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Fidelity range equality and products

Statement

For finite-dimensional density operators $ρ, σ$ , $0 \leq F (ρ, σ) \leq 1, F (ρ, σ) = 1 ⟺ ρ = σ, F (ρ, σ) = 0 ⟺ supp ρ ⊥ supp σ .$ For density operators $ρ, σ$ on $H$ and $τ, ω$ on $K$ , $F (ρ \otimes τ, σ \otimes ω) = F (ρ, σ) F (τ, ω) .$

Facts & Assumptions

Given: Nonzero finite-dimensional spaces $H, K$ and density operators on them.

[L1]

Taking the fixed auxiliary space $R = H$ , root fidelity is the maximum absolute inner product of purifications (Fidelity Purification Characterization); the QC06 theorem proves this as part of its every- $R$ statement.

[L2]

Root fidelity equals $∥ \sqrt{ρ} \sqrt{σ} ∥_{1}$ (Fidelity), the trace norm is a norm, and singular values of a tensor product are pairwise products of singular values (Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition, Trace Norm Duality And Norm Properties). The QC01 norm theorem body is present as a draft.

[F2]

Density operators are nonnegative and have trace one (Density operator).

[F1]

Cauchy–Schwarz bounds the absolute inner product of unit vectors by one, with equality exactly for linearly dependent vectors (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

[F3]

On finite product spaces, $\sqrt{ρ} \otimes \sqrt{τ}$ is nonnegative and squares to $ρ \otimes τ$ : diagonalize both nonnegative factors in orthonormal eigenbases; the product basis has eigenvalues equal to products of their nonnegative eigenvalues, and tensor-map composition gives the square identity. Uniqueness of the nonnegative square root identifies it with $\sqrt{ρ \otimes τ}$ (A non-negative operator is equivalently self-adjoint with nonnegative eigenvalues, a positive semidefinite matrix in an orthonormal basis, or an operator of the form S^*S, Ordered tensor registers, their inner product, and wire permutations, Tensor products of finitely many linear maps, Tensor maps preserve composition, identities, and adjoints, A non-negative operator has a unique non-negative square root).

[L3]

The trace norm of a tensor product of finite-dimensional maps is the product of the trace norms (Trace norm of a tensor product).

Proof

1.1givenL1L2F1F2algebra

The fidelity is nonnegative because it is a trace norm. Choose unit purifications attaining the maximum in [L1]; [F1] gives their overlap at most one, so $F \leq 1$ . If $F = 1$ , equality in [F1] makes the two purifying vectors scalar multiples of modulus one. Their partial traces are then equal, hence $ρ = σ$ . Conversely, if $ρ = σ$ , choose the same purification twice to obtain overlap one.

1.2givenL2algebra

By [L2], $F = 0$ exactly when $\sqrt{ρ} \sqrt{σ} = 0$ , since a norm vanishes only at zero. The range of $\sqrt{σ}$ is $supp σ$ , while the kernel of $\sqrt{ρ}$ is $(supp ρ)^{⊥}$ . Thus the product is zero exactly when $supp σ \subseteq (supp ρ)^{⊥}$ , which is the asserted orthogonality.

2.1givenL2F2F3L3algebra∎

By [F3] and uniqueness of the nonnegative square root, $\sqrt{ρ \otimes τ} = \sqrt{ρ} \otimes \sqrt{τ}$ . Tensor-map composition gives $(\sqrt{ρ} \otimes \sqrt{τ}) (\sqrt{σ} \otimes \sqrt{ω}) = (\sqrt{ρ} \sqrt{σ}) \otimes (\sqrt{τ} \sqrt{ω})$ . Now [L3] makes the trace norm of this product the product of the two factor trace norms, which by the root-fidelity definition is $F (ρ, σ) F (τ, ω)$ .

Depends on

Used by

Exact Mixed State Cloning Fidelity Criterion Theorem

Dependency tree · two levels

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Sources

John Watrous, The Theory of Quantum Information (standard reference, not scraped)