TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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The partial trace preserves non-negativity and total trace

Statement

For finite-dimensional complex inner-product spaces $H_{A}, H_{B}$ and a non-negative operator $R \in End (H_{A} \otimes H_{B})$ , ${Tr}_{B} (R)$ is non-negative and

$tr ({Tr}_{B} (R)) = tr (R) .$

Here $End (H)$ is the set of linear maps $H \to H$ as in Partial trace over a finite tensor factor, $I_{H}$ denotes the identity map $x \mapsto x$ , and both conclusions include zero-dimensional factors.

Facts & Assumptions

Given: Finite-dimensional complex inner-product spaces $H_{A}, H_{B}$ , a non-negative operator $R$ on their ordered tensor product, and the partial trace from Partial trace over a finite tensor factor.

[F1]

A non-negative operator is self-adjoint and has $⟨ R v, v ⟩ \geq 0$ for every vector $v$ (Non-negative and positive operators, Self-adjoint and normal endomorphisms of a finite-dimensional real or complex inner product space).

[F2]

In orthonormal bases the adjoint matrix is the conjugate transpose (In orthonormal bases, the matrix of the adjoint is the conjugate transpose of the matrix).

[F3]

The partial-trace entries are $({Tr}_{B} R)_{i i^{'}} = \sum_{j} R_{(i, j), (i^{'}, j)}$ (Partial trace over a finite tensor factor).

[F4]

The partial trace satisfies $tr (({Tr}_{B} R) \circ T) = tr (R \circ (T \otimes I_{B}))$ (The partial trace is characterized by the trace pairing and is basis-independent).

[F5]

The elementary product vectors form an orthonormal basis, including empty and zero-dimensional cases (The product basis is orthonormal and tensor dimensions multiply).

[F6]

The root pairing is linear in its first argument and conjugate-linear in its second (Real and complex inner-product spaces and their induced length).

[F7]

The tensor map is the unique linear extension of its action on elementary tensors (Tensor products of finitely many linear maps).

[F8]

Composing a linear map with the identity map on either side leaves it unchanged (Identity maps and composites of linear maps are linear).

[F9]

The trace of a finite-dimensional endomorphism is the sum of its diagonal matrix entries in any ordered basis (The basis-independent trace of an endomorphism of a finite-dimensional vector space, The trace $tr (A)$ as the sum of the diagonal entries).

[F10]

Every finite-dimensional real or complex inner-product space has an orthonormal basis, with the empty basis in dimension zero (Every finite-dimensional real or complex inner product space has an orthonormal basis).

Proof

1.1givenF1F2F3F5F10choose

By [F10], choose ordered orthonormal bases $(e_{i})$ and $(f_{j})$ for $H_{A}, H_{B}$ , and write $S = {Tr}_{B} (R)$ . Since $R$ is self-adjoint by [F1], [F2] gives $R_{(i, j), (i^{'}, j)} = \overline{R_{(i^{'}, j), (i, j)}}$ . Summing over $j$ in [F3] yields $S_{i i^{'}} = \overline{S_{i^{'} i}}$ , so $S$ is self-adjoint. The product basis is orthonormal by [F5].

2.1step 1.1F1F3F5F6algebra

For $x = \sum_{i} x_{i} e_{i}$ , [F3] and [F6] give $⟨ S x, x ⟩ = \sum_{i, i^{'}, j} R_{(i, j), (i^{'}, j)} x_{i^{'}} \overline{x_{i}}$ . For each fixed $j$ , the coefficient of $e_{i} \otimes f_{j}$ in $R (x \otimes f_{j})$ is $\sum_{i^{'}} R_{(i, j), (i^{'}, j)} x_{i^{'}}$ ; orthonormality [F5] therefore gives $⟨ R (x \otimes f_{j}), x \otimes f_{j} ⟩ = \sum_{i, i^{'}} R_{(i, j), (i^{'}, j)} x_{i^{'}} \overline{x_{i}}$ . Summing these equalities over $j$ yields $⟨ S x, x ⟩ = \sum_{j} ⟨ R (x \otimes f_{j}), x \otimes f_{j} ⟩$ . Each summand is nonnegative by [F1], hence $⟨ S x, x ⟩ \geq 0$ for every $x$ . Together with step 1.1, the definition in [F1] gives $S \geq 0$ .

3.1F4F5F7F8F9algebra∎

In [F4] set $T = I_{H_{A}}$ . By [F7], $I_{H_{A}} \otimes I_{H_{B}}$ fixes every elementary tensor; these span by the product-basis result [F5], so it is $I_{H_{A} \otimes H_{B}}$ . Therefore [F4] gives $tr (S \circ I_{H_{A}}) = tr (R \circ I_{H_{A} \otimes H_{B}})$ . By [F8] the composites are $S$ and $R$ , so their traces are equal by [F9]. Empty bases give empty traces $0$ on zero-dimensional factors, so the equality holds there as well.

Depends on

Used by

Dependency tree · two levels

39 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

John Watrous, The Theory of Quantum Information (standard reference, not scraped)
John Preskill, Quantum Computation, Chapter 2: Foundations I: States and Ensembles (standard reference, not scraped)