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X, CNOT, Toffoli, and Fredkin are involutions

Statement

On the ordered computational basis, X, CNOT, Toffoli T, and Fredkin F are involutions. Their linear extensions are unitary. With chronological order meaning the gates are applied in the written order, Fredkin is implemented by CNOT(a,b), then T(c,b,a), then CNOT(a,b).

Facts & Assumptions

Given: The displayed gate actions and the ordered tensor-product computational basis. Prove involutivity, unitarity, and the stated Fredkin implementation.

[F5]

The Pauli matrices: In the ordered computational basis, X=(0110).

[F1]

CNOT and Control-Target Order: In the ordered two-qubit basis, CNOT⁡1→2∣a,t⟩=∣a,t⊕a⟩.

[F2]

Toffoli and Fredkin gates: T∣a,b,t⟩=∣a,b,t⊕ab⟩; F∣0,a,b⟩=∣0,a,b⟩ and F∣1,a,b⟩=∣1,b,a⟩.

[F3]

The computational basis is orthonormal: For every d≥1, the ordered computational basis of Qd is an orthonormal basis for its declared inner product.

[F4]

The product basis is orthonormal and tensor dimensions multiply: The elementary tensors of specified ordered orthonormal factor bases, in lexicographic order, form an orthonormal basis of the tensor register.

[F6]

Denotational semantics of a finite layered circuit: A chronological list of layers L1,…,Ld, with L1 applied first, denotes ULd⋯UL1.

Proof

1.1givenF5algebra

The matrix in [F5] sends ∣t⟩ to ∣t⊕1⟩. Applying it twice sends t to t⊕1⊕1=t, so X2=I.

1.2givenF1algebra

By [F1], applying CNOT twice sends ∣a,t⟩ to ∣a,t⊕a⊕a⟩=∣a,t⟩, so CNOT⁡2=I.

1.3givenF2algebra

Applying T twice leaves a,b unchanged and sends t to t⊕ab⊕ab=t. Thus T2=I.

1.4givenF2choose

When the Fredkin control is 0, each application fixes (a,b). When it is 1, each application swaps a,b and leaves the control unchanged. In either case two applications restore all three bits, so F2=I.

2.1step 1.1step 1.2step 1.3step 1.4F1F2F3F4F5algebra

The four maps permute the ordered orthonormal computational basis by [F1]–[F5]. A permutation of an orthonormal basis preserves inner products, so each linear extension is unitary; by Steps 1.1–1.4 each is its own inverse.

3.1givenF1F2F6choosealgebra∎

Apply the gates chronologically as prescribed by [F6] to ∣c,a,b⟩. After CNOT(a,b) the last two bits are (a,b⊕a). If c=0, Toffoli does nothing and the last CNOT restores (a,b). If c=1, Toffoli changes its target to a⊕(b⊕a)=b, and the last CNOT changes the other bit to (b⊕a)⊕b=a. Thus the output is ∣c,a,b⟩ for c=0 and ∣c,b,a⟩ for c=1, exactly the Fredkin action.

Depends on

Used by

Cited to discharge well-definedness by Toffoli and Fredkin gates.

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources