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ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2016/paper-1/1/d/solution

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Past exam of the theoretical physics II course of the University of Cambridge / 2016 / Paper 1 / 1 / d / Solution by

Codex  0 2026-10-20

α = ℏ T / (2 I)

. The free propagator acts diagonally on each angular harmonic:

U_{T} e^{im θ} = e^{- i α m^{2}} e^{im θ} .

The kick multiplies by the Jacobi-Anger series

e^{i η c o s θ} = \sum_{ℓ \in Z} i^{ℓ} J_{ℓ} (η) e^{i ℓ θ} .

Multiply the two series and collect the output harmonic

n = m + ℓ

. The one-step coefficient update is

c_{n} (t + T) = m \in Z \sum i^{n - m} J_{n - m} (η) e^{- i α m^{2}} c_{m} (t),

where both sampling times are just after their respective kicks. Free evolution supplies the input-index phase before the kick mixes angular momenta.

The source uses the unnormalized basis functions

e^{im θ}

. Thus a unit-norm wavefunction satisfies

2 π \sum_{m} ∣ c_{m} ∣^{2} = 1

. If the basis is instead

e^{im θ} / 2 π

, its amplitudes have sum of squared moduli one and obey the same matrix update. Mixing these normalization conventions would change an energy expectation incorrectly.

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