Suppose I have a $d$-bit estimate of $\theta \in (0, 2 \pi]$, say a register of qubits $|\tilde{\theta} \rangle = |z_1 \rangle \ldots |z_d \rangle$ for $|z_i \rangle \in \{|0 \rangle, |1 \rangle \}$ where $\theta \approx \sum_{i=1}^{d} z_i \pi /2^{i}$, and I wish to implement a transformation $|\tilde{\theta} \rangle |0 \rangle \mapsto |\tilde{\theta} \rangle (cos(\theta) |0 \rangle + sin(\theta)|1 \rangle)$ using a quantum circuit consisting of two-qubit gates which can be represented by the controlled unitary transformation

$U_{c_i} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & cos(\pi / 2^i) & -sin(\pi / 2^i) \\ 0 & 0 & sin(\pi / 2^i) & cos(\pi / 2^i) \\ \end{bmatrix}$

I am trying to come up with a formula for how a sequence of these controlled operations would be applied to the ancilla qubit $|0 \rangle$ (maybe more than a single qubit?) to result in an overall rotation by $\approx \theta $ but I am not sure how to proceed. Any insights appreciated.

Edited the matrix as in the answer.


$U_{c_i}$ should not have $z_i$ in it. That information is on the input. It is not in the gate.

So instead

$$ U_{c_i} = \begin{pmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&\cos(\pi/2^i)&-\sin(\pi/2^i)\\ 0&0&\sin(\pi/2^i)&\cos(\pi/2^i)\\ \end{pmatrix} $$

If this gate acts on the $i$'th qubit of $| \tilde{\theta} \rangle$ and the target, this will give the appropriate controlled rotation.

To do all of $\tilde{\theta}$ just do all these $U_{c_i}$ on the appropriate pairs as described above. This is thanks to the fact that all of these operations are rotating along the same axis. You are just building up

$$\prod_{i=1}^d R_z (\frac{z_i}{2^i}\pi) = R_z (\sum_{i=1}^d \frac{z_i \pi}{2^i}) = R_z (\tilde{\theta})$$

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.