Manuscript I@2 · I.tex · proposition · 85631fa7

Proposition: Two-qubit nullity violation

The independent static marginal problem › Explicit failure of operational nullity · lines 2974–3002 · proof included or linked

I@2#prop:static-nullity

The family of Theorem thm:static-positive violates operational nullity on a two-qubit pure input.

Proof. Let

ψ = \frac{∣ 00 ⟩ + ∣ 01 ⟩ + ∣ 10 ⟩}{3} .

Its $11$ -amplitude vanishes. Both one-qubit reductions have off-diagonal entry $1/3$ , so both rotations have angle $θ = η /3$ .

Write $s = sin θ$ , $c = cos θ$ . Both are positive because $0 < θ < 1/3$ . Directly,

⟨ 11 ∣ (W \otimes W) ∣ ψ ⟩ = \frac{s ^{2} + 2 sc}{3} > 0.

Therefore

F_{{1, 2}} (∣ ψ ⟩ ⟨ ψ ∣)_{11} = \frac{( s ^{2} + 2 sc ) ^{2}}{3} > 0,

although $M^{{1, 2}}_{11} ψ = 0$ .

Displayed formulas in this passage

unlabeled · line 2982
$ψ = \frac{∣ 00 ⟩ + ∣ 01 ⟩ + ∣ 10 ⟩}{3} .$
unlabeled · line 2992
$⟨ 11 ∣ (W \otimes W) ∣ ψ ⟩ = \frac{s ^{2} + 2 sc}{3} > 0.$
unlabeled · line 2997
$F_{{1, 2}} (∣ ψ ⟩ ⟨ ψ ∣)_{11} = \frac{( s ^{2} + 2 sc ) ^{2}}{3} > 0,$