 Solution

ID: past-exam-of-the-computer-science-course-of-the-university-of-oxford/2020/quantum-information/1/e/i/solution

Directly applying the Hadamard gate gives

H ∣ + i ⟩ = \frac{1}{2} {(1 + i) ∣0 ⟩ + (1 - i) ∣1 ⟩} = e^{iπ /4} ∣ - i ⟩,

while

H ∣ - i ⟩ = \frac{1}{2} {(1 - i) ∣0 ⟩ + (1 + i) ∣1 ⟩} = e^{- iπ /4} ∣ + i ⟩ .

Hence the two states are interchanged up to the displayed global phases. Their relative factors

- i

and

+ i

are retained. They are the two Pauli eigenstates of the Pauli Y gate, and the Hadamard interchange of Y eigenstates is consistent with

H Y H = - Y

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