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Marker overlap contraction of the two-path cross term

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $(D, A, μ)$ be a $σ$ -finite measure space, let $L^{2} (D, μ)$ denote the quotient of square-integrable measurable functions modulo almost-everywhere equality (The space $L^{p} (μ)$ as the quotient by null functions), let $K$ be a complex Hilbert space (Hilbert space) with unit vectors $e_{1}, e_{2}$ , let $a, b \in L^{2} (D, μ)$ be unit vectors, and let $α, β \in C$ with $∣ α ∣^{2} + ∣ β ∣^{2} = 1$ . Choose measurable representatives of $a, b$ ; all displayed pointwise expressions are understood almost everywhere, and the integrals below are independent of the representatives. Put

$Ψ : = α a \otimes e_{1} + β b \otimes e_{2} \in L^{2} (D) \hat{\otimes} K .$

Then:

(a) the reduced positive trace-class weight $ρ : = {Tr}_{K} T_{Ψ, Ψ}$ on $L^{2} (D)$ is

$ρ = ∣ α ∣^{2} T_{a, a} + ∣ β ∣^{2} T_{b, b} + α \overline{β} ⟨ e_{1}, e_{2} ⟩ T_{a, b} + \overline{α} β ⟨ e_{2}, e_{1} ⟩ T_{b, a};$

(b) for every measurable $B \subseteq D$ with finite measure, writing $P_{B}$ for the multiplication operator by $1_{B}$ ,

$tr (ρ P_{B}) = \int_{B} (∣ α ∣^{2} ∣ a ∣^{2} + ∣ β ∣^{2} ∣ b ∣^{2} + 2 Re (α \overline{β} ⟨ e_{1}, e_{2} ⟩ a (x) \overline{b (x)})) d μ;$

(c) the marker overlap contracts the cross term: $∣ ⟨ e_{1}, e_{2} ⟩ ∣ \leq 1$ , and if $e_{1} ⊥ e_{2}$ the cross term in (b) vanishes, leaving the incoherent weight $∣ α ∣^{2} ∣ a ∣^{2} + ∣ β ∣^{2} ∣ b ∣^{2}$ .

The weight satisfies $tr ρ = ∥ Ψ ∥^{2} = 1 + 2 Re (α \overline{β} ⟨ a, b ⟩ ⟨ e_{1}, e_{2} ⟩)$ ; it is a density operator precisely when $∥ Ψ ∥ = 1$ . The stated assumptions do not force this normalization. These are identities for the specified vector and its unnormalized weight; they assert no measured detector law and cover no selective conditioning.

Facts & Assumptions

Given: $D, K, e_{1}, e_{2}, a, b, α, β, Ψ$ as in the Statement; a measurable $B \subseteq D$ with finite measure.

[F0]

Under the assumed Countable Choice, complex $L^{2} (D, μ)$ is complete for its $L^{2}$ norm and has the first-variable-linear inner product $⟨ f, g ⟩ = \int_{D} f \overline{g} d μ$ inducing that norm (Complex completeness, density, and inner product: the consumer interface). Thus it is a complex Hilbert space. The given unit vectors $a, b$ make it nonzero; the unit $e_{1}$ makes $K$ nonzero, so both tensor and partial-trace suppliers apply.

[F1]

The completed tensor product $L^{2} (D) \hat{\otimes} K$ carries the inner product $⟨ f \otimes u, f^{'} \otimes u^{'} ⟩ = ⟨ f, f^{'} ⟩ ⟨ u, u^{'} ⟩$ and $∥ f \otimes u ∥ = ∥ f ∥ ∥ u ∥$ (Hilbert tensor construction: inner product, completion and product bases).

[F2]

${Tr}_{K}$ is linear with ${Tr}_{K} T_{f \otimes u, f^{'} \otimes u^{'}} = ⟨ u, u^{'} ⟩ T_{f, f^{'}}$ ; $T_{f, f^{'}} g = ⟨ g, f^{'} ⟩ f$ ; and $tr (T_{f, f^{'}} P_{B}) = ⟨ f, P_{B} f^{'} ⟩ = \int_{B} f \overline{f^{'}} d μ$ for the self-adjoint multiplication operator $P_{B}$ (Separable partial trace and its basic properties, Nuclear bounded cyclicity for trace-class operators).

[F3]

$∣ ⟨ e_{1}, e_{2} ⟩ ∣ \leq 1$ (Cauchy–Schwarz: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ , with equality exactly for linearly dependent vectors); the functions $a \overline{b}$ , $∣ a ∣^{2}$ , $∣ b ∣^{2}$ are integrable on $B$ of finite measure by Cauchy-Schwarz (Cauchy-Schwarz inequality for $L^{2}$ ).

Proof

1.1givenF0F1F2algebra

[The reduced state] Expanding $T_{Ψ, Ψ}$ into its four rank-one summands and applying the rank-one rule [F2] termwise gives ${Tr}_{K} T_{Ψ, Ψ} = ∣ α ∣^{2} T_{a, a} + ∣ β ∣^{2} T_{b, b} + α \overline{β} ⟨ e_{1}, e_{2} ⟩ T_{a, b} + \overline{α} β ⟨ e_{2}, e_{1} ⟩ T_{b, a}$ , which is (a). The trace-preserving partial trace [F2] gives $tr ρ = ∥ Ψ ∥^{2}$ ; expanding the tensor norm by [F1] gives the displayed normalization formula. Positivity follows from positivity of $T_{Ψ, Ψ}$ and the positive partial trace [F2], so trace one is exactly the density condition.

2.1step 1.1F2F3algebra

[The bin weight] By [F2], $tr (ρ P_{B}) = ∣ α ∣^{2} \int_{B} ∣ a ∣^{2} + ∣ β ∣^{2} \int_{B} ∣ b ∣^{2} + α \overline{β} ⟨ e_{1}, e_{2} ⟩ \int_{B} a \overline{b} + \overline{α} β ⟨ e_{2}, e_{1} ⟩ \int_{B} b \overline{a}$ ; since the last two integrals are conjugates of one another (because $⟨ e_{2}, e_{1} ⟩ = \overline{⟨ e_{1}, e_{2} ⟩}$ and $\int_{B} b \overline{a} = \overline{\int_{B} a \overline{b}}$ ), their sum is $2 Re (α \overline{β} ⟨ e_{1}, e_{2} ⟩ \int_{B} a \overline{b})$ , giving (b).

3.1step 2.1F3algebra∎

[The overlap bound] By [F3], $∣ ⟨ e_{1}, e_{2} ⟩ ∣ \leq 1$ . Thus the modulus of the complex cross-term coefficient $α \overline{β} ⟨ e_{1}, e_{2} ⟩$ is at most $∣ α β ∣$ , and the real cross-term contribution at $x$ is bounded in magnitude by $2 ∣ α β ∣ ∣ a (x) b (x) ∣$ . A complex phase may rotate that contribution, so no comparison of its real part with the real part for overlap one is asserted. When $e_{1} ⊥ e_{2}$ , the overlap is zero and the cross term vanishes, leaving exactly the diagonal sum, as claimed in (c).

Marker overlap contraction of the two-path cross term

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources