QFT on One Qubit

The Quantum Fourier Transform (QFT) is the quantum analogue of the discrete Fourier transform (DFT). For an $n$ -qubit register it maps computational basis states as

\text{QFT}_n |j\rangle = \frac{1}{\sqrt{2^n}} \sum_{k=0}^{2^n - 1} e^{2\pi i j k / 2^n} |k\rangle.

Before building the full $n$ -qubit circuit it pays to study the smallest case: a single qubit.

The n = 1 case

Setting $n = 1$ and $N = 2^1 = 2$ the formula becomes

\text{QFT}_1 |j\rangle = \frac{1}{\sqrt{2}} \sum_{k=0}^{1} e^{2\pi i j k / 2} |k\rangle.

Evaluating on each basis state:

\text{QFT}_1 |0\rangle = \frac{1}{\sqrt{2}}\bigl(e^{0}|0\rangle + e^{0}|1\rangle\bigr) = \frac{1}{\sqrt{2}}\bigl(|0\rangle + |1\rangle\bigr),

\text{QFT}_1 |1\rangle = \frac{1}{\sqrt{2}}\bigl(e^{0}|0\rangle + e^{2\pi i / 2}|1\rangle\bigr) = \frac{1}{\sqrt{2}}\bigl(|0\rangle + e^{i\pi}|1\rangle\bigr) = \frac{1}{\sqrt{2}}\bigl(|0\rangle - |1\rangle\bigr).

In matrix form, using the columns $\text{QFT}_1|0\rangle$ and $\text{QFT}_1|1\rangle$ :

\text{QFT}_1 = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = H.

The single-qubit QFT is the Hadamard gate — exactly the gate you already know.

Why the DFT viewpoint matters

The classical DFT maps a length- $N$ vector $x$ to its frequency components $X_k = \sum_j x_j e^{2\pi i jk/N}$ . The QFT does the same thing but encodes the $2^n$ amplitudes of a quantum state, operating in $O(n^2)$ gates rather than the classical fast Fourier transform's $O(n \cdot 2^n)$ — an exponential speedup in circuit depth. For $n = 1$ there is a single amplitude "frequency component" and one Hadamard suffices.

Try it

Implement the 1-qubit QFT. The grader checks the unitary matrix of your circuit, so an empty circuit will not pass.

Run your code to see the quantum state.

After running, the statevector panel shows both amplitudes equal to $1/\sqrt{2} \approx 0.707$ when started from $|0\rangle$ — consistent with $H|0\rangle = |+\rangle$ .