= Abelian hidden-subgroup Fourier sampling
Measuring the <coset state> $|g_0+K\rangle$ in the <group shift operator> <eigenbasis> samples uniformly from the <annihilator of a subgroup of a finite abelian group> $K^\perp$. Indeed,
$$
\langle v_\chi|g_0+K\rangle
=\frac{\chi(g_0)}{\sqrt{|G||K|}}\sum_{k\in K}\chi(k),
$$
so the <character-sum cancellation lemma> gives probability $|K|/|G|$ on $K^\perp$ and zero elsewhere. The <modulus> of $\chi(g_0)$ is one, so the offset has no effect on the distribution.
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