Exercise 5.2

Following Eq.(5.5)-Eq.(5.10) in the textbook, the Fourier transformation of state \(|j\rangle\) can be written as

(1)\[ |j\rangle \to \frac{1}{2^{n/2}}\left(|0\rangle + e^{2\pi i 0.j_n}|1\rangle\right)\left(|0\rangle + e^{2\pi i 0.j_{n-1}j_n}|1\rangle\right)\cdots \left(|0\rangle + e^{2\pi i 0.j_1\dotsc j_n}|1\rangle\right) \]

According to the textbook, \(j_1\dotsc j_n\) is binary representation of integer \(j\), and

(2)\[ 0.j_{l}j_{l+1}\dotsc j_{m} = j_l/2 + j_{l+1}/4 + \dotsc + j_m/2^{m-l+1} \]

If we have a \(n-\)qubit state \(|j\rangle = |0\rangle = |00\dotsc 0\rangle\), then \(j_k = 0\) for any \(1\leq k\leq n\) in the binary representation. Also, it always have \(0.j_{k}\dotsc j_n = 0\) for any \(k\) from eq. (2). Thus, the Fourier transformation of \(n-\)qubit state \(|00\dotsc 0\rangle\) is given by

(3)\[ |00\dotsc 0\rangle \to \frac{1}{2^{n/2}}\left(|0\rangle + |1\rangle\right)\left(|0\rangle + |1\rangle\right)\cdots \left(|0\rangle + |1\rangle\right) = |+\rangle^{\otimes n} \]

where \(|+\rangle = (|0\rangle + |1\rangle)/\sqrt{2}\).