Manuscript X-Enter-ST@1 · X-Enter-ST.tex · section · f76acc1f

The encoding

Stage C: the retained finite amplitude trial › The encoding · lines 635–678

X-Enter-ST@1#eq:encoding, X-Enter-ST@1#eq:bloch, X-Enter-ST@1#eq:ABC, X-Enter-ST@1#eq:selective

Freeze $σ > 0$ , with the units of $a$ , and set

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

The constant comparison encoding is $∣ e_{0} (a)⟩ = ∣ + ⟩$ .

Let

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) .

The basis $∣ 0 ⟩, ∣ 1 ⟩$ may be the two calibrated basis states of a finite encoding clock. The values of $a$ occur in its amplitudes, not as a continuum of orthogonal clock codes.

For any fixed instrument define

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

Linearity gives the exact selective expression

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

Taking traces gives the classical law with coefficients $α_{y} = tr A_{y}, β_{y} = tr B_{y}, γ_{y} = tr C_{y}$ .

Displayed formulas in this passage

X-Enter-ST@1#eq:encoding · line 638
$p (a) = \frac{1 + tanh ( a / σ )}{2}, ∣ e (a)⟩ = 1 - p (a) ∣ 0 ⟩ + p (a) ∣ 1 ⟩, τ (a) = ∣ e (a)⟩ ⟨ e (a)∣ .$
X-Enter-ST@1#eq:bloch · line 654
$τ (a) = \frac{1}{2} (I + t (a) D + s (a) X) .$
X-Enter-ST@1#eq:ABC · line 665
$A_{y} = Φ_{y} (I /2), B_{y} = Φ_{y} (D /2), C_{y} = Φ_{y} (X /2) .$
X-Enter-ST@1#eq:selective · line 672
$Φ_{y} (τ (a)) = A_{y} + B_{y} t (a) + C_{y} s (a) .$
unlabeled · line 648
$D = ∣ 1 ⟩ ⟨ 1 ∣ - ∣ 0 ⟩ ⟨ 0 ∣, X = ∣ 0 ⟩ ⟨ 1 ∣ + ∣ 1 ⟩ ⟨ 0 ∣, t (a) = tanh (a / σ), s (a) = sech (a / σ) .$

The encoding

Displayed formulas in this passage

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