Strict fixed-entanglement advantage over pure TMSS teleportation

For every real $\lambda>0$ and every real $r\geq0$, define the following two real suprema. First, let

$$S(\lambda,r)= \sup_{\substack{\tau,y\in\mathbb R\\ 0\leq\tau,\ |1-\tau|\leq y\\ e^{-2r}(1+\tau)\leq y}} \frac{2\lambda} {2(1-\sqrt{\tau})^2+\lambda(1+y+\tau)}.$$

Second, let $H=\ell^2(\mathbb N;\mathbb C)$, let $e_j$ be its standard coordinate vector, and put

$$t_r(a,b)= \begin{cases} (-\tanh r)^a/\cosh r,&a=b,\\ 0,&a\neq b. \end{cases}$$

Define $T(\lambda,r)$ as the supremum of $F_\lambda(\Phi)$ over every state $\rho$ indexed by $\mathbb N\times\mathbb N$, every real gain $g$, and every function $\Phi$ from complex matrices indexed by $\mathbb N$ to such matrices, satisfying the conditions below. Here a matrix $X$ indexed by a set $I$ is called trace class precisely when there exist sequences $u_j,v_j\in\ell^2(I;\mathbb C)$ such that $\sum_j\|u_j\|\|v_j\|$ is summable and $X_{ab}=\sum_j u_j(a)\overline{v_j(b)}$ for every $a,b$. A state is a matrix of this kind for which, for every finite $A\subseteq I$ and every $z:I\to\mathbb C$, the number $\sum_{a,b\in A}\overline{z(a)}X_{ab}z(b)$ has imaginary part zero and nonnegative real part, and whose real diagonal entries form a summable family with sum $1$. The required conditions on $\rho,g,\Phi$ are that $\rho_{uv}=t_r(u)\overline{t_r(v)}$ for all $u,v\in\mathbb N^2$, that $g>0$, and that $\Phi$ sends trace-class matrices to trace-class matrices, is additive and complex homogeneous on trace-class inputs, and has the following finite-extension property: for every integer $d\geq0$ and every state $R$ indexed by $\mathbb N\times\{0,\ldots,d-1\}$, the matrix

$$\widetilde\Phi(R)_{(a,s),(b,t)} =\Phi\bigl((R_{(i,s),(j,t)})_{i,j\in\mathbb N}\bigr)_{ab}$$

is a state. This includes $d=0$, for which there are no states on the empty index set. To state the remaining condition and the objective, define the matrices

$$D_z(m,n)=e^{-|z|^2/2}\sqrt{m!\,n!} \sum_{j=0}^{\min(m,n)} \frac{z^{m-j}(-\overline z)^{n-j}} {j!\,(m-j)!\,(n-j)!}, \qquad \chi_X(z)=\sum_m\sum_n X_{mn}D_z(n,m).$$

For every input state $\sigma$ indexed by $\mathbb N$ and every $z\in\mathbb C$, require

$$\chi_{\Phi(\sigma)}(z) = \chi_\sigma(gz) \sum_{u\in\mathbb N^2}\sum_{v\in\mathbb N^2} \rho_{uv}\, D_{-g\overline z}(v_1,u_1)\, D_z(v_2,u_2).$$

For a matrix $K$ indexed by $\mathbb N$, let $A_K$ be a chosen bounded complex-linear operator on $H$ satisfying $(A_Ke_n)_m=K_{mn}$ for all $m,n$, if such an operator exists, and let $A_K=0$ otherwise. Let $c_\alpha\in H$ be a chosen vector with coordinates

$$c_\alpha(j)=e^{-|\alpha|^2/2}\frac{\alpha^j}{\sqrt{j!}},$$

if such a vector exists, and let it be zero otherwise; set $K^\alpha_{mn}=c_\alpha(m)\overline{c_\alpha(n)}$. The fidelity appearing in the definition of $T$ is

$$F_\lambda(\Phi)= \int_{\mathbb C}\frac{\lambda}{\pi}e^{-\lambda|\alpha|^2} \operatorname{Re} \left\langle c_\alpha, A_{\Phi(K^\alpha)}c_\alpha\right\rangle\,d\alpha,$$

where $d\alpha$ is Lebesgue measure. The infinite sums and integral here use their total definitions; the optimization conditions impose no separate summability or integrability hypotheses on these expressions. Both suprema are the total real supremum operation, whose value is $0$ for an empty set or a set unbounded above. With these definitions, the declaration asserts all of

$$T(\lambda,r)\leq S(\lambda,r), \qquad S(\lambda,0)\leq T(\lambda,r),$$ $$S(\lambda,r)=T(\lambda,r)\iff r=0, \qquad T(\lambda,r)=S(\lambda,0)\iff r=0,$$

and

$$r>0\ \Longrightarrow\ S(\lambda,0)<T(\lambda,r)<S(\lambda,r).$$

Thus the endpoint $r=0$ is included, and at that endpoint both displayed equalities hold. Finally, for each such fixed $r$,

$$\lim_{\ell\to0^+}\bigl(S(\ell,r)-T(\ell,r)\bigr)=0.$$