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Quantum theory may be formalized in several different ways. Generally, the physical discussion of different states of a quantum system distinguishes pure and mixed states, and then subsumes both in a density operator formalism where a quantum state is given by a density operator $\rho : \mathcal{H}\to\mathcal{H}$ on some Hilbert space $\mathcal{H}$ where $\rho$ is self-adjoint, positive semi-definite and is trace-class with unit trace.

However, in an alternative approach, one starts with the abstract $C^*$-algebra $\mathcal{A}$ of observables and calls any positive linear functional on $\mathcal{A}$ a state. A normal state is then one that can be represented by a density operator on some Hilbert space on which the algebra is represented (there are several equivalent definitions, but this one is the one which makes it obvious that the normal states are the usual pure+mixed states of quantum mechanics).

But, in general, there exist non-normal states at least on some admissible algebras of observables. One can find several vague statements about what those may or may not mean, but I have not yet found a unique satisfactory answer to the question:

Do the non-normal states of the algebra of observables have physical significance, that is, is there an actual quantum system where only considering the normal states leads you to different physical predictions than taking all states?

ACuriousMind
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    Consider a pair of unitarily inequivalent representations of a given $C^$-algebra induced by a pair of pure algebraic states by means of the GNS construction. Each one* of these states is not a normal state for the other GNS representation. (Otherwise it would be represented by a cyclic unit vector in the Hilbert space of the other rep. and the two reps would be unitarily equivalent via the identity operator). Unitarily inequivalent reps arise in several contexts in QFT...Spontaneous breaking of symmetry, for instance, leads to unitarily inequivalent reps... – Valter Moretti Nov 02 '15 at 15:49
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    In curved spacetime, consider a spacetime which is flat both in the future and in the past of a given Cauchy surface, but including a curved region. The $*$-algebra of the fields is fixed but there are two natural pure states on that algebra: the Minkowski vacuum of the future and the Minkowski vacuum of the past. In general, depending on the curvatures, they are not unitarily equivalent, each one is not normal for the representation of the other of the same algebra of quantum fields... – Valter Moretti Nov 02 '15 at 16:04
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    Actually there is another possiblity, different from the two cases I mentioned above, referring to non-normal states with respect to a von Neumann algebra. For a given C-algebra, fix a state, represent the C-algebra in the GNS Hilbert space and take the generated von Neumann algebra therein. There exist algebraic states for that C* algebra (the von Neumann algebra interpreted as a C* algebra), which are not normal. Maybe you were referring to this other notion of non-normal state. – Valter Moretti Nov 02 '15 at 16:12
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    @ValterMoretti: Maybe that different notion of normal state is part of why I seem to have difficulty in nailing down their meaning - people are talking about two different things! If I'm understanding you correctly, are you saying in the curved spacetime case that we would not be able render both vacua as density operators on the same Hilbert space, thus making the generalization to non-normal states of the algebra necessary? – ACuriousMind Nov 02 '15 at 16:24
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    Yes, in curved spacetime the picture is the one you described. There are states which cannot be represented in the same Hilbert space in the standard way. However, there is a class of states, called Hadamard states, which are locally unitarily equivalent i.e. referring to the subalgebras of observables restricted to bounded spacetime regions. – Valter Moretti Nov 02 '15 at 16:30
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    Non-normal states can occur if a thermodynamic limit is taken for instance for a system of particles in a box. Another case is that of scattering of Schródinger particles in the presence of bound states. – Urgje Nov 02 '15 at 20:15
  • @ValterMoretti Uh, could you give some link where the curved space example is considered in detail. At least I would like to understand the setup. If we consider e.g. the free scalar on the Friedmann-like background (with $a(t)=1$ as $t\rightarrow\pm\infty$) then we simply have creation-annihilation operators in the past and the future related through the Bogolyubov transformation (and for this $a(t)$ there are no frozen modes like in cosmology). Then the vaccum of the past doesn't look to me as a non-normal state in the future. Of course that's just the simplest setup I could think of. – OON May 29 '19 at 20:26
  • @OON I assumed that both vacua are pure algebraic states. Suppose that the first vacuum is a normal state in the GNS representation of the second one. In this case, both states must be represented by unit vectors in the GNS representation of the second vacuum, in particular, the first vacuum must be a unit vector as well since pure normal states in irreducible representations are only of that type. – Valter Moretti May 29 '19 at 20:53
  • @OON It is easy to check that the GNS representation of the second vacuum is also a GNS rep of the first vacuum (use in particular the fact in an irreducible GNS representation all vectors are cyclic). In this case the identity map is a unitary connecting the two GNS representations. This is impossible as I said. – Valter Moretti May 29 '19 at 20:54
  • @ValterMoretti Frankly, I don't understand at all how all this works at least in your language... But let me try to ask whether I get the essense of what happens the dumb way? If we consider the universe in the box, would you say that in that case everything would be ok? I mean is this simply because the unitary transformation between two vacua related by the Bogolyubov transformation becomes singular for the infinite volume and so we can't put both into one Hilbert space? – OON May 30 '19 at 11:25
  • @OON In a box (finite volume) all vacua produces equivalent representations. Indeed, the problem may arise when passing to an infinite volume as you point out. If I correctly remember, a discussion on this issue using Bogolyubov transformation appears in Fulling's textbook "aspect of QFT in curved spacetime" (or a similar title)... Try to have a look at that book. – Valter Moretti May 30 '19 at 12:30
  • @ValterMoretti ok, thanks! – OON May 30 '19 at 12:31
  • Is it not necessary in the alternative abstract approach you describe that the state when evaluated on the identity yields unity, i.e. if $\rho$ is a positive linear functional on $\mathcal{A}$, then $\rho(1) = 1$ if $\rho$ is a state? I am confused. – Apoorv Potnis Aug 13 '23 at 03:24
  • Can you please comment on the following as well? Hannabuss defines states as (normalized positive linear functionals on a *-algebra), but then at the end, he remarks that "Even this algebraic formulation of quantum theory is not the most general. Some physical observables, such as times when a photon hits a counter, cannot be easily be interpreted as elements of the algebra, and for them there are still more general approaches." Ref: Keith Hannabuss An Introduction to Quantum Theory, pp. 196--199. I am not quite sure if this is related to your question. – Apoorv Potnis Aug 13 '23 at 03:26

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