# Quantum Fourier Transform Question

Guys applying Hadamard transform on N qubits gives us an equal superposition of all states which has major applications in all quantum algorithms But

Quantum Fourier transform is said to be similar to Hadamard gate, applying QFT to a set of qubits gives us a new set of amplitudes for each state, which is a linear combination original amplitudes multiplied by e^[(2*piij)/N] How is this transformation similar to Hadamard's?

What is the use/purpose of linearly combining the state amplitudes?

How can this be used? Could you give me an example, please?

## 1 Answer

1. How is this transformation similar to Hadamard's? The effect of the QTF on the $n$-qubit empty register $|000\rangle$ is the same as applying $H^{\otimes n}$. From a algebraic point of view, an hadamard is just a QTF over $Z_2$, while the QTF works for every group.

2. What is the use/purpose of linearly combining the state amplitudes? In general: perform a change of base. :)

3. How can this be used? Could you give me an example, please? Think of the inverse of the quantum fourier transform. Is a mapping that allows you to do : $$\sum_{i=0}^N e^{-2i\pi \sigma}|i\rangle|0\rangle \to \sum_{i=0}^N e^{-2i\pi \sigma}|i\rangle|\sigma\rangle$$ It's a pretty useful transformation, because many quantum algorithms relay on the efficient creation of a relative phase. You can think of it as the "piece of software" in our software that allows to read the local phase you have created in your computation. And also, QTF itself is the thing that allows to do fourier analysis: the thing that quantum computers should be good at.