I have the following doubt. Let's assume we have two mixed states $\rho_1 = \Sigma_i a_i \omega_i^{1}$ and $\rho_2 = \Sigma_i b_i \omega_i^{2}$ on the same algebra, where the states $\omega$ are all pure states and $\Sigma_i a_i = \Sigma_i b_i = 1$.
- Could it be that the two mixed states $\rho_1$ and $\rho_2$ are disjoint despite for each pure state $i$ there is a unitary map $U_i$ such that $U_i \pi_{\omega_i^{1}} (A) U_i^{\dagger} = \pi_{\omega_i^{2}} (A), \forall A$ in the algebra?
- Let's suppose that the algebras are factors. Is the result of question 1 somehow related to the type of the factor?
Edit: Here for disjoint representations I mean that, given the GNS representations $\pi_{\rho_{1}}$ and $\pi_{\rho_{2}}$ of the mixed states $\rho_1$ and $\rho_2$, there is no *-isomorphism τ such that, for any element A of the algebra, $τ(\pi_{\rho_{1}}(A))=\pi_{\rho_{2}}(A)$. In other terms, the two representations are not quasi-equivalent. Motivation: I would like to better understand the reason why two KMS states with different temperatures are disjoint (please see Takesaki work for more details: https://link.springer.com/article/10.1007/BF01649582), and make one step more to understand equivalence between reducible representations