How would one go along to calculate the number of uses of an unitary $U$ in Iterative Phase Estimation (IPE) to compare it to the number of uses of $U$ in standard Phase Estimation (Qiskit QPE)?


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The IQPE procedure tries to reduce the extra number of ancilla qubits needed to measure the phase and replaces that with the extra number of iterations that we perform to measure our phase.

Imagine you have the phase as $\theta = 0.\theta_{0}\theta_{1}\theta_{2}\theta_{3}$. So, what IQPE does is that in the first iteration, it tries to identify the value of $ \theta_{3}$ by applying the unitary $U$ matrix, $2^{3}$ times. Why? Each phase kickback imparts a phase of $\theta$ to our ancilla. If we applied the matrix $8$ times, the phase kicked back is going to be $2^{3} \theta$ right?

Now, see what actually $e^{2 \pi i (2^{3}\theta)} $ is equal to. $2^{3}\theta$ is just $\theta_{0}\theta_{1}\theta_{2}.\theta_{3}$ as the bits get shifted 3 times left. $e^{2 \pi i (\theta_{0}\theta_{1}\theta_{2}.\theta_{3})}$ can be written as $e^{2 \pi i (0.\theta_{3})}$ as the integral part evaluates to 1. Measuring in the X basis produces the bit $\theta_{3}$ with very high probability.

Now, we see that in the estimation of the $i_{th}$ bit position (0-based), we required $2^{i}$ number of unitary applications. Summing this over to all the bits we shall have -
$$\sum_{i=0}^{n-1} 2^{i}$$ which evaluates to $2^{n} - 1$ unitary applications for n bit precision.

PS - This is exactly the same as the Standard Approach but the width of the circuit reduces :)


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