Haar-average versus entanglement fidelity

Regard complex matrices as ${\mathbb{R}}^{2d^2}$. The equations $U^{†}U=I$ define a closed set, and each column has norm one, so the set is bounded. It is a group under matrix multiplication, with inverse $U^{†}$; multiplication and inversion are continuous. It is therefore a compact Hausdorff group by [F5]. By [F4] and [A1] it has normalized left-invariant Haar probability. Write $\mu$ for this probability on the Borel sigma-algebra $\mathcal{B}(U(d))$, and write $dU=d\mu (U)$.

A continuous map between topological spaces is Borel measurable: the target sets whose inverse images are Borel form a sigma-algebra containing every open set, so they include the sigma-algebra in [F7]. Every integrand in the moment and fidelity calculations below is a finite sum of products of the coordinates of $U\mathrm{\mid }0⟩$ and their conjugates, with fixed coefficients. These functions and their moduli are continuous, and they are bounded since each coordinate has modulus at most one. For any such function $h$, choose a finite $C\ge 0$ with $\mathrm{\mid }h\mathrm{\mid }\le C$. A nonnegative simple minorant $s={\sum }_r{c}_r{\mathbf{1}}_{A_r}\le \mathrm{\mid }h\mathrm{\mid }$, after omitting empty fibers, has $c_r\le C$. Thus its simple integral satisfies ${\sum }_r{c}_r\mu (A_r)\le C\mu (U(d))=C$. Taking the supremum in [F8] gives $\int \mathrm{\mid }h\mathrm{\mid }d\mu \le C<\mathrm{\infty }$, proving integrability by [F6]. Applied to $h=1$ this bounds its integral above by one, while the admissible simple minorant $s=1$ gives the reverse inequality, so $\int 1d\mu =1$. Consequently every finite expansion integrated below is justified by [F10].

For fixed $V\in U(d)$, the left translation $T_V(U)=VU$ is a homeomorphism with inverse $T_{V^{†}}$, hence is Borel measurable by step 2.1. For every Borel $A$, left invariance gives $\mu (T_V^{-1}A)=\mu (V^{-1}A)=\mu (A)$. Thus [F9] applies, giving $\int h(VU)d\mu (U)=\int h(U)d\mu (U)$ for each integrable $h$. In particular this applies to all functions in step 2.1 and all the phase, permutation and two-coordinate unitary substitutions below.

Let $S=\{x\in {\mathbb{C}}^d:x^{†}x=1\}$ and $\mathcal{P}=\{x{x}^{†}:x\in S\}$ carry the Borel sigma-algebras of their Euclidean subspace topologies. The maps $q:U(d)\to S$, $q(U)=U\mathrm{\mid }0⟩$, and $p:U(d)\to \mathcal{P}$, $p(U)=U\mathrm{\mid }0⟩⟨0\mathrm{\mid }{U}^{†}$, are continuous and hence measurable by step 2.1. For either map $r$ with target $Y$, define its law by ${\lambda }_r(B)=\mu (r^{-1}(B))$ as in [F11]. Inverse images preserve the empty set and disjoint countable unions, so ${\lambda }_r(\mathrm{\varnothing })=0$ and, for pairwise disjoint Borel $B_n\subseteq Y$, ${\lambda }_r({\bigcup }_n{B}_n)={\sum }_n{\lambda }_r(B_n)$ by countable additivity of $\mu$. Also ${\lambda }_r(Y)=\mu (U(d))=1$. Thus these laws are probabilities. For fixed $V\in U(d)$, let $a_V(x)=Vx$ on $S$ and $a_V(P)=VP{V}^{†}$ on $\mathcal{P}$. Each $a_V$ is a homeomorphism and $r\circ T_V=a_V\circ r$. Therefore ${\lambda }_r(a_V^{-1}(B))=\mu (T_V^{-1}(r^{-1}(B)))={\lambda }_r(B)$ by step 3.1. This proves unitary invariance of the vector law and conjugation invariance of the projector law.

