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I'm learning about the standard phase estimation algorithm. Here's a diagram from Qiskit tutorial. enter image description here Suppose we use $\mathcal{C}$ to represent the entire circuit, and $|\psi\rangle$ is an eigenstate of the system we are simulating, then we can write $$ \mathcal C|0\rangle^{\otimes t}|\psi\rangle = |\psi^{(t)}\rangle|\psi\rangle $$

Where $|\psi^{(t)}\rangle$ (or $|2^t\theta\rangle$) is the $t$- bit state showing the phase information associated with the state $|\psi\rangle$. My question is if we don't know exactly the state $|\psi\rangle$, is there a way we can estimate this state based on $|\psi^{(t)}\rangle$? Suppose $|\psi\rangle = a|0\rangle + b|1\rangle$ (single qubit case), can we find $a$ and $b$?

Thanks for the help!

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Quantum Phase Estimation (QPE) estimates the phase $\theta_i$ of an eigenstate $|\phi_i\rangle$ with eigenvalue $e^{2\pi i \theta_i}$ of the unitary operator $U$.

Your arbitrary quantum state $|\psi\rangle$ is generally a superposition of the eigenstates $|\psi\rangle = \sum_i c_i|\phi_i\rangle$ of $U$. Hence, the results from the QPE will also be a superposition of the phases.

To estimate an unknown quantum state, the right algorithm to use is Quantum State Tomography.

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The first register only gives you information about the eigenvalue, not the eigenvector. However, once you have measured the first register and projected onto one of the eigenvalues (for simplicity, I'm talking about exactly finding an eigenvalue. Statements can be appropriately modified for the case where you get an extremely good approximation), the second register is projected onto the corresponding eigenvector. So, you can do tomography of the state of the second register. This is the case even if the input was an arbitrary superposition of states, although you will have to do tomography separately for each of the possible measurement outcomes.

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  • $\begingroup$ You gave a better answer than the one I gave earlier. May I ask if the method you outline has any advantage over doing tomography directly? Thanks :) $\endgroup$ Commented Feb 16, 2022 at 14:49
  • $\begingroup$ I cannot pretend to be a tomography expert. The advantage is that if you want to identify what the dominant eigenvector in an initial state which is a superposition of terms, you're doing a preparation step that isolates that terms that you want to understand. Otherwise, I imagine you'd actually need process tomography instead of state tomography, which presumably has a higher cost. The disadvantage is that you need access to controlled-$U$ instead of just $U$. $\endgroup$
    – DaftWullie
    Commented Feb 16, 2022 at 15:39

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