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Trace-class normal density states

Definition

Let $H \neq {0}$ be a separable complex Hilbert space, with inner product linear in its first argument. An operator $ρ \in B (H)$ is a density operator when it is positive, trace class, and $Tr ρ = 1$ . Here trace class means compact with summable singular values, and $Tr$ is the basis-independent trace of a trace-class operator.

The normal density state represented by $ρ$ is the functional on bounded operators $ω_{ρ} (A) : = Tr (ρ A), A \in B (H) .$ In this page, “normal density state” means this specified trace-pairing representation; no equivalence with other definitions of normal functionals is assumed. The trace pairing is used only for bounded $A$ . Unbounded observables are treated through their spectral probability measures in Trace-state cyclicity and spectral probability calculus.

Facts & Assumptions

Given: Countable Choice, a nonzero separable complex Hilbert space $H$ , a positive trace-class operator $ρ$ with $Tr ρ = 1$ , and a bounded operator $A \in B (H)$ .

[A1]

Countable Choice is the exact choice assumption used to index the nonzero eigenvalues by a finite or countable set (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

Trace class operator: a trace-class operator is compact with summable singular values; its trace norm dominates its operator norm.

[F2]

Spectral theorem for compact self adjoint operators: a positive compact operator has at most countably many positive eigenvalues with an orthonormal family spanning its support.

[F3]

Trace of a positive operator is the sum of its eigenvalues: for a positive trace-class operator, its trace is the sum of its positive eigenvalues with multiplicity and its trace norm is that sum. This gives the trace directly without treating the positive eigenfamily as a Hilbert basis.

[F4]

Trace is absolutely convergent and basis independent: the trace of a trace-class operator is absolutely defined, basis independent, linear, and bounded by its trace norm.

[F5]

Trace class is a two sided Banach operator ideal: if $T$ is trace class and $A, B$ are bounded, then $A T B$ is trace class and $∥ A T B ∥_{1} \leq ∥ A ∥ ∥ T ∥_{1} ∥ B ∥$ .

[F6]

Nuclear bounded cyclicity for trace-class operators: for a trace-class nuclear expansion $T = \sum_{j} T_{u_{j}, v_{j}}$ and bounded $A$ , $Tr (A T) = \sum_{j} ⟨ A u_{j}, v_{j} ⟩$ and $Tr (T A) = Tr (A T)$ .

Proof

1.1givenA1F1F2F3

By [F1] and [F2], write $ρ = \sum_{j \in J} p_{j} ∣ u_{j} ⟩ ⟨ u_{j} ∣$ , where $J$ is finite or countable, $p_{j} > 0$ , and the $u_{j}$ are orthonormal. The positive-operator trace formula [F3] gives $\sum_{j} p_{j} = Tr ρ = 1$ and $∥ ρ ∥_{1} = 1$ . For every finite subset $F \subseteq J$ , the remainder $ρ - \sum_{j \in F} p_{j} ∣ u_{j} ⟩ ⟨ u_{j} ∣$ is positive trace class with trace norm $1 - \sum_{j \in F} p_{j}$ . Thus these finite partial sums converge to $ρ$ in trace norm as the finite-subset net exhausts $J$ ; no basis of $\ker ρ$ is used.

2.1givenstep 1.1F4F5F6algebra

By [F5], $ρ A$ is trace class. The trace-norm expansion from step 1.1 is the nuclear representation $ρ = \sum_{j \in J} T_{p_{j} u_{j}, u_{j}}$ , since $\sum_{j} ∥ p_{j} u_{j} ∥ ∥ u_{j} ∥ = \sum_{j} p_{j} = 1$ . The basis-independent trace [F4] and bounded cyclicity and its nuclear-series formula [F6] give $Tr (ρ A) = \sum_{j \in J} p_{j} ⟨ A u_{j}, u_{j} ⟩ .$ The series is absolutely convergent because $\sum_{j} p_{j} ∣ ⟨ A u_{j}, u_{j} ⟩ ∣ \leq ∥ A ∥ \sum_{j} p_{j} = ∥ A ∥$ . Thus $ω_{ρ}$ is a well-defined bounded linear functional with $∣ ω_{ρ} (A) ∣ \leq ∥ A ∥$ .

3.1givenstep 1.1step 2.1algebra∎

For $A = I$ , step 2.1 gives $ω_{ρ} (I) = \sum_{j} p_{j} = 1$ . If $A \geq 0$ , every $⟨ A u_{j}, u_{j} ⟩ \geq 0$ , so step 2.1 gives $ω_{ρ} (A) \geq 0$ . Therefore the trace pairing is a positive normalized bounded functional, as asserted.

Trace-class normal density states

Definition

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources