In the HHL algorithm, how do you efficiently do the $\lambda-$controlled rotation on the ancillary qubit ? It seems to me after reading around some answers that this can be done in two steps :

  • First, we map $|\lambda\rangle\mapsto |\frac{1}{\pi}\arcsin(\frac{C}{\lambda})\rangle$, defining $|\frac{1}{\pi}\arcsin(\frac{C}{\lambda})\rangle$ to be a binary representation $|\frac{1}{\pi}\arcsin(\frac{C}{\lambda})\rangle$ with $m$ qubits.
  • Then perform a controlled rotation $U_y(|\theta\rangle \otimes |0\rangle)\mapsto |\theta\rangle \otimes \big(\cos(\theta)|0\rangle + \sin{(\theta})|1\rangle\big)$ where $U_y$ is simply $$ U_y(|\theta\rangle \otimes |0\rangle) = \prod_{j=1}^m (I^{\otimes^m}\otimes R_y(2\pi\theta_j/2^j)) $$ i.e. a sequence of controlled rotations where we successively halve the angle of rotation conditionally of the digits of the binary representation of $\theta$.

My question is the following how can one implement efficiently the first step in an environment such as Qiskit ?

  • $\begingroup$ Hello, did you check this tutorial from Qiskit directly? qiskit.org/textbook/ch-applications/hhl_tutorial.html $\endgroup$
    – Lena
    Dec 9, 2020 at 17:50
  • $\begingroup$ Yes, I read the all tutorial but they didn't explain how to compute eigenvalue rotation without lost of generality. In their example developed in part B, they apply a $2-$controlled rotation by $01$ and $10$ but it's a particular case. $\endgroup$
    – SRichoux
    Dec 9, 2020 at 18:00

1 Answer 1


There is a new approach that will be merged soon in qiskit terra (here for the PR) that uses polynomial approximation to compute $\arcsin(C/\lambda)$, and asymptotically this would be the efficient implementation.

In practice if you are solving a $2\times 2$ matrix or a very small system it would be better to hard code the rotations.

The theory and error analyses of this approach are explained in Section VI of this paper.


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