Manuscript Oai272-A@1 · Oai272-A.tex · lemma · a40828a0

Lemma: Finite fidelity identities

Stage 1: the operational certificate › Fidelity at singular and unnormalized blocks · lines 734–769 · proof included or linked

Oai272-A@1#lem:fidelity, Oai272-A@1#eq:fidelity-lower

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For square matrices $A, B$ ,

f (A^{†} A, B^{†} B) = ∥ A B^{†} ∥_{1} .

For positive $H, K$ ,

f (H, K) \geq \frac{Tr H + Tr K - ∥ H - K ∥ _{1}}{2} .

Also $f (H^{T}, K^{T}) = f (H, K)$ .

Proof. Extend the polar partial isometries to unitaries for the first identity. Transpose invariance follows from functional calculus and invariance of singular values under transpose and adjoint.

For (eq:fidelity-lower), put $A = H$ , $B = K$ , $D = A - B$ . Since $H - K = A D + D B$ , trace-norm duality gives

∥ H - K ∥_{1} \geq Tr (∣ D ∣ (A + B)) .

On a unit eigenvector of $D$ with eigenvalue $d$ , positivity gives $⟨ A + B ⟩ \geq ∣ d ∣$ . Thus

∥ H - K ∥_{1} \geq Tr D^{2} = Tr H + Tr K - 2 Tr (A B) \geq Tr H + Tr K - 2 f (H, K) .

Displayed formulas in this passage

Oai272-A@1#eq:fidelity-lower · line 741
$f (H, K) \geq \frac{Tr H + Tr K - ∥ H - K ∥ _{1}}{2} .$
unlabeled · line 737
$f (A^{†} A, B^{†} B) = ∥ A B^{†} ∥_{1} .$
unlabeled · line 757
$∥ H - K ∥_{1} \geq Tr (∣ D ∣ (A + B)) .$
unlabeled · line 763
$∥ H - K ∥_{1} \geq Tr D^{2} = Tr H + Tr K - 2 Tr (A B) \geq Tr H + Tr K - 2 f (H, K) .$

Lemma: Finite fidelity identities

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