 Solution

ID: past-exam-of-the-physics-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1/section-b/b7/a/solution

Invert the given ladder operators:

\overset{x}{^} = \frac{ℏ}{2 mω} (\overset{a}{^}_{+} + \overset{a}{^}_{-}), \overset{p}{^} = i \frac{m ℏ ω}{2} (\overset{a}{^}_{+} - \overset{a}{^}_{-}) .

Because the Energy eigenstate are orthonormal,

⟨ n ∣ \overset{a}{^}_{\pm} ∣ n ⟩ = 0

: each operator changes the state to one orthogonal to

∣ n ⟩

. Hence

⟨ x ⟩_{n} = ⟨ p ⟩_{n} = 0.

The result also follows from the definite parity of every oscillator energy eigenfunction, but here it follows directly from the specified operators.

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