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Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 33D / Solution Created 2026-09-24 Updated 2026-10-03

From the creation and annihilation operators commutator,

[H, A] = ℏ ω [A^{†} A, A] = - ℏ ω A .

Therefore

A ∣ n ⟩

is either zero or an energy eigenstate with energy

E_{n} - ℏ ω = E_{n - 1}

. Since each energy level of the one-dimensional quantum harmonic oscillator is nondegenerate,

A ∣ n ⟩ = c_{n} ∣ n - 1 ⟩

. Its norm determines the coefficient:

∣ c_{n} ∣^{2} = ⟨ n ∣ A^{†} A ∣ n ⟩ = n .

Choosing the conventional phases of the number states gives

A ∣ n ⟩ = n ∣ n - 1 ⟩ .

Solving the given relation for the position operator gives

X = \frac{ℏ}{2 mω} (A + A^{†}) .

Put

Q = A + A^{†}

. Repeated use of the ladder relations gives

Q ∣0 ⟩ = ∣1 ⟩, Q^{2} ∣0 ⟩ = ∣0 ⟩ + 2 ∣2 ⟩,

Q^{3} ∣0 ⟩ = 3∣1 ⟩ + 6 ∣3 ⟩, Q^{4} ∣0 ⟩ = 3∣0 ⟩ + 62 ∣2 ⟩ + 26 ∣4 ⟩ .

Hence

X^{4} ∣0 ⟩ = (\frac{ℏ}{2 mω})^{2} (3∣0 ⟩ + 62 ∣2 ⟩ + 26 ∣4 ⟩) .

In first-order nondegenerate perturbation theory, the

∣0 ⟩

component is omitted from the state correction, while

E_{0} - E_{2} = - 2ℏ ω

and

E_{0} - E_{4} = - 4ℏ ω

. It follows that

∣ 0_{λ} ⟩ = ∣0 ⟩ - \frac{ℏ λ}{4 m ^{2} ω ^{3}} (32 ∣2 ⟩ + \frac{3}{2} ∣4 ⟩) + O (λ^{2}),

which is the first-order ground state of the quartic oscillator.

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