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Finite Gibbs density and its canonical purification

Statement

Let $H$ be a nonzero finite-dimensional complex Hilbert space of dimension $d \geq 1$ , $K = K^{*}$ an endomorphism, and $β > 0$ . Define $Z_{β} = Tr (e^{- β K})$ and $ρ_{β} = Z_{β}^{- 1} e^{- β K}$ . Given an ordered orthonormal basis $(e_{j})_{j = 1}^{d}$ and its identified copy in a reference space $H_{R}$ , define $∣ Φ_{d} ⟩ = d^{- 1 / 2} \sum_{j = 1}^{d} e_{j} \otimes e_{j}, ∣ Ψ_{β} ⟩ = \sqrt{d / Z_{β}} (e^{- β K / 2} \otimes I_{R}) ∣ Φ_{d} ⟩ .$ Then $Z_{β} > 0$ , $ρ_{β}$ is a density operator, $∣ Ψ_{β} ⟩$ is a unit vector, and ${Tr}_{R} (∣ Ψ_{β} ⟩ ⟨ Ψ_{β} ∣) = ρ_{β}$ . If $K$ is interpreted as energy, the product $β K$ is dimensionless.

Facts & Assumptions

Given: $H \neq {0}$ has finite dimension $d$ ; $K$ is Hermitian; $β > 0$ ; and the ordered orthonormal bases and reference copy are as displayed.

[F1]

Finite matrix spectral and state calculus: a finite Hermitian matrix has real eigenvalues and a basis-independent functional calculus; in an eigenbasis, the trace is the sum of diagonal entries.

[F2]

Density operator: in finite dimension a density operator is positive with trace one.

[F3]

Partial trace over a finite tensor factor: in ordered product bases, $({Tr}_{R} X)_{j k} = \sum_{ℓ} X_{(j, ℓ), (k, ℓ)}$ .

[F4]

The partial trace is characterized by the trace pairing and is basis-independent: the finite partial trace is independent of the chosen reference basis and is uniquely characterized by its trace pairing.

Proof

1.1givenF1F2algebra

By [F1], choose an orthonormal eigenbasis of $K$ with real eigenvalues $λ_{1}, \dots, λ_{d}$ . Then $Z_{β} = \sum_{j} e^{- β λ_{j}}$ is a finite sum of strictly positive real numbers, so $Z_{β} > 0$ . The operator $ρ_{β}$ is diagonal in this basis with positive entries $e^{- β λ_{j}} / Z_{β}$ , whose sum is one; [F2] therefore makes it a density operator.

2.1givenstep 1.1F1algebra

Keep the arbitrary ordered orthonormal basis $(e_{j})$ and its given reference copy $(e_{j}^{'})$ from the Statement. Put $T = e^{- β K / 2}$ , so the defining vector is $Ψ_{β} = Z_{β}^{- 1 / 2} \sum_{j} T e_{j} \otimes e_{j}^{'}$ . Orthogonality of the reference vectors gives $∥ Ψ_{β} ∥^{2} = Z_{β}^{- 1} \sum_{j} ∥ T e_{j} ∥^{2} = Z_{β}^{- 1} Tr (T^{*} T) = 1$ , because [F1] gives $T^{*} = T$ and $T^{*} T = e^{- β K}$ , and step 1.1 gives its trace $Z_{β}$ . This uses the given basis, without requiring it to diagonalize $K$ .

3.1givenstep 1.1step 2.1F1F3F4algebra∎

In those same given bases, $Ψ_{β}$ has coefficient $T_{j ℓ} / \sqrt{Z_{β}}$ at $e_{j} \otimes e_{ℓ}^{'}$ . Thus [F3] gives the reduced matrix entry $Z_{β}^{- 1} \sum_{ℓ} T_{j ℓ} \overline{T_{k ℓ}} = (T T^{*})_{j k} / Z_{β} = (ρ_{β})_{j k}$ , since $T T^{*} = e^{- β K}$ by [F1]. Equivalently, the reduction is $Z_{β}^{- 1} \sum_{ℓ} ∣ T e_{ℓ} ⟩ ⟨ T e_{ℓ} ∣ = T T^{*} / Z_{β}$ . The intrinsic characterization [F4] identifies this coordinate result with the finite partial trace, so the system marginal equals $ρ_{β}$ for every allowed original basis and reference copy. No basis independence of the purification vector itself is asserted.

Finite Gibbs density and its canonical purification

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources