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Consider a separable $C^*$ algebra $\mathcal A$. The space of states is also separable in the weak* topology, let $S$ be a countable dense subset.

Denoting with $H_\omega$ the GNS representation of a state $\omega$ we retrieve a representation of $\mathcal A$ on the Hilbert space $H(S)=\bigoplus_{\omega\in S}H_\omega$. This representation is isometric and $H$ is separable.

In the context of quantum mechanics we have built a candidate for the physical Hilbert space just by knowing an algebra of observables. This construction however depends on how we chose our set $S$. My question is:

Are the representations of $\mathcal A$ on $H(S)$ and $H(S')$ unitarily equivalent for any two dense countable subsets $S,S'$ of the state space of $\mathcal A$?

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Independently from separability of the algebra, if you restrict the states to pure states the answer is generally negative provided the algebra admits a couple of algebraic pure states unitarily inequivalent with separable GNS spaces (and this is the case in nontrivial qft). Consider a such pure state $\omega$. The pure states in its folium are weakly dense in the folium of every other pure state by Fell's theorem. Suppose the Hilbert space of the GNS representation of $\omega$ is separable. You conclude that there is a countable dense set of pure states $S$ in the folium of that state which is weakly dense in all folia of all algebraic state. The GNS representations of this separable class of pure states are unitarily equivalent, since the states are vector states in the folium of a given pure state. Therefore:

In the direct sum of Hilbert spaces labeled over $S$, actually there is a unique unitary representation, that of $\omega$, acting on each summand.

If you start from a unitarily inequivalent initial pure state $\omega'$ and built up the analogous direct sum, the final direct sum of representations cannot be unitarily equivalent to the initially constructed out of $\omega$.

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  • $\begingroup$ Thank you for the answer and sorry for the late reply. I have some questions about the beginning of your reply. What do you mean with an "algebraic pure state" of the algebra? I think all positive algebraic functionals that on a unital algebra are automatically continuous, so the normal way I would understand this statement does not work. And how can a state be equivalent or inequivalent to a GNS representation? $\endgroup$ – s.harp Dec 16 '17 at 15:38
  • $\begingroup$ I mean an extremal element of the convex body of positive normalized functional on the algebra. This is equivalent to saying that its GNS representation is irreducible (and in turn, it follows that all unit vectors in the GNS space are pure states as well in the same sense). $\endgroup$ – V. Moretti Dec 16 '17 at 16:31
  • $\begingroup$ Then isn't every pure state is equivalent to a separable GNS state, namely the GNS state associated to the state itself? Provided the algebra itself is separable (I think compact operators on $\ell^2$ with $\Bbb1$ are separable, but I see that $B(\ell^2)$ in general will only be separable with a weak or strong topology). $\endgroup$ – s.harp Dec 16 '17 at 16:53
  • $\begingroup$ For a unital $C^*$-algebra, in the GNS representation of an extremal state every vector state gives rise to a GNS representation unitarily equivalent to the initial one. If the initial algebraic state is not extremal this is false in general. I am not very familiar with the $C^*$-algebra of compact operators since it is not unital... $\endgroup$ – V. Moretti Dec 16 '17 at 17:00
  • $\begingroup$ However I do not know separable $C^*$-algebras relevant in physics. The simplest one, that associated to CCR of $X$ and $P$ (finite dimensional Weyl algebra) is never separable. $\endgroup$ – V. Moretti Dec 16 '17 at 17:04

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