Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)
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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.

Hadamard preparation of a uniform binary state

Statement

For an integer n≥0, give each factor of the ordered register Vn=(C2)⊗n the basis (e0,e1) and define the one-qubit matrix H:=12(111−1). Then H is unitary and H⊗ne0n=2−n/2∑x∈{0,1}nex. For n=0, the tensor product is C, the empty word is the sole basis label, and H⊗0 is the identity. The preparation uses exactly n one-qubit H gates when H is primitive in the declared circuit model; otherwise its implementation and synthesis costs must be supplied there.

Facts & Assumptions

Given: n≥0, the ordered computational register, and the stated matrix H.

[F1]

Finite registers and their ordered computational bases: a qubit has ordered basis (e0,e1), and e0n denotes the corresponding register basis vector.

[F2]

Ordered tensor registers, their inner product, and wire permutations: tensor-register basis vectors are ordered products of the factor bases; the empty tensor product is C.

[F3]

The computational basis is orthonormal: the ordered computational basis is orthonormal.

[F4]

Circuit-family and computation cost model: a circuit gate count counts its finite one- and multi-qubit gate instances, separately from oracle-query and circuit-description costs.

[F5]

Tensor products of finitely many linear maps: the tensor map is the unique linear map that acts factorwise on elementary tensors, with the empty tensor map equal to the identity on C.

Proof

technique · direct
1.1givenF1F3algebra

Direct multiplication gives H∗H=I2. Also He0=(e0+e1)/2, so H is a unitary one-qubit operation in the specified basis.

1.2F1F2F3F5algebra

By the tensor-map rule [F5], H⊗ne0n=(He0)⊗n=2−n/2(e0+e1)⊗n=2−n/2∑x∈{0,1}nex. The last equality distributes the finite tensor product over its two-term factors, choosing exactly one of e0,e1 at each position; the ordered product basis in [F2] identifies these choices with the binary words x. For n=0, both sides are the scalar 1 in C, the empty sum-index set convention gives its single empty word, and [F5] makes the empty tensor map the identity. The basis vectors are distinct orthonormal vectors by [F3], so the right side has norm one.

2.1givenF4F5algebra∎

By [F5], the construction applies one copy of H to each of the n tensor factors and no oracle operation. It therefore has n one-qubit gate instances when H is primitive, as counted by [F4]; if it is not primitive, the declared synthesis cost is required. At n=0 the circuit has no gates.

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Sources