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In HHL, vector $\vec{b}$ is assumed to be decomposed in the eigenbasis {$u_i$}$_{i=1}^n$ of a Hermitian matrix $A$. However, as we do not calculate explicitly the eigenvectors $u_i$ in course of the algorithm, how do we load $\vec{b}$ into the quantum computer (using amplitude encoding), if we don't even know the $b_i$'s?

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We do explicitly calculate $A$'s eigendecomposition of $\vec b$, but in superposition. This is the quantum phase estimation portion of HHL.

You can provide your initial vector $\vec b$ in the computational basis (e.g., in the $|0\rangle,|1\rangle$ basis), and as long as you can simulate $A$ as a Hamiltonian you can perform controlled-versions of this simulation, with the controlling qubits being part of a $k$-qubit ancilla register. Your initial vector $\vec b$ can be anything and need not be prepared in any eigenbasis of $A$, because the QPE is what does the decomposition.

After doing the Hamiltonian simulations $2^k-1$ times, an inverse-QFT on the $k$ qubits stores the phases (eigenvalues) of $e^{-iAt}$ into the ancilla register, with probability amplitudes given by the amount of overlap to the eigendecomposition of $\vec b$. This explicitly calculates the eigenvalues of $A$ and “writes” $\vec b$ into this eigenbasis, but in superposition and indexed by the phases $|k\rangle$.

The rest of the algorithm is eigenvalue surgery on these phases (and uncomputation and post-selection if needed).

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  • $\begingroup$ Ok, so vector $\vec{b}$ can be loaded in the computational basis, the decomposition in the eigenbasis of $A$ happens during the algorithm. Can you direct me to places, where it is explained, what are the options used to load $\vec{b}$? One needs to understand how amplitude encoding is being done, what is the error of this operation, right? $\endgroup$
    – wondering
    Jun 27, 2023 at 15:05
  • $\begingroup$ @wondering I don't know of any one source that could explain how to prepare $\vec b$ as $|b\rangle$ efficiently, but yes that needs to be done and would have error associated with it perhaps, if $|b\rangle$ cannot be prepared error-free. You could for example adiabatically prepare $|b\rangle$. I guess the key-words to search for are "state preparation" or something. $\endgroup$ Jun 27, 2023 at 20:16

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