Self-testing of exact entanglement embezzlement

We consider bipartite exact entanglement embezzlement with a catalyst state vector $\psi$ 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 $(U \otimes I_d)(I_d \otimes V)(e_0 \otimes \psi \otimes e_0)=\sum_{i=0}^{d-1} \alpha_i e_i \otimes \psi \otimes e_i$, where each $\alpha_i>0$ and $\sum_{i=0}^{d-1} \alpha_i^2=1$. 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 $\text{III}_{\lambda}$ factor for some $0<\lambda \leq 1$, where $\lambda$ can be determined by an algebraic condition on the Schmidt coefficients of the state $\varphi=\sum_{i=0}^{d-1} \alpha_i e_i \otimes e_i$.