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

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A unit-vector outer product is a density operator

Statement

Let $H$ be a nonzero finite-dimensional complex inner-product space with the root pairing linear in its first argument, and let $u \in H$ satisfy $∥ u ∥ = 1$ . Define $P_{u} : = ∣ u ⟩ ⟨ u ∣$ by the root-linear-first Dirac convention. Then $P_{u}$ is self-adjoint and idempotent, has one-dimensional range, is non-negative, and has trace one. In particular, $P_{u}$ is a density operator.

Facts & Assumptions

Given: The finite-dimensional complex inner-product space $H$ and a unit vector $u \in H$ .

[F1]

Real and complex inner-product spaces and their induced length: the root pairing is linear in the first argument, conjugate-linear in the second, and conjugate-symmetric.

[F2]

The norm $∥ v ∥ = \sqrt{⟨ v, v ⟩}$ induced by a real or complex inner product: $∥ u ∥^{2} = ⟨ u, u ⟩$ .

[F3]

Dirac Notation Conversion: $P_{u} z = ⟨ z, u ⟩_{r o o t} u$ for every $z \in H$ .

[F4]

The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ : $T^{*}$ satisfies $⟨ T z, w ⟩ = ⟨ z, T^{*} w ⟩$ .

[F5]

Every linear map between finite-dimensional inner product spaces has a unique adjoint: every linear endomorphism of $H$ has a unique adjoint.

[F6]

Non-negative and positive operators: non-negative means self-adjoint with nonnegative quadratic form on every vector.

[F7]

The basis-independent trace of an endomorphism of a finite-dimensional vector space: trace is the sum of the diagonal entries in any basis.

[F8]

Every finite-dimensional real or complex inner product space has an orthonormal basis: every finite-dimensional inner-product space has an orthonormal basis.

[F9]

Bessel's inequality for a finite orthonormal list and Parseval's identity for an orthonormal basis: for an orthonormal basis $(e_{j})$ , $\sum_{j} ∣ ⟨ v, e_{j} ⟩ ∣^{2} = ∥ v ∥^{2}$ .

[L1]

Density operator: a density operator is a non-negative endomorphism of trace one on a nonzero finite-dimensional space.

Proof

1.1givenF2F3algebra

For every $z \in H$ , [F2] and [F3] give $P_{u}^{2} z = ⟨ z, u ⟩ ⟨ u, u ⟩ u = ⟨ z, u ⟩ u = P_{u} z$ , since $⟨ u, u ⟩ = ∥ u ∥^{2} = 1$ . Thus $P_{u}$ is idempotent.

1.2givenF1F3F4F5algebra

For all $z, w \in H$ , [F1] and [F3] give $⟨ P_{u} z, w ⟩ = ⟨ z, u ⟩ ⟨ u, w ⟩ = ⟨ z, ⟨ w, u ⟩ u ⟩ = ⟨ z, P_{u} w ⟩$ . The outer product is linear by [F3], so [F4] and [F5] imply $P_{u}^{*} = P_{u}$ .

2.1givenF1F2F3F6step 1.2algebra

The image of $P_{u}$ lies in $C u$ , and [F2] and [F3] give $P_{u} u = ⟨ u, u ⟩ u = u$ , so its range is exactly the one-dimensional span of $u$ . Also [F1] and [F3] give $⟨ P_{u} z, z ⟩ = ⟨ z, u ⟩ ⟨ u, z ⟩ = ∣ ⟨ z, u ⟩ ∣^{2} \geq 0$ . With self-adjointness from step 1.2, [F6] makes $P_{u}$ non-negative.

3.1givenF2F3F7F8F9L1step 2.1choosealgebra∎

Choose an orthonormal basis $(e_{j})_{j < n}$ of $H$ using [F8]. By [F3] and [F7], $tr (P_{u}) = \sum_{j < n} ⟨ P_{u} e_{j}, e_{j} ⟩ = \sum_{j < n} ∣ ⟨ e_{j}, u ⟩ ∣^{2} = ∥ u ∥^{2} = 1$ , where [F9] gives the Parseval equality. Thus [L1] makes $P_{u}$ a density operator, using non-negativity from step 2.1. The basis is finite, so no choice axiom is used.

Depends on

Used by

Dependency tree · two levels

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Sources

Ronald de Wolf, Quantum Computing: Lecture Notes, arXiv:1907.09415v6 (standard reference, not scraped)
John Preskill, Quantum Computation, Chapter 2: Foundations I: States and Ensembles (standard reference, not scraped)