# A question regarding quantum phase estimation algorithm

Why are $$U$$s raised to successive powers of two in quantum phase estimation circuit diagram when we use $$n$$ register qubits $$|0\rangle|0\rangle|0\rangle$$?

The objective of all those gates is to put the quantum state in a "nice form" to manipulate. Let me explain. Let's note $$U |\psi \rangle = e^{2i\pi\theta}|\psi\rangle$$ all the variables we are interested in, meaning we want to find $$\theta$$ here. After applying all the Hadamard gates and the $$U^{2^j}$$, the state looks like this : $$\frac{1}{2^{n/2}}\left( |0\rangle + e^{2i\pi2^{t-1}\theta}|1\rangle \right)\left( |0\rangle + e^{2i\pi2^{t-2}\theta}|1\rangle \right) ... \left( |0\rangle + e^{2i\pi2^{0}\theta}|1\rangle \right) \otimes|\psi\rangle= \frac{1}{2^{n/2}}\sum_{k = 0}^{2^n-1} e^{2i\pi\theta k} |k\rangle \otimes |\psi\rangle$$

Now if you are familiar with the Quantum Fourier Transform (if not you can check this Qiskit tutorial, I believe it is well-explained here), you'll notice that the expression looks a lot like the QFT of the state $$|2^n \theta \rangle$$. This way, since applying the inverse QFT is easily done, you can "easily" access the phase you are looking for.

I believe this Qiskit tuto (especially the first part, 1. Overview) and this book (page 221) will explain this better than I do, plus with Qiskit you can see an implementation, and in the book notice it is presented as an application of the QFT.