# Problem with controlled rotation in HHL

In HHL algorithm, for subroutine involving controlled rotation, after applying $$R_y(\theta)$$, where $$\theta=2\sin^{-1}\left(\frac{C}{\lambda}\right)$$ to the ancilla, the state changes to $$\sqrt{1-\frac{C^2}{\lambda^2}}|0\rangle+\frac{C}{\lambda}|1\rangle.$$

Question

If $$\lambda=1$$, then by $$\sqrt{1-\frac{C^2}{\lambda^2}}|0\rangle+\frac{C}{\lambda}|1\rangle$$ and choosing $$C=1$$, we get $$\theta=\pi$$. For $$\lambda=2$$, $$\theta=\frac{\pi}{3}.$$ So in general, for each $$\lambda$$, correspondingly there's a different $$\theta$$. Since we don't know eigenvalues a priori, how do we account $$\theta$$s for superpositon of eigenvalues?

With respect to the circuit here on page 5, I don't understand how the controlled rotation part works. Will this circuit work when I choose a hermitian matrix $$A_{4\times 4}$$ such that, one of it's eigenvalues, $$\lambda_j=10\neq 2^i,i \in \mathbb{Z}$$ ?

• There are too many questions here. Please ask one question per post, as it will help us give you better answers. Apr 6 '20 at 8:22
• How I remember, the circuit in the paper (Fig.4) is working only for (1) matrix from the same paper. The more generic solution/circuit is proposed in other figures of the same paper. Apr 6 '20 at 9:25

• No, it's entangled with the input state. So if the eigenvectors of $A$ are $|\lambda_i\rangle$, then the input state is decomposed as $\sum_i\alpha_i|\lambda_i\rangle$, then after phase estimation on $t$ qubits, you have $\sum_i\alpha_i\lambda_i\rangle|\theta_i\rangle$, where $\theta_i$ is the best approximation to $e^{2\pi\theta_i/2^t}=\lambda_i$. If you do a controlled rotation off the $|\theta_i\rangle$ part, that you can say "if the angle is $\theta_i$, do rotation $R_y(f(\theta_i))$ where $f$ is some extra calculation you might perform. Apr 6 '20 at 9:43
• In this case is $f(\theta_i)=arcsin(\frac{2C}{\theta_i})$? If yes, how do we compute it? Apr 6 '20 at 17:13