Entanglement fidelity on a finite logical space

Let $d\ge 1$ be an integer, let $H={\mathbb{C}}^d$ with standard orthonormal basis $\mathrm{\mid }0⟩,\dots ,\mathrm{\mid }d-1⟩$, and let $\mathcal{E}:\mathcal{L}(H)\to \mathcal{L}(H)$ be a quantum channel. Use a second copy of $H$ as a reference and set $\mathrm{\mid }{\mathrm{\Phi }}_d⟩=d^{-1\mathrm{/}2}{\sum }_{j=0}^{d-1}\mathrm{\mid }j⟩\otimes \mathrm{\mid }j⟩$. For vectors $u,v$, write $\mathrm{\mid }u⟩⟨v\mathrm{\mid }$ for the operator $x\mapsto ⟨x,v⟩u$, using the inner product linear in its first argument. The entanglement fidelity of $\mathcal{E}$ for the maximally mixed input $I\mathrm{/}d$ is

$F_e(\mathcal{E})=⟨{\mathrm{\Phi }}_d\mathrm{\mid }(\mathrm{i}\mathrm{d}\otimes \mathcal{E})(\mathrm{\mid }{\mathrm{\Phi }}_d⟩⟨{\mathrm{\Phi }}_d\mathrm{\mid })\mathrm{\mid }{\mathrm{\Phi }}_d⟩\mathrm{.}$

Here the bracket $⟨v\mathrm{\mid }A\mathrm{\mid }v⟩$ means $⟨Av,v⟩$, $I$ is the identity on $H$, and $\mathrm{i}\mathrm{d}$ is the identity map on $\mathcal{L}(H)$. The tensor map acts on elementary operators by $(\mathrm{i}\mathrm{d}\otimes \mathcal{E})(A\otimes B)=A\otimes \mathcal{E}(B)$ and extends linearly. The pure-state overlap $⟨\psi \mathrm{\mid }\mathcal{E}(\mathrm{\mid }\psi ⟩⟨\psi \mathrm{\mid })\mathrm{\mid }\psi ⟩$ for a unit vector $\psi$, and its infimum over unit vectors, are separately defined functionals; they need not equal $F_e$.

Finite-dimensional quantum channel

Density operator

Ordered tensor registers, their inner product, and wire permutations

Greatest lower bound (infimum)

Given: The finite space, standard basis, channel, rank-one-operator convention and tensor-map rule in the Definition. Write $E_{jk}=\mathrm{\mid }j⟩⟨k\mathrm{\mid }$ and $P_{\mathrm{\Phi }}=\mathrm{\mid }{\mathrm{\Phi }}_d⟩⟨{\mathrm{\Phi }}_d\mathrm{\mid }$.

1.1 By [F3], the $d$ matching product basis vectors are orthonormal, so $\mathrm{\parallel }{\mathrm{\Phi }}_d\mathrm{\parallel }=1$. The operator $P_{\mathrm{\Phi }}$ is self-adjoint and satisfies $⟨P_{\mathrm{\Phi }}x,x⟩=\mathrm{\mid }⟨x,{\mathrm{\Phi }}_d⟩{\mathrm{\mid }}^2\ge 0$, and its trace is $\mathrm{\parallel }{\mathrm{\Phi }}_d{\mathrm{\parallel }}^2=1$, so it is a density operator by [F2]. Expanding gives $P_{\mathrm{\Phi }}=d^{-1}{\sum }_{j,k}{E}_{jk}\otimes E_{jk}$. The coordinate rule [F4] yields ${\mathrm{Tr}}_1(E_{jk}\otimes E_{\mathrm{\ell }m})={\delta }_{jk}{E}_{\mathrm{\ell }m}$, since summing the first-factor diagonal entries gives ${\sum }_s{\delta }_{sj}{\delta }_{sk}={\delta }_{jk}$. Consequently ${\mathrm{Tr}}_1{P}_{\mathrm{\Phi }}=d^{-1}{\sum }_j{E}_{jj}=I\mathrm{/}d$. [given, F2, F3, F4, algebra]

1.2 By [F5], prescribing $E_{jk}\otimes E_{\mathrm{\ell }m}\mapsto E_{jk}\otimes \mathcal{E}(E_{\mathrm{\ell }m})$ defines a unique linear map $L$ on the full operator space. Expanding $A$ and $B$ in their matrix units shows $L(A\otimes B)=A\otimes \mathcal{E}(B)$, as required in the Definition. In the product basis, summing diagonal entries gives $\mathrm{tr}(A\otimes B)=\mathrm{tr}(A)\mathrm{tr}(B)$. Hence [F1] implies $\mathrm{tr}L(E_{jk}\otimes E_{\mathrm{\ell }m})=\mathrm{tr}(E_{jk})\mathrm{tr}\mathcal{E}(E_{\mathrm{\ell }m})={\delta }_{jk}{\delta }_{\mathrm{\ell }m}=\mathrm{tr}(E_{jk}\otimes E_{\mathrm{\ell }m})$; linearity extends trace preservation to every input. [given, F1, F3, F5, algebra]

