section · 8ca47088

The retained amplitude family

Stage C: readout closure of the retained coefficient › The retained amplitude family · lines 1495–1527

X-A-conv-ST@1#eq:selective

Freeze $σ > 0$ and define

p (a) = \frac{1 + tanh ( a / σ )}{2}, ∣ e (a)⟩ = 1 - p (a) ∣ 0 ⟩ + p (a) ∣ 1 ⟩, τ (a) = ∣ e (a)⟩ ⟨ e (a)∣ .

Set

D = ∣ 1 ⟩ ⟨ 1 ∣ - ∣ 0 ⟩ ⟨ 0 ∣, X = ∣ 0 ⟩ ⟨ 1 ∣ + ∣ 1 ⟩ ⟨ 0 ∣, t (a) = tanh (a / σ), s (a) = sech (a / σ) .

Then

τ (a) = \frac{1}{2} (I + t (a) D + s (a) X) .

For any fixed finite instrument ${Φ_{y}}_{y}$ , with every quantum residual included, put

A_{y} = Φ_{y} (I /2), B_{y} = Φ_{y} (D /2), C_{y} = Φ_{y} (X /2) .

Its selective output is

Φ_{y} (τ (a)) = A_{y} + B_{y} t (a) + C_{y} s (a) .

Displayed formulas in this passage

X-A-conv-ST@1#eq:selective · line 1523
$Φ_{y} (τ (a)) = A_{y} + B_{y} t (a) + C_{y} s (a) .$
unlabeled · line 1498
$p (a) = \frac{1 + tanh ( a / σ )}{2}, ∣ e (a)⟩ = 1 - p (a) ∣ 0 ⟩ + p (a) ∣ 1 ⟩, τ (a) = ∣ e (a)⟩ ⟨ e (a)∣ .$
unlabeled · line 1505
$D = ∣ 1 ⟩ ⟨ 1 ∣ - ∣ 0 ⟩ ⟨ 0 ∣, X = ∣ 0 ⟩ ⟨ 1 ∣ + ∣ 1 ⟩ ⟨ 0 ∣, t (a) = tanh (a / σ), s (a) = sech (a / σ) .$
unlabeled · line 1511
$τ (a) = \frac{1}{2} (I + t (a) D + s (a) X) .$
unlabeled · line 1517
$A_{y} = Φ_{y} (I /2), B_{y} = Φ_{y} (D /2), C_{y} = Φ_{y} (X /2) .$