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Faithful matrix modular flow and Gibbs time

Statement

Assume the Axiom of Choice. Let $n \geq 1$ , let $H = C^{n}$ , and let $ρ$ be a positive-definite Hermitian matrix, meaning $⟨ ρ v, v ⟩ > 0$ for each nonzero $v \in H$ , with $Tr ρ = 1$ . Write $H_{H S} = M_{n} (C)$ with inner product $⟨ X, Y ⟩_{H S} = Tr (Y^{*} X)$ , linear in the first argument, and let $M_{n} (C)$ act on $H_{H S}$ by left multiplication. For $Ω = ρ^{1 / 2}$ , the standard modular data are $J X = X^{*}, Δ X = ρ X ρ^{- 1}, σ_{t} (A) = ρ^{i t} A ρ^{- i t} (A \in M_{n} (C), t \in R) .$ The vector state of $Ω$ is $ω (A) = Tr (ρ A)$ . If $H = H^{*}$ has energy unit joules, $β > 0$ has unit joules $^{- 1}$ , $ℏ > 0$ has unit joule-seconds, and $ρ = e^{- β H} / Z$ with $Z = Tr (e^{- β H})$ , then the physical-time dynamics $α_{s} (A) = e^{i s H / ℏ} A e^{- i s H / ℏ} (s in seconds)$ satisfies $α_{s} = σ_{- s / (β ℏ)} .$ On the finite-dimensional C $^{*}$ -algebra $M_{n} (C)$ , $ω$ is a $β$ -KMS state for $α$ in the physical-time convention of the stated KMS definition, with strip width $β ℏ$ seconds. If $β = (k_{B} T)^{- 1}$ , then $k_{B}$ has units joules per kelvin and $T$ has units kelvins.

Statement references

Tomita graph and standard modular data

Concrete von Neumann algebra and faithful vector

Finite matrix spectral and state calculus

Trace-class normal density states

Physical-time KMS strip equilibrium condition

The Axiom of Choice

Facts & Assumptions

Given: The Axiom of Choice, $n \geq 1$ , a positive-definite Hermitian $ρ \in M_{n} (C)$ with trace one, and, for the Gibbs clause, a Hermitian $H$ , $β > 0$ , and $ℏ > 0$ with the units in the Statement.

[A1]

Assume the Axiom of Choice (The Axiom of Choice).

[F1]

Hermitian matrices have orthonormal eigenbases; the positive square root and matrix powers are defined spectrally; finite trace is cyclic; and the density pairing is $Tr (ρ A)$ (Finite matrix spectral and state calculus, Trace-class normal density states).

[F2]

A concrete von Neumann algebra is a unital star subalgebra satisfying the double-commutant condition; under Choice, weak-operator closure is equivalent to that condition (Concrete von Neumann algebra and faithful vector, The double commutant theorem for concrete von Neumann algebras).

[F3]

The physical-time KMS boundary convention requires bounded strip functions with $F (t) = ω (A α_{t} (B))$ and $F (t + i β ℏ) = ω (α_{t} (B) A)$ (Physical-time KMS strip equilibrium condition).

Proof

Given: [A1], [F1]–[F3], the finite-dimensional Hilbert space, and the matrices in the Statement.

1.1A1F1F2givenconstruct

Let $M = {L_{A} : A \in M_{n} (C)}$ on $H_{H S}$ , where $L_{A} X = A X$ . The Hilbert-Schmidt adjoint satisfies $L_{A}^{*} = L_{A^{*}}$ , so $M$ is a unital star subalgebra. The ambient operator space $B (H_{H S})$ is finite dimensional; hence $M$ is weak-operator closed. By [F2], $M$ is a concrete von Neumann algebra. Since $ρ^{1 / 2}$ is invertible, every $X \in H_{H S}$ equals $L_{X ρ^{- 1 / 2}} Ω$ , so $Ω$ is cyclic. If $L_{A} Ω = 0$ , then $A ρ^{1 / 2} = 0$ and invertibility gives $A = 0$ , so $Ω$ is separating. Finally, $⟨ L_{A} Ω, Ω ⟩_{H S} = Tr (ρ^{1 / 2} A ρ^{1 / 2}) = Tr (ρ A),$ using finite trace cyclicity.

1.2F1F3givenalgebra

For $A, B \in M_{n} (C)$ define $F_{A, B} (z) = Tr ⁣ (ρ A e^{i z H / ℏ} B e^{- i z H / ℏ}) .$ Finite matrix exponentials are entire, so $F_{A, B}$ is entire. If $0 \leq Im z \leq β ℏ$ , the real-time factors are unitary and the two imaginary-time factors have norms at most $e^{β ∥ H ∥}$ each. Since $∥ ρ ∥_{1} = 1$ , $∣ F_{A, B} (z) ∣ \leq ∥ A ∥ ∥ B ∥ e^{2 β ∥ H ∥}$ uniformly on that strip. The group $α_{s}$ is norm-continuous because its matrix entries are continuous.

2.1F1step 1.1algebra

Choose an orthonormal eigenbasis of $ρ$ , with eigenvalues $p_{j} > 0$ , and write $E_{i j} = ∣ i ⟩ ⟨ j ∣$ . Then $E_{i j} Ω = \sqrt{p_{j}} E_{i j}$ and $E_{i j}^{*} Ω = \sqrt{p_{i}} E_{j i}$ , whence $S (E_{i j}) = \sqrt{p_{i} / p_{j}} E_{j i}$ . The maps $J X = X^{*}$ and $Δ X = ρ X ρ^{- 1}$ satisfy $J Δ^{1 / 2} E_{i j} = \sqrt{p_{i} / p_{j}} E_{j i}$ and $Δ E_{i j} = (p_{i} / p_{j}) E_{i j}$ . Thus $S = J Δ^{1 / 2}$ on a basis, so uniqueness of the polar decomposition gives the stated modular data.

3.1step 2.1algebra

For $X \in H_{H S}$ , $Δ^{i t} L_{A} Δ^{- i t} X = ρ^{i t} A ρ^{- i t} X,$ because $Δ^{- i t} X = ρ^{- i t} X ρ^{i t}$ and the final right factor $ρ^{i t}$ cancels $ρ^{- i t}$ . Hence $σ_{t} (A) = ρ^{i t} A ρ^{- i t}$ .

4.1givenstep 3.1algebra

For $ρ = e^{- β H} / Z$ , functional calculus gives $ρ^{i t} = Z^{- i t} e^{- i β t H}$ ; the scalar factor cancels in conjugation. Substituting $t = - s / (β ℏ)$ into step 3.1 yields $σ_{- s / (β ℏ)} (A) = e^{i s H / ℏ} A e^{- i s H / ℏ} = α_{s} (A)$ . Here $β ℏ$ has units seconds, so the modular parameter is dimensionless.

5.1F1F3step 1.2algebra∎

At $z = t + i β ℏ$ , write $U = e^{i t H / ℏ}$ . Since $U$ commutes with $e^{\pm β H}$ , $F_{A, B} (t + i β ℏ) = Z^{- 1} Tr ⁣ (e^{- β H} A e^{- β H} α_{t} (B) e^{β H}) = Z^{- 1} Tr ⁣ (A e^{- β H} α_{t} (B)) = Tr ⁣ (ρ α_{t} (B) A) .$ The middle equality cycles $e^{β H}$ past the trace, and the last cycles $A$ past the trace. The lower boundary is $F_{A, B} (t) = Tr (ρ A α_{t} (B))$ . Thus [F3] gives the physical-time $β$ -KMS condition with width $β ℏ$ .

Faithful matrix modular flow and Gibbs time

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources