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Unique tracial KMS state is not vacuum normal

Example

Let q≥2, Γ=Zd with d≥1, and take the zero Hermitian onsite Hamiltonian h=0∈Mq(C) in joules. Fix any ℏ>0 in joule-seconds and let β>0 have units inverse joules. The onsite dynamics is αt(A)=A for t∈R in seconds, and the KMS strip has width βℏ seconds, with the convention of Unique onsite product Gibbs KMS state. For every such β, the unique quasi-local KMS state has local densities ρΛ=q−∣Λ∣IΛ. It is tracial on the full quasi-local algebra, but has no normal extension to the finite-excitation vacuum representation.

Facts & Assumptions

Given: ACω, the local spin algebra, and the zero onsite model with the energy, time, positive hbar and inverse-energy beta conventions just specified.

[F1]

Unique onsite product Gibbs KMS state: under ACω and its finite-matrix hypotheses [A2], for h=0, r=q−1I and the unique KMS state has local density q−∣Λ∣IΛ. Those finite-matrix hypotheses are the clauses in [F3] below.

[F2]

Thermal product state has no normal vacuum extension: under ACω and the same finite-matrix hypotheses retained from [F1], if q≥2 and the onsite Gibbs density is full rank, this KMS state has no normal extension to B(Hvac).

[F3]

Local spin matrix algebras and their factor inclusions, Definition: every finite region Λ has a nonzero complex Hilbert space HΛ of dimension q∣Λ∣, with inner product linear in the first argument, and AΛ=B(HΛ) with the induced operator norm. Its finite-matrix clauses give ∥T∗T∥=∥T∥2; positive complex-linear functionals are exactly T↦Tr⁡Λ(ρT) for unique ρ≥0, with functional norm Tr⁡Λρ, so normalization is equivalent to trace one. These clauses are supplied by the proved finite-matrix lemma cited in that Definition, including the empty region of dimension one. Every pair of local observables embeds into a common finite-region matrix algebra, where the finite trace is cyclic.

[F4]

Norm-completed quasi-local quantum spin algebra: the strictly local algebra is Aloc=⋃Λ⊆Γ, ∣Λ∣<∞AΛ, and the quasi-local algebra is its norm completion A=Aloc‾ ∥⋅∥; therefore strictly local observables are norm dense in A.

Verification

technique · direct

Given: [A1], [F1]–[F4], and arbitrary local observables.

1.1A1F1F3givenalgebra

For every finite region, q∣Λ∣≥1, so the finite-matrix clauses in [F3] apply and discharge exactly the C*-identity and positive-functional/density hypotheses of [F1]. Together with Countable Choice [A1], h=0=h∗, β>0 and ℏ>0, these meet its hypotheses. Since e−βh=I and Tr⁡I=q, the density is r=q−1I, so r⊗Λ=q−∣Λ∣IΛ, and [F1] identifies its extension as the unique β-KMS state.

2.1F1F3step 1.1algebra

If A,B are local, embed them into a common finite region Λ using [F3]. Then ω(AB)=q−∣Λ∣Tr⁡Λ(AB)=q−∣Λ∣Tr⁡Λ(BA)=ω(BA), by finite-matrix trace cyclicity.

3.1A1F1F4step 2.1algebra

For arbitrary A,B∈A, norm density from [F4] and Countable Choice [A1] give local sequences An→A and Bn→B. Their norms are bounded, and ∥AnBn−AB∥≤∥An−A∥ ∥Bn∥+∥A∥ ∥Bn−B∥⟶0, with the same estimate after exchanging An,Bn and A,B. Boundedness of the state then gives ω(AnBn)→ω(AB) and ω(BnAn)→ω(BA). Step 2.1 gives equality at each n, so ω(AB)=ω(BA); the state is tracial on the full algebra.

4.1A1F2F3step 1.1step 3.1givenalgebra∎

The density r=q−1I has all eigenvalues 1/q>0, hence is full rank, and its reference diagonal entry is p0=1/q∈(0,1) because q≥2. Countable Choice [A1] and the finite-matrix clauses [F3] were discharged in step 1.1 for this same onsite model. Thus every hypothesis retained in [F2] holds, and the unique tracial KMS state has no normal extension in the finite-excitation vacuum representation.

Depends on

Dependency tree · 2 levels

Sources