LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Demultiplexing a Two-Block Unitary

Statement

Let $H$ be a finite-dimensional complex inner-product space and let $U_{0}, U_{1}$ be unitary endomorphisms of $H$ . On the control-first direct sum $H \oplus H$ , the block unitary $diag (U_{0}, U_{1})$ factors as $diag (U_{0}, U_{1}) = (I \otimes V) diag (D, D^{*}) (I \otimes W)$ for unitary $V, W$ on $H$ and a diagonal unitary $D$ in an orthonormal eigenbasis; $I \otimes V$ denotes $diag (V, V)$ under $C^{2} \otimes H ≅ H \oplus H$ .

Facts & Assumptions

Given: A finite-dimensional complex inner-product space $H$ and unitary $U_{0}, U_{1}$ on $H$ .

[F1]

Complex spectral theorem: a normal endomorphism of a finite-dimensional complex inner product space has an orthonormal eigenbasis, and conversely: a normal endomorphism of a finite-dimensional complex inner-product space has an orthonormal eigenbasis.

[F2]

Every nonzero complex number has a unique polar form $r (\cos θ + i \sin θ)$ with $r > 0$ and $- π < θ \leq π$ : every unit complex number is $e^{i t}$ for some real $t$ , and therefore has a unit-modulus square root.

[F3]

Orthogonal and unitary operators form groups, and their determinants have modulus one: products and adjoints of unitary maps are unitary.

Proof

1.1givenF1F2F3choose

The product $Q : = U_{0} U_{1}^{*}$ is unitary by [F3] and hence normal. By [F1], choose a unitary $V$ and diagonal unitary $D$ with $Q = V D^{2} V^{*}$ : diagonalize $Q$ , write each eigenvalue as $e^{i t_{j}}$ by [F2], and put $e^{i t_{j} / 2}$ on the corresponding diagonal of $D$ .

2.1step 1.1algebra

Set $W : = D^{*} V^{*} U_{0}$ . Then $V D W = U_{0}$ . Also $V D^{*} W = V D^{- 2} V^{*} U_{0}$ . Taking adjoints in $U_{0} U_{1}^{*} = V D^{2} V^{*}$ gives $U_{1} U_{0}^{*} = V D^{- 2} V^{*}$ , so $V D^{- 2} V^{*} U_{0} = U_{1}$ . Thus the two diagonal blocks in the displayed factorization are respectively $U_{0}$ and $U_{1}$ .

3.1step 1.1step 2.1given∎

The factors $V, D, W$ are unitary, and the two block products agree with $U_{0}, U_{1}$ ; hence the asserted exact block factorization holds, including repeated eigenvalues. Only a finite orthonormal eigenbasis and finitely many square roots are chosen.

Depends on

Used by

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Sources

Vivek V. Shende, Stephen S. Bullock, and Igor L. Markov, Synthesis of Quantum Logic Circuits, arXiv:quant-ph/0406176v5 (standard reference, not scraped)