 Solution

ID: past-exam-of-the-computer-science-course-of-the-university-of-oxford/2019/quantum-information/1/b/ii/solution

The Pauli Y gate satisfies

Y^{†} = Y

and

Y^{2} = I

. With

c = cos (π /6) = 3 /2

and

s = sin (π /6) = 1/2

, the given matrix is Hermitian, and

M_{2}^{†} M_{2} = (c I + s Y)^{2} = (c^{2} + s^{2}) I + 2 cs Y = I + \frac{3}{2} Y \neq = I .

Therefore $M_{2}$ is not a unitary matrix. Its eigenvalues are

c + s

and

c - s

, neither of which has modulus one. An expression

cos θ I + i sin θ Y

would behave differently, but the printed matrix has no such factor of

i

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