Physics Asked on June 3, 2021
I should start by saying that I am not a physicist, and I am not really familiar with Quantum dynamics.
I am reading about the GNS representation in the book "An Introduction to the Mathematical Structure of Quantum Mechanics" by Strocchi.
What I get from the explanation is that given a $C^*$-algebra of observables $mathcal A$, and a state $omega$ then there’s a Hilbert space $mathcal H_{omega}$ and a representation $pi_{omega}:mathcal Atomathcal B(mathcal H_{omega})$ that satisfies certain conditions.
I have two doubts:
I apology I advance if the question is rather basic or ill-posed.
This questions is somehow related but it doesn’t clear all my doubts though.
I just add some further details to the other good answers.
Regarding the first issue, I think Strocchi is referring here to the explicit GNS construction of the GNS Hilbert space ${cal H}_omega$. The vectors of that Hilbert space are (limit of) vectors obtained by the action of the observables $Bin {cal A}$ on the cyclic vector throug the representation $pi_omega$: $$Psi_B := pi_omega(B) Phi_omega:.$$ In fact, $${cal H}_omega = overline{pi_omega({cal A})Phi_omega}:.$$
From the algebraic perspective, it amounts to consider the algebraic states for $B^*Bneq 0$, $$omega_B(A):= frac{omega(B^*AB)}{omega(B^*B)}:.quad Ain {cal A}.$$
This result actually holds true also in standard quantum theory in Hilbert space, when assuming that the algebra of bounded observables is made of all the selfadjoint operators in ${cal B}({cal H})$, hence in the absence of superselection rules. (In case of superselection rules, a refinement of the idea leads to the coherent sector decomposition of the Hilbert space.) In this case, if $Phi$ is any unit vector (thus representing a pure state within the said hypotheses), $$overline{{cal B}({cal H}) Phi} = {cal H}:.$$ The GNS construction is a generalization of this idea.
Regarding your second issue, yes there are states on ${cal A}$ which are related to the given $omega$, and other states which are independent. Let us first assume that $omega$ is pure (i.e., it cannot be decomposed into a proper convex combination of other states), to simplify the discussion. In this case, all statistical operators $rho$ of ${cal H}_omega$ represent other states with some relation with $omega$: $$omega_rho(A) := tr(pi_omega(A)rho):,quad A in {cal A}$$ It is possible to prove that $pi_{omega_rho}$ and ${cal H}_{omega_rho}$ can be constructed out of the corresponding objects related to $omega$. In particular, if $rho$ is pure, they coincide to them and the cyclic vector of $omega_{rho}$ is every unit vector $Psi_rho$ such that $rho = |Psi_rho rangle langle Psi_rho|$. The states of type $omega_rho$ are called normal states of the representation of $omega$ and they form the folium of $omega$.
There are however other (algebraic) states $omega'$ on ${cal A}$ which cannot be costructed as $omega_rho$ for some statistical operator acting in ${cal H}_omega$. These states and their GNS representations are called disjoint (in the general case of $omega$ that is not pure).
If $omega$ and $omega'$ are both pure, they share the same folium if and only if their GNS representations are unitarily equivalent. In other words, there is an isometric surjective operator $U : {cal H}_{omega'} to {cal H}_omega$ such that $$U pi_{omega'}(A) U^{-1} = pi_omega(A):, quad forall A in {cal A}.$$ (Notice that it is not required that $U$ maps a cyclic vector into the other of the two GNS representations and thus this requirement is quite mild.)
This fact should be viewed as an improvement of the elementary Hilbert space formulation, since there are several known cases in standard quantum theory where we have two theories with a common algebra of operators but, for some reason, "the two theories cannot be represented in the same Hilbert space". A safe approach is to see the two theories as (possibly GNS) representations unitarily inequivalent (more generally disjoint representations) of the same abstract algebra of observables.
In QFT that is the standard (Haag's theorem establishes that the free theory and the interacting one suffer of this problem). Actually, in QFT the easier approach based on $*$-algebras is more appropriate, though it is plagued by some intpretative issues when one wants to pass form the common folklore to rigorous statements.
Also the appearance of superselection rules and various aspects of the so called spontaneously breaking of symmetry can be handled this way.
Correct answer by Valter Moretti on June 3, 2021
The archetypal cyclic vector is a vacuum (or lowest energy) state - the only physically relevant states are those you can reach by starting from nothing and letting operators derived from the observables (in particular the time evolution operator) act on it. The archetypal example here is the Hilbert space of the quantum harmonic oscillator that has a basis of states labeled by a natural number $ninmathbb{N}$ where $n=0$ is the vacuum/ground state and there are creation/annihilation (or raising/lowering) operators that increase or decrease $n$ by 1 when acting on such a state.
Note that in an irreducible representation every non-zero vector is cyclic, so a GNS construction that contains the vacuum state (which is in particular always a pure state so that the representation is irreducible) is isomorphic to all the GNS constructions of the abstract states associated to the other vectors in this vacuum construction. That is, if we accept the premise that the physical Hilbert space has to include the vacuum state, then there is only a single GNS construction you need to get the "physical Hilbert space".
This isn't always true - sometimes there is no unique vacuum, but instead different "vacua" the system might jump between - but it's a good enough picture for most cases.
Answered by ACuriousMind on June 3, 2021
Just to complement the other answer, in the algebraic setting one can define pure states abstractly, as the extremal points in the cone of states (intuitvely, these are the states that can be written as a convex combination of states only if each element in the combination is the state itself).
The GNS representation of a given state is irreducible iff the state is pure. In addition, as @ACuriousMind wrote, every vector in the Hilbert space of an irreducible representation is cyclic (otherwise there would be a nontrivial invariant subspace).
Hence given a fixed pure state, all vectors in its GNS Hilbert space are pure states themselves. Unfortunately, however, in general there are also pure states that are not representable at all in a fixed GNS hilbert space.
If two states are such that one cannot be represented as a vector (or density matrix for what matters) of the other GNS Hilbert space they are called disjoint.
In non relativistic quantum mechanics (i.e. if one considers the algebra of canonical commutation relations in a finite dimensional phase space), there are no disjoint pure states. This is a reformulation of the famous Stone-von Neumann theorem: any couple of irreducible representations of the CCR are unitarily equivalent.
In relativistic theories, however, there are infinitely many possible disjoint pure states. This is due to the so-called Haag's theorem. The typical example is to consider a scalar free theory, and an interacting one. Haag's theorem says that since the two ground states (vacua) cannot be equal in this case, they must be disjoint.
Answered by yuggib on June 3, 2021
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