 Solution

ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2018/4/a/solution

Using

[x, p] = i ℏ

\overset{x}{˙} = \frac{i}{ℏ} [H_{0}, x] = \frac{p}{m}, \overset{p}{˙} = \frac{i}{ℏ} [H_{0}, p] = - m ω^{2} x .

Solving these coupled linear equations gives

x (t) p (t) = x (0) cos ω t + \frac{p ( 0 )}{mω} sin ω t, = p (0) cos ω t - mω x (0) sin ω t .

a = \frac{mω}{2ℏ} x + \frac{i}{2 m ℏ ω} p, a^{†} = \frac{mω}{2ℏ} x - \frac{i}{2 m ℏ ω} p .

They obey

[a, a^{†}] = 1

and

H_{0} = ℏ ω (a^{†} a + 1/2)

. Substitution of the position and momentum solutions, or the corresponding commutators with

H_{0}

, yields

a (t) = a (0) e^{- iω t}, a^{†} (t) = a^{†} (0) e^{iω t} .

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