Fidelity classical measurement minimum

Let $x,y$ be purifications attaining $F(\rho ,\sigma )$ as in [L1]. For a finite POVM $(E_a)$, put $x_a=(\sqrt{E_a}\otimes I)x$ and $y_a=(\sqrt{E_a}\otimes I)y$. Since ${\sum }_a{E}_a=I$, the adjoint identity gives $⟨x,y⟩={\sum }_a⟨x_a,y_a⟩$. By [L4], $\mathrm{\parallel }{x}_a{\mathrm{\parallel }}^2=\mathrm{tr}(\rho E_a)$ and $\mathrm{\parallel }{y}_a{\mathrm{\parallel }}^2=\mathrm{tr}(\sigma E_a)$. The triangle inequality and [F1] termwise give $\mathrm{\mid }⟨x,y⟩\mathrm{\mid }\le {\sum }_a\mathrm{\parallel }{x}_a\mathrm{\parallel }\mathrm{\parallel }{y}_a\mathrm{\parallel }={\sum }_a\sqrt{\mathrm{tr}(\rho E_a)\mathrm{tr}(\sigma E_a)}$. Therefore every finite POVM overlap is at least $F(\rho ,\sigma )$.

For arbitrary $\rho ,\sigma$, set ${\rho }_{\epsilon }=(1-\epsilon )\rho +\epsilon I\mathrm{/}d$ and ${\sigma }_{\epsilon }=(1-\epsilon )\sigma +\epsilon I\mathrm{/}d$ for $0<\epsilon \le 1$. Both are positive definite. The root fidelity is continuous on density operators: if $A_n\to A$ are density operators and $B_n=\sqrt{A_n}$, then [F5] and [F7] give ${\sum }_{i,j}\mathrm{\mid }(B_n{)}_{ij}{\mathrm{\mid }}^2=\mathrm{tr}(B_n^{\ast }{B}_n)=1$. Thus the coordinate matrices $B_n$ are bounded. Every convergent subsequence has a nonnegative limit $B$ by [F6] and $B^2=A$, so uniqueness in [L3] gives $B=\sqrt{A}$. If $(B_n)$ did not converge to $\sqrt{A}$, a subsequence staying a fixed positive distance away would have a convergent subsubsequence by [F2], contradicting that every such limit is $\sqrt{A}$. Applying this to both state sequences shows their square roots converge; their product converges by the finite matrix-product formula, and its trace norm converges by [L5]. Hence $F({\rho }_{\epsilon },{\sigma }_{\epsilon })\to F(\rho ,\sigma )$.

Suppose first that $\rho$ is positive definite. By [F4], choose an orthonormal eigenbasis of $\rho$; its eigenvalues are strictly positive by positive definiteness, so ${\rho }^{-1\mathrm{/}2}$ is defined by the reciprocal square roots in this basis. Define $M={\rho }^{-1\mathrm{/}2}(\sqrt{\rho }\sigma \sqrt{\rho }{)}^{1\mathrm{/}2}{\rho }^{-1\mathrm{/}2}$. Its quadratic form is nonnegative because it is the quadratic form of $(\sqrt{\rho }\sigma \sqrt{\rho }{)}^{1\mathrm{/}2}$ at ${\rho }^{-1\mathrm{/}2}x$, and direct multiplication gives $M\rho M=\sigma$. By [F4], choose an orthonormal eigenbasis of $M$, with rank-one projections $E_j$ and eigenvalues ${\mu }_j\ge 0$. If $p_j=\mathrm{tr}(\rho E_j)$, then $\mathrm{tr}(\sigma E_j)={\mu }_j^2{p}_j$, so the measurement overlap is ${\sum }_j{\mu }_j{p}_j=\mathrm{tr}(M\rho )=\mathrm{tr}(\sqrt{\sqrt{\rho }\sigma \sqrt{\rho }})=F(\rho ,\sigma )$ by [L3] and [L6]. Thus a rank-one PVM attains the lower bound when $\rho$ is faithful.

Fix an orthonormal basis of $H$ by [F3], and identify $H$ with ${\mathbb{C}}^d$. Let $Q_j$ be the rank-one projector onto its $j$th basis vector. Define ${\mathcal{P}}_d$ as ordered families $(U{Q}_j{U}^{\ast }{)}_{j=1}^d$ for $U\in U(d)$. Every ordered rank-one PVM has this form, since unit vectors spanning its mutually orthogonal one-dimensional ranges form an orthonormal basis. The map $U\mapsto (U{Q}_j{U}^{\ast }{)}_{j=1}^d$ is continuous in matrix coordinates, so ${\mathcal{P}}_d$ is compact by [F2]. For every $\epsilon >0$, step 2.1 supplies a minimizing PVM in ${\mathcal{P}}_d$. For $n\ge 1$, let $C_n$ be the closure in ${\mathcal{P}}_d$ of all such minimizing PVMs for $0<\epsilon <1\mathrm{/}n$. The $C_n$ are nonempty nested compact sets, so their intersection is nonempty; fix $P_0$ in it. The overlap function $g(\epsilon ,P)={\sum }_j\sqrt{\mathrm{tr}({\rho }_{\epsilon }{P}_j)\mathrm{tr}({\sigma }_{\epsilon }{P}_j)}$ is continuous at every $(0,P_0)$, since trace is a finite coordinate sum and scalar square root is continuous on $[0,\mathrm{\infty })$. Given $\delta >0$, choose a neighborhood of $(0,P_0)$ where $g$ differs from $g(0,P_0)$ by less than $\delta$, and choose $n$ so large that $1\mathrm{/}n$ is inside its $\epsilon$-neighborhood and, by step 1.2, $\mathrm{\mid }F({\rho }_{\epsilon },{\sigma }_{\epsilon })-F(\rho ,\sigma )\mathrm{\mid }<\delta$ for $0<\epsilon <1\mathrm{/}n$. Since $P_0\in C_n$, the definition of closure supplies a minimizing PVM $P^{\mathrm{\prime }}$ in that neighborhood for some $0<\epsilon <1\mathrm{/}n$. For this pair, $g(\epsilon ,P^{\mathrm{\prime }})=F({\rho }_{\epsilon },{\sigma }_{\epsilon })$, so $\mathrm{\mid }g(0,P_0)-F(\rho ,\sigma )\mathrm{\mid }<2\delta$. As $\delta >0$ is arbitrary, $g(0,P_0)=F(\rho ,\sigma )$. This gives an attaining rank-one PVM also in the singular case.

Step 1.1 bounds every finite POVM overlap below by $F$, and step 3.1 exhibits a finite rank-one PVM whose overlap equals $F$. This proves the stated minimum and attainment.