 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1/5/ii/solution

The absolutely convergent matrix exponential series gives

U^{†} = (e^{i H})^{†} = e^{(i H)^{†}} = e^{- i H}

for Hermitian

H

i H

and

- i H

commute, so their exponentials multiply to the exponential of their sum:

U^{†} U = e^{- i H} e^{i H} = I, U U^{†} = I .

Hence

U

is unitary. Alternatively, an orthonormal eigenbasis of

H

turns the exponential eigenvalues into

e^{i λ_{j}}

, all on the unit circle.

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