1.3 For every unit vector $\psi \in H$, put $P_{\psi }=\mathrm{\mid }\psi ⟩⟨\psi \mathrm{\mid }$. This operator is self-adjoint and satisfies $⟨P_{\psi }x,x⟩=\mathrm{\mid }⟨x,\psi ⟩{\mathrm{\mid }}^2\ge 0$, so it is non-negative. Complete positivity in [F1] and the $r=1$ case of [F6] imply that $\mathcal{E}(P_{\psi })$ is non-negative. Thus $f_{\psi }=⟨\mathcal{E}(P_{\psi })\psi ,\psi ⟩$ is real and nonnegative. Since $d\ge 1$, the unit vector $\mathrm{\mid }0⟩$ exists, so $S_{\mathcal{E}}=\{f_{\psi }:\psi \in H,\mathrm{\parallel }\psi \mathrm{\parallel }=1\}$ is a nonempty subset of $\mathbb{R}$ bounded below by $0$. By [F8] it has a real infimum. By [F7], any two infima are each at most the other and hence equal; the infimum functional in the Definition is therefore well-defined. [given, F1, F6, F7, F8, algebra]

2.1 Let $S(x\otimes y)=y\otimes x$ be the wire swap on $H\otimes H$. It permutes the orthonormal product basis and satisfies $S^2=I$, so it preserves inner products and $⟨Su,v⟩=⟨u,Sv⟩$. If $X$ is non-negative, $SXS$ is self-adjoint and $⟨SXSy,y⟩=⟨XSy,Sy⟩\ge 0$; thus conjugation by $S$ preserves non-negativity. The block rule [F6] gives ${\mathcal{E}}^{(d)}(B\otimes A)=\mathcal{E}(B)\otimes A$. Checking on the product operator basis [F5] therefore gives $L(X)=S{\mathcal{E}}^{(d)}(SXS)S$ for every $X$. Complete positivity in [F1] and [F6] makes ${\mathcal{E}}^{(d)}$ positive, so this identity proves positivity of the reference-first map $L=\mathrm{i}\mathrm{d}\otimes \mathcal{E}$. [F1, F3, F5, F6, step 1.2, algebra]

3.1 Steps 1.1, 1.2 and 2.1 show that $\rho =L(P_{\mathrm{\Phi }})$ is non-negative and has trace one, hence is a density operator by [F2]. In an orthonormal basis beginning with ${\mathrm{\Phi }}_d$, $F_e(\mathcal{E})=⟨\rho {\mathrm{\Phi }}_d,{\mathrm{\Phi }}_d⟩$ is its first diagonal entry. All diagonal entries are non-negative and sum to one, so this is a well-defined real number in $[0,1]$. [F2, F3, step 1.1, step 1.2, step 2.1, algebra]

4.1 For the comparison example take $d=2$ and define $\mathcal{R}(X)=\mathrm{tr}(X)\mathrm{\mid }0⟩⟨0\mathrm{\mid }$ on every operator $X$. This map is complex-linear and trace-preserving since $\mathrm{tr}(\mathrm{\mid }0⟩⟨0\mathrm{\mid })=1$. For any $r\ge 1$, write $X\in \mathcal{L}({\mathbb{C}}^2\otimes {\mathbb{C}}^r)$ uniquely as ${\sum }_{a,b=0}^1{E}_{ab}\otimes X_{ab}$. On elementary operator tensors the block rule [F6] acts as ${\mathcal{R}}^{(r)}(E_{ab}\otimes B)={\delta }_{ab}{E}_{00}\otimes B$; [F5] extends this to ${\mathcal{R}}^{(r)}(X)=E_{00}\otimes (X_{00}+X_{11})$. If $X$ is non-negative, each compression $X_{aa}$ is self-adjoint and satisfies $⟨X_{aa}y,y⟩=⟨X(\mathrm{\mid }a⟩\otimes y),\mathrm{\mid }a⟩\otimes y⟩\ge 0$. The displayed output is self-adjoint and, on $z=\mathrm{\mid }0⟩\otimes u+\mathrm{\mid }1⟩\otimes v$, its quadratic form is $⟨(X_{00}+X_{11})u,u⟩\ge 0$. Thus every amplification is positive, proving complete positivity by [F6]; [F1] then makes $\mathcal{R}$ a channel. Direct expansion yields $(\mathrm{i}\mathrm{d}\otimes \mathcal{R})(P_{{\mathrm{\Phi }}_2})=(I_2\mathrm{/}2)\otimes E_{00}$ and $F_e(\mathcal{R})=1\mathrm{/}4$. For a unit vector $\psi =a\mathrm{\mid }0⟩+b\mathrm{\mid }1⟩$, its pure-state overlap is $\mathrm{\mid }a{\mathrm{\mid }}^2$, equal to $1$ at $\mathrm{\mid }0⟩$ and $0$ at $\mathrm{\mid }1⟩$, so $0$ is a lower bound and every lower bound is at most the attained value $0$. By [F7], its infimum is therefore $0$. This verifies that these functionals need not equal entanglement fidelity. [F1, F3, F5, F6, F7, step 1.1, step 1.2, algebra] ∎