Self-testing of exact entanglement embezzlement
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arXiv
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| Format: | Preprint |
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2026
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| _version_ | 1866910246604111872 |
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| 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 |
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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 |