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I am trying to understand different definitions of equivalence of representations. Given an algebra $\textbf{A}$, given two different representations $\pi_1$ and $\pi_2$ on two spaces $\textbf{B}(\textit{H}_1)$ and $\textbf{B}(\textit{H}_2)$, respectively. There are the following definitions:

  • $\pi_1$ and $\pi_2$ are said to be unitarily equivalent if there is a unitary opearator U such that U $\textbf{B}(\textit{H}_1)$ $U^{-1}$ = $\textbf{B}(\textit{H}_2)$. Otherwise, they are said to be not (unitarily) equivalent
  • $\pi_1$ and $\pi_2$ are said to be almost equivalent if every normal state for $\pi_1$ is also normal for $\pi_2$ and viceversa
  • $\pi_1$ and $\pi_2$ are said to be disjoint if there is no normal state for both representations

My doubts are the following:

  1. If both $\pi_1$ and $\pi_2$ are irreducible, $\pi_1$ and $\pi_2$ are unitarily equivalent $\iff$ $\pi_1$ and $\pi_2$ are almost equivalent?
  2. $\pi_1$ and $\pi_2$ are unitarily inequivalent $\iff$ $\pi_1$ and $\pi_2$ are disjoint?

Can you please help me? Even trivial examples would be helpful. My apologies in advance if I made stupid mistakes, I am still learning C* algebras.

Update: Theorem 2.4.26 from Bratteli-Robinson vol. I proves that almost equivalence $\iff$ existence of a isomorphism relating $\pi_1$ and $\pi_2$ as pointed out by @QuantumFieldMedalist. Moreover, it also proves that $\pi_1$ and $\pi_2$ are irreducible and almost equivalent $\iff$ unitarily equivalent

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Unitary equivalence of representations is a stronger statement than almost equivalence. To see this, for two unitarily equivalent representations $\pi_{1}, \pi_{2}$, any normal state on one is unitarily related to a normal state on the other. Almost equivalence is weaker as the relation between the normal states is not necessarily implementable by a unitary.

Disjoint representations $\pi_{1}, \pi_{2}$ occur when the representations are not even quasi-equivalent. This condition means that there is no *-isomorphism between $\pi_{1}(A)'', \pi_{2}(A)''$. A unitary is only a special type of *-isomorphism, so two representations need not be disjoint if they are not unitarily equivalent.

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  • $\begingroup$ Thank you! Just to better understand what you say: in case of almost equivalence, two normal states are related by a *-isomorphism which however is not unitary? May I ask you also a physical example? It would be helpful. $\endgroup$
    – MBlrd
    Commented Jun 6, 2023 at 19:46
  • $\begingroup$ The *-isomorphism is not necessarily unitary, and when it is then you have unitary equivalence instead. For a simple example, consider the algebra of observables of a QFT defined on a causal patch of de Sitter. This has an isomorphic algebra, namely the observables localized on the antipodal patch. These are isomorphic representations but are not related by a unitary, as one cannot evolve observables from one side to the other. $\endgroup$ Commented Jun 6, 2023 at 19:51

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