Manuscript Oai276-A@1 · Oai276-A.tex · lemma · 215068af

Lemma: One-qubit output determinant

Local one-use proof › Population-resolved output entropy · lines 469–499 · proof included or linked

Oai276-A@1#lem:determinant

Every pure qubit input with excited-state probability $p$ has output determinant $v_{γ, ν} (p)$ . For every mixed input $ρ$ with the same population,

S (A_{γ, ν} (ρ)) \geq e_{γ, ν} (p) .

Proof. For a pure input, $∣ z ∣^{2} = p (1 - p)$ , so the determinant of (eq:channel) is

t (1 - t) - (1 - γ) p (1 - p) .

Expansion gives

γ ν (1 - ν) + γ (1 - γ) (p - ν)^{2} .

This is in $[0, 1/4]$ because it is the determinant of a state.

For a spectral decomposition $ρ = \sum_{a} w_{a} ∣ ψ_{a} ⟩ ⟨ ψ_{a} ∣$ , let $p_{a}$ be the component populations. Entropy concavity and Lemma lem:convex give

S (A (ρ)) \geq a \sum w_{a} e (p_{a}) \geq e (a \sum w_{a} p_{a}) = e (p) .

Displayed formulas in this passage

unlabeled · line 474
$S (A_{γ, ν} (ρ)) \geq e_{γ, ν} (p) .$
unlabeled · line 482
$t (1 - t) - (1 - γ) p (1 - p) .$
unlabeled · line 486
$γ ν (1 - ν) + γ (1 - γ) (p - ν)^{2} .$
unlabeled · line 494
$S (A (ρ)) \geq a \sum w_{a} e (p_{a}) \geq e (a \sum w_{a} p_{a}) = e (p) .$

Lemma: One-qubit output determinant

Displayed formulas in this passage

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