Fidelity Purification Characterization

Given: A nonzero finite-dimensional $H$, states $\rho ,\sigma \in \mathcal{D}(H)$, and an auxiliary space $R$ of dimension at least $\dim H$.

The root fidelity is $F(\rho ,\sigma )=\mathrm{\parallel }\sqrt{\rho }\sqrt{\sigma }{\mathrm{\parallel }}_1$ (Fidelity). The current QC06 source body is present; its fidelity convention is root fidelity.

In ordered orthonormal product coordinates, a vector with coefficient matrix $X$ has first marginal $X{X}^{\ast }$; under the root linear-first pairing the inner product of coefficient matrices $X,Y$ is $\mathrm{tr}(X{Y}^{\ast })$, whose modulus is $\mathrm{\mid }\mathrm{tr}(X^{\ast }Y)\mathrm{\mid }$ (Ordered tensor registers, their inner product, and wire permutations, Partial trace over a finite tensor factor, The partial trace is characterized by the trace pairing and is basis-independent). These identities follow by finite coordinate sums and the partial-trace trace-pairing formula.

For a rectangular coefficient map $X:R\to H$, the singular-value decomposition gives $X={\sum }_{j<r}{s}_j\mathrm{\mid }{u}_j⟩⟨v_j\mathrm{\mid }$ with $s_j>0$ and orthonormal families $(u_j)$ in $H$ and $(v_j)$ in $R$ (Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition, Dirac Notation Conversion). Then $X{X}^{\ast }={\sum }_{j<r}{s}_j^2\mathrm{\mid }{u}_j⟩⟨u_j\mathrm{\mid }$, and $B_X={\sum }_{j<r}{s}_j\mathrm{\mid }{u}_j⟩⟨u_j\mathrm{\mid }$ is non-negative because $⟨B_{X}z,z⟩={\sum }_{j<r}{s}_j\mathrm{\mid }⟨z,u_j⟩{\mathrm{\mid }}^2\ge 0$; also $B_X^2=X{X}^{\ast }$. Uniqueness of the non-negative square root gives $B_X=(X{X}^{\ast }{)}^{1\mathrm{/}2}$. Consequently $X=(X{X}^{\ast }{)}^{1\mathrm{/}2}W$ for $W={\sum }_{j<r}\mathrm{\mid }{u}_j⟩⟨v_j\mathrm{\mid }$. The SVD bases show $W{v}_j=u_j$ for $j<r$, that $W$ vanishes on the remaining basis vectors, and that $W^{\ast }$ has the reverse action; hence both are contractions. This rectangular factorization is derived from the SVD; the cited polar-decomposition theorem is used only for square endomorphisms.

Trace-norm duality gives ${\max }_{\mathrm{\parallel }C{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1}\mathrm{\mid }\mathrm{tr}(A^{\ast }C)\mathrm{\mid }=\mathrm{\parallel }A{\mathrm{\parallel }}_1$; in equal finite dimensions polar decomposition supplies an attaining unitary (Trace Norm Duality And Norm Properties, Every endomorphism has a polar decomposition T = SU with U non-negative and S an isometry on the orthogonal complement of ker T, and S is unique exactly when T is invertible).