Controlled-Phase Approximation and Global-Phase Boundary

On a basis vector $\mathrm{\mid }ab⟩$, the first CNOT maps it to $\mathrm{\mid }a,b\oplus a⟩$. The target phase contributes $e^{-i\theta (b\oplus a)\mathrm{/}2}$; the second CNOT restores $\mathrm{\mid }ab⟩$; and the two final phase gates contribute $e^{i\theta (a+b)\mathrm{/}2}$. The product is $e^{i\theta ab}$, which is $1$ on $\mathrm{\mid }00⟩,\mathrm{\mid }01⟩,\mathrm{\mid }10⟩$ and $e^{i\theta }$ on $\mathrm{\mid }11⟩$. This is exactly $C{P}_{\theta }$ from [F1].

Direct diagonal multiplication gives $P_{\alpha }=e^{i\alpha \mathrm{/}2}{R}_z(\alpha )$, hence $R_z(\alpha )=e^{-i\alpha \mathrm{/}2}{P}_{\alpha }$. Replacing $P_{-\theta \mathrm{/}2}$ on the target contributes $e^{i\theta \mathrm{/}4}$, while replacing each of the two $P_{\theta \mathrm{/}2}$ gates contributes $e^{-i\theta \mathrm{/}4}$. Their product is $e^{-i\theta \mathrm{/}4}$, so the resulting $R_z$ circuit is $e^{-i\theta \mathrm{/}4}C{P}_{\theta }$.

The circuit in step 1.1 is an exact factorization of the two-qubit $C{P}_{\theta }$ into three one-qubit phase gates and two CNOTs. Applying [F3] therefore gives a word $W$ and unit scalar $\mu$ with $\mathrm{\parallel }C{P}_{\theta }-\mu W\mathrm{\parallel }\le \epsilon$. By [F2], $C(\mu W)=(\mathrm{diag}(1,\mu )\otimes I_4)C(W)$. The difference $C(C{P}_{\theta })-C(\mu W)=\mathrm{diag}(0,C{P}_{\theta }-\mu W)$ has norm exactly $\mathrm{\parallel }C{P}_{\theta }-\mu W\mathrm{\parallel }$: on a vector $(x,y)$ it acts as $(0,(C{P}_{\theta }-\mu W)y)$, giving the upper bound, and vectors $(0,y)$ give equality. Consequently the displayed phase-corrected controlled approximation has error at most $\epsilon$. If $\mu \mathrm{\ne }1$, the correction $\mathrm{diag}(1,\mu )$ is not a scalar on the control. Omitting it is not justified by the two-qubit projective bound; using it requires both the control phase and $C(W)$ to be available.

With ${\lambda }_R=e^{-i\theta \mathrm{/}4}$ from step 2.1, [F2] gives $C({\lambda }_{R}C{P}_{\theta })=(\mathrm{diag}(1,{\lambda }_R)\otimes I_4)C(C{P}_{\theta })$. Applying $P_{\theta \mathrm{/}4}=\mathrm{diag}(1,\overline{{\lambda }_R})$ to the outer control restores the exact controlled target in the continuous phase-gate model.

The complete controlled target is the diagonal three-qubit matrix $D_{\theta }=\mathrm{diag}(1,1,1,1,1,1,1,e^{i\theta })$. For bits $a,b,c\in \{0,1\}$, put $p_{ab}=a\oplus b$, $p_{ac}=a\oplus c$, $p_{bc}=b\oplus c$, and $p_{abc}=a\oplus b\oplus c$. Checking the four possible Hamming weights $0,1,2,3$ makes the right side respectively $0,0,0,4$, so $4abc=a+b+c-p_{ab}-p_{ac}-p_{bc}+p_{abc}$. Apply $P_{\theta \mathrm{/}4}$ to each single bit. To apply $P_{-\theta \mathrm{/}4}$ to $p_{ab}$, use a CNOT with control $b$ and target $a$, apply $P_{-\theta \mathrm{/}4}$ to wire $a$, and undo that CNOT; the target wire then again contains $a$. Use the corresponding pairs for $p_{ac}$ and $p_{bc}$. To apply $P_{\theta \mathrm{/}4}$ to $p_{abc}$, use two CNOTs into wire $a$, apply $P_{\theta \mathrm{/}4}$ there, and undo both CNOTs. On each basis state these circuits restore the original bits and contribute exactly the indicated signed parity phase. The resulting phase is $e^{i\theta abc}$, so this is an exact factorization of $D_{\theta }$ with seven one-qubit phase gates and ten CNOTs. Applying [F3] to this complete three-qubit factorization gives a projective operator approximation to $D_{\theta }$ itself; its scalar is global on the three-qubit block. This supplies the complete-block compilation option in the statement.