TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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Finite abelian Fourier transform is unitary

Statement

Let $N_{R}$ be the canonical positive real image of $∣ G ∣$ from Normalized Fourier transform on a finite abelian group. Define the finite coefficient matrix $U$ , with rows indexed by $\hat{G}$ and columns by $G$ , by $U_{χ, x} = N_{R}^{- 1 / 2} \overline{χ (x)}, χ \in \hat{G}, x \in G .$ It is square and satisfies $U^{*} U = I_{G}$ and $U U^{*} = I_{\hat{G}}$ , where $U^{*}$ is the conjugate transpose and $I_{X}$ is the identity matrix indexed by $X$ , with $(I_{X})_{x, y} = 1$ when $x = y$ and $0$ otherwise. These two identities are the meaning of “unitary” here.

Facts & Assumptions

Given: The normalized finite abelian Fourier transform on $G$ .

[F1]

The transform has coefficient $N_{R}^{- 1 / 2} \sum_{x} f (x) \overline{χ (x)}$ (Normalized Fourier transform on a finite abelian group).

[F2]

$∣ \hat{G} ∣ = ∣ G ∣$ , so the finite row and column index sets have equal cardinality (A finite abelian group and its character dual have equal order).

[F3]

For additive characters $χ, ψ$ , $N_{R}^{- 1} \sum_{x \in G} ψ (x) \overline{χ (x)}$ equals $1$ if $χ = ψ$ and $0$ otherwise (Row orthogonality for additive characters of a finite abelian group).

[F4]

$N_{R}^{- 1} \sum_{χ \in \hat{G}} χ (x) \overline{χ (y)}$ equals $1$ if $x = y$ and $0$ otherwise (Dual-index character orthogonality).

[F5]

The matrix products and sums here are finite; sums over finite product index sets may be evaluated in either order (Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule).

[F6]

Positive rational powers are defined for positive reals and obey $a^{r} a^{s} = a^{r + s}$ (Rational powers $a^{r}$ of a positive base, Laws of rational exponents).

Proof

1.1F1F2

By [F1], the transform acts on a function $f$ by the coefficient formula defining $U$ . By [F2], its row and column index sets have equal finite cardinality, so $U$ is square.

1.2F4F5F6

For $x, y \in G$ , the $(x, y)$ entry of $U^{*} U$ is $N_{R}^{- 1} \sum_{χ \in \hat{G}} χ (x) \overline{χ (y)}$ , because [F6] gives $(N_{R}^{- 1 / 2})^{2} = N_{R}^{- 1}$ . By [F4], this is $1$ when $x = y$ and $0$ otherwise, exactly the entry of $I_{G}$ . Hence $U^{*} U = I_{G}$ .

1.3F3F5F6

For $χ, ψ \in \hat{G}$ , the $(χ, ψ)$ entry of $U U^{*}$ is $N_{R}^{- 1} \sum_{x \in G} \overline{χ (x)} ψ (x)$ . By [F6] and [F3], this is $1$ when $χ = ψ$ and $0$ otherwise, exactly the entry of $I_{\hat{G}}$ . Hence $U U^{*} = I_{\hat{G}}$ .

2.1F1step 1.1step 1.2step 1.3∎

By [F1], the action of this coefficient matrix on the function values of any $f \in C^{G}$ is exactly $F_{G} f$ . The two identities in steps 1.2 and 1.3 therefore prove the asserted unitarity of the transform as defined here.

Depends on

Used by

Finite cyclic discrete logarithm by Fourier sampling Algorithm
One finite abelian hidden-subgroup Fourier sample Algorithm
Finite abelian Fourier inversion Theorem
Finite abelian Parseval identity Theorem
Fourier samples are uniform in the annihilator Theorem

Dependency tree · two levels

46 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

Ronald de Wolf, Quantum Computing: Lecture Notes, version 6 (standard reference, not scraped)
Andrew M. Childs, Lecture Notes on Quantum Algorithms (standard reference, not scraped)
John Preskill, Quantum Information, Chapter 6: Quantum Algorithms (standard reference, not scraped)