Solution
= Solution
Applying the transform to the coset state gives amplitude
$$
\frac1{2\sqrt{12}}\omega^c
\sum_{j=0}^3\omega^{3jc}
$$
for $|c\rangle$. The geometric sum vanishes unless $\omega^{3c}=1$, which means $c\equiv0\pmod4$. For $c=0,4,8$ its magnitude is $4/(2\sqrt{12})=1/\sqrt3$. Thus the <quantum Fourier transform of a periodic coset state> gives
$$
\boxed{c\in\{0,4,8\},\qquad \mathbb P(c)=\frac13}.
$$