Write $b=\sum_{j=0}^{m-1}2^jb_j$. Applying the <phase gate>
$$
P\left(\frac{2\pi2^j}{Q}\right)
$$
to qubit $b_j$ contributes $\exp(2\pi i2^jb_j/Q)$. The product of these $m$ gates is therefore
$$
\boxed{U|b\rangle=\omega^b|b\rangle}.
$$
Similarly, for $a=\sum_ja_j2^j$ and $b=\sum_kb_k2^k$, apply a <controlled phase gate>
$$
CP\left(\frac{2\pi2^{j+k}}Q\right)
$$
between every pair $(a_j,b_k)$. The accumulated phase is
$$
\prod_{j,k}\exp\left(\frac{2\pi i}{Q}a_jb_k2^{j+k}\right)
=\omega^{ab}.
$$
The $m^2$ controlled phase gates implement
$$
\boxed{V|a\rangle|b\rangle=\omega^{ab}|a\rangle|b\rangle}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 324 / 3 / a / iii / Solution

{title2=$CP(\phi)$}

The controlled phase gate contributes $e^{i\phi}$ exactly when both of its computational-basis input bits are one. It is diagonal and symmetric between its two qubits.


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