 Solution

ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2014/paper-1/4/b/solution

⟨ a^{2} ⟩ = ⟨ a^{† 2} ⟩ = 0

. With

\overset{n}{ˉ} = ⟨ a^{†} a ⟩

and

a a^{†} = a^{†} a + 1

⟨ x^{2} ⟩ = \frac{ℏ}{2 mω} (2 \overset{n}{ˉ} + 1), ⟨ p^{2} ⟩ = \frac{m ℏ ω}{2} (2 \overset{n}{ˉ} + 1) .

Hence

⟨ \frac{p ^{2}}{2 m} ⟩ = ⟨ \frac{m ω ^{2} x ^{2}}{2} ⟩ = \frac{ℏ ω}{4} (2 \overset{n}{ˉ} + 1) .

Equality holds even at zero temperature, when each contribution is

ℏ ω /4

. Classical equipartition alone would miss this zero-point energy.

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