 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2016/11/b/solution

t = 0

v (0) = v_{0}

. Equate the coefficient of

t^{n}

after differentiating the series and substituting into the linear system of ordinary differential equations:

(n + 1) v_{n + 1} = - i H v_{n} .

v_{n} = (- i H)^{n} v_{0} / n!

. Summing gives the matrix exponential solution

v (t) = e^{- i H t} v_{0}

For the specified diagonal Hermitian matrix, its powers and exponential act separately on the two components. Thus

U (t) = e^{- i H t} = (e^{- i t} 0 0 e^{i t}), v (t) = U (t) v_{0} .

This unitary matrix preserves

v^{†} v

: the two components acquire opposite phases of unit modulus. The coefficient recurrence applies to any constant finite square matrix

H

; unitarity here additionally uses its Hermitian property.

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