How much information we can get about eigenstate from phase estimation?

I'm learning about the standard phase estimation algorithm. Here's a diagram from Qiskit tutorial. 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!

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.
• 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 $U$. Feb 16, 2022 at 15:39