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In this paper (last paragraph, page 3) by Wocjan and Zhang, the definition of PES requires vector/bit string b. The phase estimation problem (PE) very much inspires the definition.

I cannot understand the vector b requirement in the PES definition.

I guess it has some connection to the 'distance/measure' defined in the sample space. Perhaps vector b brings a weight factor to account for uniformity in the definition. (or, states/vectors nearby b need to be more precisely estimated than the faraway vectors to b.)

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The discussion in question appears to be discussing usage of the Quantum Phase Estimation algorithm when we do not have access to an eigenstate $|\eta_j \rangle$ of the unitary matrix $U$ in question. This is almost always the case in practice as one may assume that the task of obtaining an eigenstate is similarly as hard as finding an eigenvalue $\lambda_j = e^{2\pi\phi_j}$, which is the main purpose of QPE.

The paper you have linked suggests using a bit string-encoded state $|b\rangle$ as an approximate eigenstate $|\eta_j \rangle$, such that the probability of success now depends on the squared overlap: $P_{\text{success}} \propto |\langle b|\phi_j\rangle|^2$.

Using a bit string encoded initial state sometimes works quite well. Indeed, when using QPE to estimate an eigenvalue of a Hamiltonian H (which we encode in a unitary matrix U) in quantum chemistry, we call such a state the Hartree-Fock state. The ordered bit string corresponds to the occupation of molecular orbitals as determined by the Hartree-Fock method.

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    $\begingroup$ +1 but the paper makes no claims as to whether $|b\rangle$ is an approximate eigenstate of $U$ or not; rather, a key point of the paper is to imply that merely sampling from the probability distribution induced by $U$ and $|b\rangle$ together, with probability given by the squared overlap of $|\langle \phi_j | b\rangle|^2$, is, as a sampling problem, BQP-complete. It may be convenient for $|b\rangle$ to be a computational basis state, but that's not required for the BQP-completeness proof (only that $|b\rangle$ is easy to prepare). $\endgroup$ Mar 23 at 23:43
  • $\begingroup$ @MarkSpinelli, point taken. Now I recognise this is complementing the jsbaker's argument in the context (Wocjan's paper.) $\endgroup$ Mar 24 at 17:44

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