Self-testing of exact entanglement embezzlement

Fuente: arXiv
Saved in:
Bibliographic Details
Main Author: Harris, Samuel J.
Format: Preprint
Published: 2026
Subjects:
Online Access:
Tags: Add Tag
No Tags, Be the first to tag this record!
_version_ 1866910246604111872
author Harris, Samuel J.
author_facet Harris, Samuel J.
contents We consider bipartite exact entanglement embezzlement with a catalyst state vector $ψ$ in a Hilbert space $\mathcal{H}$ using unitaries (or more generally, contractions). If $\mathcal{M} \subseteq \mathcal{B}(\mathcal{H})$ is a von Neumann algebra and $U \in M_d \otimes \mathcal{M}$ and $V \in \mathcal{M}' \otimes M_d$ are unitaries (or more generally contractions), then such a protocol is of the form $(U \otimes I_d)(I_d \otimes V)(e_0 \otimes ψ\otimes e_0)=\sum_{i=0}^{d-1} α_i e_i \otimes ψ\otimes e_i$, where each $α_i>0$ and $\sum_{i=0}^{d-1} α_i^2=1$. We show that any such protocol must arise from a unique state on the tensor product $\mathcal{O}_d \otimes \mathcal{O}_d$ of the Cuntz algebra with itself. As a result, we prove that exact entanglement embezzlement is a self-test for a collection of $d$ Cuntz isometries for each party and a unique quasi-free state on the Cuntz algebra $\mathcal{O}_d$ in the sense of \cite{Iz93}. Moreover, we use modular theory to show that the von Neumann algebra generated by the copy of $\mathcal{O}_d$ is the unique separable approximately finite-dimensional Type $\text{III}_λ$ factor for some $0<λ\leq 1$, where $λ$ can be determined by an algebraic condition on the Schmidt coefficients of the state $φ=\sum_{i=0}^{d-1} α_i e_i \otimes e_i$.
format Preprint
id arxiv_https___arxiv_org_abs_2605_22713
institution arXiv
publishDate 2026
record_format arxiv
spellingShingle Self-testing of exact entanglement embezzlement
Harris, Samuel J.
Operator Algebras
Quantum Physics
We consider bipartite exact entanglement embezzlement with a catalyst state vector $ψ$ in a Hilbert space $\mathcal{H}$ using unitaries (or more generally, contractions). If $\mathcal{M} \subseteq \mathcal{B}(\mathcal{H})$ is a von Neumann algebra and $U \in M_d \otimes \mathcal{M}$ and $V \in \mathcal{M}' \otimes M_d$ are unitaries (or more generally contractions), then such a protocol is of the form $(U \otimes I_d)(I_d \otimes V)(e_0 \otimes ψ\otimes e_0)=\sum_{i=0}^{d-1} α_i e_i \otimes ψ\otimes e_i$, where each $α_i>0$ and $\sum_{i=0}^{d-1} α_i^2=1$. We show that any such protocol must arise from a unique state on the tensor product $\mathcal{O}_d \otimes \mathcal{O}_d$ of the Cuntz algebra with itself. As a result, we prove that exact entanglement embezzlement is a self-test for a collection of $d$ Cuntz isometries for each party and a unique quasi-free state on the Cuntz algebra $\mathcal{O}_d$ in the sense of \cite{Iz93}. Moreover, we use modular theory to show that the von Neumann algebra generated by the copy of $\mathcal{O}_d$ is the unique separable approximately finite-dimensional Type $\text{III}_λ$ factor for some $0<λ\leq 1$, where $λ$ can be determined by an algebraic condition on the Schmidt coefficients of the state $φ=\sum_{i=0}^{d-1} α_i e_i \otimes e_i$.
title Self-testing of exact entanglement embezzlement
topic Operator Algebras
Quantum Physics
url https://arxiv.org/abs/2605.22713