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Can I get the valid solution of linear problem using HHL algorithm even if the QPE is not completely correct?

My example is $A=diag(0.5, 0.2, 0.3, 0.6)$, so the solution is $[0.5, 0.2, 0.3, 0.6]$ which is the eigenvalue of $A$.

I run the QPE algorithm with 10000 shots and input the eigenstate as $b = normalized(5,4,3,2)$, so the target measurement counts are $(4630, 2963, 1667, 741)$

But I got the result $[\hat{\lambda_1},\hat{\lambda_2},\hat{\lambda_1}',\hat{\lambda_4}, \hat{\lambda_5}] = [0.5062,0.2062,0.3,0.4875,0.6]$ with measurement counts $(3180, 2045, 1654, 816, 726)$ times respectively (with other ignorable results).

I think $0.5$ eigenvalue was measured as two different corresponding $\hat{\lambda_1}, \hat{\lambda_1}'$, where one is correct and the other is wrong.


But HHL algorithm requires rotation by the parameter $\widetilde{\lambda_j}$ corresponding $b_j$. If the QPE is not perfect, is it the wrong $\hat{\lambda_k}$ multiplied to $b_j$ where $k\neq j$??

In my example, does the wrong $\hat{\lambda_1}'$ affect the solution??

The HHL algorithm result is proportional to $\sum_{j=1}^{2^2}b_j \frac{C}{\lambda_j}|u_j\rangle$. So, does the algorithm calculate $b_1 \frac{C}{\lambda_1}|u_1\rangle+b_2 \frac{C}{\lambda_2}|u_2\rangle+b_3 \frac{C}{\lambda_1'}|u_3\rangle+b_4 \frac{C}{\lambda_3}|u_4\rangle$??

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