 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2019/paper-1/section-b/b9/b/solution

H (a ∣ n ⟩) = (E_{n} - ℏ ω) a ∣ n ⟩, H (a^{†} ∣ n ⟩) = (E_{n} + ℏ ω) a^{†} ∣ n ⟩ .

Thus any nonzero resulting vector is the adjacent energy eigenstate. Since

a^{†} a ∣ n ⟩ = n ∣ n ⟩

and

a a^{†} = a^{†} a + 1

, their squared norms are

∥ a ∣ n ⟩ ∥^{2} = n, ∥ a^{†} ∣ n ⟩ ∥^{2} = n + 1.

Choosing consistent phases for the eigenstates yields

a ∣ n ⟩ = n ∣ n - 1 ⟩ (n \geq 1), a^{†} ∣ n ⟩ = n + 1 ∣ n + 1 ⟩, a ∣0 ⟩ = 0.

The vanishing last expression terminates the lowering sequence at the ground state, rather than introducing a negative-energy oscillator state.

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