From Classical Mechanics to Quantum Field Theory

186 From Classical Mechanics to Quantum Field Theory. A Tutorial

The algebraic notion of pure state is in nice agreement with the Hilbert space
formulation result where pure states are represented by unit vectors (in the absence
of superselection rules). Indeed we have the following proposition which make a
comparison between the two notions.

Proposition 2.4.7. Let ωbe a pure state on theC∗-algebra with unitAand
Φ∈Hωaunitvector.Then:

(a)the functional

Aa→〈Φ,πω(a)Φ〉ω,

defines a pure algebraic state and(Hω,πω,Φ)is a GNS triple for it. In that

case, GNS representations of algebraic states given by non-zero vectors in

Hωare all unitarily equivalent;

(b)unit vectorsΦ,Φ′∈Hωgive the same (pure) algebraic state if and only if

Φ=cΦ′for somec∈C,|c|=1, i.e. if and only ifΦandΦ′belong to the

same ray.

The correspondence pure (algebraic) states vs. state vectors, automatic in the
standard formulation, holds in Hilbert spaces of GNS representations of pure alge-
braic states, but in general not for mixed algebraic states. The following exercise
focusses on this apparent problem.

Exercise 2.4.8.Consider, in the standard (not algebraic) formulation, a physical
system described on the Hilbert spaceHand a mixed stateρ∈S(H).Themap
ωρ:B(H)A→tr(ρA)defines an algebraic state on theC∗-algebraB(H).
By the GNS theorem, there exist another Hilbert spaceHρ, a representationπρ:
B(H)→B(Hρ)an unit vectorΨρ∈Hρsuch that

tr(ρA)=〈Ψρ,πρ(A)Ψρ〉

forA∈B(H). Thus it seems that the initial mixed state has been transformed
into a pure state! How is this fact explained?

Solution. There is no transformtion from mixed to pure state because the
mixed state is represented by a vector, Ψρ, in a different Hilbert space,Hρ.More-
over, there is no Hilbert space isomorphismU:H→HρwithUAU−^1 =πρ(A),
so thatU−^1 Ψρ∈H. In fact, the representationB(H)A→A∈B(H) is irre-
ducible, whereasπρcannot be irreducible (as it would be ifUexisted), because
the stateρis not an extreme point in the space of non-algebraic states, and so it
cannot be extreme in the larger space of algebraic states.