 Solution

ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2012/paper-1/3/c/solution

Square the operator solutions, retaining the symmetrized cross term rather than treating

x^{'} (0)

and

p^{'} (0)

as commuting variables. The initial symmetrized expectation vanishes, so

⟨ x^{'2} ⟩_{t} ⟨ p^{'2} ⟩_{t} = \frac{cos ^{2} ω t}{2 α} + \frac{ℏ ^{2} α}{2 m ^{2} ω ^{2}} sin^{2} ω t, = \frac{ℏ ^{2} α}{2} cos^{2} ω t + \frac{m ^{2} ω ^{2}}{2 α} sin^{2} ω t .

The means remain zero. These are also the variances. Unless

α = mω /ℏ

, the Gaussian state breathes: its position and momentum widths exchange periodically while the mean energy remains constant.

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