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Exact Controlled One-Qubit Unitary with Two CNOTs

Statement

For every $U \in U (2)$ , the exact controlled matrix $C (U) = diag (I_{2}, U)$ is realized with two CNOTs and one-qubit gates only. The scalar phase of $U$ is implemented as a phase gate on the control and is not discarded.

Facts & Assumptions

Given: Control-first order, the exact controlled-unitary convention, and $U \in U (2)$ .

[F1]

One-Qubit Euler Decomposition: $U = e^{i δ} R_{z} (α) R_{y} (θ) R_{z} (β)$ for real parameters, with $δ = 0$ possible when $\det U = 1$ .

[F2]

Exact one-qubit gates and rotations: axis rotations are $R_{n} (t) = \exp (- i t n \cdot σ / 2)$ and $R_{z} = R_{(0, 0, 1)}$ .

[F3]

CNOT and Control-Target Order: the first-wire-control CNOT maps $∣ a, t ⟩$ to $∣ a, t \oplus a ⟩$ and is the block $C (X) = diag (I, X)$ .

[F4]

Controlled Unitary Access: $C (U) = diag (I, U)$ in control-first order.

[F5]

Denotational semantics of a finite layered circuit: later-applied gates multiply on the left of earlier gates.

[F6]

Finite circuit resource accounting: the declared gate count counts only the named gate occurrences from the operation model.

[F7]

The computational basis is orthonormal: the ordered computational basis of a register is orthonormal.

Proof

1.1givenF1F2algebra

Put $V = R_{z} (α) R_{y} (θ) R_{z} (β) \in S U (2)$ , and define $A = R_{z} (α) R_{y} (θ / 2), B = R_{y} (- θ / 2) R_{z} (- (α + β) / 2), C = R_{z} ((β - α) / 2) .$ Using additivity of same-axis rotations gives $A B C = I$ . The rotations in [F1] follow the axis convention [F2]. Direct multiplication with the matrices in [F1] gives $X R_{y} (t) X = R_{y} (- t)$ and $X R_{z} (t) X = R_{z} (- t)$ ; therefore $A X B X C = R_{z} (α) R_{y} (θ) R_{z} (β) = V$ .

2.1step 1.1F3F4F5F7algebra

The displayed matrix product $(I \otimes A) C N O T (I \otimes B) C N O T (I \otimes C)$ has chronological gate order $I \otimes C$ , CNOT, $I \otimes B$ , CNOT, $I \otimes A$ by [F5]. By [F3] and [F7], each CNOT is a unitary basis permutation, and the full product is block diagonal in the control basis. On control value $0$ its target block is $A B C = I$ ; on value $1$ the two CNOTs apply $X$ between the target factors, so the block is $A X B X C = V$ . Hence the product is exactly $C (V)$ by [F4].

3.1step 2.1F1F4F6algebra∎

Since $U = e^{i δ} V$ , block multiplication gives $C (U) = (P_{e^{i δ}} \otimes I) C (V)$ , where $P_{e^{i δ}} = diag (1, e^{i δ})$ . This is one additional one-qubit control gate and preserves the exact scalar phase. By [F6], it adds no CNOT occurrences, so the exact count remains two for every $U \in U (2)$ .

Depends on

Used by

Dependency tree · two levels

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Sources

Adriano Barenco et al., Elementary gates for quantum computation, arXiv:quant-ph/9503016v1 (standard reference, not scraped)