Proposition 38.11.4.label Let $A$ be a unital $C^{*}$ algebra, $\phi \in S(A)$ be a state, and $(H, \pi, \xi)$ be its GNS triple, then $\phi$ is a pure state if and only if $(H, \pi)$ is irreducible.
Proof. ($\Rightarrow$): Suppose that $(H, \pi)$ is reducible, then there exists a non-trivial invariant subspace $M \subset H$. Let $P \in B(H)$ be the orthogonal projection onto $M$, $\xi_{1} = P\xi/\norm{P\xi}_{H}$, and $\xi_{2} = (\xi - P\xi)/\norm{\xi - P\xi}_{H}$. For each $x \in A$, let
then $\phi_{1}$ and $\phi_{2}$ are vector states on $A$. Since $M$ and $M^{\perp}$ are invariant for $\pi$,
so $\phi$ is a strict convex combination of two states, and is not pure.
($\Leftarrow$): Suppose that $(H, \pi)$ is irreducible. Let $t \in (0, 1)$ and $\phi_{1}, \phi_{2} \in S(A)$ with $\phi = t\phi_{1} + (1 - t)\phi_{2}$. Let $T: \pi(A)H \to H$ be defined by
then since $\phi_{1} \le t^{-1}\phi$, $T$ extends continuously into a positive operator in $B(H)$. Moreover, for any $x, y, z \in A$,
so $T \in \pi(A)'$. By Schur’s Lemma, there exists $\lambda \in \complex$ such that $T = \lambda I$. However, since $\dpn{T\xi, \xi}{H}= \dpn{1_A, \phi_1}{A}= 1$, $\lambda = 1$, $T = I$, and $\phi_{1} = \phi$. By symmetry, $\phi_{2} = \phi$ as well.$\square$
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