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I have a question arising from von Neumann's C*-algebra formulation of quantum mechanics. In it, a state is a $\mathbb{C}$-linear functional on a C*-algebra $A$

$\rho:A\rightarrow\mathbb{C}$

satisfying the following conditions:

  • $\rho$ is positive: for every $a\in A$, we have $ \rho(a^*a) \geq 0 \in \mathbb{R}$;

  • $\rho$ is normalized: $\rho(\mathbf{1}) = 1$.

By the Gelfand-Naimark-Segal (GNS) construction, for each such state, there is a *-representation $\pi_{\rho}$ of $A$ in a Hilbert space $\mathscr{H}_{\rho}$ and a unit vector $\Omega\in\mathscr{H}_{\rho}$ such that $$\rho(a)=\langle\pi_\rho(a)\Omega,\Omega\rangle$$ for every $a\in A$.

Question: what conditions do two states $\rho$, $\rho'$ have to satisfy such that the associated representations be equivalent ?

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2 Answers 2

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If you want a criterion which is not tautological, that is, beyond the very definition of equivalence of *-representations, there are (at least) two situations where there is a criterion for equivalence of GNS representations, namely:

  1. $\rho$ and $\rho'$ are pure states (equivalently, the GNS representations $\pi_\rho$, $\pi_{\rho'}$ associated to $\rho$, resp. $\rho'$ are irreducible) or
  2. $\rho$ and $\rho'$ are type III factor states (i.e. the von Neumann algebras $\pi_\rho(A)''$, $\pi_{\rho'}(A)''$ are type III factors) and the GNS Hilbert spaces $\mathscr{H}_\rho$, $\mathscr{H}_{\rho'}$ are both separable.

In both cases, $\pi_\rho$ and $\pi_{\rho'}$ are equivalent iff they are quasi-equivalent, i.e. neither representation has a nonzero subrepresentation disjoint from the other. Generally, $\pi_\rho$ and $\pi_{\rho'}$ are quasi-equivalent iff there are trace-class positive linear operators $T_\rho\in\mathfrak{B}(\mathscr{H}_{\rho'})$, $T_{\rho'}\in\mathfrak{B}(\mathscr{H}_\rho)$ with unit trace (i.e. density matrices) such that $\rho(a)=\mathrm{Tr}(T_\rho\pi_{\rho'}(a))$ and $\rho'(a)=\mathrm{Tr}(T_{\rho'}\pi_\rho(a))$ for all $a\in A$. By Fell's equivalence theorem (Theorem 1.2 of J.M.G. Fell, The Dual Spaces of C*-Algebras, Trans. Amer. Math. Soc. 94 (1960) 365-403), quasi-equivalence of $\pi_\rho$ and $\pi_{\rho'}$ also amounts to requiring:

  • $\ker\pi_\rho=\ker\pi_{\rho'}=J$, and
  • The identity map of $A/J$ extends to a *-isomorphism from $\pi_\rho(A)''$ onto $\pi_{\rho'}(A)''$.

Therefore, these two conditions together may be thought of as an equivalence criterion in situations 1.) and 2.) above.

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    $\begingroup$ If $\rho$ and $\rho'$ are pure then they induce equivalent representations if and only if they are unitarily conjugate (an easy consequence of Kadison transitivity). $\endgroup$
    – Nik Weaver
    Commented Sep 3, 2016 at 21:45
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    $\begingroup$ That isn't really an issue for non-relativistic quantum mechanics due to the Stone-von Neumann uniqueness theorem: all representations of the canonical commutation relations in Weyl form are unitarily equivalent for a finite number of degrees of freedom. Inequivalent representations become important for systems with infinitely many degrees of freedom, such as thermodynamic limits of quantum statistical mechanical systems and quantum field theory. In those cases, the isometric isomorphisms among all separable Hilbert spaces need (and usually do) not intertwine different GNS representations. $\endgroup$
    – Pedro Lauridsen Ribeiro
    Commented Sep 4, 2016 at 1:43
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    $\begingroup$ Finite degrees of freedom is not the same as a finite-dimensional Hilbert space or a finite-dimensional C*-algebra. Just think of a single, free, non-relativistic quantum particle in one dimension (a system with one degree of freedom) - the Hilbert space of the system is $L^2(\mathbb{R})$, which is definitely infinite-dimensional. In that case, the C*-algebra of observables is generated by the Weyl unitaries $e^{itP}$, $e^{isX}$, where $s,t\in\mathbb{R}$ and $X,P$ are resp. the position and momentum operators. This C*-algebra is also infinite-dimensional. $\endgroup$
    – Pedro Lauridsen Ribeiro
    Commented Sep 4, 2016 at 2:01
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    $\begingroup$ Fock space is indeed one possible Hilbert space for QFT (albeit only for free field theories). There are many others, such as those of GNS representations associated to finite-temperature (i.e. KMS) states. $\endgroup$
    – Pedro Lauridsen Ribeiro
    Commented Sep 4, 2016 at 2:14
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    $\begingroup$ The number of degrees of freedom is the dimension of the classical configuration space (in the sense of classical mechanics). Harmonic oscillators are simple examples where the number of degrees of finite (it can even be 1) and the dimension of the quantum Hilbert space is infinite. $\endgroup$
    – Robert Furber
    Commented Sep 4, 2016 at 11:04
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Your title should not include normal, since that applies only to states on vN algebras (and possibly AW$*$ algebras).

Decades ago, George Elliott proved that if $A$ were a simple (and unital---all C$*$-algebras are unital here) AF C$*$-algebra, then any two extremal (aka pure) states were conjugate (via an automorphism, I think it turned out to be approximately inner). This was subsequently generalized to a much larger class, including non-simple, where the criterion applied to two extremal traces \st the kernels of the representations were equal. In this case, conjugacy is enough for equivalence (although it is not in general, consider the transposition on $\mathbf C \times \mathbf C$).

For impure states, there is not going to be much of a result, unless you know their integral decompositions.

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  • $\begingroup$ From the other comments, it looks like vN algebras come into play anyway, so may be normal is not so far out? But I'll take under advisement. Thanks. $\endgroup$
    – magya_bloom
    Commented Sep 4, 2016 at 1:28

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