 Solution

ID: past-exam-of-the-quantum-theory-course-of-the-university-of-oxford/2023/2/c/solution

[X_{j}, P_{k}] = i ℏ δ_{jk} I, [X_{j}, X_{k}] = [P_{j}, P_{k}] = 0.

c_{j} = m ω_{j} / (2ℏ)

d_{j} = 1/ 2 m ℏ ω_{j}

, direct expansion of the commutator gives

[a_{j, -}, a_{k, +}] = ℏ (c_{j} d_{k} + d_{j} c_{k}) δ_{jk} I = δ_{jk} I .

ψ_{0, 0}

is annihilated by both

a_{j, -}

. Successive creation operators generate

ψ_{n_{1}, n_{2}} \propto (a_{1, +})^{n_{1}} (a_{2, +})^{n_{2}} ψ_{0, 0} .

Hence, in the order of part (a), the six energy eigenstates are proportional to

ψ_{0, 0}, a_{1, +} ψ_{0, 0}, a_{2, +} ψ_{0, 0}, (a_{1, +})^{2} ψ_{0, 0}, a_{1, +} a_{2, +} ψ_{0, 0}, (a_{2, +})^{2} ψ_{0, 0} .

If the Fock vacuum has unit norm, dividing the general expression by

n_{1}! n_{2}!

gives normalized number states. Normalization is not required here, but using this convention makes the later change of basis unambiguous.

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