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Exact Three-CNOT Synthesis of Two-Qubit Unitaries

Statement

Every U∈U(4) has an exact circuit with no ancillas, at most three CNOT gates, and arbitrary one-qubit gates. The scalar phase is implemented as a one-qubit scalar gate and is not discarded.

Facts & Assumptions

Given: Exact matrix semantics, both CNOT orientations, and the Cartan decomposition.

[F1]

Exact Two-Qubit Cartan Decomposition: U=eiδ(A⊗B)N(a,b,c)(C⊗D), where N(a,b,c)=ei(aX⊗X+bY⊗Y+cZ⊗Z) and A,B,C,D∈SU(2).

[F2]

One-Qubit Euler Decomposition: every SU(2) factor and every scalar one-qubit unitary has an exact Euler realization using the QC7 rotation convention.

[F3]

Pauli formula for an axis rotation: Rn(t)=cos⁡(t/2)I−isin⁡(t/2)(n⋅σ) for real unit n.

[F4]

The Pauli matrices: X,Y,Z are the displayed Pauli matrices in the ordered computational basis.

[F5]

CNOT and Control-Target Order: the first-wire-control CNOT maps ∣a,t⟩ to ∣a,t⊕a⟩; its wire order is part of the definition.

[F6]

Denotational semantics of a finite layered circuit: chronological circuit layers L1,…,Ld denote the exact matrix product ULd⋯UL1.

[F7]

Finite circuit resource accounting: size and gate costs count the occurrences declared in the explicit operation model; gate access is not inferred from a matrix alone.

[F8]

The computational basis is orthonormal: the ordered two-qubit computational basis is orthonormal.

Proof

1.1givenF1F2F3F4F5F8algebra

The Pauli matrices in [F4] give commuting products XX,YY,ZZ. On the even-parity subspace with ordered basis (∣00⟩,∣11⟩), XX and YY restrict to X and −X, while ZZ restricts to I. Hence N restricts there to eicei(a−b)X, whose entries are eiccos⁡(a−b) on the diagonal and ieicsin⁡(a−b) off diagonal. On the odd-parity basis (∣01⟩,∣10⟩), XX and YY both restrict to X and ZZ restricts to −I, so the block is e−icei(a+b)X. These two blocks give N(a,b,c)=(eiccos⁡(a−b)00ieicsin⁡(a−b)0e−iccos⁡(a+b)ie−icsin⁡(a+b)00ie−icsin⁡(a+b)e−iccos⁡(a+b)0ieicsin⁡(a−b)00eiccos⁡(a−b)). In basis 00,01,10,11, the two CNOT orientations are C12=(1000010000010010),C21=(1000000100100100). Here C12 is the gate from [F5] and C21 is obtained by interchanging the wire names. Both displayed permutations are unitary because they permute the orthonormal basis in [F8]. Using [F3] with the Pauli matrices [F4] gives the QC7 rotation convention in [F2]; multiplication of those rotations and the displayed CNOT permutations gives exactly N(a,b,c)=eiπ/4(Rz(π/2)⊗I)C21(I⊗Ry(π/2−2b))C12(Rz(π/2−2c)⊗Ry(2a−π/2))C21(I⊗Rz(−π/2)). All entries agree with the two parity blocks above; thus this is an exact identity, including its scalar.

2.1step 1.1F1F2F6F7algebra∎

The factorization in step 1.1 uses three CNOTs and otherwise one-qubit rotations. By [F2], the local factors A,B,C,D and the scalar eiδ in [F1] have exact one-qubit implementations; eiδI4=(eiδI2)⊗I2. The circuit product [F6] therefore equals the written factorization exactly, and [F7] counts three CNOT occurrences and no additional wires.

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