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 $V\in {\mathcal{M}}^{\mathrm{\prime }}\otimes M_d$ 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 $0<\lambda \le 1$, where $\lambda$ can be determined by an algebraic condition on the Schmidt coefficients of the state $\phi ={\sum }_{i=0}^{d-1}{\alpha }_i{e}_i\otimes e_i$.