Optimal success probability for Gaussian processing of non-Gaussian entanglement
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- Topics
Problem
What is the maximal success probability for preparing an approximate two-mode squeezed vacuum from finitely many copies of a non-Gaussian state?
Let \(\rho\) be a non-Gaussian density operator of one bosonic mode at Alice and one at Bob. Fix an integer \(m\geq1\), squeezing \(r>0\), and error tolerance \(0<\varepsilon<1\). The target is the two-mode squeezed vacuum
Equation (1) uses the product photon-number basis.
Preparation from \(m\) copies of \(\rho\) uses local Gaussian operations and classical communication. Allowed success maps \(\Lambda\) are trace-nonincreasing completely positive maps. Each implementation is a finite sequence using local Gaussian ancillary states, local Gaussian unitaries, Gaussian measurements or vacuum-projection success branches, discarding, and classical feedforward. Gaussian measurement records may be accepted on a specified measurable set; their probabilities are integrated over that set. A nonvacuum outcome of a vacuum projection terminates that run in failure; its output cannot be reused. The input copies are the only non-Gaussian resource.
For each allowed map, write \(p_\Lambda:=\operatorname{Tr}\Lambda(\rho^{\otimes m})\). When \(p_\Lambda>0\), let \(\tau_\Lambda:=\Lambda(\rho^{\otimes m})/p_\Lambda\). Define the maximal heralding probability by
Determine Eq. (2), with \(\sup\varnothing:=0\). Here \(\|X\|_1:=\operatorname{Tr}\sqrt{X^\dagger X}\).
Source
This optimization is a precise formulation of the probability-versus-accuracy question suggested by Browne et al., Sections II–III and Figures 2–3. The source supplies particular protocols, not this universal optimum [BESP03].
Progress
For the input \(|\varphi_\lambda\rangle:=(|0,0\rangle+\lambda|1,1\rangle)/\sqrt{1+\lambda^2}\), \(0<\lambda<1\), a balanced beam-splitter tree with \(m=2^k\) input copies, \(k\geq0\), and vacuum projections on all discarded ports produces coefficients
\begin{equation} \begin{aligned} b_{m,j}&:=\lambda^j\prod_{\ell=0}^{j-1}(1-\ell/m), &Z_m&:=\sum_{j=0}^{m}b_{m,j}^2,\\ |\varphi_{\lambda,m}\rangle&:=Z_m^{-1/2} \sum_{j=0}^{m}b_{m,j}|j,j\rangle, &p_m&:=\frac{Z_m}{(1+\lambda^2)^m}. \end{aligned} \tag{3} \end{equation}Equation (3) follows by expanding the published optical iteration; the empty product equals one. [BESP03]
Consequently, with \(r=\operatorname{arctanh}\lambda\) and \(\varepsilon_m:=\sqrt{1-|\langle\psi_r|\varphi_{\lambda,m}\rangle|^2}\), the protocol gives
\begin{equation} \begin{gathered} P_G^{(m)}(|\varphi_\lambda\rangle\langle\varphi_\lambda|,r,\varepsilon_m)\geq p_m, \\ \varepsilon_m\longrightarrow0,\qquad p_m\sim\frac{(1+\lambda^2)^{-m}}{1-\lambda^2}. \end{gathered} \tag{4} \end{equation}The limits in Eq. (4) follow directly from \(0\leq b_{m,j}\leq\lambda^j\) and \(b_{m,j}\to\lambda^j\); this establishes attainability, not an optimal asymptotic yield. [BESP03], [CE12]
Comment
The optimal probability for arbitrary non-Gaussian inputs remains unresolved. The displayed convergent protocol has exponentially decreasing success probability when every branch of a fixed input tree must succeed. Convergence alone does not establish a positive asymptotic yield.