 Solution

ID: past-exam-of-the-laser-science-and-quantum-information-processing-course-of-the-university-of-oxford/2021/4/d/solution

The extra Pauli gate on the control is removed by

Z_{a} G = C_{Z}

. Conjugating the target by a Hadamard gate then gives

C_{X} = (I \otimes H) (Z \otimes I) G (I \otimes H) .

Thus, from left to right in the network, apply

H_{b}

, then

G

, then

Z_{a}

and

H_{b}

; the last two gates commute.

For a direct proof, the control-zero block of

G

I_{b}

and its control-one block is

- Z_{b}

Z_{a}

changes the latter into

Z_{b}

. Since

H_{b} I_{b} H_{b} = I_{b}

and

H_{b} Z_{b} H_{b} = X_{b}

, the result is

∣ g ⟩ ⟨ g ∣_{a} \otimes I_{b} + ∣ e ⟩ ⟨ e ∣_{a} \otimes X_{b},

which is exactly the CNOT gate on all inputs, including quantum superpositions.

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