 Solution

ID: past-exam-of-the-quantum-theory-course-of-the-university-of-oxford/2021/2/b/ii/solution

From part (a),

[N, a_{j}] = - a_{j}

and

[N, a_{j}^{*}] = a_{j}^{*}

[N, a_{1} a_{2}^{*}] [N, a_{2} a_{1}^{*}] = [N, a_{1}] a_{2}^{*} + a_{1} [N, a_{2}^{*}] = - a_{1} a_{2}^{*} + a_{1} a_{2}^{*} = 0, = [N, a_{2}] a_{1}^{*} + a_{2} [N, a_{1}^{*}] = - a_{2} a_{1}^{*} + a_{2} a_{1}^{*} = 0.

Consequently

[N, L] = 0.

Each term in

L

transfers one excitation between the two modes, preserving the total number operator eigenvalue. This is an instance of the oscillator bilinear commutator; every number operator eigenspace is an invariant subspace of

L

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