# Tag Info

### Why does the "Phase Kickback" mechanism work in the Quantum phase estimation algorithm?

A first remark This same phenomenon of 'control' qubits changing states in some circumstances also occurs with controlled-NOT gates; in fact, this is the entire basis of eigenvalue estimation. So not ...
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### Why does the "Phase Kickback" mechanism work in the Quantum phase estimation algorithm?

Imagine you have an eigenvector $|u\rangle$ of $U$. If you have a state such as $|1\rangle|u\rangle$ and you apply controlled-$U$ to it, you get out $e^{i\phi}|1\rangle|u\rangle$. The phase isn't ...

### How to describe, or encode, the input vector x of Quantum Fourier Transform?

You don't convert a classical input to the r.h.s. of Eq. (5.2). The r.h.s. of Eq. (5.2) is something you get as the output of a preceding quantum computation as a quantum state, such as in Shor's ...
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### What is the intuition of using Hadamard gate in quantum fourier transform?

The intuition, roughly speaking, is that the only way that you're going to get some difference between classical and quantum computing is if you are able to prepare qubits in a superposition. If you ...
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### Does the Quantum Fourier Transform (QFT) preserve entanglement?

TLDR: the Fourier transform is entangling. We can immediately agree on two things: if you input a computational basis state (separable) to the Fourier transform, it outputs a separable state the ...

### What is the probability to get all qubits equal zero after QFT

By looking to the circuit for the QFT presented in the M. Nielsen and I. Chuang textbook (Figure 5.1.) we can notice that all controlled rotations can be neglected because for each control rotation ...

### Application of QFT to Order-finding

Answer to question 1 There are many ways the first quantum algorithm for order finding could have been conceived and I don't know how it really happened. However, here is a plausible though entirely ...
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### How to generalize the relationship HXH = Z for higher dimensions

The appropriate $d$-dimensional analogue of $H$ turns out to be the Quantum Fourier Transform. This is obscured by the fact that even though $(1)$ is conjugation the inverse is written implicitly ...
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### Does QFT exploit entanglement?

Yes, the formula you have shows that applying QFT to a given computational basis state $|j\rangle = |j_1 j_2 \dots j_n\rangle$ results in an unentangled output state. However when applied to ...

### Implementing QFT for Shor's Algorithm

The comment made by arriopolis is correct. The output registers of these compiled circuits are important for synthesizing the circuit, but not particularly interesting to measure. As you saw already,...
In many quantum algorithms, the first step is to compute some problem on all instances at the same time -- if you wish, you compute all solutions at once. But then you are left with a state such as ...