 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2021/paper-1/section-b/b7/b/solution

⟨ a^{†} a ⟩_{n} = n

⟨ a a^{†} ⟩_{n} = n + 1

. Terms with

a^{2}

(a^{†})^{2}

have zero diagonal matrix elements, since they change the occupation number by two. Therefore

⟨ x^{2} ⟩_{n} ⟨ p^{2} ⟩_{n} = \frac{ℏ}{2 mω} ⟨ a^{2} + (a^{†})^{2} + a a^{†} + a^{†} a ⟩_{n} = \frac{ℏ ( 2 n + 1 )}{2 mω}, = - \frac{m ℏ ω}{2} ⟨(a^{†})^{2} + a^{2} - a^{†} a - a a^{†} ⟩_{n} = \frac{m ℏ ω ( 2 n + 1 )}{2} .

⟨ V ⟩_{n} = \frac{1}{2} m ω^{2} ⟨ x^{2} ⟩_{n} = \frac{1}{2} ℏ ω (n + \frac{1}{2}), ⟨ T ⟩_{n} = \frac{⟨ p ^{2} ⟩ _{n}}{2 m} = \frac{1}{2} ℏ ω (n + \frac{1}{2}) .

E_{n} = ℏ ω (n + 1/2)

. Their equality is also the virial theorem result for a quadratic potential, here derived directly from the ladder operators.

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