Write $z_i=⟨i\mathrm{\mid }\psi ⟩$ and $M_{ij,kl}=\int z_i\overline{z_j}\overline{z_k}{z}_{l}dU$. Diagonal phase invariance forces $M_{ij,kl}=0$ unless the multiset $\{i,l\}$ equals $\{j,k\}$. Permutation invariance makes all $\int \mathrm{\mid }{z}_i{\mathrm{\mid }}^4$ equal to $t$ and all $\int \mathrm{\mid }{z}_i{\mathrm{\mid }}^2\mathrm{\mid }{z}_j{\mathrm{\mid }}^2$, for $i\mathrm{\ne }j$, equal to $s$. For $d\ge 2$, apply invariance under the unitary replacing the first two coordinates by $(z_0+z_1)\mathrm{/}\sqrt{2}$ and $(z_0-z_1)\mathrm{/}\sqrt{2}$. Expanding the fourth power and using phase invariance to discard unbalanced monomials gives $t=(2t+4s)\mathrm{/}4$, so $t=2s$. Integrating $({\sum }_i\mathrm{\mid }{z}_i{\mathrm{\mid }}^2{)}^2=1$ gives $dt+d(d-1)s=1$. Consequently $s=1\mathrm{/}[d(d+1)]$ and $t=2\mathrm{/}[d(d+1)]$. For $d=1$ the same resulting formula holds directly. Hence, for all indices, $M_{ij,kl}=({\delta }_{ij}{\delta }_{kl}+{\delta }_{ik}{\delta }_{jl})\mathrm{/}[d(d+1)]$.

The definition of ${\lambda }_r$ and the simple-integral formula in [F8] give ${\int }_{Y}sd{\lambda }_r={\int }_{U(d)}s\circ rd\mu$ for every nonnegative simple $s$ on $Y$, and for signed simple $s$ by [F6] and [F10]. For a bounded real Borel $g$ and any $\epsilon >0$, partition a bounded interval containing its range into finitely many half-open intervals of width at most $\epsilon$, with the final endpoint included, and choose the lower endpoint on each inverse image. This yields a simple $s$ with $s\le g\le s+\epsilon$. The supremum definition in [F8] gives monotonicity for nonnegative functions; adding a common constant and using [F10] gives it for bounded real functions. The boundedness argument of step 2.1 applies on both probability spaces, so all these functions are integrable. Their two $g$ integrals lie between the common value of the two $s$ integrals and that value plus $\epsilon$. Since $\epsilon$ is arbitrary, ${\int }_{Y}gd{\lambda }_r={\int }_{U(d)}g\circ rd\mu$. Applying this to the real and imaginary parts proves the same identity for bounded complex Borel $g$ by [F6]. In particular $g(P)=\mathrm{tr}(P\mathcal{E}(P))$ is bounded and continuous on $\mathcal{P}$: its expression is a finite sum of products of matrix entries, and every entry of a unit-vector projector has modulus at most one. For $P=p(U)=\mathrm{\mid }\psi ⟩⟨\psi \mathrm{\mid }$, finite matrix multiplication gives $g(P)=⟨\psi \mathrm{\mid }\mathcal{E}(\mathrm{\mid }\psi ⟩⟨\psi \mathrm{\mid })\mathrm{\mid }\psi ⟩$. The Haar integral in the Statement is therefore precisely the average with respect to ${\lambda }_p$.

Put $E_{ij}=\mathrm{\mid }i⟩⟨j\mathrm{\mid }$ and $\mathcal{E}(E_{ij}{)}_{kl}=⟨k\mathrm{\mid }\mathcal{E}(E_{ij})\mathrm{\mid }l⟩$. Expanding the projector and using complex linearity, the average is ${\sum }_{i,j,k,l}\mathcal{E}(E_{ij}{)}_{kl}{M}_{ij,kl}=[\mathrm{tr}\mathcal{E}(I)+{\sum }_{i,j}\mathcal{E}(E_{ij}{)}_{ij}]\mathrm{/}[d(d+1)]$. Trace preservation gives $\mathrm{tr}\mathcal{E}(I)=d$. On the other hand, expanding the normalized maximally entangled projector in [F2] yields $F_e(\mathcal{E})=d^{-2}{\sum }_{i,j}\mathcal{E}(E_{ij}{)}_{ij}$. Substitution proves the claimed formula.