Manuscript X-Enter@1 · X-Enter.tex · section · 484b71bc

Explicit amplitude encodings

Frozen finite interface and selective formula › Explicit amplitude encodings · lines 455–504

X-Enter@1#eq:encoding, X-Enter@1#eq:state, X-Enter@1#eq:selective, X-Enter@1#eq:law

Fix a scale $σ > 0$ before the convolution witness evaluation. Freeze the two encoding trials

p_{j} (a) = \frac{1 + j tanh ( a / σ )}{2}, ι_{j} (a) = 1 - p_{j} (a) ∣ 0 ⟩ + p_{j} (a) ∣ 1 ⟩, j \in {0, 1} .

An attempt selects one trial once. The input register size is $n = 1$ . The constant trial is $ι_{0} (a) = ∣ + ⟩$ . These are mathematical preparation interfaces, not field-to-qubit implementation certificates.

Set

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

For $τ_{1} (a) = ∣ ι_{1} (a)⟩ ⟨ ι_{1} (a)∣$ ,

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

Define

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

The retained selective formula is

Φ_{y}^{P} (τ_{1} (a)) = A_{y} + B_{y} t (a) + C_{y} s (a) .

Its traces give the classical law:

Law_{P} (a) (y) = α_{y} + β_{y} t (a) + γ_{y} s (a), (α_{y}, β_{y}, γ_{y}) = (tr A_{y}, tr B_{y}, tr C_{y}) .

Equal laws need not mean equal selective records.

Displayed formulas in this passage

X-Enter@1#eq:encoding · line 459
$p_{j} (a) = \frac{1 + j tanh ( a / σ )}{2}, ι_{j} (a) = 1 - p_{j} (a) ∣ 0 ⟩ + p_{j} (a) ∣ 1 ⟩, j \in {0, 1} .$
X-Enter@1#eq:state · line 479
$τ_{1} (a) = \frac{1}{2} (I + t (a) D + s (a) X) .$
X-Enter@1#eq:selective · line 490
$Φ_{y}^{P} (τ_{1} (a)) = A_{y} + B_{y} t (a) + C_{y} s (a) .$
X-Enter@1#eq:law · line 495
$Law_{P} (a) (y) = α_{y} + β_{y} t (a) + γ_{y} s (a), (α_{y}, β_{y}, γ_{y}) = (tr A_{y}, tr B_{y}, tr C_{y}) .$
unlabeled · line 473
$t (a) = tanh (a / σ), s (a) = sech (a / σ), D = ∣ 1 ⟩ ⟨ 1 ∣ - ∣ 0 ⟩ ⟨ 0 ∣, X = ∣ 0 ⟩ ⟨ 1 ∣ + ∣ 1 ⟩ ⟨ 0 ∣ .$
unlabeled · line 484
$A_{y} = Φ_{y}^{P} (I /2), B_{y} = Φ_{y}^{P} (D /2), C_{y} = Φ_{y}^{P} (X /2) .$