\sum_{g \in G} \overline{M_{α, ij} (g)} M_{β, k l} (g) = \frac{∣ G ∣}{d _{α}} δ_{α β} δ_{ik} δ_{j l},

and

\sum_{α} d_{α}^{2} = ∣ G ∣

. Consequently the normalized vectors

∣ α ij ⟩ = \frac{d _{α}}{∣ G ∣} \sum_{g \in G} \overline{M_{α, ij} (g)} ∣ g ⟩

form an orthonormal basis. The non-abelian quantum Fourier transform is the unitary basis change

∣ α ij ⟩ ⟼ ∣ α ⟩ ∣ i ⟩ ∣ j ⟩,

up to the harmless choice of transform direction and complex-conjugation convention.

ii

Solution

Put

x = g_{0} g

in the definition of

U (g_{0}) ∣ α ij ⟩

and use

M_{α} (g_{0}^{- 1} x) = M_{α} (g_{0})^{†} M_{α} (x) .

One obtains

U (g_{0}) ∣ α ij ⟩ = k = 1 \sum d_{α} M_{α, ki} (g_{0}) ∣ α kj ⟩ .

Thus every

d_{α}^{2}

-dimensional subspace

H_{α}

is invariant; the left regular representation acts as

M_{α} (g_{0})

on the first matrix index and as the identity on the second.

Let

P_{α}

project onto

H_{α}

. Since the subspace is invariant under every

U (g)

P_{α}

commutes with those unitaries. Also

∣ g H ⟩ = U (g) ∣ H ⟩

. Therefore

Π_{H} (α) = ∥ P_{α} ∣ g H ⟩ ∥^{2} = ∥ P_{α} ∣ H ⟩ ∥^{2},

which is independent of the coset representative

g

and depends only on

H

H^{'} = x H x^{- 1}

produce the same weak irrep-label distribution. Under an irrep, the subgroup average

\sum_{h^{'} \in H^{'}} M_{α} (h^{'}) = M_{α} (x) (\sum_{h \in H} M_{α} (h)) M_{α} (x)^{- 1}

is related by unitary similarity, so its squared Hilbert-Schmidt norm, and hence every

Π_{H} (α)

, is unchanged. Weak Fourier sampling can therefore fail to identify even distinct subgroups of a non-abelian group.

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