Manuscript Psy03-A@1 · Psy03-A.tex · theorem · b30eb6e9

Theorem: Exact phase-resource performance

Exact coherence-sensitive pilots › Phase-sensitive classification under dephasing · lines 1649–1680 · proof included or linked

Psy03-A@1#thm:phase

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The binary classification success is

Q_{phase} (γ, T) = \frac{1 + e ^{- γ T}}{2} .

For the selected coherence comparison

ρ_{+} = ∣ + ⟩ ⟨ + ∣, Δ ρ_{+} = I_{2} /2, F = P_{+},

C_{F} (T) = Tr [F Λ_{λ} (ρ_{+} - Δ ρ_{+})] = \frac{λ}{2} .

The trace-distance witness has the same value.

Every diagonal preparation obtained by dephasing either task input is $I_{2} /2$ , and carries no information about the label. Consequently its optimal equal-prior task success is $1/2$ .

Proof. Equation (eq:dephchannel) preserves populations and multiplies off-diagonal entries by $λ$ . Thus the correct phase-projector probability is $(1 + λ) /2$ . The coherent-minus-dephased output is $λ X /2$ , with eigenvalues $\pm λ /2$ , proving both witness values. The two dephased task states are identical, so no output measurement can distinguish their labels.

Displayed formulas in this passage

unlabeled · line 1652
$Q_{phase} (γ, T) = \frac{1 + e ^{- γ T}}{2} .$
unlabeled · line 1656
$ρ_{+} = ∣ + ⟩ ⟨ + ∣, Δ ρ_{+} = I_{2} /2, F = P_{+},$
unlabeled · line 1660
$C_{F} (T) = Tr [F Λ_{λ} (ρ_{+} - Δ ρ_{+})] = \frac{λ}{2} .$

Theorem: Exact phase-resource performance

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