# Unitarian branching on qubits Let $|\psi\rangle=\alpha|0\rangle|\phi_0\rangle+\beta|1\rangle|\phi_1\rangle$ ,$U_0,U_1$ unitarian gates and we would like to operate $U_0$ on $|\phi_0\rangle$ and $U_1$ on $|\phi_1\rangle$. Define the controlled operation $U=|0\rangle\langle0|\otimes U_0+|1\rangle\langle1|\otimes U_1$ . **Lemma**: $U$ is unitarian operation (although the projectors $|0\rangle\langle0|, |1\rangle\langle1|$ are not unitarians) **Proof**: $U^\dagger=|0\rangle\langle0|\otimes U_0^\dagger+|1\rangle\langle1|\otimes U_1^\dagger$ . $UU^\dagger=|0\rangle\langle0|\otimes U_0U_0^\dagger+|1\rangle\langle1|\otimes U_1U_1^\dagger$ because $|0\rangle\langle0||1\rangle\langle1|$ is the zero matrix. As $U_0,U_1$ are unitarians $U_0U_0^\dagger=U_1U_1^\dagger=I$ and $UU^\dagger=|0\rangle\langle0|\otimes I+|1\rangle\langle1|\otimes I=I\otimes I=I$ . **Lemma**: $U|\psi\rangle=\alpha|0\rangle U_0|\phi_0\rangle+\beta|1\rangle U_1|\phi_1\rangle$ **Proof**: This is true as $\big(|0\rangle\langle0|\big) |1\rangle = 0$, $\big(|0\rangle\langle0|\big) |0\rangle = 1$ and $\big(|1\rangle\langle1|\big) |0\rangle = 0$, $\big(|1\rangle\langle1|\big) |1\rangle = 1$. ## Created 2026-03-15 12:13