Almost Sure

14 January 20

The GNS Representation

Filed under: Probability Theory — George Lowther @ 12:06 AM
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As is well known, the space of bounded linear operators on any Hilbert space forms a *-algebra, and (pure) states on this algebra are defined by unit vectors. Considering a Hilbert space {\mathcal H}, the space of bounded linear operators {\mathcal H\rightarrow\mathcal H} is denoted as {B(\mathcal H)}. This forms an algebra under the usual pointwise addition and scalar multiplication operators, and involution of the algebra is given by the operator adjoint,

\displaystyle  \langle x,a^*y\rangle=\langle ax,y\rangle

for any {a\in B(\mathcal H)} and all {x,y\in\mathcal H}. A unit vector {\xi\in\mathcal H} defines a state {p\colon B(\mathcal H)\rightarrow{\mathbb C}} by {p(a)=\langle\xi,a\xi\rangle}.

The Gelfand-Naimark–Segal (GNS) representation allows us to go in the opposite direction and, starting from a state on an abstract *-algebra, realises this as a pure state on a *-subalgebra of {B(\mathcal H)} for some Hilbert space {\mathcal H}.

Consider a *-algebra {\mathcal A} and positive linear map {p\colon\mathcal A\rightarrow{\mathbb C}}. Recall that this defines a semi-inner product on the *-algebra {\mathcal A}, given by {\langle x,y\rangle=p(x^*y)}. The associated seminorm is denoted by {\lVert x\rVert_2=\sqrt{p(x^*x)}}, which we refer to as the {L^2}-seminorm. Also, every {a\in\mathcal A} defines a linear operator on {\mathcal A} by left-multiplication, {x\mapsto ax}. We use {\lVert a\rVert_\infty} to denote its operator norm, and refer to this as the {L^\infty}-seminorm. An element {a\in\mathcal A} is bounded if {\lVert a\rVert_\infty} is finite, and we say that {(\mathcal A,p)} is bounded if every {a\in\mathcal A} is bounded.

Theorem 1 Let {(\mathcal A,p)} be a bounded *-probability space. Then, there exists a triple {(\mathcal H,\pi,\xi)} where,

  • {\mathcal H} is a Hilbert space.
  • {\pi\colon\mathcal A\rightarrow B(\mathcal H)} is a *-homomorphism.
  • {\xi\in\mathcal H} satisfies {p(a)=\langle\xi,\pi(a)\xi\rangle} for all {a\in\mathcal A}.
  • {\xi} is cyclic for {\mathcal A}, so that {\{\pi(a)\xi\colon a\in\mathcal A\}} is dense in {\mathcal H}.

Furthermore, this representation is unique up to isomorphism: if {(\mathcal H^\prime,\pi^\prime,\xi^\prime)} is any other such triple, then there exists a unique invertible linear isometry of Hilbert spaces {L\colon\mathcal H\rightarrow\mathcal H^\prime} such that

\displaystyle  \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle \pi^\prime(a)=L\pi(a)L^{-1},\smallskip\\ &\displaystyle \xi^\prime=L\xi. \end{array}

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