Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge 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.

Teleportation Bell-basis expansion

Statement

For arbitrary α,β∈C, put ∣ψ⟩=α∣0⟩+β∣1⟩ and VAB=(HA⊗IBC)CNOT⁡A→B, in wire order A,B,C. Then

VAB(∣ψ⟩A⊗∣Φ+⟩BC)=12∑a,b∈{0,1}∣ab⟩AB⊗XCbZCa∣ψ⟩C.

No normalization assumption on ∣ψ⟩ is needed for this vector identity.

Facts & Assumptions

Given: α,β∈C, the ordered wires A,B,C, and the operators in the Statement.

[A1]

∣ψ⟩=α∣0⟩+β∣1⟩ is an arbitrary vector; it need not be normalized.

[F1]

Qubit teleportation procedure specifies initial input ∣ψ⟩A, shared resource ∣Φ+⟩BC, wire order A,B,C, CNOT followed by HA, and measurement-record order (a,b).

[F2]

The Bell vectors defines ∣Φ+⟩=(∣00⟩+∣11⟩)/2.

[F3]

Exact one-qubit gates and rotations gives H=2−1/2(111−1).

[F4]

CNOT and Control-Target Order gives CNOT⁡1→2∣x,t⟩=∣x,t⊕x⟩ with the first wire as control.

[F5]

The Pauli matrices gives X∣0⟩=∣1⟩, X∣1⟩=∣0⟩, Z∣0⟩=∣0⟩, and Z∣1⟩=−∣1⟩.

[F6]

Tensor products of finitely many linear maps extends the rule H⊗I:x⊗y↦Hx⊗y linearly to the three-wire register.

Proof

1.1givenA1F1F2algebra

By [F2], the initial vector is 2−1/2(α∣000⟩+α∣011⟩+β∣100⟩+β∣111⟩) in wire order A,B,C.

2.1step 1.1F4algebra

Applying [F4] to the four basis vectors gives 2−1/2(α∣000⟩+α∣011⟩+β∣110⟩+β∣101⟩) after CNOT: the first two terms have control bit 0 and are fixed, while the last two have control bit 1 and their target bits flip.

3.1step 2.1F3F6algebra

By [F3] and [F6], HA sends each first-wire ∣0⟩ to (∣0⟩+∣1⟩)/2 and each ∣1⟩ to (∣0⟩−∣1⟩)/2. Regrouping the resulting eight terms by the first two wires gives 12(∣00⟩(α∣0⟩+β∣1⟩)+∣01⟩(α∣1⟩+β∣0⟩)+∣10⟩(α∣0⟩−β∣1⟩)+∣11⟩(α∣1⟩−β∣0⟩)).

4.1A1step 3.1F5algebra∎

The four receiver vectors in step 3.1 are respectively ∣ψ⟩, X∣ψ⟩, Z∣ψ⟩, and XZ∣ψ⟩ by the actions in [F5]. These are exactly XbZa∣ψ⟩ for labels 00,01,10,11, so step 3.1 is the sum in the Statement.

Depends on

Used by

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Sources