Past exam of the laser science and quantum information processing course of the University of Oxford / 2018 / 5 / b / Solution

For the lower-qubit controlled rotation, the control is the upper qubit and only the subspace $|10⟩,|11⟩$ rotates. Write $u=\cos (\theta /2)$ and $v=\sin (\theta /2)$. Unitary conjugation of its diagonal populations gives $c{u}^2+d{v}^2$ and $c{v}^2+d{u}^2$, together with off-diagonal coherences. The ideal crush gradient stipulated in the source removes those coherences without changing populations. ThereforeThe other two populations remain $a,b$. This controlled-transfer gate is not unitary as a complete operation because it includes dephasing.

For an upper-qubit target, the lower qubit is the control: the mixed subspace is $|01⟩,|11⟩$. ThusThe population $a$ of $|00⟩$ is unaffected in either direction because its control bit is zero. The dephasing step also leaves this diagonal population unchanged, so any sequence of these controlled-transfer gates preserves $a$.