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

Question

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

Answer 1

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.

If you need anything else please feel free to ask :)

Answer 2

The unitary operations of the phase estimation algorithm are mainly controlled phase rotations of an angle that must be specified in the algorithm.

This angle that you specify is the phase that you want to estimate.

So the first counting qubit will make one rotation the second will make two, qubit 3 will make 4 rotations ... up to 2 ^ t-1 rotations on the register.